Partial Differential Equations Analytical and Numerical Methods
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Differential Partial Differential Equations
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Partial Differential Differential Partial Equations Equations Analytical Analytical and and Numerical Numerical Methods Methods
Mark S. S. Gockenbach Gockenbach Mark Michigan Technological Technological University University Michigan Houghton, Houghton, Michigan Michigan
siam Society for Industrial and and Applied Applied Mathematics Mathematics Society for Industrial Philadelphia Philadelphia
Copyright © © 2002 2002 by by the Society for for Industrial Industrial and Applied Mathematics. Mathematics. Copyright the Society and Applied 1098765432 1 10987654321 All rights rights reserved. reserved. Printed Printed in in the the United United States States of of America. No part part of of this book may All America. No this book may be reproduced, reproduced, stored, transmitted in in any any manner manner without permission of without the the written written permission of be stored, or or transmitted the the Society for Industrial the publisher. publisher. For For information, information, write write to to the Society for Industrial and and Applied Applied 19104-2688. Mathematics, Mathematics, 3600 3600 University University City City Science Science Center, Center, Philadelphia, Philadelphia, PA PA 19104-2688. Library of of Congress Congress Cataloging-in-Publication Cataloging-in-Publication Data Data Library Gockenbach, Mark Mark S. Gockenbach, S. Partial differential equations S. Partial differential equations :: analytical analytical and and numerical numerical methods methods // Mark Mark S. Gockenbach. Gockenbach. p.cm. p. cm. Includes Includes bibliographical bibliographical references references and and index. index. ISBN 0-89871-518-0 ISBN 0-89871-518-0 1. 1. Differential Differential equations, equations, Partial. Partial. I.I. Title. Title. QA377 .G63 .G63 2002 2002 QA377 515'.353-dc21 515'.353-dc21 2002029411 MATLAB trademark of of The MATLAB is is a a registered registered trademark The MathWorks, MathWorks, Inc. Inc. Maple trademark of Waterloo Maple, Maple is is a a registered registered trademark of Waterloo Maple, Inc. Inc. Mathematica a registered Mathematica is is a registered trademark trademark of of Wolfram Wolfram Research, Research, Inc. Inc.
0-89871-518-0 0-89871-518-0
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Dedicated to mother, Joy Joy Gockenbach, the Dedicated to my my mother, Gockenbach, and and to to the memory of of my my father, father, LeRoy LeRoy Gockenbach. memory Gockenbach.
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Contents Contents Foreword Foreword
xiii xiii
Preface Preface
xvii xvii
11
Classification Classification of of differential differential equations equations
22
Models Models in in one one dimension dimension
2.1 2.1
2.2 2.2 2.3 2.3 2.4 2.4 33
Heat flow in Heat flow in aa bar; bar; Fourier's Fourier's law. law . . . . . . . . . . . . . . . .. Boundary 2.1.1 Boundary and and initial initial conditions conditions for for the heat equation equation fk 2.1.2 Steady-state 2.1.2 Steady-stateheat heatflow 2.1.3 Diffusion. . . . . . . . . . . . . . . . . . 2.1.3 Diffusion The bar . . . . . . . . . . . . . . . . . . . . . The hanging hanging bar 2.2.1 Boundary conditions for for the bar 2.2.1 Boundary conditions the hanging hanging bar The The wave wave equation equation for for aa vibrating vibrating string string Suggestions Suggestions for for further further reading. reading . . . . . . . . . . . .
Essential linear linear algebra algebra Essential 3.1 Linear systems 3.1 Linear systems as as linear linear operator operator equations. equations 3.2 Existence and uniqueness of to Ax = = bb 3.2 Existence and uniqueness of solutions solutions to 3.2.1 3.2.1 Existence . . . . . . . . . 3.2.2 Uniqueness . . . . . . . . 3.2.2 Uniqueness 3.2.3 The 3.2.3 The Fredholm Fredholm alternative alternative 3.3 Basis 3.3 Basis and and dimension. dimension . . . . . . . 3.4 Orthogonal 3.4 Orthogonal bases and and projections . L22 inner product . 3.4.1 The L 3.4.2 The theorem 3.4.2 The projection projection theorem 3.5 3.5 Eigenvalues and and eigenvectors eigenvectors of of aa symmetric symmetric matrix matrix 3.5.1 The 3.5.1 The transpose transpose of of aa matrix matrix and and the the dot dot product product ... 3.5.2 Special properties of 3.5.2 Special properties of symmetric symmetric matrices matrices 3.5.3 The spectral spectral method method for for solving solving Ax 3.5.3 The Ax == bb 3.6 Preview of of methods methods for for solving solving ODEs ODEs and and PDEs 3.6 Preview PDEs . 3.7 Suggestions for for further further reading reading . . . . . . . . . . . . . 3.7 Suggestions
VVII II
11 9 9 99 13 14 14 16 16 21 21 24 24 27 27 30 30
31 31 31 31 38 38 38 38 42 42 45 45 50 50 55 55 58 58 61 61 68 71 71 72 72 74 74 77 77 78
VIII viii
4 4
Contents Contents ordinary differential equations 79 Essential ordinary 79 Converting system ... 79 Converting aa higher-order higher-order equation equation to to aa first-order first-order system 79 Solutions to some simple ODEs ODEs . . . . . . . . . . . . . . . Solutions to some simple 82 82 4.2.1 The general 4.2.1 The general solution solution of of aa second-order second-order homogeneous homogeneous ODE with with constant ODE constant coefficients coefficients . . . . . . . . . . . 82 82 4.2.2 A special inhomogeneous inhomogeneous second-order ODE . 85 4.2.2 A special second-order linear linear ODE 85 4.2.3 First-order linear ODEs . . . 4.2.3 First-order linear ODEs 87 87 4.3 Linear with constant constant coefficients coefficients . . . . . . . . . . . . 4.3 Linear systems systems with 91 4.3.1 Homogeneous systems systems . . . . . . . . . . . . . . . . 4.3.1 Homogeneous 92 4.3.2 Inhomogeneous systems and and variation variation of of parameters parameters 96 4.3.2 Inhomogeneous systems 101 4.4 Numerical methods for for initial value problems . . . . . . . . . . . 101 4.4.1 Euler's method method . . . . . . . . . . . . . . . . . . . . . 102 4.4.1 Euler's 102 4.4.2 Improving on Euler's Euler's method: method: Runge-Kutta Runge-Kutta methods methods 104 104 4.4.2 Improving on 4.4.3 Numerical systems of of ODEs 4.4.3 Numerical methods methods for for systems ODEs . . . . . . 108 108 4.4.4 Automatic step control Runge-Kutta-Fehlberg and Runge-Kutta-Fehlberg 4.4.4 Automatic step control and methods 110 methods. . . . . . . . . . . . . . . 110 115 4.5 Stiff systems of of ODEs. 4.5 Stiff systems ODEs . . . . . . . . . . . . . 115 4.5.1 A simple example example of stiff system 117 4.5.1 A simple of aa stiff system 117 4.5.2 The backward backward Euler Euler method method . . . 118 4.5.2 The 118 4.6 Green's functions 123 4.6 Green's functions . . . . . . . . . . . . . . . . 123 4.6.1 The Green's function for for aa first-order first-order linear linear ODE ODE.. . 123 123 4.6.1 The Green's function 4.6.2 The Dirac Dirac delta delta function function . . . . . . . . . . . 125 4.6.2 The 125 4.6.3 The Green's function for aa second-order second-order IVP 4.6.3 The Green's function for IVP . . . . 126 126 4.6.4 Green's for PDEs 127 4.6.4 Green's functions functions for PDEs 127 4.7 Suggestions for 128 4.7 Suggestions for further further reading reading . . 128
4.1 4.1 4.2 4.2
5 5
131 Boundary value problems in statics 5.1 The analogy between between BVPs BVPs and algebraic systems 131 5.1 The analogy and linear linear algebraic systems . . . . 131 5.1.1 A note note about about direct direct integration. 141 5.1.1 A integration . . . . . . . 141 5.2 Introduction to 144 to the the spectral spectral method; method; eigenfunctions eigenfunctions . . . . 144 5.2 Introduction 5.2.1 Eigenpairs 5.2.1 Eigenpairs of of — - -j^ under under Dirichlet Dirichlet conditions conditions.. . . . 144 144 5.2.2 Representing functions functions in terms of of eigenfunctions eigenfunctions.. . 146 146 5.2.2 Representing in terms boundary conditions; conditions; other 5.2.3 Eigenfunctions under under other other boundary 5.2.3 Eigenfunctions other Fourier series 150 Fourier series . . . . . . . 150 5.3 Solving 155 5.3 Solving the the BVP BVP using using Fourier Fourier series series 155 5.3.1 A special special case case . . . . . . 155 5.3.1 A 155 5.3.2 The general general case case . . . . . 156 5.3.2 The 156 5.3.3 Other boundary conditions 161 5.3.3 Other boundary conditions 161 5.3.4 Inhomogeneous 164 5.3.4 Inhomogeneous boundary boundary conditions conditions 164 5.3.5 Summary 166 5.3.5 Summary . . . . . . . . . . . . . . . 166 5.4 Finite element element methods methods for for BVPs BVPs . . . . . . . . 172 5.4 Finite 172 5.4.1 5.4.1 The The principle principle of of virtual virtual work work and and the the weak weak form form of of 173 aa BVP BVP . . . . . . . . . . . . . . . . . . . . . . . . . . 173 5.4.2 5.4.2 The The equivalence equivalence of of the the strong strong and and weak weak forms forms of of the the BVP 177 BVP. . . 177 5.5 The Galerkin Galerkin method method . . . . . . . . . . . . . . . . . . . . . . . . 180 180 5.5 The
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Contents
5.6 5.7 5.8
IX ix
Piecewise polynomials and the finite element method . . . 5.6.1 Examples using piecewise piecewise linear finite elements . . . 5.6.2 Inhomogeneous Dirichlet conditions .. . . . . 5.6.2 BVPs. . . . . . . . . . . . . . . . . . Green's functions for BVPs 5.7.1 The Green's function and the inverse of a differential differential operator. operator . . . . . . Suggestions for further reading. reading
188 193 197 202
207 210
6
Heat flow and diffusion 211 211 6.1 211 6.1 Fourier series methods for the heat equation 211 The homogeneous 6.1.1 homogeneous heat equation 214 6.1.2 Nondimensionalization . . . . . . 217 The inhomogeneous 220 6.1.3 inhomogeneous heat equation equation 220 6.1.4 Inhomogeneous boundary conditions 222 6.1.5 Steady-state heat flow flow and diffusion diffusion 224 6.1.6 Separation variables. . . . . . . . Separation of variables 225 6.2 Pure Neumann conditions and the Fourier cosine series 229 6.2.1 One end insulated; mixed boundary conditions . . . 229 229 231 6.2.2 6.2.2 Both ends insulated; Neumann boundary conditions 231 6.2.3 Pure Neumann conditions in a steady-state BVP . . 237 237 6.2.3 6.3 Periodic boundary conditions and the full Fourier series . . . . . 245 6.3.1 Eigenpairs of -— -j^ -/l;x under periodic boundary conditions247 conditions247 249 6.3.2 Solving the BVP using the full Fourier series . . . . 249 252 6.3.3 Solving the IBVP using the full Fourier series . . . . 252 6.4 Finite element methods for the heat equation 256 6.4.1 The method of lines for the heat equation. equation . . 260 6.5 Finite elements and Neumann conditions . . . . . . . . . . 266 6.5.1 The weak form of a BVP with Neumann conditions 266 6.5.2 6.5.2 Equivalence of the strong and weak forms of a BVP with Neumann conditions . . . . . . . . . . . . . . . 267 6.5.3 Piecewise linear finite elements with Neumann conditions. ditions . . . . . . . . . . . . . . . . . . . . . . . . . 269 6.5.4 Inhomogeneous Neumann conditions . . . . . . . . . 273 6.5.4 6.5.5 The finite element method method for an IBVP with Neumann conditions . . . . . . . . . . . . . . . . . . . . 274 6.6 Green's functions for the heat equation . . . . . . . . . . . . . . 279 6.6.1 The Green's function for for the one-dimensional heat equation under Dirichlet conditions . . . . . . . . . 280 6.6.2 Green's functions 281 6.6.2 functions under other boundary conditions. conditions . 281 6.7 Suggestions for further reading. reading . . . . . . . . . . . . . . . . . . 283
7
VVaves Waves 7.1 The homogeneous wave equation without boundaries . . . . . . 7.1 7.2 Fourier series methods for the wave equation . . . . . . . . . . . 7.2.1 Fourier series solutions of the the homogeneous wave equation. equation . . . . . . . . . . . . . . . . . . . . . . . .
285 285 285 291 291
293
x
Contents
7.2.2
7.3
7.4 7.5 88
Fourier series solutions of the inhomogeneous wave equation . . . . . . . . . . . . . . equation. 7.2.3 Other boundary conditions . . . . . . . . . . Finite element methods for the wave wave equation . . . . . . 7.3.1 The wave equation with Dirichlet conditions . . . . The wave equation under other boundary conditions 7.3.2 Point sources and resonance . . . . . . . . . . . . . . 7.4.1 The wave equation with a point source . 7.4.2 Another experiment leading to resonance reading . . . . . . . . . . . . Suggestions for further reading.
296 296 301 305 306 306 312 318 318 321 324
Problems in in multiple multiple spatial spatial dimensions dimensions 327 Problems 327 8.1 dimensions . . . . .. 327 8.1 Physical models in two or three spatial dimensions. 8.1.1 8.1.1 The divergence theorem . . . . . . . . . . . . . .. 328 for a three-dimensional domain . 330 8.1.2 The heat equation for 330 8.1.3 Boundary conditions for for the three-dimensional heat 8.1.3 equation . . . . . . . . . . . . . . . . 332 equation. 8.1.4 The heat equation in a bar . . . . . . . . . . . . . . 333 8.1.5 The heat equation in two dimensions . . . . . . . . . 334 8.1.6 equation for a three-dimensional domain. domain . 334 8.1.6 The wave equation equation in two dimensions dimensions . . . . . . . . 334 8.1.7 The wave equation 8.1.8 335 8.1.8 Equilibrium problems and and Laplace's equation equation . .. .. . . 335 8.1.9 Green's identities and the symmetry of the Laplacian 8.1.9 Laplacian 336 8.2 Fourier series on on aa rectangular rectangular domain domain . . . . 339 8.2 Fourier series 339 8.2.1 Dirichlet boundary conditions. conditions . . 339 8.2.2 Solving aa boundary boundary value 345 8.2.2 Solving value problem problem 345 8.2.3 Time-dependent problems 346 Time-dependent problems. . . . . 8.2.4 Other boundary conditions for 348 for the rectangle . . . . 348 349 8.2.5 Neumann boundary conditions . . . . . . . . 8.2.6 Dirichlet and Neumann problems for Laplace's equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 8.2.7 Fourier series methods for a rectangular box in three dimensions . . . . . . . . . . . . . . 354 8.3 Fourier series on a disk 359 Fourier series on a disk . . . . . . . . . . . . . . . . . . 8.3 359 8.3.1 The Laplacian Laplacian in polar coordinates . . . . . 360 360 8.3.2 Separation Separation of variables in polar coordinates 362 362 363 8.3.3 Bessel's equation . . . . . . . . . . . . . . . 8.3.4 Properties Properties of the Bessel functions . . . . . 366 8.3.5 The eigenfunctions eigenfunctions of the negative Laplacian on the disk . . . . . . . . . . . 368 disk 8.3.6 Solving PDEs PDEs on on aa disk disk . . . . . . . . . . . . . . 372 8.3.6 Solving 372 377 Finite elements in two dimensions . . . . . . . . . . . . . . . 8.4 377 8.4.1 The weak form of a BVP in multiple dimensions . . 377 377 378 8.4.2 Galerkin's method . . . . . . . . . . . . . . . . . 378 8.4.3 Piecewise linear finite elements in two dimensions dimensions . 379 379 8.4.4 Finite elements and Neumann conditions . . . . . 388 388
Contents
8.5 8.5 99
10 10
XI xi
8.4.5 Inhomogeneous boundary conditions 8.4.5 Inhomogeneous boundary conditions Suggestions reading . . . . . . . . . . Suggestions for for further further reading
389 389 392 392
More More about about Fourier Fourier series series 9.1 The 9.1 The complex complex Fourier Fourier series series . . . . . . . . . . . . . . . . 9.1.1 Complex 9.1.1 Complex inner inner products products . . . . . . . . . . . 9.1.2 Orthogonality 9.1.2 Orthogonality of of the the complex complex exponentials exponentials 9.1.3 Representing 9.1.3 Representing functions functions with with complex complex Fourier Fourier series. series . 9.1.4 The complex Fourier series of a real-valued function 9.1.4 The complex Fourier series of a real-valued function 9.2 Fourier series series and and the the FFT FFT . . . . . . . . . . . . . . . . . . . . . 9.2 Fourier 9.2.1 Using the 9.2.1 Using the trapezoidal trapezoidal rule rule to to estimate estimate Fourier Fourier coefcoefficients .. . '. . . . . . . . . . . . . . . . . . 9.2.2 The 9.2.2 The discrete discrete Fourier Fourier transform transform . . . . . . . . . . . . 9.2.3 A 9.2.3 A note about about using using packaged packaged FFT FFT routines routines . . . . . 9.2.4 Fast transforms transforms and and other other boundary boundary conditions; conditions; the the 9.2.4 Fast discrete discrete sine sine transform. transform . . . . . . . . . . . . . . . using the 9.2.5 Computing 9.2.5 Computing the the DST DST using the FFT FFT . . . . . . . . 9.3 Relationship of of sine sine and and cosine cosine series series to to the the full full Fourier Fourier series series . 9.3 Relationship 9.4 Pointwise 9.4 Pointwise convergence convergence of of Fourier Fourier series series . . . . . . . . . . . . . 9.4.1 Modes 9.4.1 Modes of of convergence convergence for for sequences sequences of of functions functions .. . 9.4.2 Pointwise convergence convergence of of the the complex complex Fourier Fourier series series 9.4.2 Pointwise 9.5 Uniform convergence 9.5 Uniform convergence of of Fourier Fourier series series . . . . . . 9.5.1 Rate 9.5.1 Rate of of decay decay of of Fourier Fourier coefficients coefficients 9.5.2 Uniform 9.5.2 Uniform convergence convergence . . . . . . . . . 9.5.3 A note note about about Gibbs's Gibbs's phenomenon 9.5.3 A phenomenon . 9.6 Mean-square 9.6 Mean-square convergence convergence of of Fourier Fourier series series .. . 2 9.6.1 The L2( -£, £) . . . . . . . . . 9.6.1 The space space L (-l,l) 9.6.2 Mean-square 9.6.2 Mean-square convergen·ce convergence of of Fourier Fourier series series . 9.6.3 Cauchy 9.6.3 Cauchy sequences sequences and and completeness completeness 9.7 A note about 9.7 A note about general general eigenvalue eigenvalue problems problems 9.8 Suggestions 9.8 Suggestions for for further further reading. reading . . . . . .
393 393
More More about about finite finite element element methods methods 10.1 10.1 Implementation Implementation of of finite finite element element methods methods 10.1.1 Describing 10.1.1 Describing aa triangulation triangulation ... 10.1.2 Computing 10.1.2 Computing the the stiffness stiffness matrix matrix 10.1.3 Computing the the load load vector vector 10.1.3 Computing 10.1.4 Quadrature 10.1.4 Quadrature . . . . . . . . . . . 10.2 10.2 Solving Solving sparse sparse linear linear systems systems . . . . . . . . 10.2.1 Gaussian elimination elimination for for dense dense systems systems 10.2.1 Gaussian 10.2.2 Direct 10.2.2 Direct solution solution of of banded banded systems systems . . . 10.2.3 Direct solution of general general sparse 10.2.3 Direct solution of sparse systems systems 10.2.4 Iterative 10.2.4 Iterative solution solution of of sparse sparse linear linear systems systems 10.2.5 The 10.2.5 The conjugate conjugate gradient gradient algorithm algorithm 10.2.6 Convergence of of the the CG CG algorithm algorithm 10.2.6 Convergence 10.2.7 Preconditioned 10.2.7 Preconditioned CG CG . . . . . . . . .
461 461
394 394 395 395 396 396 397 397 398 398 401 401 402 402 404 404 409 409
410 410 411 411 415 415 419 419 419 419 422 422 436 436 436 436 439 439 443 443 444 444 445 445 448 448 450 450 455 455 459 459 461 461 462 462 465 465 467 467 467 467 473 473 473 473 475 475 477 477 478 478 482 482 485 485 486 486
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Contents
10.3
10.4 10.5
An outline of the convergence theory for finite element methods 488 H1( 10.3.1 TheThe Sobolev space HJ (0) . . . .space . . . . . 489 10.3.1 Sobolev 489 10.3.2 approximation in the energy norm . 491 Best approximation 491 10.3.3 491 Approximation by piecewise polynomials 491 10.3.4 L22 estimates 492 Elliptic regularity and L 492 Finite element methods for eigenvalue problems 494 Suggestions for further reading . . . . . . . . . . 499
A A
Proof Proof of of Theorem Theorem 3.47 3.47
501 501
B B
Shifting 505 Shifting the the data data in in two two dimensions dimensions 505 B.0.1 B.O.I Inhomogeneous Dirichlet conditions on a rectangle . 505 B.0.2 Inhomogeneous Neumann conditions on a rectangle 508
C C
Solutions to to odd-numbered odd-numbered exercises exercises Solutions
515 515
Bibliography
603 603
Index
607 607
Foreword Foreword Newton and differential equaNewton and other other seventeenth-century seventeenth-century scientists scientists introduced introduced differential equations physics. The The intellectual tions to to describe describe the the fundamental fundamental laws laws of of physics. intellectual and and technological technological of implications of this innovation are enormous: the differential differential equation equation description of physics is the foundation upon which all modern quantitative science and engineering ing rests. rests. Differential Differential equations equations describe describe relations relations between physical physical quantitiesquantities— forces, masses, positions, positions, etc.-and with respect to space forces, masses, etc.—and their their rates rates of of change change with respect to space and and time, i.e. their derivatives. the implications these relations time, i.e. their derivatives. To To calculate calculate the implications of of these relations requires requires the "solution" or integration integration of the differential differential equation: since derivatives approximate the the ratios ratios of of small of differential mate small changes, changes, solution solution of differential equations equations amounts amounts to to the the summing together of many small changes-arbitrarily many, many, in principle-to comcomsumming together of many small changes—arbitrarily in principle—to pute the effect effect of of aa large large change. change. Until Until roughly the middle of of the twentieth twentieth century, this meant meant the derivation derivation of of formulas formulas expressing expressing solutions solutions of of differential differential equations equations as the elementary calculus-polynomials, as algebraic algebraic combinations combinations of of the elementary functions functions of of calculus—polynomials, trigonometric functions, the exponential, extrigonometric functions, the the logarithm logarithm and and the exponential, and and certain certain more more exotic formulas are only for problems, otic functions. functions. Such Such formulas are feasible feasible only for aa limited limited selection selection of of problems, and even then often which can only be evaluated often involve infinite sums ("series") which evaluated approximately, was (until approximately, and and whose whose evaluation evaluation was (until not not so so long long ago) ago) quite quite laborious. laborious. The has changed all that. that. Fast with inexpeninexpenThe digital digital revolution revolution has changed all Fast computation, computation, with sive performing many per second, makes sive machines machines performing many millions millions of of arithmetic arithmetic operations operations per second, makes numerical the solution practical: these numerical methods methods for for the solution of of differential differential equations equations practical: these methmethby literally literally adding ods ods give give useful useful approximate approximate solutions solutions to to differential differential equations equations by adding up up the the effects effects of many many microscopic microscopic changes changes to approximate approximate the effect effect of aa large change. change. Numerical methods have been used to Numerical methods have been used to solve solve certain certain relatively relatively "small" "small" differential differential equations the eighteenth those that occur in equations since since the eighteenth century-notably century—notably those that occur in the the computacomputation of planetary orbits and other astronomical calculations. Systematic use of the computer has vastly expanded the range of differential differential equation models that can be coaxed coaxed into into making predictions predictions of of scientific scientific and and engineering engineering significance. significance. Analytical techniques the last three centuries techniques developed developed over over the last three centuries to to solve solve special special problem problem classes classes have new significance numerical methods. methods. Optimal have also also attained attained new significance when when viewed viewed as as numerical Optimal design prediction of migration, synthesis of design of of ships ships and and aircraft, aircraft, prediction of underground underground fluid fluid migration, synthesis of new numerous other tasks all rely-sometimes tacitly—upon tacitly-upon the new drugs, drugs, and and numerous other critical critical tasks all rely—sometimes the modeling phenomena by by differential modeling of of complex complex phenomena differential equations equations and and the the computational computational approximation of of their their solutions. solutions. Just Just as significantly, mathematical mathematical advances unapproximation as significantly, advances underlying derlying computational computational techniques techniques have have fundamentally fundamentally changed changed the ways ways in in which scientists and engineers think about differential differential equations and their implications.
xiii XIII
xiv XIV
Foreword
Until recently this sea change in theory and practice has enjoyed enjoyed little reflecteaching of of differential differential equations equations in undergraduate classes classes at tion in in the the teaching tion in undergraduate at universities. universities. While mention mention of of computer computer techniques techniques began began showing up in in textbooks published or or While showing up textbooks published revised in the 1970s, the view view of of the the subject propounded by by most textbooks would would revised in the 1970s, the subject propounded most textbooks have seemed conventional in the 1920s. The book you hold in your hands, along with aa few others published published in in recent recent years, years, notably notably Gil Introduction to to with few others Gil Strang's Strang's Introduction Applied Mathematics, Mathematics, represents represents aa new new approach differential equations Applied approach to to differential equations at at the the undergraduate level. level. It presents presents computation an integral integral part part of of undergraduate computation as as an of the the study study of differential equations. differential equations. It is not so so much that computational computational exercises must be part syllabus-this text of the syllabus—this text can can be used entirely entirely without any student student involvement in computation at taught that way would miss a great deal computation at all, though aa class taught that way deal of the the possible analysis and implementation possible impact. impact. Rather, Rather, the the concepts concepts underlying underlying the the analysis and implementation of of numerical methods assume an importance equal equal to to that of solutions solutions in terms of elementary functions. In series and and elementary In fact, many of these concepts are are equally effeceffective of the the series series expansion methods as This book book expansion methods as well. well. This tive in in explaining explaining the the workings workings of devotes considerable effort effort to "classical" methods, moddevotes considerable to these these "classical" methods, side side by by side side with with modern numerical numerical approaches the finite element method). method). The ern approaches (particularly (particularly the finite element The "classical" "classical" both aa means means to understand the the essential nature of of the series expansions expansions provide provide both series to understand essential nature the phenomena modeled by the equations, physical phenomena equations, and effective effective numerical methods for those special problems to which they apply. Perhaps surprisingly, some of the the most most important important concepts concepts in in the Perhaps surprisingly, some of the modern modern viewpoint on on differential differential equations equations are are algebraic: the ideas vector, vector vector space, viewpoint algebraic: the ideas of of vector, space, linear algebra algebra are are central, in the the development development of of and other components and other components of of linear central, even even in more conventional parts of the subject such as as series solutions. The The present present book book more conventional parts of the subject such series solutions. both theory theory and and computation, computation, just just as as uses linear unifying principle principle in in both uses linear algebra algebra as as aa unifying working scientists, engineers, and and mathematicians mathematicians do. do. working scientists, engineers, This book, book, along along with with aa number number of of others like it published in in recent recent years, years, difdifThis others like it published fers textbooks on differential differential equations in fers from earlier earlier undergraduate textbooks in yet yet another another respect. Especially in in the the middle of the last century, century, mathematical mathematical instrucrespect. Especially middle years years of the last instruction tion in in American universities tended to to relegate the physical context context for differential differential equations equations and other topics to the background. The "big three" differential differential equations of of science science and and engineering—the engineering-the Laplace, Laplace, wave, wave, and and heat heat equations, equations, to to which which tions the devoted—have appeared in many many texts the bulk bulk of of this this book book is is devoted-have appeared in texts with with at at most most aa curcursory nod nod to their physical physical origins meaning in in applications. applications. In In part, part, this this trend sory to their origins and and meaning trend reflected the the development the theory theory of equations as self-contained reflected development of of the of differential differential equations as aa self-contained arena of of mathematical mathematical research. research. This has been been extremely arena This development development has extremely fruitful, fruitful, and and is the the source of many many of new ideas ideas which which underlie underlie the the effectiveness effectiveness of of indeed indeed is source of of the the new modern numerical numerical methods. methods. However, However, it it has has also also led to generations modern led to generations of of textbooks textbooks which present present differential differential equations equations as as aa self-contained subject, at at most most distantly which self-contained subject, distantly related to the other intellectual disciplines disciplines in which differential equations play playaa related to the other intellectual in which differential equations crucial role. role. The The present present text, text, in in contrast, physically and crucial contrast, includes includes physically and mathematically mathematically substantial derivations of of each differential equation, equation, often often in in several several contexts, contexts, along substantial derivations each differential along with examples and homework homework problems problems which which illustrate illustrate how how differential differential equations with examples and equations really really arise arise in in science science and and engineering. engineering. With exception of the chapter chapter on on ordinary ordinary differential differential equations equations With the the exception of aa part part of of the problemswhich begins the book, this text text concerns itself itself exclusively with linear problems—
Foreword
xv
that is, problems for which scalar multiples multiples of of solutions solutions are solutions, that is, problems for which sums sums and and scalar are also also solutions, if not of of the same problem problem then be argued if not the same then of of aa closely closely related related problem. problem. It might might be argued that is odd in aa book which otherwise breaks with with recent that such such aa conventional conventional choice choice is odd in book which otherwise breaks recent tradition tradition in several several important important respects. Many if not most most of of the important important problems which which concern concern contemporary contemporary scientists scientists and and engineers engineers are nonlinear. nonlinear. For For example, example, alflow and reaction are nonlinear, and most most all all problems problems involving involving fluid fluid flow and chemical chemical reaction are nonlinear, and these these phenomena and and their simulation and and analysis of central central concern concern to modtheir simulation analysis are are of to many many modphenomena ern engineering engineering disciplines. series solutions solutions are to linear linear problems, problems, ern disciplines. While While series are restricted restricted to numerical methods are principle just just as nonlinear as as to numerical methods are in in principle as applicable applicable to to nonlinear to linear linear probproblems in making making them them work work well). However, lems (though (though many many technical technical challenges challenges arise arise in well). However, it true that that not only are equations-Laplace, it is is also also true not only are the the classic classic linear linear differential differential equations—Laplace, wave, heat-models for generic classes phenomena (potentials, wave wave, and and heat—models for generic classes of of physical physical phenomena (potentials, wave which many many nonlinear processes belong, belong, but but also somotion, motion, and and diffusion) diffusion) to to which nonlinear processes also their their sothe building blocks of methods for nonlinear problems. problems. lutions lutions are are the building blocks of methods for solving solving complex, complex, nonlinear problems seems for Therefore Therefore the the choice choice to to concentrate concentrate on on these these linear linear problems seems very very natural natural for aa first first course course in in differential differential equations. equations. The computational the theory theory and The computational viewpoint viewpoint in in the and application application of of differential differential important to equations very important equations is is very to aa large large segment segment of of SIAM SIAM membership, membership, and and indeed indeed the forefront gratifySIAM members have SIAM members have been been in in the forefront of of its its development. development. It is is therefore therefore gratifying that SIAM playa leading role bringing this this important part of of modern modern ing that SIAM should should play a leading role in in bringing important part the classroom by making the present volume available. applied mathematics into applied mathematics into the classroom by making the present volume available. William W. Symes Symes William W. Rice University University Rice
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Preface Preface This differential equations This introductory text on on partial differential equations (PDEs) (PDEs) has has several several features that that are are not not found found in other texts texts at at this this level, level, including: in other including: features
equal emphasis emphasis on on classical classical and and modern modern techniques. techniques. •• equal the explicit explicit use use of of the the language language and and results results of of linear linear algebra. algebra. •• the examples and and exercises exercises analyzing analyzing realistic realistic experiments experiments (with (with correct correct physical physical •• examples parameters parameters and and units). units). recognition that that mathematical mathematical software software forms forms aa part part of of the the arsenal arsenal of of both both •• aa recognition students students and professional mathematicians. mathematicians. In this preface, will discuss and offer getting the In this preface, II will discuss these these features features and offer suggestions suggestions for for getting the most out of the text.
Classical and and modern modern techniques Classical techniques Undergraduate courses on PDEs tend to focus on Fourier series methods and separation of variables. These techniques are still useful after after two centuries because they apply. Howthey offer insight into those problems they offer aa great great deal deal of of insight into those problems to to which which they apply. However, the subject of PDEs much of research in in both both pure pure and and applied ever, the subject of PDEs has has driven driven much of the the research applied to some more mathematics mathematics in in the the last last century, century, and and students students ought ought to to be be exposed exposed to some more modern techniques well. modern techniques as as well. The limitation limitation of of the Fourier series series technique technique is is its restricted applicability: applicability: The the Fourier its restricted it can be used only for equations equations with constant constant coefficients coefficients and only on certain certain simple geometries. To complement the classical topic of Fourier series, I present the finite element element method, method, aa modern, and flexible to solving solving PDEs. PDEs. finite modern, powerful, powerful, and flexible approach approach to Although many include some some discussion discussion of of finite finite elements elements (or Although many introductory introductory texts texts include (or finite finite differences, a competing computational computational methodology), the modern approach tends place in the exposition. tends to to receive receive less less attention attention and and aa subordinate subordinate place in the exposition. In In this this text, II have have put put equal weight on Fourier series finite elements. text, equal weight on Fourier series and and finite elements.
Linear algebra Linear algebra Both Both linear linear and and nonlinear differential differential equations equations occur occur as as models of of physical phenomena of great importance importance in in science science and and engineering. engineering. However, However, most most introductory introductory ena of great XVII xvii
xviii
Preface
texts focus equations, and mine is no exception. are several texts focus on on linear linear equations, and mine is no exception. There There are several reasons reasons is difficult, and it makes sense sense to to begin begin why this this should be so. so. The The study why should be study of of PDEs PDEs is difficult, and it makes difficult nonlinear with the simpler linear equations equations before moving on to the more difficult nonlinear equations. Moreover, linear linear equations equations are better understood. understood. Finally, Finally, much much equations. Moreover, are much much better of of what is known about about nonlinear differential differential equations equations depends on the analysis analysis so this this material material is prerequisite for moving on on to of linear differential equations, so of linear differential equations, is prerequisite for moving to nonlinear equations. nonlinear equations. Because we focus focus on on linear linear equations, equations, linear linear algebra algebra is is extremely useful. useful. InIndeed, no discussion of Fourier series or finite element methods can be complete unless it it puts the the results results in in the the proper linear algebraic algebraic framework. framework. For For example, example, both methods produce the best approximate solution certainfinite-dimensional finite-dimensional both best approximate solution from from certain subspaces, projection theorem is therefore central to both techniques. Symsubspaces, and the projection metry is another another key feature exploited by both methods. While many texts de-emphasize the linear algebraic nature of the concepts it explicit. This decision, believe, and techniques, II have to make make it and solution solution techniques, have chosen chosen to explicit. This decision, II believe, leads more cohesive better preparation preparation for for future However, leads to to aa more cohesive course course and and aa better future study. study. However, it presents certain certain challenges. challenges. Linear Linear algebra algebra does not seem seem to to receive the attention it presents does not receive the attention it deserves in in many many engineering and science programs, and so many many students students will will it deserves engineering and science programs, and so based on this text without the "prerequisites." take a course based "prerequisites." Therefore, I present a fairly complete overview overview of the necessary material in Chapter 3, 3, Essential Essential Linear Linear Algebra. Algebra. Both faculty previewing this this text text and taking aa course it will will Both faculty previewing and students students taking course from from it soon realize that that there there is is too too much much material material in 3 to to cover thoroughly in in the soon realize in Chapter Chapter 3 cover thoroughly the couple of weeks weeks it can reasonably occupy in a semester course. From experience I that conscientious dislike moving through material material that that know that know conscientious students students dislike moving so so quickly quickly through they cannot cannot master master it. However, one one of to using using this this text text is is to to avoid avoid getting they it. However, of the the keys keys to getting bogged down in Chapter 3. Students Students should should try try to get from it the "big "big picture" picture" essential ideas: and two essential •• How Howto to compute compute aa best best approximation approximation to to aa vector vector from from aa subspace, subspace, with with and and without an basis (Section (Section 3.4) without an orthogonal orthogonal basis 3.4).. •• How How to to solve solve aa matrix-vector matrix-vector equation equation when when the the matrix matrix is is symmetric symmetric and and its its eigenvalues are known known (Section (Section 3.5). eigenvalues and and eigenvectors eigenvectors are 3.5). Having Having at at least begun to grasp these ideas, students students should should move on on to Chapter Chapter 4 even if some details are not clear. The concepts from linear algebra algebra will become much clearer clearer as they are used throughout the remainder of the text. 11 II have have taught taught this this course course several several times times using using this this approach, approach, and, and, although although students find it frustrating at the beginning, beginning, the to be be good. students often often find it frustrating at the the results results seem seem to good.
Realistic problems Realistic problems The subject of PDEs is easier to grasp if one keeps in mind certain certain standard physical experiments modeled by by the the equations equations under consideration. II have have used used these ical experiments modeled under consideration. these 1
Also, Chapter Chapter 44 is is much much easier easier going going than than Chapter Chapter 3, 3, aa welcome welcome contrast! contrast!
1 Also,
Preface
XIX xix
models to introduce introduce the equations and and to to aid aid in in understanding understanding their their solutions. solutions. The The models also is worth worth studying! models also show, show, of of course, course, that that the the subject subject of of PDEs PDEs is studying! To make the the applications applications as as meaningful meaningful as as possible, possible, II have have included included many many To make examples and exercises posed in terms of meaningful experiments with realistic physical parameters.
Software Software There be used There exists exists powerful powerful mathematical mathematical software software that that can can be used to to illuminate illuminate the the material presented in material presented in this this book. book. Computer Computer software software is useful for for at least least three three reasons: reasons:
It removes removes the the need need to to do do tedious tedious computations computations that that are are necessary necessary to to compute compute •• It solutions. Just use aa table Just as as aa calculator calculator eliminates eliminates the need to to use table and and interinterpolation to to compute logarithm, aa computer algebra system polation compute aa logarithm, computer algebra system can can eliminate eliminate the need to perform perform integration by parts the need to integration by parts several several times times in in order order to to evaluate evaluate an an integral. more mechanical mechanical obstacles more time time to integral. With With the the more obstacles removed, removed, there there is is more to focus on concepts. Problems that that simply simply cannot cannot be be solved solved (in (in aa reasonable reasonable time) time) by by hand hand can can of•• Problems often be done with the assistance of a computer. This allows for more interesting ten be done with the assistance of a computer. This allows for more interesting assignments. assignments. •• Graphical Graphical capabilities capabilities allow allow students students to to visualize visualize the the results results of of their their compucomputations, improving improving understanding tations, understanding and and interpretation. interpretation. II expect or expect students students to to use use aa software software package package such such as as MATLAB, MATLAB, Mathematica, Mathematica, or Maple to the text Maple to reproduce reproduce the the examples examples from from the text and and to to solve solve the the exercises. exercises. II prefer prefer not not to to introduce introduce aa particular particular software software package package in in the the text text itself, itself, for at reasons. The the features usage of the software can at least least two two reasons. The explanation explanation of of the features and and usage of the software can detract from from the mathematics. Also, Also, if if the the book book is is based based on software detract the mathematics. on aa particular particular software package, then then it can be difficult to different package. package. For For these it can be difficult to use use with with aa different these reason, reason, my my package, text packages except However, text does does not not mention mention any any software software packages except in in aa few few footnotes. footnotes. However, since the use of software is, in my opinion, opinion, essential for a modern course, course, I have written tutorials for MATLAB, Maple that the various various written tutorials for MATLAB, Mathematica, Mathematica, and and Maple that explain explain the capabilities of these programs that that are are relevant appear capabilities of these programs relevant to to this this book. book. These These tutorials tutorials appear CD. on on the the accompanying accompanying CD.
Outline Outline The material in in this this text text is series The core core material is found found in in Chapters Chapters 5-7, 5-7, which which present present Fourier Fourier series and techniques for most important of and finite finite element element techniques for the the three three most important differential differential equations equations of mathematical Laplace's equation, equation, the the heat equation, and and the wave equaequathe wave mathematical physics: physics: Laplace's heat equation, tion. Since the concepts themselves themselves are are hard enough, these chapters are are restricted restricted tion. Since the concepts hard enough, these chapters to problems in dimension. to problems in aa single single spatial spatial dimension. Several introductory introductory chapters chapters set set the the stage stage for for this this core. core. Chapter Chapter 11 briefly briefly Several defines the basic that will will be the text. text. Chapter defines the basic terminology terminology and and notation notation that be used used in in the Chapter
xx xx
Preface
22 then then derives derives the the standard standard differential differential equations equations in in one one spatial spatial dimension, dimension, in in the the process parameters that process explaining explaining the the meaning meaning of of various various physical parameters that appear appear in in the the the associated boundary conditions equations equations and and introducing introducing the associated boundary conditions and and initial initial conditions. conditions. Chapter Chapter 3, 3, which has has already already been been discussed discussed above, above, presents presents the the concepts concepts and and techniques techniques from from linear linear algebra algebra that will will be used used in in subsequent subsequent chapters. chapters. II want want to to reiterate reiterate that that perhaps the the most most important important key key to to using this text text effectively effectively is to move through to move through Chapter Chapter 33 expeditiously. expeditiously. The The rudimentary rudimentary understanding understanding that that students obtain obtain in in going going through through Chapter Chapter 33 will will grow grow as as the the concepts concepts are are used used in students in the the rest rest of of the book. book. Chapter presents the background material material on Chapter 44 presents the background on ordinary ordinary differential differential equations equations that much easier that is is needed needed in in later later chapters. chapters. This This chapter chapter is is much easier than than the the previous previous the material material is the last one, much of one, because because much of the is review review for for many many students. students. Only Only the last two two Although sections, methods and sections, on on numerical numerical methods and stiff stiff systems, systems, are are likely likely to to be be new. new. Although the Essential Ordinary Differential Equations, the chapter chapter is is entitled entitled Essential Ordinary Differential Equations, Section Section 4.3 4.3 is is not not formally the book. give formally prerequisite prerequisite for for the the rest rest of of the book. II included included this this material material to to give students foundation for understanding stiff students aa foundation for understanding stiff systems systems of of ODEs ODEs (particularly, (particularly, the the stiff stiff system system arising arising from from the the heat heat equation). equation). Similarly, Similarly, Runge-Kutta Runge-Kutta schemes schemes and and automatic automatic step step control control are are not not strictly strictly needed. However, However, understanding aa little little about about variable variable step step size size methods methods is is useful if if one one tries tries to apply apply an an "off-the-shelf" "off-the-shelf" routine routine to to aa stiff stiff system. system. Chapter techniques developed first part Chapter 88 extends extends the the models models and and techniques developed in in the the first part of of the the book brief discussions dimensions). book to to two two spatial spatial dimensions dimensions (with (with some some brief discussions of of three three dimensions). The more in-depth The last last two two chapters chapters provide provide aa more in-depth treatment treatment of of Fourier Fourier series series (Chapter 9) and finite elements (Chapter 10). In addition to the standard theory of (Chapter 9) and finite elements (Chapter 10). In addition to the standard theory of Fourier series, Chapter 9 shows how to use the fast Fourier transform to efficiently Fourier series, Chapter 9 shows how to use the fast Fourier transform to efficiently compute compute Fourier Fourier series series solutions solutions of of the PDEs, explains explains the relationships relationships among among the the various Fourier series, the Fourier various types types of of Fourier series, and and discusses discusses the the extent extent to to which which the Fourier series series method can be extended to complicated complicated geometries and equations with nonconstant stant coefficients. coefficients. Sections Sections 9.4-9.6 9.4-9.6 present present aa careful careful mathematical mathematical treatment treatment of of the the convergence Fourier series, have aa different flavor from convergence of of Fourier series, and and have different flavor from the the remainder remainder of of the the book. In particular, they book. In particular, they are are less less suited suited for for an an audience audience of of science science and and engineering engineering have been been included the curious students, students, and and have included as as aa reference reference for for the curious student. student. Chapter Chapter 10 10 gives gives some some advice advice on on implementing implementing finite finite element element computations, computations, discusses discusses the the solution solution of of the the resulting resulting sparse sparse linear linear systems, systems, and and briefly briefly outlines outlines the convergence methods. It use finite finite the convergence theory theory for for finite finite element element methods. It also also shows shows how how to to use elements elements to solve solve general general eigenvalue eigenvalue problems. problems. The The tutorials tutorials on on the accompanyaccompanying ing CD CD include include programs implementing implementing two-dimensional two-dimensional finite finite element element methods, as as (MATLAB, described described in in Section Section 10.1, 10.1, in in each each of of the the supported supported software software packages (MATLAB, Maple). The Mathematica, and and Maple). The sections sections on on sparse sparse systems systems and and the convergence convergence thetheboth little pointing the the students toward more ory ory are are both little more more than than outlines, outlines, pointing students toward more advanced advanced concepts. Both of these topics, topics, of justify aa dedicated concepts. Both of these of course, course, could could easily easily justify dedicated semestersemesterlong long course, course, and and II had no no intention intention of of going going into into detail. detail. II hope hope that that the the material material on on implementation implementation of of finite finite elements elements (in (in Section Section 10.1) 10.1) will will encourage encourage some some students students to to experiment experiment with with two-dimensional two-dimensional calculations, calculations, which which are are already already too too tedious tedious to to carry carry out out by hand. hand. This This sort sort of of information information seems seems to to be lacking lacking from from most most books books accessible accessible to to students students at at this this level.
Preface Preface
XXI xxi
Possible course course outlines outlines Possible In In aa one-semester one-semester course course (42-45 (42-45 class class hours), hours), II typically typically cover cover Chapters Chapters 1-7 1-7 and and part part of Chapter Chapter 8. 8. II touch touch only only lightly lightly on on the material concerning concerning Green's Green's functions functions and and of the material the Dirac delta delta function function (Sections (Sections 4.6, 4.6, 6.6, 6.6, and and 7.4.1), 7.4.1), and and sometimes sometimes omit omit Section Section the Dirac 6.3, 6.3, but but cover cover the the remainder remainder of of Chapters Chapters 1-7 1-7 carefully. carefully. If part of the material material in If an an instructor instructor wishes wishes to to cover cover aa significant significant part of the in Chapters Chapters 8-10, an an obvious obvious place place to save time time is is in in Chapter Chapter 4. suggest covering to save 4. II would would suggest covering 8-10, the needed material material on the needed on ODEs ODEs on on aa "just-in-time" "just-in-time" basis basis in in the the course course of of Chapters Chapters 5-7. This This will will definitely definitely save save time, time, since since my my presentation presentation in in Chapter Chapter 44 is is more more 5-7. detailed than is really necessary. Chapter Chapter 22 can can be be given given as as aa reading assignment, detailed than is really necessary. reading assignment, particularly for the the intended particularly for intended audience audience of of science science and and engineering engineering students, students, who who will will typically be comfortable with the physical physical parameters parameters appearing appearing in in the differential the differential typically be comfortable with the equations. equations.
Acknowledgments Acknowledgments This book book began when II was visiting Rice University in in 1998-1999 1998-1999 and and taught taught aa This began when was visiting Rice University course using lecture notes of Professor William W. W. Symes. Symes. To To satisfy satisfy my my perpercourse using the the lecture notes of Professor William sonal predilections, rewrote the the notes notes significantly, significantly, and for the the convenience convenience of and for of sonal predilections, II rewrote myself and my my students, students, II typeset them in in the the form form of of aa book, book, which which was was the myself and typeset them the first first version of of this this text. text. Although the final result bears, bears, in in some some ways, ways, little little resemblance blance to to Symes's Symes's original original notes, II am am indebted to to him for for the idea idea of of recasting recasting the the undergraduate PDE undergraduate PDE course course in in more more modern modern terms. terms. His His example example was was the the inspiration inspiration for this project, project, and benefited from his advice for this and II benefited from his advice throughout throughout the the writing writing process. process. II am am also also indebted indebted to to the the students students who who have have suffered suffered through through courses courses taught taught from early early version version of of this this text. Many of of them found errors, errors, typographical typographical and and from text. Many them found otherwise, otherwise, that that might might otherwise otherwise have have found found their their way way into into print. print. II would would like like to to thank thank Professors Professors Gino Gino Biondini, Biondini, Yuji Yuji Kodoma, Kodoma, Robert Robert Krasny, Krasny, Yuan Lou, Uhlig, all whom read read part text Yuan Lou, Fadil Fadil Santosa, Santosa, and and Paul Paul Uhlig, all of of whom part or or all all of of the the text and and offered offered helpful helpful suggestions. suggestions. The various physical physical parameters used in deThe various parameters used in the the examples examples and and exercises exercises were were derived rived (sometimes (sometimes by by interpolation) interpolation) from from tables tables in in the the CRC CRC Handbook Handbook of of Chemistry Chemistry and Physics and Physics [35]. [35). The graphs graphs in in this generated with with MATLAB. MATLAB. For For MATLAB MATLAB product product The this book book were were generated information, please contact: information, please contact: The MathWorks, Inc. The MathWorks, Inc. 33 Apple Drive Apple Hill Hill Drive Natick, MA 01760-2098 01760-2098 USA Natick, MA USA Tel: Tel: 508-647-7000 508-647-7000 Fax: 508-647-7101 508-647-7101 Fax: E-mail:
[email protected] E-mail:
[email protected] Web: www.mathworks.com Web: www.mathworks.com
As mentioned above, the CD also also supports supports the the use of Mathematica and Maple. As mentioned above, the CD use of Mathematica and Maple. For For Mathematica product product information, information, contact: Mathematica contact:
xxii
Preface
Wolfram Wolfram Research, Research, Inc. Inc. 100 Drive 100 Trade Trade Center Center Drive Champaign, USA Champaign, IL IL 61820-7237 61820-7237 USA Tel: Tel: 800-965-3726 Fax: Fax: 217-398-0747 E-mail: E-mail:
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Maple product information, information, contact: For Maple contact: Waterloo Waterloo Maple Maple Inc. Inc. 57 57 Erb Erb Street Street West West Waterloo, Waterloo, Ontario Ontario Canada Canada N2L6C2 Tel: 800-267-6583 Tel: 800-267-6583 E-mail: E-mail:
[email protected] [email protected]
Chapter Chapter 1
ification of classification of differential -al
equations
Loosely speaking, a differential differential equation is an equation specifying a relation between the derivatives of a function or between one or more derivatives and the function function itself. We We will call the function appearing in such an equation the unknown Junction. function. We use this terminology because the typical task involving involving a differential differential equation, equation, that is, and the focus of this book, is to solve the differential differential is, to find a function whose derivatives are related as specified by the differential equation. In whose differential carrying out this task, everything else about the relation, other than the unknown unknown function, is regarded as known. known. Any function satisfying satisfying the differential differential equation is called a solution. In other words, a solution of a differential solution. words, differential equation is a function function that, when substituted for unknown function, causes the equation to be satisfied. for the unknown Differential Differential equations fall into several natural and widely widely used categories: 1. versus partial: 1. Ordinary Ordinary versus partial:
(a) If the unknown unknown function has a single independent variable, say t, then the equation is an ordinary ordinary differential differential equation equation (ODE). (ODE). In this case, only "ordinary" derivatives are involved. involved. Examples of ODEs are du = 3u dt
(1.1)
and rPu a dt 2
du
+ b dt + cu =
In the second example, a, b, and fJ(t) ( t ) as a known known function uu = u(t). = u(t).
J (t).
(1.2)
and c are regarded as known known constants, constants, of t. In both equations, the unknown unknown is
(b) If the unknown unknown function function has two or more independent variables, the equation is called a partial differential differential equation equation (PDE). Examples include
{Pu 8x2
11
82 u
+ 8y2
= 0
(1.3)
Chapter 1. Classification of differential equations Chapter 1. Classification
2
and (1.4)
In (1-3), the is u = u(x,y), the unknown unknown function function is = u(x, V), while while in in (1.4), (1.4), it it is is u == In (1.3), u(x, t). In In (1.4), (1.4), cc is u(x,t). is aa known known constant. constant. 2. Order: differential equation the order of the the highest highest deriva2. Order: The The order of of aa differential equation is is the order of derivative appearing the equation. Most differential science tive appearing in in the equation. Most differential equations equations arising arising in in science and or second Example (1.1) above is and engineering engineering are are first first or second order. order. Example (1.1) above is first first order, order, are second order. while Examples Examples (1.2), while (1.2), (1.3), (1.3), and and (1.4) (1.4) are second order.
3. Linear versus versus nonlinear: 3. Linear nonlinear:
(a) As the the examples examples suggest, suggest, a differential differential equation has, has, on on each each side of the the equals sign, the unknown equals sign, algebraic algebraic expressions expressions involving involving the unknown function function and and its derivatives, derivatives, and and possibly other functions functions and and constants constants regarded regarded as known. A A differential terms involving the unknown. differential equation equation is is linear if if those those terms involving the ununknown known function function contain products of known contain only only products of aa single factor of of the the unknown function or with other (known) constants constants or funcfunction or one one of of its its derivatives derivatives with other (known) or functions of the the tions of the independent independent variables. variables.22 In In linear linear differential differential equations, equations, the unknown function and its appear raised raised to to aa power power unknown function and its derivatives derivatives do do not appear other 1, or argument of of aa nonlinear nonlinear function function (like (like sin, sin, exp, exp, than 1, or as as the the argument other than log, etc.). log, etc.). For second-order ODE the form form For example, example, the the general general linear linear second-order ODE has has the
require that that a a2(t) in order that the the equation truly be be of of Here we Here we require ) 00 in order that equation truly 2 ( t f:. second order. (1.2) is special case case of this general general equation. equation. is aa special of this second order. Example Example (1.2) In linear differential differential equation, unknown or its derivatives be In aa linear equation, the the unknown or its derivatives can can be multiplied functions of the independent variable, multiplied by constants constants or by functions the independent variable, but but not by not by functions functions of of the the unknown. unknown. As another the general general linear second-order PDE in two linear second-order PDE in two indeindeAs another example, example, the pendent variables variables is is pendent
a 2u a 2u a 2u au (x, y) ax 2 +a12(x, y) axay +a22 (x, y) ay2 au au + al (x, y) ax +a2(x, y) ay
+ ao(x, y)u = f(x, V)·
Examples (1.3) (although the the independent independent Examples (1.3) and and (1.4) (1.4) are are of of this this form form (although variables are are called called x and (1.4), not and V). y). Not an, a12, a12, and tt in in (1.4), not x and Not all all of of au, variables and be zero in order order for this equation to be be second and a22 a22 can can be zero in for this equation to second order. order. A linear homogeneous if the zero function is A linear differential differential equation equation is is homogeneous if the zero function is aa solution. For For example, = 00 satisfies satisfies (1.3), (1.3), == 00 satisfies satisfies (1.1), (1.1), u(x,y) u(x,y) == solution. example, u(t) = -::-----2
2In will give precise definition In Section Section 3.1, 3.1, we we will give aa more more precise definition of of linearity. linearity.
Chapter 1. Classification differential equations equations Classification of differential
3
and u(x, t) = == 00 satisfies so these these are are examples of homogeneous homogeneous and u(x,t) satisfies (1.4), (1.4), so examples of linear differential equations. A homogeneous linear equation has another linear differential equations. A homogeneous linear equation has another important property: whenever whenever u and of the the equation, so important property: and v are are solutions solutions of equation, so is is au cm + + (3v ftv for for all real numbers a and (3. ft. For example, suppose suppose u(t) u(t) and and vet) v(t) are are solutions solutions of of (1.1) (1.1) (that (that is, is, du/dt du/dt — = 3u 3w and and dv/dt dv/dt = 3i>), 3v), and and w == au + (3v. ftv. Then Then dw
ill =
d
dt [au + (3v] du dv =a dt +(3dt = a3u + (33v = 3(au + (3v) =3w.
Thus w is is also also aa solution solution of Thus of (1.1). (1.1). A linear is not not homogeneous is called A linear differential differential equation equation which which is homogeneous is called inhoinhoFor example mogeneous. mogeneous. For example
~~
=
3u + sin (21ft)
(1.5)
is inhomogeneous, since not satisfy equation. is linear linear inhomogeneous, since u(i) u(t) = == 00 does does not satisfy this this equation. Example (1.2) might be be homogeneous homogeneous or or not, not, depending depending on on whether Example (1.2) might whether /(*) = not. J(t) == 00 or or not. It is is always always possible to group group all terms involving involving the the unknown unknown function It possible to all terms function all terms terms not not in aa linear linear differential on the the left-hand left-hand side in differential equation equation on side and and all involving the the unknown unknown function the right-hand right-hand side. For example, involving function on on the side. For example, (1.5) (1.5) is equivalent to
~~ -
3u = sin (21ft).
(1.6)
"Equivalent (1.5) and "Equivalent to" in this context means means that (1.5) and (1.6) (1.6) have exactly exactly the same solutions: if the if u(t) u(t) solves one one of these two equations, equations, it it solves the other other as as well. well. When When the the equation written this this way, way, with with all all of of the equation is is written the terms terms involving the unknown unknown on the left left and and those those not not involving involving the the involving the on the the unknown on the right, right, homogeneity homogeneity is to recognize: the equation is unknown on the is easy easy to recognize: the equation is right-hand side homogeneous if if and and only only if if the the right-hand side is identically zero. (b) Differential Differential equations which are not linear are termed nonlinear. nonlinear. For example, example, (1.7)
is nonlinear. This is is clear, clear, as as the u appears appears raised to is nonlinear. This the unknown unknown function function u raised to the second power. Another Another way way to to see that (1.7) is as as follows: the second power. see that (1.7) is is nonlinear nonlinear is follows: if the equation equation were were linear, linear homogeneous, since the linear, it it would would be be linear homogeneous, since the zero zero if the
4 4
Chapter 1. Classification equations Classification of differential equations
function is is aa solution. solution. However, However, suppose suppose that that u(t) u(t) and and v(t) v(t} are are nonzero nonzero function solutions, solutions, so so that that rPu dt 2
+u
2
rP
2
rPv = 0, dt 2
+V
2
= 0,
and w = u + v. Then rPw dt 2
+ w = dt 2
[u
+ v] + (u + v)
+
d2 v dt 2
rPu = dt 2
= ( rPu dt 2
2
2
+ u + 2uv + v
2v 2) (d + u + dt + 2
V
2
2) + 2uv
= 2uv.
Thus w w does does not not satisfy satisfy the equation, so so the the equation equation must must be nonlinear. Thus the equation, be nonlinear. Both Both linear linear and and nonlinear nonlinear differential differential equations equations occur occur as as models models of of physical However, cal phenomena phenomena of of great great importance importance in in science science and and engineering. engineering. However, linear of linear differential differential equations equations are are much much better better understood, understood, and and most most of what what is is known known about about nonlinear differential differential equations equations depends depends on on the the analysis of linear linear differential equations. equations. This This book will focus focus almost almost exclusively clusively on on the the solution solution of of linear linear differential differential equations. equations. 4. Constant versus versus nonconstant coefficient: Linear differential equations equations 4. Constant nonconstant coefficient: Linear differential like like
have constant coefficients coefficients if if the the coefficients coefficients ra, c, and and kk are are constants constants rather rather m, c, have constant than (nonconstant) functions functions of of the the independent independent variable. The right-hand right-hand side side than (nonconstant) variable. The ff(t) ( t ) may depend on the independent variable; it is only the quantities quantities which multiply the the unknown unknown function function and and its its derivatives derivatives which which must must be be constant, constant, multiply in order for for the equation to have constant constant coefficients. Some techniques that are effective effective for for constant-coefficient constant-coefficient problems problems are are very very difficult difficult to to apply apply to to are problems problems with with nonconstant nonconstant coefficients. coefficients. As will explicitly As aa convention, convention, we we will explicitly write write the the independent independent variable variable when when aa coefcoefficient ficient is is aa function function rather rather than than aa constant. constant. The The only only function function in in aa differential differential equation will omit the unknown unknown equation for for which which we we will omit the the independent independent variable variable is is the function. Thus Thus function.
d[
dU]
- - k(x)-
dx
dx
= f(x)
is is an an ODE ODE with with aa nonconstant nonconstant coefficient, coefficient, namely namely k(x). k(x).
We will will only only apply apply the "constant coefficients" coefficients" to to linear linear differential We the phrase phrase "constant differential equations. equations.
5
Chapter 1. Classification Classification of differential differential equations
5. versus system 5. Scalar Scalar equation equation versus system of of equations: equations: A A single single differential differential equaequation unknown function tion in in one one unknown function will will be be referred referred to to as as aa scalar scalar equation equation (all (all of of the the examples we have seen to this point have been scalar equations). A system of examples we have seen to this point have been scalar equations). A system of differential equations consists of several equations for one or more unknown differential equations consists of several equations for one or more unknown functions. constant-coefficient functions. Here Here is is aa system system of of three three first-order first-order linear, linear, constant-coefficient ODEs X3(t): ODEs for for three three unknown unknown functions functions Xl x i ((t), t ) , XX2(t), 2 ( t ) , and and xs(t):
dXI
dt = 2XI -
X2
+ X3,
dX2 dt
= Xl + X2,
dX3 dt
= -Xl + X2 -
X3·
It It is is important important to to realize realize that, since since aa differential differential equation equation is is an an equation equation involving the equation values of indeinvolving functions, functions, aa solution solution will will satisfy satisfy the equation for for all all values of the the independent variable(s), variable(s), or or at at least least all all values values in in some some restricted restricted domain. domain. The The equation equation pendent is is to to be be interpreted interpreted as as an an identity, such such as as the the familiar familiar trigonometric trigonometric identity cos 2 (t)
+ sin2 (t)
= 1.
(1.8)
To to say To say say that that (1.8) (1.8) is is an an identity identity is is to say that that it it is is satisfied satisfied for for all all values values of of the the 3t independent variable t. Similarly, u(t) = is (1.1) because because this this independent variable Similarly, u(t) — e3t is aa solution solution of of (1.1) function uu satisfies satisfies function du
dt (t) = 3u(t)
for for all values values of of t.t. The will note between the uses of many terms The careful careful reader reader will note the the close close analogy analogy between the uses of many terms introduced "nonlinintroduced in in this this section section ("unknown ("unknown function," function," "solution," "solution," "linear" "linear" vs. vs. "nonlinear") the uses ear" ) and and the uses of of similar similar terms terms in in discussing discussing algebraic algebraic equations. equations. The The analogy analogy between between linear linear differential differential equations equations and and linear linear algebraic algebraic systems systems is is an an important important theme theme in in this this book.
Exercises 1. the following the categories 1. ClaSSify Classify each each of of the following differential differential equations equations according according to to the categories described described in in this this chapter chapter (ODE (ODE or or PDE, PDE, linear linear or or nonlinear, nonlinear, etc.): etc.):
(a) dx dt (b)
0
au - -a2 u = (1 + t) sm. (x) at ax2
-
(c)
+ tx =
6
Chapter 1. Classification Classification of differential equations 2. Repeat Repeat Exercise 1 for for the following equations: equations:
(a) av av -+3- =x+3t at ax (b) dx
1
dt
t
- + -x =
0
(c) au _ ~ [2U au] = 0 at ax ax 3. Repeat Repeat Exercise the following 3. Exercise 11 for for the following equations: equations:
(a)
d2(} dt 2
.
+ sm ((})
= 0
(b)
au au - 2 - =l+x at ax
-
(c)
4. Repeat Repeat Exercise 1 for for the following equations: equations: (a)
d2 x dx --2-+tx=O 2 dt dt (b)
av - -a ( ~(x)av) p(x)c(x)at ax ax
= 0
(c) d2x dt 2
dx
+ 2 dt + 3x =
sin (nt)
5. Determine whether each of the functions below is a solution of the the functions the corresponddifferential equation ing differential equation in Exercise 1:
(a) x(t) = e!t (b) u(x,t) = tsin(x)
(c) w(x, t)
=
t(l - x)
6. Determine Determine whether whether each each of of the solution of of the the correspondcorrespond6. the functions functions below below is is aa solution ing differential equation ing differential equation in in Exercise Exercise 2: 2:
Chapter 1. Classification of differential equations
7
(a) v(x, t) = tx (b) x(t) =
t
(c) u(x, t) = x 2 t 7. Find that u(t) t sin (t) is is aa solution solution of the ODE ODE 7. Find aa function function ff(t) ( t ) so so that u(t) = tsm of the du 1 - - -u = f(t) t > O. dt t '
Is there only such function f? Why Why or why not? Is there only one one such function /? or why not?
8. Is Is there there aa constant constant f/ such such that that u(t) u(t] = = ee1t is of the is aa solution solution of the ODE ODE 8. d2 u dt 2
-
du 2 dt
+ 3u = f?
9. Suppose u is aa nonzero equation. 9. Suppose u is nonzero solution solution of of aa linear, linear, homogeneous homogeneous differential differential equation. What is is another nonzero solution? solution? What another nonzero solution of of 10. Suppose u is 10. Suppose is aa solution ~u
a(t) dt 2
du
+ b(t) dt + c(t)u = f(t)
and v is (nonzero) solution of and is aa (nonzero) solution of
Explain how how to to produce produce infinitely infinitely many many different of (1.9). (1.9). Explain different solutions solutions of
(1.9)
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Chapter 2
Modelass
in one one dimension in dimension
In this chapter, present several In this chapter, we we present several different different and and interesting interesting physical physical processes processes that that can be modeled by ODEs or PDEs. For now we restrict ourselves to phenomena that can be modeled by ODEs or PDEs. For now we restrict ourselves to phenomena that can dimension: can be be described described (at (at least least approximately) approximately) as as occurring occurring in in aa single single spatial spatial dimension: heat flow or mechanical mechanical vibration vibration in in aa long, long, thin thin bar, bar, vibration of aa string, string, diffusion diffusion flow or vibration of heat of chemicals chemicals in in aa pipe, and so so forth. forth. In In Chapters Chapters 5-7, 5-7, we will learn learn methods methods for we will for of pipe, and solving solving the the resulting resulting differential differential equations. equations. In Chapter Chapter 8, 8, we we consider consider similar similar experiments experiments occurring occurring in in multiple multiple spatial spatial In dimensions. dimensions.
2.1
Heat flow in law in a bar; Fourier's law
We by considering heat energy temperWe begin begin by considering the the distribution distribution of of heat energy (or, (or, equivalently, equivalently, of of temperthe cross-sections bar are ature) ature) in in aa long, long, thin thin bar. bar. We We assume assume that that the cross-sections of of the the bar are uniform, uniform, and temperature varies the longitudinal and that that the the temperature varies only only in in the longitudinal direction. direction. In In particular, particular, we assume assume that that the bar is is perfectly perfectly insulated, insulated, except except possibly at the the ends, ends, so so that that we the bar possibly at no we no heat heat escapes escapes through through the the side. side. By By making making several several simplifying simplifying assumptions, assumptions, we derive whose solution of derive aa linear linear PDE PDE whose solution is is the the temperature temperature in in the the bar bar as as aa function function of spatial position position and and time. time. spatial We when we We show, show, when we treat treat multiple multiple space space dimensions dimensions in in Chapter Chapter 8, 8, that that if if the the initial temperature is constant constant in in each each cross-section, cross-section, and and if if any any heat heat source source depends depends initial temperature is only on on the the longitudinal longitudinal coordinate, coordinate, then then all all subsequent subsequent temperature temperature distributions distributions only depend only only on on the the longitudinal longitudinal coordinate. coordinate. Therefore, Therefore, in in this this regard, regard, there there is depend is no modeling modeling error error in in adopting adopting aa one-dimensional one-dimensional model. model. (There (There is is modeling modeling error no error make.) associated associated with with some some of of the the other other assumptions assumptions we we make.) We will will now now derive derive the the model, model, which which is is usually usually called called the the heat equation. We We heat equation. We begin by defining aa coordinate coordinate system system (see (see Figure Figure 2.1). 2.1). The The variable variable xx denotes denotes begin by defining position in in the the bar bar in in the longitudinal direction, direction, and and we we assume assume that that one one end end of of the the the longitudinal position bar is The length bar will will be be f, the other £. We We bar is at at x = O. 0. The length of of the the bar £, so so the other end end is is at at x = t. denote by by A the the area area of of aa cross-section cross-section of of the bar. denote the bar. The temperature temperature of of the the bar bar is is determined determined by by the the amount amount of of heat energy in The heat energy in 99
bar; Fourier's Fourier's law 2.1. Heat flow in a bar;
11 11
The fact fact that that we we do do not not know know EQ Eo causes causes no difficulty, because because we we are are going to model The no difficulty, going to model the in the the energy energy (and hence temperature). temperature). the change change in (and hence Specifically, we examine the rate rate of change, with with respect to time, of the the heat Specifically, we examine the of change, respect to time, of heat energy in in the of the between xx and and xx + Ax. The total heat in [x, xx + Ax] energy the part part of the bar bar between ~x. The total heat in [x, ~xl is by (2.2), (2.2), so so its its rate of change is is given given by rate of change is
d dt
r+tl.x ix Apeu(s,t)ds
(the constant EQ Eo differentiates to zero). work with this expression, we need need the (the constant differentiates to zero). To To work with this expression, we the following which allows us to move the the derivative derivative past past the the integral sign: following result, result, which allows us to move integral sign:
Theorem 2.1. Let (a,b) x [c, —>• R and aF/ax dF/dx be be continuous, continuous, and define Theorem 2.1. Let F F :: (a, b) x [e, d\ d] -+ Rand and define (a, 6) ->• R R by by ¢(f>: : (a, b) -+ ¢(x)
=
ld
F(x, y) dy.
Then is continuously and Then ¢0 is continuously differentiable, differentiate, and d¢ dx (x) = That That is, is, d
dx
ld c
ld c
aF ax (x, y) dy.
F(x,y)dy=
ld c
aF ax (x,y)dy.
By Theorem Theorem 2.1, By 2.1, we we have have d dt
r+tl.x r+tl.x au ix Apcu(s, t) ds = ix Ape at (s, t) ds.
(2.3)
If we assume that there there are no internal or sinks heat, the the heat If we assume that are no internal sources sources or sinks of of heat, heat contained [x, x + Ax] ~xl can can change because heat heat flows flows through through the the crosscrosscontained in in [x, change only only because sections x and and x x + ~x. rate at which heat heat flows flows through through the the cross-section sections at at x Ax. The The rate at which cross-section at is called and is denoted q(x,t). q(x, t). It has units units of per unit at xx is called the the heat heat flux, flux, and is denoted It has of energy energy per unit area per unit time, and if heat heat energy energy is is flowing flowing in in the the positive area per unit time, and is is aa signed signed quantity: quantity: if positive x-direction, then qq is x-direction, then is positive. positive. The net heat heat entering [x, x + ~xl at time t, through through the the two two cross-sections, is The net entering [x, + Ax] at time cross-sections, is
A(q(x, t) - q(x
+ ~x, t))
= -
r+tl.x aq ix A ax (s, t) ds;
(2.4)
the last the fundamental theorem of (See Figure the last equation equation follows follows from from the fundamental theorem of calculus. calculus. (See Figure 2.2.) Equating yields 2.2.) Equating (2.3) (2.3) and and (2.4) (2.4) yields
r+tl.x au r+tl.x aq ix Ape at (s,t)ds = - ix Aax(s,t)ds
12
Chapter 2. Models in one dimension
or or
l
x
X
dX
+
{Ape au at (8, t)
q
a (8, t) } + A ax
d8 = O.
Since this all x and Since this holds holds for for all and Ax, ~x, it it follows follows that that the the integrand integrand must must be be zero: zero: au Ape at (x, t)
aq
+ A ax (x, t) = 0,
0
< x < c.
(2.5)
(The key point point here here is is that that the the integral integral above zero over over every small interval. interval. (The key above equals equals zero every small the integrand is itself zero. zero. If the integrand were positive, positive, This is only possible if the say, at some some point point in in [0, C), and and were were continuous, then the the integral integral over over aa small say, at [0,^], continuous, then small interval positive.) interval around that point would be positive.)
Figure 2.2. 2.2. The heat flux into and out of a part of of the bar. bar. Figure flow from regions of higher temperature to regions of of Naturally, heat will flow lower temperature; note that the sign of au/ax du/dx indicates whether temperature is increasing decreasing as as x increases, and hence hence indicates indicates the the direction direction of of heat increasing or or decreasing increases, and heat flow. We We now now make make aa simplifying assumption, which which is is called called Fourier's Fourier's law of of heat flow. simplifying assumption, conduction: the conduction: the heat heat flux flux q q is is proportional proportional to to the the temperature temperature gradient, gradient, du/dx. au/ax. We We the magnitude of the the constant constant of proportionality the the thermal conductivity K, r., call the and obtain the equation and obtain the equation au (2.6) q(x, t) = -r. ax (x, t) (the the units of the the heat flux and (the units of r.K can can be determined from the and those of the the temperature gradient; gradient; see Exercise 2.1.1). 2.1.1). The negative sign sign is is necessary necessary so so that that temperature see Exercise The negative heat flows flows from from hot hot regions regions to to cold. cold. Substituting Fourier's law into the the differential heat Substituting Fourier's law into differential we can eliminate eliminate q q and find a PDE PDE for u: u: equation (2.5), we equation
au a2u pc at - r. ax 2
= 0, 0 < x < C,
(2.7)
for all t.
(We common factor of A) A.) Here we have assumed that r. (We have canceled aa common K is constant, constant, the bar bar were homogeneous. homogeneous. It which would be true if the It is possible that r. K depends on x (in (in which which case case p and and ce probably probably do do as as well); well); then then we we obtain obtain
p(x)c(x)
~~ -
:x (r.(x)
~~)
= 0, 0 < x < C, for all
t.
We call call (2.7) (2.7) the the heat equation. equation. We If the bar bar contains internal sources (or sinks) heat (such (such as as chemical chemical reacreacIf the contains internal sources (or sinks) of of heat we collect all such sources into a single source function tions that produce heat), we
2.1.
Heat Heat flow flow in in a bar; bar; Fourier's Fourier's law law
13
/(x,t) (in units heat energy time per volume). Then the total heat f(x, t) (in units of of heat energy per per unit unit time per unit unit volume). Then the total heat added to + ~xl Ax] during time interval interval [t, t + At] to [x, x + during the the time ~tl is is added
l
t+.t:.t
it
x+.t:.x
Af(s, r) ds dr.
x
The time rate of change of this contribution to the total heat energy (at time t) is
l
x+.t:.x
x
(2.8)
Af(s, t) ds.
The rate change of of total total heat in [x, [#, x + ~xl Ax] is is now given by by the the sum sum of of (2.4) and The rate of of change heat in now given (2.4) and (2.8), so so we we obtain, obtain, by above reasoning, reasoning, the inhomogeneous33 heat equation by the the above the inhomogeneous heat equation (2.8), au pc at
2.1.1 2.1.1
-
a2 u "'ax 2 = f(x,t), 0< x < £, for all t.
(2.9)
Boundary and and initial initial conditions conditions for for the the heat heat equation Boundary equation
The equation by itself is not aa complete complete model model of heat flow. We must The heat heat equation by itself is not of heat flow. We must know know flows through the ends of the bar, and we we must know the temperature heat flows how heat distribution distribution in the bar at some initial initial time. flow in a bar correspond to perfect Two possible boundary conditions for for heat flow perfectly insulated, insulation and perfect thermal contact. contact. If the ends of the bar are perfectly so we have so that the heat flux across across the ends ends is is zero, zero, then then we have -K,
au ax (0, t)
= 0,
-K,
au ax (£, t)
=
°
for all t
(no heat flows flows into the left left end or out of the right end). On the other hand, ifif the ends of zero44 (through (through perfect perfect thermal the ends of the the bar bar are are kept kept fixed fixed at at temperature temperature zero thermal contact with an ice ice bath, bath, for for instance), instance), we obtain contact with an we obtain
u(0,t) =u(l,t] for all alH. u(O, t) = u(£, t) = 00 for t. Either of boundary conditions Either of these boundary conditions can be inhomogeneous (that (that is, is, have aa nonzero right-hand side), and we could, could, of course, have mixed conditions (one end insulated, the other other held at fixed temperature). the held at fixed temperature). A A boundary boundary condition condition that that specifies specifies the the value value of of the solution solution is is called called aa Dirichlet condition, while a condition derivative is called condition that specifies the value of the derivative called a Neumann condition. Neumann condition. A A problem with a Dirichlet condition condition at one one end end and and a Neuboundary conditions. conditions. As noted mann condition at at the the other is said said to to have mixed mixed boundary noted 3
The term term homogeneous homogeneous is is used used in in two two completely completely different different ways ways in in this this section section and and throughout 3The throughout the A material in the the the book. book. A material can can be be (physically) (physically) homogeneous, homogeneous, which which implies implies that that the the coefficients coefficients in differential equations equations will will be constants. On other hand, linear differential differential equation can be differential be constants. On the the other hand, aa linear equation can be is zero. zero. These uses of of (mathematically) homogeneous, which (mathematically) homogeneous, which means means that that the the right-hand right-hand side side is These two two uses the homogeneous are are unrelated confusing, but the usage is standard standard and and so the word word homogeneous unrelated and and potentially potentially confusing, but the usage is so the reader understand from from context context the the sense sense in which the the word word is used. the reader must must understand in which is used. 4 by an 4Since Since changing changing uu by an additive additive constant constant does does not not affect affect the the differential differential equation, equation, we we can can as as well use Celsius as as the the temperature rather than than Kelvin. Kelvin. (See well use Celsius temperature scale scale rather (See Exercise Exercise 6.) 6.)
14
Chapter Chapter 2. Models in one dimension
above, above, we we will will also also use use the the terms terms homogeneous homogeneous and and inhomogeneous inhomogeneous to to refer refer to to the the boundary conditions. The condition u(£) u(i] == 10 10 is called an inhomogeneous inhomogeneous Dirichlet condition, for for example. example. condition, To completely completely determine determine the temperature as function of of time time and space, we we To the temperature as aa function and space, must the temperature temperature distribution distribution at at some some initial initial time time to. to. This This is is an an initial must know know the initial rnnAH-.i.nn.' condition: u(x, to) = 1jJ(x), 0 < x < £. Putting condition, Putting together together the the PDE, PDE, the the boundary boundary conditions, conditions, and and the the initial initial condition, we obtain an initial-boundary value value problem (IBVP) for for the the heat heat equation. equation. For we obtain an initial-boundary problem (IBVP) For example, with ends of of the the bar held at at fixed we have have the the IBVP IBVP example, with both both ends bar held fixed temperatures, temperatures, we
au a2 u pc at - '" ax 2 = f(x, t), 0 < x < £, t > to, u(x, to) = 1jJ(x), 0 < x < £, u(O, t) = 0, t > to, u(£, t) = 0, t > to.
2.1.2
(2.10)
Steady-state heat flow Steady-state flow
An special case case modeled modeled by by the equation is is steady-state An important important special the heat heat equation steady-state heat heat flow—a flow--a situation in in which is constant constant with to time time (although (although not situation which the the temperature temperature is with respect respect to not necessarily with respect to this case, be necessarily with respect to space). space). In In this case, the the temperature temperature function function uu can can be thought of partial derivative thought of as as aa function function of of xx alone, alone, uu = — u(x), u(x), the the partial derivative with with respect respect to to tt is is zero, zero, and and the the differential differential equation equation becomes
J2u
-'" dx 2 = f(x), 0 < x < £. In the steady-state case, case, any any source source term must be of time In the steady-state term /f must be independent independent of time (otherwise, (otherwise, the equation equation could could not not possibly possibly be satisfied by by aa function function uu that is independent independent of the be satisfied that is of time). Boundary conditions have have the the same same meaning meaning as as in in the the time-dependent time-dependent case, case, time). Boundary conditions although the boundary although in in the the case case of of inhomogeneous inhomogeneous boundary boundary conditions, conditions, the boundary data data must must be constant. constant. On On the the other other hand, hand, it it obviously obviously does does not not make sense sense to to impose impose an temperature distribution. an initial initial condition condition on on aa steady-state steady-state temperature distribution. Collecting these these observations, observations, we see that that aa steady-state steady-state heat we see heat flow flow problem problem Collecting takes the form of of aa boundary boundary value value problem (BVP). For For example, example, if if the temperature takes the form problem (BVP). the temperature is fixed fixed at the two endpoints of the bar, we we have the following following Dirichlet problem: problem:
J2u
-'" dx 2 = f(x), 0 < x < £, u(O) = 0, u(£) = O.
(2.11)
We remark We remark that, that, when when only only one one spatial spatial dimension dimension is is taken taken into into account, account, aa steadysteadystate (or (or equilibrium) equilibrium] problem in an an ODE ODE rather rather than than aa PDE. PDE. Moreover, state problem results results in Moreover, these problems, problems, at in their be solved two intethese at least least in their simplest simplest form, form, can can be solved directly directly by by two integrations. Nevertheless, Nevertheless, we will (in Chapter 5) devote a significant significant amount of effort effort
2.1. bar; Fourier's law 2.1. Heat flow in a bar;
15
toward developing developing methods for solving solving these ODEs, since since the techniques can can be genthe techniques be gentoward methods for these ODEs, eralized multiple spatial eralized to to multiple spatial dimensions, dimensions, and and also also form form the the foundation foundation for for techniques techniques for for solving solving time-dependent time-dependent problems. problems. Example Example 2.2. 2.2. The The thermal thermal conductivity conductivity of of an an aluminum aluminum alloy alloy is is 1.5 1.5 W/(cmK). W/(cmK). We steady-state temperature We will will calculate calculate the the steady-state temperature of of an an aluminum aluminum bar bar of of length length 11 m m (insulated along along the the sides) with its its left left end end fixed fixed at at 20 20 degrees degrees Celsius Celsius and and its its right (insulated sides) with right end for the steady-state temperature, end fixed fixed at at 30 30 degrees. degrees. If If we we write write uu = u(x) u(x] for the steady-state temperature, then then satisfies the the BVP BVP uu satisfies d2 u -1.5 dx 2 = 0, 0
< x < 100,
u(O) = 20, u(100)
= 30.
(We so as (We changed changed the the units units of of the the length length of of the the bar bar to to centimeters, centimeters, so as to to be be consistent consistent with the the units units of of the the thermal thermal conductivity.) conductivity.) The The differential differential equation equation implies implies that that with 22 2 d?u/dx is zero, so, integrating once, du/dx is constant, say d u/dx is zero, so, integrating once, du/dx is constant, say du dx(x)=C1 ,0<x<100. Integrating Integrating aa second second time time yields yields
The second boundary The boundary boundary condition condition u(O) w(0) == 20 20 implies implies that that C C?2 == 20, 20, and and the the second boundary condition then then yields yields condition 1 100C1 + 20 = 30 ~ C1 = 10. Thus Thus
x u(x) = 20 + 10.
The the bar bar does The reader reader should should notice notice that that the the thermal thermal conductivity conductivity of of the does not not affect previous example. because the affect the the solution solution in in the the previous example. This This is is because the bar bar is is homogeneous homogeneous (that the bar bar is constant throughout). throughout). This (that is, is, the the thermal thermal conductivity conductivity of of the is constant This should should be contrasted be contrasted with with the the next next example. example.
Example A metal so that Example 2.3. 2.3. A metal bar bar of of length length 100 100 cm cm is is manufactured manufactured so that its its thermal thermal conductivity conductivity is is given given by by the the formula formula
II:(X) = 1.4 + 0.0025x, 0 < x < 100 (the units units of of lI:(x) K(X] are are W/(cmK)). W/(cmK)). We We will will find find the the steady-state temperature of of the the (the steady-state temperature bar under under the the same conditions as as in in the the previous example. That That is, is, we we will will solve bar same conditions previous example. solve
16
Chapter 2. Models in one dimension
the the BVP BVP -
d~
((1.4 + 0.0025x)
:~)
= 0, 0
< x < 100,
u(O) = 20, u(100)
= 30.
Integrating both sides of the the differential differential equation equation once once yields yields Integrating both sides of du
(1.4 + 0.0025x) dx (x) = 0, 0 < x < 100, where 0C is is aa constant. constant. Equivalently, we have have where Equivalently, we
0 du dx (x) = 1.4 + 0.0025x' 0
< x < 100.
Integrating second time time yields yields Integrating aa second u(x) = u(O) +
= 20 +
r
10
r
10
0
1.4 + 0.0025s ds
4000 ds 560 + s
= 20 + 4000 In ( 1 + 5:0). Applying the we obtain obtain Applying the boundary boundary condition condition u(100) w(100) = = 30, 30, we
0-
10
- 400 In (~~)"
With With this this value value for for 0, C, we we have have
_ 101n(1+~) u (x ) - 20+ (33) In
28
The previous examples in The steady-state steady-state temperatures temperatures from from the the two two previous examples are are shown shown in Figure heterogeneity of Figure 2.3, 2.3, which which shows shows that that the the heterogeneity of the the second second bar bar makes makes aa small small but but discernible temperature distribution. both cases, is discernible difference difference in in the the temperature distribution. In In both cases, heat heat energy energy is flowing right end bar has has aa flowing from from the the right end of of the the bar bar to to the the left, left, and and the the heterogeneous heterogeneous bar the right to aa higher higher temperature higher higher thermal thermal conductivity conductivity at at the right end. end. This This leads leads to temperature in the the middle middle of of the the bar. bar. in
2.1.3 2.1.3
Diffusion Diffusion
One the facts mathematics so physical pheOne of of the facts that that makes makes mathematics so useful useful is is that that different different physical phenomena can can lead lead to to essentially essentially the the same same mathematical mathematical model. In this this section section we nomena model. In we introduce another another experiment experiment that is modeled modeled by by the heat equation. equation. Of Of course, course, the the introduce that is the heat
17 17
2.1. Heat flow in a bar; bar; Fourier's law 30~--------------~------~--------------~
29
28 27 ~26 ::J
coQ5 25 a. E
224 23
constant K - _. nonconstant K
22 21
20
60
40
80
100
x
Figure steady-state temperature from Example Figure 2.3. 2.3. The The steady-state temperature from Example 2.2 2.2 (solid (solid line) line) and curve). and Example Example 2.3 2.3 (dashed (dashed curve).
meaning the quantities be different meaning of of the quantities appearing appearing in in the the equation equation will will be different than than in in the the case heat flow. flow. case of of heat We We suppose suppose that that aa certain certain chemical chemical is is in in solution solution in in aa straight straight pipe pipe having having aa uniform cross-section. has length uniform cross-section. We We assume assume that that the the pipe pipe has length I!, t, and and we we establish establish aa coordinate system system as as in in the the preceding section, with with one one end end of of the pipe at at xx = 00 preceding section, the pipe coordinate and the the other other at at xx == I!. 1. Assuming Assuming the the concentration concentration varies varies only only in in the ^-direction, and the x-direction, we can define u(x, u(x, t) to be the concentration, in units of mass per volume, volume, of the the chemical at at time time tt in in the the cross-section cross-section located located at at x. Then the the total total mass mass of of the the x. Then chemical chemical between xx and is chemical in in the the part part of of the the bar bar between and xx + ~x Aar (at (at time time t) is
l x
x+~x
Au(s, t) ds,
(2.12)
where A A is the cross-sectional where is the cross-sectional area area of of the the pipe. pipe. The The chemical chemical will will tend tend to to diffuse diffuse from regions regions of of high high concentration concentration to to regions regions of of low low concentration concentration (just (just as as heat heat from flows from regions to to cooler We will will assume that the the rate of energy energy flows from hot hot regions cooler regions). regions). We assume55 that rate of diffusion diffusion is is proportional proportional to to the the concentration concentration gradient gradient
5 5This law This assumption, assumption, which which is is analogous analogous to to Fourier's Fourier's law law of of heat heat conduction, conduction, is is called called Fick's Pick's law in in the the diffusion diffusion setting. setting.
18
Chapter 2. 2. Models in one Chapter Models in one dimension dimension
the meaning of of this assumption is is that there exists the meaning this assumption that there exists aa constant constant66 D > 00 such such that, that, at at time t, the chemical moves across the cross-section at x at a rate of of
au
-D ax (x, t).
The units ofthis (for example, g/cm22s). Since of this mass flux are mass per area per time (for 44 the units of au/ax du/dx are are mass/length coefficient D must must have the units of mass/length ,, the the diffusion diffusion coefficient have unit unit of of area per time time (for area per (for example, example, cm cm22/s). /s). With this this assumption, the equation diffusion is as With assumption, the equation modeling modeling diffusion is derived derived exactly exactly as was the heat equation, with a similar result. The total amount of the chemical contained in part of of the the pipe between x and and x + is given given by by (2.12), contained in the the part pipe between + 6.x Ax is (2.12), so so the the rate at which this total mass is is changing is
1
a [r+!::..x r+!::..x au at ix Au(s, t) ds = ix A at (s, t) ds. This same quantity can be computed fluxes at the two at xx This same quantity can be computed from from the the fluxes at the two cross-sections cross-sections at and of and x + 6.x. Ax. At At the left, mass mass is is entering at aa rate rate of
au -AD ax (x, t), of while at the right it enters at a rate of
au AD ax (x
+ 6.x, t).
We therefore therefore have
r+!::..x
ix
au au A 8t (s, t) ds = -AD 8x (x, t)
{x+!::..x
=
ix
au
+ AD 8x (x + 6.x, t)
82u AD 8x 2 (s, t) ds.
The The result result is is the the diffusion diffusion equation: equation: 2
8u = D 8 u2 0 < X < e, t > to. 8t 8x ' If accounted for for by a If the chemical chemical is added to the interior interior of the pipe, this can be accounted function /(x, t), where mass is added to the part of the and x + Ax f(x, t), where mass is added to the part of the pipe pipe between between x and 6.x function at aa rate rate of at of
r+!::..x
ix
Af(s, t) ds
(mass per unit time—/ time-f has units of mass per volume per time). We We then obtain obtain the inhomogeneous the inhomogeneous diffusion diffusion equation: equation:
8u 82 u 8t - D 8x 2 = f(x, t), 0 < x < e, t> to· 6 6This This constant constant varies varies with with temperature temperature and and pressure; pressure; see see the the CRC CRC Handbook Handbook of of Chemistry Chemistry and page 6-179. 6-179. Physics °hysics [35], [35], page
2.1.
Heat flow in a bar; Fourier's law
19
Just flow, we we can Just as as in in the the case case of of heat heat flow, can consider consider steady-state steady-state diffusion. diffusion. The The result result is is the the ODE ODE d2u -D dx 2 = f(x), 0 < x < f, with boundary conditions. with appropriate appropriate boundary conditions. Boundary Boundary conditions conditions for for the the diffusion diffusion equaequation the exercises. exercises. tion are are explored explored in in the
Exercises 1. 1. Determine Determine the the units units of of the the thermal thermal conductivity conductivity /'i,K from from (2.6). (2.6). 2. Handbook of 2. In In the the CRC CRC Handbook of Chemistry Chemistry and and Physics Physics [35], [35], there there is is aa table table labeled labeled "Heat "Heat Capacity Capacity of of Selected Selected Solids," Solids," which which "gives "gives the the molar molar heat heat capacity capacity at at constant temperature in in constant pressure pressure of of representative representative metals metals ... ... as as aa function function of of temperature the iron the range range 200 200 to to 600 600 K" K" ([35], ([35], page page 12-190). 12-190). For For example, example, the the entry entry for for iron is is as as follows: follows: Temp. Temp. (K) cc (J(mole· (J/mole • K) K)
200 21.59
250 23.74
300 25.15
350 26.28
400 27.39
500 29.70
600 32.05
table indicates, the specific As this As this table indicates, the specific heat heat of of aa material material depends depends on on its its temperatemperature. How dependence ture. How would would the the heat heat equation equation change change if if we we did did not not ignore ignore the the dependence of the the specific specific heat heat on on temperature? temperature? of 3. energy. 3. Verify Verify that that the the integral integral in in (2.1) (2.1) has has units units of of energy. 4. as 4. Suppose Suppose uu represents represents the the temperature temperature distribution distribution in in aa homogeneous homogeneous bar, bar, as discussed discussed in in this this section, section, and and assume assume that that both both ends ends of of the the bar bar are are perfectly perfectly insulated. insulated. (a) this situation? (a) What What is is the the IBVP IBVP modeling modeling this situation? (b) heat energy (b) Show Show (mathematically) (mathematically) that that the the total total heat energy in in the the bar bar is is constant constant with of with respect respect to to time. time. (Of (Of course, course, this this is is obvious obvious from from aa physical physical point point of view. view. The The fact fact that that the the mathematical mathematical model model implies implies that that the the total total heat heat energy completely energy is is constant constant is is one one confirmation confirmation that that the the model model is is not not completely divorced reality.) divorced from from reality.) 5. means of heat energy one 5. Suppose Suppose we we have have aa means of "pumping" "pumping" heat energy into into aa bar bar through through one of per second of the the ends. ends. If If we we add add rr Joules Joules per second through through the the end end at at xx == f, I, what what would the corresponding corresponding boundary boundary condition condition be? would the be? was 6. 6. In In our our derivation derivation of of the the heat heat equation, equation, we we assumed assumed that that temperature temperature was measured measured on on the the Kelvin Kelvin scale. scale. Explain Explain what what changes changes must must be be made made (in (in the the PDE, use degrees PDE, initial initial condition, condition, or or boundary boundary conditions) conditions) to to use degrees Celsius Celsius instead. instead.
7. The The thermal thermal conductivity conductivity of of iron iron is is 0.802 0.802 W W/(cm Consider an an iron iron bar bar of 7. (( cm K). Consider of length 11 m with the and length m and and radius radius 11 cm, cm, with the lateral lateral side side completely completely insulated, insulated, and assume that the the temperature temperature of the bar bar is fixed at 20 degrees assume that of one one end end of of the is held held fixed at 20 degrees Celsius, the other degrees. Celsius, while while the the temperature temperature of of the other end end is is held held fixed fixed at at 30 30 degrees.
20
Chapter 2. Models in one dimension dimension
heat energy is added to or removed from the interior of the Assume that no heat bar. bar. (a) What What is is the (steady-state) (steady-state) temperature temperature distribution distribution in the bar? flowing through the bar? (b) At what rate is heat energy flowing 8. (a) Show that the function function
u(x, t)
= e-~IPt/(pc) sin (Ox)
is is aa solution solution to to the homogeneous heat equation equation
au - aax2u = 0,
pc at
/'i,
2
0<x
< l, for all t.
(b) What What values of 0 will cause uu to to also also satisfy satisfy homogeneous homogeneous Dirichlet Dirichlet con(b) values of will cause conditions at x = f? = 0 and x = = tl 9. we consider 9. In In this exercise, exercise, we consider aa boundary boundary condition condition for for aa bar that that may be be more realistic than a simple Dirichlet Dirichlet condition. Assume that, as usual, the side of a bar is is completely insulated insulated and that the ends are placed in a bath maintained maintained at constant flows out of or into the ends constant temperature. Assume that the heat heat flows in accordance accordance with Newton's law of cooling: the heat flux is proportional proportional to to the difference difference in temperature temperature between the end of the bar and the surrounding medium. resulting boundary medium. What What are are the the resulting boundary conditions? conditions? 10. Derive the heat equation from Newton's law of cooling (cf. (cf. the previous exercise) of cise) as as follows: follows: Divide Divide the bar bar into into aa large large number number nn of of equal equal pieces, each each of length Llx. Approximate the temperature in the ith piece as a function Ui(t) Ax. temperature «th Ui(t) (thus (thus assuming assuming that that the the temperature temperature in in each each piece piece is is constant constant at at each each point point in time). Write down a coupled system of ODEs for Ul (t), U2 (t), ... Unn(t) ui(t), u^t], • • • ,,u by applying applying Newton's law of cooling cooling to each piece and its nearest neighbors. p, c, Assume that the bar is homogeneous, so that the material properties p, c, and /'i, constant. Take the the limit as Llx -+ 0, and and show that the the result is the K are constant. Aa; —> the heat equation. equation. heat 11. Suppose a chemical is diffusing diffusing in a pipe, and both ends of the pipe are sealed. sealed. What are are the the appropriate appropriate boundary boundary conditions conditions for for the diffusion equation? equation? What the diffusion What initial initial conditions conditions are are required? required? Write Write down down aa complete complete IBVP IBVP for for the the What diffusion diffusion equation equation under these conditions. conditions.
12. Suppose Suppose that chemical contained contained in in aa pipe pipe of of length length lI has has an an initial initial concenconcen12. that aa chemical tration 0) = At time time zero, the ends pipe are tration distribution distribution of of u(x, w(or,0) = 'IjJ(x). ip(x}. At zero, the ends of of the the pipe are sealed, and no mass is added to or removed from the interior of the pipe. (a) Write down the IBVP describing the diffusion diffusion of the chemical. (b) Show mathematically mathematically that the total mass of the chemical in the pipe is constant. (Derive (Derive this fact from the equations rather rather than from common sense.)
2.2. The hanging bar
21 21
(c) Describe ultimate steady-state steady-state concentration. concentration. (c) Describe the the ultimate (d) the steady-state steady-state concentration in terms terms of (d) Give Give aa formula formula for for the concentration in of 'IjJ. if). 13. 13. Suppose Suppose aa pipe pipe of of length length tf and and radius radius rr joins joins two two large large reservoirs, reservoirs, each each concontaining aa (well-mixed) the same the concentration taining (well-mixed) solution solution of of the same chemical. chemical. Let Let the concentration in one reservoir MO and in the the other ui, and assume that w(0,£) = MO in one reservoir be be Uo and in other be be Uf, and assume that u(O, t) = Uo and u(f, and u(l,t)t) = = Uf. ut. (a) the steady-state steady-state rate rate at at which the chemical chemical diffuses diffuses through (a) Find Find the which the through the the pipe (in of mass mass per per time). time). pipe (in units units of (b) does this this rate rate vary vary with the length length fi and (b) How How does with the and the the radius radius r? r?
14. Consider previous exercise, exercise, in chemical is monoxide 14. Consider the the previous in which which the the chemical is carbon carbon monoxide (CO) and the the solution is CO 0.01 and (CO) and solution is CO in in air. air. Suppose Suppose that that Uo MO = = 0.01 and Uf ui = = 0.015, 0.015, and diffusion coefficient coefficient of of CO CO in in air air is 0.208 cm /s. If If the and that that the the diffusion is 0.208 cm22Is. the bar bar is is 11 m m long and and its its radius is 22cm, cm, find find the the steady-state rate at at which which CO CO diffuses long radius is steady-state rate diffuses through the the pipe pipe (in units of per time). time). through (in units of mass mass per 15. Verify that that Theorem 2.1 holds d) = and F defined 15. Verify Theorem 2.1 holds for for (a, b) 6) xx (c, (c, d) = (0,1) (0,1) xx [0,1] [0,1] and defined F(x, y) = (xy). by F(x,y) = cos cos(xy}.
2.2 2.2
The hanging bar The hanging bar
Suppose that bar, with cross-sectional area i, hangs vertiSuppose that aa bar, with uniform uniform cross-sectional area A and and length length f, hangs vertically, and it stretches stretches due force (perhaps acting upon assume cally, and it due to to aa force (perhaps gravity) gravity) acting upon it. it. We We assume that occurs only only in in the vertical direction; direction; this this assumption assumption is is reasonreasonthat the the deformation deformation occurs the vertical able only if the the bar is long long and materials tend to contract bar is and thin. thin. Normal Normal materials tend to contract horizontally horizontally able only if when they are stretched both this contraction and coupling bewhen they are stretched vertically, vertically, but but both this contraction and the the coupling between horizontal small compared compared to to the the elongation elongation when tween horizontal and and vertical vertical motion motion are are small when the bar bar is Segel [36], [36], Chapter Chapter 12). 12). the is thin thin (see (see Lin Lin and and Segel With the the assumption assumption of of purely purely vertical vertical deformation, deformation, we we can can describe moveWith describe the the movement of the bar bar in of aa displacement function u(x, t). Specifically, suppose ment of the in terms terms of displacement function u(x,t). Specifically, suppose that the of the at xx = = 0, that the top top of the bar bar is is fixed fixed at 0, and and let let down down be be the the positive positive x-direction. x-direction. Let the cross-section cross-section of the bar bar originally move to to xx + u(x, at time time tt (see (see Let the of the originally at at xx move u(x,t)t) at Figure 2.4). 2.4). We PDE describing describing the dynamics of of the Figure We will will derive derive aa PDE the dynamics the bar bar by by applying applying Newton's second second law law of motion. Newton's of motion. We assume that the the bar bar is elastic, which which means that the the internal in the We assume that is elastic, means that internal forces forces in the bar depend on the local local relative relative of change in The deformed bar depend on the of change in length. length. The deformed length length of of the the part part of of the the bar bar originally originally between between xx and and xx + Ax 6.x (at (at time time t} t) is is (x
+ 6.x + U (x + 6.x, t))
- (x
+ u(x, t))
= 6.x
+ u(x + 6.x, t)
Since the original length is is Ax, 6.x, the the change change in length of of this this part part is is Since the original length in length u(x
+ 6.x, t) - u(x, t),
and the in length length is and the relative relative change change in is u(x+6.x,t) -u(x,t) ~ ou( ) 6.x - ax x, t .
- u(x, t).
22 22
Chapter 2.
dimension Models in one dimension
Figure 2.4. hanging bar and its Figure 2.4. The hanging its coordinate system. system.
This explains explains the as the the This the definition definition of of the the strain, strain, the the local local relative relative change change in in length, length, as dimensionless dimensionless quantity quantity au ax (x, t). As we we noted noted above, above, to to say that the the bar bar is is elastic is to to say that the the internal As say that elastic is say that internal restoring force of of the depends only only on strain. We the the deformed deformed bar bar depends on the the strain. We now now make make the restoring force further are small, internal further assumption assumption that that the the deformations deformations are small, and and therefore therefore that that the the internal forces are, are, to strain. This is equivalent to forces to good good approximation, approximation, proportional proportional to to the the strain. This is equivalent to assuming that is linearly linearly elastic, Hookean. assuming that the the bar bar is elastic, or or Hookean. Under assumption that is Hookean, expression for Under the the assumption that the the bar bar is Hookean, we we can can write write an an expression for the force acting on P, and xx + Ax. the total total internal internal force acting on P, the the part part of of the the bar bar between between xx and ~x. We We denote of proportionality proportionality denote by by k(x) k(x) the the stiffness stiffness of of the the bar bar at at x, x, that that is, is, the the constant constant of in with units units of offorce per unit unit area. the engineering engineering literature, literature, kk is is in Hooke's Hooke's law, law, with force per area. (In (In the called Young's modulus elasticity. Values materials called the the Young's modulus or or the the modulus modulus of of elasticity. Values for for various various materials can be found in as [35].) [35].) Then Then the can be found in reference reference books, books, such such as the total total internal internal force force is is
au + ~x, t) -
Ak(x + ~x) ax (x
au
Ak(x) ax (x, t).
(2.13)
The in this expression is force exerted The first first term term in this expression is the the force exerted by by the the part part of of the the bar bar below below xx + Ax above ~x on on P, P, and and the the second second term term is is the the force force exerted exerted by by the the part part of of the the bar bar above the strains positive, then bar has been x on on P. P. The The signs signs are are correct; correct; if if the strains are are positive, then the the bar has been stretched, and the is pulling the internal internal restoring restoring force force is pulling down down (the (the positive positive direction) direction) stretched, and at x + Ax and up at x+ ~x and up at at x. x.
2.2. The hanging hanging bar
23
Now, (2.13) is equal (by the fundamental theorem of calculus) to r+l:J.x a ( au ) ix A ax k(s) ax (s, t) ds.
We now assume that all external forces are lumped into a body force given by a force body force density f/ (which has units offorce of force per unit volume). Then the total external force on P (at time t) t] is {x+l:J.x ix f(s, t)Ads, and the sum of the forces acting on part P is {x+l:J.x a ( au ) ix A ax k(s) ax (s, t) ds
r+l:J.x
+ ix
Af(s, t) ds.
Newton's second law states that the total force acting on P must equal the mass mass of of P times times its its acceleration. acceleration. This law law takes takes the the form form x x r+l:J.x a ( au ) +l:J.X +l:J.X a2u ix A ax k(s) ax (s, t) ds + x Af(s, t) ds = x Ap(s) at 2 (s, t) ds,
l
l
where p(x) p(x) is the density of the bar at x (in units of mass per volume). We can rewrite this as x 2 +l:J.X [ aatu (s, t) - ax a ( k(s) a ut)) - f(s, t) ] ds = 0 x p(s) ax (s, 2
l
(note how the factor of A cancels). This integral must be zero for £) for every xx E e [0, [0,£) and every ~x Ax > O. 0. It follows (by the reasoning introduced on page 12) 12) that the integrand integrand must be identically identically zero; this gives the equation 2
a u p(x) at 2
-
a ( au) ax k(x) ax - f(x, t)
= 0,
or (2.14) The PDE equation. PDE (2.14) is called the wave wave equation. If t, and If the bar is in equilibrium, equilibrium, then the displacement does not depend on t, we u(x). In u/at22 is we can can write write uu = u(x]. In this case, case, the the acceleration acceleration ad22 u/dt is zero, zero, and and the the forcing forcing function f/ must must also also be of time. We then obtain the the following following ODE function be independent independent of time. We then obtain ODE for the equilibrium displacement of the bar: for the equilibrium displacement of the bar:
dU) = f(x).
d ( k(x)-dx dx
(2.15)
This is the same equation that governs flow! Just as in the case governs steady-state steady-state heat flow! of steady-state flow, the resulting BVPs can be solved with two integrations steady-state heat flow, integrations (see Examples Examples 2.2 and 2.3).
24 24
Chapter 2. Models in one dimension
If the bar bar is is homogeneous, homogeneous, so that pp and constants, these If the so that and k are are constants, these last last two two differential equations can be written as differential equations can be written as
and and
J2u
-k dx 2 = I(x),
respectively. respectively.
2.2.1 2.2.1
Boundary conditions conditions for the hanging bar Boundary bar
Equation (2.15) unique displacement; Equation (2.15) by by itself itself does does not not determine determine aa unique displacement; we we need need boundboundary conditions, well as as initial conditions if the problem problem is ary conditions, as as well initial conditions if the is time-dependent. time-dependent. The The statement of the the problem problem explicitly gives us us one boundary condition: condition: u(O) statement of explicitly gives one boundary u(0) == 00 (the (the top end of of the bar cannot cannot move). move). Moreover, Moreover, we second boundary top end the bar we can can deduce deduce aa second boundary condition from force force balance at the end of of the bar. If If the of the bar the other other end the bar. the bottom bottom of the bar condition from balance at is unsupported, then no contact contact force force applied applied at at xx = t. On the other hand, is unsupported, then there there is is no = e. On the other hand, the analysis that part of bar above above xx — t (which (which is the analysis that led led to to (2.13) (2.13) shows shows that that the the part of the the bar = e is all of the exerts an an internal internal force force of all of the bar) bar) exerts of
on the t. Since Since there is nothing balance this this force, have on the surface surface at at xx = = e. there is nothing to to balance force, we we must must have -Ak(f) ~~ (e) = 0,
or or simply simply
~~(f)=O. Since the the wave equation involves time derivative derivative of of u, Since wave equation involves the the second second time u, we we need need two two initial conditions to uniquely uniquely determine determine the the motion the bar: bar: the the initial initial displaceinitial conditions to motion of of the displacethe initial velocity. We We thus thus arrive the following IBVP for the wave wave ment and and the ment initial velocity. arrive at at the following IBVP for the equation: equation: 2
p(x) aatu2
-
a ( k(x) au) ax ax = I(x, t), 0
< x < e, t> to,
u(x, to) = 'I/J(x), 0 < x au at (x, to) = "Y(x), 0 < x
u(O, t) = 0, t> to, au ax (e, t) = 0, t > to·
< e,
< e,
(2.16)
2.2. The The hanging hanging bar bar 2.2.
25 25
The corresponding corresponding steady-state steady-state BVP BVP (expressing (expressing mechanical mechanical equilibrium) equilibrium) is The is
d (k(x) dx dU)
- dx
= f(x),
0< x < C, (2.17)
u(O) = 0,
~~(C) = o. There are several other other sets sets of of boundary boundary conditions interest There are several conditions that that might might be be of of interest in connection connection with with the or (2.15). For example, if both in the differential differential equations equations (2.14) or For example, if both ends of bar are are fixed (not allowed allowed to to move), move), we we have have the the boundary boundary conditions conditions ends of the the bar fixed (not
= 0,
U(O)
u(C)
=0
°
(recall that that u u is displacement, so so the condition u(C) u(i] = = 0 indicates indicates that cross(recall is the the displacement, the condition that the the crosssection at at the the end end of of the the bar corresponding to to xx == C t does not move move from from its original bar corresponding does not its original section position). If both the bar boundary conditions position). both ends ends of of the bar are are free, free, the the corresponding corresponding boundary conditions are are du du dx (0) = 0, dx (C) = o. Any of of the the above can be be inhomogeneous. inhomogeneous. For Any above boundary boundary conditions conditions can For example, example, fix one one end end of of the the bar bar at and stretch stretch the the other other to to xx = + A^. I::!.L we could could fix we at xx == 00 and =C t + This the boundary boundary conditions tJ..£. As As This experiment experiment corresponds corresponds to to the conditions u(O) w(0) = = 0, 0, u(C) u(t] = A£ another one end the bar bar (say another example, example, if if one end of of the (say xx = = 0) is is fixed fixed and and aa force force F is is applied applied to the the other the applied determines the to other end end (x == C), £), then then the applied force force determines the value value of of du/dx(C). du/dx(i}. Indeed, as as indicated indicated above, above, the restoring force force of the bar on the t cross-section cross-section Indeed, the restoring of the bar on the xx = C is is
-Ak(C) ~~ (C),
and this this must force F: F: and must balance balance the the applied applied force du -Ak(C) dx (C)
+F
= O.
This to the the boundary boundary condition This leads leads to condition
and the quantity / A has has units pressure (force per unit For mathematical and the quantity F F/A units of of pressure (force per unit area). area). For mathematical purposes, it to write an inhomogeneous boundary condition condition of this type purposes, it is is simplest simplest to write an inhomogeneous boundary of this type as as
:~ (C) = c,
but for practical problem, problem, it essential to but for solving solving aa practical it is is essential to recognize recognize that that F
C
= Ak(C)"
26
Chapter 2. Models Models in one dimension dimension
Exercises 1. the following experiment: A A bar is hanging with the the top fixed 1. Consider Consider the following experiment: bar is hanging with top end end fixed at bar is is stretched by aa pressure pressure (force per unit at xx == 0, 0, and and the the bar stretched by (force per unit area) area) pp applied uniformly free (bottom) (bottom) end. What are are the the boundary applied uniformly to to the the free end. What boundary conditions conditions describing this situation? describing this situation? 2. Suppose that aa homogeneous homogeneous bar bar (that constant stiffness stiffness k) of 2. Suppose that (that is, is, aa bar bar with with constant A;) of length top end fixed at the bar bar is stretched to to aa length length £i has has its its top end fixed at xx = 0, 0, and and the is stretched length £1+A£. + Il£ by pressure pp applied bottom end. to be the positive positive by aa pressure applied to to the the bottom end. Take Take down down to be the x-direction. x-direction. (a) Explain Explain why why pp and and Il£ have the the same (a) A^ have same sign. sign. (b) Explain why why p and and \t Il£ cannot cannot both both be be chosen chosen arbitrarily arbitrarily (even (even subject subject (b) Explain to the the requirement requirement that that they they have have the the same both physical physical to same sign). sign). Give Give both and mathematical reasons. and mathematical reasons. (c) is specified. Find Il£ terms of of p, p, k, (c) Suppose Suppose pp is specified. Find A^ (in (in terms k, and and C). i). (d) is specified. specified. Find terms of Il£, fc, k, and (d) Suppose Suppose Il£ Al is Find pp (in (in terms of A£, and C). t}. 3. A certain certain type type of of stainless stiffness of of 195 (A Pascal (Pa) is 3. A stainless steel steel has has aa stiffness 195 GPa. GPa. (A (Pa) is the standard unit of pressure, or unit area. area. The Pascal is is aa derived the standard unit of pressure, or force force per per unit The Pascal derived unit: one Pascal Pascal equals equals one square meter. meter. The the unit: one one Newton Newton per per square The Newton Newton is is the standard standard unit of force: force: one one Newton equals equals one kilogram kilogram meter per secondsecondsquared. squared. Finally, Finally, GPa GPa is is short short for for gigaPascal, or 10 1099 Pascals.) Pascals.) (a) Explain Explain in words (including units) what what aa stiffness (a) in words (including units) stiffness of of 195 195 GPa GPa means. means.
(b) Suppose Suppose aa pressure pressure of of 11 GPa GPa is is applied applied to the the end end of aa homogeneous, bar of of this this stainless steel, and and that the other is circular cylindrical cylindrical bar circular stainless steel, that the other end end is fixed. If the original original length the bar bar is m and and its its radius radius is is 11 cm, cm, what fixed. If the length of of the is 11 m what will its length length be, be, in the equilibrium after the pressure has has been been will its in the equilibrium state, state, after the pressure applied? applied? (c) the result result of of 3b by formulating the boundary boundary value value (c) Verify Verify the 3b by formulating and and solving solving the problem representing this experiment. problem representing this experiment. 4. Consider circular cylindrical cylindrical bar, length 11 m m and cm, made made from 4. Consider aa circular bar, of of length and radius radius 11 cm, from an aluminum alloy with stiffness 70 GPa. If the top end of the bar (x = 0) an aluminum alloy with stiffness 70 GPa. If the top end of the bar 0) is fixed, what what total total force must be to the the other other end end (x = = 1) is fixed, force must be applied applied to 1) to to stretch stretch the of 1.01 1.01 m? the bar bar to to aa length length of m? 5. Write the the wave wave equation equation for the bar bar of that the the density of 5. Write for the of Exercise Exercise 3, 3, given given that density of the stainless steel steel is is 7.9 7.9 g/cm g/cm33.. (Warning: Use Use consistent units!) What What must must the units the forcing be? Verify Verify that that the two terms terms on the left the units of of the forcing function function f/ be? the two on the left side the same units as side of of the the differential differential equation equation have have the same units as f./. 6. that aIm bar of the stainless stainless steel in Exercise Exercise 3, with 6. Suppose Suppose that a i m bar of the steel described described in 3, with density 7.9 g/ cm33,, is the bottom bottom but but free the top. top. Let Let the density 7.9g/cm is supported supported at at the free at at the the cross-sectional area of the the bar bar be be 0.1 m22 .• A A weight weight of placed on on cross-sectional area of O.lm of 1000 1000 kg kg is is placed
2.3. The wave wave equation for a vibrating string
27
gravitational constant top of the bar, exerting pressure on it via gravity (the gravitational is is 9.8m/s 9.8m/s22 ). ). The purpose of this problem is to compute and compare the effects effects on the bar of the mass maBS on the top and the weight of the bar itself.
(a) Write down three BVPs: (which means that i. First, take into account the weight of the bar (which gravity induces aa body body force), force), but but ignore ignore the the mass maBS on on the top (so gravity induces the top (so the the top end end of of the the bar bar is boundary condition condition is is aa homogeneous homogeneous top is free-the free—the boundary Neumann condition). condition). Neumann ii. Next, ignore the the effect effect of ii. Next, take take into into account account the the mass mass on on the the top, top, but but ignore of the weight of the bar (so there is no body force). LaBt, take take both both effects into account. iii. iii. Last, effects into account. 1.
(b) Explain why why the the third BVP can can be be solved by solving solving the the first first two two and (b) Explain third BVP solved by and adding the the results. results. adding
(c) Compare the (c) Solve Solve the the first first two two BVPs BVPs by by direct direct integration. integration. Compare the two two disdisplacements. Which is more significant, significant, the weight of the bar bar or the mass maBS on top? (d) How would would the the situation change if the the cross-sectional cross-sectional area area of of the the bar bar were were (d) How situation change changed to 0.2 0.2m changed to m 22 ?? 7. (a) Show 7. (a) Show that that the function function
u(x, t) = cos (c(}t) sin ((}x)
is solution to equation is aa solution to the the homogeneous homogeneous wave wave equation
(b) What values of 9() will cause u to also satisfy satisfy homogeneous Dirichlet conditions =0 and xx = = tl ditions at at xx = 0 and £?
2.3 2.3
The for a a vibrating string The wave wave equation equation for vibrating string
We now now present present an an argument argument that that the wave equation equation (2.14) the small small We the wave (2.14) also also describes describes the aB a guitar string). In the course transverse vibrations of an elastic string (such as unjustified assumptions which which of the following derivation, we make several aa priori unjustified are significant enough that the end result ought to be viewed viewed with some skepticism. careful analysis However, aa careful analysis leads leads to to the same model (see (see the the article article by Antman [1]). [1]). For simplicity, we For simplicity, we will will assume assume that that the the string string in in question question is is homogeneous, homogeneous, so We suppose so that any any material properties are constant constant throughout the string. string. We that string is is stretched stretched to its two endpoints are are not that the the string to length length £I and and that that its two endpoints not allowed allowed move. We We further suppose that the string occupying to move. string vibrates in in the xy-plane, xy-plane, occupying the interval interval [0, £] on on the the x-axis x-axis when when at rest, and that the the point point at at (x,O) in the the [0,1] at rest, and that (x,0) in the
28
Chapter 2. Models in one dimension
reference configuration moves to (x, u(x, t)) at time t. We We are thus postulating that configuration moves the motion motion of of the the string string is entirely entirely in the the transverse direction direction (this (this is is one one of of the the severe we mentioned previous paragraph). Granted severe assumptions assumptions that that we mentioned in in the the previous Granted this this assumption, assumption, we we now derive derive the differential differential equation equation satisfied satisfied by the the displacement displacement u. Since, by by assumption, assumption, aa string string does does not not resist internal restoring Since, resist bending, bending, the the internal restoring We force of the string under tension is tangent tangent to the string itself at every point. We magnitude ofthis will denote denote the magnitude of this restoring restoring force force by T(x, T(x, t). In In Figure 2.5, 2.5, we display a part of the deformed string, corresponding corresponding to the part of the string between x and x + 6x Ax in reference configuration, configuration, together together with the internal internal forces forces at at the the and in the the reference with the ends of this part, and their magnitudes. In the absence of any external forces, the sum of these internal forces must balance the mass times acceleration of this part of the string. To write down these equations, we we must decompose the internal force into its horizontal and vertical components.
Figure Figure 2.5. 2.5. A part of of the the deformed deformed string.
We write write n = n(x, t) for for the the force force at at the the left left endpoint and and 0 9 for for the the angle this this force vector makes with the horizontal. We We then have
ni + n~ =
T(x, t)2,
with with nl = -T(x, t) cos (0), n2 = -T(x, t) sin (0). Assuming that at every point, and and noting noting that that Assuming that lou/oxl \du/dx «
au
= 8x(x,t),
aa reasonable is reasonable approximation approximation is cos (0) == 1, sin (0) == tan (0) =
~~ (x, t).
2.3. The The wave a vibrating string 2.3. wave equation equation for for a vibrating string
We then then obtain n(x, t)
=== (
29
-T(x, t), -T(x, t) ~~ (x, t)) .
Similarly, the force at at the the right right end end of of the the string string under consideration is the part part of of the under consideration is Similarly, the force n(x + Llx, t)
===
(T(X
+ Llx, t), T(x + Llx, t) ~~ (x + Llx, t)) .
We then see that that We then see T(x
+ Llx, t) -
T(x, t)
is (an component of the total on the the part is (an approximation approximation to) to) the the horizontal horizontal component of the total force force on part of of the string, string, and the and T(x
au
+ Llx, t) ax (x + Llx, t) -
au T(x, t) ax (x, t)
is to) the the vertical vertical component. component. is (an (an approximation approximation to) By assumption, the horizontal component of the acceleration of of the string is the acceleration the string is By assumption, the horizontal component of zero, Newton's second law yields yields zero, and and so so Newton's second law T(x
+ Llx, t) -
T(x, t) = 0;
thus tension is constant throughout string at at each in time. will thus the the tension is constant throughout the the string each point point in time. We We will assume that that the length of of the string never never changes from its its length length in the assume the length the string changes much much from in the reference configuration (which (which is is f), £), and and therefore therefore that that it it makes sense to makes sense to assume assume reference configuration that T is is independent independent of of t as as well, that that is, is, that that T is is aa constant. constant. Applying Applying Newton's component yields yields second second law law to to the the vertical vertical component
t+1':!..x a 2 u
ix
au p at 2 (s, t) ds = T ax (x
+ Llx, t) -
au T ax (x, t)
{X +1':!.. x a2u
=
ix
T ax 2 (s, t) ds,
where is the the string string (in (in units per length). Since this this holds where pp is the density density of of the units of of mass mass per length). Since holds differential equation for Llx sufficiently for all all x and and Ax sufficiently small, small, we we obtain obtain the differential equation
a2 u
a2 u
p at 2 = T ax 2 ' 0
< x < £,
t
> O.
(2.18)
We this as as the wave equation. equation. It It is is usual this in We recognize recognize this the homogeneous homogeneous wave usual to to write write this in the form the form a2 u 2 a2 u at 2 - c ax 2 = 0, < x < £, t > 0,
°
2
where c = T/p. The significance significance of of the the parameter parameter cc will in Chapter Chapter where c2 = T / p. The will become become clear clear in 7. In the case that force is is applied the string string (in (in the vertical In the case that an an external external body body force applied to to the the vertical direction), the direction), the equation equation becomes becomes
aat2 u
2 -
2
a2 u
c ax 2 = I(x, t), 0
< x < £, t> O.
(2.19)
30
Chapter 2. 2. Models one dimension dimension Chapter Models in in one
Exercise asks the Exercise 11 asks the reader reader to to determine determine the the units units of of f./. The string, as above, The natural natural boundary boundary conditions conditions for for the the vibrating vibrating string, as suggested suggested above, are conditions: are homogeneous homogeneous Dirichlet Dirichlet conditions:
u(O, t)
= u(£, t) = 0, t > 0.
One can also ends of string are are allowed freely One can also imagine imagine that that one one or or both both ends of the the string allowed to to move move freely in direction (perhaps (perhaps an along aa frictionless frictionless in the the vertical vertical direction an end end of of the the string string slides slides along pole). In this this case, case, the the appropriate appropriate boundary boundary condition condition is is aa homogeneous homogeneous Neumann Neumann pole). In condition (see (see Exercise Exercise 5). condition 5).
Exercises 1. What ( x , t t) ) have (2.19)? 1. What units units must must ff(x, have in in (2.19)? 2. What What are are the the units units of of the the tension tension T in in the the derivation derivation of of the the wave wave equation equation for for 2. the string? the string?
3. are the What are the units units of of the the parameter parameter cc in in (2.18)? (2.18)? 3. What 4. Suppose the only external external force force applied gravity. the only applied to to the the string string is is the the force force due due to to gravity. 4. Suppose What form form does does (2.19) (2.19) take take in in this this case? case? (Let (Let g be be the the acceleration due to What acceleration due to gravity, and take constant.) gravity, and take g g to to be be constant.) 5. Explain why why aa homogeneous homogeneous Neumann Neumann condition condition models models an the string 5. Explain an end end of of the string that is allowed freely in in the that is allowed to to move move freely the vertical vertical direction. direction. 6. that an elastic string string is fixed at both ends, ends, as as in in this this section, section, and 6. Suppose Suppose that an elastic is fixed at both and it it sags under under the the influence influence of of an external force force ff(x) with respect sags an external ( x ) (f (f is is constant constant with respect to time). time). What What differential differential equation equation and side conditions conditions does the equilibrium equilibrium to and side does the displacement of the string satisfy? given in in units force that f/ is is given units of of force displacement of the string satisfy? Assume Assume that per length. per length.
2.4 2.4
Suggestions for reading Suggestions for further further reading
If the the If the reader reader wishes wishes to to learn learn more more about about the the use use of of mathematical mathematical models models in in the an excellent excellent place place to to start start is is the the text text by by Lin Lin and and Segel which comcomsciences, an sciences, Segel [36], which bines modeling modeling with with the the analysis analysis of of the the models models by by aa number number of of different bines different analytical analytical techniques. Lin Lin and and Segel cover the the models models that that form form the the basis basis for for this this book, book, as as techniques. Segel cover well as many others. others. A more advanced advanced text, focuses almost almost entirely the derivation derivation of of A more text, which which focuses entirely on on the differential Gurtin differential equations equations from from the the basic basic principles principles of of continuum continuum mechanics, mechanics, is is Gurtin [21]. [21].
Chapter Chapter 33
Eseential ial
linear algebra algebra linear
The techniques presented presented in in this this book be described by analogy analogy to The solution solution techniques book can can be described by to techniques for solving techniques for solving Ax=b, is an n E RllXll) bare b E where A is n x nn matrix (A (A 6 R n x n ) and and x and and b are n-vectors (x, (x,b € Rll). R n ). Recall Recall that such such a matrix-vector equation represents represents the following system system of of n linear the nn unknowns Xl, #2, X2, ... equations in equations in the unknowns x\, • • • ,, Xn: xn: anXl a2lXl
+ al2X2 + ... + alnX n + a22X2 + ... + a2nXn
bl , b2 ,
Before we discuss methods for solving differential differential equations, we review review the the fundamental about systems linear (algebraic) mental facts about systems of of linear (algebraic) equations. equations.
3.1 3.1
Linear operator equations equations Linear systems systems as as linear linear operator
To fully fully appreciate appreciate the taken in book, it it is understand To the point point of of view view taken in this this book, is necessary necessary to to understand the finitethe equation Ax = = b not just as as aa system system of of linear equations, equations, but as a finiteother words, words, we we must must view view the the matrix matrix A dimensional linear operator operator equation. dimensional linear equation. In In other A as defining an operator mapping, or simply function) from Rll operator (or mapping, Rn to Rll Rn via matrix multiplication: multiplication: A A maps x E 6 Rll Rn to to Y y = = Ax E e Rll. R n . (More generally, generally, if A is is not mxn m m square, say Rffi, Rffi say A E e RffiXll, R , then A defines a mapping from Rll Rn to R , since Ax E GR n for xe E Rll.) for each each x R .) The following following language language is useful in The is useful in discussing discussing operator operator equations. equations. Definition 3.1. 3.1. Let X be sets. A function function (^operator, X Definition X and Y be (operator, mapping,) mapping) /f from X eX X a unique yy £E Y, denoted (x). to Y is a rule for associating associating with each xX E denoted yy = = ff (x) The set set X X is called the domain range of of f is the set domain of of ff,, and the range R(A) = {f(x) E Y : x E X}. 31 31
32
Chapter 3. Essential linear algebra
We X --+ X into function from from X We write write ff :: X -> YY (lif ("f maps maps X into Y") Y") to to indicate indicate that that ff is is aa function X to Y. toY.
The reader should recognize the difference difference between the range of a function X --+ the set (which is f). The set /f :: X —)• Y Y and and the set YY (which is sometimes sometimes called called the the co-domain co-domain of of /). The set Y Y merely merely identifies identifies the the type type of of the the output output values values f(x); /(#); for for example, example, ifif YY == R, R, then then every f/(#) (x) is is a real number. On the other hand, the range of f/ is is the set of elements by f. As aa simple of of Y that that are are actually actually attained attained by /. As simple example, example, consider consider f/ : R --+ —> R defined x 22.• The but the the range range of the defined by by ff(x) (x) = =x The co-domain co-domain of of /f is is R, but of /f consists consists of of the set of nonnegative numbers: R(f) = [0,00). In many cases, it is quite difficult difficult to determine the range of a function. function. The codomain, on the other hand, must be specified as part of the definition of the function. A A set set is is just aa collection collection of of objects objects (elements); (elements); most most useful sets sets have have operations operations defined important sets vector defined on on their elements. elements. The The most most important sets used in in this this book are sue vector spaces. spaces. Definition A vector space V which two Definition 3.2. 3.2. A V is is aa set set on on which two operations operations are are defined, defined, addition (if (if u, V, then V) and and scalar scalar multiplication multiplication (if V and and aa addition u, vv 6 E V, then u u +v v e E V) (if u u € EV is scalar, then space are is aa scalar, then au cm E € V). V). The The elements elements of of the the vector vector space are called called vectors. (In (In this book, the scalars are usually real numbers, and we assume this unless otherwise this book, the scalars are usually real numbers, and we assume this unless otherwise stated. we use complex numbers stated. Occasionally Occasionally we use the the set set of of complex numbers as as the the scalars. scalars. Vectors Vectors will will denoted by always always be denoted by lower lower case case boldface boldface letters.} letters.) The satisfy the following algebraic properties: The two two operations operations must must satisfy the following algebraic properties: 1. u
+v
= v
2. (u + v)
+u
+w
for all u, v E V.
= u
+ (v + w)
for all u, v, wE V.
3. There is a zero vector 0 in V with the property that u
+0 =
4.
+ (-u)
For each u E V, there is a vector -u E V such that u
u for all u E =
v.
o.
5. a(u + v) = au + av for all u, v E V and for all scalars a. 6. (a + P)u = au + pu for all u E V and for all scalars a, p. 7. a(pu) = (ap)u for all u E V and for all scalars a, p.
8. 1u = u for all u E v.
For For every every vector vector space space considered considered in in this this book, the the verification of of these vector vector space space properties properties is is straightforward straightforward and and will will be be taken taken for for granted. granted. Example space is Euclidean Example 3.3. 3.3. The The most most common common example example of of aa vector vector space is (real) (real) Euclidean n-space: n-space:
3.1. Linear Linear systems systems as operator equations equations 3.1. as linear linear operator
33 33
Vectors in R R nn are written in Vectors in are usually usually written in column column form, form,
as is convenient times to think of G Rn as as an as it it is convenient at at times to think of u u ERn an n n x x 11 matrix. matrix. Addition Addition and and multiplication are defined scalar multiplication defined componentwise: u
+ v = (Ul, U2, ... , un) + (VI, V2, ... ,Vn ) = (Ul + VI, U2 + V2, • .. ,Un + V n ), au =
a(Ul,U2, ... ,Un)
=
(aUl,aU2, ... ,aUn).
Example Apart from from Euclidean Euclidean n-space, n-space, the the most most common common vector spaces are are Example 3.4. 3.4. Apart vector spaces function spaces -vector spaces spaces in which the the vectors vectors are are functions. functions. Functions Functions (with function spaces —vector in which (with common domains) domains) can can be be added added together and multiplied multiplied by by scalars, scalars, and and the the algebraic common together and algebraic properties of vector space space are are easily easily verified. verified. Therefore, defining aa function properties of aa vector Therefore, when when defining function space, one only check that any any desired properties of functions are are preserved preserved space, one must must only check that desired properties of the the functions by addition addition and and scalar scalar multiplication. Here are are some some important by multiplication. Here important examples: examples:
1. C[a, b] be the of all defined 1. b] is is defined defined to to be the set set of all continuous, continuous, real-valued real-valued functions functions defined on the interval [a, b]. The The sum continuous functions continuous, on the interval [a, b]. sum of of two two continuous functions is is also also continuous, as is is any any scalar of aa continuous continuous function. Therefore, C[a, b] as scalar multiple multiple of function. Therefore, b] is is aa vector space. vector space. 2. C1l [a, to be be the continuously differentiable differentiate 2. C [a, b] b] is is defined defined to the set set of of all all real-valued, real-valued, continuously functions defined defined on interval [a,b]. [a, b]. (A function is continuously continuously differenfunctions on the the interval (A function differentiable if if its derivative sum of tiate derivative exists and and is continuous.) continuous.) The The sum of two continuously continuously differentiate functions and the differentiable functions is is also also continuously continuously differentiate, differentiable, and the same same is is for a scalar multiple of of a continuously differentiable function. function. Therefore, true for continuously differentiate Therefore, C1l [a, [a, b] b] is is aa vector vector space. space. C 3. For any positive integer integer k, k, Ck[a, b] is the the space space of of real-valued real-valued functions functions defined 3. For any positive Ck[a, b] defined [a, b] b] that have k continuous continuous derivatives. on [a, derivatives. Many vector vector spaces are encountered encountered in in practice practice are are subspaces subspaces of Many spaces that that are of other othe vector spaces. spaces. vector
Definition V be and suppose W is the Definition 3.5. 3.5. Let Let V be aa vector vector space, space, and suppose W is aa subset subset ofV of V with with the following properties: following properties: 1. The The zero vector vector belongs to W. W. 2. Every Every linear vectors in also in in W. That is, if if x, yyEW 2. linear combination combination of of vectors in W is also That is, GW and a, G R, then then and a, /? j3 E ax + j3y E W.
34 34
Chapter 3. 3. Essential linear algebra algebra Chapter Essential linear
Then call W V. Then we we call W aa subspace of ofV.
A subspace of of aa vector vector space space is space in in its its own own right, as the reader A subspace is aa vector vector space right, as the reader can verify by checking that all the of aa vector space are for aa can verify by checking that all the properties properties of vector space are satisfied satisfied for subspace. Example 3.6. 3.6. We We define define Example
= {u E C 2[a, b]
C1[a, b]
: u(a)
= u(b) = o}.
The set Cb[a, subset of of C [a, b], and subset contains contains the the zero zero function The set C^ja, bj b] is is aa subset C22[a,b], and this this subset function (hopefully Also, if u,v E and (hopefully this this is obvious obvious to to the the reader). reader). Also, if u,v e Cb[a,b], 6%[a, b], a,f3 a,j3 GE R, and w = au au + f3v, then w = f3v, then 2
2
• w wE [a, bj [a, bj is aa vector and eC C2[a, b] (since (since C C 2[a,b] vector space); space); and
— au(a) a-Q + 0-Q and similarly w(b) = 0. w(a) = au (a) + /3v(a) f3v(a) = 0.·0 f3 . 0 = Q, 0, and similarly w(b) o. • w(a) Therefore, b], which which shows shows that subspace of [a, bj. Therefore, wE w € Cb[a, Cp[a,b], that Cb[a, C^a,b]bj is aa subspace of C C22[a,b].
Example 3.7. Example 3.7. We We define define C~,[a,bj={uEC2[a,bj: ~~(a)=~~(b)=O}. The set CF,-[a, subset of of C [a, bj, and and it it can shown to be aa subspace. The set C^[a,b]bj is also also aa subset C22[a,b], can be be shown to be subspace. function belongs belongs to [a, bj. If If u, [a, b], a, f3 E and Clearly zero function Clearly the the zero to C;' Cj^[a,b]. u,vv E € C;' C^[a,b], a,/? £ R, R, and w au + f3v, fiv, then then w == au dw dx (a)
du
= a dx (a) +
Similarly, Similarly,
dw f3 dx (a)
=a
·0+ f3 . 0 = o.
~: (b) = 0,
and so w w E shows that subspace of [a, bj. and so e C;'[a,bj. C^[a,b]. This This shows that C;'[a,bj C^[a,b] is aa subspace of C C22[a,b].
throughout this book. The letters The previous two examples will be used throughout "D" and "N" Dirichlet and and Neumann, Neumann, respectively respectively (see (see Section for "D" and "N" stand stand for for Dirichlet Section 2.1, 2.1, for example). The following provides an important nonexample nonexample of a subspace. subspace. Example Example 3.8. 3.8. We We define define W
= {u
E C 2 [a,bj : u(a)
= 'Y,
u(b)
= 8},
where'Y are nonzero nonzero real although W subset of of C [a, bj, where 7 and and 8 are real numbers. numbers. Then, Then, although W is aa subset C22[a,b], it is not a subspace. For example, the zero function does not belong to W, since it not a subspace. For example, the zero function does not belong to W, since it does does not satisfy the the boundary boundary conditions. conditions. Also, Also, if E Wand R, then, it not satisfy if u, u, vv e W and a, a, J3f3 E € R, then, with w = au au + f3v, flv, we with w we have have w(a)
= au(a) + f3v(a) = a'Y + f3'Y = (a + (3)"(.
3.1. Linear systems as as linear operator equations equations
35 35
Thus w(a) does does not not equal ,,(, except special case that a a + (3 j3 = Thus w(a) equal 7, except in in the the special case that = 1. I. Similarly, Similarly, w(b) does not satisfy the boundary condition at the right endpoint. w(b) does not satisfy the boundary condition at the right endpoint. The concept vector space space allows allows us us to linearity, which which describes The concept of of aa vector to define define linearity, describes many simple simple processes processes and modeling and and analysis. many and is is indispensable indispensable in in modeling analysis.
Definition X and and Yare vector spaces, spaces, and and ff :: X X -+ is an an operator Definition 3.9. 3.9. Suppose Suppose X Y are vector —>• Y Y is operator (or mapping) with range Y. if and (or function,77 or mapping) with domain X and range Y. Then f is is linear linear if only if only if f(ax + j3z) = af(x) + j3f(z) for all a, j3 E R, x, z E X. (3.1) This can be as the the following following two conditions, which which together together are are This condition condition can be expressed expressed as two conditions, equivalent to to (3.1): equivalent (3.1):
1. f(ax) = af(x) for all x E X and all a E R; 2. f(x
+ z) = f(x) + f(z)
for all x,z E X.
A linear thus aa particularly particularly simple kind of simplicity A linear operator operator is is thus simple kind of operator; operator; its its simplicity can be appreciated by comparing the (e.g. f(x f ( x ++y)y}==f(x) f ( x ) ++ can be appreciated by comparing the property property of of linearity linearity (e.g. f(y)) with with common common nonlinear nonlinear operators: operators: ^/x Jx + y ^ -:j:. -/X + v:y, sin -:j:. sin sin (x) f(y)} ^-\-^Jy-> sin (x (x + y) ^ (x) + sin(?/), sin (y), etc. etc. mxn Example The operator by aa matrix GR R ffixn via matrix-vectoi Example 3.10. 3.10. The operator defined defined by matrix A A E via matrix-vector multiplication, multiplication, ff(x) (x) == Ax, Ax,
is linear; linear; the reader should should verify this if necessary (see Exercise 7). Moreover, it is the reader verify this if necessary (see Exercise 7). Moreover, it can be shown shown that that every operator mapping mapping Rn into Rffi can be be represented represented can be every linear linear operator Rn into Rm can by aa matrix in this this way way (see (see Exercise Exercise 8). This explains why why the the study by matrix A €E Rffixn R mxn in This explains study of (finite-dimensional) (finite-dimensional) linear linear algebra algebra is of matrices. this book, book, of is largely largely the the study study of matrices. In In this upper case boldface boldface letters. matrices will be denoted by upper letters. Example show that that the sine function function is not linear, linear, we that Example 3.11. 3.11. To To show the sine is not we observe observe that sin (2i) = sin (7f) = 0,
while while
2sin(~)
=2·1=2.
Example 3.12. Differentiation operator Example 3.12. Differentiation defines defines an an operator
d~
: C 1 [a, b] -+ C[a, b],
7 7The The word word "operator" "operator" is is preferred preferred over over "function" "function" in in this this context, context, because because the the vector vector spaces spaces themselves are are often often spaces spaces of of functions. functions. themselves
36
Chapter 3. 3. Essential Essential linear linear algebra algebra Chapter
and this this operator operator is is well well known known to to be be linear. linear. For For example, and example,
d~ [2 sin (x) since since
3e X ] = 2 cos (x) - 3e x ,
d~ [sin (x)] = cos (x), d~ [eX] = eX.
In the kth kth derivative derivative operator defines aa linear linear operator operator mapping mapping Ck Ck[a, In general, general, the operator defines [a, bb] into C[a,b]. This is why linearity is so important in the study of differential equainto C[a, b]. This is why linearity is so important in the study 01 differential equations. tions. Since a matrix A E e Rnxn R n x n defines a linear operator operator from Rn Rn to Rn, Rn, the linear Ax = be regarded this point of system system Ax = bb can can be regarded as as aa linear linear operator operator equation. equation. From From this point of view, posed by Is there vector x xE view, the the questions questions posed by the the system system are: are: Is there aa vector G Rn Rn whose whose image image under A A is is the the given given vector If so, so, is is there there only only one one such such vector vector x? the next next vector b? b? If x? In In the under section, we explore these these questions. questions. section, we will will explore The point of discussing The point of view view of of aa linear linear operator operator equation equation is is also also useful useful in in discussing differential equations. equations. For For example, example, consider consider the the steady-state steady-state heat heat flow flow problem problem differential -/1,
(Pu ox 2 = I(x), 0 < x
< C,
u(O) = 0, u(e) = 0
(3.2)
from Section Section 2.1. 2.1. We We define define aa differential differential operator operator L C|,[0,^] -> C[O,C] C[Q,l] by by from LD D :: C1[0,C]-+ LDu =
~u
-/1,
dx 2 '
Then the the BVP BVP (3.2) (3.2) is is equivalent equivalent to to the the operator operator equation equation Then LDU=I
(the reader should notice Dirichlet boundary boundary conditions conditions are are enforced enforced by (the reader should notice how how the the Dirichlet by the definition definition of of the the domain domain of of LD). This and and similar similar examples examples will will be be discussed the LD)' This discussed throughout throughout this this chapter chapter and and in in detail detail in in Section Section 5.1. 5.1.
Exercises Exercises 1. R -+ 1. In In elementary elementary algebra algebra and and calculus calculus courses, courses, it it is is often often said said that that 1 / :R ->• R R is linear linear if if and only if if it it has has the the form form f(x) — ax ax + + b, &, where where aa and and bb are are is and only I(x) = constants. Does Does this this agree agree with with Definition Definition 3.9? 3.9? If If not, not, what what is is the the form form of of aa constants. linear function / : R —^ R? linear function 1 R -+ R?
2. Show explicitly explicitly that that 1 / :R R -+ -> R R defined defined by by f(x) = v'x ^/x is linear. 2. Show I(x) = is not not linear. 3. the following not it 3. For For each each of of the following sets sets of of functions, functions, determine determine whether whether or or not it is is aa vector space. space. (Define (Define addition addition and and scalar scalar multiplication multiplication in in the obvious way.) vector the obvious way.) If not, state what property to hold. If it it is is not, state what property of of aa vector vector space space fails fails to hold.
3.1.
Linear equations Linear systems as linear operator equations
(a) {J E C[O, 1] : 1(0)
37
= O}
(b) {J E C[O, 1] : 1(0) = I}
(c)
{1 E C[O, 1]
1
: 10 1(x) dx
= O}
(d) P Pnn,, the set of all polynomials of degree n or less. (e) The set of all polynomials of degree exactly n. n. 4. Prove that the [a, b] the differential differential operator operator L : C1l[a, b] -+ —> C[a, b] b] defined by du Lu=udx
is not a linear operator. operator. 5. [a, b] -+ 5. Prove Prove that that the the differential differential operator operator L : Cl1[a,b] —>• C[a, C[a,b]b] defined defined by by du 3 Lu = - +u dx
is is not not aa linear linear operator. operator. 6. Prove that the [a, b] the differential differential operator M M : C22[a, b] -+ —)• C[a, <7[a, b] b] defined by d2 u du Mu= --2-+3u 2
dx
dx
is a linear operator. operator. 7. (a) (a) Let Let A E e R2x2. R 2x2 . Prove that
A(ax + f3y) = aAx + f3Ay for all a,f3 E R, X,y E R2. (Write
and and write write out out A(ax + + f3y) /3y) explicitly.) (b) Now repeat part 7a for A E Now repeat e Rnxn. R n x n . The proof is not difficult difficult when when one uses summation notation. For example, if the (i, (i, j)-entry of A is denoted aij, then then aij, n
(Ax)i =
L
aijXj.
j=1
Write (A (A(ax /?y)) i and and (aAx (o:Ax + f3AY)i /5Ay)i in in summation summation notation, Write (ax + f3Y))i notation, and and show that the first can be rewritten show that the first can be rewritten as as the the second second using using the the elementary elementary properties of arithmetic.
38
Chapter 3. Essential linear algebra 2x2 8. is linear. linear. Prove Prove that is aa matrix matrix A R 2X2 8. (a) (a) Suppose Suppose f/ : R2 R2 -t —>• R2 R2 is that there there is AE €R 2 such that /f is by /(x) f(x) == Ax. Hint: Each Each x x E such that is given given by Ax. Hint: e R2 R satisfies satisfies xx == xiei +X2e2, + #262, where Xlel where
we have have Since Since f/ is is linear, linear, we
The desired matrix be expressed expressed in terms of the vectors (et), The desired matrix A A can can be in terms of the vectors f/(ei), f(e2). /(e 2 ). (b) Now Now show R n -t Rm m is then there there exists exists aa matrix (b) show that that if if f/ : R —>• R is linear, linear, then matrix mxn such that /(x) f(x) == Ax for AE <E RmXll R for all all x E e Rll. Rn. II
3.2
Existence Ax == b Existence and and uniqueness of solutions to Ax
We will now now discuss discuss the the linear linear system system Ax Ax = = b, where where A A eE RllXll R n x n and and bb eE Rll, R n , as as a We consider three fundamental linear operator linear operator equation. equation. We consider three fundamental questions: questions: 1. Does aa solution to the the equation 1. Does solution to equation exist? exist?
2. is it unique? 2. If If aa solution solution exists, exists, is it unique? how can compute it? 3. If 3. If aa unique unique solution solution exists, exists, how can we we compute it? It turn~ out that the first two are intimately purpose of It turns out that the first two questions questions are intimately linked; linked; the the purpose of this this to shed these two two questions and the the connection connection between between section section is is to shed some some light light on on these questions and We will will also how this this point point of can be be carried to the them. We them. also briefly briefly discuss discuss how of view view can carried over over to the case will be be deferred in case of of aa linear linear differential differential equation. equation. The The third third question question will deferred to to later later in this chapter. this chapter.
3.2.1 3.2.1
Existence Existence
The existence of to Ax b is to the condition that that b b lie in The existence of aa solution solution to Ax == b is equivalent equivalent to the condition lie in R(A), the range range of of A. A. This begs the R(A)? 72-(A), the This begs the question: question: What What sort sort of of aa set set is is 7£(A)? If y, y, w 72.(A), say say y Az, then then If wE6 R(A), y = = Ax, Ax, w w = = Az, ay + {3w
= aAx + {3Az = A(ax + (3z).
This shows that ay (3w € E R(A). Moreover, the zero vector vector lies in 7£(A), R(A), since This shows that ay + + /3w 72-(A). Moreover, the zero lies in since AO = = o. 0. It follows that R(A) 7£(A) is a subspace of Rll Rn (possibly the entire space RllRn— every vector space Every linear operator has has this every vector space is is aa subspace subspace of of itself). itself). Every linear operator this property; property; if ff :: X -» -t Y then R(f) of the if Y is is aa linear linear operator, operator, then 72-(f) is is aa subspace subspace of the vector vector space space Y. Y. (The same same need not be true for for aa nonlinear operator.) (The need not be true nonlinear operator.)
3.2. Existence and uniqueness of solutions solutions to to Ax Ax = = b 3.2. Existence and uniqueness of b
39 39
The subspaces of of Rll is particularly particularly simple: proper subspaces The geometry geometry of of subspaces Rn is simple: the the proper subspaces of R (Le. those those that are not not the the entire of RII (i.e. that are entire space) space) are are lower-dimensional lower-dimensional spaces: spaces: lines lines in R2, lines and and planes planes in in R33,, and and so forth (we visualize these these objects objects in in in R2, lines so forth (we cannot cannot visualize dimensions we can understand understand them dimensions greater greater than three, but we them by analogy). analogy). Since every must contain contain the in R for example, example, is is every subspace subspace must the zero zero vector, vector, not not every every line line in R 22 ,, for a subspace, subspace, but but only only those those passing passing through through the the origin origin are. are. n With understanding of the geometry of R Rll, we obtain obtain the following conWith this understanding , we clusions: clusions: n
• If R(A) = Rll, for each b E If 7£(A) R n , then Ax = b has a solution solution for e Rll. R n . (This is is a tautology.) tautology. ) • If for almost If R(A) 7£(A) i^ Rll, R n , then Ax = = b fails to have a solution for almost every b E e Rll. Rn. n This is because a lower-dimensional subspace comprises very little of R R II (think of a line contained contained in the the plane or in three-dimensional space). three-dimensional 2x2 Example 3.13. 3.13. We consider the the equation = b, where where A 6 ER is given by by Example We consider equation Ax = R2X2 is given
A=[~ ~]. 2 ER R2, we have For any x 6 , we
-
Xl [ 2Xl
=
(Xl
+ 2X2 + 4X2
+ 2X2)
[
~
] .
This calculation shows shows that every vector in the range of multiple of This calculation that every vector b in the range of A is is aa multiple of the the vector vector
R(A) is in the the plane plane R R22 (see Figure 3.1). Since Since this this Therefore, subspace 'R-(A) Therefore, the the subspace is aa line line in (see Figure 2 line is aa very small part part of of R the system system Ax Ax = fails to to have have a a solution solution for line is very small R 2 ,, the = b fails for 2 almost every every b bE R2. almost eR . As mentioned mentioned at at the beginning of this chapter, chapter, there there is is aa close close analogy analogy between between As the beginning of this linear (algebraic) systems and and linear linear differential equations. The reader should linear (algebraic) systems differential equations. The reader should think think carefully about carefully about the similarities between the following example and the previous one. Example 3.14. define the £] Example 3.14. We We define the linear linear differential differential operator operator LN LN :'•e1[o, C^[Q,f\£] -+ -» e[O, (7[0,£j by by
40
Chapter 3. Essential Essential linear algebra
Figure 3.1. 3.1. The A in Figure The range of the matrix A in Example 3.13. 3.13.
where where
CMO,l] =
{u E C2 [0,l] :
~~(O) = ~~(l) = O}
(as defined defined in the previous section). If If f e E 7£(Ljv), R(L N ), then there exists u €E Ch[O,l] C^[0,l] such that LNu = It follows follows that such that LNU = f.f. It that
l r rl cPu io f(x) dx = - io dx (x) dx 2
= -
dU ]1 [-(x) dx
0
= _ du (l)
dx
+ du (0) dx
= 0.
This shows that E C[O, l] cannot cannot belong belong to the range LN unless unless it it satisfies satisfies the This shows that f e C[0,^] to the range of of LN the special condition special condition
11
f(x) dx
= 0.
(3.3)
In fact, fact, 7£(Ljv) R(LN) i-iss the set of of all all such such ff,, as as the reader is asked to to show show in in Exercise Exercise In the set the reader is asked 12. 12. Because the the space space C[O, l] is is infinite-dimensional, infinite-dimensional, we we cannot visualize this situBecause C[0,^] cannot visualize this situation (as (as we could in the previous the reader reader should appreciate ation we could in the previous example). example). However, However, the should appreciate that e C[0, example, that most most functions functions ff E e[O, £] l] do do not not satisfy satisfy the the condition condition (3.3). (3.3). (For (For example,
3.2. Existence and and uniqueness of solutions to Ax Ax = b
41 41
the reader is invited to write down quadratic polynomial at random and call it down a quadratic f(x). Chances quadratic will not satisfy satisfy (3.3).) range /(#). Chances are that this quadratic (3.3).) Therefore, Therefore, the range of LN LN is only only a small part of of C[O, f], and for most most choices f], there is of C[0,i], and for choices of of f E G C[O, C[0, £], LNU = = f. no solution to to L^u f. Example 3.15. now define LD Example 3.15. We now define L C^[Q,l] ->• C[O,f] C[Q,f\ by D : Cb[O,f]-+ d2 u LDU = - dx 2 '
where
Cb[O,f] = {u E C2[0,f] : u(O) = u(f) =
O}.
The reader should recall from the previous section that the BBVP VP
°
~u
= f(x), < x < f, u(O) = 0,
- dx 2
u(f) =
°
can be operator equation Lj^u L DU = R( A) is be written as the linear operator — f .. We will show that 'R-(A) all of LDu = = f for for every every f E fl. The idea of C[O, C[Q,(]f] by showing that we can solve Lpu £ C[O, C[0,£]. of is to integrate twice and use the boundary boundary conditions to determine the constants of integration. intearation. We have ~U
dx 2 (x) = -f(x),
so
du dx (x) = We write
l
0
°< x < f,
X
f(s) ds
du dx(X) = -F(x)
where
F(x) =
+ Cb a < x < f.
°
+c1 , < x < f,
fox f(s) ds.
We then integrate to obtain u(x) =
-loX F(z) dz + C x + C = -loX Io 2
1
z
f(s) dsdz + C1 x + C2,
°< X < f.
The reader should notice the use of the dummy variables variables of integration ssand and z. z. The first boundary boundary condition, u(O) = O. w(0) = = 0, implies that C Ci2 = 0. We then have u(f) =
-10£ l
z
f(s) dsdz
+ C1 f;
u(f) = since u(i] — 0, 0, we obtain C1 =
e1111Z f(s)dsdz 0
0
42 42
Chapter 3. 3. Essential linear algebra algebra Chapter Essential linear
and so so and
r r f(s)dsdz+ x r 10r f(s)dsdz, 1010 e10 l
u(x)=-
O<x
(3.4)
The reader twice) that that this The reader can can verify verify directly directly (by (by differentiating differentiating twice) this formula formula defines defines aa solution of of the the BVP BVP represented represented by Lnu = = ff.. This shows that that we we can solve LDU Lnu = = ff solution by LDU This shows can solve E C[O,l], so K(L R(Ln) = C[O,l]. for any any ff e for C[0,4 and and so C[Q,l}. D) =
3.2.2
Uniqueness
The linearity of a matrix operator operator A implies that nonuniqueness of solutions to = b, if it occurs, has a special structure. Suppose Ax = Suppose x and and z in Rll Rn are both both solutions (i.e. Ax and Az Then solutions to to Ax Ax = =b b (Le. Ax = = b band Az = = b b both both hold). hold). Then Ax
= Az => Ax -
Az = 0 => A(x - z) = 0,
the following from linearity of of A. If x the last last step step following from the the linearity A. If x^ =I z, z, then then w w= = x x— - zz is is aa nonzero nonzero satisfying Aw Aw = = o. vector satisfying 0. = b and w is a nonzero On the other hand, suppose suppose x is a solution solution to Ax = vector 0. Then Then vector satisfying satisfying Aw Aw = = o.
A(x+w) =Ax+Aw =b+O =b, and in this case there cannot cannot be a unique solution to Ax == b. the above observations, we define the the null space of A to be Because of the
N(A) = {x E Rll : Ax = O}. Since AO AO = = 0 always holds for a linear operator we always have 0 E N(A). operator A, we e -A/"(A). Moreover, if if x,z x,z e E A/"(A) N(A) and and a, a,{3 E R, R, then then Moreover, ft € Ax = 0, Az = 0 and so and so A (ax + {3z) = aAx + {3Az = a . 0 + {3 ·0=
o.
Therefore, ax ax + {3z N(A). This shows shows that A/"(A) N(A) is a subspace of Rll. /3z E € A/"(A). Rn. If 0 is the only vector in A/"(A), N(A), we say that A/"(A) N(A) is trivial. trivial. the Our above lead conclusion: If If Ax Our observations observations above lead to to the the following following conclusion: Ax = = b b has has aa N(A) is trivial. Furthermore, nothing in the solution, it is unique if and only if wV(A) the A's being a matrix operator; the same arguments can above discussion depends on A's differential operator. be made for any linear operator, such as a differential
3.2. Existence and Ax = bb and uniqueness uniqueness of solutions to Ax
43
If is nontrivial nontrivial and b has then the has in in If N(A) JV(A) is and Ax Ax == b has aa solution, solution, then the equation equation has fact many solutions. this, suppose E RD band fact infinitely infinitely many solutions. To To see see this, suppose xx e Rn satisfies satisfies Ax Ax = = b and w e N(A), A/"(A), w w =F 7^ O. 0. Then, Then, for for each each 0: a E G R, R, we we have have wE
A(x + o:w) = Ax + o:Aw = Ax + 0:0 = Ax = b. Since the real number 0:, this shows shows Since xx + + o:w aw is is different different for for each each different different choice choice of of the real number a, this that has infinitely Moreover, it that the the equation equation has infinitely many many solutions. solutions. Moreover, it easily easily follows follows that that the the set set of of all all solutions solutions to Ax Ax = = b is, is, in in this this case, case,
x+N(A) = {x+w : w EN(A)}. Once Once again, again, the the same same properties properties hold hold for for any any linear linear operator operator equation. equation. Example Let A by Example 3.16. 3.16. Let A Ee R R4x4 be be defined defined by
Consider Ax = b, where b Consider the the equation equation Ax = b, where b 88 elimination system Ax elimination algorithm, algorithm, the the system Ax == system system Xl 13X3 4X3 + X2 +
E R4 e R4 is is arbitrary. arbitrary. Using Using the the standard standard the bb can can be be shown shown to to be be equivalent equivalent to to the 4X4 2X4
0 0
bl - 3b2, b2, b3 - b2 - 2b l b4 - b2 - bl .
,
We see that system is conditions We see that the the system is inconsistent inconsistent unless unless the the conditions
b3 - b2 - 2b 1 = 0, b4 - b2 - b1 = 0
(3.5)
hold. If If these b, then hold. these conditions conditions are are satisfied satisfied by by b. then
= bl
-
3b 2
X2 = b2
-
4X3 - 2X4,
Xl
+ 13x3 + 4X4,
where X3 and X4 can Setting #3 X3 = X4 = of where x% and x± can take take on on any any value. value. Setting = ssand and #4 = t, every every vector vector of the the form form
bl
x=
[
~2o3b
2 ]
+s
[
1~1 ] +t [ - 0~ 1
001
8 8We reduction We assume assume that that the the reader reader is is familiar familiar with with Gaussian Gaussian elimination, elimination, the the standard standard row row reduction algorithm for for solving solving Ax = b. For aa review, see any any introductory introductory text text on on linear algebra, such such as algorithm Ax = h. For review, see linear algebra, as the text by by Lay the text Lay [34]. [34].
44
Chapter 3. Essential algebra Essential linear algebra
is solution of is aa solution of the the system. system. We We have have that that
is one of Ax is one solution solution of Ax = = b, b, and and
Example 3.17. compute the space of LN defined defined in Example Example 3.17. We We compute the null null space of the the operator operator LN in Example If LNU LNU = satisfies the 3.14. 3.14- If = 0, 0, then then uu satisfies the BVP BVP d2 u
- dx 2
= 0,
du (0)
= 0,
dx
~~(f) =
°< x <
f,
0.
The differential differential equation equation implies implies that = C C\x C%2 for C\1 The that u(x] u(x) = for some some constants constants C 1x + C and C^The two boundary conditions each lead to the conclusion that C\ = 0; • and C The two boundary conditions each lead to the conclusion that C = 0,2 1 however, any value differential equation equation however, the the constant constant C C%2 can can have have any value and and both both the the differential and the two two boundary conditions will will be This shows that every every constant constant and the boundary conditions be satisfied. satisfied. This shows that function C
Example 3.18. 3.18. The null space LD defined defined in in Example Example 3.15 is is Example The null space of of the the operator operator Lp trivial. this, the reader should should note note that equation trivial. To To see see this, the reader that the the differential differential equation
d2 u dx
--= 0 2
again implies = C boundary conditions conditions w(0) u(O) = again implies that that u(x) u(x] = C\x C%; however, now now the the boundary = 1x + C 2 ,- however, u(l) — 0 force C\1 == C^ Therefore, the only solution of LDU = 00 is u(f) = force C C2 == 0. O. Therefore, the only solution of LDU = is the the zero zero function. function.
°
Since null space of LD LD is is trivial, trivial, we we see that LDU LDu = has at most one Since the the null space of see that = f has at most one solution for for any any right-hand side f./. solution right-hand side
3.2. Existence and Ax = b and uniqueness of solutions to Ax
3.2.3 3.2.3
45 45
The Fredholm The Fredholm alternative alternative
One of of the the fundamental fundamental results results of of linear linear algebra in aa certain certain sense, One algebra is is that, that, in sense, existence existence and uniqueness uniqueness are are equivalent equivalent for system. To be precise, precise, if Ax — = bb has and for aa square square system. To be if Ax has for aa solution solution for for each each bERn b G Rn (i.e. (i.e. if if R(A) 7£(A) == Rn), R n ), then then the the solution solution is is unique unique for each bERn N(A) is is trivial). trivial). On the other hand, if if the the solution solution to to Ax Ax = = b, each b 6 Rn (i.e. (i.e. A/"(A) On the other hand, b, whenever it exists, is is unique unique (i.e. (i.e. if if A/"(A) N(A) is is trivial), trivial), then then Ax Ax = has aa solution whenever it exists, = bb has solution for each bERn b 6 Rn (i.e. R(A) ft(A) = = Rn). R n ). Moreover, the case N(A) is is not not trivial, trivial, we we can can give give aa condition condition that that Moreover, in in the case that that A/"(A) b G Rn must satisfy in solution. We bERn must satisfy in order order for for Ax Ax = = bb to to have have aa solution. We collect collect these these facts facts the following following theorem. in the in theorem. Theorem 3.19. 3.19. (The (The Fredholm Fredholm alternative) Suppose A A E Theorem alternative) Suppose 6 Rnxn. R n x n . Then exactly exactly following is true. one of of the following 1. The null space of A A is trivial and for for each b b ERn, G R n , there exists a unique n n E R Ax = x G solution x R to Ax = b. b. 2. of A A is nontrivial, and the equation Ax Ax = 2. The The null space of = bb has a solution ifif only if satisfies the following following condition: and only if bb satisfies wE Rn, ATw
= 0 :::} W· b = 0,
(3.6)
or, equivalently, or, equivalently,
(3.7) satisfied, then the equation equation Ax Ax = infinitely many IfIf this condition is satisfied, = b has infinitely solutions. T In the the statement of the the Fredholm alternative, we we used used the the transpose A, A AT, In statement of Fredholm alternative, transpose of of A, , which is is the the matrix matrix whose whose rows rows are are the the columns of A A and vice versa. versa. We We also used which columns of and vice also used the product, defined defined for two vectors vectors in in Rn: the dot dot product, for two Rn:
n
w· b = LWibi. i=l
We and its its significance Section 3.4. 3.4. We will will discuss discuss the the dot dot product product and significance in in Section We will will have have more more to to say later about about the the condition (3.6). For For now now it it is We say later condition (3.6). is sufficient sufficient to = b only if if the the to understand understand that, that, if if -A/"(A) N(A) is is not not trivial, trivial, then then Ax Ax = b has has aa solution solution only right-hand side vector certain compatibility right-hand side vector b b satisfies satisfies aa certain compatibility condition. condition. Example 3.20. In 3.16, J\f(A) is nontrivial, nontrivial, and Example 3.20. In Example Example 3.16, N(A) is and we we saw saw directly directly that that Ax = only ifb if b satisfies satisfies conditions {3.5}. Ax = bb has a solution if if and only (3.5). These conditions can be be written written as as can WI'
b = 0,
W2'
b = 0,
46
Chapter 3. Essential linear algebra
where where
is straightforward straightforward to show show that that It is (3.8)
Example 3.21. 3.21. In In Examples Examples 3.14 3.14 and and 3.17, we showed that that A/"(L/v) N(LN) is is nontrivial nontrivial Example we showed LNu == ff has has aa solution solution if if and and only only if the compatibility compatibility condition condition (3.3) (3.3) is is and that that L^u and if the as stated stated in in Theorem satisfied. Although Although the the Fredholm Fredholm alternative, alternative, as satisfied. Theorem 3.19, 3.19, does does not not apply analogous statement We will apply to to this this situation, situation, an an analogous statement can can in in fact fact be be made. made. We will see see how close the analogy is is when when we we define define aa "dot product" for for functions junctions (see Sections how close the analogy "dot product" (see Sections 3.4.1 and and 6.2.3). 3.4.1 6.2.3). nxn If A E Rnxn and N(A) is trivial, trivial, then then A A is is called called nonsingular (or invertible), If A eR and JV(A) is nonsingular (or invertible), nxn and there there exists matrix A" A- 11 G ER R nxn (the inverse of A), with with the the property property that that and exists aa matrix (the inverse of A),
The matrix II is is the the n n x xn n identity matrix; it it has has the the property property that that Ix = x x for all The matrix identity matrix; Ix = for all
xx eERn Rn.
1 If we know know A" A -1, we wish wish to to solve solve Ax Ax = b, b, we we need need only only compute If we , and and we compute aa matrix-vector calculation shows: matrix-vector product, product, as as the the following following calculation shows:
::::} ::::}
Ax A-lAx Ix
::::}
x
Thus = b is is x = Thus the solution solution of of Ax = = A -1x b. 1 Since A" defines aa linear b Since A -1 defines linear operator, operator, we we see see that that the the solution solution x x to to Ax Ax = = b depends on the side b. fact, we can demonstrate depends linearly linearly on the right-hand right-hand side b. In In fact, we can demonstrate this this for for any are vector any linear linear operator operator equation. equation. We We assume assume that that X X and and Y Y are vector spaces spaces and and f :X X -t -» YY is is aa linear linear operator operator with with aa trivial trivial null null space. space. Then, Then, for for each each yy E € R(f), 7?.(f), f: there solution x (x) = operator there is is aa unique unique solution x e EX X to to ff(x) = y. y. Let Let us us define define the the solution solution operator S :: R(f) 7£(f) -t -» X X by by the the condition condition that that xx = Sy Sy is is the the solution solution to to f(x) f (x) = y. y. (If S (If 1 R(f) = that ff is is invertible, invertible, then then S nothing more more than than f" f-1.) We wish wish to K(f) — Y, so so that S is is nothing .) We to show that that S is linear. show S is linear. We will will show show that that if if y, y, zZ e E 7£(f) R(f) and R, then We and a, (3 ft E € R, then S(ay + (3z) = aSy
+ (3Sz.
3.2. Existence and uniqueness of solutions solutions to to Ax Ax = = bb 3.2. Existence and uniqueness of
47
We define define xx = Sy Sy and and w Sz; then then We w = Sz; f(x)
= y,
f(w)
= z.
By f, we By the the linearity linearity of of f, we have have f(ax
+ (3w)
= af(x)
+ (3f(z)
= ay
+ (3z.
This last last equation equation is is equivalent equivalent to to This S(ay + (3z) = ax
+ (3w =
aSy
+ (3Sz.
Thus, the linearity necessarily linear. Thus, the linearity of of f implies implies that that the the solution solution operator operator is is also also necessarily linear. To put it another another way, the solution solution to linear operator operator equation equation depends depends linearly put it way, the to aa linear linearly To on the right-hand side side of of the equation. In In the the context context of of differential differential equations, equations, this this on the right-hand the equation. property is usually called the the principle of superposition. superposition. principle of property is usually called
Example 3.22. 3.22. The The BVP BVP Example
d2 u -/1,
dx 2
= f(x), 0 < x < f,
u(O) = 0, u(f) = 0 can be be written written as as the the linear linear operator operator equation equation LDu Lpu == /, as explained explained in in Examples can f, as Examples 3.15 and and 3.18, 3.18, and and we we showed showed in in those those examples examples that that there there is is aa unique unique solution solution uu 3.15 for each fl. Since Since integration Formula (3.4) shows for each ff E e C[O, C[0,£]. integration is is aa linear linear operation, operation, Formula shows that uu depends depends linearly linearly on f. that on f. For we cannot find an for For many many linear linear differential differential equations, equations, we cannot find an explicit explicit formula formula for the makes this this linear those cases, we the solution solution that that makes linear dependence dependence obvious. obvious. Even Even in in those cases, we know know from from the the above above argument argument that that the the solution solution depends depends linearly linearly on on the the right-hand right-hand side. side.
Exercises Exercises 1. Let Let 1.
A=[
Graph R(A) H(A) in plane. Graph in the the plane. 2. 2.
(a) the missing missing steps (a) Fill Fill in in the steps in in Example Example 3.16. 3.16.
(b) Let Let A A by by the matrix in in Example Example 3.16. 3.16. Compute Compute the solution set set of of the the the matrix the solution (b) equation AT ATw = 0, 0, and and show show that that the is (3.8). equation w = the result result is (3.8). 3. For For each each of of the the following following matrices A, determine determine if if Ax Ax == b b has matrices A, has aa unique unique 3. solution for for each each b, b, that that is, determine if if A is nonsingular. For each each matrix matrix is, determine A is nonsingular. For solution A which which is is singular, singular, find vector bb such such that that Ax Ax = =b has aa solution solution and and aa b has A find aa vector vector cc such such that that Ax = cc does does not have aa solution. solution. Ax = not have vector
48
Essential linear algebra Chapter 3. Essential
(a) A =
[ 1 -1] -2
4
~]
(b) A =
[!
(c) A =
[~ ~]
4. For For each of the following matrices matrices A, = bb has has aa unique unique 4. each of the following A, determine determine if if Ax Ax = solution b, that that is, determine if if A For each each matrix matrix solution for for each each b, is, determine A is is nonsingular. nonsingular. For A is singular, singular, find vector b such that that Ax Ax = = bb has has aa solution solution and and aa A which which is find aa vector b such vector such that — cc does does not solution. vector cc such that Ax Ax = not have have aa solution.
(e) A
~[
1 2
-11
2 -1 4 -2 -2 1
1
A €E Rnxn and b bERn, b '" the solution of the 5. Suppose Suppose A R nxn and € Rn, b ^ O. 0. Is Is the solution set set of the equation equation Ax = b, Ax=b, {x ERn : Ax = b} , aa subspace subspace of of R Rnn ?? Why Why or or why why not? not? 6. Let Let D : CI[a, b] be be the the derivative is the null space 6. Cl[a,b]b] -+ —»• C[a, b] derivative operator. operator. What What is the null space of Dl D? 7. Let L : C22[a, [a, b] b] be be the the second second derivative derivative operator. operator. What is the the null null 7. Let b] -+ —» C[a, b] What is space space of L? L? 8. As shown Chapter 2, 2, differential differential equations involving the the second 8. As shown in in Chapter equations involving second derivative derivative operator are are sometimes paired with with mixed mixed boundary boundary conditions. Define the operator sometimes paired conditions. Define the set set C![a,b] = {u E C2 [a,b] : u(a) = ~~(b) = O} and define Lm : C![a, b]-+ C[a, b] by
The reader note that that C![a, b] is of C22[a, [a, b]. The reader should should note C^[a, b] is aa subspace subspace of (a) the null null space space of of Lm. (a) Determine Determine the Lm.
3.2. Existence Existence and and uniqueness uniqueness of of solutions solutions to = b b 3.2. to Ax Ax =
49 49
(b) Explain Explain how how to solve L for u, w, when is aa given given function function in (b) to solve Lmu when f/ is in mu = f for C[a,b]. (Hint: Integrate Integrate twice.) Is this this differential differential equation equation solvable solvable for C[a, b]. (Hint: twice.) Is for every f/ E £ C[a, C[a, b]? &]? every 9. Repeat Exercise with 9. Repeat Exercise 8, 8, but but with C~[a,b] = {u E C 2 [a,b]
~~(a) = u(b) =
o}
and Lm : C~.[a, b] -+ C[a, b] defined by
cPu
Lm u = -
dX2·
10. Consider Consider the following system system of of nonlinear nonlinear equations: equations: 10. the following
xi + x~ = 1, X2 - xi = o. This be written b, where R22 is by This system system can can be written as as f(x) f (x) = = b, where f: f : R2 R2 -+ —> R is defined denned by f(x) =
and and
2
2
Xl X~ [ X2 - Xl
+
b=[~].
(a) (a) Show Show that that f(x) f (x) == bb has has exactly exactly two solutions. solutions. (Hint: (Hint: A A graph graph is is useful.) useful.)
(b) Show Show that that the only solution solution to to f(x) f(x) == 00 is is xx = = o. 0. (Yet, (Yet, as as Part lOa (b) the only Part lOa shows, unique.) shows, the the solution solution to f(x) f (x) = = b is is not not unique.) This example example illustrates illustrates that that the properties of of linear linear systems systems do do not not necessarnecessarThis the properties ily carryover carry over to to nonlinear ily nonlinear systems. systems. 11. Let Let D D :: C CIl[a, b] -+ -> C[a, C[a,b]b] be be the differentiation operator: operator: 11. [a, b] the differentiation Df
= df. dx
(a) Show Show that that the the range range of of D D is is all all of of C[a, b]. (a) (b) Part lla is is equivalent equivalent to to saying saying that, that, for for every every f/ E e C[a, b], b], the (differen(b) Part 11a the (differential) equation equation Du = = f has has aa solution. solution. Is Is this this solution solution unique? Why or tial) unique? Why or why not? why not? 12. Let Let LN LN be defined as as in in Example Example 3.14, 3.14, and and suppose suppose f/ E e C[O, C[Q,l]l] satisfies 12. be defined satisfies
1£ f(x) dx Show that that f/ E e R(Ln). n(Ln}. Show
=
o.
Chapter 3. Essential linear algebra
50
3.3
Basis and dimension dimension Basis
The matrix-vector product product Ax is equivalent to aa linear the columns columns The matrix-vector Ax is equivalent to linear combination combination of of the of A. If A has VI, V V2,2 ,... of the the matrix matrix A. If A has columns columns YI, . . . ,, Vvnn ,, then then
The reader should write out is not not clear (see Exercise The reader should write out aa specific specific example example if if this this is clear (see Exercise x 1). XI,X2, ••• ... ,Vn vectors; an ex1). The The quantities quantities £i,#2, • • • ,Xn i n are are scalars scalars and and VI,V2, YI, V2,...., vn are are vectors; an expression such as as XiVi is called called aa linear linear combination combination of of the the pression such Xl Vl + #2V2 X2 V2 + cldots cldots + x Xn Vnn is nv vectors Vl, Vnn because using the the linear vectors YI , V2, v 2 , ... . . . ,, v because the the vectors vectors are are combined combined using linear operations operations of addition addition and and scalar scalar multiplication. multiplication. of E RDXD way When A £ R n x n is is nonsingular, each b E£ RD Rn can be written in a unique way as aa linear linear combination combination of of the the columns columns of of A A (that (that is, is, the equation Ax Ax = = bb has has aa the equation as unique is related. related. unique solution). solution). The The following following definition definition is Definition Let V space, and V2, , Vnn are vectors Definition 3.23. 3.23. Let V be be aa vector vector space, and suppose suppose Vl, vi,V 2 ,... . . .,v are vectors in the property property that be written linin V V with with the that each each Vv E6 V V can can be written in in aa unique unique way way as as aa linear combination combination of of {Vl, (YI, V2,···, v 2 , . . . , vvn }. Then {Vb {vi, V . . . ,, v is called called aa basis basis of of V. V. }. Then V2,2 ,... v nn )} is ear Moreover, we say say that that nn is is the the dimension dimension of of V. V. Moreover, we A be shown A vector vector space space can can have have many many different different bases, bases, but but it it can can be shown that that each each contains the same number of well-defined. contains the same number of vectors, vectors, so so the the concept concept of of dimension dimension is is well-defined. We now now present present several We several examples examples of of bases. bases.
Example standard basis for RD ... ,e Example 3.24. 3.24. The The standard basis for Rn is is {el,e2, (ei,e2,... ,en}},; where where every every entry entry n of ej is zero except the jth, which is one. Then we obviously have, for any x E RD, of GJ is zero except the jth, which is one. Then we obviously have, for any x 6 R ,
and see that For example, for xx E R 33;, and it it is is not not hard hard to to see that this this representation representation is is unique. unique. For example, for £R
Example An alternate for R3 V2, V3},; where Example 3.25. 3.25. An alternate basis basis for R3 is is {Vb {vi,V2,V3J where
It may be obvious why one want to use this of It may not not be obvious to to the the reader reader why one would would want to use this basis basis instead instead of the much much simpler basis {el, {61,62,63}. it is is easy easy to to check check that the simpler basis e2, e3} . However, However, it that Vl . V2 = 0, Vl . V3 = 0, V2· V3 = 0,
3.3. Basis Basis and and dimension 3.3. dimension
51 51
and the basis basis {VI, {vi,V2,vs} easy to use as {61,62,63}. and this this property property makes makes the V2, V3} almost almost as as easy to use as {el, e2, e3}. We explore this topic in the the next section. We explore this topic in next section. Example The set the vector of degree Example 3.26. 3.26. The set P P n is is the vector space space of of all all polynomials polynomials of degree n n or or n less (see (see Exercise 3.1.3). The The standard basis is {l,x,x22,... To see standard basis is {1,x,x ,xn}. see that that this this less Exercise 3.1.3). , ... ,x }. To is indeed first note that every every polynomial € Pn can is indeed aa basis, basis, we we first note that polynomial ppEP can be be written written as as aa 2 n 2 linear combination of 1, x, x , . . . , x : linear combination of 1, X, x , ... , xn: p(x) = co' 1 + CIX + C2X 2 + ... + cnx n (this is is just definition of that this (this just the the definition of polynomial polynomial of of degree degree n). n). Showing Showing that this representarepresentation is unique is a little subtle. If we also had tion is unique is a little subtle. If we also had p(x) = do' 1 + dlx
+ d2x2 + ... + dnxn,
then, subtraction then, subtraction would would yield yield (eo - do)
+ (Cl
+ ... + (c n -
- ddx
dn)x n
=0
for every x. can have at most for every x. However, However, a a nonzero nonzero polynomial polynomial of of degree degree n n can have at most n n roots, roots, so it must be the case that that (CQ 4- (GI + ... ••• + + (c (cnn -— dn)x dn)xnn is is the the zero so it must be the case (eo — - do) do) + (CI — - di)x ddx + zero polynomial. That is, CQ=—do, do,Clci == dd1,i ... , . . ., ,Ccnn== ddnn must must hold. hold. polynomial. That is, Co Example 3.27. 3.27. An Example An alternate alternate basis basis for for P^ P 2 is is { I , x-~2' X2_X+~} 6
(the advantage advantage of this basis (the of this basis will will be be discussed discussed in in Example Example 3.39 3.39 in in the the next next section). section). To To 2 show this is we must that, given any p(x] — Co CQ+cix+C2X show that that this is indeed indeed aa basis, basis, we must show show that, given any p(x) = +CIX+C2X2,, there is aa unique unique choice choice of of the the scalars scalars ao,ai,fl2 ao, aI, a2 such such that that there is ao . 1 + al (x -
~) + a2 (X2 - X + ~)
= Co
+ CIX + C2X2.
This equation equation is is equivalent equivalent to to the the three This three linear linear equations equations 1
ao - '2a1
1
+ 6'a2
al -
= Co,
a2 =
CI,
a2 = C2·
The reader regardless of the The reader can can easily easily verify verify that that this this system system has has aa unique unique solution, solution, regardless of the values values 0/co,ci,C2. of eo,Cl,C2· Example 3.28. Yet another basis forP? is {L {1/1,1/2,1^3}, Example 3.28. Yet another basis for P 2 is where 1, L 2, L 3 }, where LI(X) = 2 (x
-~) (x -1),
L2(X) = -4x(x - 1), L3 (x) = 2x (x -
~) .
52
Chapter Chapter 3. 3.
Essential linear Essential linear algebra algebra
// write x\ Q, # = 1/2, 1/2, and and x% I , then the property property If we we write Xl = 0, X22 = X3 = 1, then the 1, i
L i (Xj ) = { 0,
i
= j, ¥- j
(3.9)
holds. the properties of aa basis can be be verified (see Exercise 5). holds. From From this this property, property, the properties of basis can verified (see Exercise 5). There are are two two essential of aa basis basis {v {vi, v2, of aa vector space There essential properties properties of 1,V ... , V n} vector space 2 ,...,v n } of every vector in V can can be represented as as aa linear linear combination of the V. First, First, every vector in be represented combination of the basis basis is unique. two definitions definitions provide provide vectors. vectors. Second, Second, this this representation representation is unique. The The following following two concise ways to two properties. concise ways to express express these these two properties. Definition 3.29. 3.29. Let be aa vector and suppose {YI, V2, v 2 , ... . . . ,, v Definition Let V be vector space, space, and suppose {Vl, V n} is aa colcoln } is lection in V. The The span span of {YI, V2, v 2 , ... . . . ,, v is the the set lection of of vectors vectors in of {Vl, v nn} } is set of of all all linear linear combicombinations of of these these vectors: nations vectors: spon{vi,v 2 ,...,v n } = {o!iVi+Q!2V2 +
that that
h anvn : c*i, a 2 , • • •, an e R}
Thus, one one of of the of aa basis {YI, v . . . ,, v of aa vector space V is Thus, the properties properties of basis {v!, V2, V n} vector space is 2 , ... n } of V = span{vi,v 2 ,...,v n }.
The reader should also also be aware that, for any vectors vVl, i,V 2 ,... . . . ,, Vn v n in The reader should be aware that, for any vectors V2, in aa vector vector space V, span span{vi, V 2 ,... . . . ,, v } is a subspace of (possibly the entire space as in {Vl' V2, V n} is a subspace of V (possibly the entire space V, as in space n the of aa basis). the case case of basis). Definition 3.30. A set of of vectors vectors {Vl' {YI, v . . . ,, v is called linearly independent Definition 3.30. A set V2, V n} called linearly independent ifif 2 , ... n } is the only . . . ,,Ccnn satisfying satisfying the only scalars scalars c\, Cl, c C2, 2 ,•••
are
Cl
= C2 = ... = Cn = o.
It can shown that the uniqueness part of the definition definition of of aa basis is equivaIt can be be shown that the uniqueness part of the basis is equivalent to independence of basis vectors. vectors. Therefore, for aa vector lent to the the linear linear independence of the the basis Therefore, aa basis basis for vector space is is aa linearly independent spanning spanning set. set. space linearly independent A third the number A third quality quality of of aa basis basis is is the number of of vectors vectors in in it-the it—the dimension dimension of of the the vector any two these properties properties imply the third. vector space. space. It It can can be be shown shown that that any two of of these imply the third. That has dimension two of of the following statements That is, is, if if V has dimension n, n, then then any any two the following statements about about {vi, V2, v 2 , ... . . . ,, vvd } imply the third: {Vl, imply the third: fc • k =n; •
{Vl, V2, ... ,
vd is linearly independent;
3.3. Basis and dimension
53 53
Thus if {vi, v . . . ,, Vfc} satisfy two of the above properties, it is Thus if {Vl' V2, vd isis known known to to satisfy two of the above properties, then then it is aa 2 , .•. basis V. basis for for V. Before the topic remind the fact Before leaving leaving the topic of of basis, basis, we we wish wish to to remind the reader reader of of the the fact indicated of this section, which which is is so so fundamental indicated in in the the opening opening paragraphs paragraphs of this section, fundamental that that we it formally formally as we express express it as aa theorem. theorem. Theorem Let A A be n x x n n matrix. A is Theorem 3.31. 3.31. Let be an an n matrix. Then Then A is nonsingular nonsingular if if and and only only ifif the columns of A form form aa basis basis for for R R nn .. the columns of
Thus, when columns form form aa basis for Rn, R n , and Thus, when A A is is nonsingular, nonsingular, its its columns basis for and solving solving Ax equivalent to finding the weights that that express express b as aa linear combination Ax = =b b is is equivalent to finding the weights b as linear combination of this basis. This answers the Suppose we of this basis. This fact fact answers the following following important important question. question. 99 Suppose we n n have . . . ,, v for Rn R and and aa vector b G R . Then, of course, course, b b is have aa basis basis YI, Vl, v V2, Vn vector bERn. Then, of is aa 2 , ... n for linear combination combination of find the weights in linear linear of the the basis basis vectors. vectors. How How do do we we find the weights in this this linear combination? How How expensive is it to do so (that is, how much work is required)? the scalars Xl, #2, X2, ..• the equation To find find the scalars #1, • • • ,X , xn in in the equation
10 we define definelO
and solve solve and Ax=b
via Gaussian elimination. The expense computing xx can countvia Gaussian elimination. The expense of of computing can be be measured measured by by counting the operations—the number of additions, subtractions, ing the number number of of arithmetic arithmetic operations-the number of additions, subtractions, multiplications, and divisions—required. The total number of of operations operations required required multiplications, and divisions-required. The total number to solve solve Ax Ax = and it convenient to leading to = bb isis aa polynomial polynomial in in n, and it is is convenient to report report just just the the leading term in in the the polynomial, polynomial, which which can can be be shown to be be term shown to 2
-n
3
3
(the lower-order terms terms are not very very significant large). We We usually usually express (the lower-order are not significant when when nn is is large). express this saying saying that this that the the operation operation count count is is
("on the the order order of of (2/3)n (2/3)n33"). "). In section, we discuss aa certain type of of basis for which In the the next next section, we discuss certain special special type basis for which it it is is much easier to in terms of the the basis. basis. much easier to express express aa vector vector in terms of 9
91f If the the importance importance of of this this question question is is not not apparent apparent to to the the reader reader at at this this point, point, it it will will be be after after he or or she reads the the next two sections. he she reads next two sections. 10 This notation is the columns are are the vectors VI, vi, V 2 ,..• . . . ,, v notation means means that that A A is the matrix matrix whose whose columns the vectors V2, Vn. lOThis n.
54 54
Chapter Essential linear Chapter 3. 3. Essential linear algebra algebra
Exercises Exercises 1.
(a) Let A
~ -~ -~ [
-n, ~ -n x
[
Compute boteAx ax and and Compute both
and are equal. equal, and verify verify that that they they are nxn (b) Let 6 R x n and e R n , and and suppose suppose the columns of of A (b) Let A A ERn and x x ERn, the columns A are are
so that j)-entry of of A Compute both and (XlVI (#iVi + so that the the (i, (i,j)-entry A is is (vj)j. (vjk Compute both (Ax)j (AX)i and xX2V2 1- x equal. XnVn)i, and verify verify that that they they are are equal. 2V2 + ... + nvn)i, and 2. Is 2. Is
basis for for R3? R3? (Hint: (Hint: As As explained explained in in the the last last paragraphs paragraphs of of this this section, section, the the aa basis three Rn if and only = b three given given vectors vectors form form aa basis basis for for Rn if and only if if Ax Ax = b has has aa unique unique solution for for every 6 R n , where columns are solution every b b ERn, where A A is is the the 33 x x3 3 matrix matrix whose whose columns are the the three vectors.) three given given vectors.) 3. Is 3. Is
basis for for R3? R3? (See (See the the hint hint for for the the previous previous exercise.) exercise.) aa basis 4. Show that 4. Show that {X2
+ 1, X + 1, X2 -
X
+ I}
is space of less. (Hint: Verify is aa basis basis for for P%, P2, the the space of polynomials polynomials of of degree degree 22 or or less. (Hint: Verify directly directly that that the the definition definition holds.) holds.) 5. Show that {L1? L22 ,,L L33}, defined in in Example is aa basis for P 5. Show that {L }, defined Example 3.28, 3.28, is basis for P22-. (Hint: (Hint: Use Use I ,L (3.9) to show that (3.9) to show that
holds for every p E P2 .)
projections 3.4. Orthogonal Orthogonal bases and projections
55
6. Let V be be the the space of all continuous, complex-valued complex-valued functions functions defined on the 6. Let space of all continuous, defined on the real line: line: real V = {f : R -+ C : f is continuous}. lx spanned by e ei:v and e-i:v, where ii = = A. Define W to be the subspace of V spanned e lx, where ^/^l. Show that {cos sin (x) another basis basis for for W. (Hint: Use Euler's Euler's formula: Show that {cos (x), (x), sin (x}}} is is another (Hint: Use formula: eiIJ = cos (0) + i sin (0).)
3.4 3.4
Orthogonal and projections projections Orthogonal bases bases and
At the the end end of of the the last last section, section, we we discussed discussed the the question question of expressing aa vector vector in in At of expressing terms of of aa given basis. This This question is important important for for the the following following reason, reason, which which terms given basis. question is describe in general general terms at problems that are we can only describe at the moment: Many problems are admit a special posed in vector spaces admit special basis, in in terms of which the problem is is easy easy to solve. solve. That That is, is, for for many many problems, problems, there there exists exists aa special basis with with the the property to special basis property that if if all all vectors vectors are in terms terms of that basis, basis, then then aa very very simple simple calculation that are expressed expressed in of that calculation will produce produce the the final final solution. For this this reason, reason, it it is is important important to to be be able able to to take take aa will solution. For vector (perhaps in terms terms of of aa standard basis) and and express in terms terms of of aa vector (perhaps expressed expressed in standard basis) express it it in different basis. In the latter part of this section, we we will study different study one one type of problem problem advantageous to use aa special another such for which it is advantageous special basis, and we will discuss another the next next section. problem in in the problem section. It is easy to to express vector in terms of of aa basis basis if if that that basis basis is orthogonal. is quite quite easy express aa vector in terms is orthogonal We wish wish to to describe describe the the concept concept of of an an orthogonal orthogonal basis basis and and show some important We show some important examples. Before Before we we can can do do so, we must must introduce introduce the the idea of an product, examples. so, we idea of an inner inner product, which is is aa generalization generalization of of the the Euclidean Euclidean dot dot product. which product. 3 R22 and R R3. The dot product plays a special role in the geometry of R . The 2 3 reason for this is is the the fact that two two vectors vectors x,y x, y in R2 or R3 are are perpendicular perpendicular ifif reason for this fact that in R or R and only ifif and only x·y =0. Indeed, one can show that Indeed, one can show that
x . y = Ilxllllyll cos (0), where 09 is is the the angle angle between between the the two two vectors vectors (see (see Figure Figure 3.2). 3.2). where From elementary elementary Euclidean Euclidean geometry, geometry, we we know know that, that, if if x x and and y y are are perpenperpenFrom dicular, then dicular, then
Ilx + yW = IIxl1 2+ IIyl12
(the Pythagorean Pythagorean theorem). theorem). Using Using the the dot dot product, product, we we can can give give aa purely purely algebraic (the algebraic proof of the the Pythagorean Pythagorean theorem. By definition, definition, proof Ilull=~,
so
Ilx + Yl12 = (x + y) . (x + y) =x,x+x·y+y·x+y·y =x·x+2x·y+y·y = IIxl12 + 2x . y + IIYI12.
56 56
Chapter Chapter 3. Essential Essential linear algebra y
e
x
Figure 3.2. The The angle two vectors. Figure 3.2. angle between between two vectors. This calculation shows that This calculation shows that
holds O. holds if if and and only only if if x .• y = = 0. Seen this theorem is is an an algebraic algebraic property can be Seen this way, way, the the Pythagorean Pythagorean theorem property that that can be deduced in in RD, Rn, n 3, even in those spaces we cannot visualize vectors or deduced n > 3, even though though in those spaces we cannot visualize vectors or what it to use the word word orthogonal it means means for for vectors vectors to to be be perpendicular. perpendicular. We We prefer prefer to use the orthogonal what instead of of perpendicular: Vectors xx and in RD Rn are are orthogonal orthogonal if = 0. instead perpendicular: Vectors and y y in if x x·• y y = O. In course of of solving solving differential differential equations, deal with with function function spaces spaces In the the course equations, we we deal in to Euclidean and our methods are heavily dependent dependent on in addition addition to Euclidean spaces, spaces, and our methods are heavily on the the existence product-the analogue dot product product in in more more general general existence of of an an inner inner product—the analogue of of the the dot vector spaces. spaces. Here Here is vector is the the definition: definition:
Definition 3.32. V be be aa real (real) inner inner product product on V is is aa Definition 3.32. Let Let V real vector vector space. space. A A (real) on V function, usually .) or or (., taking two two vectors from V producing function, usually denoted denoted (', (•,•) ( - , -.)) vv,> taking vectors from V and and producing aa real function must following three properties: real number. number. This This function must satisfy satisfy the the following three properties: 1. (u, (v, u) vectors uu and and Vi v; 1. (u, v) v) = = (v, u) for for all all vectors 2. (cm + + fiv, /?v, w) = a(u, a(u, w) + /?(v, (w, au cm + + fiv) /?v) = a(w, u) + fi(w, /?(w, v) v) for 2. (au w) = w) + fi(v, w) w) and and (w, = a(w, u) + for all vectors u, u, v, v, and and w, and all all real real numbers and fii all vectors w, and numbers aa and /3;
3. all vectors vectors u, u, and and (u, (u, u) 0 if if and if uu is 3. (u, (u, u) u) > 2': 00 for for all u) = = 0 and only only if is the the zero zero vector. vector. It should should be easy easy to check check that these properties hold for for the ordinary ordinary dot dot product on Euclidean product on Euclidean n-space. n-space. Given an an inner inner product space (a (a vector an inner product), we define Given product space vector space space with with an inner product), we define orthogonality just as in in Euclidean space: two are orthogonal orthogonal if if and and only only ifif just as Euclidean space: two vectors vectors are orthogonality their zero. It their dot product is zero. It can can then then be shown shown that that the the Pythagorean Pythagorean theorem holds holds (see Exercise Exercise 3). 3). (see
57
Orthogonal bases and projections 3.4. Orthogonal projections
An orthogonal orthogonal basis an inner space V V is is aa basis {vi, v An basis for for an inner product product space basis {VI, V2, ... , v v nn}} 2 ,..., with the property that that with the property i =I- j
'* (Vi,Vj) =
0
(that is, basis is to every every other other vector vector in in the the basis). basis). (that is, every every vector vector in in the the basis is orthogonal orthogonal to We now now demonstrate the first first special special property property of of an an orthogonal orthogonal basis. basis. Suppose Suppose We demonstrate the {VI, V2, is an an orthogonal basis for an inner inner product product space x is is any any {vi, v . . . ,, vvn }} is orthogonal basis for an space V and and x 2 , ... exist scalars scalars aI, a2, ... ann such that vector in in V. Then vector Then there there exist ai, 0:2, • • • ,, a such that
(3.10) To deduce deduce the we take take the inner product (3.10) with To the value value of of Q:J, ai, we the inner product of of both both sides sides of of (3.10) with Vi: (Vi,X) = (Vi,aIVI + a2v2 + ... + anv n ) = al (Vi, vd + a2 (Vi, V2) + ... + an(Vi, Vn ) = ai(Vi, Vi). The step follows that every product (Vi, Vj) vanishes vanishes except The last last step follows from from the the fact fact that every inner inner product (v^, YJ) except (vj,Vj). then obtain obtain (Vi, Vi). We We then
(Vi, x) ai = ( )' i = 1,2, ... , n, Vi, Vi and so and so
X=
(VI,X) VI (VI, VI)
+
(V2,X) V2 (V2, V2)
+ ... +
(vn,x) Vn. (V n , v n )
(3.11)
This formula shows shows that that it it is is easy easy to to express express aa vector terms of of an an orthogonal orthogonal vector in in terms This formula basis. Assuming Assuming that that we we compute compute (VI, vt), (v2, (V2, V2), basis. (vi, vi), V 2 ) ... , . . .,,(V (vnn,,Vvnn)) once once and and for forall, all, itit n inner products to requires just n to find the weights in in the linear linear combination. In the the - 11 arithmetic arithmetic operations case n-vectors, aa dot dot product product requires requires 2n — case of of Euclidean Euclidean n-vectors, operations (n multiplications and and nn -— 1I additions), additions), so the total total cost cost is multiplications so the is just just
If n is large, this this is less costly the O(2n O(2n33/3) /3) operations operations required for aa If is large, is much much less costly than than the required for nonorthogonal basis. basis. We We also remark that that if the basis basis is is orthonormal-each nonorthogonal also remark if the orthonormal—each basis basis vector is is normalized normalized to have have length one-then one—then (3.11) simplifies simplifies to to (3.12)
Example 3.33. The basis {vi,v 2 ,v 3 } forH3, where
58
bra Chapter3.3. Essential Essentiallinear linear algebra Chapter algebra
is verified directly. If is orthonormal, as as can can be be verified If
then
3.4.1 3.4.1
L22 inner product The L
We have seen that functions functions can can be be regarded regarded as We have seen that as vectors, vectors, at at least least in in aa formal formal sense: sense: functions can be added together together and multiplied by scalars. (See Example 3.4 in Section 3.1.) functions are not so different 3.1.) We will now show more directly that functions different we show that aa suitable product can can from vectors. In In the the process, from Euclidean Euclidean vectors. process, we show that suitable inner inner product be defined defined for for functions. functions. be Suppose we have have aa function b]-a continuous function defined Suppose we function 9g E G era, C[a, b]—a continuous function defined on on the the interval [a, b]. we can produce a vector that approximates interval [a, &]. By sampling 9g on a grid, we approximates the function g. Let Xi Xi = i~x, A# ~x = = (b - a)/N, a)/N, and define a vector vector G E = a + iAx, (b — G RN RN by
Gi=g(Xi), i=O,l, ... ,N-1. Then G can can be be regarded regarded as as an an approximation approximation to to g 9 (see (see Figure 3.3). Given Given another another Then Figure 3.3). function f(x) and the corresponding corresponding vector vector FERN, we have function f(x) and the F € R N , we have N-l
F·G=
L
FiGi
i=O N-l
=L
f(xi)g(x;).
i=O
Refining the discretization leads to that obviously obviously Refining the discretization (increasing (increasing N) leads to aa sampled sampled function function that original function more accurately. Therefore, we we ask: What happens represents the original to as N -t to F F .• G G as ->• oo? oo? The The dot dot product product N-l
F .G =
L
f (Xi) 9 (Xi)
i=O
does not converge to any value as as N -t but aa simple modification induces does not converge to any value —>• 00, oo, but simple modification induces convergence. We replace the ordinary dot product by the following scaled dot we introduce new notation: notation: product, for which we product, for which introduce aa new N-l
(F, G) =
L
i=O
FiGi~X,
59
3.4. Orthogonal bases and projections projections
o
0.2
0.6
0.4
0.8
x
Approximating a a function function g(x] g(x) by by a a vector vector G ERN. Figure 3.3. Approximating € RN.
we have Then, when F and G are sampled functions as above, we N-l
(F, G) =
L
1 b
f(xi)g(xi)6.x -+
f(x)g(x) dx as N -+
00.
a
i=O
Based on on this this observation, observation, we we argue argue that that aa natural natural inner inner product product (•, (.,.)•) on on Based C[a, b] is b] is
era,
(f,g) =
lb
I(x)g(x) dx.
(3.13)
Just dot product on Euclidean Euclidean n-space (\\x\\ = • x], so the Just as the the dot product defines defines aa norm norm on n-space (11xll = ^/x y'X-X), so the functions: inner product (3.13) defines a norm for functions:
11111 = v(f, f) = foC II(x)1 2 dx.
(3.14)
the definition of norm. Norms Norms measure the the size or magniFor completeness, we give the tude of of vectors, vectors, and and the the definition definition is is intended intended to to describe describe abstractly the properties tude abstractly the properties reasonable notion of size ought to have. that any reasonable Let V be a a vector vector space. space. A A norm on is aa real-valued real-valued function Definition 3.34. 3.34. Let V be on V V is function with V, usually usually denoted denoted by by II\\ .• II\\ or or II\\ .• Ilv, \\v, and and satisfying satisfying the the following following with domain domain V, properties: properties:
60
Chapter 3. Essential linear algebra
1. Ilvll ~ 0 for all v E V and Ilvll = 0 if and only if v =
o.
2. Ilavll = lalllvil for all scalars a and all v E V. 3. Ilu + vii ~ Ilull
+ IIvll
for all u, v E V.
The last property property is called called the The last the triangle inequality.
For For Euclidean vectors in the plane, the triangle inequality expresses expresses the fact fact that one side side of a triangle cannot be longer than the sum of the other two sides. 11 The inner product defined by (3.13) is the so-called L L22 inner product.l1 Two b] are g) = 0. O. This condition does functions in in era, C[a, b] are said said to be orthogonal if if (f, (/,#) does not have a direct geometric meaning, as the analogous condition does for Euclidean 3 R22 or R R3, we argued above, orthogonality is still important vectors in R , but, as we important algebraically. we measure norm in the L L22 sense, we we say that functions f/ and g 9 are When we close (for (for example, good approximation approximation to that g 9 is is aa good to /) f) ifif close example, that
2 g(X))2 is small for for every x €E [a, [a, b]b] is small. This does not mean that (f(x) (/(#) - g(x}) is 2 ((f(x) _g(X))2 can be large in places, as long as this difference ((f(x) —g(x)) difference is large only over very but rather it implies that (f(x) - g(x))2 the average over small intervals), but (f(x) — g(x)}2 is small on the the interval [a, 6]. b]. For this reason, we often often -use use the the term "mean-square" "mean-square" in the interval [a, in referring referring L22 norm (for we might say to the the L (for example, example, we say "g "# is close close to f/ in in the mean-square mean-square sense"}. sense").
Example If /f ;: [0,1] R is defined defined by by f(x) x(1 — - x), then Example 3.35. 3.35. // [0,1] --t —> R f(x) = = x(l then
Ilfll =
t x2(1- x)2 dx = V35 fT = _1_ == 0.1826. v'3O
10
With 1] -» --t R defined by With g9 ;: [0, [0,1] defined by
g(x) = 83 sin (1I"x) , 11"
we have we have Ilf-gll=
fo
1
(X(1-X)-:3 sin (1I"X)r dx=
(11"6 - 960) == 0.006940. 3011"6
These two functions functions differ differ by by less than 4% in in the mean-square sense sense (cf. These two less than the mean-square (cf. Figure Figure
3·4)· 3.4).
11 llThe refers to to the the French French mathematician mathematician Lebesgue, the "2" to the the exponent exponent in in the The "L" "L" refers Lebesgue, and and the "2" to the formula for L22 norm norm of of aa function. symbol L L22 is is read read "L-two." formula for the the L function. The The symbol "L-two."
61 61
3.4. Orthogonal bases and projections 0.3r-------..---~---_r_;::=:::::::::;::3r:=;;===iI
.......
-- .......
0.1
0.2
0.4
0.6
0.8
x Figure 3.4. functions of Example 3.35. 3.35. Figure 3.4. The The functions of Example
3.4.2 3.4.2
The projection theorem
The is about approximating aa vector vector vv in vector space by The projection projection theorem theorem is about approximating in aa vector space V by aa vector vector from from aa subspace subspace W. W. More More specifically, specifically, the the projection projection theorem theorem answers answers the the question: Is Is there £W W closest closest to (the best best approximation approximation to to vv from from w E to vv (the question: there aa vector vector w W), and and if if so, so, how can it it be computed? Since Since this this theorem theorem is so important, important, and its W), how can be computed? is so and its proof we will prove the the theorem. proof is is so so informative, informative, we will formally formally state state and and prove theorem. Theorem Let V be aa vector space with with inner inner product product (-,-), (', .), let Theorem 3.36. 3.36. Let V be vector space let W W be be aa finite-dimensional subspace subspace of and let v E V. finite-dimensional o f VV,, and letv^V. 1. EW such that 1. There There is is aa unique unique uu G W such that
Ilv - ull = min Ilv - wll· wEW
That is, there there is unique best to v v from from W. That is, is aa unique best approximation approximation to W. We We also also call call uu the projection o/v onto W, W, and the projection of v onto and write write u = projwv.
2. A A vector vector u best approximation from W only ifif 2. u E EW W is is the the best approximation to to vv from W if if and and only
(v - u, z) = 0 for all z E W.
(3.15)
62
Chapter Chapter 3. 3. Essential Essential linear linear algebra algebra 3. If If {WI, ... , w for W, 3. {wi, W2, w 2 ,..., wnn}} is is aa basis basis for W, then then n
projw v =
(3.16)
LXiWi, i=1
where where Gx = b,
(Wj, Wi),
G ij =
(Wi, v).
bi =
(3.17)
The equations equations represented represented by by Gx Gx = = bb are are called called the the normal normal equations, equations, and The and the matrix G is is called called the the Gram Gram matrix. matrix. the matrix G 4- If (w1; W2,···, w2,..., w wnn}} is is an an orthogonal orthogonal basis basis for W, then then the the best best approximation 4· If {Wl, for W, approximation to vv from from W W is to is
.
proJw v =
~ (Wi'V) ~ ( . .)Wi. i=1
(3.18)
W" W,
IfIf the the basis basis is is orthonormal, orthonormal, this this simplifies to simplifies to n
projw v
= L(Wi, V)Wi.
(3.19)
i=l
Proof. We prove the uE any Proof. We will will prove the second second conclusion conclusion first. first. Suppose Suppose that that u e W, and and zz is is any other vector in W. Then, Then, since since W is is closed closed under under addition addition and and scalar scalar multiplication, multiplication, other vector in we that uu + + tz tz E €W W for for all all real real numbers numbers t. On On the the other other hand, hand, every every other other we have have that vector W w in in W can can be written as as uu + + tz tz for for some some zz E £ W and and some some tt E €R R (just (just take take vector be written w -— uu and and tt == 1). 1). Therefore, Therefore, uu E €W W is is closest closest to to vv if if and and only only if zz =— W if Ilv - ull
s: IIv -
(u + tz)11 for all z
E
W, t E R.
(3.20)
2 2
Since IIxl1 ||x|| == (x,x), (x,x), this this last last inequality inequality is is equivalent equivalent to to Since (v - u, v - u)
s: (v -
(u + tz), v - (u + tz)) = ((v - u) - tz, (v - u) - tz) = (v - u, v - u) - 2t(v - u,z) + e(z,z)
or to to or t 2 (z, z) - 2t(v - u, z)
2': 0 for all z E W, t E R.
If we regard If we regard zz as as fixed, fixed, then then
e(z,z)+2t(v-u,z) O. is is aa simple simple quadratic quadratic in in t, and and the the inequality inequality holds holds if if and and only only if if (v (v -- u, u, z) z) = = 0. It holds for for all It follows follows that that the the inequality inequality holds all zz and and all all tt if if and and only only if if (3.15) (3.15) holds. holds. In addition, addition, provided / 0, 0, (3.20) (3.20) holds holds as as an an equation equation only only when when tt = = 00 (since (since In provided zz i(z, z) > 00 for (z, z) for zz i^ 0). 0). That That is, is, if if W wE € W and and W w i^ u, u, then then
IIv-ull < IIv-wlI·
Orthogonal bases and projections projections 3.4. Orthogonal
63
Thus, if the the best best approximation problem has has aa solution, Thus, if approximation problem solution, it it is is unique. unique. We now prove prove the remaining conclusions conclusions to to the the theorem. theorem. Since finiteWe now the remaining Since W is is aa finitedimensional subspace, has aa basis basis {WI, W2, ... , w }. A vector u E W solves the dimensional subspace, it it has {wi, w , . . . , w }. A vector u e solves the 2 n best approximation approximation problem problem if if and and only only if holds; moreover, moreover, it it is is straightforbest if (3.15) (3.15) holds; straightforthat (3.15) (3.15) is is equivalent equivalent to ward to ward to show show that to (v - u, Wi) = 0 for i = 1,2, ... , n
(3.21)
(see Exercise Exercise 5). Any Any vector vector uu E can be be written written as (see G W can as n
u= LXjWj.
(3.22)
j=1
Thus, is aa solution if and and only if (3.22) holds and Thus, uu 6E W W is solution if only if (3.22) holds and
(
V-
Wi) = 0, i = 1,2, ... ,n,
(3.23)
= (Wi,V), i = 1,2, ... ,n.
(3.24)
"tXjWj, )=1
which simplifies to to which simplifies n L(Wj,Wi)Xj j=1
nxn If we define G Gy = (Wj, (w j? Wj) 6 Rn by (3.24) If we define G 6 ER R nxn by by G Wi) and and b bERn by bit = (w (Wi,i? v), then then (3.24) ij = is is equivalent equivalent to to Gx=b.
It can be be shown shown that that G is nonsingular nonsingular (see (see Exercise Exercise 6), the unique unique best best approxIt can G is 6), so so the approximation W is is given given by (3.22), where solves Gx Gx = b. to v v from from W by (3.22), where x x solves = b. imation to If the the basis basis is orthogonal, then then (Wj, Wi) = unless jj = In this this case, G is is If is orthogonal, (wj, w;) = 00 unless = i. i. In case, G the diagonal matrix with diagonal entries entries
and Gx Gx — = b b is is equivalent the n simple and equivalent to to the simple equations equations
that is, is, to that to
(v, Wi)
Xi
This completes completes the proof.
= (Wi,Wi )'
i = 1,2, ... ,n.
0
We now present present two two examples examples of of best best approximation approximation problems problems that that commonly commonly We now computational practice. occur in scientific scientific and computational practice.
64
Chapter Chapter 3. 3. Essential Essential linear algebra
Example 3.37. 3.37. We assume that that data data points Example We assume points
(xl,yd (xi,yi) (X2, Y2) (z 2 ,!fe) (z 3 ,2/Y3) (X3, 3) (0:4,2/4) (X4, Y4) (X5,Y5) (a*,2/5)
= (0.10,1.7805), = = (0.30,2.2285), = (0.40,2.3941), = (0.40,2.3941), = (0.75,3.2226), = (0.75,3.2226), = (0.90,3.5697) -(0.90,3.5697)
have collected in in the laboratory, and and there there is is aa theoretical theoretical reason reason to have been been collected the laboratory, to believe believe that yi = = axi ought to of a, that Yi aXi + bb ought to hold hold for for some some choices choices of a, b b6 E R. R. Of Of course, course, due due to to measurement error, this relationship is unlikely to hold exactly for any any choice choice of of aa measurement error, this relationship is unlikely to hold exactly for and b, b, but but we we would to find and bb that come as close as as possible possible to and would like like to find aa and that come as close to satisfying satisfying it. If we we define it. If define
Y=[H~:~ ,x~ 3.2226 3.5697
0.10 0.30 0.40 0.75 0.90
I
,e =
1 1 1 1 1
then one to pose this problem is to + be is as then one way way to pose this problem is to choose choose aa and and bb so so that that ax ax + be is as close close as possible possible to to yy in the Euclidean Euclidean norm. norm. That That is, we find the best best approximation approximation to to as in the is, we find the y W= = span{x, e). y from from W span{x, e}. The matrix is is The Gram Gram matrix G
= [ x· x x .e
e· x ] e· e
= [1.6325 2.45
2.45] 5 '
and the right-hand of the the normal normal equations equations is is and the right-hand side side of b = [ x· y ] == [ 7.4370 ] e· y 13.196· Solving the the normal yields Solving normal equations equations yields
== [ 2.2411 ] [ a] b 1.5409' The resulting linear model is is displayed, together with with the original data, data, in in Figure Figure The resulting linear model displayed, together the original 3.5.
Example 3.38. One One advantage advantage of working with with polynomials polynomials instead Example 3.38. of working instead of of transcentranscendental functions like eXx is is that that polynomials polynomials are are very very easy evaluate~only the basic dental functions like e easy to to evaluate—only the basic arithmetic required. For For this it is often desirable to approxarithmetic operations operations are are required. this reason, reason, it is often desirable to approximate imate aa more more complicated complicated function function by by a a polynomial. polynomial. Considering Considering f(x] f(x) = = eX ex as as aa function function in in (7[0,1], C(O, 1], we we find find the the best best quadratic quadratic approximation, approximation, in in the the mean-square mean-square sense, basis for the space of polynomials {l,x,x sense, to to ff.. A A basis for the space V^ P2 of polynomials of of degree degree 22 or or less less is is {I, x, x 22 }. }.
3.4. Orthogonal bases and projections
65
4~------~------~------~~------~------~
x
Figure 3.5. 3.5. The The data 3.37 and the approximate approximate linear linear reFigure data from from Example Example 3.37 and the relationship. lationship. It is for the problem of best It is easy easy to to verify verify that that the the normal normal equations equations for the problem of finding finding the the best approximation from P2 matrix-vector form) approximation to to f from p2 are are (in (in matrix-vector form)
Ga = b, Ga=b, where where
G
=[
1~2
1/2 1/3 ] [ e- 1 ] 1 . 1/3 1/4 ,b = 1/5 e 2 1/3 1/4
Since Since G-1b ==
1.013] 0.8511 , [ 0.8392
the best best quadratic quadratic approximation approximation is is the eX
== JJ2{x)
= 1.013 + 0.8511x + 0.8392x2.
The function and quadratic approximation approximation are graphed in The exponential exponential function and the the quadratic are graphed in Figure Figure 3.6. 3.6.
Example 3.39. An orthonormal for p2 P 2 (on the interval Q2, Q3}, 3.39. An orthonormal basis for interval [0,1]) [0,1]J is {ql, {^1,^25^3}; where where
66
Chapter 3. Essential linear algebra
2.8 2.6
x
y=e - _. y=P2(X)
2.4 2.2 2 >-
1.8 1.6 1.4 1.2 1
0
0.2
0.4
0.6
0.8
x
Figure 3.6. The function and the Figure 3.6. The function f(x) f(x) — = eeXx and the best best quadratic quadratic approximation approximation yy — = Pi(x) p2(X) (see (see Example Example 3.38). 3.38). The best best approximation approximation p^ to eeXx (which (which was the previous The P2 to was calculated calculated in in the previous exercise) exercise) can can be computed by the formula be computed by the formula P2(x) = (qij)qi(x) + (q2,f}qi(x) + (tfs,/)^). A A direct direct calculation calculation shows shows that that the the same same result result is is obtained. obtained. (See (See Exercise Exercise 10.) 10.) The approximation is The concept concept of of best best approximation is central central in in this this book, book, since since the the two two main main solution techniques techniques we we discuss, discuss, Fourier series and finite elements, elements, both both produce produce aa solution Fourier series and finite best approximation to of aa BVP. BVP. best approximation to the the true true solution solution of
Exercises 1. (a) Show that {vi,vV2, from Example Example 3.33 orthonormal 1. (a) Show that the the basis basis {VI, V3} from 3.33 is is an an orthonormal 2 ,V3} basis R33.. basis for for R (b) Express Express the (b) the vector vector
as combination of VI,, V2, V2, \s. V3. as aa linear linear combination of YI 2. Is the {1/1,1/2,1/3} (on the interval [0,1]) [0,1]) given 3.28 2. Is the basis basis {L for P P2 the interval given in in Example Example 3.28 2 (on I ,L 2 ,L 3 } for an orthogonal orthogonal basis? basis? an
67
Orthogonal bases and projections 3.4. Orthogonal projections E V satisfy 3. Let V be an inner product space. Prove that x, yy e
and only if (x, (x, y) = = 0. O. ifif and only if
4. Use the results of of this section section to show show that any any orthonormal orthonormal set containing containing n n vectors in R Rnn is a basis for for R Rn. . (Hint: Since the dimension of R Rnn is is n, it suffices orthogonal set spans Rn suffices to show either that the orthogonal Rn or that it is linearly independent. Linear independence is probably easier.) 5. Let W be be aa subspace subspace of inner product product space and let W2, w n }} 5. Let of an an inner space V and let {Wi, {wi, w ..., w 2 , ... E V, be a basis for for W. Show that, for for y e V, all zz E (y, z) z) = 00 for for all GW
holds if and and only ifif
(y,Wi) =0, i=1,2, ... ,n holds. holds. 6. W2, w n} } be a linearly independent 6. Let {Wi, {w1; w ..., w independent set set in an an inner inner product space space 2 , ... nxn V, and define G E R nxn by y, and eR by G = (Wj,Wi), = 1,2, Gijij = (wj-jWj), i,j i,j = 1 , 2 ,... . . . ,,n. n.
Prove that G is invertible. Hint: By the Fredholm alternative, alternative, it suffices suffices to = O.0. Assume that x EE Rn show that the only solution to Gx = 0 is is x = Rn satisfies Gx = = O. 0. This implies that x·Gx = O. Show that that x X·• Gx Gx = = (u, u), where where u E is given given by by Show € V is
that u u cannot cannot be be the the zero zero vector vector unless unless x x= o. and show and show that = 0. 7. Consider the following data: x X 1.0000 1.2500 1.4000 1.5000 1.9000 2.1000 2.2500 2.6000 2.9000
yy
2.0087 2.4907 2.8363 2.9706 3.9092 3.9092 4.1932 4.1932 4.5057 4.5057 5.2533 5.8030
68
Chapter 3. Essential linear algebra
Find the the relationship = mx these data. data. Plot Plot the Find relationship yy = mx + Cc that that best best fits fits these the data data and and the best fit line. 8. the following following data: 8. Consider Consider the data: X x 1.0000 1.0000 1.2500 1.2500 1.4000 1.4000 1.5000 1.9000 2.1000 2.1000 2.2500 2.6000 2.9000
y
Y
1.2475 1.2475 1.6366 1.6366 1.9823 1.9823 2.2243 3.4766 4.2301 4.8478 6.5129 8.2331
Find the the relationship relationship yy = = C2X c^x22 + +C\X data. If If possible, Find C1X + CQ Co that that best best fits fits these these data. possible, plot the data and the best fit fit parabola. parabola. 9. 9. Using the the orthonormal orthonormal basis basis {Ql,Q2,q3} {^1,^2,^3} for P2 P% given given in in Example 3.39, find find the quadratic quadratic polynomial p(x] p(x) that the that best approximates approximates g(x) = — sin sin (nx), (TTX), in the the mean-square sense, over the interval [0,1]. Produce a graph of 9g and the quadratic approximation. quadratic approximation. 10. Verify that in Example Example 3.39, is an 10. (a) (a) Verify that {Ql, (<7i, Q2, #2) Q3}, #3}, defined defined in 3.39, is an orthonormal orthonormal basis basis for for P2 on on the interval interval [0,1]. [0,1]. (b) Using this this orthonormal basis, find find the the best to ff(x) (b) Using orthonormal basis, best approximation approximation to (x) = = eeXx (on the the interval interval [0,1]) [0,1]) in in the sense, and and verify same (on the mean-square mean-square sense, verify that that the the same result is is obtained obtained as as in in Example 3.38. 3.38.
3.5
Eigenvalues and eigenvectors of a symmetric matrix matrix
of The eigenvectors eigenvectors of of a matrix matrix operator operator are are special special vectors for for which which the action action of the we discuss of the matrix is is particularly simple. simple. In In this section, section, we discuss the the properties of eigenvalues and and eigenvectors of symmetric symmetric matrices. this topic eigenvalues eigenvectors of matrices. Understanding Understanding this topic will will set the stage method for for solving set the stage for for the the Fourier Fourier series series method solving differential differential equations, equations, which which takes advantage advantage of of the the simple differential operator operator on its eigenfunctions. takes simple action action of of aa differential on its eigenfunctions. nxn Definition 3.40. 3.40. Let A E Rnxn. of A Definition 6R . We say that the scalar A is an eigenvalue of if there exists nonzero vector such that that if there exists aa nonzero vector xx such
Ax = AX. As we discuss below, below, the the scalar scalar A may be complex complex even even if has only only real entries; As we discuss may be if A has real entries; if A is complex, then then xx must complex entries. entries. if is complex, must have have complex
3.5. Eigenvalues Eigenvalues and eigenvectors eigenvectors of a symmetric symmetric matrix
69
As we eigenvalues are are naturally expressed as the As we are are about about to to demonstrate, demonstrate, eigenvalues naturally expressed as the roots of polynomial, with the coefficients the polynomial polynomial computed roots of aa polynomial, with the coefficients of of the computed from from the the entries in in the matrix. A with real coefficients can roots. entries the matrix. A polynomial polynomial with real coefficients can have have complex complex roots. For complex eigenvalues and eigenvectors eigenvectors in the is natural natural to to allow allow complex eigenvalues and in the For this this reason, reason, it it is above definition. However, we will show below that a symmetric matrix can have above definition. However, we will show below that a symmetric matrix can have only real will simplify matters. Moreover, Moreover, the and only real eigenvalues, eigenvalues, which which will simplify matters. the eigenvalues eigenvalues and eigenvectors of of aa symmetric other useful which we also eigenvectors symmetric matrix matrix have have other useful properties, properties, which we also explore below. explore below. The following calculation is is essential for understanding understanding eigenvalues eigenvalues and The following calculation essential for and eigeneigenvectors: vectors:
Ax= .>.x, x:f:.O {:} .>.x - Ax = 0, x :f:. 0 {:} (.>.1 - A)x = 0, x:f:. o. This last condition is is possible if 'AI is aa singular This last condition possible if if and and only only if >'1 -— A A is singular matrix; matrix; that that is, is, if if and only only ifif and det (.>.1 - A) = 0, 12 where is the determinant12 of the square matrix matrix B. det(B) is the determinant of the square B. In In principle, principle, then, then, we we where det(B) can find find the the eigenvalues eigenvalues of by solving solving the the equation det (AI can of A A by equation det (.>.1 -— A) A) = 0. O. It is hard to PA (A) = det (.>.1 (AI — is aa polynomial of degree degree nn It is not not hard to show show that that PA('>') = det - A) A) is polynomial of (the characteristic polynomial of A), and so A has n eigenvalues (counted according (the characteristic polynomial of A), and so A has n eigenvalues (counted according to as roots of PA (A)). Any all of of the eigenvalues can can be to multiplicity multiplicity as roots of PA('>')). Any or or all the eigenvalues be complex, complex, even if real entries, entries, since coefficients can have complex even if A A has has real since aa polynomial polynomial with with real real coefficients can have complex roots. roots.
Example 3.41. Let
Then Then
PA('>') = det (.>.1 - A) -1 '>'-1 -1 .>. - 2 1
= (.>. - 1)('>' - 2) -1 =.>.2 _ 3'>' + 1. Therefore, eigenvalues are are Therefore, the the eigenvalues
3± v'5 2 12 12We that the the reader reader is is familiar familiar with with the the elementary properties of of determinants determinants (such (such as as We assume assume that elementary properties the computation of of determinants determinants of matrices). The The most most important matrix the computation of small small matrices). important of of these these is is that that aa matrix is singular singular if if and its determinant introductory text on linear is and only only if if its determinant is is zero. zero. Any Any introductory text on linear algebra, algebra, such such as as [34], can can be details. [34], be consulted consulted for for details.
70
Chapter Chapter 3. 3. Essential linear algebra algebra
Example 3.42. 3.42. Let Example Let
Then Then PA (A) = det (AI - A)
=1
A-I
-1 A-I
1
+1 A2 - 2A + 2.
= (A - 1)2 =
Therefore, the the eigenvalues eigenvalues are Therefore, are 1 ± i,
where i = ^f—^.. wherei=A· Example Example 3.43. 3.43. Let Let
Then Then PA(A)
= det (AI -
A) -1 A-I
o
-1 -1 A-I
Therefore, the only only eigenvalue is 1, of multiplicity multiplicity 3. In In this Therefore, I, which is an eigenvalue of example, We must it is is instructive instructive to to compute compute the the eigenvectors. eigenvectors. We must solve solve (XI (AI — - A.)x A)x = = 0 example, it for We have for A A= = 1, 1, that that is, is, (I (I — - A)x A)x = = 0. O. We have I- A
=
01 01 00 [ 001
1-
III = [00
[10 1 1 001
-10 -1 -1 000
1,
and and aa straightforward straightforward calculation calculation shows shows that that the the solution solution space space (the (the eigenspace) eigenspace) is is
{(a,O,O) : a
E
R} = span{(I,O,O)}.
of the fact fact that that the eigenvalue has multiplicity multiplicity 3, 3, there is is only only one Thus, in spite of linearly independent independent eigenvector corresponding to the eigenvalue. linearly The fact fact that that the the matrix matrix in in the the previous previous example example has has only only one one eigenvector eigenvector for for The an eigenvalue eigenvalue of of multiplicity multiplicity 33 is is significant. significant. It It means means that that there there is is not not aa basis basis of of an 3 R consisting A. R3 consisting of of eigenvectors eigenvectors of of A.
3.5. 3.5.
Eigenvalues and and eigenvectors eigenvectors of matrix Eigenvalues of a a symmetric symmetric matrix
3.5.1 3.5.1
71 71
The of a a matrix matrix and dot product The transpose transpose of and the the dot product
The eigenvalues and eigenvectors is greatly simplified The theory theory of of eigenvalues and eigenvectors is greatly simplified when when aa matrix matrix A Ae E nxn Rnxn is R is symmetric, symmetric, that is, is, when AT AT == A. In order to appreciate appreciate this, we we need to of the of aa matrix inner the transpose transpose of matrix to to the the Euclidean Euclidean inner to understand understand the the relationship relationship of product (or dot dot product). product (or product). mxn operator defined x n,, and perform the following We consider aa linear linear operator defined by A E eR Rm following calculation: calculation:
n
=
m
LLAijViUj j=l i=l
n
j=l
(The above manipulations are elementary properties (The above manipulations are just just applications applications of of the the elementary properties of of arithmetic-basically that numbers numbers can be added in any any order, order, and and multiplication arithmetic—basically that can be added in multiplication mxn distributes over addition.) Thus the transpose AT A E Rmxn satisfies the AT of A € R following fundamental property: following fundamental property: (3.25) nxn For aa symmetric matrix A A E R nxn this simplifies simplifies to to For symmetric matrix eR ,, this
(Au) . v = u· (Av) for all u, vERn.
When for complex complex scalars and vectors entries, we we must When we we allow allow for scalars and vectors with with complex complex entries, must y E then we we define modify the product. If modify the dot dot product. If x, x, y 6 en, C n , then define n
x·y
= LXiYi. i=l
The second vector in product must conjugated; this this is is necessary so The second vector in aa dot dot product must thus thus be be conjugated; necessary so that defined. We same notation notation that x-x X·X will will be be real, real, allowing allowing the the norm norm to to be be defined. We will will use use the the same product whether the vectors are in or C en.n . The The complex complex dot product for for the the dot dot product whether the vectors are in R Rnn or dot product has following properties, form the definition of of an an inner on aa has the the following properties, which which form the definition inner product product on complex vector space: complex vector space: 1. ~ 0 for all x E 1. xX·• x > 6 en, C n , and x .• x = 0 if and only if x = = o. 0. n 2. x X·• y Y= = y Y .• x x for en. 2. for all all x, x, yy E 6C .
72
Essential linear algebra Chapter 3. Essential
(ax + f3y) /3y) .• zz = ax· ax • zz + + f3y 0y .• zz for for all all x, x,y, 6 en, C n , a, f3ft E £ e. C. Together Together with 3. (ax y, zz E with the the second second property, property, this this implies implies that that z . (ax
+ f3y)
=
az . x + f3z . y
for all x,y,z £ E en, all x,y,z C n , a,f3 a,/3 E £ e. C. Complex products are Complex vector vector spaces spaces and and inner inner products are discussed discussed in in more more detail detail in in Section Section 9.1. 9.1.
3.5.2 3.5.2
Special symmetric matrices matrices Special properties properties of of symmetric
We We can can now now derive derive the special special properties of of the the eigenvalues eigenvalues and and eigenvectors eigenvectors of of aa symmetric symmetric matrix. matrix. Theorem If A E Theorem 3.44. 3.44. //A e Rnxn R n x n is symmetric, symmetric, then then every every eigenvalue eigenvalue of of A is real. real. Moreover, each eigenvector. Moreover, each eigenvalue eigenvalue corresponds corresponds to to aa real real eigenvector.
Proof. Ax = AX, xx =P where for the moment moment we we do the Proof. Suppose Suppose Ax = Ax, / 0, 0, where for the do not not exclude exclude the possibility that be complex. possibility that A A and and xx might might be complex. Then Then (Ax) . x = (AX) . x = A(X . x)
and and x· (Ax)
= x . (AX) = X(x . x).
But But (Ax) (Ax). •xx== xx. •(Ax) (Ax) when whenAA isissymmetric, symmetric,soso A(X . x)
= X(x . x).
Since x .•xx i:-/ 0,0,this this yields yields
A = A, which implies that A is real. real. Let xx = =u iv, where v £ R n . Then Then Let U + iv, where u, u, vERn. Ax
= AX <=> A(u + iv) = A(U + iv) <=> Au + iAv = AU + iAV <=> { Au = AU, Av = AV.
Since i:- 0, both), so v (or Since xx 7^ 0, we we must must have have either either U u =P / 00 or or vv f:^ 00 (or (or both), so one one of of u, u, v (or both) both) real eigenvector A corresponding A. 0 must be must be aa real eigenvector of of A corresponding to to A. we will From this point Prom this point on, on, we will only only discuss discuss eigenvalues eigenvalues and and eigenvectors eigenvectors for for real real to the theorem, then, then, we we will will not to use use symmetric matrices. matrices. According According to symmetric the last last theorem, not need need to complex numbers or vectors. Theorem be eigenvectors Theorem 3.45. 3.45. Let Let A A E £ Rnxn R n x n be be symmetric, and and let let Xl, xi, X2 x2 be eigenvectors of of AA corresponding AI, A2. Xl and and X2 orthogonal. corresponding to to distinct distinct eigenvalues eigenvalues \i, \2- Then Then xi X2 are are orthogonal.
73 73
Eigenvalues and eigenvectors eigenvectors of of a symmetric symmetric matrix matrix 3.5. Eigenvalues
Proof. We have Proof.
But and and Therefore Therefore
Al(XI . X2) = A2(Xl . X2), Al ^ i- AA2,2 , this implies implies that (xi,x (Xl,X2) = 0. 0. and since AI 2) =
D
Example 3.46. 3.46. Consider Example Consider
A =
1/3 1/3 1/3] 1/3 1/3 1/3 . [ 1/3 1/3 1/3
straightforward calculation shows that A straightforward
so that so the the eigenvalues eigenvalues of of A A are are 0,0,1. 0,0, 1. Another Another straightforward straightforward calculation calculation shows shows that there are are two two linearly linearly independent independent eigenvectors eigenvectors corresponding corresponding to to X ,\ — = 0, 0, namely, there namely,
and (independent) eigenvector A= = 1, 1, and aa single single (independent) eigenvector for for ,\
°
We can can verify verify by by observation observation that that the the eigenvectors eigenvectors for X= = 0 are are orthogonal orthogonal to to the the We for A eigenvector for A = eigenvector for = 1. Here is another special special property property of symmetric matrices. matrices. Example 3.43 3.43 shows that this result is not true for for nonsymmetric nonsymmetric matrices. matrices. nxn Theorem 3.47. 3.47. Let Let A A E 6 R Rnxn be symmetric, symmetric, and and suppose suppose A A has has an an eigenvalue Theorem be eigenvalue IJL of of (algebraic) (algebraic) multiplicity multiplicity kk (meaning (meaning that that J1fj, is is aa root root of of multiplicity multiplicity kk of of the the J1characteristic polynomial of A). A). Then Then A A has has kk linearly linearly independent independent eigenvectors eigenvectors characteristic polynomial of . corresponding corresponding to J1-. [L.
74
Chapter Chapter 3. Essential linear algebra
The proof of this theorem theorem is rather involved involved and and does not generalize to differThe proof of this is rather does not generalize to differential operators operators (as above). We therefore relegate it to ential (as do do the the proofs proofs given given above). We therefore relegate it to Appendix Appendix A. A. If /L is is an an eigenvalue of multiplicity multiplicity k, as in the previous previous theorem, theorem, then then we we can If n eigenvalue of k, as in the can choose the the kk linearly linearly independent independent eigenvectors corresponding to to // /L to be orthonorchoose eigenvectors corresponding to be orthonor13 mal. We obtain the corollary. mal. 13 We thus thus obtain the following following corollary.
Corollary 3.48. 3.48. (The spectral theorem for symmetric symmetric matrices) Corollary (The spectral theorem for matrices) Let Let A A GE nxn Rnxn be U2,2 ,... ... ,un} Rnn R be symmetric. symmetric. Then Then there there is is an an orthonormal orthonormal basis basis {Ul' {ui,u ,u n ) of of R consisting of of A. consisting of eigenvectors eigenvectors of A.
3.5.3 3.5.3
The spectral spectral method method for solving Ax The for solving Ax = = bb
E Rnxn When A e R nxn is symmetric symmetric and and the eigenvalues eigenvalues and eigenvectors eigenvectors of A are known, 14 is aa simple method 14 for Ax = there is there simple method for solving solving Ax = b. b. Let A A be be symmetric symmetric with with eigenvalues AI,, \A2, ... ,An Let eigenvalues AI An and and orthonormal orthonormal eigeneigen2,..., vectors ui, Ul, U U2, For any b ERn, we can write vectors 2 ,... . . . ,,Un' un. For any b e R n , we can write
We can can also write We also write
(Of course, ai o,i = are thinking cannot = Uj Ui .• x, x, but, but, as as we we are thinking of of xx as as the the unknown, unknown, we we cannot (Of course, compute o>i from from this compute ai this formula.) formula.) We We then then have have Ax
= b::::} A (alul + ... + anun ) =
+ ... + (un' b)un ::::} alAul + ... + anAun = (Ul . b)UI + ... + (un' b)u n ::::} alAlUl + ... + anAnUn = (Ul . b)Ul + ... + (un' b)un . (Ul . b)Ul
Since b can have only expansion in terms of of the basis {Ul, ... ,u , un}, we must Since b can have only one one expansion in terms the basis {ui,... must n }, we have have aiAi
= (Ui' b), i = 1,2, .. . ,n,
that is, that is,
Thus we solution we obtain obtain the the solution Thus (3.26) 13 13It is always always possible possible to to replace replace any linearly independent independent set with an an orthonormal orthonormal set It is any linearly set with set spanning spanning the same same subspace. The technique technique for doing this this is procedure; it it is is the subspace. The for doing is called called the the Gram-Schmidt Gram-Schmidt procedure; explained in in elementary texts such such as explained elementary linear linear algebra algebra texts as [34]. 14 14This method is is not not normally normally taught taught in in elementary linear algebra algebra courses or books, books, because because it This method elementary linear courses or it is more difficult to to find find the the eigenvalues and eigenvectors eigenvectors of of A A than than to to just just solve solve Ax Ax = = b b by by other is more difficult eigenvalues and other means. means.
75 75
3.5. Eigenvalues Eigenvalues and eigenvectors of of a a symmetric 3.5. and eigenvectors symmetric matrix matrix
We see see that that all all of of the the eigenvalues eigenvalues of of A A must must be be nonzero nonzero in in order order to to apply We apply this this method, which which is is only only sensible: if 00 is is an an eigenvalue eigenvalue of of A, A, then then A A is is singular method, sensible: if singular and and Ax = bb either either has has no no solution or has has infinitely many solutions. Ax solution or infinitely many solutions. If already have the eigenvalues method for If we already eigenvalues and and eigenvectors eigenvectors of A, then this method solving Ax = = b is simpler and less less expensive expensive than than the the usual usual method method of of Gaussian solving Ax b is simpler and Gaussian elimination. elimination. 3.49. Let Let Example 3.49.
A
=
11 -4 [ -1
-4 14 -4
-1 -4 11
1 =[ 1 ,b
12 1
A direct direct calculation calculation shows shows that that the the eigenvalues eigenvalues of of A A are are A A1 = 6, A2 = 12, A3 = 18,
corresponding (orthonormal) (orthonormal) eigenvectors are and the corresponding
(According to Theorem the eigenvectors eigenvectors are are automatically automatically orthogonal orthogonal in in this this (According to Theorem 3.45, 3.45, the we had to ensure that each eigenvector was normalized.) The solution case; however, case; however, we had to ensure that each eigenvector was normalized.) The solution to Ax = = bb isis to Ax
Exercises 1. 1. Let
164 A = [ -48
-48] [ 116 ] 136 ,b = 88 .
Compute, by by hand, hand, the the eigenvalues eigenvalues and eigenvectors of of A, A, and use them them to Compute, and eigenvectors and use to solve for xx (use (use the "spectral method"). solve Ax Ax — = bb for the "spectral method"). 2. Repeat Exercise for 2. Repeat Exercise 11 for
A=
3 -1
-1 11 ] . 3 ] ,b = [ -9
76
Chapter 3. Essential linear algebra
3. Repeat Repeat Exercise Exercise 11 for for 3.
4. Repeat Repeat Exercise Exercise 11 for for 4.
[ 7 -2 1]
A =
-2 1
10 -2
-2 7
,b =
[ 6] 12 18
.
nxn 5. Let Let A A E R nxn be symmetric, symmetric, and and suppose suppose the the eigenvalues eigenvalues and 5. €R be and (orthonormal) (orthonormal) eigenvectors of of A A are known. How How many many arithmetic eigenvectors are already already known. arithmetic operations operations are are required to to solve Ax = = b b using using the the spectral spectral method? required solve Ax method? nxn 6. A A symmetric matrix A A E Rnxn 6. eR is called positive definite definite ifif
=I 0 ~
x
(Ax) . x > O.
the spectral Use the spectral theorem to show show that A is positive definite if and only if all of the eigenvalues A are are positive. of the eigenvalues of of A positive.
7. Let defined by 7. Let L L be be the the n n x nn matrix matrix defined by the the condition condition that that 2
p, Lij
-b,
= {
0,
i
= j,
li-jl=1,
otherwise,
where l/(n + 1). 1). For example, with where h = = 1j(n For example, with n = = 5, 5, 72 -36 L =
0
[
o o
-36 72 -36 0 0
°
-36 72 -36 0
0 0 -36 72 -36
o
o . 01 -36 72
(a) For For each each jj = = 1,2, ... ,n, define the the discrete discrete sine wave s^ s(j) of frequency jj (a) 1,2,..., n, define sine wave of frequency by s~) = sin (jk7rh). J Show that that s( s(j) an eigenvector eigenvector of of L, L, and and find find the the corresponding corresponding eigenShow ) is is an eigenJ value Xj. Aj. (Hint: (Hint: Compute Ls(j) and apply apply the the addition addition formula for the value Compute Ls( ) and formula for the sine function.) sine function.)
(b) What is is the the relationship relationship between between the the frequency and the the magnitude magnitude of of (b) What frequency j and Aj? A,-?
3.6. Preview of methods for solving solving ODEs and PDEs
77
(c) (c) The The discrete sine sine waves are orthogonal orthogonal (since (since they are the eigenvectors eigenvectors of of aa symmetric symmetric matrix matrix corresponding to distinct eigenvalues) eigenvalues) and and thus thus form an orthogonal basis for for Rn. be shown that every every form an orthogonal basis R n . Moreover, Moreover, it it can can be shown that s(j) s(^ has the same norm: 1 I s 11-- V2h' (j)
J. -- 1, 2 , .. . ,n.
Therefore, {I V2hs(1) V2hs^l\, V2hS(2), ^/2hs^2\ . .... ,,V2hs(n)} \/2hs^ > is an orthonormal basis for n n R . We We will will call call aa vector vector xx E e Rn R smooth smooth or or rough rough depending depending on on whether Rn. whether its in the the discrete sine wave wave basis are heavily toward its components components in discrete sine basis are heavily weighted weighted toward that the the solution x of of the low low or high frequencies, the or high frequencies, respectively. respectively. Show Show that solution x Lx = b is Lx =b is smoother smoother than than b. b.
3.6 3.6
Preview Preview of of methods methods for for solving solving ODEs ODEs and and PDEs PDEs
The close formal relation concepts of of linear differential equation equation and The close formal relation between between the the concepts linear differential and that ideas linear operators operators might role linear algebraic system system suggests suggests that ideas about about linear might playa play a role linear algebraic in solving former, as the case the latter. In order this parallel in solving the the former, as is is the case for for the latter. In order to to make make this parallel the context explicit, we will will apply apply the the machinery machinery of of linear algebra algebra in the context of of differential differential equations: view view solutions solutions as as vectors vectors in in aa vector space, space, identify identify the the linear linear operators operators which define linear differential equations, equations, and and understand facts such such as as the the which define linear differential understand how how facts Fredholm alternative appear Fredholm alternative appear in in the the context context of of differential differential equations. equations. In chapters to follow we we will will accomplish this. In the process, process, we we will will In the the chapters to follow accomplish all all of of this. In the develop three classes of methods for for solving ODEs and and PDEs, each develop three general general classes of methods solving linear linear ODEs PDEs, each one closely closely analogous analogous to method for for solving solving linear linear algebraic algebraic systems: systems: to aa method one 22 2 1. The The method method of of Fourier series: Differential operators, like like dd?/dx can have have / dx ,, can 1. Fourier series: Differential operators, eigenvalues and eigenfunctions; eigenfunctions; for example eigenvalues and for example
;:2
[sin(wx)] = _w 2 sin(wx).
The Fourier series using the the eigenfunctions eigenfunctions The method method of of Fourier series is is aa spectral spectral method, method, using of the the differential differential operator. operator. 2. method of Green's functions: A Green's Green's function function for differential 2. The The method of Green's functions: A for aa differential equation equation is is the solution solution to to aa special special form form of of the the equation equation (just (just as as A-I A"1 is is the the solution J) that down the the solution solution to to AB = /) that allows allows one one to to immediately immediately write write down solution -1 to the equation equation (just could write write down x= A -lb). to the (just as as we we could down x =A b). 3. method of method that 3. The The method of finite finite elements: elements: This This is is aa direct direct numerical numerical method that can used when fails (or intractable). It can can be be compared can be be used when (1) (1) or or (2) (2) fails (or is is intractable). compared with Gaussian elimination, numerical method method for for solving solving with Gaussian elimination, the the standard standard direct direct numerical finite element not produce Ax = h. b. Like Like Gaussian Gaussian elimination, elimination, finite element methods methods do do not produce formula for for the the solution; solution; however, however, again again like like Gaussian Gaussian elimination, elimination, they they are are aa formula broadly applicable. applicable.
t78
Chapter 3. Essential linear algebra
In this this book, book, we we concentrate concentrate on Fourier series and finite finite element We In on Fourier series and element methods. methods. We will will also also explain explain the the idea idea of of aa Green's Green's function function and and derive derive aa few few specific specific Green's Green's functions that we will find useful.
3.7 3.7
Suggestions Suggestions for for further further reading reading
A good good introductory introductory text, text, which which assumes assumes no no prior prior knowledge knowledge of of linear linear algebra algebra and and A is is written written at at aa very very accessible accessible level, level, is is Lay Lay [34]. [34]. An An alternative alternative is is Strang Strang [44]; [44]; this this book also also assumes no background but it is written book assumes no background in in linear linear algebra, algebra, but it is written at at aa somewhat somewhat more demanding more demanding level. level. It It is is noteworthy noteworthy for for its its many many insights insights to to the the applications applications of of linear algebra for its tone. A A more advanced text text is is Meyer Meyer [40]. linear algebra and and for its conversational conversational tone. more advanced [40]. Anyone seriously mathematics must must become become familiar Anyone seriously interested interested in in applied applied mathematics familiar with with the computational algebra. The the computational aspects aspects of of linear linear algebra. The text text by by Strang Strang mentioned mentioned above above includes the numerical the subject. subject. There are also also more more includes material material on on the numerical aspects aspects of of the There are specialized references. A good introductory introductory text is Hager [23], [23], while more advanced advanced treatments include Bau [49]. encyclopedic treatments include Demmel Demmel [13] [13] and and Trefethen Trefethen and and Bau [49]. An An encyclopedic reference is Golub and Van Loan [19].
Chapter 4 Chapter 4
tial ordinary ordinary Jsseiitial differential eal
equations
In an ordinary differential differential equation equation (ODE), there is a single independent variable. Commonly ODEs model change over time, so the independent variable is tt (time). Our interest in ODEs derives from the following fact: both the Fourier series method and the finite element method reduce time-dependent time-dependent PDEs into systems of ODEs. In the Fourier In the the case case of of the Fourier series series method, method, the the system system is is completely completely decoupled, decoupled, so so the the "system" is really just a sequence of scalar ODEs. In we learn how to In Section 4.2, we solve the scalar ODEs that arise in the Fourier series method. The finite element method, on the other hand, results in coupled systems of of ODEs. In Section 4.3, we we discuss the solution of linear, coupled systems of firstorder ODEs. Although we we present an explicit solution technique in that section, the emphasis is is really on the properties of the solutions, as the systems that arise in practice are destined to be solved by numerical rather than analytical analytical means. In Sections 4.4 and 4.5, we we introduce some simple numerical methods that are adequate adequate for our purposes. We close this chapter chapter by interpreting our simple solutions in terms of Green's functions. Although we do not emphasize the method of Green's function in this book, the basic basic idea 4.6. book, we we do do explain explain the idea in in Section Section 4.6.
4.1 4.1
Converting a a higher-order higher-order equation equation to to a Converting a first-order system first-order system
We begin our discussion of ODEs with a simple observation: It is always possible to convert a single ODE of order two or more to a system of first-order first-order ODEs. We illustrate the following illustrate this this on on the following second-order second-order equation: equation:
cPu
a dt 2
du
+ b dt + cu = f(t).
We define Xl (t)
= u(t),
79
(4.1)
80
Chapter 4. Essential Essential ordinary differential equations
Then we have Then we have
dXl
dt
=X2,
dX2 ~u C b du 1 dt = dt2 = --;;,u - -;;, dt + -;;,f(t) =
c --;;,Xl -
b 1 -;;,X2 + -;;,f(t).
We can can write this as as aa first-order first-order system system using using matrix-vector matrix-vector notation: notation: We write this dx =Ax+f(t), A= [ 0c -d t - -a
(4.2)
In the vector-valued In (4.2), (4.2), xx is is the vector-valued function function
The The same same technique technique will will convert convert any any scalar scalar equation equation to to aa first-order first-order system. system.
If the the unknown unknown is is u and and the equation is is order order m, m, we we define define If the equation Xl
du dm-lu = U, X2 = -dt ' ... , Xm = -t md 1 . -
The The first first m -— 1 equations equations will will be be
dx l dX2 dt = X2, dt = X3, ... ,
dXm-l - - - =X m , dt
and the original the new If and the the last last equation equation will will be be the original ODE ODE (expressed (expressed in in the new variables). variables). If the original scalar scalar equation equation is is linear, linear, then then the the resulting system will will also also be be linear, the original resulting system linear, and be written written in and it it can can be in matrix-vector matrix-vector form form if if desired. desired. An mth-order typically has many solutions; An mth-order ODE ODE typically has infinitely infinitely many solutions; in in fact, fact, an an mthmthorder has an order linear linear ODE ODE has an m-dimensional m-dimensional subspace subspace of of solutions. solutions. To To narrow narrow down down aa unique unique solution, solution, m m auxiliary auxiliary conditions conditions are are required. required. When When the the independent independent variable is is time, time, the auxiliary conditions conditions are are initial initial conditions: conditions: variable the auxiliary
du ~-lU u(to) = ao, dt (to) = al,···, dt m- l (to) = am-l· These the new new These initial initial conditions conditions translate translate immediately immediately into into initial initial conditions conditions for for the unknowns Xl, x\, x%,..., xm: X2, ... , Xm: unknowns
We can can apply apply aa similar similar technique technique to to convert convert an an mth-order mth-order system system of of ODEs We ODEs to aa first-order first-order system. system. For For example, example, suppose suppose x(t) x(i) is is aa vector-valued vector-valued function, function, say to say x(t) x(£) ERn. 6 R n . Consider Consider aa second-order second-order system: system:
~x dt
2 = f
(
dX) t, x, dt .
(4.3)
4.1. a higher-order to a a first-order first-order system system l.l. Converting Converting a higher-order equation equation to
81 81
We define yet)
= x(t), dx
z(t) = dt (t). We then have have We then dy dt
dz dt
= z, = f(t,y,z).
The original system (4.3) consists of n second-order ODEs in n unknowns (the components of x( t)). We We have have rewritten rewritten this original system system as 2n first-order ODEs components of x(i)). this original as In first-order ODEs in 2n unknowns unknowns (the n components of y(t) yet) and in In (the n components of and the the nn components components of of z(t)). The fact that any any ODE ODE can can be be written written as system has has the the followThe fact that as aa first-order first-order system following benefit: Any Any theory theory or or algorithm algorithm developed ODEs is is ing benefit: developed for for first-order first-order systems systems of of ODEs automatically applicable applicable to to any This leads to aa considerable in automatically any ODE. ODE. This leads to considerable simplification simplification in the study the study of of this this subject. subject.
Exercises 1. Write the the ODE ODE 1. Write
rPu
2 dx 2
+u =0
as system of first-order ODEs. as aa system of first-order ODEs. 2. Write the ODE
rPu
dx 2
+ csin (u)
= 0
as aa system system of first-order first-order ODEs. 3. 3. Write Write the the ODE ODE
d4 u
dt 4
d2 u
-
2 dt 2 + u = sin (t)
as system of first-order ODEs. as aa system of first-order ODEs. 4. Write the ODE
d2 u du 2 - 2 -2u- +t u=O dt dt as system of first-order ODEs. as aa system of first-order ODEs.
5. Write the following system of second-order ODEs as a system of first-order first-order ODEs: ODEs:
82
Chapter 4. Essential ordinary equations ordinary differential equations
Here ml m2 are are constants, constants, and and II and fz real-valued functions functions of of four Here mi and and m^ j\ and /2 are are real-valued four variables.
4.2
Solutions to some simple ODEs Solutions
In this section, we show how to to solve some simple simple firstand second-order second-order ODEs ODEs In this section, we show how solve some first- and that will will arise arise later in the the text. text. that later in
4.2.1 4.2.1
The general general solution solution of of a second-order homogeneous The a second-order homogeneous ODE ODE with constant coefficients with constant coefficients
In the ensuing ensuing chapters, often encounter encounter the the second-order In the chapters, we we will will often second-order linear linear homogehomogeconstant coefficients, neous ODE ODE with constant coefficients, d2u a dt 2
du
+ b dt + cu =
(4.4)
0,
we will present present aa simple so we simple method method for for computing computing its its general general solution. solution. The The method method is based based on on the faced with with aa differential differential equation, equation, one one can is the following following idea: idea: When When faced can sometimes guess guess the general form form of the solution solution and, by substituting substituting this general form into the equation, determine the specific specific form. form. (4.4) is of the In this case, In this case, we assume that that the the solution solution of (4.4) the form form
Substituting into (4.4) (4.4) yields yields Substituting into
since the exponential exponential is never zero, this equation equation holds holds if if and only ifif since the is never zero, this and only
ar2
+ br + c = O.
This quadratic is is called called the the characteristic characteristic polynomial polynomial of of the the ODE ODE (4.4), and its This quadratic (4.4), and its roots roots are ODE. are called called the the characteristic characteristic roots roots of of the the ODE. formula: The characteristic characteristic roots are given by the the quadratic quadratic formula: rl =
-b - Vb 2 2a
-
4ac
, r2 =
-b + v'b 2
-
4ac
2a
We three cases. We distinguish distinguish three cases. - 4ac 1. The characteristic roots are are real real and and unequal unequal (i.e. (Le. 6 b22 — 4ac > In 1. The characteristic roots > 0). 0). In rit r2t this since the the equation equation is any linear combination of and eer2t this case, case, since is linear, linear, any linear combination of e er1t and (4.4) can is also also aa solution solution of of (4.4). In In fact, as as we now show, show, every every solution solution of of (4.4) can be written written as as be
(4.5)
4.2. Solutions to some some simple ODEs 4.2. Solutions to simple ODEs
83
for choice of ci, C2In fact, initial for some some choice of Cl, C2. In fact, we we will will show show that, that, for for any any &i, kl' £2, k2, the the initial value problem (IVP) (IVP) d2 u a dt 2
du
+ b dt + cu =
0,
u(O) = kl'
(4.6)
du(O) = k dt
2
has a unique solution solution of the form (4.5). (4.5). (4.5), we have With uu given by (4.5),
= CI + C2,
u(O)
du dt (0) =
rlCI
+ r2 C2,
and and we we wish wish to to choose choose c\, CI, c^ C2 to to satisfy satisfy CI rlCI
+ C2 = kl'
+ r2C2
= k2'
that is, is, that [
ll]C_k r2
rl
The matrix in this equation equation is is obviously nonsingular (since rl / :j:. r-2), r2), The coefficient coefficient matrix in this obviously nonsingular (since r\ and there is is aa unique unique solution CI, C2 for for each each fci, kl' fo k 2•. and so so there solution c\,C2
Since every in the (4.5), we the Since every solution solution of of (4.4) (4.4) can can be be written written in the form form (4.5), we call call (4.5) (4.5) the general solution of of (4.4). (4.4). qeneral solution - 4ac 2. The characteristics roots roots are are complex complex (Le. 4ac < 0). In 2. The characteristics (i.e. b22 — < 0). In this this case, the characteristics characteristics roots roots are are also also unequal; unequal; in in fact, they form complex case, the fact, they form aa complex conjugate pair. The analysis of the previous case applies, and we could write the general the form form (4.5) (4.5) in this case as well. well. However, However, with with 7*1, rl, r<2 r2 the general solution solution in in the in this case as ri r2 t ,eT2t functions, and this is undesirable. complex, ee T1 *,e * are complex-valued functions, With rl = f.L - Ai, r2 = f.L + Ai, we have (by Euler's formula) e T1t
=
el-'t
(cos (At) - i sin (At)),
T2t
=
el-'t
(cos (At)
e
+ i sin (At)),
and and so so 1
2"
(eTlt+eT2t)
~ (e T1t _
e
T2t
)
=
el-'t
cos (At),
= el-'t sin (At).
This shows equation is the equation is linear) linear) that that This shows (because (because the el-'t
cos (At),
el-'t
sin (At)
84
Chapter 4. Essential ordinary differential differential equations
are also solutions of (4.4) (4.4) in this case. We will write the general solution in the form Cle Pt cos (At) + C2ept sin (At). we can show that every solution As in the previous case, we solution of (4.4) can be written in this form (see (seeExercise Exercise 2). 2). 3. polynomial has single (repeated) real root root (i.e. (Le. 3. The The characteristic characteristic polynomial has aa single (repeated) real b2 -— 4ac 4ac = 0). In this case the root is is r =— -b/(2a). —b/(2a). We We cannot write the we try teTt general solution solution with the single solution eeTtrt;; by an inspired guess, we tert rt as the second solution. Indeed, with u(t) we have u(t) = — teTt, te , we
=
du dt (t)
= (1 + rt)e Tt ,
~u
dt 2 (t)
= (2 + rt)reTt ,
and so ~u
du
+ b dt (t) + cu(t) = a(2 + rt)re Tt + b(1 + rt)e Tt + cte Tt = (ar2 + br + c) teTt + (2ar + b) eTt = 0 (since ar2 + br + c = 0 and 2ar + b = 2a( -b/(2a)) + b = 0). a dt 2 (t)
rt teTt is also a solution of (4.4), and we This shows that te we write the general solution in this case as
of where r = = -b/(2a). —b/(2a). Once again, it is not hard to show that every solution of (4.4) can be written in this form (see (see Exercise Exercise3). 3). Example 4.1. characteristic polynomial of of 4.1. The characteristic ~u du - -2 - + u = O dt dt
is rr22 -— r + 1, I, which which has roots 1 rl = 2
+ -v'3. z
2'
1 v'3. r2 = - - -z. 2 2
Therefore this this example falls in case case 22 above, above, and and the the general solution of the ODE ODE is Therefore falls in of the u(t)
t . (v'3 = (Cl cos (2v'3 ) + C2 sm -2- ) ) t
Example 4.2. 4.2. The The ODE ~u du --2-+u=0 2 dt dt
et /
2
•
4.2. Solutions to some simple ODEs
85
has characteristic polynomial rr22 — 2r + + 1. 1. The characteristic roots are - 2r ri = r2 = 1,
general solution is is so case 3 above applies. The general
4.2.2 4.2.2
A special special inhomogeneous A inhomogeneous second-order second-order linear linear ODE ODE
In we will will encounter encounter the the following inhomogeneous IVP: IVP: In Chapter Chapter 7, 7, we following inhomogeneous
d2 u
+ ()2U = f(t),
dt 2
u(to) = a, du dt (to) = {3.
(4.7)
The coefficient is aa positive positive real this section, we will will present present aa The coefficient ()2 02 is real number. number. In In this section, we surprisingly simple for the surprisingly simple formula formula for the solution. solution. By the principle principle of of superposition, we can solve the the two two IVPs IVPs By the superposition, we can solve ~u
+ ()
dt 2
2
u = 0,
u(to) = a, du dt (to) = {3
(4.8)
and and d2 u
dt 2
+ (Pu = f(t), u(to) = 0, du dt (to) = 0
(4.9)
and add the solutions to get the solution to (4.7). to apply and add the solutions to get the solution to (4.7). It is is straightforward straightforward to apply the the techniques of of Section to derive the solution techniques Section 4.2.1 4.2.1 to derive the solution of of (4.8): (4.8): Ul (t)
= a cos (()( t -
to))
+ ~ sin (()( t - to))
(4.10)
(see Exercise Exercise 12). (see 12). The of (4.9) is more more difficult to derive, the final final answer The solution solution of (4.9) is difficult to derive, although although the answer has has aa pleasingly pleasingly simple simple form: form:
llt
U2(t) = (j
to
sin (()(t - s))f(s) ds.
(4.11)
86
Essential ordinary ordinary differential differential equations Chapter 4. Essential equations
The of this formula is is outlined outlined in in Exercise section. The derivation derivation of this formula Exercise 4.3.10 4.3.10 of of the the next next section. For now, we we will will content ourselves with with verifying (4.11) really solves (4.9). For now, content ourselves verifying that that (4.11) really solves (4.9). To solution, we differentiate U2(t), W2(i), which which is since the the we must must differentiate is aa bit bit tricky tricky since To verify verify the the solution, independent variable appears both integrand and and as integration limit. limit. To independent variable appears both in in the the integrand as an an integration To make derivative clear, clear, we make the the computation computation of of the the derivative we write write
h(x,y)
11
X
= (j
sin (O(y - s))f(s)ds,
to
and note note that that U2(t) u^t) = h(t, h(t,t). By the chain rule, rule, then, then, and t). By the chain
dU2 ah dt di(t) = ax (t, t) dt ah
= ax (t, t)
ah
dt
+ ay (t, t) dt ah
+ ay (t, t).
By the of calculus, calculus, By the fundamental fundamental theorem theorem of
~~ (x, y) = ~ sin (O(y and, by Theorem 2.1, 2.1, and, by Theorem
ah a(x,y) y
x))f(x),
11 o l x
= (j
cos (O(y - s))f(s)ds
to
x
cos (O(y - s))f(s) ds.
=
to
Therefore, Therefore,
dU2 1. dt (t) = (j sm (O(t - t))f(t)
rt cos (O(t - s))f(s) ds
+ lto
= it cos (O(t - s))f(s) ds. to
A similar similar calculation calculation then then shows shows that A that
~u22 (t) dt
= cos (O(t - t))f(t) - 0
it sin (O(t -
s))f(s) ds
to
= f(t) - 0
t sin (O(t - s))f(s) ds.
lto
We have We then then have
~~2 (t) + 02 U2 (t)
= f(t) - 0
t
rt sin (O(t - s))f(s) ds + 0 ltort sin (O(t - s))f(s) ds
lto
= f(t),
U2(tO)
= (jlito sin (O(t -
s))f(s) ds
= 0,
to
du
2
dt
(to) =
ito cos (O(t to
s))f(s) ds = O.
4.2. Solutions to simple ODEs 4.2. Solutions to some some simple ODEs
87
Thus U2 solves (4.9), which Thus U2 solves (4.9), which is is what what we we wanted wanted to to show. show. We have have now now shown shown that that the the solution to (4.7) (4.7) is is We solution to
u(t) = a cos (O(t - to))
+ 7i(3 sin (O(t - to)) + (jlit sin (O(t - s))f(s) ds.
(4.12)
to
EXaJIlple 4.3. 4.3. The to Example The solution to ~u
dt 2
+u =
cos (t),
u(O) = 0, du(O) = 0 dt i
u(t)
= lot sin (t -
s) cos (s) ds
= ~ sin (t)
(the integml can can be be computed using the the subtraction subtmction formula formula for for sine sine and (the integral computed using and simplified simplified using other other trigonometric using trigonometric identities). identities).
4.2.3 4.2.3
First-order linear linear ODEs First-order ODEs
Another type ODE for general solution first-order Another type of of ODE for which which we we can can find find the the general solution is is the the first-order linear ODE: linear ODE: du (4.13) dt - au = f(t). (We could could treat treat the the case case of ofaa nonconstant nonconstant coefficient coefficient similarly; similarly;however, however,wewehave havenon (We need need to.) to.) The solution The solution is is based based on on the the idea idea of of an an integmting integratingfactor: factor: we wemultiply multiplythe th equation equation by by aa special special function functionthat that allows allowsus ustotosolve solveby bydirect directintegration. integration.We Wenote not that that Therefore, we have Therefore, we have
du - - au dt
= f(t)
=}
du e- at _ - ae-atu dt d
=} -
dt
=}
= e-atf(t)
[e-atu(t)] = e- at f(t)
e-atu(t) = e-atOu(to)
+ it e- as f(s) ds to
=}
u(t) = ea(t-to)u(to)
+ it ea(t-s) f(s) ds. to
This immediately solution to This immediately gives gives us us the the solution to the the IVP IVP du
dt - au = f(t), u(to) = Uo.
(4.14)
88
Chapter 4. Essential ordinary ordinary differential equations equations
Example 4.4. 4.4. The The IVP Example IVP
du . () dt +u = sm t, u(O) = 2 has solution has solution u(t)
= 2e- t + lot e-(t-s) sin (s) ds = 2e- t =
1
+ '2 (sin (t)
- cos (t)
~e-t + ~ (sin (t) -
+ e- t )
cos(t))
(the integral can be computed computed by by two applications of by parts). (the integral can be two applications of integration integration by parts). It should should be be appreciated that, although although we we have have an an explicit explicit formula appreciated that, formula for for the the solution of of (4.13), it will will be be impossible impossible in cases to to explicitly explicitly evaluate evaluate the solution (4.13), it in many many cases the integrals involved involved in terms of of elementary elementary functions. functions. integrals in terms
Exercises 1. be the the solution (4.4). 1. Let Let S 5 be solution set set of of (4.4). (a) Show that S (R), the the set of twice-continuously twice-continuously differ(a) Show that S is is aa subspace subspace of of C22 (R), set of differentiable defined on R. entiable function function defined on R. (b) Use this section section to show S is any values (b) Use the the results results of of this to show is two-dimensional two-dimensional for for any values of a, a, 6, b, and provided only that a ¥ 0. O. of and c, c, provided only that a^
2. Suppose (4.4) characteristic roots \i, where R and A / 0. Show Show 2. Suppose (4.4) has has characteristic roots /z± /-l±>..i, where /i, /-l, A >.. e E Rand>.. ¥ O. that, for kl' k2, there is is aa unique unique choice choice of of Cl, C2 such that the the solution of that, for any any ki, £2, there ci, C2 such that solution oi (4.6) (4.6) is CleP,t cos (>..t) + C2eP,t sin (>..t). 3. Suppose has the the single root r = = -bj(2a). that, for for 3. Suppose (4.4) (4.4) has single characteristic characteristic root -b/(2a). Show Show that, any kl' k2' there is is aa unique unique choice of ci,C2 Cl, C2 such such that that the the solution of (4.6) (4.6) is is any fei, fo, there choice of solution of
Clert
+ C2 tert .
4. each of following IVPs, general solution solution of ODE and and use 4. For For each of the the following IVPs, find find the the general of the the ODE use it the IVP: IVP: it to to solve solve the (a)
d2 u dt 2
du
+ 2 dt + u = 0, u(O)
= 1,
du(O) = 0 dt
4.2. Solutions to some simple ODEs ODEs 4.2. Solutions to some simple
89
(b)
dlu 3 dt 2
du
+ 2 dt + u =
0,
= -1,
u(o)
du(O) = 3 dt
(c) dlu 2 dt 2
du
+ 2 dt + u = u(O)
0,
= 0,
du (0) = 1 dt
(d)
d2 u dt 2
du
+ 3 dt + u =
0,
u(O) = 1, du (0) = 2 dt
5. following differential differential equations equations are accompanied by by boundary boundary conditionsconditions— 5. The The following are accompanied auxiliary conditions that of aa spatial spatial domain auxiliary conditions that refer refer to to the the boundary boundary of domain rather rather than to to an an initial initial time. time. By using the the general general solution of the ODE, determine than By using solution of the ODE, determine to the value problem problem (BVP) exists, and whether aa nonzero whether nonzero solution solution to the boundary boundary value (BVP) exists, and if so, whether whether the the solution solution is is unique. unique. if so, (a)
d2 u dx 2
-
2u
= 0,
u(O) = 0, u(l) = 0 (b)
d2 u dx 2
+ 2u =
0,
u(O) = 0, u(l) = 0
(c) dlu dx 2
+ 7r
2
U
= 0,
u(O) = 0, u(l)
=0
90
Chapter ordinary differential Chapter 4. 4. Essential Essential ordinary differential equations equations
6. Determine Determine the values of A E e R such that the BVP d2 u dx 2
+ AU =
0,
u(o) = 0, u(l) =
°
has aa nonzero has nonzero solution. solution.
7. Prove directly (that (that is, is, by by substituting substituting uu into into the the differential differential equation 7. Prove directly equation and and initial condition) that initial condition) that u(t) = ea(t-to)uo
+ i t ea(t-s) /(s) ds to
solves solves
du
dt - au = /(t), u(to )
= Uo·
8. the following following IVPs: 8. Solve Solve the IVPs:
(a)
du dt
= 2u -
0.1,
u(O) = 1.0 (b)
du
dt = -2u + O.lt,
u(O) = 0
(c)
du
= t, u(O) = 0
dt +u
9. 9. Find the solution solution to the the IVP ~u
dt 2
+ 4u =
1,
u(o) = 0,
du
dt (0)
= O.
91 91
4.3. Linear systems with constant coefficients
10. 10. Find the the solution solution to the IVP IVP d2 u
dt 2
+ 9u = t, u(O) = 0,
du(O) dt
= O.
11. Find the the solution to the the IVP IVP 11. Find solution to
d2 u
-t
dt 2 +u = e ,
= 1, du(O) = O. u(O)
dt
12. Use Use the of Section Section 4.2.1 4.2.1 to to derive derive the the solution (4.8), and verify 12. the techniques techniques of solution to to (4.8), and verify that you you obtain obtain (4.10). way to to do do this is to write the the general that (4.10). (Hint: (Hint: One One way this is to write general solution of of the as u(t) u(t) = c\ cos (Ot) (9t] + + C2 02 sin sin (Ot) and and then solve for for Cl c\ the ODE ODE as Cl cos then solve solution and C2C2. If you do this, you will then have to apply trigonometric identities identities to put the solution in the form given in (4.10). It is simpler to recognize at the beginning that the general solution could just as well be written as u(t) = Cl ci cos (0(t -- to)) t0})++C2c2sin sin(O(t (0(t- -to)).) t0)).) u(t) cos (O(t
4.3
coefficients Linear systems with constant coefficients
We now consider first-order linear system of with constant coefficients. We now consider aa first-order linear system of ODEs ODEs with constant coefficients. Such be written Such aa system system can can be written
dXl
dt = anXl + a12 X2 + ... + alnXn + /1 (t), dX2
dt = a21 Xl + a22 x 2 + ... + a2n Xn + h(t),
(4.15)
dX n (it = anlXl + an2X2 + ... + annXn + fn(t). As algebraic system, system, there is aa great advantage in in using using matrix-vector As with with aa linear linear algebraic there is great advantage matrix-vector notation. notation. System System (4.15) (4.15) can be written as
dx - =Ax+f(t), dt
(4.16)
where
a21
a12 a22
al n a2n .
anl
an2
a~n
an
A=
1, x(t) = [ X2(t) x,(t) 1 [ h(t) j,(t) 1 . , f(t) = . . xn·(t)
fn·(t)
92
Essential ordinary differential equations Chapter 4. Essential
Although it it is is possible possible to to develop develop aa solution solution technique technique that that is is applicable applicable for for any any Although matrix A whkl1 there matrix AE € Rnxn, R n x n , it it is is sufficient sufficient for for our our purposes purposes to to discuss discuss the the case case in in which there is aa basis basis for for R Rnn consisting consisting of of eigenvectors eigenvectors of of A, A, notably notably the the case case in in which which A is A is is symmetric. we can spectral method, much the spectral symmetric. In In this this case, case, we can develop develop aa spectral much like like the spectral 15 method for solving solving Ax Ax = = bb in in the case that that A A is is symmetric. symmetric.15 method for the case As we develop reader should As we develop an an explicit explicit solution solution for for (4.16), (4.16), the the reader should concentrate concentrate on on the qualitative qualitative properties properties (of (of the the solution) solution) that that are are revealed. revealed. These These properties turn properties turn the out to to be be more more important important (at (at least least in in this this book) book) than than the the formula formula for the solution. solution. out for the
4.3.1 4.3.1
Homogeneous systems Homogeneous systems
We begin the homogeneous version of of (4.16), (4.16), We begin with with the homogeneous version
dx dt = Ax.
(4.17)
We will look We will look for for solutions solutions to to (4.16) (4.16) of of the the special special form form
x(t)
= a(t)u,
where uf:.O u ^ 0 is is aa constant constant vector. vector. We We have have where
dx da x(t) = a(t)u, dt = Ax => dt u = aAu. This last last equation equation states states that two vectors vectors are are equal, equal, which which means means that that Au Au must must This that two be multiple of of uu (otherwise, (otherwise, the the two two vectors in different different directions, directions, be aa multiple vectors would would point point in which is is not possible if if they they are are equal). equal). Therefore, Therefore, uu must must be be an an eigenvector eigenvector of of A: A: which not possible Au=)"u.
We obtain We then then obtain
da = ),,0: => a(t) = Ce),(t-to) dt ' where C C is is aa constant. constant. Therefore, Therefore, if if )", A,uu isis any any eigenvalue-eigenvector eigenvalue-eigenvector pair, pair, then then where x(t)
= e),(t-to)u
is aa solution solution of of (4.17). (4.17). is Example Let Example 4.5. 4.5. Let A
= ~ [-: -: 6
1
! ].
4-5
eigenvalues of of A A are \i = = 0, 0, A2 = -1, —1, and),,3 and AS = = -2, —2, and the corresponding corresponding Then the eigenvalues are),,1 ),,2 = eigenvectors are
15 15See See Section Section 3.5.3. 3.5.3.
4.3. Linear Linear systems systems with constant coefficients coefficients 4.3. with constant
93
We therefore therefore know three independent We know three independent solutions solutions of of dx dt
= Ax,
namely, namely, X1(t)=U1, X2(t) = e-(t-t O )U2, X3(t) =
e- 2 (t-t O)U3.
If, as in the previous previous example, R nn consisting of eigenvectors If, example, there is a basis for R {ui, U2, u 2 , ... . . . ,, u of A, A, with corresponding eigenvalues we can {U1' un} with corresponding eigenvalues AI, AI, A A2,"" An, then we can n} of 2,..., A n , then write of (4.17). Indeed, Indeed, it it suffices suffices to show that write the the general general solution solution of to show that we we can can solve solve
dx -A dt x,
x(to)
(4.18)
= Xo·
n Since {U {ui, un} is basis for Rn, , there exists aa vector Rn such such that Since 1, 112,..., U2, ... , Un} is aa basis for R there exists vector cc G ERn that
The solution to then The solution to (4.18) is is then
(4.19) This is verified by computing dx/ dt:
~~ (t)
+ c2A2eA2(t-to)U2 + ... + CnAneAn(t-to)un cle),dt-to) AUI + c2e),2(t-t O) AU2 + ... + cne),n(t-t o) AUn A (C1e),1(t-to)U1 + c2e A2 (t-t o )U2 + ... + CneAn(t-tO)un)
= cIAle A1 (t-t o)UI =
=
= Ax(t).
We also have have We also by construction. by construction. Example 4.6. 4.6. Let Let A A be be the matrix Example matrix in in Example 4.5, 4-5, and let
94
Chapter 4. Essential Essential ordinary ordinary differential equations differential equations
Since the basis of {111,112,113} is orthonormal, orthonormal, as as is is easily checked, Since the basis of eigenvectors eigenvectors {Ul, U2, U3} o/A of A is easily checked, we have we have
We We can then immediately write down the the solution to to
dx
-=Ax, dt
x(O) = xo. is It is
as in in the the previous previous example, example, A A is is symmetric, the eigenvectors eigenvectors If, as symmetric, then then the
{ui,u 2 ,...,u n } can be chosen chosen to to be be orthonormal, orthonormal, in in which which case case it it is particularly simple to compute can be is particularly simple to compute the coefficient coefficient Cl, C2, ...- ,,c Cn expressing in terms terms of of the the eigenvectors. We obtain the ci,C2,.. expressing Xo x0 in eigenvectors. We obtain the following the solution (4.18): the following simple simple formula formula for for the solution of of (4.18): n
x(t) = 2)xo . ui)e"i(t-to)Ui. i=l
We have have derived derived the the solution to (4.18) in the the case that A n linearly inWe solution to (4.18) in case that A has has n linearly independent We remark that formula formula (4.19) (4.19) might might involve involve complex complex dependent eigenvectors. eigenvectors. We remark that numbers if A A is is not not symmetric. we have have aa complete complete and and satisfactory numbers symmetric. However, However, we satisfactory the case that A A is is symmetric-there n linearly eigensolution in in the solution case that symmetric—there are are n linearly independent independent eigenthe eigenvalues to be be real, real, and the basis of eigenvectors can vectors, the vectors, eigenvalues are are guaranteed guaranteed to and the basis of eigenvectors can be chosen chosen to be orthonormal. be to be orthonormal. the significance The interpretation interpretation is very We now now discuss discuss the We significance of of solution solution (4.19). (4.19). The is very The solution to (4.18) (4.18) is the sum components, one simple: simple: The solution to is the sum of of nn components, one in in each each eigendireceigendirection, and is tion, and component component ii is • exponentially exponentioallyincreasing increasing(if (if Ai yi > 0); 0);
exponentioallydecreasing decreasing(if(ifAiyi<< 0); 0); or or • exponentially constant (if (if Ai yi == 0). • constant 0).
4.3.
95 95
Linear systems with constant coefficients coefficients
Moreover, Moreover, the the rate rate of of increase increase or or decrease of of each each component component is is governed governed by the the magnitude magnitude of of the the corresponding corresponding eigenvalue. eigenvalue. We can therefore draw the following following conclusions: conclusions: 1. the eigenvalues of A are negative, then every solution x of (4.18) 1. If all of the
satisfies satisfies Ilx(t)11 -t 0 as t -t
00.
2. If all of the eigenvalues of A are positive, then every solution x of (4.18) (except the zero solution) satisfies Ilx(t)11 -t
00
as t -t
00.
3. If any of the the eigenvalues of A is positive, then, for most initial values XQ, xo, the the solution x of (4.18) satisfies solution x of (4.18) satisfies
Ilx(t)11 -t
00
as t -t
00.
The only statement which requires justification is the third. Suppose an eigenvalue Aj of of A positive, and \j A is is positive, and suppose suppose A A has has k linearly linearly independent independent eigenvectors eigenvectors correcorresponding to \j. Aj. Then, Then, unless Xo lies the (n )-dimensional subspace sponding to unless XQ lies in in the (n -— kk)-dimensional subspace of of R Rnn spanned by the other n -— kk eigenvectors, the solution x of (4.18) will contain a Aj O ), guaranteeing the factor of eeAj(t-t (*~*°), the
||x(£)|| —> 00 oo as as tt —> oo. Ilx(t)ll-t -t 00. - k)-dimensional Even if kk == 1, 1, an (n — fc)-dimensional subspace of R Rnn is a very small part of Rn Rn 3 (comparable to a plane or a line in R3). R ). Therefore, most initial vectors do not lie in this subspace.
Example 4.7. 4.7. Let Let Example 1 -1 1
Jl
Then A are Al = 1, A2 = = -1, A3 = -2, Then the the eigenvalues eigenvalues of of A are AI 1, A2 —1, and and \s —2, and and the the corresponding corresponding eigenvectors are eigenvectors are
The solution of The solution of (4.18) (4-18) is is
where where
96
Essential ordinary differential equations Chapter 4. Essential
The only not grow without bound only initial values that lead to a solution x that does not are are Xo E S = span{U2' U3}. But S is plane, which very small part of Euclidean 3-space. almost But is aa plane, which is is aa very small part of Euclidean 3-space. Thus Thus almost every every initial initial value leads leads to to a solution solution that grows exponentially. exponentially.
Another can draw draw is is somewhat somewhat more more subtle subtle than those Another conclusion conclusion that that we we can than those given but it 6. given above, above, but it will will be be important important in in Chapter Chapter 6. 4. has eigenvalues very different then solutions (4.18) have 4. If If A A has eigenvalues of of very different magnitudes, magnitudes, then solutions of of (4.18) have components whose whose magnitudes change at very different different rates. Such solutions can be difficult difficult to compute efficiently efficiently using numerical methods. methods. We We will will discuss discuss this this point point in in more more detail detail in in Section Section 4.5. 4.5.
4.3.2 4.3.2
Inhomogeneous and variation variation of parameters In homogeneous systems systems and of parameters
We can now explain how to solve the inhomogeneous inhomogeneous system system
dxdt = Ax + f(t),
(4.20)
again only considering the case in which which there is a basis of Rn Rn consisting of eigenvectors of A. The method is a spectral method, and the reader may wish to review Section 3.5.3. 3.5.3. Section If ... , u un} R nn can be written If {Ul' {ui, U2, u 2 ,..., for R R nn,, then every vector in R n} is a basis for uniquely as a linear combination of these vectors. In particular, for each t, £, we we can Ul, U2, ... , u Un: write write f(t) as aa linear linear combination combination of ui, 112,..., n:
Of the vector t, so ... ,en(t). Of course, course, since since the vector f(t) f (t) depends depends on on £, so do do the the weights weights Cl c\ (t), (£), C2(t), C2(t),..., cn(t). These weights can be computed explicitly from f, f, which of course is considered to be be known. We can also write the solution of (4.20) in terms of the the basis vectors: (4.21)
,ann(t). However, However, when Since x(t) x(£) is unknown, unknown, so are are the the weights al(t),a2(t), ai(i),«2(*), ... • • • ,a, unknown weights. these basis vectors are eigenvectors of A, it is easy to solve for the unknown Indeed, substituting (4.21) (4.21) in in place Indeed, substituting place of of x x yields yields
dxdt
d[n
- - A x = - "'a·udt~" i=l
n
= L i=l
dt
~ i=l
d
~Ui
1-A (n"'a-u) •• n
-
LaiAui i=l
4.3.
Linear Linear systems with constant coefficients
The ODE The ODE
dx
- - Ax dt then implies then implies that that n
'"' L...J
{d~ } dt - A·a· ~
i=1
~
97
= f(t)
U·~
n
= '"' L...J c·(t)u· ~ ~, i=1
which can only only hold hold if which can if
~i
_
Aiai = Ci(t), i = 1,2, ... ,no
(4.22)
Thus, by by representing representing the the solution solution in in terms terms of of the the eigenvectors, eigenvectors, we we reduced reduced the the Thus, system of of ODEs ODEs to to nn scalar scalar ODEs ODEs that can be solved independently.16 independently.16 The The comsystem that can be solved computation of of the the functions functions C1 ci(£), C 2(t), ( t ) ,... . . . ,, C cnn(t) is simplest simplest when when A is symmetric, symmetric, so putation (t), C2 (t) is A is so that the eigenvectors eigenvectors can that case, that the can be be chosen chosen to to be be orthonormal. orthonormal. In In that case,
Ci(t) = f(t) . Ui, i = 1,2, ... ,n. We will will concentrate concentrate on on this case henceforth. We this case henceforth. The x(to) unknowns The initial initial condition condition x(£ = Xo x0 provides provides initial initial values values for for the the unknowns 0) = ai(t],a-2,(t},... Indeed, we we can can write a1 (t), a2(t), ... ,a , an(t). write n(t}. Indeed,
so x(to) so x(£ = Xo XQ implies implies that that 0) =
or or
= bi , i = 1,2, ... ,no
ai(to)
When the basis is orthonormal, the coefficients coefficients b&1i ,, &b22 ,j .•• - - - >, &bnn can be computed in the usual the usual way: way: bi = Xo . Ui, i = 1,2, ... ,n. Example Example 4.8. 4.8. Consider Consider the the IVP IVP
dx
= Ax + f(t), x(O) = 0, dt
16 16This just what what happened happened in the spectral b-the system This is is just in the spectral method method for for solving solving Ax Ax == b—the system of of nn simultaneous equations equations was was reduced to nn independent independent equations. equations. simultaneous reduced to
98 98
Chapter 4. Essential ordinary ordinary differential equations Chapter equations
where
A= [11 1] l'
f(t)
= [ C?S (t) sm (t)
] .
2 The matrix A is symmetric, and an orthonormal basis for R R2 consists of vectors consists of the vectors
U1
l'
= [ _;,
U2 = [ ; ,
l·
The corresponding corresponding eigenvalues are are \i A1 = = 0, A2 = = 2. 2. 0, X% f(t) in terms of of the eigenvectors: We now express express f(t) f(t) = Cl(t)Ul + C2(t)U2, 1 . Cl(t) = f(t) . U1 = J2(cos (t) - sm (t)),
C2(t)
= f(t) . U2 =
1
J2(cos (t)
.
+ sm (t)).
The solution is where 1
dal
dt = J2 (cos (t) da2
. - sm (t)), a1(0)
1
= 0,
.
dt = 2a2 + J2 (cos (t) + sm (t)) , a2(0) = O. We obtain obtain 1 aI(t) = J2 =
and
a2(t)
1 = J2
it 0
{CDS(S) -sines)} ds
~ (sin (t) + cos (t) -
it 0
1)
e2(t-s) {cos (s) + sin (s)} ds
1 2t = 5y2 In (3e -
.
3cos(t) - sm(t)).
Finally, x(t)
= a1 (t)U1 + a2(t)u2 =
/0 (2cos(t) +4sin(t) +3e 2t -5) [ 110 (-8 cos (t) _ 6 sin (t) + 3e2t + 5)
1 .
The technique for for solving presented in this section, section, which which we we have have The technique solving (4.20) (4.20) presented in this described as aa spectral spectral method, method, is is usually usually referred as the described as referred to to as the method method of of variation of of parameters.
4.3.
99
constant coefficients Linear systems with constant
Exercises 1. Consider the the IVP IVP (4.18), (4.18), where where A A is is the the matrix matrix in in Example Example 4.5. 4.5. Find Find the 1. Consider the solution solution for for
2. Consider Consider the the IVP IVP (4.18), (4.18), where where A A is is the the matrix matrix in in Example Example 4.5. 2. 4.5.
(a) matter what what the the value value of of x xo, the solution solution x(£) x(t) converges converges (a) Explain Explain why, why, no no matter 0 , the to constant vector to aa constant vector as as tt —> -+ oo. 00. (b) Find Find all all values values of of x Xo0 such such that that the the solution x is is equal equal to to aa constant constant vector (b) solution x vector for for all all values values of of t.t.
3. to 3. Find Find the the general general solution solution to
dx
dt = Ax,
where where
A=[~ ~]. 4. Find Find the the general general solution solution to to 4.
dx
dt = Ax,
where where
5. Let Let A A be be the the matrix matrix in in Exercise Exercise 3. 3. Find Find all all values values of Xo such such that that the the solution 5. of XQ solution exponentially to zero. to (4.18) decays exponentially 6. Let Let A A be be the the matrix matrix in in Exercise Exercise 4. 4. Find Find all all values values of of XQ Xo such such that that the the solution solution 6. to (4.18) (4.18) decays exponentially to to zero. zero. to decays exponentially 7. Let Let A A be be the the matrix matrix of of Example Example 4.5. 4.5. Solve Solve the the IVP 7. IVP
dx
= Ax + ret), x(O) = 0, dt
where where
ret) = [
~
sin (t)
l.
100 100
Chapter 4. Essential Essential ordinary differential equations
8. Let Let 8.
-4 2 -4
-4]
2 -4 -4 -4 2
.
Solve the the IVP IVP Solve
dx
= Ax + f(t),
dt
x(o) = xo,
where where
(t) ] 1 , sin (t)
COS
f(t) = [
Xo
=
[ 1] 1 . 1
9. following system been proposed proposed as the population population dynam9. The The following system has has been as aa model model of of the dynamics two species animals that that compete ics of of two species of of animals compete for for the the same same resource: resource:
dx dt
= ax -
~~
= -ex + dy, y(O) = yo.
by, x(O)
= Xo,
Here a, b, positive constants, x(t) is the population population of of the the first first species Here o, 6, e, c, dd are are positive constants, x(t] is the species at time £, and and y(t] is the corresponding population of the the second second species species at time t, yet) is the corresponding population of (x measured in convenient units, units, say thousands or millions (x and and yy are are measured in some some convenient say thousands or millions of The equations increases of animals). animals). The equations are are easy easy to to understand: understand: either either species species increases (exponentially) if the other is not present, the two compete is not present, but, but, since since the two species species compete (exponentially) if the other for resources, the the presence one species species contributes negatively to to for the the same same resources, presence of of one contributes negatively the growth rate of the growth rate of the the other. other.
(a) Solve the IVP with with 6b = = ec = o= = 1, 1, and (a) Solve the IVP = 2, 2, a = dd == l1,, x(0) x(O) = = 2,2, and and j/(0) yeO) = and explain (in (in words) to the of the the two two species explain words) what what happens happens to the populations populations of species in the long term. term. in the long (b) With the the values values of of a, a, 6, c, dd given given in 9a, is is there there an an initial initial condition (b) With b, e, in part part 9a, condition which will outcome? which will lead lead to to aa different different (qualitative) (qualitative) outcome? 10. The The purpose of this exercise is to derive derive the solution (4.11) (4.11) of of the IVP 10. purpose of this exercise is to the solution the IVP
J2u dt 2
2
+ B u = J(t), u(to) du dt (to)
The solution solution The u(t)
=~
= 0, = o.
(4.23)
rt sin (B(t - 8»J(8) d8
ito
can using the the techniques techniques of the computations can be be found found using of this this section, section, although although the computations are more any of of the presented. than any the examples examples we we have have presented. are more difficult difficult than
4.4.
methods for initial value problems Numerical methods
101 101
(a) Rewrite Rewrite (4.23) in the the form (a) (4.23) in form
dx dt
= Ax + F(t),
x(to)
= O.
matrix A is not symmetric. Notice that the matrix Al, A2 A2 and and eigenvectors eigenvectors Ui,u Ul, U22 of (b) Find Find the the eigenvalues eigenvalues AI, (b) of A. A. (c) Write Write the the vector-valued vector-valued function function F(t) in the form (c) in the form
F(t)
= Cl(t)Ul +C2(t)U2.
Since the eigenvectors eigenvectors ui Ul and and u U2 not orthogonal, orthogonal, this this will will require Since the are not require 2 are solving 2) system system of of equations to find find Cl(t) and C2(t). solving aa (2 (2 xx 2) equations to c\ (t} and c2 (t). (d) Write Write the the solution solution in in the the form (d) form
x(t) = al(t)ul
+ a2(t)u2,
and solve solve the scalar IVPs get ai (t}, a2(t). a2 (t}. and the scalar IVPs to to get al(t), (e) The desired Show that is (4.11). (e) The desired solution solution is is u(t] u(t) = xi(t). Xl(t). Show that the the result result is (4.11).
4.4
Numerical methods methods for initial value problems problems
So far simple classes it is So far in in this this chapter, chapter, we we have have discussed discussed simple classes of of ODEs, ODEs, for for which which it is However, most differential possible to produce an explicit formula for the solution. However, differential equations in this sense. Moreover, sometimes the that equations cannot cannot be be solved solved in this sense. Moreover, sometimes the only only formula formula that difficult to evaluate, involving integrals that cannot cannot be computed in can be found is difficult functions, eigenvalues and eigenvectors that cannot cannot be found terms of elementary functions, exactly, so forth. exactly, and and so forth. In cases it may to that the the only only way way to to investigate investigate the the solution solution is is to In cases like like these, these, it may be be that approximate it using approximate it using aa numerical method, method, which which is is simply simply an an algorithm algorithm producing producing an approximate solution. We an approximate solution. We emphasize emphasize that that the the use use of of aa numerical numerical method method always always implies that that the the computed computed solution will be be in in error. error. It is essential essential to to know know something implies solution will It is something about the the magnitude magnitude of of this this error; one cannot cannot use use the the solution solution with with any any about error; otherwise, otherwise, one confidence. confidence. Most numerical methods for IVPs in ODEs are designed for first-order first-order scalar almost without change, to first-order systems, and equations. These can be applied, almost hence to to higher-order higher-order ODEs (after they they have have been been converted converted to to first-order hence ODEs (after first-order systems). systems). the general first-order scalar IVP, which is of Therefore, we begin by discussing the first-order scalar of the the form form du (4.24) dt = !(t, u), u(to) = Uo· We shall discuss time-stepping methods, which seek seek to find approximations approximations to to U(tl), U(t2),"., u(tnn},), where where to < ti tl < t^ t2 < ... define the the grid. grid. The The quantities n(ti),^(£2), • • • ,u(t • • • < ttnn define quantities ti — to,t2 tn-i are it - to, t2 -— ti,..., tl,···, ttnn -— tn-l are called called the the time steps. The basic basic idea idea of of time-stepping time-stepping methods methods is is based based on on the the fundamental fundamental theorem The theorem if of calculus, which implies that if
du dt (t)
= !(t,u(t)),
102
Chapter equations Chapter 4. Essential Essential ordinary ordinary differential equations
then then (4.25) we cannot Now, we cannot (in general) hope to evaluate evaluate the integral integral in (4.25) (4.25) analytically; analytically; however, any numerical method for approximating the value of a definite integral (such methods are often quadrature rules) can be adapted to form the often referred to as quadrature basis of a numerical method of IVPs. By the way, way, equation (4.25) explains why the process often referred process of of solving solving an an IVP IVP numerically numerically is is often referred to to as as integrating integrating the the ODE. ODE.
4.4.1
Euler's Euler's method
The simplest left-endpoint rule: rule: simplest method of integrating integrating an ODE is based based on the left-endpoint
lb
f(x) dx == f(a)(b - a).
(4.25), we Applying this to (4.25), we obtain obtain (4.26) Of course, course, this is not a computable formula, because, except possibly on the first step (i = 0), we we have an approximation, we do not know U(ti) u(ti) exactly. Instead, we approximation, Ui == = U(ti)' u(ti). Formula Formula (4.26) (4.26) suggests suggests how how to to obtain obtain an an approximation approximation Ui+1 m+\ to to U(ti+1): u(ti+i): Ui (4.27) The reader should notice that (except for ii == 0) 0) there are two sources of error in the using formula the estimate estimate UHl. MJ+I. First First of of all, all, there there is is the the error error inherent inherent in in using formula (4.26) (4.26) to to advance the integration by one time step. Second, Second, there is the accumulated error due we do ti) in due to to the the fact fact that that we do not know know u( u(ti] in (4.26), (4.26), but but rather rather only only an an approximation approximation to to it. it. The method (4.27) is referred to as Euler's Euler's method. method. Example Example 4.9. 4.9. Consider Consider the the IVP IVP du
dt The The exact exact solution solution is is
u
I
+ t2'
u(O) = 1.
(4.28)
1
u(t) = etan - t, and formula to and we we can can use use this this formula to determine determine the the errors errors in in the the computed computed approximation approximation to u(t). u(t). Applying Applying Euler's method with with the the regular regular grid t{ = iLlt, iAt, Ai 10/n, we to Euler's method grid ti Llt = lOin, we have have
103 103
4.4. Numerical Numerical methods methods for value problems 4.4. for initial initial value problems
For example, with UQ = 11 and = 100, 100, we For example, with Uo and nn = we obtain obtain Ul ~
1.1000,
U2 ~
1.2089,
U3 ~
1.3252, ....
In Figure Figure 4-1, 4.1, we graph the exact solution solution and and the approximations computed using In we graph the exact the approximations computed using Euler's method method for for nn = = 10,20,40. As we we should should expect, produced Euler's 10,20,40. As expect, the the approximation approximation produced as At IJ..t (the step) decreases. In fact, fact, Table 4.1, by Euler's Euler's method method gets gets better by better as (the time time step) decreases. In Table 4-1, we collect suggests that global where where we collect the the errors errors in in the the approximations approximations to to u(10), w(10), suggests that the the global of steps) steps) is is O(At). O(lJ..t). The error (which comprises the total error error after after aa number error (which comprises the total number of The symbol symbol of At") IJ..t") denotes denotes aa quantity quantity that proportional to to or smaller than O(lJ..t) 0(At) ("big-oh ("big-oh of that is is proportional or smaller than IJ..t as At IJ..t ~ At as ->• O. 0. 5~------~-------r------~--------~------,
4.5
, ..
~.
4
~
o
".!,.
~.
o
'! . . . . .
I!. ~ .
I! . . . .
3.5 :::J
3 2.5
2 -
exact
o It
2
4
6
8
n=10 n=20 n=40 10
Figure 4.1. (4-28). Figure 4.1. Euler's Euler's method method applied applied to to (4.28).
Proving that the order method is is really really O(lJ..t) is beyond beyond the the scope scope Proving that the order of of Euler's Euler's method O(At) is of this this book, book, but but we we can easily sketch essential ideas the proof. proof. It be of can easily sketch the the essential ideas of of the It can can be proved that the the left-endpoint left-endpoint rule, rule, applied to an an interval interval of of length length proved that applied to of integration integration of IJ..t, has an error error that that is is O(At 0(lJ..t 22 ). ). In In integrating an IVP, we apply apply the the left-endpoint At, has an integrating an IVP, we left-endpoint rule repeatedly, in in fact, 0(1/ IJ..t) times. times. Adding Adding up up 1/ IJ..t errors order At IJ..t 22 rule repeatedly, fact, nn = = 0(l/At) I/At errors of of order gives total error that this this heuristic heuristic reasoning reasoning is gives aa total error of of O(lJ..t). 0(At). (Proving (Proving that is correct correct takes takes some rather rather involved but elementary be found most numerical numerical some involved but elementary analysis analysis that that can can be found in in most analysis textbooks.) analysis textbooks.)
104 104
Chapter 4. Essential ordinary ordinary differential differential equations equations
n n
Error Error in in u(10) w(10)
Error Error in in u(10) w(io)
10 10 20 40 80 80 160 320 640
-3.3384.10- 11
-3.3384. -3.3384 • 101Q-11 -3.7014.10-3.7014 -10-11 -4.0218. 10- 11 -4.0218 -HT -4.2232 .- 10HT11 -4.3378. -4.3378 • 10lO"11 -4.3992. -4.3992 • 101Q-11 -4.4310 .• 101Q-11
-3.3384 • 1Q-1.8507.10-1.8507 -KT 11 -1.0054.10-1.0054 -HT 11 -5.2790· -5.2790 • 101Q-22 -2.7111.10-2.7111 -1Q-22 -1.3748.10-1.3748 -HT 22 -6.9235 .• 1010~33
At At
Table 4.1. Global error error in in Euler's method for (4-28). Table 4.1. Global Euler's method for (4.28).
4.4.2 4.4.2
on Euler's method: Runge-Kutta methods Improving on
If method can If the the results results suggested suggested in in the the previous previous section section are are correct, correct, Euler's Euler's method can be used to to approximate approximate the the solution to an an IVP to any any desired desired accuracy accuracy by be used solution to IVP to by simply simply making tlt Although this this is might making At small small enough. enough. Although is true true (with (with certain certain restrictions), restrictions), we we might hope for that would as many many time hope for aa more more efficient efficient method-one method—one that would not not require require as time steps steps to achieve aa given given accuracy. accuracy. to achieve To method, we To improve improve Euler's Euler's method, we choose choose aa numerical numerical integration integration technique technique that that is more accurate than the is more accurate than the left-endpoint left-endpoint rule. rule. The The simplest simplest is is the the midpoint midpoint rule: rule:
lar
b
(a+b)
f(x) dx == (b - a)f -2- .
If tlt At = = bb -— aa (and (and the the integrand integrand f/ is is sufficiently sufficiently smooth), smooth), then the error error in in the the If then the midpoint ). Following the reasoning the previous we expect midpoint rule rule is is O(tlt O(A£33). Following the reasoning in in the previous section, section, we expect that the the corresponding that corresponding method method for for integrating integrating an an IVP IVP would would have have O(tlt O(A£22 )) global global error. error. It use the rule for It is is not not immediately immediately clear clear how how to to use the midpoint midpoint rule for quadrature quadrature to to integrate is integrate an an IVP. IVP. The The obvious obvious equation equation is
However, after after reaching time ti ti in in the integration, we we have an approximation approximation for However, reaching time the integration, have an for u(ti), but but none u(ti + A£/2). use the the midpoint midpoint rule requires that that we we first U(ti), none for for U(ti tlt/2). To To use rule requires first generate an an approximation approximation for for U(ti u(ti + tlt/2); Ai/2); the the simplest simplest way way to to do do this is with an generate this is with an Euler Euler step: step: U (ti
+ ~t) == u (ti) + ~t f
(ti,U (ti)).
Putting Putting this this approximation approximation together together with with the the midpoint midpoint rule, rule, and and using using the the approxapproximation Ui Ui == = U(ti), u(ti), we we obtain obtain the improved Euler method: imation the improved Euler method:
105 105
4.4. Numerical methods for initial value problems
nn
Error in in u(10) it (10) Error
Error in in2 u(lO) «(io) Error
10 10 20 20 40 40 80 80 160 160 320 320 640 640
-2.1629.10-2.1629 • 10~22 -1.2535 .-Mr 10- 22 -4.7603 .• 101Q-333 -1.4746.10-1.4746 -nr 43 -4.1061 -4.1061 .• 10-4 10~ -1.0835 .• 10-4 10-4 -2.7828.10-2.7828 • 10~55
-2.1629.10-2.1629-1Q-22 -5.0140.10-5.0140 -1Q-22 -7.6165.10-7.6165 -1Q-22 -9.4376. -9.4376 • 1010~22 -10.512.10-10.512 -1Q-22 -11.095.10-11.095 -i(r 222 -11.398.10-11.398 -10- 2
lJ.t At 2
Table 4.2. 4.2. Global Global error error in in the the improved improved Euler Euler method method for (4-28). Table for (4.28).
results of with Figure Figure 4.2 4.2 shows shows the the results of the the improved improved Euler Euler method, method, applied applied to to (4.28) (4.28) with n = by — 10, 10, n = 20, 20, and and n = 40. 40. The The improvement improvement in in accuracy accuracy can can be be easily easily seen seen by comparing Figure 4.1. We can Table 4.2-when comparing to to Figure 4.1. We can see see the the O(~t2) O(At2) convergence convergence in in Table 4.2—when At is is divided divided by by two two (Le. (i.e.nnisisdoubled), doubled),the theerror errorisisdivided dividedby byapproximately approximately four. four. ~t
4.5~------~-------r------~--------r-----~
-
exact
o •
n=10 n=20 n=40
8
10
Figure Improved Euler Euler method Figure 4.2. 4.2. Improved method applied applied to to (4.28). (4-28).
The improved improved Euler Euler method method is is aa simple simple example example of of aa Runge-Kutta The Runge-Kutta (RK) (RK)
106
Chapter Chapter 4. 4. Essential Essential ordinary ordinary differential differential equations equations
method. An (explicit) (explicit) RK RK method method takes the following following form: form: method. An takes the m
ui+1 = Ui
+ D.t L (}:jkj , j=l
kl
= 1 (ti' Ui) ,
k2 = 1 (ti k3 = 1 (ti
km =
+ 'Y2D.t, Ui + f321D.tkl) ' + 'Y3D.t, Ui + f331D.tk l + f332D.tk2) ,
(4.29)
1 (ti + 'YmD.t,ui + ~ f3mlD.tkl) ,
f3kl, and "0 with certain restrictions on the values of the parameters parameters (}:j, ctj, PM, 7^ (e.g. (e.g. (}:l a\ + ... +h (}:m am = —1). !)• Although (4.29) looks complicated, it is not hard to understand the the idea. A general form for a quadrature rule is
W2, ... ,W X2, ... ,X [a, b] b) are where WI, wi,W2,.--, wmm are the quadrature quadrature weights and Xl, xi, x%,..., xmm E G [a, axe the quadrature nodes. In the weights weights are quadrature nodes. In the the formula formula for for Ui+l Ui+i in in (4.29), (4.29), the are
and values • • • ,,kkm are estimates estimates of of fI(t, ( t , uu(t)) ( t ) ) at in the the interval and values fci, kl' £2, k2, ... at ra m nodes nodes in interval m are [ti,ti We will the general general formula formula (4.29), (4.29), but but the should apap[ti' ti+1)' will not not use use the the reader reader should +i]. We preciate the following point: there are are many many RK RK methods, obtained by by choosing preciate the following point: there methods, obtained choosing various values for the parameters in (4.29). This fact is used in designing algorithms rithms that that attempt to to automatically control the the error. We We discuss discuss this this further further in Section 4.4.4 below. Section 4.4.4 below. The most popular RK method is analogous to Simpson's rule for quadrature:
i
b
I(x) dx
== b ~
a(f(a) + 41 (a; b) + f(b)) .
Simpson's Simpson's rule rule has has an an error error of of O(D.t O(A£5)) (D.t (At = = bb -— a), and and the the following following related related method method for for integrating integrating an an IVP has global global error error of of O(D.t4): O(A£ 4 ): Ui+l
= Ui
D.t
+6
(k l
+ 2k2 + 2k3 + k4) ,
kl
=f
k2
D.t D.t) = 1 ( ti + 2' Ui + 2 kl ,
(ti' Ui) ,
k3 =
D.t D.t) 1 ( ti + 2' Ui + 2 k2 ,
k4 =
1 (ti+l,Ui + D.tk 3).
(4.30)
4.4.
107 107
Numerical Numerical methods methods for initial value problems
n n
Error Error in in u(10) w(10)
Error in in4 u(io) u(10) Error
10 20 40 80 160 320 640
1.9941 .• 1010~22 1.0614. 10-33 1.0614 -Hr 6.5959 .• 1010~55 4.0108 .• 1010~66 2.4500. 2.4500 • 101Q-77 1.5100. 10-88 1.5100 -l(r 10 9.3654. 9.3654 • 10- 10
1.9941 .• 1010~22 1.6982 .• 1010~22 1.6885. 1.6885 • 1010~22 1.6428 .• 1010~22 1.6057. 10-22 1.6057 -101.5833 .-1010- 22 1.5712 .-ID" 10- 22
.6.t At4
Table 4.3. Global error in the fourth-order Runge-Kutta method (RK4) Table 4.3. (RK4) for (4.28). for (4.28).
We call this the RK4 method. Figure 4.3 and Table 4.3 demonstrate demonstrate the accuracy of D..t by by two two decreases decreases the the error by (approximately) (approximately) aa factor factor of of 16. 16. of RK4; RK4; dividing dividing A£ error by This is typical typical for for O(D..t4) convergence. This is O(At 4 ) convergence. 4.5r-------~-------r------~--------r_----__,
-
exact
o ..
8
n=10 n=20 n=40
10
Figure Fourth-order Runge-Kutta Runge-Kutta method (RK4) Figure 4.3. Fourth-order (RK4) applied applied to (4.28). (4-28).
We must must note note here here that that the the improvement these higher-order higher-order We improvement in in efficiency efficiency of of these methods is not as dramatic as it might appear appear at first glance. For instance, comcomwe notice notice that that just just 10 of RK4 RK4 gives error paring Tables 4.1 and 4.3, we paring Tables 4.1 and 4.3, 10 steps steps of gives aa smaller smaller error than method. However, the improvement in efficiency efficiency is is not not aa than 160 160 steps steps of of Euler's Euler's method. However, the improvement in factor steps of of Euler's Euler's method use 160 while factor of of 16, 16, since since 160 160 steps method use 160 evaluations evaluations of of ff(t, ( t , uu), ) , while
108 108
Chapter 4. Essential Essential ordinary ordinary differential equations
10 RK4 use evaluations of problems of realistic complexity, 10 steps steps of of RK4 use 40 40 evaluations of ff(t, ( t , uu). ) . In In problems of realistic complexity, the is the the most part of the calculation is the evaluation evaluation of of ff(t,u) ( t , u ] is most expensive expensive part of the calculation (often (often /f is defined by aa computer computer simulation rather than so defined by simulation rather than aa simple simple algebraic algebraic formula), formula), and and so aa more more reasonable reasonable comparison comparison of of these these results results would would conclude conclude that that RK4 RK4 is is 44 times times as efficient Euler's method method for level of of error. error. HigherHigherand level as efficient as as Euler's for this this particular particular example example and order methods tend tend to that they order methods to use use enough enough fewer fewer steps steps than than lower-order lower-order methods methods that they are efficient even they require more work step. are more more efficient even though though they require more work per per step.
4.4.3 4.4.3
Numerical methods for systems systems of ODEs
The numerical methods methods presented presented above can be of The numerical above can be applied applied directly directly to to aa system system of ordinary differential differential equations. equations. Indeed, the system system is is written written in in vector ordinary Indeed, when when the vector form, form, the notation notation is identical. We We now now present an example. example. the is virtually virtually identical. present an
Example 4.10. species of share aa habitat, and suppose Example 4.10. Consider Consider two two species of animals animals that that share habitat, and suppose one prey to to the the other. Let Xl population of predator species species one species species is is prey other. Let x\ (t) (t) be be the the population of the the predator at time and let let X2(t) be the the population of the the same time. The The LotkaLotkaat time t, t, and X2(t) be population of the prey prey at at the same time. Volterra predator-prey model for these populations is Volterra predator-prey model for these two two populations is dXl dt = e2 e l XI X2 - qXI, dX2 dt
= rX2 -
elXIX2,
where ei, el, 62, e2, q, are positive positive constants. constants. The parameters have have the following interprewhere #, rr are The parameters the following interpretations: tations:
BI • el
describes the attack rate rate of the predators (the rate at which which the the prey describes the attack of the predators (the rate at prey are are killed bv the the predators vredators is is elxlx2); e^ x-\ x? ); killed by
the growth rate of the predator the number number of • 62 e2 describes describes the growth rate of the predator population population based based on on the of prey killed efficiency of predators at at converting predators to prey killed (the (the efficiency of predators converting predators to prey); prey); is the the rate rate at at which which the the predators • qq is predators die; die; is the the intrinsic intrinsic growth growth rate rate of of the the prey population. prey population. • rr is The the rate rate of population growth growth (or of each each species. The equations equations describe describe the of population (or decline) decline) of species. predator population population is the rate of The The rate rate of of change change of of the the predator is the the difference difference between between the rate of growth due due to feeding on prey and rate of of death. death. The rate of change of of growth to feeding on the the prey and the the rate The rate of change the prey population population is is the natural growth growth rate would the prey the difference difference between between the the natural rate (which (which would govern in absence of of predators) predators) and and the death rate due to govern in the the absence the death rate due to predation. predation. 2 Define -» R22 by Define ff :: R x R 2 -+ by
f(t,x) = [ e2 e l Xl X2 rX2 -
-
qXl ]
elXIX2
(f is is actually actually independent independent of of t; t; however, include tt as as an independent variable (f however, we we include an independent variable so general form form discussed discussed above). given initial so that that this this example example fits fits into into the the general above). Then, Then, given initial
4.4.
109 109
methods for initial value problems Numerical methods
values #1,0 Xl,O and X2,O of the predator predator and and prey prey populations, populations, respectively, respectively, the IVP of of values and #2,0 of the the IVP interest is is interest dx dt
= f(t,x), x(O) = xo,
(4.31)
where where
= [ Xl,O
xo
X2,O
] .
Suppose Suppose
e = 0.01, 0.01, e2 e2 == 0.2, 0.2, rr == 0.2, 0.2, qq == 0.4, 0.4, Xl,O XI,Q == 40, 40, X2,O x2,o == 500. 500. ell =
usmg the we estimated estimated the of (4.31) (4-31) on on the the Using the RK4 RK4 method method described described above, above, we the solution solution of time interval [0,50]. [0,50]. The The implementation is exactly described in in (4.30), (4-30), except except time interval implementation is exactly as as described that the various various quantities (lcj,Uj,f(tj,Ui)J The time time step Ui, f(ti' Ui)) are are vectors. vectors. The step used used was was that now now the quantities (ki, Llt = 0.05 (for steps). In In Figure Figure 4-4> 4.4, wwe plot the the two two populations At = 0.05 (for aa total total of of 1000 1000 steps). & plot populations versus time; graph suggests suggests that populations vary versus time; this this graph that both both populations vary periodically. periodically. 600 I
"',
I
500
I
I
400 :
I
c
I
I
0
I
~300 c..
I I
0
c..
I
I
I
I I
200 100
I
,,
.. .. 10
I
,,
I
, ,,
I
.. 20
,,
I I
/
40
50
Figure 4.4. The variation of populations with with time Figure 4.4. The variation of the the two two populations time (Example (Example 4.10). 4-10). Another way way to to visualize visualize the results is graph x^ X2 versus versus Xl; such aa graph Another the results is to to graph x\; such graph is meaningful ODE (one (one in appear in which which tt does does not not appear is meaningful because, because, for for an an autonomous autonomous ODE explicitly), the curve curve (xi(t)jX^(t}} is determined be the the initial XQ (see (Xl (t), X2 (t)) is determined entirely entirely be initial value value xo (see explicitly), the for aa precise precise formulation formulation of this property). property). In In Figure Figure 4-5, 4.5, we graph x% X2 Exercise 55 for Exercise of this we graph versus x\. This This curve curve is clockwise direction (as can be determined determined versus Xl. is traversed traversed in in the the clockwise direction (as can be
110 110
Chapter Chapter 4. Essential ordinary ordinary differential differential equations equations
by Figure 4-4)4.4}. F° Forr example, when the prey population population is by comparing comparing with with Figure example, when the prey is large large and predator population population is small (the left part part of predator and the the predator is small (the upper upper left of the the curve), curve), the the predator As it does, the prey population population begins population will will begin grow. As population begin to to grow. it does, the prey begins to to decrease decrease (since more prey prey are Eventually, the prey population population gets gets small enough that that (since more are eaten). eaten). Eventually, the prey small enough cannot support support aa large large number number of predators (far part of the curve), curve), and itit cannot of predators (far right right part of the and the the predator population population decreases rapidly. 'When the predator predator population population gets gets small, predator decreases rapidly. When the small, the the grow, and the again. prey population begins to grow, the whole cycle cycle begins again. 600r---~----'---~~---r----~--~----~--~
10
20
30
40
50
60
70
80
predator population
Figure 4.5. The variation of populations (Example Figure 4.5. The variation of the the two two populations (Example 4.10). 4-10).
4.4.4 4.4.4
and Runge-Kutta-Fehlberg Automatic step control and methods
As As the above examples examples show, show, it it is possible to design numerical numerical methods which which give aa predictable improvement predictable improvement in in the error error as as the the time step step is is decreased. decreased. However, However, these these to choose in order desired methods do not not allow allow the the user methods do user to choose the the step step size size in order to to attain attain aa desired level D..t from level of of accuracy. accuracy. For For example, example, we we know that that decreasing decreasing At from 0.1 0.1 to 0.05 0.05 will will decrease decrease the the error in in an O(D..t4) 0(At 4 ) method method by approximately approximately aa factor of 16, 16, but this this does does not tell tell us that either either step step size size will will lead lead to a global global error error less than than 1010~33.. It It would would be desirable to have have aa method .that, that, given given aa desired desired level level of of accuracy, accuracy, could could choose the step step size size in an an algorithm so so as as to attain attain that accuracy. accuracy. While While no method guaranteed to do method guaranteed do this is is known, known, there there is is aa heuristic technique that that is is usually successful successful in in practice. The The basic idea idea is is quite quite simple. simple. When an algorithm wishes to to tn+!, uses two two different D..tfek and and to integrate integrate from from tn tn to tn+i, it it uses different methods, methods, one one of of order order Ai fc+1 k another D..t +!.. It then regards the more accurate another of order A£ accurate estimate estimate (resulting from from
4.4. Numerical methods for initial value problems
111 111
fe+1 the exact value, it with from the O(At O(~tk+l)) method) method) as as the the exact value, and and compares compares it with the the estimate estimate from the lower-order lower-order method. method. If the difference difference is is sufficiently sufficiently small, then it the If the small, then it is is assumed assumed that the the step is sufficiently the step accepted. (Moreover, (Moreover, if if the that step size size is sufficiently small, small, and and the step is is accepted. the difference is too too small, small, then then it it is assumed that that the the step than it it needs needs difference is is assumed step size size is is smaller smaller than to be, be, and and it it may may be be increased increased on on the the next next step.) step.) On On the the other other hand, hand, if if the the difference to difference is too too large, then it it is is assumed that the the step step is inaccurate. The The step step is rejected, and is large, then assumed that is inaccurate. is rejected, and ~t, the the step size, is is reduced. reduced. This This is is called automatic step step control, control, and and it it leads leads to At, step size, called automatic to an approximate approximate solution solution computed on an since At can vary an computed on an irregular irregular grid, grid, since ~t can vary from from one one step to This technique error below step to the the next. next. This technique is is not not guaranteed guaranteed to to produce produce aa global global error below the desired level, since the only controls error (the the desired level, since the method method only controls the the local local error (the error error resulting resulting from aa single single step step of and from of the the method). method). However, However, the the relationship relationship between between the the local local and global error is is understood in principle, and this global error understood in principle, and this understanding understanding leads leads to to methods methods that achieve the desired accuracy. accuracy. that usually usually achieve the desired A popular popular class of methods methods for control consists consists of of the the RungeRungeA class of for automatic automatic step step control Kutta-Fehlberg (RKF) (RKF) methods, methods, which which use use two two RK methods together together to to control Kutta-Fehlberg RK methods control the local local error error as in the the previous previous paragraph. paragraph. The The general form of of RK RK the as described described in general form methods, given shows that that there there are are many many possible possible RK RK formulas; formulas; indeed, indeed, methods, given in in (4.29), (4.29), shows there are many different RK methods methods that that can be derived. derived. for given order order !ltk, for aa given At fc , there are many different RK can be is used used in the RKF methodology to to choose pairs of of formulas that evaluate This fact fact is This in the RKF methodology choose pairs formulas that evaluate /, the function defining as few few times as possible. f, the function defining the the ODE, ODE, as times as possible. For For example, example, it it can can be be proved that that every method of of order order !lt requires at least six evaluations of of f./. proved every RK RK method At55 requires at least six evaluations It is (i.e. the ti,ti 72/1,... + 16h ^h in It is possible possible to to choose choose six six points points (Le. the values values ti, ti + 12h, ... ,, ttij + in (4.29)) (4.29)) so can be in an points so that that five five of of the the points points can be used used in an O(At O( !lt 44 )) formula, formula, and and the the six six points together define an O(At 55 )) formula. (One such of formulas formulas is the together define an O(!lt formula. (One such pair pair of is the the basis basis of of the popular efficient implementation implementation of automatic popular RKF45 RKF45 method.) method.) This This allows allows aa very very efficient of automatic step control. step control.
Exercises 1. The purpose purpose of this exercise exercise is is to to estimate estimate the the value value of u(0.5), where where u(t) u(t) is is 1. The of this of u(0.5),
the solution of IVP the solution of the the IVP du
t
dt = u + e , u(O) = O.
(4.32)
The solution is and so = ee 11//22 /2 /2 = The solution is w(t) u(t) = = te*, te t , and so w(0.5) u(0.5) = == 0.82436. 0.82436. (a) steps of method. u(0.5) by by taking taking 44 steps of Euler's Euler's method. (a) Estimate Estimate w(0.5) (b) Estimate steps of u(0.5) by by taking taking 22 steps of the the improved improved Euler's Euler's method. method. (b) Estimate w(0.5) (c) Estimate Estimate w(0.5) u(0.5) by by taking taking 11 step the classical classical fourth-order RK method. (c) step of of the fourth-order RK method.
Which each evaluate evaluate ff(t, ( t , u) = = is more more accurate? accurate? How How many many times times did did each Which estimate estimate is u + eetet ?? u+ 2. 2. Reproduce Reproduce the the results results of of Example Example 4.10. 4.10.
112
3.
Chapter 4. Essential Essential ordinary differential equations 17
17
The system of of ODEs ODEs The /-LI(X - /-L2) T2(X,y)3
(4.33)
/-LlY T2(X,y)3'
where
+ /-LI)2 + y2, = V(X - /-L2)2 + y2,
TI (X,y) = V(X T2(X,y)
models the the orbit orbit of of aa satellite satellite about two heavenly heavenly bodies, which we we will will assume assume earth and to be the earth and the moon. In these these equations, equations, (x(t),y(t)) (x(t),y(t)) are the the coordinates coordinates of of the the satellite satellite at at time time t.t. The origin origin of of the the coordinate coordinate system system is the center of mass of the the earth-moon earth-moon The is the center of mass of system, and and the the x-axis ar-axis is is the the line line through the centers centers of and the the system, through the of the the earth earth and moon. The center center of of the is at the point point (1 (1 — the center center of moon. The the moon moon is at the - /^i,0) /-LI,O) and and the of the is at at ((—//i, 0), where 1/82.45 is is the the ratio ratio of of mass of the the earth earth is /-LI, 0), where IJL\ /-LI = = 1/82.45 mass of the moon moon to the mass of of the The unit unit of length is is the distance from from the center to the mass the earth. earth. The of length the distance the center earth to the moon. We write ^2 /-L2 = /-L1.• of the the earth to the the center of the = 11 -— A*i If known If the satellite satellite satisfies the following initial conditions, then its orbit is known to be be periodic with with period T = 6.19216933: 6.19216933:
dx x(O) = 1.2, dt (0) = 0, y(O) = 0,
~~ (0) =
-1.04935751.
(a) Convert Convert the the system system (4.33) (4.33) of of second-order second-order ODEs ODEs to first-order system. system. (a) to aa first-order (b) (b) Use Use aa program that implements implements an an adaptive (automatic (automatic step step control) orbit for one period. Plot the method18 (such as RKF45) RKF45) to follow the orbit orbit in plane, and make sure the tolerance tolerance used by the orbit in the the plane, and make sure that that the used by the adaptive adaptive algorithm algorithm is small small enough that that the orbit orbit really appears in in the plot plot to to be periodic. periodic. Explain Explain the the variation variation in in time time steps steps with with reference to the the motion of the satellite. (c) fixed step (c) Assume Assume that, in in order to obtain obtain comparable accuracy accuracy with with aa fixed step size, it it is use the the minimum step chosen chosen by the adaptive adaptive size, is necessary necessary to to use minimum step by the algorithm. How many steps steps would would be by an an algorithm algorithm. How many be required required by algorithm with with 19 fixed step size? How algorithm?19 fixed How many steps were used by the adaptive algorithm? 17 17 Adapted pages 153-154. Adapted from from [16], [16], pages 153-154. 18 18Both and Mathematica provide provide such respectively. Both MATLAB MATLAB and such routines, routines, ode45 ode45 and and NDSolve, NDSolve, respectively. 19 to determine number of routine. It It is possible 19It It is is easy easy to determine the the number of steps steps used used by by MATLAB's MATLAB's ode45 ode45 routine. is possible but tricky the number number of of steps used by by Mathematica's Mathematica's NDSolve NDSolve routine. but tricky to to determine determine the steps used routine.
methods for initial value problems 4.4. Numerical methods
113
4. Let A=![ll 48] 5 48 39 . The problem is to produce a graph, on the interval [0,20], of the two components of of the the solution nents solution to to du dt = Au,
(4.34)
= 110,
u(O)
where
Llt = = 0.1 (a) Use Use the RK4 method with a step size of At 0.1 to compute the solution. Graph both components on the interval [0,20].
(b) Use the techniques of the previous section to compute the exact exact solution. Graph Graph both components components on on the interval [0,20]. [0,20]. difference. (c) Explain the difference.
b] and and f: b]-t 5. (a) Suppose tto0 E G [a, &] f : Rll Rn -t ->• Rll, R n , u : [a, 6] ->• Rll Rn satisfy du dt = f(u),
u(t o) = Uo· Show that v : [a + (it - to), b + (h - to)] -t Rll defined by
v(t) = u(t - (tl - to)) satisfies satisfies
dv
dt = f(v),
v(td = 110· Moreover, show that Moreover, show that the the curves curves
{u(t)
tE[a,b]}
and and
{v(t)
t
E
[a + (tl - to), b + (tl - to)]}
are the same. (b) Show, by an explicit explicit example (a scalar scalar IVP the (b) Show, by producing producing an example (a IVP will will do), do), that that the differential equation equation that depends above property does not hold for a differential explicitly explicitly on on tt (that (that is, is, for for aa nonautonomous ODE). ODE).
114 114
Chapter Chapter 4. Essential ordinary differential equations equations
6. 6. Consider Consider the the IVP IVP dx dt
x(O)
= cos (t)x, = 1.
Use both Euler's method and the improved improved Euler method to to estimate x(lO), Use both Euler's method and the Euler method estimate #(10), using step sizes 1,1/2,1/4,1/8, and and 1/16. 1/16. Verify in Euler's Euler's using step sizes 1,1/2,1/4,1/8, Verify that that the the error error in 2 method is O(A£), while the error in in the improved Euler Euler method is O(Ai ). method is O(~t), while the error the improved method is O(~t2). the IVP IVP 7. Consider Consider the
-dx = 1+ lx-II ' dt x(O) = O. (a) Use the the RK4 RK4 method and 1/128 (a) Use method with with step step sizes sizes 1/4,1/8,1/16,1/32,1/64, 1/4,1/8,1/16,1/32,1/64, and 1/128 to x(1/2) and verify that that the the error is O(~t4). to estimate estimate x(l/2) and verify error is O(Af 4 ).
(b) Use the RK4 RK4 method method with step sizes sizes 1/4,1/8,1/16,1/32,1/64, 1/4,1/8,1/16,1/32,1/64, and 1/128 (b) Use the with step and 1/128 to estimate estimate x(2). x(2). Is the error case as to Is the error O(~t4) O(A£ 4 ) in in this this case as well? well? If If not, not, speculate speculate as to to why why it it is is not. as not. 8. Let Let 8.
A~
-31
1 o -11 2 0 1 -1 0 -1 -1 1 0 0 2 -1 0-2
l
[ -2
-3
.
The matrix A has an eigenvalue eigenvalue A = -0.187; —0.187; let corresponding eigeneigenThe matrix A has an A == let x x be be aa corresponding problem is is to to produce produce aa graph the interval of the five vector. vector. The The problem graph on on the interval [0,20] [0,20] of the five components components of of the the solution solution to to du dt
= Au,
u(O) = x.
(a) Solve Solve the the problem method (in arithmetic) (a) problem using using the the RK4 RK4 method (in floating floating point point arithmetic) and graph the results. and graph the results. (b) What is is the the exact solution? (b) What exact solution?
(c) Why is there discrepancy between the computed computed solution solution and exact there aa discrepancy between the and the the exact (c) Why is solution? solution? (d) this discrepancy be eliminated eliminated by by decreasing the step step size (d) Can Can this discrepancy be decreasing the size in in the the RK4 method? method? RK4
115 115
4.5. Stiff systems of of ODEs
4.5
Stiff systems of ODEs
The numerical methods described described in in the the last last section, section, while for many IVPs, The numerical methods while adequate adequate for many IVPs, can be unnecessarily unnecessarily inefficient for some We begin begin with comparison of can be inefficient for some problems. problems. We with aa comparison of and the the performance performance of state-of-the-art automatic two IVPs, IVPs, and two of aa state-of-the-art automatic step-control step-control algoalgorithm on on them. rithm them. Example following two Example 4.11. 4.11. We We consider consider the the following two IVPs: IVPs:
dx dt
= Alx,
(4.35)
x(O) = Xo,
dx
dt = A 2 x,
(4.36)
x(O) = Xo. For both IVPs, the is the the same: same: For both IVPs, the initial initial value value is
We matrices AI Al and A22 by spectral decompositions. decompositions. The We will will describe describe the the matrices and A by their their spectral The two two are symmetric, symmetric, with the same same eigenvectors eigenvectors but different eigenvalues. matrices matrices are with the but different eigenvalues. The The eigenvectors eigenvectors are are
while the eigenvalues of —I, A = -2, —1, XA33 = —3 and and the eigenvalues of while the eigenvalues of AI Al are are AI Al = = -1, A22 = = -3 the eigenvalues of A are AI -I, A -10, A3 A3 = -100. -100. A22 are Al = -1, A22 = -10, Since Since the the initial initial value value x Xo0 is is an an eigenvector eigenvector of of both both matrices, matrices, corresponding corresponding to to the eigenvalue eigenvalue -1 —I in in both cases, the the two two IVPs exactly the the same solution: the both cases, IVPs have have exactly same solution:
We solved both systems using MATLAB command which implements We solved both systems using the the MATLAB command ode45, ode45, which implements aa 20 state-of-the-art automatic automatic step-control step-control algorithm algorithm based fourth-fifth-order scheme. scheme. 2o state-of-the-art based on on aa fourth-fifth-order The The results results are are graphed graphed in in Figure Figure 4-6, 4.6, where, where, as as expected, expected, we we see see that that the the two two comcomputed solutions solutions are are the the same same (or at least least very very similar). puted (or at similar). performed by ode45 reHowever, examining the the results results of of the the computations performed by ode45 veals surprise: the algorithm used used only steps for for IVP IVP (4-35) (4.35) but but 1124 steps for veals aa surprise: the algorithm only 52 52 steps 1124 steps for (4-36)1 (4·36)! 20 20The ode45 was was implemented by Shampine and Reichelt. Reichelt. See See [42] for details. The routine routine ode45 implemented by Shampine and [42] for details.
116 116
Chapter 4. 4. Essential differential equations Chapter Essential ordinary ordinary differential equations
2
4
6
8
10
2
4
6
8
10
Figure The computed computed solutions (4.35) (top) (top) and and IVP (4-36) Figure 4.6. 4.6. The solutions to to IVP IVP (4.35) IVP (4.36) (bottom). (Each (Each graph of the the solution; however, the the (bottom). graph shows shows all all three three components components of solution; however, exact solution, solution, which which is is the same for for the the two two IVPs, IVPs, satisfies satisfies xi(t] Xl(t) = = X3(t).) exact the same = £X2(t) (£) = xs(t)•) 2 A detailed explanation explanation of of these results is is beyond beyond the the scope scope of of this this book, but A detailed these results book, but we the reason comments are are illustrated illustrated we briefly briefly describe describe the reason for for this this behavior. behavior. These These comments by an an example below. Explicit Explicit time-stepping time-stepping methods methods for for ODEs, ODEs, such by example below. such as as those those that form the basis basis of have an stability region for the time step. that form the of ode45, ode45, have an associated associated stability region for the time step. fc This means means that the expected O(~tk)) behavior behavior of of the the global global error error is is not not observed observed This that the expected O(At unless is small small enough enough to in the stability region. values of ~t is to lie lie in the stability region. For For larger larger values of At, ~t, unless A£ the method method is is unstable produces computed that "blow up" as as the the unstable and and produces computed solutions solutions that "blow up" the integration proceeds. stability region and gets gets is problem problem dependent dependent and integration proceeds. Moreover, Moreover, this this stability region is 21 smaller the eigenvalues of the the matrix matrix A A get get large and negative. negative. 21 smaller as as the eigenvalues of large and If A has large eigenvalues, then system being has some If A has large negative negative eigenvalues, then the the system being modeled modeled has some transient the presence of the behavior transient behavior behavior that that quickly quickly dies dies out. out. It It is is the presence of the transient transient behavior 21 If 21 If
the of ODEs consideration is the system system of ODEs under under consideration is nonlinear, nonlinear, say say
= f(x,t), then it the eigenvalues the Jacobian Jacobian matrix / ax that that determine behavior. then it is is the eigenvalues of of the matrix 8f df/dx determine the the behavior. x
ODEs 4.5. Stiff systems systems of ODEs
117
that requires requires aa small small time time step—initially, step-initially, the the solution over aa very very small that solution is is changing changing over small time scale, the time time step model this this behavior behavior accurately. time scale, and and the step must must be be small small to to model accurately. If, the same time, there there are small negative negative eigenvalues, then the the system If, at at the same time, are small eigenvalues, then system also also models components out more Once the out, that die die out more slowly. slowly. Once the transients transients have have died died out, models components that one ought to able to increase the step to more slowly slowly varying varying to be be able to increase the time time step to model model the the more one ought components of the the solution. However, this this is is the the weakness weakness of explicit methods methods like like components of solution. However, of explicit the transient transient behavior behavior inherent Euler's method, RK4, Euler's method, RK4, and and others: others: the inherent in in the the system system haunts the numerical even when when the the transient transient components solution haunts the numerical method method even components of of the the solution have A time time step that ought to be be small enough to to accurately have died died out. out. A step that ought to small enough accurately follow follow the the slowly varying components produces instability in the numerical method and ruins slowly varying components produces instability in the numerical method and ruins the the computed computed solution. solution. A system with negative magnitudes (that A system with negative eigenvalues eigenvalues of of widely widely different different magnitudes (that is, is, aa that models models components different rates) rates) is is sometimes sometimes system system that components that that decay decay at at greatly greatly different 22 stiJJ.22 Stiff problems require require special numerical methods; we give give examples called called stiff. Stiff problems special numerical methods; we examples we discuss discuss aa simple for which which the below in however, we below in Section Section 4.5.2. 4.5.2. First, First, however, simple example example for the ideas presented above can be be easily easily understood. ideas presented above can understood.
4.5.1 4.5.1
A simple example of a stiff system
The IVP
d~l
= -107Ul, Ul(O) = 1,
(4.37)
dU2 dt = -U2, U2(0) = 1
has solution has solution (t) u
= [
107t
ul(0)eu2(0)e- t
]
'
and the the system qualifies as as stiff according to the description description given and system qualifies stiff according to the given above. above. We We will will apply method, since is simple enough that apply Euler's Euler's method, since it it is simple enough that we we can can completely completely understand understand its behavior. However, similar similar results be obtained another its behavior. However, results would would be obtained with with RK4 RK4 or or another explicit explicit method. method. the behavior on (4.37), (4.37), we we write exTo understand understand the To behavior of of Euler's Euler's method method on write it it out out explicitly. We have plicitly. We have with with
that is, that is, U~n+1) = u~n) _ 107 ~tu~n) = (1 - 107 ~t) u~n), u~n+1) = u~n) _ ~tu~n) = (1 - ~t) u~n). 22
22Stiffness of Stiffness is a surprisingly subtle concept. The definition we follow here does not capture all of this subtlety; moreover, moreover, there is is not aa single, single, well-accepted well-accepted definition definition of of a stiff stiff system. system. See See [33], Section 6.2, for for aa discussion discussion of of the the various various definitions definitions of of stiffness. Section 6.2, stiffness.
118 118
Chapter 4. 4. Essential Essential ordinary ordinary differential differential equations equations Chapter
It It follows follows that that u~n)
= (1 - 107 b.t) U~n-l) = (1 - 107 b.t)2 U~n-2) = (1-10 7 b.t)3 U~n-3)
= ... = (1 -
107b.tt uiO)
= (1 - 107 b.tt
and, similarly, similarly, and, Clearly, the the computation computation depends depends critically critically on on Clearly, 11 - 107 b.tl, 11 - b.tl·
If one one of of these these quantities quantities is is larger larger than than 1, 1, the the corresponding corresponding component component will will grow grow If exponentially as as the the iteration iteration progresses. progresses. At At the the very very least, least, for for stability stability (that (that is, is, to to exponentially avoid spurious spurious exponential exponential growth), growth), we we need need avoid 11 -10 7 b.tl :::; 1, 11- b.tl :::; 1.
The first first inequality inequality determines determines the the restriction restriction on on b.t, At, and and aa little little algebra algebra shows shows that that The we must must have have we o < b.t :::; 2 . 10- 7 . Thus aa very very small small time time step step is is required required for for stability, stability, and and this this restriction restriction on on b.t Af is is Thus imposed by by the the transient transient behavior behavior in in the the system. system. imposed As we we show show below, below, it it is is possible possible to to avoid avoid the the need need for for overly overly small small time time steps steps As in aa stiff stiff system; system; however, however, we we must must use use implicit implicit methods. methods. Euler's Euler's method method and and the the in RK RK methods methods discussed discussed in in Section Section 4.4 4.4 are are explicit, explicit, meaning meaning that that the the value value u(n+l) w(n+1) is is defined by by aa formula formula involving involving only only known known quantities. quantities. In In particular, particular, only only urn) u^ apapdefined pears pears in in the the formula; formula; in in some some explicit explicit methods methods (multistep (multistep methods), methods), u(n-I), u^n~l\ u(n-2), u^n~2\ n+1 ... ... may may appear appear in in the the formula, formula, but but u(n+l) w( ) itself itself does does not. not. On the other hand, an implicit method defines On the other hand, an implicit method defines u(n+l) u^n+l^ by by aa formula formula that that ininn+1 volves volves u(n+l) w( ) itself itself (as (as well well as as urn) u^ and and possibly possibly earlier earlier computed computed values values of of u). u). This This means that, that, at at each each step step of of the the iteration, iteration, an an algebraic algebraic equation equation (possibly (possibly nonlinear) nonlinear) means must must be be solved solved to to find find the the value value of of u(n+l). u( n+1 ). In In spite spite of of this this additional additional computational computational expense, implicit implicit methods methods are are useful useful because because of of their their improved improved stability stability properties. properties. expense,
4.5.2 4.5.2
The backward backward Euler Euler method method The
The simplest simplest implicit implicit method method is is the the backward backward Euler Euler method. method. Recall Recall that that Euler's Euler's The method for for the the IVP IVP method du dt = f(t, u), u(to) = Uo
4.5. Stiff of ODEs 4.5. Stiff systems systems of ODEs
119 119
was derived from the equivalent formulation
by applying the left-endpoint left-endpoint rule for quadrature:
lb
hex) dx == h(a)(b - a).
The result is
= Un + Atf(tn, Un)
U n +1
(see Section 4.4.1). To obtain obtain the backward Euler method, we we apply the right-hand (see right-hand rule, rule,
lb
hex) dx == h(b)(b - a),
instead. The result is
(4.38) This This method is indeed indeed implicit. implicit. It is necessary to solve the equation equation (4.38) for In the the case case of of aa nonlinear nonlinear ODE ODE (or (or aa system system of of nonlinear nonlinear ODEs), ODEs), this this may may be be In difficult (requiring aa numerical numerical root-finding root-finding algorithm algorithm such such as as Newton's Newton's method). method). difficult (requiring However, not difficult difficult to backward Euler However, it it is is not to implement implement the the backward Euler method method for for aa linear linear system. We We illustrate illustrate this for the linear, constant-coefficient system:
unn+ i. U +1.
dx dt = Ax + f(t),
(4.39)
x(O) = Xo. The backward Euler method takes the form
+ At(Axn+1 + f(t n+ 1 )) =} X n +1 - AtAxn+1 = Xn + Atf(tn+d =} (1 - AtA)xn+l = Xn + Atf(tn + 1 ) =} Xn+1 = (1 - AtA)-l (x n + Atf(tn+d) . X n +1
= Xn
Example the backward method to example (4.37) (4-37) Example 4.12. 4.12. Applying Applying the backward Euler Euler method to the the simple simple example methods. We gives some insight into into the the difference difference between between the implicit implicit and explicit explicit methods. We following iteration for the first obtain the the following first component: (n+1) _ (n) _ U1 - U1
+
107 At)
=}
(1
=}
ui
=}
uin ) = (1
n
+1)
107 ~ AtU (n+1) 1
u~n+1) = uin)
= (1 + 107 At) -1 uin ) + 107 At) -n .
120
Chapter 4. Essential ordinary differential equations
Similarly, for for the second component, component, we we obtain obtain Similarly, the second
any positive value For any value of of ilt, A£; we we have
and so so no no instability instability can step size size ilt will still still have be small enough and can arise. arise. The The step A£ will have to to be small enough accurate, but we do small to for the solution to for the solution to be be accurate, but we do not not need need to to take take ilt Ai excessively excessively small to avoid spurious growth. avoid spurious exponential exponential growth. Example 4.13. Euler's method method and the backward backward Euler Euler method method Example 4.13. We We will will apply apply both both Euler's and the to the IVP IVP
dx
dt = A 2 x,
(4.40)
x(O) = xo, where A2 is is the matrix from from Example Example 4-11 4.11 and where A.% the matrix o,nd
We integrate over over the the interval with aa time step of We integrate interval [0,2) [0,2] with time step of ilt At =— 0.05 0.05 (40 (40 steps) steps) in in both both 4.7. Euler's Euler's method algorithms. The The results are are shown in in Figure 4-7method "blows "blows up" up" with this step, while while the Euler method produces aa reasonable reasonable approximation this time time step, the backward backward Euler method produces approximation to true solution. to the the true solution. Remark higher-order methods for use use with with stiff ODEs, nonoRemark There There are are higher-order methods designed designed for stiff ODEs, tably method. It is also also possible possible to to develop develop automatic automatic step-control tably the the trapezoidal trapezoidal method. It is step-control 23 methods for methods for stiff stiff systems. systems.23
Exercises 1. 1. Let Let
A=[
and the IVP and consider consider the IVP
dx
dt = Ax,
x(O) = xo. (a) Find the the exact x(t). (a) Find exact solution solution x(£). 23
23MATLAB includes such routines. MATLAB includes
121 121
4.5. Stiff systems of ODEs
0 x
-1 -2
-3
0
0.5
1.5
2
0.5
1.5
2
2
x
1l 0, -1
0
Figure 4.7. The computed solutions to IVP IVP (4.40) method (4-40) using Euler's method (top) (top) and the backward backward Euler method (bottom). (Each graph shows all three components of the solution.) solution.) (b) Estimate x(l) using using 10 of Euler's method and the norm (b) Estimate x(l) 10 steps steps of Euler's method and compute compute the norm of the of the error. error. (c) using 10 and compute (c) Estimate Estimate x(l) x(l) using 10 steps steps of of the the backward backward Euler Euler method method and compute the norm norm of of the the error. error. the 2. for 2. Repeat Repeat Exercise Exercise 11 for
3. 3. Let Let
A =
~
-113 -80 [ -107
-80 -140 -80
-107] -80 . -113
Find largest value of At Ilt such that Euler's method is numerical Find the the largest value of such that Euler's method is stable. stable. (Use (Use numerical experimentation. ) experimentation.)
122 122
Chapter Essential ordinary ordinary differential differential equations Chapter 4. 4. Essential equations
4. Let Let ff:: R R x xR R22 -+ R22 be be defined by 4. ->• R defined by
The nonlinear nonlinear IVP The IVP
dx dt
= f(t,x),
x(o)
=[~ ]
has solution has solution
te-t ] x(t) = [ e- t . (a) by example example that that Euler's Euler's method method is unstable unless unless At D..t is chosen (a) Show Show by is unstable is chosen small Determine (by (by trial trial and and error) error) how how small D..t must must be be in in small enough. enough. Determine small At that Euler's Euler's method method is is stable. order order that stable.
(b) (b) Apply Apply the the backward backward Euler Euler method. method. Is Is it it stable stable for for all all positive positive values values of of D..t? At? 5. Let 5. Let
A = [-50.5 49.5
49.5 ] -50.5 .
(a) Find the the exact solution to (a) Find exact solution to
~: = Ax,
x(O)
= [ ~ ].
(b) Determine Determine experimentally experimentally (that (that is, is, by by trial trial and and error) error) how how small small At D..t must (b) must be in in order that Euler's Euler's method method behaves behaves in in aa stable manner on this IVP. be order that stable manner on this IVP. (c) (c) Euler's Euler's method method takes takes the the form form Xi+1
= Xi + D..tAxi, i = 0,1,2, ....
Let the the eigenpairs of A A be be .A1, U1 and and A2, .A2, 112. U2. Show Show that that Euler's Euler's method Let eigenpairs of AI, ui method is equivalent equivalent to is to
y~i+l) = (1 y~i+l) where where
+ D..t.Ady~i), = (1 + D..t.A2)y~i), (i) _
(i) _ YI
-
Xl . UI,
Y2
-
Xi . U2·
y~i), y% y~i),, and derive an upper bound on At D..t that Find explicit formulas for y^', guarantees guarantees stability stability of of Euler's Euler's method. method. Compare Compare with with the the experimental experimental rpsiilt. result.
4.6. Green's functions
123
(d) Repeat with the backward Euler Euler method (d) Repeat with the backward method in in place place of of Euler's Euler's method. method. nxn 6. Let €R be symmetric with with negative ..., A 6. Let A A E R nxn be symmetric negative eigenvalues eigenvalues AI, Al, A A2,2 ,... An. n.
(a) (a) Consider Consider Euler's Euler's method method and and the the backward backward Euler Euler method method applied applied to to the the homogeneous homogeneous linear linear system system
dx
dt = Ax.
Show that Euler's method produces produces aa sequence sequence xo,xi,X2,... XO, Xl, X2, ... where Show that Euler's method where
the backward backward Euler method produces produces aa sequence Xo, Xl, X2, ... while while the Euler method sequence x 0 ,xi,x 2 ,... where where Xi = (I - AtA)-ixo.
(b) What are the the eigenvalues eigenvalues of of 1I + + AtA? (b) What are AtA? (c) What What condition condition must satisfy in in order order that that Euler's method be stable? (c) must At satisfy Euler's method be stable? previous exercise.) (Hint: (Hint: See See the the previous exercise.) What are AtA)-l? (d) (d) What are the eigenvalues eigenvalues of of (I -— At A)"1? (e) that the the backward backward Euler all positive positive values values of of (e) Show Show that Euler method method is is stable stable for for all At. At.
4.6
Green's functions functions
The Green's Green's function an IVP, IVP, BVP, BVP, or or IBVP IBVP has has aa fairly fairly simple simple meaning, meaning, alThe function for for an although details can can get get complicated, complicated, particularly for PDEs. PDEs. A A homogeneous the details particularly for homogeneous though the always has has the zero solution; nonzero solutions arise only only when when the linear ODE always the zero solution; nonzero solutions arise the iniinilinear not zero, the differential has aa nonzero nonzero forcing tial tial conditions conditions are are not zero, or or when when the differential equation equation has forcing function (that is, is initial values values and function (that is, is inhomogeneous). inhomogeneous). The The initial and the the forcing forcing function function can can be called the data of problem. The function represents represents the contribution be called the of the the problem. The Green's Green's function to of the the datum datum at at each each point in time. time. to the the solution solution of point in In the this section, we explain for two IVPs that In the remainder remainder of of this section, we explain this this statement statement for two IVPs that we first 4.2. We explain how how the we first considered considered in in Section Section 4.2. We also also explain the Green's Green's function function can can be interpreted interpreted as to the the differential differential equation, and introduce introduce the be as aa special special solution solution to equation, and the concept of the Dirac Finally, we comment on the form form of the concept of the Dirac delta delta function. function. Finally, we comment on how how the of the Green's function function will change when PDEs. Green's will change when we we consider consider PDEs.
4.6.1 4.6.1
The Green's function for a a first-order first-order linear linear ODE ODE The Green's function for
The The solution solution to to
du dt - au = f(t), t> 0,
u(to) = uo,
(4.41)
124
Chapter ordinary differential Chapter 4. 4. Essential Essential ordinary differential equations equations
as derived in Section 4.2, is
u(t)
= ea(t-to)UO +
We We define define
(j(t;s)
={
e
rt ea(t-s) f(s) lto
a(t-s)
, 0,
t> t
ds.
s,
< s.
(4.42)
Then we can write formula for for the u as Then we can write the the formula the solution solution u as
u(t) = (j(t; to)uo + {'XI (j(t; s)f(s) ds.
(4.43)
lto
The function (causal) Green's function for (4.41), and and formula formula (4.43) The function G (j is is called called the the (causal) for (4.41), (4.43) already hints at the significance of (j. The effect already hints at the significance of G. The effect of of the the initial initial datum datum Uo UQ on on the the solution at time time tt is while the the effect the datum datum ff(s) on the the solution solution u at is (j(t; G(t] to)uo, while effect of of the ( s ) on solution 24 at time time tt is is (j(t; G(t] s)f(s)ds. s)/(s)ds.24 u at As aa concrete concrete example, example, the the IVP the growth growth of population (of As IVP (4.41) (4.41) models models the of aa population (of aa country, where a is is the the natural natural growth growth rate rate and is the rate of country, say), say), where and ff(t) ( t ) is the rate of immigration immigration 25 For the sake of definiteness, suppose the population is measured in at time t. 25 at time For the sake of definiteness, suppose the population is measured in millions of and tt is is measured in years. Then Uo UQ is at time millions of people people and measured in years. Then is the the population population at time to (in rate, in millions of of people people per per year, (in millions) millions) and and ff(t) ( t ) is is the the rate, in millions year, of of immigration immigration at at time that u(t) to) is is the following IVP: time t. We We can can see see that u(t) = (j(t; G(t;to) the solution solution to to the the following IVP:
du - - au dt u(to)
= 0' = 1.
t > to,
(4.44)
That is, u(t) = to) is the population resulting from initial populaThat is, = (j(t; G(t;to) is the population function function resulting from an an initial population of 11 (million) and no immigration (here consider only only times times tt after to). We We can tion of (million) and no immigration (here consider after to). can also recognize u(t) population function to aa certain also recognize u(t) = = (j(t; G(t; s) as as the the population function corresponding corresponding to certain pattern of immigration, namely, namely, when when 11 million million people people immigrate instant pattern of immigration, immigrate at at the the instant si Of Of course, course, this this is is an an idealization, idealization, but but there there are are many many situations situations when when we tt = s! we wish to to model model aa phenomenon that takes takes place place at time or at aa sinwish phenomenon that at an an instant instant in in time or at single point point in in space point force mechanics). We We now now explore gle space (such (such as as aa point force in in mechanics). explore this this idea idea further. further. We whose population population is by the the differential We consider consider aa country country whose is modeled modeled by differential equation equation
du dt - au = f(t). We assume that initially initially there country (u(to) = 0), million We assume that there are are no no people people in in the the country (u(to) = 0), and and 11 million over the interval [s,s [s, s + 6.tj (at aa constant rate, during people immigrate people immigrate over the time time interval At] (at constant rate, during that that 24 24The represented by by the has different units than represented by by Uo. Indeed, The data data represented the function function f/ has different units than that that represented UQ. Indeed, an examination of differential equation shows that that the of f/ are are the of Uo UQ divided divided an examination of the the differential equation shows the units units of the units units of is aa rate, rate, and it must be multiplied by the to being being by time. time. Therefore, by Therefore, f/ is and it must be multiplied by the time time "interval" "interval" ds ds prior prior to
multiplied by by G(t; multiplied G(t; s). 25 25This particularly good assumes that that the This model model of of population population growth growth is is not not aa particularly good one, one, since since it it assumes the growth rate remains constant time. In In reality, reality, the the growth growth rate changes due growth rate remains constant over over time. rate changes due to to changing changing birth birth rates, life life spans, rates, spans, etc. etc.
4.6. Green's Green's functions
125
interval, D..t people per year). year). We interval, of of 1/ I/At people per We a.ssume assume that that ss > to. to. The The resulting resulting population population satisfies the IVP
du dt - au = dl'>.t(t - s), t> to, u(to) = 0, where where
dto,.t(t)
= {
It,
0< t < D..t,
0, t <
°
or t > D..t.
The is The solution solution is
(t; T) the interval As we This This la.st last expression expression is is the the average average of of G G(t; r] over over the interval [s, [s, s + D..t]. At]. As we take take D..t At smaller smaller and and smaller, smaller, we we obtain obtain u(t) -+ G(t; s) as D..t -+ 0. But the limit? limit? But what what forcing forcing function function do do we we obtain obtain in in the
4.6.2
The Dirac delta function
The has the properties, the the first first two just the The function function dto,.t d&t has the following following properties, two of of which which are are just the definition: definition:
• dto,.t(t) = 0, t ~ [0, D..t]. • dto,.t(t) = l/D..t, t • J: dto,.t(t) dt J: dto,.t(t) dt
=
E
[O,D..t].
It
Joto,.t dt = 1 for all D..t > 0, provided [0, D..tj c [a, b], and if D..t < a or > b.
=°
°
• If 9 is a continuous function, and [0, D..tj
c [a, bj, then
{b 1 (to,.t ia dto,.t(t)g(t) dt = D..t io g(t) dt, which is the the average average value of 9g on on the the interval interval [0, [0, D..tj. At]. which is value of Since D..t -+ value of Since the the limit, limit, as as At ->• 0, 0, of of the the average average value of 9g over over [0, [0, D..t] At] is is g(O)
°
• 6"(t) = • 8(0) =
°
if t :j:. 0.
00.
126
ordinary differential equations Chapter 4. Essential ordinary
• J: 8(t) dt
= 1 if 0 E [a, b], and
J: 8(t) dt
= 0 if 0
~ [a, bj .
• If 9 is a continuous function and 0 E [a, bj, then
o ~ [a, b], then
J: 8(t)g(t) dt
J: 8(t)g(t) dt
= g(O). If
= O.
There is is no no ordinary ordinary function function 6 8 satisfying the above properties; for for example, There satisfying the above properties; example, if 8(t) all t :j; / 0, 0, then 6(t) dt = 00 should should hold. However, it if 8(t) = 00 for for all then J 8(t) hold. However, it is is useful useful to to consider 6, as defined defined by It is consider 8, as by the the above above properties, properties, to to be be aa generalized function. It is called the the Dirac delta function, and and sometimes referred to to as as aa unit point source or or called sometimes referred impulse. The last property property implies that, when when 9g is continuous, unit impulse. The last implies that, is continuous,
(b 8(t _ s)g(t) dt
Ja
=
{g(s), 0,
8 8
E
[a, bj,
~ [a, bj
(4.45)
is formally formally an an even even function, function, 6(—t) 8( -t) = 8(t) for so (see Exercise Exercise 6). Moreover, 86 is (see 6). Moreover, = 6(t] for all all t, so viro also a1cr» Vicnro we have (b 8(8 _ t)g(t) dt = {9(8), 8 E [a, b], (4.46) 0, 8~[a,bj.
Ja
sifting property of the Dirac delta This is is called the sifting delta function. Incidentally, Incidentally, aa sequence functions such such as as {d {di/ converges to the delta delta function, is called }, which which converges to the function, is called quence of of functions 1 / nn}, delta-sequence. aa delta-sequence. Our analysis analysis now now suggests suggests that that the the IVP Our IVP du - - au = 8(t - 8) dt ' u(to) =
°
has solution has solution u(t) u(t) == G(t; 8). s). Indeed, this this follows follows from the solution solution of 0) Indeed, from our our formula formula for for the of (4.41) (4.41) (with (with Uo UQ == 0) and from the the sifting sifting property property of the delta and from of the delta function: function:
Since G(t; G(t',s) for tt < 8s anyway, obtain simply simply u(t) u(t) = = G(t; G(t]s). This 8) is is zero zero for anyway, we we obtain 8). This Since again meaning of 8): it it is is the the response response (of the system under again emphasizes emphasizes the the meaning of G(t; G(t;s): (of the system under at time to aa (unit) point source time s > 2: to. consideration) consideration) at time t to (unit) point source at at time to-
4.6.3
The Green's Green's function function for a a second-order second-order IVP
In Section we saw that the the solution In Section 4.2.2, 4.2.2, we saw that solution to to d2 u
dt 2
+ ()2U = !(t), u(to) = 0, du dt (to) = 0
(4.47)
4.6. Green's functions
127
is is
u(t) =
~
t sin (O(t - s))f(s) ds.
ito
Therefore, the (4.47) is Therefore, the (causal) (causal) Green's Green's function function for for (4.47) is sin (1I(t-s))
G(t;S) =
4.6.4 4.6.4
II
{
0'
,
t> s, t < S.
Green's functions for PDEs
We can can now now briefly briefly preview preview the the form form of the Green's for an an IVP IVP or IBVP We of the Green's function function for or IBVP defined by aa PDE. PDE. We We will will assume that any any boundary boundary conditions defined by assume that conditions are are homogeneous homogeneous for simplicity. Both Both the the initial initial values values and and the forcing function function will will be be prescribed prescribed over over for simplicity. the forcing the spatial the problem, problem, and and we we will will have have to to "add "add up" up" the the contributions the spatial domain domain of of the contributions well as as across across time. time. Therefore, the Green's function will will have have the across space across space as as well Therefore, the Green's function the form £, s), giving influence of of the at the the space-time space-time point s) on t;~, giving the the influence the datum datum at point (£, (~, s) on form G(x, G(x, £; the solution at (x, (x, t). Moreover, the solution formula will involve involve an an integral integral over over the solution at t}. Moreover, the solution formula will we must must add up the the contributions contributions across the spatial spatial space as well as as time, time, since since we space as well add up across the domain as well as the time time interval. interval. A A similar similar notion notion of domain as well as over over the of Green's Green's function function applies in the case of static static problems, problems, that PDEs in in which which the the independent applies in the case of that is, is, PDEs independent G(x;~) defines the influence of the datum variables are only space variables; then variables are only space variables; then G(x; £) defines the influence of the datum at the the space space point point £ ~ on solution at x. This This reasoning reasoning also hints at at one one of of the at on the the solution at x. also hints the l analogies mentioned in Section 3.6. The inverse matrix A-I defines the influence analogies matrix A~ bj in in the the data data (right-hand of the system Ax. Ax = = b b on on of each component bj of each component (right-hand side) side) of the linear linear system component Xi Xi of of the the solution: bj contributes contributes (A-I )ijbj to to Xj, Xi, and we get the full component solution: bj (A~1)ijbj and we get the full of the the datum datum component. component. This is in in result by by adding adding up up over the position position jj of result over the This story story is precise correspondence with the the way way in in which which the the Green's precise correspondence with Green's function function G(t; G(£; s) s) defines defines the influence influence of the datum datum at at "position" "position" (time) (time) ss on on the the solution solution at at "position" the of the "position" So G G is is presumably presumably in in some sense an inverse operator; we will will develop develop (time) t. So (time) some sense an inverse operator; we this point point of view further further in in Chapter 5. this of view Chapter 5.
Exercises 1. Find the following following IVP: IVP: 1. Find the the Green's Green's function function for for the du dt
+ 2u = f(t), u(to) = Uo·
2. certain population population grows grows according according to to the the ODE ODE 2. Suppose Suppose aa certain
dP
dt
= 0.02P + f(t),
where /f is is the the immigration. immigration. If the population population at at the the beginning beginning of 55.5 where If the of 1990 1990 is is 55.5 million immigration during during the 10 years is given function million and and immigration the next next 10 years is given by by the the function = 1 - 0.2t (in millions), millions), where where tt = = 0 represents the the beginning beginning of ff(t) (t) = 1— Q.2t (in 0 represents of 1990, 1990,
128
Chapter 4. Essential ordinary differential differential equations
what 10)? Use the what will will the the population population be be at at the the beginning beginning of of 2000 2000 (t (t = 10)? Use the P(t) and then P(10). Green's function to find P(t) 3. A certain radioactive isotope decays exponentially according to the law dm dt
-=-km
'
where k = = In In (2)/2. Here Here m(t) m(t) is is the the mass mass of of the the isotope isotope at time tt (seconds), (seconds), where at time and the differential differential equation equation indicates indicates that that aa constant constant fraction fraction of of the the atoms atoms are are and the disintegrating at each each point point in in time. time. The The above above differential differential equation holds, of of disintegrating at equation holds, course, when none none of of the the isotope is being being added added or or taken taken away by other other means. course, when isotope is away by means. Suppose that initially of the the isotope isotope are present, and and mass mass is added from Suppose that initially 10 10 gg of are present, is added from an external source source at the constant rate of of 0.1 0.1 g/s. Solve the IVP IVP modeling modeling this this an external at the constant rate Solve the situation, using the the Green's Green's function, function, and and determine determine the the long-time long-time behavior behavior situation, using of of m(t). m(t). 4. Find Find the the Green's Green's function function for for the the IVP 4. IVP
rf2u dt 2
+ 4u =
f(t),
u(O) = 0,
~~(O) = O. 5. Use the Green's function from the previous exercise to solve d2 u
dt 2
+ 4u = u(O)
cos (t),
= 0,
~~ (0) =
O.
6. Use the property property
o E [a, b]:::} lab 8(t)g(t) dt =
g(O)
and a change of variables to justify justify (4.45), (4.45).
4.7 4.7
Suggestions for further further reading reading Suggestions for
There such as as the and There are are many many introductory introductory textbooks textbooks on on ODEs, ODEs, such the text text by by Goldberg Goldberg and Potter [18]. An introductory introductory text text with with aa particularly particularly strong Potter [18]. An strong emphasis emphasis on on applied applied mathematics is Boyce and DiPrima DiPrima [5]. A more advanced book that focuses on mathematics the theory theory of of ODEs ODEs (and (and the the necessary necessary linear linear algebra) algebra) is is Hirsch Hirsch and and Smale [26]. the Smale [26]. The reader reader is is referred referred to to Hirsch Hirsch and and Smale for aa complete complete description of the the theory The Smale for description of theory of linear, constant-coefficient systems. systems. (We covered only only aa special special case, case, albeit of linear, constant-coefficient (We covered albeit an an important one, in Section important one, in Section 4.3.) 4.3.)
4.7. Suggestions for further reading
129
For more information on numerical methods for ODEs, the reader is referred to Shampine or Lambert Lambert [33]. Most books books on on numerical numerical analysis, analysis, such such as Atkinson Shampine [43] [43] or [33]. Most as Atkinson [2] Kincaid and and Cheney Cheney [30], cover numerical numerical methods methods for for ODEs, ODEs, as as well well [2] and and Kincaid [30], also also cover as the the background background material material on on quadrature and interpolation. as quadrature and interpolation. The [38] gives the The text text by by Mattheij Mattheij and and Molenaar Molenaar [38] gives an an integrated integrated presentation presentation of of the theory and numerical analysis of ODEs, as well as their use for modeling physical physical systems.
This page intentionally This page intentionally left left blank blank
Chapter 5 Chapter 5
Boundary dary value problems In statice
Our will arise of Our first first examples examples of of partial partial differential differential equations equations (PDEs) (PDEs) will arise in in the the study study of static phenomena in heat flow. flow. To static (equilibrium) (equilibrium) phenomena in mechanics mechanics and and heat To make make our our introducintroduction the subject possible, we begin with tion to to the subject as as simple simple as as possible, we begin with one-dimensional one-dimensional examples; examples; that is, is, all all of of the the variation variation is is assumed assumed to to occur occur in in one one spatial spatial direction. direction. Since Since the the that phenomena are static, static, time time is is not involved, and and the the single single spatial spatial variable variable is is the phenomena are not involved, the only only independent variable. variable. Therefore, Therefore, the "PDEs" are are actually actually ODEs! ODEs! Nevertheless, Nevertheless, the the independent the "PDEs" techniques for the problems generalize techniques we we develop develop for the one-dimensional one-dimensional problems generalize to to real real PDEs. PDEs.
5.1
The analogy between BVPs and and linear algebraic systems
As mentioned As mentioned in in Chapter Chapter 3, 3, the the solution solution methods methods presented presented in in this this book book bear bear strong strong resemblance, at at least least in in spirit, spirit, to to methods for solving solving aa linear linear system system of of the the methods useful useful for resemblance, form of form Ax Ax == b. b. In In the the exercises exercises and and examples examples of of Chapter Chapter 3, 3, we we showed showed some some of the similarities similarities between linear systems systems and and linear linear BVPs. We now now review review these these the between linear BVPs. We similarities, similarities, and and explain explain the the analogy analogy further. further. We will We will use use the the equilibrium equilibrium displacement. displacement uu of of aa sagging sagging string string as as our our first first example. example. In In Section Section 2.3, 2.3, we we showed showed that that u satisfies satisfies the the BVP BVP
d2u
-T dx 2 = f(x), 0 < x < £, u(O)
= 0,
u(£)
= 0,
(5.1)
where T is is the the tension tension in in the the string string and and f/ is is an an external external force force density density (in (in units where units of of force per per length). length). force If If A A E e RIDxn, R m x n , then then A A defines defines an an operator operator mapping mapping Rn Rn into into RID, R m , and and given given m b e RID, R , we can ask ask the following questions: questions: bE we can the following mapped to by A? • Is Is there there an an xx ERn e Rn which which is is mapped to bb by A? 131 131
132
Chapter 5.
Boundary value value problems problems in statics
• If so, is is x x unique? unique? If so,
• How How can can such such an x be be computed? an x computed? In much much the the same same way, way, In
d2 -T-2
dx defines referred to to as as aa differential differential operator. For For convenience, convenience, defines an an operator-usually operator—usually referred we will will refer refer to to this this operator operator as as L. It It maps maps one one function function to to another, another, and and the we the differential ermation differential equation d2 u (5.2) - T dx 2 = f (x), < x < £
°
can be question: Given on the can be interpreted interpreted as as the the following following question: Given a a function function /f defined defined on the interval does there u which interval [0,^], [0,£], does there exist exist aa function function u which is is mapped mapped to to /f by by L? L? That That is, is, does there exist exist aa function function u satisfying = f? does there satisfying Lu = /? The analogy analogy between between Ax Ax = b and strengthened by by the the fact fact that that The =b and Lu = = f is is strengthened the is linear, A. the differential differential operator operator L L is linear, just just as as is is the the operator operator defined defined by by the the matrix matrix A. Indeed, if if u and and v are are two two functions functions and and a a and (3 are Indeed, and {3 are scalars, scalars, then then L(au + (3v)
~
= -T dx 2 [au + (3v] ~u
~v
=-aT--(3Tdx 2 dx 2 = aLu + (3Lv. Underlying this this calculation calculation is the fact fact that that aa space of functions can be be viewed viewed as as Underlying is the space of functions can aa vector vector space-with space—with functions functions as as the the vectors. vectors. The The operator operator L L is is naturally naturally defined defined on (722[0,£| Thus L takes as input on the the space space C [0,£] introduced introduced in in Section Section 3.1. 3.1. Thus takes as input aa twicetwicecontinuously differentiable differentiable function u and and produces produces as as output output aa continuous continuous function continuously function u function Tjii.. Lu.
Example Consider the Example 5.1. 5.1. Consider the function function uu :: [0,1] [0,1] —> ---+ R R defined defined by by
°< x < ~, ~<x<1.
The function [0,1] (we check that the two two cubic The function uu is is continuous continuous on on [0,1] (we need need only only check that the cubic pieces pieces which they do). do). We have have the same value at x = — 1/2, 1/2, which du { ~ (x - x ) , dx(x)= ~(x2_X+~), 2
°< x < ~, ~
< x < 1,
and to see 1/8 at and it it is is easy easy to see that that both both quadratic quadratic pieces pieces defining defining du/dx du/dx have have value value 1/8 at xx — = 1/2. 1/2. Therefore, Therefore, du/dx du/dx is is continuous continuous on on [0,1]. Finally, Finally,
~u2 (x) = { dx
l-x 2
x-
1
'
2'
°<
x < ~, ~ < x < 1,
133 133
5.1. The analogy between BVPs and systems 5.1. and linear algebraic systems
that is, is, that
::~(x) = Ix - ~I· Thus [0, 1]. Infact, C 22[0,1], [0, 1], Thus cPu/dx d?u/dx is is also also continuous continuous and anduu E EC C22[0,1]. In fact, uu is is "exactly" "exactly" in in C k in the the sense sense that, that, because because the the third third derivative derivative of of uu is is discontinuous, discontinuous, uu $. C [Q, 1] for for in f/. Ck [0, 1] any kk > 2: 2: any 22
3
d u X dx 3 ( )
_
{-I, 0< x < ~, 1,
-
~ < x < 1.
Figure 5.1 graphs of Figure 5.1 shows shows the the graphs of uu and and its its first first three three derivatives. derivatives.
0.15
0.1 0.1
0.05
0 0
0.5
1
0.5
X
1
X
0.5 0 -0.5 0.1
-1
0.5
1
0
X
0.5
1
X
Figure 5.1. (722[0,1] = u(x) u(x) (top left), Figure 5.1. A A function function in in C [0, 1] (see (see Example Example 5.1): 5.1): yy = (top left), the second derivative derivative of the first first derivative derivative of of uu (top (top right), right), the the second of uu (bottom (bottom left), left), and and the the third third derivative derivative of of uu (bottom (bottom right). right). The should notice notice that Example 5.1 is the the The reader reader should that the the function function uu defined defined in in Example 5.1 is solution of of Lu Lu == f for for aa certain certain f/ E 6 C[O, C[0,1], namely, solution 1], namely, f(x) =
-Tlx - ~I·
134
Chapter 5. Boundary Boundary value problems problems in statics
On the the other other hand, hand, if we define define vv = = du/dx, solution of of Lu On if we du/dx, then then vv is is not not the the solution Lu = = fJ 22 for any J E C[O, 1] (because rPv/dx is discontinuous). Indeed, Lv is not defined, for any / £ C[0,1] (because d?v/dx is discontinuous). Indeed, Lv is not defined, strictly dv/dx is is not the entire interval [0,1]. strictly speaking, speaking, since since dv/dx not differentiable differentiable on on the entire interval [0,1]. Thus Thus 22 it makes sense to regard regard the the domain domain of Las [0, 1]: it makes sense to of L as C (7 [0,1]: L: C 2 [0, 1]-t C[O, 1]. It is possible possible to to take take the the analogy analogy of of Lu Lu = f with with Ax = b b even even further. further. If If It is A E G Rffixn, R m x n , then then Ax = = bb represents m equations equations in in n unknowns, unknowns, each each equation equation having the having the form form
In the — jJ represents collection of of equations, equations, namely, namely, the In the same same way, way, Lu Lu = represents aa collection the conditions conditions that, the the functions functions that _T rPu dx 2 and /J must be equal each xx E fl. Here, and must be equal at at each e [0, [0,£J. Here, though, though, we we see see aa major major difference: difference: Lu = represents an infinite number number of Lu = f represents an infinite of equations, equations, since since there there is is an an infinite infinite number number x E fl. Also, Also, there there are infinitely many many unknowns, namely, the of points x of points e [0, [0,£j. are infinitely unknowns, namely, the values values u(x), x E [O,f]. u(x),xe [0,4 The role of the boundary conditions is not not difficult to explain The role of the boundary conditions is difficult to explain in in this this context. context. Just matrix A E have aa nontrivial nontrivial null null space, space, in in which which case Just as as aa matrix 6 Rffixn R m x n can can have case Ax == = b cannot cannot have have aa unique unique solution, operator can can have b solution, so so aa differential differential operator have aa nontrivial nontrivial null to verify verify that that the null space space of of all null space. space. It is is easy easy to the null of L consists consists of all first-degree first-degree polynomial functions: polynomial functions: N(L)=={u: u(x)=ax+b, a,bER}.
Therefore, Lu = /J cannot unique solution, since if if u u is is one one solution solution and and Therefore, Lu cannot have have aa unique solution, since w(x) ax + b, b, then then uu + w w is is another solution. On other hand, hand, Lu Lu = does w(x] == = ax another solution. On the the other = ff does have solutions for for every e C[O,f]; C^O,^]; for for example, example, have solutions every f/ E u(x)
1rr
= -Iji 10 10
J(z)dzds
solves Lu Lu = This is equivalent to to saying saying that that the the range range of of L L is is all solves = f. f . This is equivalent all of of C[O,£]. C[0,4 Example 5.2. Let ff E 1] be == x. x. Then, for every every a, R, Example 5.2. Let G C[O, C[0,1] be defined defined by by ff(x) (x) — Then, for a, bE b e R, 1 u(x) == - 6Tx3
+ ax + b
Lu = Lu — == f consists of of is is a solution of of Lu = ff.. That That is, the the general solution of of Lu u(x) -x33/(6T) 43.) u(x) = = —x /(6T) plus any any function from the the null null space of of L. (See (See page 43.)
The that the two-dimensional suggests The fact fact that the null null space space of of L is is two-dimensional suggests that that the the operator operator equation Lu Lu == == b, b, where where equation = f is is closely closely analogous analogous to to aa linear linear algebraic algebraic equation equation Ax = A is an (n-2) A (n — 2) x n matrix and R(A) H(A) == = Rn-2 R n ~ 2 (that is, the columns of A A span Rn-2). R n ~ 2 ).
5.1. systems 5.1. The The analogy analogy between between BVPs BVPs and and linear linear algebraic algebraic systems
135
Such has the for every Such aa matrix matrix A A has the property property that that Ax Ax = = bb has has aa solution solution for every bb E € Rn-2, Rn 2 , but N(A) is is two-dimensional, but the the solution solution cannot cannot be be unique. unique. Indeed, Indeed, in in this this case, case, M(A) two-dimensional, and therefore therefore so so is is the the solution solution set set of of Ax Ax == bb for for every every bb ERn. e R n . In In order order to obtain and to obtain aa unique unique solution, solution, we we must must restrict restrict the the allowable allowable vectors vectors xx by by adding adding two two equations. equations. 2x4 Example 5.3. 5.3. Let eR R2x4 be defined defined by Example Let A A E be by
_[-1° 2 -1 ° ]
A-
-1
2
-1
.
Then it it is is easy easy to to show show that that N(A) N (A) = = span{vl' span{vi,v<2}, Then V2}, where where
linearly (It (It is is also also easy easy to to see see that that R(A) "^-(A) == R2, R2; since since the the first first two two columns columns of of A A are are linearly 4 independent.) independent.) Thus, Thus, if «/x e R4 R satisfies satisfies Ax Ax = = 0, 0, then then there there exist exist s, s,tt E €R R such such that that x E x =
SVl
+ tV2.
(5.3)
In order In order to to narrow narrow down down aa unique unique solution, solution, we we must must impose impose additional additional conditions conditions that force ssand If we preserve the that force and tt to to both both be be zero. zero. If we are are to to preserve the linear linear nature nature of of the the problem, these problem, these conditions conditions will will simply simply be be two two more more linear linear equations, equations, of of the the form form w·x=o,
(5.4)
z·x = 0, where W, z are work; where w,z are two two vectors vectors in in R R44.• There There are are many many vectors vectors wand w and z that that will will work; indeed, as as long long as as w, w; z, and and the the two two rows rows of of A form linearly independent independent set, form aa linearly set, indeed, then these additional equations equations will will result result in unique solution. solution. then these additional in aa unique For example, example, let let w w= = el ei = = (1,0,0,0) (1,0,0,0) and and zz = = e4 64 = = (0,0,0,1). (0,0,0,1). Then Then (5.3) (5.3) and and For (5.4) (5-4) together together imply imply s = 0, s + 3t = 0,
R4 and system is and the the only only solution solution to to this this system is ss == t == 0. 0. Therefore, Therefore, the the only only vector vector xx E GR satisfying satisfying (5.3) (5.3) and and (5.4) (5.4) is is xx = = o. 0. x4 The defines The matrix matrix A AE eR R22x4 defines an an operator operator mapping mapping
°
(Kx. = — Ax). Ax). One One way way to to understand understand the the role role of of the the additional additional equations equations el ei .• xx = = 0 (Kx 4x4 and and R44 and e4·X 64 -x = = 00 is is that that they they allow allow us us to to define define aa matrix matrix BE B eR R4X4 and aa vector vector cc E 6R
136
Boundary value problems in statics Chapter 5. Boundary
such = cc has has aa unique unique solution in aa simple such that that By By = solution related related in simple way way to to a a solution solution of of The matrix Ax = = b. The matrix B would would be be
l
-~
° °2 °° ° ° However, this way of looking at the situation does not generalize in a natural way to B=
2
-1
-1
However, this way of looking at the situation does not generalize in a natural way to differential equations. equations. Therefore, we interpret interpret the side equations differential Therefore, we the two two side equations as as restricting restricting the domain of of the operator. The equations The equations el . x = 0, e4· x =
(5.5)
°
are satisfied by vectors of are satisfied of the the form form
(and only by by vectors vectors of of this this form), form), where where e2 = (0,1,0,0) and and e3 (and only 62 = 63 = = (0,0,1,0). Therefore, we are allowing only only solutions solutions to to Ax Ax == Therefore, by by imposing imposing the the equations equations (5.5), we are allowing belong to to the the subspace bb that that belong subspace S = span{e2,e3}. We define aa new operator KS Ks : SS -t R22 by Ksx = = Kx Kx (so (so KS Ks is is We define new operator —> R by restriction: KS*. except that the same as K, except that we we restrict restrict the set of allowable allowable inputs). We We have then then Ks has space-the only Ksx = shown that the the operator KS has a trivial trivial null space—the only solution to to KSX = 00 is x= bE Ksx = = o. 0. By By the the same token, token, for any any b € R2, R 2 , there there is is aa unique unique solution to to KS~X. = b. b. The final final step step in in this example is is to the operator operator equation Ksx = The this example to interpret interpret the equation K§x = bb matrix-vector equation. After all, all, KS Ks is a linear as a matrix-vector equation. After linear operator defined defined on finitefinitedimensional vector spaces, and and it it therefore therefore must must be be representable representable by by aa matrix. dimensional vector spaces, matrix. We We write e 5 in their coordinates {62, e write vectors vectors x xES in terms terms of of their coordinates with with respect respect to to the the basis basis {e2, e3}: 3}: xES:::} x = Ule2
+ U2e3.
2 We can thus identify SS with with R R2: We can thus identify :
x = Ul e2
+ U2e3
E
S B u E R 2.
The matrix revresentinq representing K$ Ks is and the reader can can show show2266 that that this The matrix is therefore therefore 2 xx 22,, and the reader this matrix is is matrix
c=
[ 2 -1] -1
2
.
26
26The reader should should recall recall from Exercise 3.1.8 that there there is is aa canonical method for for computing computing The reader from Exercise 3.1.8 that canonical method the matrix matrix representing representing aa linear operator on on finite-dimensional finite-dimensional spaces: The columns columns of the matrix the linear operator spaces: The of the matrix are operator of of the standard basis If uu = = (1,0), (1,0), then then K$u —1), are the the images images under under the the operator the standard basis vectors. vectors. If Ksu = = (2, (2, -1), and so the first first column column of C must must be and so the of C be
[ -i ].
The second second column column of C is is determined by computing the image image of of (0,1) (0,1) under under Ks. The of C determined by computing the KS-
5.1. between BVPs 5.1. The The analogy analogy between BVPs and and linear linear algebraic algebraic systems systems
137
By inspection, we can see that nonsingular, as as expected, By inspection, we can see that C is is nonsingular, expected, and, and, as as an an added added bonus, C C is is aa symmetric symmetric matrix, matrix, whereas whereas the the matrix matrix B, B; obtained obtained from the other from the other bonus, point of above, is not symmetric. point of view view described described above, is not symmetric.
We now show that the the role role of in (5.1) is exactly We now show that of the the boundary boundary conditions conditions in (5.1) is exactly analogous to the role of the extra equations ei • x = 0, 62 • x = 0 in the above analogous to the role of the extra equations el . x = 0, e2 . x = in the above example. We view boundary conditions example. We view the the boundary conditions
°
u(o)
= u(£) =
°
as restricting the the domain of the L, and as restricting domain of the operator operator I/, and define define LD : C1[0,£]-+ C[O,£],
where where c1[0,£] = {u E C 2 [0,£] : u(O) = u(£) =
by by
O},
d2 u LDu = Lu = -T dx 2 '
This LD has has aa trivial trivial null Indeed, if [0, £] satisfies This new new operator operator LD null space. space. Indeed, if u E e C22[0,l] satisfies
-T
~u
dx2
=0,
then then u(x) u(x) = ax ax + + bb for for some some a, a, bb E e R. If If uu also also satisfies satisfies u(O) w(0) = = u(£) u(t] == 0, 0, then then L D , like L, has aa == bb == 0, 0, and and uu is is the the zero zero function. function. The The operator operator LD, like L, has the the property property that LDu = f has has aa solution £] (that the range range of LD is that LDU solution for for each each f/ E € C[O, (7[0,1] (that is, is, the of LD is C[O, C[0, £]), and is unique unique because LD is and this this solution solution is because the the null null space space of of LD is trivial. trivial. We next LD is symmetric operator, as such, of We next show show that that LD is aa symmetric operator, and, and, as such, it it has has many many of the the properties of of aa symmetric symmetric matrix. Before we we can can demonstrate demonstrate this, of of course, we what symmetry means for operator. In we must must explain explain what symmetry means for aa differential differential operator. In Section Section 3.5, 3.5, we if and we saw saw that that aa matrix matrix A A €E Rnxn R n x n is is symmetric symmetric if and only only ifif
(Ax)· y =
X·
(Ay) for all x,y ERn.
We can analogous definition definition for for aa differential operator, using using the inner We can make make the the analogous differential operator, the L L22 inner product in place of the dot product in place of the dot product. product. Definition Let 5 bj, and K :: 5 linear Definition 5.4. 5.4. Let S be be aa subspace subspace of of Ck[a, Ck[a, b], and let let K S -+ —>• C[a, C[a, bj b] be be aa linear operator. say that K is operator. We We say that K is symmetric symmetric ifif (Ku, v)
= (u,Kv)
for all u,v E 5,
where L22 inner product on [a, b]. where (.,.) (-,-) represents represents the the L inner product on the the interval interval [a,b]. symmetric w = Ku, zZ = Kv, then symmetric if, if, whenever whenever u,v u,v E € 5, S, and and w = Ku, = Kv, then
lb
w(x)v(x) dx
=
lb
u(x)z(x) dx.
(5.6) That K is That is, is, K
138 138
Chapter 5. Chapter
Boundary value problems in statics Boundary
In working with with differential integration by by parts parts is is In working differential operators operators and and symmetry, symmetry, integration an essential essential technique, technique, as the following following examples examples show. an as the show. Example 5.5. 5.5. Let Let LD LD be be the defined above. if u, f], we we Example the operator operator defined above. Then, Then, if u,vv E € C1[0, Cf^O,^], have have
{l d2u
(LDu, v) = -T
10
dx 2 (x)v(x) dx
du
l
]
{l du
= [ -T dx (x)v(x) 0 + T 10 ~
= -T dx (f)v(f) = T =
{i
10
dv dx (x) dx (x) dx
~
r~
+ T dx (O)v(O) + T 10
~
dx (x) dx (x) dx
du dv dx (x) dx (x) dx (since v(O) = v(f) = 0)
dV]l r ~v [Tu(x) dx (x) 0 - T 10 u(x) dx 2 (x) dx
dv du = Tu(f) dx (f) - Tu(O) dx (0) - T = -T
(l
10
~v
(l
10
u(x) dx 2 (x) dx (since u(O)
~v u(x) dx 2 (x) dx
= u(f) = 0)
= (U,LDV).
LD is is a operator. Therefore, Therefore, LD a symmetric symmetric operator.
Example 5.6. Example 5.6. Define Define c1[0,ij = {u E C 1 [0,f] : u(O) = u(f) =
O}
and MD : C1[0, f) -+ C[O, f) by
Then Then
(l
(MDu,v) = =
10
du dx(x)v(x)dx
u(x)v(x)l~
-
r
10
dv u(x) dx (x) dx
= u(f)v(f) - u(O)v(O) = -
(i
10
(l
10
dv u(x) dx (x) dx
dv u(x) dx (x) dx (since u(O) = u(f) = 0)
= -(U,MDV).
5.1. The analogy between BVPs and linear algebraic systems systems
139
Therefore, MD is is not Therefore, MD not symmetric. symmetric.
A symmetric matrix matrix A A E G Rnxn R n x n has the following following properties properties (see Section 3.5): All eigenvalues eigenvalues of of A real. • All A are are real. Eigenvectors of corresponding to eigenvalues are are orthogonal. orthogonal. to distinct distinct eigenvalues • Eigenvectors of A corresponding There exists exists aa basis of R consisting of of eigenvectors eigenvectors of of A. A. • There basis of R nn consisting
properties exist for a symmetric differential operator. In fact, the first Analogous properties exist for differential operator. 27 symmetric matrices. two properties can be proved proved exactly as they were for symmetric matrices. 27 k In the the following discussion, S S is a subspace of Ck[a, C [a,b]b] and and K : S S —> C[a,b] -t C[a, b] is a symmetric linear operator. eigenvalue of K K if there exists a operator. A scalar scalar A is an eigenvalue nonzero function function u such that Ku = AU. Just as in the case of matrices, we we cannot cannot assume aa priori that A is real, or that the the eigenfunction u is real-valued. When working with complex-valued functions, the eigenfunction complex-valued functions, L22 inner product on [a, b] L 6] is defined by (f,g) =
lb
f(x)g(x) dx.
(5.7)
The properties properties
(al,g) =a(f,g), (f,ag) =a(f,g) hold, just as for the complex dot product. product. We immediately show that there is no need to allow complex numbers when operators. Indeed, suppose u is a nonzero function and AA working with symmetric operators. is a scalar satisfying Ku = AU.
Then, since (Ku,u) (Ku,u) = = (u,Ku) complex-valued functions as well as real(u,Ku) holds for complex-valued valued functions functions (see Exercise 4), we have
(Ku,u) = (u,Ku)
= (U,AU)
=?
(AU,U)
=?
A(U,U) = X(u,u)
=?A=X =?AER. It is easy to show that there must be a real-valued eigenfunction corresponding corresponding to AA (the proof is the same as for matrices; see Theorem Theorem 3.44). Therefore, we do not need to consider complex numbers any longer when we are dealing with symmetric operators. operators. 27 27The third property property is more difficult. difficult. We discuss an an analogous property for for symmetric symmetric difdifThe third is more We discuss analogous property ferential operators in the subsequent sections and and delay delay the the proof proof of of this this property until Chapter ferential operators in the subsequent sections property until Chapter 9. 9.
140
Chapter 5. Boundary Boundary value problems in statics statics Chapter
We now now assume that AI, A1, A2 A2 E the operator operator K, We assume that € R R are are distinct distinct eigenvalues eigenvalues of of the K, with corresponding corresponding eigenfunctions eigenfunctions U1, U2 E S. We We then then have have with wi,W2 G S.
= (A1U1,U2)
A1(U1,U2)
= ( KU1,U2)
= (U1, Ku 2) = (U1' A2U2) = A2(Ut, U2).
Since AI ^ A2, A2, this if (U1' (^1,^2) u\ and u^ are are orthogorthogSince A1 =I this is is only only possible possible if U2) == 0, 0, that that is, is, if if U1 and U2 onal. Therefore, aa symmetric differential operator orthogonal eigenfunctions. eigenfunctions. onal. Therefore, symmetric differential operator has has orthogonal The symmetric has aa special property not shared by by The symmetric differential differential operator operator LD has special property not shared every operator. Suppose every symmetric symmetric operator. Suppose Aisisan an eigenvalue eigenvalueof ofLD LD and and Uu isisaacorresponding corresponding that (u,u) = eigenfunction, normalized so eigenfunction, normalized so that — 1. 1. Then Then A = A(U,U) = (AU,U) = (LDU,U)
= -T =T =T
l
0
£d2u
dx 2 (x)u(x) dx
1£ ~~ ~~ 1£ (~~(x)) (x)
(5.8)
(x) dx (integration by parts) 2
dx
> 0. Therefore, every of LD is is positive. positive. (This explains why why we we prefer prefer to to work work Therefore, every eigenvalue eigenvalue of (This explains with the the negative operator-the eigenvalues then positive. positive. with negative second second derivative derivative operator—the eigenvalues are are then Actually, in in the the computation computation above, it is is only only obvious that Actually, above, it obvious that
and therefore therefore that that A > 2: 0. 0. How How can can we we conclude conclude that, that, in in fact, the inequality and fact, the inequality must must be strict? Exercise 5.) 5.) be strict? See See Exercise
Summary Summary We seen that differential operator operator We have have now now seen that the the differential
subject to to Dirichlet Dirichlet conditions, conditions, is is symmetric, which means that any any eigenvalues subject symmetric, which means that eigenvalues the eigenfunctions eigenfunctions will will be be orthogonal. We have seen directly directly must be be real real and must and the orthogonal. We have also also seen that any must be be positive. positive. Assuming Assuming we we can can find find the the eigenvalues that any eigenvalues eigenvalues must eigenvalues and and
5.1. The analogy between BVPs and linear algebraic algebraic systems
141 141
eigenfunctions of of LD, and that that there there are are "enough" "enough" eigenfunctions eigenmnctions to to represent represent u eigenfunctions L D , and u and f, /, we can contemplate contemplate aa spectral spectral method for solving and we can method for solving
tPu
-T dx 2 = f(x),
0< x < £,
u(o) = 0, u(£) = 0.
(5.9)
We next two two sections. We develop develop this this method method in in the the next sections.
5.1.1 5.1.1
A note about direct integration
The how The BVP BVP (5.9) (5.9) is is quite quite elementary; elementary; indeed, indeed, we we already already showed showed in in Chapter Chapter 22 how to 2.3). to solve solve this this and and similar similar problems problems by by integrating integrating twice twice (see (see Examples Examples 2.2 2.2 and and 2.3). There are are two two reasons why we we spend spend this this chapter chapter developing developing other other methods methods for for solvThere reasons why solvfinite elements, elements, ing ing (5.9). (5.9). First First of of all, all, the the methods methods we we develop, develop, Fourier Fourier series series and and finite form the basis form the basis of of methods methods for for solving solving time-dependent time-dependent PDEs PDEs in in one one spatial spatial variable. variable. Such admit direct seSuch PDEs PDEs do do not not admit direct solution solution by by integration. integration. Second, Second, both both the the Fourier Fourier series method method and and the the finite element method method generalize generalize in in aa fairly fairly straightforward straightforward way ries finite element way to problems in multiple spatial dimensions, dimensions, whereas whereas the the direct direct integration integration method method to problems in multiple spatial does does not. not.
Exercises 1. MD be 1. Let Let MD be the the operator operator defined defined in in Example Example 5.6. 5.6.
(a) Show Show that that if if f/ Ee C[O, C[0,£] is such such that that Mpu has aa solution, solution, then that (a) £] is MDu = f has then that solution is unique. unique. solution is (b) Show Show that that MDu Mpu =— f has has aa solution solution only only if if f/ Ee C[O, C[0,^] satisfies aa certain certain (b) £] satisfies constraint. is that that constraint? constraint. What What is constraint? l
2. Define Define C][O, C}[Q,t]£] == {u {u E€ C C 1[Q,i] 2. [0, £]
u(0) == O} 0} and and MI M7 :: C][O, C)[0,£] ->• C[0, C[0,£] by : u(O) £] --+ £] by du
M1u = dx'
(a) Show Show that has aa unique unique solution solution for for each each f/ E6 C[O,£]. C[0,^]. (Hint: (a) that M/w M1u == f/ has (Hint: Use fundamental theorem of calculus.) calculus.) This This is is equivalent equivalent to Use the the fundamental theorem of to showing showing that that N(M JV(M/) is trivial trivial and and R(M ft(M/) = C[O,£]. C[Q,f\. 1 ) is 1) = (b) symmetric. (b) Show Show that that MI M/ is is not not symmetric.
3. Consider Consider again again the the differential differential operator operator equation equation Lu = — f discussed discussed in in Section 3. Section 5.1. Suppose Suppose an an equation equation Lsu LSU = = f/ is is produced restricting the the domain domain of 5.1. produced by by restricting of L to 5). to aa subspace subspace S of of C C22[0,£] [0,£] (that (that is, is, by by defining defining LSU Lsu = = Lu for for all all u € E S). (a) Show Show that LSU == f/ has has at at most one solution solution for for each each f/ Ee C[O, (7[0, £] t] provided (a) that Lsu most one provided N(L) n S = N(L)nS = {O}. {Q}.
142
Chapter 5.
problems in statics Boundary value problems
(b) Show Show that L$u = — fj has unique solution solution for for each each f/ E e C[O,f] C[0,1] for for either either (b) that Lsu has aa unique of the following choices of S: of the following choices of i. S={UEC2[0,f]:
~~(~)=O, f~u(x)dx=O}.
ii. S = {u E C 2[0,f] : u(O)
= ~~(O) = O}.
4. Suppose 4. Suppose S 5 is is aa subspace subspace of of Ck[a, Ck[a, b] b] and and K : S S -+ —>• C[a, b] 6] is is aa symmetric symmetric linear operator. Suppose u, v are complex-valued functions, with u linear operator. Suppose u, v are complex-valued functions, with u = = f/ + ig, ig, vv == w w + iz, iz, where where /, g, w, z EG S. 5. Show Show that that f,g,w,z (Ku,v) = (u,Kv)
holds, where (.,.) (•, •) is defined by (5.7). 5. Explain why, calculation (5.8), the integral why, in the last step of the calculation
must be positive. (Hint: (Hint: If du/dx is must be constant. What If du/dx is zero, zero, then then u must be constant. What must be positive. constant functions functions belong C£)[0,^]?) constant belong to to C1[0, f]?) 6. Define 6. Define C![O,f]
= {u E C2[0,f]
u(O)
= ~~(£) = O}
and Lm : C~ [0, f] -+ C[O, f] by
(a) (a) Show Show that that the the null null space space of of Lm Lm is is trivial. trivial. of C[o,f]. (b) (b) Show Show that that the the range range of of Lm Lm is is all all of (7[0,^]. (c) Show is symmetric. (c) Show that that Lm Lm is symmetric. (d) Show that all of the eigenvalues of Lm Lm are positive. 7. Repeat Exercise 6 with C~[O,f] and Lift : C~[O,f]-+ C[O,f] defined by C~[O,f]
= {u E C2[0,f]
: ~~(O)
= u(f) = O},
J2u
Liftu = - dx 2' 8. 8. Repeat Repeat Exercise Exercise 66 with with K: K : C1[0,f]-+ C^[Q,f\ ->• C[O,f] C[Q,f\ defined defined by by Ku=-a where where aa and and bb are are positive positive constants. constants.
J2u dx2
+bu,
5.1. between BVPs 5.1. The The analogy analogy between BVPs and and linear linear algebraic algebraic systems systems
143 143
9. Consider the 1) defined by the differential differential operator operator M M :: C1[0, C^O, 1)-+ 1] —>• C[O, C[0,1]
Mu=
cPu
du
--+ -+5u. dx 2 dx
Show that M M is is not symmetric by producing two functions u, u, v E 6 C1[0,1] (7^(0,1] such that such that
=I (u,Mv).
(Mu,v) 10. Define B : C44[0,l] [0, £) -+ £) by -> C[O, C[Q,£\ Bu
d4 u
= dx 4 .
(a) Determine the null space of B. B. (b) Find a set of boundary conditions such that, if 5 S is the subspace of of C [0, £] consisting of functions satisfying these boundary conditions, then (744[0,1] B, restricted S, is is symmetric and has a trivial trivial null space. B, restricted to 5, 11. 11. In this exercise, we we consider a new boundary condition, more complicated than Dirichlet or Neumann conditions, that provides a more realistic realistic model of certain physical phenomena. This boundary condition is called a Robin condition; we flow in a we will introduce it in the context context of steady-state heat flow one-dimensional bar. Suppose the ends of a bar are uninsulated and the heat flux through each end is proportional proportional to the difference between the temperature temperature at the end of the bar and the the surrounding surrounding temperature (assumed constant). constant). If we let aa > 0 be be the the and temperature (assumed we let constant temperatures surrounding the constant of proportionality proportionality and To, TO, T£ TI be the the temperatures the ends of of the the bar bar at at xx = = 0 and and xx = £, i, respectively, respectively, then then the the resulting resulting boundary boundary ends conditions are
°
°
du K,
dx (0) = -a (To - u(O)) ,
du K,
dx (£) = a (T£ - u(£)).
These can be rewritten as
du -K,
dx (0)
du K,
dx (£)
+ £lu(O)
= £lTD,
+ £lu(£)
= £IT£.
Define
c1[0, £] = {u and let
LR :
E
C2 [0, £] :
-K,
~~ (0)
+ £lu(O)
CMO, £] -+ C[O, £] be defined by LRU =
cPu -K,
dx 2 ·
= 0,
K,
~~(£)
+ £lu(£)
= O},
144 144
Chapter Boundary value problems in statics Chapter 5. Boundary
(As usual, we we define the operator terms of of the the homogeneous (As usual, define the operator in in terms homogeneous version version of of the the boundary conditions.) boundary conditions.)
(a) Prove Prove that is symmetric. that LR LR is symmetric. (a) (b) Find Find the space of of LLR. (b) the null null space R.
5.2 5.2
Introduction to to the spectral method: method; Introduction the spectral eigenfunctions eigenfunctions
In this this section the next, next, we we develop method for for the the BVP In section and and the develop aa "spectral" "spectral" method BVP -T
J2u
=/(x), O<x<£, dx2 u(O) = 0,
(5.10)
u(£) = O.
This BVP can can be be written simply as as Lpu = /, is the symmetric linear linear This BVP written simply LDu = /, where where LD LD is the symmetric differential in the differential operator operator defined defined in the last last section. section. The spectral spectral method symmetric linear linear system system Ax Ax = = b is based The method for for aa symmetric b is based on on the fact matrix A A E orthonormal eigenvectors, and the fact that that aa symmetric symmetric matrix e RllXll R n x n has has nn orthonormal eigenvectors, and therefore any any vector vector in in Rll the right-hand b and the solution can therefore Rn (including (including the right-hand side side b and the solution x) x) can be written, written, in in aa simple way, as be simple way, as aa linear linear combination combination of of those those eigenvectors. eigenvectors. (The (The reader may may wish wish to to review review Section 3.5 at at this time.) We have seen, at the end of of reader Section 3.5 this time.) We have seen, at the end the last of LD LD must must be be real real and and positive that the last section, section, that that any any eigenvalues eigenvalues of positive and and that eigenfunctions LD corresponding to distinct eigenvalues must must be be orthogonal. We eigenfunctions of of LD corresponding to distinct eigenvalues orthogonal. We now find the the eigenvalue-eigenfunction eigenvalue-eigenfunction pairs pairs (eigenpairs (eigenpairs for short) of of LD and explore explore now find for short) LD and the following we represent represent the the right-hand right-hand side /(x) and and the the following question: question: Can Can we side f(x) the solution solution u(x) terms of u(x) in in terms of the the eigenfunctions? eigenfunctions?
5.2.1 5.2.1
:1:2
Eigenpairs of — - -j^ under under Dirichlet conditions
We will will show show that that the the operator operator LD infinite collection eigenpairs. To We LD has has an an infinite collection of of eigenpairs. To we may may as well assume is an simplify the calculations, simplify the calculations, we as well assume that that T T = = 1, 1, since since if if A, A,wu is an eigenpair of eigenpair of d2 dx 2 (subject to conditions), then then T).., T\,u is an an eigenpair eigenpair of (subject to Dirichlet Dirichlet conditions), u is of
J2 -T-2 dx
under the the same boundary conditions. under same boundary conditions. To find the need to to solve To find the eigenpairs, eigenpairs, we we need solve d2 u - dx 2 = AU, u(O) = 0, u(£)
= O.
(5.11)
5.2. Introduction Introduction to to the the spectral spectral method; method; eigenfunctions eigenfunctions 5.2.
145 145
"Solving" this identifying those the "Solving" this problem problem means means identifying those special special values values of of A A such such that that the BVP (5.11) has u. (For BVP (5.11) has aa nonzero nonzero solution solution u, u, and and then then determining determining u. (For any any value value of of A, the the zero zero function function is is aa solution solution of just as as x = = 00 isis aa solution of Ax Ax = = Ax AX A, of (5.11), (5.11), just solution of for any In both cases, the lead to for any A. A. In both cases, the eigenvalues eigenvalues are are the the scalars scalars which which lead to aa nonzero nonzero solution.) we know, know, from the previous previous section, section, that that all all such A are solution.) Since Since we from the such A are positive, positive, we write write A A= (J2.2. We We can use the the results results of Section 4.2 4.2 to to write write the the general we = O can use of Section general solution solution the ODE of the of ODE cf2u
2
+0 U
dx 2
= 0,
(5.12)
and then satisfy the conditions. and then try try to to satisfy the boundary boundary conditions. The The characteristic characteristic roots roots of of (5.12) (5.12) are are ±0i, ±Oi, and and the the general general solution solution is is u(x) =
Cl
cos (Ox)
+ C2 sin (Ox).
Therefore, Therefore,
u(O) =
Cl,
and the c\ =— 0. u(x] — and the first first boundary boundary condition condition implies implies Cl O. We We then then have have u(x) = C2sin(0x), C2 sin (Ox), and the and the second second boundarv boundary condition condition becomes becomes C2
sin (Of)
= O.
This in two cases: This equation equation only only holds holds in two cases: 1. value of of C2 zero. But But then then u(x) u(x) is the zero zero function, which cannot 1. The The value C2 is is zero. is the function, which cannot be be
an eigenfunction. eigenfunction. an 2. The value O22 is such that = 0. cases 2. The value of of A A= = 0 is such that sin sin (Of.) (Of) = O. This This holds holds in in the the cases Of = ±n1f, n = 1,2, ....
Solving for Solving for A, A, we we have have n 2 1f2
A=~,
n= 1,2, ...
(we discard = 0, function). It It turns out (we discard the the case case that that A A= 0, for for then then uu is is the the zero zero function). turns out that the the value value of of c^ C2 is is immaterial; immaterial; these these values values of of A, A, and and no no others, others, produce produce that eigenfunctions with with the the correct correct boundary boundary conditions. eigenfunctions conditions. We thus thus see that the the operator operator LD LD has has infinitely many eigenpairs: We see that infinitely many eigenpairs: (5.13)
We We already already know know that, that, since since the the eigenvalues eigenvalues of of LD LD are are distinct, distinct, the the eigenfunceigenfunctions listed listed above above must must be be orthogonal. orthogonal. This This can can also be verified verified directly directly using using the tions also be the trigonometric identity trigonometric identity 1 sin Qsin.B = "2 (cos (Q - .B) - cos (Q + .B))
146
Chapter 5.
Boundary Boundary value value problems problems in statics
(see Exercise Exercise 1). Also, for for each (see 1). Also, each n,
that that is, is,
II1/1nll
[f,
=
n = 1,2, ....
Thus are orthogonal, Thus the the eigenfunctions eigenfunctions are orthogonal, and and they they can can be be normalized normalized by by multiplying multiplying each by by ^/2/l. each The first first four four eigenfunctions eigenfunctions 1/11,1/12,1/13,1/14 are6 graphed graphed in in Figure Figure 5.2. 5.2. The V>i, V>2, V's, V>4 a^
-/27"f.
1
-\""'-'W"'" - ......... ,,'
:,
,- ,.,
,: i ,'
",; "
,
'
,,
,,
, :', , "
,:,'
0.5
\ \
\
\
, :' , ,::.', :;
\
\
\
:" jl
,
\
\
,
I
\ \
\
"
\
\
\
.
\t
\
\'
: \ '.,'
,
"
,
,
'
...... ,,;
"
0.6
0.4
, , , ,, :
: : ,:
\
\
0.2
, ,, , ,-' , ,, \
\
\
o
, ,
\ \
-1
\
,
,: :"
\
\
"
,
\
\
-0.5
\
,', '"
,
:
:
:'
, "',' "
..
' ---;
:
,-
.-
",'
0.8
1
x Figure 5.2. The first first four eigenfunctions 1/11, '1/12,1/13,1/14 ft(£ — = 1JI), ^15^2,^85^4
5.2.2 5.2.2
Representing functions in terms Representing functions in terms of of eigenfunctions eigenfunctions
We operator —d?/dx subject to condiWe have have shown shown that that the the operator -d? / dx22,, subject to Dirichlet Dirichlet boundary boundary conditions, an infinite eigenpairs, and eigenfunctions are orthogtions, has has an infinite number number of of eigenpairs, and that that the the eigenfunctions are orthogonal. answer the crucial question: the onal. We We now now must must answer the crucial question: is is it it possible possible to to represent represent both both the solution of (5.10) in terms of the solution and and right-hand right-hand side side of (5.10) in terms of the eigenfunctions? eigenfunctions?
5.2. Introduction to the the spectral spectral method; method; eigenfunctions 5.2. Introduction to eigenfunctions
147 147
Let us any function <7[0,^]. Section 3.5, Let us take take any function f/ e E e[o, fl. We We know, know, from from Section 3.5, how how to to find find the best best approximation approximation to to f/ from from aa finite-dimensional finite-dimensional subspace. define the subspace. We We define
e
. (7fX) ,sm . (27fX) . (N7fX)} FN = span { sm -f- , ... ,SIn -f.
The approximation to The best best approximation to f/ from from FN FN is is
Since (1/Jn,1/Jn) = f/2 for all all n, n, we we have Since (?/> ^/2 for have n ,V>n) =
r
l
(1/Jn, f) 2 ( ) . (n7fX) Cn = (1/Jn,1/Jn) ="iJo fx sm -f- dx, n=1,2, ... ,N. 28 We now now show show several several examples examples of of this this approximation. approximation. 28 In the the following following examples, We In examples, f1 == 1. 1.
Example 5.7. Let Let Example 5.7. f(x)
= x(l -
x).
Then Then
1
4(1+ (_l)n+1)
1
cn =2
o
x(1-x)sin(n7fx)dx=
33
n 7f
.
Figure errors in 10,20,40,80. Figure 5.3 5.3 shows shows the the errors in approximating approximating /(#) f(x) by by /N(X] fN(X) for for N N = = 10,20,40,80. (The approximations so good if we same (The approximations are are so good that, that, if we were were to to plot plot both both ff and and /AT fN on on the the same graph, curves would Therefore, we the graph, the the two two curves would be be virtually virtually indistinguishable. indistinguishable. Therefore, we plot plot the difference this example, small difference f(x) f(x) — - /N(X) fN(x) instead.) instead.) In In this example, the the approximation approximation error error is is small and gets gets increasingly increasingly small small as as N N increases. increases. It It is easy to and is easy to believe believe that that fN(x) -+ f(x) as N -+
00.
Example Let Example 5.8. 5.8. Let f(x) = I-x.
Then Then
r (1 Jo 1
Cn
=2
x) sin (n7fx) dx
= ~. n7f
Figure 5.4 shows shows f(x) f(x) together together with with the the approximations approximations /jv(ar) fN(X) for for N N = = 10,20,40,80. Figure 10,20,40,80 28 28Many specific integrals integrals must must be be computed to compute these approximations. approximations. It It is is beyond beyond the Many specific computed to compute these the scope often allows scope of of this this book book to to discuss discuss methods methods of of integration; integration; moreover, moreover, modern modern technology technology often allows us us to avoid avoid this this mechanical mechanical step. Computer algebra software, such such as as Mathematica, Mathematica, MATLAB, MATLAB, Maple, to step. Computer algebra software, Maple, etc., as as well well as as the the more more powerful powerful handheld handheld calculators, calculators, will will compute compute many many integrals integrals symbolically. symbolically. etc.,
148 148
Chapter Chapter 5. 5. Boundary Boundary value value problems problems in in statics statics -4
1.5 x 10 1 0.5
o
0 -0.5 -1
-5~------------------~
o
0.5
1
0
2
X1O
0.5
1
-6
1
2
-2~------------------~ 0.5 1 o
-2~------------------~
o
0.5
1
Figure 5.3. 5.3. Error Error in in approximating approximating f(x} f(x) = = x(l—x) xCI-x) by by /N(X) fN(X) (see (see Example Example Figure 5.7). The The value value of of N N is is 10 10 (top (top left), left), 20 20 (top (top right), right), 40 40 (bottom (bottom left), left), and and 80 80 (bottom (bottom 5.7). right). right). is worst worst near near xx = = 00 is is due due to to the the fact fact that that The fact fact that that the the approximation approximation is The does not not satisfy satisfy the the Dirichlet Dirichlet condition condition there. there. Indeed, Indeed, this this example example shows shows that that ff does fN need need not not converge converge to to ff at at every every xx eE [0,€j; [0,£]; here here /(O) f(O) == 1, but but /Ar(0) fN(O) == 00 for for /AT every N. However, However, itit seems seems clear clear from from the the graphs graphs that that fw fN approximates approximates ff in in some some every N. meaningful meaningful sense, sense, and and that that this this approximation approximation gets gets better better as as N N increases. increases. The The oscillation oscillation near near xx = = 00 is is known known as as Gibbs's Gibbs's phenomenon. phenomenon.
In In Chapter Chapter 9, 9, we we justify justify the the following following fact: fact: /jv fN converges converges to to /f as as N N —> -+ oo, 00, in in 2 the sense sense that that ||/ Ilf —- /AT|| fNIl —> -+ 00 (that (that is, is, the the L L2 norm norm of of the the error error /f — - /AT fN goes goes to to zero). zero). the We We write write 00
f(x) = Lcnsin (n;x) n=l
to to indicate indicate this this fact. fact. As As we we saw saw in in Example Example 5.8, 5.8, this this does does not not necessarily necessarily imply imply the ( x ) for the /N(X) fN(X) ->• -+ ff(x) for all all xx eE [0,^]. [0,£]. ItIt does does imply, imply, however, however, that that /N fN gets gets arbitrarily arbitrarily
149
5.2. Introduction Introduction to to the spectral method; method; eigenfunctions eigenfunctions 5.2. the spectral
1.5r----------...,
1.5r----------...,
1
1
0.5
0.5
1.5r----------...,
1
1.5r----------...,
1
1
-0.5'-------------' 0.5 o 1
-0.5'----------~
o
0.5
1
Figure Approximating /(#) f(x) = fN(X) (see Example 5.8). The The Figure 5.4. Approximating = 1I-x — x by by /N(X) (see Example value of 20 (top value of N is is 10 10 (top (top left), left), 20 (top right), right), 40 40 (bottom (bottom left), left), and and 80 80 (bottom (bottom right). right).
close the entire interval, in type of convergence close to to f, /, over over the entire interval, in an an average average sense. sense. This This type of convergence (in the L L22 norm) referred to to as mean-square convergence. (in the norm) is is sometimes sometimes referred as mean-square convergence. The The resulting resulting representation, representation,
=
f(x) =
L n=l
Cn
sin (n;x),
Cn
=
~
1 l
f(x) sin (n;x) dx,
(5.14)
0
is referred to Fourier sine series of the scalars C2, called the is referred to as as the the Fourier sine series of /,f, and and the scalars Cl, ci,C 2 ,••. . . . are are called the Fourier sine coefficients coefficients of of f./. In the next next section, will show how to use the to solve In the section, we we will show how to use the Fourier Fourier sine sine series series to solve the the BVP First, however, however, we boundary conditions the resulting BVP (5.10). (5.10). First, we discuss discuss other other boundary conditions and and the resulting Fourier Fourier series. series.
150
5.2.3 5.2.3
problems in statics Chapter 5. Boundary value value problems
Eigenfunctions under other boundary boundary conditions; other Fourier series
We consider the We now now consider the BVP BVP with with mixed mixed boundary boundary conditions, conditions, -T
d'2u
dx2 u(O)
=f(x), O<x<£,
= 0,
(5.15)
~~(£)=o, which models, models, for example, aa hanging hanging bar bar with with the the bottom bottom end end free. free. In In order order to which for example, to consider spectral method, method, we we define define the the subspace consider aa spectral subspace c![O,£] = {u E C 2[0,£] : u(O) = ~~(£) = O}
and C[0,l], and the the operator operator L Lm C;, [0, £] —> --+ C[O, £], where where m :: C^[0,€]
d'2u
Lm u = - dx 2 ' The operator Lm is symmetric with positive positive eigenvalues, eigenvalues, it it has has aa trivial trivial null null space, The operator L symmetric with space, m is and the range range of Lm is all all of of C[0,^] C[O, £] (see Exercise 5.1.6). 5.1.6). Therefore, Therefore, L Lmu = f has and the of L (see Exercise has m is mu = unique solution solution for for every every f/ E e C[O,£], C[0,^], and and we we will will show, show, in in the the next next section, section, how how aa unique to compute the solution solution by by the the spectral method. First, First, however, however, we we must must find find the to compute the spectral method. the eigenvalues and and eigenfunctions eigenfunctions of of Lm. eigenvalues Lm. 2 Since Lm has only only positive positive eigenvalues, we define define A A= where 9B > 0, 0, and Since L eigenvalues, we = 0(J2, , where and m has solve solve
d'2u
dx 2
2
+B u =
0,
°<
x
< £,
u(O) = 0, du dx (£) = 0.
(5.16)
The general of the the differential differential equation equation is is The general solution solution of u(x)
= Cl cos (Bx) + C2 sin (Bx).
The c\ — any nonzero The first first boundary boundary condition, condition, u(Q) u(O) = 0, 0, implies implies that that Cl = 0, 0, and and hence hence any nonzero eigenfunction eigenfunction must must be be aa multiple multiple of of u(x) = sin (Bx).
The second boundary The second boundary condition. condition, du dx (£) = 0,
yields yields the the condition condition that that Bcos (B£)
= 0.
5.2. Introduction Introduction to to the method; eigenfunctions 5.2. the spectral spectral method; eigenfunctions
151
Since by assumption, this is possible only only if if Since 0 > 00 by assumption, this is possible cos (OC)
=0
or or
(2n - 1)7r OC = ~, 37r , .... 2 2'···' 2 Solving for obtain eigenvalues eigenvalues AI, . . . ,' with with Solving for A, A, we we obtain A1, A A2'2 ,...
An =
(2n - 1)27r 2 4C2 ' n = 1,2,3, ....
The corresponding eigenfunctions eigenfunctions are The corresponding are . ((2n - l)7rX) cPn(x)=sm 2C ' n=1,2,3, .... The as is is guaranteed guaranteed since Lm is symmetric, symmetric, The eigenfunctions eigenfunctions are are orthogonal, orthogonal, as since L m is and as be verified (0 n ,^>cPm). The best to verified directly directly by by computing computing (cPn, best approximation approximation to and as can can be TO ). The C], using 0i,02,...,0jv, cP1, cP2, . .. , cPN, is /f E G e[O, C[0,£], (5.17)
Since Since
(cPn, cPn)
r£.2((2n-1)7rX) sm 2C dx
= 10
£
= 2'
(5.17) becomes (5.17) becomes N
b . ((2n - l)7rX) f( x ) == '"' ~ n sm 2C ' n=l
where where
~
bn =
1£
f(x) sin ((2n
;c
1
)7rx) dx, n = 1,2,3, ....
The representation, The resulting resulting representation, . ((2n-1)7rX) f( x ) -~b - ~ n sm 2£ ' n=l
is referred as the It is valid in in the mean-square is referred to to as the Fourier Fourier quarter-wave quarter-wave sine sine series. series. It is valid the mean-square sense. sense. Example Let f(x) f(x) = and £i = Example 5.9. 5.9. Let = x and = 1. 1. Then
21
1
o
.
xsm
((2n-1)7rx)d _ 8(-1)n+1 x- (2 1)22' n=1,2,3, ... , 2 n7r
152
Chapter 5. Boundary Boundary value problems problems in statics
and the to ff using the first eigenfunctions is and the best best approximation approximation to using the first N eigenfunctions is N
b . ((2n -1)7fX) f N (x ) -- ""' ~ n sm 2 . n=l
The in approximating approximating ff on on [0,1] fN, for for N = is graphed graphed in The error error in [0,1] by by /N, = 10,20,40,80, 10,20,40,80, is in Figure 5.5. 5.5. Figure
0.03,....----------,
15
0.02
10 .... 0 .... 5 .... Q)
....
g
X10-3
0.01
Q)
0 -0.01 '---------~ 0.5 o
-5 0
0.5
x
X
3 4
2
2
.... ....0.... 1 Q)
a
0
....
g Q)
-1
-2~------------------~ 0.5
a
X1O
-3
..., .&.
a
x
0.5 x
Figure 5.5. 5.5. Approximating Approximating ff(x) by aa quarter-wave quarter-wave sine sine series series (see (see Figure (x) = — xx by Example 5.9). The value of of N N is is 10 left), 20 20 (top right), 40 40 (bottom left), and and Example The value 10 (top (top left), (top right), (bottom left), 80 (bottom right). right). 80 (bottom
In the reader is is asked asked to series, including In the exercises, exercises, the the reader to derive derive other other Fourier Fourier series, including the appropriate series series for conditions the appropriate for the the boundary boundary conditions du dx (0)
= 0,
u(£)
= 0.
In Chapter we will present Fourier Fourier series series for Neumann conditions conditions In Chapter 6, 6, we will present for the the Neumann du (0)
dx
= 0,
du (£)
dx
=
°
5.2. Introduction to spectral method; eigenfunctions 5.2. Introduction to the the spectral method; eigenfunctions
153 153
and conditions and the the periodic periodic boundary boundary conditions u( -P)
= u(P),
du dx (-P)
du
= dx (P)
(for the the interval interval [-P, Generally speaking, (for [ — i , P]). f \ ) . Generally speaking, given given aa symmetric symmetric differential differential operaoperator, sequence of of orthogonal eigenfunctions exists, exists, and it can can be used to represent and it be used to represent tor, aa sequence orthogonal eigenfunctions functions in in aa Fourier However, it may not easy to it may not be be easy to find find the the eigenfunctions eigenfunctions functions Fourier series. series. However, in some cases cases (see (see Exercise 4). in some Exercise 4).
Exercises 1. Use the trigonometric identity 1. Use the trigonometric identity sin a sin,8 =
1
2 (cos (a -
,8) - cos (a + ,8))
to verify verify that, that, if if nn / ¥- m, m, then then ('¢n, where '¢l, '1/J2, '¢3, ... are given given in in to (V>n? '¢m) V'm) = — 0, 0, where ^1,^25^3? • • • are (5.13). (5.13).
2. Define Define K K as in Exercise (from the exercise) that has 2. as in Exercise 5.1.8. 5.1.8. We We know know (from the earlier earlier exercise) that K K has only real, real, positive eigenvalues. Find all of of the eigenvalues and eigenvectors of the eigenvalues and eigenvectors of only positive eigenvalues. Find all K K. 3·. differential operator Lin defined in Exercise 3-. Repeat Repeat Exercise Exercise 22 for for the the differential operator Lfh defined in Exercise 5.1.7. 5.1.7. The resulting eigenfunctions the quarter-wave cosine functions. functions. The resulting eigenfunctions are are the quarter-wave cosine
4. (Hard) Define 4. (Hard) Define LR LR as as in in Exercise Exercise 5.1.11. 5.1.11. (a) Show that eigenvalues. that LR LR has has only only positive positive eigenvalues. (a) Show (b) Show Show that sequence of of positive (b) that LR LR has has an an infinite infinite sequence positive eigenvalues. eigenvalues. Note: Note: The equation that that determines the positive positive eigenvalues be solved The equation determines the eigenvalues cannot cannot be solved analytically, graphical analysis analysis can can be be used used to that but aa simple simple graphical to show show that analytically, but they exist and and to estimate their they exist to estimate their values. values. (c) K= = 1, 1, find find the finding accurate = 1£ the first first two two eigenpairs eigenpairs by by finding accurate estimates estimates (c) For For o: a = of the two smallest smallest eigenvalues. eigenvalues. of the two 5. (Hard) (Hard) Consider Consider the C[0,1] defined by by 5. the differential differential operator operator M : C^[0,1] C~[O, 1] ->• -+ C[O, 1] defined cFu dx 2
du dx
Mu=--+~+5u 2 (recall that u(0) = g(l) 0}). AnalyzetheeigenAnalyze the eigen(recall that C£[0,1] C~[O, 1] = {u e E C 2[0,1] [0, 1] :: u(O) ~~(1) = o}). pairs of of M pairs M as as follows: follows:
(a) Write the characteristic polynomial of the ODE (a) Write down down the characteristic polynomial of the ODE d2 u dx 2
-
du dx
+ (A -
5)u = O.
Using quadratic formula, formula, find find the roots. Using the the quadratic the characteristic characteristic roots.
(5.18)
154
Chapter 5. Boundary value value problems in statics
(b) There There are are three three cases to consider, consider, depending depending on the discriminant (b) cases to on whether whether the discriminant in the quadratic quadratic formula formula is is negative, negative, zero, or positive (that is, depending in the zero, or positive (that is, depending on whether characteristic roots roots are complex conjugate, conjugate, or on whether the the characteristic are complex or real real and and repeated, or real and write down the general repeated, or real and distinct). distinct). In In each each case, case, write down the general solution of of (5.18). (5.18). solution (c) real roots there is is (c) Show Show that that in in the the case of of real roots (either (either repeated repeated or or distinct), distinct), there no nonzero nonzero solution. no solution. that there is an infinite sequence values of A, leading (d) (d) Show Show that there is an infinite sequence of of values of A, leading to to complex complex conjugate roots, yielding aa nonzero nonzero solution (5.18). Note: conjugate roots, each each yielding solution of of (5.18). Note: The The equation that the eigenvalues eigenvalues cannot cannot be be solved solved analytically. analytically. equation that determines determines the However, a simple graphical graphical analysis is sufficient sufficient to show the existence of the sequence of eigenvalues. of the sequence of eigenvalues.
(e) Find Find the the first two eigenpairs (that is, is, those those corresponding corresponding to the two (e) first two eigenpairs (that to the two smallest smallest eigenvalues), using some numerical method to compute the first two eigenvalues two eigenvalues accurately. accurately. (f) Show that the first two eigenfunctions (f) eigenfunctions are not orthogonal. orthogonal.
6. Define Define
and define Lb : GUO, £] -+ G[O, f] by
Find conditions on a12, a21, #22 a22 for for Find necessary necessary and and sufficient sufficient conditions on the the coefficients coefficients au, an,012,021? the operator L^ to be symmetric. the operator Lb to be symmetric.
7. Compute Compute the sine series, series, on on the [0,1], for each of the the following following 7. the Fourier Fourier sine the interval interval [0, 1], for each of 29 functions: 29 (a) g(x) = x (b) hex)
=! -Ix-!I
(c) m(x)=x-x 3 (d) k(x)
= 7x -
10x 3 + 3x5
Graph the error in in approximating each function function by 10 terms terms of of the the Fourier Graph the error approximating each by 10 Fourier sine series. series. sine 29 29The Mathematica or program to to compute the necessary necessary integrals recomThe use use of of Mathematica or some some other other program compute the integrals is is recommended. mended.
5.3. Solving BVP using using Fourier Fourier series series 5.3. Solving the the BVP
155 155
8. Explain why why aa Fourier Fourier sine converges to to aa function function on defines 8. Explain sine series, series, if if it it converges on [0, [0,e], £], defines an odd function € R. R. (A function function /f : R ->• is odd odd if an odd function of of x E -+ R is if f(—x) f( -x) = = - —f(x) f(x) for all x e E R.) for all
9. is the the Fourier sine series series of of /(#) sin (37fx) (3-rrx) on on the interval [0, [0,1]? 9. What What is Fourier sine f(x) == sin the interval I]? 29 10. Repeat Exercise using the the quarter-wave quarter-wave sine 5.2.3).29 10. Repeat Exercise 77 using sine series series (see (see Section Section 5.2.S). 29 11. Repeat Exercise using the the quarter-wave quarter-wave cosine cosine series (see Exercise Exercise 3). 3).29 11. Repeat Exercise 77 using series (see
5.3 5.3
Solving BVP using using Fourier Fourier series Solving the the BVP series
We now now return the problem of solving solving the BVP We return to to the problem of the BVP
°< x < e,
cPu
-T dx 2 = f(x), u(o) = 0, u(e) = 0,
(5.19)
which can be expressed simply simply as as LDU LDU = = f. f. which can be expressed
5.3.1 5.3.1
A special case A
We begin begin with with aa special special case that is easy to to solve. solve. Suppose the form We case that is easy Suppose f/ is is of of the form N
f(x)
=L
Cn
sin (n;x),
(5.20)
n=l
where the coefficients 2 ,••• . . . ,, CN then look look for in the the where the coefficients ci, Cl, C C2, CN are are known. known. We We then for the the solution solution in form form N
u(x) =
L
bn sin (n;x).
n=l
By G C1[0, C|j[0,£], so Uu satisfies satisfies the the boundary (regardless By construction, construction, Uu E e], and and so boundary conditions conditions (regardless of the the values values of of b b2 , ..• N ). We We therefore therefore choose choose the coefficients b N so of bi, • • • ,, b&w)the coefficients &i, 622, ,.•• . . . ,, b&AT so l ,b l , 62, that the the differential is satisfied. satisfied. that differential equation equation is Since in terms terms of the eigenfunctions, the expression expression for LDu is is Since uu is is expressed expressed in of the eigenfunctions, the for LDU very simple: very simple:
The equation LDU == f then then becomes The equation LDu becomes N
2
"" Tn 7f2 . (n7fx) ~~bnsm -en=l
N
. (n7fx) = "" ~cnsm -e- . n=l
156
Chapter 5. Boundary value problems Chapter 5. Boundary value problems in in statics statics
It to find the coefficients coefficients b It is easy to find the &i, 62, • • • ?, bN &AT satisfying satisfying this equation; we need 1,b 2 , ...
Tn 2 7r 2
~bn =
or or
cn , n = 1,2, ... ,N,
C2 Cn
bn = - T 2 2' n = 1,2, ... ,N. n7r We then then have have the We the following following solution: solution: N
u (x)
=
02
'"' ~
<- Cn
Tn 2 7r 2
. 8m
(n7rX) -C- .
n=l
Example = 1, and Example 5.10. 5.10. Suppose Suppose C t— I, T T = 1, I, and
I(x) = sin (7rx) - 2 sin (27rx)
+ 5 sin (37rx).
The solution (5.19), with The solution to to (5.19), with this this right-hand right-hand side side fI,, is is
+ b2 sin (27rx) + b3 sin (37rx),
u(x) == b1 sin (7rx) where where
b1
125 b2 = - 47r2' b3 = 97r2 '
= 7r2'
u. In Figure 5.6, 5.6, we display display both the right-hand right-hand side If and the solution u.
When I/ is of of the special special form form (5.20), (5.20), the solution technique described described above above is nothing more than the spectral method explained for for matrix-vector equations in Section 3.5. 3.5. Writing Section Writing 2
2
Tn 7r An =~, '¢n(X)
= sin (n7rx) -C- ,
we simply the right-hand right-hand side terms of the (orthogonal) we simply express express the side /I in in terms of the (orthogonal) eigenfunctions, eigenfunctions,
and then divide the coefficients by the to get get the the solution: solution: and then divide the coefficients by the eigenvalues eigenvalues to
5.3.2
The general case Now suppose that I/ is the special 5.2 Now suppose that is not not of of the special form form (5.20). (5.20). We We learned learned in in Section Section 5.2 that by that we we can can approximate approximate /I by
157 157
5.3. Solving Solving the the BVP BVP using using Fourier Fourier series 5.3. 10r--------T--------~------~--------~------_,
x 0.25 0.2 0.15 0.1 0.05 0.2
0.4
0.6
O.B
x
Figure 5.6. The right-hand right-hand side side f (top) and solution solution uu (bottom) Figure 5.6. The (top) and (bottom) from from Example 5.10. 5.10. Example
where Cn
=
~
1£
f(x) sin (n;x) dx, n = 1,2,3, ....
This approximation approximation gets better and and better better (at (at least least in in the the mean-square mean-square sense) sense) as as This gets better TV oo. N -»• --+ 00. We know know how how to to solve solve Lpu LDu = fN; let us call the solution We — /AT; let us call the solution UN: UNf Cn . (n'JTx) =~ ~ Tn 2'JT2 sm -g- . 2
UN(X)
n=l
It is reasonable reasonable to to believe believe that, that, since fN gets gets closer and closer to /f as as N —> --+ oo, 00, the It is since /N closer and closer to the function UN will get get closer and closer closer to to the the true true solution solution U as N N —> --+ oo. 00. function UN will closer and u as
Example 5.11. 5.11. We consider the BVP Example BVP
d2 u
- dx 2 =
X,
0 < x < 1,
u(O) = 0, u(l) = O.
(5.21)
The exact solution solution is is easily easily computed computed (by (by integration) integration) to to be be The exact
u(x)
=
1
6x (1 - x 2)
(5.22)
158
Chapter 5. Boundary value problems in in statics statics Chapter
(see Exercise 5). We We begin begin by by computing computing the the Fourier sine coefficients coefficients of of f(x] = x. x. (see Exercise 5). Fourier sine f(x) = We have have We 1 . 2(_1)n+1 cn =2 xsm(n7fx)dx= , n=1,2, ....
1 o
n7f
The solution solution to to LDu L&U = — /N, with The fN, with N
fN(X)
= L Cn sin (mrx), n=1
is is then then UN(X) =
LN n=1
~n 2 sin (n7fx)
LN
=
n 7f
2(_1)n+1 sin (n7fx).
3 3
n 7f
n=1
We can can now compare UN UN with with the the exact exact solution u. In 5.7, we We now compare solution u. In Figure Figure 5.7, we show show the graphs of U u and anduw; on this this scale, scale, the the two two curves curves are are indistinguishable. indistinguishable. Figure the graphs of UlOi on Figure 5.8 shows the 5.8 shows the approximation approximation error error u(x) u(x) -— UlO(X). UIQ(X). The fact that UN UN approximates approximates U u so closely is is no no accident. accident. If we compute compute the the The fact that so closely If we Fourier coefficients a1, ai, a2, 02, a3, as,... of u directly, we find find that that Fourier sine sine coefficients ... of directly, we an
=2
1 1
o
1 2. -6x (1- x ) sm (n7fx) dx
=
2( _1)n+1 3 3
n 7f
'
n
= 1,2,3, ....
These These are are precisely the the coefficients coefficients of of UN! UN'- Therefore, Therefore, UN UN is is the the best approximation approximation to to uu using using the the first first N N eigenfunctions. eigenfunctions. The example is typical: By By solving LDU = /N, fN, where fN is the best The last last example is typical: solving LDU where /AT is the best apapproximation to to f/ using using the the first N eigenfunctions, eigenfunctions, we we obtain obtain the best approximation approximation proximation first N the best UN to to the solution uu of of LDU We can can demonstrate demonstrate this this directly. directly. We We assume assume UN the solution LDU = /. f. We that Uu satisfies satisfies u(O) w(0) = u(f) u(C) == 0, 0, and and we we write write al, ai,012,03,... for the (sine) that a2, a3, ... for the Fourier Fourier (sine) coefficients of of u: u: coefficients an
= ~ foi u(x) sin (n;x) dx,
n
= 1,2,3, ....
We can then compute the sine coefficients coefficients of We can then compute the Fourier Fourier sine of
_T~U dx 2
by integration by parts: by using using integration by parts: 2
d u -21£ -T-(x) sin (n7fx) dx £ 0 dx 2 £ -2T{[dU . (n7fx)]l = -(x)sm -
£
dx
£
",=0
- -n7f1ldU -(x)cos (n7fx)} dx £ 0 dx £
-2T {dU. du. n7f dx (£) sm (n7f) - dx (0) sm (0) - f
= -£-
(n7fx)} 10t· du dx (x) cos -£- dx
159 159
5.3. Solving the BVP BVP using Fourier Fourier series O.OB.------r-----r------r-------r------..
0.07
Figure Figure 5.7. 5.7. The The exact exact solution uu of of {5.21} (5.21) and and the the approximation approximation U10. UIQ.
2
-2~---~----~---~----~---~
o
0.2
0.4
0.6
O.B
1
Figure 5.8. 5.8. The The error error in in approximating approximating the the solution of {5.21} (5.21) with with U10' UIQ. Figure solution u U of
160
Chapter 5. Boundary value value problems in statics statics
2Tmr
r du i
(nnx)
.
.
= ~ 10 dx (x) cos -e- dx (smce sm (0) = sin (nn) = 0) =
(5.23)
2~~n { [u(x) cos (n;x)] :=0 + nen fo£ u(x) sin (n;x) dX} 2 2
Tn n 2 ( = -e2 -g 10
(nnx) u(x) sin -e- dx (since u(O) = u(e) = 0)
Tn 2 n 2
=~an'
We of We thus thus see see that that the the Fourier Fourier sine sine series series of
_T~u dx 2 is is
2 Since -Td —Td?u/dx = ff by assumption, the Fourier sine sine series series of of the the two two functions functions Since u/dx22 = by assumption, the Fourier must be be the the same, same, and and so so we we obtain obtain must
Tn 2 n 2
~an
or or
= en, n = 1,2,3, ... ,
e2 cn
an = - T 2 2' n= 1,2,3, .... nn This This is is exactly exactly what what we we obtained obtained by by the the reasoning reasoning presented presented earlier. earlier. Example 10 N, Example 5.12. 5.12. Consider Consider an an elastic elastic string string that, that, when when stretched stretched by by aa tension tension of of10 N, has aa length length 50 50 cm. cm. Suppose Suppose that that the the density density of of the the (stretched) (stretched) string string is is pp == 0.2 0.2 g/ cm. has g/ cm. sags under gravity, what IfIf the the string string is is fixed fixed horizontally horizontally and and sags under gravity, what shape shape does does it it assume? assume* We let u(x), the vertical displacement (in the string. We let u(x), 00 < xx < 50, 50, be be the vertical displacement (in cm) cm) of of the string. 2 To use consistent units, we convert 10 cm/?). 10 Newtons to 10 1066 dynes dynes (g (gcm/s ). Then u satisfies satisfies the the BVP BVP ~u
-T dx 2 = -pg, 0
< x < e,
= 0, u(e) = 0,
u(O)
or or 6~U
= -196, 0 < x < 50, u(O) = 0, u(50) = 0
-10 dx 2
161 161
Fourier series 5.3. Solving the BVP using Fourier series
2 (using 980 980 cm/ cm/s82 for the gravitational To solve this, we we first first compute compute the the (using for the gravitational constant). constant). To solve this, side: Fourier sine Fourier sine coefficients coefficients of of the the right-hand right-hand side:
21
= -
Cn
50
0
50
. (n1rx) -196sm - - dx = 392((-1)n -1) , n = 1,2,3, .... 50 n1r
We can then find the Fourier sine coefficients coefficients of the solution u: u: 50 2 cn 49(( -l)n - 1) an = 106 2 2 = 3 3 ' n = 1,2,3, .... n 1r 50n 1r
We Figure 5.9. 5. g. The maximum deflection of We plot the the approximate approximate solution U20 u^o in in Figure The maximum deflection of the string is quite small, small, less of aa millimeter. millimeter. This to be be expected, expected, since since the string is quite less than than half half of This is is to the the tension tension in in the the string is is much much more more than than the the total total gravitational force acting acting on it. it. 0.5 0.4 0.3 0.2
---
0.1 :::J
0 ___ -0.1 -0.2 -0.3 -0.4 -0.5 0
10
20
30
40
50
x
Figure shape of sagging string string in Example 5.12. 5.12. Figure 5.9. 5.9. The The shape of the the sagging in Example
5.3.3
Other boundary boundary conditions
We can apply apply the the Fourier Fourier series method to to BVPs boundary conditions We can series method BVPs with with boundary conditions other other than Dirichlet conditions, provided provided the the differential differential operator operator is under the than Dirichlet conditions, is symmetric symmetric under the boundary and provided provided we boundary conditions, conditions, and we know know the the eigenvalues eigenvalues and and eigenfunctions. eigenfunctions. For consider the the BVP For example, example, we we consider BVP
162
Chapter 5. Boundary statics Boundary value problems in statics
cPu -T dx2 = I(x), 0 < x < C, u(O) = 0,
(5.24)
du (C) = O. dx Written as as aa differential differential operator operator equation, equation, this takes the form L Lm Written this takes the form Lmu I, where where Lm mu = /, is defined defined as as in in Section Section 5.2.3, 5.2.3, but including the the factor factor of of T: T: is but including Lm : C;. [0, C] -+ C[O, CJ, cPu Lm u = -T dx 2 • We saw, in that section, that the eigenvalues Lm eigenvalues and eigenfunctions eigenfunctions of L m are
An =
. ((2n-1)7rx) T(2n-1)27r 2 4C2 ' ¢n(X) = sm 2f. ' n = 1,2,3, ....
Therefore, write Therefore, we we write
. ((2n - 1)7rX) I( x ) -- ~ ~cnsm 2f. ' n=l where C
. ((2n-1)7rX)
2 (
n = flo I(x) sm
2f.
dx, n
= 1,2,3, ....
We represent the solution as We represent the solution as
~
. ((2n - 1)7rX)
u(x)=~ansm
2£
'
where a2, a3, . .. are of where the coefficients coefficients aI, 01,02,03,... are to to be be determined. determined. The The Fourier Fourier series series of L Lmu -TcPu/dx22 is is then then mu = —Td?u/dx
_Td2u( )_~T(2n-1)27r2 . ((2n-1)7rX) dX2 x - ~ 4C2 an sm 2C . n=l The that these these are are the the correct correct Fourier Fourier coefficients can be be verified by aa calculation The fact fact that coefficients can verified by calculation exactly like like (5.23) (5.23) (see (see Exercise Exercise 7) 7) and is also below in Section 5.3.5 5.3.5 (see exactly and is also justified justified below in Section (see (5.29)). (5.29)). The The BVP BVP (5.24) (5.24) can can now be written as as
2 . ((2n -l)7rX) _ ~ ~ T(2n - 1)27r . ((2n -1)7rX) ~ 4C2 an sm 2C - ~ en sm 2f. . n=l From From this this equation, equation, we we see see that that
n=l
5.3. Solving the BVP using Fourier series
or or
163
4£2Cn an = T(2n _ 1)27f2 , n = 1,2,3, ....
Thus the solution to (5.24) (5.24) is
~ 4£2Cn . ((2n-1)7fX) u(x) = ~ T(2n _ 1)27f2 sm 2£ . Example circular cylindrical m and Example 5.13. 5.13. Consider Consider aa circular cylindrical bar, bar, of of length length 11 m and radius radius 1 cm, made from an stiffness 70 gj cm33 .. cm, made from an aluminum aluminum alloy alloy with with stiffness 70 CPa GPa and and density density 2.64 2.64g/cm Suppose bar (x force acting bar Suppose the the top top end end of of the the bar (x = = 0) Oj is is fixed fixed and and the the only only force acting on on the the bar is force due gravity. We will find the displacement displacement u of of the is the the force due to gravity. We will find the the bar bar and and determine determine the in length length of of the the bar just u(l)). the change change in bar (which (which is is just u(\}). The B VP describing u is BVP ~u
-k dx 2 = pg, 0 < x < 1, u(O)
= 0,
(5.25)
~~(1)=0, where kk = = 77 .• 1010, 2640 kgj m33,, and Fourier quarter-wave where 1010, P p= — 2640 kg/m and 9g = = 9.8 9.8 mj m/s~2.. The The Fourier quarter-wave of the sine coefficients coefficients of the constant constant function pg are Cn
= 2
1 1
0
.
pg sm
((2n - l)7rX) 2
4pg
dx = (2n _ 1)7f' n = 1,2,3, ... ,
and differential operator are and the eigenvalues of of the differential k(2n - 1)27f2 4 ' n = 1,2,3, .... Therefore, is Therefore, the the solution solution is
() = L oo
ux
n=1
.
16pg . ((2n - l)7fX) SIn k(2n - 1)37f3 2
= (5.9136· 10
-6
~ 1 . ((2n -l)7fX) ) ~ (2n _ 1)37f3 sm 2 .
We We then then have
. (
u(l) = 5.9136· 10
-6) ~ -6
= (5.9136·10 ~ 1.85 . 10- 7 .
1
. ((2n - 1)7f) 2
~ (2n _ 1)37f3 sm 00
(_l)n+I
) ~ (2n _ 1)37r3
164
Chapter 5.
problems in statics Boundary value problems
Thus the bar stretches stretches by only about 1.85· m, or less than two ten-thousandths 1.85 • 1010~77 m, of aa millimeter. millimeter. 01 The The Fourier Fourier series series method method is is aa fairly fairly complicated complicated technique technique for for solving solving such such simple problem problem as as (5.13). (5.13). As As we we mentioned mentioned in in Section Section 5.1.1, 5.1.1, we we could could simply aa simple simply integrate the equation twice and use use the the boundary boundary conditions integrate the differential differential equation twice and conditions to to deterdetermine the of integration. The Fourier its worth worth mine the constants constants of integration. The Fourier series series method method will will prove prove its when apply it it to to PDEs, PDEs, either either time-dependent time-dependent problems (Chapters 6 and 7) we apply problems (Chapters 6 and 7) or or when we problems multiple spatial spatial dimensions dimensions (Chapter (Chapter 8). 8). We We introduce introduce the the Fourier problems with with multiple Fourier series method method in in this simple context context so so that reader can its essence series this simple that the the reader can understand understand its essence before dealing with more complicated applications. before dealing with more complicated applications.
5.3.4 5.3.4
In homogeneous boundary boundary conditions Inhomogeneous
We now now consider consider the the following with inhomogeneous boundary conditions: We following BVP BVP with inhomogeneous boundary conditions:
d?u -T d,x2
= I(x), 0 < x < e, u(O) = a,
(5.26)
u(e) = b
(T constant). constant). To apply the Fourier series series method problem, we (T To apply the Fourier method to to this this problem, we make make aa transformation of the dependent variable variable uu to obtain aa BVP BVP with with homogeneous homogeneous transformation of the dependent to obtain boundary conditions. that the the linear boundary conditions. We We notice notice that linear function function
b-a
p(x) = a + -e-x
satisfies the the boundary (p(0) = a, b). If If we define vex) v(x) = u(x) — satisfies boundary conditions conditions (P(O) a, p(t) pee) = b). we define #(V), then then p(x), d?v d?u d2p -T dx 2 = -T d,x2 + T d,x2 = I(x) + 0 = I(x), since the second derivative of since the second derivative of aa linear linear function function is is zero. zero. Also, Also,
v(O)
= u(O) -
p(O)
=a-
a
= 0,
vee)
= u(e) -
pee)
= b - b = O.
Thus vex) (5.10), so so we we know how to to compute then have have u(x) Thus v(x) solves solves (5.10), know how compute vex). v(x). We We then u(x) = = v(x] +p(x). vex) + p(x). This technique, of transforming transforming the to one one with boundThis technique, of the problem problem to with homogeneous homogeneous boundary conditions, referred to to as as the the method method of shifting the data. ary conditions, is is referred of shifting data.
Example 5.14. Example 5.14. Consider Consider d?u - dx 2 = x, 0 < x u(O) = 1, u(l) = O.
< 1, (5.27)
5.3. Solving Solving the BVP BVP using Fourier series
165
Let p(x] p(x) = - x, x, the linear function function satisfying satisfying the the boundary conditions, and Let = 1I — the linear boundary conditions, and define define vex) p(x). Then v(x) solves {5.21}, v(x) =— u(x) u(x] -—p(x). Then v(x) (5.21), and so so 2( _l)n+l
L 00
v(x) =
3
n7r
n=l We then have have We then
2(_1)n+l
L 00
u(x) = 1- x+
sin (ml'x).
3
3 3
n7r
n=l
sin (n7rx).
Example of Example Example 5.13, and and suppose now aa Example 5.15. 5.15. Consider Consider again again the the bar bar of suppose that that now mass of of 1000 kg is from the the bottom bottom end the bar. Assume that mass is mass 1000 kg is hung hung from end of of the bar. Assume that the the mass is evenly distributed over the end of evenly distributed over the end of of the the bar, bar, resulting resulting in in a pressure of _ (1000kg) (9.8mj ?) _ 9.8· 107 Pa p 7r (10- 2 )2 m2 7r -
BVP satisfied satisfied by the displacement u is on the the end of of the the bar. bar. The The BVP the displacement is now now d2 u - k dx 2 = pg, 0
< x < 1,
u(O) = 0,
(5.28)
dU(l) = E dx k' 10 where lO lD kgj m33,, and and gg == 9.8 9.8 m/s mj?2. To solve this this problem problem using using where kk =— 7· 7 • 10 ,, pP == 2640 2640 kg/m To solve the shift the function satisfying the Fourier series series method, we we shift the data data by by finding finding aa linear linear function satisfying the the boundary boundary conditions. conditions. The The function function
q(x) =
tx
satisfies satisfies dq p q(O) = 0, dx (1) = 'k' so we we define q. It It then turns out out that, that, since since d~qjdx2 = 0, 0, that so define vv = uu -— q. then turns q/dx = that vv satisfies satisfies {5.25}, so (5.25), so vex)
~ (5.9136.10- 6 )
f(
n=l and and
. P u(x) = 'kx We We then then obtain obtain
+ (5.9136·10 u(1) ~
-6
t+
1)3 3 sin ((2n -l)7rX) 2n - 1 7r 2
~ 1 . ((2n - 1)7rX) ) ~ (2n _ 1)37r3 sm 2 .
1.8 .10- 7 ~ 1.4 .10- 3 •
This duedue to gravity {1.85· 10-•410millimeters) millimeters} This isis 1.4 1.4 millimeters. millimeters. The Thedisplacement displacement to gravity ("1.85 is of is insignificant insignificant compared compared to to the the displacement displacement due to to the the mass mass hanging hanging on on the the end end of the bar. bar. the
166 166
Chapter Boundary value value problems problems in in statics statics Chapter 5. 5. Boundary
5.3.5 5.3.5
Summary Summary
We We can now summarize the method of Fourier series series for solving aa BVP. BVP. We assume assume Ku — = f, that the BVP has been written in the form Ku /, where, as usual, the boundary boundary definition of the domain of the differential differential operator operator K. conditions form part of the definition K. For the Fourier series series method to be applicable, it it must be the case that
• K K is is symmetric; symmetric; ...,, with eigenfunc• K has has aa sequence sequence of of eigenvalues eigenvalues AI, A1, A2, A2, AS, A3, ... with corresponding corresponding eigenfunctions '¢1, Vi? 1/J2, V% 1/J3, V'S) ... • • • (which are are necessarily orthogonal); function 9 g in f] can in (7[0, e[O, £] can be be written written in in aa Fourier Fourier series series • any any function 00
g(x) =
L dn1/Jn(x) , n=l
where
It then then follows follows that that the the function function Ku K u is is given given by by the the series series 00
(Ku)(x) =
L
Anan1/Jn(X) ,
n=l
where ai, 02,03,... are the u. This follows from where al, a2, a3,'" are the Fourier Fourier coefficients coefficients of of u. This follows from the the symmetry symmetry K and the the definition of the the Fourier coefficients coefficients b b2 , b3 , ..• of KKu: u: of K &i, l , 62,63,... b = (K u, 1/Jn) n (1/Jn,'¢n) (u, K1/Jn) =-'c--'_--'--'-'':(1/Jn, 1/Jn) (u, An1/Jn) (1/Jn,1/Jn)
= An
(u, 1/Jn) (1/Jn,1/Jn) = Anan·
The BVP Ku form The BVP Ku = = f then then takes takes the the form 00
00
n=l
n=l
L An an1/Jn(X) = L cn1/Jn(x), whence obtain whence we we obtain
Cn
an = An' n = 1,2,3, ... ,
(5.29)
5.3. Solving using Fourier series 5.3. Solving the the BVP BVP using Fourier series
167
and and 00
u(x) =
2: ~: 1Pn(x). n=l
Here is is aa summary summary of the Fourier Fourier series series method method for solving BVPs. Here of the for solving BVPs. O. Pose the BVP a.s differential operator operator equation. 0. as aa differential equation.
1. Verify the operator operator is eigenvalue/eigenfunction 1. Verify that that the is symmetric, symmetric, and and find find the the eigenvalue/eigenfunction pairs. pairs. 2. the unknown unknown function as aa series series in in terms terms of the eigenfunctions. eigenfunctions. 2. Express Express the function u as of the The are the the unknowns unknowns of the problem. problem. The coefficients coefficients are of the 3. the right-hand right-hand side side /f of of the a.s aa series series in in terms terms 3. Express Express the the differential differential equation equation as of the eigenfunctions. Use the the formula formula for the projection projection of onto the the eigeneigenof the eigenfunctions. Use for the of f/ onto to compute compute the the coefficients. functions to functions coefficients. of 4. 4. Express the the left-hand side side of the differential differential equation equation in aa series series in in terms of the is done by simply simply multiplying multiplying the Fourier coefficients the eigenfunctions. eigenfunctions. This This is done by the Fourier coefficients by the the corresponding corresponding eigenvalues. eigenvalues. of of u by 5. Equate the the coefficients coefficients in the series for the left- and right-hand sides of the unknowns. differential differential equation, and and solve for the unknowns. As mentioned mentioned above, above, the Fourier series is analogous to what what we we called As the Fourier series method method is analogous to called (in 3.5) the method for for solving solving Ax Ax = This method method is is only only (in Section Section 3.5) the spectral spectral method = b. b. This applicable if the is symmetric, symmetric, so that the eigenvectors are are orthogonal. the eigenvectors orthogonal. In In applicable if the matrix matrix A is so that the same way, the the method described above the first first step if the the eigenfunctions eigenfunctions the same way, method described above fails fails at at the step if are not orthogonal, orthogonal, that is, if if the differential differential operator operator is is not symmetric. that one one rarely rarely uses uses the spectral method An more pertinent pertinent observation observation is An even even more is that the spectral method to solve solve Ax Ax = b, for for the simple reason reason that that computing computing the the eigenpairs eigenpairs of of A A is is more more to = 6, the simple costly, in most cases, cases, than solving Ax Ax = b directly directly by by other means. It It is is only only in in costly, in most than solving = b other means. way, it special cases that one one knows knows the eigenpairs eigenpairs of a matrix. In the same same way, it is only only for very special differential differential operators that the the eigenvalues and eigenfunctions can be be for very special operators that eigenvalues and eigenfunctions can found. Therefore, Therefore, the the method method of of Fourier Fourier series, which works works very very well well when when it it can found. series, which can be used, is applicable applicable to only only aa small set set of of problems. On most problems, such as the BVP BVP the
d (k(x) dx dU) = f(x),
- dx
0
< x < £,
u(O) = 0, u(£) = 0
when it is eigenfunctions than when k(x) is is nonconstant, nonconstant, it is more more work work to to find find the the eigenfunctions than to to just just solve the the problem problem using using aa different method. For For this this reason, reason, we we discuss discuss aa more more solve different method. finite element method, beginning beginning in in Section 5.4. broadly applicable method, the broadly applicable method, the finite element method, Section 5.4.
168
Chapter 5. Chapter
Boundary Boundary value problems in statics statics
Example conditions) We We will the folfolExample 5.16. 5.16. (Homogeneous (Homogeneous Dirichlet Dirichlet conditions) will solve solve the BVP, applying applying the the procedure procedure given given above: above: lowing BVP, lowing d2U _ - 3 dx 2 -
3 X
,
0< x
< 2,
u(O) = 0, u(2) = O.
(5.30)
1. have already already seen seen that that the the eigenfunctions of the the negative negative second second derivative 1. We We have eigenfunctions of derivative operator, the interval interval [0,2] and to Dirichlet and subject subject to Dirichlet conditions, conditions, are are operator, on on the
sin (n;x), n = 1,2,3, .... The effect of operator by by 3 is to multiply multiply the the eigenvalues The effect of multiplying multiplying the the operator is to eigenvalues by by 3, 3, so the eigenvalues are are so the eigenvalues 3n 2 7f2
An = -4-' n = 1,2,3, .... 2. write the unknown solution solution uu as as 2. We We write the unknown 00
u(x) =
L an sin (n;x). n=l
The problem now reduces computing 0,1,0,3,0,3, — The problem now reduces to to computing al, a2, a3, .... 3. Fourier sine sine coefficients of f(x] f(x) = are 3. The The Fourier coefficients of = xx33 on on the the interval interval [0,21 [0,2] are
and so
2 4. The Fourier sine sine coefficients coefficients of of —3d?u/dx -3~u/dx2 are are just just 4The Fourier
Thus Thus
5. Finally, Finally, the the differential differential equation equation can can be be written as 5. written as
5.3. 5.3. Solving Solving the the BVP BVP using using Fourier Fourier series series
169 169
so we have
This yields
and and so so
is the solution solution to (5.30). (5.30)
Example 5.17. 5.17. (Inhomogeneous Dirichlet conditions) Next Next we we solve a BVP with inhomogeneous inhomoqeneous boundary conditions: conditions:
d2 u
-= I-x 0< x dx 2 '
1
< 2'
u(O) = 3, u
(5.31)
(~) = -1.
We We must must first first shift shift the the data to to transform transform the the problem to to one one with with homogeneous homogeneous boundary function p(x] p(x) = boundary conditions. The function — 3 -— 8x Sx satisfies satisfies the boundary conditions, conditions, so we define define v = VP = u -— p and solve the BBVP
d2 v --= l - x 0< x ' dx 2
v(O) v
1
< 2'
= 0,
(~) = O.
1. 1. The eigenfunctions eigenfunctions are now
sin (2mfx) , n = 1,2,3, ... ,
and the eigenvalues are
2. 2. We write the solution solution v in the form form 00
v(x) =
L an sin (2mfx). n=l
170
Chapter Chapter 5. Boundary value problems in statics
3. The Fourier sine coefficients coefficients of ff(x) (x) = = 11 -— x are
1
1/2
4
o
(1 - x) sin (2mrx) dx
and so
L00
1- x =
=
2-(-1)n nn
, n = 1,2,3, ... ,
2 - (_1)n
n=l
nn
sin (2nnx).
2 4. _~V/dX2 are 4- The Fourier sine coefficients coefficients of of —d?v/dx
and so
~~ (x) = f: 4n n
2 2
-
a n sin(2nnx).
n=l
5. We thus obtain
L00 4n n a 2
2
n
n=l
002_(_1)n sin(2nnx) = L sin (2nnx) , n=l nn
which implies that 22
4n n an =
or or an =
2-(-1)n , n = 1,2,3, ... , nn
2-(-1)n 4 3 3 ,n = 1,2,3, .... nn
The solution is then then The solution vv is v(x) =
002_(_1)n. 4 3 3 sm (2nnx).
L n=l
n n
This yields 002_(_1)n u(x)=3-8x+L 4 33 sin (2nnx). n=l n n
Exercises In In Exercises 1-4, 1-4, solve solve the the BVPs using the method method of of Fourier series, series, shifting shifting the the data if necessary. If possible,3o produce aa graph by plotting if necessary. possible,30 produce graph of of the the computed computed solution solution by plotting aa partial (The partial Fourier Fourier series series with with enough enough terms terms to to give give aa qualitatively qualitatively correct correct graph. graph. (The number of terms can can be trial and plot no number of terms be determined determined by by trial and error; error; if if the the plot no longer longer changes, changes, qualitatively, when more more terms are are included, included, it it can can be assumed assumed that the the partial partial series contains contains enough enough terms.) terms.) series 30 30That if you have the the necessary That is, is, if you have necessary technology. technology.
5.3.
1.
Solving Solving the BVP BVP using using Fourier series
171
(a) - ~:~ = 1, u(O) = u(l) = 0 (b) - ~:~ = 1, u(O) = 0, u(l) = 1
2.
(a) -~ = eX, u(O) = u(l) = 0 (b) -~:~ = eX, u(O) = u(l) = 1
3. The results will be be useful useful for for these these problems: 3. The results of of Exercises Exercises 5.2.2 5.2.2 and and 5.2.3 5.2.3 will problems: (a) -
2
d u = X dx2'
u(O) = 0 ,
du dx
(1) = 0
(b) -~ = x, ~~(O) = 0, u(l) = 1
(c) -~+2u=l, u(O)=u(l)=O (d) -
2
d u dX2
+u - 0,
u(O) - 0, u(l) -- 1
The results of Exercises 5.2.2 4. The 5.2.2 and 5.2.3 will be useful for these problems: (a) -~:~ = x 2 , u(O) = u(l) = 0 (b) -~ + 2u = ~ - x, u(O) = u(2) = 0 (c) -~:~ =x+sin(1fx), u(O) = ~~(1) =0 (d) -~:~ = f(x), ~~(O) = u(l) = 0, where
f(x)
=
{I, 0<< xx << ~, 0,
~
1.
5. 5. Solve Solve (5.21) (5.21) to to get get (5.22). (5.22). 6. satisfies 6. Suppose Suppose uu satisfies du u(O) = 0, dx (f) = 0, and the Fourier Fourier sine is and the sine series series of of uu is
Show not possible possible to to represent represent the the Fourier Fourier sine dx 22 Show that that it it is is not sine coefficients coefficients of of -~u/ —d?u/dx . .. This shows that it is not possible to use the "wrong" in terms of b 61,6 &3, 1 , b22 ,, b 3 , •— eigenfunctions (that corresponding to eigenfunctions (that is, is, eigenfunctions eigenfunctions corresponding to different different boundary boundary conditions) to to solve solve aa BVP. 7. Suppose u satisfies du u(O) = 0, dx (£) = 0,
and has Fourier Fourier quarter-wave quarter-wave sine sine series
~
. ((2n -2£1)1fX) .
u(x) = ~ansm
172
Chapter 5. Boundary value problems in in statics Chapter 5. Boundary value problems statics Show the Fourier Fourier series -Td 2 ufdx22 is is Show that that the series of of —Td?u/dx
2 _Td u( ) _ ~ T(2n - 1)27f2 . ((2n -l)7fX) dx2 x - L..J 4£2 an sm 2f . n=l
2 (Hint: The The computation the Fourier coefficients of of -Td —Td?u/dx similar to to (Hint: computation of ofthe Fourier coefficients ufdx22 is is similar (5.23)). (5.23)).
radius 11 cm. that the 8. 8. Consider Consider an an aluminum aluminum bar bar of of length length 11 m m and and radius cm. Suppose Suppose that the side the bar bar is is perfectly perfectly insulated, the bar bar are are placed placed in side of of the insulated, the the ends ends of of the in ice ice baths, baths, throughout the interior of bar at at aa constant and heat and heat energy energy is is added added throughout the interior of the the bar constant rate of 0.001 0.001 W/cm thermal conductivity conductivity of aluminum alloy alloy is rate of W fcm 33.. The The thermal of the the aluminum is 1.5W/(cmK). Find the steady-state steady-state temperature temperature of the bar. bar. Use 1.5W/(cmK). Find and and graph graph the of the Use the Fourier Fourier series the series method. method.
9. Repeat exercise, assuming assuming that that the right end end of the bar 9. Repeat the the previous previous exercise, the right of the bar is is perperthe left left is bath. fectly insulated insulated and fectly and the is placed placed in in an an ice ice bath. 10. of Example the right the string is 10. Consider Consider the the string string of Example 5.12. 5.12. Suppose Suppose that that the right end end of of the string is free to move vertically (along pole, for pressure of of free to move vertically (along aa frictionless frictionless pole, for example), example), and and aa pressure 2200 per centimeter, upward direction, applied along 2200 dynes dynes per centimeter, in in the the upward direction, is is applied along the the string string unit of dyne equals equals one (recall (recall that that aa dyne dyne is is aa unit of force-one force—one dyne one gram-centimeter gram-centimeter per square second). second). What What is is the the equilibrium equilibrium displacement the string? string? How per square displacement of of the How does the answer answer change change if if the the force due to gravity is is taken into account? account? force due to gravity taken into does the
5.4 5.4
Finite element element methods methods for BVPs Finite for BVPs
As we we mentioned mentioned near near the end of of the the last last section, section, the primary utility of the As the end the primary utility of the Fourier Fourier is for problems with coefficients, in in which which case case the the eigenpairs series method is series method for problems with constant constant coefficients, eigenpairs can be found problems in in two order for can be found explicitly. explicitly. (For (For problems two or or three three dimensions, dimensions, in in order for the the eigenfunctions to explicitly computable, computable, it it is is also also necessary the domain domain eigenfunctions to be be explicitly necessary that that the on which the to be be solved be geometrically geometrically simple. We discuss on which the equation equation is is to solved be simple. We discuss this this further in in Chapter 8.) For For problems with nonconstant nonconstant coefficients, coefficients, it to further Chapter 8.) problems with it is is possible possible to perform analysis that the the Fourier method applies applies in perform analysis to to show show that Fourier series series method in principle-the principle—the eigenfunctions they are orthogonal, and consult eigenfunctions exist, exist, they are orthogonal, and so so forth. forth. The The reader reader can can consult Chapter of Haberman Haberman [22] [22]for foran anelementary elementaryintroduction introductiontotothis thiskind kindofofanalysis, analysis, Chapter 55 of which we we also also discuss further in in Section Section 9.7. 9.7. However, However, without explicit formulas formulas for discuss further without explicit for which the it is easy to the Fourier Fourier series the eigenfunctions, eigenfunctions, it is not not easy to apply apply the series method. method. to the the BVP These remarks apply, apply, for These remarks for example, example, to BVP -
d~ (k(X)~:)
= f(x), 0
< x < £,
u(O) = 0, u(£) = 0
(5.32)
(where kk is is aa positive The operator operator K C?JO,£] -t —> C[O,£] C[Q,l] defined defined by by (where positive function). function). The K:: Cb[O,£]
d(
dU)
Ku = - - k(x)dx dx
(5.33)
5.4.
Finite element methods for BVPs BVPs
173 173
is symmetric symmetric (see Exercise 1), 1), and and there there exists exists an an orthogonal orthogonal sequence sequence of of eigenfunceigenfuncis (see Exercise tions. However, when when the the coefficient coefficient k(x) k(x] is constant, there is no simple way is not not aa constant, there is no simple way tions. However, to eigenfunctions. Indeed, Indeed, computing computing the requires much to find find these these eigenfunctions. the eigenfunctions eigenfunctions requires much more work work than more than solving solving the the original original BVP. BVP. the limitations approach, we now introduce Because of the Because of limitations of of the the Fourier Fourier series series approach, we now introduce the the finite element the most most powerful powerful methods finite element method, method, one one of of the methods for for approximating approximating solutions solutions to PDEs. The The finite element method variable coefficients coefficients and, and, in to PDEs. finite element method can can handle handle both both variable in multiple spatial dimensions, dimensions, irregular irregular geometries. geometries. We We will will still still restrict restrict ourselves ourselves to to multiple spatial symmetric although it it is is possible possible to to apply finite element symmetric operators, operators, although apply the the finite element method method to to nonsymmetric nonsymmetric problems. problems. The finite element based on on three The finite element method method is is based three ideas: ideas: 1. BVP is in its weak or form, which 1. The The BVP is rewritten rewritten in its weak or variational variational form, which expresses expresses the the
problem infinitely many scalar equations. equations. In In this this form, form, the problem as as infinitely many scalar the boundary boundary conconditions are are implicit implicit in in the the definition definition of of the the underlying underlying vector vector space. space. ditions
2. The The Galerkin Galerkin method is applied "solve the the equation equation on on aa finite-dimensional 2. method is applied to to "solve finite-dimensional subspace." This in an an ordinary ordinary linear linear system system (matrix-vector (matrix-vector equation) equation) subspace." This results results in that be solved. that must must be solved. piecewise polynomials polynomials is 3. basis of 3. A A basis of piecewise is chosen chosen for for the the finite-dimensional finite-dimensional subspace subspace the matrix matrix of linear system sparse (that has mostly mostly zero zero so so that that the of the the linear system is is sparse (that is, is, has entries). entries).
We describe describe each each of of these the following following sections, sections, using using the BVP (5.32) (5.32) as as the BVP We these ideas ideas in in the our model We always always assume assume that the coefficient coefficient k(x) k(x] is is positive, positive, since that the since our model problem. problem. We it represents represents aa positive positive physical (stiffness or or thermal conductivity, for it physical parameter parameter (stiffness thermal conductivity, for example). example).
5.4.1 5.4.1
The principle of virtual work and and the weak form of a BVP BVP
When potential energy due to to internal When an an elastic elastic material material is is deformed, deformed, it it stores stores potential energy due internal elastic forces. It not obvious first principles principles how how to to measure elastic forces. It is is not obvious from from first measure (quantitatively) (quantitatively) this elastic elastic potential potential energy; we can the correct from the this energy; however, however, we can deduce deduce the correct definition definition from the equations of of motion and the of conservation conservation of of energy. equations motion and the principle principle of energy. Suppose bar, with fixed, is is in in motion, motion, and Suppose an an elastic elastic bar, with its its ends ends fixed, and its its displacement displacement function satisfies the wave equation: equation: function u(x, t) t) satisfies the homogeneous homogeneous wave 2
a u -Aax a ( k(x)ax au) =0, O<x
= 0, t > 0,
t>o, (5.34)
= 0, t > 0.
We now following calculation calculation (a trick—multiply both sides of the wave We now perform perform the the following (a trick-multiply both sides of the wave equation by du/dt and integrate): and integrate): equation by au/at
174 174
Chapter 5. Chapter
Boundary value problems problems in statics
Applying integration by parts, parts, Applying integration by
-
1£
2 £ -a ( k(x)au) -dx au au a u = - [auau]£ k(x)-- + k(x)---dx o ax ax at ax at 0 0 ax axat 2 r£ au a u ( . au au = 10 k(x) ax atax dx smce at (0, t) = at (l, t)
1
=0
)
Also, Also,
we obtain Therefore, we Therefore, obtain
1
[1
£ Ap(x)a - (au)2] dx + at 2 at
o
1£ Ak(x)-ata [1- (au)2] dx =0. ax 0
2
(5.35)
Applying 2.1 to Applying Theorem Theorem 2.1 to (5.35) (5.35) yields yields
a [ 1 r£ (a ) at "210 Ap(x) a~
2
1 rl (a ) dx + "210 Ak(x) a~
2
]
dx = 0, t > o.
(5.36)
This equation implies that the sum of the to This equation implies that the sum of the two two integrals integrals is is constant constant with with respect respect to time, that is, is, that that time, that 1 r£ (a ) 2 1 r£ (a ) 2 "210 Ap(x) a~ dx + "210 Ak(x) a: dx
is conserved. conserved. Obviously, Obviously, the the first integral represents energy (one-half is first integral represents the the kinetic kinetic energy (one-half mass times times velocity velocity squared), squared), and and we we define the second integral to to be be the the elastic mass define the second integral elastic potential energy. energy. The The elastic potential energy the internal arising from potential elastic potential energy is is the internal energy energy arising from the strain strain au/ax. du/dx. the We equilibrium, that to the satisfying the the We now now return return to to the the bar bar in in equilibrium, that is, is, to the bar bar satisfying BVP (5.32). (5.32). We bar: BVP We have have an an expression expression for for the the elastic elastic potential potential energy energy of of the the bar: 1 r£ (d)2 "210 Ak(x) d~ (x)
dx.
The bar possesses possesses another another form form of potential energy, energy, due due to to the the external The bar of potential external force force f/ acting upon it. The proper proper name name of of this this energy the nature nature of of the acting upon it. The energy depends depends on on the the force; force; if it due to for example, gravitational if it were were the the force force due to gravity, gravity, for example, it it would would be be called called the the gravitational
.
5.4. Finite element methods for BVPs
175 175
potential will call the external potential energy. energy potential energy. energy. We We will call this this energy energy the external potential energy. This This energy potential energy: is way from is different different in in aa fundamental fundamental way from the the elastic elastic potential energy: it it depends depends on on the the absolute the bar, not on position of bar to to other absolute position position of of the bar, not on the the relative relative position of parts parts of of the the bar other parts. For this we do not know potential energy in its parts. For this reason, reason, we do not know the the external external potential energy of of the the bar bar in its zero elastic elastic potential potential "reference" (zero displacement) displacement) position position (whereas "reference" (zero (whereas the the bar bar has has zero energy in in the We define define Co £Q to to be external potential potential energy energy of the reference reference position). position). We be the the external of energy the reference position, between the bar bar in in the the reference position, and and we we now now compute compute the the change change of of energy energy between the bar its reference reference position position and and the bar in in its equilibrium position. The change change the bar in in its the bar its equilibrium position. The in to the to the work done by f/ when when in potential potential energy energy due due to the external external force force f/ is is equal equal to the work done by the its reference reference position the bar bar is is deformed deformed from from its position to to its its equilibrium equilibrium position. position. The work work done done by by f/ acting acting on on the the part of the the bar originally between between xx and The part P of bar originally and approximately xx + Ax ~x is is approximately volume ,.,.-"-..
A~x
f(x)
u(x)
~
~
force per unit volume
distance
u(x) is the displacement bar originally (recall (recall that that u(x) is the displacement of of the the cross-section cross-section of of the the bar originally at at x-thus u(x) is is the the distance distance traveled traveled by by that that cross-section). cross-section). To To add add up up the the work work x—thus u(x) the external external force all little parts of the bar, we integrate: due to due to the force acting acting on on all little parts of the bar, we integrate:
l'
(5.37)
Af(x)u(x) dx.
We now have an total potential We now have an expression expression for for the the total potential energy energy of of the the bar bar in in its its equilibrium position: position: equilibrium
Cpot(u)
=~
l'
Ak(x)
(:~ (X))
2
dx
-1'
Af(x)u(x) dx + co.
If f/ and and uu are are both both positive, for instance, instance, then the bar less potential energy in If positive, for then the bar has has less potential energy in its position (think second its equilibrium equilibrium position (think of of gravity). gravity). This This accounts accounts for for the the sign sign on on the the second integral. integral. Next we wish to to prove prove the the following following fact: fact: the the equilibrium equilibrium displacement Next we wish displacement uu of of the displacement with potential energy. w ^ ¥- uu is the bar bar is is the the displacement with the the least least potential energy. That That is, is, if if w is any any other other displacement displacement satisfying satisfying the the Dirichlet Dirichlet conditions, conditions, then then
Cpot(w) > Cpot(u). We can can write write any any displacement displacement w as as w = u + v (just (just take w), and and We take v = w -— u), the conditions conditions u(O) u(Q) = — 0, 0, w(O) w;(0) = = 00 imply imply that that v(O) v(0) =— 0, 0, and and similarly similarly at at x = = £.t. the Thus vv must must belong to c1 C|)[0, ^]. For brevity, we we shall shall write = C1 Cf^O,^], the set set of Thus belong to [0, fl. For brevity, write V = [0, f], the of allowable wish to allowable displacements displacements v. v. Therefore, Therefore, we we wish to prove prove that that
Cpot(u + v) > Cpot(u) for all v E V. We We have have
Cpot(u
+ v)
=
1 r' (dU dV) 2 dx - 10( "210 k(x) dx (x) + dx (x)
f(x) (u(x)
+ vex)) dx + Co
176
Chapter 5. Boundary value problems problems in statics
III
=2
0
dv (dV )2) dx k(x) ((dU)2 dx(X) +2 du dx (X)dx(X)+ dx(X)
- fot f(x)u(x) dx - fot f(x)v(x) dx + [0
It ( +"2 It
="21
0
1
k(x)
0
(dU) 2 dx (x) dx dv (x) ) dx
k(x)
2
+
It lf
du dv k(x) dx (x) dx (x) dx
0
dx -
0
f(x)u(x) dx -
lf 0
f(x)v(x) dx
+ [0;
that is, that is, [pot(U + v) = [pot(u)
11t
+"2
0
+ fof k(x) ~~ (x) ~: (x) dx k(x)
fol f(x)v(x) dx
(d~(x) )2 dx.
(5.38)
Using integration by by parts, have Using integration parts, we we have
fof k(x) ~~ (x) ~: (x) dx - fot f(x)v(x) dx =
[k(X)~~(X)V(X)]: - fof (d~ (k(x)~~(x)))V(X)dX- fof f(x)v(x) dx
=-
fot (d~ (k(X)~~(X)) + f(X)) v(x)dx.
(The boundary term disappears disappears in in the the above above calculation calculation since since v(O) v(0] = v(£) v(t] = 0.) (The boundary term 0.) The equilibrium equilibrium displacement displacement Uu satisfies satisfies The
d ( k(x) dx du (x) ) dx
+ f(x)
= 0,
the differential differential equation equation from from (5.32), (5.32), as as we derived in in Section Section 2.2. 2.2. Therefore, Therefore, the the the we derived integral above vanishes, and we have integral above vanishes, and we have
[pot(U
+ v)
= [pot(U)
11£
+2
0
k(x) (dv)2 dx (x) dx.
If not identically zero, then / dx (only If v is is not identically zero, then neither neither is is dv dv/dx (only
v(O) v(£) == 0). v(Q) = =v(l) G). Since Since
11t
2
0
k(x)
(dd:(x) )2 dx>O
whenever ^ 0 (the (the integral integral of of aa nonzero, function is is positive), whenever vv:/;O nonzero, nonnegative nonnegative function positive), we we have obtained have obtained the the desired desired result: result:
5.4. Finite element element methods methods for BVPs
177 177
This result gives us aa different different understanding understanding of of the the equilibrium of the This result gives us equilibrium state state of the bar—the state of energy. This This complements complements our bar-the bar bar assumes assumes the the state of minimal minimal potential potential energy. our earlier all forces forces balance. balance. earlier understanding understanding that that the the bar bar assumes assumes aa state state in in which which all The potential function mapping space of The potential energy energy £ [pot defines aa function mapping the the space of physically physically poi defines meaningful R, and the equilibrium displacement u is is its its miniminimeaningful displacements displacements into into R, and the equilibrium displacement mizer. A A standard result from from calculus is that that the the derivative derivative of real-valued function function mizer. standard result calculus is of aa real-valued must be zero zero at at aa minimizer. minimizer. Above Above we we computed must be computed
+ v)
[pot (u
=
+
[pot (u)
a term linear in v
+
a term quadratic in v
(see (5.38)). The linear term term in in this this expression expression must must be be D£ D[pot(u)v, the directional (see (5.38)). The linear directional pot(u)v, the [pot at in the direction of The minimality of the potential energy derivative of £ derivative of at u in the direction of v. The minimality of the potential energy pot then implies implies that that D£ D[pot(u) or, equivalently, then = 0, 0, or, equivalently, poi(u} =
D[pot(u)v = This yields This yields
i
f
du
dv
o k(x) dx (x) dx (x) dx -
°
if 0
for all v E V.
f(x)v(x) dx =
°
for all v E V
(as we saw which is is called called the the principle principle of of virtual virtual work. work. (The (as we saw above), above), which (The term term linear linear in u dominates the work displacement v is arbitrarily in dominates the work £ [pot v) — - £[pot when the the displacement is arbitrarily pot (u + v} pot (u) when small and is called the virtual work. The principle principle says the virtual small ("virtual"), ("virtual"), and is called the virtual work. The says that that the virtual work is is zero zero when when the the bar bar is is at equilibrium.) work at equilibrium.) The principle principle of virtual work work is is also also called called the the weak weak form form of of the the BVP; BVP; it is The of virtual it is the form form that that the BVP would would take take if basic principle principle were were the the minimality minimality of of the the BVP if our our basic potential energy rather at equilibrium. equilibrium. By By contrast, potential energy rather than than the the balance balance of of forces forces at contrast, we we the original original BVP, BVP, will call call the will
d ( k(x) du - dx dx (x) ) =f(x),O<x
(5.39)
u(f) = 0, the can show show directly are equivalent, the strong strong form. form. We We can directly that that the the two two forms forms of of the the BVP BVP are equivalent, in the for f/ E6 C[O, C[0, I], form if if and and only in the sense sense that, that, for £], uu satisfies satisfies the the strong strong form only if if it it satisfies satisfies the form. the weak weak form.
5.4.2 5.4.2
The equivalence of of the strong and and weak forms of of the BVP The equivalence the strong weak forms the BVP
The precise precise statement of the the weak weak form form of of the the BVP BVP is: is: The statement of find u E V such that
i
0
f
du dv k(x) dx (x) dx (x) dx =
if 0
f(x)v(x) dx for all v E V. (5.40~
We have have already already seen the proof proof of of the the following following fact: satisfies (5.39), (5.39), the We seen the fact: If If uu satisfies the strong form (5.40), the form. This This was strong form of of the the BVP, BVP, then then uu also also satisfies satisfies (5.40), the weak weak form. was proved proved
178
Chapter 5.
Boundary value problems in statics statics Boundary
deriving just below (5.38), using integration integration by parts. Indeed, the simplest way of deriving differential equation the weak form is to start with the differential
d ( k(x) dx du (x) ) = f(x), 0 < x < f, - dx function" vv e E V to get multiply by an arbitrary arbitrary "test "test function"
d ( k(x) dx du (x) ) v(x) = f(x)v(x), 0 < x < f, - dx and f: and then integrate integrate both sides from 00 to i:
- 10{l dxd
(
dU) (l k(x) dx (x) v(x) dx = 10 f(x)v(x) dx.
Integrating Integrating by parts on the left left and applying the boundary conditions yield the weak form. weak form. satisfies (5.40), (5.40), the the weak form of the BVP. Then, by definition Now suppose suppose Uu satisfies definition = u(i] u(f) = 0. O. We We now now use integration integration by parts to show of V, u(O) w(0) = show that u satisfies the the strong We have have strong form form of of the the differential differential equation. equation. We
{i
10
du dv (l k(x) dx (x) dx (x) dx - 10 f(x)v(x) dx = 0 for all v E V dU
=> [ k(x)dx(x)v(x)
]i
0 -
10(- dxd
(
du ) k(x) dx (x) v(x)dx
- foi f(x)v(x) dx = 0 for all v E V =>-
foi {d~ (k(X)~~(x)) +f(X)}V(X)dx=oforallvEv.
The boundary terms vanish because of the boundary conditions v(O) = v(f) O. v(Qi) = v(i] = 0. We have have We
fol {d~ (k(X) ~~ (X)) + f(X)} v(x) dx = 0 for all v E V.
(5.41)
This condition can can only only be be true This condition true ifif
d ( k(x) dx du (x) ) dx
+ f(x) = 0
(5.42)
on the interval [0, [0, fl. i]. The The proof proof isis simple, simple, given given the the following following fact: fact: If If 00 << ec << dd << f,I, then function V[c,dj V[c^] £ that V[c,dj(X) v^c^(x) > 0 for all (c,d) and then there there exists exists aa function E V such such that 0 for all x € E (e,d) and v V[c,dj(x) = 00 for for all all xx in in [0,c] [O,e] or or [d,i\. [d,f]. (Exercise (Exercise 33 asks asks the the reader reader to to construct construct such such [c,d](x) = function V[c,dj.) v\c &.} If aa function If d ( k(x)dx(x) du ) +f(x) dx
5.4. Finite element element methods methods for BVPs
179
were positive positive on on aa small interval [c, [c, d]d]CC [0, £], then were small interval [0,£], then
would be positive on [c, and zero zero on on the of [0, [0, £], ^], and and hence would be positive on [c, d\ d] and the rest rest of hence
would be positive. Since this is zero zero for for all £ V, this shows that positive. Since this integral integral is all v E this shows that would be d ( k(x)dx(x) du ) +f(x) dx cannot be positive on on any the same cannot be be cannot be positive any small small interval. interval. By By the same reasoning, reasoning, it it cannot negative on any small that (5.42) must hold. negative on any small interval. interval. It It follows follows that (5.42) must hold. to solve to solve solve the the weak weak form Thus, Thus, to solve (5.39), (5.39), it it suffices suffices to weak form form (5.40). (5.40). The The weak form consists of infinitely many equations, equations, since there are infinitely many many "test functions" consists of infinitely many since there are infinitely "test functions" v E £ V. The The original original PDE PDE also also implies implies an an infinite infinite number number of of equations, equations, since since the the equation must hold hold for for each of the the infinitely many values equation must each of infinitely many values of of x E £ [0, [0, fl. i\. In In either either case, case, we find the the solution directly, since since any any practical practical computational procedure we cannot cannot find solution directly, computational procedure must to aa finite finite number number of of steps. steps. The admits the the use of the must reduce reduce to The weak weak form form admits use of the Galerkin method, which which reduces reduces the the infinitely infinitely many many equations to aa finite finite collection Galerkin method, equations to collection to the the BVP. of equations equations whose whose solution solution provides provides an of an approximate approximate solution solution to BVP.
Exercises 1. the differential 1. Show Show that that the differential operator operator
K : c1 [0, £] -+ C[O, f] defined by defined by
Ku
d (k(x)dU)
= --
dx
dx
is symmetric: symmetric: is
(Ku,v)
=
(u,Kv)
for all
U,V
E
C1[0,£].
°
2. Suppose Suppose that for all all xx £E [0, [0,1]. Show that if the operator K, defined in 2. that k(x) k(x) > 0 for fl. Show that if the operator K, defined in the previous previous exercise, exercise, has an eigenvalue, eigenvalue, then then it it must must be be positive. positive. (Hint: Let the has an (Hint: Let u, u) AA be be an an eigenvalue eigenvalue and and u aa corresponding corresponding eigenfunction. eigenfunction. Compute Compute (K (Ku,u) two ways, ways, once once using using the the fact that u is an an eigenfunction eigenfunction and using two fact that u is and again again using parts once.) once.) integration by integration by parts
°
3. d < t£ hold. function V[c,d] V[CJ(Q that hold. Find Find aa function that is is twice-continuously twice-continuously 3. Let Let 0 < cc < d differentiable on on [0, differentiate [0, i]£] and and satisfies satisfies the the conditions conditions that that V[c,dJ(X)
°
> for all x
E
(c, d)
180
Chapter Chapter 5. 5. Boundary Boundary value value problems problems in statics statics
and and
°
for all x E [0, c] U [d, fl.
V[c,d](X) =
Hint: Let Let p(x) p(x] be be aa polynomial such that that Hint: polynomial such p(c)
d2 p
dp
= dx (c) = dx 2 (c) = 0,
p(d)
dp
~p
= dx (d) = dx 2 (d) = o.
Then Then define define V[c,d]
() {P(X), c<x
4. Show Show that that both both of of the the integrals integrals 4.
r
f
1 "210 Ap(x)
(aa~ ) dX'"2 10r Ak(x) (aa: ) dx 2
1
f
2
have units have units of of energy energy (force (force times times distance). distance). 5. the damped 5. Consider Consider the damped wave wave equation equation 2
a u Ap(x) at2
+ c au at
a ( au) - A ax k(x) ax = 0, 0 < x < f, t> 0, u(O, t) = 0, t > 0, u(f,t) = 0, t > 0,
where cc > 00 is modifying the beginning on where is aa constant. constant. By By modifying the calculation calculation beginning on page page 173, that the the total total energy energy) 173, show show that energy (kinetic (kinetic energy energy plus plus elastic elastic potential potential energy) is decreasing decreasing with with time. time. is
5.5 5.5
The method The Galerkin Galerkin method
As we stated stated earlier, earlier, the Galerkin method method defines defines an an approximate approximate solution solution to the As we the Galerkin to the weak form form of of the the BVP BVP by by restricting restricting the the problem problem to to aa finite-dimensional subspace. weak finite-dimensional subspace. This the infinitely This has has the the effect effect of of reducing reducing the infinitely many many equations equations contained contained in in (5.40) (5.40) to to aa finite finite system system of of equations. equations. To To describe describe the the Galerkin Galerkin method, method, it it is is convenient convenient to to introduce the the following following notation. notation. We We define define the the symmetric symmetric bilinear bilinear form a(-, •) by by introduce form a(·,·)
r k(x)dx(x)dx(x)dx. du dv f
a(u,v)
= 10
The bilinear because, because, holding holding one fixed, the the function is The function function is is called called bilinear one argument argument fixed, function is linear linear in in the the other: other:
a(alul +a2U2,V) a(u, al Vl + a2v2)
= ala(Ul,V) +a2a(u2,V) = ala(u, Vl) + a2a(u, V2)
for all al,a2 E R, Ul,U2,V, for all al, a2 E R, U, Vl, V2.
5.5. 5.5. The The Galerkin Galerkin method method
181 181
The bilinear bilinear form symmetric because The form is is called called symmetric because
a(u,v) = a(v,u) for all u,V. Bilinearity and symmetry two of the properties properties of Bilinearity and symmetry are are two of the of an an inner inner product product (see (see Section Section 3.4). third property property also holds if we restrict 3.4). The The third also holds if we restrict a(·,·) a(-, •) to to vectors vectors in in V = = Cb[O,f]: Cf^O,^]:
a( u, u)
~
0 for all u E V
and and
a(u,u) > 0 for all u E V,U
i- O.
From the definition From the definition of of a(·, a(-, .), •), we we have have
a(u,u) =
1£
k(x)
(~~(X)r dx.
Since the integrand nonnegative, we we clearly a( u, u) > ~ 0 all u. Moreover, Since the integrand is is nonnegative, clearly have have a(u, 0 for for all u. Moreover, the nonnegative function can only be zero function (the the integral integral of of aa nonnegative function can only be zero when when the the function (the integrand) the zero function. Thus, integrand) is is the zero function. Thus,
a(u,U)=0=?k(X)(~~(X))2 =0, O<x
~~ (x)
=?
u(x) =
< x < f (since k(x) > 0), C, 0 < x < f,
= 0, 0
where C is But we know that that u(O) is continuous on where is aa constant. constant. But we already already know w(0) = 00 and and u is continuous on [O,f]. Therefore, zero, which is what we wanted wanted to to prove. prove. Thus [0,1]. Therefore, C must must be be zero, which is what we Thus a(·,·) o(-, •) defines an product on on V. defines an inner inner product V. We recall recall that that We
11
f(x)v(x) dx
= (f,v),
where (.,.) is the the L L22 inner product. Thus, Thus, the weak form (5.40) of of the the model where (-, •) is inner product. the weak form (5.40) model BVP BVP (5.39) can be written as (5.39) can be written as find u E V satisfying a(u, v) = (f, v) for all v E V.
(5.43)
The Galerkin Galerkin idea finite-dimensional subspace The idea is is simple. simple. Choose Choose aa finite-dimensional subspace Vn Vn of of V, and and reduce (5.43) (5.43) to the the subspace: subspace: find Vn E Vn satisfying a(vn,V) = (f,v) for all v E Vn.
(5.44)
Below we show that (5.44) (5.44) has has aa unique unique solution that can be found found by by solving Below we show that solution that can be solving aa matrix-vector equation equations). First we matrix-vector equation (that (that is, is, aa system system of of linear linear algebraic algebraic equations). First we show show that solving solving (5.44) (5.44) is is useful: useful: the Galerkin approximation is the best approxiapproximation, from from V to the mation, Vnn,, to the true true solution. solution.
182
Chapter 5. Boundary Boundary value problems in statics
Suppose U uE 6 V is is the solution to to (5.43), (5.43), Vn Vn is is aa finite-dimensional subspace of Suppose the solution finite-dimensional subspace of V, and E Vn the solution to (5.44). have and Vn vn e Vn is is the solution to (5.44). Then Then we we have
UEV, a(u,v) vnEVn , a(vn,v)
= (f,v) = (f,v)
forallvEV, forallvEVn ·
Subtracting Subtracting the the second second equation equation from from the the first, first, we we see see that that
a(u,v) - a(vn,v) = 0 for all v E Vn a(·, .), or, using the bilinearity of a(-,-),
a(u - vn,v) = 0 for all v E Vn . Since a(·,·) o(-, •) defines defines an an inner inner product product on on V, the theorem (see (see Theorem Theorem Since the projection projection theorem 3.36 when 3.36 in in Section Section 3.4) 3.4) shows shows that that Vn vn is is the the best best approximation approximation to to u from from V Vnn, when "best" "best" is is defined defined by the the norm norm induced induced by a(·, o(-, .): •):
Ilu -
vnllE :::;
Ilu - wilE
for all w E Vn,
where where
IlvIIE=va(v,v)=
1\(X)(~~(X))2 dx
forallvEV.
We We usually usually refer to II|| .• liE \\E as as the energy energy norm and and to to a(·,·) «(-,-) as as the the energy energy inner product. product.
To compute V ¢2, ... ¢n} is a basis for V vnn, suppose {(/>1, {i,02, • • • ,,^n} Vnn. Then (5.44) (5.44) is equivalent equivalent to to find Vn E Vn satisfying a(vn,¢i) = (f,¢i),i = 1,2, ... ,n.
(5.45)
belongs to to V written as basis vectors: Since Since Vn vn belongs Vnn,, it it can can be be written as aa linear linear combination combination of of the the basis vectors: n
Vn =
I: Uj¢j'
(5.46)
j=l
(We have two the symbol the exact (We now now have two meanings meanings for for the symbol "u." "u." The The function function uu is is the exact solution while the solution to to the the BVP, BVP, while the components components of of the the vector vector uu are are the the weights weights in in the the representation of the approximation basis ¢l, representation of the approximation Vn vn in in terms terms of of the the basis 0i, ¢2, 02, ... • • • , ¢n' (f>n- We We will will to standard live with this live with this ambiguity ambiguity in in order order to to adhere adhere to standard notation.) notation.) . .. , Un. (5.46) Finding Finding Vn vn is is now now equivalent equivalent to to determining determining Ul, HI, U2, 1*2,..., un. Substituting Substituting (5.46) into yields into (5.45) (5.45) yields
or, or, using the the bilinearity bilinearity of of a(·, o(-, .), •), n
I:a(¢j,¢i)Uj j=l
= (f,¢i), i = 1,2, ... ,n.
5.5. The Galerkin method
183
This This is is aa system system of of n linear linear equations equations for for the the unknowns unknowns Ul, wi, U2, U2, ... • • • ,, Un. un. We We define define aa matrix matrix K K E € Rnxn R n x n and a vector fERn f e Rn by
(We (We now now also also have have two two meanings meanings for for the the symbol symbol "f": "f": the the function function f/ is is the the right-hand right-hand side
i).) side of of the the differential differential equation, equation, while while the the vector vector ff has has components components Ii fi = — (I, (/,&).} Finding Vn vn is is now now equivalent equivalent to solving the the linear linear system system Finding to solving Ku=f.
The problems The matrix matrix K K is is usually usually called called the the stiffness stiffness matrix, matrix, since, since, when when applied applied to to problems in mechanics, the entries in K depend on the stiffness k(x) of the material. For in mechanics, the entries in K depend on the stiffness k(x) of the material. For analogous analogous reasons, reasons, the the vector vector ff is is usually usually called called the the load load vector.31 vector.31 The that, in best approximation, The reader reader will will recall recall that, in computing computing aa best approximation, it it is is advanadvantageous the stiffness is tageous to to have have an an orthogonal orthogonal basis, basis, since since then then the stiffness (or (or Gram) Gram) matrix matrix is diagonal, products. However, diagonal, and and the the projection projection is is computed computed by by nn inner inner products. However, the the basis basis for be orthogonal respect to for Vn Vn must must be orthogonal with with respect to the the energy energy inner inner product, product, and and with with aa 32 nonconstant coefficient usually difficult to find find an nonconstant coefficient k(x), k ( x ) , it it is is usually difficult32 to an orthogonal orthogonal basis. basis. finite element method, we nonorthogonal basis. Therefore, Therefore, in in the the finite element method, we will will use use aa nonorthogonal basis. We now to two with We now illustrate illustrate the the Galerkin Galerkin method method by by applying applying it it to two BVPs, BVPs, one one with constant constant coefficients coefficients and and the the other other with with nonconstant nonconstant coefficients. coefficients. Example Let FN Example 5.18. 5.18. Let FN be be the the subspace subspace of ofVV spanned spanned by by the the basis basis {sin (7fx) , sin (27fx) , ... ,sin (N 7fx)} ; This L22 inner product on interval This basis basis is is orthogonal orthogonal with with respect respect to to the the ordinary ordinary L inner product on the the interval
[0,1]. Consider Consider the the BVP BVP
d2 u - dx 2
= X, u(O) = 0, u(l) = 0
0
< x < 1, (5.47)
(this is (this is is (5.39) (5.39) with with k(x) k(x) == I}. I ) . The The exact exact solution solution is X
u(x) = 31
x3
6 - 6'
31The the Gram see The stiffness stiffness matrix matrix is is precisely precisely the Gram matrix matrix appearing appearing in in the the projection projection theorem; theorem; see page we apply the projection we need to compute page 62. 62. If If we apply the projection theorem theorem directly directly to to compute compute Vn, vn, we need to compute the the right-hand-side are a(u, a(u, ¢i)j allows us compute these these right-hand-side vector vector ff whose whose components components are i); the the weak weak form form allows us to to compute components ¢i) = components without without knowing knowing u, u, since since a(u, a(it,(fo) = (j, (/, ¢;). i). 32 32That That is, is, in in the the sense sense that that it it is is computationally computationally expensive expensive to to compute compute an an orthogonal orthogonal basis. basis. principle to to apply orGiven Given any any basis, basis, it it is is simple simple in in principle apply the the Gram-Schmidt Gram-Schmidt procedure procedure to to obtain obtain an an orexplanation thogonal thogonal basis. basis. The The reader reader can can consult consult any any introductory introductory book book on on linear linear algebra algebra for for an an explanation of of the the Gram-Schmidt Gram-Schmidt procedure. procedure.
184
Chapter 5. Boundary value problems in statics
Since Since k(x) k(x) = = 1, I , we we see see that that
(1 du dv a(u,v) = 10 dx(x)dx(x)dx and and
Therefore, Therefore,
and basis turns product as as and the the basis turns out out to to be be orthogonal orthogonal with with respect respect to to the the energy energy inner inner product well. have well. We We also also have (x, ¢i)
=
1
1 o
(_1)i+1
x sin (i1fx) dx
)
= -'----,. :-.Z1f
These form the we These form the entries entries of of the the matrix matrix K and and the the vector vector f. f. Since Since K is is diagonal, diagonal, we can solve solve the can the system system K Kuu == ff immediately immediately to to obtain obtain
Ui
=
2(-1)i+1 ·3 3
2 1f
The for N N = Figure 5.10. The resulting resulting approximation, approximation, for = 20, 20, is is displayed displayed in in Figure 5.10. The The energy energy inner product, for for a a constant constant function function k(x), just the product applied inner product, k ( x ) , is is just the £2 L2 inner inner product applied to to we perform the and so the derivatives, derivatives, and so the the calculations calculations are are intimately intimately related related to to those those we perform in result is in computing computing aa Fourier Fourier series series solution; solution; indeed, indeed, the the final final result is the the same! same!
Example Example 5.19. 5.19. Consider Consider the the BVP BVP
d(
dU) =x, °< x < 1,
- - (l+x)dx
dx
u(o) = 0, u(l) = 0. The solution is is The exact exact solution
_ ~ _ x2 ( ) u x - 2 4
_
(5.48)
In (1 + x) 4In2'
We apply the basis We will will apply the Galerkin Galerkin method method with with the the same same approximating approximating subspace subspace and and basis as previous exercise. as in in the the previous exercise. The The calculations calculations are are similar, similar, but but the the bilinear bilinear form form changes changes to to (1 du dv a(u,v) = 10 (1 + x)dx(x)dx(x) dx, and functions are are no energy inner inner product. and the the basis basis functions no longer longer orthogonal orthogonal in in the the energy product.
185 185
5.5. The Galerkin method 5.5.
We have have We a(¢>j, ¢>i)
= ij7r2
11
(1
+ x) cos (i7rx) cos (j7rX) dx
={ and and
(_I)i+l
1
(x, ¢>i) =
1 o
x sin (i7rx) dx = -'---:-.-
Z7r
These form the entries the matrix matrix K K and and the the vector vector f. We cannot solve Ku Ku == ff These form the entries of of the f. We cannot solve explicitly; that that is, is, we cannot derive derive aa useful the coefficients Ui,ii = explicitly; we cannot useful formula formula for for the coefficients Ui, = 1,2,..., in the the formula formula 1,2, ... , N, N, in N
VN(X) =
LUi
sin (i7rx).
i=1
However, can solve the unknowns unknowns numerically numerically using using Gaussian Gaussian elimination, elimination, However, we we can solve for for the result, for for N N = is and produce the from FN. FN . The and thereby thereby produce the approximation approximation w w from The result, = 20, 20, is shown Figure 5.10. shown in in Figure 5.10. 0.08.-----------.
0.06.------------.
-5
10 x 10
2
5
-5~---~----j
o
0.5
0.5
x
x
Figure 5.10. 5.10. Solving (5.47) {left} (left) and and {5.48} (6.48) {right} (right) using using the Figure Solving BVPs BVPs {5.47} the Galerkin method. method. The The solutions are graphed in the the top top figures and the the errors errors in in Galerkin solutions are graphed in figures and the solutions in the the bottom. bottom. the solutions in
186
Chapter 5. Boundary Boundary value problems in statics
The results these two two examples are of of roughly roughly the the same showing The results in in these examples are same quality, quality, showing that the Galerkin method can the Fourier Fourier method. method. The that the Galerkin method can be be as as effective effective as as the The most most is the need to form the matrix K in significant difference in these two methods significant difference in these two methods is the need to form the matrix K in the the second and solve the N x x N system = f.f. This time-consuming second example example and solve the system Ku Ku = This is is very very time-consuming to the the computations required in in the the first first example example (where the system of compared compared to computations required (where the system of linear linear equations equations is is diagonal). diagonal). This question question of of efficiency efficiency is particularly particularly important important dimensions; in aa problem of when we we have have two or three spatial spatial dimensions; of realistic size, it may may take long to to solve solve the the resulting resulting linear linear system the coefficient matrix is is take impossibly impossibly long system if if the coefficient matrix dense. A in which all of entries are are nonzero. dense. A dense dense matrix matrix is is aa matrix matrix in which most most or or all of the the entries nonzero. A sparse matrix, on on the other hand, has mostly mostly zero zero entries. entries. A sparse matrix, the other hand, has The finite finite element method is is simply simply the method with choice The element method the Galerkin Galerkin method with aa special special choice the subspace subspace and its basis; basis; the the basis basis leads to aa sparse for the for and its leads to sparse coefficient coefficient matrix. matrix. The The ultimate sparse, nonsingular is aa diagonal diagonal matrix. Obtaining aa diagonal diagonal ultimate sparse, nonsingular matrix matrix is matrix. Obtaining matrix for the subspace be be chosen chosen to matrix requires requires that that the the basis basis for the approximating approximating subspace to be be As mentioned mentioned earlier, earlier, it is too orthogonal to the orthogonal with with respect respect to the energy energy inner inner product. product. As it is too difficult, for for aa problem variable coefficients, coefficients, to The to find find an an orthogonal orthogonal basis. basis. The difficult, problem with with variable finite method uses basis in in which which most pairs of functions are orthogonal; finite element element method uses aa basis most pairs of functions are orthogonal; the resulting matrix matrix is not diagonal, it is is quite quite sparse. the resulting is not diagonal, but but it sparse.
Exercises 1. whether the the bilinear bilinear form 1. Determine Determine whether form
product on on each the following [0, fl. If defines defines an an inner inner product each of of the following subspaces subspaces of of C (722[0,^]. If it it does not, show show why. why. does not,
(a) {v E C 2 [0,f] : v(f) =
O}
2
(b) {v E C [0,f] : v(O) = v(f)}
(c) C 2 [0,f] (the entire space) 2. that if the subspace 5.18, and and the the Galerkin Galerkin 2. Show Show that if FN FN is is the subspace defined defined in in Example Example 5.18, method is to the the weak weak form of method is applied applied to form of
d2 u -k dx 2 = f(x), 0
< x < f,
u(O) = 0, u(f) = 0, with FN as the the approximating approximating subspace, subspace, the the result result will will always always be be the the partial with FJV as partial Fourier sine series terms) of of the solution u. u. Fourier sine series (with (with N N terms) the exact exact solution 3. S to to be be the the set polynomials of of the the form ax + bx considered as as 3. Define Define 5 set of of all all polynomials form ax bx22,, considered functions the interval interval [0,1]. functions defined defined on on the [0,1]. (a) Explain Explain why [0, 1]. (a) why S 5 is is aa subspace subspace of of C22[0,1].
5.5. The Galerkin method method
187
(b) Explain form (b) Explain why why the the bilinear bilinear form
10r
l
a(u,v) =
du dv dx (x) dx (x) dx
defines defines an inner product on on S. S.
(c) Compute Compute the from S, in in the the energy energy norm, = (c) the best best approximation approximation from norm, to to f(x] f(x) = eX. ex. (d) Explain why in the to (d) Explain why the the best best approximation approximation from from P-2, P2, in the energy energy norm, norm, to = eeXx isis not not unique. unique. (The P2 is defined in in Exercise ff(x) (x) = (The subspace subspace PI is defined Exercise 3.l.3d.) 3.1.3d.) 4. Repeat Repeat Exercise Exercise 33 with
s=
{v E C 2 [0, 1]
v(x) = ax+ bx 2
5. 5. Define Define V2 = span { x(l - x), x
+ cx 3
(~ -
for some a,b,c E
R}.
x) (1 - x) }
and regard regard V as aa subspace subspace of of C|j[0,1]. C1[0, 1]. Apply Apply the the Galerkin Galerkin method, method, using and ¥22 as using V the approximating approximating subspace, to estimate the solution solution of of ¥22 as as the subspace, to estimate the
d ((1 + x) dx dU) = x,
- dx
0 < x < 1,
u(O) = 0, u(l) = O. Find the the exact the exact and approximate approximate solutions solutions totoFind exact solution solution and and graph graph the exact and gether. gether. 6. Exercise 55 using using the the subspace subspace 6. Repeat Repeat Exercise
V3
= span {X(l- x),x (~- x)
(1- x),x
(~- x) (~- x) (1- x)}.
7. The standard Gaussian Gaussian elimination elimination algorithm solving Ax Ax = b requires requires 7. The standard algorithm for for solving =b
nxn Rnxn. cubic polyarithmetic operations when A E eR . (The exact number is a cubic 3 3 2n /3 the leading leading term. term. When When n is is large, the lower lower degree degree nomial in nomial in n, and and 2n /3 is is the large, the terms terms are negligible in in comparison to the leading term.)
(a) Suppose that, on certain computer, takes 11 second to solve (a) Suppose that, on aa certain computer, it it takes second to solve aa 100 100xx100 100 it take take to dense by Gaussian long will will it dense linear linear system system by Gaussian elimination. elimination. How How long to solve solve a 1000 1000 x 1000 1000 system? system? A 10000 10000 x 10000 10000 system?
188
Chapter 5. Boundary value value problems in statics nxn (b) Now suppose ER Rnxn is tridiagonal, Aij = whenever (b) Now suppose A e tridiagonal, that is, is, suppose suppose AIJ —a 0 whenever how many many arithmetic arithmetic operations does it it take Ii\i -—ilj\ > 1.1. About About how operations does take to to solve solve Ax = by Gaussian that the the algorithm Ax = bb by Gaussian elimination, elimination, assuming assuming that algorithm takes takes advantage of the fact of the zero? advantage of the fact that that most most of the entries entries of of A are are already already zero? (The example by by hand hand and count the (The student student should should work work out out aa small small example and count the number of of operation taken at number operation taken at each each step. step. A A short short calculation calculation gives gives this this operation as aa function operation count count as function of of n.) n.) (c) Finally, suppose that, on on aa certain certain computer, it takes takes 0.01 to solve (c) Finally, suppose that, computer, it 0.01 ss to solve aa 100 by Gaussian elimination. How How long long 100 xx 100 100 tridiagonal tridiagonal linear linear system system by Gaussian elimination. will it it take take to to solve x 1000 system? will solve aa 1000 1000 x 1000 system? system? A A 10000 10000 xx 10000 10000 system?
5.6 5.6
Piecewise the finite finite element Piecewise polynomials polynomials and and the element method method
To the Galerkin to solve and efficiently, we must must To apply apply the Galerkin method method to solve aa BVP BVP accurately accurately and efficiently, we [0, l] and basis for for Vn the following choose aa subspace choose subspace Vn Vn of of C C22[0,^] and aa basis Vn with with the following properties: properties: 1. K and and the the load load vector vector ff can be assembled 1. The The stiffness stiffness matrix matrix K can be assembled efficiently. efficiently.
the basis basis for consist of that are easy to This means that This means that the for Vn Vn should should consist of functions functions that are easy to manipulate, in particular, differentiate manipulate, in particular, differentiate and and integrate. integrate. 2. for Vn be as close to to orthogonal orthogonal as as possible possible so K will 2. The The basis basis for Vn should should be as close so that that K will be sparse. Although expect every every pair functions to to be be be sparse. Although we we do do not not expect pair of of basis basis functions matrix), we we want orthogonal (which orthogonal (which would would lead lead to to aa diagonal diagonal stiffness stiffness matrix), want as as many pairs pairs as possible to be orthogonal will be be as to diagonal many as possible to be orthogonal so so that that K K will as close close to diagonal as as possible. possible.
3. The The true true solution solution u of of the BVP should should be be well well approximated from subspace subspace the BVP approximated from 3. V^, with the approximation becoming arbitrarily good as n —> oo. , V with the approximation becoming arbitrarily good as n -+ 00. n We will will continue continue to to use the BVP We use the BVP d ( k(x) dx du (x) ) - dx
= f(x),
0
< x < /!,
u(O) = 0, u(/!) = a
(5.49)
as our as our model model problem. problem. Finite element methods methods use use subspaces subspaces of wise polynomials; polynomials; for for simplicity, Finite element of piece piecewise simplicity, we will on piece piecewise piecewise we will concentrate concentrate on wise linear linear functions. functions. To To define define aa space space of of piecewise linear functions, mesh (or (or grid) on the the interval interval [0, [0,^]: we begin begin by by creating creating aa mesh grid) on l]: linear functions, we
o = Xo < Xl < ... < Xn = l. The points points XQ are called called the the mesh. choose aa regular regular The Xo,, Xi,..., Xl, ... , x Xn the nodes nodes of of the mesh. Often Often we we choose n are mesh, with xi — ih, ih, hh = lin. t/n. A A function [0,^] —>• R R is is piecewise linear (relative (relative mesh, with Xi = function pp :: [0, l] -+ piecewise linear to given mesh) for each each i = l , 2 , ... . . . ,,n, n , there d{, bb{, to the the given mesh) if, if, for = 1,2, there exist exist constants constants ai, with i , with p(x)
=
aiX
+ bi
for all
X
E (Xi-l,Xi).
5.6. Piecewise polynomials and the finite element element method
189 189
See Figure 5.11 5.11 for for an an example continuous piecewise piecewise linear linear function. function. See Figure example of of aa continuous
0.9
x Figure 5.11. 5.11. A A piecewise piecewise linear linear function function relative relative to the mesh mesh XQ Xo = Figure to the = 0.0, Xl = = 0.25, x X2 = 0.75, z X4 = 1.0. 0.0,31 = 0.5, xX33 = 1.0. 2 = 4 =
We define, define, for for aa fixed fixed mesh mesh on £], We on [0, [0,£],
Sn = {p: [0,£]-+ R : p is continuous and piecewise linear, p(O) = pee) = O}. (5.50)
We argue that subspace S satisfies the for an approxthat the the subspace Snn satisfies the three three requirements requirements for an approxWe now now argue imating subspace that are listed above. above. Two Two of of the the properties properties are are almost almost obvious: obvious: imating subspace that are listed 1. the functions belonging to to 1. Since Since the functions belonging
Snn are piecewise polynomials, polynomials, they they are are easy easy S are piecewise to manipulate—it manipulate-it is is easy easy to to differentiate differentiate or or integrate integrate aa polynomial. polynomial. to
3. can be be well-approximated well-approximated by by piecewise piecewise linear linear functions. In3. Smooth Smooth functions functions can functions. Inthe days days before before hand-held hand-held calculators, the elementary deed, in the deed, in calculators, the elementary transcendental transcendental functions like like sin sin (x) or eX were given given in in tables. tables. Only number of of funcfunctions (x) or ex were Only aa finite finite number function values values could could be be listed listed in in aa table; table; the the user user was was expected to use use piecewise piecewise tion expected to linear interpolation listed in table. to estimate estimate values values not not listed in the the table. linear interpolation to The fact that smooth functions can can be be well-approximated well-approximated by by piecewise piecewise linear The fact that smooth functions linear functions can also illustrated in graph. Figure shows aa smooth smooth funcfuncfunctions can also be be illustrated in aa graph. Figure 5.12 5.12 shows 1] with with two two piecewise piecewise linear the first first corresponding corresponding tion on on [0, tion [0,1] linear approximations, approximations, the to 10 and approximation can can be made to n = 10 and the the second second to to n = 20. 20. Clearly Clearly the the approximation be made arbitrarily good by by choosing choosing n arbitrarily good n large large enough. enough. In order order to to discuss discuss Property Property 2, that the the stiffness stiffness matrix matrix should should be be sparse, sparse, In 2, that we must must first first define define aa basis basis for Sn. The idea idea of of piecewise piecewise linear linear interpolation we for S interpolation n. The
190
Chapter 5.
problems in statics Boundary value problems
x
0.2
x
Figure 5.12. A smooth smooth function function and and two two piecewise piecewise linear approximations. Figure 5.12. A linear approximations. is based based on nodes, while bottom is based on on 21 21 nodes. nodes. The The top top approximation approximation is on 11 11 nodes, while the the bottom is based
suggests the natural basis: a piecewise linear function is completely determined by its nodal values (that (that is, is, the the values values of the function function at the nodes nodes of of the the mesh). mesh). For For its nodal values of the at the n= - 1, ¢i £E Sn be that piecewise linear function satisfying satisfying n = 1,2, 1,2, ... . . . ,, nn — 1, define define fa Sn to to be that piecewise linear function .+. ( )
{I, 0
=
j = i, . ..J.. , J r~·
Snn satisfies satisfies Then Then each each pp E eS n-l
p(x) = ~p(Xi)¢i(X).
(5.51)
i=l
To prove this, we merely need to show that p and n-l
~P(Xi)¢i
(5.52)
i=l
have have the the same same nodal nodal values, values, since since two two continuous continuous piecewise piecewise linear linear functions functions are are equal equal if we substitute if and and only only if if they have the the same same nodal values. Therefore, Therefore, we substitute x == Xj into into (5.52): (5.52): n-l
~P(Xi)¢i(Xj) i=l
= p(xd· 0 + ... + p(xj)·l + ... + p(xn-d· 0 = p(Xj).
5.6. 5.6.
191
Piecewise polynomials and the finite element method method
This shows holds. This shows that that (5.51) (5.51) holds. Snn can can be be written written as as aa linear linear combination combination of (/J!, 02, ¢2, ... Thus Thus every every pp E 6 S of {{
Then, Then, in in particular, particular, n-1 LCi¢i(Xj) =
Cj
= 0, j = 1,2, .. . ,n-l.
i=1
This that all all of of the the coefficients C1, C2, ... , C zero, so ¢1, 0 ¢2, This implies implies that coefficients c\, c^,..., cnn -1 _i are are zero, so {{(f>i, • • • ,? ¢n-1} (f>n-i} 2 , ... is aa linearly A typical typical basis basis function is displayed displayed in in Figure is linearly independent independent set. set. A function ¢i(X) (f)i(x) is Figure 5.13. 5.13.
0.9
0.8 0.7
0.6 0.5 0.4
0.3 0.2
0.6
0.2
0.8
x
Figure 5.13. A typical typical basis function for for Sn. Figure 5.13. A basis function Sn. The entries matrix are The entries of of the the stiffness stiffness matrix are Kij=a(¢j,¢i), i,j=1,2, ... ,n-l.
Here is the observation concerning element method: Here is the fundamental fundamental observation concerning the the finite finite element method: since since each fa is interval [O,fJ, [0,£], most inner products a(0j,^>j) is zero zero on on most most of of the the interval most of of the the inner products a(¢i,¢j) each ¢i are zero zero because product are because the the product
192 192
Chapter 5. 5. Boundary Boundary value value problems problems in in statics statics Chapter
turns out to to be be zero [0,^]. turns out zero for for all all xx €E [0, fl· To be specific, fa zero except except on on the the interval interval [Xi-l, [ori_i,£i+i]. (We say say that the To be specific, ¢i is is zero Xi+l]. (We that the 33 support of fa is the interval [xj_i,Xj+i].) From this we see that support 33 of ¢i is the interval [Xi-l, Xi+1].) From this we see that
can be nonzero, but can be nonzero, but
Ii - il > 1 =:? a(¢j, ¢i) =
0,
since in that that case, f], either since in case, for for any any xx E G [0, [O,/], either d¢ix () o r d¢j ( ) -x dx dx
is zero. matrix K K turns is, all all of entries is zero. Therefore, Therefore, the the matrix turns out out to to be be tridiagonal, tridiagonal, that that is, of its its entries are zero except (possibly) (possibly) those three diagonals (see Figure Figure 5.14). 5.14). are zero except those on on three diagonals (see
o •• 5
10 15
20 25
30 35
... ... ... ... ... ... ... ... ... ... ... ......... ••• ... ... ... ... •••
... ...
......
••• •••
... ... ...
40L-________~________~__________~________~ 10 20 30 40 o
Figure 5.14. A schematic schematic view the only nonzeros Figure 5.14. A view of of aa tridiagonal tridiagonal matrix; matrix; the only nonzeros are are on on the the main main diagonal diagonal and and the the first first subsub- and and super-diagonal. super-diagonal.
We Snn of linear functions We now now see see that that the the subspace subspace S of continuous continuous piecewise piecewise linear functions satisfies requirements for for aa good However, the satisfies all all three three requirements good approximating approximating subspace. subspace. However, the attentive Sn: it attentive reader may may have been been troubled by one one apparent apparent shortcoming shortcoming of of S it n is not aa subset of C1[0, f], because most of of the the functions Snn are not continucontinuis not subset of Cf>[0,^], because most functions in in S are not ously twice-continuously differentiable. differentiable. This This would would seem seem ously differentiable, differentiate, much much less less twice-continuously 33 33The of aa continuous continuous function function is the closure of the the set set on on which the function function is nonzero, The support support of is the closure of which the is nonzero, that is, the set set on on which which the the function nonzero, together together with with its boundary. In ci>;, that is, the function is is nonzero, its boundary. In the the case case of of i, the function is nonzero nonzero on of this interval is the closed interval. the function is on (Xi-I,Xi+ll. (xi-i,Xi+i). The The closure closure of this open open interval is the closed interval.
polynomials and and the finite element element method 5.6. Piecewise polynomials
193
to invalidate invalidate the the entire context on on which which the the Galerkin Galerkin method method depends. In fact, fact, to entire context depends. In however, presents no no difficulty-the difficulty—the important important fact fact is is that that equations equations defining defining however, this this presents the the weak weak form form are are well-defined, well-defined, that that is, is, that that we we can can compute compute
Since piecewise linear, piecewise constant hence Since cPi
5.6.1 5.6.1
Examples using piecewise piecewise linear finite elements
Let the subspace previous Let us us apply apply the the Galerkin Galerkin method, method, with with the subspace Sn Sn defined defined in in the the previous section, to to (5.49). (5.49). As As we we showed showed in in Section Section 5.5, 5.5, the the Galerkin Galerkin approximation approximation is is section, n-l
L UicPi(X),
vn(x) =
i=l
where satisfy where the the coefficients coefficients Ul, MI, U2, U2, ... • • • ,, Un-l w n -i satisfy
Example 5.20. 5.20. We We apply apply the the finite finite element element method method to to the the BVP BVP Example d2 u - dX2
= eX, 0 < x < 1,
u(O) = 0, u(l) = O.
(5.53)
The n The exact exact solution solution is is u(x) u(x] == -ex —ex + (e (e -— l)x l)x + 1. 1. We We use use aa regular regular mesh mesh with with n subintervals and h = lin. l/n. We We then then have subintervals and h= have ~(x - (i -l)h),
cPi(X) =
{
< x < Xi, --k(x - (i + l)h), Xi < x < Xi+l,
and d¢i(x) dx
0,
~{
1
Xi-l
otherwise
-II'
< X < Xi, Xi < X < Xi+1,
0,
otherwise.
II' 1
Xi-l
194 194
Chapter Chapter 5. Boundary Boundary value problems in statics
Therefore, Therefore, Kii =
l
a(¢i' ¢i) =
(Hl)h
(i-l)h
dx h2
2
=h and and r(Hl)h
= a(¢Hl, ¢i) = J
K i ,Hl
ih
1( 1)
h -h
dx
1
-h' (These are critical, the reader reader should make sure sure he she thoroughly thoroughly (These calculations calculations are critical, and and the should make he or or she understands them. them. The The basis basis function is nonzero only on on the the interval interval [( [(ii-I) — l)h, ¢i is nonzero only h, (i(i+ + understands function fa l)h], so the the interval reduces to subinterval. The l)h], so interval of of integration integration reduces to this this subinterval. The square square of of the the of ¢i I1h22 on subinterval. This gives the the result for KH. K ii . As derivative derivative of fa is is l/h on this this entire entire subinterval. This gives result for As for the Ki ,Hl, the the only part of 1] on which both both ¢i and ¢Hl are for the computation computation of of Ki^+i, only part of [0, [0,1] on which fa and fa+i are nonzero is is [ih,(i l)h]. On On this this subinterval, the derivative fa is is -l/h —l/h and nonzero [ih, (i + l)h]. subinterval, the derivative of of ¢i and the the derivative of ¢Hl fa+i is is Ilh.) l/h.) derivative of This gives us on the super-diagonal ofofK; K; This gives us the the entries entries on the main main diagonal diagonal and and first first super-diagonal since K symmetric and tridiagonal, we deduce the of the entries. We since K is is symmetric and tridiagonal, we can can deduce the rest rest of the entries. We also have also have
(I, ¢i) =
l~h
(,-I)h eih
= -
(~(X - (i -
(e h + e- h
-
l)h)) eX dx
+
(i+l)h
Jih
(-~(X - (i + l)h)) eX dx
2) .
h It remains remains only assemble the It only to to assemble the tridiagonal tridiagonal matrix matrix K K and and the the right-hand-side right-hand-side vector vector f, and solve Ku f, and solve Ku = = ff using using Gaussian Gaussian elimination. elimination. 4x4 For example, for nn = 5, we have K E R 4X4 For example, for 5, we have K GR ;, 10 -5
=
K
-5 10
0 -5
[
o f
==
0
0 -5 10 -5
0.2451 ] 0.2994 0.3656 ' [ 0.4466
and
== u
0.1223] 0.1955 0.2089' [ 0.1491
0 ] 0
-5 10
'
195
5.6. Piecewise the finite element method 5.6. Piecewise polynomials polynomials and and the finite element method
The corresponding piecewise piecewise linear with the exact solution, solution, The corresponding linear approximation, approximation, together together with the exact is displayed displayed in Figure 5.15.
0.25r-----.....,...-----r-------,----.....,------,
0.6
0.8
1
x -3
15 X 10
-5~------~--------~--------~--------~------~
o
0.2
0.6
0.4
0.8
1
X
Figure Exact solution solution and piecewise linear linear finite element approximaFigure 5.15. 5.15. Exact and piecewise finite element approximation w(l) = (top). Error in piecewise linear approximation approximation tion for for -£-% -~ = = eeX,x,u(Q) u(O) = u(l) = 00 (top). Error in piecewise linear (bottom). (bottom). Example approximate solution solution to the BVP Example 5.21. 5.21. We We now find find an approximate BVP
d ((l+x)-(x) dU)
-dx
dx
=1, 0< x
< 1,
u(O) = 0, u(l) = O.
(5.54)
The solution is + x)/ln2 x)/In2 -— x. We The exact exact solution is u(x) u(x) = = In In (1 (1 + We again again use use a regular regular mesh mesh with with n subintervals subintervals and and hh =— lIn. l/n. The bilinear form form a(·,·) now takes The bilinear a(-, •) now takes the the form form a(u,v) =
(l
10
du dv (1 + x)dx(x)dx(x)dx.
196 196
Chapter Boundary value value problems Chapter 5. 5. Boundary problems in statics statics
We We obtain obtain a(¢i, ¢i)
l
=
(i+I)h (
(i-1)h
1) (1 + x) h2 dx
2(ih + 1) h
and a(¢i+1' ¢i)
=
1+
(i+1)h ( ih
h
1( 1))
(1 + x)h
-h
dx
2ih + 2 2h
As in previous example, calculations suffice suffice to also have have As in the the previous example, these these calculations to determine determine K. We We also
(1 l
(1, ¢i) = .ih
h(x - (i - l)h)
(.-1)h
)
1
dx
+ 1(i+ .
)h (
1
-h(x - (i + l)h)
)
dx
.h
=h. remains only only to assemble the tridiagonal matrix K and the right-hand-side vector It remains f, f, and solve Ku Ku = = f using using Gaussian Gaussian elimination. elimination. 5, the For example, if if n = — 5, the matrix matrix K K is 44 xx 4, 4, 13 12 - 2 0 13 15 14 - 2 K-2 15 0 -2 16 [ 17 o 0 -2
)), 18
and ff is is aa 4-vector, and 4-vector, f=
1/5 1/5 ) 1/5 . [ 1/5
The resulting uu is is The resulting
== u
0.0628] 0.0851 0.0778' [ 0.0479
The piecewise linear approximation, approximation, together together with with the the exact exact solution, is shown shown in in The piecewise linear solution, is Figure 5.16. 5.16. Figure Fundamental issues for the finite finite element method are: Fundamental issues for the element method are: • How How fast fast does the approximate to the the true does the approximate solution solution converge converge to true solution solution as as nn -+ -» oo? oo?
197
5.6. and the finite element 5.6. Piecewise Piecewise polynomials polynomials and the finite element method method 0.15..--------r----,.------,----...,-------. 0.1
0.4
x
0.6
O.B
x
of
Figure linear finite finite element element approximation solution Figure 5.16. 5.16. Piecewise Piecewise linear approximation to to the the solution 1,u(O) = u(l) = O.
-/x ((x + l)~~) =
•• How Howcan can the the stiffness stiffness matrix matrix K K and and the the load load vector vector ff be be computed computed efficiently efficiently and automatically automatically on on aa computer? and computer?
Howcan can the the sparse sparse system system Ku Ku == ff be be solved solved efficiently? efficiently? (This (This is is primarily primarily an an •• How issue when when the problem involves two or or three issue the problem involves two three spatial spatial dimensions, dimensions, since since then then K can be very K can be very large.) large.) will be be important higher-dimensional problems problems is Another issue Another issue that that will important for for higher-dimensional is the the descripdescripmesh-special data structures are needed to to allow tion of the mesh—special tion of the data structures are needed allow efficient efficient computation. computation. treatment of but we will A careful careful treatment of these questions questions is is beyond beyond the scope scope of of this this book, book, but provide answers in provide some some answers in Chapter Chapter 10. 10.
5.6.2 5.6.2
Inhomogeneous Dirichlet conditions
The inhomogeneous inhomogeneous Dirichlet problem,
d (k(x) dU) dx = f(x),
- dx
u(O) =
u(£)
0:,
0 < x < £, (5.55)
= /3,
can be solved method of the data" In this can be solved using using the the method of "shifting "shifting the data" introduced introduced earlier. earlier. In this method, we we find g(x) satisfying boundary conditions conditions and method, find aa function function g(x] satisfying the the boundary and define define aa
198
Chapter Chapter 5.
value problems problems in statics Boundary value
new BVP for the unknown w(x) w(x) = u(x) u(x] -— g(x). g(x). We We can apply this idea to the weak form of of the the BVP. BVP. form A calculation of calculation similar to that of Section 5.4.1 shows that the weak form of (5.55) is a(u,v) = (f,v) for all v E V. Here, Here, however, we we do do not not look look for for the the solution solution uu in in V, but rather in in g + +V = {g + +v : vE € V}. We We now now write write u = gg + + w, w, and and the the problem problem becomes becomes {g find wE V such that a(w
+ g,v) = (f'v)
for all v E V.
(5.56)
But a(w + g, v) = = a(w, v) + + a(g, v), v ) , so so we can write write (5.56) (5.56) as as But we can find w E V such that a(w,v) = (f'v) - a(g,v) for all v E V.
(5.57)
It turns turns out out that that solving solving (5.57) (5.57) is is not not much much harder harder than than solving solving the the weak weak formulation formulation we modify of of the the homogeneous homogeneous Dirichlet Dirichlet problem; problem; the the only only difference difference is is that that we modify the the right-hand-side vector f, f, as as we we explain explain below. below. right-hand-side vector Shifting because we Shifting the the data data is is easy easy because we can can delay delay the the choice choice of of the the function function gg until until we find aa finite finite element function we apply apply the the Galerkin Galerkin method; method; it it is is always always simple simple to to find element function g gnn which which satisfies satisfies the the boundary boundary conditions conditions on on the the boundary boundary nodes nodes (actually, (actually, for for aa one-dimensional find such one-dimensional problem problem like like this, this, it it is is trivial trivial to to find such aa function function gg in in any any case; case; however, this this becomes while finding however, becomes difficult difficult in in higher higher dimensions, dimensions, while finding aa finite finite element element function in function to to satisfy satisfy the the boundary boundary conditions, conditions, at at least least approximately, approximately, is is still still easy easy in higher dimensions). dimensions). The Galerkin problem is:
find Wn E Sn such that a(w n , v)
= (f, v) -
a(gn, v) for all v
E
Sn-
(5.58)
We choose choose g to be piece wise linear linear and and satisfy satisfy gn(XO) gn(xo} = a, a, gn(xn) gn(xn) = (3; /3- the the simplest simplest We gnn to be piecewise such function function has value zero zero at at the the other other nodes of the mesh, mesh, namely namely g = a¢o+(3¢n. afio+ftfin. nodes ofthe gnn = such has value Substituting (5.58), we we obtain, obtain, as as before, before, the the linear linear system system Substituting Wn wn == l:~:ll ^ii=i aai¢i i^i minto *° (5-58), Ku = = f. f. The The matrix matrix K does not not change, change, but becomes Ku K does but ff becomes i = 2,3, ... , n - 2 (since a(gn, ¢i) i = 1, i=n-1.
= 0), (5.59)
Example Example 5.21; specifExample 5.22. 5.22. We We now now solve solve an an inhomogeneous inhomogeneous version version of of Example specifi.rnll.ii we IIJP apply n.nnln the the finite finite element pl.pm.pni. method mpthntt to t.n ically,
d ((1 + x) dx dU) (x) = 1,
- dx
u(O) = 2, u(l) = 1.
0
< x < 1, (5.60)
199 199
5.6. Piecewise polynomials polynomials and the finite element method
As we just explained, the approximate by As we have have just explained, the approximate solution solution is is given given by n-l
=L
Vn(X)
UirPi(X),
i=l
where coefficients iti, Ul, U2, where the the coefficients «2,.... • • ,, Un-l un-i satisfy satisfy Ku=f.
The matrix K K is is identical identical to to the the one in Example while the the The stiffness stiffness matrix one calculated calculated in Example 5.21, 5.21, while load modified as as indicated 59}. We load vector vector ff is is modified indicated in in (5. (5.59). We obtain obtain
i = 2,3, ... , n - 2 (since a(gn, rPi) i = 1, i=n-l.
= OJ,
Now, Now, a(rPo,rPd =
lh ((1 +x)~ (-~))
dx
h+2 2h ' a(rPn,rPn-d
= l~h
((l+X)~ (-~))
dx
h-4
2h' so so
II
h+2 =h+2----v;-
and and fn-l = h-
h-4
2h"".
With n = 5, obtain 5, we we obtain With n =
_
f -
56/5] 1/5 1/5.
r 97/10
The results are shown in solution equals The results are shown in Figure Figure 5.17. 5.17. In In this this case, case, the the computed computed solution equals the the exact solution (the error error shown shown is is Figure 5.17 is is only only due due to to round-off round-off error error in in the the exact solution (the Figure 5.17 computer computer calculations) calculations) (see (see Exercise Exercise 2). 2).
200 200
Chapter 5. Boundary value problems in statics 2~------~--------~--------~------~--------~
1.8
1.6 1.4
1.2 1L-------~~------~--------~--------~------~~
o
0.2
0.4
0.6
0.8
x
0.2
0.8 x
Figure 5.11. 5.17. Piecewise linear finite finite element approximation, with with nn = 5 Figure Piecewise linear element approximation, 5 subintervals in the the mesh, the solution ((x + + 1) l)ff) = 1, l,w(0) to the solution of of — -^ ((x ~~) = u(O) = = 2,2,w(l) u(1) = = 1.1. subintervals in mesh, to The error in the the bottom bottom graph is due to round-off round-off error—the "approxiThe error shown shown in graph is due only only to error-the "approximation" to the exact solution. mation" is in fact equal to solution.
-lx
Exercises 1. (a) (a) Using direct integration, exact solution solution to to the BVP from from ExamExam1. Using direct integration, find find the the exact the BVP ple 5.21. ple 5.21. increasing sequence 5.21, using an increasing (b) Repeat the calculations from Example 5.21, sequence n. Produce graphs similar of values of n. similar to Figure Figure 5.16, or otherwise measure the errors in the approximation. Try to identify the size of the error 34 as a function of h. 34 2. the exact 5.22. 2. (a) (a) Compute Compute the exact solution solution to to the the BVP BVP from from Example Example 5.22.
(b) Explain BVP, the the finite finite element element method method computes computes (b) Explain why, why, for for this this particular particular BVP, the exact, exact, rather than than an an approximate, solution. solution. 3. Use piecewise linear finite elements elements and and aa regular to solve solve the 3. Use piecewise linear finite regular mesh mesh to the following following 34 34To perform these much larger the use use of To perform these calculations calculations for for nn much larger than than 55 will will require require the of aa computer. computer. One could, of write aa program, in aa language language such such as Fortran or or C, C, to to do do the the calculations. calculations. One could, of course, course, write program, in as Fortran there exist powerful interactive programs that However, However, there exist powerful interactive software software programs that integrate integrate numerical numerical calculations, calculations, programming, and convenient to to use. MATLAB is is particularly programming, and graphics, graphics, and and these these are are much much more more convenient use. MATLAB particularly suitable for the finite element for this designed to to facilitate facilitate suitable for the finite element calculations calculations needed needed for this book, book, since since it it is is designed matrix Mathematica and and Maple Maple are are other possibilities. matrix computations. computations. Mathematica other possibilities.
5.6. the finite 5.6. Piecewise Piecewise polynomials polynomials and and the finite element element method method
201 201
problem:
J2u
- dx 2 =
2 X
,
U(O) = 0, u(1) = O. Use nn = 10,20,40, ... elements, Use 10,20,40,... elements, and and determine determine (experimentally) (experimentally) an an exponent exponent p» so is so that that the the error error is
o
(:p) .
Here error be the the maximum between the Here error is is defined defined to to be maximum absolute absolute difference difference between the exact exact and you and approximate approximate solutions. solutions. (Determine (Determine the the exact exact solution solution by by integration; integration; you will need it to compute the error in the approximations.) 4. 4. Repeat Repeat Exercise Exercise 33 for for the the BVP BVP
d ( (1 + x 2 ) dU) dx - dx
= 1, 0
< x < 1,
u(O) = 0, u(1) = O.
The exact exact solution solution is The is
u(x)
2ln2 = -7rtan- 1 (x) -
1 2ln (1
+ x 2 ).
5. weak form 5. Derive Derive the the weak form of of the the BVP BVP
d ( k(x) dU) - dx dx
+ p(x)u = I(x), u(O) u(f)
0
< x < f,
= 0, = 0,
where and p satisfy satisfy k(x) > 0, 0, p(x] for x G [0,^]. is the the bilinear where k and p(x) > 00 for E [0, fl. What What is bilinear form BVP? form for for this this BVP? 6. Repeat Exercise 3 for for the BVP
J2u
1
- -2 +2u= --x dx 2' u(O) = 0, u(1) = O. The the last is The results results of of the last exercise exercise will will be be required. required. The The exact exact solution solution is
u(x) = clev'2x
+ cze-v'2x + ~ - ~,
202
Chapter 5. Boundary value problems problems in statics Chapter
where
Cl
= - 4 (e-v'2 _ ev'2) , ev'2
5.7 5.7
+1
Green's functions functions for for BVPs Green's BVPs
The zero zero function is always solution of of aa linear linear homogeneous equation. The function is always aa solution homogeneous differential differential equation. Suppose we we have a BVP, defined by a linear differential equation and Dirichlet or differential equation Neumann conditions, satisfying satisfying the property. Neumann boundary boundary conditions, the existence existence and and uniqueness uniqueness property. Then the BVP has a nonzero solution only if the differential differential equation equation is inhomogeThen neous inhomogeneous (or focus on the boundary boundary conditions conditions are are inhomogeneous (or both). both). We We will will focus on neous or or the the differential differential equation, assuming the boundary conditions are homogeneous, and present aa formula that expresses expresses the the solution terms of of the right-hand present formula that solution explicitly explicitly in in terms the right-hand side of differential equation. side of the the differential equation. As our our first first example, consider the BVP As example, we we consider the BVP
d (k(x) dx dU) = f(x),
- dx
0
< x < l,
u(O) = 0, du dx(l) =0.
(5.61)
The "data" for (5.61) the forcing f, which the spatial The "data" for (5.61) is is the forcing function function /, which is is aa function function of of the spatial variable x. x. By By two two integrations, integrations, we we can determine aa formula formula for in terms of f:/: variable can determine for u in terms of
d ( k(x) dx du (x) ) - dx du ~ k(x) dx (x) =
= f(x)
1xr fey) dy (since i
~~(l) = 0)
ri fey) dy r 1 rt ~ u(x) = 10 k(z) 1z f(y)dydz (since u(O) = 0) du
1
~ dx (x) = k(x)
~ u(x) =
r
i
1x
{ rmin{x,y} (
10 10
1) }
k(z)
dz
fey) dy.
(To verify the the reader reader should the domain integration for (To verify the last last step, step, the should sketch sketch the domain of of integration for the the double integral, integral, and the order integration.) We We can write the the solution double and change change the order of of integration.) can thus thus write solution as as
u(x)
= Io
i
g(x;y)f(y)dy,
(5.62)
5.7. Green's functions for BVPs
203
where rmin{x,y} (
g(x;y) = Jo
1 ) k(z) dz.
(5.63)
By we see is By inspection inspection of of formula formula (5.62), (5.62), we see that, that, for for aa particular particular x, the the value value u(x) u(x) is weighted sum sum of of the the data data f/ over over the the interval interval [O,.€]. [0,£]. Indeed, Indeed, for for aa given given x and and y, y, aa weighted g(x; g ( x ; yy)) indicates the effect effect on the solution u at x of the data f/ at y. The function function g is Green's function (5.61). 9 is called called the the Green's function for for BVP BVP (5.61). To the Green's To expose expose the the meaning meaning of of the Green's function function more more plainly, plainly, we we consider consider aa right-hand-side that is point xx =~. right-hand-side function function f/ that is concentrated concentrated at at aa single single point = £. We We define define d dAx Ax by x+Ax -Llx < x < 0, AX2 , dAu-X (x) = _x-Ax 0< x < Llx, AX2 , { otherwise 0,
(see Figure Figure 5.18) 5.18) and and let let f& d^x(x — ^). The The solution solution of of (5.61), (5.61), with with f/ = /A (see fAx(x) fAx, x(x) = dAx(X-~). X) is is
Since
l
{+AX
d Ax (y - ~) dy
=1
{-Ax
for we see for all all Llx, Ax, we see that that u(x) u(x) is is aa weighted weighted average average of of g(x; y) (considered (considered as as aa function function of y) over the interval ~£ -— Llx Ax ::; < yy ::; < ~£ + Llx. Ax. As As Llx Ax -+ —>• 0, 0, this this weighted weighted average average of y) over the interval converges to
fot Ahx (y) dy is the total force exerted on the entire bar. Since
fot hx(Y) dy = 1 for all Llx > 0 fAx is zero except on the interval [~~ + Llx], we see that the total pressure and /AX [£ — Llx, Ax,£+Ax], exerted on on the the bar bar is is A and and it it is is concentrated concentrated more and more more at at xx = ~£ as as Llx Ax -+ -> 0. exerted more and O. In unit acting In the the limit, limit, the the external external force force on on the the bar bar consists consists of of aa pressure pressure of of 11 unit acting on x = =~. response of the bar, is on the the cross-section cross-section of of the the bar bar at at x £. The The response of the bar, in in the the limit, limit, is u(x) g(x; ~). w(x) ==g(x]£).
204
Chapter 5. Boundary value problems in statics 10~------~-------r------~------~
9 8 7
6 5 4 3
2
O~------~-----L~~----~------~
-1
-0.5
0
0.5
x
Figure Figure 5.18. 5.18. The The function d/:;.x d&x (!lx (Ax = 0.1). O.I). Thus that, for fixed £, ~, g(x; g(x;~) Thus we we see see that, for aa fixed £) is is aa solution solution to to (5.61) (5.61) for for aa special special rightrighthand forcing function ~. However, this hand side-a side—a forcing function of of magnitude magnitude 11 concentrated concentrated at at x = = £. However, this forcing forcing function function is is not not aa true true function function in in the the usual usual sense. sense. Indeed, Indeed, the the function function d/:;.x d^\x has properties: has the the following following properties:
• d~x(x) = 0, x ft [-!lx, !lx]. • I: d/:;.x(x) dx = I~~x d/:;.x(x) dx = 1 for all !lx > 0, provided [-!lx, !lx] [a, b], and I: d/:;.x(x) dx = 0 if !lx ::; a or -!lx 2: b. • If 9 is a continuous function, and [-!lx, !lx]
c
c [a, b], then
is weighted average is aa weighted average of of 9g on on the the interval interval [-!lx, [—Ax, !lx]. Ax]. Since g(O) Since the the limit, limit, as as !lx Ax ~ —>• 0, 0, of of the the weighted weighted average average of of 9g over over [-!lx, [—Ax, !lx) Ax] is is #(0) when properties suggest that we d/:;.xx as when 9g is is continuous, continuous, these these properties suggest that we define define the the limit limit 86 of of d<\ as !lx Ax ~ —>• 00 by by the the following following properties: properties:
• 8(x)
= 0 if x # O.
• 8(0) =
00.
• J: 8(x)dx =
1 if 0 E [a,b], and I: 8(x)dx
= 0 if 0 rf- [a, b).
5.7. Green's functions for BVPs
205
If 9 g is is aa continuous continuous function function and and 0 E [a, b], then • If
o (j [a, b], then
J: 8(x)g(x) dx
J: 8(x)g(x) dx
= g(O). If
o.
=
The the property The "function" "function" 88 is is not not aa function function at at all-any all—any ordinary ordinary function function d has has the property that if is nonzero then the the integral integral of that if dd is nonzero only only at at aa single single point point on on an an interval, interval, then of dover d over that zero. However, useful to to regard generalized function. function. The that interval interval is is zero. However, it it is is useful regard 88 as as aa generalized The particular generalized function function 88 is delta function (or simply simply the the particular generalized is called called the the Dirac Dirac delta junction (or delta by the the siftinq sifting property: property: delta function) function) and and is is defined defined by
J:
If 9 g is is aa continuous continuous function function and and 00 E e [a, [0,6], 8(x)g(x}dxdx = If b], then then Ja 8(x)g(x) =
J:
c/(0). If If 00 (j <£ [a, 6], S(x}g(x) dx dx == O. 0. g(O). bj, then then fc 8(x)g(x) By we see By aa simple simple change change of of variables, variables, we see that that the the following following property property also also holds: holds:
J:
If 9 g is is aa continuous continuous function function and and ~£ E G [a, b], 6], then then J 8(x)g(x 8(x)g(x — £} dx dx == If -~) g(~). 0(0Since -x) = 8(x)), Since 8S is is formally formally an an even even function function (8( (8(—x) 8(x)), we we also also have have
J:
If 9g is is aa continuous continuous function function and and ~£ Ee [a, 6], then 8(x)g(£ -— x}dx = If [a, b], then J 8(x)g(~ x) dx = g(~). 0(0.
Since we regard regard the Since 86 is is not not aa function, function, but but aa generalized generalized function, function, we the last last two two properties properties as assumptions. assumptions. as From this discussion, discussion, we we see see that that the Green's function function 9g for for (5.61) (5.61) has the the Green's has the From this property that that u(x) u(x) = = g(x;~) g(x; £) is is the the solution solution of of the the BVP BVP property
-!
(k(X) ~~) = 8(x -
~),
0
u(O) = 0,
< x < £, (5.64)
dUe£) =0. dx We can verify this result result using using physical reasoning; this this is is particularly We can verify this physical reasoning; particularly simple simple in the the case case of of aa homogeneous homogeneous bar. consider the the following following thought thought experiment: experiment: in bar. We We consider We unit pressure the positive positive direction) the cross-section We apply apply aa unit pressure (in (in the direction) to to the cross-section at at x = = £~ E6 (0, e). (It not obvious pressure could be applied (0,^). (It is is not obvious how how such such aa pressure could actually actually be applied in in aa physical physical why we this aa "thought the bar is experiment, experiment, which which is is why we term term this "thought experiment.") experiment.") Since Since the bar is fixed at at xx = 0, 0, and and the the end end at at xx = — £t is free, it it is hard to to see see what what the the resulting fixed is free, is not not hard resulting displacement will be. part of the bar between xx = will displacement will be. The The part of the bar originally originally between = 00 and and xx = — ~£ will stretch from from its its original original length length of of £~ to to aa length length of of £~ + + ~~, A£, where stretch where
k~~ ~
= 1
(the pressure, pressure, 1, 1, is is proportional proportional to change in in length length (A£/£)). (the to the the relative relative change (~~/~)). Therefore, Therefore, which is the displacement u(~), u(£), is is ~/k. £/fc. Since Since the bar is the which is the displacement the bar is homogeneous, homogeneous, the displacement in displacement in the the Dart part of of the the bar bar above above xx = = £~ isis A£, ~~,
u(x) =
x
k'
0 ~ x ~ ~.
Chapter Chapter 5. 5. Boundary Boundary value value problems problems in in statics statics
206
What bar originally between x = ~£ and part of of What about about the the part part of of the the bar originally between and x = = l? 11 This This part the bar is not stretched at all; it merely moves because of the displacement in the displacement part of the bar above it. We We therefore obtain
Combining two formulas Combining these these two formulas gives gives
u(x)
={
~' k'
or simply
u(x) = min{x,~} k
(see Figure 5.19). Thus we ~), where 9 we see that u(x) u(x) = g(x; g(x;£), g is given by (5.63) (specialized to the case in which k is constant).
-
=g(x;O.6)
4.5
4 3.5
x
0.8
Figure 5.19. 5.19. The The Green's Green's function ( x ; yy), ] , yy = = 0.6, 0.6, in in the the case case of of aa Figure function gg(x; constant constant stiffness stiffness (k (k = = 200, 200, £ t == 1). I). Example = 1, k(x) k(x) = - x, and We Example 5.23. 5.23. Consider Consider BVP (5.61) (5.61) with with £1=1, = 21-x, and ff(x) (x) = = 1. 1. We have have g(x;y) =
l
dz
ffiin {X,Y}
o
-
2-z
5.7. Green's Green's functions functions for for BVPs BVPs 5.7.
207
= -log (2 _ z)l~in{x,y}
= log 2 -log (2 -
min{x, y})
_ {IOg2-10g (2- y ), log 2 - log (2 - x),
O
Therefore, Therefore, u(x) = =
fo1 g(x; y) dy
(since fey) = 1)
fox {log 2 -log (2 -
= x(1 + log 2) + (2 = x
5.7.1 5.7.1
v)} dy
+
11
{log 2 -log (2 - x)} dy
x) log (2 - x) - log 4 + (1 - x) (log 2 - log (2 - x))
+ log (1 - ~).
The Green's function and and the inverse of a differential operator operator
We write We write C![O,i] = {u
E
C 2 [O,i]
u(O) = ~~(i) = o}
and K : c;' [0, i] -+ C[O, i] by
d ( dU) .
Ku = - - k(x)dx dx
Our results above show show that, for each each f/ E € C[O, C[0,£], there is is aa unique solution to to Our results above that, for i], there unique solution Ku In other other words, an invertible operator. Moreover, Moreover, the solution to to Ku == ff.. In words, K K is is an invertible operator. the solution Ku = f is is u == Mf, C[0,l] -> C;'[O,i] C^[0,£] is is defined denned by Mf, where where M M:: C[O,i]-+ by (Mf)(x)
=
1£
g(x;y)f(y) dy.
Therefore, KM shows that that M == K Therefore, K M ff = /, f, which which shows K- x1,, the the inverse inverse operator operator of of K. It It follows that that follows (5.65) MKu = u for all u E C![O,i) must the differential must also also hold. hold. Thus Thus the the Green's Green's function function defines defines the the inverse inverse of of the differential operator. operator.
Exercises x 1. Use Use the the Green's Green's function function to solve (5.61) (5.61) when when it == 1, 1, /(#) 1, and and k(x) = =e 1. to solve f(x) == 1, eX..
2. into the the differential 2. Prove Prove directly directly (by (by substituting substituting u into differential equation equation and and boundary boundary conditions) that that the the function function uu defined defined by (5.62) and and (5.63) (5.63) solves solves (5.61). (5.61). by (5.62) conditions)
208
Chapter 5.
Boundary value problems problems in statics
3. Consider Consider the the BVP 3. BVP d ( k(x)dx dU) =/(x),O<x
:: (0) = 0, u(f)
= O.
(a) Using Using physical physical reasoning reasoning (as (as on on page page 205), 205), determine determine aa formula formula for for the the (a) Green's Green's function function in in the the case case that that k is is constant. constant. (b) Derive Derive the the Green's Green's function function for for the the BVP BVP (with (with aa possibly possibly nonconstant nonconstant (b) k} by by integrating integrating the the differential differential equation twice and and interchanging interchanging the the k) equation twice order order of of integration integration in in the resulting resulting double double integral. integral. (c) (c) Simplify Simplify the the Green's Green's function function found found in in Exercise Exercise 3b 3b in in the the case case that that the the stiffness stiffness kk is is constant. constant. Compare Compare to to the the result result in in part part 3a. 3a.
(d) Use Use the the Green's Green's function function to to solve solve (5.66) (5.66) for for f1=1, f ( x ) == x, k(x) = = l+x. 1 + x. (d) = 1, I(x) x, k(x) 4. 4. Consider Consider the the BVP BVP
d ( k(x) dU) - dx dx = I(x), 0 < x < f, u(O) = 0, u(f) = O.
(a) Using Using physical physical reasoning reasoning (as (as in in the the text), text), determine determine aa formula formula for for the the (a) Green's Green's function function in in the the case case that that kk is is constant. constant. (Hint: (Hint: Since Since the the bar bar is homogeneous, homogeneous, the the displacement displacement function function u(x) u(x) == g(x;y) g(x; y) (y fixed) fixed) will will is be piecewise piecewise linear, linear, with with u(O) be w(0) = = u(f) u(i) = = O. 0. Determine Determine the the displacement displacement of the the cross-section cross-section at at xx = yy by by balancing balancing the the forces forces and and using using Hooke's of Hooke's law, law, and and then then the the three three values values u(O), w(0), u(y), u(y], and and u(f) u(i] will will determine determine the the displacement function.) displacement function.) (b) the Green's the BVP, (b) Derive Derive the Green's function function for for the BVP, still still assuming assuming that that kk is is constant, constant, by integrating the the differential differential equation equation twice twice and interchanging the the order order by integrating and interchanging of integration integration in in the the resulting resulting double double integral. integral. Verify that you you obtain obtain the the of Verify that same formula formula for for the the Green's Green's function. same function. (c) (c) (Hard) Derive Derive the the Green's Green's function function in in the the case case of of aa nonconstant nonconstant stiffness stiffness k(x). k(x). 5. holds, and 5. Since Since the the BVP BVP (5.61) (5.61) is is linear, linear, the the principle principle of of superposition superposition holds, and the the solution to to solution
d ( k(x) dU) - dx dx = I(x), 0 < x u(O) = 0, kef) du (f) = p dx
< f,
209
5.7. Green's functions functions for BVPs
is is u(x) u(x) = — w(x) + + vex), v(x), where w solves solves (5.61) (5.61) (hence (hence w is is given given by by (5.62)) (5.62)) and and vv solves
d (k(x) dx dU) = 0, °< x < f,
- dx
u(o) = 0, du kef) dx (f) = p.
(5.67)
Solve (5.67) (5.67) and show that the solution can be written in terms of the Green's function function (5.63). 6. Prove directly that (5.65) holds. (Hint: Write
g(x; y) =
{
};Y dz fOx k~~)' Jo k(z)'
y < x, x
and rewrite
fot g(x; y)(Ku)(y) dy as the the sum of two integrals, integrals,
fox g(x; y)(Ku)(y) dy +
it
9(Xiy)(Ku)(y) dy.
Then apply apply integration integration by by parts parts to integral, and and simplify.) simplify.) Then to the the first first integral, 7. the Green's 7. Let g(x; y) be be the Green's function function for for the the BVP BVP
tPu
-k dx 2 = I(x),
°< x <
f,
u(o) = 0, u(f) = 0. The purpose of this problem g(x; y) in The purpose of this problem is is to to derive derive the the Fourier Fourier sine sine series series of of g(x] in three ways. Since y, its three different different ways. Since 9g depends depends on on the parameter parameter y, its Fourier Fourier sine coefficients coefficients will be functions functions of y. y. (a) (a) First First compute compute the the Fourier Fourier sine sine series series of of g(x; y) directly directly (that (that is, is, using the the formula formula for for the the Fourier Fourier sine sine coefficients coefficients of of aa given function), function), using the the formula formula for for 9g as as determined determined in in Exercise Exercise 4. 4. (b) Next, derive the the Fourier sine series of g(x; y) as follows: Write down the Fourier series solution to the BVP. Pass the summation summation through the I. Identify Identify integral the Fourier Fourier coefficients integral defining defining the coefficients of of the the right-hand right-hand side side /. the Fourier sine series of the Green's function function g(x; y). y).
210
Chapter 5. Boundary Boundary value problems in statics statics
(c) Find the Fourier Fourier sine sine series series of of gg(x;y) ( x ; y ) by by solving solving the BVP (c) Find the the BVP
cPu
= J(x u(O) = 0,
-k dx 2
y), 0 < x < J!.,
u(J!.) = 0 using the Fourier series method. Use Use the sifting property of the delta function to determine Fourier coefficients. determine its Fourier coefficients.
Verify that all three methods give the same result.
5.8
Suggestions for further reading reading
We have now introduced introduced the two primary topics of this book, Fourier series and finite elements. Fourier analysis has been important important in applied mathematics mathematics for 200 years, and there are many books on the subject. A classic introductory textbook textbook is Churchill and Brown [10], [10], which has been in print for over 60 years! A classic reference the theory Fourier series series is [53]. Two Two modern textbooks to the theory of of Fourier is Zygmund Zygmund [53]. modern textbooks reference to that methods are Bick that focus focus on Fourier Fourier series methods Bick [4] [4] and Hanna Hanna and Rowland Rowland [24]. [24]. One can can also also look look to books on on PDEs PDEs for for details about Fourier Fourier One to introductory introductory books details about series, series, in addition addition to other other analytic analytic and and numeric numeric techniques. techniques. Strauss Strauss [46] [46] is is aa wellwritten text that concisely surveys analytic methods. It contains a range of topics, from elementary to relatively advanced. Another introduction to PDEs that covers Fourier series well as as many many other topics, is Fourier series methods, methods, as as well other topics, is Haberman Haberman [22]. [22]. The finite element The finite element method method is is much much more more recent, recent, having having been been popularized popularized written at more demanding in the 1960s. Most textbooks textbooks on in the 1960s. Most on the the subject subject are are written at aa more demanding than is is ours. ours. Brenner Brenner and and Scott the theory of level than level Scott [6] [6] is is aa careful careful introduction introduction to to the theory of and Strang finite problems, while finite elements elements for for steady-state steady-state problems, while Johnson Johnson [27J [27] and Strang and and Fix Fix [45J both steady-state and time-dependent Ciarlet [11J [45] treat treat both steady-state and time-dependent equations. equations. Ciarlet [11] is is another another careful treatment of the the theory of finite elements for careful finite elements for steady-state problems. There is also the six-volume six-volume Texas Finite Element Element Series by Becker, Carey, Carey, and and Oden Oden [3], is also the by Becker, [3], which level. which starts starts at at an an elementary elementary level.
Chapter 6
Heat flow ow
and diffusion diffusion and
We now turn our attention to to time-dependent problems, still restricting ourselves ourselves We now turn our attention time-dependent problems, still restricting to problems in one spatial dimension. Since both time (t) and space (x) (x) are indedifferential equations will be PDEs, and we we will need initial pendent variables, the differential conditions as well as boundary conditions. We begin with with the the heat heat equation, equation, We begin au pc at -
/'i,
a2u ax 2 = f(x, t),
0< x < e, t > to·
This PDE PDE models models the the temperature temperature distribution distribution u(x,t) u(x, t) in in aa bar bar (see 2.1) or or This (see Section Section 2.1) the concentration u(x, t) of aa chemical chemical in solution (see 2.1.3). Since the concentration w(ar, t] of in solution (see Section Section 2.1.3). Since the the first first the unknown unknown appears in the need aa single time derivative derivative of of the time appears in the equation, equation, we we need single initial initial condition. condition. Our first first model model problem problem will will be be the the following initial-boundary value value problem problem Our following initial-boundary (IBVP): au pc at -
a 2u ax 2 = f(x, t),
0< x < e, t> to, u(x, to) = 'I/J(x), 0 < x < e, u(O, t) = 0, t > to, /'i,
u(e, t) = 0, t
(6.1)
> to.
method. We apply apply the the Fourier series method first and then turn to the the finite element element method We close close the the chapter look at Green's functions functions for for this and related related problems problems. We chapter with with aa look at Green's this and Along the the way way we we will will introduce new combinations of boundary boundary conditions. Along introduce new combinations of conditions.
6.1
Fourier Fourier series series methods methods for the heat equation
We now solve (6.1) using Fourier series. Since the unknown u(x, u(x,t]t) is a function function represented as a Fourier sine of of both time and space, it can can be represented sine series series in which which the the In other words, for each t, we have sine Fourier depend on on t. In other words, for each £, we have aa Fourier Fourier sine Fourier coefficients coefficients depend 211 211
212
Chapter 6. Heat Heat flow and and diffusion diffusion
series representation representation of of u(x, u(x,t)t) (regarded (regarded as as aa function function of of x). The The series series takes takes the the series form form
. (n1l'x) 2 (i . (n1l'x) u(x, t) = ~ ~ an(t) sm -,,- , an(t) = 10 u(x, t) sm -,,- dx.
e
n=l
0
The The function function u will will then then automatically automatically satisfy satisfy the the boundary boundary conditions conditions in in (6.1). (6.1). We We must must choose choose the the coefficients coefficients an(t) an(t) so so that that the the PDE PDE and and initial initial condition condition are are satisfied. satisfied. We way: We represent represent the the right-hand right-hand side side f(x, f ( x , t t) ] in in the the same same way: 00
f(x, t)
=
L cn(t) sin (n;x),
cn(t) =
~l
~
1 i
f(x, t) sin (n;x) dx.
0
We We then then express express the the left-hand left-hand side side of of the the PDE PDE as as aa Fourier Fourier series, series, in in which which the the (t), a2 (t), . ... To Fourier Fourier coefficients coefficients are are all all expressed expressed in in terms terms of of the the unknowns unknowns al ai(i), a^t], To do do this, this, we we have have to to compute compute the the Fourier Fourier coefficients coefficients of of the the functions functions
au
pc at (x, t)
and and
a2 u
-1'0 ax 2 (x, t). Just as as with with the the function function uu itself, itself, we we are are using using the the Fourier Fourier series series to to represent represent the the Just spatial spatial variation variation (Le. (i.e. the the dependence dependence on on x) x} of of these these functions. functions. Since Since they they also also depend on on t,t, the resulting Fourier coefficients will will be functions functions of of t.t. To compute compute the Fourier coefficients of
au 2
-1'0 ax2(x, t), we we integrate integrate by by parts parts twice, twice, just just as as we we did did in in the the previous previous chapter, chapter, and and use use the the boundary conditions conditions to eliminate the boundary terms: terms:
21'0 {i cPu
(n1l'x)
-7 10 ax2 (x, t) sin -e- dx
=
([au (x, t) sin (n1l'x)] £ _ n1l' {l au (x, t) cos (n1l'x) dX) ax " 0 " 10 ax " 21'On1l' {i au (n1l'x) ~ 10 ax (x, t) cos -,,- dx
~
2;:, ([
= _ 21'0
e
=
u(x, t) 008
21'On 2 1l'2 (i
,,3 10
I'On 21l'2
=
C;X) 1: + ne, [u(x, t) sin C;X) dx )
(n1l'X) u(x, t) sin -,,- dx
i 2 an (t).
6.1. Fourier series methods for the heat equation 6.1.
213
We used used the the conditions conditions sin (0) = = 0 0 and u(O, t]t) == u(l,t) u(£, t) == 0 0 in canceling We sin (0) = sin sin (n1f) (n?r) = and w(0, in canceling the boundary in the calculation. the boundary terms terms in the above above calculation. We now now compute the Fourier Fourier sine pc8uj8t. Using We compute the sine coefficient coefficient of of pcdu/dt. Using Theorem Theorem 2.1, 2.1, we have we have
d [2 [f. . (n1fx) 1 e210t· 8uat (x, t) sm. (n1fx) -f.- dx = dt e10 u(x, t) sm -f.- dx d
= dt[an(t)]
_ dan (t ) . dt
-
Thus of pcdu/dt pc8uj8t is is Thus the the nth nth Fourier Fourier coefficient coefficient of
We now We now see see that that
Since Since
CXl
f(x, t)
= L Cn(t) sin (n;x), n=l
the implies the PDE PDE implies
the coefficients al(t),a2(t),a3(t), .... we This is ODEs for This is aa sequence sequence of of ODEs for the coefficients ai(£),a2(£),a3(£), — Moreover, Moreover, we have have
u(x, to) = 'ljJ(x), 0 < x < f.
: :} I: n=l
an(to) sin (n;x) = I:bnsin(n;x), n=l
where b we obtain where 61,62; • • • are are the the Fourier Fourier sine sine coefficients coefficients of of 'ljJ(x). ijj(x). Therefore, Therefore, we obtair 1 , b2, ...
an(to) = bn , n = 1,2,3, .... These conditions conditions provide provide initial for the the ODEs. ODEs. To These initial conditions conditions for To find find an(t), an(t), we we solve solve the IVP the IVP dan K,n 21f2 (6.2) pCTt + ~an = cn(t),
an(to) = bn .
214
Chapter Chapter 6. Heat flow and diffusion
The coefficients ccn(t) and b bnn are computable from the given given functions functions ff(x, The coefficients are computable from the ( x , t t) ) and and n(t] and 'lj;(x). $(%)• The IVP IVP (6.2) (6.2) is is of of the the type type considered considered in 4.2.3, and and we we have have aa formula formula. The in Section Section 4.2.3, for the the solution: for solution: (6.3)
6.1.1 6.1.1
The homogeneous heat equation equation
We will will begin begin with with examples of the the homogeneous homogeneous heat heat equation. We examples of equation. Example 6.1. 6.1. We consider an iron bar, bar, of of length 50 cm, with with specific specific heat heat cc — = Example We consider an iron length 50cm, 0.437 J/(gK), density density pp = 7.88 g/cm g/ cm33,, and and thermal thermal conductivity conductivity KK, = = 0.836 0.836 W/(cmK). 0.437 J/(gK), = 7.88 Wf(cmK). We except at We assume that the bar is insulated insulated except at the the ends and and that that it it is (somehow) (somehow) given the initial temperature the initial temperature 'lj;(x)
1
= 5 - Six -
251,
where'lj;(x) is given given in in degrees degrees Celsius. Celsius. Finally, Finally, we we assume assume that, that, at time tt = 0, the the where ^(x) is at time — 0, ends of of the the bar bar are are placed placed in in an an ice degrees Celsius). will compute compute the ends ice bath bath (0 degrees Celsius). We We will the temperature distribution after after 20, 20, 60, and 300 300 seconds. seconds. temperature distribution 60, and We must solve solve the the IBVP We must IBVP au pc at -
a 2u ax 2 = 0, 0
< x < 50, t > 0, u(x,O) = 'lj;(x), 0 < x < 50, u(O, t) = 0, t > 0, K,
u(50, t) = 0, t > O.
The The solution solution is is 00
x u(x,t) = Lan(t) sin (n:a ) , n=l
where the the coefficient an(t) satisfies the where coefficient a the IVP IVP n(t) satisfies dan dt
+
K,n 2 7r 2 502pc an = 0, an(O)
and and
bn
= ~ (50 "1.( £ 10
The coefficient a ann is is given given by by The coefficient
'I' X
= bn
) . (n7rx) d sm 50 x
= 40 sin (n7r /2) 7r 2 n 2
'
215
6.1. Fourier series methods for the heat equation
where Ib, p, p, and have the given above. gives us explicit formula formula for where K, and cc have the values values given above. This This gives us an an explicit for the solution u, the solution u,
L bne 00
u(x, t) =
_
"n 2 ,,2 t 2 50 pc
n7fX
sin
(50)'
(6.4)
n=l
approximated by which can can be approximated by a finite finite series series of of the the form form
In 6.1, we we display "snapshots" of of the the temperature temperature distributions distributions at at the the times times In Figure Figure 6.1, display "snapshots" tt = 20, tt = 60, junction u(x,20), 60, and and tt = = 300; that that is, is, we we show show the the graphs graphs of of the the function w(x,20), u(x,QQ), and and u(x,3QO). This is is often often the the preferred way to to visualize visualize aa function u(x,60), u(x,300). This preferred way junction of of space and and time. time. space Sr-----~----~--~--~--~======~ u(x,O) u(x,20) 4.S u(x,60) 4 u x,300)
3.S ~ ~
3
til
Cii2.S c.. E
2
2
1.S
10
20
30
40
so
x
Figure 6.1. The The solution solution u(x, u(x,t)t) from 6.1 at times 0, 0, 20, 60, and Figure 6.1. from Example Example 6.1 at times 20, 60, and 300 (seconds). (seconds). Ten curves. 300 Ten terms terms of of the the Fourier Fourier series series were were used used to to create create these these curves. We now make several observations concerning the homogeneous heat equation equation and its solution.
1. For the Fourier coefficients an(t) an(t] decay decay exponentially exponentially as as nn -+ —>• 1. For aa fixed fixed time time tt > 0, 0, the Fourier coefficients
00 the second the homogeneous oo (cf. (cf. equation equation (6.3), (6.3), in which which the second term term is is zero zero for for the homogeneous n correspond heat equation). Since Since larger values of of n correspond to higher frequencies, frequencies, this this shows that the solution solution u(x, u(x,t)t) is is very smooth as as aa function function of for any any tt > 0. shows that the very smooth of xx for O.
216
Chapter 6. Heat flow and diffusion
This be seen the temperature This can can be seen in in Figure Figure 6.1; 6.1; at at tt = = 0, 0, the temperature distribution distribution has has aa 25. This singularity singularity (a (a point point of of nondifferentiability) nondifferentiability) at at xx == 25. This singularity singularity is is not not seen even after 10 10 seconds. seconds. In In fact, fact, it shown mathematically mathematically that the it can can be be shown that the seen even after solution u(x, u(x, t) t) is is infinitely differentiate as as aa function function of of xx for for every every tt > 0, infinitely differentiable 0, solution and this holds 'ljJ E fl. and this holds for for every every initial initial temperature temperature distribution distribution ty £ e[o, (7[0,^]. The between the the smoothness and the of decay The relationship relationship between smoothness of of aa function function and the rate rate of decay of be explained thoroughly in will of its its Fourier Fourier will will be explained thoroughly in Section Section 9.5.1. 9.5.1. For For now, now, we we will content with aa qualitative relationship. A content ourselves ourselves with qualitative description description of of this this relationship. A "rough" "rough" function with x) must have high frequency function (one (one that that varies varies rapidly rapidly with x] must have aa lot lot of of high frequency (mrx / f) with with nn large-change content (because high high frequency frequency modes-sin content (because modes—sin (rnrx/l) large—change rapidly rapidly with with x), and and therefore therefore the the Fourier coefficients coefficients must decay decay to zero slowly as as nn -t -» 00. oo. On On the other hand, hand, aa smooth smooth function, function, one one that changes the other that changes slowly must be be made slowly with x, must slowly with made up up mostly mostly of of low low frequency frequency waves, waves, and and so so the the Fourier x. Fourier coefficients coefficients decay decay rapidly rapidly with x. For example, a function we function with a discontinuity is "rough" in the sense that we are Fourier coefficients A are discussing, discussing, and and its its Fourier coefficients go go to to zero zero only only as as fast fast as as lin. 1/n. A function that is is continuous whose derivative function that continuous but but whose derivative is is discontinuous discontinuous has has Fourier Fourier coefficients the function coefficients that that decay decay to to zero zero like like 1/n 1/n22 (assuming (assuming the function satisfies satisfies the the same the eigenfunctions). continues: same boundary boundary conditions conditions as as the eigenfunctions). This This pattern pattern continues: Each to an Each additional additional degree degree of of smoothness smoothness in in the the function function corresponds corresponds to an addiadditional factor of n in the denominator of the Fourier coefficients. For examples, tional factor of n in the denominator of the Fourier coefficients. For examples, reader is the reader is referred to to Exercise Exercise 13 13 and, and, for for aa complete complete explanation, explanation, to Section Section 9.5.1. 9.5.1. The that the the solution the homogeneous homogeneous heat is very very The fact fact that solution u(x, u(x, t) t} of of the heat equation equation is smooth be understood physically as as follows: smooth can can be understood physically follows: If If somehow somehow aa temperature temperature distribution has aa lot the temdistribution that that has lot of of high high frequency frequency content content (meaning (meaning that that the temrapidly with x) is induced, and perature changes changes rapidly and then the the temperature disdistribution is will tribution is allowed allowed to to evolve evolve without without external external influence, influence, the the heat heat energy energy will quickly flow from high temperature regions to to low quickly flow from high temperature regions low temperature temperature regions regions and and smooth smooth out out the the temperature temperature distribution. distribution. 2. 00. 2. For each each fixed n, n, the the Fourier coefficient coefficient an(t) an(t) decays decays exponentially exponentially as as tt -t —>• oo. Moreover, the of Moreover, the rate rate of of decay decay is is faster faster for for larger larger n. This This is is another another implication implication of the fact temperature distribution the fact that that the the temperature distribution u becomes becomes smoother smoother as as tt increases, increases, and means that, as tt grows, grows, fewer and fewer terms in the Fourier series are to produce the solution t). W'hen required required to produce an an accurate accurate approximation approximation to to the solution u(x, u(x,t}. When tt is from the excellent is large large enough, enough, the the first first term term from the Fourier Fourier series series provides provides an an excellent approximation to approximation to the the solution. solution.
As an an illustration illustration of of this, we graph, graph, in in Figure Figure 6.2, 6.2, the difference between As this, we the difference between u(x, first term Fourier series 20, and u(x, t) t) and and the the first term of of its its Fourier series for for tt == 10, 10, tt == 20, and tt = = 300. 300. (We used used 10 10 terms of the the Fourier Fourier series series to approximate the the exact exact solution. solution. (We terms of to approximate Figure Figure 6.2 6.2 suggests suggests that this this is is plenty.) 3. The material characteristics characteristics of the bar appear appear in the formula formula for the solution
217 217
6.1. Fourier series methods for the heat equation
t=10 - _. t=60 .- - t=300
...
e
W
20
10
30
x
40
50
Figure approximating, at (secFigure 6.2. 6.2. The The error error in in approximating, at times times 0, 0, 20, 20, 60, 60, and and 300 300 (seconds), the u(x, t) from from Example with only the first Fourier onds), the solution solution u(x,t) Example 6.1 6.1 with only the first term term in in its its Fourier series. series. only in the the combination onlv in combination ",
pC£2·
For the case bar of have For example, example, in in the case of of the the iron iron bar of Example Example 6.1, 6.1, we we have
If the made of of aluminum instead, with with pp = — 2.70, 0.875, and and If the bar bar were were made aluminum instead, 2.70, cc = 0.875, ", = 2.36, 2.36, and and still had length length 50cm, 50 cm, we we would would have have K= still had 02'" 1;
pc
~ 3.996 . 10-4 .
An examination of of the solution (formula (formula (6.4)) (6.4)) shows shows that that the the temperature temperature An examination the solution u(x,t)t) will decay to to zero zero faster faster for for aa bar bar made of aluminum aluminum than for aa bar bar u(x, will decay made of than for made of Figure made of iron. iron. Figure Figure 6.3 6.3 demonstrates demonstrates this; this; it it is is exactly exactly analogous analogous to to Figure 6.1, 6.1, except except the the solution solution is is for for an an aluminum aluminum bar bar instead instead of of an an iron iron bar. bar. 6.1.2 6.1.2
Nondimensionalization
Point we cannot explicPoint 33 above above raises raises the the following following question: question: Suppose Suppose we cannot solve solve aa PDE PDE explicwe still identify the parameters (or itly. itly. Can Can we still identify the critical critical parameters (or combination combination of of parameters) parameters)
218
Chapter 6. Heat flow and diffusion diffusion 5r-----~----~--~--~--_r====~~
u(x,O) u(x,20) u(x,60) u(x,300)
4.5
4 3.5 ~
::J
3
0;2.5 0-
E
.sa
2 1.5
10
20
x
30
40
50
Figure 6.3. The temperature distribution distribution u(x,t) u(x, t) for for an an aluminum aluminum bar bar at at Figure 6.3. The temperature and 300 of the Fourier series series were were used times 0, times 0, 20, 20, 60, 60, and 300 (seconds). (seconds). Ten Ten terms terms of the Fourier used to to create these curves. Example 6.1, except the the iron iron create these curves. The The experiment experiment is is exactly exactly as as in in Example 6.1, except bar. bar is replaced replaced by an aluminum bar.
in the problem? problem? In the case of the the homogeneous heat equation, the parameters parameters p, p, in the In the case of homogeneous heat equation, the c, K, tr" and lt are not significant individually, but rather c, rather in the combination tr,/(pC£2). K/(pcl2). Can we deduce this from the equation itself, itself, rather from its solution? Can we deduce this from the equation rather than than from its solution? which It turns out out that this is is often often possible through through nondimensionalization, which is the process process of of replacing replacing the the independent (and sometimes the dependent is the independent variables variables (and sometimes the dependent variable) with nondimensional variables. variables. We We continue to consider flow in variable) with nondimensional continue to consider heat heat flow in aa bar. The independent variables variables are are xx and The spatial spatial variable has dimensions of bar. The independent and t. The variable has dimensions of cm, it is is rather rather obvious how to to nondimensionalize x-we describe cm, and and it obvious how nondimensionalize x—we describe the the spatial spatial location in terms terms of of the the overall overall length length tl of of the the bar. is, we we replace replace x with location in bar. That That is, with
y=
e·x
Since both x and units of of centimeters, centimeters, y is dimensionless. Since both and tl have have units is dimensionless. It is more difficult difficult to to nondimensionalize nondimensionalize the the time time variable, but there It is more variable, but there is is aa general general technique for creating nondimensional nondimensional variables: List all parameters appearing in technique for creating variables: List all parameters appearing in the with their and find combination of the parameters parameters the problem, problem, together together with their units, units, and find aa combination of the that has the same units as as the variable you you wish wish to to nondimensionalize. nondimensionalize. In that has the same units the variable In this this problem, the the parameters parameters and and their their units units are are as as follows: problem, follows:
6.1. 6.1.
Fourier series methods methods for for the the heat Fourier series heat equation equation
parameter parameter
units 3 g/cm g/cm3 J/(gK) J/(scmK) J/(s cmK) cm cm
Pp cc K
'"
" t
Then the ratio
219 219
pd2
'"
is have units units of of seconds, so we we define dimensionless time time variable variable ss by by is seen seen to to have seconds, so define aa dimensionless t
",t
s = -2 - = - -2. pd /", pd
= u(x, t), where (y, s) and (x, We then define v(y, v ( y , ss)) = u(x,t), (y,s) ( x , tt)) are related by ",t
x
Y=
Then, by the chain rule, AU
at and and
Therefore, the PDE PDE
is equivalent to to is equivalent
which simplifies to
e'
s = pd2 •
avos as at
220
Chapter 6. Heat flow and diffusion
Since xx = £y and and xx = corresponds to to yy = = 1, the IBVP IBVP (6.1), (6.1), with with ff(x, = 0, 0, Since = iy — t£ corresponds 1, the ( x , t) = becomes becomes
av a2 v as - ay2
= 0,
0< y < 1, s > so,
= 'ljJ(£y), 0< y < 1, v(O,s) = 0, s > so, v(1,s) =0, s > so,
v(y, so)
(6.5)
2 where sSo0 = = Kt /'1,tO/(pC£2). The solution of (6.5) (6.5) is is computed computed as above; the the result result is is where solution of as above; 0/(pcl ). The 00
v(s, y)
=L
bne-n27r2S
sin (mry).
n=l
The Fourier sine sine coefficients of 'ljJ(£y) considered as function of the interval The Fourier coefficients of if)(ly), , considered as aa function of yyon on the interval [0,1], are same as as the sine coefficients of i})(x) as the [0,1]' are the the same the Fourier Fourier sine coefficients of'ljJ(x) as aa function function of of xx on on the interval [0, [0,1] (see Exercise 6). interval £] (see Exercise 6). What did did we we gain gain from from nondimensionalization? nondimensionalization? We We identified identified the the natural What natural time scale, namely, pcl pC£22/K, / /'1" for the experiment, experiment, and and we we did this without without solving time scale, namely, for the did this solving the equation. equation. Whether Whether we we have have an an iron iron or bar, the the problem problem can can be be the or aluminum aluminum bar, written as as (6.5). This shows shows that that the the evolution of the the temperature temperature will will be be exactly written (6.5). This evolution of exactly the same same for the two two bars, bars, except except that that the the temperature temperature evolves different time the for the evolves on on different time scales, depending on the material properties of the bars. Moreover, we see exactly scales, depending on the material properties of the bars. Moreover, we see exactly how this this time time scale the various various parameters, parameters, and and this this can can give give some how scale depends depends on on the some information that that is not immediately immediately obvious obvious from the equation. For example, it information is not from the equation. For example, it the equation that, holding holding p p constant, bar with with specific specific heat heat cc is obvious is obvious from from the equation that, constant, aa bar and thermal thermal conductivity conductivity K/'1, will will behave behave the the same same as bar with with specific heat 2c 2c and and as aa bar specific heat and thermal conductivity 2/'1,. 2«. However, at all and cc thermal conductivity However, it it is is not not at all obvious obvious that, that, holding holding pp and constant, aa bar bar of of length length t£ with with thermal thermal conductivity conductivity K /'1, will will behave behave just just as as aa bar bar of of constant, 35 with thermal thermal conductivity conductivity 4«. 4/'1,.35 length 2£ length 21 with
6.1.3 6.1.3
The inhomogeneous heat equation
We now now consider consider an an inhomogeneous inhomogeneous problem. problem. We
Example 6.2. coefficient for for carbon carbon monoxide monoxide (CO) in air air is D = Example 6.2. The The diffusion diffusion coefficient (CO) in is D = 0.208cm2 consider aa pipe pipe of of length cm, filled filled with with air, air, that that initially 0.208cm 2//s.s. We We consider length 100 100cm, initially of the the pipe pipe open open onto onto large large reservoirs reservoirs contains no no CO. assume that that the two ends contains CO. We We assume the two ends of of air, air, also also containing containing no no CO, that CO produced inside inside the the pipe pipe at at aa constant constant of CO, and and that CO is is produced 3 rate 10~77 g/cm We write t}, 00 < 100, for rate of of 10g/ cm3 per per second. second. We write u(x, u(x, t), < xx < 100, for the the concentration concentration 3 of in air, model uu as of the IBVP of CO in air, in in units units of of g/cm g/ cm3,, and and model as the the solution solution of the IEVP
co
au a2 u -7 at - D ax 2 = 10 ,0 < x < 100,
t
> 0,
u(x,O) = 0, 0 < x < 100, 35
35For thorough discussion nondimensionalization and scale models, see [36], Chapter For a thorough discussion of nondimensionalization scale models, see [36], Chapter 6.
6.1. Fourier series methods for the heat equation
221 221
u(O, t) = 0, t > 0, u(100, t) = 0, t > O. We wish wish to to determine determine the the evolution evolution of of the the concentration concentration of of CO CO in in the the pipe. pipe. We We write the u in We write the solution solution u in the the form form 00
u(x, t) =
L an(t) sin (~~~), n=l
so that that so
We have We have 10
-7
~ Cn sm . (mrx) 100 '
= ~
n=l
where where 2
(IOO
en = 100 io
_ .
mrx 2.1O- 7 (1-(-1)n) 10 7 sm ( 100) dx = mr ' n = 1,2,3, ....
To find find the the coefficients an(t) of of u(x, u(x,t), we must must solve solve the the IVPs IVPs To coefficients an(t) t), we
dan dt
+
Dn 2 Jr2 1002 an = cn ,
an(O) =
o.
Using (6.3), we we obtain Using obtain
In Figure Figure 6.4, we show show the pipe at 3600 (1 In 6.4, we the concentration concentration of of CO CO in in the the pipe at times times tt == 3600 hour), tt = = 7200 7200 (2 (2 hours), hours), tt = = 14400 14400 (4 (4 hours), = 21600 (6 hours), = 360000 360000 hour), hours), tt = 21600 (6 hours), andt andt = (100 hours). The approaches steady state (the (the (100 hours). The graph graph suggests suggests that that the the concentration concentration approaches steady state reader should notice little the the concentration concentration changes changes between between 44 hours and 100 hours and 100 reader should notice how how little hours steady state state below hours of of elapsed elapsed time). time). We We discuss discuss the the approach approach to to steady below in in Section Section 6.1.5. 6.1.5.
222
Chapter 6. Heat flow and diffusion
t=3600 - _. t=7200 ._.- t=14400 t=21600 t=360000
6 5
----- .......
2
20
40
60
80
100
x
Figure pipe of Figure 6.4. 6.4. The The concentration concentration of of co CO in in the the pipe of Example Example 6.2 6.2 after after 1, 1, 2, 4, 6, and these and 100 100 hours. hours. Ten Ten terms terms of of the the Fourier Fourier series series were were used used to to create create these curves. curves.
6.1.4 6.1.4
Inhomogeneous conditions In homogeneous boundary boundary conditions
Inhomogeneous boundary conditions in in aa time-dependent time-dependent problem problem can can be handled Inhomogeneous boundary conditions be handled by the method by the method of of shifting shifting the the data, data, much much as as in in the the steady-state steady-state problems problems discussed discussed in with an in the the last last chapter. chapter. We We illustrate illustrate with an example. example. Example suppose that Example 6.1 to aa constant Example 6.3. 6.3. We We suppose that the the iron iron bar bar of of Example 6.1 is is heated heated to constant temperature placed in temperature of of 44 degrees degrees Celsius, Celsius, and and that that one one end end (x (x = = O) 0) is is placed in an an ice ice bath bath (0 (0 degree degree Celsius), Celsius), while while the the other other end end is is maintained maintained at at 4 4 degrees. degrees. What What is is the the temperature temperature distribution distribution after after 55 minutes'? minutes ? The IEVP is The IBVP is {)u
{)2u
pc {)t -
K,
{)x 2
= 0, 0 < x < 50,
u(x,O) = 4, 0
= 0,
t > 0,
< x < 50,
> 0, u(50, t) = 4, t > O. u(O, t)
t
Since = 4x/50 4#/50 satisfies satisfies the the boundary boundary conditions, conditions, we we will will define define v(x, v(x,t)t) == Since p(x) p(x) = u(x,t) -— p(x). Then, as as is is easily easily verified, verified, vv satisfies satisfies u(x,t) p(x). Then, {)v
pc {)t -
{)2v K,
{)x 2 = 0, 0
< x < 50,
t
> 0,
6.1. 6.1.
Fourier series Fourier series methods methods for for the the heat heat equation equation
223
4 v(x,O) = 4 - 50x, 0< x < 50,
v(O, t) = 0, t > 0, v(50, t) = 0, t > O. The The initial initial temperature temperature satisfies satisfies oo
4 4- -x= 50 where where
L b sm (nnx) -50' .
n=l
n
8 bn = - , n = 1,2,3, .... nn
The The solution 00
v(x, t)
=L
an(t) sin (n;x)
n=l is determined by the IVPs is determined by the IVPs dan dt
+
K,n 2n 2 pc502 an = 0, an(O) = bn ,
so the solution so the solution is is
L -nn8e _ ~ sin (nnx -). 50 «n 2 ,,2
00
v(x,t) =
t
n=l by The solution u u is given by The solution is then then given 4
L 00
8
u(x, t) = -x + -e 50 n=l nn
_"n 2 .. 2 t pc50
2
nnx sin ( - ) . 50
The temperature temperature distribution is shown The distribution at at tt = 300 300 is shown in in Figure Figure 6.5. 6.5. In aa time-dependent time-dependent problem, problem, inhomogeneous inhomogeneous boundary boundary conditions conditions can can themIn themselves be time time dependent; selves be dependent; for for example, example,
u(O, t) = g(t), u(£, t) = h(t), t > 0,
are valid The function function are valid inhomogeneous inhomogeneous Dirichlet Dirichlet conditions. conditions. The p(x, t) = g(t)
x
+ C(h(t)
- g(t))
satisfies these conditions. The reader reader should should notice notice that the function v(x, t) = satisfies these conditions. The that the function v(x,i) u(x, t) -—p(x, p(x, t) will not not satisfy the same same PDE as does does u; the right-hand right-hand side will be u(x, t) t] will satisfy the PDE as w; the side will be different. See Exercise Exercise 5. 5. different. See
224 224
Chapter 6. Heat flow and diffusion diffusion
5 4.5
~
:::J
,,
3
~ (1)2.5 c.. E 2 2
,
,
,,
I
40
50
~
--- ---
4 3.5
t-O 1 - _. t=300
.. ,.
I
I
I I
I I
1.5
I I I I I I I
0.5
I I I
00
10
20
x
30
Figure 6.5. The temperature Example 6.3). temperature distribution after after 300 seconds (see (see Example
6.1.5 6.1.5
Steady-state Steady-state heat flow flow and diffusion
We discuss the the special special case or diffusion diffusion in the source source function We now now discuss case of of heat heat flow flow or in which which the function f(x, t) appearing the right-hand right-hand side /(#,£) appearing on on the side of of the the differential differential equation equation is is independent independent of tt (this (this is is the the case case in Example 6.2). 6.2). In In this case, the the Fourier Fourier coefficients coefficients of of f/ are are in Example this case, of constant with and formula constant with respect respect to to t, and formula (6.3) (6.3) can can be be simplified simplified as as follows: follows:
As —>• oo, term tends tends to to zero zero for for each each n, n, and and so As tt -+ 00, the the first first term so
This This implies implies that 00
u(x,t) -+ Us(x)
=
L n=l
£2
",~27f2 sin (n;x).
6.1. Fourier Fourier series series methods methods for for the the heat heat equation equation 6.1.
225
The limit us(x) the steady-state steady-state temperature not difficult The limit us(x) is is the temperature distribution. distribution. It is is not difficult to to show that us is is the the solution solution to show that Us to the the BVP BVP
cPU
-r;, daP
= f(x), 0
< x < f,
= 0, u(f) = 0
(6.6)
u(O)
(see be noted that the the initial initial temperature is (see Exercise Exercise 2). It It should should be noted that temperature distribution distribution is irrelevant in determining the the steady-state temperature distribution. irrelevant in determining steady-state temperature distribution. Similar results results hold hold for for the the diffusion diffusion equation, equation, which which is is really no different different from from Similar really no parameters. Indeed, 6.2 the heat equation except meaning of the heat equation except in in the the meaning of the the parameters. Indeed, Example Example 6.2 is an illustration; illustration; the concentration u steady-state concentration, is an the concentration u approaches approaches aa steady-state concentration, as as suggested by 6.4 (see Exercise 1). 1). (see Exercise suggested by Figure Figure 6.4 In the the next next section, section, we we will will turn turn our our attention attention to to two two new sets of of boundary In new sets boundary conditions the heat Fourier series that apply conditions for for the heat equation equation and and derive derive Fourier series methods methods that apply to to the new new boundary however, we briefly discuss the boundary conditions. conditions. First, First, however, we briefly discuss another another derivation derivation of the Fourier series series method. method. of the Fourier
6.1.6 6.1.6
Separation variables Separation of of variables
There is is another another way way to to derive derive Fourier series solutions solutions of of the the heat heat equation, which There Fourier series equation, which is called the method method of of separation separation of of variables. in some more elementary is called the variables. It It is is in some ways ways more elementary than the point of have presented, presented, but but it it is than the point of view view we we have is less less general. general. use the the following To the method of variables, variables, we To introduce introduce the method of of separation separation of we will will use following IBVP: IBVP: {)u
{)2u
pc {)t -
r;, {)x 2
u(x, to) u(O, t) u(f, t)
< x < f, t > to, = 'l/J(x), 0 < x < f, = 0, t > to, = 0, t > to. = 0, 0
(6.7)
The reader reader should should notice notice that that the PDE is is homogeneous; this is is necessary necessary and The the PDE homogeneous; this and aa significant this technique. technique. significant limitation limitation of of this Here is the the key idea of of the method of of separation separation of of variables: look for for Here is key idea the method variables: we we look separated solutions: t) = X(x)T(t) X(x)T(t) (X are functions functions of separated solutions: u(x, t) (X and and T are of aa single single variable). variable). Substituting; uu into into the the PDE PDE vields Substituting yields {)u U
= XT, pc {)t -
{)2U r;, {)x 2
= 0
d2 X dx
dT dt
=> pcX- - r;,T--2 = 0
cP X => -r;,T-2 => -r;,X
dT = -pcXdt
dx _ld2 X
d,x2 = -peT
_ldT
di·
226 226
Chapter Chapter 6. 6. Heat Heat flow flow and and diffusion diffusion
The XT. The The last last step step follows follows upon dividing dividing both both sides sides of of the the equation equation by by XT. The crucial crucial observation the following: observation is is the following:
is aa function function of of x alone, alone, while is while dT
-pcT- 1 dt is is aa function function of of tt alone. alone. The The only only way way these these two two functions functions can can be be equal equal is is if if they they are are both both constants: constants: -1 J2 X -1 dT -liX dx 2 = -pcT dt = A. We X(x) and We now now have have ODEs ODEs for for X(x) and T(t): T(t) -Ii
d2 X dx 2
= AX,
-pc
dT
dt = AT.
Moreover, boundary conditions Moreover, the the boundary boundary conditions conditions for for uu imply imply boundary conditions for for X: X:
u(O, t) u(l, t)
= 0 :::} X(O)T(t) = 0 :::} X(O) = 0, = O:::} X(l)T(t) = O:::} X(l) = 0
(the function function T(t) T(t) cannot cannot be zero, for for otherwise otherwise u is is the the trivial trivial solution). solution). (the be zero, We see see that that the the function function X must satisfy satisfy the the eigenvalue eigenvalue problem problem We X must
J2x -Ii
dx 2 = AX, 0 < x < l,
X(O) = 0, X(l) = O. From we know From our our earlier earlier work, work, we know that that there there is is aa sequence sequence of of solutions: solutions:
n21f2
An =~, Xn(x)
= sin (n1fx) -l- , n = 1,2,3, ....
For each each value value of of A we can can solve solve the ODE For An, the ODE n , we dT
-pc- = AT
dt
to find corresponding function function Tn: Tn: to find aa corresponding Tn(t)
= bne
_ An(t-·o) pc
,
n
= 1,2,3, ...
(bn is is an an arbitrary arbitrary constant). constant). (b We have now to the the homogeneous We have now found found aa sequence sequence of of separated separated solutions solutions to homogeneous heat heat equation, namely, namely, equation, un(x,t)=bne
_"'n(t-to) pc
•
(n1fX)
sm -l- ,n=1,2,3, ....
6.1. Fourier series methods for the heat equation
227
By the homogeneous Dirichlet By construction, each each of of these these functions functions satisfies satisfies the Dirichlet conditions. tions. To initial condition, condition, we superposition (which (which To satisfy satisfy the the initial we use use the the principle principle of of superposition applies because the the heat heat equation to form form the the solution applies because equation is is linear) linear) to solution <Xl
U(X, t) =
~
~
An(t-tO) pc
bne-
sin
(nnx)
-c- .
n=l
We have We then then have
<Xl
Lb
u(x, to) =
n
sin (n;x).
n=l
If we choose choose 61, b1 , b%, b2 , 63,... b3 , ••• to to be be the the Fourier Fourier sine If we sine coefficients coefficients of'lj;, of i/»,
bn =
~
1i
'lj;(x) sin (n;x) dx,
then u(x, to) == 'lj;(x) will hold. hold. then u(x,to) i^(x] will We the solution We thus see see that the method of separation separation of of variables variables yields the solution we obtained However, the method obtained earlier earlier (see Section Section 6.1.1). 6.1.1). However, method requires that the PDE PDE and the the boundary boundary conditions conditions be homogeneous, and For and be homogeneous, and this this is is aa serious serious limitation. limitation. For to aa problem problem with with inhomogeneous, example, one cannot cannot apply apply separation of variables example, one separation of variables to inhomogeneous, time-dependent boundary conditions. The time-dependent The method does not apply directly, and one the trick trick of the data, it would would result result in inhomogeneous cannot use cannot use the of shifting shifting the data, since since it in an an inhomogeneous PDE. PDE. We will find variables useful useful in when we we derive We will find separation separation of of variables in Section Section 8.2, 8.2, when derive the the eigenvalues and on aa rectangle. eigenvalues and eigenfunctions eigenfunctions for for aa differential differential operator operator defined defined on rectangle.
Exercises 1. What is is the the steady-state of Example 6.2? What What BVP does it 1. What steady-state solution solution of Example 6.2? BVP does it satisfy? satisfy?
2. Find Fourier sine sine series series of of the solution of (6.6) and show that equals 2. Find the the Fourier the solution of (6.6) and show that it it equals us(x). u8(x). 3. Solve the following using the the Fourier Fourier series series method: 3. Solve the following IBVP IBVP using method: {)u fPu {)t - {)x2
= 0,
°< °<
u(x,O) = x, u(O, t) = 0, t
u(I, t) =
< 1, t > 0, x < 1,
x
> 0, 0, t > 0.
Produce an Produce an accurate accurate graph graph of of the the "snapshot" "snapshot" u(·,O.I). w(-,0.1). 4. Solve the following IBVP using using the the Fourier 4. Solve the following IBVP Fourier series series method: method: {)u
{)2u
.
{)t - {)x2 = sm (t),
°<
x
< 1, t > 0,
228
Chapter 6. Heat flow and diffusion
°
= 0, < x < 1, = 0, t > 0, = 0, t > 0.
u(x,O) u(O, t) u(l, t)
Produce an accurate graph of the "snapshot" "snapshot" u(·,O.I). w(-,0.1). 5. boundary conditions: 5. Consider Consider the the heat heat equation equation with with inhomogeneous inhomogeneous boundary conditions:
au pc at
a2 u
- ~ ax2
°< < °< <
= f(x, t),
x
u(x, to) = ¢(x), x u(O, t) = g(t), t > to, u(f, t) = h(t), t > to. Define Define
p(x, t) = g(t)
x
+ e(h(t) -
f, t> 0,
f,
g(t))
and v(x, v(x,i)t) = = u(x, u(x,t)t) -— pp(x, ( x , tt). ) . What IBVP does v satisfy? satisfy?
f), and 1> 1) is defined by 1>(y) 6. Suppose ¢ VE e C[O, C[0,4 (f> E e C[O, C[0,1] (j>(y) = = ¢(fy). ^(iy). Show that the sine coefficients of of 1> as the of ¢ the sine coefficients <j> on on [0,1) [0,1] are are the the same same as the sine sine coefficients coefficients of ty on on [0, f). [0,4 7. Use the method of shifting the data to solve the following IBVP with inhomogeneous Dirichlet conditions: 2
°°
au _ a u2 = < x < 1, t > 0, ' at ax u(x, 0) = x, 0 < x < 1, u(O, t) = 0, t > 0, u(l, t) = cos (t), t > 0. If If possible, produce an accurate accurate graph of the "snapshot" "snapshot" u(·, w(-, 1.0).
8. an iron iron bar, of diameter cm and and length m, with with specific 8. Consider Consider an bar, of diameter 44cm length 11m, specific heat heat 0.437 J/(gK), J/(gK), density density pp = 7.88g/cm 7.88g/cm33,, and and thermal thermal conductivity conductivity ~K = cc = 0.437 0.836 W/(cm W / (cm K). Suppose that the bar is insulated except at the ends, is heated Celsius, and the ends are placed heated to a constant constant temperature temperature of 5 degrees Celsius, in an an ice degrees Celsius). Celsius). Compute Compute the (accurate to in ice bath bath (0 (0 degrees the temperature temperature (accurate to 33 digits) at the midpoint of the bar after after 20 20 minutes. (Warning: Be sure to use consistent consistent units.)
9. Repeat Repeat Exercise with aa copper copper bar: specific heat = 0.379J/(gK), 0.379 J/(gK), density density 9. Exercise 88 with bar: specific heat cc = pp = = 8.97g/cm 8.97 g/cm33,, thermal thermal conductivity conductivity K ~= = 4.04 4.04 W/(cmK). W /(cmK). 10. 10. Consider the bar of Exercise 8. Suppose the bar is completely insulated except at the ends. The bar is heated heated to a constant constant temperature temperature of 5 degrees Celsius, and then the right end is placed in an ice bath (0 degrees Celsius), Celsius), while the other end is maintained maintained at 5 degrees Celsius.
6.2.
Neumann conditions and the Fourier cosine series Pure Neumann
229
(a) What What is is the the steady-state steady-state temperature temperature distribution distribution of of the bar? (a) the bar? (b) How How long long does does it it take to reach steady-state (to (to within within 11%)? (b) take the the bar bar to reach steady-state %)? 11. Repeat the bar of Exercise 9. 9. 11. Repeat the preceding preceding exercise exercise with with the the bar of Exercise
12. Consider Consider aa 100cm 100 cm circular circular bar, with radius radius 44cm, cm, designed designed so so that thermal 12. bar, with that the the thermal conductivity is is nonconstant nonconstant and and satisfies satisfies II;K(X) = 11 + + o::x ax W/(cmK) for some (x ) = W / (cm K) for some conductivity a > O. 0. Assume Assume that that the the sides sides of of the the bar bar are are completely completely insulated, insulated, one one end 0:: end (x == 0) 0) is at 00 degrees degrees Celsius, Celsius, and and heat is added added to to the other end end at at (x is kept kept at heat is the other rate of of 44 W. W. The The temperature temperature of of the the bar bar reaches reaches aa steady steady state, state, uu == u(x), u(x), aa rate and the the temperature at the end xx == 100 100 is is measured measured to be and temperature at the end to be
u(100) == 7.6 degrees Celsius. Estimate Estimate 0::. a. 13. 13. Compute Compute the the Fourier Fourier sine sine coefficients coefficients of of each each of of the the following following functions, functions, and and verify the the statements statements on on page page 216 concerning the the rate rate of of decay decay of of the the Fourier verify 216 concerning Fourier coefficients. the interval be [0,1].) coefficients. (Take (Take the interval to to be [0.11.) (a) f(x) = {
x,
Os x < 1/2,
2 - 2x,
1/2 S x S 1.
(b)
6.2 6.2
{
s
0 x < 1/2, 1'-x,I/2SxS1.
X
f(x)=
Pure Neumann Neumann conditions conditions and and the the Fourier Fourier cosine Pure cosine . series series
In this section section we we will consider the effect of of insulating insulating one one or or both both ends ends of of aa bar In this will consider the effect bar in in which is flowing. flowing.3636 With the new new boundary boundary conditions conditions that result, we must use which heat heat is With the that result, we must use different eigenfunctions in in the the Fourier Fourier series series method. Of particular particular interest interest is is the different eigenfunctions method. Of the case in in which ends of of the the bar bar are are insulated, insulated, as as this this raises some mathematical mathematical case which both both ends raises some questions that we we have yet encountered. encountered. questions that have not not yet 6.2.1 6.2.1
One end insulated; mixed boundary conditions
We begin with the case in in which which one one end end of of the the bar is insulated, insulated, but the other other is is not. not. We begin with the case bar is but the In Section Section 2.1, 2.1, we saw that an insulated insulated boundary boundary corresponds corresponds to to aa homogeneous In we saw that an homogeneous Neumann condition, which indicates that no heat energy energy is is flowing through that Neumann condition, which indicates that no heat flowing through that end of of the consider aa bar bar of of length length C, t, perfectly perfectly insulated insulated on on the sides, and and end the bar. bar. We We consider the sides, assume that that one one end end (x = = C) 1} is is perfectly perfectly insulated insulated while while the the other other end end (x = 0) assume 0) 36 360r both ends pipe in Or closing closing one one or or both ends of of aa pipe in which which aa chemical chemical is is diffusing. diffusing.
230
flow and diffusion Chapter 6. Heat flow
is uninsulated uninsulated and and placed in an an ice ice bath. If the the initial initial temperature temperature distribution distribution is is placed in bath. If is ^(x), then then the the temperature temperature distribution distribution u(x, u(x,t)t) satisfies satisfies the the IBVP IBVP 'l/J(x), au pc at -
/l,
°< < £, °< < £,
a2u ax 2 = f(x, t),
u(x, to) = 'l/J(x) , u(O, t) = 0, t au ax (£, t) = 0, t
x
t> to,
(6.8)
x
> to, > to·
The negative second The eigenvalue/eigenfunctions eigenvalue/eigenfunctions pairs pairs for for L Lmm,, the the negative second derivative derivative operator, operator under conditions (Dirichlet (Dirichlet at at the the left, left, Neumann at the the right) right under the the mixed mixed boundary boundary conditions Neumann at are are (2n-1)21f2 . ((2n-1)1fx) 4£2 ' sm 2£ ' n = 1,2,3, ... (see we represent represent the (see Section Section 5.2.3). 5.2.3). If If we the solution solution as as
~ . ((2n - l)1fX) u(x, t) = ~ an(t) sm 2£ ' then uu will satisfy the the boundary boundary conditions. conditions. Just Just as as in in the the previous section, we then will satisfy previous section, we can derive derive an an ODE ODE with an initial initial condition condition for for each each coefficient coefficient an(t) an(t) and and thereby thereby can with an determine u. u. We We will with an an example. example. determine will illustrate illustrate with Example 6.4. the temperature distribution in Example Example 6.4. We We consider consider the temperature distribution in the the iron iron bar bar of of Example 6.1, with the 6.1, with the experimental experimental conditions conditions unchanged unchanged except except that that the the right right end end (x (x = = £) I) of the bar perfectly insulated. of the bar is is perfectly insulated. The The temperature temperature distribution distribution u(x, u(x,t)t) then then satisfies satisfies the the IBVP IBVP au pc at -
/l,
a 2u ax 2 = 0,
°< < °< x
50, t > 0,
u(x,O) = 'l/J(x), x u(O, t) = 0, t > 0, du dx (50, t) = 0, t > 0,
< 50,
3 where = 7.88 7.88g/cm = 0.437 QA37J/(gK), 0.836 W/ W/(cmK), where p p= g/ cm3,, cc = J/ (g K), K/l, == 0.836 (cm K), and and
'l/J(x) We have We have
1
= 5 - Six -
251·
~ . ((2n-1)1f) 'l/J(x) = ~cnsm 100 '
where where cn
_ ~ (50.,,( ) . ((2n -l)1fX) d 50 J0 'f' X sm 100 x
-
=-
80 (-\1'2 sin (n1f /2) + \1'2 cos (n1f/2) - (-l)n) 2()2 ' n 1f 2n-1
= 1,2,3, ....
6.2.
Pure Neumann Neumann conditions and the Fourier Fourier cosine series
231 231
We write the solution u u in in the form We write the solution the form
. ((2n 100 - 1)1fX) .
~
u(x, t) = ~ an(t) sm
Then, same computation computation as as on on page page 212, we have have Then, by by almost almost exactly exactly the the same 212, we
The PDE and and initial condition for for u u then an(t) The PDE initial condition then imply imply that that the the coefficient coefficient a must satisfy satisfy n(t) must the following IVP: the following IVP: dan pc dt
+
~(2n
- 1)21f2 4.50 2 an = 0, an(O) = cn ·
The solution is The solution is
t). which determines determines u(x, which u(x,t). In Figure Figure 6.6, we show show the temperature distributions distributions at at times In 6.6, we the temperature times tt = = 20, 20, tt = = 60, 60, and tt = should be with Figure Figure 6.1. insulating and = 300; 300; this this figure figure should be compared compared with 6.1. The The effect effect of of insulating the end is is clearly clearly seen. seen. the right right end
6.2.2 6.2.2
Both ends insulated; Neumann boundary boundary conditions
In order to treat the case heat flow flow in in aa bar both ends insulated, we we must In order to treat the case of of heat bar with with both ends insulated, must first first analyze analyze the negative second derivative derivative operator operator under Neumann conditions. conditions. We We define define e1[0, e) = and LN : eF,r[o, £) -+
{u E e 2 [0, e) :
~~(O) = ~~(e) =
o},
e[o, £} by
negative second derivative operator under the the other Unlike the Unlike the negative second derivative operator under other sets sets of of boundary boundary conditions we we have considered, LN LN has null space; indeed, LNu L^u = have considered, has aa nontrivial nontrivial null space; indeed, = 0 ifif conditions and only only if constant function function (see 3.17). Therefore, Therefore, AQ an and if uu is is aa constant (see Example Example 3.17). AD = = 0 isis an eigenvalue with eigenfunction straightforward to to show eigenvalue of of LN LN with eigenfunction 'Yo(x) 70 (x) == 1. 1. It It is is straightforward show that that LN is symmetric, its eigenvalues eigenvalues are real and eigenfunctions corresponding LJV is symmetric, so so that that its are real and eigenfunctions corresponding to to distinct from AQ, AD, the the other distinct eigenvalues eigenvalues are are orthogonal. orthogonal. Moreover, Moreover, apart apart from other eigenvalues eigenvalues of of LN Z/jv are are positive (see (see Exercise Exercise 3). We We now determine determine these other other eigenvalues eigenvalues and and the corresponding eigenfunctions. the corresponding eigenfunctions.
°°
232
Chapter 6. Heat flow and diffusion 5r-----~----~--~--~--~:%====~
4.5 4
3.5 ~
::J
3
Cil Q52.5
a. E .$
2 1.5
10
20
x
30
40
50
Figure 6.6. 6.6. The The solution at times times 0, 0, 20, 20, 60, 60, and Figure solution u(x,t) u(x, t) from from Example Example 6.4 at and 300 {seconds}. (seconds). Ten Ten terms terms of of the the Fourier Fourier series series were were used used to to create create these these curves. 300 curves. Since we we are axe looking looking for for positive positive eigenvalues, eigenvalues, we we write A = (j2 02 and and solve Since write>. solve
(6.9)
The general general solution solution of of the the ODE ODE is The is
Therefore, Therefore,
and and
du dx (0) = c2 fJ,
so the first so the first boundary boundary condition condition implies implies C2 C2 = = O. 0. We We then then have have u(x) u(x] = = Cl c\ cos cos (Ox) (Ox) and and the second second boundary boundary condition condition becomes the becomes
6.2. Pure Neumann conditions and the Fourier cosine cosine series
233
When When A > > 0, 0, this holds only only ifif
B£=±n7l", n=1,2,3, ... , that that is, is, ifif
n27l"2
>..= T ' n= 1,2,3, ... (we (we disregard disregard the the possibility possibility that that Cl c\ == 0, 0, as as this this does does not not lead lead to to an an eigenfunction). eigenfunction). The of the is irrelevant. irrelevant. The value value of the constant constant Cl c\ is We thus see that the We thus see that the operator operator
has the following following eigenpairs eigenpairs under under Neumann conditions: conditions
(6.10) in the eigenpair in addition addition to to the eigenpair >"0 AQ = = 0, 0, 'Yo(x) 7o(#) = = l. 1A direct direct calculation calculation shows shows that that
("(m,'Yn)
={
m", n,
0, ~,
m=n>O,
£,
m
= n = 0.
The eigenfunctions are orthogonal, we knew knew they would be be due due to to the symmetry The eigenfunctions are orthogonal, as as we they would the symmetry of of LN. I/ATBy projection theorem, theorem, the the best £] from By the the projection best approximation approximation to to f/ E G e[O, C*[0,£] from the the subspace subspace spanned spanned by { 1, cos
(7l"£X), ... , cos (N;X) }
is is
where where
bo =
l e1 10r f(x) dx, b = e2 10rl f(x) cos (n7l"X) -£- dx, n = 1,2, .... n
It can can be shown that It be shown that
Ilf - fN"
-t
°
as N
-t
00;
that is, that is,
L bncos C;X), 00
f(x) = bo +
n=l
234 234
Chapter 6. Heat flow and diffusion
where mean-square sense necessarily in the pointwise where the the convergence convergence is is in in the the mean-square sense (not (not necessarily in the pointwise sense-the will be sense—the distinction distinction will be discussed discussed in in detail detail in in Section Section 9.4.1). 9.4.1). This This is is the the the function Fourier cosine series of of the function f, /, and and the the coefficients coefficients b 6o0 ,,&b1i ,, &b22, , •- •• - - are are the the Fourier cosine coefficients coefficients of of f./. We IBVP for We can can now now solve solve the the following following IBVP for the the heat heat equation: equation:
°< x < £, u(x, 0) = 'I/J(x), °< x < OU
oU 02U pc ot - '" ox2 = f(x, t),
t
> to, (6.11)
£,
ox (0, t) = 0, t> to, au ox (£, t) = 0, t > to·
We write write the the unknown unknown solution solution u(x, t) t) in in aa Fourier cosine series series with timeWe Fourier cosine with timedependent Fourier Fourier coefficients: coefficients: dependent 00
u(x, t) = bo(t)
+ 2: bn(t) cos (n;x), n=l
1 (-
bo(t)
= i io
bn(t)
= i2 io(l u(x, t) cos (n7rx) -c- dx,
u(x, t) dx, n
= 1,2, ....
The problem problem of of determining determining u(x, u(x,t)t) now now reduces to the the problem of solving solving for The reduces to problem of for bo(t), b1 (t), b22(t),.... (t), .... We b0(t),bi(t),b We write write the the PDE PDE
in this requires requires determining in terms terms of of Fourier Fourier cosine cosine series; series; this determining formulas formulas for for the the Fourier Fourier cosine coefficients coefficients of of ou/ot(x, du/dt(x,t)t) and and -o2u/ax —d2u/dx22(x,t) in terms of the the Fourier Fourier coefficoefficosine terms of (x, t) in cients we have cients of of u(x, t). By By Theorem Theorem 2.1, 2.1, we have
1
(n7rx) d [2 {l (n7rx) db i2io{lou ot (x, t) cos -c- dx = dt i io u(x, t) cos -£- dx = dt(t) n
(and Therefore, (and similarly similarly for for nn = = 0). 0). Therefore,
au db o ~ db n (n7rx) ot (x, t) = (it(t) + ~ dt(t) cos -c- . n=l
Computing the the Fourier coefficients of of -02u/ox —d2u/dx22(x, requires two two applications applications Computing Fourier coefficients (x, t) requires of integration integration by parts, as as in in Section Section 6.1. 6.1. In In the the following following calculation, calculation, the the boundary boundary of by parts,
cosine series 6.2. Pure Neumann conditions and the Fourier cosine
235
terms vanish because of the boundary boundary conditions: terms vanish because of the conditions:
l
a2u(x, t) cos (n7rx) -""i2 ior ax2 -e- dx =
n7r (au . (n7rx) -""i2 ([au ax (x, t) cos (n7rx)]l -e- 0 + T io ax (x, t) sm -e- dx )
2n7r rl au . (n7rx) io ax (x, t) sm -e- dx
= -~
l 2n7r ([u(x, t) sm . (n7rx)]l n7r io u(x, t) cos (n7rx)) -e- 0 - T -e- dx 2n27r 2 l (n7rx) ~ io u(x, t) cos -e- dx n 2 7r 2
r
= -~
=
r
= ----g2bn (t).
To compute the the n we just just use use direct To compute n= = 00 Fourier Fourier coefficient, coefficient, we direct integration: integration:
(since 0). We obtain aujax(e, t) — = du/dx(Q,t] aujax(O, t) == 0). We obtain (since du/dx(l,t)
Therefore, Therefore, (6.12)
have have
be represented represented by by aa cosine we Now, since since any any function function in Now, in C[O,e] C[0,£] can can be cosine series, series, we
f(x, t) =
=
eo(t)
+ L cn(t) cos (n;x),
(6.13)
n=l where the the coefficients cO(t),Cl(t),C2(t), be computed explicitly because because f/ is is where coefficients CQ (t}, c\ (t), c^ ( t ) , ... . . . can can be computed explicitly given: given:
rl f(x, t) dx, cn(t) = ""i2 iorl f(x, t) cos (n7rx) -e- dx, n = 1,2, ....
co(t) = ""i1 io
236
and diffusion Chapter 6. Heat flow and
We equate equate the the two two series, series, given given in in (6.12) (6.12) and and (6.13), obtain We (6.13), and and obtain
We can can then bo(t),b\(t),b<2,(t),... solving these these ODEs, ODEs, together together with with We then compute compute bo(t), bl (t), b2 (t), ... by by solving the the initial the initial initial values values obtained obtained from from the initial condition condition u(x, to) = ,¢(x), 0< x <
e.
Example 6.5. 6.5. We We continue continue to to consider the iron iron bar 6.1, now both Example consider the bar of of Example Example 6.1, now with with both ends insulated. insulated. We We must IBVP ends must solve solve the the IB VP au pc at -
K,
°
a2u ax 2 = 0,
< x < 50, t
°
u(x,O) = ,¢(x), < x au ax (0, t) = 0, t > 0, au ax (50, t) = 0, t
> 0,
< 50,
(6.14)
> 0,
where K, and and '¢ ip are are as as given before. where p, p, c, c, K" given before. Before u, we that it it is is easy easy to to predict the long-time long-time behavior Before solving solving for for u, we note note that predict the behavior o f uu,, the the temperature. temperature. Since the bar bar is is completely completely insulated insulated and and there there is is no internal of Since the no internal source of heat, heat, the heat energy energy contained in the the bar bar at time 00 (represented (represented by by the the the heat contained in at time source of initial value value of of u) u) will will flow flow from to colder colder regions and eventually eventually reach from hot hot regions regions to regions and reach initial equilibrium at constant temperature. temperature. The The purpose modeling this this phenonemon of modeling phenonemon equilibrium at aa constant purpose of with heat equation equation is to obtain obtain quantitative quantitative information information about about this is to this process—the process-the with the the heat process is already already understood understood qualitatively. qualitatively. We write write We 00
u(x, t) = bo(t)
+ 2..: bn(t) cos (n:ox) n=l
and and then then
Since the right-hand right-hand side we obtain of the the PDE PDE is is zero, zero, we obtain Since the side of
Also, Also, 00
,¢(x) = do
+ 2..: dn cos (n:ox) , n=1
6.2. Pure Neumann conditions and the Fourier Fourier cosine series
237
where where 1 (50 do = 5010
2 (50
dn = 5010
=
5
2'
'l/J(x) dx = 'l/J(x) cos
(n1fX)
50
dx
20 (2 cos (n1f /2) - 1 - (-1)n) 22 ,n=1,2,3, .... n1f
Since u(x,O) we have have Since w(#,0) = = 'l/J(x), i£>(x), we
that is, is, that bo(O)
= do,
bn(O)
= d n , n = 1,2,3, ....
Thus we solve solve Thus we to to get get
bo(t) =do ,
and and to get
where where
Therefore, Therefore, 00
u(x,t)
= do + 2: dne->'nt cos (n1fx) ,
50 with ^0,^1,^2,... do ,d1 ,d2 , ... given given above. above. Figure Figure 6.7 is analogous analogous to to Figures Figures 6.1 6.1 and and 6.6, with 6.6, showing snapshots snapshots of of the the temperature temperature distribution distribution at same times times as as before. showing at the the same before. It It also shows shows the the equilibrium equilibrium temperature; temperature; an an inspection formula for for u(x, t) also inspection of of the the formula u(x,t) makes it it clear clear that that u(x, --+ do do == 5/2 5/2 as as tt —> --+ oo. 00. makes u(x,t}t) -> n=l
6.2.3 6.2.3
conditions in a steady-state steady-state BVP Pure Neumann conditions
The BVP The BVP ~u
-r;, dx 2
= f(x), 0 < x < e,
~~ (0) = 0,
~~(e) = 0,
(6.15)
Chapter 6. 6. Heat Heat flow flow and and diffusion Chapter diffusion
238 238
5r-----~----~~~--~----r:==~~
- - t=O - _. t=20 ._.- t=60 t=300 - - t=oo
4.5 4 3.5
~3
~
~
.........
.,../
w2.5~------------~~---------------------------~~------------~
0..
E $
2
10
20
30
x
40
50
Figure 6.7. 6.7. The The solution u(x,t]t) from 6.5 at at times times 0, 0, 20, 20, 60, 60, and Figure solution u(x, from Example Example 6.5 and 300 (seconds). (seconds). Ten Ten terms terms of of the the Fourier series were were used used to to create create these these curves. Fourier series curves. 300 describes the steady-state temperature distribution in aa completely describes the steady-state temperature distribution u(x) u(x) in completely insulated insulated bar (time-independent) heat source. Up Up to this point, each BVP BVP or or IBVP IBVP bar with with aa (time-independent) heat source. to this point, each in Chapters Chapters 55 and and 66 has has had had aa unique solution for for any any reasonable choice of the unique solution reasonable choice of the in "ingredients" of the the PDE: the coefficients the stiffness the heat con"ingredients" of PDE: the coefficients (such (such as as the stiffness or or the heat conductivity), boundary conditions, and the the right-hand-side function. ductivity), the the initial initial and and boundary conditions, and right-hand-side function. We have have not emphasized this point; rather, rather, we concentrated on on developing We not emphasized this point; we have have concentrated developing solution techniques. However, we (does solution techniques. However, we must must now now address address the the issues issues of of existence existence (does the problem have have aa solution?) solution?) and and uniqueness uniqueness (is (is there there only only one one solution?). solution?). The The the problem class of of PDEs PDEs for for which can demonstrate demonstrate existence existence and and uniqueness by explicitly class which we we can uniqueness by explicitly producing the is very is, there many PDEs PDEs for we producing the solution solution is very small. small. That That is, there are are many for which which we cannot derive derive aa formula formula for for the solution; we we can only approximate approximate the the solution solution by cannot the solution; can only by numerical methods. For those those problems, problems, we that aa unique numerical methods. For we need need to to know know that unique solution solution exists before before we try to approximate it. it. we try to approximate exists Indeed, this is an issue finite element method, although Indeed, this is an issue with with the the finite element method, although we we did did not not acknowledge it it when we first first introduced introduced the the technique. showed in in Section Section 5.5 acknowledge when we technique. We We showed 5.5 that the Galerkin Galerkin method, on which element method method is that the method, on which the the finite finite element is based, based, produces produces the best approximation approximation (in (in the the energy energy norm, and from from aa certain certain subspace) subspace) to the the best norm, and to the true solution. However, assumed in that argument argument that that aa true true solution solution existed, existed, true solution. However, we we assumed in that and that there there was was only one! and that only one! The question of existence existence and and uniqueness is not academic. The steadyThe question of uniqueness is not merely merely academic. The steadystate is aa physically physically meaningful problem that state BVP BVP (6.15) (6.15) demonstrates demonstrates this-it this—it is meaningful problem that aa scientist may to solve, solve, and and yet, as we we are about to show, either either it it does does not have yet, as are about to show, not have scientist may wish wish to
6.2. Pure Neumann Neumann conditions and the Fourier cosine series
239 239
aa solution solution or or the the solution solution is is not not unique! unique! We We already already discussed discussed this this fact fact in in Examples Examples 3.14 and 3.17 in Section 3.2. For emphasis, we the justification from we will repeat the several points of view. view. Physical reasoning Physical reasoning
The source function ff(x) represents the rate at which (heat) energy is ( x ) in (6.15) represents added to the the bar bar (in units of of energy per volume). volume). It It depends on x (the added to (in units energy per per time time per depends on (the position in the bar) we allow the possibility possibility that different different parts of the bar position bar) because because we bar receive different different amounts of heat. it is independent of of time, indicating that that is independent time, indicating receive amounts of heat. However, However, it the rate at which added is is constant. constant. Clearly, for aa steady-state steady-state solution solution to to the rate at which energy energy is is added Clearly, for exist, zero; if exist, the the total total rate at at which energy energy is is added added must be zero; if energy energy is added added to to one part of of the it must taken away from another Otherwise, the the total total heat heat part the bar, bar, it must be be taken away from another part. part. Otherwise, energy energy in in the bar would be constantly constantly growing or or shrinking, shrinking, and the the temperature temperature could not not be respect to to time. time. All of this this implies implies that that (6.15) (6.15) may may not not All of could be constant constant with with respect have whether it it does does depends depends on the properties f(x). have aa solution; solution; whether on the properties of of /(#). On other hand, hand, if (6.15) does cannot be be unique. KnowOn the the other if (6.15) does have have aa solution, solution, it it cannot unique. Knowing only that no net energy is being added to or taken from the bar does not tell us how much energy is in the bar. That is, the energy, and hence the temperature, temperature, is not uniquely determined determined by BVP. is not uniquely by the the BVP. Mathematical reasoning
Recall ( x ) has has units of energy per unit time; the the total Recall that that ff(x) units of energy per per unit unit volume volume per unit time; total rate rate at which energy is added to at which energy is added to the the bar bar is is
1i
f(x)Adx,
where A is where is the the cross-sectional cross-sectional area areaof ofthe the bar. bar. By Byphysical physicalreasoning, reasoning,weweconcluded conclude that that
1£
or simply or simply
f(x)A dx = 0,
1£
f(x) dx = 0,
(6.16)
must in order for aa solution solution to to exist. exist. But from the the But we we can can see see this this directly directly from must hold hold in order for BVP. Suppose Suppose uu is aa solution solution to (6.15). (6.15). Then Then
{l io f(x) dx
=
(l dlu -I),
io dx 2 (x) dx
d Ii = -I),~(X) dx
0
du du = 1),-(0) - /'i,-(£) dx dx = 0-0 = O.
240
Chapter 6. Heat flow and diffusion Chapter 6. Heat flow and diffusion
Thus, solution exists, exists, then compatibility condition condition (6.16) (6.16) holds. holds. Thus, if if aa solution then the the compatibility Now suppose that solution u of (6.15) (6.15) does does exist. exist. Let Let C be any constant, constant, Now suppose that aa solution u of be any and function v by u(x] + C. C. Then Then we and define define aa function by v(x) v(x) == u(x) we have have ~v
-K,
dx 2 (x) =
~u -K,
dx 2 (x) + 0 =
~u
-K,
dx 2 (x) = f(x)
and and
dv (0) = du (0) + 0 = du (0) = 0, dv (£) = du (£) + 0 = du (£) = O. dx dx dx dx dx dx Therefore, another solution solution of of (6.15), (6.15), and and this any real C. Therefore, v is is another this is is true true for for any real number number C. This shows that if (6.15) has a solution, then it has infinitely many solutions. The Fourier series method The Fourier series method
Suppose that, rather rather than reasoning, we we just just set to solve solve Suppose that, than engage engage in in the the above above reasoning, set out out to (6.15) by by the the Fourier of the the boundary boundary conditions, conditions, we we write (6.15) Fourier series series method. method. Because Because of write the solution as as aa cosine the solution cosine series: series: 00
u(x)
= bo + 2:: bncos (n;x). n=l
2 _~U/dX2(x) the usual way, have Then, computing the the cosine coefficients coefficients of —d?u/dx (x) in the way, we have
~u ~ K,n 7r (n7rx) dx 2 (x) = ~ ~bn cos -£2 2
-K,
n=l
(note that the the constant term b vanishes when when u(x) is differentiated). Also, (note that constant term 60 u(x) is differentiated). Also, o vanishes 00
f(x) =eo+ 2:cnCOs(n;x), n=l
where where 1
eo = f
10{l f(x) dx,
Cn
2
=f
10(l f(x) cos (n7rx) -£- dx,
n = 1,2, ....
Equating the the series series for for -—Kd?u/dx and /(#), find that that the following equations equations Equating K,d2 u / dx22(x) (x) and f (x), we we find the following must must hold: hold: K,n 2 7r 2 0= Co, ~bn = Cn, n = 1,2, .... Since coefficient eo,Cl,C2, ... are given and the Since the coefficient co,ci,C2,... the coefficients coefficients bo,bl,b O o ? 0 i 5 022, , ... - - - are are to to be we can can choose be determined, determined, we choose
and and thereby satisfy satisfy all all but the first equation. equation. However, However, the equation
eo = 0
6.2. 6.2. Pure Pure Neumann Neumann conditions conditions and and the the Fourier Fourier cosine cosine series series
241 241
is not solution w, rather on on the right-hand-side function function f./. is not aa condition condition on on the the solution u, but but rather the right-hand-side Indeed, Indeed, no choice of u can can satisfy satisfy this equation unless Co =
l1£ f(x)dx
=
o.
Thus we see again again that condition (6.16) Thus we that the compatibility compatibility condition (6.16) must must hold in order order for (6.15) to have have aa solution. (6.15) to solution. Moreover, = 00 does see that that CQ Co = does hold, hold, we we see that the the value value of of 60 bo is is Moreover, assuming assuming that undetermined by by the the BVP. BVP. Indeed, with CQ Co = undetermined Indeed, with =0,0,the the general generalsolution solutionofofthe the BVP BVP isis
2 ~ C cn (n1fx) u(x) = bo + L..J /1,n 21f2 cos -C- , n=l
where b can take take on value. Thus, Thus, when when aa solution exists, it it is not unique. unique. where 60 on any any value. solution exists, is not o can Relationship to the alternative for for symmetric symmetric linear systems Relationship to the Fredholm Fredholm alternative linear systems
If If A A €E Rnxn R n x n is symmetric and singular (so that N(A) A/"(A) contains nonzero vectors), then the the Fredholm Fredholm alternative alternative (Theorem (Theorem 3.19) implies that that Ax b has solution then 3.19) implies Ax = =b has aa solution if and and only b is orthogonal to N(A). is the the compatibility compatibility condition condition for if only if if b is orthogonal to AT (A). This This is for aa
symmetric, singular The compatibility compatibility condition condition (6.16) is precisely precisely symmetric, singular linear linear system. system. The (6.16) is we now now show. show. analogous, as as we analogous, The operator operator LN LN (the (the negative second derivative derivative operator, operator, restricted to funcfuncThe negative second restricted to tions satisfying homogeneous conditions) is symmetric. Moreover, the tions satisfying homogeneous Neumann Neumann conditions) is symmetric. Moreover, the null space space of of the the operator operator contains contains nonzero nonzero functions, namely, all all nonzero nonzero solutions null functions, namely, solutions to to ~u
-/1, dx 2
= 0,
0 < x < C,
:: (0) = 0,
(6.17)
du(C) = o. dx It to show, by direct integration, that that the the solution solution set It is is easy easy to show, by direct integration, set consists consists of of all all constant constant Cl. Therefore, the suggested by the functions defined on [0, [0,£]. the compatibility compatibility condition condition suggested the finite-dimensional situation situation is that (6.15) only if is orthogonal finite-dimensional is that (6.15) has has aa solution solution if if and and only if f/ is orthogonal to every every constant constant function. function. But to But
(c, f) is equivalent equivalent to is to
or simply simply to to or
This is is precisely precisely (6.16). This (6.16).
= 0 for all c E R
1£ cf(x) dx = 0 for all c E R, 1£ f(x) dx = o.
242 242
Chapter 6. 6. Heat Heat flow Chapter flow and and diffusion diffusion
Summary As with pure pure As the the above above example example shows, shows, aa steady-state steady-state (time-independent) (time-independent) BVP BVP with Neumann property that either Neumann conditions can have the property either there is no solution solution or there there is however, that this is more more than one solution. solution. It is is important important to note, note, however, this situation situation depends on the differential well as on the boundary conditions. For differential operator operator as well example, the BVP
d2 u -/'i,
dx 2 + au = f(x), 0< x < i, du (0) dx
= 0,
du (i) dx
=0
(6.18)
(/'i" a > 0) there is is no condition imposed imposed (K, 0) has has aa unique unique solution, solution, and and there no compatibility compatibility condition on difference between on f. /. The The difference between (6.18) (6.18) and and (6.15) (6.15) is is that that the differential differential operator operator in (6.18) involves the the function not just just the the derivatives unlike (6.18) involves function uu itself, itself, not derivatives of of u. u. Therefore, Therefore, unlike in problem (6.15), (6.15), the addition of a constant constant to u is not "invisible" to the PDE.
Exercises 1. Solve the IBVP
au a 2u at - ax2
= 0, °< x < 1, t > 0, u(x,O) = x(I- x), 0< x < 1, au ax (0, t) = 0, t > 0, au ax (1, t)
= 0,
t > 0,
and graph the solution at at times steady-state and graph the solution times 0,0.02,0.04,0.06, 0,0.02,0.04,0.06, along along with with the the steady-state solution. solution. 2. the IBVP 2. Solve Solve the IBVP
au a 2u 1 at - ax2 = "2 - x,
°<
x
< 1, t > 0,
u(x,O) = x(I - x), 0< x < 1, au ax (0, t) = 0, t > 0, au ax(I,t) =0, t>o
and graph the the solution solution at at times times 0,0.02,0.04,0.06, 0,0.02,0.04,0.06, along with the the steady-state steady-state and graph along with solution. solution. 3. (a) Show Show that LN Ljy is symmetric.
6.2.
conditions and the Fourier cosine series series Pure Neumann conditions
243
(b) (b) Show Show that that the the eigenvalues eigenvalues of of LN LN are are nonnegative. nonnegative. 4. Section 6.1.4, 6.1.4, show show how how to to solve solve the the following following IBVP IBVP with with inhomo4. Following Following Section inhomogeneous Neumann conditions: geneous Neumann conditions:
au pc at -
K,
a2u ax 2 = J(x, t), 0 < x to,
u(x, to) = 'l/J(x), 0 < x < !!, au ax (0, t) = aCt), t> to, au ax (!!, t)
= bet), t > to·
5. (a) (a) Consider Consider the following BVP (with inhomogeneous inhomogeneous Neumann 5. the following BVP (with Neumann conditions): conditions): ~u
-K,
dx 2 = J(x), 0 < x < !!,
du dx (0) = a, du (!!) dx
= b.
Just as as in in the the case case of of homogeneous homogeneous Neumann Neumann conditions, conditions, if if there there is is one one Just solution solution u(x), w(x), there there are are in in fact fact infinitely infinitely many many solutions solutions u(x)+C, u(x}+C, where where C is aa constant. constant. Also Also as as in in the the case case of of homogeneous homogeneous Neumann conditions, is Neumann conditions, there is aa compatibility right-hand side side /(#) must there is compatibility condition condition that that the the right-hand J(x) must satisfy in in order order for for aa solution solution to to exist. exist. What What is is it? it? satisfy (b) (b) Does Does the the BVP BVP ~u
-K,
dx 2
+ u = J(x),
0
< x < !!,
du dx (0) = a, dUe!!) = b
dx
(note the difference in in the differential equation) equation) require compatibility (note the difference the differential require aa compatibility condition on on /(#), or is is there solution for for every every /I E E C[O,!!]? C[0, ^]? Justify condition I(x), or there aa solution Justify your answer. answer. your 6. Consider Consider aa copper copper bar = 8.96 8.96 g/ g/cm = 0.385 0.385 JJ/(gK), 4.01 W W/(cmK)) 6. bar (p = cm 33,, cc = / (g K), K,« == 4.01 / (cm K)) of length length 11 m and cross-sectional cross-sectional area area 22 cm cm22.. Suppose Suppose the the sides sides and and the the left left of m and end (x = = 0) 0) are are perfectly insulated and and that is added added to (or taken taken from) from) end perfectly insulated that heat heat is to (or the the other other end end (x = = 100, 100, x in in centimeters) centimeters) at at aa rate rate of of fJ(t) ( t ) =— 0.1 0.1 sin sin (607rt) (QQirt) W. W. If temperature of Celsius, If the the initial initial (t == 0) 0) temperature of the the bar bar is is aa constant constant 25 25 degrees degrees Celsius, find the the temperature temperature distribution distribution u(x, u(x,t)t) of of the the bar bar by by formulating and solving formulating and solving find an IBVP. IBVP. an
244
Chapter 6. 6. Heat flow and Chapter Heat flow and diffusion diffusion
7. Consider the IBVP IBVP 7. Consider the
au a u 2
= 0, 0 < x < l, t > 0, u(x,O) = 'lj;(x), 0< x < l,
at -
{)x2
{)u
ax (0, t) = 0, t > 0, {)u
ax(l,t)
= 0, t > 0.
Show that the the solution satisfies Show that solution u(x, u(x,t)t) satisfies lim u(x, t) = C for all x E (0, l),
t-+oo
for some some constant constant C, (7, and and compute compute C. C. for 8. in Exercise that, after 8. Consider Consider the the experiment experiment described described in Exercise 6.1.8. 6.1.8. Suppose Suppose that, after the the 20 20 minutes are are up, up, the ends ends of the bar are are removed removed from the the ice bath bath and and are are insulated. Compute the eventual steady-state the insulated. Compute the eventual steady-state temperature temperature distribution distribution in in the bar. bar. the copper copper bar described in in Exercise 6.1.9. 9. Repeat Exercise Exercise 8 8 for for the 9. Repeat bar described Exercise 6.1.9. 3 10. Consider Consider an an iron iron bar = 7.87 0.449 J/(gK), KK, = = 0.802 0.802 W/(cm K)) 10. bar (p = 7.87 g/cm g/cm3 ,, cc = = 0.449J/(gK), W /(cmK)) of length area 2 cm22.• Suppose Suppose that the bar is heated to of length 11 m m and and cross-sectional cross-sectional area 2 cm that the bar is heated to aa constant perfectly constant temperature temperature of of 30 30 degrees degrees Celsius, Celsius, the the left left end end (x = 0) 0) is is perfectly insulated, and and heat heat energy energy is is added from) the end (x — insulated, added to to (or (or taken taken from) the right right end = 100) 100) at the of 11W (the variable Suppose further further at the rate rate of W (the variable xx is is given given in in centimeters). centimeters). Suppose that heat heat energy is added added to to the the bar bar at at aa rate of 0.1 /cm 33.• that energy is the interior interior of of the rate of 0.1 W W/cm
(a) steady state. (a) Explain Explain why why the the temperature temperature will will not not reach reach steady state. (b) the temperature temperature distribution distribution u(x, t) by formulating formulating and solving (b) Compute Compute the t] by and solving the appropriate IBVP, IBVP, and and verify that u(x, t) does does not approach aa steady steady the appropriate verify that not approach state. state. (c) that, instead the left left end's being insulated, insulated, heat heat energy is (c) Suppose Suppose that, instead of of the end's being energy is removed from the the left left end end of of bar. bar. At At what rate must must energy be removed removed from what rate energy be removed if is to to reach steady state? state? reach aa steady if the the temperature temperature distribution distribution is that your answer to to the the previous previous part part is is correct correct by by formulating (d) Verify (d) Verify that your answer formulating and solving the IBVP and and showing that u(x, and solving the appropriate appropriate IBVP showing that u(x,t)t) approaches approaches a limit. limit. (e) that aa steady is to to be be achieved achieved by by holding (e) Suppose Suppose that steady state state is holding the the temperature temperature at the the left left end end at at aa fixed the value Explain. at fixed value. value. What What must must the value be? be? Explain. The purpose purpose of this exercise is to to show that if if u(x, homogeneous 11. 11. The of this exercise is show that u(x,t)t) satisfies satisfies homogeneous Neumann conditions, conditions, and is represented by aa Fourier Neumann and u(x, u(x,t)t) is represented by Fourier sine sine series, series, then then aa
6.3. the full Fourier 6.3. Periodic Periodic boundary boundary conditions conditions and and the Fourier series series
245 245
2 formal calculation the Fourier of —8 -()2u/ax (x, t) is formal calculation of of the Fourier sine sine series series of u/dx22(x, is invalid. invalid. That That is, suppose is, suppose
. (nnx) -e- , an(t) = "£2 io(l u(x, t) sm. (nnx) -e- dx
~
u(x, t) = ~ an(t) sm
0
n=l
and u satisfies
au ax (0, t)
au
= ax (e, t) = 0 for all t > O.
We We wish to show show that (6.19) 2 2 does not represent —d u/dx2 2(x, (x,t). does represent -a u/ax t).
(a) that the the method of Sections 6.2.2 (two (two applications applications of (a) Show Show that method of Sections 6.1 6.1 and and 6.2.2 of inteintegration by parts) parts) cannot used to to compute compute the the Fourier Fourier sine gration by cannot be be used sine coefficients coefficients 2(x, t) in of -d22u/dx in terms of al ai(t),O2(t),— u/ax2(x,t) terms of (t), a2(t), .... of -a (b) Let u(x, t) = 1. the Fourier Fourier sine sine coefficients (b) Let u(x, t} == tt and and e t — \. Compute Compute the coefficients
of of u. Then compute the series series (6.19), and and explain explain why it it does does not not equal equal 2 -d22u/dx -a u/ax2(x,t). (x, t).
6.3 6.3
Periodic Fourier Periodic boundary boundary conditions conditions and and the the full full Fourier series series
We of aa circular circular (instead of straight) straight) bar, specifically, aa set set We now now consider consider the the case case of (instead of bar, specifically, of the the form of form
(see 6.8). We can model in the by the the heat equation We can model the the flow flow of of heat heat in the ring ring by heat equation (see Figure Figure 6.8).
au
pc at
-
K,
a2 u ax 2 = 0;
however, the variable x, the cross-section cross-section of the bar, is now to however, the variable x, representing representing the of the bar, is now related related to mathematical convenience, convenience, we we will assume that the the polar angle O. 9. Purely for mathematical length of of the the bar and we will label label the the cross-sections cross-sections by by xx €E (-e, (—£,^j, bar is is 21, 2£, and we will el, where where length = Oi/7r. oe/n. In In particular, particular, x — = —t -e and and x = = te now now represent represent the the same same cross-section cross-section of of x= the the bar. bar. There additional level of approximation involved in flow in in There is is an an additional level of approximation involved in modeling modeling heat heat flow aa circular circular bar bar (as (as compared compared to to aa straight straight bar) bar) by by the the one-dimensional one-dimensional heat heat equation. equation.
246 246
Chapter Chapter 6. 6. Heat Heat flow flow and and diffusion diffusion
Figure 6.8. bar. Figure 6.8. A circular circular bar.
We mentioned earlier earlier that, that, in in the the case case of of the the straight bar, if the initial initial temperature We mentioned straight bar, if the temperature distribution heat source distribution and and the the heat source depend depend only only on on the the longitudinal longitudinal coordinate, coordinate, then then so does the temperature temperature distribution for for any subsequent time. This is not the case for aa circular circular bar, as the suggests (for (for example, example, the the distance distance around around the the the geometry geometry suggests for bar, as ring varies varies from from 2C It -— 2m5 2-jrS to It + + 21[(5, 2-jrS, depending depending on on the path). Nevertheless, the ring to 2C the path). Nevertheless, the modeling involved in using the not large modeling error error involved in using the one-dimensional one-dimensional heat heat equation equation is is not large bar is when 86 is when when the the bar is thin thin (Le. (i.e. when is small small compared compared to to f). t). We will We will consider consider an an initial initial condition condition as as before: before: u(x, to) = 'l/J(x), -f < x ~ f.
A ring does not have physical boundaries as a straight straight bar does (a ring has a lateral boundary, which which we we still still assume assume to to be be insulated, insulated, but it does does not not have have "ends"). "ends"). boundary, but it However, since xx == -f — t represents represents the the same same cross-section cross-section as as does does xx == f, l,we still have have However, since we still boundary conditions: u(-f,t)
= u(f,t),
ou ox (-f,t)
ou
= ox(f,t),
t > to
(the temperature and temperature gradient must be (the temperature and temperature gradient must be the the same same whether whether we we identify identify the or by x = f). to as periodic the cross-section cross-section by by xx = -f —lorbyx l). These These equations equations are are referred referred to as periodic boundary conditions. boundary conditions. We therefore wish to solve the following IBVP: ou pc 8t -
82 u ox 2 = f(x, t), -f
< x < f, t > to, u(x, to) = 'l/J(x), -f < x < f, fi,
u( -f, t) ou ox (-f, t)
= u(f, t), t> to, ou = 8x (f, t), t> to·
(6.20)
6.3. Periodic boundary boundary conditions and the full Fourier series 6.3.
247 247
As we we have have seen seen several several times the first first step to develop Fourier series series As times now, now, the step is is to develop aa Fourier method for the the related related BVP method for d2 u dx 2
= lex), -f::; x < f, u( -f) = u(f), du (-f) = du (f). -/'i,
dx
6.3.1 6.3.1
(6.21)
dx
**j
Eigenpairs of of -— -£-5 under periodic periodic boundary boundary conditions Eigenpairs d~2 under conditions
We define Lp: C;[-f,fJ-t C[-f,fJ by We define
where C;[-f,f]
= {u E C 2 [-f,f]
: u(-f)
= u(f),
~~(-f)
= ~~(f)}.
Using now familiar, should be demonstrate Using techniques techniques that that are are by by now familiar, the the reader reader should be able able to to demonstrate that that Lp is and • Lp is symmetric, symmetric, and • Lp has no negative eigenvalues eigenvalues (see Exercise 5). We first observe observe that an eigenvalue eigenvalue of of L corresponding We first that AQ AO = 00 is is an L pp,, and and aa corresponding eigenfunction is ggo(x) = 1. straightforward show that the only eigenfunction (x) — I. Indeed, it is straightforward to Q solutions to solutions to d2 u - dx 2
= 0,
-f
< x < f,
u( -e) = u(£),
dUe_f) = dUCf) dx dx
constant functions. are the constant = (j2, We must solve We next consider consider positive eigenvalues. eigenvalues. Suppose Suppose A = 02, 09 > O. 0. We ~u 2 --=Ou -f<x
(6.22)
u( -E) = u(E), du C-E) dx
= du (E). dx
differential equation equation is The general solution to the differential u(x) =
Cl
cos (Ox)
+ C2 sin (Ox).
248
Chapter 6. Heat flow Chapter 6. Heat flow and and diffusion diffusion
The first boundary boundary condition condition yields yields the the equation The first equation Cl
cos (OC)
+ C2 sin (OC) = Cl cos (OC) -
C2
sin (OC),
while while the the second second yields
0, these these equations equations reduce reduce to Since 0 8 > 0, Since to C2
=
Cl
=
and and
° °
or sin (8C)
=
°
or sin (OC) = 0,
respectively. (6.22) correspond respectively. It follows that the only nonzero solutions solutions of (6.22) correspond to to
n21l"2
A= T' n
= 1,2,3, ... ,
that with with one these values function of of the the form and, moreover, moreover, that and, one of of these values for for A, A, every every function form cos (Ox) or sin (Ox) satisfies (6.22). satisfies (6.22). Now, Now, cos
(n;x)
and sin
(n;x)
are two independent are linearly linearly independent, which means means that there are are two independent eigenfunceigenfuncwe have have not not seen before. The The complete tions for for each positive eigenvalue, eigenvalue, aa situation tions each positive situation we seen before. complete list the negative derivative operator, periodic list of eigenpairs eigenpairs for for the negative second second derivative operator, subject subject to to periodic boundary conditions, conditions, is is boundary Ao = 0, 'Yo(x) = 1;
2
An
n 1l"2' 'Yn(X) = cos (n1l"x) =T -C- , n = 1,2,3, ... ,
n21l"2 (n1l"x) An = T ' '¢n(X) = sin -C- ,
n
(6.23)
= 1,2,3, ....
We know that, that, since Lpp is symmetric, eigenfunctions corresponding to to distinct distinct eigenWe know since L is symmetric, eigenfunctions corresponding eigenvalues orthogonal. That is, if m =f. n, n, then values are orthogonal. That is, m^
m1l"X)
cos ( -Cis orthogonal to both both is orthogonal to
Similarly, Similarly,
n1l"X) and sm . (n1l"x) cos ( -C-C- . . (m1l"x) sm -C
6.3. Periodic boundary conditions and and the full Fourier series 6.3.
is is orthogonal orthogonal to to both both
nKX) and sm . cos ( -e-
249
(nKX) -e- .
There is guarantee that the two two eigenfunctions eigenfunctions corresponding corresponding to There is no no aa priori priori guarantee that the to A Amm are are orthogonal to each direct calculation to be true. We We orthogonal to each other. other. However, However, aa direct calculation shows shows this this to be true. can also also calculate calculate can (1,1)
= 2e, bn, 'Yn) = ('l/Jn, 'l/Jn) = e, n = 1,2,3, ....
(6.24)
The lull full Fourier function I/ E € C[—t,(\ given by C[-e, £] is is given by The Fourier series series of of aa function
where where
ao
1 = 2£
ji I(x) dx, -i
an =£1 ji _/(x)cos (nKX) -£- dx, n=1,2,3, ... , bn =
~ ii/(X) sin (n;x) dx,
(6.25)
n = 1,2,3, .. ..
We will show in Section that the C[ -e, I]£] function conWe will show in Section 9.6 9.6 that the full full Fourier Fourier series series of of aa C[—l, function converges to to it the mean-square mean-square sense (that is, that the the sequence partial Fourier Fourier verges it in in the sense (that is, that sequence of of partial series converges to to the the function function in the I/ L 22-norm). series converges in the
6.3.2 6.3.2
BVP using the full Fourier series Solving the BVP
We now show how how to to solve solve the BVP (6.21) using the full Fourier Fourier series. series. Since Since the the We now show the BVP (6.21) using the full method should by familiar, we we leave the intermediate steps to to the the exercises. exercises. method should by now now be be familiar, leave the intermediate steps We the (unknown) (unknown) solution solution of of (6.21) (6.21) as as We write write the 00
u(x)
=
ao +
2:= {an cos (n;x) + bn sin (n;x) } ,
(6.26)
n=l
aI, a2, b1 , b&2, to be be determined. Then u auwhere the the coefficients coefficients ao, ai, 02, ... • • • ,, &i, • • • are are to au2, ... tomatically periodic boundary boundary conditions. The full Fourier series for tomatically satisfies satisfies the the periodic conditions. The full Fourier series for —d22ujdx u/dx22 is is given given by by -d 2
2
~u 00 {K,n K2 (nKX) K,n K2 (nKX)} -K,dx2(X)=~ an~cos -e- +bn~sin -e(see 6). (see Exercise Exercise 6). We assume that that the the source source function function I/ has full Fourier series We now now assume has full Fourier series 00
I(x) = C{) +
2:= {en cos (n;x) + dnsin (n;x) } , n=l
(6.27)
250 250
Chapter Heat flow and diffusion diffusion Chapter 6. 6. Heat flow and
where the the coefficients coefficients CQ, Co, ci, Cl, e2, d2 , •.. can be determined determined explicitly, explicitly, since since f/ C 2 ..• , . . . ,d , d 1i ,,fife? • • • can is Then the differential equation equation is known. known. Then the differential ~u
-r;, dx 2
= f(x)
(6.28)
implies implies that that the the Fourier Fourier series series (6.27) (6.27) and and (6.28) (6.28) are are equal equal and and hence hence that that 0= Co, r;,n 2 7r 2 ~an=en'
r;,n 2 7r 2
~bn=dn.
From we deduce that the From these these equations, equations, we deduce first first of of all all that the source source function function f/ must must satisfy satisfy the the compatibility compatibility condition condition
i£/(X)dX = 0
(6.29)
in order order for for aa solution solution to to exist. exist. If If (6.29) (6.29) holds, holds, then then we we have have in
but do is is not determined by by the the differential differential equation equation or or by by the the boundary boundary conditions. conditions but ao not determined Therefore, for any Therefore, for any ao, CIQ,
is aa solution solution to (6.21) (again, (again, assuming assuming that holds). is to (6.21) that (6.29) (6.29) holds). Example 6.6. 6.6. We We consider consider aa thin thin gold ring, of of radius radius 1em. 1cm. In the above above notation, notation, Example gold ring, In the then, fI = = 7r. TT. The The material material properties of gold are properties of gold are then, 3 p=l9.3g/cm c = 0.129 Jf(gK), r;,=3.17W/(cmK). K = 3.17 W/(cmK). p=19.3g/cm3,, e=0.129J/(gK),
We of We assume assume that that heat heat energy energy is is being being added added to to the the ring ring at at the the rate rate of
Since Since
i:
f(x) dx = 0,
and the full Fourier series 6.3. Periodic boundary conditions and
251 251
the BVP BVP
J2u
-I\, dx 2 = f(x), -7f
u( -7f) du dx (-7f)
< x < 7f,
= u(7f), du
= dx (7f)
has aa solution. solution. We We write write has 00
u(x) =
L (an cos (nx) + bn sin (nx)) n=l
(we take take ao ao == 0, 0, thus thus choosing choosing one one of of the the infinitely infinitely many many solutions). We then then have {we solutions}. We have -I\,
~~ (x) =
f:
(I\,n 2 a n cos (nx)
+ I\,n 2 bn sin (nx))
.
n=l
Also, Also, 00
f (x)
=L
(c n cos (nx)
+ dn sin (nx)) ,
n=l
where where
= -1111"
f(x) cos (nx) dx = 0, n = 1,2,3, ... , 7f -11" 1111". 6(-1)n dn = f(x) sm (nx) dx = 3 ' n = 1,2,3, .... 7f -11" 5n Cn
(The value of Co CQ is is already already known known to be zero zero since since f satisfies satisfies the the compatibility compatibility con(The value of to be condition.) Equating the series yields dition.) Equating the last last two two series yields 2
Cn = Cn :::} an = -I\,n 2 = 0,
2
n = dn :::} bn = -I\,n2 =
I\,n an I\,n bn
d
n = 1,2,3, ... ,
12(-1)n 10 5 ' n I\,n
= 1,2,3, ....
Thus Thus 12(-1)n . L 10 sm (nx). n=l I\,n 00
u(x) =
5
A graph graph of of uu is is shown shown in in Figure Figure 6.9. 6.9. We We emphasize emphasize that that this this solution solution really really only A only shows for some shows the the variation variation in in the the temperature-the temperature—the true true temperature temperature is is u(x) u(x) + +C C for some real real number number C. C. The The assumptions assumptions above above do do not not give enough enough information information to to determine determine C; we we would would have have to to know know the the amount amount of of heat heat energy energy in in the the ring ring before before the the heat C; heat source source ff is is applied. applied.
252
Chapter 6.
diffusion Heat flow and diffusion
0.5,.,.-----,,.-----.---r----r--.....----n 0.4 0.3
~ 0.1
ca
Iii
a.
0
E
2-0.1 -0.2 -0.3 -0.4
-0.5 ........----''---.........- - ' - - - - ' - - - . & . - -......... -3 -2 -1 0 2 3 x
Figure temperature distribution in Example 6.6. Figure 6.9. 6.9. The tempemture 6.6. This graph gmph was computed of the Fourier series. computed using 10 terms of
6.3.3 6.3.3
Solving using the Fourier series Solving the the IBVP I BVP using the full full Fourier series
Now we we solve the IVP IVP (6.20) (6.20) for for the the heat heat equation, equation, using using aa full full Fourier Fourier series series Now solve the representation of the solution. As representation of the solution. As in in the the previous previous section, section, the the technique technique should should by by now be be familiar, familiar, and we will will leave leave many many of the supporting computations as as exercises. now and we of the supporting computations exercises. Let u(x, t) (6.20). We as aa full full Fourier Let t) be be the the solution solution to to (6.20). We express express u as Fourier series series with with time-varying coefficients: time-varying Fourier Fourier coefficients: 00
u(x, t) = ao(t) +
L
{an(t) cos (n;x) + bn(t) sin (n;x) } .
(6.30)
n=l
The for —Kd?u/dx The series series for -l'i.J2u/dx22 is is (6.31)
As apply Theorem obtain As for for the the time time derivative, derivative, we we can can apply Theorem 2.1 2.1 to to obtain
. (n1fx)} au dao ~ {dan (n1fx) db n at (x, t) = (jt(t) + ~ dt (t) cos -e- + (jt(t) sm -e(see (see Exercise Exercise 8). 8).
(6.32)
6.3.
Periodic boundary boundary conditions and the full Fourier series
253
We series, We then then write write the the source source term term /(#,£) f(x, t) as as the the (time-varying) (time-varying) full full Fourier Fourier series, 00
f(x, t)
= co(t) + L
{cn(t) cos (n;x)
+ dn(t) sin (n;x)},
(6.33)
n=l
where the Fourier coefficients coefficients can be computed explicitly from
co(t)
= ;e il/(X, t) dx,
eIII f(x, t) cos (n7rx) -e- dx, n = 1,2,3, ... , dn(t) = ~ ill f(x, t) sin (n;x) dx, n = 1,2,3, ... . cn(t) =
_£
Substituting (6.32), and and (6.33) (6.33) into into the the heat heat equation equation and and equating Substituting (6.31), (6.31), (6.32), equating coefficients coefficients yield
dao pCTt(t) 2
r;,n 7r
= eo(t),
2
dan pc dt (t)
+ ~an(t) = cn(t), n = 1,2,3, ... ,
db n pcdt(t)
+ ~bn(t)
r;,n 2 7r 2
(6.34)
= dn(t), n = 1,2,3, ....
We can solve these ODEs for the the unknown coefficients once we have derived derived initial conditions. These These come from the initial condition for the PDE, just just as in Section 6.1. the initial say 6.1. We write the initial value 1jJ(x) ib(x] in aa full Fourier series, sav 00
1jJ(x) = Po
+ L {Pn cos (n;x) + qn sin (n;x)} .
(6.35)
n=l
We also have have 00
1jJ(x)
= u(x, 0) = ao(O) + L
{an(O) cos (n;x)
+ bn(O) sin (n;x) } .
n=l
This yields the initial initial conditions
ao(O) = Po, an(O) = Pn, n = 1,2,3, ... , bn(O) = qn, n = 1,2,3, ... . Solving the IVPs defined by (6.34) and (6.36) yields
ao(t),al(t),a2(t), ... ,b 1 (t),b 2(t), ... , and u(x,t)t) is and then then u(x, is given given by by (6.30), (6.30).
(6.36)
254
Chapter 6. Heat flow and and diffusion
Example consider again Example 6.6. will assume Example 6.7. 6.7. We We consider again the the ring ring of of Example 6.6. We We will assume that that the temperature temperature in in the the ring is initially initially 25 25 degrees degrees Celsius, Celsius, and at time time tt = the ring is and that that at =0 (t added according to the function ff given given in (t measured measured in in seconds), seconds), heat heat energy energy is is added according to the function in that example. We wish to u(x, t), which which describes describes the evolution of of temperature that example. We wish to find findu(x,t], the evolution temperature in do this, solve the in the the ring. ring. To To do this, we we solve the IBVP IBVP
°
{)u
{)2U
< x < Jr, t> 0, u(x,O) = 25, -Jr < x < Jr, u( -Jr, t) = u(Jr, t), t > 0,
pc {)t -,.;, {)x 2 = f(x),
{)U
{)t
-Jr
(6.37)
{)u
(-Jr,t) = {)t (Jr,t), t > 0.
We solution as as We write write the the solution 00
u(x, t) = ao(t)
+L
{an(t) cos (nx)
+ bn(t) sin (nx)}.
n=l
We already already have have the the full and the the initial initial temperature temperature distribution distribution We full Fourier Fourier series series of of ff,, and is given by aa full with exactly exactly one one nonzero nonzero term, constant is given by full Fourier Fourier series series with term, namely, namely, the the constant term solve the term 25. 25. We We must must therefore therefore solve the IVPs IVPs pCTt
da o
= 0,
ao(O)
pc dan dt +,.;,n2 an
= 0,
an (
= 25,
°= )
0, n
= 1,2,3, ... ,
db n 2 ( ) pCTt+,.;,nbn=dn, bnO =0, n=1,2,3, ....
The are The solutions solutions are ao(t) = 25, an(t) = 0, n = 1,2,3, ... , bn(t) =
12(-1)npC ( _I
Therefore, solution is Therefore, the the solution is u(x, t)
= 25 + ~ ~
12(-1)n ( 1<,,2 ) 10,.;,n5 1 - e-pc-t sin (nx).
n=l
Snapshots of shown in Figure 6.10. reader Snapshots of this this temperature temperature distribution distribution are are shown in Figure 6.10. The The reader should notice relationship between steady-state temperature and should notice the the relationship between the the steady-state temperature of of the the ring ring and the previous example. the solution solution to to the the previous example.
6.3.
255
conditions and the full Fourier series Periodic boundary boundary conditions 25.5 t=O - _. t=O.25 t=O.5 t=1 o t=oo
25.4 25.3 25.2 ~25.1
:::J
~ Q)
2
0.
~,~.......
E 224.9
0.'"
ao·
24.8
°
,":t) ....... __
0
--
.. '" "'-,,'~ v
....:. "," -,- ,~'~".'.~·l
',....
-
-
-
-
....
24.7 24.6 24.5 -3
-2
-1
o x
2
3
Figure 6.10. 6.10. The solution u(x, t) to Example Example 6.7 at times 0, 0, 0.25, Figure w(x,t) 0.25, 0.5, estimated and 11 (seconds), (seconds), along with the steady-state solution. These solutions were estimated using 10 terms in the Fourier series.
Exercises 1. Repeat Example Example 6.6, the ring ring is is made made of of silver instead of gold: 1. Repeat 6.6, assuming assuming the silver instead of gold: pp = g/cm33, , cc = J/(gK), «K, == 4.29 W/(cm K). How steady= 10.5 10.5g/cm = 0.235 O.235J/(gK), 4.29W/(cmK). How does does the the steadytemperature compare to that of state temperature of the gold ring? Repeat Example 6.7, assuming the the ring is made of silver instead instead of gold: 2. Repeat 3 = 10.5g/cm IO.5g/cm3, = 0.235 O.235J/(gK), K, = = 4.29W/(cmK). 4.29W/(cmK). pp = , cc = J/(gK), K 3. cm in radius, made of lead (which has physical properties 3. Consider aa ring, 55cm properties = 11.3 11.3g/cm = 0.129 O.129J/(gK), = 0.353 O.353W/(cmK)). Suppose, by by heating heating pp = g/cm33,, cc = J/(gK), «K, = W/(cm K)). Suppose, one side of the the ring, ring, the the temperature temperature distribution one side of distribution
heat source is induced. Suppose Suppose the heat source is then removed and the ring is allowed to cool. an IBVP IBVP describing describing the the cooling of the the ring. (a) Write Write down down an (a) cooling of ring. (b) Solve the IBVP. (b) Solve the IBVP. (c) Find the steady-state temperature of of the the ring. ring. (c) Find the steady-state temperature (d) How How long long does does it it take take the the ring ring to to reach reach steady steady state 1%)? (d) state (within (within 1%)?
256
Chapter 6. Heat flow and diffusion diffusion
4. the lead the temperature is 4. Consider Consider the lead ring ring of of the the previous previous exercise. exercise. Suppose Suppose the temperature is a constant constant 25 25 degrees Celsius, Celsius, and an (uneven) (uneven) heat source is is applied to the the of ring. If If the heat source delivers heat heat energy to the ring at a rate of
f(x) = 1-
~~ Wjcm 3 ,
temperature at the hottest hottest part of the ring to how long does it take for for the temperature rpa.rh 30 degrees HPCTTPPS Celsius? ("Iplsiiifi? reach 5. Lpp is is symmetric. symmetric. 5. (a) (a) Show Show that that L Lpp does not have any negative eigenvalues. (b) Show that L
6. Assuming Assuming that that u is is aa smooth smooth function function defined on [-£, [—£, £], full Fourier series defined on £l, the the full Fourier series 6. of uu is given by (6.26), and uu satisfies periodic periodic boundary boundary conditions, conditions, show that 2 the Fourier series ujdx22 is is given given by the full full Fourier series of of -d —d^u/dx by (6.27). (6.27). 7. Justify the compatibility compatibility condition condition (6.29) (a) by physical reasoning (assume that (6.21) models a steady-state steady-state temperature distribution in in aa circular circular ring); ature distribution ring); (b) by using the differential differential equation equation ~u
- dx 2 (x)
= f(x),
-£
< x < £,
and the periodic boundary conditions to compute
i:
f(x)dx;
(c) by analogy to the compatibility condition given in the Fredholm Predholm alternative. 8. Show that the Fourier series representation representation (6.32) (6.32) follows from (6.30) (6.30) and Theorem 2.1.
6.4 6.4
Finite element element methods methods for the heat heat equation equation Finite for the
To apply the Fourier series method to the heat equation, we we used the familiar eigenfunctions to represent represent the spatial variation of the solution, while while allowing the Fourier coefficients coefficients to on time. values of We then then found found the the values of these these Fourier Fourier Fourier to depend depend on time. We coefficients by coefficients by solving solving ODEs. ODEs. We can can use the finite element element method in an an analogous analogous fashion. We We use finite element element functions functions to approximate the the spatial spatial variation of the the fashion. use finite to approximate variation of solution, solution, while while the the coefficients coefficients in in the the representation representation depend depend on on time. time. We end end up up with system of of ODEs ODEs whose whose solution solution yields coefficients. with aa system yields the the unknown unknown coefficients.
6.4. 6.4. Finite Finite element element methods methods for for the the heat heat equation equation
257
We again the following IBVP: We will will consider consider again the following IBVP: {)u
°< x < e, t > to, u(x, to) = 'Ij!(x), °< x < e, {)2U
pc {)t -
/'i,
{)x 2
I(x, t),
=
(6.38)
u(O, t) = 0, t > to, u(e,t) = 0, t > to.
We begin begin by deriving the weak form form of of the IBVP, following following the the pattern of Section Section We by deriving the weak the IBVP, pattern of 5.4.2 (multiply (multiply the the differential differential equation equation by by aa test test function function and and integrate integrate by parts 5.4.2 by parts on the left). In In the the following following calculation, calculation, V V is is the the same same space space of of test test functions functions used on the left). used for the the problem in statics: statics: for problem in V = cb[o,e]. We have We have {)u
pc {)t
{)2u
-
/'i,
{)x 2 = I(x, t),
{)u
=}
=}
{)2U
pc {)t (x, t)v(x) -
1£
°< x < t, t > to
/'i,
{)x 2 (x, t)v(x) = I(x, t)v(x),
{pc ~~ (x, t)v(x) -
/'i,
~:~ (x, t)v(x)} dx =
°< x < t, t> to, v
1£
EV
I(x, t)v(x) dx, t> to, v E V.
We by parts parts in boundary term We integrate integrate by in the the second second term term on on the the left; left; the the boundary term from from integration by parts parts vanishes the test integration by vanishes because because of of the the boundary boundary conditions conditions on on the test function function v(x), and and we we obtain obtain vex),
1£
{pc
~~ (x, t)v(x)
~~ (x, t) :: (x) } dx
+
/'i,
=
10
£
I(x, t)v(x) dx, t> to for all v
E
V.
This This is is the the weak weak form form of of the the IBVP. IBVP. To To get get an an approximate approximate solution, solution, we we apply apply the the Galerkin method with approximating approximating subspace subspace Galerkin method with
Sn p(O) = pee) = O} Sn = {p: {p '• [0, [0, e] f\ -+ —> R R :: p is is continuous continuous and and piecewise piecewise linear, linear, p(0) = p(t] 0} = ... ,CPn-d, = span{¢h,CP2, span{0i,0 2 ,...,0n-i},
°
where the piecewise piecewise linear linear functions functions are are defined defined on on the grid 0 = = XQ x\ < ... ••• < where the the grid Xo < Xl xn = t and and {0i, {CPI, 02? CP2, •... CPn-l} is is the the standard standard basis basis defined defined in in the the previous previous chapter. xn = • • ,, 0n-i} chapter. For each each t,t, the the function function u(·, u(-,t]t) must must lie lie in in S that is, For Sn, is, n, that
n-l un(x, t) =
L D:i(t)cpi(X). i=l
(6.39)
Chapter Chapter 6. 6. Heat Heat flow flow and and diffusion diffusion
258 Galerkin's Galerkin's method method leads leads to to
n
10£ {PC a;t
(x, t)v(x)
+ '" ~:n (x, t) ~~ (X)} dx
= 10£ !(x, t)v(x) dx for all v E Sn,
t> to,
or or
i
f {
o
aU n ( ) ( ) pC7it x, t cPi X
n ( )} ) dcPi + '" aU ax x, t dx (x dx
=
Ioi !(x, t) cPi (x) dx, t > to, i = 1,2, ... , n -1.(6.40)
Substituting n -— 1,1, Substituting (6.39) (6.39) into into (6.40) (6.40) yields, yields, for for tt > to to and and ii = = 1,2, 1,2, ... . . . ,, n
If by If we we now now define define the the mass matrix matrix matrix M M and and the the stiffness stiffness matrix K K by
r
r '" dcP' K—^-(x)^ (x)dx, a:: (x) dcPi dx (x) dx,
i
f
Mij I pc(f)j(x)(f)i(x}dx, I Mij == 10 pecPj (X)cPi(X) dx, Kij Kij == 10 Jo Jo
±
dx
dx
and by and the the vector-valued vector-valued functions functions f(t) f (t} and and a(t) a(t) by
f(t) =
then write (6.41) then we we can can write (6.41) as as da M dt
+ Ka = f(t).
(The names "mass matrix" and "stiffness "stiffness matrix" for for M and K, respectively, respectively, arise from the interpretation for similar matrices appearing in mechanical models. We have already seen the stiffness stiffness matrix in Chapter 5; 5; we will see the mass matrix in 7. In the context of the heat equation, these Chapter Chapter 7. In the context of the heat equation, these names names are are not not particularly particularly meaningful, but the usage usage is meaningful, but the is well well established.) established.) to) == 'I/J(x), impleThe The initial initial condition condition u(x, u(x,to) ip(x), 00 < x < lt can can be be approximately approximately implemented mented as as
un(x, to) =
n-l
n-l
i=l
i=l
L ai(to)cPi(X) = L 'I/J(Xi)cPi(X),
6.4. 6.4. Finite element methods for the heat equation
259
that is, as ai(to) = '¢(Xi). if>(xi). This is reasonable, because the function function n-l
,(f;(x) =
L '¢(Xi)(Pi(X) i=l
satisfies n-l
,(f;(Xj) =
L ,¢(Xi)(Pi(Xj) = '¢(Xj). i=l
We call ,(f; V> the the piecewise linear linear interpolant interpolant of ,¢(x) tf}(x) (see Figure 6.11 for an example).
interpolant. Figure 6.11. 6.11. ,¢(x) ^(x) = = sin (37fx) (STTX) and its piecewise linear linear interpolant. We therefore therefore arrive at the following system of ODEs and initial conditions:
dO' M dt
+ Ka = f(t), t > to, aCto)
where
= 0'0,
260
Chapter 6. Heat flow and diffusion diffusion Chapter
1 To solve this, we we can can multiply multiply the the ODE ODE on on both both sides by M M~l to to obtain obtain To solve this, sides by
that that is, is, do: dt == Ao:
+ get),
A == _M~lK, get) == M~lf(t).
We recognize recognize this this as an inhomogeneous inhomogeneous system system of of linear, We as an linear, constant constant coefficient coefficient firstfirstorder ODEs ODEs for for the the unknown unknown a(t) o:(t) (if the coefficients the PDE PDE are nonconstant, order (if the coefficients in in the are nonconstant, then so will in that case, the system of then so will be be the the matrices matrices M M and and K K and and hence hence A; A; in that case, the system of 37 ODEs will not not have have constant constant coefficients). ODEs will coefficients).37 We have have now now discretized discretized the the spatial variation of of the solution (a (a process process referred We spatial variation the solution referred to as semidiscretization system of of ODEs. apply space) to to obtain obtain aa system ODEs. We We can can now now apply to as semidiscretization in in space) aa numerical numerical method method to to integrate integrate the the ODEs. ODEs. This This general general technique technique of of solving solving aa time-dependent PDE by integrating the the system system of of (semidiscrete) called by integrating (semidiscrete) ODEs ODEs is is called time-dependent PDE the of lines. the method method of lines.
6.4.1 6.4.1
The method of lines for the heat equation
We now now illustrate illustrate the the method method of via an example, which which will will show show that We of lines lines via an example, that the the system ODEs arising the heat stiff. system of of ODEs arising from from the heat equation equation is is stiff. Example 6.8. Suppose Suppose an iron == 7.88, 7.88, c = == 0.437, K. — == 0.836,) 0.836) is chilled to Example 6.8. iron bar (p = 0.437, K to a constant Celsius, and then heated internally with both ends constant temperature of of 00 degrees Celsius, ends maintained at degrees Celsius. Suppose further further that that the the bar bar is is 100 length maintained at 00 degrees Celsius. Suppose 100 cm cm in in length of and heat energy energy is added at rate of and heat is added at the the rate
We the temperature temperature distribution distribution in in the the bar bar after after 33 minutes. We wish wish to to find find the minutes. The temperature distribution u(x,t) u(x, t) is of the IBVP The temperature is the the solution of IBVP
au
a2 u
< x < 100, t > 0, u(x,O) == 0, 0 < x < 100, u(O, t) == 0, t > 0, u(100, t) == 0, t > O.
pc at -
K.
ax 2 == I(x, t), 0
(6.42)
We will approximate the linear finite finite elements regular We will approximate the solution solution using using piecewise piecewise linear elements with with aa regular mesh 100 subintervals. We write 100, h = ih, ih, ii == = 0,1,2, 0,1,2,..., n. of 100 subintervals. We write n n= == 100, h= == 100/n, 1001n, Xi Xi == ... , n. mesh of 37 37In practice, we we do do not not actually actually compute A. When implementing numerical numerical In practice, compute either either M-l M"1 or or A. When implementing it is is rarely rarely efficient efficient to to compute compute an inverse matrix, matrix, particularly particularly when when the the matrix matrix is is algorithms, algorithms, it an inverse this case. case. Instead, Instead, the presence of of M" M- 11 is signal that that aa linear system with with coefficient sparse, in this sparse, as as in the presence is aa signal linear system coefficient matrix M must be be solved. solved. This This is is explained explained below below when when we we discuss discuss the the implementation implementation of matrix M must of Euler's Euler's method and and the the backward backward Euler Euler method. method method.
6.4. 6.4.
Finite element element methods methods for heat equation equation Finite for the the heat
261 261
As usual, {
1tOO
100
Mii = Mi,i+l = Kii= Ki i+l = ,
o
2hpc pC(
3
hpc
10
PC
10{lOO
K;
1
100
0
(d
2
6'
. z = 1,2, ... ,n - 2,
2K;. dX=h' z=1,2, ... ,n-l,
d
K;-
symmetric). We also have have (both (both M and and K are are tridiagonal tridiaqonal and and symmetric). We also
(100
fi(t) =
10
= t
f(x, t)
h2(60000i - 200h - 1200i 2 h + 6h 2i 3 6.10 8
+ 3ih2 ) '
i = 1,2, ... , n - 1.
We first first try try taking N = 180 180 steps (At = 11 s) of Euler's however, the We taking N steps (.D.t s) of Euler's method; method; however, the result result 40 is meaningless, with with temperatures degrees is meaningless, temperatures on on the the order order of of 10 1040 degrees Celsius! Celsius! Clearly Clearly Euler's method method is unstable with this choice A little experimentation shows shows that Euler's is unstable with this choice of of .D.t. At. A little experimentation that At cannot cannot be be much more than than 0.7 0.7 seconds, instability will will result. result. The The temperature temperature .D.t much more seconds, or or instability distribution at attt = = 180 180 seconds, computed using using 260 time steps (At = 0.69J, is is shown shown seconds, computed 260 time steps (.D.t == 0.69), distribution in Figure 6.12. in Figure 6.12. Before leaving this this example, example, we explain how method is is impleimpleBefore leaving we should should explain how Euler's Euler's method mented in practice. The system of ODEs ODEs is is mented in practice. The system of da dt
= M- 1 (-Ka + f(t)),
and so Euler's and so Euler's method method takes takes the the form form
As we we mentioned before, it since M M is is As mentioned before, it is is not not efficient efficient to to actually actually compute compute MM l ;, since 1 l Instead, we tridiagonal tridiagonal and and MM" is is completely completely dense. dense. Instead, we implement implement the the above above iteration iteration as as found by where sci) sW is is found by solving MSCi)
= -Ka C;) + f(ti)·
262
Chapter Chapter 6. 6. Heat Heat flow flow and and diffusion diffusion
x
Figure secFigure 6.12. 6.12. The temperature distribution in Example 6.8 after after 180 seconds (computed using using the the forward method, with with At t:..t == = 0.69). 0.69J. onds (computed forward Euler Euler method,
Therefore, Therefore, one one tridiagonal tridiagonal solve solve is is required required during during each each iteration, iteration, at at aa cost cost of of O(8n) O(8n) which costs O(2n2) operations, instead of of a multiplication by a dense matrix, which O(2n2) operoper38 ations. ations.38
Although for Although the the time time step step of of t:..t At = = 0.7 0.7 seconds, seconds, required required in in Example Example 6.8 6.8 for it is is easy to show by numerical stability, stability, does does not not seem seem excessively excessively small, small, it easy to show by numerical experiexperimentation mentation that that aa time time step step t:..t At on on the the order order of of hh22 is is required required for for stability. stability. (Exercise (Exercise 10 to verify we wish to refine refine the 10 asks asks the the reader reader to verify this this for for Example Example 6.8.) 6.8.) That That is, is, if if we wish to the spatial grid grid by factor of of 2, 2, we we must must reduce reduce the the time time step step by by aa factor factor of of 4. 4. This This spatial by aa factor is is the the typical typical situation situation when when we we apply apply the the method method of of lines lines to to the the heat heat equation: equation: The resulting system The resulting system of of ODEs ODEs is is stiff, stiff, with with the the degree degree of of stiffness stiffness increasing increasing as as the the spatial grid is refined. understand on on The reason for the heat heat equation The reason for the the stiffness stiffness in in the equation is is easy easy to to understand an that stiffness an intuitive intuitive level. level. The The reader reader should should recall recall from from Section Section 4.5 4.5 that stiffness arises arises when a (physical) system has components that decay at very different different rates. In heat flow, the the temperature heat flow, the components components of of interest interest are are the the spatial spatial frequencies frequencies in in the temperature distribution. High frequency components components are damped damped out very quickly, as we we saw from the Fourier series solution (see Section 6.1.1). Low Low frequency components, on the more slowly. refine the the spatial the other other hand, hand, decay decay more slowly. Moreover, Moreover, as as we we refine spatial grid, grid, we we can can 38 38We bit better, factoring the matrix M We can can even even do do aa bit better, by by factoring the matrix M once once (outside (outside the the main main loop) loop) and and then the factors to solve the cost to O(6n) iteration. then using using the factors in in the the loop loop to solve for for s(i). s^). This This reduces reduces the cost to O(6n) per per iteration. Factorization briefly in 10.2. Factorization of of matrices matrices is is discussed discussed briefly in Section Section 10.2.
6.4. Finite element element methods for the heat equation
263
represent represent higher higher frequencies, frequencies, and and so so the the degree degree of of stiffness stiffness worsens. worsens. Example Example 6.9. 6.9. We We now now apply apply the the backward backward Euler method method to to the the previous example. example. Using nn = 100 100 subintervals subintervals in in space space and and Ilt At = = 22 (seconds), (seconds), we we obtain obtain the the result Using result shown for which shown in in Figure 6.13. 6.13. Using Using aa time time step step for which Euler's method method is is unstable, unstable, we produced aa result correct. produced result which which is is qualitatively qualitatively correct. The Euler method The backward backward Euler method takes takes the the form form a(i+l) =
a(i)
+ IltM- 1 ( _Ka(i+1) + f(tHd)
.
Multiplying through Multiplying through by by M M and and manipulating, manipulating, we we obtain obtain
(M + IltK)a(Hl) = Ma i
+ Iltf(ti+1).
We solving aa system system with We therefore therefore compute compute a(Hl) a^+1) by by solving with the the (tridiagonal) (tridiagonal) matrix matrix M M+ AtK as the the coefficient coefficient matrix. matrix. IltK as
Br--------r--------r--------r--------r---------,
x
Figure Example 6.8 secFigure 6.13. 6.13. The The temperature temperature distribution distribution in in Example 6.8 after after 180 180 seconds (computed (computed using using the the backward backward Euler method, with with At = 2). 1). onds Euler method, Ilt =
Exercises Show that that the the mass mass matrix M is is nonsingular. nonsingular. (Hint: (Hint: This This follows follows from from the the 1. Show matrix M general result that if if {Ul' {ui,U2,... is aa linearly linearly independent independent set set in in an an ingeneral result that U2, ... ,,u un} inn } is ner product product space, space, then then the Gram matrix matrix G G defined defined by by G Gij = (Uj, (uj, Ui) Uj) is the Gram is ner ij = nonsingular. by nonsingular. To To prove prove this, this, suppose suppose Gx Gx = = O. 0. Then Then (x, (x, Gx) Gx) = = 0, 0, and and by
264
Chapter 6. Heat flow and diffusion
expanding Ui), it is possible to expanding (x, (x, Gx) Gx) in in terms terms of of the the inner inner products products (Uj, (iij,iii), to show that that (x, (x, Gx) Gx) = = 00 is is only only possible if xx = = 0. This suffices suffices to to show show that that G G show possible if O. This is nonsingular.) is nonsingular.)
2. Consider Consider aa function function /f in C[0,^]. 2. in e[O, l). (a) the projection Sn. (a) Show Show how how to to compute compute the projection of of f/ onto onto the the subspace subspace S Be sure sure n. Be to mass matrix matrix M in to observe observe how how the the mass M and and load load vector vector f arise arise naturally naturally in this this problem. problem. (b) find an an approximation approximation to to /f from Snn by the piece(b) We We can can also also find from S by computing computing the piecewise projecwise linear linear interpolant interpolant of of f. /. Show Show by by aa specific specific example example that that the the projecSnn and not tion tion of of f/ onto onto S and the the piecewise piecewise linear linear interpolant interpolant of of f/ from from Sn Sn are are not the be chosen to make make the the computations the same. same. (The (The number number nn can can be chosen small small to computations simple.) 3. Apply the to the the IBVP 3. Apply the method method of of lines lines to IBVP
au a2 u at - ax2 = x(l u(x, 0)
x) cos (t), 0
< x < 1, t > 0,
= 1, 0 < x < 1,
u(O, t) = 0, t > to, u(l, t) = 0, t
> to.
Use the finite finite element S3 (and Use the element method, method, with with the the approximating approximating subspace subspace 83 (and aa regular regular grid), grid), to to do do the the discretization discretization in in space. space. (a) Explicitly compute the mass mass matrix (a) Explicitly compute the matrix M, M, the the stiffness stiffness matrix matrix K, and and the the load load vector vector f(t). f(t). (b) Explicitly set up the system of ODEs resulting from applying the method of of lines. 4. bar, that bar in 4. Consider Consider aa heterogeneous heterogeneous bar, that is, is, aa bar in which which the the material material properties properties /9, c, and K are not constants but rather functions of space: p = p(x), c(x), p, c, and '" are not constants but rather functions of space: = p(x), cc == c(x), K'" = K(X). = "'(x). the appropriate the heat (a) (a) What What is is the appropriate form form of of the heat equation equation for for this this bar? bar? the appropriate weak form the resulting resulting IBVP (assuming (b) (b) What What is is the appropriate weak form of of the IBVP (assuming homogeneous homogeneous Dirichlet Dirichlet conditions)? conditions)? (c) the system (c) How How does does the system of of ODEs ODEs change? 5. An advantage therefore of fi5. An advantage of of the the weak weak formulation formulation of of aa BVP BVP (and (and therefore of the the finite element the equation makes sense nite element method) method) is is that that the equation makes sense for for discontinuous discontinuous cm. AsAscoefficients. coefficients. Consider Consider aa metal metal bar bar of of length length 1m 1m and and diameter diameter 44cm. sume that that one-half one-half of of the is made of copper copper (c (c = = 0.379J/(gK), 0.379 J/(gK), pP = = the bar bar is made of sume 8.97g/cm K = 4.04 W/(cmK)), and the the other other half is made made of of iron iron (spe8.97 g/cm 33,, '" 4.04 W /(cmK)), and half is (spe/ (g K), density g/ cm33,, thermal cific heat cc = cific heat = 0.437 0.437 JJ/(gK), density pp == 7.88 7.88 g/cm thermal conductivity conductivity K= Suppose the is initially degrees Celsius '" = 0.836 0.836 W/(cmK)). W / (cmK)). Suppose the bar bar is initially heated heated to to 55 degrees Celsius and then then (at (at time time 0) 0) its its ends ends are are placed placed in in an an ice bath (0 (0 degrees degrees Celsius). Celsius).
6.4. Finite Finite element element methods methods for for the heat equation equation 6.4. the heat
265 265
(a) (a) Formulate Formulate the IBVP describing describing this this experiment. experiment.
(b) Apply element method method and and the of lines lines to obtain the the (b) Apply the the finite finite element the method method of to obtain resulting system of of ODEs. ODEs. Use only aa uniform uniform grid an even even number resulting system Use only grid with with an number of midpoint of of subintervals subintervals (so (so that that the the midpoint of the the bar bar is is always always aa gridpoint). gridpoint). the stiffness matrix K explicitly. Give mass matrix Give the the mass matrix M M and and the stiffness matrix K explicitly. 6. 6. Consider Consider the the IBVP IBVP with with inhomogeneous Dirichlet Dirichlet conditions: conditions:
au pc at -
a2 u
< x < £, t > to, u(x, to) = ,¢(x), 0 < x < £, Ii ax 2
u(O, t) u(£,t)
= f(x, t), 0
= a(t), = b(t),
(6.43)
t> to, t> to.
(a) Formulate Formulate the the weak weak form form of of the IBVP. (a) the IBVP. (b) Show Show how apply the the finite element method by representing the apapto apply finite element method by representing the (b) how to proximate solution solution in in the form un(x, un(x,t)t) = vn(x, vn(x,t)t) + g gn(X, t), where where proximate the form n(x,t), n-l
vn(x, t) =
L Qi(t)¢i(X) i=l
and and
gn(X, t) = a(t)¢o(x)
+ b(t)¢n(x).
(c) Illustrate Illustrate by by solving p, c, to the material constants (c) solving (6.43) (6.43) with with /9, c, IiK equal equal to the material constants for iron, £t = 100cm, ^)(x) degrees Celsius, Celsius, a(t) a(t] = 0, 0, and = 100cm, ,¢(x) = 00 degrees and b(t) = = for iron, sin (601ft) (607r£).. 7. the IBVP 7. Consider Consider the IBVP from from Examples Examples 6.8 6.8 and and 6.9. 6.9.
(a) Using method of of Fourier series, find find the exact solution solution u(x, u(x,t). t). (a) Using the the method Fourier series, the exact (b) Using enough terms in the the Fourier Fourier series series to obtain aa highly accurate (b) Using enough terms in to obtain highly accurate solution, regular grid = 100 used in in solution, evaluate evaluate u(x, 180) 180) on on the the regular grid with with nn = 100 used Examples 6.8 6.8 and 6.9. Examples and 6.9. (c) Reproduce the the numerical numerical results results in by comcom(c) Reproduce in Examples Examples 6.8 6.8 and and 6.9, 6.9, and, and, by paring to the the Fourier series result, accuracy of each result. result. paring to Fourier series result, determine determine the the accuracy of each
8. Repeat Repeat Example Example 6.8, 6.8, with the following following changes. changes. First, First, assume assume that the bar bar that the 8. with the 33, c = 0.385 J/(gK), K = 4.01 W/(cmK)) is made of copper (p = 8.96g/cm is made of copper (p = 8.96 g/ cm , c = 0.385 J / (g K), Ii = 4.01 W / (cm K)) instead of of iron. iron. Second, Second, assume assume that that the "source" is given by instead the heat heat "source" is given by
f(x, t) = 10- 7 tx(60 - x)(100 - x) (note adds energy it away an(note that that f/ adds energy over over part part of of the the interval interval and and takes takes it away over over another the temperature Does the the temperature other part). part). Graph Graph the temperature after after 180 180 seconds. seconds. Does temperature approach aa steady steady state state as as tt -+ ->• oo? oo? approach
266
Chapter 6. Heat flow and diffusion diffusion
9. Consider a heterogenous bar of length 100 cm whose material material properties properties are 100cm given by by the the following formulas: given following formulas:
p(x) + O.01x O.Ola; g/cm g/cm33,, 0 100, p(x) = = 7.5 7.5 + 0 < xx < 100, c(x) = = 0.45 O.OOOlx J/(gK), 100, c(x) 0.45 + + O.OOOlx J/(gK), 0 0 < xx < 100, K(X) = 2.5 + 0.05z g/(cmK), 0 100. II;(X) = 2.5 + 0.05x g/(cmK), 0 < xx < 100. Suppose that the the initial initial temperature temperature in in the the bar bar is uniform 5 5 degrees degrees Celsius, Celsius, Suppose that is aa uniform and that that at = 00 both both ends are placed placed in in ice ice baths baths (while (while the the lateral side of of and at tt = ends are lateral side the bar bar is is perfectly perfectly insulated). the insulated). (a) Formulate Formulate the the IBVP IBVP describing describing this this experiment. (a) experiment. (b) Formulate Formulate the the weak weak form form of of the the IBVP. (b) IBVP. (c) (c) Use the finite element element method with backward Euler integration integration to to estiestimate the temperature after mate the temperature after 22 minutes. minutes. 10. Example 6.8, and suppose that the number of elements in the mesh 10. Consider Example is increased increased from 200, so so that that the the mesh mesh size size h is is cut in half. half. Show, by is from 100 100 to to 200, cut in Show, by numerical experimentation, that the the time time step in the the forward forward Euler Euler method numerical experimentation, that step tlt At in method be reduced reduced by by aa factor of approximately approximately four preserve stability. must be must factor of four to to preserve stability.
6.5 6.5
Finite and Neumann conditions Finite elements elements and Neumann conditions
So far we have only used finite element methods for problems with Dirichlet boundary conditions. conditions. The weak form of the BVP or IBVP, on which the finite element method is based, incorporates incorporates the Dirichlet conditions in the definition definition of V, the space test functions. When the the weak weak form form is discretized via the Galerkin method, space of of test functions. When is discretized via the Galerkin method, the boundary boundary conditions conditions form part of of the the definition definition of of S Sn, the approximating subthe form part approximating subn, the space space (see (see (5.50)). (5.50)). It turns turns out that Neumann Neumann conditions even easier easier to to handle handle in in the the finite finite out that conditions are are even element method. As we we show below, aN eumann condition condition does does not not appear appear explicitly explicitly element method. As show below, a Neumann in the weak weak form form or or in in the the definition the approximating in the definition of of the approximating subspace subspace (the (the analogue analogue is often called a boundary of Sn). For this reason, a Neumann condition of S )For this reason, a Neumann condition is often called a natural boundary n condition (since (since it automatically by by aa solution of the the weak weak form), form), while while it is is satisfied satisfied automatically solution of is aa Dirichlet Dirichlet condition condition is is referred referred to to as as an an essential essential boundary boundary condition condition (since (since it it is essential to include the condition explicitly in the weak form). essential
6.5.1
conditions The weak form of a BVP with Neumann conditions
(time-independent) BVP We will first first consider consider the (time-independent) BVP
d ( dx dU) = lex), 0< x < £,
- - k(x)dx
du (0) = 0 dx ' du dx(£) =0.
(6.44)
6.5. 6.5. Finite Finite elements elements and and Neumann Neumann conditions conditions
267
To derive the sides of To derive the weak weak form, form, we we multiply multiply both both sides of the the differential differential equation equation by by aa test function, and integrate by parts. We take for our space of test functions test function, and integrate by parts. We take for our space of test functions
that the the boundary boundary conditions conditions do not appear in this this definition). (note (note that do not appear in definition). Assuming (6.44), we for each € V, V, that u solves solves (6.44), we have, have, for each v E Assuming that
d ( k(x) dx du (x) ) v(x) = f(x)v(x), 0 < x < £ - dx (x) v(x) dx 1£ f(x)v(x) dx 1 dxd (k(x) dxdU) 0
=} -
=
£+ 1£
du ] [ k(x) dx (x)v(x) 0
=} -
=}
£
0
0
du dv k(x) dx (x) dx (x) dx =
1£ f(x)v(x) dx 0
1£ £ du dv 1 k(x) dx (x) dx (x) dx = f(x)v(x) dx. 0
0
In the the last last step, step, we we use use the the fact that In fact that
du (0) = du (£) = dx dx
o.
The (6.44) is The weak weak form form of of (6.44) is find u E V such that
1£
k(x)
~~ (x) ~~ (x) dx =
1£
f(x)v(x) dx for all v E V. (6.45)
In the the context context of (6.45), we we refer refer to to (6.44) (6.44) as the strong strong form of the the BVP. BVP. In of (6.45), as the form of
6.5.2 6.5.2
Equivalence the strong strong and weak forms forms of a BVP Equivalence of of the and weak of a BVP with with Neumann conditions conditions Neumann
The derivation derivation of of the the weak weak form form (6.45) (6.45) shows shows that that if if uu satisfies the original original BVP The satisfies the BVP (6.44) (the (the strong strong form), then u also also satisfies the weak weak form. form. We We now now demonstrate (6.44) form), then satisfies the demonstrate the converse: converse: If satisfies the the weak weak form form (6.45) (6.45) of of the the BVP, BVP, then then uu also also satisfies satisfies the If uu satisfies (6.44)----'including the the boundary boundary conditions! conditions! This is quite quite surprising at first first sight, sight, (6.44)—including This is surprising at since the the boundary boundary conditions mentioned in in (6.44) (6.44) are not mentioned mentioned in in (6.45). (6.45). since conditions mentioned are not We assume that u E GV V and that
1
£ du dv o k(x) dx (x) dx (x) dx
=
1£ 0
-
f(x)v(x) dx for all v E V.
Integrating by parts on the left left side Integrating side of of this equation equation yields
du ]£ [ k(x) dx (x)v(x) 0
-
1£ 0
d [ du ] (£ dx k(x) dx (x) v(x) dx = 10 f(x)v(x) dx for all v E V. (6.46)
268
Chapter Heat flow diffusion Chapter 6. 6. Heat flow and and diffusion
Now, V C where V is is defined defined as as the the space space of of test test functions used for for aa Dirichlet Dirichlet Now, C V, where functions used problem: problem: V = c1[0, R] = {v E V : v(O) = v(R) = O} .
Therefore, (6.46) holds for all V. Since Since the Therefore, in in particular, particular, (6.46) holds for all vv e E V. the boundary boundary term term in in (6.46) v(0) = v(R) v(t} = = 0, 0, we obtain (6.46) vanishes vanishes when when v(O) we obtain -
or or
i
0
f
d[
dU]
dx k(x) dx (x) vex) dx =
if 0
f(x)v(x) dx for all v E V,
fat {! [k(X) :~ (x)] + f(X)} vex) dx = 0 for all v E V.
Using in Section Using the the same same argument argument as as in Section 5.4.2, 5.4.2, we we see see that that d [ k(x)dx(x) du ] +f(x) =0, O<x
d [ k(x) du - dx dx (x) ] = f(x), 0 < x < £. Thus, if if u U satisfies satisfies the the weak weak form the BVP, BVP, it it must must satisfy satisfy at at least least the the differential Thus, form of of the differential equation appearing form. equation appearing in in the the strong strong form. It is The condition It is now now easy easy to to show show that that the the boundary boundary conditions conditions also also hold. hold. The condition (6.46) is equivalent equivalent to (6.46) is to
]i - [ k(x) du dx (x)v(x) 0
rt {dxd [k(x) dU]} dx (x) + f(x) vex) dx
+ 10
= 0 for all v E
_ V,
and, and, since since we we have have already already shown shown that that d [ k(x) dx du (x) ] +f(x) =0, O<x
du ]i [ k(x) dx (x)v(x) 0
_
= 0 for all v E V.
(6.47)
We v(x) = 1I — x/l, this We now now take, take, for for example, example, vex) x/R, which which certainly certainly belongs belongs to to V. With With this choice (6.47) becomes choice of of v, v, (6.47) becomes -k(O): (0)
=0
(since = 1, 1, v(i) = 0). 0). Since Since fc(0) (since v(0) v(O) = v(R) = k(O) > 0 0 by by assumption, assumption, this this can can hold hold only only if if du dx (0) = O.
6.5. conditions 6.5. Finite Finite elements elements and and Neumann Neumann conditions
269
Similarly, xli shows that Similarly, choosing choosing v(x) v(x) = = x/i shows that
du (i) = 0 dx must hold. Thus, necessarily satisfy must hold. Thus, when when uu satisfies satisfies the the weak weak form form (6.45), (6.45), it it must must necessarily satisfy the the Neumann Neumann boundary boundary conditions. conditions. This This completes completes the the proof that the strong strong and weak forms weak forms of of BVP BVP (6.44) (6.44) are are equivalent. equivalent.
6.5.3 6.5.3
Piecewise linear finite elements with Neumann conditions
Now that we have have the weak form we choose Now that we the weak form of of the the BVP, BVP, we choose the the appropriate appropriate subspace subspace of to obtain finite of piecewise piecewise linear linear functions functions and and apply apply the the Galerkin Galerkin technique technique to obtain aa finite weak form, we element boundary conditions element method. method. Since Since no no boundary conditions are are imposed imposed in in the the weak form, we augment the space Snn defined basis functions augment the space S defined in in Section Section 5.6 5.6 by by adding adding the the basis functions 4>0 o and and 0n (see (see Figure Figure 6.14). 6.14). We We denote denote the the resulting resulting subspace subspace by by Sn: Sn: 4>n
Sn = i] -+ continuous and and piecewise piecewise linear} linear} — {p :'• [0, [Oj^j —*• R R- :: Pp isis continuous = span span{o,0i,...,0 = {4>0, 4>1, ... , 4>n} n }..
j
._._
0.8 0.7
,
~
0.9
,! ,
\
,
0.6
,
0.5
i
0.4
,
,
0.3 0.2 0.1 00
\
,
, 0.2
0.4
0.6
0.8
Figure piecewise linear functions o 4>0 and Figure 6.14. 6.14. The The piecewise linear basis basis functions and 4>n. (f)n.
Having defined defined the the approximating approximating subspace subspace Sn, 5n, we apply the Galerkin method Having we apply the Galerkin method to the problem to the the weak weak form form (6.45), (6.45), yielding yielding the problem
-
r
l
r
l
dVn dv find Vn E Sn such that 10 k(x) dx (x) dx (x) dx = 10 f(x)v(x) dx for all v E Sn,
Chapter 6. 6. Heat Heat flow flow and and diffusion Chapter diffusion
270
or, or, equivalently, equivalently, find Vn E Sn such that
lot k(x) ~; (x) dxi (x) dx = 10£ !(X)tPi(X) dx, i = 0,1, ... , n. (6.48)
For simplicity simplicity of of notation, we will again again write write
a(v, w)
=
(t dv dw io k(x) dx (x) dx (x) dx,
so that (6.48) can be be written written as as so that (6.48) can (6.49)
We write the vn as We then then write the unknown unknown Vn as n
Vn = LUjtPj j=O
and substitute substitute into (6.49). (6.49). The result is n
La(tPj,tPi)Uj = (J,tPi), i =O,l, ... ,n. i=O
n+ in the the n + 1I unknowns unknowns uo, UI, ... , Un. We can can This is is aa system system of of n + 11 equations equations in uo,ui,...,u This n. We write the system system as the matrix-vector matrix-vector equation = f by by defining defining write the as the equation Kii =
Kij
= a(tPj, tPi),
i,j
= 0,1, ... , n,
ii = (J,tPi)' i = O,l, ... ,n. The calculations calculations involved involved in in forming and the the vector vector f are are exactly The forming the the matrix matrix K and exactly the the same as for problem, except that 0 tPo0 and tPnn are are qualitatively same as for aa Dirichlet Dirichlet problem, except that and 4> qualitatively different different tPI, tP2, ... , tPn-1 (tPo and and 0 tPnn are nonzero on on aa single from from 0i, 02,..., 0n-i (0o are each each nonzero single subinterval, subinterval, namely namely [XQ, and [x respectively, while while each each 0j, 1,2,..., 1, is is nonzero nonzero on on [xo, Xi] Xl] and [Xn-l, xn], tPi, ii = = 1,2, ... , n n -— 1, n _i, x n], respectively, [Xi-i,X [Xi-I, Xi+l]). i+i}}.
Having now now reduced reduced the the problem problem to to the the linear linear system = f, where where now now Having system Kii = n+i 5 we and ^ ii, ff eE RRn+1, weface facethe theDifficulty difficultythat thatKKisisa asingular singular and we know that (6.44) does matrix. This is not surprising, surprising, since since we that the original BVP BVP (6.44) does not not have a unique unique solution. We We can can show show directly directly that that K is is singular, and understand understand the If the nature of of the singularity, as as follows. follows. If
K E 6 R(n+1)x R(n+i)x(n+i) (n+1)
271 271
6.5. Finite elements and Neumann conditions
then then
But But (6.50) j=O
is piecewise linear value one mesh node. node. It follows that is is piecewise linear and and has has value one at at each each mesh It follows that (6.50) (6.50) is the constant function derivative zero. zero. Therefore Therefore CKUe)i (Ku c )j = 0 for the constant function one, one, and and hence hence has has derivative = 0 for each i, ij which Ku ce = 0 and and hence singular. Moreover, Moreover, it it can each which shows shows that that KU = 0 hence that that K K isis singular. can = 0, then ii be shown shown that Uuec spans the the null space space of K; that is, is, if Kii Ku = u is is a multiple of Uc Exercise 5} of u (see Exercise 5]. e (see The fact that K is singular means that we we must give special attention to the Kii == f. In particular, particular, if we process of solving Ku we ignore the singularity of K K and Kii == f using using computer computer software, we will will get get either solution solve Kii software, we either aa meaningless meaningless solution solve or an error message. To solve this this singular we must must add or an error message. To solve singular system system correctly, correctly, we add another another equation sufficient, because the equation (one (one additional additional equation, correctly chosen, chosen, will will be sufficient, the null space of is one-dimensional). simple choice choice is is the equation Un un = = 0; one-dimensional). A A simple the equation 0; this this is is null space of K is equivalent equivalent to choosing, choosing, out out of of the the infinitely infinitely many solutions solutions to (6.44), (6.44), the one with u(i] =— O. 0. Moreover, Moreover, we additional equation equation by simply removing u(C) we can can impose i~pose this this additional by simply removing; the last last row and column column from and the the last last entry The last of K the row and from K, K, and entry from from f. f. The last column column of K un; insist consists of the the coefficients coefficients of the terms in the the equations involving u ; if we n that Un = 0, we can remove these these terms terms from from all We then that un = 0, then then we can remove all nn + 11 equations. equations. We then have n + 1l equations in the the n unknowns unknowns uo, This is one more more equation have equations in WQ, ut, W i ,... . . . ,,Un-l. w n _i. This is one equation we need, so obtain aa square system. than we so we we remove remove one one equation, equation, the the last, to obtain system. It can be proved proved that that the the resulting resulting n xx n system system is is nonsingular nonsingular (see 11). can be (see Exercise Exercise 11). Of have removed removed aa different row and remove Of course, course, we we could could have different row and column. column. If If we we remove the column, we are selecting selecting the approximate solution solution Vn vn satisfying the iih ith row row and and column, we are the approximate satisfying Vn(Xi) Vn(Xi) = = O. 0. Example 6.10. 6.10. Consider the Neumann Neumann problem problem Example
d2 u 1 -- = x - - 0 dX2 2'
< x < 1,
~: (0) = 0,
(6.51 )
~: (1) = 0. It is easy to show (by direct integration) that X3
u(x) =
X2
-"6 + 4" + C
272
Chapter 6. Heat flow and diffusion
is aa solution solution for any constant and that that is for any constant C, and x3 x2 1 u(x) = - - + - - 6 4 12 is the the solution solution satisfying u(l) = 0. is satisfying u(l) 0. We will the finite finite element We define define xi = ih, ih, Xi = We will use use aa regular regular grid grid with with the element method. method. We = 0,1, 0,1,...,n, l/n. With With piecewise piecewise linear linear finite finite elements, elements, as as we we have have seen seen ii = ... , n, hh =— lin. before, the stiffness matrix matrix K is tridiagonal before, the stiffness K is tridiagonal since since
Ii - il > 1 :::}
ixi a:; (x)
°
on (0, f).
(x) =
We compute (as suggested suggested above, above, the the computations involving <^o and CPn (f)n We compute as as follows follows (as computations involving CPo and must be handled separately from those involving CPl, CP2, ... n-i)-' - .. _l(Hl)h (dCPi ) 2 _l(Hl)h ~ _ ~ K .. d (x) dxh2dx - h' i=1,2, ... ,n-1, (i-l)h x (i-l)h h (dCPO ) 2 tIl Koo = io dx (x) dx = io h2 dx = h'
- r
_ /1
Knn =
I-h
(dCPn ) dx (x)
2 /1 dx =
1
I-h h2 dx =
1
h'
1)
l(i+1)h dCPi dCPHI l(i+1)h ( K i ,i+l = -(x)--(x)dx = -2" ih dx dx ih h 1 . = -
h'
Z
dx
= 0, 1, ... , n - l.
Since symmetric and of K. Next, Since K is is symmetric and tridiagonal, tridiagonal, this this completes completes the the computation computation ofK. Next, we have we have
h=
l(HI)h (x - ~) CPi(X) dx (i-I)h 2
= lih (x _ ~) (x - (i - l)h) dx (i-I)h 2 h
+ I~HI)h = ih2 io =
Io
~)
(1 _ ~ x
~
4'
h
h2
(x _
(x -
~) (1 - ~) dx
h
6-4' in=
!~h(x-~)X-(~-h)dX h 4
h2 6
---
ih) dx
6.5. Finite elements and and Neumann conditions
273
As suggested As suggested above, above, we compute a solution ii u to Kii Ku = f by removing the last row and column from K and the last component from ff,, and solving the resulting square, square, nonsingular system. result, for for n n = nonsingular system. The The result, = 10, 10, is is shown shown in in Figure Figure 6.15. 6.15.
0.1 r----r-----"T"""""---....---,=:::;;::;;:::=:::;,
o
-0.1~------~--------~------~--------~------~
o
0.2
0.4
0.6
0.8
x -3
1 x 10
-1~---~---~---~----~--~ o 0.2 0.4 0.6 0.8
x
Figure Figure 6.15. 6.15. The The exact exact and computed computed solution (top) (top) and and the the error error in in the computed solution (bottom) for (6. 51}. The computed solution was found using (6.51). piecewise linear finite finite elements with a regular grid of of 10 10 subintervals. subintervals.
6.5.4 6.5.4
Inhomogeneous conditions Inhomogeneous Neumann Neumann conditions
The method requires The finite finite element element method requires only only minor minor modifications modifications to to handle handle the the inhomoinhomogeneous geneous Neumann Neumann problem problem
d (k(x)dx dU) =f(x),o<x
- dx
du dx (0) = a,
~~ (f) =
(6.52)
b.
In the weak weak form the BVP, BVP, the boundary terms terms from from the the integration by In deriving deriving the form of of the the boundary integration by parts parts do do not not vanish: vanish:
- ior£ dxd
[
du ] k(x) dx (x) v(x) dx
274
Heat flow and diffusion Chapter 6. Heat diffusion
The weak weak form form becomes The becomes find u E V such that a(u, v)
= (f, v) + k(R)v(R)b -
k(O)v(O)a for all v E V. (6.53)
When we we apply apply the the Galerkin Galerkin method, method, we we end end up up with with aa slightly slightly different different rightrightWhen hand-side vector vector f. We We have hand-side have
ii
= (f, rPi) + k(R)rPi(R)b -
k(O)rPi(O)a
_ { (f, rP~)' ~ :: 1,2,.... , n - 1 ~ince rPi(R~ = rPi(O) (f, rP~) - k(O)a, ~ - 0 (smce rPo(R) - 0, rPo(O) - 1), (f, rPi) + k(R)b, i = n (since rPn(R) = 1, rPn(O) = 0).
= 0), (6.54)
Example 6.11. now apply method to the inhomogeneous Example 6.11. We We now apply the the finite finite element element method to the inhomogeneous Neumann problem Neumann problem d2 u - dx 2
du (0) dx
=X
1
-
2'
0
< x < 1,
= 1,
(6.55)
dU(l) = 1. dx
The exact solution solution is The exact is
x3
u(x)
x2
13
= -6 + '4 +x - 12
(again choosing choosing the solution with = 0). calculations are are exactly the same (again the solution with u(l) u(l] = Q). The The calculations exactly the same as 6.10, except first and last components as for for Example Example 6.10, except that that the the first and last components of of f are are altered altered as as given in (6.54). We therefore have have given in (6.54)We therefore
h2 fo = 6
h
- 4-
h h2 1, fn = 4 - 6
+ 1.
The results results are = 10, 10, in The are shown, shown, for for n n = in Figure Figure 6.16. 6.16.
6.5.5
The finite element method for an an IBVP with Neumann conditions
We now now briefly briefly consider consider the the following We following IBVP: IBVP:
au
a
2 u pc at - '" ax 2
= f(x, t), 0 < x < R,
t> to,
6.5. Finite elements and Neumann conditions
275 275
_1.51-_ _ _-'-_ _ _---'......._ _ _-'-_ _ _--L_ _ _- - J o 0.2 0.4 0.6 0.8
x
_1~------~---------L--
o
0.2
______ ________ ______ ~
0.4
0.6
~
~
0.8
x
Figure computed solution {top} Figure 6.16. 6.16. The The exact exact and computed (top) and and the the error in in computed solution was found found using using the computed computed solution {bottom} (bottom) for {6.55}. (6.55). The The computed subintervals. piecewise linear finite finite elements with a regular grid of of 10 subintervals.
u(x, to) = 'l/J(x), 0< x du dx (0, t) = 0, t > to, du dx (e, t) = 0, t
< e,
(6.56)
> to·
Finite element methods for. the heat equation with Neumann conditions are derived as in Section 6.4. arise: as in Section 6.4. The The following following differences differences arise: •• In functions In the the derivation derivation of of the the weak weak form form of of the the IBVP, IBVP, the the space space V V of of test test functions is replaced replaced by by if. is V.
When the the Galerkin Galerkin method method is is applied, applied, the the approximating approximating subspace subspace Sn Sn is is re•• When replaced by by Sn. placed Sn. • The The mass matrix M ME e R(n-l)x(n-l), R( n - 1 ) x ( n - 1 ), stiffness stiffness matrix K K E€ R(n-l)x(n-l), RC11-1)^"-1), and 11 1 load vector f(t) Rn-l ME x (n+l) ?,K R(n+1)x (n+1) }, f(t) E GR " are replaced by M € R(n+l) R(n+i)x(n+i) £E e R(n+i)x(n+i) n+1 f(t) G E Rn+l, respectively. The Secand f(t) R , respectively. The new stiffness stiffness matrix K was was derived derived in Secthe mass mass matrix, matrix, M, derived analogously. tion 6.5.3, tion 6.5.3, and and the the new new form form of of the M, is is derived analogously. In contrast contrast to of aa steady-state BVP, the In to the the case case of steady-state BVP, thesingularity singularityof ofthe thematrix matrixKK causes causes IBVP, since since we we are are not not required required to to solve linear system no particular problem problem in in an no particular an IBVP, solve aa linear system whose matrix is is K. whose coefficient coefficient matrix K.
276
Chapter Chapter 6. Heat flow and diffusion
Moreover, the the source not required required to to satisfy Moreover, source function function fI(x, ( x , t t)) is is not satisfy aa compatibility compatibility condition condition in in order order for for (6.56) (6.56) to have aa solution. solution. Physically, Physically, this makes sense. sense. ThinkThinking flow, for the temperature distribution is changing, ing of of heat heat flow, for example, example, the temperature distribution is in in general general changing, so so there there is no no inconsistency inconsistency in having having aa net gain or loss of of total total heat heat energy energy in in the the bar. the steady-state the temperature be bar. On On the the other other hand, hand, for for the steady-state case, case, the temperature could could not not be independent of time time if the total energy in in the bar were were changing changing independent of if the total amount amount of of heat heat energy the bar with with time. time. This also be be seen This can can also seen directly. directly. If If uu is is the the solution solution of of (6.56), (6.56), then then
fof I(x, t) dx = fof pc ~~ (x, t) dx _fof ~ ~:~ (x, t) dx =
I
f
au
pC- (x, t) dx o t
a
= fof pc ~~ (x, t) dx = foe pc
[au]e ~-a (x, t) x
x=o
~ (~~ (C, t) - ~~ (0, t))
~~ (x, t) dx (by the Neumann conditions)
= %t [fof pcu(x, t)
dx].
This last expression to) the the time rate of of change of the total heal heat This last expression is is (proportional (proportional to) time rate change of the total energy contained in the (see (2.2) (2.2) in in Section Section 2.1). 2.1). energy contained in the bar bar (see We leave the detailed formulation of aa piecewise piecewise linear linear finite finite element method We leave the detailed formulation of element method to the the exercises for for (6.56) (6.56) to exercises (see (see Exercise Exercise 6). 6). Exercises Exercises
1. bar of length 11 m m and 1. Consider Consider an an aluminum aluminum bar of length and radius radius 11 cm. cm. Suppose Suppose that that the the sides the bar bar are are perfectly perfectly insulated, heat energy sides and and ends ends of of the insulated, and and heat energy is is added added to to the interior interior of of the the bar bar at rate of of the at aa rate
I(x) = 10- 7 x(25 - x) (100 - x)
1
+ 240 W /cm3
(x given in in centimeters, the usual used). The (x is is given centimeters, and and the usual coordinate coordinate system system is is used). The thermal conductivity W / (cm K). Find and graph thermal conductivity ofthe of the aluminum aluminum alloy alloy is is 1.5 1.5W/(cmK). Find and graph temperature of the steady-state steady-state temperature of the bar. Use Use the finite element method. method. 2. Repeat the previous 2. Repeat previous exercise, exercise, except except now assume assume that the the bar is heterogeheterogeneous, with thermal conductivity conductivity neous, with thermal
~(x) =
1.0 +
(1~0) 2
3. copper bar bar (thermal 4.04 W/(cm W / (cm K)) oflength 3. Consider Consider aa copper (thermal conductivity conductivity 4.04 K)) of length 100 100 cm. cm. Suppose that heat is added to the the bar Suppose that heat energy energy is added to the interior interior of of the bar at at the the constant constant
6.5. Finite elements and Neumann conditions
277
rate of ff(x) (x) = = 0.005W/cm 0.005 W/cm33,, and and that that heat heat energy energy is is added added to to the the left left end rate of end (x = = 0) 0) of of the the bar rate of W/cm22.. (x bar at at the the rate of 0.01 0.01 W/cm (a) what rate rate must heat energy be removed removed from right end 100) (a) At At what must heat energy be from the the right end (x = = 100) that aa steady-state or just in in order order that steady-state solution solution exist? exist? (See (See Exercise Exercise 6.2.5a, 6.2.5a, or just use common common sense.) sense.) use
(b) Formulate Formulate the the BVP BVP leading leading to to aa steady-state steady-state solution, solution, and and approximate approximate (b) the using the the finite finite element the solution solution using element method. method. 4. Suppose Neumann problem problem (6.44), the comcom4. Suppose that, that, in in the the Neumann (6.44), f/ does does not not satisfy satisfy the patibility condition patibility condition
lo
f-
f(x)dx =
o.
Suppose we just just ignore ignore this this fact try to Suppose we fact and and try to apply apply the the finite finite element element method method anyway. the method break down? with the Neumann anyway. Where Where does does the method break down? illustrate Illustrate with the Neumann problem problem d2 u
- dx 2 = x, O<x
~~(O) = 0, ~~(1) =0. 5. purpose of exercise is to provide of the proof that that the null 5. The The purpose of this this exercise is to provide aa sketch sketch of the proof the null space of is spanned spanned by is the the stiffness stiffness matrix matrix arising in aa by u U c ,, where where K is arising in space of K is Neumann problem problem and the vector vector with equal to Neumann and Uucc is is the with each each component component equal to one. one.
(a) Suppose Suppose Ku = 0, 0, where where the the components of u are are UO,Ul, UQ, ui,..., un. Show Show Ku = components of ..• ,Un. (a) that this this implies that a( v, v) =0, = 0, where where that implies that a(v,v) n
V
=
L
Ui¢>i·
i=O
(Hint: (Hint: Compute Compute U· u • Ku.) Ku.)
(b) Explain Explain why why a( a(v, = 00 implies implies that is aa constant constant function. (b) v, v) = that v is function. (c) Now explain why implies that is aa multiple multiple of of Uucc .• (c) Now explain why Ku Ku = 00 implies that u is 6. Formulate the the weak weak form (6.56). 6. (a) (a) Formulate form of of (6.56). to use use piecewise finite elements elements and the method lines (b) how to (b) Show Show how piecewise linear linear finite and the method of of lines to system of of ODEs. ODEs. to reduce reduce the the weak weak form form to to aa system
(c) the backward Euler method method for for the the time integration, estimate estimate the the (c) Using Using the backward Euler time integration, solution solution of of the the IBVP IBVP 2 au _ aax2 u = 0' at
0 <.x < 1, t
> 0,
278
Chaoter Chapter 6.
Heat flow and difFusior diffusion
u(x, 0) = x(l - x), 0 < x < 1,
au ax (0, t) = au (1, t) = ax
0, t
> 0,
0, t
>0
and graph graph the the solution at times times 0,0.02,0.04,0.06, 0,0.02,0.04,0.06, along along with with the the steadyand solution at steadystate 6.2.1 and state solution. solution. (See (See Exercise Exercise 6.2.1 and Figure Figure C.12.) C.12.) 7. (a) (a) Formulate Formulate the the weak weak form form of the following following BVP BVP with with mixed mixed boundary boundary 7. of the conditions: conditions:
d ( k(x) dU) - dx dx
= I(x), 0 < x < £,
u(O) = 0,
~~(£) =
(6.57)
O.
(b) (b) Show Show that that the the weak weak and and strong strong forms forms of of (6.57) (6.57) are are equivalent. equivalent. 8. Repeat Repeat Exercise Exercise 77 for for the the BVP 8. BVP
d ( k(x)dx dU) =/(x),O<x<£, -dx du (0) = 0 dx ' u(£) = O. 9. (p = 8.97g/cm 0.379 J/(gK), '" K= 9. Consider Consider a a copper copper bar bar (p = 8.97 g/cm 33,,cc = = 0.379J/CgK), = 4.04 4.04 W/(cmK)) W/CcmK)) of length 11 m m and cross-sectional area area 2 2 cm cm22.• Suppose the bar bar is is heated heated to to aa of length and cross-sectional Suppose the uniform temperature temperature of of 88 degrees degrees Celsius, Celsius, the the sides sides and and the the left left end 0) uniform end (x (a; = = 0) = 100) are perfectly insulated, and the right end end (x = 100) is is placed in an ice bath. of the the bar bar by by formulating formulating and Find the the temperature temperature distribution distribution u(x, Find u(x,t)t) of and solvsolving ing an IBVP. IBVP. Graph u(x, 300) (t in in seconds). seconds). Use the finite element method method with backward backward Euler Euler time time integration. with integration. 10. Consider aa chromium chromium bar bar of oflength 25 cm. cm. The The material material constants constants for for chromium chromium 10. Consider length 25 are are pp = «= 0.937W/(cmK). = 7.15g/cm 7.15g/cm33,, cc = = 0.439J/(gK), 0.439J/(gK), '" = 0.937W/(cmK). Suppose that the the left left end end of the bar bar is is placed placed in in an an ice ice bath, bath, the the right right end end is is Suppose that of the held fixed fixed at at 88 degrees degrees Celsius, Celsius, and the bar bar is allowed to to reach reach aa steady-state held and the is allowed steady-state Then (at = 0) 0) the the right right end is insulated. Find the the temperature temperature. Then temperature. (at t = end is insulated. Find temperature distribution using using the the finite finite element element method method and and backward backward Euler Euler integration. distribution integration. Graph u(x, 100). Graph n 1 x n 1 11. R(n+1)x(n+1) matrix for for the Neumann problem, and 11. Let K E 6 R,( + ) ( + ) be the stiffness stiffness matrix nxn nxn let E R R be the the matrix matrix obtained by removing removing the the last row and column let K e be obtained by last row and column
6.6. Green's functions for the heat equation equation
279
of nonsingular. (Hint: of K. Prove Prove that that K is is nonsingular. (Hint: K is is the the stiffness stiffness matrix matrix for for the the BVP with with mixed mixed boundary boundary conditions, conditions, Neumann at the the left left and and Dirichlet Dirichlet at at BVP Neumann at the right.) the right.)
6.6 6.6
Green's Green's functions functions for for the the heat heat equation equation
We now discuss the heat We will will now discuss Green's Green's functions functions for for the heat equation. equation. Our Our treatment treatment of of this this topic topic is is intentionally intentionally rather rather superficial, superficial, and and our our goals goals are are modest. modest. We We simply simply want want to time-dependent PDE show to introduce introduce the the concept concept of of aa Green's Green's function function for for aa time-dependent PDE and and show an more would main topics. an example. example. To To do do anything anything more would distract distract us us from from our our main topics. present any Before Before we we present any specific specific Green's Green's functions, functions, we we describe describe the the form form that that the the Green's two examples functions. Green's function function must must take. take. We We have have so so far far seen seen two examples of of Green's Green's functions. First First of of all, all, in in an an IVP IVP for for an an ODE, ODE, the the solution solution u == u(t) u(t) is is of of the the form form
u(t) = (JO G(t; s)f(y) ds,
ito
where the right-hand where f/ is is the right-hand side side of of the the ODE ODE (see (see Section Section 4.6 4.6 for for an an example). example). This This time ss on the solution formula formula shows shows the the contribution contribution of of the the data data at at time on the solution at at time time t t— 39 this The the value is this contribution contribution is is G(t; G(t] s)f(s) s)f(s) ds. ds.39 The reader reader should should notice notice how how the value u(t) u(t) is found by "adding found by "adding up" up" the the contributions contributions to to u(t) u(t) of of the the data data from from the the time time interval interval oo). (The (The value value of of G(t; s) is is zero zero when when s > t, t, so so in in fact fact only only the data from from the the [to, 00). the data time the value time interval interval [to, [tQ,t]tl can can affect affect the value of of u(t).) u(t).) We We have have also also seen seen an an example example of of aa Green's Green's function function for for aa steady-state steady-state BVP BVP (see where the is (see Section Section 5.7), 5.7), where the data data (the (the right-hand right-hand side side of of the the differential differential equation) equation) is given fl. In value u(x) given over over aa spatial spatial interval interval such such as as [0, [0,1]. In this this case, case, the the value u(x) is is determined determined points in by adding the contributions the data by adding up up the contributions from from the data at at all all points in the the interval. interval. The The solution solution formula formula takes takes the the form form
u(x) =
10£ g(x;y)f(s) dy.
In t) (the of In the the case case of of aa time-dependent time-dependent PDE, PDE, the the data data !(x, /(x,t) (the right-hand right-hand side side of the PDE) prescribed over the PDE) is is prescribed over both both space space and and time. time. Therefore, Therefore, we we will will have have to to add add up the the contributions contributions over over both both space space and and time, time, and and the the formula formula for for the the solution solution will will involve involve two two integrals: integrals:
u(x, t) =
roo
ito
r G(x, t; y, s)f(y, s) dy ds. io l
39 39By mean that is By this this we we mean that the the contribution contribution of of the the data data over over aa small small interval interval (8 (s -— £,8 e, s + + £) e) is
8+<
1
8-<
G(tj s)f(s) ds.
Chapter 6. 6. Heat Heat flow flow and and diffusion diffusion Chapter
280 280
6.6.1 6.6.1
The function for The Green's Green's function for the the one-dimensional one-dimensional heat heat equation equation under under Dirichlet Dirichlet conditions conditions
We will will compute We compute the the Green's Green's function function for for the the IBVP IBVP
au a 2u pc at - "'ax 2 = f(x,t), 0< x < £, t> to,
°
u(x, to) = 0, < x < £, u(O, t) = 0, t > to, u(£, t) = 0, t > to.
(6.58)
The to (6.58) is The solution solution to (6.58) was was derived derived in in Section Section 6.1; 6.1; it it is 00
u(x, t) =
L an(t) sin (n;x), n=l
where where
and and
Cn(s) =
~ 1£ fey, s) sin C?) dy.
We can derive derive an an expression expression for for the Green's functions functions by manipulating the the formula formula We can the Green's by manipulating for u(x, u(x,i): for t):
If we define If
G(x , t ;Y, s) =
~ ~oo e-t
{
£
pel L...n=l
0,
£'
to S sst, s > t, (6.59)
then we we have then have u(x,t) =
t ltD
XJ
(t G(x,t;y,s)f(y,s)dyds.
10
Therefore, G(x, G(x,t;y,s) is the the Green's Green's function function for for (6.58). Therefore, t; y, s) is (6.58).
281 281
6.6. 6.6. Green's Green's functions for the the heat heat equation
Having derived the Green's ought to to ask the question: What is is it Having derived the Green's function, function, we we ought ask the question: What it good for? It be obvious the reader reader that that the the formula is no good for? It may may be obvious to to the formula for for uu is no more more useful, useful, for the original Fourier series Therefore, for computational computational purposes, purposes, than than the original Fourier series solution. solution. Therefore, the of the Green's function function is as aa computational computational tool. Rather, the the the value value of the Green's is not not as too1. 40 Rather, Green's function is valuable because of the insight it gives us into the behavior of Green's function is valuable because of the insight it gives us into the behavior of solutions to the the IBVP. IBVP. Indeed, Indeed, the the Green's Green's function function can can be be regarded regarded as solution solutions to as aa solution to IBVP with special right-hand right-hand side. side. The The solution to to the the IBVP with aa special solution to
au pc at
a2 u
- ~ ax2
= 6(x - yo)6(t - so), 0
< x < e, t > to,
u(x, to) = 0, 0 < x < e, u(O, t) = 0, t > to, u(e,t) = 0, t> to
(6.60)
is is
u(x, t)
= G(x, t; Yo, so).
The right-hand side at the The right-hand side 6(x-yo)6(t-s 6(x—yo)6(t—SQ)o) represents represents aa source source concentrated concentrated entirely entirely at the space-time point (Yo, so). The IBVP (6.60) the following following experiment: A space-time point (yo,so). The IBVP (6.60) describes describes the experiment: A bar is initially initially at temperature 0, time So to, A Joules energy, bar of of length length t is at temperature 0, and and at at time SQ > to, Joules of of energy, where is the the cross-sectional the bar,41 added to to the the cross-section cross-section at where A is cross-sectional area area of of the bar,41 are are added at xx — = y Yo. The heat heat energy energy then then flows flows in in accordance accordance with with the the heat heat equation. equation. By By 0. The examining snapshots snapshots of of G(x, G(x,t;yo,so), an idea idea of of how temperature examining t; Yo, so), we we can can get get an how the the temperature changes with time. changes with time.
e
e
Example 6.12. 6.12. We We consider an iron bar of of length length t =— 100 100cm (p = 7.8&g/cm Example consider an iron bar cm (p = 7.88 g/ cm33,, QA37J/(gK), 0.836 W/(cmK)). W/(cmK)). We We graph graph the the solution solution to to (6.60) (6.60) with cc = 0.437 J/(gK), ~K = 0.836 with to = 0, 0, So SQ = 60, 60, and and yo 75 (time (time measured measured in in seconds) 6.17, which to Yo = 75 seconds) in in Figure Figure 6.17, which shows how how the spike of heat energy energy diffuses diffuses over over time. shows the spike of heat time. Formula (6.59) (6.59) has its shortcomings; for example, wishes to to compute compute aa Formula has its shortcomings; for example, if if one one wishes snapshot of of G G(x,t;y , s ) for t greater than but very close to SQ, then many terms (x, t; Yo, so) for t greater than but very close to so, then many terms snapshot 0 0 to obtain obtain an result (see is of the Fourier required to of the Fourier series series are are required an accurate accurate result (see Exercise Exercise 1). 1). It It is possible to obtain aa more more computationally computationally efficient efficient formula formula for for use when tt is possible to obtain use when is very very close to SQso. However, to do do so require aa lengthy lengthy digression. close to However, to so would would require digression.42
6.6.2
Green's functions under other boundary boundary conditions
The method can can be be used to derive derive the the Green's Green's function function for the one-dimensional one-dimensional The above above method used to for the heat under other boundary conditions, conditions, such as Neumann Neumann or mixed boundboundheat equation equation under other boundary such as or mixed ary conditions. The calculations are are left to the the exercises. exercises. ary conditions. The calculations left to 40 40That useful as a computational tool. We encounThat is, this particular particular Green's function is not useful We have encountered other Green's functions, particularly those in Section 4.6, are useful 4.6, that are useful computationally. 41 41The from Section 2.1 including the The reader should recall from 2.1 that each term in the heat equation, including right-hand side, is multiplied by A in the original derivation. 42 42The The interested reader can consult Haberman [22], Chapter 11.
282
Chapter 6. Heat flow and and diffusion Graph of G(x,t;75,60) for various values of t
0.02
Q)
- _. ._....... -
t=120 t=240 t=360 t=480 t=600
0.015
::; "§ Q)
c..
.ill
0.01
0.005
20
40
80
x
Figure 6.17. 6.17. Snapshots Snapshots of of the the Green's Green's function in Example 6.12 at at tt == Figure function in Example 6.12 120,240,360,480,600 120,240,360,480,600 seconds. seconds. Twenty Twenty terms terms of of the the Fourier Fourier series series were were used used to to create these these gmphs. graphs. create Exercises Exercises 1. Let Let G(x, G(x,t]yo,so) be the the Green's Green's function function from from Example Example 6.12 6.12 (yO (yo = 75, 75, So SQ == L t; Yo, so) be 60). Suppose Suppose one one wishes wishes to to produce produce an an accurate accurate graph graph of of G(x, G(x,t;yo, 60). t; Yo, SQ) so) for for various values of of tt > so. SQ. How How many many terms terms of of the the Fourier Fourier series series are are necessary necessary various values to the heat is to obtain obtain an an accurate accurate graph graph for for tt = 6060 6060 (10 (10 minutes minutes after after the heat energy energy is added)? For For tt == 61 61 (1 (1 second second afterward)? afterward)? (Hint: (Hint: Using Using trial trial and and error, error, keep added)? keep adding until the visibly.) adding terms terms to to the the Fourier Fourier series series until the graph graph no no longer longer changes changes visibly.) 2. 2. Compute Compute the the Green's Green's function function for for the the IBVP IBVP
au
pc at -
/'i,
02u ox 2 = f(x, t), 0
u(x, to)
< x < f, = 0, 0 < x < f,
au ax (0, t) = au
0, t
t
> to,
> to,
ox(f,t) = 0, t> to·
Produce values of p, c, yo, and Produce aa graph graph similar similar to to Figure Figure 6.17 6.17 (use (use the the values of p, c, /'i"K, yo, and So SQ from Exercise from Exercise 6.12). 6.12).
6.7. Suggestions for further reading
283
3. 3. Compute Compute the the Green's Green's function function for for the the IBVP IBVP
au - 1£ aax2 u =
pc at
2
f(x, t), 0< x < f, t > to,
°<
u(x, to) = 0, u(O, t) = 0, t
au
x
< f,
> to,
ax(f,t) = 0, t > to·
Produce aa graph graph similar similar to to Figure 6.17 (use (use the values of of p, c, 1£, K, Yo, yo, and and 80 SQ Produce Figure 6.17 the values p, c, from Exercise Exercise 6.12). 6.12). from 4. 4. Compute Compute the the Green's Green's function function for for the the IBVP IBVP
au
pc at
a2 u
- 1£ ax 2 = f(x, t), u(x, to) = 0,
au ax (0, t) = 0,
°<
°< x < x
f, t > to,
< f,
t > to,
u(f, t) = 0, t
> to.
Produce aa graph graph similar similar to to Figure Figure 6.17 6.17 (use (use the the values values of of p, c, 1£, «, Yo, yo ? and and 80 SQ Produce p, c, from Exercise Exercise 6.12). 6.12). from
6.7 6.7
Suggestions for for further further reading reading Suggestions
In the development, In this this section, section, we we have have continued continued the development, begun begun in in the the previous previous chapter, chapter, of both both Fourier Fourier series series and and finite finite element element methods. methods. The The references references noted noted in in Section Section of 5.8 are, are, for for the revelant for this chapter as as well. well. The The books books [10, [10, 4, 46, 22], 22], the most most part, part, revel ant for this chapter 4, 46, 5.8 along with with many many others, others, introduce introduce Fourier Fourier series series via via the the method of separation separation of along method of of variables perhaps unique variables briefly briefly described described in in Section Section 6.1.6. 6.1.6. Our Our approach approach is is perhaps unique among among introductory texts. introductory texts. To gain gain aa deep deep understanding understanding of of solution solution techniques, techniques, it it is is essential essential to To to know know the which is texts. A underthe qualitative qualitative theory theory of of PDEs, PDEs, which is developed developed in in several several texts. A basic basic understanding can can be be gained gained from from introductory introductory books books such such as as Strauss Strauss [46] [46] or or Haberman Haberman standing [22]. [22]. An An alternative alternative is is the the older older text text by by Weinberger Weinberger [52]. [52]. For For aa more more advanced advanced and and complete study, study, one one should should consult consult McOwen McOwen [39] [39] or or Folland Fblland [14] [14] or or Rauch Rauch [41]. [41]. complete
This page intentionally This page intentionally left left blank blank
Chapter 7
waves
We now now treat treat the the one-dimensional one-dimensional wave wave equation, equation, which which models models the the transverse We transverse vibrations of an elastic string or the longitudinal vibrations of a metal bar. We will concentrate on modeling a homogeneous medium, medium, in which case the wave equation concentrate takes the form form 2 82u 28 u 8t2 - C 8x 2 = f(x, t).
Our first order order of of business to understand Our first business will will be be to understand the the meaning meaning of of the the parameter parameter c. c. We We then Fourier series and finite finite element methods for for solving the wave wave equation. then derive derive Fourier series and element methods solving the equation. It is straightforward straightforward to extend the finite element methods to handle heterogeneous media coefficients in the PDE). media (nonconstant (nonconstant coefficients in the PDE).
7.1
homogeneous wave equation without The homogeneous boundaries
We of the equation by are no We begin begin our our study study of the wave wave equation by supposing supposing that that there there are no boundaries— boundariesthat equation holds oo < xx < oo. seem to be that the the wave wave equation holds for for — -00 00. Although Although this this may may not not seem to be very realistic realistic problem, problem, it it will will provide provide some some useful useful information information about about wave wave motion. motion. aa very We We therefore therefore consider consider the the IVP IVP
82 u
2 2 f) u 8t 2 - c f)x 2 = 0, -00 < x < 00, t > 0, u(x,O) = 'l/J(x), -00 < x < 00, 8u 8t (x,O) = ')'(x), -00 < x < 00.
(7.1)
(To simplify simplify the follows, we in this section that time (To the algebra algebra that that follows, we assume assume in this section that the the initial initial time With aa little cleverness, it it is is possible possible to to derive derive an an explicit is is to = = 0.) 0.) With little cleverness, explicit formula formula for for the the solution in the initial i^(x) and and ')'(x). ^(x). The in deriving solution in terms terms of of the initial conditions conditions 'l/J(x) The key key point point in deriving this formula formula is to notice operator this is to notice that that the the wave wave operator
82
2
f)t2 -
C
285
82 f)x2
286
Chapter 7. Waves
can be factored: factored:
For example, For example,
The mixed derivatives cancel cancel because, according to to aa theorem theorem of of calculus, calculus, if The mixed partial partial derivatives because, according if the second second derivatives derivatives of of aa function function u(x, t) are are continuous, continuous, then then the 82 u 8t8x
82 u 8x8t·
It follows follows that that any any solution solution u(x, w(x, t) of of either either It 8u+ c 8u=O 8t 8x
or or 8u_ c 8u=O 8t 8x will also also solve solve the the homogeneous wave equation. equation. If If u(x, u(x,t)t) == f(x will homogeneous wave f(x -— ct), where where /f :: R -t —t R R is is any any twice twice differentiable differentiate function, function, then then R 8u 8t (x, t)
8u
+ c 8x (x, t)
= -c!'(x - ct)
+ c!,(x -
ct) = O.
Similarly, if if g : R R -t —> R R is is twice twice differentiable, differentiate, then then u(x, u(x,t)t) = = g(x ct) satisfies satisfies Similarly, g(x + ct) 8u 8u 8t (x, t) - c 8x (x, t) = cg'(x + ct) - cg'(x + ct) = O.
Thus, by by linearity, linearity, every every function function of of the the form form Thus, u(x, t)
= f(x -
ct)
+ g(x + ct)
(7.2)
is is aa solution solution of of the the homogeneous homogeneous wave wave equation. equation. We We now now show show that that (7.2) (7.2) is is the the general solution. solution. general the homogeneous To To prove prove that that (7.2) (7.2) really really represents represents all all possible possible solutions solutions of of the homogeneous wave equation, it it suffices suffices to to show show that we can can always always solve solve (7.1) (7.1) with with aa function function of wave equation, that we of
7.1. The The homogeneous homogeneous wave wave equation equation without without boundaries 7.1. boundaries
287
the the form form (7.2) (7.2) (since (since any any solution solution of of the the wave wave equation equation satisfies satisfies (7.1) (7.1) for for some some 'lipjJ and and ')'). 7). Moreover, Moreover, by by linearity linearity (the (the principle principle of of superposition), superposition), it it suffices suffices to to solve solve
fPu at 2
-
2 fPu _ c ax 2 -
°
< x < 00, t > 0, 'ljJ(x), -00 < x < 00, , -00
u(x,O) = au at (x,O) = 0,
-00
(7.3)
< x < 00,
and 2
u at 2
a
-
2 2a U _ c ax 2 -
°
, -00
< x < 00, t > 0, < x < 00,
u(x,O) = 0, -00 au at (x,O) = ')'(x), -00 < x <
(7.4)
00,
separately separately and and add add the the solutions. solutions. To find solution to (7.3), we we assume assume that that u(x, u(x,t)t) = f(x f ( x -- ct) ct) + g(x g(x + ct). ct}. To find aa solution to (7.3), Then uu solves solves the the PDE. PDE. We We want want to choose f/ and and 9g so so that that u(x,O) u(x,Q) == 'ljJ(x) if)(x) and Then to choose and au/at(x,O) But du/dt(x,Q} == 0. 0. But u(x,O) = f(x) + g(x) and
au at (x,O) = c(g'(x) - !'(x)),
so so if if we we take take f(x)
1
= "2'ljJ(x),
g(x)
1
= "2'ljJ(x),
the is, the required required conditions conditions are are satisfied. satisfied. That That is, u(x, t)
= "21 ('ljJ(x -
ct) + 'ljJ(x + ct))
is (7.3). is the the solution solution of of (7.3). Finding aa solution solution to (7.4) is is aa bit bit harder. harder. With With u(x, u(x,t)t) = f(x-ct) f(x-ct}+g(x + ct), Finding to (7.4) + g(x+ct), we want u(x,O) two equations: we want u(x,0) = 00 and and au/at(x, du/dt(x,0]0) = ')'(x). 7(#)- This This yields yields two equations: f(x) + g(x) = 0, -c!'(x) + cg'(x) = ')'(x).
The first equation that f(x) The first equation implies implies that /(#) == -g(x); —g(x); substituting substituting this this into into the the second second equation yields equation yields 2cg'(x) = ')'(x).
A solution is A solution to to this this is 1 g(x) = 2c
10r
')'(s) ds.
288
Chapter 7. Waves
With g(x}, we we obtain obtain With /(re) f(x) = — -g(x),
u(x, t)
= --21
c
l
x ct - ')'(8)
d8
1 + -2
l
C
0
0
x ct
+ ')'(8) d8
l
1 = -2
x ct
+ ')'(8) d8.
x-ct
C
This is the the solution solution to (7.4). Adding we obtain obtain This is to (7.4). Adding the the two, two, we
u(x, t)
= "21 (¢(x -
ct)
1 + ¢(x + ct)) + -2
c
l
x ct
+ ')'(8) d8.
(7.5)
x-ct
This is d'Alemberl's solution This is d'Alembert's solution to to (7.1). We can significance of the constant also why the We can now now understand understand the the significance of the constant c, c, and and also why the PDE is is called first consider consider aa function function of of the form ff(x ( x -— ct). ct}. PDE called the the wave equation. equation. We We first the form Regarding u(x, t} t) = = ff(x ct) as fixed t, we see that each Regarding ( x -— ct} as aa function function of of x for for each each fixed t, we see that each u(x,t) is aa translate of u(x, w(x,0) = /(#). That is, "time snapshots" snapshots" of function u(x, t) is translate of 0) = f(x). That is, the the "time of the the function u(x, t) t) all all have same shape; as time time goes goes on, this shape shape moves have the the same shape; as on, this moves to to the the right right (since (since cc > 0). 0). We We call call u(x, u(x,t)t) == f(x ct} aa right-moving right-moving wave; wave] an an example example is is shown shown in in f(x -— ct) Figure 7.1. Similarly, Similarly, u(x,t} + ct) ct} is left-moving wave. u(x, t) == g(x + is aa left-moving wave. Moreover, Moreover, c is is just just Figure 7.1. the speed. It It is easy to d'Alembert's solution solution to to the the the wave wave speed. is therefore therefore easy to understand understand d'Alembert's wave equation—it is is the sum of of aa right-moving left-moving wave, both wave equation-it the sum right-moving wave wave and and aa left-moving wave, both moving at speed speed c. moving at c.
1.Sr--..------..------..------..------,
y=u(x,O)
y=u(x,S)
y=u(x, 10)
O.S
O~~----~--~--~~--~~~~~--~--~--~
-S
0
S
10
Figure 7.1. A right-moving wave u(x,t} u(x, t) == ff(x ct) (with c = = 1). Figure 7.1. ( x -— ct} I).
Example 7.1. 7.1. The solution to Example E)2u
8t 2
-
28
2u
c 8x 2
= 0,
-00
< x < 00, t > 0,
7.1.
289
The wave equation The homogeneous homogeneous wave equation without boundaries U (X,
0) = e
_,,2
au at (x, 0) = 0,
,-00 -00
< x < 00,
(7.6)
< x < 00
is is
Seveml snapshots snapshots of of this this solution solution are gmphed in in Figure Figure 7.2. Notice how how the the initial initial Several are graphed 7.2. Notice "blip" splits into two parts parts which which move move to the right right and "blip" splits into two to the and left. left.
(\
uLO) u(·,1 ) u(·,2) u(·,3)
0.9 0.8 0.7
0.6 0.5 0.4
0.3
0.2 0.1
-5
0
5
10
Figure 7.2. Snapshots of of the the solution solution to to (7.6) at at times = 0,1,2,3. 0,1,2,3. Figure 7.2. Snapshots times tt =
D' Alembert's solution to the the wave wave equation equation shows shows that that disturbances disturbances in in aa D'Alembert's solution to medium modeled modeled by by the the wave wave equation equation travel travel at at aa finite finite speed. We can can state state this medium speed. We this precisely as as follows. Given any any point point x in space and and any any positive positive time time t, the interval precisely follows. Given in space £, the interval [x [x -— ct, ct,xx + ct\ ct] consists consists of of those those points points from from which which aa signal, signal, traveling traveling at at aa speed speed of of c, c, would would reach reach xx in in time time t or or less. less.
Theorem 7.2. 7.2. Suppose Suppose u(x, solves the the IVP IVP (7.1), xx is any real real number, number, and and Theorem u(x,t)t) solves is any
t> for all t > o. 0. If // 1jJ(x) ifi(x) and and 'Y(x) 'j(x) are are both both zero zero for all xx satisfying satisfying x -— ct ::; < xx ::; <X x+ + ct, then then u(x,t]t) = = 0. u(x, O.
290 290
Chapter 7. 7. Waves Chapter Waves
Proof. We We have Proof. have -
u(1', t) =
21 ['IjJ(1' -
ct)
l
x d
+ 'IjJ(1' + ct)] + 21 _ +_ ')'(s) ds.
c
x-ct
If all three If 'IjJ(x) tfj(x) and and ')'(x) 7(x) are are both both zero zero for for all all x satisfying satisfying 1'x —d ct :::; < x :::; < 1'+ x + d, ct, then then all three terms in = 0.0. D0 terms in this this formula formula for for u(1', u(x,t)t) are are zero, zero, and and hence hence u(1', u(x,i)t) =
We call the the interval ct,xx + d] ct\ the the domain of dependence dependence of of the the We therefore therefore call interval [x -— d, domain of space-time illustrate the the domain domain of dependence in in Figure space-time point point (x, t). We We illustrate of dependence Figure 7.3. 7.3.
1.5,----------.,
1.5,---------~
1
0.5
0.5
--=-.....J 1500
OL.......:_-... 500
1000
O~------~~~--~
500
1000
1500
X
Figure 7.3. Left: The domain domain of dependence of of the the point (x, t) == (1000,1) (1000,1) Figure 7.3. Left: The of dependence point (x, with dependence is shaded on x-axis. Right: Right: with cc == 522. 522. The The domain domain of of dependence is the the interval interval shaded on the the x-axis. The of influence of the point (xo,O) The domain domain of influence of the point (#o,0) == (1000,0). (1000,0). We can can look at this this result from another another point of view. The initial initial data at aa We look at result from point of view. The data at point xo, and ')'(xo), can only the solution enough, point #o, 'IjJ(xo) VKxo) and 7(^0), can only affect affect the solution u(x, u(x,t)t) if if t is is large large enough, specifically, if specifically, if t
~ \x ~ Xo \.
The set The set {(x, t) : ct ~ Ix - xol}
is therefore therefore called the point point (xo,O). is called the the domain domain of of influence influence of of the (xo,0). An An example example of of aa domain influence is in Figure Figure 7.3. domain of of influence is given given in 7.3. From the the preceding discussion, we we see see that if aa disturbance disturbance is is initially initially confined confined From preceding discussion, that if to the bounded region, region, the boundary cannot the solution to the interior interior of of aa bounded the boundary cannot affect affect the solution until until
7.2. Fourier series methods for the wave equation
291 291
sufficient time has has passed. passed. For this reason, reason, the the solution to the the wave wave equation in an sufficient time For this solution to equation in an infinite medium provides provides aa realistic realistic idea of what what happens happens (in (in certain certain experiments) infinite medium idea of experiments) bounded region, region, at for aa finite period of time (namely, (namely, until until enough enough in in aa bounded at least least for finite period of time time has elapsed for the disturbance to reach the boundary). However, there no time has elapsed for the disturbance to reach the boundary). However, there is is no simple for the the solution solution of the wave equation subject to boundary boundary conditions. conditions. simple formula formula for of the wave equation subject to To attack must use Fourier series, finite elements, attack that that problem, problem, we we must use Fourier series, finite elements, or or other other less less To elementary elementary methods. methods.
Exercises 1. (7.1) with with 1. Solve Solve (7.1)
0, x < -2 or x > 2, 3, 5· 1Q-4 + 10-2 :s; x :s; 0, x ,,(x) = { -5· 1Q-4 x + 10- 3 , 0 < X :s; 2, 43 ,(x) == 00 and and c = 522.43 Graph several of the the solutions 7(0;) — 522. Graph several snapshots snapshots of solutions on on aa common common plot. plot.
2. (7.1) with with 'IjJ(x) 0, 2. Solve Solve (7.1) ip(x) = 0, 0,
,(x) = { -1,
x < -0.5 or x > 0.5, Ixl :s; 0.5,
and 522. Graph Graph several the solutions solutions on on aa common and cc = = 522. several snapshots snapshots of of the common plot. plot. 3. the IVP IVP (7.1) ,(x) = = 0 0 and 3. Consider Consider the (7.1) with with j(x) and
The initial displacement consists of two "blips," which, according according to to d' Alembert 's The initial displacement consists of two "blips," which, d'Alembert's right-moving and and aa left-moving wave. What formula, will will each each split formula, split into into aa right-moving left-moving wave. What happens when the the right-moving right-moving wave wave from from the the first first blip blip meets meets the the left-moving left-moving happens when wave from the second blip? wave from the second blip? the previous previous exercise exercise with with 4. 4. Repeat Repeat the
7.2
Fourier series methods for the wave equation
Fourier series series methods methods for the wave wave equation equation are just as for the the heat heat Fourier for the are developed developed just as for equation-the key is is to to represent the solution in aa Fourier series in in which which the Fourier equation—the key represent the solution in Fourier series the Fourier We begin begin with with the the IBVP IBVP coefficients coefficients are are functions functions of of t. We
aat2 u 2
43
2
a2 u
c ax 2 = f(x, t), 0
< x < £, t > to,
43The value 522 is physically physically meaningful, we will will see the next next section. The value 522 is meaningful, as as we see in in the section.
292 292
Chapter 7. Waves Chapter 7. Waves
u(x, to) = 1/J(x) , au at (x, to) = 1'(x),
°< x < e,
°<
x
< e,
(7.7)
u(O, t) = 0, t > to, u(e,t) = 0, t> to, which small displacements displacements of of aa vibrating string. We We write write the solution which models models the the small vibrating string. the solution in form in the the form . (n7fx) u(x, t) = ~ ~ an(t) sm -e- , n=l
where where an(t) =
~
1i
u(x, t) sin (n;x) dx, n = 1,2,3, ....
f(x, t) in We also also expand expand the right-hand side side /(#,£) in aa sine sine series: series 00
f(x,t) = LCn(t) sin (n;x), n=l where where
Cn(t) =
~
1i
f(x, t) sin (n;x) dx, n = 1,2,3, ....
Using the the same same reasoning as in in Section Section 6.1, 6.1, we we can can express express the left-hand side side of of the the Using reasoning as the left-hand PDE PDE in in aa sine sine series: series: 2 2 2 2 ) a u a u 00 (d2a c n 7f2 (n7fx) at 2 (x, t) ax 2 (x, t) = ~ dt 2n (t) + ~an(t) sin -e- .
c
Setting this equal to the the series for /(#,£), f(x, t), we we obtain the ODEs Setting this equal to series for obtain the ODEs ~an 2n 27f2 dt 2 + -e-2-an = cn(t), n = 1,2,3, ....
Initial conditions conditions for for these these ODEs ODEs are are obtained obtained from from the the initial initial conditions conditions for for the the Initial If wave equation wave eauation in in (7.7). (7.7). If . (n7fx) ~. (n7fx) 1/J(x) = ~ ~ bn sm -e- , 1'(x) = ~ dn sm -e- , n=l n=l
then we we have have then
dan an(to ) = bn, dt (to)
= dn.
We Fourier sine sine coefficients coefficients of of the the solution solution u(x, u(x, t) by solving the the IVPs IVPs We thus thus find find the the Fourier by solving ~an
dt 2
c2 n 2 7f2
+ -e-2-an = an(to)
cn(t),
= bn ,
dan (to ) =d , --;It n
(7.8)
7.2. Fourier series methods for the wave equation
293
for n n= .... In In Section 4.2.2, we we presented presented an an explicit explicit formula formula for for the the solution for = 1,2,3, 1,2,3, — Section 4.2.2, solution The use of this formula is illustrated in Example 7.4 below. of (7.8). of (7.8). The use of this formula is illustrated in Example 7.4 below.
7.2.1 7.2.1
Fourier series solutions solutions of the homogeneous homogeneous wave wave equation equation Fourier series of the
When the the wave wave equation equation is is homogeneous homogeneous (that (that is, is, when when f/ = == 0), the the Fourier Fourier series When series solution is is especially especially instructive. We then must solve
tPa n
r?n 2 7f2
dt 2 +~an=O, an(to)
= bn ,
dan ( ) dt to
= dn,
which yields
Therefore, the the solution to Therefore, solution to
a2 u at 2
a
-
2 u 2 c ax 2 = 0, 0 < x
< i, t > to, 'IjJ(x) , 0 < x < i,
u(x, to) = au at (x, to) = 'Y(x), 0 u(O, t) = 0, t
< x < i,
(7.9)
> to,
u(i, t) = 0, t > to,
is
~{ bncos (m7f( u(x,t ) = ~ -i- t-to) )
dni. sm (m7f + m7f --yet-to) )}. sm (n7fx) -f- .
This formula shows the the essentially essentially oscillatory oscillatory nature nature of of solutions to the the wave wave equaThis formula shows solutions to equation; indeed, indeed, the the Fourier Fourier sine sine coefficients coefficients of of the the solution are periodic: tion; solution are periodic: an (t
+ ~!)
= an(t) for all t.
44 frequencies cn! the natural frequencies frequencies of the string. string.44 freThe frequencies en/ (2l) (21} are called the of the These frequencies are are called natural because because any any solution of the the homogeneous homogeneous wave wave equation quencies called natural solution of equation combination of the the normal modes of the the string, is an (infinite) combination
cn7f (t - to) ) _ _ _ _--'A_n cos ( -f.(
. (cn7f . (n7fx) + Bn sm -f.- (t - to) )) sm -f.- .
44 44The units of of frequency frequency are literally l/second, which is is interpreted interpreted as as cycles cycles per per second, second, or or The units are literally I/second, which Hertz. Sometimes Sometimes the the natural natural frequencies are given given as as Hertz. frequencies are
cn cn7l' 271'-=-
2£
£'
in which which case case the the units units can can be be thought thought of of as in as radians/seconds. radians/seconds.
294 294
Chapter 7. Waves
Each normal normal mode is aa standing wave with cn/(2£) Each mode is with temporal temporal frequency frequency en/ (21) cycles cycles per per second. Several examples are displayed in Figure 7.4.
0.5
0.5 O~---____::.:----__:JI
:::J
-0.5
-0.5
-1L-----~~~----~
a
-1~--~~------~&---~
a
0.5
x
0.5
x
1
0.5
-0.5 -1~~----~--~----~
-1~--~----~------~~
a
a
0.5
x
0.5
1
x
Figure = If = to == 0). Figure 7.4. 7.4. The first four normal modes (c = = 1, 1, to 0). Displayed Displayed are multiple time snapshots of wave. of each standing wave.
In In particular, particular, since since 2t/c 2£/c is is an an integer integer multiple multiple of of 2^/(cn), 2£/(cn), we we have have
an
(t + 2cf )
= an(t) for all
t,
for all n = 1,2,3, ....
It follows follows that that the the solution u(x, t) satisfies solution u(x,t) satisfies f
u (x, t + 2c ) = u(x, t) for all x
E
(0, f), for all t > 0;
that is, is, the the solution solution is periodic periodic with period 2f/c. 2£/c. The The frequency frequency c/(2£) c/(2l) is is called called the the fundamental of the string—it is string fundamental frequency frequency of the string-it is the the frequency frequency at at which which the the plucked plucked string vibrates. An question is string that is the the following: following: What What happens happens to to aa string that is is An interesting interesting question subjected to an oscillatory transverse pressure with a frequency equal to one of the
7.2. 7.2.
Fourier series series methods methods for Fourier for the the wave wave equation equation
295
natural frequencies of of the string? The answer is that resonance resonance occurs. occurs. This This is natural frequencies the string? The answer is that is illustrated A. illustrated in Exercise Exercise 8 and also in Section 77.4. We now give an example example of using the Fourier series method to solve an IBVP IBVP for for the wave equation. Example 7.3. = 1, c = = 0, and Example 7.3. Consider Consider (7.7) with with £1=1, = 522, 522, ff(x, ( x , t t)) = 0, 'Y(x) 7(0;) = = 0, 0, and 004 < x < 0.5, 0.5 < x < 0.6, 0 < x < 004 or 0.6
004, -(x - 0.6), 0, X -
'IjJ(x) = {
(7.10)
< x < 1.
The fundamental fundamental frequency frequency of of the string is is then C 522 - = =261 Hz 2£ 2 ' which is is the frequency of of the the musical musical note note called conditions which the frequency called "middle "middle C." C." The The initial initial conditions indicate string is is ''plucked'' center. Using indicate that that the the string "plucked" at at the the center. Using the the notation notation introduced introduced above, we have cn(t) cn(t] = = 00 and and ddn = = 00 for all n, n, and we have for all and above,
bn = 2 101 'IjJ(x) sin (mfx) dx
= -2 -2 sin(I/2mf)
+ sin(2/5mf) + sin(3/5mf). n 2 7f2
We solve We must must solve
~an dt 2
+c
2 2
n
2 7f
an = 0,
an(O) = bn ,
d:
tn (0) = O.
The solution is easily seen seen to be The solution is easily to be an(t) = bn cos(cn7ft), n = 1,2,3, ... , and and therefore therefore 00
u(x,t) = L:)n cos (cn7ft) sin (n7fx). n=l
graphed in should Several snapshots of of the the solution solution are are graphed in Figure 7.5. 7.5. The The reader reader should observe that when a wave wave hits a boundary where there is boundary where is a Dirichlet condition, condition, the wave inverts for example, 3T /8) == -u(x, T /8) the wave inverts and and reflects. reflects. Therefore, Therefore, for example, u(x, w(x,3T/8) —u(x,T/8] and T /2) = period corresponding corresponding to and u(x, u(x,T/2) = -u(x,O), —u(x,ty, where where T T is is the the period to the the fundamental fundamental frequency: T = — 2/522 2/522 = = 1/261. 1/261. Also, u(x,T) = w(x,0), as expected. expected. frequency: T Also, u(x, T) = u(x, 0), as
The reader should recall, from Section 2.3, that the wave speed c is determined determined by the the tension and the density p of the string: cc22 = = T T/p. fundamental by tension T and the density p of the string: / p. The The fundamental frequency the string is frequency of of the string is
296
Chapter 7. Chapter 7. Waves
0.2r------.-----r-------.---;:=======il y=u(x,O) - -' y=u(x,T/8) y=u(x,3T/8) y=u(x,T/2) o =U x,T
, ,,
, ,. , ,
, , ... , , , , ,
,,
-0.2'-----.......---'-----'-----.&...-----' o 0.2 0.4 0.6 0.8 x
Figure The solution u(x,t) to to {7.7} (7.7) at at times times 0, 0, T18, T/8, 3T/8, T/2, and Figure 7.5. 7.5. The solution u(x,t) 3T18, T12, and T. For this example, example, we took I£ == 1, 1, cc = 522, /(#,£) = 0, 0, "((x) 7(2) = = 0, 0, and the initial initial T. For this we took = 522, f(x,t) = and the condition'IjJ given in of the Fourier series series were were used. used. condition ijj given in (7.10). (7.10). One One hundred hundred terms terms of the Fourier which shows that tension and and density of the the string string stay stay the the same, same, but but the the which shows that if if the the tension density of string is shortened, then explains how string is shortened, then the the fundamental fundamental frequency frequency is is increased. increased. This This explains how aa guitar guitar works-the works—the strings strings are are shortened shortened by by the the pressing pressing of of the the strings strings against against the the "frets," and thus different notes are sounded. "frets," and thus different notes are sounded.
7.2.2 7.2.2
Fourier series solutions of the inhomogeneous wave equation
We now show two two examples examples of of the wave equation. equation. We now show the inhomogeneous inhomogeneous wave Example Consider metal string, string, such as aa guitar guitar string, string, that can be be Example 7.4. 7.4. Consider aa metal such as that can attracted magnet. Suppose Suppose the string in question is 25 cm cm in length, with with attracted by by aa magnet. the string in question is 25 in length, cc = 522 522 in in the the wave wave equation, equation, its its ends ends are are fixed, fixed, and and it it is is "plucked" "plucked" so so that that its its initial displacement displacement is initial is
'I/J(x)
=
25 ~ (1 ~ Ix _ 1) 10 25 2
is zero). zero). Suppose Suppose further further that and released velocity is and released (so (so that that its its initial initial velocity that aa magnet magnet of 100 dynes. We wish to find the motion exerts a constant upward force exerts a constant upward force of 100 dynes. We wish to find the motion of of the the string.
7.2. 7.2.
297 297
Fourier Fourier series methods for the the wave equation
We must must solve the IBVP IBVP
cPu
2
a2 u
°< x < 1jJ(x), °< x < °< x <
at 2 - c ax 2 = 100,
25, t
u(x, 0) = au at (x, 0) = 0,
> 0,
25,
25,
u(O, t) = 0, t > 0, u(25, t) = 0, t > 0, 522 and ^ where cc — = 522 1jJ is given above. We write the solution as 00
u(x, t)
=L
an(t) sin (n;x)
n=l determine the coefficients coefficients ai al(t),a2(t),a3(t), ... by solving the IVPs IVPs {7.8}. and determine (£), 02 (£), 03 (£),... (7.8). We have 00
100
=L
cnsin (n;x),
n=l with 2
Cn
[25
= 25 10
. (n7fx)
100 sm
25
dx
=
200(1- (_l)n) n7f ' n
= 1,2,3, ... ,
and
with
bn
=~ [25 ",.( ) . (n7fx) d = 4 sin (n7f/2) 25 10 'f/ X sm 25 x 5n 7f2 , 2
n = 1,2,3, ....
an(t], therefore, must solve To find an(t), therefore, we must solve d2 a n
dt 2
c2 n 2 7f2
+ ~an
=
Cn,
an(O) = bn , n
d;t (0) = 0. Using the results of Section 4-2.2, Using 4.2.2, we see that the solution is
298
Chapter Chapter 7. Waves
The solution is periodic with period The solution is periodic with period 2n cn /25
=
50 c
Twenty-five snapshots of the the solution solution (taken (taken during during one one period) are shown shown in in Figure Twenty-five snapshots oj period) are Figure 7.6. The The effect effect oj of the the external external Jorce force on on the the motion motion oj of the the string string is is clearly clearly seen. seen. (The (The 7.6. reader reader should should solve solve Exercise Exercise 11 iJ if it it is is not not clear clear how how the the string string would would move move in in the the absence oj of the the external external Jorce.) force.) absence 0 . 1~------r-------~--~--~------~------~
0 .08
"E Q) E Q) ~
a. !Jl '6
10
5
20
15
25
x
Figure string from from Example Example 7.4. Figure 7.6. 7.6. Twenty-five Twenty-five snapshots snapshots oj of the the vibrating vibrating string 7.4We which one We now now consider consider an an example example in in which one of of the the boundary boundary conditions conditions is is ininhomogeneous. As usual, data. homogeneous. As usual, we we use use the the method method of of shifting shifting the the data. Example Example 7.5. 7.5. Consider a string oj of length 100 100 cm that has one end fixed at x == 0, 0, with the the other other end end free to move move in in the the vertical vertical direction direction only only (tied (tied to to aa frictionless frictionless with free to 45 S.45 Suppose Suppose the pole, for example). Let the wave speed speed in the string be be c == 2000 2000 cm/ cm/5. free end is is "flicked," "flicked," that that is, is, moved moved up up and and down down rapidly, rapidly, according according to to the the formula formula free end u(100,t) = /(*), t) = J(t), where where u(100, f(t) = { co-it -
3c 2 t 2 + 3c 3 t 3 - c
4 4
t
,
---,-----45
45The then The fundamental fundamental frequency frequency is is then
_c_ = 10Hz.
2 ·100 By comparison, comparison, the lowest tone of aa piano 27.5 Hz. By the lowest tone of piano has has frequency frequency 27.5 Hz.
0 < t < c, t > c,
7.2. Fourier Fourier series methods for the wave equation 7.2. series methods for the wave equation
where
E
299
= 0.01 (see Figure 7.7). 0.12..-----....-----...-----,..----~--.....,..--___,
0.1
c
0.08
Q)
E Q)
al 0.06 a. ({)
15
0.04
0.02
\ 0.02
0.04
0.06
0.08
0.1
0.12
t
free end Figure 7.7. 7.7. The The "flick" "flick" of the the free end in Example 7.5.
To determine determine the the string, must solve the IBVP IBVP To the motion motion of of the string, we we must solve the
cPu c2 2{PU =o !!~ ^ =_ °' o<xo> ' ot2
-
ox 2
C
-
,0 < x < 100, t
0,
u(ar,0) = 00,, 0< Q<xx < 100, u(x,O) < 100, du ou — ( x ,0) 0 )== 00,, 00< < xx<<1 100, 00, ot (x,
(7.11) (7.11)
w(0,t) u(O, t) = 0, 0, tt > > 00,, u(lOO, t) = f(t), tt>Q. > 0. u(100,*) = /(*), Since the boundary at the the right end of of the the string Since the boundary condition condition at right end string is is inhomogeneous, inhomogeneous, we we will will shift the data to to obtain problem with with homogeneous homogeneous boundary conditions. Define shift the data obtain aa problem boundary conditions. Define v(x,t) = u(x,t) ( x , tt), ) , where v(x, t) = u(x, t) -— pp(x, where p(x, t) =
xf(t) 100'
Then satisfies the VP Then vv satisfies the IB IBVP 02v
202v _
ot 2 -c ox2 v(x,O)
OV
-
X d2f
-100 dt 2 (t), 0< x < 100, t > 0,
= 0,
0< x < 100, x
ot (x,D) = -IOOE'
°<
X
< 100,
Chapter 7. 7. Waves Chapter Waves
300
v(O, t) v(100, t)
= 0, = 0,
t > 0, t > O.
In computing the initial initial conditions for v, we used
df 1 f(O) = 0, -d (0) = -.
t
E
The new right-hand side of PDE is of the PDE is g(x, t)
= -1~0 ~{ (t) = { ~ l~O
2 (-6E-
3
+ 18c t -
4
12c t2) ,
~;:.:::; E,
Since g is defined piecewise, so so are its Fourier coefficients: coefficients: is defined 00
g(x, t) = I:>n(t) sin (mrOx) , 10
n=l
(lOO
2
cn(t) = 10010 = {
(n1fx) g(x, t) sin 100 dx
2(~;t ~(t), 0:::; t :::; E, t > E.
0,
The initial is given by initial velocity for v is x
-100E =
~
. (n1fx)
~dnsm 100 '
n=l
2
{lOO (
dn = 100 10
X)
- 100E sin
(n1fx) 100
dx
2(-1)n
IfIf we write 00
v(x, t)
= ~ an(t) sin (~~~),
coefficient an(t) an (t] is IVP then the coefficient is determined determined bv by the IVP ~an
dt 2
+
c2n 21f2
= cn(t), an(O) = 0, ~tn(O) = dn·
1002 an
Again using the results of Section 4-2.2, 4.2.2, we see that Again of Section 100 ( . (cn1ft) an(t) = cn1f dn sm 100
) ) + 10{t.sm (cn1f 100 (t - S) cn(s) ds .
301 301
7.2. 7.2. Fourier series methods for the wave equation
We note note that Cn(t)
,pf = Edn dt 2 (t),
simplify the formulas formulas slightly. We will write which allows us to simplify
Then we have an(t) = d; (Sin (Ot) For t
E
lot sin (O(t -
s)) (-6E- 2 + 18C 3 s - 12C 4 s 2) dS) .
lot sin (O(t -
s)) (-6E- 2 + 18E- 3 S
+E
[0, EJ,
an(t)
= dOn
(Sin (Ot)
+E
= dn (. (()) _ 6
(}smt
-1
(1 - cos (Ot) _ 3
E
()
~
E
Ot - sin ((}t)
E
2 _2(}2t2_2+2COS((}t)))
+
-1
12E-4 S2) dS)
-
(}2
,
while for t > e, E, an(t) = d(}n (sin (Ot) n -_ -d
o
3E
2E
+ E10' sin (O(t - S))
-2
+ 18E- 3 S -
12E- 4S2) dS)
sm (ll) ut - 6E -1 (COS ((}(t - E)) - COS (Ot)
(.
-1
(-6E- 2
sin ((}(t -
()
E))
+ E(} COS ((}(t - S)) ~
2
sin ((}t)
E 02 COS (O(t - E)) - 2 cos (O(t - E)) (}3
+
+ 2EO sin (O(t -
E))
+ 2 cos (Ot)))
•
Several snapshots snapshots of the solution solution are shown in Figure 7.8. 7.8. For each t, the Fourier coefficients of u(x, t) decay to zero relatively slowly in n, so a good approximation coefficients of u(x,t] approximation requires many Fourier series. To produce Figure 7.8, many terms in the Fourier 7.8, we used 200 terms. terms. The reader should observe the behavior of the wave motion: the wave proend travels from right to left, from left duced by the flick of of the end left, reflects, reflects, travels from left to 46 At each reflection, the wave inverts. 46 right, reflects reflects again, and so on. At
7.2.3 7.2.33
Other boundary conditions
Using Fourier series to solve the the wave with aa different different set set of of boundary Using Fourier series to solve wave equation equation with boundary conditions is is not provided the the eigenvalues and eigenfunctions conditions not difficult, difficult, provided eigenvalues and eigenfunctions are are known. known. We now now do do an with mixed boundary conditions. conditions. We an example example with mixed boundary 46
46The reader can this experiment with aa long rope. The reader can approximate approximate this experiment with long rope.
302
Chapter 7. Waves Waves 0.1 5
O. 1 ,,',\
,,
cSb o o0
\
0
\
,
\
0.05 '
\
,
c
0
\ \
Q)
\
E Q)
,
u
ttl
,
\
, , ,
\
-0.0 5
-0. 1
\..
,
"
C. C/) 'C
\ \
, , , ; \
-- _. t=0.025 t=0.05 ,-,- t=0.06
,
',J
0
-0.1 5
o
0 0 0 0 0
t=0.1 t=0.11
40
20
x
60
80
100
Figure from Example Example 7.5. Figure 7.8. 7.8. Snapshots Snapshots of of the the string string from 7.5.
Example steel bar Example 7.6. 7.6. Consider Consider aa steel bar of of length length 11 m, m, fixed fixed at at the the top top {x (x =— O} 0) with with the bottom bottom end free to to move. Suppose the the bar bar is is stretched, by means the end {x (x == I} I) free move. Suppose stretched, by means of of aa downward pressure applied free end, released. downward pressure applied to to the the free end, to to aa length length of of 1.002 1.002 m, m, and and then then released. We We wish wish to to describe describe the the vibrations vibrations of of the the bar, bar, given given that that the the type type of of steel steel has has stiffness stiffness 3 cm3 •. kk = = 195 195 GPa GPa and and density density pp == 7.9 7.9 g/ g/cm The The longitudinal longitudinal displacement displacement u(x, u(x,t)t) of of the the bar bar satisfies satisfies the the IBVP IBVP fpu p at 2
-
k
a2 u ax 2 =
u(x,O) =
< x < 1, t > 0, 0.002x, 0 < x < 1,
0, 0
au at (x,O) =0, u(O, t)
O<x
(7.12)
= 0, t > 0,
au ax (1, t) =
0, t
>0
(see Section Section 2.2}. 2.2). The The eigenpairs eigenpairs of of the the negative negative second second derivative derivative operator, operator, under under {see these are these mixed mixed boundary boundary conditions, conditions, are
An
= (2n -4(21)21T2 '
tPn(x)
. ((2n - 1)1TX) = sm 2f ' n = 1,2,3, ...
7.2. 7.2.
303 303
Fourier series methods for the wave equation
(see = 1. We therefore represent the solution solution of (7.12) as (see Section Section 5.2.3), 5.2.3), with f1=1. We therefore of (7.12)
~ an(t) sin ((2n ~ l)7rX).
u(x, t) = The left of the PDE PDE is left side of
2 a u
2 a u
2 ~ {J2a n k(2n - 1)27r } . ((2n - l)7rX) P dt 2 (t) + 4 an(t) sm 2 ;
p at 2 (x, t) - k ax2 (x, t) = ~
since the right side is is zero, we obtain the the ODEs ODEs
J2a n
+
P dt 2
k(2n - 1)27r2 4
an = 0, n = 1,2,3, ....
The initial displacement is given by The initial
0.002x
= ~ bn sin ((2n ~ l)7rX),
with with bn = 2
1 1
o
.
0.002x sm
((2n-1)7rX) 2
dx
= 1252(-1)n+l (2n - 1)2 7r 2' n = 1,2,3, ....
This, together with the fact fact that the initial is zero, yields initial initial velocity of the bar is conditions conditions for the the ODEs: ODEs:
d2a n
P dt 2
+
k(2n - 1)27r 2 an = 0,
4
an(O) = bn ,
~tn (0) =
O.
The solution The solution is is
an(t) (with c =
Jk/P),
= bncos ( C(2n -2
l)7rt)
' n
= 1,2,3, ...
so
~ bnCOS (C(2n -l)7rt) sm . ((2n -l)7rX) . u ( x,t ) = L....n=1 2 2 We speed in We see that the the wave wave speed in the the bar bar is is 1.95.10 11 7.9.103
.
= 4970 mj s
(we consistent units before computing (we convert k and and p to to consistent units before computing c), c), and the fundamental fundamental frequency is frequency is
~ ==
49;0
==
1240 Hz.
of the Several snapshots of of the the displacement of the bar are shown in in Figure 7.9. 7.9.
304
Chapter 7. Waves
"E Q) E
~1
0.. Ul '5 ..........................................................................
t=O - _. t=5e-05 ._.- t=0.0001 t=0.00015
0.2
0.4
0.6
0.8
1
x
Figure Snapshots of displacement of the bar bar in in Example Example 7.6. 7.6. Figure 7.9. 7.9. Snapshots of the the displacement of the
Exercises 1. Find the the motion Example 7.4 external force 1. Find motion of of the the string string in in Example 7.4 in in the the case case that that no no external force and produce produce aa graph graph analogous to Figure Figure 7.6. Compare. Compare. is applied, applied, and is analogous to
2. Consider string of of length length 50 50 cm, cm, with with cc = 400 400 cm/s. cm/s. Suppose Suppose that the string string 2. Consider aa string that the is rest, with with both both ends fixed, and that it is initially initially at at rest, ends fixed, and that it is is struck struck by by aa hammer hammer at at its center so to induce induce an an initial of its center so as as to initial velocity velocity of "(x)
={
-20, 0,
24 < x < 26, 0::; x ::; 24 or 26 ::; x ::; 50.
(a) long does it take take for the resulting wave to to reach the ends ends of of the (a) How How long does it for the resulting wave reach the the string? string? (b) and solve the IBVP IBVP describing the motion the string. (b) Formulate Formulate and solve the describing the motion of of the string. (c) Plot several the string, the wave wave first (c) Plot several snapshots snapshots of of the string, showing showing the first reaching reaching the the ends Verify your your answer 2a. ends of of the the string. string. Verify answer to to 2a. 3. Find the the motion motion of in Example 7.4 if the only change in the experexper3. Find of the the string string in Example 7.4 if the only change in the imental that the the right right end is free free to vertically. imental conditions conditions is is that end (x == 25) 25) is to move move vertically. (Imagine the right is held at uu = 00 and and then along with (Imagine that that the right end end is held at then released, released, along with the the rest of of the the string, at tt = = 0.) 0.) rest string, at 4. Repeat Repeat Exercise 2, assuming assuming that right end end of of the the string is free to move move ver4. Exercise 2, that right string is free to vertically. tically.
7.3. Finite element methods for the wave wave equation equation
305
5. Find the motion ofthe of the string in Example 7.4 in the case that both ends of the vertically. (All remain string string are are free free to to move move vertically. (All other other experimental experimental conditions conditions remain the same.) the same.) 6. is 0.25 6. Consider Consider aa string string whose whose (linear) (linear) density density (in (in the the unstretched unstretched state) state) is 0.25 g/cm g/cm and whose longitudinal N (meaning that aa force N is is and whose longitudinal stiffness stiffness is is 6500 6500 N (meaning that force of of 13p 13pN required to stretch the string by p%). If the unstretched (zero tension) length (zero tension) of the string is is 40 40 cm, cm, to must the string be stretched in in order order to what what length length must the string be stretched of the string that its fundamental fundamental frequency frequency be be 261 261 Hz Hz (middle (middle C)? C)? 7. bar in Example 7.6 if the the force 7. How How does does the the motion motion of of the the bar in Example 7.6 change change if force due due to to gravity is taken into account? gravity is taken into account?
8. Consider Consider aa string string that that has one end end fixed, the other other end end free free to to move has one fixed, with with the move 8. in the the vertical vertical direction. direction. Suppose Suppose the string is is put into motion motion by virtue of put into by virtue of in the string the free end's being manually moved up and down periodically. An IBVP down periodically. describing this motion is
a2 u at2
2 2 au c ax 2 = 0, 0 < x < i, t > to,
-
u(x, to) = 0, 0 < x < i,
au
at (x, to) = 0, 0 < x < i, u(O, t) = 0, t > to, u(i, t) = E sin (27TWt) , t > to.
Take = 1, c = 4. Take it=l,c = 522, 522, tto0 == 0, 0, and and Ee == Ie le -- 4. (a) (a) Solve Solve the IBVP IBVP with w u not not equal equal to aa natural frequency of of the string. string. Graph Graph the the motion of of the the string string with w uj equal equal to to half the fundamental frequency. frequency. (b) Solve the IBVP with w uj equal to a natural frequency of the string. Show that the motion motion of with u) w equal that resonance resonance occurs. occurs. Graph Graph the of the the string string with equal to to the fundamental the fundamental frequency. frequency.
7.3
Finite element methods for the wave wave equation equation
The wave equation presents presents special special difficulties difficulties for for the design of of numerical The wave equation the design numerical methods. methods. These These difficulties difficulties are are largely largely caused caused by the the fact fact that waves waves with with abrupt abrupt changes jump discontinuities) with (even (even singularities, singularities, such such as as jump discontinuities) will will propagate propagate in in accordance accordance with the wave equation, the waves. waves. (Of has aa the wave equation, with with no no smoothing smoothing of of the (Of course, course, if if u(x, u(x,t]t) has that is, is, it it fails be twice then it solution singularity, singularity, that fails to to be twice differentiable, differentiable, then it cannot cannot be be aa solution of the the wave wave equation equation unless unless we expand our our notion notion of of what constitutes aa solution. solution. we expand what constitutes of But ways to to do But there there are are mathematically mathematically consistent consistent ways do this, this, and and they they are are important important for modeling modeling physical phenomena phenomena exhibiting exhibiting discontinuous discontinuous behavior.) behavior.)
Chapter Chapter 7. 7. Waves
306
For example, example, the the function 1] -t defined by For function if)'IjJ : [0, [0,1] -> R R defined by
Ix - ~I < 0.05, Ix - ~I ~ 0.05
x _ {I, 'IjJ( ) 0,
is discontinuous. discontinuous. We We solve solve both both the the heat heat equation, is equation,
au a2 u at - ax2 =
0, 0 < x < 1, t
> 0,
u(x,O) = 'IjJ(x) , 0< x < 1, u(O,t) = 0, t > 0, u(1, t) = 0, t > 0,
(7.13)
and the wave wave equation, and the equation,
a2 u a2 u at2 - ax 2 =
°< x < t > u(x,O) = 'IjJ(x), 0< x < 1, au at (x, °< x < 0,
1,
0) = 0,
1,
0,
(7.14)
u(O, t) = 0, t > 0, u(l, t) = 0, t > 0,
with if)(x} as the initial second initial initial condition the wave with 'IjJ(x) as the initial condition condition (we (we need need aa second condition in in the wave equation; we take the initial velocity equal to zero). The results are shown in Figures equation; we take the initial velocity equal to zero). The results are shown in Figures and 7.11. 7.11. The The graphs graphs show that the the discontinuity is immediately immediately "smoothed "smoothed 7.10 7.10 and show that discontinuity is away" by by the the heat heat equation, equation, but but preserved preserved by by the the wave wave equation. away" equation. The standard standard method method for for applying applying finite finite elements to the the wave wave equation, equation, which which The elements to we now now present, present, works works well well if the true true solution is smooth. Methods that that are we if the solution is smooth. Methods are effective effective when the the solution solution has has singularities singularities are beyond the the scope of this this book. when are beyond scope of book.
7.3.1 7.3.1
The wave equation with Dirichlet conditions
We as we suppose that We proceed proceed as we did did for for the the heat heat equation. equation. We We suppose that u(x, i) t) solves solves
a2 u at 2
2
-
a2 u
c ax 2 = f(x, t),
u(x, to)
= 'IjJ(x),
au at (x, to) = l'(x), u(O, t) u(£, t)
= 0, = 0,
°< x < £,
t
> to,
0< x < £, 0< x < £,
t> to, t> to.
We sides by function We then then multiply multiply the the PDE PDE on on both both sides by aa test test function vEV
= 01[0,£],
(7.15)
7.3. Finite element methods for the the wave wave equation equation Finite element methods for
307 307
1.2 y=u(X,O) - _. y=u(x,0.0005) y=u(x,0.001 ) ,-,- =u(x,0.0015)
",
0.8
>-0.6
0.4
0.2
00
0.2
0.4
0.6
0.8
x
Figure 7.10. equation with a discontinuous initial Figure 7.10. The solution to the heat equation are four time snapshots, including t = = 0. O. condition (see (see (7.13}). (7.13)). Graphed Graphed are
integrate by parts in the second term on the left left to obtain and integrate obtain
foe {~:~ (x, t)v(x) + c ~~ (x, t) ~~ (x)} 2
dx
= foe f(x, t)v(x) dx, t > to,
v
E
V.
(7.16) This is the weak form. We now apply the Galerkin technique, approximating u(x, t) This is the weak form. We now apply the Galerkin technique, approximating by n-l
vn(x, t) =
L Ui(t)CPi(X),
(7.17)
i=l
where i, CPl, CP2, are the the standard standard piecewise piecewise linear linear finite element basis basis funcwhere ^ 2 , ... • • • ,,CPn-l <$>n-\ are finite element functions, equation (7.16) for all tions, and and requiring requiring that that the the variational variational equation (7.16) hold hold for all continuous continuous piecewise piecewise linear linear test test functions: functions:
n ( ) () r£ {{Pv at x, t CPi X + c
io =
2
foe f(x, t)CPi(X) dx,
2
aV n ( ) dCPi ( )} ax x, t dx x dx
t> to, for all i = 1,2, ... ,
n- 1.
Substituting (7.17) (7.17) into Substituting into (7.18) (7.18) yields yields n-l
L
j=l
d2u. dt2J (t)
r£
io 0
CPj (X)CPi ex) dx
n-l
r£
j=l
0
dcp·
dcp'
+ L Uj(t) io c2-t-ex) ax' (x) dx X
(7.18)
308 308
Chapter 7. Waves 1.2
y=u(x,O) -- _. y=u(x,0.15) . _. - y=u(x,0.3) - -. y=u(x,0.45)
0.8
>'0.6
._ .. _'. •
- -0.4
- - --
- --
'-"-"
• • • • • • •i •
• • • • • •i
0.2
•
0.2
0.6
0.4
- --
0.8
x
Figure 7.11. The The solution wave equation equation with with aa discontinuous Figure 7.11. solution to the the wave discontinuous initial initial four time snapshots, including condition condition (see (see (7.14)). Graphed Graphed are are four time snapshots, including tt = = O. 0.
=
lot f(x, t)¢i(X) dx, t "2: to, i = 1,2, ... ,
n - l.
(7.19)
If now define matrices M and K K (the (the mass respectively) If we we now define matrices M and mass and and stiffness stiffness matrices, matrices, respectively) by bv 2 d¢j d¢i Mij = ¢j (X)¢i (x) dx, Kij = c dx (x) dx (x) dx,
rt
rt
10
10
I [~~~:~
and the f (t) and u(£) by and the vector-valued vector-valued functions functions f(t) and u(t) by
f(t) =
I~ ~~:: :~:~~:~~:
[f~
, u(t) =
f(x, t)¢n-l (x) dx
.
1 ,
Un-·l(t)
we can write we can write (7.19) (7.19) as as ~u
M dt 2
+ Ku = f(t).
We now linear ODEs with constant We now have have aa system system of of second-order second-order linear ODEs with constant coefficients. coefficients. from the the initial condiWe two initial We need need two initial conditions, conditions, and and they they are are easily easily obtained obtained from initial conditions in have tions in (7.15). (7.15). We We have u(x, to) = 1jJ(x), 0
< x < £,
7.3. Finite element methods for the wave wave equation equation
309
and by its and 'IjJ(x) T/>(X) can can be be approximated approximated by its linear linear interpolant: interpolant: n-l
L 'IjJ(Xi)¢>i(X).
'IjJ(x) ==
i=l
We therefore therefore require require that We that Similarly, we require Similarly, we require dUi
dt(t o) = ,(Xi)'
These lead lead to the initial initial conditions These to the conditions
1 du
'IjJ(xd 'IjJ(X2)
u(to) = Yo = [
-(to) = Zo = dt
'IjJ(x~-d
'
,(Xl) ,(X2) [
'(X~-l)
1 .
We therefore therefore obtain obtain the the IVP IVI We ~u
M dt 2 +Ku = f(t), t> to, u(t o) = Yo, u'(to) = Zoo
(7.20)
To apply apply one one of of the the numerical methods for for ODEs ODEs that that we we studied studied in in Chapter Chapter To numerical methods 4, we must first first convert convert (7.20) (7.20) to to aa first-order first-order system. system. We will define define 4, we must We will y(t) = u(t), du z(t) = dt (t) (y,z ERn-I). € R11"1). Then dy
dt = z, dz
~u
dt
dt 2 = M- I (-Ku + f(t)) = -M-1Ky + M-1f(t) = Ay
47 47
where
+ g(t),
1
1 = M M-lf(t). We also also have have initial initial conditions: conditions: A == -M-1K -M K and g(t) = f(t). We
y(to)
= u(to) = Yo,
du z(to) = dt (to) = zoo 47 47As As we matrix MInstead, we discussed discussed in in Section Section 6.4, 6.4, we we do do not not actually actually form form the the inverse inverse matrix M 11.. Instead, when action of of M-l M^ 1 is is needed, needed, we solve the appropriate linear linear system. system. the action we solve the appropriate when the
310 310
Chapter 7. 7. Waves Chapter Waves
We obtained the the (2n (In -— 2) 2) x (In 2) system system of of IVPs IVPs We have have thus thus obtained (2n -— 2)
z,
y(t o)
Yo,
dz
Ay + g(t),
z(to)
zoo
dt
dt
(7.21)
We can now now apply apply any any of of the the numerical numerical methods methods that that we we have have previously previously learned, learned, We can such as as Euler's Euler's method, method, RK4, RK4, or or an an adaptive adaptive scheme. scheme. such Just as as in in the case of of the heat equation, equation, stability stability is is an an issue. issue. If If the the time time step step Just the case the heat At is is chosen chosen to to be be too too large large relative relative to to the the spatial spatial mesh size h, h, then the approximate approximate !:l.t mesh size then the solution the time solution computed computed by by the time stepping stepping method method can can "blow "blow up" up" (i.e. (i.e. increase increase without without bound). For the heat using Euler's we needed needed !:l.t bound). For the heat equation, equation, using Euler's method, method, we At = = O(h2). O(h2). This time steps that it to use use This required required such such aa large large number number of of time steps that it was was advantageous advantageous to special methods (such as as the the backward Euler method) method) for for the the sake sake of of efficiency. efficiency. The The special methods (such backward Euler requirement not so the wave wave equation; just need requirement on on !:l.t At is is not so stringent stringent when when solving solving the equation; we we just need At == O(hlc). O(h/c}. For reason, we we usually usually use explicit methods methods to solve the the system system of !:l.t For this this reason, use explicit to solve of ODEs arising arising from from the the wave wave equation; equation; the improved stability stability properties properties of of implicit implicit ODEs the improved 48 methods We mind, methods are are not not really really needed. needed.48 We must must keep keep the the stability stability requirement requirement in in mind, though. If, to approximate the wave we though. If, in in applying applying (7.20) (7.20) to approximate aa solution solution to to the wave equation, equation, we notice that the notice that the approximate approximate solution solution is is growing growing unreasonably unreasonably in in amplitude, amplitude, then then the time time step step must must be be decreased. decreased. the Example Example 7.7. 7.7. We We will will solve solve the the lBVP IBVP f) 2 u f)t 2
2
-
f) 2 u
°
= 0, < x < 1, t> 0, u(x,O) = 'IjJ(x), 0< x < 1, c f)x 2
f)u f)t (x,O) = 0, 0< x < 1,
(7.22)
> 0, u(l, t) = 0, t > u(O, t) = 0, t
°
with cc = 522 and and with = 522 'IjJ(x)
= O.01x(l -
x).
The resulting resulting solution is quite quite smooth, smooth, so so the the finite finite element element method method should work The solution is should work well. We We use use aa regular regular grid with nn = 20 20 subintervals (h = lin), 1/n), and and apply apply the the RK4 RK4 subintervals (h well. grid with method to integrate integrate the The fundamental fundamental frequency frequency of the string method to the resulting resulting ODEs. ODEs. The of the string is is 48 48However, finite element between an implicit However, in in aa sense, sense, the the use use of of the the finite element obscures obscures the the distinction distinction between an implicit and timeand an an explicit explicit method. method. The The mass mass matrix matrix couples couples the the derivatives, derivatives, and and so, so, to to advance advance aa timestepping necessary to means stepping method, method, it it is is necessary to solve solve aa linear linear system, system, even even in in an an explicit explicit method. method. This This means that of that explicit explicit methods methods are are no no more more efficient efficient than than implicit implicit methods. methods. The The same same was was true true in in the the case case of the heat equation, equation, with, with, however, however, aa key key difference. difference. For For the the heat heat equation, equation, the the stability stability requirement the heat requirement introduced is is such such that that it it is is not not advantageous advantageous to to use use explicit explicit methods methods anyway, anyway, and and so so the the coupling coupling introduced by the finite finite element wave equation, by the element method method is is not not important. important. For For the the wave equation, explicit explicit finite finite difference difference methods do do not not couple couple the derivatives, and and there there is is no need to to solve solve aa linear linear system system to to take take aa time time methods the derivatives, no need step. reason, finite geometry step. For For this this reason, finite difference difference methods methods are are preferred preferred for for the the wave wave equation equation if if the the geometry is simple. have advantages is simple. Finite Finite element element methods methods still still have advantages for for complicated complicated geometries. geometries.
311 311
7.3. Finite element methods methods for the wave equation
c/(1i} == 261, making the the fundamental c/(2f) 261, making fundamental period period
T = 261 1. = 0.003814. We motion of the string string over over one period, using 50 steps steps {!:1t We compute compute the the motion of the one period, using 50 (At = = T/50}. T/50J. The results are The results are shown in in Figure 7.12. 7.12. 0.01 , - - - - - - . . - - - - - . - - - - . . . , . - - - - . , . - - - - - ,
o. 0.006 0.004
~ 0.002 E Q)
:\l a.
:6 - 0.002 - 0 .004 - 0 .006 - 0 .008 0.2
0.4
x
0 .8
0.6
Figure 7.12. The The solution to the the wave wave equation equation in in Example Figure 7.12. solution to Example 7.7. 7.7. Shown Shown are taken over over one of the the motion. motion. are 25 25 snapshots snapshots taken one period period of The time step step taken taken in in the the previous previous example, example, The time AT. 7
Uot
= 50 =
.663· 10
-5
,
seems rather small. Moreover, Moreover, it it can can be be shown that it cannot be much larger seems rather small. shown that it cannot be much larger without without instability appearing in the computation. computation. On On the the other other hand, hand, as mentioned above, instability appearing in the as mentioned above, the h to preserve stability. the time time step step !:1t At must be decreased decreased only in in proportion to h stability. the heat heat equation, where !:1t is This to the This is is in in contrast contrast to the situation situation with with the equation, where At = = O(h2) O(/i 2 ) is required. reauired. 7.S. We now solve solve {7.22} with Example 7.8. We will will now (7.22) with ¢(x)
={
O.Ol(x - 0.4), -O.Ol(x - 0.6), 0,
0.4 < x < 0.5, 0.5 < x < 0.6, 0 < x < 0.4 or 0.6
<x <1
312 312
Chapter 7. Waves
(compare Example 1.3). Since the initial displacement displacement '¢ is not very smooth smooth (V> ('¢ is (compare Example 7.3). Since the initial if} is not very is continuous but but its its derivative derivative has has singularities), our comments comments at at the the beginning beginning of of continuous singularities), our the section section suggest suggest that that it it may may be be difficult difficult to to accurately accurately compute compute the the wave wave motion motion the using finite elements. using finite elements. We use use the the same spatial mesh mesh (h (h = = 1/20), 1/20,), time-stepping time-stepping algorithm algorithm (RK4) (RK4),, We same spatial and / 50) as in the the previous previous example. In Figure Figure 7.13, show and time-step time-step (ilt (£±t == T T/50J as in example. In 7.13, we we show three solution. We solution should look: three snapshots snapshots of of the the solution. We already already know know how how the the solution should look: The splits into pieces, one The initial initial "blip" "blip" splits into two two pieces, one moving moving to to the the right right and and the the other other to to the left, left, preserving the original original shape (see Figure Figure 7.5). 7.5). The The computed computed solution, the preserving the shape (see solution, while shapes are well while demonstrating demonstrating the the basic basic motion, motion, is is not not very very accurate-the accurate—the shapes are not not well represented. represented.
E
Ql
E Ql
o(1J 0. !I) '6
-1
-20~------~-------L--------~------~------~
0.2
0.4
0.6
0.8
1
x
Figure Example 7.8, Figure 7.13. 7.13. The The solution solution to to the the wave wave equation equation in in Example 7.8, computed computed with mesh size of hh = 1/20 1/20 and and aa time time step step of of ilt A£ = T/50. T/50. Shown are 33 snapshots snapshots with aa mesh size of Shown are corresponding tt = 0, Q,tt = T/50, T/50; tt = 2T/50. 2T/50. corresponding
To get an grid. We To get an accurate accurate computed computed solution solution requires requires aa fine fine grid. We repeat repeat the the calculations using using hh = 1/100 1/100 and and ilt At = = TT/250, and obtain obtain the the results results shown shown in in calculations /250, and Figure 7.147.14. Even Even with grid, there distortion in in the Figure with this this finer finer grid, there is is aa distortion the waves. waves.
7.3.2 7.3.2
The wave equation under other boundary conditions
If an an IBVP for the the wave equation involves involves aa Neumann Neumann condition, condition, we apply the the If IBVP for wave equation we can can apply finite weak form finite element element method method by by adjusting adjusting the the weak form and and the the approximating approximating subspace, subspace, as explained explained in in Section Section 6.5. 6.5. By By way way of of example, example, we we will will consider consider the the following following IBVP IBVP as
7.3.
313 313
Finite element methods methods for the wave equation
c
Q)
E
Q) (.)
ttl
Ci
en '5
-1
-2~------~------~--------~------~------~
o
0.2
0.4
0.8
0.6
x
Figure 7.14. 7.14. The solution to the wave equation in Example 7.B, computed Figure 7.8, computed /250. Shown are are 3 with a mesh size of h = = 1/100 1/100 and a time step of 6..t At = = T T/250. snapshots corresponding t = 0, 0, t — 2T/50. snapshots corresponding = TT/50, /50, t = 2T /50.
with mixed mixed boundary boundary conditions: with conditions:
aat2 u 2
2
a2 u
c ax 2 = f(x, t), 0< x <
e,
t> to,
u(x, to) = 'I/J(x), 0< x < e, au at (x, to) = l'(x), 0< x < e,
(7.23)
u(O, t) = 0, t> to, au ax (e, t) = 0, t > to.
Since is essential, it appears in the the definition Since the the Dirichlet Dirichlet condition condition is essential, it appears in definition of of the the space space of test test functions; condition is and does appear of functions; however, however, the the Neumann Neumann condition is natural natural and does not not appear explicitly. Therefore, the the weak weak form form is is explicitly. Therefore,
1t {~:~
(x, t)v(x)
+ c2 ~: (x, t) :~ (x) }
dx =
l'
f(x, t)v(x) dx, t > to, v
E
ii,
(7.24) where where
ii =
{v E C 2 [0,e] : v(O) = O}.
The reader should should note note that that this this is is exactly e~actly the the same same as the weak weak form form derived for The reader as the derived for the Dirichlet Dirichlet problem, problem, except for the space of of test test functions functions used. the except for the space used.
314 314
Chapter 7. Waves
To we establish To apply the Galerkin method, method, we establish aa grid
o=
Xo
< Xl < X2 < ... < Xn
= C
and choose choose the the finite finite element element subspace consisting of of all all continuous continuous piecewise piecewise linear and subspace consisting linear to the the given given mesh) satisfying the the Dirichlet condition at at the the left functions (relative (relative to functions mesh) satisfying Dirichlet condition left endpoint. This subspace can be represented as
difference between Sn Sn, for the Dirichlet problem, The only difference Sn and S n, the subspace used for is the of the basis function corresponding to the right For is the inclusion inclusion of the basis function ¢n (f)n corresponding to the right endpoint. endpoint. For simplicity, will use use aa regular in the the following following calculations. simplicity, we we will regular grid grid in calculations. The then proceeds proceeds just just as the Dirichlet problem, except The calculation calculation then as for for the Dirichlet problem, except that that there is is an extra basis basis function function and hence an an extra unknown and corresponding there an extra and hence extra unknown and corresponding equation. We write equation. We write n
vn(x, t)
=L
Ui(t)¢i(X)
i=l
for the to the the solution weak form form yields yields for the approximation approximation to solution u(x, u(x, t). t). Substituting Substituting into into the the weak the equations _ d2 u _ _ M dt 2 + Ku = f(t),
ret)
nxn nxn where M ME Rnxn, ER Rnxn, 6R ,K€ , and f(t) ERn. € Rn. The mass matrix M is is tridiagonal, with the following following nonzero with the nonzero entries: entries:
i = 1,2, ... ,n - 1,
MHI,i = Mi,HI, i = 1,2, ... , n - 1.
The stiffness stiffness matrix K is also tridiagonal, with the following nonzero entries:
7.3. Finite element methods for the wave equation
315
Finally, the the load vector f(t) by Finally, load vector f (t) is is given given by
!i(t) =
fal f(x,t)¢i(x)dx, i = 1,2, ... ,n.
Having derived derived the the formulation formulation of the initial Having the system system of of ODEs, ODEs, the of the initial conditions conditions just as before: is is just as before:
The IVP The IVP
- cPu
M dt 2
+ Ku = f(t), t > to, u(to) = Yo, u'(to) = Zo
is then then converted converted to first-order system system is to the the first-order dy
dt
= z,
y(t o)
= Yo,
dz dt = Ay + g(t), z(to) = Zo,
where A = _M~l:K -M^K and and g(t) = M~1f(t). M~lf(t). Exalllple motion of Example 7.9. 7.9. We will compute the motion of the string in Example 7.7, 7.7, with the only change change being that the the right right end end of of the the string string is is now to be being that now assumed assumed to be free free to to move move only IBVP is is vertically. The IBVP
a2 u at 2
-
2 2au c ax 2
.
° °<
= 0, < x < 1, t > 0,
u(x,O)=x(l-x),O<x 0, au ax (1, t) = 0, t > 0,
316 316
Chapter 7. Waves Chapter 7. Waves
with 522. The The only change from the earlier calculations is is that that there there is is an with cc = = 522. only change from the earlier calculations an extra row and and column in the and stiffness stiffness matrix. matrix. We will again again use 20 extra row column in the mass mass and We will use nn == 20 subintervals mesh (h (h = 1/20,). The The fundamental subintervals in in the the spatial spatial mesh = 1/20). fundamental period period is is now now T =
~= c
0.007663.
We RK4 algorithm algorithm and steps to compute the motion over over one one We use use the the RK4 and 100 100 time time steps to compute the motion period (Llt = TIlOO). results are are shown shown in in Figure Figure 7.15. 7.15. period (At = T/lOOj. The The results 0.01 0.008 0.006 0.004
C 0.002 Q) E Q)
0......-
0
cQ
a.
~
~
~~
~
--=::::::::::-
:6 - 0.002
-
-
- 0.004 - 0.006 - 0.008 - 0.01
0
0.2
0.4
0.6
0.8
x
Figure 7.15. The The solution solution to to the in Example Example 7.9. 7.9. Shown Figure 7.15. the wave wave equation equation in Shown are over one of the the motion. are 25 25 snapshots snapshots taken taken over one period period of motion.
Exercises 1. piecewise linear finite elements elements to to solve solve the the IBVP 1. Use Use piecewise linear finite IBVP
fPu at 2
2 fpu
-
c ax 2 = 0, 0
< x < 100, t > 0,
= 1O- 6 x(100 - X)2, 0 < X < 100, au at (x, 0) = 0, 0 < x < 100, u(x,O)
u(O, t) = 0, t > 0, u(lOO, t) = 0, t > 0 on the interval 00 < tt < T, T, where where T T is the fundamental fundamental period. period. Take Take cc = = 1000. 1000. on the interval is the Graph several several snapshots. snapshots. Graph
7.3. Finite Finite element element methods methods for the wave wave equation equation 7.3. for the
317 317
2. Repeat the previous exercise, assuming the PDE is inhomogeneous with righthand side to the the constant constant -100. hand side equal equal to —100.
3. (Cf. of length length 50cm, 3. (Cf. Exercise Exercise 7.2.2.) 7.2.2.) Consider Consider aa string string of 50cm, with with cc = = 400cm/s. 400cm/s. Suppose that the the string string is is initially initially at at rest, rest, with with both both ends fixed, and and that that it it Suppose that ends fixed, is of is struck by a hammer at its center so so as as to induce an initial velocity of -20,
'Y(x) = { 0,
24 < x < 26, 0 :::; x :::; 24 or 26 :::; x :::; 50.
How long does it take for (a) How for the resulting wave to reach the ends of the the string? string? (b) Formulate and solve the IBVP describing the motion of the string. Use (b) Formulate other numerical method for the finite element method with RK4 or or some other solving system of of ODEs. solving the the system ODEs. snapshots of the string, showing the wave first (c) Plot several snapshots first reaching the the ends string. Verify 3a. to 3a. ends of of the the string. Verify your your answer answer to that the the right right end end of of the the string free to to move move 4. Repeat Repeat Exercise Exercise 3, assuming that 4. 3, assuming string is is free vertically. vertically. try to reproduce Figure 5. Solve (7.14) using the finite element method, and try 7.11. (the length in the to the mesh) mesh) be be to 7.11. How How small small must must h h (the length of of the the subintervals subintervals in obtain aa good good graph? obtain graph?
6. Consider aa steel at the 6. (Cf. (Cf. Example Example 7.6.) 7.6.) Consider steel bar bar of of length length 11 m, m, fixed fixed at the top top (x = = 0) 0) with the the bottom bottom end end (x = = 1) free to to move. move. Suppose the bar bar is is stretched, stretched, by by with 1) free Suppose the means of a downward pressure applied to the free end, to a length of 1.002 m, of and then released. Using the finite element method, compute the motion of the bar, bar, given that the the type type of of steel has stiffness = 195 GPa and and density the given that steel has stiffness kk = 195 GPa density pp = = 7.9g/cm 7.9g/cm33.. Produce Produce aa graph graph analogous analogous to to Figure Figure 7.9. 7.9. 7. Consider Consider aa heterogeneous heterogeneous bar bar with with the the top top end end fixed fixed and and the the bottom bottom end end free. 7. free. According Section 2.2, of the bar the derivation derivation in in Section 2.2, the the displacement displacement u(x,t) u(x, t) of the bar According to to the satisfies the the IBVP satisfies IBVP 2
p(x) aatu2
-
a ( k(x) au) ax ax
=
f(x, t), 0 < x < e, t > to,
u(x, to) = 'lj!(x), 0 < x < e, au at (x, to) = 'Y(x), 0 < x < e, u(O, t) = 0, t > to, au 8x(e,t) =0, t>to. (a) Formulate the the weak weak form form of of (7.25). (7.25). (a) Formulate
(7.25)
318 318
Chapter 7. Waves Chapter
(b) Apply Apply the the Galerkin Galerkin method method to to obtain obtain aa system of ODEs with initial (b) system of ODEs with initial the mass mass and and stiffness stiffness matrices matrices change the case conditions. How How do conditions. do the change from from the case of bar? of aa homogeneous homogeneous bar? 8. Repeat Exercise Exercise 6, assuming now now that that the the bar bar is is heterogeneous heterogeneous with with density density 8. Repeat 6, assuming p(x)
= 7500 + 1000x kg/m3
and stiffness and stiffness
k(x) = 1.9.10 11
+ 101O xN/m2 •
(The coordinate is measured measured in in meters: meters: 0 0 < x < 1.) 1.) (The coordinate x is
7.4 7.4
Point and resonance Point sources sources and resonance
We now now consider consider two two special special experiments experiments that that can can lead lead to to resonance: We resonance: fixed, is •• A A metal metal string string (such (such as as aa guitar guitar string), string), with with both both ends ends fixed, is subjected subjected to an an oscillatory oscillatory electromagnetic electromagnetic force force generated generated by by aa small to small electromagnet. electromagnet. the frequency frequency of of the the external external force force equals equals one one of of the the natural natural frequencies frequencies When the When the string, string, resonance resonance occurs. occurs . of the of down: •• A A string string has has one one end end fixed, fixed, and and the the other other is is mechanically mechanically moved moved up up and and down: u(l, t) = = sin(27ru;£). sin (21fwt). Again, Again, resonance resonance occurs occurs when when the the applied applied frequency, frequency, w, u(i,t) w, equals one of the the natural natural frequencies frequencies of of the the string. equals one of string. 49 In the the first first experiment, experiment, the the modeling modeling involves involves the the Dirac Dirac delta delta function. In function.49
7.4.1 7.4.1
The wave equation with a point source
We following IBVP equation: We begin begin with with the the following IBVP for for the the wave wave equation: fpu
at 2
-
2fPu c ax 2 = f(t)8(x - a),
u(x, to)
= 0,
ou at (x, to) = 0,
°< x < £, t > to,
0 < x < £,
< x < £, 0, t > to, 0
(7.26)
u(O,t) = u(£, t) = 0, t > to,
where string, where we assume assume that that 0 0
~
1£
f(t)8(x - a) sin (n;x) dx
49 The reader Section 4.6 Dirac 49The reader is is assumed assumed to to have have read read Section 4.6 and/or and/or Section Section 5.7; 5.7; the the material material on on the the Dirac delta found in in those sections is is prerequisite section. delta function function found those two two sections prerequisite for for this this section.
7.4. Point sources and 7.4. Point sources and resonance resonance
319
- 2f(t) --
e
1£
!i:( ). dx ux-a SIn (n1fX) -
e
0
= 2 sin ~T) f(t).
We then must solve the following IVP to find the unknown unknown coefficients coefficients an(t): an(t):
in Section According to the the results derived derived in Section 4.2.2, 4.2.2, the solution solution to this IVP is
an(t)
it
= 2 sin (n1ra) £ c:n1f
to
(c:n1f ) sin -----n(t - s) f(s) ds. ~
Oscillatory Oscillatory forcing forcing and and resonance resonance
interested in the case in which which f(t) convenience, we take We are interested f ( t ) is oscillatory. For convenience, Then Then the the solution solution to to the the IBVP IBVP is is
eI — = 11 and and to to — = 0, 0, and and let let /(t) f(t) = = sin sin (2,-jrut). (21fwt). 00
u(x, t)
=
L an(t) sin (mfx), n=I
where an(t) = 2 sin (mfa) c:n1f
rt sin (21fws) sin (cn1f(t - s)) ds.
10
A lengthy but elementary calculation calculation shows shows that, u; =I / c:n/2, cn/2, then then A lengthy but elementary that, if if w an (t ) =
2 sin (n1fa) (c:n sin (21fwt) - w sin (c:n1ft)) . cn1f2(c2n 2 - 4w2)
(7.27)
One the other hand, if frequencies of the string, then if w cj = c:m/2, cm/2, one of the natural frequencies _ sin (m1fa) (sin (c:m1ft) -c:m1ftcos(c:m1ft)) () amt222 cm1f
(7.28)
(formula still holds, o> = cm/2, for the coefficients coefficients an(t) an(t) with n =I / m). (formula (7.27) (7.27) still holds, with with w = c:m/2, for the with n m). In the case that the external frequency a; w is a natural frequency of the string, the result (7.28) (7.28) shows that the string oscillates with an ever-increasing amplitude. (The reader should should notice the the factor of t, which shows shows that the amplitude amplitude increases without bound.) This phenomenon is called resonance. resonance. When the exterexternal frequency is close to but not equal to a natural frequency, formula (7.27) shows that the the nearby nearby natural natural frequency amplified in in the the solution (though it remains that frequency is is amplified solution (though it remains bounded)-this is because the the denominator in (7.27) is very small. bounded)—this
320
Chapter 7. Waves Waves Chapter
Example 7.10. We We now now solve = 522 522 and = 11 (this (this corresponds corresponds Example 7.10. solve (7.26) with with c = and l1 = to aa fundamental of middle middle C, C, as as we we saw in Example 7.3), to == 00 and to fundamental frequency frequency of saw in Example 7.3), and ff(t) ( t ) = sin s'm(27rujt) various values values of ofui. We locate locate the the point source at at aa = 1/0. l/\/2(27rwt) for for various w. We point source Figure 7.16 7.16 shows shows the the solutions corresponding to to w u = 260, 260, w LJ = 520, 520, w u; = = 780, 780, and Figure solutions corresponding and u} = 1040, 1040, close close to natural frequencies 0/261, 522, 783, 783, and and 1044, 1044, respectively. respectively. to the the natural frequencies of 261, 522, w= The notice how how the four normal The reader reader will will notice the solutions solutions resemble resemble the the first first four normal modes modes of of the the string. 7.17, we display some some snapshots of the the solution u = = 1044, 1044, the In Figure Figure 7.17, we display snapshots of solution for for w the string. In fourth natural frequency. The effect effect of of resonance resonance is is clearly clearly seen seen in in the the increasing increasing fourth natural frequency. The amplitude. amplitude.
4
X
10-
5
2 0
-2 -4
4
0 .5 x
0 X
10-
1
6
-2~------------------~
o
1 0 .5 x 10- 6 __________________
2~ x~~
~
1
- 1 -4 ~--------------------~
o
0 .5
x
1
-2~--------------------~
o
0 .5
x
Figure 7.16. Solutions Solutions to to (7.26) with with an oscillatory forcing term (see Figure 7.16. an oscillatory forcing term (see Example 7.10). The The four solutions shown shown correspond toujw = 260, 260, w w= = 520, 520, w u; = = 780, Example 7.10). four solutions correspond to 780, and w uj = = 1040; 1040; since these frequencies are close to the the natural natural frequencies of 261, and since these frequencies are close to frequencies of 261, 522, 783, solutions are 783, and and 1044, 1044) the the solutions are almost almost equal equal to to the the corresponding corresponding normal normal modes. modes.
7.4.
321 321
Point sources and resonance Point
1 . 5 r-------~-------,--------~------~--------,
0.2
0 .8
0 .6
0.4
x
Figure 7.17. 7.17. Solution Solution to to {7.26} with an an oscillatory oscillatory forcing forcing term term {see Figure (7.26) with (see ExExample 7.10). The the forcing term term is the fourth natural frequency o/the/orcing is 1044, 1044, the/ourth natural frequency ample 7.10}. The frequency frequency of 0/ the string, string, and and the the solution solution exhibits resonance. of the exhibits resonance.
7.4.2 7.4.2
Another experiment leading to resonance
Next we we solve the following following IBVP Exercise 7.2.8): 7.2.8): Next solve the IBVP (cf. (cf. Exercise a
2
2 2a U
u
at2
-
< x < £, t > to, 0, 0 < x < £,
c ax 2 = 0, 0
u(x, to) = au at (x, to) = 0, 0
< x < £, 0, t > to,
(7.29)
u(O, t) = u(£, t) = lOsin (211'wt), t> to. We will will continue continue to to take take £ O. We I = = 11 and and to t0 = = 0. Since the Dirichlet Dirichlet condition condition at at the the right right endpoint is inhomogeneous, we will will Since the endpoint is inhomogeneous, we shift the the data. data. We We define shift define p(x, t) =
lOX
sin (211'wt)
and v(x, t) — = u(x,t) u(x, t) -— p(x,t}. p(x, t). Then Then aa straightforward calculation shows shows that that vv and v(x,t) straightforward calculation satisfies the IBVP IBVP satisfies the
aatv 2
2 -
a
2
2 v 2 2 • ( ) c ax 2 = 41T W EX sm 21TWt , 0
v(x,O) = 0, 0 < x
< 1,
< x < 1, t> 0,
322
Chapter 7. Waves Waves
°
av at (x, 0)
= -27fWEX, < X < 1,
v(O, t) = 0, t v(l, t) = 0, t
(7.30)
> 0, > 0.
We We write write the the solution solution as as 00
v(x, t) =
L an(t) sin (n7fx). n=1
Then, by the Then, by the usual usual calculation, calculation, the the coefficient coefficient an(t) an(t) satisfies satisfies the the IVP IVP n rPa 2 2 2 dt 2 + C n 7f an
( )
= Cn t ,
an(O) = 0, dan (0) = b dt n,
where where Cn(t)
= 211 47f2w2Exsin (27fwt) sin (n7fx) dx =
87fW 2E( _l)nH
n
sin (27fwt)
= en sin (27fwt) and and bn = 2
1 1
o
.
(-27fWEX) sm (n7fx) dx =
4WE( _l)n n
.
By the the results of Section Section 4.2.2, we have have By results of 4.2.2, we an(t) = =
~ sin (en7ft) + _1_ cn7f 4WE( _l)n
+
en7f
2 sin (cn7ft) en 7f 8W 2E( _1)n+l
en
2
11
10rt sin (cn7f(t -
s))cn(s) ds
sin (cn7f(t - s)) sin (27fws) ds.
(7.31)
0
If W u =I7^ en/2 cn/2 for for all all nn (that (that is, is, if if W u; is is not not one one ofthe of the natural natural frequencies frequencies of of the string), If the string), then then we we obtain obtain
rt sin (en7f(t - s)) sin (27fws) ds = 7f (c n22_ 4W2) c; sin (27fwt) - Wsin (en7ft)) .
10
2
(7.32)
323
7.4. Point sources and resonance
On the hand, if w= = c:n/2, en/2, we obtain On the other other hand, if W we obtain
i
t .
o
sm (c:n7f(t - s)) sin (27fws) ds
(cmrt) = sin2c:n7f - -2t cos (c:n7ft).
(7.33)
The results are very similar to those for for the point source. If the externally applied applied frequency frequency is is not not aa natural natural frequency frequency of of the the string, string, then then no no resonance resonance occurs. occurs. Formula (7.32) (7.32) shows, shows, though, that the amplitude of the vibrations gets larger as W (jj approaches one of the the natural natural frequencies. If If the external frequency is a natural frequency, then resonance occurs, as indicated by the factor of tt in the formula for an(t). an(t}. Example 7.11. Suppose Suppose c = 522 and and W the fundamental fundamental frequency Example 7.11. = 522 u) = = c/2 c/2 = = 261, the frequency of the string. string. With we obtain Figure 7.18. 7.18. of the With ef = = 0.001, 0.001, we obtain the the motion motion shown shown in in Figure
Exercises 1. 1. Suppose a string with total mass of 10 g is stretched to a length of 40 cm. By
varying the frequency frequency of an oscillatory forcing term until resonance occurs, it is determined that the fundamental fundamental frequency of the string is 500 Hz. (a) How How fast do waves waves travel along the string? (b) What is the tension in the string? (Hint: (Hint: See See the derivation of the wave equation 2.3.) equation in in Section Section 2.3.) (c) At what other frequencies will the string resonate? bar of length 1 m, with a stiffness 2. Consider an iron bar stiffness of kk = — 90 GPa GPa and a 7.2 g/cm33 .. Suppose bottom end bar is is density density of of pp == 7.2g/cm Suppose that that the the bottom end (x = = 1) 1) of of the the bar fixed, fixed, and the top end is subjected to an oscillatory pressure: k ~~ (0, t) = B sin (27rwt), t
> o.
(a) What What is the w that causes resonance? Call this frefrethe smallest value of u; quency w(Jrr.. (b) Find and graph the displacement of the bar for w w= =W wrr .. (c) (c) Find Find and and graph graph the the displacement displacement of of the the bar bar for for W a; = = Wr w r /4. /4. 3. If a string is driven by an external force whose whose frequency equals a natural natural frequency, does resonance always occur? Consider the string of Example 7.10, and suppose W u = = 1044, a == 1/2. 1/2. Produce a plot like Figure 7.17. Does resonance occur? resonance occur? Why Why or or why why not? not?
4. Solve the IBVP 2
2
8t
8x
-8 u2 - c2 _8 u2
= sin (27fwt)6
u(x, 0) = 0, 0
(2) x --
< x < 1,
3
'
0< x < 1, t> 0,
324
Chapter 7. Waves
0.01 r-----------,
0.01 r-----------,
0.005
0.005
-0.005
-0.005
-0.01 '----------~ 1 o 0.5
-0.01 '--_ _ _ _ _ _ _ _...J o 0.5 1
x
x
0.01 ,...---------""""1
0.01 , . . . - - - - - - - - - . . . . ,
-0.01 '--_ _ _ _ _ _ _ _...J 1 o 0.5
-0.01 '----------~ o 0.5 1
x
x
Dirichlet condition (see Figure 7.18. 7.18. Solution to (7.29) with an oscillatory Dirichlet Example 7.11). 7.11). The 261, the fundamental frequency Example The frequency of of the forcing term term is is 261, the fundamental frequency of fundamental period (upper of the string. Shown are 10 snapshots from the first first fundamental (upper left), left), 10 fundamental period (upper 10 from the second fundamental (upper right), right), 10 10 from the the third third (lower (lower left), left), and 10 from the lower right (lower right). The increase in the amplitude is due to resonance.
au at (x,O) = 0, u(O, t) u(l, t)
0< x < 1,
= 0, t > 0, = 0, t > 0,
where c = 55 and w = 120. Graph several snapshots of the solution.
7.5 7.5
Suggestions further reading Suggestions for for further reading
The reader can consult Sections information about PDEs, The reader can consult Sections 5.8 5.8 and and 6.7 6.7 for for further further information about PDEs, Fourier Fourier series, and finite element element methods. methods. There is another computational technique technique that can be viewed as an alternative
Suggestions for further reading 7.5. Suggestions
325
to elements, namely, of finite finite differences. This method method is to finite finite elements, namely, the the method method of differences. This is very very popular for the wave equation equation and and similar for the reason briefly mentioned popular for the wave similar PDEs, PDEs, for the reason briefly mentioned in the footnote on Page 310. 310. Strikwerda Strikwerda [47] [47] is is an an advanced advanced treatment of the the finite finite in the footnote on Page treatment of difference method method for for PDEs. PDEs. Celia and Gray [9] covers covers both both finite differences and and difference Celia and Gray [9] finite differences finite elements elements (including (including finite element methods are not methods). finite finite element methods that that are not Galerkin Galerkin methods).
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Chapter 8 Chapter 8
Jgihlems in multiple multiple • menslons spatial dimensions
Most applications involve multiple spatial leading to partial Most practical practical applications involve multiple spatial dimensions, dimensions, leading to partial to four independent variables: variables: #1, Xl, #2 X2 or or #1, Xl, xz, X2, tt differential differential equations equations involving involving two two to four independent or xi,X2,x$,t. this chapter chapter is is to extend the the techniques last or Xl, X2, X3, t. The The purpose purpose of of this to extend techniques of of the the last elements-to PDEs involving involving two or more three chapters-Fourier chapters—Fourier series and finite elements—to spatial dimensions. We begin by developing developing the fundamental physical physical models in in two two and three three dimensions. We then Fourier series methods; as these techdimensions. We then present present Fourier series methods; as we we saw saw earlier, earlier, these techniques are applicable only in the case of constant-coefficient differential differential equations. Moreover, higher dimensions, we can the eigenfunctions explicitly when when Moreover, in in higher dimensions, we can only only find find the eigenfunctions explicitly the computational we treat the case of a rectangle and a circular computational domain is simple; simple; we disk in two dimensions. We then turn to finite element methods, focusing on two-dimensional problems and piecewise linear finite elements defined on a triangulation triangulation of the computational computational domain. domain.
8.1
Physical models in in two or three spatial spatial dimensions
The derivation of the physical models in Chapter 2 depended on the fundamental theorem of of calculus. we were were able to relate relate aa quantity defined on on theorem calculus. Using Using this this result, result, we able to quantity defined the boundary boundary of interval (force (force acting acting on on aa cross-section cross-section or or heat energy flowing the of an an interval heat energy flowing across for example) the interior interior of of the across aa cross-section, cross-section, for example) to to aa related related quantity quantity in in the the interval. interval. In higher higher dimensions, analogue of of the is dimensions, the the analogue the fundamental fundamental theorem theorem of of calculus calculus is the divergence theorem, which relates a vector field acting on the boundary of the divergence theorem, which relates a vector field acting on the boundary of aa domain the divergence divergence of of the the vector field in in the the interior. We begin begin by by explaining domain to to the vector field interior. We explaining the the divergence divergence theorem theorem and and proceed to apply apply it it to to a derivation derivation of of the the heat equation. equation. We first establish We must must first establish some some notation. notation. 327
Chapter Chapter 8. Problems in multiple spatial dimensions dimensions
328
B.1.1 8.1.1
The divergence theorem
An advantage advantage of of using using vector vector notation notation is is that that many many results results of of calculus calculus take take the An the same form for for two two and and three three dimensions dimensions when when expressed expressed in in vector vector form. form. Therefore, Therefore, same form we will will treat treat both both two-dimensional two-dimensional and three-dimensional problems problems in in this this section, we and three-dimensional section, 3 and much of the background calculus requires no distinction between R R22 and R R3. . 2 3 Let f2 0 be be aa domain domain (a open set) in R R2 or or R R3 with with aa piecewise piecewise Let (a connected connected open set) in 50 smooth boundary 80. R33 using either vector notation <9fi.50 We will denote points in R notation 2 notation as R2. or coordinate notation as is convenient, and similarly for points in R . Thus a point R33 can can be be denoted denoted as as xx or or as as (Xl,X2,X3). in in R (xi,X2,xs). An important concept related to aa domain domain fi 0 is is that that of of the the (outward) (outward) unit An important concept related to at which the boundary is smooth, there is a unique normal. At each point x of 80 dfl 80 at x (to be precise, n(x) is orthogonal unit vector n(x) that is orthogonal orthogonal to dtl orthogonal to the the line or plane plane tangent tangent to to 80 at x) x) and and points points toward toward the the exterior exterior of to line or dft at of O. fi. It It is customary customary to to suppress suppress the the dependence dependence of of the the normal normal vector vector n n on on x, x, writing writing nn is instead of n(x), but but it is important important to keep this dependence in mind. Figure 8.1 instead shows an example in two dimensions. nix)
Figure 8.1. 8.1. A two-dimensional domain J7 0 with a sample unit normal vector. Figure
A (volume) (volume) integral over 0fJ in R3 R3 will be denoted denoted as as
In f(x)dx or simply by or simply by 50 SOFor our purposes, purposes, it it does does not not seem seem worthwhile worthwhile to to define define precise precise conditions conditions on on n and its For our fi and its boundboundary that that allow allow the the application application of the divergence theorem. We We will will assume assume that that the the reader reader has has an an ary of the divergence theorem. intuitive idea idea of of the the meaning meaning of of "smooth "smooth boundary" boundary" and "piecewise smooth smooth boundary." boundary." For For examintuitive and "piecewise exam2 2 2 ple, the the unit unit ball, ball, S S = = {(Xl, X2, X3) G ER R33 :: x X2 + y y2 + z z2 < 1 I}, ple, {(xi,X2,xs) j, has hasaa smooth smooth boundary, boundary, while while the the 3 unit cube, R= = {(Xl, ER R3 :: 0 O:S Xl :S X2 :S X3 :S I}, has has a a boundary boundary that that is is unit cube, R { ( x i , XX2, 2 , xX3) z) e < x\ < 1,0 1,0 :S < #2 < 1,0 1,0 :S < #3 < ij, piecewise smooth. piecewise smooth.
8.1. Physical Physical models in two or three spatial dimensions 8.1.
329
where £) --+ -> R. R. If If aa domain domain n £) in in R3 R/5 can can be be described described as as where f/ : n {(Xl, X2, X3) :
a
< Xl < b, gl (xI) < X2 < g2(XI), h1 (Xl, X2) < X3 < h 2 (X1, X2)} ,
for for example, example, then the volume volume integral integral can can be be rewritten rewritten as as an an iterated iterated integral: integral:
A (surface) (surface) integral integral over over an <9fJ will will be be denoted denoted by by A
r g(x) dO",
i 8 1l
where 9g :: an d£l --+ —»• R R and and da an infinitesimal infinitesimal surface surface area. area element, element, or or simply where dO" represents represents an simply
hll
g.
Both Both volume volume and and surface surface integrals integrals can can have have vector-valued vector-valued integrands, integrands, in in which which case case each component component is is integrated. integrated. each If n Jl is is aa domain domain in in R R 22 ,, then then If
is an an area area integral integral that can, for for many many regions fi, be as aa doubly doubly iterated iterated is that can, regions n, be rewritten rewritten as this case, integral. integral. Since Since an dQ, is is aa curve curve in in this case.
hll
9
51 is is aa line line integral. integral.51 A vector vector field field defined defined on on n £) is is just vector-valued function-a function—a mapping mapping of of the the A just aa vector-valued form form f(x) = [ il (x) ]
hex)
22
if n fi is is in in R R or or if
f(x)
= [ ;~~:j fs(x)
1
if n isis in if f) in R R 33 .. The The divergence divergence of of a.a vector vector field field ff is is denoted denoted by by V' V .•ff and and isis defined defined by by
V' . f(x)
ail (x) + -a ah (x) = -a Xl X2
in in two two dimensions dimensions and and
V' . f(x) =
aa il (x) + aa h (x) + aa Xl
51
X2
fs
(x)
X3
Line integral integral is is aa misnomer, since the the domain domain of of integration integration is is generally generally aa curve, curve, not not aa (straight) 51 Line misnomer, since (straight) line. line.
330 330
Chapter 8. Problems in multiple spatial dimensions dimensions
in three dimensions. The The following following theorem, theorem, which in both and three three in three dimensions. which holds holds in both two two and dimensions, this combination combination of of derivatives derivatives is is so so significant. dimensions, explains explains why why this significant. Theorem 8.1. 8.1. (The (The divergence divergence theorem) theorem) Let bounded open set with Theorem Let 17 n be be aa bounded open set with aa piecewise smooth smooth boundary boundary an. Assume that smooth vector vector field defined on piecewise $17. Assume that ff is is aa smooth field defined on 17 d£l, and that nn is is the the (outward (outward pointing) unit normal normal to 017. Then n uU an, and that pointing) unit to an. Then
r\7 . ianr
in
f =
f· n.
In the case in R R 22,, the the divergence In the case of of aa rectangular rectangular domain domain in divergence theorem theorem follows follows immediately from from the the fundamental fundamental theorem theorem of of calculus calculus (see 2). immediately (see Exercise Exercise 2).
8.1.2 8.1.2
The heat equation for a three-dimensional domain
We solid occupying occupying aa domain fi in in R We now now consider consider aa three-dimensional three-dimensional solid domain n R 33.. (We (We will often often call call the solid n, 17, although, although, in fact, n 17 is is aa mathematical mathematical description description of will the solid in fact, of the location of of the solid.) For simplicity, assume assume that solid is is homogeneous, the location the solid.) For simplicity, that the the solid homogeneous, p, specific heat c, c, and thermal conductivity K,. Analogous Analogous to onewith density density p, with specific heat and thermal conductivity K. to the the onedimensional case, if if the domain w a; is is aa subset subset of of n, 17, then then the total heat heat energy the domain the total energy dimensional case, contained in o> w is contained in is Eo + pc(u(x, t) - To) dx,
L
where w(x,t) is the the temperature 6n 17atattime timet,i,ToT0isisthe thereference referencetemperature, temperature, where u(x, t) is temperature at at xx E and EQ is the energy contained in the solid at at temperature and Eo is the thermal thermal energy contained in the solid temperature T To. 0. The rate of change change of energy contained contained in is given given by The rate of of the the energy in u; w is by
!
[Eo
+
L
pc(u(x, t) - To) dX] =
!L
pcu(x, t) dx.
Moving the derivative derivative through through the the integral sign, we we obtain obtain the the following following expression Moving the integral sign, expression for of change total heat heat energy energy contained for the the rate rate of change of of total contained in in w: uj:
iwr pc au at (x, t) dx.
(8.1)
We can can also also describe this rate of change computing the the rate rate at at which which heat We describe this rate of change by by computing heat 3 energy across ow. duj. Let Let q(x, q(x,£) €R at x 17atattime timet.t. The The energy flows flows across t) E R3 be be the the heat heat flux flux at x e En heat flux is aa vector-its vector—its direction direction is the direction direction of of the of energy, energy, and its heat flux is is the the flow flow of and its magnitude gives the rate at at which which energy energy is is flowing across the the plane at x magnitude gives the rate flowing across plane at x to to which which q(x, t) £) is is orthogonal, in units of energy energy/(time-area). At each each point duj, the q(x, orthogonal, in units of /(time·area). At point of of ow, the rate rate at which which energy is flowing into w u; is is at energy is flowing into -q-n.
(This follows follows from from the fact that that qq is is aa vector and hence hence can be decomposed (This the fact vector quantity quantity and can be decomposed into the the direction is -q —q .• n, n, and and the the component into the component component in in the direction of of —n, -n, which which is component
8.1. Physical models in two or three spatial dimensions
331 331
orthogonal to -— n.) flowing into is then orthogonal to n.) The The total total amount amount of of energy energy flowing into w uj is then
- Jawr
q.n.
We now make make the heat flux flux is proWe now the assumption assumption called called Fourier's Fourier's law, law, that that the the heat is pro52 portional to temperature gradient: gradient:52 portional to the the temperature
q(x, t) = -1\;\7u(x, t). The proportionality, I\; is the thermal conductivity, just as in The constant constant of of proportionality, K > 0, 0, is the thermal conductivity, just in one one spatial dimension. dimension. We We therefore obtain the the following following expression expression for for the the rate rate of spatial therefore obtain of change of heat energy contained in a;: w:
We wish to equate this expression with (8.1) and derive derive a PDE. To do so, we we must apply the divergence theorem convert the volume integral: apply the divergence theorem to to convert the boundary boundary integral integral to to aa volume integral:
We thus have
or or
i
pc ~~ (x, t) dx
i (pc ~~
=
i
1\;\7 . \7u(x, t) dx,
(x, t) - 1\;\7 . \7u(x,
t)) dx = O.
(8.2)
Since (8.2) (8.2) holds holds for for every every subdomain subdomain w uj of of n, £), by by exactly exactly the the same same reasoning Since reasoning one dimension, the integrand must be be identically identically zero. We therefore therefore obtain obtain as in in one dimension, the integrand must zero. We the PDE
pc au at -
n, t > to·
1\;\7 . \7u = 0, x E
The just such such The combination combination \7 V .• \7 V of of differential differential operators operators arises arises frequently frequently due due to to just an the divergence have an application application of of the divergence theorem. theorem. We We have
\7u =
gx~ 1
[ ~:
'
aX3
52
52 The of several variables points points in in the the direction which The gradient gradient of of aa scalar-valued scalar-valued function function of several variables direction in in which the rapidly. the function function increases increases most most rapidly.
332
Chapter Chapter 8. Problems in multiple spatial dimensions
and and so so
The differential operator The differential operator
a
2 82 82 6.=8 Xl2 +8X 2 +aX32 2
is called the is called the Laplacian. Laplacian. equation: equation:
53 53
Using obtain the the three-dimensional three-dimensional heat heat Using this this notation, notation, we we obtain
au pc 8t - ",6.u = 0, x
E
.11, t > to·
(8.3)
If f/ : n fi xx (to, 00} R is heat source source (or (or sink), sink), with of energy energy per If 00) ->• -+ R is aa heat with units units of per volume volume per time, then we we obtain obtain the the inhomogeneous inhomogeneous three-dimensional heat equation: per time, then three-dimensional heat equation:
8u pc 8t - ",6.u = f(x, t), x E .11, t> to·
(8.4)
It be obvious obvious to to the the reader reader how how to to modify modify this this equation equation if the material It should should be if the material properties (p, c, c, "') are not not constants constants but but rather of xx E (see Exercise Exercise 1). properties K) are rather functions functions of Gn 0 (see 1).
8.1.3 8.1.3
Boundary Boundary conditions for the three-dimensional heat equation
Two common common types types of boundary conditions conditions for for the heat equation Dirichlet and Two of boundary the heat equation are are Dirichlet and Neumann conditions. An equation of form Neumann conditions. An equation of the the form u =
°
on
an
is Dirichlet condition; it corresponds corresponds to to an an experiment experiment in in is called called aa homogeneous homogeneous Dirichlet condition; it the temperature temperature of the boundary held fixed fixed at o. Inhomogeneous Dirichlet which the which of the boundary is is held at 0. Inhomogeneous Dirichlet conditions are are also also possible, possible, and and boundary boundary data can be conditions data can be time-dependent: time-dependent:
u(x, t)
= g(x),
x E
8n, t > to,
or u(x, t)
= g(x, t),
x E
an, t > to.
It is is also also possible possible to the boundary, boundary, which which leads to the the condition condition that to insulate insulate the leads to that the heat heat flux flux across across the is zero: zero: the the boundary boundary is
\7u(x, t) . n = 0, x E an, t > to.
The derivative Vw is often referred to derivative and and The directional directional derivative \7 u .• nn is often referred to as as the the normal normal derivative denoted denoted au 53 53 Another Another
an·
common symbol symbol for for the '\7 22 .• common the Laplacian Laplacian is is V
8.1. Physical two or spatial dimensions 8.1. Physical models models in in two or three three spatial dimensions
333 333
Just as as in in the the one-dimensional case, we have mixed conditions, Just one-dimensional case, we can can have mixed boundary boundary conditions, it is possible to to treat treat different parts of of the boundary differently. since it is possible different parts the boundary differently. Suppose Suppose dtl of the into two disjoint sets. sets. Then Then it an == FI rl U U F2 r 2 isis aa partition partition of the boundary boundary into two disjoint it is is possible the following following mixed conditions: possible to to pose pose aa problem problem with with the mixed boundary boundary conditions:
u(x,t) = 0, x au an (x, t) = 0, x
8.1.4 8.1.4
E
rl, t > to,
E
r 2 , t > to·
The heat equation in in a The heat equation a bar bar
In Section Section 2.1 and in in subsequent subsequent sections, sections, we we discussed heat flow in aa bar In 2.1 and discussed heat flow in bar with with insuinsulated sides. sides. We following fact: fact: If If the initial temperature distribution lated We mentioned mentioned the the following the initial temperature distribution in bar depends the longitudinal longitudinal coordinate coordinate (that is, if it is in in aa bar depends only only on on the (that is, if it is constant constant in each cross-section), and if any heat source also depends only on the longitudinal each cross-section), and if any heat source also depends only on the longitudinal coordinate, then the the temperature temperature at at all all subsequent times also also depends depends only coordinate, then subsequent times only on on aa now justify justify this this statement. single spatial coordinate. coordinate. We single spatial We can can now statement. Let the domain domain n ft in in R R33 be Let the be defined defined by by
n=
{(XI,X2,X3) :
°<
< l, x~ +x~ < r2},
Xl
where li and and rr are are given positive constants. The set set n is aa circular circular cylinder where given positive constants. The fHs cylinder centered centered on the interval [0, [0,^1 Consider the following IBVP: IBVP: the interval l] on on the the xi-axis. Xl -axis. Consider the following on
au
= I(Xl, t), x En, t> to, u(x, to) = '¢(xt), x E n,
pc at - ",flu
au
an (x, t) = 0, u(x,t)
= 0,
x E
rt,
(8.5)
t> to,
x Ere, t
> to,
where ends of of the the bar sides: where F r ee represents represents the the ends bar and and T r t the the transverse transverse sides: re = {(O,X2,X3) : x~ +x§ < r2} U {(f,X2,X3) : x~ +x§ < r2},
rt ={(Xl,X2,X3):
O<xl
The reader reader should notice that that both both the the heat heat source source /I and the initial initial temperature The should notice and the temperature distribution '¢ if) are are independent independent of of x-2 and X3. x3. distribution X2 and The solution to (8.5) is w(xi,x 2 ,x 3 ,i) = V(Xl,t), v(xi,t), where The solution to (8.5) is U(Xl,X2,X3,t) where v is is the the solution solution of of the the IBVP IBVP OV 02V pc--;:;- - "'>:. 2 = I(Xl,t), < Xl < l, t> to, ut
° V(XI, to) = '¢(xd, °< uX l
Xl
< l,
(8.6)
V(O, t) = 0, t > to, v(i, t) = 0, t > to. The proof of this of the equations in (8.5) and and is left to an The proof of this is is aa direct direct verification verification of the equations in (8.5) is left to an exercise. exercise.
334 334
8.1.5 8.1.5
Chapter 8. 8. Problems Problems in in multiple multiple spatial dimensions Chapter spatial dimensions
The heat equation in in two dimensions dimensions
It to restrict the heat heat equation equation to to two two dimensions. heat flows flows in in It is is straightforward straightforward to restrict the dimensions. If If heat solid, such such as as aa thin thin plate, plate, in in such such aa way way that that the the temperature temperature is is constant constant in in one one aa solid, dimension, the the derivative derivative with with respect to the the third spatial variable, say #3, dimension, respect to third spatial variable, say X3, vanishes. vanishes. The result result is two-dimensional heat equation, which which is is usually usually written written exactly exactly The is the the two-dimensional heat equation, as in in (8.3), (8.3), since since we we will A to denote the the Laplacian Laplacian in in two independent will also also use use A to denote two independent as variables: variables:
The distinction distinction between between the the twotwo- and and three-dimensional The three-dimensional Laplacian Laplacian will will be be underunderstood stood from from context. context.
8.1.6 8.1.6
The wave equation for a three-dimensional domain
The the wave wave equation dimensions is more comcomThe derivation derivation of of the equation in in three three dimensions is considerably considerably more the heat equation. A A complete treatment begins begins with plicated than plicated than that that of of the heat equation. complete treatment with an an elastic elastic solid, Newton's law forces, both both internal external, acting on solid, and and applies applies Newton's law to to the the forces, internal and and external, acting on the solid. The The result is aa system system of three (coupled) (coupled) PDEs PDEs for three components result is of three for the the three components the solid. of displacement displacement that that describes the vibration of the the solid. solid. (Each (Each point in the of describes the vibration of point in the solid solid can move move in in three three dimensions, dimensions, so so there are three dependent variables.) can there are three dependent variables.) This This system system is is one one form form of of the the wave wave equation. equation. Much physical modeling modeling can can be under apparently apparently severe severe simMuch useful useful physical be performed performed under simplifying Specifically, assuming assuming that solid is is aa fluid (meaning that that plifying assumptions. assumptions. Specifically, that the the solid fluid (meaning the only only stress stress supported supported is and that that the consideration is the is aa pressure) pressure) and the motion motion under under consideration is aa small small perturbation perturbation of of an an equilibrium equilibrium state, state, with with the the motion motion induced induced by by aa force force density F(x, F(x, t), £), the the result is the acoustic wave wave equation equationfor forthe the pressure pressureperturbation perturbation result is the acoustic density u(x.t): uu == u(x, t): (8.7)
In this equation, equation, the the forcing forcing function function f/ is negative divergence divergence of the body body force: In this is the the negative of the force: f(x, t)
= - \7 . F(x, t).
An example physical phenomenon by the three-dimensional acoustic acoustic An example of of aa physical phenomenon modeled modeled by the three-dimensional wave equation is is the of sound sound waves waves in in air. air. wave equation the propogation propogation of
8.1.7 8.1.7
The wave equation in in two dimensions dimensions
The (acoustic) wave wave equation equation models models the the small small transverse The two-dimensional two-dimensional (acoustic) transverse vibravibrations of of an an elastic elastic membrane. An elastic elastic membrane analogous to to an an elastic elastic string, tions membrane. An membrane is is analogous string, it does bending. The The form wave equation equation is is exactly in in that in that it does not not resist resist bending. form of of the the wave exactly as as in (8.7); however, however, the the meaning meaning is quite different. The dependent dependent variable is the the (8.7); is quite different. The variable uu is vertical component of displacement, the right-hand right-hand side is the the transverse vertical component of displacement, while while the side f/ is transverse pressure. 8.2 illustrates illustrates the the small small deflection of aa square square membrane. pressure. Figure Figure 8.2 deflection of membrane.
8.1. Physical models in two or three spatial dimensions
335
0.1
o
0
Figure small vertical vertical deflection deflection of of aa square square membrane under Figure 8.2. 8.2. The The small membrane under Dirichlet conditions. conditions. Dirichlet Dirichlet the vibrating vibrating membrane Dirichlet boundary boundary conditions conditions for for the membrane indicate indicate that that the the boundary (one can can picture, example, aa circular circular drumhead boundboundary is is fixed fixed (one picture, for for example, drumhead with with its its boundary to the conditions indicate that the the boundary is free free ary attached attached to the drum). drum). Neumann Neumann conditions indicate that boundary is to move in in aa vertical vertical direction. direction. to move
8.1.8 8.1.8
Equilibrium problems problems and and Laplace's Laplace's equation Equilibrium equation
The differential operator ~ appears in both both the The differential operator A appears in the heat heat equation equation and and the the wave wave equaequation, therefore in in the versions of example, tion, and and therefore the equilibrium equilibrium versions of these these equations. equations. For For example, steady-state heat is modeled by Poisson's steady-state heat flow flow is modeled by Poisson's equation, equation, -K,~u =
f(x), x E
n,
or or by by its its counterpart counterpart for for aa heterogeneous heterogeneous material, material, -~. (K,(x)~u)
= f(x), x E
n.
This be applied two or three dimensions. This equation equation can can be applied in in either either two or three dimensions. An elastic membrane subject to aa steady tranverse pressure f/ has has aa vertical An elastic membrane subject to steady tr an verse pressure vertical deflection u approximately satisfying deflection approximately satisfying -k~u =
f(x), x E
n.
The heterogeneous version of of this this equation equation is The heterogeneous version is -~. (k(x)~u) =
f(x), x
E
n.
Chapter Problems in multiple spatial spatial dimensions Chapter 8. 8. Problems in multiple dimensions
336 336
The homogeneous homogeneous version version of Poisson's equation is referred to as Laplace's Laplace's equation: -~u = 0, x EO. Of course, this equation equation is interesting interesting only if the boundary conditions are inhomoOf geneous. geneous.
8.1.9 8.1.9
Green's identities and the the symmetry symmetry of the Laplacian Laplacian Green's identities and of the
expect to be able to extend the methods from the preceding chapters We should expect chapters -~ is a symmetric operator. The inner product is only if —A
In fg,
(f,g) = and we we must show that
(-~u,v) = (u,-~v)
u, v satisfying the desired boundary conditions. We We define for all u,v
c(722(J7) (n) = = {u : n -+ R :: u and its partial partial derivatives derivatives up up to to order order 2 2 are are continuous} continuous} {w:(]->R and its (n the closure of of Q, 0, that that is, is, fi 0 together together with its boundary: boundary: n = 0 u ao) (£1 is is the with its J7 = i7 U 9fi) and and cb(n)
= {u E C 2 (n)
:
x E ao =? u(x) =
o}
and LD : cb(n) -+ C(O), LDu = -~u.
In one one dimension, fundamental manipulation manipulation underlying underlying the the development development In dimension, the the fundamental of both Fourier series and finite element methods is integration integration by parts. The anawe now derive. logue in higher dimensions is Green's (first) identity, which we derive. It It is helpful to to recall recall that by parts parts is based on on the product rule rule for helpful that integration integration by is based the product for differendifferentiation and the fundamental theorem of calculus, as follows:
d dv dx [u(x)v(x)] = u(x) dx (x)
r
b
r
b
d
du
+ dx (x)v(x)
r du b
dv
=?
fa dx [u(x)v(x)] dx = fa u(x) dx (x) dx + fa dx (x)v(x) dx
=?
u(x)v(x)l:
=?
l
r u(x) dxdv (x) dx + far dudx (x)v(x) dx b
b
a
b
dv
= fa
U(X) dx (x) dx
= u(x)v(x)l: -
lb a
du dx (x)v(x) dx.
To derive derive Green's analogous product for higher To Green's identity, identity, we we need need an an analogous product rule rule for higher dimensions dimensions.
8.1. 8.1.
Physical models in two or three spatial dimensions
337
It is not not completely obvious how how to to obtain obtain the the needed needed product product rule, rule, but we It is completely obvious but we can be guided by by the formula involving can be guided the fact fact that that we we want want to to obtain obtain aa formula involving v~u
= v\7 . \7u.
The desired product rule is u; \7 . (v\7u) = \7u· \7v
+ vAu.
(8.8)
is proved proved directly; Exercise 5. Combining (8.8) with the the divergence theorem This This is directly; see see Exercise 5. Combining (8.8) with divergence theorem gives the desired desired result: gives the result: \7 . (v\7u)
=>
l
= \7u· \7v + vAu
\7. (v\7u) =
=> { v\7u· n
ian
=> { vAu
in
=
l
\7u· \7v +
l
vAu
= { \7u· \7v + r v~u in in
r
ian
vaau - { \7u· \7v. n in
This is is Green's first This first identity: identity: { vAu = {
v au - ( \7u· \7v.
ian an in
in
(8.9)
We can now prove the symmetry of the Laplacian Laplacian under under Dirichlet Dirichlet conditions. conditions. We can now prove the symmetry of the If If u, w ,vv E € Cb(n), 07^(0), then then (LDu, v)
=
-l
vAu
= { \7u· \7v -
in
=
l
= {
n
\7u· \7v (since v = 0 on an)
ian
= -
{ vaau (Green's first identity)
ian
u
In
OV - { uAv (Green's first identity) a n in
uAv (since v = 0 on an)
= (U,LDV),
It is no more difficult difficult to prove the symmetry of -A —A under Neumann conditions conditions (see Exercise Exercise 6). (see 6).
Exercises 1. How does does the the heat heat equation (8.3) change if p, p, c, functions of 1. How equation (8.3) change if c, and and K,K are are functions of x? x?
Chapter 8. Problems in multiple spatial dimensions
338
2. C R22 be domain 2. Let Let 0 nCR be the the rectangular rectangular domain
n = {x E R2 22
: a
< Xl < b, c < X2 < d} ,
22
and smooth vector and let let F F :: R —»• -+ R be be aa smooth vector field. field. Prove Prove that that
r\7 . F lenr F· n. =
ln
Hint: Rewrite Rewrite the the area area integral integral on on the the left an iterated iterated integral integral and apply Hint: left as as an and apply the fundamental theorem of the result result to to the the boundary the fundamental theorem of calculus. calculus. Compare Compare the boundary integral integral on on the right. right.
3. Let n £7 c C R2 R2 be the unit square: square:
n = {x E R2
: 0
< Xl < 1,
0
< X2 < I} .
-+ R2 by Define F : n ft -> R2 by F(x)
= [ Xl X+2 X2
] •
Verify divergence theorem fl and and this vector Verify that that the the divergence theorem holds holds for for this this domain domain n this vector fieldF.F. field 4. Let il c CR 4. Let n R22 be be the the unit unit disk: disk:
n = {x E R2
: xi
+ x~ < I} .
-+ R2 by Define F : n ft -» R2 by F(x)
= [ Xl X2+ X2
] .
Verify divergence theorem Q and and this Verify that that the the divergence theorem holds holds for for this this domain domain n this vector vector field field F.F. 5. as follows: follows: Write 5. Prove Prove (8.8) (8.8) as Write \7. (v\7u) =
L -0 3
.
.=1
OXi
[ v~ 0 ] OXi
and apply apply the right. and the ordinary ordinary (scalar) (scalar) product product rule rule to to each each term term on on the the right. 6. 6. Define Define
and LN :
C~(n) LNU
Show that Show that LN LN is is symmetric: symmetric:
-+ C(n),
= -~U.
8.2.
Fourier rectangular domain domain Fourier series on a rectangular
339
7. Suppose that the the boundary boundary of of £7 n isis partitioned partitioned into into two two disjoint disjoint sets: sets: dtl an —= 7. Suppose that . rFI1 u r Define UF 2 . Define
and and Lm : C~01) -t C(O), Lmu
= -~u.
Show that that L Lm is symmetric: Show symmetric: m is (Lmu,v) = (u,Lmv) for all
U,V
2 -
E Cm(n).
8. Verify Verify that that the the solution to (8.6) (8.6) is is also also the the solution to (8.5). (8.5). 8. solution to solution to
8.2
Fourier series on a rectangular domain
We now now develop develop Fourier Fourier series series methods methods for for the the fundamental fundamental equations equations (Poisson's We (Poisson's equation, the the heat heat equation, equation, and and the the wave wave equation) equation) on on the the two-dimensional two-dimensional rectequation, rectangular angular domain domain
(8.10) We will will begin begin by by discussing discussing Dirichlet Dirichlet conditions, conditions, so the operator operator is is L We so the Lp, as defined defined D , as at the section. at the end end of of the the last last section.
8.2.1 8.2.1
boundary conditions Dirichlet boundary
As should expect development in 5, 6, crux of As we we should expect from from the the development in Chapters Chapters 5, 6, and and 7, 7, the the crux of the matter is to determine determine the the eigenvalues and eigenfunctions of Lp. L D . We We have the matter is to eigenvalues and eigenfunctions of have that LD LD is is symmetric, symmetric, so so we we know know that that eigenfunctions eigenfunctions corresponding already already seen seen that corresponding to distinct distinct eigenvalues eigenvalues must must be be orthogonal.' orthogonal: Moreover, Moreover, it it is to show show directly to is easy easy to directly that LD LD has has only only positive positive eigenvalues. For suppose is an an eigenvalue eigenvalue of LD and and u u that eigenvalues. For suppose A is of LD is corresponding eigenfunction, (the norm is aa corresponding eigenfunction, normalized normalized to to have have norm norm one one (the norm is is derived derived from ||w|| = Then the inner inner product: product: Ilull = ^(u,u)}. v(u,u)). Then from the A = A(U,U) = (AU,U) = (LDu,u) = -lLlUU = l V'u· V'u,
the last from Green's Green's first first identity identity and the fact that u vanishes with the with last step step following following from and the fact that vanishes on the the boundary boundary of of n. Since on f). Since V'u·
V'u = IIV'ul1 2 :2: 0,
this certainly shows > O. 0. Moreover, that A A :2: Moreover, this certainly shows that
illV'u W=
0
340
Chapter 8. Problems dimensions Problems in multiple spatial spatial dimensions
only only if if \7u Vw is is identically identically equal equal to to the the zero zero vector. vector. But But this this holds holds only only if Uu is is aa constant constant function, function, and, and, due due to to the boundary boundary conditions, conditions, the the only only constant constant function in C1(n) Cpfil) is is the zero function. function. By By assumption assumption Ilull ||w|| = = 1, 1, so so uu is is not the zero zero function. function. the zero not the in Thus Thus we we see see that, in in fact, fact, A A > 00 must hold. hold. This This proof proof did did not not use use the the particular particular form form of of n, fJ, and and the the result result is is therefore therefore true true for for nonrectangular domains. domains. Thus Thus we wish wish to find all all positive values values of of A such such that that the the BVP BVP -Llu = AU, x E n, u = 0, x E an
(8.11)
has now faced has aa nonzero nonzero solution. solution. We We are are now faced with with aa PDE PDE for for which which we we have have no no general general solution time-honored approach: solution techniques. techniques. We We fall fall back back on on aa time-honored approach: make make an an inspired inspired guess guess as as to the the general general form form of of the solution, solution, substitute substitute into into the the equation, equation, and and try try to to determine determine specific specific solutions. solutions. We will look look for for solutions solutions of of the form form
that is, functions functions of two variables that can be written as the product of two funcfunctions, tions, each each with with only only one one independent independent variable. variable. Such Such functions functions are are called called separated, separated, and this technique variables. and this technique is is called called the the method method of of separation separation of of variables. We therefore suppose sunnose that that u(x) w,(x1 = = U1 wi (XdU2(X2) (x-i }u^(x^} satisfies We therefore satisfies -Llu = AU, x E
n.
We We have have
so the PDE becomes
Dividing through by U2 yields yields Dividing through by U1 u\ui -1 ~U1 -U 1 - d 2 Xl
-1 U2
~U2 -d 2 = A,
(X1,X2)
X2
En.
If rewrite this this as If we we rewrite as
(8.12) we obtain the conclusion that that both both we obtain the conclusion -1 U1
d2 u1 -d 2 and Xl
-1 U2
~U2
_d 2 X2
341 341
rectangular domain 8.2. Fourier series on a rectangular
must must be be constant constant functions. functions. For For on on the the left left side side of of (8.12) (8.12) is is aa function function depending depending only Xl, and X2. If we differentiate differentiate only on on x\, and on on the the right right is is aa function function depending depending only only on on X2. If we with respect to to Xl, 0:1, we we see see that the derivative derivative of of the the first function must be zero, that the first function must be zero, with respect that the the function be constant. function and and hence hence that function itself itself must must be constant. Similarly, Similarly, the the second second function be constant. must be must constant. We have therefore shown the the following: If A is positive and and -~u
= AU, x E
n
X2) = Ul (Xl)U2(X2), then u\ Ul and U2 has a solution solution of the form U(Xl, w(#i,£2) = ^(x^u^x^)-, u% satisfy 2 -1 ~Ul -1 d u2 -d 2 = (h, Xl
-U l
-U 2
-d 2 = X2
()2,
where ()1 ()2 = A. These where 9\ + 62 = A. These we we can can rewrite rewrite as as 2 d u1 --d 2 = ()l U l, Xl
and so so we we obtain obtain ODEs ODEs for for U1 u\ and and U2. u^. and Moreover, find boundary boundary conditions Moreover, we we can can easily easily find conditions for for the the ODEs, ODEs, since since the the boundary conditions conditions on on the PDE also also separate. separate. The The boundary of n 0 consists consists of of four boundary the PDE boundary of four line line segments segments (see (seeFigure Figure 8.3): 8.3): f1 =
{(X1,X2)
X2 = 0, 0::; Xl ::; £t} (bottom),
f2 =
{(X1,X2)
f3 =
{(X1,X2)
0::; X2 ::; £2} (right), X2 = £2, 0::; Xl ::; £1} (top), Xl = 0, 0::; X2 ::; £2} (left). Xl =
f4 = {(X1,X2)
£1,
(8.13)
On On fFI1,, for for example, example, we we have have U = 0 =} Ul(Xt}U2(0)
= 0 =} U2(0) = O.
(There is also the possibility (xt) == but then then U zero function, and (There is also the possibility that that U1 u\(x\) = 0, 0, but u is is the the zero function, and we obtain we are are only only interested interested in nontrivial nontrivial solutions.) solutions.) By similar similar reasoning, reasoning, we obtain U1(0)
= u1(£d = 0,
U2(0)
= U2(£2) = o.
Our reduces to to finding finding nonzero to the the BVPs Our problem problem now now reduces nonzero solutions solutions to BVPs ~Ul
--d 2 = ()lUl, 0 Xl
< Xl < £1, (8.14)
U1(0) = 0, U1(£1) =0
and and ~U2
--d 2 = X2
U2(0)
()2 U 2,
= 0,
U2(£2) = 0,
0
< X2 < £2, (8.15)
Chapter Chapter 8. 8. Problems Problems in in multiple multiple spatial spatial dimensions dimensions
342 342
r,
Figure Figure 8.3. 8.3. The domain n fJ and its boundary. boundary.
where (h 9\ and (h #2 can be any real numbers adding to A. A. But we we have already solved these these problems problems (in (in Section Section 5.2). 5.2). For For (8.14), (8.14), the the permissible permissible values values of of (h B\ are are
and and the the corresponding corresponding eigenfunctions eigenfunctions are are . 'ljJ1(m) (Xl ) -_ sm
For (8.15), we have
(m1l' XI) , m -_ 1,2,3, .... -------;;;--
(n) _ n 21l'2 (}2
-
e~
,n=
1,2,3, ... ,
and the corresponding eigenfunctions eigenfunctions are
_ . (n1l'X2) _ 'ljJ2(m)( X2 ) -sm T ' n-1,2,3, .... we Any solution to (8.14) times any solution to (8.15) forms a solution to (8.11), so we obtained a doubly indexed sequence of solutions to (8.11): obtained doublv
m21l'2
n 21l'2 Amn=-r+T' I
'ljJmn(XI,X2)
(8.16)
2
= sin (m;IXI) sin (n;:2),
m,n = 1,2,3, ....
We have succeeded in in computing computing all all the the eigenvalue-eigenfunction eigenvalue-eigenfunction pairs pairs of of LD LD We have succeeded in which which the eigenfunctions eigenfunctions are separated. We We have no guarantee guarantee (at least, not from from our analysis so far) far) that there are not other eigenpairs with nonseparated nonseparated eigenfunceigenfunctions. particularly important, tions. However, However, it it turns turns out out that that this this question question is is not not particularly important, because because
8.2. Fourier Fourier series on a rectangular domain
343
sufficient for our we can can show that the eigenpairs (8.16) are sufficient our purposes. Indeed, Indeed, it is difficult to not difficult to argue argue that that every every function Uu £E CeO) C(ft) should should be representable as
where the convergence is in the mean-square sense. We will call such a series a Fourier (double) sine sine series. We consider any U necessarily satisfysatisfyFourier (double) series. We consider any u = U(Xl,X2) u(xi,x<2) (not (not necessarily ing the the Dirichlet Dirichlet boundary boundary conditions). Regarding x% X2 €E (0,^) (0'£2) as as aa parameter, parameter, we we ing conditions). Regarding can write can write
where bm (X2) =
£~ 1£1 U(Xl,X2) sin (m~Xl) dXl.
X2 E £2) that can be expanded in a Fourier sine But now bmm is a function of #2 € (0, (0,^2) corioc as QC iirollseries well:
where
amn =
~ 1£2 bm (X2) sin (n;:2 )
dX2.
Putting these these results results together, together, we we obtain obtain Putting (8.17) with with (8.18) Equation (8.17) is valid in the mean-square sense. Equation formulas (8.17), (8.18) It is is straightforward to show that formulas (8.18) are are exactly what the the that is, is, projection theorem theorem produces; produces; that projection
~~
. (m7rXl) . (n7rX2) --y;- sm T
~l ~ amn sm
is the best approximation L22 norm, from the subspace spanned by approximation to u, in the L {~mn:
(see Exercise 6).
m=1,2, ... ,M, n=1,2, ... ,N}
344
Chapter 8. Problems Problems in multiple spatial dimensions dimensions
Example Let n R22 :: 0 < xi Xl < 1, let Example 8.2. Let Q == {(X1,X2) {(xi,x2) E eR 1, 0 < xX22 < I}, l}, and and let
Then Then 1
amn
= 410 10
1 X1X2 sin (mnx1) sin (nnx2)
4(-1)m(-1)n mnn 2 In Figure Figure 8.4, gmph the partial series series approximation to u, M = = 20 20 In 8.4, we we graph the resulting resulting partial approximation to u, using using M and N = 20 20 (for (for aa total total of of 400 400 terms). terms). Gibbs's Gibbs's phenomenon is clearly clearly visible, visible, as and N = phenomenon is as would be be expected expected since only satisfies satisfies the the Dirichlet boundary conditions conditions on on part since uu only Dirichlet boundary part of of would the boundary. the boundary.
1.5
1
0.5
o 1 1
o
0
Figure A partial partial Fourier Fourier (double) sine series series (400 Figure 8.4. 8.4. A (double) sine (400 terms) terms) approxiapproximating mating U(X1,X2) ll(xi,X2) = = X1X2. XiX-2.
8.2. Fourier Fourier series series on a rectangular rectangular domain domain
8.2.2 8.2.2
345
Solving a boundary value problem Solving a boundary value problem
It no more difficult to to apply apply the the Fourier Fourier series method to BVP in in two two or It is is no more difficult series method to aa BVP or three three dimensions that we we know dimensions than than it it was was in in one one dimension, dimension, provided, provided, of of course, course, that know the the eigenvalues and eigenfunctions.
Example following BVP: Example 8.3. 8.3. We We will will solve solve the the following -~U U
n, = 0 on an,
= 1 in
(8.19)
where n £1 is is the the unit unit square: where square:
We first first write constant function = 11 as We have We write the the constant junction /(x) f(x) = as aa Fourier Fourier sine sine series. series. We have 00
00
1 = L L Cmn sin (m1fxd sin (n1fX 2), m=l n=l
where where Cmn
=
411 11
sin (m1fxd sin (n1fX2) dX2 dX1
4((-1)n(-1)m - (_l)n - (_l)m mn1f 2
+ 1)
We next next write the solution u as 00
U(X1,X2)
00
=L
L a mn sin (m1fX 1) sin (n1fX2). m=ln=l
straightforward to satisfies homogeneous It is is straightforward to show that, since Uu satisfies homogeneous Dirichlet conditions, conditions, 00
00
-~U(X1,X2) = L LAmnamnsin(m1fxdsin(n1fx2), m=l n=l
where where Amn
= (m 2 +n2)1f2,
m,n
= 1,2,3, ...
(see Exercise The PDE therefore implies implies that that (see Exercise 8). The PDE therefore 00
00
00
00
L L Amnamn sin (m1fxd sin (n1fX 2) = L L Cmn sin (m1fxd sin (n1fX 2), m=l n=l m=l n=l
and therefore and therefore Amnamn = Cmn ' m, n = 1,2,3, ....
Chapter dimensions Chapter 8. 8. Problems Problems in in multiple multiple spatial spatial dimensions
346 This in turn implies that
We can approximate the solution by a partial Fourier series of of the form M
N
L La
mn
sin (m1rxt) sin (n1rX2).
m=l n=l
In Figure 8.5, we graph the partial Fourier series with M = — 10, 10, N = 10.
1
o
0
x,
BVP (8.19), approximated Figure 8.5. 8.5. The solution to the BVP approximated by a Fourier series with 100 terms. terms.
8.2.3 8.2.3
Time-dependent problems Time-dependent problems
It is also straightforward straightforward to extend extend the one-dimensional one-dimensional Fourier series techniques techniques for time-dependent time-dependent problems to two- or three-dimensional three-dimensional problems, again again assuming assuming that eigenvalues eigenvalues and eigenfunctions are known. We We illustrate with an example.
Example 8.4. We We will solve the the following equation on on the the unit unit Example 8.4. will solve following IBVP IBVP for for the the wave wave equation square:
cPu
{)t 2 -
c2 du = 0, x E
n, t > 0,
347
8.2. Fourier series on a rectangular domain
u(x,O)
= 0,
x
E
au at (x, 0) = ,,((x),
n,
(8.20)
x E n,
u(x, t) = 0, x E an, t > O.
The initial velocity function will be be the function velocity function be taken to be function "(
(x) =
2
- 1,
{
5
< Xl <
otherwise.
0,
3
2
5' 5
< X2 <
3
5'
We take c = = 261V5. 261.;2. This IBVP IBVP models a (square) square hammer. (square) drum struck by a square We write the solution u as 00
00
U(XI,X2,t) = L m=l
where
1
amn(t)
= 410 10
Lamn(t)sin(m7fxdsin(n7fX2), n=1
1 U(Xl>X2,t) sin (m7fXI) sin (n7fX2) dX2 dx 1.
The PDE PDE then takes the form
~ ~ { ~~r;n (t) + c Amnamn(t)} sin (m7fX1) sin (n7fX2) = 0, 2
2 2 Amn = (m 2 + )7f2.. The initial conditions become where X +nn22)7r mn = 00
U(X1,X2,0)
00
=L
L amn(O) sin (m7fX 1) sin (n7fX2)
=0
m=1 n=l and
We have
00
"((X1,X2)
=L
00
L bmnsin (m7fxt} sin (n7fX2),
m=l n=l
where 1
bmn =
410 10
1 "((X1,X2) sin (m7fxt) sin (n7fX2) dX2 dX1
= -4 [3/5 [3/5 sin (m7fxd sin (n7fX 2) dX2 dx 1
12/5 12/5
4(cos (3n7f /5) - cos (2n7f /5)(cos (3m7f /5) - cos (2m7f /5)) mn7f 2
Chapter 8. Problems spatial dimensions Problems in multiple spatial dimensions
348
Putting together PDE and conditions, we we obtain Putting together the the PDE and the the initial initial conditions, obtain the the following following of IVPs: sequence of IVPs:
m, n — 1,2,3,. 1,2,3, — The solutions m, n= ... The solutions are are
amn(t)
= dmn sin (CVm2 + n 2 7ft) ,
m, n
= 1,2,3, ... ,
where where
We graph selected snapshots of u in Figure 8.6.
8.2.4 8.2.4
Other boundary conditions for the rectangle
Using separation of variables, variables, it find the the eigenvalues eigenvalues and and eigeneigenUsing separation of it is is straightforward straightforward to to find functions for for the Laplacian, on on aa rectangle, conditions or the negative negative Laplacian, rectangle, under under Neumann Neumann conditions or functions some combination combination of Dirichlet and and Neumann conditions. For example, if the some of Dirichlet Neumann conditions. For example, if £) n isis the rectangle (8.10) and and an dtl = with the the F; as in in (8.13), (8.13), we rectangle (8.10) = TI r 1UU Tr 22 UU Tr33 UU rT44,, with r i defined defined as we can consider the the following following IBVP: can consider IBVP: au pc at - K,Llu = f(x, t), x E
n, t> 0, u(x,O) = 'I/J(x), x E n, u(x, t) = 0, x E rl, t> 0,
au an (x, t) = 0, x E r 2 , t > 0, au an (x, t)
= 0, x
E
(8.21 )
r 3 , t > 0,
u(x,t) =0, xEr 4 , t>O.
By the method method of of variables, we can can determine By applying applying the of separation separation of variables, we determine the the appropriate Fourier series to represent the solution: appropriate Fourier series to represent the solution:
(see Exercise 14). The The determination determination of of the the coefficients amn(t) follows follows the the nowcoefficients amn(t) now(see Exercise 14). familiar pattern. familiar pattern.
8.2. Fourier series 8.2. Fourier series on on a a rectangular rectangular domain domain
X 10-
349
4
X 10-4
2 0
0
-2
-1 1
1
1
0 0
0 0
X 10-4
X 10-4
1
2
0
0
-1 1
-2 1
1
0 0
1
0 0
Figure 8.6. 8.6. Snapshots Snapshots of the vibrating vibrating membrane of Example Example 8.4: Figure of the membrane of 8.4'- tt == 5 .• 101CT44 (upper (upper left), left), t == 101(T33 (upper (upper right), t == 2· 2 • 101(T33 (lower left), left), t == 3· 3 • 101(T33 (lower right). right). (lower
8.2.5 8.2.5
Neumann boundary boundary conditions Neumann conditions
Neumann conditions are more difficult handle than Dirichlet or or mixed mixed Neumann conditions are slightly slightly more difficult to to handle than Dirichlet boundary conditions. This simply because conditions lead boundary conditions. This is is simply because Neumann Neumann conditions lead to to two two difdifferent constant function function 1, the cosines cosines of ferent kinds kinds of of eigenfunctions, eigenfunctions, the the constant 1, and and the of increasing increasing frequencies. To To be be specific, the reader reader will will recall recall from from Section 6.2 that that the the eigenpairs eigenpairs frequencies. specific, the Section 6.2 of the second derivative operator, under conditions, are are of the negative negative second derivative operator, under Neumann Neumann conditions,
Ao = 0, ro(x) = 1, 2
An
2
n 7r =T '
rn(X)
= cos (n7rx) -e- , n =
1,2,3, ....
Just Just as as in in the the case case of of Dirichlet Dirichlet or or mixed mixed boundary boundary conditions, conditions, the the eigenfunctions eigenfunctions of Laplacian, under and on on the rectangle, are are of the the negative negative Laplacian, under Neumann Neumann conditions conditions and the rectangle, products of the one-dimensional one-dimensional eigenfunctions. eigenfunctions. Since there are are two two formulas formulas for for products of the Since there the one-dimensional one-dimensional eigenfunctions, eigenfunctions, this this leads leads to to four four different different formulas formulas for the for the the
Chapter 8. 8. Problems Problems in in multiple spatial dimensions Chapter multiple spatial dimensions
350
two-dimensional (see Exercise 12): two-dimensional eigenfunctions eigenfunctions (see 12): Aoo = 0, ')'oo(x) = 1, (8.22)
By the the same same reasoning reasoning given given in in Section Section 8.2.1, 8.2.1, this this collection collection of of eigenfunction eigenfunction should By should representation is be sufficient sufficient to represent any function in C(IT)j (7(fi); the series representation
m7rXl
+ fl~amnCOS l ; 0000
( )
cos
(
n7rX2 )
T
'
where where 11
aoo = £:£21 amo =
aO a
112 f(Xl,X2) dX2 dx 1,
£~2 foil fol2 f(Xl, X2) cos (m~Xl )
dX2 dx 1,
n= £12£2 foil fol2 f(Xl, X2) cos (n;:2 ) dX2 dXl,
mn = £1~2 foil fol2 f(Xl, X2) cos (m~Xl ) cos (n;:2 )
dX2 dXl.
A steady-state steady-state Neumann Neumann problem imposes aa compatibility compatibility condition condition on on the the A problem imposes right-hand side of of the the PDE, PDE, as as might expected from from the the existence existence of of aa zero right-hand side might be be expected zero picrpnva.liip Suppose Snnnnsp u 11. is is a a. solution solution of nf t.hp RVP eigenvalue. the BVP
-.:lu au
= f(x)
an =
0 on
in
n,
(8.23)
an.
The compatibility follows from from the the divergence theorem: The compatibility condition condition follows divergence theorem:
f f =- f
Jo
Jo
.:lu
=- f
Jo
V' . V'u
=- f
Jao
If f/ satisfies satisfies the the compatibility compatibility condition condition If
If=O,
V'u. n
=_ f
au
Jao an
= o.
351 351
8.2. Fourier Fourier series series on on a a rectangular rectangular domain domain 8.2.
then Neumann problem has infinitely not then the the Neumann problem (8.23) (8.23) has infinitely many many solutions, solutions, while while if if f/ does does not satisfy there is satisfy the the compatibility compatibility condition, condition, then then there is no no solution. solution.
Example 8.5. We We will the Neumann Example 8.5. will solve solve the Neumann problem problem 1
2
-~U=XIX2-6
au
an =
(8.24)
inn,
an,
0 on
square. We write the solution as where n fi is the unit square. 00
00
+L
U(Xl' X2) = aOO
amo cos (m1rxt)
m=l 00
+L
aO n cos (n1rX 2)
n=l
00
+L
L a mn cos (m1rxd cos (n1rX 2).
m=l n=l
Then Then 00
-~U(Xl' X2)
00
=L
2 2 m 1r a mOcos (m1rxt)
+L
m=l
n21r2aOn cos (n1rX2)
n=l
00
00
+L
2:= (m21r2 + n 21r2) amn cos (m1rxd cos (n1rX2).
m=l n=l
We also have 00
f(Xl, X2)
00
= Coo + L
Cmo cos (m1rX l)
m=l 00
+L
+ 2:= COn cos (n1rX2) n=l
00
L Cmn cos (m1rxd cos (n1rX 2) ,
m=ln=l
where COO
= 0,
2(-1)m Cmo= COn = Cmn
22' m1r
2((-1)n-1) 3n1r 2 2 ' 8 (( -1) m+n + (-1) m+1 )
=
2
2
mn1r
4
.
The The reader reader should should notice that that COO CQO = = 00 is is the the compatibility compatibility condition, condition, and and so so there are infinitely for -~U infinitely many solutions. Equating the series for —Aw and f yields CmO
amo
= ----:!2' m1r
aO n =
Con 22' n1r
352
Chapter 8. 8. Problems in multiple Chapter Problems in multiple spatial spatial dimensions dimensions
withttQQ aoo undetermined. The graph of of u, with aoo == 0, 0, is shown shown in Figure 8.7. 8.7.
0.06 0.04 0.02
o - 0.02 - 0.04 1
o 0
8.7. The solution to the BVP BVP in Example 8.5. This graph was Figure 8.7. produced using a total of of 120 terms of (double) Fourier cosine cosine series. produced of the (double)
8.2.6 8.2.6
Dirichlet and Neumann problems for for Laplace's Dirichlet and Neumann problems Laplace's equation equation
The PDE
°
(8.25) -~U = in n is called called Laplace's Laplace's equation. equation. This This equation, equation, with with inhomogeneous inhomogeneous boundary boundary condicondiis tions, commonly encountered encountered in applications. For example, aa steady-state steady-state heat tions, is is commonly in applications. For example, heat flow source, in in aa homogeneous Laplace's flow problem problem with with no no heat heat source, homogeneous domain, domain, leads leads to to Laplace's equation, equation, which which can can be be paired paired with with inhomogeneous inhomogeneous Dirichlet Dirichlet or or Neumann Neumann conconditions. Dirichlet conditions conditions indicate that the the temperature temperature is is fixed fixed on on the ditions. The The Dirichlet indicate that the that the the heat heat flux flux is boundary, while while the the Neumann Neumann conditions conditions indicate boundary, indicate that is prescribed. prescribed. As another example, stretched on As another example, if if aa membrane membrane is is stretched on aa frame frame described described by by aa curve ( x , y , g ( x i , X 2 ) ) , (xi,x
U
°
in n, = 9 on an.
-~U =
(8.26)
If we solve (8.26) of Fourier series, then desirable If we wish wish to to solve (8.26) using using the the method method of Fourier series, then it it is is desirable to shift the data data to to obtain obtain homogeneous homogeneous boundary boundary conditions. conditions. The The reader reader should should to shift the
353
8.2. Fourier series on a rectangular domain
recall the method of shifting the data from Section 5.3: Given a smooth function function pp defined on ft satisfying p(xi,#2) P(X1,X2) == 5(^1,^2) g(X1,X2) on an, 5ft, we we define vv == uu -— p, where uu satisfies satisfies (8.26). (8.26). Then, Then, in in n, ft,
n
-Llv
= -Llu + Llp = 0 + Llp = Llp,
and, on an, 5ft, v
=u -
p
=g-
g
= O.
Therefore, v satisfies the BVP -Llv = Llp in
v = 0 on
n,
(8.27)
an.
For aa general general domain domain n, ft, it it may may be be difficult difficult or or impossible impossible to to find For find (explicitly) (explicitly) smooth function function p, p, defined defined on on all all of of ft and and satisfying satisfying P(X1' p(x\,x<2) g(xi,xz] for all all aa smooth X2) == g(X1' X2) for (Xl, X2) E (rci, #2) € an. 5ft. However, However, for for such such aa general general domain, domain, it it is is probably probably impossible impossible to to find the eigenvalues and and eigenfunction eigenfunction of of the the negative negative Laplacian, Laplacian, also. also. For the simple the eigenvalues For the simple domains domains (rectangles (rectangles and and disks) disks) for for which which we we can can explicitly explicitly find the the eigenpairs eigenpairs of of the the Laplacian, we we can can also also find the function function pp explicitly. explicitly. Laplacian, find the For For the rectangle (8.10), (8.10), the boundary boundary data data g will will typically be be described described as follows: follows: (X1,X2) E r 1 , (X1,X2) E r 2 , (X1,X2) E r 3 , (X1,X2) E r 4 ,
n
where an 9ft is partitioned as in (8.13). There is a completely mechanical technique for finding pp:: n ft -+ —> R R satisfying satisfying P(XI,XZ) = g(X1,X2) g(xi,#2) on on an; 5ft; however, this technique technique P(X1,X2) = however, this finding requires which we Appendix B. the following requires aa lengthy lengthy explanation, explanation, which we relegate relegate to to Appendix B. In In the following example, example, we we use use the the results results from from that that Appendix, Appendix, which which interested interested reader reader can can consult consult for the details. Example Example 8.6. 8.6. We We assume assume that that an an elastic elastic membrane membrane of of dimensions dimensions 10 cm cm by by 15 cm cm occupies occupies the the set set
n=
{x E R2 : 0
< Xl < 10, 0 < X2 < 15},
and frame, so and the the edges edges of of the the membrane membrane are are fixed fixed to to aa frame, so that that the the vertical vertical deflection deflection of the the membrane membrane satisfies satisfies the the following boundary conditions: of following boundary conditions: ~
200' 1
2" - 30' Jc. _ ~ 10
r 1, E r 2, E r3, E r4·
(X1,X2) E X2
100'
~
(X1,X2) (X1,X2) (Xl, X2)
15 '
Then Then uu satisfies satisfies -Llu = 0 in u
n,
= g on an.
(8.28)
354 354
Chapter 8. Problems in multiple spatial dimensions
smooth function function pp satisfying satisfying pp = gg on on d£t an isis A smooth _ 20X2 - 2X1X2 - Xi(X2 - 15) P( X1,X2 ) 3000 ' and we we have have and We solve We therefore therefore solve
-b.v = f(x1,x2) in 0., v = 0 on an, where /(xi,^) f(x1,x2) = (#2 (X2 — -15)/1500, by the Fourier series series method. eigenpairs of of where 15)/1500, by the Fourier method. The The eigenpairs -b. on on 0. are —A fi are Amn =
m
2
'Jr2
100
+
2 n 'Jr2
225 ' '¢mn(X1,X2)
Ts '
(n'Jrx) = sin (m'Jrx) ~ sin
m,n = 1,2,3, ....
The Fourier coefficients coefficients of of ffare The Fourier are Cmn
= 1:0 fo10 fo15 f(x1,x2) sin (m;;l) sin =
(-I)m-l 25mn'Jr
2
'
(n~:2) dX2dxI
m,n = 1,2,3, ... ,
and so the solution v is given given by
where where
a mn
Cmn = ~,
m,n
"mn The desired solution solution u u is given by by The desired is then then given
= 1,2,3, ....
The graph graph of of uu is is shown shown in Figure 8.8. B.B. The in Figure
The Neumann problem for Laplace's equation can be handled in much the same way. way. An example is given in Appendix B (Example B.2).
8.2.7 8.2.7
Fourier series series methods methods for a rectangular rectangular box box in in three Fourier for a three dimensions dimensions
It is straightforward, straightforward, in principle, principle, to extend the Fourier series method to problems posed on a rectangular box in three dimensions. For example, ifif
0. = {(Xl, X2, X3) E R 3 : 0 < Xl
< £I, 0 < X2 < £2, 0 < X3 < £3} ,
355
rectangular domain 8.2. Fourier series on a rectangular
0.5 0.4
0.3 0.2 0.1
o 15 10
o 0
Figure 8.8. solution to BVP in in Example Example 8.6. graph was was Figure 8.8. The The solution to the the BVP 8.6. This This graph produced using the (double) Fourier sine sine series. produced using 100 100 terms terms of of the (double) Fourier series. -b. on f) n and under Dirichlet then the the eigenpairs eigenpairs of of —A Dirichlet conditions conditions are are k2~2
Akmn 1/Jkmn (Xl,
m2~2
n2~2
= T + ----er + T'
X2, X3) = sin
(k~~l )
X2 sin ( m;2 ) sin
(n~:3 ) ,
k,m,n = 1,2,3, ....
Again, Again, having determined determined these eigenpairs, it is is straightforward straightforward to to solve solve Poisson's Poisson's equation, equation, or or the the wave equation under under Dirichlet conditions. The The equation, the the heat heat equation, wave equation Dirichlet conditions. reader note, however, however, that that the the resulting are rather reader should should note, resulting solutions solutions are rather expensive expensive to to evaluate, because they are given given as triply indexed For example, example, to evaluate, however, however, because they are as triply indexed series. series. For to include the lowest dimension means means working working with include the lowest 10 10 frequencies frequencies in in each each dimension with 10 1033 = = 1000 1000 terms! Fast Fast transforms, the fast fast Fourier Fourier transform, transform, can be used used to to reduce reduce terms! transforms, such such as as the can be the cost 9.2). the cost (see (see Section Section 9.2).
Exercises 2 1. R2, 1. Let n fJ be the the unit square in R ,
n = {x E R2
0 < Xl < 1, 0 < X2 < 1} ,
and ~ R R by by and define define f/ : n fi —> f(x) = XIX2
(~ -
Xl) (1 - xd(l - X2)'
Chapter 8. Problems in in multiple multiple spatial dimensions
356 356
(a) Compute the (double) Fourier sine series (a) Compute the (double) Fourier sine series 00
00
L LC
mn
sin (m1TxI) sin (n1Tx2)
m=ln=l
of /f (that is, compute compute the the coefficients coefficients ccmn ). of (that is, mn ). (b) Graph Graph the the error error in in approximating approximating f/ by by (b) M
N
L LC
mn
sin (m1Tx l) sin (n1Tx 2)
m=l n=l
for M M =N N =2 and again for 2 and again for for M M = =N N = 5. 5. 2. Repeat Repeat the the previous previous exercise with 2. exercise with
the BVP 3. Solve Solve the 3. BVP -Llu = f(x) in u = 0 on
n,
an,
R22 and f/ is the function 1. where n fJ is the unit square in R function in Exercise 1. 4. Solve Solve the the BVP 4. BVP
n, u =0 on an,
-Llu = f(x) in
where £) n isis the the unit unit square square in in R and f/ is is the the function in Exercise Exercise 2. 2. where R22 and function in 5. Suppose Suppose that that aa square iron plate plate of of side side length length 50cm 50 cm initially initially has has constant 5. square iron constant temperature of of 55 degrees Celsius and its its edges edges are placed in an ice bath (0 degrees Celsius) (the plate is perfectly insulated insulated on the top top and and bottom). bottom). At At degrees Celsius) (the plate is perfectly on the the same same time time that that the the edges are placed placed in in the the ice ice bath, bath, we we begin begin adding the edges are adding heat energy in the the interior interior of of the the plate plate at at the the (constant) (constant) rate rate heat energy everywhere everywhere in of = 7.88g/cm 7.88 g/cm33,, cC = of 0.02W/cm 0.02W/cm33.• The The material material constants constants for for iron iron are are p p = = 0.437 and « = 0.437 J/(gK), J/(gK), and". = 0.836W/(cmK). 0.836 W/(cm K). (a) Formulate an IBVP that describes the temperature u(x, t). u(x,£). (b) u. (b) Find the (double) (double) Fourier sine sine series of u.
(c) Find the Fourier series of the steady-state steady-state temperature us. ua. What BVP does Us does us satisfy? satisfy? (d) (d) How How close close is is u u to to Us us after after 10 10 minutes? minutes? Graph Graph the the difference difference in in the the two two temperature distributions. distributions.
8.2. Fourier series on a rectangular domain domain
357
6. Let u E be given, define FMN FMN to to be be the the subspace subspace of of C(f1) 6. Let € C(f1) C(fJ) be given, and and define C(f)) spanned spanned by {~mn : m = 1,2, ... , M, n = 1,2, ... , N},
where
~mn(Xl' X2) =
sin
(m~Xl
)
sin (n;:2 ).
Show that the best approximation L22 norm) from FMN FMN is given approximation to u (in the I/ by (8.17), (8.17), (8.18). (8.18). 7. the iron iron plate plate of Exercise 5. Suppose that that plate plate initially has constant 7. Consider Consider the of Exercise 5. Suppose initially has constant temperature 2 2 degrees 0 the the edges edges are are instantly temperature degrees Celsius Celsius and and that that at at time time tt = 0 instantly to temperature temperature 55 degrees degrees and and held held there. there. How How long is take take until until brought to brought long does does is
the entire entire plate plate has has temperature temperature at degrees? (Hint: deal with the at least least 4 degrees? (Hint: To To deal with the the inhomogeneous boundary condition, just shift shift the the data. data. It is trivial in in this this inhomogeneous boundary condition, just is trivial is constant.) case, because because the the boundary boundary data case, data is constant.) n is the rectangle 8. Suppose fi
u is differentiable function and that that u is aa twice-continuously differentiate function defined defined on n. fL Let the Fourier sine sine series series of of u be be
Suppose u u satisfies u(x) u(x) = 0 for x e E dfl. an. Prove Prove that the Fourier Fourier series of -.6.u Suppose =0 —Aw is is
where
9. In In Example Example 8.4, does it take for for the the leading leading edge edge of of the wave to 9. 8.4, how how long long does it take the wave to reach the the boundary boundary of of 0? n? Is Is the the computed computed time time consistent consistent with with Figure Figure 8.6? reach 8.6? 10. n is a domain in R R22 or R R33 with a piecewise smooth boundary an, 10. Suppose £) dft, and suppose suppose Oft an is is partitioned partitioned into U rF2Define . and into 22 disjoint disjoint subsets: subsets: an dti, = = rFI1 U Define 2
Let L Lm be defined by L Lmu -.6.u. Show that L Lm is Let C^(fl) -t —>• C(f1) C(fi) be defined by = —Aw. Show that m :: C!(f1) mu = m is symmetric and has has only positive eigenvalues. eigenvalues.
358 358
Chapter 8.
Problems in multiple spatial dimensions
11. square drumhead the unit unit square, 11. Consider Consider aa square drumhead occupying occupying the square,
o=
{(Xl,
X2) E R2 : 0 <
Xl
< 1, 0 < X2 < I} ,
when at at rest, rest, and and suppose suppose the the drum is "plucked" so that that its its initial initial (vertical) (vertical) when drum is "plucked" so displacement is is the function displacement the (piecewise (piecewise linear) linear) function
(X1,X2) (Xl, X2) (Xl, X2) (X1,X2)
E 0 1, E O2 , E 03, E 04.
f3 = 10and Here 0 1(T44 and 01 =
{(X1,X2)
E
R2
02 = {(X1,X2) E R2
0 3 = {(X1,X2) E R2 04 =
{(X1,X2)
E R2
< !, X2 < Xl < 1 - X2} , ! < Xl < 1, 1 - Xl < X2 < xI}, ! < X2 < 1, 1 - X2 < Xl < X2} , o < Xl < !, Xl < X2 < 1 - Xl }
o < X2
(see Figure 8.9). Find the the resulting resulting vibrations the drumhead drumhead by by solving solving (see Figure 8.9). Find vibrations of of the (Pu
8t2
c ~u = 0, x E 0, t 2
-
> 0,
u(x,O) = ¢(x), x E 0,
8u
8t (x, 0) = 0, x E 0,
u(x, t) = 0, x E 80, t >
o.
12. Let ft fl be be the the rectangle rectangle (8.10). (8.10). Define 12. Let Define
civ(n) = {u
E C 2 (n)
: ~~(x)
= 0,
x E aO}.
Let LN -+ C(O) be defined defined by by L^u LNu = = —Aw. -~u. Use Use separation of Let LN :'• Civen) Cpf(ft) —> C(ft) be separation of variables to find eigenfunctions. variables to find the the eigenpairs eigenpairs of of LN LN with with separated separated eigenfunctions. 54 Suppose Suppose an an elastic elastic membrane membrane occupies occupies the the unit unit square, square, 13. 13. 54
0= {(XI,X2) ER2 : O<xI
and something pushes pushes up up at the center of the the membrane membrane (a (a pin pin or the tip and something at the center of or the tip of aa pencil, pencil, for for example). purpose of of this this question question is of example). The The purpose is to to determine determine the the resulting the membrane. solving the resulting shape shape of of the membrane. This This can can be be done done by by solving the BVP -~u
= 8(XI -
1/2)8(x2 - 1/2) in fl, u = 0 on 00.
Graph solution. Note: Fourier coefficients be Graph the the solution. Note: The The Fourier coefficients of of the the point point source source can can be found using the the sifting property of the Dirac Dirac delta delta function. found using sifting property of the function. 54 Section 4.6 4.6 or Section 5.7 is needed this 5 4The The material material on on the the delta delta function function found found in in Section or Section 5.7 is needed for for this exercise. exercise.
359
8.3. Fourier series on a disk
Figure Figure 8.9. 8.9. The partition of the unit square used in Exercise Exercise 11.
14. n be be the the rectangle rectangle (8.10), is partitioned in (8.13). 14. Let Let (7 (8.10), and and suppose suppose an d£l is partitioned as as in (8.13). Define Define
Let C^(Q) -+ —>• C(O) C(fi) be defined by —Aw. Use Use separation separation of Lm be defined by L Lmu of Let L m :: C!(O) mu = -~u. variables eigenpairs of of L with separated eigenfunctions. to find find the the eigenpairs Lm with separated eigenfunctions. variables to m 15. but assuming that the r 2and r3 are 15. Repeat Repeat Exercise Exercise 5, 5, but assuming now now that the edges edges F2 and Fa are insulated, insulated, while edges edges FI and F4 are placed in an an ice bath (an (dft is is partitioned partitioned as as in in (8.13)). (8.13)). while r 1 and r 4 are placed in ice bath (You will need need the Exercise 14.) (You will the eigenpairs eigenpairs determined determined in in Exercise 14.)
8.3 8.3
Fourier series series on on a a disk Fourier disk
We now now discuss discuss another another simple simple two-dimensional domain for for which which we can derive We two-dimensional domain we can derive Fourier methods, namely, the case Fourier series series methods, namely, the case of of aa circular circular disk. disk. We We define define
where is aa constant. constant. We We wish to develop develop Fourier Fourier series series methods for the the A > 00 is wish to methods for where A following Dirichlet Dirichlet problem: problem: following
= f(x), x E n, u(x) = 0, x E an.
-~U
(8.29)
360
Chapter 8. Problems in multiple spatial spatial dimensions dimensions
The find the the eigenvalues eigenfunctions of the Laplacian, The first first step step is is to to find eigenvalues and and eigenfunctions of the Laplacian, subject subject to to the the Dirichlet Dirichlet conditions conditions on on the circle. differential operator We have only one technique for finding eigenfunctions eigenfunctions of a differential operator in two namely separation separation of of variables. However, this this technique is not not in two variables, variables, namely variables. However, technique is very promising in rectangular rectangular coordinates, coordinates, since since the the boundary does not not very promising in boundary condition condition does separate. For this reason, we we first first change to polar coordinates. If we we define
v(r,O) =
U(Xl,X2),
where where (r,O) (r, 9) are are the polar coordinates coordinates corresponding corresponding to the rectangular rectangular coordinates coordinates (Xl,X2), then the Dirichlet condition becomes simply (#1,0:2), then the Dirichlet condition becomes simply
v(A,O) = 0, -7r::; 0 < 7r. If we apply separation of 0) = R(r)T(0), R(r)T(O), then If we of variables and write write v(r, v(r,0) then the Dirichlet Dirichlet condition is is R(A)T(O) R(A}T(9] = for all all 0, which which implies implies (if v is is nontrivial) nontrivial) that R(A] = 0. = 00 for that R(A) O. condition The The use of polar coordinates introduces introduces periodic boundary conditions conditions for for T:
T(-7r) = T(7r), dT dT dO (-7r) = dO (7r). There also aa boundary condition for at r = although it it is is not we have There is is also boundary condition for R at = 0, 0, although not one one we have We simply need that be aa finite finite number. number. seen before. We seen before. simply need that R(O) R(Q) be
B.3.1 8.3.1
coordinates The Laplacian Laplacian in in polar coordinates
Before we we can apply separation separation of variables to the PDE, we must change variables to determine the form of the (negative) Laplacian in polar coordinates. This is an exercise in in the the chain chain rule, rule, as as we we now The chain rule implies, implies, for example, exercise now explain. explain. The chain rule for example, that that OU OV or OV 00 = + . OXl or OXl 00 OXl This allows replace the derivaThis allows us us to to replace the partial partial derivative derivative with with respect respect to to x\ Xl by by partial partial derivatives with respect to new variables provided we we can can compute tives with respect to the the new variables rr and and 0, 0, provided compute
or
00
OX1' OX1· X2, we will To deal deal with with derivatives derivatives with with respect to to #2, will also also need need
ar ao OX2' aX2·
It to compute compute the needed derivatives derivatives from It is is straightforward straightforward to the needed from the the relationship relationship between rectangular and polar coordinates: between rectangular and polar coordinates: Xl
= r cos (0),
r =
VXi + x~,
X2
= r sin (0), X2
tan (0) = - . Xl
8.3. Fourier series on a disk
361 361
We have Xl
y'xi + x~
= r cos (61) = cos (61)
r
'
and, similarly, and, similarly, :r = sin (61). UX2 Also,
X2 tan(61)=Xl ::::} sec2 (61) 861 = _ X2 = _ r sin (61) 8XI xi r2 cos2 (61) 861 sin (61) ::::}----8XI r' and, and, similarly, similarly, 861 cos (61) 8X2 r formulas for future reference: We collect these formulas :r =cos(61), UXI 861 sin (61)
:r = sin (61), UX2 861 cos (61) r
We can now compute the We can now compute the Laplacian Laplacian in in polar polar coordinates. coordinates. We We have have
8u
8v 8r
8v 861
=8r -8XI - +861-8XI 8XI = cos (61) 8v _ sin (61) 8v 8r r 80' and so
2 8 2u (8 2v 8r 8 v 861) . 861 8v 8xI = cos (61) 8r2 8XI + 8618r 8XI - sm (61) 8XI 8r _ sin (61) (8 2v 8r + 8 2v 861 ) _ r cos (B) sin (B) r 8r861 8XI 8612 8XI r2 2 2 2 = (l1) ( (l1) 8 v _ sin (61) 8 v) sin (61) 8v cos cos 8r2 r 8618r + r 8r 2 2 _ sin (61) ( (l1) 8 v _ sin (61) 8 v) 2 sin (61) cos (61) 8v r cos 8r8B r 861 2 + r2 80 2 2 2 2 2 = cos2 (61) 8 v _ 2 sin (61) cos (B) 8 v + sin (61) 8 v + sin (B) 2 r 8r2 r 808r r2 861 2 sin (B) cos (61) 8v + r2 861·
It -
Q
'*
8v 8B
Q
Q
8v 8r
362 362
Chapter Chapter 8. Problems in multiple spatial dimensions
A similar similar calculation calculation shows shows that that A (Pu = 8x~
.
2 2 (0) 8 2v 2 sin (0) cos (0) 8 2v cos 2 (0) 8v + - - + cos (0) 8 v2 + -----'--'8r2 r a08r r2 80 r 8r 2 sin (0) cos (0) 8v r2 80'
SIn
2
Adding results and and using using the sin22 (9) = 1, 1, we we obtain obtain Adding these these two two results the identity identity cos cos22 (0) (0) + sin (0) =
8 2v
-Apv
= - 8r2
1 8 2v 18v - r2 802 - -:;. 8r'
where we — Ap for for the in polar polar coordinates. coordinates. where we write write -Ap the Laplacian Laplacian in
8.3.2 8.3.2
Separation Separation of of variables variables in in polar polar coordinates coordinates
We now apply separation separation of of variables, and look look for for eigenfunctions eigenfunctions of —A p (under (under We now apply variables, and of -Ap Dirichlet Dirichlet conditions) conditions) of of the the form form
v(r, 0) = R(r)T(O). That of That is, is, we we wish wish to find all all solutions solutions of
-Apv = Av in n, v = 0 on an
(8.30)
that have the the form form v(r,9) Substituting the separated function function v == RT RT that have v(r,O) == R(r}T(0). R(r)T(O). Substituting the separated into into the PDE PDE yields the following: following:
-Apv = Av 8 2v 1 8 2v 18v ::::}--------=Av 8r2 r2 80 2 r 8r d2R 1 d2T 1 dR ::::} - - T - -R- --T= ART 2 dr r2 d0 2 r dr 2 ::::} _R-1 d R _ ~T-I d2T _ ~R-I dR = A dr2 r2 d0 2 r dr 2 -1 d2T 2 -1 2 d R -1 dR ::::} - T d0 2 = Ar + R r dr 2 + R r dr' Since the left left side of this this equation while the Since the side of equation is is aa function function of of 00 alone, alone, while the right right side side is is aa function We will write function of rr alone, both both sides sides must must be constant. We write
or or
363 363
8.3. Fourier series on a disk
We then have or or
eF R + ~ dR + R = o. dr 2 r dr r2 This last equation equation can be written written as eigenvalue problem problem for but it is not not one one This last can be as an an eigenvalue for R, R, but it is we have studied before. before. We We will study it in detail below, below, but but first we deal with the easier eigenvalue problem problem for easier eigenvalue for T. T. As we we mentioned above, the angular variable 0 introduces periodic boundary conditions for T, so we we must solve the eigenvalue problem
(A _ 2)
eFT - d(}2
='YT, -1f
T( -1f) = T(1f), dT dT dO (-1f) = d(} (1f).
(8.31)
we saw As we saw in in Section Section 6.3, 6.3, the eigenvalues are 0 and and n22 ,, n = 1,2,3, 1,2,3, .... — The The constant constant function 11 is is an to '7Y = = 0, 0, and each positive function an eigenfunction eigenfunction corresponding corresponding to and each positive eigenvalue eigenvalue = n 22 has has two two independent independent eigenfunctions, eigenfunctions, cos cos (nO) and and sin sin (nO). (nO). '7Y =
8.3.3
Bessel's equation
We consider the We now now consider the problem problem
eFR dr 2
+ ~ dR + (A _ 2) R = r dr
0,
r2 R(O) E R, R(A) =
(8.32)
o.
We know A must be positive, positive, since the Laplacian Laplacian has positive eigenvalues eigenvalues We know that that A must be since the has only only positive under Dirichlet Dirichlet conditions under conditions (see (see Section Section 8.2.1). 8.2.1). We We can can considerably considerably simplify simplify the the analysis by by the the change change of of variables variables analysis s =
We define We define
S(s) = R
(Jx)
v>"r.
¢}
R(r) = S (VXr) ,
so tnat so that
dR(r) = V>..dS (V>..r) = V>..dS (s) dr ds ds and and
364
Chapter 8. Problems in multiple spatial dimensions
We then have
The ODE
dl S ds 2
+ ~ dS + S
ds
(1 _n2) 0 S=
S2
n. In the new variables, the BVP (8.32) becomes is called Bessel's equation of of order n.
n2)
-dl8 + -1 -d8 + ( 1 - -s2 8 ds 2 S ds
= 0,
8(0) E R,
(8.33)
S(v'XA) = O. Bessel's of Bessel's equation does not have solutions that can be expressed expressed in terms of we can use a power elementary functions. To find find a solution of Bessel's equation, we 55 series expansion of the solution. We suppose55 that (8.33) has a solution of the form form 00
00
8(s) = r Laksk = LakSk+a, ao f:. O. k=O k=O The value of a, as well as the the coefficients aO,al,a2, ... The coefficients 005^15^2, • • •,, must be determined from the differential and boundary boundary conditions. We have have the differential equation equation and conditions. We a
and so and so (8.34)
Also,
dlS ds 2 (s)
00
= L(k + a)(k + a -
1) ak sk+ a-2.
(8.35)
k=O
55
55This This is yet another another example example of an an inspired guess guess at at the form of a solution. solution. It It is is based on the the fact that fact that Euler's Euler's equation, r
r) = rO;a. has solutions solutions of the form form R( R(r) =r
2d 2R -
dr 2
+ r -dR dr
-
>'R = 0,
365
8.3. Fourier series on a disk
It to write write It is is helpful helpful to 00
00
S(s) = Laksk+a = Lak_2Sk+a-2. k=O k=2
(8.36)
Substituting (8.36), (8.34), (8.34), and and (8.35) (8.35) into into Bessel's yields Substituting (8.36), Bessel's equation equation yields 00
00
L(k k=O
1) ak sk+ a -2
+ o:)(k + 0: -
+ L(k + 0:)akSk+a-2 k=O
00
00
+ L ak_2 Sk +a - 2 - n 2 L ak sk+a - 2 = 0, k=2 k=O
or or
00
+ L {((k k=2
+ o:)(k + 0: -
1) - n 2)ak
+ ak-d Sk+ a- 2 = 0.
This equation equation implies implies that that the the coefficient power of must vanish, vanish, and we This coefficient of of each each power of s must and so so we obtain the the equations obtain equations
(0:(0: - 1) + 0: - n 2 )ao = 0, ((0: + 1)0: + (0: + 1) - n 2 )al = 0, ((k + o:)(k + 0: - 1) + (k + 0:) - n 2)ak + ak-2 = 0, k
= 2,3,4, ... ,
which simplify simplify to which to (0:
2
-
n
2
)
ao
= 0, (8.37)
((0:+1)2-n2)al=0, ((k
+ 0:)2 -
n
2
)
ak
+ ak-2 = 0,
k
= 2,3,4, ....
We have assumed assumed that 0, so so the in (8.37) (8.37) implies We have that ao / ::j:. 0, the first first equation equation in implies that that 0:= ±n.
Moreover, the condition condition that be finite finite rules out the the possibility possibility that that a 0: = Moreover, the that S(O) 5(0) be rules out = -n, —n, so we conclude that o:=n
must together with with the equation in in (8.37), (8.37), immediately must hold. hold. This, This, together the second second equation immediately implies implies that that al = 0. The third (8.37) can can be as the relation The third equation equation in in (8.37) be written written as the recursion recursion relation ak = - (k
ak-2
+ n)2
_ n2
ak-2
= - k(k + 2n)'
k = 2,3,4,....
(8.38)
366
Chapter 8. 8. Problems multiple spatial spatial dimensions Chapter Problems in in multiple dimensions
Since 0, (8.38) (8.38) implies that all coefficients are Since al a\ = = 0, implies that all of of the the odd odd coefficients are zero: zero:
a2i+1
= 0, j = 0, 1, 2, ....
On the hand, we we have have On the other other hand,
a 2 -
-
ao ao 2(2 + 2n) - - 22(n + 1)'
a2 ao a4 = - 4(4 + 2n) = 2!24(n + 1)(n + 2)' a4 ao a6 = - 6(6 + 2n) = - 3!2 6 (n + l)(n + 2)(n + 3)'
and, in in general, general, and, (-I) iao
a2i = j!22i (n
+ 1)(n + 2) ... (n + j)'
. J
= 0,1,2, ....
(8.39)
Since any any multiple multiple of of aa solution solution of of (8.33) (8.33) is is again again aa solution, solution, we as well well choose choose Since we may may as the value of possible for For this this reason, we the value of ao to to give give as as simple simple aa formula formula as as possible for a2i' a 2 j • For reason, we choose choose 1 ao = 2nn! and obtain obtain and
(-I)i . a2i = 22i+nj!(n + j)!' J = 0,1,2, ....
We obtain the the solution solution We then then obtain 00
S(s) =
(
-1)i
(S ) 2i+n
~ j!(n + j)! 2
.
This function is aa solution of order n. It usually written This function is solution of of Bessel's Bessel's equation equation of order n. It is is usually written
(-I)i
(S)2i+n
= ~ j!(n + j)! 2 00
In(s)
(8.40)
and Bessel function function of note that that the and called called the the Bessel of order order n. n. The The reader reader should should note the above above calculations are are valid for nn = = 0 as as well asn = 1,2,3, l,2,3, .... — calculations valid for well as n =
°
8.3.4 8.3.4
Properties of the the Bessel functions Properties of Bessel functions
The Bessel functions studied because because of of their applied The Bessel functions have have been been extensively extensively studied their utility utility in in applied 56 mathematics, and their their properties are well known.56 We will following properties are well known. We will need need the the following mathematics, and properties of the the Bessel Bessel functions: functions: properties of 1. If If nn > 0, 0, then = 0. 0. This This follows 1. then JIn(O) follows directly directly from from (8.40). (8.40). n (0) = 56 56Standard Standard
software packages, including including MATLAB, MATLAB, Mathematica, Mathematica, and Maple, implement Bessel software packages, and Maple, implement Bessel functions just they do elementary functions functions such such as or sine. functions just as as they do elementary as the the natural natural logarithm logarithm or sine.
367
8.3. Fourier series on a disk
2. The Bessel functions satisfy 2. The Bessel functions satisfy (8.41)
This recursion relation relation can be verified This recursion can be verified directly directly by by computing computing the the power power series series equals the of the right-hand right-hand side of the side and and simplifying simplifying it it to to show show that that it it equals the left-hand left-hand side side (see (see Exercise Exercise 10). 10).
3. If, for for aa given given value value of of a, a, JIn(aA) — 0, 0, then then 3. If, n(aA) = (8.42)
We will sketch sketch the the proof of this this result. Since IJn satisfies satisfies Bessel's We will proof of result. Since Bessel's equation, equation, we we have have 2 d Jn 1dJn ( n2) dr2 + dr + 1- r2 I n = 0,
r
and multiplying through through by yields and multiplying by rr yields
2 d Jn r dr 2 or or
!
n
-1 ( 2 2) + dJ dr + r r - n In =
[r d~n (r)]
+ r- 1 (r2 - n 2) In(r)
0,
= O.
Next, we through by 2rdJnn/dr manipulate the /dr and Next, we multiply multiply through by 2rdJ and manipulate the result: result:
2r d~n (r)! [r d~n (r)]
~!
[
(r
d~n (r)) 2] + !
+ 2 (r2 - n 2) In(r) d~n (r) = 0 [(r2 _ n 2) (In(r))2] = 2r (In (r))2 .
Integrating both both sides sides yields Integrating yields
last step, we used used the the fact that n 22 J nn(0) (0) = = 00 since or In In the the last step, we fact that since either either n = 00 or JIn(O) (0) = 0. We can now evaluate the integral = O. We can now evaluate the integral n
Chapter spatial dimensions dimensions Chapter 8. 8. Problems Problems in in multiple multiple spatial
368
by or. The The result by making making the the change change of of variables variables s8 = aT. result is is
with the last step following following from from the the assumption assumption that that JIn(aa) = 0. Finally, last step O. Finally, with the n(o:a) = applying (8.41), (8.41), we we obtain obtain (8.42). (8.42). applying 4. ... , has has an we 4. Each Each IJnn,, nn = = 0,1,2, 0,1,2,..., an infinite infinite number number of of positive positive roots, roots, which which we label no simple the roots .) label s8 n i1 < s8 n22 < < 8snn33 < < .... • • •• (There (There is is no simple formula formula for for the roots 8snm nm.) We not prove prove this this fact. fact. Figure Figure 8.10 8.10 shows the graphs Jo, Ji, J 1, and We will will not shows the graphs of of Jo, and J2. J^. 57 For here some the roots roots 57 of the Bessel Bessel functions: functions: For future future reference, reference, we we give give here some of of the of the 80m
81m
82m
83m
8.3.5 8.3.5
= 2.404825557690968,5.520078109856846,
8.65372791291017,11. 79153381314112, ... , = 3.831704472655219,7.015586669057602, 10.17346813505608,13.32368935305259, ... , = 5.135622247449045,8.41724413443633, 11.61984117182147,14.79595178232688, ... , = 6.380161894536216,9.76102306245301, 13.01520072163691,16.22346616026436, ....
(8.43)
Laplacian on on the disk The eigenfunctions of the negative laplacian
Since infinite sequence positive roots, roots, there there are are infinitely Since IJn has has an an infinite sequence of of positive infinitely many many positive positive solutions to to solutions namely, namely, \ Amn
2
snm
= A2 ' m = 1,2,3, ....
Therefore, for for each each value of n, 0,l,2,..., there are are infinitely infinitely many many solutions solutions to to Therefore, value of n, nn == 0,1,2, ... , there (8.32). For For each each value of n, n, there are two solutions to (8.31): (8.32). value of there are two independent independent solutions to (8.31): cos (nO), sin (nO). 57
These estimates estimates were computed in in Mathematica, function and 57These were computed Mathematica, using using the the BesselJ BesselJ function and the the FindFindRoot command. Root command.
369
8.3. Fourier series on a disk
- y=Jox __ . y=J (X) 1 _._ y=J (X) 2
0.5
_0.51....-----'------'--------1 10 15 o 5 x
Figure function JQ, Jo , J3\, Figure 8.10. 8.10. The The Bessel Bess el function and J^. 1 , andh. We the following -~p We therefore therefore obtain obtain the following doubly doubly indexed indexed sequence sequence of of eigenvalues eigenvalues for for — Ap (on the disk with one eigen(on the disk and and under under Dirichlet Dirichlet conditions), conditions), with one or or two two independent independent eigenfunctions functions for for each each eigenvalue: eigenvalue: 2
_ SOm (1) ( B) _ A2' 'PmO r, - Jo (Snmr) ----;r-
AmO -
_ , m -1,2,3, ... ,
2
\ (1) ( B) _ T (Snmr) Amn -_ Snm A2 ' 'Pmn r, - I n ----;r- cos (B) n , m, n -_ 1,2,3, ... ,
'P~~(r,O)
(8.44)
= I n Cn;r) sin (nO), m,n = 1,2,3, ....
Since -~p under Dirichlet we know Since — Ap is is symmetric symmetric under Dirichlet conditions, conditions, we know that that eigenfunctions eigenfunctions to distinct turns out that 'Pgh and corresponding corresponding to distinct eigenvalues eigenvalues are are orthogonal. orthogonal. It turns out that (pmn an d /2"\ 'P~~ are as aa direct shows: (Pmn are orthogonal orthogonal to to each each other other as as well, well, as direct calculation calculation shows:
('PgL'P~~) = !~IoA I n Cn;r) cos (nO)Jn Cn;r) sin (nB)r dr dO =
(1:
=0· = O.
cos (nO) sin (nO) dO)
(loA Jncn;rfrdr)
(loA I n (Sn;rf rdr)
Chapter 8. 8. Problems in multiple spatial dimensions
370 For convenience, convenience, we will write
Snm amn=T' n=0,1,2, ... , m=1,2,3, ....
(8.45)
now have aa sequence We now sequence of of eigenfunctions eigenfunctions of of --~p Ap on on n fJ (the (the disk disk of ofradius radius AA centered at the origin), and these eigenfunctions are orthogonal on n. Just in centered at the origin), and these eigenfunctions are orthogonal on £7. Just as as in the case in which which n f) was rectangle, it it is is possible possible to any function function in in Cen) C(0) the case in was aa rectangle, to represent represent any in terms of these eigenfunctions: eierenfunctions: <Xl
<Xl
m=l
<Xl
m=l n=l (8.46)
The coefficients coefficients are are determined determined by the usual usual formulas: The by the formulas:
C
mn
=
(I, ~~~) (1)
n = 0,1,2, ... , m
(1) ) ,
= 1,2,3, ... ,
( ~mn,~mn
(J,~~~)
dmn = -;(,.....:...(2-)--(2~)7")' m, n = 1,2,3, .... ~mn,~mn
Using the of the the Bessel Using the properties properties of Bessel functions functions developed developed above, above, we we can can simplify simplify 2
):
(!) (!) \ (((pmrnVmn): (2) ( ) ^\ ~~~, ~~~) andA (~~h, ~~~ (((VmniVmn)
(~~b, ~~b) =
i:
= 27f (
loA (Jo(amor))2 r dr dB
~2 (J1(a m oA))2)
= 7fA2 (J1 (a moA))2 ,
(~~~,~~~) = 1:loAcos2(nB)(Jn(amnr))2rdrdB =
(1:
2
cos (nB) dB) (loA (In(amnr))2 r dr)
7fA2
2
= -2- (In+1(a mn A)) ,
and, and, similarly, similarly, (2) (2) ) _ ( ~mn'~mn -
7fA2
2
2 (In+1(amnA)) .
We can now represent a function l(r,B) /(r, 0) as in (8.46), (8.46), where (1) ) ( I,~mn Cmo
=
7fA2 (J1(a moA))
2'
m = 1,2,3, ... ,
8.3. disk 8.3. Fourier series on a disk
2 (f, Cmn
=
n = 1,2,3, ... , m = 1,2,3, ... ,
2'
n = 1,2,3, ... , m = 1,2,3, ....
2
1fA (In+1(a mn A)) 2
dmn =
371 371
(1,
1fA2 (In+1 (a mn A))
(8.47)
Given that that software software routines routines implementing implementing the the Bessel Bessel functions functions are are almost almost as readily Given as readily the trigonometric trigonometric functions, functions, it it would would appear that these these available as routines routines for for the available as appear that formulas are are as as usable usable as the analogous analogous formulas formulas on on the the rectangle. rectangle. However, However, we we formulas as the must in mind Therefore, must keep keep in mind that that we we do do not not have have formulas formulas for for the the values values of of a a mn . Therefore, to do aa certain to to actually actually use use Bessel Bessel functions, functions, we we must must do certain amount amount of of numerical numerical work work to obtain the needed needed values values of of a a mn and and the the eigenvalues eigenvalues \),mn a~n" obtain the mn = c? mn. Example 8.7. 8.7. Consider function f E defined by Example Consider the function € C(IT) C(£l) defined
f(x) = 1 -
xi - x~,
where n fi is = 1). I ) . The corresponding function expressed expressed in polar is the unit disk (A = coordinates is g(r,O) = 1- r2. approximation to g using the eigenfunctions A p . This We wish to compute an approximation eigenfunctions of of — -~p. is aa particularly 0, is particularly simple simple example, example, since, since, for for any any nn > 0,
(g,
i: 11
rr 1-7r 10
g(r,
O)
1
=
=
(i:
=0·
(1 _ r2) In(amnr) cos (nO)r dr dO
cos (nO) dO)
(1
(1
1
(l-r2) In(amnr)rdr)
1
(1-r
2
)
In(amnr)rdr)
= 0. Similarly, Similarly,
°
(g, 0. O. It follows that g can be 00
g(r,O)
=L
cmOJo(amor).
m=1
not surprising; surprising; since g is independent 0, it only the This is not independent of 9, it is reasonable that only eigenfunctions independent eigenfunctions independent of of 09 are needed needed to to represent g. Since we must compute numerical numerical estimates of the eigenvalues \),mn, we will mn, we content eigenfunctions, those corresponding to the eigenvalues content ourselves with using 3 eigenfunctions, ),10, ),20, ),30.
Chapter spatial dimensions Chapter 8. 8. Problems Problems in in multiple multiple spatial dimensions
372
To six digits, we have SlO
== 2.40483, S20 == 5.52008, S30 == 8.65373
{see Smn s~n' (see {8.43}}. (8.43)). Since A = 1 in this example, we we have Q amn = s and \Amn = ^ mn and mn = mn58 The needed needed coefficients coefficients are are58 ClO
1
=
rr(Jl(QlO)) C20
=
C30 =
1
rr(Jl(Q20)) 1
rr(Jl(Q30))
2111"
(l
-1I"io
g(r,())Jo(QlO r )rdrd() == 1.10802,
2111" (l g(r, ())JO(Q20r)r dr d() == -0.139778, -1I"io 2111" (l g(r,())Jo(Q30 r )rdrdfJ == 0.0454765. -1I"io
of the Fourier series for We now have the first three terms of for g. The graph of 9g is shown in Figure 8.11, 8.11, while the approximation approximation is shown in Figure 8.12. 8.12. Exercise 1 approximation by using more terms. terms. asks the reader to improve the approximation
0.8 0.6 0.4 0.2
o 1
-1 -1
2 Figure 8.11. function g(r g(r,: ()) 8.11. The The function 9} = = 11 - rr2. .
8.3.6 8.3.6
Solving PDEs on a disk
Having found the the disk and under the eigenvalues eigenvalues and eigenfunctions eigenfunctions of -Ll — App on the Dirichlet we can now apply Fourier Dirichlet conditions, conditions, we Fourier series methods to solve any of the equation. familiar equations: the the Poisson Poisson equation, the the heat equation, and the wave wave equation. 58 58The were computed Mathematica, using The integrals integrals were computed in in Mathematica, using the the Integrate Integrate command. command.
373
8.3. 8.3. Fourier Fourier series on a disk
-1 -1
2 Figure approximation to to function function g(r,0] g(r, 0) = = 1I — - r r2, using three Figure 8.12. 8.12. The The approximation , using three of the Fourier series series (see Example 8.7). 8.7). terms of the Fourier (see Example
Example will solve solve (approximately) (approximately) the Example 8.8. 8.8. We We will the BVP BVP
-.:lpu
=1-
r2
u = 0 on
in
n,
(8.48)
an,
where disk. We have already already seen seen that that where n 0, is is the the unit unit disk. We have 00
1 - r2 =
L
cmOJo(amor),
m=l
where coefficients cCmO be computed as in in the the previous previous example. example. We where the the coefficients can be computed as We now now mo can write write 00
u(r,B) =
L
bmoJo(amor),
m=l
where the the coefficients coefficients b bmo are to to be be determined. determined. (Since (Since the the forcing forcing function function is is radial radial where mo are (i.e. independent independent of the solution be radial. we do (i.e. of 0), B), the solution will will also also be radial. Therefore, Therefore, we do not not need need the eigenfunctions that depend on 9.) Since each Jo(a Qr) is an eigenfunction of the eigenfunctions that depend on B.) Since each Jo(amor) is an eigenfunction of m —A p under under the boundary conditions, have -.:lp the given given boundary conditions, we we have 00
-.:lpu(r,O)
=L m=l
AmObmOJo(amor),
374 374
Chapter Chapter 8. 8. Problems Problems in in multiple multiple spatial spatial dimensions dimensions
where the eigenvalue A AmO AmO PDE implies that = a~o. #mo • Therefore, Therefore, the PDE m o is given by A mo = AmObmO = Cmo, m = 1,2,3, ... , and so the solution is
00
L
c~o Jo(amor). m=1 a mO We computed ClO, C\Q, C20, c^o, C30, c^o, alO, «iO; a20, &2Q, and o,nd a30 0,3$ in the previous example, so we see that u(r, B) =
u(r,B) = 0.191594Jo(alOr) - 0.00458719Jo(a2or)
+ 0.000607268Jo(a30r) + ....
Example Example 8.9. 8.9. We will now solve the wave equation on a disk and compare the fundamental frequency of fundamental of a circular drum to that of of a square drum. We consider the IBVP IBVP
(Pu
at 2
-
c2 Ll p u = 0, (r,B)
E
fl, t
> 0,
(8.49)
u(r,B,O) = ¢(r,B), (r,B) E fl,
au at (r, B, 0) = 'Y(r, B), u(A,B, t)
= 0,
-7r
(r, B) ~
E
(8.50)
fl,
(8.51)
B < 7r, t> 0,
where fl f) is the disk of radius A, A, centered at the origin. We write the solution as 00
u(r, B, t)
=L
amo(t)Jo(amor)
m=1 00
+L
00
L
(amn(t) cos (nB)
+ bmn(t) sin (nB)) In(amnr).
m=1 n=1
argument, Then, by the usual argument, mo aat2 u(r, B, t) - C2 Llpu(r, B, t) =;;;:1 ~ (dla 2 ) ~(t) + CAmOamO(t) 2
+ ~1 ~
(
(,pd~;n (t) +
Jo(amor)
2 C Amnamn(t))
cos (nB)
+ (,p!r;n (t) + C2 Amnbmn(t)) sin (nB)) In(amnr). The wave equation then implies the following ODEs: ,pamn
~ 2
d bmn
~
2
+C Amnamn 2
= 0, m = 1,2,3, ... ,n = 0,1,2, ... ,
+ C Amnbmn = 0, m,n = 1,2,3, ....
375
8.3. Fourier series on a disk
From obtain From the the initial initial conditions conditions for for the the PDE, PDE, we we obtain amn(O) = cmn' damn dt (0) = d mn, m = 1, 2, 3, ... , n = 0 , 1 , 2, ... , bmn(O) db mn (0) --;]i'"
= emn , = fmn,
m, n
= 1,2,3, ... ,
where (generalized) Fourier fmn where c Cmn d mn are are the the (generalized) Fourier coefficients coefficients for for , l/J, and and e emn mn', d mn,, fmn are (generalized) Fourier 7. By the results of Section Section are the the (generalized) Fourier coefficients coefficients for for 'Y. By applying applying the results of 4-7 and using the fact 4.7 and using the fact that that \Amn = o^ O:~n' we obtain obtain the the following following formulas formulas for for the the mn — nn, we coefficients solution: coefficients of of the the solution: amn(t) = Cmn cos (co:mnt)
+
mn d sin (Co:mnt) , m = 1,2,3, ... , n = 0, 1,2, ... , CO: mn
bmn(t) = e mn cos (co:mnt)
+
fmn sin (Co:mnt) , m, n = 1,2,3, .... CO: mn
The of any of these these coefficients of aw: 0,1$: The smallest smallest period period of any of coefficients is is that that of Tw=
~ = ~=
2A1l'.
Co: 10
CSOl
csodA
The corresponding is the this circular The corresponding frequency, frequency, which which is the fundamental fundamental frequency frequency of of this circular drum, drum, is is F = CSOl = ~ SOl == 0 382740~ 10 2A1l' A 21l" A. It would of radius (diameter 1A) It would be be reasonable reasonable to to compare compare aa circular circular drum drum of radius A A (diameter 2A) with with drum of of side side length length 2A. 1A. (Another (Another possibility is to to compare compare the the circular aa square square drum possibility is circular drum the same This is is Exercise fundamental drum with with aa square square drum drum of of the same area. area. This Exercise 9.) The The fundamental frequency is frequency of of such such aa square square drum drum is
The square the circular drum. The square drum drum sounds sounds aa lower lower frequency frequency than than the circular drum.
Exercises Exercises 2
1. Let 1— the 1. Let g(r,9) g(r, B) = = 1 - rr2,, as as in in Example Example 8.7. 8.7. Compute Compute the the next next three three terms terms in in the series for g. series for g. 2. Let Let f(r,9] f(r,B) = = 1Compute the the first first nine nine terms terms in in the the generalized generalized Fourier 2. I — r. Compute Fourier series for /f (that (that is, is, those those corresponding corresponding to to the the eigenvalues series for eigenvalues
Take to be be the the unit unit disk. Take n fJ to disk.
376
Chapter 8.
Problems spatial dimensions dimensions Problems in multiple spatial
3. Solve the BVP
= 1 - Vx 2 + y2 in 0, u = 0 on a~,
-~u
0 is the unit disk. Graph the the solution. (Hint: Change to polar polar coordiwhere £7 nates and use the results of the previous exercise.) 4. Let f(r,O) the first first nine nine terms terms in in the the generalized generalized Fourier Fourier series 4. Let /(r, 9} = r. Compute Compute the series for /f (that is, those those corresponding corresponding to to the the eigenvalues for (that is, eigenvalues
Take f2 0 to be the unit disk. the BVP 5. Solve the -~u u
= VX2 +y2 in 0, = 0 on a~,
0 isisthe where £7 theunit unit disk. disk. Graph Graphthe the solution. solution. (Hint: (Hint: Change Changetotopolar polar coordicoordinates and use the results of the previous exercise.) 6. Let Let /(x) f(x) = = (1 (1xi -— Xz)(xi X~)(XI + xz). X2). Convert Convert to to polar polar coordinates coordinates and and compute 6. — #1 compute the first 16 16 terms in the (generalized) Fourier series (that (that is, those corresponding to Amn n — = 0,1,2,3). Graph approximation and its 1,2,3,4, n Graph the approximation mn ,, m = 1,2,3,4, error. error. 7. Consider aa disk disk made made of of copper 8.97 g/cm 33,, c = = 0.379J/(gK), 0.379 J/(gK), K, 7. Consider copper (p = 8.97g/cm K = (cm K)), of radius 10 cm. Suppose that the temperature temperature in the disk 4.04 WI W/(cmK)), 10cm. initially 4>(r,6) ¢(r, 0) = = rcos r cos (0)(10 — - r)/5. What What is the temperature temperature distribution is initially distribution after 30 30 seconds seconds (assuming (assuming that that the the top top and and bottom bottom of the disk disk are are perfectly after of the perfectly insulated, and the edge of the disk is held at 0 degrees Celsius)?
8. Consider Consider an an iron 0.836W/(cmK)) 8. iron disk disk (p (p == 7.88 7.88 g/cm gl cm33,, cc = = 0.437 0.437 J/(gK), JI (g K), « K, = = 0.836 WI (em K)) of radius radius 15 cm. Suppose that heat heat energy energy is is added added to to the the disk disk at at the the constant of 15 cm. Suppose that constant of rate of f(r,O) = I~Or(sin (0) + eos (0)) W lem 3 , where the disk occupies the set 0= {(r,O) : r<15}.
bottom of the disk are perfectly insulated Assume that the top and bottom insulated and that degrees Celsius. the edge of the disk is held at 0 degrees
(a) steady-state temperature What is is the the steady-state temperature of of the the disk? disk? (a) What (b) disk is degrees Celsius (b) Assuming Assuming the the disk is initially initially 00 degrees Celsius throughout, throughout, how how long long does for the steady state 1%)? does it it take take for the disk disk to to reach reach steady state (to (to within within 1%)?
elements in two dimensions dimensions 8.4. Finite elements
377
9. Which Which has has aa lower fundamental frequency, square drum drum or or aa circular drum 9. lower fundamental frequency, aa square circular drum of equal area? of equal area? 10. Show Show that holds by the power power series series for for the side that (8.41) (8.41) holds by deriving deriving the the right-hand right-hand side 10. and showing that it the power power series the left-hand left-hand side. side. and showing that it simplifies simplifies to to the series for for the
8.4
Finite elements in two two dimensions dimensions
We now now turn turn our attention to finite element element methods methods for for multidimensional multidimensional problems. problems. We our attention to finite the sake we will restrict our to problems in two two spatial For the For sake of of simplicity, simplicity, we will restrict our attention attention to problems in spatial dimensions and, as one dimension, to piecewise piecewise linear linear finite finite elements. elements. dimensions and, as in in one dimension, to The The general general outline outline of of the the finite finite element element method method does does not not change change when when we we move to to two-dimensional two-dimensional space. space. The The finite finite element method is obtained via via the move element method is obtained the following following steps: steps:
1. Derive form of of the given BVP. 1. Derive the the weak weak form the given BVP. 2. Apply the method to weak form. 2. Apply the Galerkin Galerkin method to the the weak form.
3. Choose aa space space of approximating subspace subspace in 3. Choose of piecewise piecewise polynomials polynomials for for the the approximating in the method. the Galerkin Galerkin method. The first step is similar to in one one dimension; one might The first step is very very similar to the the derivation derivation in dimension; as as one might expect, the the main main difference difference is is that that integration integration by by parts parts is replaced by Green's first first expect, is replaced by Green's The Galerkin Galerkin method method is unchanged when when applied to aa two-dimensional two-dimensional identity. identity. The is unchanged applied to The most most significant two-dimensional problems problems is is problem. problem. The significant difference difference in in moving moving to to two-dimensional in defining defining the element spaces. spaces. We We must create aa mesh on the the computational computational the finite finite element must create mesh on in domain and and define on the We will restrict ourselves ourselves to to define piecewise piecewise polynomials polynomials on the mesh. mesh. We will restrict domain triangulations and and piecewise linear functions. functions. triangulations piecewise linear 8.4.1 8.4.1
The weak form of a BVP BVP in in multiple dimensions
We begin begin with with the following Dirichlet Dirichlet problem: problem: We the following
- \7 . (k(x)\7u) u(x)
= f(x), x E 0, = 0, x E ao.
(8.52)
2 3 Here 0f) is domain in there is difference in derivation. Here is aa domain in R R2 or or R R3;; there is no no difference in the the following following derivation. We space V V of of test functions by We define define the the space test functions by
V=
eben) =
{v E
e 2 (n)
: v(x) =
0,
x E
ao}.
We original PDE an arbitrary arbitrary vv E € V V and and apply apply Green's Green's We then then multiply multiply the the original PDE by by an identity: identity: -\7. (k(x)\7u)(x) = f(x), x E 0 ::::} -\7. (k(x)\7u(x))v(x) = f(x)v(x), x E 0, v E V
: : } -In
\7 . (k(x)\7u)v
=
In
fv for all v E V
Chapter 8. Problems in multiple spatial dimensions
378
=> - {
ien
=>
In
k(x)v ~u
+ {
un
k(x)\7u . \7v =
in
k(x)\7u· \7v = ( Jv for all v E V
In
in
Jv for all v E V.
The last last step from the the fact that vv vanishes The step follows follows from fact that vanishes on on an. dtl. We define define the the bilinear bilinear form We form a(·, a(-, .)•) by by a(u,v) =
In
k(x)\7u· \7v.
We then obtain form of of the BVP (8.52): (8.52): We then obtain the the weak weak form the BVP Find u E V such that a( u, v) = (f, v) for all v E V.
(8.53)
The proof that that aa solution solution of of the the weak also satisfies the original original BVP BVP follows The proof weak form form also satisfies the follows the same same pattern as in in Section Section 5.4.2 5.4.2 (see (see Exercise Exercise 6). 6). the pattern as
8.4.2 8.4.2
Galerkin's method Galerkin's method
To apply Galerkin's method, we we choose choose aa finite-dimensional finite-dimensional subspace of V To apply Galerkin's method, subspace Vn Vn of V and and ¢2,···, a basis {(1)1, {^1,^2, • • • , ¢n} n} of ofVVnn. · We then pose the the Galerkin problem: Find u E Vn such that a(u,v) = (f,v) for all v E Vn .
We as We write write the the solution solution as
(8.54)
n Vn
= LUi¢i
(8.55)
i=1
and note note that that (8.54) (8.54) is is equivalent equivalent to: to: and Find U E Vn such that a( u, ¢i) = (f, ¢i), i = 1,2, ... , n.
(8.56)
Substituting (8.55) (8.55) into into (8.56), (8.56), we we obtain the following following equations: Substituting obtain the equations:
n
=> La(¢j,¢i)Uj
= (f,¢i), i = 1,2, ... ,n
j=1
=> Ku = f. The stiffness stiffness matrix K and the load load vector ff are are defined by
Kij = a(¢j,¢i), i,j = 1,2, ... ,n, Ji = (f,¢i), i = 1,2, ... ,no The reader reader will notice that that the the derivation of the the equation Ku = exactly as in The will notice derivation of equation Ku = ff is is exactly as in Section since Galerkin's method is is described described in in aa completely completely abstract fashion. Section 5.5, 5.5, since Galerkin's method abstract fashion.
8.4. 8.4.
379
Finite Finite elements in in two dimensions dimensions
The specific specific details, details, and and the the differences differences from from the the one-dimensional one-dimensional case, case, arise arise only only The 59 when the the approximating approximating subspace when subspace Vn Vn is is chosen. chosen.59 Moreover, as in form a(·,·) a(-, •) defines defines Moreover, just just as in the the one-dimensional one-dimensional case, case, the the bilinear bilinear form an inner inner product, product, called called the the energy inner product, product, and the Galerkin Galerkin method method proan energy inner and the prothe best best approximation, approximation, in in the the energy norm, to to the the true true solution u from from the duces the duces energy norm, solution u the approximating subspace approximating subspace Vnn..
8.4.3 8.4.3
Piecewise finite elements elements in in two two dimensions dimensions Piecewise linear linear finite
The graph graph of degree polynomial polynomial in in two two variables variables is is aa plane, plane, and and three three points The of aa first first degree points it algebraically, algebraically, the the equation equation of of aa first determine aa plane. plane. Or, Or, looking looking at determine at it first degree degree polynomial is is determined determined by by three three parameters: polynomial parameters:
z = a+bx+cy. For this this reason, reason, it it is is natural natural to to discretize discretize aa two-dimensional two-dimensional domain 0 by by defining For domain fU defining aa triangulation on on O-that fi—that is, is, 0f) is is divided divided into into triangular triangular subdomains. subdomains. (If 0il is is not not aa polygonal polygonal domain, domain, then then 0f) itself itself must must be be approximated approximated by by aa polygonal polygonal domain domain at at 60 the cost of some approximation error. 60) See Figure 8.13 for examples of triangular the cost of some approximation error. ) See Figure 8.13 for examples of triangular Figure 8.14 8.14 for for the the graph graph of of aa piecewise piecewise linear meshes defined defined on various regions regions and meshes on various and Figure linear The reader reader should notice how how the the graph graph of of aa piecewise piecewise linear is function. The function. should notice linear function function is made up of of triangular "patches." "patches." 0, we we write write n for for the the number number of "free" nodes, nodes, For aa given given triangulation triangulation 7T of of il, For of "free" that is, is, nodes nodes that that do do not not lie lie on on the the boundary boundary and and hence hence do do not not correspond to aa that correspond to Dirichlet boundary boundary condition. condition. We We will will denote denote aa typical typical triangular triangular element element of of T 7 by by Dirichlet T, and and aa typical typical node node by by z. z. Let Let Vn be the the following following approximating approximating subspace Vn be subspace of of V:
Vn == {v {v E€ C(O) C(fi) :: vv isis piecewise piecewise linear linear on on 7, T, v(z) v(z) == 00 for for all all nodes nodes zz E6 aO} dfl} .. Vn To apply apply the the Galerkin Galerkin method, method, we we must must choose choose aa basis basis for This is is done done exactly To for Vnn.. This exactly as . . . ,, zZn. We number number the the free free nodes nodes of of T 7 as as zZ1, Z2, Then, since since as in in one one dimension. dimension. We i5z 2 , ... n . Then, a piecewise piecewise linear linear function function is is determined determined by by its its values values at at the the nodes nodes of of the the mesh, mesh, we define fa ~i e E Vn by the the condition condition define Vn by ,j, ( )
'l'i Zj
=
{I, 0
,
i = j, . ...t.. Z I J.
(8.57)
~i is the one-dimensional A typical i, {~1' 02, ~2' ... ~n} is is aa basis basis for Vn and and that, that, for for any any vv E6 VVnn,, Vn M
v(x)
= 2.: V(Zi)~i(X). i=1
59 59Just Just as as in Section 5.5, the symbols symbols "u" "f" has has two two meanings: meanings: u is is the the true true in Section 5.5, each each of of the "u" and and "f" solution of of the the BVP, BVP, while while u u G ER Rnn is the vector vector whose whose components components are are the the unknown unknown weights weights in in solution is the the expression expression (8.55) (8.55) for the approximate approximate solution solution Vn. Similarly, /f is is the the forcing forcing function function in in the the the for the vn. Similarly, BVP, while while ffERn the load load vector vector in in the the matrix-vector matrix-vector equation Ku = f. BVP, e Rn is is the equation Ku f. 60 60Another Another way way to to handle handle this this is is to to allow allow "triangles" with aa curved curved edge. edge. "triangles" with
380 380
Chapter 8. Problems Problems in in multiple multiple spatial dimensions Chapter
-0.5 -1~----~~~~~
-1
o
1
1r-----~~----~
1
0.5
0.5 0
-0.5
- 0.5 -1 -1
-1~----~~-----J
0
-1
1
o
1
Figure 8.13. Triangular Triangular meshes defined on on (1) (1) aa square (upper left), left), (2) (2) aa Figure 8.13. meshes defined square (upper union union of of three three squares (upper (upper right), right), (3) (3) an an annulus annulus (lower (lower left), left), and and (4) (4) aa rhombus rhombus (lower Since the polygonal, the (lower right). right). Since the annulus annulus is is not not polygonal, the triangulated triangulated domain domain is is only only an an approximation approximation to to the the original original domain. domain. It now straightforward It is now straightforward (albeit quite tedious) to assemble the stiffness stiffness matrix K and the the load load vector vector f-it f—it is is just just aa matter matter of of computing computing the the quantities quantities a( o(0j,0f) K and
Example For simplicity, Example 8.10. 8.10. For simplicity, we we take take as as our our example example the the constant constant coefficient coefficient problem problem -~U=Xl,
xEn,
u(x) = 0, x E
where n £) is is the the unit unit square: square: where
an,
(8.58)
381
8.4. Finite elements in two two dimensions dimensions 8.4. Finite elements in
0.4
0.2
o -0.2 1
-1 -1
Figure 8.14. A defined on triangulation of Figure 8.14. A piecewise piecewise linear linear function function defined on the the triangulation of the annulus Figure 8.13. 8.13. the annulus from from Figure
o
0
Figure 8.15. One One of the standard basis functions. functions. Figure 8.15. of the standard piecewise piecewise linear linear basis
382
Chapter Chapter 8. Problems in multiple spatial dimensions
We triangles, 25 nodes, and and 9 free nodes. We choose choose aa regular regular triangulation triangulation with with 32 32 triangles, 25 nodes, 9 free nodes. The shown in Figure 8.16, with the free nodes nodes labeled from 11 to The mesh mesh is is shown in Figure 8.16, with the free labeled from to 9. 9.
Figure 8.16. 8.16. The mesh for for Example Example 8.10. Figure The mesh 8.10.
We with the computation of of We begin begin with the computation
K11
= a((/h,¢>d =
k
\l¢>l' \l¢>l'
The support!H1 of of ¢>l is shown shown in Figure 8.17, which also triangles in The support? fa is in Figure 8.17, which also labels labels the the triangles in the the mesh, TI1 ,, TTI2 ,..., Ts2. This support triangles; on mesh, T support is is made made up up of of six six triangles; on each each of of these these , •.. , T 32 • This triangles, ¢>l fa has has aa different different formula. We therefore therefore compute by adding the triangles, formula. We compute KH K11 by adding up up the contributions contributions from from each each of of the the six six triangles; triangles:
kr\l¢>l . \l¢>l = inr \l¢>l' \l¢>l + hr \l¢>l' \l¢>l + hr \l¢>l' \l¢>l +
r \l¢>l . \l¢>l + irnl \l ¢>l . \l ¢>l + in2r \l¢>l . \l¢>l .
~o
On TI,1 , V0i(x) (l//i, 0), where 1/4 (observing fa changes changes along On T \l¢>l(X) = = (l/h,O), where h h = 1/4 (observing how how ¢>l along the horizontal and vertical vertical edges edges ofT ofTi1 leads leads to to this conclusion). The The area the horizontal and this conclusion). area ofTi ofT1 (and (and 2 of the other other triangles triangles in in the the mesh) mesh) is is /i /2. Thus, Thus, of all all the h 2 /2.
hi 61
\l ¢>l . \l ¢>l =
hi :2 (:2) (~2) =
=
~.
61 As As we we explained Section 5.6, the support of aa function function is set on function is is explained in in Section 5.6, the support of is the the set on which which the the function nonzero, together together with the boundary boundary of that set. nonzero, with the of that set.
8.4. two dimensions 8.4. Finite Finite elements elements in in two dimensions
383
On T T-2, V^i(x) = = (0, (0, l/h), and we obtain On I/h), and we obtain 2 , V'4>1(X)
I/h), and and On On TT33,, V'4>1(X) V^i(x) = = (-l/h, (-1/fc, l/h),
Figure support of 4>1 (Example Figure 8.17. 8.17. The The support of 4>i (Example 8.10). mesh, mesh, T TI1 ,, TT-2,..., T32 are are also also labeled. labeled. 2 , •.• ,T
It should be be easy easy to to believe, believe, from symmetry, symmetry, that that
Adding up from the six triangles, we obtain obtain Adding up the the contributions contributions from the six triangles, we
Kll =
k
V'4>1 . V'4>1 = 4.
The The triangles triangles of of the the
384
Chapter 8. Problems in multiple spatial dimensions dimensions
Since the PDE (and the the basis the same Since the PDE has has constant constant coefficients coefficients (and basis functions functions are are all all the same, up to to translation), translation), it it is is easy easy to to see that up see that Kii
= 4, i = 1,2, ... ,9.
We now now turn turn our our attention attention to to the the off-diagonal off-diagonal entries. entries. For what values values of We For what of jj ^ ? By fa, we fa, =I- ii will will Kij Kij be be nonzero nonzero? By examining examining the the support support of of (PI, we see see that that only only
We now now compute compute the the first first row row of ofK. We already already know that KH 4. ConsultConsultWe K. We know that Kll == 4. ing Figure Fiaure 8.17, 8.17, we see see that
OnT3,
\7
On T 12 12,,
\7
Thus K12 Ki2 == -1. -l. Thus The following calculations: calculations: The reader reader can can verify verify the the following
385 385
8.4. Finite elements in two dimensions
= -1,
K 15 =
1
\l¢5' \l¢l
+
Tn
= (0)
1
\l¢5' \l¢l
T12
(~2) + (0) (h;)
=0. The rest rest of are similar, and the the result result is is the stiffness The of the the calculations calculations are similar, and the following following stiffness matrix: 4 -1 0 -1 0 0 0 0 0 4 -1 -1 0 -1 0 0 0 0 4 0 0 0 0 -1 0 0 -1 0 0 4 -1 0 0 -1 0 -1 4 -1 0 -1 0 -1 0 0 -1 K= 4 0 0 -1 0 0 -1 0 -1 4 -1 0 0 0 -1 0 0 0 4 -1 0 0 0 0 -1 0 -1 4 0 0 0 0 -1 0 0 -1 To load vector vector f, must evaluate integrals To compute compute the the load f , we we must evaluate the the integrals
L
f(x)¢i(X) dx,
which is is quite quite tedious tedious when when done done by by hand hand (unlike (unlike \l¢i, V0j, ¢i 0j itself itself is not piecewise which is not piecewise constant). constant). We We just show one one representative representative calculation. calculation. We We have Xl
X
E
T1 ,
X
E
T2 ,
X
E
T3 ,
X
E
T lO ,
,
X
E
Tu ,
,
X
E
T 12 ,
h' X2 h' ¢1(Xl,X2) =
X2-X1+h h Xl-X2+h h 2h-X2 -h2h-X1 -h-
0, Therefore, Therefore,
, ,
otherwise.
386
Chapter 8. Problems dimensions Problems in multiple spatial dimensions
Continuing in in this this manner, manner, we we find Continuing find h3 2h 3
f
=
3h3 h3 2h 3 3h3 h3 2h 3
3h3 62 We now solve Ku obtain We can can now solve62 Ku = = ff to to obtain
u =
0.015904 0.027344 0.027065 0.020647 0.035156 0.034040 0.015904 0.027344 0.027065
The piecewise linear displayed in Figure 8.18. 8.18. The resulting resulting piecewise linear approximation approximation is is displayed in Figure
The calculations calculations in the previous example are are elementary, elementary, but extremely timetimeThe in the previous example but extremely consuming, and and are are ideal ideal for for implementation implementation in in aa computer computer program. When these these consuming, program. When not organized the same as calculations are programmed, programmed, the calculations are the operations operations are are not organized in in the same way way as in the above hand For example, example, instead instead of of computing one entry entry in K in the above hand calculation. calculation. For computing one in K at time, it to loop the elements the mesh to compute at aa time, it is is common common to loop over over the elements of of the mesh and and to compute all contributions to to the various entries entries in in K K and from each each element element in turn. all the the contributions the various and ff from in turn. Among other other things, this simplifies simplifies the data structure structure for for describing describing the mesh. (To Among things, this the data the mesh. (To automate know, given a automate the calculations calculations presented above, it would be necessary to know, certain node, are adjacent adjacent to it. This is avoided avoided when one loops loops over certain node, which which nodes nodes are to it. This is when one over the elements over the Also, the the various various integrations are rarely the elements instead instead of of over the nodes.) nodes.) Also, integrations are rarely carried out but rather by using rules. These carried out exactly, exactly, but rather by using quadrature quadrature rules. These implementation implementation details are are discussed discussed in in Section Section 10.l. 10.1. details As we have have discussed before, an finite element As we discussed before, an important important aspect aspect of of the the finite element method method is the the linear linear systems that must is the fact fact that that the systems that must be be solved solved are are sparse. sparse. In In Figure Figure 8.19, 8.19, we we display display the sparsity sparsity pattern of of the stiffness stiffness matrix for for Poisson's Poisson's equation, equation, as as in Example banded, that Example 8.10, 8.10, but with with a finer finer grid. The The matrix K K is banded, that is, all of its its entries the main main diagonal. entries are are zero zero except except those those found found in in aa band band around around the diagonal. Efficient Efficient solution solution of of banded banded and and other other sparse sparse systems systems is is discussed discussed in in Section Section 10.2. 10.2. 62 6 2 It It
is that aa computer computer program program will be used used to linear system is expected expected that will be to solve solve any any linear system large large than than 22 xx 22 or 33 xX 3. 3. We We used used MATLAB MATLAB to solve Ku = ff for for this this example. or to solve Ku = example.
387 387
8.4. 8.4. Finite Finite elements elements in in two two dimensions dimensions
0.01
o
0,.
0
Figure 8.18. The The piecewise linear approximation approximation to to the the solution of {8.58}. (8.58). Figure 8.18. piecewise linear solution of
10~i_. 20 30
40 50
, ••••••••
-.-. ••••
_.
e.
,
-.-. ••••
-.••••••
_.
,,
-.•••••• -...... -...... •••• •••• •.....• •.....• •••• ... ••••... ..•.... ..•.... -. -. .~... :,~~ .... ,
e.
~
70 80
o
~
,, _.
~
~.
m
~
~
00
nz = 369
Figure 8.19. sparsity pattern of Figure 8.19. The The sparsity of the the discrete discrete Laplacian {200 (200 triangular triangular elements}. elements).
Chapter in multiple multiple spatial Chapter 8. 8. Problems Problems in spatial dimensions dimensions
388
8.4.4 8.4.4
Finite elements elements and and Neumann Neumann conditions Finite conditions
We will close Neumann conditions conditions are are handled We will close by by describing describing briefly briefly how how Neumann handled in in twotwodimensional elements. For For the sake of of generality, generality, we we will consider aa problem dimensional finite finite elements. the sake will consider problem with possibly mixed mixed boundary boundary conditions. n isisaadomain with possibly conditions. So So suppose suppose £7 domaininineither eitherR2 R2 oror 3 R3, R , and assume that an dtl has been partitioned partitioned into two two disjoint sets: an d£l == rI\1 ur LJ T^. 2. We consider the BVP: We consider the following following BVP: -\7 . (k(x)\7u) = I(x), x E n,
u(x)
= 0,
au an (x) = 0,
x E r 1,
(8.59)
x E r2.
As we discussed in Section 6.5, 6.5, Dirichlet Dirichlet conditions conditions are are termed essential boundary boundary As we discussed in Section termed essential conditions because because they be explicitly in the method, conditions they must must be explicitly imposed imposed in the finite finite element element method, while Neumann conditions are are called called natural natural and and need not be mentioned. We while Neumann conditions need not be mentioned. We therefore define the space space of of test functions functions by by
v = {u E C
2
(O) : u(x)
= O,X E rt}.
The weak form of of (8.59) is is now derived exactly exactly as as on on page page 378; the the boundary boundary The weak form now derived integral integral
au = r k(X)V au + r k(x)V au r k(X)V a an Jr2 an Jon n Jrl
now vanishes = 00 on and du/dn — 00 on on rF22.. Thus now vanishes because because vv = on rFI1 and au/an = Thus the the weak weak form form of of is: (8.59) is: Find u E V such that a(u, v) = (f, v) for all v E V. (8.60) The bilinear form a(-, is defined defined exactly exactly as as before (the only from (8.53) a(·,·)•) is before (the only difference difference from The bilinear form is in in the space of of test functions). is the space test functions). polygonal doWe now now restrict restrict our our discussion discussion once once more more to to two-dimensional two-dimensional polygonal mains. To To apply apply the the finite element method, method, we must choose choose an an approximating approximating mains. finite element we must there are least two two subspace subspace of of V. Since Since the the boundary boundary conditions conditions are are mixed, mixed, there are at at least Neumann. We We will points where where the the boundary conditions conditions change change from from Dirichlet to Neumann. make the assumption assumption that is chosen chosen so so that that all all such such points are nodes nodes (and (and make the that the the mesh mesh is points are that all such such nodes nodes belong that is, that rFI1 includes its "endpoints"). "endpoints"). We We can that all belong to to rF11;, that is, that includes its can then choose the approximating subspace subspace of of V V as as follows: then choose the approximating follows:
Vn
= {v E C(O') : v is piecewise linear on /, v(z) = 0 for all nodes z E
rt}.
A basis for for Vn Vn is is formed formed by by including including all all basis basis functions functions corresponding corresponding to to interior interior A basis nodes (as case), as as well the basis basis functions nodes (as in in the the Dirichlet Dirichlet case), well as as the functions corresponding corresponding to to boundary nodes that belong to to rFI. For an an example example of function boundary nodes that do do not not belong 1. For of aa basis basis function corresponding see Figure corresponding to to aa boundary boundary node, node, see Figure 8.20. 8.20. Example Example 8.11. 8.11. We We consider consider the BVP BVP -~U=X1,
xEn,
8.4.
Finite elements elements in two dimensions
389
0.8 0.6 0.4
0.2
o 1
o
0
Figure 8.20. A standard standard piecewise piecewise linear function corresponding corresponding to to Figure 8.20. A linear basis basis function aa boundary node. boundary node. u(x) 0, x x E r 3 ,, u(x) = = 0, €T du —(x) an (x) = = o, 0, x x eE rril uUrr22uUrr44,, an
au
(8.61)
where Jl 0 is is the the unit unit square, square, as as in in the the previous previous example, example, and fl' 1^, r 2 , T$, r 3 , and r 4 are are where and TI, and T± defined as in Figure 8.3. We use the same regular triangulation of 0 as in Example defined 8.3. of ft, 05 Example 8.10. Now, however, however, the to r and F r 4 are free nodes 8.10. Now, the boundary boundary nodes nodes belonging belonging to TI,l , r22,, and are free nodes (but not the the corners corners (1,1) and and (0,1) of of OJ. It follows follows that has dimension dimension 20. (but not Ci). It that Vn Vn has 20. The free nodes are labeled labeled in in Figure Figure 8.21. 8.21. The free nodes are The calculation of of the the new new stiffness stiffness matrix matrix K K and and load load vector vector ff proceed proceed as as The calculation before, and the linear approximation to the true solution is disbefore, and the resulting resulting piecewise piecewise linear approximation to the true solution is displayed in Figure 8.22. 8.22. played in Figure If stiffness matrix will If the BVP includes only Neumann conditions, then the stiffness be singular, singular, reflecting reflecting the the fact fact that that BVP BVP either not have solution or or has has be either does does not have aa solution infinitely many solutions. Special care must be taken taken to compute a meaningful meaningful solution to Ku = = f. f. solution to Ku
8.4.5
Inhomogeneous boundary conditions conditions
In a two-dimensional problem, inhomogeneous boundary conditions are handled conditions are addressed just as in one dimension. Inhomogeneous Dirichlet Dirichlet conditions addressed via the method of shifting shifting the data (with a specially chosen piecewise linear function), while
390
Chapter 8. Problems in dimensions in multiple spatial dimensions
20
15
10
Figure Figure 8.21. 8.21. The The mesh for Example 8.11. 8.11.
0.3
0.2
0.1
o 1
Figure 8.22. Figure 8.22. The piecewise linear approximation to the solution of (8.61). (8.61).
391
8.4. Finite elements in in two two dimensions 8.4. Finite elements dimensions
inhomogeneous Neumann Neumann conditions conditions are are taken taken into into account account directly directly when when deriving inhomogeneous deriving the weak form. Both types the load the weak form. Both types of of boundary boundary conditions conditions lead lead to to aa change change in in the load vector. Exercises and 77 ask ask the reader to to derive derive the the load load vectors vectors in in these these cases. cases. vector. Exercises 55 and the reader
Exercises 1. 1. Consider the BVP
-V'. (k(x)V'u)
= I(x),
x EO,
u(x) = 0, x E a~,
where is the the unit square, k(x) k(x) == 1I + x\x^ where 0£) is unit square, X1X2 piecewise linear finite element (approximate) with 18 18 triangles.
and I(x) /(x) = = 1. 1. and solution using a
Produce the the Produce regular grid
2. Repeat Repeat Exercise 1, but but suppose suppose that that the the boundary boundary conditions conditions are are homogeneous 2. Exercise 1, homogeneous Neumann conditions conditions on on the the right right side side f2 F2 of of the square and and homogeneous Neumann the square homogeneous Dirichlet conditions conditions on the other three sides. sides. 3. Repeat Exercise 1, 1, but suppose that the boundary conditions are homogeneous Neumann conditions conditions on on every every part part of of the the boundary of O. fi. Let Let the the right-hand right-hand boundary of Neumann side be /(x) = = #1 1/2 (so (so that that the compatibility condition condition is is satisfied satisfied and and side be I(x) Xl -—1/2 the compatibility solution exists). exists). aa solution
4. If If Uu does does not belong to to V Vnn,, then then 4. not belong n W
= LU(Zi)¢i i=l
is not equal to u. Instead, linear interpolant is not equal to Instead, w w is is the the piecewise piecewise linear interpolant of of Uu (relative (relative to to the given is one the given mesh mesh T). The The interpolant interpolant of of Uu is one piecewise piecewise linear linear approximation approximation to U; to aa BVP, BVP, then then the finite element computes to u] if if Uu is is the the solution solution to the finite element method method computes another. The purpose of this exercise is to compare the two. (a) Verify X1X2(1 -— #i)(l xd(1 -— #2) X2) is the the solution to the Verify that u(x) w(x) = = #1X2(1 the Dirichlet problem problem -6.u = 2Xl(1 - Xl) u(x) = 0, x E a~,
+ 2x2(1 -
X2), x E fl,
where 0fi is is the unit square. square. (b) (b) Let Let T be be the regular triangulation of of 0£) with 18 18 triangles. Compute Compute the the finite element approximation to u using piecewise linear functions on finite element approximation to u using piecewise linear functions on T. T. (c) relative to to T. (c) Compute Compute the the piecewise piecewise linear linear interpolant interpolant of of uu relative T. (d) better approximation finite element or (d) Which Which is is aa better approximation of of u, the the finite element approximation approximation or the your answer numerically. the interpolant? interpolant? Justify Justify your answer both both theoretically theoretically and and numerically.
392 392
Chapter 8. Problems in multiple spatial dimensions
5. (a) (a) Explain Explain how how to solve an an inhomogeneous inhomogeneous Dirichlet 5. to solve Dirichlet problem problem using using piecepiecewise linear linear finite finite elements. elements. wise (b) Illustrate the procedure on the following BVP: -~u
= 0 in 0,
u(x) =
xi, x E a~.
Let be the unit square use aa regular regular grid grid with Let 0J7 be the unit square and and use with 18 18 triangles. triangles. 6. 6. Suppose uu is a solution to (8.53). (8.53). Prove that uu also solves (8.52). (8.52).
7. (a) (a) Explain Explain how to solve solve an inhomogeneous Neumann Neumann problem 7. how to an inhomogeneous problem using using piecepiecewise linear finite elements. (b) the compatibility compatibility condition Neumann (b) What What is is the condition for for an an inhomogeneous inhomogeneous Neumann problem? problem? (c) the procedure procedure on (c) Illustrate Illustrate the on the the following following BVP: BVP:
-~u = X1X2 + ~ in 0,
au
an(x)
3xi
= -5'
x E a~.
Let fi 0 be the unit square and use a regular grid with 18 18 triangles.
8.5 8.5
Suggestions for for further further reading Suggestions reading
The foundation of PDEs in two or more spatial dimensions is advanced calculus. Kaplan [29] introduction; an alternative at the same is Kaplan [29] gives gives aa straightforward straightforward introduction; an alternative at the same level level is Greenberg advanced treatment can be found in and 'Tromba Tromba Greenberg [20]. A A more more advanced treatment can be found in Marsden Marsden and [37]. All of the references references cited PDEs in All of the cited in in Sections Sections 5.8 5.8 and and 6.7 6.7 deal deal with with PDEs in multiple multiple spatial dimensions. Another source source of of information Bessel function is Folland Folland spatial dimensions. Another information about about Bessel function is [15].
Chapter 9
about f.+~,Mope iI?!ri~'~I¢.~~ bout Fourier Fourier series series
In the preceding chapters, we we introduced several kinds of Fourier series: series: the Fourier sine series, cosine series, quarter-wave sine series, quarter-wave cosine series, and the full Fourier series. These series were primarily used to represent the solution to differential differential equations, and their usefulness was based based on two facts: facts: 1. with the property that con1. Each Each is is based based on on an an orthogonal orthogonal sequence sequence with the property that every every continuous function can be represented represented in terms of this sequence. sequence.
2. The terms in the series represent eigenfunctions of certain simple differential differential operators (under various boundary conditions). conditions). This accounts for the fact that it is computationally tractable representation of of tractable to determine a series representation the solution to the corresponding differential differential equation. equation. In this chapter we we will go deeper into the study of Fourier series. Specifically, Specifically, we will consider the following questions: 1. 1. What is the relationship among the various kinds of Fourier series?
2. partial Fourier 2. How How can can aa partial Fourier series series be be found found and and evaluated evaluated efficiently? efficiently? conditions and in what sense can a function be represented by its 3. Under what conditions Fourier series? 4. the Fourier be generalized to more 4. Can Can the Fourier series series method method be generalized to more complicated complicated differential differential equations nonconstant coefficients equations (including (including nonconstant coefficients and/or and/or irregular irregular geometry)? geometry)? book Our justify many many of Our discussion discussion will will justify of the the statements statements we we made made earlier earlier in in the the book concerning the convergence of Fourier series. It also introduces the fast fast Fourier convergence transform (FFT), (FFT), which which is is an an exciting exciting and and recent recent development development (from (from the last half transform the last half 63 of the twentieth century63) of ) in the long history of Fourier series. The calculation calculation of 2 coefficients would appear appear to require O(JV O(N2) operations (for (for reasons that N Fourier coefficients ) operations 63 63The The FFT FFT was was popularized popularized in in the the 1960s, 1960s, by by the the paper paper [12]. [12]. However, However, the the method method was was known to Gauss long long before; before; see see [25]. [25]. to Gauss
393
Chapter 9. 9. More More about about Fourier Fourier series series Chapter
394
we explain explain in in Section Section 9.2). 9.2). The The FFT FFT reduces operation count count to to O(NlogN), we reduces the the operation O(NlogN), aa considerable when N considerable savings savings when TV is is large. large. We begin introducing yet yet another type of the complex complex We begin by by introducing another type of Fourier Fourier series, series, the Fourier series, series, that that is is convenient convenient both for analysis analysis and and for for expressing expressing the the FFT. FFT. both for Fourier
9.1 9.1
The complex complex Fourier Fourier series series The
We saw saw in in Chapter Chapter 44 that that problems problems apparently apparently involving involving only only real can We real numbers numbers can sometimes be addressed with with techniques techniques that that use use complex complex numbers. For exsometimes be best best addressed numbers. For example, if if the the characteristic characteristic roots and r2 r-2 of of the ODE roots r\ rl and the ODE ample, rPu
du
+ b dt + cu = 0
a dt 2
are distinct, the general general solution is are distinct, then then the solution is
u(t) =
Cl e
Tlt
+ c2e
T2t
•
This formula formula holds even if if rl r\ and and r2 r% are are complex, complex, in which case case uu is is complex-valued complex-valued This holds even in which for most most choices choices of of Cl c\ and and C2. c^. (We (We also also saw, saw, in in Section Section 4.2.1, 4.2.1, how how to to recover recover the the for general real-valued real-valued solution solution if if desired.) desired.) general In to using using complex-valued In aa similar similar way, way, there there are are some some advantages advantages to complex-valued eigenfunceigenfunctions, even even when solving differential differential equations equations involving involving only only real-valued real-valued functions. functions. tions, when solving Using Euler's formula, formula, Using Euler's eiO = cos e + i sin e, we see see that we that
d~
[eiW
=
!
+ i sin (wx)] = -wsin (wx) + iwcos (wx) = iw (cos (wx) + i sin (wx)) [cos (wx)
= iweiw
and therefore therefore and 2
, ] = -(iw) 2 ' _d_ [e'wx e'wx
dx 2
This the negative This suggests suggests that that the negative second second derivative derivative operator operator has has complex complex exponential exponential eigenfunctions. eigenfunctions. We leave to the exercises (see to show show that solutions to to We leave it it to the exercises (see Exercise Exercise 4) 4) to that nonzero nonzero solutions the BVP BVP the rPu
- dx 2
= AU,
-£ < x < £,
u( -f) = u(£), du (_£) = du (£), dx dx
(9.1)
9.1. 9.1. The complex complex Fourier series
395 395
22 2 exist 0, 1f2 2/£2, ... ,n 7r /£2, ... The has eigenfuncexist only only for for A == 0,7r /7 2 ,..., n 22?r /£ , .— The eigenvalue eigenvalue A == 00 has eigenfunction (the constant function), while each eigenvalue eigenvalue A has two two linearly linearly tion 11 (the constant function), while each A= = n 22ir7r 22/I/£22 has independent eigenfunctions, independent eigenfunctions,
2 2 constant function 1 and A = 1f2/£2 When n == 0, ei7fnx/l e™nx/1 reduces to the constant = n22?r /7 reduces we will of to A = to = O. 0. Therefore, Therefore, to to simplify simplify the notation, notation, we will write the the complete complete list of eigenvalue-eigenfunction pairs as eigenvalue-eigenfunction pairs as
2 2
n 1f i7fnx/l ,n= 0,±, 1 ±2 , .... T,e
(9.2)
In this section we use to form In this section we use the the eigenfunctions eigenfunctions given given in in (9.2) (9.2) to form Fourier Fourier series series the interval (-£, f). Since Fourier series representing functions on representing functions on the interval (—£,(.). Since Fourier series calculations calculations are are we must complex vector vector space based on based on orthogonality, orthogonality, we must take take aa slight slight detour detour to to discuss discuss complex space and products. Some the following have been been used used earlier the text, text, and inner inner products. Some of of the following results results have earlier in in the but operator has only but only only briefly in the course course of demonstrating demonstrating that aa symmetric symmetric operator real eigenvalues. real eigenvalues.
9.1.1 9.1.1
Complex inner products products
As discussed in in Section As we we discussed Section 3.1, 3.1, aa vector vector space space is is aa set set of of objects objects (vectors), (vectors), along along with two operations, and scalar multiplication. To this point, point, we we with two operations, (vector) (vector) addition addition and scalar multiplication. To this have used real real numbers as the however, the can also have used numbers as the scalars; scalars; however, the complex complex numbers numbers can also be be we use use complex numbers as as scalars, we refer used emphasis, when when we used as as scalars. scalars. For For emphasis, complex numbers scalars, we refer to the the vector vector space as aa complex vector space. to space as complex vector space. The vector space The most most common common complex complex vector space is is en, C n , complex complex n-space: n-space:
Just as as for for R addition and scalar multiplication multiplication are are defined defined componentwise. componentwise. Just R nn ,, addition and scalar The only only adjustment adjustment that must be working with complex vector vector The that must be made made in in working with complex spaces is in the inner product. If u vector spaces is in the definition definition of of inner product. If u is is aa vector vector from from aa complex complex vector space, then it is is permissible permissible to to multiply multiply uu by to get get iu. iu. But space, then it by the the imaginary imaginary unit unit ii to But ifif there inner product product (.,.) and if properties of there is is an an inner (•, •) on on the the space, space, and if the the familiar familiar properties of inner inner product hold, product hold, then then
(iu,iu) = i2(u,U) = -(u,u). This suggests suggests that that either (w,u) or or (iu,iu) is is negative, negative, contradicting contradicting one one of of the rules This either (u,u) the rules of inner inner products (and making making it impossible to define aa norm norm based inner of products (and it impossible to define based on on the the inner product). product). For this reason, reason, the inner product is modified modified for for complex complex vector For this the definition definition of of inner product is vector spaces. spaces.
Chapter 9. More More about about Fourier Fourier series series
396
Definition be a complex vector space. An An inner product on V is is aa Definition 9.1. 9.1. Let V be vectors from V and producing a complex complex number, number, in such a way function taking two vectors that the following following three properties properties are satisfied: satisfied: 1. 1. (u,V) (w,t>) = = (v,u) (v,u) for all vectors u and V; v;
2. (au + j3v, — a(u, a(u, w) (w, au au + /3v) fiv) = — a(w, a(w, u) fi(w, v) 2. (au /3v, w] w) = w) + {3(v, /3(v, w) w) and and (w, u) + 73(w, v) for for all all v, and w wand numbers a 0: and /3; vectors u, v, and all all complex numbers 0;
2: 00 for for all vectors u, u, and (u,u) = if and and only only ifu vector. 3. (u,u) (w, w) > — 00 if if u is the zero vector. In the the above above definition, definition, z denotes the the complex complex conjugate conjugate of In z denotes of z E € C. C. (If z = = x+ + iy, with with x, x,yy €E R, R, then then ~zz = = x — - iy.) For C the inner product is is suggested suggested by by the the fact that For Cnn ,, the inner product fact that
IZI=# for We define for z €E C. C. We define
n
(U,V)cn
= U· V = L
UiVi·
j=1
It the properties satisfied. It is is straightforward straightforward to to verify verify that that the properties of of an an inner inner product product are are satisfied. In particular, In particular, n
(u,u)c n = LUi'lli j=1 n
=
L
IUil 2 2: O.
j=1
If we of complex-valued functions defined interval If we now now consider consider the the space space of complex-valued functions defined on on an an interval [a, b], bj, the the same same reasoning reasoning as in Section 3.4 leads leads to to the the complex complex L L22 inner inner product, product, [a, as in Section 3.4 (J,g) =
i
b
f(x)g(x) dx.
(We will will use use the the same notation as the real real L L22 inner inner product, product, since, when /f and (We same notation as for for the since, when and g9 are real-valued, real-valued, the the presence presence of of the complex conjugate in the the formula formula has has no no effect, are the complex conjugate in effect, and the the ordinary ordinary L L22 inner product is is obtained.) and inner product obtained.)
9.1.2 9.1.2
Orthogonality of complex exponentials Orthogonality of the the complex exponentials
A direct that the (9.2) are A direct calculation calculation now now shows shows that the eigenfunctions eigenfunctions given given in in (9.2) are orthogonal orthogonal with respect respect to to the L22 inner product: with the (complex) (complex) L inner product: (ei1fnX/t, em1fix/i)
=
rt ei1fnx/£e-m1fix/£ dx
i-I
397
9.1. The complex Fourier series 9.1.
1£
=
e(n-m)1rix/£ dx
-£
= [e(n-m)1rix/£ ] £
(n - m)7ri/C
=
C
_£
(e(n-m)1Ti _ e-(n-m)1Ti)
(n - m)7ri = 0, m f. n.
n m 1 The last step follows because e^ e(n-m)Oi is 27r-periodic. 27r-periodic. We We also also have have ^ is
(ei1TnX/l,ei1rnX/l) =
1£ 1£
ei1rnx/fe-i1Tnx/fdx
-l
=
dx
-£
= 2£. This shows that that the eigenfunction is is VU. y/%1. This shows the L L22 norm norm of of the the eigenfunction
9.1.3 9.1.3
Representing functions complex Fourier Fourier series Representing functions with with complex series
If f/ is complex-valued function denned on on [[-C, — I , IC], ] , then complex If is aa continuous, continuous, complex-valued function defined then its its complex Fourier is Fourier series series is CXl
'""' cn e i1Tnx /£ , ~ n=-CXl
where where
1 = 2C
1f
.
f(x)e-'1rnx/£ dx.
-£
These complex according to theoThese complex Fourier Fourier coefficients coefficients are are computed computed according to the the projection projection theorem; rem; it it follows follows that that N
L
cnei1Tnx/£
n=-N
is element of subspace is the the element of the the subspace 1 e i1rx, /l span {e -i1rNX/£ , e-i1r(N-l)x/£ , ... , e- i1rx / L" .. "' ei1rNX/L} closest to to /f in the L L22 norm. norm. Also, Also, we we will will show show in in Section 9.6 that the complex complex closest in the Section 9.6 that the in the the L L22 norm norm under under mild mild conditions conditions on on f/ (in Fourier series Fourier series converges converges to to f/ in (in particparticular, if f/ is is continuous). continuous). ular, if
Chapter 9. More about Fourier series
398
9.1.4 9.1.4
The The complex complex Fourier Fourier series series of of a a real-valued real-valued function function
If E C[ -e, e] isis real-valued, If f/ € C[—i,i] real-valued, then, then, according according to to our our assertions assertions above, above, it it can can be represented now show represented by its its complex complex Fourier Fourier series. series. We We now show that, that, since since f/ is is real-valued, each partial sum (n = ... , N) = -N, —TV, -N — JV + 1, 1,..., AT) of the the complex Fourier series is real, and and moreover moreover that that the the complex complex Fourier Fourier series series is is equivalent equivalent to to the the full full Fourier Fourier series series in this case. So we we suppose suppose f/ is is real-valued real-valued and and that that Cn, cn, nn = 0, 0, ±1, ±1, ±2, ±2,... are its its complex complex So ... are we will write the full Fourier coefficients. For reference, we full Fourier series of f/ (see Section Section 6.3) 6.3) as as 00
L
ao +
(an cos (n;x) + bn sin (n;x)) ,
n=l where
ao an
= 2f1
1£
f(x) dx,
-l
11£ f(x)cos (n7rx) -e- dx, n =
="£
-£
bn
11£
="£
-£
.
f(x)sm (n7rx) -e- dx, n
Now, Now, 1
Co = 2f
II
1,2,3, ... ,
= 1,2,3, ....
f(x) dx = ao·
-l
For nn > 0, 0, For C
n=
2~ ill f(x)e-i7rnx/l dx
= ;e ill f(x)
(cos (n;x) - i sin (n;x)) dx
= ;e ill f(x) cos (n;x) dx 1
;e ill f(x) sin (n;x) dx
i
= 2an - 2bn . Similary, with n > 0, we we have C-
;e [ll f(x)ei7rnx/l dx = ;e [i/(X) (cos (n;x) +isin (n;x)) dx = ;e [le f(x) cos (n;x) dx + ;e [: f(x) sin (n;x) dx
n=
9.1. 9.1. The The complex complex Fourier Fourier series series
399 399
Therefore, Therefore, cnei7rnx/£
+ c_ne-i7rnx/l = (~an _ ~bn )
(cos
+ (~an+~bn)
(n;x) + i sin (n;x))
(n;x) _isin(n;x)) = an cos (n;x) + bn sin (n;x). (cos
quantities sum to zero in this calcula(The reader should notice how all imaginary quantities tion.) tion.) 2 0, 0, Therefore, for for any Therefore, any N > N
L
N
Cnei7rnx/l = ao
n=-N
+L
(an cos
(n;x) + bn sin (n;x)) .
n=1
This shows that every (symmetric) (symmetric) partial sum of the complex Fourier series of a is real real and and also also that the complex series is is equivalent equivalent to real-valued function real-valued function is that the complex Fourier Fourier series to the full series. the full Fourier Fourier series.
Exercises
f:
f(x)
x
2
1. Let Let / : [-1,1] [-1,1] ->• 1. -+ R R be be defined defined by by f ( x ) = = 1I -— x2 .. Compute Compute the the complex complex Fourier series Fourier series
of f, /, and of and graph graph the the errors errors N
f(x) - L
cnei7rnx
n=-N
for = 10,20,40. 10,20,40. for TV N =
f:
f(x) x.
2. Let [-71",71"] -+ R be be defined by f ( x ) = the complex complex Fourier Fourier 2. Let / : [—7r,7r] —>• R denned by = x. Compute Compute the coefficients Cn, c n , nn = 0, of /, the errors errors coefficients 0, ±1, ±1, ±2,..., ±2, ... , of j, and and graph graph the N
f(x) - L
cne inx
n=-N
for = 10,20,40. for N TV =
f:
f(x) e
lx 3. be defined by f ( x ) = the complex Fourier 3. Let Let / : [-1,1]-+ [—1,1] —> C be defined by = eix .. Compute Compute the complex Fourier coefficients Cn, cn, nn — = 0, 0, ±1, ±1, ±2,..., ±2, ... , of of f, /, and and graph graph the the errors errors coefficients N
f(x) - L cne7l"nx i
n=-N
Chapter 9. Chapter 9. More More about about Fourier Fourier series series
400
for N 10,20,40. (Note: (Note: You imaginary for N = = 10,20,40. You will will have have to to either either graph graph the the real real and and imaginary separately or just graph the modulus of parts of of the error error separately of the error.) 4. Consider the operator under 4. Consider the negative negative second second derivative derivative operator under periodic periodic boundary boundary conditions on [-f,f]. conditions on [—£,•£]. (a) Show Show that is an eigenpair. (a) that each each pair pair listed listed in in (9.2) (9.2) is an eigenpair. (b) that (9.2) (9.2) includes includes every every eigenvalue. eigenvalue. (b) Show Show that (c) Show Show that in terms given in in that every every eigenfunction eigenfunction is is expressible expressible in terms of of those those given (c) (9.2). (9.2). (d) Section 6.3, 6.3, and to those those derived derived in in Section and explain explain why why (d) Compare Compare these these results results to they consistent. they are are consistent. 5. Suppose that /f:: [—•£,•£] [-f,f] —> ~ C is defined by ff(x) = g(x] g(x) +ih(x), where 9g and (x) = + ih(x), where hh are are real-valued real-valued functions functions defined defined on on [-f,f]. [—i,£]. Show Show that that the the complex complex Fourier Fourier coefficients can can be be expressed in terms terms of of the the full full Fourier Fourier coefficients coefficients of of 9g and coefficients expressed in and h. h. 6. Prove that, that, for for any real numbers numbers 0 () and >., 6. Prove any real and A, ei(II+>') = ei () e i >..
(Hint: Use Euler's Euler's formula formula and trigonometric identities.) 7. Let TV 1, and and 7. Let N be be aa positive positive integer, integer, jj aa positive positive integer integer between between 11 and and N N — - 1, by define vectors u ERN e RN and v ERN e RN by Un
.
= cos ( jn~) N ,Vn = SIn (jn~) N ,n = 0,1, ... ,N \
\
/
1.
/
(For this exercise, it is convenient to index the components of xx E as (For € RN RN as 0,1,..., TV — 1, instead 1,2,..., TV as The purpose 0, 1, ... ,N - 1, instead of of 1,2, ... , N as is is usual.) usual.) The purpose of of this this exercise exercise is to is to prove prove that that
(a) is, u orthogonal. (a) u u·• v v= = 0, 0, that that is, u and and v v are are orthogonal. (b)
Ilull = IIvll = IN/2.
trick: The sum These results are based on the following trick: N-l
L
2
(e1fijn/N)
n=O
N-I
=
L (e
21fij N /
f
n=O
is aa finite formula. On On the other is finite geometric geometric series, series, for for which which there there is is aa simple simple formula. the other hand. hand, N-I
L
2
(e1fijn/N)
(9.3)
n=O 7r l n N can be ' J / by +i sin can be rewritten rewritten by by replacing replacing ee1fijn/N by cos cos (jnir/N) (jn~/N)+i sin (jrnr/N), (jn~/N), expandexpandthe square, square, and and summing. Find the the sum sum (9.3) (9.3) both both ways, ways, and and deduce deduce the ing the ing summing. Find the results above. results given given above.
401
9.2. Fourier series and the FFT FFT
9.2
Fourier series and the FFT FFT
We now show that there is a very efficient efficient way to estimate coefficients and estimate Fourier coefficients evaluate We will will use use the the BVP evaluate (partial) (partial) Fourier Fourier series. series. We BVP cf2u
-K,< x < £, dx 2 = f(x) ,-£ (9.4)
u( -e) = u(£), du(_£) = du(£) dx
dx
as our To solve solve this this by by the the method method of u as our first first example. example. To of Fourier Fourier series, series, we we express express u and f/ in in complex complex Fourier say and Fourier series, series, say 00
u(x) = n=-oo
(en, (cn, n n
= = 0, 0, ±1, ±1, ± ±2, 2 , ... . . . unknown) unknown) and and 00
f(x) =
L
dnei7rnx/f
n=~oo
(dnn,n differential equation equation can then be (d = 0,0,±l,±2,... ±1, ±2, ... known). known). The The differential can then be written written as as ,n = 22
00
'""' K,n ~
1f
£2
00
c
n
ei7rnx/f
n=-oo
= '""' d ei7rnx/f ~n,
(9.5)
n=-oo
and obtain and we we obtain (9.6) The The reader will will notice the similarity to how the the problem was solved solved in in Section Section 6.3.2. (The are simpler when using using the the complex Fourier 6.3.2. (The calculations, calculations, however, however, are simpler when complex Fourier series of the the full Fourier series.) the coefficient series instead instead of full Fourier series.) As As in in Section Section 6.3.2, 6.3.2, the coefficient do must must be zero, zero, that is,
1 f
f(x) dx = 0
~f
must hold, hold, in for aa solution to exist. When this this compatibility holds, must in order order for solution to exist. When compatibility condition condition holds, the value value of Co is not determined by the the equation infinitely many many solutions solutions exist, the of CQ is not determined by equation and and infinitely exist, differing the choice differing only only in in the choice of of cCo. 0. For analytic analytic purposes, purposes, the the formula formula (9.6) need. For For example, For (9.6) may may be be all all we we need. example, as as we discuss in Section 9.5.1, 9.5.1, (9.6) (9.6) shows that the solution uu is is considerably smoother we discuss in Section shows that the solution considerably smoother than the the forcing (since the the Fourier to zero zero faster than forcing function function f/ (since Fourier coefficients coefficients of of u decay decay to faster than those those of of /). f). However, However, in many cases, cases, we we want want to to produce produce aa numerical estimate than in many numerical estimate of the solution We may may wish, example, to to estimate the values values of on aa grid grid of the solution u. We wish, for for example, estimate the of u on so that we we can the solution solution accurately. requires three three steps: so that can graph graph the accurately. This This requires steps:
402
Chapter about Fourier series Chapter 9. 9. More More about Fourier series
1. the Fourier Fourier coefficients ±1, ±2,..., ±2, ... , TV, N, of by evaluating 1. Compute Compute the coefficients ddnn,, nn == 0, 0, ±1, of /f by evaluating
the the appropriate appropriate integrals. integrals.
2. ±1, ±2,..., ±2, ... , N, 2. Compute Compute the the Fourier Fourier coefficients coefficients en, cn, nn = 0, 0, ±1, TV, of of uu from from formula formula (9.6). (9.6). 3. Evaluate the partial sum sum 3. Evaluate the partial N
un(x) =
L
cnei7rnx/£
n=-N on Xj = jh, jj = ±2, ... ,,±JV, ±N, on aa grid grid covering covering the the interval interval [-f, [—I, f], I], say say Xj — jh, = 0, 0, ±1, ±1,±2,... /ih == £/JV. fiN. If produce accurate = If N JV is is chosen chosen large large enough, enough, this this will will produce accurate estimates estimates of of u(Xj), u(xj), jj — 0,±1,±2,...,±JV. 0, ±1, ±2, ... , ±N. Until this point, point, we necUntil this we have have implicitly implicitly assumed assumed that that all all of of the the calculations calculations necbe done various integration essary essary to to compute compute Uu would would be done analytically analytically (using (using various integration rules rules to to compute the necessary for example). This is is not not always compute the necessary Fourier Fourier coefficients, coefficients, for example). This always possipossinot always ble (some ble (some integrals integrals cannot cannot be be computed computed using using elementary elementary functions) functions) and and not always possible (because be more more efficient We desirable desirable when when it it is is possible (because there there may may be efficient methods). methods). We will therefore therefore consider will consider how how to to estimate estimate u. u.
9.2.1 9.2.1
Using the trapezoidal rule to estimate Fourier coefficients
We begin begin by We by estimating estimating the the integrals integrals defining defining the the Fourier Fourier coefficients coefficients of of f:/:
dn =
:f ill
f(x)e-i7rnx/l dx.
A A simple simple formula formula for for estimating estimating this this integral integral is is the the so-called so-called (composite) (composite) trapezoidal trapezoidal rule, which replaces the the integral by aa discrete be interpreted rule, which replaces integral by discrete sum sum that that can can be interpreted as as the the eiim rvf arooc nf trar^ovnirlc' sum of areas of trapezoids:
J{b g(x) dx = 2h a
(
g(xo)
+2
f;
N-l
g(Xj)
)
+ g(XN) + O(h2),
(9.7)
where hh = (b (b -— a)IN a)/N and = 0,1,2, 0,1,2,..., Figure 9.1 9.1 shows the where and Xj Xj = a a + jh, jh, jj = ... , N. N. Figure shows the geometric the trapezoidal trapezoidal rule is geometric interpretation interpretation of of the trapezoidal rule. rule. Although Although the the trapezoidal rule is generally the error for generally only only second-order second-order (that (that is, is, the error is is 0O (h2)), (h2}), it it is is highly highly accurate accurate for 64 periodic functions ,.64 periodic functions To rule to the computation we define To apply apply the the trapezoidal trapezoidal rule to the computation of of ddnn,, we define the the grid grid 0, ±1, ±1, ±2, ±2,..., ±N, hh = fiN. i/N. Then Then Xj = jh, jh, jj = 0, ... , ±N, Xj d n
. 1h
= 2f2
6 4 For
details, see any book on numerical analysis, for example, [2].
9.2. Fourier Fourier series and the FFT
403
0.2
0.6
0.4
0.8
x
Figure 9.1. 9.1. Illustration fllustration of of the the trapezoidal rule for for numerical numerical integration; Figure trapezoidal rule integration; the (signed) (signed) area area under under the the curve curve is is estimated estimated by by the the areas areas of of the the trapezoids. trapezoids. the
Since Since and and
h
2f
1 _ i7fnxj _ i7fnj 2N' l -]\i'
we can can simplify this to we simplify this to d n ..:.. - F n --
~
N-l
'~ " f·Je -i7rnj/N , n -- -N, -N +, I... O N , , ... , - 1, 2N.J=- N
(9.8)
where where
Ii
=f(xj), j=-N+l,-N+2, ... ,O, ... ,N-l
and and f-N
1
= 2 (/(-l) + f(l)).
The reader should take note note of the special _N, which which reduces reduces to simply The reader should take of the special definition definition of of ff-N, to simply f_ = /(-£) is 2^-periodic = f(l)). f(t}}. f-N = f(xf(X-N) f(-l) when when /f is 21-periodic (so (so that that f(-l) f(-l) = N = N) = The sequence F-N, F_JV+I, F_ N +1,"" F N - 1 is essentially The sequence of estimates estimates F-N, • • • , FN-I essentially produced by by discrete Fourier Fourier transform transform to to the the sequence f - N , f-N+i, f - N +1, ... f N -1, as applying the discrete applying the sequence /-AT, • • • ,, /jv-i, as we now now show. show. we
404
Chapter 9. More about Fourier series
9.2.2 9.2.2
The discrete transform The discrete Fourier Fourier transform
[-i, -f] Just as a function defined on \—i, —H\ can be synthesized from functions, each having a distinct distinct frequency, frequency, so so a finite sequence can be synthesized from sequences with distinct distinct frequencies. Definition 9.2. 9.2. Let 0,0,0,1,..., OM-I be a sequence numbers. Definition Let ao, al, ... , aM -1 be sequence of of real or complex numbers. -1 to the sequence sequence Then the discrete Fourier transform (DFT) maps ao, ao, al, a i , ... . . . ,, aM OM-I
where where An
=~ M
M-1 "a·e-27rinj/M L...J J , n
0 1, ... , M - 1. =,
(9.9)
j=O
The original sequence can be recovered by applying the inverse DFT: M-l
- 1. aJ. -- " L...JA n e27rinj/M , J. -- 0 , 1 , 2 , ...M ,
(9.10)
n=O
relationship is a direct calculation calculation (see (see Exercise 5). We We will often often The proof proof of this relationship to the the sequence A o0,A ,AM-1 as the refer to sequence A ,Ai,... the DFT DFT of ao,al, ao>«i, ... • • • ,aM-1, >«M-I, although 1, ... ,AM-I way to express the relationship relationship is the correct way is that A A o0,,Ai,..., A 1, ... , AM-I AM -1 is the image under DFT of of ao, OQ, a1, a i , ... . . . ,, aM-l OM-I (the (theDFT DFT isisthe the mapping, mapping, not notthe the result result ofofthe the under the the DFT mapping). This abuse of terminology is just a convenience, and is quite common. 1 1 we will write the sequence ao, a1, ... ,,OM-I aM -1 as {aj} ~01, As another another convenience, we 00,01,... {aj} ^ , or even even just {a,j} (if the of jj are are understood). understood). or just {aj} the limits limits of 1 an now show that the relationship relationship between the sequences {fJ }.f=--.!-N We will now {/jj^T. ^ and d can e {Fn};;==-~N can be expressed in terms of the DFT. (The reader will recall that [Fn}n=-N b expressed in terms of the DFT. (The reader will recall that i () is Fn = Fn == d d nn,, the the nth nth Fourier Fourier coefficient coefficient of of the the function function /f on on [—I,i}.} [-i,i].) Since Since eelQ is 27r-periodic, we have e-7ri(j+2N)n/N
= e-7rijn/N-27rin = e-7rijn/N e- 27rin = e-7rijn/N.
We can therefore write, for for nn = = -N, —N, -N —N + + 1, 1,..., N — 1, We can therefore write, ... , N - 1,
9.2. Fourier series and and the FFT
2~
=
405
N-l {
L
/je-1rijn/N
j=O
+
2N-l
L
} /j_2Ne-1rijn/N
.
j=N
If {jj}~:O-1 by If we we define a sequence {fj}?= *^ 0
J. =
{ Ij,
)
/j-2N,
j=0,1, ... ,N-1, = N, N + 1, ... , 2N - 1,
j
then have then we we have 2N-l
1 Fn -- 2N
"1~ je -1rijn/N ,
n -- - N , - N
+ 1, ... , N
- 1.
)=0 ze Moreover, by the periodicity of write Moreover, by the periodicity of eeifJ ,, we we can can write
2N-l
Fn
1rij (n+2N)/N =~ I, -l . 2N " ~ 1-·eJ , n = - N ,- N + , ... j=O
Defining the sequence (Fn}^0 l by
= O,l, ... ,N - 1, n = N, N + 1, ... , 2N - 1,
n
we we have have
F:n
2N-l
= _1_ 0, 1, ... , 2N - 1. 2N "~ f-·e-1rijn/N J , n = j=O
Thus {Fn} is is the DFT of of {]j}. {/j}. With Thus {Fn} the DFT With this this rearrangement rearrangement of of terms terms understood, understood, we we can say say that that {Fn} {Fn} is is the the DFT DFT ofJ/j}. of {fj}. can To actually compute {Fn}n~~N we perform perform the the following {Fn}n=-N from f rom {fj}f=-lN' {/jljL^-V' we three steps: three steps: 1. the sequence sequence 1. Replace Replace the !-N'···'
1-1, to, It,···, IN-l
with with
10, It,···, !N-I, I-N' ... ' 1-1,
.
l and label the latter latter sequence sequence as as {]j} {fj}2j~~:0-1 . 0
- 22N-l - 2N-l 2. Compute Compute the the DFT DFT of of {!j}j=o {fj} ^1 to to get get {Fn}n=O {F 2. n}^-i..
3. The The desired desired sequence sequence {F 3. {Fn};;==-~N is n}^~^N is
Chapter Chapter 9. More about Fourier series
406
The representation of the the original original sequence sequence in in terms terms of of its its DFT DFT and and the the comThe representation of complex exponentials viewed as plex exponentials can can be be viewed as trigonometric interpolation, interpolation, since since it it provides provides aa combination combination of of complex complex exponentials exponentials (and (and hence hence sines sines and and cosines) cosines) that that interinterf -N , f -N+1, ...• ,,/JY-I fN -1 and therefore polates the sequence sequence /_AT,/_JV+I,-• therefore the function /.f. To To be by precise, precise, define define aa function function I/ :: R R --t -» C C by N-1
L
I(x) =
(9.11)
n=-N
where Fn, ... is satisfies where F n == 0, 0, ±1, ±1, ±2, ±2,... is defined defined by by (9.8). (9.8). Then Then I/ satisfies n, n I (Xj)
= 1 (Xj) , j = - N + 1, - N + 2, ... ,N -
1
(9.12)
and and (9.13) (see (see Exercise Exercise 4). 4). 3 Example f(x) == x X3. Example 9.3. 9.3. We define define f : [-1,1] [—1,1] --t —>• R by /(#) . We will compute the interpolating function of the the previous with N 3. The The sequence {fj} previous paragraph paragraph with N = 3. sequence {h} interpolating function II of is (approximately) (approximately)
0, -0.29630, -0.037037,0,0.037037,0.29630
(recall f(X3))/2 = (f(-1) f(1))/2 = (-1 The (recall that that 1-3 /_ 3 = - (f(X-3) ( f ( x - 3 ) + fM)/2 (/(-I) + /(l))/2 (-1 + 1)/2 l)/2 = 0). Q). The 65 sequence {F {Fn} y65 by n} is then given b F_ 3 == 1.9466.10- 17 + 1.5519. 1O- 17 i, F-2 == 3.7066.10- 17
-
F-1 == -1.0606.10- 16
7.4842· 1O- 2 i,
+ 9.6225· 1O-2 i,
17
Fo == 3.7007.10- , Fl == 1.4501· 10- 17 - 9.6225· 1O- 2 i, F2 == -4.6788.10- 17 + 7.4842· 1O- 2 i. In Figure 9.2, 9.2, we we display display the the function f , the the sequence sequence {h}, {fj}, and and the the interpolating In Figure function f, interpolating function function 2
I(x)
=
L
Fnei7rnx/l.
n=-3
We We now now have have two two important important facts: facts: 1. 1. By By computing computing aa DFT, DFT, we we can can estimate estimate 2N IN of of the the complex complex Fourier Fourier coefficients coefficients of aa known of known function. function. 65 65The using aa built-in The calculation calculation was was performed performed in in MATLAB MATLAB using built-in function function for for computing computing the the DFT. DFT.
9.2. Fourier series and the FFT FFT
407
0.8
0.6 0.4
0.2
o x
0.5
Figure 9.2. The (x) = —x {fj}, and Figure 9.2. The function function ff(x) x 33,, the the sequence sequence {Ii}, and the the trigonotrigonotwnx metric (x) = __ F e (see Example 9.3). metric interpolating interpolating function function Ilex) = X) L~=-3 Fnei1rnx (see Example 9.3). n 3 n 2. coefficients of func2. Conversely, Conversely, given given the the first first 2AT 2N complex complex Fourier Fourier coefficients of aa periodic periodic function, estimate the + 11 nodes) we can can estimate the function function (on (on aa regular regular grid grid with with IN 2N + nodes) by by tion, we computing DFT. computing an an inverse inverse DFT. These significant because is aa fast fast algorithm algorithm for DFT; These facts facts are are significant because there there is for computing computing the the DFT; the fast fast Fourier Fourier transform (FFT). this algorithm algorithm is is called, called, appropriately appropriately enough, enough, the this transform (FFT). There similar fast simply called the There is is aa similar fast algorithm algorithm for for the the inverse inverse DFT, DFT, which which is is simply called the inverse inverse FFT. FFT. A direct implementation implementation of (9.9) requires O((M) 2 ) arithmetic operA direct of formula formula (9.9) requires O((M)2) arithmetic operations compute aa DFT. on the other hand, can require ations to to compute DFT. The The FFT FFT algorithm, algorithm, on the other hand, can require as as little O(Mlog the little as as OeM log22 (M)) (M)) operations operations to to obtain obtain the the same same result. result. The The efficiency efficiency of of the FFT M; if FFT depends depends on on the the prime prime factorization factorization of of M; if M M is is the the product product of of small small primes, primes, then situation is then the the FFT FFT is is very very efficient. efficient. The The ideal ideal situation is that that M M be be aa power power of of 2; 2; in in this (M)) operation applies. There this case, case, the the O(Mlog OeM log22 (M)) operation count count applies. There are are many many books books that that explain how how the the FFT FFT algorithm algorithm works, works, such such as as Kammler Kammler [28], [28], Chapter 6. We We will will explain Chapter 6. not concern ourselves implementation of of the algorithm in but not concern ourselves with with the the implementation the algorithm in this this text, text, but 66 just regard regard the the FFT FFT as as aa black black box box for for computing computing the the DFT. DFT.66 just 66
66The FFT is of the the most most important important computer computer algorithms algorithms for for computational computational science science (indeed, (indeed, The FFT is one one of the cited mathematmathematthe original original paper paper [12] [12] announcing announcing the the algorithm algorithm is is reputed reputed to to be be the the most most widely widely cited ical paper paper of of all all time—see time-see [28], page 295). 295). For For this this reason, reason, implementations implementations of of the the FFT FFT exist exist on on ical [28], page every is also every major major computer computer platform platform and and it it is also aa feature feature of of computer computer packages packages such such as as MATLAB, MATLAB, Mathematica^ Maple. Mathematica, and and Maple.
408
Chapter 9. More about Fourier series
Using the the above above results, we can can devise solving (ap(apUsing results, we devise an an efficient efficient algorithm algorithm for for solving proximately) BVP (9.4): (9.4): proximately) the the BVP 1. FFT, estimate the Fourier Fourier coefficients coefficients d-w,d-N+i, d-N, d_ N +1 , ... , dN -1 of 1. Using the FFT, estimate the • • •,djv-i of f./.
2. Use (9.6) 2. Use (9.6)totoestimate estimatethe thecorresponding corresponding Fourier Fourier series seriesofofthe thesolution solution u.u.
3. to estimate of uu on FFT to estimate the the values values of on the the grid grid —i, -i, —l+h,..., -i+h, . .. , i—h, i-h, 3. Use Use the the inverse inverse FFT h = iiN. t/N. h Example solution of of Example 9.4. 9.4. We We use the the above algorithm algorithm to to estimate estimate the solution
d2 u
- -dx 2 = x 3 ' -1
< x < 1,
-
u( -1) = u(l),
(9.14)
du (-1) = du (1).
dx
dx
The forcing forcing function function f(x] f(x) = satisfies the the compatibility as can The = xx33 satisfies compatibility condition, condition, as can easily easily be verified. An An exact exact solution solution is )/20, which to choosing choosing be verified. is u(x) u(x) = = (x (x -— xx55)/2Q, which corresponds corresponds to Co the Fourier Fourier series CQ = = 00 in in the series 00
u(x) =
L
cn ei 7l'nx.
n=-oo We will use = 2277,; which which makes the FFT FFT particularly particularly efficient. efficient. We will use N N = 128 128 = makes the implemented by by the the FFT, FFT, we we produce produce the the estimates (9.8), implemented estimates
Using Using
(Recall f-128 is average of f(-l) and /(I); f(l); in case, this aver(Recall the the /_i28 i$ taken taken as the average of f(—1) in this this case, age is indicate accuracy of the the computed computed Fourier Fourier coefficients, coefficients, we we graph graph the age is 0.) To To indicate accuracy of the logarithm of the the absolute absolute error error in interpolating function logarithm of in the the resulting resulting trigonometric trigonometric interpolating function as given given by Figure 9.3. 9.3. As As this graph shows, shows, the error is II(x), ( x ] , as by (9.11), (9.11), in in Figure this graph the error is essentially essentially zero at interpolation nodes nodes and and very very small small in in between, between, except near the zero at the the interpolation except near the endpoints endpoints where Gibbs's Gibbs's phenomenon is is evident (because (because f is is not not periodic). (We (We graph the the of the magnitude of logarithm logarithm of the error error because because of of the the great disparity in the magnitude of the error near the endpoints and and in in the interior of of the the interval. interval. If If we we graphed graphed the the error error itself, near the endpoints the interior itself, only the of the only the Gibbs's Gibbs's phenomenon would be be discernible on on the scale of the graph.) graph.) We next compute the the estimates ... ,0127.' We next compute estimates 0/c_i28,C-i27 of C-128, C-127, 5... ,C127: _ Cj
Fn = 22' n = ±1,±2, ... , n7f
and we we take Co = Finally, we we use the inverse inverse FFT FFT to produce the the estimates and take c$ — O. 0. Finally, use the to produce estimates 127
u- J.
--
'"'" ~
n=-128
i7l'jn/N , J. -- -128 , -127 , ... , 127. cne
9.2.
409
Fourier series and the FFT FFT
Log error in trigonometric interpolation
-15
-.:--20
g -5;
..Q
-25
-30
-4~~1---~0.8::-----:-0:-::.6---~0.~4--:-0~.2:----:0~-~0.2~~0~.4-~0.~6--:0~.8:--"""
Fi~ure 9.3. 9.3. Logarithm Logarithm of of error the trigonometric trigonometric interpolating interpolating function Figure error in in the function
I(x)
= 2:~~-128 Fnei1rnx/f
for f(x)
= x3 .
(See Example 9.4.)
We then have have We then
Uj == u C;8) , j = -128, -127, ... ,127. We graph the the error in the computed solution solution in Figure 9-49.4. We graph error in the computed in Figure
9.2.3 9.2.3
packaged FFT A note about using packaged FFT routines
There is is more more than than one one way way to to define define the the discrete discrete Fourier Fourier transform, transform, and and therefore therefore There various implementations implementations of of the the FFT FFT may may implement implement slightly slightly different formulas. For For various different formulas. formulas for the DFT and the inverse DFT are example, the formulas are asymmetric in in that the the factor of of 1/M 11M appears appears in in the the DFT DFT (9.9) (9.9) but but not not in in the the inverse DFT (9.10). (9.10). However, However, factor inverse DFT some software packages put put the the factor factor of in the the inverse inverse DFT DFT instead. instead. 67 It is some software packages of 11M 1/M in It is also possible to to make make symmetric symmetric formulas by putting putting aa factor factor of in each each of of also possible formulas by of 1/VM 1/vM in 68 the DFT DFT and and the the inverse inverse DFT. DFT.68 One can define the the DFT DFT while while indexing indexing from the One can also also define from -N to to N as in (9.8). -N N - 1, 1, as in (9.8). Given the the diversity diversity in definitions of of the the DFT DFT and therefore of the FFT FFT (which, (which, Given in definitions and therefore of the the reader reader should should bear bear in in mind, mind, is is just just aa fast fast algorithm algorithm for computing the the DFT), as the for computing DFT), it essential that that one one knows knows which which definition definition is is being being used used before before trying trying to to apply apply it is is essential aa packaged packaged FFT FFT routine. routine. 67
57MATLAB one such such package. MATLAB is is one package. 58Mathematica Mathematica does this by by default. does this default.
68
410
Chapter 9. about Fourier series Chapter 9. More More about Fourier series Error in computed solution
-1~--------~----------~--------~--------~
-1
-0.5
0 x
0.5
Figure 9.4. Error Error in computed solution solution in in Example Example 9.4. Figure 9.4. in computed 9.4-
9.2.4 9.2.4
Fast transforms and and other Fast transforms other boundary boundary conditions; conditions; the the sine transform transform discrete sine discrete
Fast transform transform methods methods are not restricted restricted to to problems problems with with periodic periodic boundary Fast are not boundary be defined defined that that are are useful useful for working with conditions. Discrete transforms can conditions. Discrete transforms can be for working with sine, cosine, cosine, quarter-wave and quarter-wave quarter-wave cosine series, as as well well as as with with the sine, quarter-wave sine, sine, and cosine series, the full Fourier series. A comprehensive comprehensive source software for for the the corresponding corresponding fast full Fourier series. A source of of software fast transforms is is [48]. transforms [48]. As an an example, example, we discuss the discrete sine transform (DST) and and its use in solving problems with Dirichlet conditions. The DST of a sequence ft, !N-1 solving problems with Dirichlet conditions. The DST of a sequence /i, 12, /2, •..• ·, • , /AT-I is defined by is defined by Fn
(.) = 2 N-1 ~ Ii sin 7r;J ,n = 0,1, ... ,N -
1.
(9.15)
)=1
It be shown shown that that the the DST DST is its own own inverse, inverse, up up to to aa multiplicative multiplicative constant: It can can be is its constant: , F 1 , ••. applying N -1 produces the original sequence multiplied by applying the DST DST to F Fo0,FI, ...,, F FN-I 2N (see Exercise 9). 27V 9). In solving a BVP or an an IBVP with Dirichlet conditions, the main calculations calculations are:
1. computing computing the function; 1. the Fourier Fourier sine sine coefficients coefficients of of aa given given function; 2. computing computing the solution from its Fourier coefficients. coefficients.
411 411
9.2. Fourier series and the FFT
As the DFT Fourier series, series, we can solve solve these these problems approxAs with with the DFT and and complex complex Fourier we can problems approximately using DST. If If f/ e (?[(),£], then the Fourier sine sine coefficients coefficients of imately using the the DST. E e[O, el, then the Fourier of f/ are are cnvpn V>v given by 2
{l
cn=eJo f(x)sin
(n7rX) -e- dx, n=1,2,3, ....
Using — 0,1,2, 0,1,2,..., and the the trapezoidal trapezoidal Using the the regular regular grid grid Xj Xj == jh, jh, jj = ... , A/", N, h == t/N, elN, and rule, we we obtain rule, obtain
. C2" f (Xj ).sm (n7rXj) -e- h N-1
en =
~
j=1
where ( x j ) . The The trapezoidal rule simplifies simplifies due due to fact that sin(O) where fj fJ == ff(xj). trapezoidal rule to the the fact that sin (0) = sin (n7r) O. We We therefore see that the approximate (n?r) = = 0. approximate Fourier sine coefficients coefficients are f(xd, /(a^), f(X2), ... just a multiple of the DST of the sequence /(#i), ••••>, f(XN-d. f(%N-i)On the other hand, suppose the Fourier sine sine series series of On the other hand, suppose the Fourier of uu is is 00
u(x) = L
Cn
sin
(n;x)
n=1
and C2, .•. N -1 (or and we know know C1, Ci,C2,. - . ,, Ccjv-i (or approximations approximations to them). We then have, for jj = , 2 , .... . . ,, JNV-- l1,, = l1,2, N-l
.~ " Cn sm . (n7rXj) u (Xj ) = -en=1 N-l
.
7rnJ
(.)
= ~Cnsm N
'
and we see that can be estimated on on aa regular grid by by another another application application of of the the and we see that u u can be estimated regular grid DST. DST.
9.2.5 9.2.5
Computing the the DST DST using using the FFT Computing the FFT
As mentioned above, above, there are special special programs available for for computing As mentioned there are programs available computing transforms, transforms, such as as the the DST, DST, using using FFT-like FFT-like algorithms. algorithms. However, However, these are more more such these programs programs are 69 specialized and therefore less widely widely available FFT.69 Fortunately, available than the FFT. Fortunately, the FFT) and few additional DST can can be be computed using using the DFT DFT (and hence hence the the FFT) and aa few additional manipulations. We will will now explain how to do do this. this. manipulations. We now explain how to 69 69For MATLAB has but no no fast command. For example, example, MATLAB has an an FFT FFT command, command, but fast DST DST command.
412
Chapter 9. More about Fourier series
l We is We assume assume that that {Ii {fj}f}f=11 is aa sequence sequence of of real real numbers numbers and and that that we we wish wish to to =l 1 compute the the DST DST of of {Ii}. {fj}. We We define define aa new new sequence sequence {ij}~:0-1 {fj}™^ by by compute
h=
j=1,2, ... ,N-1,
Ij, {
-hN-j,
0,
~=N+1,N+2, ... ,2N-1, J = O,N.
We then define {Fn}2^ to be the DFT of {fj}:
2~
Fn =
2N-1
L
he-1rijnIN
j=O
1 = 2N
?= Ii cos
2N-1
-
z.
(7rJn .)
N
2N-1
- 2N
J~
L
-.
fJ sm
(7rJn . )
N
.
J~
Since we are interested interested in in the the DST, DST, we we will look at at the part of of F Since we are will look the imaginary imaginary part Fn, which n, which is (using the fact that /0 = = /AT is (using the fact that }o }N = = 0) 0) 1 Im(Fn) = - 2N
?= Ii- sin
2N-1
(7rJn . )
N
J=O
= _ 2~
{~}j sin (7r~n) + 2f1 }j sin (7r~n)} }=o
J=N+l
~-2~ {X;fjsm(~tn) - X;fN_jSin(~(j~N)n)} (using /N-J for 1,2,..., 1). We have (using the the fact fact that that fj+N jJ+N = = /ZN-N-J hN-N-j = IN-j for jj = = 1,2, ... , TV N — - 1). We have N-1 ~
- L.....-
I
N-j
.
sm
(
7r(j
+ N)n ) N
=-
j=1
N-1 ~
L.....-
I. j
sm
(
7r ( 2N - j)n ) N
j=1
= -
L
N-1
(
Ij
.
sin 2n7r _ 7r~n
3=1
=
?=
N-l
Ij
sin
7r~n ,
(.
)
J=1
and and so so
Im(Fn) It sequence It follows follows that that the the sequence
~ - 2~ { 2 X; /; sin ( ~tn) } .
)
413 413
Fourier series and the 9.2. Fourier the FFT
is is the the DST DST of of {lj}. {/_,-}. The reader reader should should note note that, that, when when using using the the FFT to compute compute the the DST, DST, N N The FFT to should product of should be be aa product of small small primes primes for for efficiency. efficiency. A similar technique technique allows allows one one to to compute compute the the discrete discrete cosine cosine transform transform (DCT) A similar (DCT) using 10. using the the FFT. FFT. The The DCT is explored explored in in Exercise 10.
Exercises 1. Define Define aa sequence sequence {Ii {fj}}= by 1. g~o by 0
fj
j (37r 10 ) .
. = sm
Compute the the DFT DFT {Fn} {Fn} of of {Ii} {fj} and and graph graph its its magnitude. magnitude. Compute 2. [—7r,7r] -+ —> R R be defined by = x Use the DFT to estimate 2. Let Let /f :: [-7r,7r] be defined by /(#) f(x) = x 22 .• Use the DFT to estimate complex Fourier Fourier coefficients coefficients Ccnn of of f, /, nn == -8, — 8 ,-7, — 7 ,... . . . ,7. , 7 . Compare Compare with complex with exact values: CQ =— 7r 7r22/3, /3, Ccnn = = 2(—l) /n 22 ,, n n == ±1, — 2( -I)nn/n ±I, ±2, .... exact values: Co
the the the the
3. Let Let f/ : [-1,1] [—1,1] ^ R be be defined defined by ( x ) = x(l The complex complex Fourier 3. -+ R by ff(x) x(I -- x22}. ). The Fourier coefficients of / are coefficients of f are Co =
0,
Cn
6i( -I)n
=
3 3
'
n7r
n = ±I, ±2, ....
DFT to estimate Use the the inverse inverse DFT estimate f/ on on aa grid, grid, and and graph graph both both f/ and and the computed estimate estimate on on [-1,1]. [—1,1]. Use Use C-16, c_ie, C-15,· c _ i 5 , .... . ,, C15. Ci5. computed
4. Show that, that, with with I/ defined defined by by (9.11), (9.11), the the interpolation interpolation equations equations (9.12) (9.12) and 4. Show and (9.13) hold. 5. Let Let ao, ao, aI, « i , ... . . . ,, aN CLN-I be aa sequence sequence of of real or complex complex numbers, numbers, and and define define 5. -1 be real or N-l
An
=
~ " a 'e-21Tinj/N ,n=" 0 1 ... , N- . 1 N~J j=O
that Prove Prove that
N-l
A n e21Tinj / N , J. -- 0 , 1 , ... , N - 1 . aJ. -- " ~ n=O
Hint: Substitute Substitute the formula for for A into Hint: the formula An n into N-l
L
Ane21Tinj/N
n=O
and interchange the order of of summation. summation. (A (A dummy dummy index, index, say say m, must be be and interchange the order m, must used in in place place of of jj in in the formula for for A Look for for aa geometric geometric series series and and use use used the formula An.) n.) Look
~ rn = ~
n=O
K
1
r +
-
r-I
1.
414
Chapter Chapter 9. 9. More More about about Fourier Fourier series series
6. Suppose uu satisfies periodic boundary £], and boundary conditions conditions on the interval [-£, [—•£,•£],
L <Xl
cnei1fnx/l
n=-(XJ
is the the complex complex Fourier Fourier series series of of u. Prove Prove that that is
is the the complex complex Fourier Fourier series series of of -cPu/ —d?u/dx is dx 22.• Use the DST to to estimate estimate the first 15 15 Fourier Fourier coefficients coefficients of of the function J(x) f(x} = 7. Use the DST the first the function = x(l -— x) on on the interval [0,1]. Compare Compare to to the exact values x(1 the interval the exact values an = -
4(-1 + (-I)n) 3 3 n7r
n = 1,2,3, ....
'
8. The The Fourier Fourier sine sine coefficients coefficients of of J(x) f ( x ) == x(l x(l -— x) on on [0,1] are are an = -
4(-1 + (_l)n) 3 3 n7r
n = 1,2,3, ....
'
Use Use the the DST DST and and a1, ai, a2, 02, ... • • • ,, a63 «es to estimate estimate /J on on aa grid grid with 63 63 evenly evenly spaced spaced points. points. 9. ... , IN-1 real numbers, 9. Let Let Jo,ft, /o, /i,..., /jv-i be be aa sequence sequence of of real numbers, and and define define Fo0,,F FI1, .... . . , FFN-I N- 1 by ,9N-1 by by (9.15). Then define 90,91, # 0 ,£i, ... • • • ,9N-i by 9j
N-1
=2L
(.)
= 0,1, ... , N
Fn sin 7r;; , j
- 1.
n=l
I[
Show that that QJ = 2N 2JV/j, 0,1,..., I. Hint: Hint: Proceed according to to the the Show 9j = fJ, jj == 0,1, ... , N N -— 1. Proceed according hint the fact vectors hint in in Exercise Exercise 5. 5. You You can can use use the fact that that the the vectors
(~)
sin sin (2~"') .
sm
((N~1)m1f) N
,
.
sin ( 1V ) sin (~)
sm
((;-1)j1f) N
N l1 - for are orthogonal in for j,j, m m= = 1,2,..., 1,2, ... , N N— -1, j :j:. m (see Exercise Exercise 3.5.7). in R RN l,j^m Also, each each of of these these vectors vectors has has norm norm \/N/1 (see Exercise Exercise 9.1.7). 9.1.7). IN/2 (see Also,
10. The The DCT DOT maps maps aa sequence sequence {h}f=o {fj}^=0 of of real real numbers to the the sequence sequence {Fn};;=o, {Fn}^=0, 10. numbers to where where N-l
(.)
Fn=Jo+2~fJcos 7r;} + (-l)nJN' n=O,l, ... ,N. 3=1
9.3. Relationship Relationship of sine and cosine series to the full full Fourier series
415
(a) Reasoning as in Section 9.2.4, 9.2.4, show show how can be estimate (a) Reasoning as in Section how the the DCT neT can be used used to to estimate the C[0,£j. the Fourier cosine coefficients coefficients of a function in C[O, fl. Section 9.2.4, show how the DCT estimate (b) Reasoning Reasoning as in Section neT can be used to estimate aa function function from from its its Fourier Fourier cosine cosine coefficients. coefficients.
(c) Modifying the technique presented presented in Section 9.2.5, show how to compute neT using the DFT nFT (and hence the FFT). FFT). (Hint: Given {f;}.f=o, the DCT {fj}?=0, - N-l define {&*£„ by define {fj} j=-N by jj=fljl, j=-N,-N+1, ... ,N-1,
and treat {jj}f=-l-N {fj}^-^N by the three step process on page 405.) (d) Show that the DCT neT is its own own inverse, up to a constant multiple. To be precise, show that if the DCT neT is applied to a given sequence and then the neT is applied to the result, one obtains DCT obtains 2N IN times the original sequence.
9.3 9.3
Relationship of of sine sine and and cosine cosine series series to the full Relationship to the full Fourier series Fourier series
In Section 9.1, we showed showed that that the complex and and full full Fourier series are are equivalent equivalent In Section 9.1, we the complex Fourier series function. We We will now show that both for a real-valued function. both the Fourier Fourier cosine and the Fourier sine series can be recognized as special cases of the the full Fourier series and hence of the complex Fourier series. This will show that the complex Fourier series is the most general concept. relationships between the various Fourier Fourier series for realTo understand understand the relationships valued must understand following terms: terms: we must understand the the following valued functions, functions, we Definition Let f : R R -+ Definition 9.5. 9.5. Let ->• R. R. Then Then f is 1. odd if f( -x) = f(x) for for all 1. odd if f(-x) = -—f(x) all xx E G R; R;
2. if f(—x) ( x ] for all xx E 6 R; R; 2. even even if f( -x) — = jf(x) for all 3. with period T if if T T > 0, ff(x (x + + T) T) = all xx E e R, R; and and this 3. periodic periodic with period T = f(x] f(x) for for all this T. condition condition does not not hold hold for any smaller positive value of ofT. Examples Examples of odd functions include polynomials with only odd powers and (x). Polynomials Polynomials with only even powers and and cos (x) are examples of even funcfuncsin (x). tions, while sine and cosine are the prototypical prototypical periodic functions (both have period 2?r). functions imply that the graph 27f). The algebraic properties properties defining odd and even functions of an odd function is symmetric through the origin, while the the graph of an even function is symmetric across the y-axis (see Figure 9.5). We will will show show that the full of an an odd odd function function reduces sine We that the full Fourier Fourier series series of reduces to to aa sine series, and that the full Fourier series of an even function reduces to a cosine series. We this preliminary preliminary result: result: We need need this
416
Chapter More about about Fourier Fourier series series Chapter 9. 9. More 2~---------r--------~
1 ~
O~----~~-r--+-----~
-1
-1
o
1
-2~--~----~----~--~
-2
2
-1
o
2
x
x
Figure Figure 9.5. 9.5. Examples of odd (left) (left) and even (right) (right) junctions. functions. Lemma 9.6. Lemma 9.6. 1. Suppose 1 -+ R is odd. odd. Then 1. Suppose f : R —>•
1ft
I(x) dx
= o.
2. f :R R —>• R is 2. Suppose Suppose 1 -+ R is even. Then
( I(x) dx
1-t
=2
(f(x) dx.
10
Proof. Suppose is odd. Then Proof. Suppose 1 / is Then
rt I(x) dx = 1f(x) dx + 10r£ f(x) dx. 1- t 0
l
Making variables x = the first integral integral on the right, we Making the change change of of variables = -s — s in the we obtain
rt I(x) dx = -1£ f( -8) ds + 10(f(x) dx 1-£ 0
=
-1£
I(s) ds
+
1£
I(x) dx (since I( -s) = - I(s))
= O. The result result for for even 1 / is proved by making the same change of variables. We can now derive the main result.
D
9.3.
Relationship of sine and cosine cosine series to the full Fourier series
417
2 Theorem L2(-e,e), suppose that Theorem 9.7. 9.7. Let f E 6L (—t,I), and suppose
f. is the full full Fourier series of f·
1. IfIf ff is odd, odd. then an
and bn =
= 0,
n
= 0,1,2, ...
~ 10£ f(x) sin (n;x) dx.
That is, the full full Fourier series (on [—1, [-e, f]) function is the same as i]) of of an odd function its Fourier sine series (on [0, el). [0,f\). If f is even, then 2. If bnn == 0, ... 6 0, nn == 1,2,3, l,2,3,...
and ao = an =
~ 10£ f(x) dx,
2 r£ C 10 f(x) cos (n7rx) -f- dx.
That is, the full full Fourier series (on [—1,1]) [-e, e]) of of an even function function is the same as its Fourier cosine series (on [0, fl). [Q,i]). Proof. If Proof. If f/ is is odd, odd, then then n7rx) f(x) cos ( -eis is odd odd and and
f(x) sin (n;x) is This, together with the yields the first result. result. is even. even. This, together with the previous previous lemma, lemma, yields the first If f/ is is even, even, then then If n7rx) f(x) cos ( -eis even and is even and
f(x) sin (n;x) is odd. From we obtain the second conclusion. Prom this we
0
We can can use use the preceding result in the we wish wish We the preceding result in the following following fashion. fashion. Suppose Suppose we to understand understand the the convergence the Fourier sine series of f/ : (0, (0, f) -+ R. to convergence of of the Fourier sine series of i) —>• R. Define Define fodd :'•(-f, f) —>• -+ R, the odd odd extension of of /,f, by fodd (—£•,£) R-5 the by
f(x), o<x<e, fOdd(X) = { _ f( -x), -e < x < 0.
418
Chapter 9.
More about Fourier series
Then, by the previous theorem, the (full) (full) Fourier series of ffodd /, 0dd is the sine series of f, so the convergence of of the the sine sine series series can can be understood in in terms of the the convergence so the convergence be understood terms of convergence of a related related (full) Fourier series. Similarly, f) --t we define : (-f, Similarly, given given /f :: (0, (0,^) -> R, R, we define ffeven (—£,£)f) --t -> R, R, the the even even f, by extension of /, f(x), 0 < x < f, feven(x) = { f( -x), -f < x < O. The Fourier series of ffeven f, so, of so, again, the convergence of even is the cosine series of /, the cosine series can be examined in terms of the convergence of a related related Fourier series. Figure 9.6 9.6 shows the odd and even extensions extensions of /f : [-1,1] [-1,1] --t ->• R defined by ff(x) ( x ) = 11 + x. x.
v
2 1
,,
2
1
,,
V
,
>- 0
>- 0 -1
-1
~' ~ ~ ~
-2 -2
~.
-1
o X
1
2
-2 -2
-1
o
1
2
x
Figure 9.6. The The function = 1l + xx and and its its odd odd (left) (left) and and even even (right) (right) Figure 9.6. function f(x} f(x) = extensions.
Exercises 1. = x on 1. Compute Compute the the Fourier Fourier sine sine series series of of ff(x) (x) = on the the interval interval [-1,1] [—1,1] and and graph graph
the sum of the first 50 terms on the interval [-3,3]. [—3,3]. To what function does the series appear appear to converge? What is the period of this function? function?
2. Compute the Fourier cosine series of ff(x) (x) = = x on the interval [-1,1] [—1,1] and graph it it on on the the interval interval [-3,3]. [—3,3]. To To what what function function does does the the series series appear appear to to graph converge? What What is is the the period period of of this this function? function? converge?
9.4.
419 419
Pointwise convergence of series of Fourier series
3. Suppose Suppose I/ : [0, [0,^] —>• R is continuous. continuous. 3. f] -+ R is (a) Under Under what what conditions f0dd continuous? continuous? conditions is is lodd (a) (b) Under Under what what conditions feven continuous? continuous? (b) conditions is is leven 4. Consider aa function function I/ : [—•£,•£] —>• R, and let cn, n = 0, 0, ±1,±2,... 4. Consider [-f, f] -+ R, and let Cn, n = ±1, ±2, ... be be its its complex Fourier coefficients. coefficients. complex Fourier (a) Suppose Suppose I/ is is odd. cn, nn = odd. What What special special property property do do the the coefficients coefficients Cn, = (a) 0, ±1, ±2,... ±2, ... have? have? 0, ±1, (b) Suppose I/ is is even. even. What special property do the cn, n What special property do the coefficients coefficients Cn, n = = (b) Suppose 0, ±1, ±2,... 0, ±1, ±2, ... have? have? 5. Show Show how how to relate the quarter-wave sine sine series series of of I/ : [0, [0,£| -> R full to relate the quarter-wave f] -+ R to to the the full 5. Fourier series series of function. (Hint: (Hint: This function will be denned Fourier of aa related related function. This other other function will be defined on the interval [-2^,2^].) on the interval [2£, 2£].) 6. Show Show how to relate relate the cosine series series of of I/ : [0, [0, t]f] —>• R to full 6. how to the quarter-wave quarter-wave cosine -+ R to the the full Fourier of aa related related function. will be be defined Fourier series series of function. (Hint: (Hint: This This other other function function will defined the interval interval [2f, 2£].) on the on [-2^,2^].)
9.4 9.4
Pointwise convergence convergence of Fourier series Pointwise of Fourier series
Given the relationships relationships that that exist exist among the various various Fourier Fourier series, series, it Given the among the it suffices suffices to to discuss the of complex complex Fourier Fourier series. series. The convergence of of other other series, series, discuss the convergence convergence of The convergence such the Fourier will then then follow follow directly. such as as the Fourier sine sine series, series, will directly. If real- or complex-valued function defined on [-f, f], the partial partial If I/ is is aa realor complex-valued function defined on [—.£, £], then then the sums sums w N
IN(X) =
L
cnei1fnx/l
n=-N
form sequence of of functions. functions. Before discussing the convergence of these partial form aa sequence Before discussing the convergence of these partial sums specifically, we describe different functions can sums specifically, different ways in in which which a sequence sequence of of functions can converge. converge.
9.4.1 9.4.1
Modes sequences of functions Modes of of convergence convergence for for sequences of functions
Given any any sequence sequence {IN {/AT}^ of functions defined defined on Given }]V'=1 on [a, 6] b] and and any any (target) (target) function function =I offunctions I, also on [a, [a, b], we define define three three types types of of convergence convergence of to I./. /, also defined defined on 6], we of the the sequence sequence to That we assign three different different meanings -+ /I as -+ oo." 00." That is, is, we assign three meanings to to "IN "/AT ->• as N N —>• 1. We say converges pointwise to to I/ on on [a, b] b] if, for each each x E [a, b], we 1. We say that that {IN} {/AT} converges if, for £ [a, 6], we have /jv(ar) ->• fI(x) ( x ) as as N ->• oo as N have IN(X) -+ N -+ 00 (i.e. (Le. |/(ar) If(x) -- /AT(X)| fN(X) I -)• -+ 0 as N -> -+ oo). (0).
°
2. We We say say that that {IN} {/AT} converges to /I in (or in in the 2. converges to in L L22 (or the mean-square mean-square sense) ifif ll/ ~ INII /wll —>• as ./V oo. The the (L2) (L 2 ) norm norm of real111-+ 0 as N —>• -+ 00. The reader reader will will recall recall that that the of aa real-
°
420
Chapter 9. Chapter
More about about Fourier series
or g on is or complex-valued complex-valued function function 9 on [a, b] b] is
Therefore, fN —> -+ f in in the the mean-square mean-square sense means that that Therefore, /TV sense means
lb
If(x) - fN(X)12 dx -+ 0 as n -+
00.
3. We say say that that UN} converges to to /f uniformly uniformly on on [a, [a, b] b] ifif 3. We {/AT} converges max {If(x) - fN(X)1 : x E [a, b]} -+ 0 as N -+
00.
Actually, this this definition definition is is stated stated correctly if all all of of the the functions involved Actually, correctly only only if functions involved are so that that the the maximum maximum is is guaranteed guaranteed to to exist. exist. The The general are continuous, continuous, so general definition is: {/AT} converges converges to on [a, if, given given any any te >> 0, 0, definition is: UN} to /f uniformly uniformly on [a, b]b] if, there exists exists aa positive positive integer integer N N,e such such that, that, if if N N > ~ N N,e and and x E [a, 6], b], then there e [a, then \f(x) any If(x) -— /TV(X)| fN(X)1 < < e. t. The The intuitive intuitive meaning meaning of of this this definition definition is is that, that, given given any small tolerance e, t, by by going going far far enough enough out in the the sequence, sequence, /N fN will will approximate small tolerance out in approximate to within within this this tolerance tolerance uniformly, that is, is, on the entire /f to uniformly, that on the entire interval. interval. The following theorem shows that the the uniform uniform convergence convergence of of aa sequence sequence imimThe following theorem shows that plies its its convergence in the the other other two two senses. senses. This This theorem theorem is is followed followed by by examples plies convergence in examples that show show that that no no other conclusions can can be be drawn drawn in in general. that other conclusions general. Theorem 9.8. 9.S. Let Let {fN} sequence of of complex-valued defined on Theorem {/AT} be a sequence complex-valued functions defined [a,6]. -» C, it also [a,b]. //If this sequence converges uniformly uniformly on [a,b] to f : [a,b] [a,b]-+ C, then it mean-square sense. converges to f pointwise and in the mean-square
Proof. See Exercise 5. 5. Proof. See Exercise
0
Example 9.9. 9.9. We define Example define
Nx,
gN(X) =
{
1- N
0,
(x -
*') ,
O::;X<*"
< < N' 2 i.r::;x::;1. 1
N - x
9.7 shows shows the the graphs of of g§, g5, giQ, g1O, and and #20 g20·• Then {gN} converges pointwise Figure 9.7 zero function. Indeed, pAr(O) gN(O) = = 0 0 for all N, gN(O) -» -+ 0, 0, and ifif to the zero N, so clearly clearly pAr(O) 0 < xx ::; < 1, 1, then then for N sufficiently sufficiently large, large, 2/N 2/N < x. x. Therefore Therefore gN(X) gN(x) = 00 for all N N 0< for N for all sufficiently —> O. 0. sufficiently large, large, and and so so gN(x) gN(X) -+ 2 calculation shows that gN —>• -+ 00 in in L L2: A direct calculation :
9.4. 9.4.
421 421
Pointwise Pointwise convergence of Fourier series
Y=9 5 X - _. Y=9 10 (x) . _. _ Y=9 20 (X)
II
0.9 0.8 0.7
I
. if
i I
~ J~
0.6 !
,I
>-0.5 0.4
0.1 0.2
0.6
0.4
0.8
x
The functions functions 95,910,920 Example 9.9). 9.9). Figure 9.7. The g$, gw, #20 (see (see Example The convergence is uniform, since since The convergence is not not uniform,
max {19N(x) - 01 : x E [0, I)} = g
(~)
=
1 for all N.
Example Example 9.10. 9.10. This This example example is is almost almost the the same same as as the the previous previous one. one. We We define define
osx<:kr, < < fij, 2 ~ Sx S1
1 _ X fij
(see Figure 9.8). 9.8). We have HN hN -> -+ 0 pointwise pointwise on on [0,1], and and {hN} does not not converge converge (see Figure We have {h^} does uniformly to to 0 on on [0,1], by by essentially the same same arguments arguments as in the previous examuniformly essentially the as in the previous examIn this example, however, sequence also also fails fails to converge in in the L22 norm. norm. ple. In ple. this example, however, the the sequence to converge the L We have We have IlhN - 011£2 = =
11 f!:
Ih N (x)12 dx
-+
00.
Chapter 9. 9. More More about about Fourier Fourier series series Chapter
422
18
16
Ii
14
I I
I I
I
I
12
I I
I
>-10
I
I"
~I ,
8
II
I
~ 'i
\ \
,
6! " ~I
0.2
0.6
0.4
0.8
x
Figure 9.8. The The functions (see Example 9.10). Figure 9.S. functions h^, h5; HIQ, hlOJ H^Q h 20 (see Example 9.10). We saw saw in in Theorem Theorem 9.8 9.8 that that aa uniformly uniformly convergent convergent sequence sequence also also converges converges We pointwise mean-square sense. that neither pointwise and and in in the the mean-square sense. Example Example 9.9 9.9 shows shows that neither pointpointwise nor nor mean-square convergence implies implies uniform uniform convergence, convergence, while while Example Example 9.10 9.10 wise mean-square convergence shows that convergence does does not imply mean-square mean-square convergence. convergence. We We will shows that pointwise pointwise convergence not imply will see, the context context of of Fourier Fourier series, that mean-square convergence does does not see, in in the series, that mean-square convergence not imply imply pointwise convergence convergence either. either. pointwise 9.4.2 9.4.2
complex Fourier series Pointwise convergence of the complex
We function conconWe now now begin begin to to develop develop conditions conditions under under which which the the Fourier Fourier series series of of aa function verges pointwise, uniformly, and and in in the the mean-square mean-square sense. sense. We We begin with pointwise verges pointwise, uniformly, begin with pointwise convergence. convergence. The partial partial Fourier Fourier series series as integration against against a a kernel kernel The as integration
Our point is partial Fourier Our starting starting point is aa direct direct calculation calculation showing showing that that aa partial Fourier series series of of aa function f/ can can be be written written as as the the integral integral of of f/ times times aa term term from from aa delta delta sequence. sequence. function The part of that this The difficult difficult part of the the proof proof will will be be showing showing that this sequence sequence really really is is aa delta delta that it property. (Delta sequence; sequence; that that is, is, that it satisfies satisfies the the sifting sifting property. (Delta sequences sequences and and the the sifting property were discussed discussed in in Sections Sections 4.6 4.6 and and 5.7, 5.7, but the essence essence of of those those sifting property were but the discussion will be be repeated necessary to have studied discussion will repeated here, here, so so it it is is not not necessary to have studied the the earlier earlier sections. those section be modified sections. The The concepts concepts from from those section must must be modified slightly slightly here here anyway, anyway, to deal deal with with periodicity.) to periodicity.)
9.4. Pointwise convergence of Fourier series
423
We assume that Co, C±1, C±2, . .. are the complex complex Fourier Fourier coefficients coefficients of of 1 We assume that co,c±i,c±2,... are the / :
[-e,e]-+ c: [-*,<]->C:
C
n
= ;e [LL l(s)e-i7rns/L ds,
n
= 0, ±1, ±2, ....
We write write We n=-N
and then then substitute the formula formula for for Ccnn and and substitute the and simplify: simplify: IN(X) = ntN {
= [LL {
;e [LL l(s)e-i7rns/L dS} ei7rnx/L
;e ntN ei7rn(x-s)/f } I(s) ds.
We already already see see that that the the /N IN can can be be written written as as We (9.16)
Dirichlet kernel kernel KN KN is is defined defined by by where the the Dirichlet where
With the the formula finite geometric geometric sum, sum, With formula for for aa finite
and using using some clever manipulations, manipulations, we we can can simplify the formula formula for for KN KN considand some clever simplify the considerably: erably:
1
U
(e i7rli / i )N+1 _ e i7rIi / L -
(ei7rli / l
rN
1
(e i7rli / l (+1/2 _ (ei7rIi/L) -(N+1/ 2) U (e i7rl!/f) 1/2 _ (e i7rli/ f ) -1/2 1
424
Chapter about Fourier Chapter 9. More about Fourier series
1 ei n:(N+l/2)9/l
= 2l
_
e- i n:(N+l/2)(J/i
ei n:(1/2)(J / l _ e-in:(1/2)£J / l
., ((N+~)n:(J) 1 2~sm i 2i sin ( ~~ )
2l • sm
-
((2N+l)n:(J) 2l
2lsin (~n
The N = = 20, 20, is graphed graphed in Figure 9.9. The kernel KN, for TV 9.9. 25~---------r----------~--------~----------~
20 15
10
5
Figure 9.9. 9.9. The kernel K220 Q.• Periodic convolution
The convolution of two functions g9 : R -t and h : R -t The —> C and -» C is defined by
(g*h)(x) =
f
g(x-s)h(s)ds.
The integral integral sign without limits limits means means integration integration over over the the entire entire real real line line (we (we do do The sign without guarantee that these integrals integrals exists). not discuss here conditions on g9 and hh that guarantee The formula for convolution, except IN is is similar similar to to aa convolution, except that that the the integration integration is is over over The formula for /jv the fundamental property its the finite finite interval interval [—1,1]. [-f, fl. A A fundamental property of of an an ordinary ordinary convolution convolution is is its symmetry; of variables symmetry; the the change change of variables that that replaces replaces xx — - ss by by ss yields yields
f
g(x - s)h(s) ds =
f
g(s)h(x - s) ds.
9.4.
Fourier series Pointwise convergence of of Fourier series
425
A similar holds for that in provided the functions A similar formula formula holds for integrals integrals such such as as that in (9.16), (9.16), provided the functions involved are involved are periodic. periodic. A that if A direct direct calculation calculation shows shows that if g, #, hare h are 2C-periodic 2^-periodic functions functions defined defined on on R, R, then then
£ 1
_/(x - s)h(s) ds
= =
1x+£ x-£ g(s)h(x -
1£-l
s) ds
g(s)h(x - s) ds.
The last last step step follows follows from from the the fact fact that >-)• g(s)h(x is 2^-periodic, so its its The that s f-t g(s)h(x -— s) s) is 2C-periodic, and and so integral over any interval 2C is is the the same. We refer refer to integral over anv interval of of length length 21 same. We to
as as the the periodic convolution of of 9g and and h. We can extend the periodic convolution convolution to to nonperiodic nonperiodic functions by We can extend the idea idea of of aa periodic functions by C], then then its its periodic periodic using using the the periodic extension of of aa function. function. If If 9 g is is defined defined on on [-C, [—£, £], extension is gper ER R by by is the the function function g defined for for all all x G per defined
gper (X )
_ { g(x), g(x _ 2kC),
-
-C < x < C, (2k - l)C < x
< 2kC.
For example, we graph gper, g(x) = x on For example, in in Figure Figure 9.10 9.10 we graph the the function function g where g(x] on [-1,1]. [—1,1]. per, where For any 9 and h defined on [-C, C], periodic or not, we interpret the periodic For any g and defined on [—i, ^], periodic or not, we interpret the periodic convolution of and hh to be to be convolution of 9g and
We will will often often write convolution as as simply We write this this convolution simply
[ll g(x - s)h(s) ds, but 9g and are assumed to be be replaced their periodic but and hh are assumed to replaced by by their periodic extensions extensions if if necessary necessary (that is, if they they are are not (that is, if not periodic). periodic). 2C-periodic. Provided Provided the complex The kernel KN KN is is 2^-periodic. complex Fourier series series of of /f conconZ7rnx /£ verges, verges, the the limit limit is obviously obviously aa 2C-periodic 2£-periodic function function (because (because each each function function ei1fnx ^ is 2C-periodic). Therefore, extension of is 2^-periodic). Therefore, it is is natural natural to to consider consider ffpen , the periodic extension of per f, then have /, and and to to interpret interpret (9.16) (9.16) as as aa periodic periodic convolution. convolution. We We then have (9.17)
aa fact below. fact which which we we will will use use below.
426
Chapter 9.
More about Fourier series
3r-----~------~------~------~----~------~
2
-1
-2 -3~----~----~~----~----~~----~----~
-3
-2
-1
0
2
3
x Figure The periodic extension extension of [—1,1] -> R, g(x) x. Figure 9.10. The of g9 :: [-1,1)-+ g(x) — = x. KN as a a delta delta sequence KN as sequence
the kernel KN of An examination examination of the graph of the K^ (see Figure 9.9) shows that most of concentrated on a small interval O. The effect effect of this on its "weight" is concentrated interval around ()0 = = 0. the periodic convolution
integral produces a weighted average of the values of f, of is that the integral /, with most of the values of fI(s) effect is accentuated accentuated as the weight on the the ( s ) near s = — x. Moreover, this effect N -+ oo, 00, so that if 1 regular enough, we obtain obtain pointwise convergence. Thus KN KN N —> / is regular acts like a delta sequence (see Section 4.6). / be in order for for convergence to occur? It turns out that How regular regular must 1 continuity is not sufficient, there are continous functions on [-€, [-l, i]l) whose mere continuity sufficient, as there Fourier series fail to converge at an infinite number of points. Some differentiability differentiability necessary to guarantee we do not really want of 1 / is necessary guarantee convergence. On the other other hand, we we often often encounter discontinuities discontinuities in practical to require /f to be continuous, since we practical problems, of /, fper. I, then then of of fper. problems, if if not not of regularity for our purposes is that of piecewise smoothness. A suitable suitable notion of regularity We make the the following definitions. Definition 9.11. Suppose f is a real- or complex-valued complex-valued function function defined defined on (a,b), 9.11. Suppose (a,b), except possibly at a finite say that 1 except possibly finite number of of points. We say f has a jump discontinuity
427
9.4. Pointwise convergence convergence of Fourier series
at XQ (a, b) if at Xo EG (a, b) if lim f(x) and lim f(x) x-tx;-
x-txci
both both exist exist but but are are not not equal. equal. (Recall (Recall that that if if both both one-sided one-sided limits limits exist exist and and are are equal equal to ff(xo), ( x o ) , then then ff is is continuous continuous at at xx = — xo.) XQ.) to We piecewise continuous We say say that that ff is is piecewise continuous on on (a, (a, b) b) ifif 1. points in b); 1. ff is is continuous continuous at at all all but but aa finite finite number number of of points in (a, (a,b); 2. jump discontinuity; 2. every every discontinuity discontinuity of of ff in in (a, (a, b) b) is is aa jump discontinuity;
3. limx-ta+ lima._>a+ ff(x) ( x ) and and limx-tb\im.x_^b- f(x] exist. 3. f(x) exist. Finally, we we say that ff is is piecewise smooth on on (a, (a, b) b) if if both both ff and and df df/dx say that piecewise smooth Idx are are Finally, piecewise continuous piecewise continuous on on (a, (a, b). b). An example of of aa piecewise piecewise smooth smooth function function is is given given in in Figure 9.11. An example Figure 9.11. 10~------------------~------------------~
5
-
5
Figure 9.11. A function. Figure 9.11. A piecewise piecewise smooth smooth function. Before result, we we will Before we we give give the the main main result, we give give two two lemmas lemmas that that we will find find useful. useful. Lemma 9.12. 9.12. Lemma
(9.18)
Proof. This proved by by direct 4). Proof. This is is proved direct calculation calculation (see (see Exercise Exercise 4).
o
428
Chapter 9. More about Fourier series
The next lemma is called Bessel's inequality and follows directly from the projection theorem.
Lemma 9.13. 9.13. Suppose {0j ;: jj = = 1,2, l,2,...} is an an orthogonal orthogonal sequence sequence in in an Lemma Suppose {cPj ... } is an {infinite-dimensional} (infinite-dimensional) inner product space V, and f E € V. Then
Proof. N, Proof. By the the projection theorem, for for each each JV,
is N = span cP1 , cP2, N} closest N is is is the the element element of of the the subspace subspace V VN span {{i, < / > 2... , . . . , cP 0jv} closest to to f, /, and and f/AT orthogonal N· Therefore, by the the Pythagorean theorem, orthogonal to to f/ -— f/N. Therefore, by Pythagorean theorem,
This implies that IlfNII~
::;
Ilfll~ for all N = 1,2,3, ....
cP1, c/>2, Moreover, since 0i, 2, cP3, ^>s, ... • • • are are mutually orthogonal, we have (also by by the the Pythagorea Pythagorea theorem) theorem) n
JVf
Therefore,
t
l(f, cPj)12 ::; j=l (cPj, cPj)
holds for each N, TV, and and the the result follows.
Ilfll~
0
Since a Fourier series is based on an orthogonal series, Bessel's inequality can be applied to to the the sequence sequence of of Fourier Fourier coefficients. coefficients. For For instance, instance, if if en, cn, n n = = can be applied 2 0, ±2, ... are Fourier coefficients L2( -f, f), then 0, ±1, ±1, ±2,... are the the complex complex Fourier coefficients of of f/ E eL (—l,l), then
(I, ei7rnx/l) and Therefore, Therefore,
9.4.
429
Pointwise convergence convergence of Fourier series
Bessel's inequality yields 00
n=-oo
Similar Fourier sine Similar results results hold hold for for Fourier sine or or cosine cosine coefficients. coefficients. We can and prove prove the main result pointwise converconverWe can now now state state and the main result concerning concerning the the pointwise gence Fourier series. that the Fourier gence of of Fourier series. The The reader reader should should note note that the convergence convergence of of the the Fourier series terms of properties of those of of series is is naturally naturally expressed expressed in in terms of the the properties of ffper rather than than those per rather /• f· Theorem 9.14. 9.14. Suppose Suppose f : (-i, -+ C smooth. Then the complex Theorem (—l,€)i) -» C is piecewise smooth. Fourier series of f,f, 00
L
cnei1fnx/l,
n=-oo
if converges to ffper(x) if x is a point of of continuity of of ffper, to per(x) per, and to 1 -2 [
lim fper(s)
s--+x -
+ s--+x+ lim fper(S)]
if fper has a jump jump discontinuity at x. if fper atx. Proof. As before, we write Proof.
L
=
1 £
N
fN(X)
cnei1fnx/l
=
n=-N
KN(S)f(x - s) ds.
-l
If ffper is continuous continuous at at x, then then If per is
fper(x)
1
= -2
[lim fper(s) s--+x-
+ s--+x+ lim fper(X)] .
From now on, for for conciseness, will write write /f rather than ffper and remember remember to to From now on, conciseness, we we will rather than per and interpret interpret ff(s) ( s ) in in terms terms of of the periodic periodic extension extension of of /f whenever s is outside outside of of the the infprval (-i, f — P i). t\ Also, Alsn -rarp writ.p we write interval
f(x-)
= s--+xlim f(s),
f(x+)
= s--+x+ lim f(s).
With this understanding, our our task is to to show show that, for each [—f.,1], With this understanding, task is that, for each x G E [-i,i], lim fN(X) =
N--+oo
~2 [f(x-) + f(x+ )].
To do do this, it suffices suffices to show that To this, it to show that 0
lim N--+oo
1 -l
1
KN(S)f(x - s) ds = - f(x+) 2
(9.19)
430
Chapter 9. More about Fourier series
and and
1
l
lim
N-+oo
1 0
KN(S)f(x - s) ds
= -2 f (x-).
(9.20)
We will that (9.20) is similar. Equation (9.20) is We will prove prove that (9.20) holds; holds; the the proof proof of of (9.19) (9.19) is similar. Equation (9.20) is equivalent eauivalent to
By (9.18), (9.18),
fl
1
'2 f (x-) = f(x-) 10
KN(S) ds
ft
= 10
KN(S)f(x-) ds,
so so
fl KN(S)f(x - s) ds -
10
! f(x-) =
fl KN(S) (f(x - s) - f(x-)) ds
10
2
~ fl f(x - s) ~/(x-) sin ((2N +
=
2i10
sin(u)
l)7rS) ds.
2i
We can recognize this integral integral as as 1/(2i) l/(2£) times the L inner product (on the interval We can recognize this times the L22 inner product (on the interval (0,£)) of the the functions functions (0, £)) of f(x-s)-f(x-) F(x)(s) = . ('IrS) sm 2l and and . ((2N + sm 2i .
l)7rX)
The sequence The sequence .. N - 0" 1, 2 ... } . ((2N+l)11'X) 2l { sln
is an orthogonal orthogonal sequence sequence with the L inner product (see, for for example, example, is an with respect respect to to the L22 inner product (see, Exercise Bessel's inequality Exercise 5.2.2). 5.2.2). Bessel's inequality then then implies implies that that
~
~o
1(
.
F(x),sm
((2N
+ l)7rS)) 12 < 00 2i
and therefore and therefore
~ f1f(x-s)-f(x-). ((2N+1)7rs)d 2£
10
sin (~;)
sm
2l
_(D . ((2N+1)7rS)) 2£
s-
.l'(x),sm
-t
O.
This is the desired result. result. This is the desired However, there is is one problem with (Lemma However, there one problem with this this argument. argument. Bessel's Bessel's inequality inequality (Lemma 9.13) requires F^ and and the the orthogonal sequence belong to aa common 9.13) requires that that F(x) orthogonal sequence belong to common inner inner product space. take this to be of all piecewise continuous product space. We We will will take this space space to be the the space space V of all piecewise continuous functions defined defined on on [0, [0,£|. Certainly the the functions functions functions £J. Certainly
. ((2N + sm 2£
l)7rX) ' N = 0,1,2, ...
431 431
9.4. Pointwise convergence of series of Fourier Fourier series
piecewise continuous). proof, therefore, therefore, belong belong to to V (continuous (continuous functions functions are are piecewise continuous). The The proof, to showing that F(x) E V. reduces reduces to showing that F^ e V. Since has only jump discontinuities, F(x) jump discontinuSince ffper only jump discontinuities, F( also has has only only jump discontinuper has x) also there is is aa zero the denominator. we show show ities, ities, except except possibly possibly at at s == 0, 0, where where there zero in in the denominator. If If we that that lim F(x)(s) s-+o+
exists number, this will show F(x) is is piecewise exists as as aa finite finite number, this will show that that F^ piecewise continuous, continuous, and and hence hence in V. Since Since /f and and df/dx each has at most of jump discontinuities, in df / dx each has at most aa finite finite number number of jump discontinuities, interval (O,E) and there is there is an an interval (0,e) such such that that the the function function s I-t !->• ff(x ( x - s) is is continuous continuous and differentiable E). It follows value theorem differentiate on on (0, (0,e). follows from from the the mean mean value theorem that, that, for for each each s €E (0,e), there exists 7 such that (O,E), there exists ISS Ee (0,1) such that
df f(x - s) - f(x-) = - dx (x - fSS)S. Also, by the the mean mean value value theorem, theorem, there Ass E Also, by there exists exists A € (0,1) such such that that
. (7rS) 7rS sm 2£ = cos (AS7rs) U 2£ .
°
Thus, for for all all s > 0 near zero, we have Thus, near zero, we have F(x)(s) = =
f(x-s)-f(x-)
. (1[8)
sm
2f
-1x(x - ISS)S cos (>'.1[S) 1[S
--"""--':7"'-,---'--
2l
2f
1x(x - ISS) cos (>'o1[S).1!:.. 2l
-+ -
~(x-) dx
1[
2£
as s -+ 0+ .
2l
This shows shows that that lims-+o+ lims_>0+ F( exists, which completes the the proof. This F(:r;)(s) which completes proof. x)(s) exists,
0
Example 9.15. R Example 9.15. We will compute the complex Fourier series of 9g :: [-1,1] [—1,1] -+ -> R defined by g(x) = x. 9.10 shows shows that thatggper is piecewise continuous defined by g(x) = x. Figure Figure 9.10 piecewise smooth, smooth, and and continuous per is except at integral of x. therefore expect expect that the Fourier series of of 9g will x. We therefore except integral values values of At the points of discontinuity, the converge to ggper(x) except when x is an integer. integer. At per(x} except average of of the leftleft- and right-hand limits is 0, 0, which which will be be the limit of of the series at those points. The The Fourier coefficients coefficients of of 9g are Co =
~ ill X dx = 0,
111 .
i(-I)n xe-1[mx dx = - - - , n = ±1, ±2, ....
en = -
2
-1
n7r
432 432
Chapter about Fourier Fourier series Chapter 9. More about
The partial Fourier series N
L
cn e7rin",
n=-N
for N = 10,20,40, are shown in Figure 9.12. for 10,20,40, are 1.5~-----r------r------r------r------r----~
-1 .5 I..-_ _ _ _--L._ _ _ _ _ _.L...-_ _ _ _--L._ _ _ _ _ _.L...-_ _ _ _--L._ _ _ _----l
-3
-2
-1
0
2
3
x
Figure 9.12. The partial Fourier series /4o f40 for Example 9.15. Figure 9.12.
The following results results can be derived the relationship relationship be beThe following can be derived immediately immediately from from the tween the the complex Fourier series various other other forms forms of Fourier series. tween complex Fourier series and and various of Fourier series.
Corollary 9.16. 9.16. Corollary 1. Suppose Suppose f : (-I, I) -t of (—£, i) -> R is piecewise smooth. Then the full Fourier series of j, f,
converges to !per(x) fper(x} if if x is a point continuity of of ffper, point of of continuity and to per, ana
~2 [lim fper(s) + lim jper(s)] s---+"s---+,,+ if jump discontinuity discontinuity at x. if ffper per has a jump atx.
9.4.
Pointwise convergence of Fourier series
433
2. Suppose ff :: (0, (0, £) t) -t —> R R is is piecewise and let let ffodd be the the periodic, 2. Suppose piecewise smooth, smooth, and periodic, odd odd 0dd be extension of of ff to to R (that (that is, is, the the periodic extension to to R of of the the odd odd extension extension extension periodic extension of °f f to t° (-£,£)). (—£•>£))• Then the Fourier sine series of of f,f, 00
Lb
n
sin (n;x),
n=l
converges x is point of converges to to ffOdd(X) if % is aa point of continuity continuity of of ffodd' and to to 0dd(%) if 0dd, and
if fodd has jump discontinuity if fodd has aa jump discontinuity at at x. x.
3. Suppose (0,£) —> R R is is piecewise and let let ffeven be the the periodic, even 3. Suppose ff :: (0, £) -t piecewise smooth, smooth, and periodic, even even be R (that periodic extension extension extension of of ff to toU (that is, is, the the periodic extension to to R of of the the even even extension extension of ff to to (-£, (—£,1)). Then the the Fourier cosine series series of of f,f, of £)}. Then Fourier cosine
converges to to leven(x) feven(x) if if xx is is aa point of continuity continuity of of leven, feven, and and to to converges point of
if leven feven has has aa jump discontinuity at atx. if jump discontinuity x. Proof. See See Exercise Exercise 6. 6. Proof.
0
ExalTIple R be periodic, odd odd Example 9.17. 9.17. Let Let If : (0,1) (0,1) -t —> R be defined defined by by ff(x) ( x ) = 1. 1. Then Then the the periodic, extension extension of of ff,, lodd, f0dd, is is defined defined by by f
() = {1, odd x -1,
if x E (2k, 2k + 1) (k an integer), if x E (2k -1, 2k) (k an integer)
(the so-called "square wave"). wave"). Thus Thus ffodd is discontinuous discontinuous at at every every integer integer value value of of (the so-called "square 0dd is x; at those those points of dicontinuities, dicontinuities, the the average average of of the the left left and and right right endpoints endpoints is is Xi at points of zero. partial sine sine series zero. Figure Figure 9.13 9.13 shows shows lodd f0dd together together with with its its partial series having having 5, 5, 10, 10, 20, 20, and 40 40 terms. terms. As the above above corollary corollary guarantees, guarantees, the the sine sine series converges to to ff at and As the series converges at every point of continuity continuity (every (every nonintegral nonintegral point, in this this case) case) and and to to zero at every every every point of point, in zero at point of discontinuity. discontinuity. point of We can can draw draw some some further further conclusions conclusions from from Theorem Theorem 9.14 9.14 and and Corollary Corollary 9.16: 9.16: We
434
Chapter 9. More Chapter 9. More about about Fourier Fourier series series
2~------------------~
2
- -
- -
6 ........ ...A
06 ......... 06
o -1
A
-1
-2 -2
-1
o
1
2
-2 -2
V ....
-1
A
~
.... V
o
A
A
V .... - .... V
2
x
X
2
2~------------------~
1
1 ....-
0
o
..
-1
-1
-2 -2
-1
o
x
1
2
-2~----------------~ -1 -2 o 1 2
x
Figure 9.13. square-wave function function from Example 9.17, 9.17, together with Figure 9.13. The square-wave its Fourier sine series with 5, 5, 10, 10, 20, and 40 terms. terms. Corollary Corollary 9.18. 9.18.
1. If If f : (-f, ~ C smooth, then ffper (—£,£)f) —> C is is continuous and and piecewise smooth, is continuous per is everywhere except ±3f, . ... It follows that the except {possibly} (possibly) at the points ±f, ±£,±31,— converges to f(x) f(x) for for all x €E (-f, f). Fourier series of of f converges (—i,i). If f : [—t,£\ [-f,f] ->• ~ C and f(—t] f( -f) = 2. If C is is continuous and and piecewise smooth, and = f(f), f(l), then ffper is everywhere, and so the of converges is continuous everywhere, Fourier series of f converges per to f(x] f(x) for for all x E [-f,f] {including € [—•£,•£] (including the endpoints}. endpoints). If f : [0, ~ R R is is continuous and periodic 3. If [0,1]f] —>• and piecewise smooth, then ffodd (the periodic 0dd {the of ff}) is extension of of the odd extension of is continuous everywhere everywhere except except {possi(possibly} 0, ±f, ... It follows that the Fourier sine series of f bly) at the points 0, ±£, ±2£, ±21, .— f(x) for all x E converges to f(x) e (O,f). (0,^).
4· If f :'• [O,f] ~ R R is continuous and f(O) == ff(f) 4- If [0,^] —>• and piecewise smooth, and and /(O) ( i ) == 0, then ffodd is continuous everywhere. of everywhere. It follows that the Fourier sine series of 0dd is converges to to f(x] f(x) for for all all xx €E [0,£] [0,£] (including {including the the endpoints). endpoints}. ff converges
9.4.
435 435
Pointwise Pointwise convergence of Fourier series
5. If If ff : [0, R is continuous and piecewise smooth, then ffeven periodic [0, t]£] -+ —>• R and piecewise even (the periodic extension of the even extension of of ff)) is continuous everywhere. everywhere. It follows that of f converges to f(x) [0,^] (including f(x) for for all x EG [0,£] the Fourier cosine cosine series of the endpoints). Proof. The directly from Corollary 9.16, 9.16, except Proof. The conclusions conclusions all all follow follow directly from Corollary except for for the the [0, £] -+ is following: following: If If /f : [0,1] —> R R is is continuous, continuous, then then ffeven is continuous, continuous, and and ffodd even is 0dd is continuous provided /(O) f(O) = f(£) = = 0.0. The reader is to verify verify these these conclusions conclusions continuous provided — f(i] The reader is asked asked to in D in Exercise 7. 7. D When ffper feven) is continuous everywhere, everywhere, we we can When (or ffodd, or feven) is continuous can draw draw aa stronger stronger per (or 0dd, or conclusion, as as we show in in the the next section. The The reader should note that, from from the the conclusion, we show next section. reader should note that, point of view of approximating approximating aa given given function, function, the series is powerful point of view of the cosine cosine series is more more powerful than everywhere for than the the sine sine series, series, since since ffeven is continuous continuous everywhere for any any continuous continuous /f : even is [0, Therefore, for Gibbs's phenomena phenomena cannot cannot occur with the [Q,£]£] -+ -» R. R. Therefore, for example, example, Gibbs's occur with the cosine function. cosine series series approximating approximating such such aa function.
Exercises 1. [-1f,1f]-+ R be be defined defined by by f(x) f(x) == x33 .. Does Does the the full full Fourier of 1. Let Let /f:: [-7r,7r] ->• R Fourier series series of f/ converge on R? is the the limit of the the series? converge (pointwise) (pointwise) on R? If If so, so, what what is limit of series? 3 2. Define Define f/ : [-2,2]-+ by f(x] f(x) = = x x3 + of 2. [-2,2] ->• R R by + 1. 1. Write Write down down the the limits limits of
(regarded as as aa function function defined defined on on [0,2]), (a) the the Fourier Fourier sine sine series of f/ (regarded (a) series of [0,2]), (b) the Fourier cosine cosine series series of of f/ (regarded (regarded as as aa function defined on on [0,2]), [0,2]), (b) the Fourier function defined and ((c) c) the the full full Fourier Fourier series series of of f./.
3. Find an example example of function f/ : [0, [0,1] —>• R such that that it it is is not the that 3. Find an of aa function 1] -+ R such the case case that the Fourier Fourier cosine cosine series series of of 1 / converges converges to / at at every every xx E € [0, [0,1]. the to 1 1]. 4. Using Using the 4. the original original formula formula for for KKN, N , K
N
(()) =
~ 2£
N
"'" ~
ei1fnO/l
,
n=-N
prove that (9.18) holds. 5. Prove Theorem Theorem 9.8. For simplicity, of the the functions are continuous. 5. Prove 9.8. For simplicity, assume assume all all of functions are continuous. 6. Prove Corollary 6. Prove Corollary 9.16. 9.16. 7. £] -+ R is continuous everyevery7. Suppose Suppose /f : [0, [0,1] —>• R is continuous. continuous. Prove Prove that that feven is is continuous where, and f(£) = where, and that that if if 1(0) /(O) == f(£) — 0, 0, then then lodd f0dd is is continuous continuous everywhere. 8. 8. (a) (a) Extend Extend Corollaries Corollaries 9.16 9.16 and and 9.18 9.18 to the the case case of of the the quarter-wave quarter-wave cosine series. series. (Hint: (Hint: see see Exercise 9.3.6.) 9.3.6.)
Chapter 9. More about Fourier series
436 436
by f(x] f(x) = x. To (b) Define /f :: [0,1] -+ -> R by = 11 + x. To what function does the quarter-wave cosine series of f/ converge? 9. (a) Extend Corollaries 9.16 and 9.18 to the case of the quarter-wave sine Extend Corollaries series. (Hint: (Hint: see Exercise 9.3.5.) (b) Define Define /f :: [0,1] [0,1] -+ —> R by = 11 + To what what function function does does the by f(x] f(x) = + x. x. To the (b) quarter-wave quarter-wave sine sine series series of of f/ converge?
9.5 9.5
Uniform Uniform convergence convergence of of Fourier Fourier series series
In the previous section, we the pointwise convergence of the Fourier series we proved the of of aa piecewise smooth smooth function. function. We We will now consider consider conditions conditions under under which which the the convergence uniform. The property of uniform convergence very dedeconvergence is is actually actually uniform. The property of uniform convergence is is very sirable, sirable, since since it it implies implies that that aa finite finite number number of of terms terms from the Fourier series series can approximate approximate the function accurately accurately on the entire interval; in particular, particular, uniform convergence rules out the possibility possibility of Gibbs's phenomenon, which was introduced in Section Section 5.2.2 5.2.2 and and will will be discussed further further below. in be discussed below. The rapidity with which Fourier series converges, and, and, in in particular, particular, whether The rapidity with which aa Fourier series converges, whether it converges or not, is intimately intimately related related to to the the rate at which which its its coefficients coefficients it converges uniformly uniformly or not, is rate at converge to zero. We We address this question first. first.
9.5.1
Rate of decay of Fourier coefficients coefficients
In we will n = In the following following theorems, theorems, we will show show that that the Fourier Fourier coefficients coefficients Ccnn,, n = 0, ±1, ±2, ... , of satisfy 0,±1,±2,..., o f f/ satisfy (9.21) determined by the degree of smoothness of f. where the exponent exponent kk is determined /. Recall that (9.21) means that there is is a constant constant M M > 0 such that M
Icnl :::; Inlk
for all
n.
The same condition condition that guarantees pointwise pointwise convergence convergence of of Fourier Fourier series series The same that guarantees also ensures that that the Fourier coefficients coefficients decay-converge decay—converge to zero-at zero—at least least as fast as as l/n, 1/n, n n == 1,2,3, 1,2,3, .... — fast Theorem £) -+ piecewise smooth, en, Theorem 9.19. 9.19. Suppose Suppose that that f : (-£, (—£,1) -> C C is is piecewise smooth, and and let let c n, nn = ±1, ±2, ... , be Fourier coefficients = 0, 0, ±1, ±2,..., be the the complex complex Fourier coefficients of of f. f . Then Then
Proof. finite number of discontinuities of of Proof. Since f/ is piecewise piecewise smooth, there are a finite f/ and/or We label points as as and/or dfldx. df/dx. We label these these points -£ <
Xl
< X2 < ... < Xm-l < £
437
9.5. Uniform convergence convergence of Fourier series
and write XQ Xo = xm We then then have have and write — -e, —t, x = e. t. We m =
Cn = ;e [ff !(x)e-i7rnx/f dx =
~ 21!
f= i j=l
X ; f(x)e- i7rnx /£ dx. X;-1
It then suffices to for each each jj = = 1,2, 1,2,..., It then suffices to show show that, that, for ... , m, m, x ; f(x)e-i7rnx/£ dx = 0 (~) . X;-1 Inl
i
Since f/ is is continuously continuously differentiable on (x ( x jj --1 i , X, Xj )j ,), and and the Since differentiable on the limits limits of of both both f/ and and df / dx exist the endpoints of this this interval, we can can apply apply integration integration by by parts parts to df/dx exist at at the endpoints of interval, we to obtain obtain x; X; df 'e [ -i7rnx/£] x; f(x)e- i7rnx /£ dx = f(x) e . __z_ _(x)e- i7rnx /£ dx X;_1 -z7fn/e X;_1 7fn X;-1 dx
i
i
=
!!... (!(x, - )e- i7rnx ;/f n7f J
i
f(x
'-1 J
+ )e- i7rnX ;_l/f)
X ; -df (x )e -i7rnx/f dx. 7fn X;_1 dx
- ie -
Since each of three terms terms is is bounded bounded by by aa constant constant times times l/lnl, Since each of these these three l/|n|, the the result result follows. We have have used used the the fact that! df /dx, being being piecewise piecewise continuous, continuous, are are follows. We fact that / and and df/dx, both bounded e). 0 both bounded on on (-e, (—1,1). When f/ has has some some additional its Fourier Fourier coefficients coefficients can be shown shown When additional smoothness, smoothness, its can be to decay more rapidly. Theorem 9.20. 9.20. Suppose Suppose f : (-e, C has the property property that its its periodic periodic extenTheorem (—(.,£)e) -+ —>• C sion and the first k — 2 derivatives of f are continuous, where kk >2, and the first k 2 derivatives of fper are continuous, where ~ 2, and and sion ffper per per the (k -— l)st derivative of f is piecewise smooth. smooth. If If c Cn, = 0, ±1, ±1, ±2,..., ±2, ... , are the l)st derivative n, n = Fourier coefficients complex Fourier coefficients of f,f, then
continuous and its derivative smooth, then particular, if In particular, if ffper derivative is piecewise smooth, per is continuous
Proof. Suppose / dx is piecewise smooth. smooth. Then, Then, by by inteinteProof. Suppose ffper is continuous continuous and and df df/dx is piecewise per is gration by by parts, parts, we have gration we have
I
t f(x)e-i7rnx/f dx
_£
= [-i7rnX/t]f f(x) e .
-Z7fn/e -t
__ z'e
1£
dlf i7rnx /£ dx _(x)e7fn -f dx
Chapter 9. More about Fourier series
438
iTfn since eel9 iIJ Since fjper -£+) = /(^—), j(£-), and e™ eiTfnn = ee-I7rn is 27r-periodic. per is continuous, fj((—(.+) Therefore, we obtain C
. Ie
n = __z_ 27m
-l
dj _(x)e-iTfnx/l dx dx
--~d n,
-
7rn
where d d n,, n n = 0, 0, ±1, ±I, ±2, ... , are the Fourier coefficients coefficients of dj /dx. By Theorem 9.19, ±2,..., df/dx. 9.19, we have s * \
and so and so
desired. as desired. The proof proof of the general Fourier coefficients coefficients general case is similar; one shows that the Fourier 11 k k1 1 k k2 2 j /dX Fourier coefficients j /dX of - f/dx of d kk~ ~ - are of order order I/Inl, l/|n|, the Fourier coefficients of d kk~- 22f/dx ~ - are of order l/|n| , and so forth forth to the desired to obtain obtain the desired result. result. 0 order 1/lnI2, and so Example 9.21. Example 9.21. 1. Define Define f : (-1,1) = x. Then fjper has jump discontinuities (—1,1) -t —> R R by fj(x) (x) = discontinuities per (at (2k -- 1), ±2, .. .), and, and, by 9.20, the (at x = (2k 1), kk = 0, 0, ±I, ±1,±2,...), by Theorem Theorem 9.20, the Fourier Fourier be of of order I/Inl. In fact, the Fourier coefficients coefficients coefficients of of f must be l/|ri|. In coefficients are Cn
i(-I)n 7rn
= - - - , n = ±1,±2, ... ,
with Co CQ = = 0. 0. with 2. Define Define f : (-1,1) x 22.. Then ffper is continuous, but its (—1,1) -t —>• R R by ff(x) (x) = = x per derivative has jump discontinuities discontinuities (at .. .J. By derivative has jump (at xx = = (2k (2k -— 1), kk = = 0, ±1, ±2, ±2,...). By Theorem Fourier coefficients In fact, Theorem 9.20, 9.20, the the Fourier coefficients of of ff must must be be of of order order 1/lnI2. l/|n|2. In fact, the Fourier coefficients are the Fourier coefficients are Cn
=
2(-I)n 2 2 ' n = ±I, ±2, ... , 7rn
with Co CQ = — 1/3. with 1/3. 3. Define ff :: (-1,1) by ff(x) = xx 55—2x -2x 33+x. are 3. Define (—1,1) -t —>• R by (x) = +x. Then Then ffper and its its derivative derivative are per and continuous, but its its second derivative has has jump discontinuities (at (atxx = (2k -1), —I), continuous, but second derivative jump discontinuities = (2k
Fourier series 9.5. Uniform convergence of Fourier series
439
kk = — 0, ±1, ±1, ±2, ±2,...). By Theorem Theorem 9.20, 9.20, the the Fourier Fourier coefficients coefficients of of ff must must be be of .. .). By of order 1/lnI3. l/|n| 3 . In the Fourier Fourier coefficients coefficients are order In fact, fact, the are Cn
=
-8i(n 2 7f2
15)( _1)n
-
55
'
7fn
n=±1,±2,... ,
with Co CQ =— 0. 0. with
4. Define ff :: (-1, 1) ~ f(x) = and 4- Define (—1,1) —>• R by by f(x] = X4 x4 -— 2X2 2x2 + + 1. 1. Then Then ffper and its its first first two two per jump discontinuities derivatives derivatives are are continuous, continuous, but but its its third third derivative derivative has has jump discontinuities {at (at = (2k (2k — - 1), = 0,±1,±2,...J. 0, ±1, ±2, .. .). By By Theorem Theorem 9.20, the the Fourier Fourier coefficients xx — 1}, kk = coefficients of ff must must be be of of order order 1/lnI4. l/\n 4 . In of In fact, fact, the the Fourier Fourier coefficients coefficients are are Cn
=
-24( -1)n
4 4 ' n=±1,±2,... ,
7fn
with Co CQ == 8/15. 8/15. with In In Figure Figure 9.14 9.14 we we graph graph the the error error in in approximating approximating each each of of the the above above four four functions functions by Fourier series series with with 41 lowest by aa truncated truncated Fourier 41 terms terms (the (the constant constant term term and and the the 20 20 lowest frequencies). The results results are are as as expected; expected; as as fIper gets smoother, the convergence of frequencies). The gets smoother, the convergence of per the Fourier Fourier series to 1 f becomes becomes more more rapid. rapid. the series to
9.5.2 9.5.2
Uniform convergence Uniform convergence
It now now follows follows immediately immediately that, if fIper is sufficiently sufficiently smooth, smooth, then its Fourier Fourier series series It that, if then its per is converges converges uniformly uniformly to to f./. Theorem Suppose piecewise Theorem 9.22. 9.22. Suppose ff :: (-l,l) (—1,1) ~ —> C C is is continuous, continuous, dfldx df/dx is is piecewise smooth, and and f(—i) = 1f ((l). l } . Then Then the the partial sums of of the the complex complex Fourier smooth, f (-l) = partial sums Fourier series series of 01 ff converge on RJ. converge uniformly uniformly to to f1 on on (—i,t) (-l,l) (and (and therefore therefore to to fIper on R). per
Proof. The boundary boundary conditions conditions imply imply that that fIper is continuous, continuous, so so by Theorem Proof. The by Theorem per is 9.14, we know that that the Fourier series series of of 1 / converges converges pointwise point wise to to !per fper on on R. R. We 9.14, we know the Fourier We must therefore therefore show show that that the convergence is is uniform. uniform. Since Since must the convergence 00
fper(x) =
L
cnei1fnx/£,
n=-oo
we we have have N
Iper(x) -
-N-l
L
cnei1fnx/l =
n=-N
L
00
cnei1fnx/£
+
n=-oo
L
cnei1fnx/£.
n=N+l
Therefore, Therefore, by by Theorem Theorem 9.20, 9.20, there there is is aa constant constant M such such that that -N-l
N
Iper(x) -
L n=-N
cnei1fnx/£
L
n=-oo
00
cnei1fnx/£
+
L n=N+l
cnei1fnx/£
440
Chapter 9. More about Fourier series
0.01 r--------------a
0.5
-0.01
-0.5 -1L------------~
-1
-0.5
o
-0.02 .......- - - - - - - - - - - - " -1 -0.5 o 0.5 1
0.5
X
X
10-4
5~X--------------------__,
_2L------------~
-5L-------------~
-1
-0.5
o
0.5
o
-0.5
-1
0.5
1
X
X
Figure 9.14. function from from Example Figure 9.14. The The errors errors in in approximating approximating each each function Example its partial partial Fourier 9.21 9.21 by by its Fourier series series with with 41 terms: terms: ff(x) ( x ) = xx (upper (upper left), left), ff(x) ( x ) = xx22 f(x) = x x44 -- 2X2 (upper (upper right), right), ff(x) ( x ) = xx 55 -— 2x 2x33 + + xx (lower (lower left), left), f(x] 2ar2 + + 11 (lower (lower right). right). -N-l
00
: :; 2:
ICnei1rnx/il
n=-oo
2:
ICnei1rnx/ll
n=N+l 00
:::; 2M
+
2:
1 n2
n=N+l
i1rnx / i I = = 1). This This holds holds for all all xx €E R, R, and so so (using the fact that Ieei7rnx/l N
fper(x) -
2:
00
cnei7rnx/i :::; 2M
n=-N
2: n=N+l
Since the infinite series series 00
1
Ln
n=l
2
1 2" for all x E R.
n
9.5. 9.5.
441 441
Uniform convergence convergence of Fourier series Uniform of Fourier series
is convergent, convergent, the of the series tends tends to zero, which which shows is the tail tail of the series to zero, shows that that N
fper(x) -
L
cnei'lmxll
n=-N
tends to zero at aa rate is independent independent of of xx E 6 R. This proves desired result. result. tends to zero at rate that that is R. This proves the the desired D
As of other As with with pointwise pointwise convergence, convergence, the the uniform uniform convergence convergence of other types types of of Fourier series can can be be deduced from their complex Fourier Fourier series. Fourier series deduced from their relationship relationship to to the the complex series. Corollary 9.23. 9.23. Corollary 1. Suppose (—t,i]I!.) -+ —>• R is continuous df/dx is piece1. Suppose ff :: (-I!., R is continuous and and piecewise piecewise smooth, smooth, elf / dx is piecewise Then the the full of ff converges converges wise smooth, smooth, and and f(—£) f( -I!.) = = /(£)• f(I!.). Then full Fourier Fourier series series of uniformly to ff on (—1,1) (and (and therefore therefore to on R). R,). uniformly to on (-I!.,I!.) to ffper per on 2. Suppose (0,^) -+ -> R is is continuous, = 2. Suppose ff :: (O,I!.) continuous, df/dx df/dx is is piecewise piecewise smooth, smooth, and and /(O) f(O) = f(l!.) = Fourier sine sine series series converges uniformly to to ff on on (-I!., I!.) f(i] = O. 0. Then Then the the Fourier converges uniformly (—£,£} (and hence to the periodic, periodic, odd of f,f, on (and hence to ffodd, odd extension extension of on all all ofR). ofH). 0dd, the 3. Suppose ff :: (0, I!.) —> -+ R is is continuous and df/ is piecewise piecewise smooth. smooth. Then Then 3. Suppose (0,£) continuous and df/dxdx is the Fourier converges uniformly (—i-,K)I!.) (and (and hence to hence to the Fourier cosine cosine series series converges uniformly to to ff on on (-I!., feven, the the periodic, periodic, even extension of on all all ofofH). R). feven, even extension of f,f, on
Proof. Proof· 1. This follows follows immediately from the the equivalence the full Fourier series 1. This immediately from equivalence of of the full Fourier series and and the Fourier series. series. the complex complex Fourier
2. This follows fact that sine series series of of f/ is full Fourier series of 2. This follows from from the the fact that the the sine is the the full Fourier series of fodd and from fact that that the of /f and and the conditions fodd and from the the fact the continuity continuity of the boundary boundary conditions /(O) = f(t] =a 0 guarantee guarantee that f(l!.) = that ffodd is continuous. continuous. f(O) = 0dd is 3. follows from series of series 3. This This follows from the the fact fact that that the the cosine cosine series of f/ is is the the full full Fourier Fourier series of from the the fact of /f guarantees guarantees that of feven feven and and from fact that that the the continuity continuity of that Seven feven is is continuous. continuous. D
The special property cosine series series is important; no no boundary conditions The special property of of the the cosine is important; boundary conditions must satisfied in order for convergence to Indeed, the relevant must be be satisfied in order for the the convergence to be be uniform. uniform. Indeed, the relevant boundary for aa cosine cosine series series are conditions, which boundary conditions conditions for are Neumann Neumann conditions, which involve involve df/dx. The Fourier cosine coefficients of /f are are 0O (1/lnI3) (l/|n| 3 ) or or 0O (1/lnI2) (l/|n| 2 ) according according df /dx. The Fourier cosine coefficients of to or does to whether whether f/ satisfies satisfies or does not not satisfy satisfy homogeneous homogeneous Neumann Neumann conditions conditions (see (see Exercise in either either case, series converges converges uniformly Exercise 4), 4), but but in case, the the Fourier Fourier series uniformly to to f./. Example 9.24. Let ff :: [0,1] -+ R be be defined f(x) = x 22.• Since Since ff does does not Example 9.24. Let —> R defined by by f(x) = x not satisfy the Dirichlet the sine on [0, [0,1]. satisfy the Dirichlet conditions, conditions, the sine series series does does not not converge converge uniformly uniformly on 1].
Chapter 9. More 'about Fourier Chapter 9. More-about Fourier series series
442 442
However, does not the Neumann conditions either, However, even even though though ff does not satisfy satisfy the Neumann conditions either, the the cosine cosine series does converge uniformly on illustrates these these results. results. series does converge uniformly on [0,1]. Figure Figure 9.15 illustrates Odd, periodic extension of x
2
Even, periodic extension of;
=
fi
: :
1\
1\
I
:!
0.8
0.5
ii
0.6 0 0.4 -0.5
0.2
-1
-4
-2
0
2
4
o~-~~-~--~-~
-4
x2 together with partial sine series 1.Sr-----------~
-2
o
2
4
x2 together with partial cosine series
1~--------------------~ 0.8 0.6
0.5 0.4 0 -0.5
0.2 0
0.5
0
0
0.5
Figure 9.15. (x) = — xx 22 by by sine cosine series (40 terms Figure 9.15. Approximating Approximating ff(x) sine and and cosine series (40 terms in each each series). series). in
Example 9.21, we showed several progressively smoother In Example several functions with progressively smoother periodic extensions. The smoother periodic extension, extension, the faster faster the Fourier periodic smoother the periodic Fourier coefficients decay to zero, and the better the partial Fourier Fourier series approximates the coefficients decay original original function. We now give an extreme example example of this. cos 27ra; Example 9.25. Define ff:: (-1,1) -t Then ffper and all 9.25. Define -»• R by ff(x) (x) = = eecos ( (21rJ;). ). Then and per of its derivatives are are continuous. By Theorem 9.20, the the Fourier Fourier coefficients coefficients of of ff of its derivatives continuous. By Theorem 9.20, of l/lnl. It is not possible possible to converge than any any power l/\n\. It is not to compute compute aa converge to to zero zero faster faster than power of formula for for these these coefficients that must be computed computed are are intractable), formula coefficients (the (the integrals integrals that must be intractable), but we we can discussed in 9.2. Figure Figure 9.16 9.16 but can estimate estimate them them using using the the techniques techniques discussed in Section Section 9.2. in estimating terms in in its its Fourier Fourier series. series. This This graph shows the error error in estimating ff with with 41 terms graph should be compared with with Figure Figure 9.14. should be compared 9.14-
9.S. 9.5.
Uniform convergence of Fourier Fourier series
443
2
o
-2
- 1
- 0 .5
o x
0 .5
cos(27ra; Figure 9.16. 9.16. The The errors errors in in approximating approximating f(x) ) by (27rx) Figure f(x) = eecos by its its partial partial Fourier series series with with 41 41 terms terms (see (see Example 9.25). Example 9.25). Fourier
9.5.3 9.5.3
A note about Gibbs's phenomenon
A standard standard theorem theorem in in real real analysis analysis is is the the following: following: If If aa sequence sequence of of continuous continuous A functions converges converges uniformly, uniformly, then then the limit function function is is also also continuous. continuous. Every Every parparfunctions the limit tial Fourier Fourier series series is is aa continuous continuous function, function, so so if if ffper is discontinuous, it is impossible is discontinuous, it is impossible tial per for its its Fourier Fourier series series to to converge converge uniformly. uniformly. Gibbs's Gibbs's phenomenon phenomenon is is the the nonuniform for nonuniform convergence of of the the Fourier Fourier series series near near aa jump discontinuity of of ffper. . Near the end end of jump discontinuity Near the of convergence per near aa jump jump discontinuity, Fourier the the 19th 19th century, century, Gibbs Gibbs showed showed that, that, near discontinuity, the the partial partial Fourier 70 series the jump jump70(cf. series overshoot overshoot the the true true value value of of f/ by by about about 9% 9% of of the (cf. Exercise Exercise 3). 3).
Exercises 1. Consider Consider the following functions functions defined defined on on [-1,1]: [—1,1]: 1. the following f(x) = lxi, g(x) = x - x 3 , h(x)=l+x. Rank f, /, g, g, and and hh in in order order of of the the speed speed of of convergence convergence of of their their Fourier Fourier series series Rank and illustrate illustrate with with graphs. eraohs. and 70
The 70 The
history is is briefly briefly described described by by Kammler Kammler [28], [28], page page 45. history 45.
Chapter 9. 9. More about Fourier Fourier series series Chapter More about
444
2. Find an example of a function -+ R whose Fourier coefficients coefficients are function /f:: [-1,1] [—1,1] —>• are
o
C:14)
but but not not
o C:15). (Hint: to be be a a fifth-degree fifth-degree polynomial.) (Hint: Choose Choose f/ to polynomial.)
3. Consider [— 1,1] —>• defined by Consider the the function function /f : [-1,1] -+ R defined by f(x) =
{O, -1::; x < 0, 1,
0::; x ::; 1.
Verify numerically numerically that that the the overshoot overshoot in in Gibbs's phenomenon is is about about 9% of Verify Gibbs's phenomenon 9% of the at xx = = 0. the jump jump at O.
4. Suppose f/ :: (0,^) is continuous continuous and smooth. 4. Suppose (0, £) -> -+ R R is and df/dx df / dx is is piecewise piecewise smooth. that the the periodic, periodic, even even extension extension of is continuous continuous and and therefore (a) Prove Prove that (a) of /f is therefore that / are 0O (1/lnI2). (l/|n| 2 ). that the the Fourier Fourier cosine cosine coefficients coefficients of of fare
(b) Prove (l/|n| 3 ) if (b) Prove that that the the Fourier Fourier cosine cosine coefficients coefficients of of /f are are 0 (1/lnI3) if and and only only ifif
! =! (0)
9.6 9.6
(£)
= O.
Mean-square convergence convergence of Fourier series Mean-square of Fourier series
We show that, of functions C, the the We now now wish wish to to show that, for for aa large large class class of functions f/ : (—1,1) (-£, £) —> -+ C, complex Fourier series of of I, /, complex Fourier series n=-oo
converges to to f/ in in the the mean-square mean-square sense. Mean-square convergence convergence means means that that ifif converges sense. Mean-square we write we write N
L
IN(X) =
cnei1rnx/f,
n=-N
then then
III - INII£2
-+ 0 as N -+
00.
In contrast contrast to to pointwise pointwise or uniform convergence, convergence, it it turns turns out out that that the the mildest mildest asasIn or uniform mean-square convergence convergence of of the the Fourier Fourier series. sumption about about I/ guarantees sumption guarantees mean-square series. In order order to to discuss discuss mean-square mean-square convergence, convergence, it it had had better better be be the the case that I/ In case that finite L £22 norm: has a finite
iff a
IIIII =
I/(xW dx
< 00.
9.6.
Mean-square convergence of Fourier series Fourier series
445
It turns out that essential assumption assumption is guarIt turns out that this this essential is the the only only assumption assumption required required to to guarantee to f/ in in the sense. To To justify justify antee the the convergence convergence of of the the Fourier Fourier series series to the mean-square mean-square sense. this statement, we some technical technicalities this statement, we must must begin begin with with some technical preliminaries. preliminaries. The The technicalities are subtle as as to have are sufficiently sufficiently subtle to be be beyond beyond the the scope scope of of this this book—the book~the reader reader will will have to accept certain assertions on faith. to accept certain assertions on faith.
9.6.1
The -.e,.e) The space L2( L2(-t,£)
We identify the the space space of functions f/ : (-i, (—(., i) i] —)• C such such that the complex We wish wish to to identify of all all functions -+ C that the complex Fourier series series of of f/ converges converges to mean-square sense. sense. Accepting the truth Fourier to f/ in in the the mean-square Accepting the truth of of the statement statement in define the in the the previous previous paragraph, paragraph, it it would would be be natural natural to to define L2(_£,£)
= {f: (-i,i)
-+ C :
f~l If(x)j2dx < oo}.
(9.22)
There is is more than meets more to to this this definition definition than meets the the eye, eye, however. however. There First of all, all, one one of essential properties norm is is the the following: following: First of of the the essential properties of of aa norm
11/11=0=*1=0.
(9.23)
When / is / = = 00 means / is the zero (x) = —0 for all When 1 is aa function, function, 1 means that that 1 is the zero function: function: fI(x) 0 for all 71 E (—1,1). (-i, i). As long as we restrict restrict ourselves to continuous functions, functions, (9.23) holds. holds.71 xe 2 However, norm equal to However, aa discontinuous discontinuous function function can can be be nonzero nonzero and and yet yet have have I/ L2 norm equal to zero. For example, consider the the function / : (-1, (—1,1) zero. For example, consider function 1 1) -> -+ R R defined defined by by
I(x) = {
I x = 0, '
0,
x
I:- O.
This function is the zero zero function, but This function is not not the function, but
since the zero everywhere everywhere except since the integrand integrand is is zero except aa single single point. point. Therefore, Therefore, in in order order for (9.23) (9.23) to agree that / and regarded as same for to be be satisfied, satisfied, we we have have to to agree that 1 and 9g are are regarded as the the same function provided function provided ill If(x) - g(x)j2 dx = O.
We say about We will will have have more more to to say about this this below. below. There is difficulty with with (9.22). Since our mean-square There is another another difficulty (9.22). Since our interest interest is is mean-square convergence convergence of functions, functions, the the following following property is desirable: If If UN} {/N} is a sequence 71 If f/ is is continuous continuous and and not not the function, then, an interval 71 If the zero zero function, then, by by continuity, continuity, there there must must exist exist an interval (a, 6) (—1,1) on which is nonzero, nonzero, and and so (a, b) C C (-t, t) on which /f is so
i:
must be positive. positive. must be
2
If(x)1 dx
~
lb
2
If(x)1 dx
Chapter 9. 9. More More about Fourier series Chapter about Fourier series
446
of functions in and /AT in the the mean-square sense, then L2((—£,l). offunctions in L L22(—l,l) (-f, f) and IN —>• --+ /I in mean-square sense, then /I 6 E L2 -f, f). Is satisfied by t) a.s as defined defined by (9.22)? The The answer answer is, is, it it depends depends Is this this property property satisfied by L?(-t, L2( -f, f) by (9.22)? on the the definition definition of of the integral used used in (9.22). For example, any any function with an an on the integral in (9.22). For example, function with infinite be Riemann Riemann integrable; integrable; a.s example, consider consider the infinite singularity singularity fails fails to to be as an an example, the function I/ : [-1, [— 1,1] defined by 1] —>• --+ R defined by function 1
I(x)
= IxI 1 / 4 '
This function function is is not on (-1,1) (—1,1) because of the discontiThis not Riemann Riemann integrable integrable on because of the infinite infinite discontinuity at x = 0. \f(x}\2 has has aa finite finite area area under its graph, as the following nuityat O. However, However, I/(x)12 under its graph, a.s the following calculation shows: calculation shows: lim { €---+o+
r-€ I/(xW dx + 1-1
/1 I/(xW dX} f
= lim 4(1 -
-IE)
= 4.
f---+O+
Therefore, we would to include we must Therefore, we would like like to include I/ in in L L22 (—1,1), (-1, 1), which which means means that that we must interpret the integral integral a.s as an an improper improper Riemann integral when has aa finite interpret the Riemann integral when I/ has finite number number of singularities. of singularities. However, allowing allowing improper since it it is is posposHowever, improper Riemann Riemann integrals integrals is is not not enough, enough, since sible to to construct with the sible construct aa sequence sequence {/N} {IN} with the following following properties: properties: 1. IN has has N infinite infinite discontinuities; 1. /jv discontinuities;
2. /TV IN
is square-integrable square-integrable when when the integral is is interpreted as an an improper Rieis the integral interpreted as improper Riemann integral; integral; mann
3. {IN} {/AT} converges converges in in the the mean-square mean-square sense sense to function with with an an infinite infinite number 3. to aa function number of discontinuities. of discontinuities. A function with with an an infinite infinite number of discontinuities discontinuities cannot cannot be Riemann integrable, integrable, A function number of be Riemann even in in the the improper improper sense. sense. even Faced with these these difficulties, difficulties, mathematicians eventually concluded concluded that that aa betbetFaced with mathematicians eventually ter notion notion of of integration integration was which was Lebesgue (the (the "L" ter was needed, needed, which was created created by by Henri Henri Lebesgue "L" 2 L2). integral is beyond in L ). The definition of the Lebesgue Lebesgue integral beyond the scope scope of this book, but we its important begins with with aa measure measure for for but we can can describe describe its important features. features. The The theory theory begins subsets of of R that with our intuition for for simple simple sets sets (for (for example, example, the measure subsets that agrees agrees with our intuition the mea.sure of The Lebesgue Lebesgue measure mea.sure is defined in fa.shion of an an interval interval [a, b] b] is is b-a). b — a). The is defined in aa consistent consistent fashion even for for very very complicated subsets of of R, but not for subsets. Sets Sets whose Lebesgue even complicated subsets but not for all all subsets. whose Lebesgue measure is is defined defined are are called called (Lebesgue) (Lebesgue) measurable. measurable. Any set with measure zero zero is mea.sure Any set with mea.sure is then certain sense. sense. Every set containing containing aa finite number of of points points ha.s has then neglible neglible in in aa certain Every set finite number mea.sure measure zero.72 zero.72 The Lebesgue Lebesgue integral defined for functions that that are the proppropThe integral is is defined for functions are measurable; the erty of of measurability measurability is like continuity continuity or or differentiability, differentiability, but but erty is aa regularity regularity property property like 72 72Some number of also have measure zero. A countable Some sets sets having having an an infinite infinite number of points points also have measure zero. A countable set set (a (a set be put put in in one-to-one with the the integers integers 1,2,3, ... ) has zero. set that that can can be one-to-one correspondence correspondence with 1,2,3,...) has measure measure zero. For example, example, the the set set of of all rational numbers measure zero. zero. Some Some uncountable sets also also have For all rational numbers has has measure uncountable sets have measure occupies aa neglible neglible part part of the real line. measure zero, zero, although, although, obviously, obviously, such such aa set set occupies of the real line.
9.6. 9.6.
Mean-square Mean-square convergence convergence of of Fourier Fourier series series
447
much much weaker-functions weaker—functions with with an an infinite infinite number number of of discontinuities discontinuities can can be be meameaWe cannot the precise these conceptssurable, for example. surable, for example. We cannot give give the precise definitions definitions of of these concepts— measurable measurable function, measurable set, set, measurable function, Lebesgue Lebesgue integral-in integral—in this this text. text. The The definidefinithe following consequences: tions, however, have tions, however, have the following consequences:
1. Two Two measurable measurable functions functions f/ and and 9a satisfy satisfy 1.
if if and and only only if f/ and and 9g differ differ only only on on aa set set of of measure measure zero. zero. It is is in in this this sense sense of that a set set of measure measure zero is negligible: functions differing differing only on aa set set of the same. measure regarded as being the measure zero zero are are regarded as being same. 2. If f/ is 2. is Riemann Riemann integrable, integrable, then then it it is is Lebesgue Lebesgue integrable, integrable, and and the the two two integrals integrals have have the the same same value. value. that whenever whenever we or This property assures us that This last last property assures us we are are dealing dealing with with aa continuous continuous or piecewise continuous function, function, we we can can compute compute its its integral integral using using the the familiar familiar techtechpiecewise continuous niques of caculus. caculus. On On the other hand, hand, the the Lebesgue Lebesgue integral integral is is sufficiently sufficiently powerful powerful the other niques of to handle pathological cases that arise mean-square convergence convergence to handle pathological cases that arise inevitably inevitably when when mean-square is properties of Lebesgue integration L22 ((—£,(.) -f, f) is studied. studied. We We will will discuss discuss other other properties of Lebesgue integration and and L below. below. We can now define L L22(—l,l) ( -f, f) more precisely as the set of all measurable measurable functions tions f/ : (-£, (—t,K)f) --t —> C C satisfying
with with the the understanding that that two two functions functions that that differ differ only only on on aa set set of of measure measure zero represent represent the same same element element of of the the space. space. In understand the we need two more more facts In order order to to understand the L2 L2 theory theory of of Fourier Fourier series, series, we need two facts 2 about L L2(—l.,l}, ( -f, f), facts that we we cannot prove here, as their development would take us too too far work. us far astray. astray. Even Even stating stating the the second second fact fact takes takes aa certain certain amount amount of of work. The The first first fact, fact, Theorem Theorem 9.26, 9.26, allows allows us us to to prove prove that that the the Fourier Fourier series series of of every every L2 L2 function function converges converges to to the the function function in in the the mean-square mean-square sense. sense. 2 Theorem EL L2( -£, e), and Theorem 9.26. 9.26. Let f e (—l,l), and let let Ee be be any any positive real real number (no (no matter matter how function 9g :: [-£, £] --t how small). small). Then Then there there is is an an infinitely infinitely differentiable differentiate function [—•£,•£] —>• C such that that
Moreover, 9g Moreover,
Ilf - gll£2 < E. can satisfy g( -f) = g(f) can be be chosen chosen to to satisfy g(—t] g(i] = O. 0.
This theorem simply says that any L L22(—t, (-£, £) (,} function function can be approximated approximated arusing aa smooth Dirichlet bitrarily well, bitrarily well, in in the the mean-square mean-square sense, sense, using smooth function function satisfying satisfying Dirichlet conditions. conditions.
448 448
9.6.2 9.6.2
Chapter 9. More about Fourier series
Mean-square convergence convergence of Fourier series
2 Accepting L2( -£, f), we now prove desired Accepting the the facts facts presented presented above above about about L (—£,£), we can can now prove the the desired result. result.
Theorem II /I E€ L2(_£,£), Fourier series series converges Theorem 9.27. 9.27. // L2(—l,t), then then its its Fourier converges to to fI in in L2. L2. That is, is, let That let n=-oo
be the the complex complex Fourier of fI,, and and define define be Fourier series series 01 N
IN(X) =
L
cneiomx/l.
n=-N
Then Then
III -
INII£2 --+ 0 as N --+
00.
Proof. We must must show show that for any any e€ > 0, 0, there there exists exists aa positive integer N 7Vfe such Proof. We that for positive integer such that that III - INIIL2 < € for every every N N :::: >N The proof proof is is aa typical argument—we show show that that /N is close close to to for N e•.. The typical e/3 €/3 argument-we IN is I/ by by using be specific, using the the triangle triangle inequality inequality with with two two intermediate intermediate functions. functions. To To be specific, let 9g be be aa smooth smooth function function satisfying satisfying g(—I) = g(i] and let g( -f) = g(£) and €
III - gll£2 < a' We will will show IN is to I/ (for N sufficiently by showing IN is is We show that that /N is close close to (for TV sufficiently large) large) by showing that that fw close gN (where the corresponding gN is close to to QN (where gN QN is is the corresponding partial partial Fourier Fourier series series for for g) and and QN is close to g- we we then then use use the the fact fact that that 9g is is close close to to I/ by by construction. construction. close to g; Let n ==: 0 0,, ±1, ±2, Let ddnn ,, n ±l,± 2 , ... . . . bbee the the Fourier Fourier coefficients coefficients of of g, p, and and define define N
gN(X) =
L
dnei7rnx/l.
n=-N
By Corollary Corollary 9.23, 9.23, QN on [-£, [—1, £], and and therefore positive By gN —>• --+ g9 uniformly uniformly on therefore there there exists exists aa positive integer N such that that integer Nf f such
It It follows follows that that
9.6.
449
Mean-square convergence convergence of Fourier series
and so and so
f
IIg - gNII£2 < 3' Finally, we have that iN gN is the partial the function g, and Finally, we have that /AT -— QN is the partial Fourier Fourier series series of of the function if -— g, and hence that that iN the projection onto the the space hence /AT -— 9N QN is is the projection onto space spanned spanned by by ei7rNx/l . e -i7rNx/l , e-i7r(N-l)x/l , ... , It that It follows follows that
Thus Thus we we obtain obtain
This This completes completes the the proof. proof.
0
we have this fundamental we can draw some Now Now that that we have obtained obtained this fundamental result, result, we can draw some further further conclusions. First, question of of terminology: the above above theorem theorem shows shows that that the the conclusions. First, aa question terminology: the nx 1 i7rnx complex exponentials ee™ = 0, ±1, ±1, ±2,..., ±2, ... , are complete complete in L2( -C, C). //£,, n = I/ 2 (—£,£).
Definition 9.28. 9.28. Let be an an (infinite-dimensional) (infinite-dimensional) inner inner product space. We We say Definition Let V be product space. say }i=l is that that an an orthogonal orthogonal sequence sequence {t/>j {j}^i is complete complete if if every every vv £E V V satisfies satisfies
sense of norm on where where the the convergence convergence is is in in the the sense of the the norm on V. V.
The inner The completeness completeness of of an an orthogonal orthogonal sequence sequence in in an an infinite-dimensional infinite-dimensional inner product thus analogous property of of product space space is is thus analogous to to the the property of spanning spanning for for aa finite finite collection collection of vectors finite-dimensional vector vectors in in aa finite-dimensional vector space. space. Moreover, Moreover, orthogonality orthogonality implies implies linear linear independence, infinite-dimensional vector independence, so so aa complete complete orthogonal orthogonal sequence sequence for for an an infinite-dimensional vector space is analogous to a basis for a finite-dimensional vector space is analogous to a basis for a finite-dimensional vector space. space. A that is useful is A result result that is sometimes sometimes useful is the the following. following.
Theorem 9.29. 9.29. Let Let {t/>j {(f)j : jj = — 1,2, 1,2,...} be aa complete complete orthogonal orthogonal sequence sequence in in an an Theorem ... } be inner product space space V. inner product V. Then Then v E V,
(v,t/>j)v = 0 for all j = 1,2,3, ...
~ v
= O.
(9.24)
Proof. By ... }, we have, for Proof. By the the completeness completeness of of {t/>j {j : j = 1,2, 1,2,...}, we have, for any any vv E e V, V, v =
f:
(v,t/>j)v t/>j. j=l (t/>j, t/>j)v
It follows follows immediately immediately that that if if (v,t/>j)v (v, 0j)y = = 00 for for all all jj = = 1,2,3, 1,2,3,..., = 0. It ... , then then vv = O.
0
450 450
Chapter 9. More about Fourier series
This result justifies justifies aa fact fact that that we we have have used used constantly constantly in in developing developing Fourier Fourier series This result series they have have the the same methods for for BVPs: BVPs: Two Two functions functions are are equal equal if and only only if if they methods if and same Fourier Fourier coefficients. coefficients. Corollary 9.30. 9.30. Let Let {¢j ... } be complete orthogonal orthogonal sequence sequence in Corollary {fij :: j = 1,2, 1,2,...} be aa complete in an an inner product product space space V. for any any u, u, v v E inner V. Then, Then, for e V, V, u = v {:} (u, ¢j)v = (v,¢j)v for all j = 1,2,3, ....
Proof. if u 0j)y = = (v, 0j)v for for all the other hand, Proof. Obviously, Obviously, if u = = v, v, then then (u, (u, ¢j)v (v, ¢j)v all j. j. On On the other hand, v (u, ¢j)v ¢j)v for for all all j, j, then then w w == uu — - vv satisfies satisfies if (u, if (f)j)v == ((v, > 0j)v (w,¢j)v =
°
for all j = 1,2,3, ....
By the implies that that w 0, that v. By the preceding preceding theorem, theorem, this this implies w = = 0, that is, is, that that u u = = v.
0
Finally, because because all all other Fourier series (sine, cosine, cosine, full, full, quarter-wave quarter-wave sine, sine, Finally, other Fourier series (sine, quarter-wave cases of the complex complex Fourier Fourier series, series, all of the the above quarter-wave cosine) cosine) are are special special cases of the all of above to these these other Fourier series well. For For example, we define define L L22 (0,^) (0, £) results extend extend to results other Fourier series as as well. example, we to be the the space space of of all all measurable measurable functions £) -+ R such such that that to be functions f/ : (0, (0,^) -> R
Then, if if /f e E£ L22(0, (0, £), i), the the Fourier Fourier sine of /f converges to /f in in the the mean-square Then, sine series series of converges to mean-square sense (see Exercise Exercise 6). sense (see 6). We have have now now completed completed the the basic basic theory theory used used in in the earlier chapters chapters of We the earlier of this this the next next section, section, we we discuss point for the sake of tying up aa book. In In the book. discuss one one further further point for the sake of tying up loose loose end. end.
9.6.3 9.6.3
Cauchy Cauchy sequences and completeness
We assume assume that that {¢j }~l is an orthogonal orthogonal sequence sequence in an inner inner product product space We {^j}^! is an in an space V and without loss loss of has norm norm one one (so the sequence sequence is is and also, also, without of generality, generality, that that each each ¢j (f>j has (so the orthonormal). v 6 E V and and orthonormal). If If v ,/,) (v,¢j)v aj= (v,'/'j v= (¢j,¢j)v' j=1,2,3, ... ,
then, by Bessel's Bessel's inequality, then, by inequality, DO
Llajl2 <
00.
j=l
We now now consider consider the converse: If a2, a3, . .. is is any of complex complex We the converse: If al, 01,02,03,... any sequence sequence of numbers satisfying satisfying numbers DO
9.6.
451
Mean-square convergence of of Fourier Fourier series
does does the series series
00
Lajj
(9.25)
j=1
converge VI To consider this question, we write converge to to aa vector vector in in V? To consider this question, we write n Un
=
Lajj, j=1
and m > n, then then and note note that, that, if if m m
n Un -
Um
m
= L:ajj - Laj¢j = L j=1
j=1
ajj.
j=n+l
It follows that that 2
m
IIUn
-
umll~ =
L
ajj
j=n+1
v
00
::; L
lajl2
j=n+1
¢3, ... (using the orthogonality orthogonality of {¢1, {0i, ¢2, 02,0s, • • • }})).• Since we are assuming that (9.25) we conclude that holds, we 00
L
lajl2 -+ 0 as n -+ 00.
j=n+1
It follows follows that that
Ilun
-
umllv -+ 0 as m,n -+ 00.
This is is the mark mark of of a sequence sequence that "ought" "ought" to converge. Definition 9.31. Let V normed vector vector space, space, and and suppose suppose {un} sequence Definition 9.31. Let V be be aa normed {un} is is aa sequence in V. We say that that {un} for any any ef > there exists exists aa positive positive integer integer in We say {un} is is Cauchy Cauchy if, if, for > 0, there N such such that that m,n 2: N => Ilun - umllv < f,
or, in in brief, or, brief,
Ilun
-
umllv -+ 0
as m,n -+
00.
The terms of a Cauchy sequence "bunch up," up," as do the terms of a convergent sequence. Indeed, it is easy to show that every convergent sequence is a Cauchy
452 452
Fourier series Chapter 9. More about Fourier
sequence. the other hand, the sequence. On On the other hand, the converse converse (does (does every every Cauchy Cauchy sequence sequence converge?) converge?) depends on the space well as the norm norm of the space. depends on the space as as well as on on the of the space.
Example 9.32. 9.32. Define numbers {xn} {xn} by by the that Xn xn Example Define aa sequence sequence of of rational rational numbers the rule rule that is the the number number obtained obtained by by truncating truncating the the decimal decimal expansion expansion of of 11" IT after after nn digits digits {so (so xi = 3, 3, X2 3.14, and and so on). Then Then {xn} {xn} is is certainly certainly Cauchy Cauchy {ifm (ifm > n, n, Xl = X2 = 3.1, 3.1, #3 X3 = 3.14, so on}. then /x m —x -xnn\/ < lO-n+1}. However, the then \x W~n+1). However, the question question of of whether {xn} {xn} converges converges depends depends on the If the of rational rational numbers, on the space space under under consideration. consideration. If the space space is Q, Q, the the set set of numbers, then then {xn} fails to to converge. hand, if the space R, then then {xn} converges, {xn} fails converge. On On the the other other hand, if the space is R, {xn} converges, and number 11". and the the limit limit is the the irrational irrational number TT. example may seem This last last example seem rather trivial; it it just points out out the fact that that the real number system numbers that not rational, rational, and that without the real number system contains contains numbers that are are not and that without the irrational numbers there are "holes" in the system. The next example is more relevant to our study. relevant Example Define ff :: [0, 1] -+ by Example 9.33. 9.33. Define [0,1] -»• R R by f(x) =
{I, E[~,~) , X
0,
XE[2,1].
Let /AT fN :: [0,1] -+ function defined of Let —>• R R be be the the function defined by by the the partial Fourier sine series series of so with N N terms. terms. Then Then \\f IIf — - /N\\L* fNIIL2 ~* -+ 0 as as nn —> -+ °°j 00, so {fn} {In} isis convergent convergentand and f with hence Cauchy. However, if space under under consideration e[O,I], the the space space of all hence Cauchy. However, if the the space consideration is is (7[0,1], of all and the L22 norm, continuous functions defined defined on continuous functions on the the interval interval [0,1], and the norm norm is the the L norm, then not convergent convergent (since 1]). then {IN} {/AT} is is still Cauchy Cauchy and and yet is is not (since f ¢ 0 C[O, C[0,1]).
°
We see that the second example is analogous analogous to the first. 1], first. The space e[o, (7[0,1], 2 L2(0, 1), has some "holes" in it, at least which is a subspace of L (0,1), least when the norm is L22 norm. taken to be the L We have a name for a space in which every Cauchy sequence converges. Definition 9.34. Let V be normed vector vector space. space. If If every sequence in in V Definition 9.34. Let be aa normed every Cauchy Cauchy sequence converges to then we we say that V is complete. converges to an an element element of of V, then say that We now have two different different (and (and unrelated) uses for the word "complete." An orthogonal sequence in an inner product orthogonal product space can be complete, and a normed vector space can be complete. The above examples 1] under examples show that Q is not complete, and neither is e[O, C[0,1] 73 L22 norm. On the other is complete (as is Rn). the L norm.73 other hand, R is R n ). Also, the space 2 L2(0,1) L (0,1) is complete, the proof of which result is beyond the scope of this book. 73 73A A standard that e[O,I] under aa different different norm, norm, namely, namely, the standard result result from from analysis analysis is is that C[0,1] is complete complete under the norm of norm of uniform uniform convergence: convergence:
11/1100 =
max {If(x)1 : x E [0, In.
This from the if aa sequence sequence of continuous functions functions converges uniformly to to This result result follows follows from the fact fact that that if of continuous converges uniformly aa function, function, the the limit limit function function must must also also be be continuous. continuous.
453
9.6. Mean-square convergence convergence of of Fourier Fourier series 9.6. Mean-square series 2 Theorem Both real real and and complex L2(a, b) are complete spaces. Theorem 9.35. 9.35. Both complex L (a,b) are complete spaces.
We can now answer the the question question we at the of this section: If We can now answer we posed posed at the beginning beginning of this section: If is an an orthonormal sequence in in aa complex complex inner inner product space V and {a,j} is {{0j}
Llajl2 < 00, j=1
does does
00
L aj
converge an element element of of V? VI The The answer answer is if V is complete, the converge to to an is that that if is complete, the convergence convergence is guaranteed. is guaranteed. Theorem 9.36. 9.36. Let V be be aa complete (complex) inner inner product Theorem Let V complete (complex) product space, space, and and let let 02,03, • • be be an in V. V. IfIf 01,02,03,... is aa sequence of complex complex numbers numbers satisfying satisfying 00
then then
V. converges to an an element element of converges to ofV.
Proof. Since V is complete, it it suffices suffices to to show sequence {v {vn} Proof. Since is complete, show that that the the sequence of partial partial n} of sums, sums, n
Vn
=
L
aj
j=1
is Cauchy. Cauchy. If If m then is m > n, n, then m
Vm -
Vn
L
=
aj
j=n+l
and, {j} is orthonormal, we have have
Since the series
m
00
j=n+l
j=n+l
00
454
Chapter 9. More about Fourier series
is must converge which shows is convergent, convergent, its its tail tail must converge to to zero, zero, which shows that that
Ilvm This proof. This completes completes the the proof.
-
vnllv
-t
°as
n,m -t
00.
0
The The conclusion conclusion of of all all this this is is that that if if aa (complex) (complex) inner inner product product space space contains contains complete orthonormal orthonormal sequence sequence {cPj}, {0j}, then then there there is is aa one-to-one one-to-one correspondence correspondence aa complete between between V and and the the space space
The vector v E 6 V corresponds to {(v, cPj)v} (f>j)v} E G f2. ^ 2 - In certain regards, then, V and 2 f2 the same space, with the elements given different different names. We say that I are really the V and and £2 I2 are are isomorphic. isomorphic. In particular, particular, L2( -£, £) and £2 £) is isomorphic to L2(—l,l) I2 are isomorphic, and L2(0, I/ 2 (0,^) real £2.
Exercises 1. 1. Consider Consider the the sine sine series series
1
2: ;;: sin (mfx). 00
n=l 2 (a) L2(0, 1). (a) Explain Explain why why this this series series converges converges to to aa function function f/ E 6L (0,1).
(b) Graph the the partial sine series with N TV terms, for various values of N, TV, and and guess guess the the function function f. /. Then Then verify verify your your guess guess by by calculating calculating the the Fourier Fourier sine coefficients coefficients of of f./. sine
2. Consider Consider the the series series 2.
00
L
cnein7rx,
n=-oo
where Ccnn = lin 1/n for for n i' ^ 0, 0, and and Co CQ = — 0. where 0. 2 (a) this series L2( -1,1). (a) Explain Explain why why this series converges converges to to aa function function f/ in in (complex) (complex) £ (-l, 1).
(b) Graph Graph the the partial partial series series with 2TV + + 11 terms, terms, for for various various values values of of TV, and (b) with 2N N, and guess the the function function f. /. Then Then verify verify your your guess guess by by calculating calculating the the Fourier Fourier guess coefficients coefficients of of f./. 3. Does the sine series series 00
~
1
vn sin (mfx)
2 converge in the mean-square L2(0, I)? Why why mean-square sense to a function f/ E €£ (0,1)? Why or why not?
455 455
9.7. A note about about general general eigenvalue eigenvalue problems 9.7. A note problems
4. f(x) = X1//4(1 4. Consider Consider the the function function /f :: [0,1] [0,1] -t —»• R R defined defined by by /(#) =x (l -— x), and and its its Fourier Fourier sine sine series series 00
L an sin (mrx). n=l
(a) the convergence f? What What (a) Which Which of of the convergence theorems theorems of of this this chapter chapter apply apply to to /? kind of of convergence convergence is is guaranteed: guaranteed: pointwise, mean-square? kind pointwise, uniform, uniform, mean-square? (b) a2, ... ,a63 (One (b) Estimate Estimate aI, 01,02,... , OBS using using some some form form of of numerical numerical integration. integration. (One option to use possioption is is to use the the DST, DST, as as described described in in Section Section 9.2.4. 9.2.4. Another Another possibility is to the formula formula to use use the bility is an =
210
1
Xl/4(1_ x) sin (mrx) dx
and numerical integration. because of and some some form form of of numerical integration. In In this this case, case, because of the the singularity to change singularity at at xx =— 0, 0, it it is is aa good good idea idea to change variables variables so so that that the the integrand is is smooth. smooth. Use Use x = — s8 44.) integrand .) (c) (c) Graph Graph f/ together with with its its partial Fourier Fourier sine sine series. series. Also Also graph graph the the two functions. functions. difference between the difference between the two 5. values of 1]-t by ff(x) 5. (a) (a) For For what what values of k does does the the function function /f :: [0, [0,1] -» R R defined defined by (x) = = xk xk 2 belong L2(0, 1)? (Consider belong to to L (0,1)? (Consider both positive positive and and negative negative values of of k.) 1 4 1 4 (b) X- / . (b) Estimate Estimate the the first 63 63 Fourier sine sine coefficients coefficients of of ff(x) ( x ) = x" (c) Graph /(#) f(x) == x x--11// 44 ,, together together with its partial Fourier sine series (with 63 63 terms). terms). Also Also graph graph the the difference difference between between the the two two functions. functions. 2
6. Prove if f/ E 6L (Q,€), then the the Fourier Fourier sine sine series series of of f/ converges converges to to f/ in 6. Prove that that if L2(0, f), then in the the mean-square mean-square sense. sense. 7. 7. Let {Vj} {vj} be a sequence of vectors in a normed vector space V, and suppose Vj -t -> vv E E V. Prove Prove that that {Vj} {vj} is is Cauchy. Cauchy.
9.7 9.7
A note problems A note about about general general eigenvalue eigenvalue problems
We will will now We now briefly briefly discuss discuss the the general general eigenvalue eigenvalue problem problem -\7. (k(x)\7u) = AU in
u=
n,
(9.26)
°on an.
We assume assume that that n fJ is is aa bounded bounded domain domain in in R2 R2 with with aa smooth smooth boundary, and and that the coefficient kk is smooth function function of of xx satisfying satisfying fc(x) > ko ko > 0 for for all all xx E € n. f). the coefficient is aa smooth k(x) :2: We have have already already seen, seen, for for aa constant constant coefficient coefficient and and the case in in which which £) We the case n isis a rectangular rectangular domain or a disk, that (9.26) (9.26) has an infinite infinite sequence of solutions. solutions. Moreover, the the eigenpairs eigenpairs had certain properties properties that could verify directly: the the that we we could verify directly: Moreover, had certain eigenvalues were positive, forth. eigenvalues were positive, the the eigenfunctions eigenfunctions were were orthogonal, orthogonal, and and so so forth. The to the The properties properties we we observed observed in in Sections Sections 8.2 8.2 and and 8.3 8.3 extend extend to the general general eigenvalue eigenvalue problem (9.26), (9.26), although although they they cannot cannot be verified verified by directly directly computing computing
°
456
Chapter 9. 9. More about Fourier Fourier series Chapter More about series
the eigenpairs. eigenpairs. Instead, Instead, the the results results can can be be deduced deduced from fairly sophisticated the from some some fairly sophisticated that are are beyond beyond the the scope scope of of this this book. book. Therefore, Therefore, we we will will mathematical arguments mathematical arguments that not attempt attempt to to justify justify all of our statements in in this this section. not all of our statements section. KD : e1(n) We define Kr> C&(ti) by KDu
= -\7. (k(x)\7u) ,
(9.26) can be written as simply so that (9.26) KDU = AU.
The following results results hold: hold: The following 1. All eigenvalues eigenvalues of of KD are real and and positive. positive. To To show this, we we assume that 1. All are real show this, assume that KDU (u,u) = 1. I. Then KDU = \u AU and and (u,u) Then A = A(U,U) = (AU,U) = (KDu,u) =
L L
(KDU)U
\7 . (k(x)\7u) U
= -
= = =
rk(x)\7u. \7u - Janr k(x)u~U
Jn
L L
un
k(x)\7u . \7u k(x)
II\7u112 .
(The reader reader should should notice notice that, that, since we cannot cannot assume assume aa priori priori that that A and (The since we and uU are are real, real, we we use use the the complex complex L2 I/2 inner inner product.) product.) This This shows shows that that A A is is aa nonnegative nonnegative real real number. number. Moreover, Moreover, A = 0 =?
L
k(x)
II\7u112 = 0
\7u = 0 (since k(x) > 0) u(x) = constant =? u(x) = 0, =?
=?
boundary conditions. the last last step following from the Dirichlet boundary conditions. Therefore, KD has only only positive positive eigenvalues. has eigenvalues. 2. Eigenfunctions Eigenfunctions of of KD corresponding to distinct distinct eigenvalues are orthogonal. 2. corresponding to eigenvalues are orthogonal. As we we have have seen seen before, before, the the orthogonality depends only only on As orthogonality of of eigenfunctions eigenfunctions depends on the easy to is easy to verify verify using using integration integration by by the symmetry symmetry of of the the operator, operator, which which is parts (see (see Exercise Exercise 1). parts 1). 3. operator KD has has an an infinite infinite sequence sequence {\ {An}~=l of eigenvalues eigenvalues satisfying satisfying 3. The The operator n}^L1 of
457 457
9.7. about general general eigenvalue problems 9.7. A A note note about eigenvalue problems
and and An
-+
00
as n -+
00.
In the the sequence sequence {An}, {An}, each each eigenvalue eigenvalue is according to to the In is repeated repeated according the dimension dimension of of the associated associated eigenspace, eigenspace, that that is, is, according according to the the number of of linearly linearly independent eigenfunctions eigenfunctions associated associated with that eigenvalue. eigenvalue. In In particular, each particular, each dependent with that eigenvalue eigenvalue corresponds corresponds to only only finitely finitely many many linearly linearly independent independent eigenfunctions. tions. It is always possible to replace a basis for for a finite-dimensional subspace by an 74 orthogonal basis. Therefore, all all of of the the eigenfunctions eigenfunctions of of KD KD can taken orthogonal basis. 74 Therefore, can be be taken to We will will assume assume that an orthogonal orthogonal sequence sequence that {^n}^Li {~n}~=l is is an to be be orthogonal. orthogonal. We satisfying satisfying
4. The The set {~n} 4. set of of eigenfunctions eigenfunctions {ip is aa complete complete orthogonal orthogonal sequence sequence in in L2(0): Z/ 2 (fi): n} is 2 For EL L2(0), For each each f/ 6 (ft), the series series (9.27) 2 converges in in the sense to /. The space L (J7) is is defined defined as The space L2(0) as converges the mean-square mean-square sense to f. 2 was (a, &)—informally, it is is the space of square-integrable functions functions defined defined was L L2(a, b)-informally, it the space of square-integrable on 0, il, with with the that if if two functions differ differ only only on on aa set set of on the understanding understanding that two functions of measure measure zero, zero, then then they are are regarded regarded as as the the same. same. The The series series (9.27) (9.27) is is called called aa generalized generalized Fourier Fourier series series for for f./.
For For specific specific domains, it it may may be be more convenient convenient to enumerate enumerate the the eigenvalues and and than aa singly eigenfunctions indexed list list rather eigenfunctions in in aa doubly doubly indexed rather than singly indexed indexed list list as as suggested suggested above. For For example, example, the the eigenvalue/eigenfunction eigenvalue/eigenfunction pairs of the the negative negative Laplacian Laplacian above. pairs of on the unit unit square on the square are are
It is is possible possible (although (although not not necessarily necessarily useful) order the the \Amn and ~mn ^mn in in (singlyIt useful) to to order (singlymn and indexed) sequences. sequences. indexed) The of the the above above facts facts for for many many computational computational tasks tasks is limited, The usefulness usefulness of is limited, since, k(x), it is not possible to since, for for most most domains domains 0f) and and coefficients coefficients fc(x), it is not possible to obtain obtain the the eigenvalues However, as eigenvalues and and eigenfunctions eigenfunctions analytically analytically (that (that is, is, in "closed "closed form"). form"). However, as we have have seen seen before, before, the eigenpairs eigenpairs give some some information information that that can can be be useful in its its useful to own own right. right. It It may may be be useful to expend expend some some effort effort in in computing computing aa few few eigenpairs eigenpairs numerically. We We illustrate numerically. illustrate this this with with an an example. example. Example 9.37. 9.37. Consider Consider aa membrane membrane that that at at rest rest occupies the domain domain 0, tl, and Example occupies the and suppose small transverse of the satisfy the suppose that that the the (unforced) (unforced) small transverse vibrations vibrations of the membrane membrane satisfy the 74 74The technique for the Gram-Schmidt is explained elemenThe technique for doing doing this this is is called called the Gram-Schmidt procedure; procedure; it it is explained in in elementary linear linear algebra algebra texts texts such tary such as as [34].
458 458
Chapter 9. More about about Fourier Fourier series series Chapter 9. More
IBVP IB VP
aat2 u
2
2 -
n, t > 0, x E n,
c Au = 0, x E u(x,O) = ¢(x),
au at (x,O) = ,),(x),
(9.28)
x E n,
u(x, t) = 0, x E an, t
> O.
The question we wish to answer is: What is the lowest frequency at which the membrane will resonate? This could be be an important important consideration in designing a device that contained such a membrane. membrane. The key key point point is that, if if we knew the eigenvalues and eigenvectors of of the negafi (subject (subject to Dirichlet IBVP Dirichlet conditions), then we could solve the IBVP tive Laplacian on n just as in the Fourier series method. Suppose the eigenpairs eigenpairs are An, 'l/Jn(x) , n = 1,2,3, .... Then we can write the solution w(x,£) u(x, t) as 00
u(x, t) =
l: an (t)'l/Jn(x). n=l
Substituting into the PDE PDE yields Substituting
an(t) satisfies satisfies and so an(t)
Therefore, Therefore, an(t)
= bn cos (cAt)
+ Cn sin (cAt),
with the the coefficients and c determined by initial conditions. conditions. Therefore, Therefore, with coefficients bbnn and enn determined by the the initial the fundamental fundamental frequency—the frequency-the smallest smallest natural frequency—of frequency-of the membrane is cx/V^TT). cv'):; / (2n). This example example shows that we gain some useful information computing This shows that we would would gain some useful information by by computing the smallest smallest eigenvalue eigenvalue of of the the operator operator on on n. £). We will revisit revisit this this example example in Section the We will in Section lOA, where we use finite methods to to estimate estimate the the smallest of 10.4, where we use finite element element methods smallest eigenvalue eigenvalue of various various domains. domains.
Exercises Exercises 1. Prove Prove that that the operator K defined defined above above is is symmetric. symmetric. 1. the operator
9.8. Suggestions Suggestions for 9.8. for further further reading reading
459 459
2. Let 2. Let the the eigenvalues eigenvalues and and eigenvectors eigenvectors of of KD KD be be
.An, '¢n(x), n=1,2,3, ... , as the text. text. Suppose E C1(0) represented in as in in the Suppose that that uu € Of,(17) is is represented in aa generalized generalized Fourier Fourier series as as follows: follows: series 00
U(X) =
L an'¢n(x). n=l
Explain generalized Fourier Fourier series series of KDU is equal to to Explain why why the the generalized of KDu is equal 00
L .An an '¢n (x). n=l
3. Consider Consider an an iron iron plate plate occupying occupying aa domain £) in R2. Suppose Suppose that that the the plate 3. domain 0 in R2. plate is heated to degrees Celsius, Celsius, the and bottom bottom of of the the plate is heated to aa constant constant 55 degrees the top top and plate are perfectly insulated, and and the edges of of the the plate are fixed at 00 degrees. degrees. Let are perfectly insulated, the edges plate are fixed at Let u(x, temperature at Using the the first first eigenvalue w(x, t) be be the the temperature at xx E6 0fJ after after t seconds. seconds. Using eigenvalue and the negative negative Laplacian Laplacian on of and eigenfunction eigenfunction of of the on 0, 0, give give aa simple simple estimate estimate of w(x,£) that is valid for for large large t. What What must in order order that that aa single u(x, t) that is valid must be be true true in single eigenpair suffices suffices to good estimate? estimate? (The (The physical physical properties of iron eigenpair to define define aa good properties of iron 3 are p = 7.88g/cm , c = 0.437 J/(gK), AC = 0.836 W/(cmK).) are p = 7.88 g/cm3, c = 0.437 J/(gK), ,.. = 0.836W /(cm K).)
9.8 9.8
Suggestions for further Suggestions for further reading reading
The as dealing with Fourier present the the conconThe books books cited cited in in Section Section 5.8 5.8 as dealing with Fourier series series all all present vergence theory. More specialized specialized books, books, both both with wealth of about vergence theory. More with aa wealth of information information about Fourier series series and and other aspects of of Fourier Fourier analysis, analysis, are are Kammler [28] and and Folland Folland Fourier other aspects Kammler [28) [15]. Folland's Folland's book book is classical in Kammler delves delves into into aa number number of [15]. is classical in nature, nature, while while Kammler of applications applications of modern interest. Briggs and Henson [7] gives comprehensive introduction introduction to to the discrete Briggs and Henson [7] gives aa comprehensive the discrete Fourier transform transform and and covers covers some some applications, applications, while while Van Van Loan Loan [50) [50] provides an provides an Fourier in-depth treatment treatment of of the the fast fast Fourier transform. Other Other texts texts in-depth the mathematics mathematics of of the Fourier transform. that explain the the FFT FFT algorithm algorithm include include Walker Walker [51] [51] and and Brigham Brigham [8]. [8]. that explain
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Chapter 10 Chapter
about finite finite element element More about methods
In this this chapter, chapter, we we will will look look more more deeply deeply into into finite finite element element methods for solving In methods for solving steady-state BVPs. BVPs. Finite Finite element element methods methods form form aa vast vast area, area, and and even even by the end steady-state by the end of this this chapter, chapter, we we will will have have only only scratched scratched the the surface. surface. Our Our goal goal is is modest: modest: We of We wish to give the reader a better idea of how how finite element methods are implemented in in practice, practice, and and also also to to give give an an overview overview of of the the convergence convergence theory. theory. The main tasks involved in applying the finite element method are Defining aa mesh mesh on on the the computational computational domain. domain. •• Defining Computing the the stiffness stiffness matrix matrix K K and and the the load load vector vector f. f. •• Computing Solving the the finite finite element element equation equation Ku Ku == f. f. •• Solving We will mostly ignore the first question, except to provide examples for simple domains. right, and mains. Mesh Mesh generation generation is is an an area area of of study study in in its its own own right, and delving delving into into this this subject begin by by addressing involved subject is is beyond beyond the the scope scope of of this this book. book. We We begin addressing issues issues involved in computing computing the the stiffness stiffness matrix matrix and and load load vector, vector, including including data data structures structures for in for representing and and manipulating manipulating the the mesh. mesh. We will concentrate concentrate on on two-dimensional two-dimensional representing We will problems, triangular triangular meshes, and piecewise piecewise linear finite elements, as these are sufficient Next, in ficient to illustrate illustrate the the main main ideas. ideas. Next, in Section Section 10.2, 10.2, we we discuss discuss methods methods for solving the the finite element equation equation Ku Ku == f, f, specifically, specifically, on on algorithms algorithms for for taking taking solving finite element advantage we provide advantage of of the the sparsity sparsity of of this this system system of of equations. equations. In In Section Section 10.3, 10.3, we provide aa brief brief outline outline of of the the convergence convergence theory theory for for Galerkin Galerkin finite finite element element methods. methods. Finally, Finally, we close the book with a discussion of finite element methods for solving eigenvalue problems, such such as as those those suggested suggested in in Section Section 9.7. 9.7. problems,
10.1 10.1
Implementation of finite element methods
There are several issues that must be resolved in order to efficiently efficiently implement the finite element element method method in in aa computer computer program, program, including finite including
Data structures structures for for representation representation of of the the mesh. mesh. •• Data
461 461
462
Chapter 10. 10. More about finite element methods
•• Efficient Efficient computation computation of of the the various various integrals integrals that that define define the the entries entries in in the the stiffness matrix matrix and and load load vector. vector. stiffness •• Algorithms Algorithms for for assembling assembling the the stiffness stiffness matrix matrix and and the the load load vector. vector.
We discuss these these issues. issues. We will will now now discuss
10.1.1 10.1.1
Describing a a triangulation triangulation Describing
Before we can describe describe data data structures structures and and algorithms, algorithms, we must develop develop notation Before we can we must notation for the the computational computational mesh. First of of all, all, we we will will assume assume that that the domain n 0 on for mesh. First the domain on which the BVP BVP is is posed is aa bounded bounded polygonal polygonal domain domain in in R2, R2, so so that that it it can can be which the posed is be 75 triangulated.75 Let triangulated. Let Th = {Ti : i = 1,2, ... , L} be the collection collection of of triangles triangles in the mesh mesh under under consideration, consideration, and and let let be the in the
Nh
=
{llj :
j = I,2, ... ,M}
be the nodes of triangles in notation in be the set set of of nodes of the the triangles in Th. Th- (Standard (Standard notation in finite finite element element methods labels each mesh h, the the length of the of any any triangle triangle in methods labels each mesh with with /i, length of the longest longest side side of in the mesh. Similarly, the the space space of of piecewise piecewise linear linear functions functions relative to that that mesh mesh is the mesh. Similarly, relative to is denoted Vh, Vh, and and an an arbitrary arbitrary element element of of Vh Vh as as Vh. Vh- We We will will now now adopt adopt this standard denoted this standard notation.) notation.) To use as To make make this this discussion discussion as as concrete concrete as as possible, possible, we we will will use as an an example example aa regular triangular mesh, mesh, defined defined on the unit 32 triangles triangles and regular triangular on the unit square, square, consisting consisting of of 32 and 25 25 nodes. This mesh is shown in Figure IO.1. 10.1. In order order to to perform the necessary calculations, we must know which nodes nodes are are In perform the necessary calculations, we must know which associated with with which which triangles. Of course, course, each each triangle has three three nodes; we need need associated triangles. Of triangle has nodes; we to know these three nodes in ...- ,, llM. We define to know the the indices indices of of these three nodes in the the list list lll, HI, ll2, 112,.. HM- We define the the mapping ee by by the property that of triangle Tj are are lle(i,1),lle(i,2),lle(i,3). ne(i;1), n e (j ;2 ), ne(i,3) • mapping the property that the the nodes nodes of triangle Ti (So ee is is aa function function of of two two variables; the first must take an integral integral value value from from 11 to to (So variables; the first must take an L, the second L, and and the second one one of of the the integers integers 1,2,3.) 1,2,3.) In our our sample sample mesh, mesh, there are 32 32 triangles triangles (L = 32), 32), and and they they are are enumerated enumerated In there are 25 nodes are enumerated as shown which as in Figure 10.2. 10.2. The M == 25 shown in Figure 10.3, which explicitly the indices the nodes nodes belonging to each triangle. For explicitly shows shows the indices of of the belonging to each triangle. For example, example, e(IO, 1) == 6, e(IO,2) == 7, e(1O,3) = 12,
expressing the fact that vertices of of triangle 10 are are nodes 6, 7, 7, and and 12. expressing the fact that the the vertices triangle 10 nodes 6, 12. Each of of the standard basis basis functions functions for for the the space space of of continuous continuous piecewise piecewise Each the standard linear linear functions functions corresponds corresponds to to one one node in in the triangulation. triangulation. The The basis functions functions 75 751f boundaries, then, the context piecewise linear elements, If the the domain domain n fi has has curved curved boundaries, then, in in the context of of piecewise linear finite finite elements, the must be approximated by polygonal curve curve made made up of edges edges of of triangles. triangles. This This the boundary boundary must be approximated by aa polygonal up of introduces using higher-order higher-order finite introduces an an additional additional error error into into the the approximation. approximation. When When using finite elements, elements, for for example, piecewise piecewise quadratic quadratic or or piecewise cubic functions, then it it is is straightforward straightforward to to approximate piecewise cubic functions, then approximate example, the piecewise polynomial polynomial curve finite element element the boundary boundary by by aa piecewise curve of of the the same same degree degree as as is is used used for for the the finite functions. This This technique technique is is called isoparametric method, and it it leads significant reduction reduction called the the isopammetric method, and leads to to aa significant functions. in the error. error. We We will ignore all all such such questions questions here here by by restricting restricting ourselves ourselves to will ignore to polygonal polygonal domains. domains. in the
10.1. Implementation Implementation of finite element methods
Figure 10.1. 10.1. A triangulation of the unit square. Figure square.
Figure 10.2. Enumerating the triangles in the the sample sample mesh. mesh. Figure 10.2. Enumerating the triangles in
463
464
Chapter 10. about finite finite element element methods methods Chapter 10. More More about 25
20
15
10
0.1 5
0.2
0.4
Figure 10.3. Figure 10.3. Enumerating the nodes in the sample mesh.
correspond in the approximation. However, correspond to to the the degrees degrees of of freedom freedom in the finite finite element element approximation. However, not every node in in the triangulation need give rise to a degree of of freedom; freedom; some some may be associated with Dirichlet boundary conditions which determine the weight may be associated with Dirichlet boundary conditions which determine the weight given function. For is not sufficient to to to the the corresponding corresponding basis basis function. For this this reason, reason, it it is not sufficient given to identify the set of the triangulation; we must also determine identify of all nodes in the determine the free free nodes—those do not condition. We free nodes-those that that do not correspond correspond to to aa Dirichlet Dirichlet condition. We will will label label the the free nodes n/ 2 ,..., n/^. ll/l,ll/2, ... ,llfN. nodes as as n^, Referring again to the the sample mesh of Figure 10.1, if we we assume that this mesh will be be used used in in aa Dirichlet Dirichlet problem, problem, only only the the interior nodes correspond to degrees will interior nodes correspond to degrees = 99 and of freedom. Thus N N = and the free free nodes can can be enumerated enumerated in the same same order order as the of all all nodes (left-to-right and as the set set of nodes (left-to-right and bottom-to-top). bottom-to-top). This This yields yields
It = 7, h = 8, h = 9, /4 = 12, 15 = 13, 16 = 14, h = 17, Is = 18,
fg
= 19.
We is aa We now now define define the the standard standard basis basis function function fa ¢>i by by the the conditions conditions that that fa ¢>i is continuous 7ft, and and continuous piecewise piecewise linear linear function function relative relative to to the the mesh mesh 1h, A. ( ) 'l'i lli
= Vii = J:
{I, 0
,
i = j, . ...J.. Z r J.
The finite element space-the space—the space of continuous piecewise linear functions (relative 7ft) that any given Dirichlet conditionsconditions— (relative to to the the mesh mesh 1h) that satisfy satisfy any given homogeneous homogeneous Dirichlet is then is then
10.1. finite element methods 10.1. Implementation Implementation of of finite element methods
10.1.2
465
Computing the stiffness matrix
We We assume assume that that the the weak weak form form of of the the BVP B VP in in question question has has been been written written in in the the form form u E V, a(u,v) = (f,v) for all v E V and that the Galerkin problem takes the form
NxN We We need need to to compute compute the the stiffness stiffness matrix matrix K K E €R R NXN ,, where where
By be zero; fixed ii between between 11 and By design, design, most most of of these these values values will will be zero; let let us us focus focus on on aa fixed and N, and values of nonzero? AT, and ask ask the the question: question: For For which which values of jj will will Kij KIJ be be nonzero? As As an an example, example, consider consider ii =— 55 for for our our sample sample mesh. mesh. The The support support of of CP13 ^13 == cP0/165 is basis functions is shown shown in in Figure Figure lOA. 10.4. The The question question we we must must answer answer is: is: What What other other basis functions CPt; (f>fj have have aa support support that that intersects intersects of of CP13 ^13 (the (the intersection intersection must must have have aa positive positive area; area; if if the the intersection intersection is is just an an edge edge of of aa triangle, triangle, this this will will not not lead lead to to aa nonzero contribution contribution to to K). K). It It is is easy easy to to see see that that the the desired desired basis functions functions correspond correspond to to nodes nodes of of triangles triangles of of which which n13 1113 is is itself itself aa node, node, that that is, is, to to the the nodes nodes of of the the triangles triangles shaded in Figure lOA. 10.4.
Figure CP13. Figure 10.4. 10.4. The The support support of 0/^13.
It is is possible to store store all all of of the the necessary necessary connectivity connectivity information information (that (that is, is, to to It possible to store, along along with with each each node, node, aa list list of of nodes belonging to to aa common common triangle). triangle). With With store, nodes belonging
Chapter More about methods Chapter 10. 10. More about finite finite element element methods
466
this information, information, one one can can compute compute the the matrix K by by rows, according to following this matrix K rows, according to the the following algorithm: algorithm: for ii = = 1,2, 1 , 2 ,... . . . ,, N N for for each each jj = l,2,...,N such that that the the supports supports of of cPlj (f)fj and and cPl. (f>ft overlap, overlap, for = 1,2, ... , N such compute Kij = aa (cPlj, (<j>f., cPl.)· 0^). compute Kij = However, this this is is probably not the the most most common common approach, approach, for for the the following following reason: reason: However, probably not order to to compute compute the the necessary integrals, one one must must know form the the In order necessary integrals, know the the nodes nodes that that form vertices of of each each triangle. out that that the the stiffness stiffness matrix can be formed with It turns turns out matrix can be formed with vertices triangle. It only this without explicitly storing the connectivity information that is, is, without explicitly storing the connectivity information only this information, information, that mentioned above. The integral integral represented represented by (0^., cPf;) <}>f.) is is naturally naturally decomposed decomposed into into the the sum sum The by aa (cP/;, of several several integrals, integrals, one one for for each each triangular triangular element element making making up the domain domain of of inteinteof up the gration. It It is is convenient convenient to to make make this decomposition, as as cPf; <j>f4 and and cPlj 0^. are are given given by this decomposition, by gration. different formulas formulas on on the different elements. elements. It It is then natural to compute, compute, for for each is then natural to each different the different element T Tk, stiffness matrix K. K. k , the contributions to various entries of the stiffness For example, example, consider consider i%==5,5,j j ==11ininour oursample samplemesh. mesh. Then Thenaa(cPlj ((f>f^, cPl;) 0/J isis For the sum sum of of two two integrals, integrals, one one over over triangle triangle T12 Ti2 and and the the other other over over T13: Ti3: the
There is reason that that these these two integrals must be computed computed together, together, as as long long as There is no no reason two integrals must be as both contributions to to K are eventually eventually accumulated. accumulated. both contributions K51 5i are Following the above reasoning, standard algorithm for computing computing K is is to to Following the above reasoning, aa standard algorithm for loop over over the the elements elements (Le. (i.e. the the triangles) in the (rather than than over over the loop triangles) in the mesh mesh (rather the nodes) nodes) and compute compute the contributions to to K K from from each each element element in in turn. Since each each triangle triangle and the contributions turn. Since has three nodes, aa given element contributes to up to 99 entries in K. K. There is one piece of information, information, however, is not available unless unless we make aa point to one piece of however, that that is not available we make point to record it in in the the data data structure: structure: the the "inverse" "inverse" of of the the mapping mapping ii f-+ t-t Ii. fa. In In the record it the course course nj is of assembling the stiffness stiffness matrix, we need to be able to decide if aa given node n^ free node node (that (that is, is, if if jj = = fa for some some i) i) and, and, if if so, so, to to determine determine i. i. So So let let us us define define aa free Ii for mapping jj f-+ •->• gj gj by by aa mapping i, if j 0, if j
= Ii :# Ii for all i =
1,2, ... , N.
The algorithm algorithm then then takes takes the following form: form: The the following 2, ... , L for for A?k = = 1, 1,2,...,L for f o r »i = 1,2,3 for j = 1,2,3 forj=1,2,3 if P = ge(k,i) 9e(k,i) :# ^ 00 and and q ± 00 if p = q = = g ge(k,j) e(kij) :# Add the to Kpq Kpq obtained obtained by integrating over over Tk. Tk. by integrating Add the contribution contribution to
The condition condition that ge(k,i) :# / 00 indicates indicates that the ith ith vertex (i = = 1,2,3) 1,2,3) of of triangle triangle The that ge(k,i) that the vertex (i is aa free free node, node, and and similarly similarly for for ge(k,j).
Tk n
10.1. 10.1. Implementation of finite element methods
467
of this algorithm is the contribution to K Kpq (tPe(k,j) , tPe(k,i)) from = aa (0 fr°m the element pq = e (fc,j)>0e(M)) T TV Also, it should be pointed out that, in the case that K is symmetric (as it is in k . Also, the examples considered considered in in this this book), of the the work work can be eliminated: eliminated: we we just just the examples book), part part of can be compute the the entries entries K with q>p, and then assign the value of K to K . compute Kpq pq with q 2 p, and then assign the value of Kpq pq to Kqp. qp We have described above the information needed to carry out this algorithm. It is easy to store this information use in in aa computer program. A A It is easy to store this information in in arrays arrays for for use computer program. 76 simple data structure would consist of four arrays: arrays:76
1. NodeList: NodeList: An x 22 array; array; the consists of of the the xx and and yy coordinates coordinates 1. An M M x the iih ith row row consists of node ni. of node n^. 2. consists of 2. ElementList: ElementList: An An L xx 33 array; array; the the kth row row consists of e(k,1),e(k,2),e(k,3). e(fc,l),e(fc,2),e(fc,3).
3. FreeNodePtrs: FreeNodePtrs: An An N x 11 array; array; the the ith iih entry entry is is k /j. 3. N x 4. Mx array; the iih entry entry is is gi. 4. NodePtrs: NodePtrs: An An M x 11 array; the ith gi. By By way way of of example, example, Table Table 10.1 10.1 shows shows these these arrays arrays for for the sample sample mesh of Figure Figure 10.1. The information recorded in these arrays will suffice suffice for straightforward finite element problems with homogeneous Dirichlet or conditions element problems with homogeneous Dirichlet or Neumann Neumann boundary boundary conditions (or even even for for inhomogeneous inhomogeneous Dirichlet Dirichlet problems—see Exercise 5). 5). For problems with (or problems-see Exercise For problems with inhomogeneous Neumann conditions, it may also be necessary to record whether a given edge belongs to the boundary. The ElementList ElementList array can be augmented by columns containing the edges edges lie lie on boundary. (Or, (Or, of on the the boundary. of columns containing flags flags indicating indicating whether whether the course, these flags flags can be stored in independent arrays, if desired.) This extension is left left to reader. is to the the reader.
10.1.3
Computing the load vector vector
Computing the load vector ff is similar to computing the stiffness stiffness matrix, only easier. We loop over over the the elements elements and, and, for for each each node of aa given given triangle, compute the the We loop node n/ nt.{ of triangle, compute contribution i . The sum of integrals over or more more contribution to to fIi. The value value fi Ii will will be be the the sum of integrals over one one or we loop triangles, and and this sum sum is is accumulated accumulated as as we loop over over the the triangular triangular elements. elements.
for kk = 1,2, 1 , 2 ,... . . . ,, LL for for ii = 1,2,3 1,2,3 for if p = if P = ge(k,i) 9e(k,i) f:/0 Add the contribution Add the contribution to to fIp obtained by by integrating integrating over over T T^. k· p obtained
10.1.4 Quadrature Quadrature Up to this point, we we have computed all integrals exactly, perhaps with the help of a computer package such as Mathematica Mathematica or Maple. Maple. However, However, this would be difficult difficult to incorporate incorporate into a general-purpose computer program for finite finite elements. Moreover, it to compute be more more it is is not not necessary necessary to compute the the various various integrals integrals exactly, exactly, and and it it may may be 76
76Using modern techniques, such as programming, one one Using modern techniques, such as modular modular programming programming or or object-oriented object-oriented programming, might define define aa structure structure or or aa class class to to represent might represent aa mesh. mesh.
Chapter 10. More about finite element methods Chapter
468
ElementList ElementList 11 66 7 7 2 2 11 2 2 7 7 2 3 3 2 7 7 88 4 2 4 2 33 8 8 55 3 3 8 8 9 9 4 6 4 6 3 3 99 4 77 4 10 99 10 4 10 88 4 55 10 11 12 12 9 9 66 11 12 10 10 66 7 7 12 11 7 12 13 11 13 7 12 12 7 12 7 88 13 13 13 88 13 14 13 14 13 14 14 14 88 99 14 14 15 15 99 14 15 15 16 16 9 9 10 15 10 15 11 16 17 11 16 17 17 17 11 12 12 17 18 11 18 17 12 17 19 12 19 17 18 18 12 12 20 13 20 13 18 18 21 21 13 13 18 18 19 19 22 13 22 14 19 13 14 19 14 19 23 14 23 19 20 20 24 24 14 14 15 15 20 20 22 25 16 16 21 25 21 22 22 26 17 22 16 17 26 16 27 27 17 22 23 23 17 22 28 28 17 17 18 18 23 23 24 29 23 24 29 18 18 23 30 24 18 19 19 24 30 18 31 19 24 25 25 31 19 24 32 32 19 19 20 20 25 25 11
11
2 2 3 3 4 4 5 5 66 7 7 88 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25
NodeList NodeList 0 0 0 0 0.2500 0.2500 0 0 0.5000 0 0.5000 0 0.7500 0.7500 0 0 1.0000 1.0000 0 0 0 0 0.2500 0.2500 0.2500 0.2500 0.5000 0.5000 0.2500 0.7500 0.7500 0.2500 1.0000 1.0000 0.2500 0 0 0.5000 0.2500 0.5000 0.5000 0.5000 0.5000 0.7500 0.5000 0.7500 1.0000 1.0000 0.5000 0 0 0.7500 0.2500 0.2500 0.7500 0.7500 0.5000 0.7500 0.7500 0.7500 0.7500 0.7500 1.0000 1.0000 0.7500 0 1.0000 0 1.0000 0.2500 1.0000 1.0000 1.0000 0.5000 0.5000 1.0000 0.7500 1.0000 0.7500 1.0000 1.0000 1.0000 1.0000 1.0000
NodePtrs NodePtrs 11 0 0 2 2 0 0 3 3 0 0 4 4 0 0 55 0 0 66 0 0 7 7 11 2 88 2 9 9 3 3 10 0 10 0 11 0 11 0 12 12 4 4 13 13 55 14 6 14 6 15 15 0 0 16 16 0 0 17 77 17 18 18 8 8 19 19 9 9 20 0 20 0 21 21 0 0 22 22 0 0 23 0 23 0 24 24 0 0 25 25 0 0
FreeNodePtrs PreeNodePtrs 11 7 7 22 8 8 3 99 3 44 12 12 13 55 13 14 66 14 77 17 17 8 8 18 18 99 19 19
Table 10.1. data structure structure for for the mesh of of Figure Figure 10.1. 10.1. Table 10.1. The The data the mesh
efficient to to use use numerical numerical integration integration (quadrature). (quadrature). It is only necessary to to choose efficient It is only necessary choose aa quadrature quadrature rule rule that that introduces introduces an an error error small small enough enough that that it it does does not not affect affect the the rate solution to rate of of convergence convergence of of the the error error in in the the approximate approximate solution to zero zero as as the the mesh mesh is is refined. refined.
10.1. methods 10.1. Implementation Implementation of of finite finite element element methods
469
A quadrature quadrature rule is an approximation approximation of the form (10.1)
where W1, ... , Wk, the the weights, w\, W2, w?,..., weights, are real numbers and (Sl' (si, t1), £1), (S2' ($2, t2),"" £ 2 ) , . . . , (Sk' (fifc, tk), £&), the nodes, are are points points in in the the domain domain of of integration integration R. R. Equation Equation (10.1) (10.1) defines defines aa kk the point quadrature quadrature rule. Choosing Choosing a a quadrature quadrature rule rule
There are two issues that must be resolved: quadrature rule and its resolved: the choice of a quadrature implementation for for an an arbitrary arbitrary triangle. triangle. We We begin by discussing discussing the quadrature begin by the quadrature implementation rule. rules is by their precision. While While rule. A A useful useful way way to to classify classify quadrature quadrature rules is by their degree of of precision. it may not first, it it may not be be completely completely obvious obvious at at first, it is is possible possible to to define define rules rules of of the the form form (10.1) (10.1) that give the exact value when applied to certain polynomials. A rule has degree of precision d if the the rule is exact exact for all polynomials of degree d or less. Since both integration and and aa quadrature quadrature rule of the the form form (10.1) (10.1) are are linear linear in in f, /, it it suffices suffices both integration rule of to consider only monomials. As aa simple simple example, example, consider consider the the rule As rule
[11 f(x) dx
~ 2f(0).
(10.2)
This This rule rule has has degree degree of of precision 1, 1, since
[11 p(x) dx = 2 = 2p(0), p(x) = x::::} [11 p(x) dx = = 2p(0), p(x)
= 1 ::::}
0
p(x)
f1
2
= x 2 ::::} J_/(x) dx = 3 I=- 2p(O).
Equation (10.2) Equation (10.2) is the one-point Gaussian quadrature quadrature rule. For one-dimensional one-dimensional integrals, the n-point Gaussian quadrature quadrature rule is exact for polynomials of degree degree 2n 2n -— 1 or less. For example, the two-point Gauss quadrature rule, which has degree degree of of precision 3, 3, is is
The Gaussian quadrature 1]; to apply quadrature rules are defined on the reference reference interval [-1, [-1,1]; the rules rules on requires aa simple variables. the on aa different different interval interval requires simple change change of of variables. In multiple dimensions, In multiple dimensions, it it is is also also possible possible to to define define quadrature quadrature rules rules with with aa given given degree degree of of precision. precision. Of Of course, course, things things are are more more complicated complicated because because of of the the variety possible. We will exhibit triangles with with variety of of geometries geometries that that are are possible. We will exhibit rules rules for for triangles degree triangle degree of of precision 11 and and 2. 2. These These rules rules will will be be defined defined for for the the reference reference triangle
Chapter 10. 10. More More about about finite finite element element methods Chapter methods
470
TR TR with with vertices vertices (0,0), (0,0), (1,0), (1,0), and and (0,1) (0,1) (see (see Figure Figure 10.5). 10.5). The The following following one-point one-point rule has degree of precision 1: (10.3)
(see Exercise Exercise 1), 1), while the following following two-point two-point rule rule has has degree degree of of precision (see while the precision 2: 2: (lOA)
(see Exercise 2). (0,1)
(1,0)
(0,0)
Figure 10.5. 10.5. The The reference reference triangle triangle TR. TR. Figure What is necessary the finite finite element What degree degree of of precision precision is necessary in in applying applying the element method? method? Here Here is is aa rule rule of of thumb: thumb: The The quadrature quadrature rule rule should should compute compute the the exact exact entries entries in in the the stiffness matrix in the case of a PDE with constant coefficients. Such a quadrature stiffness constant quadrature rule will will then be accurate accurate enough, enough, even even in in aa problem problem with with (smooth) nonconstant rule then be (smooth) nonconstant coefficients, that that the the accuracy accuracy of of the the finite element method not be coefficients, finite element method will will not be significantly significantly degraded. When using using piecewise linear functions, functions, the the stiffness stiffness matrix matrix is is assembled assembled from from When piecewise linear integrals of of the the form form integrals
L
k"VrPj . "VrPi,
and the integrand is is constant constant when is constant constant (a (a linear linear function function has has aa conand the integrand when k is constant gradient). gradient). Therefore, Therefore, the the one-point one-point rule rule will give exact exact results results for for aa constant constant stant will give coefficient coefficient problem,77 problem,77 and, and, by our our rule of of thumb, thumb, the the same same rule rule is is adequate adequate for nonconstant coefficient coefficient problems. nonconstant problems. If use piecewise piecewise quadratic then "V rPj .• "V rPi would If we we were were to to use quadratic functions, functions, then V0j V0j would be be piecewise quadratic would need need aa quadrature of piecewise quadratic as as well, well, and and so so we we would quadrature rule rule with with degree degree of precision 2. 2. The The three-point three-point rule rule (lOA) (10.4) would be be appropriate appropriate in in that that case. case. 77
77Actually, Actually,
rule with degree of of precision would suffice; suffice; however, however, aa rule rule cannot cannot use use less less than aa rule with degree precision 00 would than one quadrature quadrature node, node, and and so so we we use above one-point one-point rule, rule, which which has degree of of precision precision 1. one use the the above has degree 1.
471 471
10.1. element methods 10.1. Implementation of finite element methods Integrating over over an arbitrary triangle
There is still still the quadraThere is the following following technical technical detail detail to to discuss: discuss: How How can can we we apply apply aa quadrature TR to an integral triangle ture rule rule defined defined on on the the reference reference triangle triangle TR to an integral over over an an arbitrary arbitrary triangle T? We assume that T has has vertices vertices (PI,P2), (ql, q2), and (rl, r2). We We can can then then map map T? We assume that (pi,p2), (tfi^), and (ri,r2). R (ZI,Z2) E TR to (XI,X2) E T by the linear mapping (^1,^2) G T to (xi,#2) € by the linear mapping Xl = PI X2 = P2
+ (ql - pt}ZI + (rl - PI)Z2, + (q2 - P2)ZI + (r2 - P2)Z2.
(10.5)
The mapping mapping (10.5) sends (0,0), to (PI,P2), (ql,q2), and and (rl,r2), The (10.5) sends (0,0), (1,0), (1,0), and and (0,1) (0,1) to (pi,p2), (tfi,
J = [ qi - PI q2 - P2
rl - PI ] . r2 - P2
78 We can now apply the rule rule for for aa change change of of variables variables in in aa multiple multiple integral: integral: 78 We can now aonlv the
{ f(F(y))ldet(J)1 dy. iTr f(x)dx= iTR As the the determinant determinant of of JJ is is the the constant As constant
it is easy to apply apply the the change change of of variables variables and and then then any any desired desired quadrature quadrature rule. it is easy to rule. For example, the three-point three-point rule rule (10.4) (10.4) would would be be applied as follows: For example, the applied as follows:
h
f ==
I~I
(! (X~l), X~I)) + f (X~2) , X~2)) + f
(X~3) , X~3)))
,
(X~I), X~I)) = (PI + (qi - PI)/6 + (ri - PI)/6,P2 + (q2 - P2)/6 + (r2 - P2)/6), (x~2), x~2)) = (PI + 2(ql - pt}/3 + (ri - PI)/6,P2 + 2(q2 - P2)/3 + (r2 - P2)/6), (X~3), X~3))
= (PI + (ql
- pd/6 + 2(rl - Pt}/3,P2 + (q2 - P2)/6 + 2(r2 - P2)/3).
Exercises 1. Verify that that (10.3) (10.3) produces produces the the exact the monomials monomials 1, X, y. 1. Verify exact integral integral for for the 1, x, y. 2 2. Verify Verify that that (10.4) (10.4) produces produces the the exact exact integral integral for for the the monomials monomials 1, X, ?/, y, x x2, 2. 1, x, , 2 xy, y2. xy, y.
3. (1,0), (2,0), (2,0), and (3/2,1), and and let 3. Let Let T T be be the the triangular triangular region region with with vertices vertices (1,0), and (3/2,1), let /f :: TT -> -+ R R be be defined defined by by /(x) f(x) = = x\x\. XIX~. Compute Compute
78 78This rule is is explained explained in calculus texts texts such as Gillett Gillett [17], or more more fully fully in in advanced This rule in calculus such as [17], or advanced calculus calculus texts such such as as Kaplan Kaplan [29]. texts [29].
472 472
Chapter 10. More about finite element methods Chapter R directly, and and also also by by transforming transforming the the integral to an an integral over T TR, as directly, integral to integral over , as suggested in in the result. that you you obtain obtain the the same same result. suggested the text. text. Verify Verify that
4. domain n il is coarse 4. A A common common way way to to create create aa mesh mesh on on aa domain is to to establish establish aa coarse mesh, suitable. A simple way it is is suitable. A simple way to to refine refine aa mesh, and and then then to to refine refine it it until until it triangular each triangle connecting is to to replace replace each triangle with with four four triangles triangles by by connecting triangular mesh mesh is the midpoints midpoints of of the the three three edges edges of the original original triangle. triangle. An An example example is shown the of the is shown in Figure 10.6, in which which aa mesh mesh with with two two triangles triangles is is refined refined to to aa mesh mesh with with in Figure 10.6, in triangle with eight triangles by this method. This technique of of replacing a triangle with four triangles can be enough. is fine fine enough. four triangles can be repeated repeated several several times times until until the the mesh mesh is (a) Consider the the coarsest coarsest mesh mesh in in Figure Figure 10.6. Describe the the mesh mesh using using (a) Consider 10.6. Describe the NodeList, EleElethe notation notation in in the the text, text, and and also also compute compute the the arrays arrays NodeList, mentList, FreeNodePtrs, mentList, FreeNodePtrs, NodePtrs. NodePtrs. Assume Assume that that the the mesh mesh will will be be used for a Dirichlet problem. used for a Dirichlet problem. (b) Repeat Repeat for for the the finer finer mesh mesh in in Figure Figure 10.6. (b) 10.6.
(c) 10.6 one one more (c) Refine Refine the the (right-hand) (right-hand) mesh mesh from from Figure Figure 10.6 more time, time, and and comcompute the the arrays arrays describing describing the the refined refined mesh. mesh. pute Note: since the is not not aa unique unique right right answer answer to to this this exercise, exercise, since the nodes nodes and and Note: There There is elements can elements can be be ordered ordered in in various various ways. ways.
Figure 10.6. A A coarse coarse mesh mesh (left), refined in the standard standard fashion fashion (right). Figure 10.6. (left), refined in the (right). (Note: mesh is is aa triangulation of the the rhombus rhombus in both figures; the circumscribed circumscribed (Note: The The mesh triangulation of in both figures; the box just box just indicates indicates the the axes.) axes.)
10.2. 10.2. Solving sparse linear systems
473
Explain how how to use the information stored in the data structure suggested 5. Explain suggested in the text text to to solve inhomogeneous Dirichlet the solve an an inhomogeneous Dirichlet problem. problem. 6. What information 6. Consider Consider the data structure structure suggested suggested in in the the text. What information is is lacking that is needed to solve an inhomogeneous Neumann problem?
10.2 10.2
Solving sparse sparse linear systems Solving linear systems
As we we have mentioned several times, the finite element method is a practical practical numerical method, method, even even for two- and and three-dimensional three-dimensional problems, problems, because because the merical for large large twothe matrices that it it produces produces have have mostly zero entries. matrix is is called called sparse sparse if matrices that mostly zero entries. A A matrix if aa we wish to large percentage of its entries are known to be zero. In this section, we briefly survey the subject of solving sparse linear systems. Methods for for solving linear systems systems can be divided into into two categories. A method that will produce the exact exact solution in a finite number of steps is called a direct method. (Actually, (Actually, when implemented on a computer, even a direct method round-off error. To be will not compute the the exact exact solution solution because of unavoidable round-off be precise, a direct method will compute the exact solution in a finite number of steps, provided it is implemented in exact arithmetic.) The most common direct methods we discuss modifications of of Gaussian are variants variants of of Gaussian are Gaussian elimination. elimination. Below, Below, we discuss modifications Gaussian elimination for sparse matrices. linear system only as An algorithm that computes computes the exact solution solution to a linear the limit of an infinite sequence of approximations is called an iterative method. the approximations we will will discuss the most most popular: popular: the There many iterative iterative methods methods in in use; use; we There are are many discuss the the conjugate gradient gradient algorithm. We also touch on the the topic of preconditioning, the conjugate the art of of transforming transforming a linear system system so as as to to obtain obtain faster convergence convergence with an an iterative method. iterative
10.2.1 10.2.1
Gaussian elimination for dense systems Gaussian elimination for dense systems
In order to have some standard of comparison, we first briefly discuss the standard Gaussian elimination elimination algorithm algorithm for the solution of dense systems. Our interest is Gaussian in the operation operation count—the count-the number of arithmetic necessary to to solve in the number of arithmetic operations operations necessary solve an an n x n x nn dense dense system. system. The The basic basic algorithm algorithm is is simple simple to to write write down. down. In In the the following following description, we we assume that no row interchanges are required, as these do not use 79 anyway. 79 any arithmetic arithmetic operations anyway. The following pseudo-code solves Ax = = b, nxn n E Rnxn and b bERn, where A € R 6 R , overwriting overwriting the values of A and b: b: for i = 1,2, ... , n - 1 for j = i + 1, i + 2, ... , n k· t- ~ ______ 3'_ Aii 79 79In general, Gaussian elimination is is numerically numerically unstable unstable unless unless partial pivoting is is used. used. Partial In general, Gaussian elimination Partial pivoting pivoting is is the the technique technique of of interchanging interchanging rows rows to to get get the the largest largest possible possible pivot pivot entry. entry. This This ensures ensures that all all of of the the multipliers multipliers appearing appearing in in the algorithm are are bounded bounded above above by by one, and in in virtually virtually that the algorithm one, and every case, case, that that roundoff roundoff error error does increase unduly. unduly. There There are classes of matrices, every does not not increase are special special classes of matrices, most notably notably the the class class of positive definite definite matrices, matrices, for which Gaussian elimination is is most of symmetric symmetric positive for which Gaussian elimination provably stable stable with with no no row interchanges.
Chapter 10. 10. More about finite element methods
474
bj
+- bj - Ajibi
for k = i Ajk
+ 1, i + 2, ... , n
+- Ajk - AjiAik
for i
= n, n -
1, ... ,1
bi
+- (bi -
E7=i+1 Aijb j ) /Aii
The first (nested) (nested) loop loop is is properly Gaussian elimination; systematiThe first properly called called Gaussian elimination; it it systematically eliminates variables to produce an upper triangular triangular system that is equivalent to back substitution; substitution; it to the original system. system. The second second loop is called back it solves solves the the last equation, for # last equation, which which now now has has only only one one variable, variable, for X n ,, substitutes substitutes this this value value in in the preceding solves for _i, and so forth. The pseudo-code above the preceding equation equation and and solves for or X nn-1, and so forth. The pseudo-code above overwrites the solution x. x. overwrites the right-hand-side right-hand-side vector vector b b with with the the solution The Gaussian Gaussian elimination part of of the the algorithm algorithm is equivalent (when, (when, as we The elimination part is equivalent as we have assumed, are required) have assumed, no no row row interchanges interchanges are required) to to factoring factoring A into into the the product product of of an upper unit lower lower triangular triangular matrix matrix (that (that is, is, aa lower lower an upper triangular triangular matrix matrix U U and and aa unit triangular matrix all ones L: A = LU. LU. As algorithm triangular matrix with with all ones on on the the diagonal) diagonal) L: A = As the the algorithm is written is overwritten above is written above, above, the the matrix matrix A A is overwritten with with the the value value of of U U (on (on and and above the of the diagonal) and L (below the diagonal-there diagonal—there is no need to store the diagonal of known to consist of all ones). At the same time, the first loop above L, since it is known -1 effectively replaces b by L L -1 b. The back substitution phase is then equivalent to effectively b. x 1 1 -1 1 1 replacing L=A A -lb. A = = LU is simply L~ b by U-1LU~ L~ b = b. The factorization A simply called the the LU factorization. factorization. It is is not not difficult difficult to to count count the the number number of arithmetic operations operations required required by by It of arithmetic Gaussian elimination elimination and and back back substitution. substitution. It turns turns out out to to be be Gaussian
L U (the computation of L -1x b most of the operations used to factor with most factor A into LU 11 1 2 L -1 bb requires only O(2n2) and then UU~ L~ O(2n ) operations). (Exercise 1 asks the reader to results.) to verify verify these these results.) 80 (SPD), then one can When symmetric and When the the matrix matrix A A is is symmetric and positive positive definite definite80 (SPD), then one can take factor the LLTT,, where to factor the matrix matrix as as A A = = LL where L L is is lower lower take advantage advantage of of symmetry symmetry to triangular. This This factorization factorization is is called called the the Cholesky Cholesky factorization. factorization. (The (The Cholesky triangular. Cholesky factorization-in particular, factor L is not the same matrix L appearing in the LU factorization—in particular, it does not have have all all ones ones on on the the diagonal—but diagonal-but it it is closely related.) related.) The symmetry it does not is closely The symmetry of LU factorization; of A makes makes the the Cholesky Cholesky factorization factorization less less expensive expensive than than the the LU factorization; the count is O(n33/3). For the operation operation count is O(n For simplicity simplicity (because (because the the Cholesky Cholesky factorization factorization harder to describe than the LU we will discuss is less familiar and harder LU factorization), we the direct direct solution solution of of linear terms of of Gaussian elimination and and the the LU the linear systems systems in in terms Gaussian elimination factorization. However, should bear in mind stiffness matrix K factorization. However, the the reader reader should bear in mind that that the the stiffness matrix K is SPD, and so so the the Cholesky Cholesky factorization factorization is is preferred preferred in in practice. is SPD, and practice. 80 80The The
nXn reader should should recall recall that that aa symmetric symmetric matrix matrix A A e ER Rnxn is called called positive positive definite reader is definite ifif
x G Rn,, x ^ 0 =» x • Ax > 0. xER x,tO=}x·Ax>O. The matrix matrix A A is positive definite definite if if and and only only if if all of its its eigenvalues eigenvalues are are positive. positive. The is positive all of
475
10.2. Solving Solving sparse sparse linear linear systems 10.2. systems
n n 100 200 200 400 400 800 800 1600
time (s) (s) 0.0116 0.107 0.107 1.57 12.7 102
Table 10.2. Time required required to to solve solve an an nn xx nn dense dense linear linear system on aa Table 10.2. Time system on personal computer. computer. personal The size The operation operation counts counts shows shows how how the the computation computation time time increases increases as as the the size of the the system system increases--doubling increases—doubling the the size size of of the system causes causes an an 8-fold 8-fold increase increase of the system in computation computation time. time. As As aa frame frame of of reference, Table 10.2 10.2 gives gives the the time time required required to to in reference, Table solve an an nn x linear system system on on aa reasonably good personal computer at at the the time time solve x nn linear reasonably good personal computer 81 of this this writing. writing.81 of With the rapid improvements hardware of two decades, With the rapid improvements in in computer computer hardware of the the last last two decades, it might might seem seem that that concerns concerns about about algorithmic algorithmic efficiency efficiency are are less less significant significant than than it they used used to to be. be. Indeed, Indeed, many mainframe computers computers 25 they many problems problems that that required required mainframe 25 years ago ago can can now now easily easily be be solved solved on on inexpensive inexpensive personal personal computers. computers. However, years However, 3 with not difficult to encounter problems that that with an an operation operation count count of of O(2 0(2n /3), it it is is not difficult to encounter problems n 3/3), exceed available computing power. power. For For example, example, consider consider the the discretization discretization of of aa exceed available computing three-dimensional N divisions then the of three-dimensional cube. cube. If If there there are are N divisions in in each each dimension, dimension, then the size size of 3 the the order N33 x N3, the system system will will be be on on the order of of N xN , and, and, if if aa dense dense linear linear system system of of this this order order is be solved by aa dense method, that is to to be solved (or (or even even if if aa sparse sparse linear linear system system is is solved solved by dense method, that is, not take then the is, aa method method that that does does not take advantage advantage of of sparsity), sparsity), then the operation operation count count will be be O(N9). to refine refine the the mesh mesh by by aa factor will O(N9). Merely Merely to factor of of 22 in in each each dimension dimension will will increase the the computation computation time time by factor of of 512. 512. (We (We are are not even considering considering the the increase by aa factor not even memory requirements.) Thus, Thus, given given the times in in Table Table 10.2, 10.2, to to solve solve aa problem problem on memory requirements.) the times on 10 xx 10 10 xx 10 10 grid grid might might take take aa minute; minute; to to solve solve the the same same problem problem on on aa 20 20 xx 20 20 xx 20 aa 10 20 grid would would take more than than 88 hours! hours! grid take more that algorithmic is critical This This discussion discussion should should convince convince the the reader reader that algorithmic efficiency efficiency is critical for solving solving some some realistic We now now discuss discuss the the solution solution of of sparse sparse systems; systems; for realistic problems. problems. We aa comparison comparison of of operation operation counts counts will will show show the the advantages advantages of of aa method, method, such such as as finite finite elements, elements, that that leads leads to to sparse sparse linear linear systems. systems.
10.2.2
Direct solution of banded systems
When the employed in element method has aa regular structure, the the When the mesh mesh employed in the the finite finite element method has regular structure, resulting stiffness matrix tends to to be be banded. The entries entries of of aa banded matrix are are resulting stiffness matrix tends banded. The banded matrix zero except for for those those in in aa band band close close to the main diagonal. The The precise precise definition definition zero except to the main diagonal: is the following. is the following. Definition Definition 10.1. 10.1.
nxn We say that A A is is banded banded with with half-bandwidth Let A A e ER Rnxn.. We say that half-bandwidth Let
81 81The MHz Pentium rate of improvement of The CPU CPU is is aa 450 450 MHz Pentium II. II. Of Of course, course, the the rate of improvement of CPUs CPUs being being what what it this will will no no longer good CPU by the the time time this this book book is it is, is, this longer be be considered considered aa reasonably reasonably good CPU by is published! published!
Chapter 10. 10. More about finite element methods
476
Pp ifif li-jl >p=>Aij =0.
the stiffness arising from from solving Poisson's As an an example, example, consider consider the stiffness matrix matrix arising solving Poisson's equation: equation: -~u
= f(x)
u = 0 on
in
n,
(10.6)
an.
2 Suppose is the unit square ER R2 :: 0 0 < x\ Xl < 1,0 X2 < I}), we Suppose n £7 is the unit square (n (J7 = = {x {x € 1,0 < #2 l})> and and we method with a regular triangulation of n apply the the finite element method f) and piecewise linear finite element functions. Dividing the xX and y intervals, [0,1], [0,1], into n subin2 2 2 tervals and (n + 1)2 I) nodes. Only the interior nodes nodes we obtain obtain 2n elements elements and nodes. Only the interior tervals each, each, we the stiffness - 1) - 1)2. are free, so the stiffness matrix matrix is (n — I) 22 xX (n — I) 2 . As we showed in Example 8.10, aa typical typical free rifi iteracts with nodes 8.10, free node node n/; iteracts with nodes
(also with n//;-n coefficients, the corresponding and n/ n /;+n' but, with with constant constant coefficients, the corresponding (also with nodes nodes n i-n and i+n, but, entries of of K to be zero due due to cancellation), so so that subdiagonals entries K turn turn out out to be zero to cancellation), that the the sub diagonals of of A —n + + 1, 1, -1 — 1 and and the superdiagonals indexed indexed by by 1, 1, n -— 1, 1, along along with A indexed indexed by by -n the superdiagonals with banded with half-bandwidth the main diagonal, contain nonzero entries. Thus A is banded n n -- 1. 1. See See Figure Figure 10.7 10.7 for for the the sparsity sparsity pattern pattern of of A. A. It is completely straightforward to apply elimination with with back back subis completely straightforward to apply Gaussian Gaussian elimination substitution banded system. stitution to to aa banded system. Indeed, the typical step of the algorithm algorithm is just as in in the the dense dense case, case, except that the inner inner loops run over a limited limited range of of indicesindices— keeping within the the band. band. Here Here is is the the algorithm: keeping the the computation computation within algorithm: for i = 1,2, ... , n - 1 for j = i + 1, i + 2, ... , mini i k· t- ~ J~ Ai;
+ p, n}
bj t- bj - Ajibi
for k = i + 1,i + 2, ... ,min{i + p,n} Ajk t- Ajk - AjiAik
for i = n, n - 1, ... ,1 bi t- (bi -
Ej=i+l Aijbj )
jAii
we assume that no row interchanges interchanges are required.) (Here, again, we It is easy to show that the above algorithm requires O(2p22n) n) operations operations (see Exercise 2)—a considerable considerable savings if pp «n. <£ n. For For example, example, consider consider solving savings if solving PoisPoisExercise 2)-a son's equation in two dimensions, as discussed above. The discrete Laplacian (under son's equation in two dimensions, as discussed above. The discrete Laplacian (under Dirichlet conditions), is an (n - 1)2 (n — I) 2 x (n -— 1)2 I) 2 banded matrix with half-bandwidth n -— 1. 1. The cost of solving such a banded system is O(2(n -— 1)4), I) 4 ), as opposed to a cost cost of of
10.2. 10.2. Solving sparse linear systems
477
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..
10 _.
,
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_.
20 30
e. _.
.....•
,,
-.••••
e.
,
_. e.
-..•...
40
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_.
50
,,
-....•. -.-. -...•... -.-.-.•..•..
60 70 80~
o
______
~
______
~
nz
, _.
•• __ _____________
40
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e.
.... -.-. -. ...-. -....-. -.-.-. -.-.-. ....
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~
60
80
Figure pattern of Laplacian (200 triangular Figure 10.7. 10.7. The The sparsity sparsity pattern of the the discrete discrete Laplacian (200 triangular elements). elements). if Gaussian elimination used. (For these if the the standard standard Gaussian elimination algorithm algorithm is is used. (For SPD SPD matrices, matrices, these operation be divided by two.) two.) operation counts counts can can be divided by The cost cost of performed on The of aa direct direct method method performed on aa sparse sparse matrix matrix is is controlled, controlled, not not merely by the in the the matrix, matrix, but but by by the the fill-in fill-in that merely by the number number of of nonzeros nonzeros in that occurs occurs during during the course course of of the algorithm. Fill-in is said said to to have occurred whenever an entry entry that the the algorithm. Fill-in is have occurred whenever an that is originally zero becomes nonzero during the course of the algorithm. In factoring aa banded fill-in occurs banded matrix, matrix, fill-in occurs within within the the bands, bands, as as can can be be seen seen by by inspecting inspecting the the Land L and U U factors factors (see (see Figure Figure 10.8). 10.8). Entries Entries that that become become nonzero nonzero during during the the course course of the algorithm be included included in the cost. of the algorithm must must be in subsequent subsequent calculations, calculations, increasing increasing the cost.
10.2.3
Direct solution of general sparse systems
When A is but not more difficult the cost When A is sparse sparse but not banded, banded, it it is is more difficult to to predict predict the cost of of aa direct direct matrix is method. The cost cost is governed governed by the degree of of fill-in, fill-in, and and when the the matrix is not not banded, the the fill-in fill-in is unpredictable. Figure banded, is unpredictable. Figure 10.9 10.9 shows shows aa symmetric symmetric sparse sparse matrix matrix A with with the the nonzeros nonzeros placed placed randomly, randomly, and and also also the the resulting fill-in. fill-in. The The original original matrix triangular factor matrix has has aa density density of of 9.46%, 9.46%, while the lower lower triangular factor of of A has has aa density density of 61%. of over over 61%. Sparse matrices matrices whose whose sparsity sparsity pattern unstructured are are particularly suited pattern is is unstructured particularly suited Sparse to iterative methods. methods only only require require the the ability to compute to iterative methods. Iterative Iterative methods ability to compute matrixmatrixvector products, products, such possibly, for nonsymmetric matrices, vector such as as Ap Ap (and (and possibly, for nonsymmetric matrices, the the product product T AT p). Therefore, the particular sparsity pattern is relatively unimportant; the A p). Therefore, the particular sparsity pattern is relatively unimportant; the cost cost
Chapter 10. 10. More about finite element element methods
478
20 40
40
60
60
80"'-_ _ _ __
80"'-_ _ _ _ __
o
20
40 nz = 737
60
80
o
20
40
60
80
nz =737
Figure sparsity pattern pattern of and U U {right} factors of of Figure 10.8. 10.8. The The sparsity of the the L L {left} (left) and (right) factors the discrete Laplacian (200 triangular elements). the discrete Laplacian {200 triangular elements}.
of the the matrix-vector is mostly mostly governed governed by density of of the matrix (the matrix-vector product product is by the the density the matrix (the of percentage of of nonzero nonzero entries). entries). percentage 10.2.4 10.2.4
Iterative solution sparse linear linear systems Iterative solution of of sparse systems
Even when A A is be too in terms or Even when is sparse, sparse, it it may may be too costly, costly, in terms of of arithemtic arithemtic operations operations or memory (or (or both), both), to solve Ax Ax = directly. An An alternative alternative is is to to use use an an iterative memory to solve = bb directly. iterative algorithm, which which produces sequence of of increasingly increasingly accurate accurate approximations approximations to to the the algorithm, produces aa sequence true Although the the exact be obtained in true solution. solution. Although exact solution solution can can generally generally not not be obtained except except in the number of get the the limit limit (that (that is, is, an an infinite infinite number of steps steps is is required required to to get the exact exact solution), solution), in many cases that in many cases aa relatively relatively few few steps steps can can produce produce an an approximate approximate solution solution that is we should keep in that, in of is sufficiently sufficiently accurate. accurate. Indeed, Indeed, we should keep in mind mind that, in the the context context of the "exact" Ax = b b is not really solving solving differential differential equations, equations, the "exact" solution solution to to Ax is not really the the exact the true exact solution-it solution—it is is only only an an approximation approximation to to the true solution solution of of the the differential differential iterative algorithm no larger equation. equation. Therefore, Therefore, as long long as the the iterative algorithm introduces introduces errors errors no larger than the discretization perfectly satisfactory. than the discretization errors, errors, it it is is perfectly satisfactory. beyond the of Many Many iterative iterative algorithms algorithms have have been been developed, developed, and and it it is is beyond the scope scope of to survey them. We will content ourselves with outlining the conjugate this book this book to survey them. We will content ourselves with outlining the conjugate gradient (CG) (CG) method, most popular algorithm for for solving solving SPD SPD systems. systems. (The the most popular algorithm (The gradient method, the stiffness matrix from aa finite element problem, being aa Gram Gram matrix, matrix, is is SPD-see SPD—see stiffness matrix K K from finite element problem, being Exercise 6.) 6.) We We will will also also briefly briefly discuss discuss preconditioning, method for for accelerating accelerating Exercise preconditioning, aa method convergence. convergence.
479
10.2. 10.2. Solving Solving sparse linear systems
o
50
100
nz =3096
nz =946
Figure A random sparse matrix its lower triangular factor factor Figure 10.9. 10.9. A random sparse matrix (left) (left) and and its lower triangular
(right). (right). The method is is actually actually an an algorithm algorithm for for minimizing quadratic form. If The CG CG method minimizing aa quadratic form. If n nxn igisSPD A eE RRnxn SPDand and0¢: :RRn-»--tRRisisdenned definedbyby 1 ¢(x) = "2x. Ax - b· x,
(10.7)
then calculation (see (see Exercise Exercise 3) that then aa direct direct calculation 3) shows shows that "\l¢(x) = Ax - h.
(10.8)
Therefore, the the unique unique stationary stationary point point of of ¢ is Therefore, 0 is
x = A-lb.
Moreover, aa consideration of the second derivative shows that that this stationary Moreover, consideration of the second derivative matrix matrix shows this stationary point is is the the global global minimizer minimizer of quadratic form defined by by an matrix is is point of ¢0 (a (a quadratic form defined an SPD SPD matrix analogous to scalar quadratic quadratic ax > 0—see 10.10). Therefore, to aa scalar ax22+bx+c +bx+c with with a a> O-see Figure Figure 10.10). Therefore, analogous solving minimizing ¢ are equivalent. equivalent. solving Ax = =b b and and minimizing 0 are We can can thus thus apply apply any minimization algorithm algorithm to to ¢, and, assuming assuming We any iterative iterative minimization 0, and, it works, converge to desired value x. A class of it works, it it will will converge to the the desired value of of x. A large large class of minimization minimization algorithms are Such algorithms are based based algorithms are descent descent methods methods based based on on aa line line search. search. Such algorithms are idea of of aa descent direction: Given an estimate estimate x^ x(i) of the solution, solution, aa descent on on the the idea Given an of the descent from x^, x(i), ¢ in the the direction direction p is is aa direction direction p direction such such that, that, starting starting from 0 decreases decreases in direction p. This means that, that, for for all a > 00 sufficiently sufficiently small, small, of of p. This means all a ¢(x(i)
+ ap) < ¢(x(i))
Chapter 10. methods 10. More about finite element methods
480
6
y
x
Figure graph of form defined positive definite Figure 10.10. 10.10. The The graph of aa quadratic quadratic form defined by aa positive definite matrix. The The contours contours of of the the function are also also shown. junction are shown. matrix. holds. this means means that that the holds. Equivalently, Equivalently, this the directional directional derivative derivative of of ¢0 at at x(i) xW in in the the direction of of pp is is negative, that is, direction negative, that is, V'¢(x(i)) . p
< O.
Given to minimize Given aa descent descent direction, direction, aa line line search search algorithm algorithm will will seek seek to minimize ¢0 along along the the ray 20} which is ray {x(i) (x^ + ap ap :: aa > 0} (that (that is, is, it it will will search search along along this this "line," "line," which is really really aa ray). ray). Since ¢0 is is quadratic, quadratic, it it is is particularly easy to the line line search-along search—along Since particularly easy to perform perform the one-dimensional subset, subset, ¢(j> reduces reduces to to aa scalar scalar quadratic. quadratic. Indeed, Indeed, aa one-dimensional ¢(x(i)
+ ap)
=
~(x(i) + ap) . A(x(i) + ap) 2
- b· (x (i)
= ~x(i) . Ax(i) + ap' Ax(i) + a 2
2
2
2
=
~
=
~2 p.Ap-ap. (b-Ax(i))
p. Ap
+a
+ ap)
p. Ap - b· x(i) - ab· p
(p. Ax(i) - P . b)
+ ¢(x(i))
(the symmetry of A Ap/2 and p. Ax(i) /2). A was used to combine the terms x(i) x^ .• Ap/2 p • Ax^/2)The minimum is easily easily seen seen to occur at at The minimum is to occur p. (b - Ax(i)) p·Ap .
a- --'----.:....
(10.9)
481 481
10.2. 10.2. Solving sparse linear systems
How should the descent direction is the steepest How should the descent direction be be chosen? chosen? The The obvious obvious choice choice is the steepest descent direction descent direction the directional derivative of ¢ x(i) is as negative as possible in this direction. since the at x^ The resulting algorithm (choose in the the The resulting algorithm (choose aa starting starting point, point, move move to to the the minimum minimum in steepest descent descent direction, calculate the steepest descent at that point, steepest direction, calculate the steepest descent direction direction at that new new point, and is called algorithm. It is guaranteed converge to to and repeat) repeat) is called the the steepest steepest descent descent algorithm. It is guaranteed to to converge the the minimizer minimizer x x of of 0, ¢, that that is, is, to to the the solution solution of of Ax Ax = = b. h. However, However, it it can can be be shown shown that the the steepest descent method method converges converges slowly, especially when when the the eigenvalues eigenvalues that steepest descent slowly, especially differ greatly in of A differ in magnitude (when (when this is is true, the matrix A is said said to be ill-conditioned).. ill-conditioned) For an example of search in steepest descent see Figure For an example of aa line line search in the the steepest descent direction, direction, see Figure 10.11 7. 10.11 and and Exercise Exercise 7. 5~--~----------~--~----~--~
4
3
o -10~--~----~2----~3~---4~--~5----~6
x
Figure 10.11. contours of 10.10. The Figure 10.11. The contours of the the quadratic form from Figure 10.10. The steepest direction from (marked by the "0") "o") is is indicated, along steepest descent descent direction from xx = = (4,2) (4,2) (marked by the indicated, along with the the steepest descent direction direction (marked (marked by "o"). desired with the minimizer minimizer in in the steepest descent by "<:> "). The The desired (global) minimizer is marked by "x." (global) minimizer is marked by
Example 10.2. test the the algorithms described in in this section, we we will will use use the the Example 10.2. To To test algorithms described this section, BVP BVP -~u
= f(x)
u = 0 on
in
n,
an,
(10.10)
482
Chapter 10. methods 10. More about finite element methods
where fi is is the the unit unit square where n square and and f(x) = -2
(-xi + x~
- (-1
+ X2)X2 + 3X1(-1 + X2)X2).
The exact solution solution is is The exact u(x) = xix2(1- x1)(1- X2). We mesh on dividing the Xl and X2 intervals subinWe establish establish aa regular regular mesh on n £7 by by dividing the x\ and #2 intervals into into 64 subintervals each. each. This in 22 .• 64 triangles and = 3969 3969 free free nodes. nodes. The tervals This results results in 6422 triangles and 63 6322 = The finite finite = f, isis therefore therefore 3969 3969 x 3969 3969 (a (a fairly fairly large large system, system, but but very element equation, equation, K element Kuu = very sparse). sparse). We apply apply the the steepest to solve Ku = f. f. One One hundred hundred steps, steps, We steepest descent descent algorithm algorithm to solve Ku starting with the the zero as the the initial estimate, produces estimate of starting with zero vector vector as initial estimate, produces an an estimate of u u that that by about in the Euclidean norm. error in differs from the exact value value uu by differs from the exact about 76% in the Euclidean norm. The The error in the corresponding linear function, to the the exact exact solution u, is is the corresponding piecewise piecewise linear function, compared compared to solution u, about 72% in in the the energy energy norm. norm. By comparison, the the exact Ku = By comparison, exact solution solution uu of of Ku = ff about 72% corresponds corresponds to to aa piecewise linear linear function that that has has an an error of about 3% in in the the energy energy norm. If we take 1000 Clearly Clearly these results results are not not very good. If 1000 steps instead of of 100, 25% in the Euclidean Euclidean norm and 23% 23% in the energy the errors become about 25% energy norm. It appears accurate answer answer using steepest descent, descent, but but only It appears that that we we can can obtain obtain an an accurate using steepest only by of steps. steps. by taking taking aa large large number number of
10.2.5
The conjugate gradient algorithm
The conjugate gradient gradient (CG) another descent algorithm that is usually usually aa The conjugate (CG) algorithm algorithm is is another descent algorithm that is method. The problem with with the the steepest great great improvement improvement over over the the steepest steepest descent descent method. The problem steepest descent while the steepest descent method is that, that, while steepest descent direction direction is locally optimal, optimal, from from aa global global point point of of view, view, the the search search directions directions are are poorly poorly chosen. chosen. Indeed, Indeed, it it can can be be shown that successive are orthogonal orthogonal (see (see Exercise Exercise 4). shown that successive search search directions directions are 4). It It is is not not efficient to approach the desired via orthogonal all, the the shortest efficient to approach the desired solution solution via orthogonal steps steps (after (after all, shortest path between two two points points follows straight line). path between follows aa straight line). The defines the successive search search directions pleasThe CG CG algorithm algorithm defines the successive directions to to satisfy satisfy aa pleasing global property—basically property-basically that that each step preserves preserves the the optimality property of of ing global each step optimality property previous steps. To To be after kk steps steps of estimated solutions solutions is the of CG, CG, the the estimated is the previous steps. be precise, precise, after minimizer the /^-dimensional k-dimensional subset by the the first direcminimizer of of ¢0 over over the subset spanned spanned by first k search search directions. tions. It is rather rather difficult difficult to algorithm-the final final form It is to derive derive the the CG CG algorithm—the form results results from from several nonobvious simplifications. We will content ourselves ourselves with with showing several nonobvious simplifications. We will content showing the the critical the computation the search direction. We We will will assume critical step: step: the computation of of the search direction. assume that that the the x(O) = 0, that the first kk search directions are initial estimate estimate of the solution is x^°^ p(i) ? p(2) p(fc) ^ ^d • • ,,ak &k so p(l), p(2),^ ... ^,p(k), and tthat after kk steps steps we we have have determined determined ai, a1, 0:2, 0:2, •... so that that nat after k
x(k)
=L
aiP(i)
i=l
solves
min {¢(x)
x E span{p(1), ... ,p(k)}} .
Solving sparse linear systems 10.2. Solving
483
fe+11 p(k+ If We now wish to find a new search direction p( )) with the following property: If
x(k+1) = x(k)
+ ak+lp(k+1) ,
where ak+l ctk+i solves fc+11 then x( x(k+ )) solves
min {¢>(x) : x E span{p(1), ... ,p(k+1)}} .
(10.11)
fc+1 It is not clear that such a p( p(k+1) however, it can be, as we we now now show. ) can be found; however, To (10.11), we /?i, /92, ... • • • ,,f3k+1 0k+i such that To solve (1O.11), we must must find f31,/h that
is as small as possible. (We (We separate the last term, f3k+1p(k+ 0k+ip(k+11)\ , from the sum because we already because already know how to make
algebra shows that as small as possible.) Some straightforward straightforward algebra ¢>
(t,
f3iP(i)
+ f3k+lP(k+1))
= ¢>
(t,
f3iP(i))
1 +f3k+lP(k+ ) . A
(t,
f3iP(i))
+ ¢> (f3k+ 1P(k+1))
.
fc+1 the crucial observation: observation: If p(k+1)) so that Here is the If we can choose p(
is zero, then
minimization problem problem is then "decoupled." The minimization "decoupled." That is, we can independently choose 8-1. 8<> 81. to minimize choose f31, f32, ... ,f3k to minimize
(10.12)
484
10. More about finite element methods Chapter 10.
and minimize and Pk+I @k+i to minimize
¢' (Pk+ I P(k+1))
,
and the resulting resulting PI,P2, ,Pk+1 will will be be the the solution This is is what what we we and the /?i,$2, ... • • • ,flk+i solution of of (10.11). (10.11). This want, since we of (10.12). want, since we already already have have computed computed the the minimizer minimizer of (10.12). Our problem then then reduces reduces to to finding finding p(k+ to satisfy Our problem p(fe+1I )) to satisfy k
L
(PiP(k+1) . Ap(i))
= O.
i=1
It is It is certainly certainly sufficient sufficient to to satisfy satisfy p(k+1) . Ap(i)
= 0, i = 1,2, ... , k.
We can assume by induction that p(j) . Ap(i)
= 0,
i,j
= 1,2, ... ,k,
i ¥-j.
(10.13)
p(1), p(2), ... ,, p(k) We can can recognize condition (10.13) as stating stating that the vectors p^,p^,... p^ 82 respect to the inner product82 are orthogonal with respect
(X,Y)A = X· Ay. search direction ), then, we we just take aa descent To compute the search direction p(k+ p( fc+1I ), descent direction direction and subtract subtract off off its component lying lying in in the the subspace and its component subspace Sk =
span {p(1), p(2), ... , p(k) }
;
the vectors p(1), p(2), ... the result will be orthogonal to each of the p^\p^,. - - ,p(k). , p^- We will use steepest descent direction r = = — -\7¢,(x(k)) the steepest V^(x( fc )) to generate the new search direction. To achieve achieve the the desired orthogonality, we To desired orthogonality, we must must compute compute the the component component of of rr in in SSkfc projecting r onto Sk Sk in the inner product product defined by A. We We therefore take by projecting P
(k+1) _ _ ~ r· Ap(i) (i) - r ~ p(i) . Ap(i) P . • =1
For reasons that that we we cannot cannot explain explain here, here, most most of of the the inner inner products products rr·• ApW Ap(i) are For reasons are zero, and the result is (k+I) _
P
- r
_
r· Ap(k) p(k). Ap(k)
(k)
P
(10.14)
We can then formula from from (10.9) direction We can then use use the the formula (10.9) to to find find the the minimizer minimizer ak+i Ctk+1 in in the the direction fe+1 ) , and p(k+ I ), and we will have found x(k+ ). of p( x(fe+1I ). By taking taking advantage of the the common By advantage of common features features ofthe of the formulas formulas (10.9) (10.9) and and (10.14) (10.14) (and simplifications), we can express express the in the the (and using using some some other other simplifications), we can the CG CG algorithm algorithm in fc efficient form. (The vector b -— Ax( Ax(k)) is called the the residual in the the equation following efficient Ax Ax = = b—it b-it is is the the amount amount by by which which the the equation equation fails fails to to be be satisfied.) satisfied.) 82 82It be shown shown that, that, since A is positive definite, definite, x X·• Ay Ay defines defines an an inner inner product product on see It can can be since A is positive on Rnj R n ; see Exercise 5. Exercise 5.
10.2. Solving sparse sparse linear systems 10.2.
485 485
= bb -— Ax Ax (* (* Compute Compute the the initial initial residual residual *) rr = *) pp «— r (* Compute the initial search direction *) +- r (* Compute the initial search direction Ci <- r • r Cl+-r·r for k = for fc = 1,2, l , 2 , ... ... v «— Ap Ap v+c2 <- p • v C2+-P'V o: +<— ~ ^ (* (* Solve Solve the the one-dimensional one-dimensional minimization minimization problem problem *) 0: *) xx +-f- xx + o:p ap (* (* Update Update the the estimate estimate of of the the solution solution *) rr +«— rr -—o:v CKV (* Compute Compute the the new new residual residual *) c3 «- rr ·• rr c3+f3 +- ~ £<-S Cl p -fftp + rr (* (* Compute Compute the the new new search search direction direction *) P +- f3p *) Ci <-C Cl +- C33 The reader reader should should note that only only aa single single matrix-vector matrix-vector product product is is required required at at each each The note that step making it step of of the the algorithm, algorithm, making it very very efficient. efficient. We We emphasize emphasize that that we we have have not not 83 derived all all of of the the steps steps in in the the CG CG algorithm. algorithm.83 derived The name "conjugate gradients" gradients" is is derived derived from from the the fact fact that that many many authors authors The name "conjugate refer the orthogonality the inner refer to to the orthogonality of of the the search search directions, directions, in in the inner product product defined defined by A, as as A-conjugacy. A-conjugacy. Therefore, Therefore, the the key key step step is is to to make make the the (negative) (negative) gradient gradient by A, direction conjugate conjugate to to the the previous previous search search directions. directions. direction Example 10.3. 10.3. We We apply apply 100 100 steps steps of of the the CG CG method method to to the the system system Ku Ku == ff from from Example Example 10.2. The The result result differs differs from the exact exact solution by about about 0.001% 0.001% in in the the Example 10.2. from the solution uu by Euclidean norm, piecewise linear function is just as Euclidean norm, and and the the corresponding corresponding piecewise linear function is just as accurate accurate from solving (error (error of of about about 3% in in the the energy energy norm) norm) as as that that obtained obtained from solving Ku Ku = f exactly. exactly.
10.2.6
CG algorithm Convergence of the CG
The kth estimate, The CG CG algorithm algorithm was was constructed constructed so so that that the the fcth estimate, x(k), x( fc ), of of the the solution solution 2 k-dimensional subspace p(2), ... ,p(k)}. xx minimizes minimizes 4> (j> over over the the fc-dimensional subspace Sk == span{p(1), spanjp^^p^ ),... ,p^}. minimizes 4> all Therefore, Therefore, x(n) x^ n ) minimizes 0 over over an an n-dimensional n-dimensional subspace subspace of of Rn, R n , that that is, is, over over all of of Rn. Rn. It It follows follows that that x(n) x(n) must must be be the the desired desired solution: solution: x(n) x(n) == x. x. Because of of this this observation, observation, the the CG CG algorithm algorithm can can be be regarded regarded as as aa direct direct Because method-it method—it computes computes the the exact exact solution solution after after aa finite finite number number of of steps steps (at (at least least when performed in point arithmetic). why when performed in floating floating point arithmetic). There There are are two two reasons, reasons, though, though, why this property is this property is irrelevant: irrelevant: 1. point arithmetic, 1. In In floating floating point arithmetic, the the computed computed search search directions directions will will not not actually actually
round-off errors), fact, be be A-conjugate A-conjugate (due (due to to the the accumulation accumulation of of round-off errors), and and so, so, in in fact, x(n) x^ n ) may may differ differ significantly significantly from from x. x. 2. used as 2. Even Even apart apart from from the the issue issue of of round-off round-off errors, errors, CG CG is is not not used as aa direct direct method for the the simple simple reason reason that that nn steps steps is is too too many! many! We We look look to to iterative iterative method for 83 83For For aa complete complete derivation derivation and and discussion discussion of of the the CG CG algorithm, algorithm, see see [19], [19], for for example. example.
Chapter More about finite element element methods methods Chapter 10. 10. More about finite
486
methods elimination too expensive. methods when n is very large, making Gaussian elimination In such aa case, iterative method method must must give reasonable approximation in In such case, an an iterative give aa reasonable approximation in much less than n iterations, or it also is too expensive. The CG algorithm much less than n iterations, or it also is too expensive. The CG algorithm useful precisely because it can give very good results in a relatively small is useful number of iterations. iterations. The rate rate of of convergence is related related to to the the condition number of which The convergence of of CG CG is condition number of A, which is defined as as the the ratio ratio of of the the largest largest eigenvalue of A to is denned eigenvalue of to the the smallest. smallest. When When the the condition number number is is relatively relatively small (that is, is, when when A is is well-conditioned), well-conditioned), CG CG will will condition small (that converge rapidly. rapidly. The algorithm also works well well when when the the eigenvalues eigenvalues of converge The algorithm also works of A are are clustered into aa few groups. In In this case, even the largest largest eigenvalue is much much clustered into few groups. this case, even if if the eigenvalue is larger than the the smallest, will perform perform well. well. The The worst worst case case for for CG CG is larger than smallest, CG CG will is aa matrix matrix A whose eigenvalues eigenvalues are out over wide range. A whose are spread spread out over aa wide range.
10.2.7 10.2.7
Preconditioned CG Preconditioned CG
It is is often It often possible possible to to replace replace aa matrix matrix A A with with aa related related matrix matrix whose whose eigenvalues eigenvalues are quickly. This technique is called are clustered, and and for for which CG CG will converge converge quickly. called preconditioning, it requires find aa matrix that is preconditioning, and and it requires that that one one find matrix M M (the (the preconditioned preconditioner) that is somehow similar to A (in of its its eigenvalues) eigenvalues) but simpler to somehow similar to (in terms terms of but is is much much simpler to invert. invert. At At each step gradient (PCG) it is is necessary each step of of the the preconditioned preconditioned conjugate conjugate gradient (PCG) algorithm, algorithm, it necessary to solve an an equation equation of form Mq r. to solve of the the form Mq == r. Preconditioners can be in many Preconditioners can be found found in many different different ways, ways, but but most most require require an an intimate knowledge of the there are are few few general-purpose general-purpose intimate knowledge of the matrix matrix A. A. For For this this reason, reason, there methods. One One method method that that is used is is to to define preconditioner from from an methods. is often often used define aa preconditioner an incomplete factorization of A. An An incomplete factorization (like (like incomplete factorization of A. incomplete factorization factorization is is aa factorization fill-in is preCholesky) in in which fill-in is limited by fiat. fiat. Another method for constructing preconditioners replace A by aa simpler simpler matrix matrix (perhaps (perhaps arising from aa simpler simpler conditioners is is to to replace A by arising from PDE) inverted by methods. PDE) that that can can be be inverted by FFT FFT methods.
Exercises nxn 1. A E Rnxn. Determine the number of 1. Suppose Suppose A 6 R . Determine the exact exact number of arithmetic arithmetic operations operations = LU required for the computation of of A = LU via via Gaussian elimination. elimination. Further operations required to compute L-1b count the number of operations L-1b and U-1L-1b. U~ 1 L~ 1 b.
Verify the text. formulas will useful: Verify the the results results given given in in the text. The The following following formulas will be be useful:
ti = ti i=l
2
i=l
n(n + 1), 2
= n(n+1)(2n+1).
6
nxn 2. Suppose A A E Rnxn is banded with half-bandwidth half-bandwidth p. p. Determine Determine the exact € R number of arithmetic arithmetic operations operations required into L LU. to factor factor A A into U. number of required to
487
10.2. 10.2. Solving sparse linear systems
nxn Rnxn 3. Let A E 6 R be symmetric, and suppose bbERn. € R n . Define ¢(f> as in (10.7). holds. (Hint: method is write out Show Show that that (10.8) (10.8) holds. (Hint: One One method is to to write out
and then show that and then show that
8¢ n -8 (x) = "L...J AijXj X· Z j=i
bi ·
-
Another method is to show that ¢(x + y)
= ¢(x) + (Ax -
+ 0 (1IYI12) .
b) . y
In either case, the symmetry symmetry of A is essential.) nxn Rnn be given, 4. Let ER Rnxn be SPD, let b,y eE R Let A e given, and suppose a* a* solves
min¢(y - aV'¢(y)),
'" where ¢(x)
1 = -x· Ax 2
b . x.
Show that Show that V'¢(y - a*V'¢(y)) is orthogonal to V' ¢(y ). V0(y). nxn 5. Suppose A eE R Rnxn is SPD. Show that
(X,y)A = X· Ay defines an inner product on R R nn.. 6. V2, 6. Let {Vi, {vi, v . . . ,, vvnn}} be be aa linearly linearly independent set set of of vectors in an an inner inner prodprod2 , ... uct space, and let G E Rnxn R n x n be the corresponding Gram matrix:
Prove G is SPD. (See (See the the hint for Exercise 3.4.6.) Prove that that G is SPD. hint for Exercise 3.4.6.) 7. The quadratic quadratic form shown in Figures 10.10 10.10and and 10.11 10.11is is ¢(x)
where where
1 = 2'x, Ax -
A=[i
b .x
+ 20,
~],b=[~].
Chapter 10. 10. More about finite element methods
488
x(O) = = beginning at x(°) (a) Perform one step step of the steepest descent algorithm, beginning x(O) and the minimizer minimizer (4,2). Compute the steepest descent direction at x(°) along the line defined by the steepest steepest descent direction. Compare your results to Figure 10.1I. 10.11. (b) the second steepest descent algorithm. Do you (b) Now compute the second step of the steepest arrive solution of b? arrive at at the the exact exact solution of Ax Ax = = b? (c) Perform two steps of CG, beginning with x(°) x(O) = = o. 0. Do you obtain the exact solution? exact solution? iterative method, a stopping stopping criterion is essential: essential: One 8. When applying an iterative must decide, on the the basis basis of of some some computable quantity, when when the the computed must decide, on computable quantity, computed solution is is accurate accurate enough enough that that one one can can stop stop the the iteration. iteration. A common common stopsolution stopis to to stop when the the relative relative residual, ping criterion criterion is ping stop when residual, iiAx(k) -
bii
Ilbll E. falls below some some predetermined tolerance e.
(a) Write a program implementing the CG algorithm with the above stopstopping criterion. ping criterion. (b) Use Use your your program program to to solve the finite element equation equation Ku = = f arising (b) solve the finite element arising the BVP (10.10) for meshes of various sizes. For a fixed value of of from the E, how how does the number number of of iterations iterations required required depend the number number of of e, does the depend on on the unknowns (free nodes)? Explain how the CG algorithm algorithm must be modified if the initial estimate estimate x(°) x(O) 9. Explain Think of using the CG algorithm to compute is not the zero vector. (Hint: Think compute - x(°). x(O). What linear linear system does y satisfy?) y =x —
10.3 10.3
An outline the convergence for finite finite An outline of of the convergence theory theory for element methods element methods
present an overview the convergence theory for Galerkin finite eleWe will now present overview of the ment for BVPs. describing ment methods methods for BVPs. Our Our discussion discussion is is little little more more than than an an outline, outline, describing the kinds of analysis that are required to prove that the finite element method method converges to the true solution as the mesh is refined. We will use the Dirichlet problem problem
-V'. (k(x)V'u) = f(x) in 0, u = 0 on ao as our our model model problem. as problem.
(10.15)
10.3. 10.3. An outline of the convergence theory for finite element methods
489
10.3.1 The Sobolev space space H~(n) H*(n) 10.3.1 The Sobolev We begin addressing the the proper for the the weak form of the We begin by by addressing the question question of of the proper setting setting for weak form of the the weak form as BVP. In Section Section 8.4, 8.4, we derived derived the as find
U
E eben) such that a(u,v) = (f,v) for all v E eben).
(10.16)
However, as we pointed out in Section 5.6 (for (for the one-dimensional case), the choice (n) for the space of of eb Cf^rj) space of of test functions is not really appropriate for for our purposes, since the finite element method uses functions functions that are not smooth enough to belong Moreover, the the statement of the weak form does not require that the the test to Cb(n). (£)) • Moreover, functions be so so smooth. smooth. We can define the appropriate space of test functions by posing the following question: the test test functions functions have variational question: What What properties properties must must the have in in order order that that the the variational equation equation a(u,v) = (f,v)
L22 inner product of of be well-defined? The energy inner product is essentially the L OU OV OXl OXl
ou
OV
+ OX2 OX2 .
(The the PDE, PDE, fc(x), k(x), is is also involved. However, we assume assume that that kk (The coefficient coefficient of of the also involved. However, if if we is smooth and bounded, bounded, then, then, for the purposes purposes of this discussion, discussion, it it may may as well is smooth and for the of this as well ignored.) It be constant and therefore can be ignored.) It suffices, therefore, therefore, that the partial 2 derivatives of of u and belong to L2(O), is natural natural to derivatives and v belong to L (t)}, and and it it is to define define the the space space (10.17)
An immediately arises arises is definition of derivatives of u. An issue issue that that immediately is the the definition of the the partial partial derivatives of u. Need at every every point If so, so, then Need du/dxi, OU/OX1, du/dx^ OU/OX2 exist exist at point of of £)? O? If then the the use use of of piecewise piecewise linear functions functions is still ruled out. The of partial partial derivatives derivatives (as (as presented presented in in calculus calculus books books The classical classical definition definition of and courses), in terms of limits of difference difference quotients, is purely local. There is another way to define derivatives that is more global in nature, and therefore more another tolerant of certain kinds of singularities. We begin by defining the the space C8"(O) Co°(tl) to 84 be the set of all infinitely differentiable differentiable functions whose support84 does not intersect O. (That the boundary of of fi. (That is, the support of of a e8"CO) (70°(fJ) function function is strictly in in the the any smooth function defined on fK 0 and interior of 0.) fi.) If If f/ is any and ¢ (f> E G C8"(O), (70° (fJ), then, by integration by by parts, integration parts, (10.18)
(the boundary integral must vanish since ¢ 0 and all of its derivatives are zero on the 2 boundary). We can define define aa derivative derivative of f/ E G LL2(O) (f2) by this equation: equation: If If there is is aa 2 L2(O) function g E €L (17) such that
r
in
g¢ = -
rf ~¢ for all ¢ E ego(o),
in
UXl
84 84The reader will will recall recall that that the the support support of is the the closure closure of of the the set set on on which The reader of aa function function is which the the function is nonzero. function is nonzero.
Chapter 10. More about finite element methods
490
respect to x\, Xl, and denoted then 9g is called the (weak) (weak) partial derivative of of 1 / with respect denoted a 1/ aXI. The definition of 1/ aX2 in df/dxi. The definition of a df/dx2 in the weak sense sense is entirely analogous. analogous. notation for we do for the We use the same notation for the weak partial derivatives derivatives as as we the classical classical partial derivatives classical partial derivatives; derivatives; it can be proved that if if the classical derivatives of / exist, so do the weak the same. same. The exist, then then so do the weak partial partial derivatives, derivatives, and and the the two two are are the The of 1 interpreted in terms of weak derivatives. The space HI (0) definition (10.17) is to be interpreted Hl(tl) is Sobolev space. is an example of of aa Sobolev space. It (0), It can can be shown shown that continuous continuous piecewise linear functions functions belong to HI /f1(J7), while, for example, discontinuous piecewise linear functions do not. Exercise 2 explores question in one dimension; complete justification the explores this this question in one dimension; aa more more complete justification is is beyond beyond the scope of scope of this this book. book. The standard product for for HI(O) is The standard inner inner product Hl(tl) is (f,g)Hl
=
i( n
Ig
a 1 ag a 1 ag ) + -a -a + -a -a Xl Xl X2 X2
'
and induced norm and the the induced norm is is
A g is is aa good good approximation / in in the H1 sense sense if if and and only only if g is is aa A function function 9 approximation to to 1 the HI if 9 good approximation to / and to V#. If we to good approximation to 1 and Vg \7 9 is is aa good good approximation approximation to \7 g. If we were were to L22 norm, on \7 9 need not be close measure closeness in the L on the other hand, V# close to to \71. V/. For an (in one is easier For an example example of of the the distinction distinction (in one dimension, dimension, which which is easier to to visualize), visualize), see Figure 10.12. see Figure 10.12. l Just as makes sense as long long as as u and vv belong belong to to H HI(O), Just as the the expression expression a(u, a(u, v) v) makes sense as u and (0), the right-hand right-hand side (/, (f, v) of the variational equation well-defined as long as 1/ equation is well-defined belongs L 2 (f2). It in this that the = (f'v) (f,v) is belongs to to L2(0). It is is in this sense sense that the variational variational equation equation a(u,v) a(u,v) = is under relatively relatively weak called the weak form of the BVP: the equation makes sense under assumptions on functions involved. assumptions on the the functions involved. l We now now define the subspace of H HI(O) by We define the subspace HJ(O) #o(fJ) of (ft) by HJ(O)={uEHI(O): u=OonaO}.
Then, so-called trace trace theorem, HQ (17) is is aa well-defined subspace Then, by by the the so-called theorem, HJ(O) well-defined and and closed closed subspace 1 HI(O). of H ^). Moreover, it can be shown that HI(O) Hl(fl) (and hence HJ(O)) #Q(^)) is complete under the HI Hl norm. (The notion of completeness was defined in Section 9.6.) equivalent to the HI Moreover, the energy inner product is equivalent H1 norm on the subspace HQ($I), which implies that that HJ(O) HQ(O,) is is complete energy norm It HJ(O), which implies complete under under the the energy norm as as well. well. It Riesz representation representation theorem theorem that the weak form of then follows from the the Riesz of the BVP, which as which we we now now pose pose as find u E HJ(O) such that a(u,v) = (f,v) for all v E HJ(O),
(10.19)
2 for each each 1 L2(0). and there is has aa unique solution solution for / E e L (f2). When When /f is is continuous continuous and is a classical solution classical solution uu to the the BVP, then uu is also the unique solution solution to to the weak form form of the the BVP.
10.3. An outline of the convergence theory for finite element element methods
1.5,....------"""1
491
5-------"""1
), 0
-0.5'---------' o 0.5
-5'"-------
o
0.5
X
X
Figure Figure 10.12. 10.12. Two functions belonging to to HI(O, Hl(Q, 1) 1) (left) (left) and their derivatives (right). The difference difference in the two functions is less than 5% in the L2 L2 norm, norm, 1 but it % in the HI it is more than 91 91% H norm. norm.
The various analytical analytical results mentioned in this section, such as the trace l theorem, completeness of the Sobolev space H HI(O), (£t), and the Riesz representation representation theorem, books on theorem, are are discussed discussed in in advanced advanced books on finite finite elements, elements, such such as as [6]. [6].
10.3.2 10.3.2
Best norm Best approximation approximation in in the the energy energy norm
We now We now know know that that the the weak weak form form of of the the BVP BVP (10.15) (10.15) has has aa unique unique solution solution uu E € HQ($}) for for each each f/ E e L2(0). L 2 (f)). Moreover, Moreover, given given aa triangulation triangulation T of of fl, fi, the the space space HJ(fl) Vh of of continuous continuous piecewise linear functions functions relative to T, satisfying satisfying the the Dirichlet Vh piecewise linear relative to Dirichlet HJ(O). We can therefore boundary conditions, is a finite-dimensional finite-dimensional subspace of -ffg(fl). apply the Galerkin method to compute the best approximation approximation from V Vh, h , in the energy energy norm, norm, to to the the true true solution solution u. u.
10.3.3
Approximation polynomials Approximation by by piecewise piecewise polynomials
We We have have seen seen (at (at least least in in outline) outline) that that the weak form form (10.19) (10.19) has has aa unique solution solution u, and and that that we we can can compute compute the the best best approximation approximation to to u from from the finite element element space Vh. Vh. The The next question is: is: How How good good is is the the best best approximation approximation from from Vh? V/j,? next question space This is a question of piecewise polynomial approximation, approximation, and the simplest way way to answer answer it it is is to to consider consider the the piecewise linear linear interpolant interpolant of of the the true true solution, solution, which
492
Chapter 10. 10. More about finite element methods
is is (using (using the the usual usual notation) notation) N
UI(X) =
L
U(XfJ¢i(X),
i=l
where x^, X/I, X/2,' XfN mesh. If where x / 2 , . .. . . ,, ~x.f are the the free free nodes nodes of of the the mesh. If we we can can determine determine how how well well N are approximates approximates Uu in in the the energy energy norm, norm, then, then, since since UI uj E € Vh Vh and and the the finite finite element element approximation approximation Uh Uh is is the the best best approximation approximation to to U u from from Vh, Vh, we we know know that that the the error error in Uh Uh is is no no greater. greater. in It can can be be shown shown that, that, if if Uu has has some some extra extra smoothness smoothness (in (in particular, particular, if if Uu is is aa solution lution of of the the strong strong form form of of the the BVP, BVP, so so that that it it is is twice twice continuously continuously differentiable), differentiable), then then UI HI
where h is is the the length length of of the the largest largest side side of of any any triangle triangle in in the the mesh mesh 1h Th and and C is is aa where 85 constant constant that depends on the particular particular solution Uu but is independent of the mesh. mesh.85 It follows that (10.20) also holds, holds, since since \\u - Uh\\E ~ \\u - UI\\E' also This completes the outline of the basic convergence theory for piecewise piecewise linear Galerkin finite finite elements. elements. Galerkin
10.3.4 10.3.4
Elliptic estimates Elliptic regularity regularity and and L2 L2 estimates
It would L22 norm. This can would be desirable to have an estimate estimate on the error in the L be obtained obtained by by aa standard standard "duality" "duality" argument, argument, provided provided "elliptic "elliptic regularity" regularity" holds. holds. be Assuming that the the coefficient coefficient kk is is smooth, smooth, when when the the domain domain 0£) has has aa smooth smooth boundboundAssuming that 86 ary, or or when when 0£) is is convex convex86 with aa piecewise smooth boundary, boundary, then solution of ary, with piecewise smooth then the the solution of (10.15) is smoother than the right-hand side f/ by two differentiability. two degrees of differentiability. 2 Thus, L2, Thus, if if f/ lies lies in in L , then Uu and and its its partial derivatives derivatives up up to to order order two two lie lie in in L2. L2. This property (that the solution is smoother than the right-hand side) referred side) is referred to as elliptic regularity. As As long as elliptic regularity holds, it can be shown (by the duality argument mentioned above) that
so so that that In In each each of of the the three three previous previous inequalities, inequalities, the the constant constant C represents represents aa generic generic conconstant stant (not (not necessarily necessarily the the same same in in each each inequality) inequality) that that can can depend depend on on the the particular particular solution solution Uu and and the the domain domain 0f) but but is is independent independent of of h. h. 85 85There that the There are are some some restrictions restrictions on on the the nature nature of of the the meshes meshes Th TH as as hh -+ —>• 0, 0, basically basically that the triangles become arbitrarily "skinny." triangles do do not not become arbitrarily "skinny." 86 86 A the endpoints then the the entire entire A set set 0fi is is called called convex convex if, if, whenever whenever the endpoints of of aa line line segment segment lie lie in in 0, J7, then line line segment segment lies lies in in O. fi.
10.3. An An outline outline of of the theory for for finite finite element element methods methods 10.3. the convergence convergence theory
493
Exercises 1. Suppose Suppose ¢> 0E € CoCO) Co*(ti) and and I/ : n fi -+ ->• R R is is smooth. smooth. Prove 1. Prove that that
for = 1,2. What would the boundary integral be be in by parts for i = 1,2. What would the boundary integral in this this integration integration by parts formula if it were not not assumed that ¢> is zero zero on formula if it were assumed that 0 is on aO? dfi?
2. The The purpose of this this exercise exercise is is to why continuous continuous piecewise linear 2. purpose of to illustrate illustrate why piecewise linear functions to HI (0,1), but piecewise linear do functions belong belong to /^(O,!), but discontinuous discontinuous piecewise linear functions functions do not. (a) by (a) Define Define I/ : [0,1] [0,1] -+ -» R R by
x, I(x) = { I-x,
°: ;
x ::;
~,
~<x::;1.
Then is differentiable differentiate (in (in the the classical classical sense) sense) everywhere everywhere except except at at Then I/ is xx = = 1/2, 1/2, and and dl (x) = ~ < x < ~, dx -1, 2' < x < 1.
{I,
Show that that I/ belongs belongs to to HI H1 (0,1) (0,1) and and that that its its weak derivative is as defined defined Show weak derivative is as above above by by verifying verifying that that
{I dl
io
(I
dx (x)¢>(x) dx = -
io
d¢> I(x) dx (x) dx
for ¢> E with the for every every 0 6 Co(O, C™(Q, 1). 1). (Hint: (Hint: Start Start with the integral integral on on the the right, right, rewrite the sum integration by rewrite it it aB as the sum of of two two integrals, integrals, and and apply apply integration by parts parts to to each.) each.)
(b) Define Define 9g :: [0,1] [0,1] -+ ->• R by R by (b) g(x)
={
x, 2-x,
°: ;
x ::; ~,
~<x::;1.
Then g, like /, differentiate (in (in the the claBsical classical sense) sense) except except at at Then g, just just like I, isis differentiable xx = 1/2, and = 1/2, and 0< x < ~, dg (x) = { 1, dx -1, ~<x<1. However, 1). Show that However, 9g ¢£ HI(O, ^1(0,1). Show that that it it is is not the the CaBe case that
{I dg
io
dx (x)¢>(x) dx = -
for every every ¢> 0e C^°(0,l). for E CoCO, 1).
(I
io
d¢> g(x) dx (x) dx
494
Chapter 10. 10. More about finite element element methods
l 3. Let Let 0f] be the unit square and and define define /, €H (ty by 3. be the unit square I,gg E HI(O) by
I(x) = 1 + Xl + X2, g(x) = I(x) + sin (m1fXI) sin (n1fX2). Compute L22 and HI1 norms. Compute the the relative relative difference difference in in 1 / and and 9 g in in the the L and H norms. 4. 4. The The exact exact solution solution of of the the BVP BVP
-cPu - = 12x2 dx 2 u(O) = 0, u(l) = 0
6x 0 '
< X < 1,
is u(x) u(x) = xx33(l x). Consider Consider the mesh with three elements, elements, the the is (1 -— x). the regular regular mesh with three subintervals [0,1/3]' [0,1/3], [1/3,2/3], [1/3,2/3], and and [2/3,1]. [2/3,1]. subintervals (a) Compute Compute the the piecewise piecewise linear linear finite element approximation approximation to to u, u, and and call call (a) finite element it it v. v. (b) piecewise linear (b) Compute Compute the the piecewise linear interpolant interpolant of of u, w, and and call call it it w. w. (c) Compute the error in v and w (as approximations approximations to u) in the energy norm. Which is is smaller? smaller? norm. Which
10.4 10.4
Finite element element methods methods for for eigenvalue eigenvalue problems Finite problems
In we described In Section Section 9.7, 9.7, we described how how eigenvalues eigenvalues and and eigenfunctions eigenfunctions exist exist for for nonconnonconstant coefficients coefficients problems problems on on (more (more or or less) less) arbitrary arbitrary domains, domains, but but we we gave gave no no stant We will will now now show how finite techniques them. We techniques for for finding finding them. show how finite element element methods methods can can be be used used to estimate estimate eigenpairs. eigenpairs. This This will will be be aa fitting fitting topic topic to to end end this this book, book, as it it ties together together our our two two main main themes: (generalized) Fourier Fourier series series methods methods and and finite ties themes: (generalized) finite element element methods. methods. We consider the following model model problem: problem: We will will consider the following
= Au in 0, u = 0 on ao.
-V'. (k(x)V'u)
(10.21)
We We follow follow the the usual usual procedure to to derive derive the the weak weak form: form: multiply multiply by by aa test test function, function, and integrate integrate by parts: -V'. (k(x)V'u) = AU, x E 0 ~ -V'. (k(x)V'u) v
= AUV,
x E 0, v E H6(O)
-In V' . (k(x)V'u) v AIn uv, v H6(O) ~ In k(x)V'u· V'v = AIn uv, v HJ(O). ~
=
E
E
10.4. 10.4. Finite element element methods for eigenvalue problems
495
In In the the last last step, step, the boundary term term vanishes vanishes because of of the boundary conditions conditions satisfied satisfied by the the test function function v. The The weak form form of of the eigenvalue eigenvalue problem is is therefore
U E HJ(O), a(u,v) = A(U,V) for all v
E HJ(O).
(10.22)
We We now now apply apply Galerkin's Galerkin's method method with with piecewise piecewise linear linear finite finite elements. elements. We We write of continuous piecewise linear functions, satisfying write V Vhh for for the the space space of continuous piecewise linear functions, satisfying Dirichlet Dirichlet conditions, to a given conditions, relative relative to given mesh mesh 'Th. 7ft. As usual, {n} will will represent represent the basis for now reduces the standard standard basis for Vh. Vh- Galerkin's Galerkin's method method now reduces to to
or or to to
n
Uh =
2:
Clj¢j,
a(uh' ¢i) = A(Uh' ¢i), i = 1,2, ... , n.
j=l
Substituting for Uh, we we obtain Substituting the the formula formula for obtain
=>
n
n
j=l
j=l
2: a(¢j, ¢i)CXj = A 2:(¢j, ¢i)CXj, i = 1,2, ... , n
=> Ku= AMu, nxn where K E Rnxn are and mass matrices, respectively, 6 Rnxn R n x n and ME M eR are the stiffness stiffness and u satisfying = AMu, then then A and encountered before. before. If encountered If we we find find A and and u satisfying Ku Ku = and the the corresponding piecewise piecewise linear corresponding linear function function Uh Uh will will approximate approximate an an eigenpair eigenpair of of the the differential operator. differential operator. We not have the space to discuss discuss the the accuracy We do do not have the space here here to accuracy of of eigenvalues eigenvalues and and eigenfunctions approximated approximated by this method; the analysis depends on the theory that was previous section. eigenvalues comcomthat was only only sketched sketched in in the the previous section. Only Only the the smaller smaller eigenvalues estimates of the true eigenvalues. To puted by the above method will be reliable estimates keep discussion on on an an elementary elementary level, level, we restrict our attention to to the the keep our our discussion we will will restrict our attention computation of (The reader will recall that computation of the the smallest smallest eigenvalue. (The that this this smallest smallest eigenvalue can be quite important, for example, in mechanical vibrations. It defines the fundamental frequency of of aa system by the the fundamental frequency system modeled modeled by the wave wave equation.) equation.) nxn nxn Given R nxn and M M E the problem problem Given two two matrices matrices K K Ee R and eR Rnxn ,, the
u ERn,
U
1:- 0, A E C, Ku = AMu
generalized eigenvalue eigenvalue problem. problem. It can be converted to an ordinary eigenis called a generalized is invertible, invertible, then then multivalue problem problem in example, when when M value in aa couple couple of of ways. ways. For For example, M is multiplying both both sides of the the equation by Myields plying sides of equation by M""11 yields
496
Chapter 10. More about about finite methods Chapter 10. More finite element element methods
1 which is the ordinary ordinary eigenvalue problem for for the the matrix matrix M M-1K. However, this this which is the eigenvalue problem K. However, -11 that MK will transformation has has several several drawbacks, drawbacks, not not the the least least of of which which is transformation is that M K will usually not not be be symmetric symmetric even both K K and and M (The reader reader should usually even if if both M are are symmetric. symmetric. (The should recall that symmetric matrices matrices have have many properties with with respect respect to recall that symmetric many special special properties to the the eigenvalue orthogonal eigenvectors, etc.) eigenvalue problem: problem: real real eigenvalues, eigenvalues, orthogonal eigenvectors, etc.) The preferred method for for converting converting Ku AMu into eigenvalue The preferred method Ku = = .AMu into an an ordinary ordinary eigenvalue problem uses uses the the Cholesky Cholesky factorization factorization.. Every Every SPD matrix can be written written as as problem SPD matrix can be the product of of aa nonsingular nonsingular lower triangular matrix times its its transpose. transpose. The the product lower triangular matrix times The mass matrix, SPD (see Exercise 10.2.6), 10.2.6), so so there exists aa mass matrix, being being aa Gram Gram matrix, matrix, is is SPD (see Exercise there exists T nonsingular lower lower triangular triangular matrix matrix L L such that M = LL LLT. can then then rewrite rewrite nonsingular such that M = . We We can the as follows: follows: the problem problem as
Ku=.AMu :::} Ku = .ALLT U :::} L -lKu = .ALT u :::} L -lKL -T (LT u) = .A (LT u) :::} Ax = .Ax, A = L -lKL -T, x = LT u.
This is is an an ordinary ordinary eigenvalue problem for A, and because K K is. is. This eigenvalue problem for A, and A A is is symmetric symmetric because This shows that the the generalized generalized eigenvalue eigenvalue problem problem Ku Ku = = .AMu has real real eigenvaleigenvalThis shows that AMu has ues. The The eigenvectors corresponding to to distinct distinct eigenvalues when ues. eigenvectors corresponding eigenvalues are are orthogonal orthogonal when x, but but not necessarily in the original expressed in in the not necessarily in the original variable variable expressed the transformed transformed variable variable x, u. given distinct eigenvectors uu u. However, However, given distinct eigenvalues eigenvalues A .A and and // J.1. and and corresponding corresponding eigenvectors and v, v, the the piecewise piecewise linear linear function function Uh with nodal values given by u u and and vv and Uh and and Vh Vh with nodal values given by are orthogonal orthogonal in the L L22 norm are in the norm (see (see Exercise Exercise 1). 1). There is drawback to to using using the the above above method method for for converting the generalized There is aa drawback converting the generalized eigenvalue problem to an ordinary is no reason why why A A should should be be sparse, eigenvalue problem to an ordinary one: one: There There is no reason sparse, even though K K and M are are sparse. For this reason, the the most most efficient efficient algorithms algorithms for for even though and M sparse. For this reason, the generalized instead of of using the the generalized eigenvalue eigenvalue problem problem treat treat the the problem problem directly directly instead using the above is useful one has above transformation. transformation. However, However, the the transformation transformation is useful if if one has access access only only to software for problem. to software for the the ordinary ordinary eigenvalue eigenvalue problem. Example 10.4. will estimate the smallest smallest eigenvalue the corresponding Example 10.4. We We will estimate the eigenvalue and and the corresponding eigenfunction of the negative Laplacian (subject to unit the negative Laplacian (subject to Dirichlet Dirichlet conditions) conditions) on on the the unit eigenfunction of square n. reader will will recall recall that the exact eigenpair is square fi. The The reader that the exact eigenpair is
We establish aa regular regular mesh on n, with 2n 2n22 triangles, nodes, and and (n-1)2 We establish mesh on £), with triangles, (n+1)2 (n+1)2 nodes, (n—I)2 free free matrix K K and and the the mass mass matrix matrix M, and solve solve the the nodes. We compute the stiffness M, and nodes. We compute the stiffness matrix function eig eig from from MATLAB. MATLAB. The results, generalized eigenvalue problem using the the function The results, genemlized eigenvalue problem using of n, are shown in 10.B. The reader should should notice notice that that the for various various values for values ofn, are shown in Table Table 10.3. The reader the eigenvalue estimate estimate is is converging faster than eigenfunction estimate. is eigenvalue converging faster than the the eigenfunction estimate. This This is typical. As the error in the is O(h O(h22),), while the error As the the results results suggest, suggest, the error in the eigenvalue eigenvalue is while the error typical. in eigenfunction is in the the energy in the the eigenfunction is O(h) O(h) (measured (measured in energy norm). norm).
10.4. 10.4. Finite element methods for eigenvalue problems
n n 22 44 88 16 16 32 32
Error in in AH Error Au 12.261 3.1266 3.1266 0.76634 0.76634 0.19058 0.19058 0.047583 0.047583
497
Error in '¢u t/>n Error in 1.5528 1.5528 0.84754 0.84754 0.43315 0.21772 0.21772 0.10900 0.10900
Table Table 10.3. 10.3. Errors in the finite finite element estimates of the smallest eigensquare. value and corresponding corresponding eigenfunction eigenfunction of of the negative negative Laplacian on the unit unit square. The eigenfunction are computed The errors errors in in the the eigenfunction computed in the the energy energy norm.
The The reader will recall recall from from Section Section 9.7 9.7 that that the the fundamental fundamental frequency of aa membrane occupying occupying aa domain domain n 17isisc";>::;/(27r), ci/A7/(27r),where whereAlAI isisthe the smallest smallesteigenvalue eigenvalue membrane of the the negative Laplacian on n 17(under (underDirichlet Dirichletconditions) conditions)and andccisisthe the wave wavespeed. speed. Therefore, Therefore, the fundamental fundamental frequency frequency of aa square square membrane of of area area 11 is cv'27r = .~ == 0.7071c. 27r v2 2
Using finite elements, we can compute the fundamental frequency frequency for membranes of we consider an equilateral of other other shapes. shapes. In In the the next example, example, we equilateral triangle with area area 1. 1. Example Example 10.5. 10.5. Consider the triangle n 17 whose vertices are
This triangle triangle is equilateral and has area 1. We use the finite finite element method to estimate the smallest eigenvalue domain, using eigenvalue of the negative negative Laplacian on this domain, 5 successive successive regular grids. The coarsest has 16 16 triangles, triangles, and is shown shown in Figure 10.13. The The remaining are obtaining obtaining by refining the the original original grid the standard 10.13. remaining grids grids are by refining grid in in the standard fashion (see (see Exercise 10.1.4). 10.1.4). The The finest finest mesh has has 4096 triangles. triangles. The The estimates estimates of of the smallest eigenvalue, as obtained on the successively successively finer meshes, are 27.7128, 23.9869, 23.0873, 22.8662, 22.8662, 22.8112. 27.7128, 23.9869, 23.0873, 22.8112. We conclude that the smallest eigenvalue fundamental freeigenvalue is Al AI == = 22.8, and the fundamental quency is quency is cA27r == 0.7601c.
In In Exercise Exercise 3, 3, the the reader reader is is asked asked to to study study the the fundamental fundamental frequency frequency of of aa membrane in membrane in the the shape shape of of aa regular area 1. regular n-gon n-gon having having area 1.
Chapter 10. about finite finite element methods Chapter 10. More More about element methods
498
1.4.-------,------r-----.----....., 1.2
0.8 x'" 0.6
0.4 0.2 0
-1
-0.5
0
0.5
x1
Figure 10.13. Example 10.5. Figure 10.13. The The coarsest mesh from Example
Exercises 1. 1.
(a) Suppose Suppose u, e Rn contain of piecewise functions (a) u, v v ERn contain the the nodal nodal values values of piecewise linear linear functions 2 Uh, Vh E how to to compute the L L2 inner inner product Uh,Vh € Vh, Vh, respectively. respectively. Explain Explain how compute the product of Uh and and Vh Vh from from u v. u and and v. of Uh (b) Show Show that that if (b) if u u and and v v satisfy satisfy Ku
= AMu,
Kv
= p,Mv,
where // and and M M are are the stiffness and and mass respecwhere A A/ =I- JL and K K and the stiffness mass matrices, matrices, respectively, the corresponding functions Uh, Uh,Vh Vh are are then the corresponding piecewise piecewise linear linear functions Vh € E Vh tively, then 2 orthogonal in the the I/ inner product. product. orthogonal in L2 inner 2. The results of this section to one-dimensional one-dimensional problems. 2. The results of this section are are easily easily specialized specialized to problems. Consider the eigenvalue eigenvalue problem problem Consider the
d (k(x) dx dU) = AU, 0 < x < 1,
- dx
u(O) = 0, u(1) = 0,
where k(x] = = 11 + + x. Use finite elements elements to smallest eigenvalue A. where k(x) Use finite to estimate estimate the the smallest eigenvalue A. 3. (Hard) (Hard) This exercise requires that you to assemble stiff3. This exercise requires that you write write aa program program to assemble the the stiffness and mass for -..6. —A on on an arbitrary triangulation of aa polygonal polygonal ness and mass matrix matrix for an arbitrary triangulation of
10.5. Suggestions Suggestions for for further further reading 10.5. reading
499
domain fJ. D. You You will also need access to a routine to solve the generalized eigenvalue problem.
(a) Repeat Repeat Example 10.5, replacing replacing the the triangular domain by domain (a) Example 10.5, triangular domain by aa domain bounded by area 1. 1. bounded by aa regular regular pentagon pentagon with with area (b) Let A~n) -~ on a region having area 1 and AI be the smallest eigenvalue of —A bounded by a regular n-gon. Form a conjecture about
· 11m
n-+oo
dn)
Al
.
(c) Test your hypothesis by repeating Example 10.5 for a regular n-gon, choosing nn to to be the largest largest integer is practical. (Whatever value choosing be the integer that that is practical. (Whatever value of of nn you choose, you have to create one or more meshes on the corresponding n-gon.) n-gon.) _1)2. 4. Repeat Repeat Exercise 2 using k(x) = = 1+ + (x — I) 2 .
10.5 10.5
Suggestions for further Suggestions for further reading reading
An excellent to the the theory of finite finite element element methods is the An excellent introduction introduction to theory of methods is the book book by by Brenner and Scott [6] [6] mentioned mentioned earlier. earlier. Strang Strang and Fix [45] [45] also also discusses discusses the and Fix the conconBrenner and Scott time-dependent PDEs and eigenvalue vergence theory for for finite elements applied to time-dependent problems. Most finite element element references references discuss discuss the computer implementation implementation of of finite finite Most finite the computer elements in only general terms. Readers wishing to learn more about this issue can consult the the Texas Finite Element Series [3] [3] and and Kwon Kwon and and Bang Bang [32]. [32]. consult As early in in this chapter, mesh mesh generation generation is is an an important important area As mentioned mentioned early this chapter, area of of study in its own right; it is treated by Knupp and Steinberg [31].
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Appendix A
of Theorem 3.47 3.47 Proof of
Theorem A.I. A.I. Let A E be symmetric, and suppose suppose A has has an eigenvalue Theorem 6 Rnxn R n x n be of (algebraic) (algebraic) multiplicity multiplicity kk (meaning (meaning that that A A is is aa root root of of multiplicity multiplicity kk of of the XA of the characteristic polynomial polynomial of of A). Then has kk linearly linearly independent independent eigenvectors characteristic Then A has eigenvectors corresponding to A. corresponding to A.
Proof. We We argue by induction on n. n = = 1. We Proof. n. The The result holds trivially trivially for for n 1. We nxn ER Rnxn assume it holds for symmetric (n -— 1) x (n -— 1) matrices, and and suppose A e is symmetric symmetric and and has an eigenvalue eigenvalue A of (algebraic) multiplicity k. There exists exists xi ^ 00 such such that = Axi. We can can assume assume that that Ilxlll ||xi|| = = 11 (since Xi/||xi|| is is also also Xl ::fthat Axi AXl = AXl. We (since xdlixlil an eigenvector of of A), A), and and we extend xi Xl to an orthonormal basis {Xl,X2, •.. ,x ,Xn}. }. an eigenvector we extend to an orthonormal basis {xi,x 2 ,... Then Then
XTAX=
[ Xl Ax, xfAxl
xf Ax 2 xf Ax 2
x[Axn xfAxn
X;AXl
X;AX2
x;Axn
.
.
1
nxn where X e ER Rnxn is X = = [xllx21·· ]. We have where X is defined by X [xi|x2| • • ·Ix • |xn].
and and xi AXi
= (AXlf Xi
(since A is symmetric)
= Axi Xi = AJil .
501 501
,
502
Theorem 3.47 Appendix A. Proof of Theorem
(The j.) Therefore, (The symbol symbol 8ij <% is is 11 if if ii == jj and and 00 if if ii ¥^ j.) Therefore,
~ XTAX=
xI
~X2
xIAxn
o
0:
[ X;AX2
:::
x nT
lx
1
n
=[~ ~], where BE B <E R(n-1)x(n-1) RCn-1)*^-1) is [s defined by
The matrix matrix B is symmetric. Also, det (..xl - XT AX)
= det (XT (..xl - A)X) = det (XT) det (..xl - A) det (X) = det (..xl - A),
l since X X TT == X" X- 11 and det (X) Itfollows follows that that and therefore det det (XT) (XT) = = det (X)-1. . It
det (..xl - A)
=
l
..x-..x 0 ..xl 0_ B I
= (..x -..x) det (,\1 - B).
Since B of of Since ,\ A is is an an eigenvalue eigenvalue of of A of of multiplicity multiplicity k, Ar, it it must must be be an an eigenvalue eigenvalue of of B multiplicity kk -- 1, therefore, by by the the induction induction hypothesis, multiplicity 1, and and therefore, hypothesis, B B has has kk -— 11 linearly linearly Y2, Y3, ... independent eigenvectors y2,ys, independent • • • ,,Yk yfc corresponding to to A. We We define
for ii = = 2,3, 2,3,..., k, and and define define Ui Uj == XYi' XyV Then Then ... , k, for AUi
= X [
~
= X [
~
Appendix A. Proof of Theorem Theorem 3.47 3.47
503 503
T T XTX XTX dis(the columns of X are orthonormal, so X X= = I; this is why the factor of X X disappeared U2, U3, ... , Uk eigenvectors appeared in the above calculation). This shows shows that u ,113,..., u/t are 2 of A to >.. Moreover, of A corresponding corresponding to A. Moreover,
(0 the zero from which which we we can can deduce deduce that that {ui, {Ul,U2, ... ,Uk} is (0 is is the zero vector vector in in Rn-l), Rn 1 ] , from 112,..., u^} is linearly independent independent set set of of eigenvectors eigenvectors of of A A corresponding corresponding to to >.. A. This This completes completes aa linearly the proof. D
This page intentionally This page intentionally left left blank blank
Appendix B Appendix
eng the data in two Shifting dmensionss
B.O.l B.O.I
Inhomogeneous Inhomogeneous Dirichlet Dirichlet conditions conditions on on a a rectangle rectangle
The method of shifting the data can be used to transform an inhomogeneous Dirichlet problem to a homogeneous Dirichlet problem. This technique works just as it did for a one-dimensional problem, although although in two dimensions it is more difficult difficult to function satisfying satisfying the the boundary boundary conditions. conditions. We We consider consider the the BVP BVP to find find aa function -~u =
f(x), x
En,
gl(xd, u(x) = g2(X2), { g3(xd, g4(X2),
r 1, E r 2, E r 3, E r 4,
x E
x x x
(B.1)
the rectangle rectangle defined in (8.10) and r 1U Tr 22 U Tr33 U rT4,4, as in (8.13). where n Q, is the and an dtt = = TI We will assume that the so We will assume that the boundary boundary data data are are continuous, continuous, so gl(fd = g2(O), g2(f 2) = g3(fI), g3(O) = g4(f2), g4(O) = gl(O).
Suppose Suppose we we find find aa function p defined defined on n J7 and and satisfying satisfying p(x) == g(x) #(x) for for all xE e an. dfL We We then then define define vv == uu -— pp and and note note that that -~v
and and
= -~u + ~p = f(x) + ~p(x)
v(x) = u(x) - p(x) = 0 for all x E an
(since p satisfies the same Dirichlet conditions that uu is to satisfy). We We can then solve -~v
= j(x),
v(x) = 0, x
n, E an,
j(x) = f(x)
+ ~p(x).
where
RDR 505
xE
506
Appendix B. Shifting the data in two dimensions
The result result will will be converging series series for for v, v, and The be aa rapidly rapidly converging and then then uu will will be be given given by by u = v + p. u v +p. We describe aa method tedious) for for computing funcWe now now describe method (admittedly (admittedly rather rather tedious) computing aa function p that satisfies the given Dirichlet conditions. We We first first note that there is a polynomial of the form polynomial of the form q(X1, X2) = a + bX1 + CX2 + dx1X2
which corners: which assumes assumes the the desired desired boundary boundary values values at at the the corners: q(O,O) = 91(0) = 94(0), q(£l,O) = 91(£t) = 92(0), q(£1,£2) = 92(£2) = 93(£t), q(0'£2) = 93(0) = 94(£2).
A direct calculation calculation shows shows that that a=91(0), b = 91(£1) - 91(0) £1 ' 94(£2) - 94(0) c= , £2 d = 92(£2) + 91(0) - 91(£t} - 94(£2). £1£2
We then define h 1(X1)=91(Xl)- ( 91(0)+ h 2(X2)
= 92(X2) -
( 92(0)
+
91(£d - 91(0) ) £1 Xl, 92(£2) - 92(0) ) £2 X2,
h3(Xt} = 93(Xt} - ( 93(0)
) + 93(it) i1- 93(0) Xl,
h 4(X2) = 94(X2) - ( 94(0)
+
94(i2) - 94(0) ) £2 X2·
We replaced each function hi differs from from gi linear We have have thus thus replaced each gi 9i by by aa function hi which which differs 9i by by aa linear function, and which has value zero at the two endpoints:
Finally, we define we define Finally,
Appendix B. in two Appendix B. Shifting Shifting the the data data in two dimensions dimensions
507
The should notice notice how how the the second second term interpolates between between the the boundary boundary The reader reader should term interpolates values on rFI1 and F , while the third term interpolates between the boundary values on and r 3, values 3 while the third term interpolates between the boundary values on and rIT^. 4. In order for these two two terms terms not interfere with with each is on rF22 and In order for these not to to interfere each other, other, it it is necessary necessary that the the boundary data data be zero at at the corners. It It was was for for this this reason that that we transformed transformed the into the term in in the the formula formula for we the g^s gi's into the h^s. hi'S. The The first first term for pp undoes undoes this transformation. transformation. It is straightforward that p satisfies the desired boundary conditions. It is straightforward to to verify verify that satisfies the desired boundary conditions. For example, example, For P(X1'0)
= a + bX1 + cO + dx 10 + h1(xd +
h3(xd - h1(xd £2 0
h2(0)-h4(0)
= =
+h4 (0) + £1 Xl a + bX1 + h1(xd (since h4(0) = h2(0) gl(O) + bX1 + gl(xd - (gl(O) + bX1)
= 0)
=gl(xd· Thus the boundary is satisfied. calculations show Thus the boundary condition condition on on rFI1 is satisfied. Similar Similar calculations show that that pp satisfies the the desired desired boundary conditions on on the the other other parts of the satisfies boundary conditions parts of the boundary. boundary. Example Example B.1. B.I. We will solve the BVP BVP
x x x x
E r 1, E r 2, E r 3, E f4
by the the above above method, where n fJ is is the the unit unit square: by method, where square:
To compute p, we define define a=91(0)=0, b = 91(1) - gl(O) = 1 1
'
c = 94(1) - 94(0) = 1 1
d = g2(1) h1 (Xl) h2(X2) h3(xd h4(X2)
+ gl(O) -
'
gl(l) - g4(1) = -2 1·1 ' = xi - (0 + xd = xi - Xl, = 1- x~ - (1 + (-I)x2) = X2 - x~, = (1- X1)2 - (1 + (-I)X1) = xi - Xl, = X2 - (0 + X2) = o.
(B.2)
B. Shifting the data in in two dimensions Appendix B.
508 We can can then then define define We P(XI,X2)
= Xl + X2 =
X2 -
XIX2
2XIX2
2
+ Xl
+ xi -
Xl
+ o· X2 + 0 + (X2
-
X~)XI
2
- XIX2·
Having computed computedp, we define define vv = uu—p, so that that Having p, we - p, so -Llv = -Llu + Llp = 0 + Llp = 2 - 2XI, X En
and and v(x)
= 0,
xE
an.
We then then see see that that We
where where
We then then obtain obtain the the following u: We following formula formula for for u:
The solution B.I. in Figure Figure B.1. The solution is is shown shown in B.0.2 8.0.2
Inhomogeneous Inhomogeneous Neumann Neumann conditions on on a rectangle
We We can can also also apply apply the the technique technique of of shifting shifting the the data data to to transform transform aa BVP BVP with with inhomogeneous inhomogeneous Neumann Neumann conditions conditions to to aa related related BVP BVP with with homogeneous homogeneous Neumann Neumann conditions. Dirichlet the details details are are somewhat somewhat more more involved involved than than in in the the Dirichlet conditions. However, However, the case. case. Suppose Suppose the the Neumann Neumann conditions conditions are are x x x x
We We first first note note that that this this is is equivalent equivalent to to
rl , r 2, r 3, E r 4. E
E E
Appendix B. B. Shifting Shifting the the data data in in two two dimensions Appendix dimensions
o
509
0
Figure B.t. B.I. The The solution to the the BVP BVP (B.2), (B.2), approximated approximated by by aa Fourier Figure solution to Fourier series with 400 terms. series with 400 terms.
We make following observation: observation: If If there is aa twice-continuously twice-continuously differentiable differentiable We make the the following there is function uu satisfying satisfying the given Neumann Neumann conditions, conditions, then, then, since since the given function
we have have we
82 u
82 u
8:lh 8X2
8X28xl '
510
Appendix B. B. Shifting the data two dimensions dimensions Appendix Shifting the data in in two
and and
which together which together imply imply that that dg 1 (0) dx 1
= dg4 (0). dx 2
the following conditions: By similar similar reasoning, we have all of the conditions: dg 1 (0) = dg4 (0), dX1 dX2 1 _ dg (£1) = dg2 (0), dx 1 dX2 dg 2 (£ ) = dg 3 (£ ) 2 d 1, dx 2
Xl
_ dg 4 (£2)
= dg3 (0).
dx 2
(B.3)
dx 1
We will will assume We assume that that (B.3) (B.3) holds. holds. now explain how to compute a function pp that satisfies We now explain how to compute satisfies the desired desired Neumann conditions. conditions. The method is is similar similar to that used to shift shift the data in aa Dirichlet problem: we will "interpolate" "interpolate" between the the Neumann conditions conditions in in each dimension and so that that the two interpolations not interfere other. and arrange arrange things things so the two interpolations do do not interfere with with each each other. We use use the that We the fact fact that (BA)
satisfies satisfies (B.5)
The step is is to data g\ function h\ The first first step to transform transform the the boundary boundary data gl to to aa function h1 satisfying satisfying dh1 (0) = dh1 (£1) = 0, dX1 dX1
and similarly similarly for for 92,93,94 and h/i2,/is,/i4. Since these these derivatives of the and g2, g3, g4 and derivatives of the boundary boundary 2, h3, h4. Since data the corners corners are are (plus or minus) minus) the of the data at at the (plus or the mixed mixed partial partial derivatives derivatives of the dedesired function function at it suffices suffices to to find function Q(X1, q(xi,xz) the at the the corners, corners, it find aa function X2) satisfying satisfying the sired conditions
511 511
Appendix B. B. Shifting the data data in two dimensions Appendix Shifting the in two dimensions
We can can satisfy satisfy these conditions with with aa function of the the form We these conditions function of form
The reader can verify verify that the necessary coefficients coefficients are are
If is to to satisfy satisfy the the desired Neumann conditions, conditions, then If pp is desired Neumann then
where where
hl(xd = gl(Xl) + aXl + bxi, h2(X2) = g2(X2) - (a + 2i1 b)X2 - (c + 2ild)x~, ha(xd = ga(xd - (a + 2i2C)Xl - (b + 2i2d)xi, h4(X2) = g4(X2) + aX2 + cx~. We can now define define pp — - q q by by the the interpolation described by by (BA), (B.5): We can now interpolation described (B.4), (B.5):
Then the original Neumann conditions, conditions, as reader can can verify Then pp satisfies satisfies the original Neumann as the the interested interested reader verify We will directly. We will illustrate this this computation computation with with an example. example.
Example B.2. B.2. We will solve the BVP Example BVP
x E fl' X E f2' X E fa,
x E f4
by the above method, where f) is is the the unit unit square: by the above method, where n square:
(B.6)
512
Appendix B. Shifting the data in two dimensions
To compute p, p, we define define dg l a= --(0) = 0, dx l b = ! (d9l (0) _ dg l (1)) = 0, 2 dXl dXl ga l C = ! (d (0) + dg (0)) = 0, 2 dXl dXl d = ! (d92 (1) + dg l (1) _ dga (0) _ dg l (0)) 4 dX2 dXl dXl dXl hI (xd = gl(Xl) = -1, h2(X2) = g2(X2) - x~ = 0, ha(Xl) = ga(Xl) = 1,
= !, 2
xr 2 h4(X2) = g4(X2) = -a' We can can then then define define We P(Xl,X2)
= q(Xl,X2) 1
hl (Xl)X2
+
2
2 2
ha(xd
1
= 2XlX2 + X2 + aXl -
+ hl(xl)
2£2
2
X2 - h4(X2)Xl
+
h2(X2)
2
aXl'
define v == U Having computed computed p, we define u -— p, so that 2 3
-Av=-Au+Ap=O+Ap=xr+x~--, XEO
.
and and
ov on (x) =
0, x E 80.
We can can write write the the solution solution V(Xl,X2) v(x\,xi} in in the the form form We 00
U(Xl, X2) =
aOO
+L
00
amo
cos (mnxd
+L
m=l 00
+
aO n
cos ( nnx2)
n=l
00
L La
mn
cos (mnxl) cos (nnx2).
m=ln=l
With With we have we have 00
f(xl,x2)
= Coo + L
00
CmO
m=l 00
+L
cos (mnxd
+
L
Con
cos (nnx 2)
n=l
00
L
m=l n=l
Cmn
cos (mnxd cos (nnx2),
+ h4(x2)
2£1
2
Xl
Appendix B. Shifting the data in two dimensions
513
where where 1
Coo =
10 10
1 f(X1,X2)dx2dx1 =0,
111 4(-1)m f(x1,x2) cos (m1fx1)dx2 dx1 = 2 2 ' m = 1,2,3, ... , m 1f 1o 0 111 4(-1)n Con = 2 f(x1,x2) cos (n1fX2) dX2dx1 = 2 2 ' n = 1,2,3, ... , n 1f 1o 0 1 1 Cmn = f(x1, X2) COS (m1fxd COS (mrX2) dX2 dX1 = 0, m, n = 1,2,3, .... CmO
=2
410 10
It then follows It then follows that that
X2) = X2) + P(X1' X2). -^ A graph of the solution u, and U(X1' u(xi,X2) = V(X1, ^(^1,^2) + p(a?i,#2)u, with aOO aoo == 0, is B.2. shown in Figure B.2. shown in Figure
2 1.5
0 .5
o - 0 .5 1
o
0
Figure B.2. The solution to BVP in Example B.2. B.2. This graph was was Figure B.2. The solution to the the BVP in Example This graph produced using of the Fourier cosine produced using aa total total of of 40 terms terms of the (double) (double) Fourier cosine series. series.
This page intentionally This page intentionally left left blank blank
Appendix C Appendix
Solutions ions to to
• odd-numbered exercises bered exercises
Chapter 1 1 1. 1.
(a) (a) First-order, First-order, homogeneous, homogeneous, linear, linear, nonconstant-coefficient, nonconstant-coefficient, scalar scalar ODE. ODE. (b) (b) Second-order, Second-order, inhomogeneous, inhomogeneous, linear, linear, constant-coefficient, constant-coefficient, scalar scalar PDE. PDE. (c) (c) First-order, First-order, nonlinear, nonlinear, scalar scalar PDE. PDE.
3. 3.
(a) (a) Second-order, Second-order, nonlinear, nonlinear, scalar scalar ODE. ODE. (b) (b) First-order, First-order, inhomogeneous, inhomogeneous, linear, linear, constant-coefficient, constant-coefficient, scalar scalar PDE. PDE. (c) (c) Second-order, Second-order, inhomogeneous, inhomogeneous, linear, linear, nonconstant-coefficient, nonconstant-coefficient, scalar scalar PDE. PDE.
5. 5.
(a) No. No. (a) (b) Yes. (b) Yes. (c) (c) No. No.
7. f: ff(t) cos (t). 7. There There is is only only one one such such /: ( t ) = ttcos (t).
9. w(t) = Cu(t) 9. For For any any constant constant C f:. ^ 1, 1, the the function function w(t) Cu(t) is is aa different different solution solution of of the the differential differential equation. equation.
2.1 Section 2.1 1. per temperature, temperature, for J / (cm ss K) 1. The The units units of of '"K are are energy energy per per length length per per time time per for example, example, J/(cm K) == W/(cmK). W/(cmK).
3. Apcu~x are 3. The The units units of of Apcu&x are cm
2
g J cm gK
- - 3 -Kcm
= J.
5. ~~(f,t)=ll<,t>to. 5- &&*) = £ , « > *>. 7. Measuring xx in 7. Measuring in centimeters, centimeters, the the steady-state steady-state temperature temperature distribution distribution is is u(x) u(x] = O.lx + + 20, 20, the the solution solution of 0.1x of
d2 u - dx 2
= 0,
u(O) 515 515
= 20,
u(100)
= 30.
516
Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises Appendix The heat heat flux The flux is is -K,
du dx (x)
= -0.802 ·0.1 = -0.0802 W /cm 2 .
Since the the area area of of the the bar's bar's cross-section cross-section is is 1r vr cm cm22,, the the rate rate at at which which energy energy is is flowing Since flowing through the bar W (in direction). through the bar is is 0.08021r 0.08027T ~ = 0.252 0.252 W (in the the negative negative direction). 9. Consider Consider the the right right endpoint endpoint (x = f). I). Let Let the temperature of of the the bath bath be be Ul, ue, so so that that 9. the temperature the difference difference in in temperature between the the bath bath and and the end of of the the bar bar is is u(f, u(i, t} ut. the temperature between the end t) -—Ul. The heat heat flux at xx = fi is, is, according according to Fourier's law, The flux at to Fourier's law,
so the the statement statement that that the heat flux is proportional proportional to to the is written so the heat flux is the temperature temperature flow flow is written
where aa > 00 is is aa constant constant of of proportionality. proportionality. This This can can be be simplified simplified to to where
The condition condition at at the end is is similar: The the right right end similar: au(O, t) -
K,
au ax (0, t) = auo·
(The sign sign change change appears appears because, because, at at the the left left end, end, aa positive positive heat heat flux that (The flux means means that heat flows into the bar, bar, while heat flows into the while at at the the right right end end the the opposite opposite is is true.) true.) 11. 11. The IBVP IBVP is
au au at - D ax = lex, t), 2
2
u(x, to) = tf;(x),
< x < f, < x < f,
0
°
t
> to,
au ax (0, t) = 0, t > to, au ax (f, t) = 0, t > to. 13. 13.
(a) The The rate rate is (a) is o) - f u-. -1rT 2D (Ul
(b) (b) The The rate rate of of mass mass transfer transfer varies varies inversely inversely with with ft and and directly directly with with the the square square of of T. r.
15. We We have 15. have
1
1 cos (xy)dy = o
and and
.!:... [SinX] dx
x
[sin~Xy)]1
= xcosx - sin x x2
y=O
= cosx X
sin x x _ sinx x2 •
517 517
Appendix C. Solutions Solutions to to odd-numbered exercises Appendix C. odd-numbered exercises On the the other other hand, hand, On
11
d o dx [cos (xy)] dy= -
11 0
ysin(xy)dy
1-11
= [YCOS (X Y)] x y=o
0
cos (xy) dy x
[Sin (X y )] 1
_ cosx
- -x- -
-x-2-
cos x
y=o
sinx
= -x--7·
Section 2.2 1. The The sum sum of of the forces on cross-section originally originally at 1. the forces on the the cross-section at x = t£ must must be be zero. zero. As As derived in in the the text, text, the (elastic) force force acting acting on on this derived the internal internal (elastic) this cross-section cross-section is is -Ak(£) ~: (£),
while the external traction traction results results in total force pA (force per unit unit area area times while the external in aa total force of of pA (force per times area). Therefore, Therefore, we we obtain area). obtain du -Ak(£) dx (£)
or or
+ pA = 0,
du k(£) dx (£)
= p.
The other boundary condition, of of course, course, is u(Q) = — 0. The other boundary condition, is u(O) O. 3. 3.
(a) The stiffness stiffness is the ratio internal restoring restoring force relative change (a) The is the ratio of of the the internal force to to the the relative change in length. length. Therefore, bar of of length length £ t and is stretched stretched in Therefore, if if aa bar and cross-sectional cross-sectional area area A A is to aa length length of +p (0 (0 < pp < 1) 1) times I, then the bar with aa force force to of 11 +p times f, then the bar will will pull pull back back with of this is is the the force that must must be be applied of 195pA 195pA gigaNewtons. gigaNewtons. Equivalently, Equivalently, this force that applied to to the of the the bar bar to bar to length of p)£. the end end of to stretch stretch the the bar to aa length of (1 (1 + +p)i. (b) 196/195 196/195 m, or approximately approximately 1.005 1.005 m. m. (b) m, or (c) BVP is is (c) The The BVP 2
-195 . 10 9 d u2 = 0 0 ' dx u(O) = 0,
< x < 1,
195. 10 9 du (1) = 109 . dx It by direct direct integration the solution is u(x) x/195, It is is easy easy to to show show by integration that that the solution is u(x] = o;/195, and therefore u(l) = 1/195. Since Since u is is the the displacement displacement (in (in meters), and therefore u(l) = 1/195. meters), the the final final position 1/195 m, is, the stretched bar position of of the the end end of of the the bar bar is is 11 + + 1/195 m, that that is, the stretched bar is is 196/195 m. 196/195 m.
5. The The wave wave equation equation is 5. is 2 (7.9 . 103) a 8tu2
-
(
a2 = !(x, t),
1.95· 10 11) axu2
0<x
< 1.
518 bib
Appendix C. C. Solutions to odd-numbered exercises
The units units of of 1 are N/m N/m33 (force (force per per unit unit volume). volume). The The units units of the first first term term on on the The / are of the the left are left are kg m kgm/s2 N m 3 - m3 ' m 3 S2 and the the units units of the second on the the left left are and of the second term term on are N 1
m2 m
7. 7.
(a) (a) We We have have
a{)t22 u (x, t) = -c2 () 2 u(x, t),
while while Therefore, Therefore,
N
= m3 ·
2 aax2 u 2 (x, t) = -f} u(x, t).
a{)t22 u -
2 C
aax2 u = -c2 f} 2 u + c2 () 2 u = o. 2
(b) Regardless of of 0, f}, u(O, for all all t. The way that that (b) Regardless of the the value value of u(Q, t) = = 00 holds holds for The only only way u(l, can hold all tt is u(l, t) t) — = 00 can hold for for all is if if sin «(}l)
= 0,
so f}9 must one of of the so must be be one the values values
() = ~7r,
n
= 0, ±1, ±2, ....
Section Section 2.3 2.3 1. Units acceleration (length (length per squared). 1. Units of of acceleration per time time squared). 3. Units (length per time). 3. Units of of velocity velocity (length per time). 5. The The internal internal force on the end of the string string at, 5. force acting acting on the end of the at, say, say, xx = = Il,, isis au
T ax (l, t).
(C.1)
If this end can can move the vertical vertical direction, (C.I) If this end move freely freely in in the direction, force force balance balance implies implies that that (C.1) must be zero. zero. must be
Section Section 3.1 3.1 1. the form form f(x) I(x) = linear if and only only if if b = = 0. O. Indeed, Indeed, if if 1. A A function function of of the = ax ax + + b is is linear if and
/I:: R —» R linear, then a= /(I). R -t R is is linear, then f(x) I(x) — = ax, ax, where where a = 1(1). 3. (a) Vector Vector space e[O, 1)). 3. (a) space (subspace (subspace of of C[0,1]). (b) Not space; does contain the the zero zero function. function. (Subset (Subset of of e[O, C[0,1], but (b) Not aa vector vector space; does not not contain 1], but not aa subspace.) subspace.) not
(c) space (subspace (subspace of of e[O, C[Q, 1]). (c) Vector Vector space (d) Vector space. (d) Vector space. (e) Not Not aa vector vector space; space; does does not not contain contain the zero polynomial. polynomial. (Subset P nn,, but (e) the zero (Subsei of of P but not aa sub subspace.) not space.)
519 519
Appendix C. C. Solutions to odd-numbered odd-numbered exercises Appendix Solutions to exercises
€ Cl1[a,b] 5. Suppose u E [a, b] is any nonzero nonzero function. Then
L(2u)
3 du 3 = 2 du dx + (2u) = 2 dx + 8u ,
while
2Lu 7.
du
= 2 dx + 2u
3
.
L(2u) =1= Since L(1u) ^ 2Lu, 2Lu, L is not linear. (a) The proof is a direct calculation:
A(ax + ,Bz)
=
= [:~~
:~~] [ ~:~! ~~~ ]
all (axl + ,BZl) + a12(ax2 + ,BZ2) ] [ a21(axl + ,Bzl) + a22(ax2 + ,Bz2) a(allxl + a12x2) + ,B(allzl + a12z2) ] [ a(a21xl + a22x2) + ,B(a21z1 + a22z2)
=a
[ allXl a21Xl
+ a12X2 + a22X2
] +,B [ allZl a21Z1
+ a12Z2 + a22Z2
]
= aAx+,BAz. This shows that the operator defined by A is linear. (b) We have n
(A(ax+,BY))i
= La;j(axj +,BYj) j=l
and
n
(aAx + ,BAY)i
=a L j=l
aijXj +,B L a;jYj· j=l
Now, Now, n
(A(ax+,By)); = Laij(aXj +,BYj) j=l n
=L
(aaijXj
j=l n
n
= Laaijxj j=l n
=a L
+ ,BaijYj) + L,BaijYj j=l
n
aijXj
+ ,B L
aijYj
j=l j=l = (a Ax + ,BAy);, as desired. The third equality follows from the the commutative and associative the fourth equality follows from the the distributive while the properties of addition, while property of multiplication multiplication over addition.
Appendix odd-numbered exercises Appendix C. C. Solutions Solutions to to odd-numbered exercises
520
Section 3.2 1. The range A is is 1. The range of of A
{o:(l,-l) {a(l,-l) : O:ER}, a€R}, which is the plane. plane. See Figure C.I. C.l. which is aa line line in in the See Figure 3~--~----~----r---~~--~----~-----r--~
2
)(N
0
-1
-2 -3~--~----~----~--~~--~----~----~--~
-3
-2
-1
0
2
3
Xl
Figure C.l. range of Figure C.I. The The range of A A (Exercise (Exercise 3.2.1). 3.2.1). 3. 3.
(a) A is nonsingular. (a) A is nonsingular. (b) A is nonsingular. (b) A is nonsingular. (c) A is is singular. in the the range (c) A singular. Every Every vector vector in range is is of of the the form form
Ax
= [~
~]
= [ ~~!~:
[ :: ]
].
That is, y E whose first That is, every every y €R R22 whose first and and second second components components are are equal equal lies lies in in the the range of Thus, for is solvable for range of A. A. Thus, for example, example, Ax Ax = = bb is solvable for
b=[~], but not for but not for
b=[;].
5. The The solution solution set set is is not subspace, since since it it cannot contain the the zero zero vector vector (AO (AO = 5. not aa subspace, cannot contain =
). oOi=^ bb).
521
Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises Appendix exercises
The null space of L the set of all first-degree polynomials: 7. The L is the
N(L) 9.
= {u: [a,b]-+ R
: u(x) = mx + c for some m,c E R}.
The only functions functions in CZ[a, b] that satisfy (a) The C2[a, 6]
dZu dx
--= z 0
are functions Cix + C C?. condition are functions of of the the form form u(x) u(x) = C1x The Neumann Neumann boundary boundary condition z . The at the left left endpoint implies that C\ Cl = = 0, 0, and the Dirichlet boundary condition at the right endpoint implies that €2 C2 = = O.0. Therefore, only the zero function is u = 0, a solution to Lin L^u 0, and so so the null space of Lin Z/m is trivial.
E C[a, b] is given and u € E C^[a, C~ [a, b] b] satisfies (b) Suppose /f G C[a,6] d2 u - dx 2 (x)
= f(x),
a < x < b.
Integrating Integrating once yields, by the the fundamental theorem of calculus,
- iar dx~ (s) ds = iar f(s) ds d2
du
du
=> d:(x) = -
iar
=> - dx (x) + dx (a) d
(X
= ia
f(s) ds
f(s)ds.
The integrate The last last step follows from from the the Neumann Neumann condition condition at at x = a. a. We We now now integrate again:
d
d: (x)
=>
l
X
=-
bd
iar
d:(z)dz
f(s) ds
=-
Ibx iar f(s)dsdz
-lb 1 i1 z
=> u(b) - u(x) = a
=> u(x) =
f(s) ds dz
Z
f(s)dsdz.
This shows shows that solution for This that L/mU Linu = = f has has aa unique unique solution for each each /f € E C[a, b].
11. By the the fundamental theorem of calculus,
u(x)
= IX f(s)ds
Du == ff.. However, However, this solution is not unique; unique; for C, satisfies Du for any constant constant (7, u(x) is is another another solution. solution.
= IX f(s) ds + C
Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises Appendix exercises
522
Section 3.3 1. 1.
(a) Both Both equal (14,4, -5). (a) equal (14,4, -5).
(b) Both equal (b) Both equal
2:7=1 (Vj)iXj.
3. The The given given set not aa basis. basis. If A is the matrix matrix whose whose columns columns are the given given vectors, vectors, 3. set is is not If A is the are the then N(A) is not not trivial. then A/"(A) is trivial. 5. We know that P% P2 has has dimension dimension 3, 3, and therefore it suffices to to show show either either that that the 5. We know that and therefore it suffices the given set set of three vectors vectors spans P2 or or that that it linearly independent. E Pz, P2, then given of three spans Ti it is is linearly independent. If If pp e then we write write we Cl = p(Xl), C3 = P(X3), C3 = p(X3) and define define the the polynomial polynomial and
Then Then
and, similarly, q(X2) = pp(X2), q(X3) = p(%s)p(X3)' But But then and, similarly, q(x-z) ( x z ) , q(xz) then p and and q are are secondseconddegree polynomials polynomials that that agree distinct points, points, and three points points determine degree agree at at three three distinct and three determine Therefore, aa quadratic quadratic (just (just as as two two points points determine determine aa line). line). Therefore,
and we have that every every pp € E Pi P2 can be written combination of of and we have shown shown that can be written as as aa linear linear combination L 1 , 1/2,£3. L 2 , L 3 • This the proof. proof. LI, This completes completes the
Section 3.4 1. 1.
(a) The verification is is aa direct direct calculation of Vi Vi •• Vj for the combinations of i, j.j. (a) The verification calculation of Vj for the 66 combinations of i, For example, example, For 1
1
1
1
1
1
Vl'Vl=--+--+--
V3 V3
V3 V3 3" + 3" + 3" = 1,
V3 V3
111
=
Vl'V2 =
~ ~ ~O ~ +
1
=
+
( -
~)
1
v'6 - v'6
= 0,
and so so forth. forth, and
(b) x
= (VI' X)Vl + =
~
(V2 . X)V2 + (V3 . X)V3
VI + OV2 +
( -
~)
V3.
Appendix to odd-numbered odd-numbered exercises exercises Appendix C. C. Solutions Solutions to
523
3. We have have
Ilx+yl12
= (x+y,x+y)
= (x, x + y) + (y, x + y) = (x,x) + (x,y) + (y,x) + (y,y) = (x,x) + 2(x,y) + (y,y) = IIxll 2 + 2(x,y) + Ilyll2. Therefore, Therefore,
if and and only if (x,y) (x, y) = = 0. only if O. if 5. First all, if if 5. First of of all,
(y, z) = 0 for all z E W, then since since Wi each i, i, we we have, have, in in particular, then vn E e W for for each particular,
(y, Wi)
= 0, i = 1,2, ... , n.
Suppose, the other that Suppose, on on the other hand, hand, that
(y, Wi)
= 0, i = 1,2, ... ,n.
If zz is any vector vector in in W, then, since since {Wi, for W, If is any W, then, {wi, W2, W 2 ,... . . . ,, w wnn }} is is aa basis basis for W', there there exist exist a2, ... such that that scalars scalars ai, 0:2, • • • ,,an otn such n Z
= Laiwi. i=l
We then then have We have
n
= Lai (y,Wi) i=l
i=l
=0.
This proof. This completes completes the the proof. 7. best approximation approximation to to the the given given data - 0.0018984. 7. The The best data is is y = — 2.0111x 2.0111x — 0.0018984. See See Figure Figure C.2. 9. The onto 7*2 P2 is 9. The projection projection of of g9 onto is
2 -Ql(X) 7r or or See C.3. See Figure Figure C.3.
+ OQ2(X) +
2V5(7r 2 - 12) 3 Q3(X) 7r
524 524
Appendix C. exercises Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises
Figure data from from Exercise Exercise 3.4.7 Figure C.2. C.2. The The data 3-4-7 and and the the best best linear linear approximation. approximation.
Figure function g(x] g(x) = Figure C.3. C.3. The The function = sin sin (1fx) (TTX) and and its its best best quadratic quadratic approxiapproximation over mation over the the interval interval [0,1].
Appendix C. Solutions Solutions to to odd-numbered Appendix C. odd-numbered exercises exercises
525
Section 3.5 1. The The eigenvalues of A 200 and and >'2 A2 = 100, and and the corresponding (normalized) 1. eigenvalues of A are are AI >'1 = = 200 = 100, the corresponding (normalized) eigenvectors are are eigenvectors U1 =
[
-0.8 ] 0.6'
U2 =
[ 0.6 ] 0.8 '
respectively. is respectively. The The solution solution is b . U1
b . U2
X=~U1+~U2=
[
1 ] 1
.
3. The and eigenvectors eigenvectors of 1, >'3 AS == —1 and 3. The eigenvalues eigenvalues and of A A are are AI >'1 = = 2, 2, A2 >'2 = = 1, -1 and
The of the The solution solution of the Ax Ax = b b is is
[
-3/2] 1/2 .
5/2
5. of u^ operations, so computing all ra5. The The computation computation of Ui .• b/Aj b/>.; costs costs 2n 2n operations, so computing all nn of of these these ra2n22 operations. tios costs costs 2n operations. Computing Computing the linear linear combination is is then equivalent equivalent to to another so the the total total cost cost is another n dot dot products, products, so is 2n 2 7. 7.
+ n * (2n -1) = 4n 2 -
n
= O(4n 2 ).
(a) For For k = -1, have (a) = 22,3, , 3 ,... . . . ,,n n— 1, we we have (Ls(j)) k
;2 (+ = ;2 (= ;2 = ;2 (2-2cos(j7rh))s~). =
sin ((k - l)j7rh)
2 sin (kj7rh) - sin ((k
sin (kj7rh) cos (j7rh)
+ l)j7rh))
+ cos (kj7rh) sin (j7rh) + 2 sin (kj7rh)
- sin (kj7rh) cos (j7rh) - cos (kj7rh) sin (j7rh)) (2 - 2 cos (j7rh)) sin (kj7rh)
Thus Thus (Ls(j)) k
=
;2
(2 - 2 cos (j7rh)) s~), k
= 2,3, ... , n -
l.
Similar Similar calculations show show that this formula holds holds for for kk = = 1I and and k = = nn also. also. Therefore, s(j) s^' is an eigenvector of L with eigenvalue \. _ 2 - 2 cos (j7rh) h2 •
AJ -
(b) The eigenvalues \j >.j are all positive and are increasing with the frequency j. j.
Appendix to odd-numbered Appendix C. C. Solutions Solutions to odd-numbered exercises exercises
526
(c) as (c) The The right-hand right-hand side side bb can can be be expressed expressed as
b = (8{1) . b) 8(1) + (8(2). b) 8(2) + ... + (8(n). b) 8(n), while the the solution solution xx of of Lx Lx = =b while b is is X
=
8(1) . x (1) ---8
Al
8(n) . X (n) + -8(2)- -. x8 (2) + ... + --8
A2
An
Since with j, j, this frequency Since Aj Xj increases increases with this shows shows that, that, in in producing producing x, x, the the higher higher frequency b are components of of b are dampened dampened more more than than are are the the lower lower frequency frequency components components components of b. Thus x is smoother smoother than than b. b. of b. Thus x is
4.1 Section 4.1 1. Define Define 1. Xl
= U,
du
X2
= dt'
Then Then dXl
dt
=X2,
dX2
dt
=
1
-2" X l.
3. Define Define
Then Then dXl
dt
=X2,
dX2 --a.:;:=X3, dX3 --a.:;:=X4, dX4
Tt
5. 5. Define Define
= 2X3 dp
Xl
. + sm t.
Xl = p, X2 = dt' X3 = q, X4
Then the the first-order first-order system system is Then is dXl
Tt =X2, dX2
mITt
= /!(Xl,X3,X2,X4),
dX3
dt dX4 m2Tt
dq
= dt'
=X4,
= h(Xl,X3,X2,X4).
Appendix C. C. Solutions Appendix Solutions to to odd-numbered odd-numbered exercises exercises
527
Section 4.2 1. 1.
(a) We first zero function so Sis nonempty. If If (a) We first note note that that the the zero function is is aa solution solution of of (4.4), (4.4), so S is nonempty. u, € S and and a, a, /3fJ E G R, satisfies u, v E R, then then w = au + /3v fJv satisfies
d2w a dt 2
d2
dw
d
+ bdt + cw = a dt 2 [au + fJv] + b dt
[au + fJv]
+ c(au + fJv)
=a ( a ~:: + fJ ~:~) + b ( a ~~ + fJ ~:) + c( au + fJv)
~:: + b~~ + cu) + fJ (a ~:~ + b ~: + cv)
=a
(a
=a
. 0 + fJ . 0
= O.
Thus w is is also (4.4), which shows that that S S is Thus w also aa solution solution of of (4.4), which shows is aa subspace. subspace. text, no matter what what the the values of a, b, c (a (b) explained in in the (b) As As explained the text, no matter values of a, 6, (a ::j:. ^ 0), 0), the the solution by two two functions. solution space space is is spanned spanned by functions. Thus Thus S S is is finite-dimensional finite-dimensional and and it has dimension at most 2. Moreover, it is to see each of the sets it has dimension at most 2. Moreover, it is easy easy to see that that each of the sets {eTlt,eT2t} } isislinearly {e rit ,e r2 *} (rl::j:. (n / rr2), {e"* cos cos (At),el£tsin(At)}, (A*), eM* sin (At)}, and and {eTt,te {er\tertrt} linearly inin2 ), {el£t two functions is linearly and only dependent, since aa set dependent, since set of of two functions is linearly dependent dependent if if and only if if one one of multiple of the other. Thus, in has aa basis of the the functions functions is is aa multiple of the other. Thus, in every every case, case, S S has basis with two two functions, functions, and with and so so S is is two-dimensional. two-dimensional. that the the characteristic polynomial of has aa single, repeated root root rr = 3. 3. Suppose Suppose that characteristic polynomial of (4.4) (4.4) has single, repeated = —6/(2a) (so b62 -— 4ac = 0). Then, shown in in the -b/(2a) (so = 0). Then, as as shown the text, text,
u(t)
= Clert + c2te Tt
is aa solution solution of of (4.4) for each each choice choice of of Cl, ci, C2. 02- We We wish show that, given fci, ki € R, is (4.4) for wish to to show that, given kl, k2 E R, there is is aa unique choice of of CI, ci, ci such that -u(O) = = ki, du/dt(0) = k^. We have have there unique choice C2 such that u(O) kl, du/dt(O) k2. We
du dt (t ) = rCle Tt
+ C2e Tt + rC2te Tt .
Therefore, the equations •u(O) simplify to to Therefore, the equations u(O) == ki, kl' du/dt(0) du/dt(O) — = k<2 k2 simplify
~]
[;
C
= k.
Since coefficient matrix matrix is is obviously nonsingular (regardless the value of r), Since the the coefficient obviously nonsingular (regardless of of the value of r), there is aa unique unique solution CI, C2 for k2• there is solution Ci,C2 for each each kl, ki,kz. 5. 5.
(a) The only solution is is u(x) O. (a) The only solution u(x) = 0.
(b) The The only only solution solution is is u( u(x) (b) x) = = 0.O. (c) The The function function u( u(x) sin (-KX) is aa nonzero solution. It It is any (c) x) = sin (7rx) is nonzero solution. is not not unique, unique, since since any multiple of it another solution. solution. multiple of it is is another 7. By By the the product product rule rule and the fundamental theorem of calculus, 7. and the fundamental theorem of calculus,
:t
[1:
1: = it = it
ea(t-S)f(S)dS] = :t [eat ae at
e-asf(S)dS]
e -as f(s) ds
+ eate- at f(t)
to
ea(t-s) f(s) ds + J(t).
a
to
Appendix C. Solutions exercises Appendix C. Solutions to to odd-numbered odd-numbered exercises
528 Therefore, with
u(t)
= uoeu(t-t o) +
[t
eu(t-.) f(s) ds,
to
we have
du (t) dt
= auoeu(t-to) + a [t eu(t-.) f(s) ds + f(t) to
= f(t). = au(t) au(t) + /(*)• Thus uu satisfies the differential differential equation. We also have
u(to)
= uoeu(to-t o) + [
to
eu(t-s) f(s) ds
to
=Uo +0 =uo, so the the initial condition holds as well.
= ~ sin 2 (t) u(t) = ~ (e- t + cos (t) + sin (t))
9. u(t) 11.
Section 4.3 1. The solution is 1. The solution is X
(t )
= -1U l v'3
1 -t - e U2
V2
+ -1e -2t ll.3
v'6
3. The general solution is
5. the vector vector (1, —1). 5. XQ Xo must must be be aa multiple multiple of of the (1, -1). 7. 7.
x(t) 9.
=
t 2t 1 [ 55 - 3e- - 45e- + 20t - 7 cos (t) - 11 sin (t) 2t 60 10 + 6e- + 20t - 16 cos (t) - 8 sin (t) -5 - 3e- 2t + 45e- t + 20t - 37 cos (t) + 19 sin (t)
1
(a) The solution is
3 -t 1 M 3 -t 1 M x(t) ="2 e +"2 e , y(t) ="2 e -"2e. the first species (x (x(t)) The population of the (t)) increases exponentially, while the the second species (y( t)) goes to zero in finite time (y( t) = 0 at population of the (y(t)) (y(t) t = In ln(3)/4). while the first first species species (3)/4). Thus the second species becomes extinct, while increases without bound.
C. Solutions to odd-numbered exercises Appendix C. to odd-numbered exercises
529
(b) If If the the initial initial populations populations are x(0) = r, y(0) = s, s, then the solution solution to to the the IVP IVP (b) are x(O) r, yeO) then the is is x(t) =
~ (r + s)e- t +
(r - s)e 3t ) , yet) =
~
((r + s)e- t + (s - r)e 3t ) .
Therefore, if rr = s, both both populations Therefore, if = s, populations decay decay to to zero zero exponentially; exponentially; that that is, is, both both species species die die out. out.
Section 4.4 4.4 Section 1. 1.
(a) Four Four steps steps of yield an an estimate of 0.71969. (a) of Euler's Euler's method method yield estimate of 0.71969. (b) Two Two steps steps of of the the improved improved Euler Euler method yield an estimate of 0.80687. (b) method yield an estimate of 0.80687. (c) the RK4 RK4 method 0.82380. (c) One One step step of of the method yields yields an an estimate estimate of of 0.82380. Euler's method gives no correct digits, digits, the the improved Euler method method gives gives one one correct correct Euler's method gives no correct improved Euler gives three three correct digits. Each methods evaluated evaluated ff(t, digit, digit, and and RK4 RK4 gives correct digits. Each of of the the methods ( t , u) u) four four times. times.
3.
(a) Let
Ul
= X, U2 = dx/dt, U3 = y, U4 = dy/dt. dUl
dt
Then the system is
=U2,
duz _ Ul + 2U4 _ 1-'2(Ul + I-'d _ 1-'1(Ul - 1-'2) dt rl (Ul, U3)3 r2( Ul, U3)3 ' dU3 dt =U4, dU4 = - 2U2 dt
-
+ U3 -
I-'zU3 ----:'-------'----::-::rl(Ul,U3)3
(b) routine ode45 421 steps (b) The The routine ode45 from from MATLAB MATLAB (version (version 5.3) 5.3) required required 421 steps to to produce produce aa graph graph with with the the ending ending point point apparently apparently coinciding coinciding with with the the initial initial value. value. The of y(t) yet) versus x(t) is in Figure Figure C.4. The graph graph of versus x(t) is given given in C.4. The The graphs graphs of of x(t) x(t) and the times show, and yet), y ( t ) , together together with with the times steps, steps, are are given given in in Figure Figure C.5. C.5. They They show, not surprisingly, that that the time step step is is forced forced to to be small precisely when the the not surprisingly, the time be small precisely when coordinates are rapidly. coordinates are changing changing rapidly. minimum step taken by was 3.28.10(c) (c) The The minimum step size size taken by ode45 ode45 was 3.28 • 10~44,, and and using using this this step step entire interval interval of require almost length over over the length the entire of [0, [0,T] would would require almost 19000 19000 steps. steps. This This is to to be to the the 421 the adaptive is be compared compared to 421 steps steps taken taken by by the adaptive algorithm. algorithm. (Note: (Note: The The exact for minimum step size, size, etc., etc., will will differ differ according according to the algorithm algorithm exact results results for minimum step to the and tolerances used.) and tolerances used.)
5. 5.
(a) We first note that a+(tl-tO) a+(ti— to) < b+(ti — to) if if and and only a < t-(lI t—(t\ — to) S
(just replace replace tt — (ti -— to) to) by in the last step). step). Gust - (tl by tt in the last
530 530
Appendix C. Solutions to odd-numbered exercises exercises
0.8 0.6 0.4 0.2 >.
0 -0.2 -0.4 -0.6 -0.8 =t5
-1
-0.5
0
0.5
x
1.5
Figure satellite in Figure C.4. C.4. The The orbit orbit of of the the satellite in Exercise Exercise 4.4.3. 4-4-3-
~JS;;:~ 1 ~_:~l f:~~:;; 1 o
1
2
3
4
5
6
7
o
1
2
3
4
5
6
7
o
1
2
3 4 Step index
5
6
7
Figure Figure C.5. C.5. The The coordinates coordinates of of the the satellite satellite in in Exercise Exercise 4.4.3 4-4-3 (top (top two two graphs) with graphs) with the the step step lengths lengths taken taken (bottom (bottom graph). graph).
531
Appendix C. C. Solutions Solutions to odd-numbered exercises We have We have dv -(t)
dt
= -dtd [u(t -
(fI - to))]
du = -(t - (h dt =f(u(t - (t1 = f(v(t)),
to)) to)))
so the ODE. so vv satisfies satisfies the ODE. Finally, Finally, V(t1)
= u(h -
(t1 - to))
= u(to) = Uo.
Therefore v is is aa solution the given given IVP. Therefore v solution to to the IVP. (b) (b) Consider the the IVP IVP
du
dt u(O)
=t
'
= 1,
which has /2 + 1. IVP which has solution solution u(t) u(t) = e t2/2 l. The The IVP
dv =t
dt
'
v(1)
=1
which is is not equal to has solution has solution v(t) = tt 22 /2 + + 1/2, 1/2, which not equal to 2 1 u(t - (1 - 0)) = u(t - 1) = -(t -1) + 1.
2
7. is 7. The The exact exact solution solution is x(t)
={
2-
2~-:'
2e,
0
< In 2,
t > In 2.
(a) = 0.5: (a) The The following following errors errors were were obtained obtained at at t =
At 6.t 1/4 1/4 1/8 1/8 1/16 1/16 1/32 1/32 1/64 1/64 1/128 1/128
Error 2.4332e-05 1.3697e-06 8.1251e-08 4.9475e-09 3.0522e-10 1.8952e-ll
By that as as 6.t in half, half, the error is reduced by by aa factor factor By inspection, inspection, we we see see that At is is cut cut in the error is reduced of of approximately approximately 16, as expected expected for 0(6.t4) O(At 4 ) convergence. errors were 2.0: (b) The (b) The following following errors were obtained obtained at at tt = 2.0:
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
532
I:l.t At
1/4 1/4 1/8 1/8 1/16 1/16 1/32 1/32 1/64 1/64 1/128 1/128
Error 5.0774e-03 3.2345e-03 3.1896e-04 3.1896e-04 1.0063e-04 1.0063e-04 8.9026e-06 3.4808e-06
The The error error definitely definitely does does not not exhibit exhibit O(l:l.t4) O(At 4 ) convergence convergence to to zero. zero. The The reason reason is is the the lack lack of of smoothness smoothness of of the the function function 11 + + Ix \x -11; — 1|; the rate rate of of convergence given in in the the text text only only applies applies to to ODEs ODEs defined defined by by smooth smooth functions. functions. When When given integrating integrating from from t = = 00 ttoo t = = 0.5, 0.5, the the nonsmoothness nonsmoothness of of the the right-hand right-hand side side is why we observed not not encountered encountered since since x(t) x(t) < 11 on on this this interval. interval. This This explains explains why we observed problem. good good convergence convergence in in the the first first part part of of this this problem.
Section 4.5 1. 1.
(a) is (a) The The exact exact solution solution is
(b) norm of 0.089103. (b) The The norm of the the error error is is approximately approximately 0.089103. (c) norm of the error 0.10658. (c) The The norm of the error is is approximately approximately 0.10658. 3. 0.04. 3. The The largest largest value value is is I:l.t At == 0.04. 5. 5.
Ca) Applying the the methods we find (a) Applying methods of of Section Section 4.3, 4.3, we find the the solution solution xCt)
t
lOOt
3e- t
+ e- lOOt
1 [ 3e- _ e-
= 2"
] .
(b) By By trial trial and and error, error, we find that that I:l.t At $< 0.02 0.02 is is necessary necessary for for stability. stability. Specifically, (b) we find Specifically, integrating from from tt = = 00 ttoo tt = = 1, l,n = 49 49 (i.e. (i.e. I:l.t At = = 1/49) 1/49) yields yields n= integrating
yields while while n ::::: > 50 50 yields l) (c) With With Xj+i = Xi + I:l.tAXi AtAxi and and y[ ui .• Xi, Xj, we we have Xi+l = Xi + yii) = Ul have (c)
=
Ul . Xi+l Ul . Xi + I:l.tUl . AXi 1 yii+ ) = yii) + I:l.t(Aut) . Xi 1 yii+ ) yii) + I:l.tAIUl . Xi 1 yii+ ) = yii) + I:l.tAlyii)
=> => = => => yii+ 1) = (1 + I:l.tAt)yii). A A similar similar calculation calculation shows shows that that
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
533
For For stability, stability, we we need need 11+~t'\11::;
1, 11+~t'\21::; 1.
Since the the eigenvalues eigenvalues of of A A are are -1, —1, -100, —100, it it is is not hard to to see see that that the the upper upper Since not hard bound for for ~t At is ~t At ::; < 0.02, 0.02, just as was was determined determined by experiment. (d) For the backward Euler method, a similar calculation calculation shows that y~i+l) y~i+l)
= (1- ~t,\l)-ly~i), = (1 _ ~t,\2)-ly~i).
Stability is is guaranteed for any any ~t, At, since since Stability guaranteed for
for every every ~t At > 0. for O.
Section Section 4.6 4.6 1. G(t; s) = e- 2 (t-s).
3. The IVP is
dm - + km
dt
= 0.1,
k
m(O)
= 10.
s
The Green's function is G(t; s) = and the the solution solutionisis — e-k(t-s), e ^ \ and 0.1) -kt ( ) -_ (10 -T mt e
0.1 +T·
The The mass converges, as tt -t —>• 00, oo, to
0~1 ~ 0.289 g. 5. The solution is u(t)
.2(t) = 3"2 (1 + 2 cos (t)) sm "2 = 3"1 (cos (t) -
cos (2t» .
Section Section 5.1 5.1 1. 1.
(a) the null If (a) Since Since Mn MD is is linear, linear, it it suffices suffices to to show show that that the null space space of of Mn MD is is trivial. trivial. If Mnu then, du/dx MDU = = 0, 0, then, du/dx = — 0, 0, that that is, is, u is is aa constant constant function: function: u(x) u(x) = — c. c. But But then the condition u(O) u(0) = 00 implies implies that that cc = 0, 0, and and so so u is is the the zero zero function. function. then the condition Thus N(Mn) M(MD] = {O}. {0}. Thus
(b) If If uu E € c1[o, CofO, f] f] and and Mnu MDU = — f, /, then then we we have have (b)
du dx(x)=f(x), O<x
u(x)
=
1"
f(s)ds.
Solutions to to odd-numbered odd-numbered exercises Appendix C. Appendix C. Solutions exercises
534 534
But have u(i) — 0, But we we also also must must have u(f) = 0, which which implies implies that that
it
I(s) ds
= o.
=
If 1 / €E C[O, C[Q, f] I] does does not satisfy this condition, then is impossible for MDU f If not satisfy this condition, then it it is impossible for MDU = 1 to have solution. to have aa solution.
3. 3.
(a) If v,w are both solutions of of Lu linearity) (a) If v, w G E S S are both solutions Lu = /, I, then then Lv Lv = Lw, Lw, or or (by (by linearity) L(v Therefore vv — € A/"(Z>). But, since since S is is aa subspace, subspace, vv — L(v -— w) w) = = 0. O. Therefore - w wE N(L). But, - w w is is also in If the only function and SS is is the then also in S. S. If the only function in in both both Af(L) N (L) and the zero zero function, function, then v -— w be the function. w must must be the zero zero function, function, that that is, is, v and and w w must must be be the the same same function. Therefore, if if J\f(L) {0}, then at most one solution N(L) n n S = {O}, then Lu Lu = = 1f can can have have at most one solution for for Therefore, any/. f. any (b) We have that Lu f has has aa solution solution for for any any 1 / € C[0, f] f\ (see the (b) We have already already seen seen that Lu = 1 E C[O, (see the discussion immediately immediately preceding We will from discussion preceding Example Example 5.2). 5.2). We will use use the the result result from the the first part of of this this exercise exercise to show show that the the solution solution is unique. unique. The L is the the space The null space of of L space of of all first first degree degree polynomials: N(L)={u:[O,f]--tR: u(x)=ax+bforsomea,bER}.
i. Suppose Suppose u — ax for some €R and i. u€ E Af(L) N(L) n S. S. Then Then u(x) u(x) = ax + b b for some a, a, b bE R and du dx
(~) = 0 =} a = o. 2
Then u(x) = b, 6, and and Then u(x) =
it
u(x) dx
= 0 =} bf = 0 =} b = o.
Therefore uu is the zero zero function, and so so A/"(L) H SS = {O}. {0}. The uniqueness Therefore is the function, and N(L) n The uniqueness property then then follows follows from from the the first first part part of of this property this exercise. exercise, ii. Suppose uu E € A/"(L) S. Then u(x) = ax ax + b for some some a, o, b eR ii. Suppose N(L) n S. Then u(x) b for bE R and and u(O)
= 0 =} b = O.
Then u(x) — ax, ax, and and Then u(x) = du dx (f) = 0
=}
a = O.
Therefore is the the zero zero function, N(L) n S = The uniqueness Therefore uu is function, and and so so N(L} C\S = {O}. {0}. The uniqueness then follows follows from from the the first first part part of of this property then property this exercise. exercise. 5. The condition 5. The condition
(C.2) implies that that du/dx du/dx is that u u is is constant. constant. The boundary conditions implies is zero, zero, and and hence hence that The boundary conditions on u u would would then then imply that uu is the zero function. But, But, by by assumption, nonzero on imply that is the zero function. assumption, uu is is nonzero that (u,u) cannot hold. hold. (we assumed that (we assumed (u, u) = 1). 1). Therefore, Therefore, (C.2) cannot
7. 7.
(a) If LmU — 0, the form u(x) = ax + b. condition, (a) If Lihu = 0, then then u u has has the form u(x) = ax b. The The first first boundary boundary condition, du/dx(Q) = 0, implies that — 0, and then second boundary condition, du/dx(O) 0, implies that aa = 0, and then the the second boundary condition, u(£) = 0, O. Therefore, Therefore, uu is the zero function and N(Lih) is u(i] = 0, yields b6 = = 0. and M(Lm] is trivial. trivial.
Appendix C. Solutions Solutions to odd-numbered exercises exercises Appendix C. to odd-numbered
535 535
(b) For For any any f/ E <E C[O, C[Q,£\, the function function (b) .e], the
=
1£ l'
d2 u - dx 2 (x)
= f(x),
u(x)
f(s)dsdz
belongs to c~ C%, [0, i] and satisfies belongs to [O,.e] and satisfies 0
< x < .e.
This that L Lmu any /f € E C7[0,^], C[O, .e], and This shows shows that has aa solution solution for for any and therefore therefore mu = f has R(Lm) K(Lto) = = C[O,.e]. C[0,f\. (c) (c) Suppose Suppose u,v E € C~[O,.e]. Cft[Q,£\. Then Then
(Lmu,v)
I
=-
t d2
dx~(x)v(x)dx
0
]1 du =- [dx(X)V(X) 0 = =
+
Ii 0
du dv dx(X)dx(x)dx
i du dv . 0 dx (x) dx (x) dx (SInce v(.e)
I
-lot U(X)::~(X)dX
[U(X):~(X)]:
I = =-
0
t
= 0, du/dx(O) = 0)
d2
u(x) dx~ (x) dx (since u(.e)
= 0, dv/dx(O) = 0)
(u, Lmv).
Thus Thus Lm is is symmetric. symmetric.
(d) Suppose Suppose A is an an eigenvalue eigenvalue of of Lm with with corresponding eigenfunction u, u, and and corresponding eigenfunction (d) A is assume that has been been normalized normalized so so that that (u, (u,u) — 1. I. Then Then assume that u has u) = A = A(U,u)
= (AU,U) = (Lmu,u) i
d2
=-
I dx~(x)u(x)dx
=-
[::(x)u(x)]:
=
0
1c: i
0
d
+
li
(::(x)f dx
2
(x)) dx (since u(.e)
= 0, du/dx(O) = 0)
> O.
The C~[O,.e] has aa nonzero The last last step step follows follows because because every every nonzero nonzero function function in in CjjO, •£] has nonzero derivative. derivative.
9. u(x) = = x(lx(l — x), v(x) = x22(l — x). Then Then aa direct calculation shows shows that 9. Define Define u(x) v(x) = (1direct calculation that
(Mu,v) but but
7
= 30' 4
(u,Mv) = 15'
Appendix Solutions to Appendix C. C. Solutions to odd-numbered odd-numbered exercises exercises
536 536
11. 11.
(a) Then (a) Suppose Suppose u,v u, v €E C|[0,^]. C~[O, fl. Then
Applying conditions on obtain Applying the the boundary boundary conditions on u, u, v, v, we we obtain du -K dx (f)v(f)
du
dv
+ Kdx (O)v(O) + U(f)K dx (f) -
= au(f)v(f) + au(O)v(O) -
dv U(O)K dx (0)
au(f)v(f) - au(O)v(O)
= 0,
so (LRU,V) = (U,LRV), as desired. first degree polynomial: polynomial: u(x) = ax ++ b.b. The (b) If If LRU LRU = = 0, 0, then u must be a first u(x) = boundary conditions imply that a and b must satisfy satisfy the following equations: -Ka +ab
(af
= 0,
+ K)a + ab = O.
The computed as follows: The determinant determinant is is computed as follows: n- K
I a<- +
a K
a
I = -a(2K + af.) < O.
= 0, is, u = so N(LR) N(LR) is trivial. The only solution is a = b = 0, that is, = 0, so
Section 5.2 5.2 Section 1. If n =I- m, then
= !li{ ((n-m)1rx) 2 cos f
_cos ((n+m)1rx)} dx f =! [ f sin ((n-m)1rx) _ f sin ((n+m}7rx)]i. 2 (n - m)1r f (n + m)1r f o
0
This last expression simplifies simplifies to four terms, each including the sine function evaluated at an integer multiple of 1r, zero. Thus (Un, TT, and hence each equal to zero. (un,uUrn) m) = 0 for n 7^ TO. =I- m. for
Appendix C. Solutions to to odd-numbered odd-numbered exercises exercises Appendix C. Solutions
537
3. The eigenpairs are
2n -l)1['X)
cos (
2f
' n = 1,2,3, ....
5. The - rr + (X (A — - 5), so the the characteristic characteristic roots are The characteristic characteristic polynomial is rr22 — rl =
1 - -/21 - 4A 1 + -/21 - 4A 2 ' r2 = 2
(C.3)
Case 1: 1: A 1/2,1/2 and the A= = 21/4. 21/4. In In this this case, case, the the characteristic characteristic roots roots are are rr = = 1/2,1/2 and the Case general of the the ODE is general solution solution of ODE is
u(x)
= cle
X /
2 + c2xe X / 2.
The boundary condition u(O) u(Q) = — 00 yields = 0, 0, and and then condition The boundary condition yields c\ CI = then the the boundary boundary condition u(l) C2 = = O. u(l) = = 1 implies that c
The boundary conditions lead to the system system
clne
r
!
CI + C2 = + c2r2e r2 =
0,
O.
This system has the unique solution C1 = ci C2 = = 0, 0, so there is no nonzero solution for for c\ = < 21/4. Hence no no A A< 21/4 is eigenvalue. AA < 21/4. Hence < 21/4 is an an eigenvalue. A> Case 3: 3: A > 21/4. In this case, the roots are complex conjugate: r1
= "21 -
8'
t, r2
= "21 + 8'z,
8= -/4A-21
2'
The general solution of the ODE is
u(x) = (Clcos(8x) +C2 sin (8x)) eX / 2. The boundary condition u(0) 0, so so any any eigenfunction eigenfunction is is of form The boundary condition u(O) = = 0 0 yields yields c\ C1 = = 0, of the the form
u(x) = sin (8x )e x / 2. The Neumann Neumann condition condition at is equivalent equivalent to to the the equation The at x = = 11 is equation tan (8) = -28, 8 > O.
Although this equation equation cannot be solved explicitly, a simple graph shows that there are infinitely many solutions 0 < 81 < 8622 < < ... are < 6\ • • •,, with 2k -12k + 1 ) 8k E ( -2-1[', -2-1[' , k = 1,2, ....
Define by Define A& Ak by 8 _ -/4A k - 21 k 2 '
538 538
Appendix C. Solutions to to odd-numbered odd-numbered exercises Appendix C. Solutions exercises
that is, is, that 2
21
= 8k + 4'
Ak
k
= 1,2, ....
Ak, A; k= and the the corresponding eigenfunctions eigenfunctions are Then A*,, = 1,2, 1, 2 , ... . . . , are eigenvalues, and
Using find that we find that Using Newton's Newton's method, method, we 81 == 1.8366, 82 == 4.8158,
which yields yields which Al
== 8.6231,
A2
== 28.4423.
The eigenfunctions V2, as given by by the the above above formula. The eigenfunctions are are VI, vi,vz, as given formula. shows that that A direct direct calculation calculation shows
so vi and and V2 v? are are not orthogonal. so VI not orthogonal. 7.
(a)
,",00
(b) ( c)
2( _l)n+l
.
(
)
.
(
)
mr
sm n7rX
,",00
L....n=1
4sin(Vf) n2".2
SIn n7rX
,",00
12(_I)n+l n3".3
L....n=1
L....n=l
.
(
sIn n7rX
)
720( _I)n+l . ( ) (d) ,",00 L....n=l n5".o SIn n7rX the original functions using 10 the Fourier sine The errors in approximating the 10 terms of the graphed in Figure C.6. series are graphed
9. sin (37rx) (That is, all of the the Fourier sine coefficients coefficients are zero, except the the third, which which (STTX) (That is one.) 11. The series have the the form
~ ((2n -1)7rX) L...Ja n cos 2 ' n=l
where where
(b) an --
(d) an
32cos «2n-I)"./4) sin 2 «2n-I)"./8). ".2(2n_l)2 ,
=-
8 5760+2880( _l)n (2n-I)".+240(2n-l)2".2+ 7 (2n_l)4".4 (2n-l)
".
The errors in approximating approximating the the original functions using 10 10 terms of the Fourier Fourier quarter-wave cosine series are graphed in Figure C.7.
539 539
Appendix C. C. Solutions odd-numbered exercises exercises Appendix Solutions to to odd-numbered
0 . 0 3 , . . . . - - - - - - - - -.....
1~------------------~
0.02
0.5
0.01
-0.5 0
0.5
-0.01
1
0
0.5
x
x
-3
1.5 x 10
5
X1O
-5
1
-1~----------------~
o
0.5
-5~----------------~ 0.5 1 o x
1
x
Figure in approximating approximating four 10 terms terms of their Figure C.6. C.6. Errors Errors in four functions /unctions by 10 of their g(x) = Fourier sine sine series (see (see Exercise 5.2.7). 5.2.7). Top Top left: left: g(x) = x; Top Top right: right: h(x) h(x) = = 3 5 •3 | —x | ; Bottom left:= m(x] — x3; Bottom right: k(x)- =7x —3x 10x ;x ~ ~ — Bottom left: m(x) x - x= Bottom right: k{x) = 7x lOx 3 +
- Ix - I;
Section Section 5.3 5.3 ~oo
2(1_~_1)n) . ( ) n .".3 Sln n7rX
1.
(a)
3
2(1-C-l)n) . ( ) (b) x + ~oo L..-n=l n3.".3 sm n7rX (a) ~oo 32(_1)n+l sin (2n-l).".X)
.
L..-n=l
L..-n=l (2n_l)4.".4
(b) 1
(c)
+
L
16(-1)n.".+2(-1)n+l n .".-2) L..-n=l (2n-l)411'4
OO
2(1-(-1)n)
n=l nn-(2+n27r2)
5. We have
2
~oo
(2n-l)"X) 2
sin (n7rx)
d2 u
dX2 (x)
so so
cos
= -x,
0
< x < 1,
Appendix C. exercises Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises
540
0.04,....----------"
0.5
-0.02 -0.5'----------~
o
-0.04 0
1
0.5
0.5
x
0.01 , . . . . - - - - - - - - - - "
0.05
o
0
-0.01
-0.05
-0.02
-0.1
-0.03'----------~
o
1
x
0.5
-0.15 0
1
0.5
x
x
Figure Errors in four functions junctions by Figure C.7. C.7. Errors in approximating approximating four by 10 10 terms terms of of their their X; Top Top Fourier Exercise 5.2.11). Fourier quarter-wave quarter-wave cosine cosine series series (see (see Exercise 5.2.11). Top Top left: left: g(x) g(x) = = x; right: h(x) h(x) == !\- —Ixx- —\ ;Bottom Bottomleft: left: m(x) m(x) ==x x- —x x3 3; ;Bottom Bottomright: right: k(x) k(x] == right: 5 7x - lOx33 + 3xt. 7x-Wx +3x .
! I;
du = -(0) -
1'"
= ~:(O) _
x22.
dx
sdx
0
Write C C for for du/dx(O); du/dx(Q); then then another another integration integration yields Write yields u(x)=u(O)+
=0 +
l
0
"'d d;(s)ds
1'" {c -
=Cx-
s; } ds
x3
6·
Finally, we we choose choose C C so so that that u(l) u(l) = = 0, 0, which which yields C= = 1/6. 1/6. This This yields yields Finally, yields C u(x) =
~ (x-x 3 ).
541 541
Appendix C. Solutions Solutions to to odd-numbered odd-numberedexercises exercises Appendix C. 7. The The Fourier Fourier quarter-wave quarter-wave sine sine coefficients coefficients of of -Td —Td2u/dx u/dx22 are are 7. bn
= _ 2T £
1i
2 d u( ) . ((2n -l)'JrX) d 0 dx2 x sm 2£ x,
while while those those of of u are are
an
2
=£
1 0i
. u(x)sm
(2n - l)7rx ) dx. 2£
(
Using by parts parts twice, in Using integration integration by twice, almost almost exactly exactly as as in in (5.23), (5.23), we we can can express express b&„n in terms of of a terms an: n:
_ 2T £
1t 0
2 d u( ). ((2n-1)7rx) d dx 2 x sm 2£ x
[dU (x) sin ((2n -
= _ 2T {
£
dx _ (2n -1)7r n 2"
= 2T(2n -1)7r
2£2
1i
ii 0
l)7rX)] i 2f x=O
((2n -l)7rX) d } ddu() X cos n X 2"
X
du() ((2n -l)7rX) d odxxCOS 2f x
= du/dx(£) = 0)
(since sin (0)
= 2T(2n2£2-1)7r {[u () ((2n -l)7rX)] i X cos 2£ +
(2n -1)7r 2f
1i ().
2 _ T(2n _1)27r ~ 4£2 £
1t ().
uxsm
o
ux sm
,,=0
((2n -l)7rX) d } 2£ x
((2n - l)7rX) d 2£ x
o (since u(O)
= cos ((2n - 1)7r/2) = 0)
This the desired This gives gives the desired result. result. (Actually, (Actually, since since it it is is known known that that the the negative negative second second under the mixed boundary boundary conditions, we can just derivative derivative operator operator is is symmetric symmetric under the mixed conditions, we can just appeal which is appeal to to (5.29), (5.29), which is the the above above calculation calculation written written abstractly.) abstractly.)
9. Let ai, a2, 0,2,as,... Fourier quarter-wave quarter-wave sine sine coefficients coefficients of of U; u; then then 9. Let aI, a3, . .. be be the the Fourier 2
_ d u( ) ,.. dX2 x
= ~ ,..(2n _1)27r 2 ~
2002
an s
in ((2n -l)7rX) 200 '
n=l
where K = 3/2. 3/2. We also have where,.. We also have 0.001
=~
0.004
~ (2n - 1)7r
n=l
sin ((2n-1)7rX). 200
Appendix C. to odd-numbered exercises Appendix C. Solutions Solutions to odd-numbered exercises
542
Setting the two two series series equal solving for we find find Setting the equal and and solving for an, a n , we
~ 3.2.10 2 . (2n-1)7rx) u(x) = ~ 3(2n _ 1)37r3 sm 200 . The temperature distribution distribution is is graphed C.S. The temperature graphed in in Figure Figure C.8. 3.5r-----r-----r-----~---...__--__,
3
2.5 Q)
:s
2
~ Q)
c. ~ 1.5
0.5
40
20
x
60
80
100
Figure C.S. distribution (in degrees Celsius) Exercise 5.3.9. Figure C.8. The The temperature temperature distribution (in degrees Celsius) in in Exercise 5.3.9.
Section 5.4 1. €J, so 1. Suppose Suppose fI,, g9 E <E C1[0, Cr>[0,<], so that that
= f(€) = g(O) = g(€) = O.
1(0)
Then Then
(-! (k(X):~) ,g) =-li
d~ (k(x):~(x))9(X)dX
df =-k(x)dx(x)g(x)
= =
1 i
0
Ii + 1i k(x)dx(x)d;(x)dx df d 0
0
dl d k(x) dx(x) d~(x)dx (using g(O)
k(x)/(x)~;(x)l:
-It
f(x)!
(integration by parts)
= g(€) = 0)
(k(X)~;(X)) dx
(integration by parts)
Appendix C. Solutions to to odd-numbered odd-numbered exercises exercises Appendix C. Solutions
=
-11
I(x)! (k(x)
~~ (x))
543 543
dx (using 1(0)
= I(t) = 0)
= (/,- d~ (k(x)~~)). 3 3. Take p(x) p(x) = = (x — - c) c)3(d x)33 and and V[c,d] suggested in in the the hint. 3. Take (d -— x) V[c>d] as as suggested hint.
5. Repeating the the calculation calculation beginning beginning on on page page 173, we obtain 5. Repeating 173, we obtain
! [~lt
1 £
= -
0
c(
Ap(x)
au at)
(~f dx+ ~
2
dx,
1t
Ak(x)
(~;f dX]
t > O.
the total energy is always negative, This shows that derivative with respect to time of the and hence that that the the total total energy is always and hence energy is always decreasing. decreasing.
Section Section 5.5 5.5 1. 1.
(a) Yes. (b) No, o(/,f)/) ==00for forI(x) /(x)==1,1,but butI / i=/O.0. (b) No, a(f, (c) No, a(f, (c) No, o(/,f)/) ==00for forI(x) f ( x ) ==1,1,but butI / i=/O.0.
3. 3.
(a) The set S is of {x, {x, x subspace. (a) The set is the the span span of x 22 }} and and therefore therefore is is aa subspace. (b) As discussed the text, text, a(-,-) a(·,·) automatically satisfies two two of of the the properties (b) As discussed in in the automatically satisfies properties of an an inner product. Every Every function function p in in S satisfies p(O) = = 0,0, so of inner product. satisfies p(Q) so a(·,·) a(-, •) also also the third third property property (a(u,u) 0 implies implies that that u = = 0) 0) for for exactly exactly the satisfies the satisfies (a(u, u) = 0 the same reason as given in in the text for the subspace V (see page 181). 181). The best approximation is p(x) p(x) = = (9 — - 3e)x 3e)x22 + - 10)x. (c) The + (4e — (d) The bilinear bilinear form form is not an inner product product on P2, since since a(l, a(l, 1) = 00 but but 11i= (d) The is not an inner on 7>2, 1) = / 00 (a constant is to the the energy energy inner inner product product and and norm). norm). Therefore, Therefore, (a constant is "invisible" "invisible" to polynomial of the the form (9 — - 3e)o; 3e)x22 + + (4e -— 10)x ++ C E every polynomial € S is equidistant equidistant x energy norm. from from fI(x) ( x ) = eX e in the the energy
5. Write p(x] p(x) — = x(l x(l -— x) x) and q(x) — = x(l/2 x(1/2 — - x)(l x)(l -— x). x). The The approximation will be 5. Write and q(x) approximation will be v(x) = mp(x) + U2q(X), u-2q(x), where vex) = U1P(X) where Ku Ku = = f.f. The The stiffness stiffness matrix matrix K K is is K
= [ 1:(1 +x)*(x)*(x)dx 1:(1 +x)~(x)*(x)dx
101(1 + x)*(x)~(x) dx 101(1 + x)~(x)~(x) dx
=
[-~
-t],
and the the load vector ff is is and load vector f=
[f~XP(X)dX]
10 xq(x)dx
= [
12 1
1
-120
Therefore, 0.16412 ] -0.038168 .
]
.
1
Appendix C. Solutions to to odd-numbered odd-numbered exercises Appendix C. Solutions exercises
544 544
The exact is The exact solution solution is 1 2 u(x) = --x 4
1 + -x 2
1 - - I n (1 4ln2
+ x) .
The exact and approximate solutions solutions are are graphed in Figure Figure C.9. C.g. The exact and approximate graphed in 0.045r-----r-----r-----r-----r----~
0.04 0.035 0.03 0.025 0.02 0.015 exact approximate
0.01
0.2
0.4
0.6
0.8
Figure C.9. exact and and approximate approximate solutions solutions from from Exercise Exercise 5.5.5. Figure C.9. The The exact 5.5.5. 7.
long, or (a) It will take about about 10 1033 times as long, or 1000 1000 seconds (almost (almost 17 17 minutes), to to solve aa 1000 1000 xx 1000 1000 system, 10033 times 1066 seconds seconds (about 11.5 solve system, and and 100 times as as long, long, or or 10 (about 11.5 days) to to solve days) solve aa 10000 10000 xx 10000 10000 system. system. (b) Gaussian elimination elimination consists of aa forward forward phase, phase, in which the diagonal entries (b) Gaussian consists of in which the diagonal entries are used to eliminate and in are used to eliminate nonzero nonzero entries entries below below and in the the same same column, column, and and aa backward phase, phase, in in which which the the diagonal entries are used to to eliminate nonzero backward diagonal entries are used eliminate nonzero entries and in in the the same the forward phase, at at aa typical entries above above and same column. column. During During the forward phase, typical is only only 11 nonzero nonzero entry below the the diagonal, diagonal, and and only only 5 5 arithmetic step, step, there there is entry below arithmetic operations are are required to eliminate eliminate it it (1 (1 division division to to compute compute the the multiplier, multiplier, operations required to the current row row and 22 multiplications and 2 additions to add add a multiple of of the to the next). Thus the forward phase requires O(5n) O(5n) operations. A typical the backward backward phase phase requires requires 3 3 arithmetic arithmetic operations operations (a (a multiplication multiplication step of the step of and an an addition adjust the side and the and addition to to adjust the right-hand right-hand side and aa division division to to solve solve for for the unknown). Thus Thus the the backward backward phase phase requires requires O(3n) The grand unknown). O(3n) operations. operations. The grand total is O(Sn) total is O(8n) operations. operations. (c) will take take about about 10 times as long, or to solve 1000 (c) It It will 10 times as long, or 0.1 0.1 seconds, seconds, to solve aa 1000 1000 xx 1000 and 100 times as or 11 second, to solve 10000 tridiagonal system, tridiagonal system, and 100 times as long, long, or second, to solve aa 10000 10000 xx 10000 tridiagonal tridiagonal system. system.
Appendix C. C. Solutions to to odd-numbered Appendix odd-numberedexercises exercises
545
Section Section 5.6 5.6 1. 1.
the errors for for n = = 10,20,40,80: (b) Here are the 10, 20,40, 80: maximum error 1.6586 . 10 10~-0>3 4.3226 . 10 10~4 1.1036 . 10 io--'l4 2.7881.10 10~-05 2.7881
n 10 10 20 40 80 80
-q
We see that, when n is doubled, the error decreases decreases by a factor of approximately four. Thus error
= 0 (~2 ) .
exact solution is u(x) = x(l x(l3. The exact u(x) — — x 3 )/12. Here are the the errors for n = 10,20,40,80: 10, 20,40,80: n n 10 10 20 40 40 80 80
maximum error 1.1286 .• 10 10~3 2.9710 .• 10 10~-'l4 7.6186 .• 10 10~-0& 1.9288 .• 10 10~-0& -;j
descreases by a factor of approximately We see that, when n is doubled, the error descreases approximately four. Thus error = 0
(~2 ) .
5. The The weak form is
1
£{
find u G E V such that
o
=
1£
du dv k(x) dx (x) dx (x)
+ p(x)u(x)v(x) }
dx
f(x)v(x)dx for all v E V.
The bilinear form is The a(u,v)=
1 0
£{
du dv } k(X)dx(X)dx(X)+P(X)U(X)V(X) dx.
Section 5.7 1. u(x)
3.
= xe- x .
(a) This BVP models a bar whose top end (originally at x = = 0) is free and whose bottom end is fixed fixed at x = i. £. If we we apply a unit force to the cross-section bottom cross-section at x = s, then the part of the bar originally between x = s and x = £ t will compress, and the part of the bar originally above x = s will just move rigidly with u(x) u(x) — u(s) for 0 ::; < x < s. The compression of the bottom the = u(s) bottom part of the bar will satisfy Hooke's law: bar u(x) () =£-x k- = 1 ~ux i-x k
Appendix C. odd-numbered exercises Appendix C. Solutions Solutions to to odd-numbered exercises
546 546
(the (the uncompressed uncompressed length length of of the the part part of of the the bar bar between between xx = = ss and and xx = =£ t is is x). Therefore Therefore we we obtain obtain £i -—x). i-s
< s,
0< x
k'
u(x) = {
i-x
s<x<£.
-k-'
The is The Green's Green's function function is
G(XjS)
i-s -k-'
={
< s, < x < e.
0< x
i-x
S
-k-'
(b) The is (b) The Green's Green's function function is
G(Xj s)
=
dz
1 £
k(z)'
max{x,s}
(c) (c) If If kA; is is constant, constant, then then
G(XjS) = e-ma;{x,s}
l-s
={
k' i-x
-k-'
0< x < s, S
< x < e.
(d) (d)
u(X)
= In2 _! _ X2 2
4
+ ~ _In(l+x)
4
2
2'
5. we obtain obtain the the following to (5.67): (5.67): 5. By By direct direct integration, integration, we following solution solution to
u(x) =p
r k(s)' ds
10
Since Since xx = min{x, min{x, i} i\ for for xx E € [0, [0,il, £.], we wecan can write write _
ds
{minix,s}
U(X) - p
10
that is, is, that
u(X)
k(s)'
= g(Xje)p.
7. Green's function is 7. The The Green's function is
g(Xj s)
. (n7rs) . (n7rx) =~ L.J kn2l 27r 2 sm -e- sm -e- . n=l
Section 6.1 6.1 1. 1. The The steady-state steady-state solution solution of of Example Example 6.2 6.2 satisfies satisfies the the BVP BVP
-D::~ =
10-
u(O) = 0, u(lOO) = O.
7
,
0
< X < 100,
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numberedexercises exercises
547
Either the limit of Either by by solving solving this this BVP BVP or or by by taking taking the limit (as (as t ---+ —> 00) oo) of of the the solution solution of Example 6.2, 6.2, we we find find that that the the steady-state steady-state solution solution is is Example (
Us X
. (n7rx) ) _~2.1O-3(1-(-1)n) ~ Dn 7r SIll 100 . -
3
3
n=l
3. is 3. The The solution solution is 2(_1)n+l n 2 2 e- 7r tsin(n7rx). n7r
00
u(x,t)
= '"' ~ n=l
A graph A graph of of u(·, u(-, 0.1) 0.1) is is given given in in Figure Figure C.lO. C.10. 1~====~~--~------~----~----~ u(·,O) 0.9 u(·,O.1 )
0.8 0.7
0.6 0.5 0.4
0.3 ~.~ ~.~
0.2
_.-.-,-,-,-,-
-'~,~
"'
.
0.1
... ... 0.2
0.4
,
0.8
0.6 x
Figure C.1O. 0.1), together C.10. The The snapshot snapshot u(·, w(-,0.1), together with the initial temperature temperature distribution, for Exercise 6.1.3. distribution, Exercise 6.1.3. 5. 5. With With
p(x, t) and v(x, t)
= u(x, t) pcov ot
x = g(t) + C(h(t) -
g(t))
p(x, t), we have
_",02 V
=pcou
ox 2
at
_",02 U _
ox 2
(pc OP ot
= f(x t) - pc (d9 (t) dt ,
_",02 p )
ox 2
+~
(dh (t) _ dg (t))) . "dt dt
We We also also have have
v(x, to)
= u(x, to) -
p(x, to)
= 1j;(x) -
x
g(to) - C(h(to) - g(to))
Appendix C. Solutions to Appendix C. Solutions to odd-numbered odd-numbered exercises exercises
548 and and
= u(O, t) = u(l, t) -
v(O, t) vel, t) We We define define
g(x, t)
= I(x, t) -
and and ifJ(x)
pc
= 1j;(x) -
p(O, t) = get) - get) pel, t) = h(t) - h(t)
= 0, = o.
7
(~! (t) + (~~ (t) - ~! (t)) ) x g(to) - l(h(to) - g(to)).
Then Then v satisfies satisfies {)v
{)2V
pc {)t -
K
{)x 2 = g(x, t), 0 < x < l, t
> 0,
v(x,to) = ifJ(x), 0 < x < l, v(O, t) = 0, t > to, vel, t) = 0, t > to. 7. vex, t) = = u(x, t) -— xcos x cos (t). homo7. Define Define v(x, (t). Then Then v satisfies satisfies the the following following IBVP IBVP (with (with homogeneous boundary conditions, but but with with aa nonzero source term): term): nonzero source geneous boundary conditions, {)v
{)2V
.
at - {)x2 = x sm (t), 0 < x < 1, t > 0, v(x,O) = 0, 0 < x < 1, v(O, t) = 0, t > 0, v(l,t) = 0, t > O.
This This IBVP IBVP has solution 00
v(x,t)
= Lan(t) sin (mrx), n=l
where where 2( _l)n
an (t) = mr (n4 7r 4
+ 1)
(
()
cos t - e
_n 2 ,..2 t
2
2 •
(
)
- n 7r sm t) .
The solution solution to the original original IBVP IBVP is is then then u(x, u(x,t)t) = vex, v(x,t)t) + xcos(t). A graph graph of The to the x cos (t). A of u(·,1.0) Figure C.11. «(•, 1.0) is is given given in in Figure C.ll. 9. temperature u(x, t) satisfies IBVP 9. The The temperature satisfies the the IBVP {)u
pc at
{)2U
- K {)x 2
< x < 100, t > 0, 0 < x < 100, t > 0, t > O.
= 0, 0
u(x,O) = 5, u(O, t) = 0, u(100, t) = 0, The solution solution is The is u(x t) ,
_"n ,,2 t (n7rx) . = '~ \:""'" 10(1-(-1)n) e 1002 sin n7r 100 00
2
pc
n=l
The temperature temperature at at the midpoint after after 20 20 minutes minutes is The the midpoint is ti(50,1200) = 1.58 1.58 degrees degrees Celsius. u(50, 1200) == Celsius.
Appendix C. C. Solutions odd-numbered exercises Appendix Solutions to to odd-numbered exercises
549
Figure C.ll. The snapshot u(-, 1.0), together with the initial temperature snapshot u(·, initial temperature distribution, distribution, for Exercise 6.1.7. 11. 11.
(a) The steady-state temperature temperature is is u u.(x) x/20. (a) The steady-state = re/20, s (x) = (b) The The temperature temperature u(x,t) u( x, t) satisfies satisfies the the IBVP IBVP (b) pc
AU
at -
02U
= 0, u(x,O) = 5, K,
ox 2
u(O, t) u(lOO, t)
= 0,
< x < 100, 0 < x < 100, t > 0,
0
t
> 0,
= 5, t > o.
The solution solution is is The 10e - ''''' " t sin (nlrx) = -20X + L -nlr -- . 100 00
u(x t) ,
2
2
l002pc
n=l
Considering the the results results of Exercise 9, is obvious obvious that that at least several thousand Considering of Exercise 9, it it is at least several thousand seconds will elapse before the the temperature temperature is is within within 1% of steady so we we seconds will elapse before 1% of steady state, state, so can accurately estimate using aa single single term term of of the the Fourier Fourier series: can accurately estimate u(x, u(x,t)t) using series:
x
u(x t) == , 20
10 + -e 11'
",,2
lO02pc
t
We want want to to find find tt large enough that that We large enough
lu(x, t) - u.(x)1 < 0.01
lu.(x)1
-
(
lrX )
sin 100 .
Appendix C. Solutions to to odd-numbered odd-numbered exercises exercises Appendix C. Solutions
550 550
for all all x E this is for e [0,100]. [0,100]. Using Using our our approximation approximation for for u, u, this is equivalent equivalent to to 200 sin ( ffo-) 7rX
A A graph shows shows that 200 sin (;o~) 7rX
~2
for we need for all x, so so we need
This yields t
>
-100 pcln 0.005 ~ 4520. 2
K.7r 2
-
75 minutes 20 seconds About 75 minutes and 20 seconds are are required. required.
Section 6.2 1. The solution is 1. The solution is 00
u(x, t)
= do + L
dne-n21r2t cos (n7rx),
n=1
where do =
1 1 o
1 x(1-x)dx=-6' d n =2
11
x(l-x)cos(n7rx)dx=
0
2 ((-It+
The graphs are given given in in Figure Figure C.12. The graphs are C.I2. /I 3. 3.
(a) Ifu,v E C1[0,f], (a) Ifu,v£ C^[0,4 then then (LNU,V)=-
10t
d
2
dx~(x)v(x)dx
du
= - dx(x)v(x)
=
l
Ii0+ 10t
du dv dx(x)dx(x)dx
i du dv . -(x)-(x)dx (smcedU(O)=du(£)=O) odx dx do; do;
dv = u(x)dx(x)
= - 10t
Ii0- 10t 2
2
d v u(x)dx 2 (x)dx
d u(x) dx~ (x) dx (since
= (u, LNV).
This shows that LN is symmetric.
1
22 n 7r
~~ (0) = ~~ (£) = 0)
-1) •
551
Appendix Solutions to to odd-numbered odd-numbered exercises Appendix C. C. Solutions exercises 0.25,....----"T""""----r-~-~__r---~---___,
"
----.,
..,-. ..... - -'-'- -'- ............ ... " ...
0.2
'!:-:::: ............................... ,:.':"'... ~~ 0.15. __ , ....
"-0.1 u(x,O) - - - u(x,0.02) ,-, - u(x,0.04 u(x,0.06) u(x,oo) 0.2
0.8
0.6
0.4
x
Figure C.12. The The solution w(x,t) 6.2.1 at times 0, 0, 0.02, Figure C.12. solution u(x, t) from from Exercise Exercise 6.2.1 at times 0.02, 0.04, and and 0.06, 0.06, along the steady-state solution. These These solutions were estimated estimated 0.04, along with with the steady-state solution. solutions were using the Fourier Fourier series. series. using 10 10 terms terms in in the (b) an eigenvalue LN and is aa corresponding eigenvector, nor(b) Suppose Suppose A A is is an eigenvalue of of LN and U it is corresponding eigenvector, normalized so that (u, (u. u) = 1. 1. Then A = A(U, u) = (AU, u)
= (LNU,U) =
-it ::~
(x)u(x) dx
= - ~:(x)u(x)l: + £
= ~
1 0
d
C:(x))
1£ (::(X))2
2
dx (since
dx
~~(O) = ~~(f) = 0)
O.
Thus LN cannot cannot have any negative eigenvalues. 5. 5.
(a) U is is aa solution to the the BVP. BVP. Then Then (a) Suppose Suppose u solution to
if
I(x)dx
= -t>
it ::~(X)dX
= t> (::(0) - ::(f)) = t>(a -
This is is the the compatibility This compatibility condition: condition:
1£ I(x) dx = t>(a - b).
b).
C. Solutions to to odd-numbered odd-numbered exercises Appendix C.
552
The operator [0, f] -t C[O, f] defined defined by (b) The operator K : c1 C^[0,£\ ->• C[Q,t\
has eigenpairs eigenpairs has
>'0 =
~n2~2 (n~x) 1, 'Yo(x) = 1, and >'n = 1 + -;y-, 'Yn(x) = cos -f- , n = 1,2,3, ....
The method of Fourier series can be applied to show that a unique solution C[O, fl. (The key is is that that 0 is is not not an an eigenvalue eigenvalue of of K, as as it it exists for for each each /f E exists G C[0,^]. (The key is LN.) is of of LN.) 7. As shown in this section, section, the solution solution to the IBVP is
u(x, t)
~ dne- n 1T = do + L...J 2
2
t
/l2 cos (n~x) -f- ,
n=l
dl, c?2, d2, ... the Fourier cosine coefficients coefficients of of"p. We have where do, di, • • • are are the if). We
L dne-n21T2t/l2 cos (n;x) ~ L JdnJ e-n21T2t/l2Icos (n;x) I 00
00
n=l
n=l
n=l
L JdnJ e-(n 2-1)1T 2t/l 2. 00
~ e-1T2t/l2
n=l
This last last series series certainly certainly converges converges (as (as can be proved, proved, for for example, using the the coracomThis can be example, using narisrm anr\ parison t.pst^ test), and
This shows that 00
do
+L
dne-n21T2t/£2 cos (n;x) -t do as t -t
00.
n=l
The limit do The limit do is is
~
1£
"p(x) dx,
the average of the initial temperature temperature distribution. the 0.992 degrees 9. The steady-state temperature is about 0.992 degrees Celsius. Celsius. 11. 11.
the Fourier sine series of u(x,t) u(x, t) on (0,^) (0, f) is (a) Suppose that the
u(x, t)
= L an(t) sin (n;x), 00
n=l
and and u satisfies satisfies
an (t)
=~
it
u(x, t) sin (n;x) dx
0
au ax (0, t) = au ax (f, t) = 0 for all t > o.
AnnpnHiv Appendix C C.
553
Solutions to r»HH-nu inhered exercises exerr.ises Solutions to odd-numbered
2 The of —d u/dx22 is follows: The nth nth Fourier Fourier sine sine coefficient coefficient of -02U/OX is computed computed as as follows:
2
r 02U
. (n1l"x)
-"llo ox 2 (x, t) sm -e- dx i
. (n1l"x)ll ou (n1l"x)] = --2 [ou -(x,t)sm - -n1l"1 -(x,t)cos dx
e
2n1l"
e
Ox
r
ou
e
0
e
ox
0
(n1l"x)
=£2 10 ox(x,t)cos -e- dx 2n1l" = £2 2n1l"
[ u(x, t) cos (n1l"x) -e-
= £2 ((-1)
n
Ii + f 10r n1l"
0
u(e, t) - u(O,t))
.
u(x, t) sm (n1l"x) -e- dx ]
n 211"2
+ f2an(t).
Since the the values values w(0, u(O, t) and u(e, t) are are unknown, unknown, we we see see that that it it is not possible possible Since and u(l, is not 2 to express express the sine coefficients coefficients of u/dx22 in — to the Fourier Fourier sine of —d -02U/OX in terms terms of of oi(t),a2(t), al (t), a2(t), .... (b) The Fourier Fourier sine = t isis (b) The sine series series of of u(x, t) = 00
'"'
2 (1 - (_1)n) t
~
n1l"
n=l
sin (1mx),
2 and the the formal formal calculation of the the sine sine series of —d -02U/OX yields and calculation of series of u/dx22 yields 00
L 2 (1 -
(-1t) n1l"tsin (n1l"x).
n=l
However, However,
02U - ox 2 (x, t)
= 0,
2 and so all of of the the Fourier Fourier sine sine coefficients coefficients of of —d -02U/OX be zero. Thus and so all u/dx22 should should be zero. Thus the calculation is the formal formal calculation is wrong. wrong.
Section 6.3 6.3 Section 1. The formula formula for for the the solution solution u the same as in in Example Example 6.6, with aa different 1. The u is is exactly exactly the same as 6.6, with different
value for for Kfl, (4.29 instead of 3.17). This This implies that the the amplitude amplitude of of the the solution value (4.29 instead of 3.17). implies that solution is Therefore, there temperature is reduced reduced by by about about 26%. 26%. Therefore, there is is less less variation variation in in the the temperature distribution the silver ring as opposed to to the the gold gold ring. ring. distribution in in the silver ring as opposed 3. 3.
The IBVP is (a) fa) The IBVP is
ou pc at -
02U ox 2 = 0, -511"
< X < 511", t > 0, u(x,O) = "p(x), -511" < X < 511", u( -511", t) = u(511", t), t > 0, fl,
ou ou ox (-511", t) = ox (511", t), t > O.
554 554
Appendix Solutions to Appendix C. C. Solutions to odd-numbered odd-numbered exercises exercises (b) The solution is (b) The solution is
+L 00
u(x, t) =
Co
"n2
t
ene-25P""C cos
(nx) ""5 '
n=l
where where
71"4
Co
= 25 + 9'
en
=
10( _l)n+l n4
'
n
= 1, 2,3, ....
4 (c) is the /9 degrees Celsius. (c) The The steady-state steady-state temperature temperature is the constant constant u Uss = = 25 25 + + 7T 71"4/9 degrees Celsius.
(d) We so that (d) We must must choose choose t so that Ius - u(x, t)1
lusl
< 0.01
-
holds for every every xx E € [-571",571"]. [—STT, STT]. By error, we find that holds for By trial trial and and error, we find that about about 360 360 seconds (6 minutes) minutes) are are required. seconds (6 required. 5. 5.
(a) Lp is (a) To To show show that that Lp is symmetric, symmetric, we we perform perform the the now now familiar familiar calculation: calculation: we we form integral (Lpu, (Lpu, v) v) and (u, Lpv). Lpv). The The the integral and integrate integrate by by parts parts twice twice to to obtain obtain (u, form the boundary boundary term term from from the the first first integration integration by by parts parts is is du - [ dx (x )v(x)
]l
-t
du
= dx (-f)v( -f) -
du dx (f)v(f).
Since both satisfy periodic Since both u and and v satisfy periodic boundary boundary conditions, conditions, we we have have v( -f)
= v(f),
du dx (-f)
du
= dx (f),
so the second integraintegraso the boundary boundary term term vanishes. vanishes. The The boundary boundary term term from from the the second tion by by parts parts vanishes vanishes for for exactly exactly the the same same reason. tion reason. (b) Suppose (u, u) = 1. 1. Then Then Lpu AU, where where u has has been been normalized: normalized: (u, (b) Suppose L pu = \u,
A = A(U, u)
= (Au, u) = (Lpu, u) = == ~
1 £ _£
d2U dx 2 (x)u(x) dx
[~:(x)u(x)rl +
I: (~:(x)f
I: (~:(x)f
dx
dx
O.
The conditions: The boundary boundary terms terms vanishes vanishes because because of of the the periodic periodic boundary boundary conditions: du dx (-f)u( -f) 7. 7.
du
= dx (f)u(f).
(a) the ring ring is completely completely insulated, insulated, aa steady-state temperature distribution (a) Since Since the steady-state temperature distribution cannot exist unless the the net net amount amount of heat being being added to the the ring ring is is zero. zero. cannot exist unless of heat added to This is is exactly exactly the the same situation as bar with with the the ends, ends, as as well well as This same situation as aa straight straight bar as the insulated. the sides, sides, insulated.
555
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
(b) (b) Suppose Suppose Uu is is aa solution solution to to (6.21). (6.21). Then Then
I
t
f(x) dx
=
-K,
1£
-i
02
ax~ (x, t) dx
-i
=
-K,
(
=K,
aU ]l [-(x,t) ax -l aU aU) ax(-l,t)- ax(l,t)
=0.
The follows from the periodic periodic boundary The last last step step follows from the boundary conditions. conditions. (c) The (c) The negative negative second second derivative derivative operator, operator, subject subject to to boundary boundary conditions, conditions, has has aa nontrivial null space, namely, namely, the the space space of of all all constant constant functions functions on (—i,i). on (-l, l). In In nontrivial null space, to the Fredholm alternative for symmetric matrices, we we would would expect analogy analogy to the Predholm alternative for symmetric matrices, expect aa solution solution to to the the boundary boundary value value problem problem to to exist exist if if and and only only if if the the right-handright-handside side function function is orthogonal to this null space. This condition is
I:
cf(x) dx
= 0 for all c E R,
or or simply simply
t
J- l
f(x)dx =
o.
Section 6.4 =
1. give the the proof the general Gram matrix where 1. We We give proof for for the general case case of of aa Gram matrix G. Suppose Suppose Gx Gx = 0, 0, where and x ERn. € R n . Then Then (x, (x, Gx) = 00 must must hold, hold, and n
n
(x, Gx)
= Z)GX)iXi = L i=1
n
L GijXjXi
i=1 j=1 n
n
i=1 j=1
=~
(tXjUj,Ui) Ui
(t XjUj, ~ XiUi) n
=
LXjUj j=1
Thus Gx = = 00 implies that the the vector Thus implies that vector n
LXjUj j=1
2
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
556
is the zero Un} this in is the zero vector. vector. But, But, since since {U1, {wi, U2, W 2 ,... . . . ,,u is linearly linearly independent, independent, this in turn turn n} is implies that x\ xi ••• = x 0, that is, that x = 0. Since the only vector implies that Xl = = X2 = ... = Xn = 0, that is, that x = 0. Since the only vector n x €R Rnn satisfying satisfying Gx Gx = = 0 is is the the zero zero vector, vector, G G is is nonsingular. nonsingular. x E 3. 3. (a) (a)
°
_ [
M-
2/9 1/18
=[
1/18] K 2/9'
-3] f __1_ [ 11 cos (t) ] 6 ' (t) - 162 11 cos (t) .
6 -3
(b) The system is (b) The system of of ODEs ODEs is
do:
= -Ko: + f(t),
M dt
that is, that is, 2 d0: 1
9" d.t
1 d0:2
.!.. d0:1 18 dt
5. 5.
11 cos (t) 162 '
+ 18 d.t = -60:1 + 30:2 + ~ d0:2 _ 3
+9
dt -
_ 6 0:2
0:1
+
11 cos (t) 162 .
(a) Write pi = 8.97, 8.97, P2 p2 = (a) Write P1 = 7.88, 7.88,
°50<< x<<50,100, X
p(x) = {
and similarly similarly for c(x) and and K(X). the IBVP IBVP is and for c(x) ,.(x). Then Then the is aU p(x)c(x) at
a ( aU) - ax ,.(x) ax = 0,
= 5,
°< °<
x
u(x,O) u(O,t)
x = 0, t > 0,
u(100, t)
= 0, t > 0.
< 100, t > 0,
< 100,
(b) (b) The The mass mass matrix matrix M M is is tridiagonal tridiagonal and and symmetric, symmetric, and and its its nonzero nonzero entries entries are are Mi,i+1=
and
M;;
={
~ 6 ' ~
{
6
'
~ 3 ' (Pi C1 +P2C2 )h 3
~ 3'
i = 1,2, ... , ~ - 1,
i
= ~,~ + 1, ... , n -
= 1,2, ... , ~ n 2 = 2' i
1,
1,
.
'-n+1n+2 2 - 2 '2 , ...
1 ,n-.
The stiffness matrix K is tridiagonal its nonzero The stiffness matrix K is tridiagonal and and symmetric, symmetric, and and its nonzero entries entries are are -~ i = 1,2, ... , ~ - 1, h' K"+l = 1.,1. ~2 i = ~, ~ + 1, ... , n -1, -h' and ~ i = 1,2, ... , ~ - 1, h '
{
K;;
={
1<1+1<2
h
'
~
h '
i =~,
i = ~
+ 1, %+ 2, ... , n -
1.
Appendix C. C. Solutions Solutions to odd-numbered exercises
557
7. The solution is 00
u(x,t)
= I>n(t) sin C~;), n=l
where
4 (
2
( 10 pc e -~ "n ,,2t -1 ) an (t) __ 400(2+(-1)n) 2 7 7 K 7r n
+ Kn 27r 2) t .
The errors in Examples 6.8 6.8 and 6.9 6.9 are graphed in Figure C.13. 0.05,.----..-----.....----....----...,......----,
____~
°l~ -0.05
-0.1 \
\\
-0.15
\ \ \ \
""
-0.2
"
....
_-_
20
....
"
40
. "'
,,
, ,,
"
'
,,
, ,,
,
, ,,
, ..
. .. .
l-
.. .
Euler - _. Backward Euler
x
60
I
80
100
Examples 6.8 and 6.9 (see Exercise Exercise 6.4.7). Figure C.I3. C.13. The errors in Examples 6.4-V9. 9.
(a) ThelBVPis The IBVP is p(x)c(x) au at
-
a ( K(X) aU) ax ax = 0, 0
< x < t', u(x,O) = 5, 0 < x < t', u(O, t) = 0, t > 0, u(t', t) = 0, t > O.
t
> 0,
(b) The weak form is to find u satisfying
l l{ o
p(x)c(x)
& ~ } at (x, t)v(x) + K(X) & ax (x, t) dx (x) dx = 0, t > 0,
\Iv E V.
(c) The temperature distribution after after 120 120 seconds is shown in Figure C.14.
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
558
t = 120 (seconds) 5r-----~----~====~~~----~----__,
4.5
4 3.5 ~
::::I
3
T!!Q)2.5 c..
E
2
2
20
40
60
80
100
x
after 120 seconds in Exercise Exercise 6.4.9. Figure C.14. The temperature temperature distribution after 6.4-9.
Section 6.5 1. The BVP to be be solved is 1. The BVP to solved is
d2 u -k dx 2
= f(x),
0
< x < 100,
~:(O) = 0, du (100) dx
= 0,
with kk = 1.5 (25-x)(100-x) with 1.5and andf(x) f(x) ==1O-7 10~x7x(25 - z)(100 - z)+1/240. +1/240.The Thesteady-state steady-statetempertemperature is with u(100) = 00 isis shown ature is not not unique; unique; the the solution solution with ii(100) = shown in in Figure Figure C.15. C.15. 3. total amount heat energy added to bar is W, where 3. (a) (a) The The total amount of of heat energy being being added to the the bar is 0.51A 0.51A W, where AA is (O.OlA W through W in is the the cross-sectional cross-sectional area area (0.01 AW through the the left left end end and and 0.5A 0.5AW in the the interior). W must interior). Therefore, Therefore, 0.51A 0.51 AW must be be removed removed through through the the right right end; end; that that is, is, rate of W jcm22 through the right right end. heat energy must be be removed heat energy must removed at at aa rate of 0.51 0.51 W/cm through the end. (b) is (b) The The BVP BVP is d2 u dx 2
= 0.005,
du k dx (0)
= -0.01,
du k dx (100)
= -0.51.
-I>,
0
< x < 100,
The temperature is unique; the with u(100) The steady-state steady-state temperature is not not unique; the temperature temperature with w(100) = = 00 is graphed in in Figure C.16. is graphed Figure C.16.
559 559
Appendix Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises 6r-------~------_r------~--------r_------~
5 4 Q)
::; 3
"§ Q)
a. E 2
2
o -1~------~------~------~--------~----~
o
20
40
x
60
80
100
Figure The computed computed steady-state fron Figure C.I5. C.15. The steady-state temperature temperature distribution distribution from J^.fOTfnoo fi6.5.1. t\ 1 Exercise 7~------~------~--------~------~-------,
6
5
2
°o~------~------~------~------~----~ 20 40 60 80 100 x
Figure The computed from Figure C.I6. C.16. The computed steady-state steady-state temperature temperature distribution distribution from Exercise 6.5.3. Exercise 6.5.3.
560 5.
Appendix C. Appendix C. Solutions Solutions to odd-numbered odd-numberedexercises exercises (a) We have n
n
u·Ku= LLK;jUjU; i=O j=O
n
=L
n
La (4)j, 4>i) UjU;
i=O j=O
~
t,a (t, .;f;,f}
~a
(t, .;~;, t, .;~;)
=a(v,v). Therefore,
Ku = 0 ~ U· Ku = 0 ~
a(v,v)
= o.
(b) Suppose v E and a(v, a( v, v) == 0. O. By definition of a(-, .), G iT V and •), this is equivalent to
Since the integrand integrand is nonnegative, this implies that the integrand integrand is in fact K(X) is positive, we conclude that zero, and, since K(X)
dv(X) dx
is the zero function. Therefore, v is a constant function. function, 0, it follows that (c) Thus, if Ku = 0, n
v(X) = L
U;4>i(X)
;=0
the components of u. But But then is a constant function, where Uo, MO, U2, 1*2, •. .. • • ,, Un un are are the 0,1,2, there is a constant constant C such that V(Xi) v(xi) = — C, ii = 0,1, 2 , ... . . . , n. These nodal Un, = CUe, values of v are precisely the numbers Uo, UQ, Ul, MI, ... ...,u we see see that uu = Cuc, n , so we we wanted which is what we wanted to prove. 7. 7.
(a) Define v(O)
= O}.
Then the the weak form of the the BVP is
find U E V such that a(u, v)
= (I, v) for all v E V.
(CA)
(b) The fact that a solution of the strong form is also a solution of the weak form is proved by the the usual argument: multiply the differential differential equation by an arbitrary test function v € E V, integrate by parts. V", and then integrate
561
Appendix C. C. Solutions odd-numbered exercises Appendix Solutions to to odd-numbered exercises
The form The proof proof that that aa solution solution of of the the weak weak form form is is also also aa solution solution of of the the strong strong form is similar to the the argument given on on page page 268. 268. Assuming Assuming that that Uu satisfies is similar to argument given satisfies the the weak (C.4), an integration by some simplification simplification shows weak form form (C.4), an integration by parts parts and and some shows that that
k(£)
~: (£)v(£)
-1£
{:x [k(X)
~: (x)] + f(x)} v(x) dx = 0 for all v E V.
Since C V, this this implies implies that that the the differential differential equation equation Since V C
d [k(x)dx(x) du ] +f(x)=O, O<x<£ dx holds, and we have holds, and we then then have
du k(£) dx (£)v(£)
= 0 for all v E V.•
Choosing any v € shows that condition holds holds Choosing any E V with with v(i} v(£) ^ f::. 00 shows that the the Neumann Neumann condition at at xx — = t. £. 9. The The temperature seconds is C.I7. 9. temperature distribution distribution after after 300 300 seconds is shown shown in in Figure Figure C.17. t=300 (seconds) 8r-----------==~~-----r------~------,
7 6
2
20
60
40
80
100
x
Figure C.I7. The The temperature Figure C.17. temperature distribution distribution from from Exercise Exercise 6.5.9 (after {after 300 300 seconds). seconds}. 11. Here Here is sketch of of the If Ku = 11. is aa sketch the proof: proof: If = 0,0, then then
u· Ku = 0 => a(v, v) = where where
n-l
V
= LUi¢>i. ;=0
0,
Appendix Solutions to to odd-numbered Appendix C. C. Solutions odd-numbered exercises exercises
562
But the usual must be constant function, and V(Xn) v(xn) = But then, then, by by the usual reasoning, reasoning, vv must be aa constant function, and = 00 (since 0 for Thus v is the zero (since CPi(X (j)i(xnn) ) = 0 for ii = 0,1,2, 0,1, 2 , ... . . . ,, n-l). n — 1). Thus is the zero function, function, which which implies implies that the nodal values of vv are all zero. zero. Therefore uu = = 0, 0, and so so K is nonsingular.
Section 6.6 1. For t = only 2 2 terms terms are are required required for for an an accurate graph, while 61, 1. For = 6060, 6060, only accurate graph, while for for t = = 61,
about 150 150 terms are required. required. about terms are 3. Green's function function G(x, G(a;,t;r/,s) is given by 3. The The Green's t; y, s) is given by 2
~
~~e
-K(2n-l)2,,2(t-8)/(4pcl 2 ) .
~
((2n -1)1rY )
U
. ((2n -1)1rX)
~
U
n=l
for to < s < t, while 0 for for s > t. Selected Selected snapshots snapshots of of G(x, t; t] 75, 75,60) for while G(x, t; y, y, s) = 0 60) are C.18. are shown shown in in Figure Figure C.I8. Graph Graph of G(x,t;75,60) G(x,t;75,60) for for various values values of of t
0.02
- _. ._,'"'' -
t=120 t=240 t=360 t=480 t=600
~ 0.015 :::l
"@
Cl.
E 0.01 $
0.005
20
60
40
80
100
x
Figure C.IS. Snapshots of the function in Exercise 6.6.3 at tt = Figure C.18. Snapshots of the Green's Green's function in Exercise 6.6.3 at = 120,240,360,480,600 seconds. Twenty Twenty terms terms of of the the Fourier to 120, 240, 360, 480, 600 seconds. Fourier series series were were used used to create these graphs. create these graphs.
Section 7.1 7.1 Section 1. the solution 1. Some Some snapshots snapshots of of the solution are are shown shown in in Figure C.19. C.19.
3. two waves add constructively, and then then separate separate again. See Figure Figure 3. The The two waves join join and and add constructively, and again. See C.19.
Appendix C. C. Solutions to odd-numbered odd-numbered exercises exercises Appendix Solutions to
563
1.5.--.-----r--.....---.....--.----.-r========i1 t=o t=2.5 t=5
, "
1\
"
" ,," ,, ,, , , 0.5
~ _,
,
...
, I
,"
'
I
~i
I
I I
I I
~ ..
,- I
I
-I
,.'
-f
I;~
i
I
I
1&
,~"-
I
I
~~,
~
I),
, , • -, ~ '\. .,•
"
I
l~!
~"
..
I
I
I.
~
I.
/'j'
j\ '\
1\ I I I
,ii
~!~
r' I
~
.
."
, __~____~ I I OL-____~__6_~~/~\~~-~J_'\~/_~,-~I_\~_'~~~~~~ -20
-15
-10
-5
0
5
10
15
20
x -3
1.5 x 10 t=O t=0.02 t=0.04
0.5
I
!
i'
"
,
I
,• -~O
-30
-20
I
,"
""I
, ,,
,•
-10
,-i
"
'I • I I I I I
I
,
0
I I I I I
10
,,! -I
i I i ! i !
,
I
20
30
40
x
Figure C.19. Some snapshots snapshots of of the solution to Exercise 7.1.1. Figure C.19. (Top) (Top) Some the solution to Exercise 7.1.1. (Bottom) Constructive interference in in Exercise Exercise 7.1.3. center is is (Bottom) Constructive interference 7.1.3. The The "blip" "blip" in in the the center the result of of two waves combining combining temporarily. temporarily. (The (The velocity velocity in in this this example is the result two waves example is l.) cc = 1.)
Section Section 7.2 1. is found found by by setting setting c enn = 0 0 in in Example Example 7.4: 7.4: 1. The The solution solution is
. (n7rx) u(x,t) = ~ ~bncos (cn7rt) ~ sm 25 . n=l
Appendix C. C. Solutions to odd-numbered odd-numbered exercises exercises Appendix Solutions to
564
A graph analogous to Figure 7.6 7.6 is given in Figure C.20.
o
10
5
15
20
25
x
Figure C.20. C.20. Fifty Fifty snapshots snapshots of of the the vibrating vibrating string string from from Example Example 7.4, Figure 7-4, with no no external with external force. force. The solution is 3. The
u(X,t)
= Lan(t)sm (2n-1)1I'X) 50 ' 00
•
n=l
where
and and
8 (v'2sin (n1l'/2) - v'2cos (n1l'/2) + (_1)n) 400 ,en= (2n - 1)11' . 5(2n - 1)211'2 The fundamental frequency frequency is snapshots are The fundamental is now now c/100 instead instead of of c/50. Fifty Fifty snapshots are shown shown in Figure C.21. C.2l. bn
=
5. The The solution is 1 +50t u(x,t) = ( 20
2) + ~ ~bncos (cn1l't) 25 cos (n1l'X) 25 ' n=l
where
= 2(2cos(n1l"2~ ~ 1- (-It),
n = 1,2,3, .... n1l' Fifty Fifty snapshots of the the solution are shown in Figure C.22. The solution gradually moves are free both ends ends are free to to move move vertically. vertically. moves up; up; this this is is possible possible because because both bn
565 565
Appendix C. Solutions to odd-numbered exercises
0.2
c
CD
E CD
~
'i5.. In 'i5
- 0.2
o
5
15
10
20
25
x
Figure C.21. Fifty snapshots vibrating string string from from Example Example 7.4, 7.4, Figure C.21. Fifty snapshots of of the the vibrating with right end with the the right end free. free.
'E CD E CD
<..>
a. 0.2
5
15
10
20
25
x
Figure C.22. C.22. Fifty Fifty snapshots snapshots of of the the vibrating string from Figure vibrating string from Example Example 7.4, 7.4, with both ends with both ends free. free.
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
566 7. The The solution solution is 7. is
(2n-1)1rX)
u(x, t) = ~ an(t) sm 00
•
2
'
where where
C2(2n4~1)21r2)cos(C(2n;1)1rt) + c2(2n4~1)21r2'
an(t) = (bn-
2(_1)n+l bn
= 125(2n -
en
= -,-----"--:-(2n - 1)1r
1)21r 2 '
4g
Snapshots of the the solution are graphed graphed in Figure C.23, be compared compared to Snapshots of solution are in Figure C.23, which which should should be to Figure 7.9. Figure 7.9.
;
,- -
;
,,' .... .;.,.;:~:~ .. -;~ .... ""
..
0°
,.
'
. . .>;>' ;
;
;
-,
;
-,_._,- - _._,-,-,-
-,
;
........................................ . t=O - _. t=5e-05 t=0.0001 t=0.00015
.-:>;"
.0,'.,, .--';'
;
;{'-
°o~------~------~------~------~------~ 0.2 0.4 0.6 0.8 x
Figure C.23. displacement of the bar in Example Example 7.6, C.23. Snapshots Snapshots of the displacement taking into account the effect effect of of gravity. gravity.
Section 7.3 7.3 Section 1. The 100/20 = = 5, = T/60, and the the 1. The fundamental fundamental period period is is li/c 2l/c = 0.2. 0.2. Using Using h h = 100/20 5, Dt Dt = T/60, and RK4 we obtain solution shown shown in in Figure C.24. RK4 method, method, we obtain the the solution Figure C.24.
3. 3.
(a) Since Since the the initial disturbance is is 24cm from the the boundary boundary and and the wave speed speed (a) initial disturbance 24cm from the wave is 400cm/s, it will take 24/400 24/400 = = 0.06 0.06ss for for the reach the the boundary. boundary. will take the wave wave to to reach is 400 cm/s, it
Appendix C. Solutions Solutions to odd-numbered exercises exercises Appendix C. to odd-numbered
567
O . 2 r-------~-------,--------~------~------~
'E Q) E
g
Ci
'6
-0 . 2 ~------~--------~------~--------~------~
o
0 .2
0.4
0.6
x
0.8
Figure C.24. The computed computed solution of the wave equation equation in Exercise Exercise 7.3.1. Shown are 30 snapshots, snapshots, plus the initial displacement. displacement. Twenty 7.3.1. Shown Twenty subintervals subintervals space and 60 time steps were used. in space used. (b) The IBVP is
aatu 2
2
2 -
a u = 0, 2
< x < 50, u(x,O) = 0, 0 < x < 50, c ax 2
0
t
> 0,
au at (x,O)
= ,(x), 0 < x < 50, u(O, t) = 0, t > 0,
u(50, t)
= 0, t > 0,
with c = 400 statement of the exercise. 400 and ,7 given in the statement = 50/80 (c) Using piecewise linear finite elements and the RK4 method (with h = and .6.t = 6 .• 10), we ~ t < ~ 0.06. 0.06. At = 10~44 ), we computed the solution over the interval 0 < snapshots are shown in Figure C.25, which shows that it does take 0.06 0.06 Four snapshots C.25, which wave to reach the boundary. (The reader should notice the seconds for for the wave spurious "wiggles" "wiggles" in the computed solution; solution; these are due to the fact that the true solution is not smooth.) 5. It is not possible to obtain a reasonable numerical solution using finite elements. In Figure C.26, = 2.5.10= 3.75-10~ 3.75.10- 44 .• C.26, we we display the result obtained obtained using h — 2.5-10~33 and .6.t At = Three snapshots are shown. shown. 7. (a) The The weak form is
{a u au dV} It f(x, t)v(x) dx p(x) (x, t)v(x) + k(x) ax (x, t) dx (x) dx =
1 £
o
2
(Jt2
0
Appendix C. C. Solutions Solutions to to odd-numbered exercises Appendix odd-numbered exercises
568
0.1r-----~------~------~------~----__,
'E Q) E Q) u
CIl
a. rn '0
10
20
x
40
30
50
Figure Figure C.25. C.25. The computed solution solution of of the wave equation in Exercise 7.3.3. 7.3.3. 1.2~------~------~------~------~------~
0.8
'E Q) E Q) u
CIl
a. rn '0
0.4
0.2
0.6
0.8
x
Figure C.26. Figure C.26. The computed solution solution of of the wave equation in Exercise 7.3.5. 7.3.5.
for > to to and and vv E € V, where for t 2: where
v = {v E C
2
[O, £]
v(O)
= O} .
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
569
homogeneous bar: (b) The system of ODEs has the same form as in the case of a homogeneous _ d2 u _ _ M dt 2 + Ku = f(t), except that the entries in the mass matrix are now now
1t = 1t
Mij
=
p(x)cPi(x)(/Ji(x)dx, i,j = 1,2, ... , n,
now and the entries in the stiffness stiffness matrix are now Kij
k(x)
d:: (x) ~!i (x) dx,
i,j
= 1,2, ... , n.
The components of are unchanged (see Section Section 7.3.2). 7.3.2). The components of the the vector vector f are unchanged (see
Section 7.4 1. 1.
=
(a) c/80. Therefore (a) The The fundamental frequency frequency is c/(2i) el(2Ji.) = e180. e 80 = 500 c = 40000 cm/s
'*
(or 400m/s). (b) T/p, where T is the tension. The is 10/40 = = 0.25g/cm, 0.25 g/cm, (b) We We have have ec2 = T/p, whereTisthetension. The density density p pis so T is 4 • 1088 dynes (gcm/s 22 ). 4.10 ). (c) The other resonant frequencies are the integer multiples of the fundamental fundamental frequency: 1000, 1500, 2000,... 2000, ... Hz. 1000,1500, 3. Snapshots of the solution are shown in Figure C.27. C.27. The reason that resonance does not occur is that the point source is placed at a point of the string that does point). (In the formula (7.28), not move under the fourth standing wave (a fixed point). sin (rmra) — sin (27r") (2vr) = 0.) = sin = 0.) sin (m7ra)
Section Section 8.1 8.1 1. The heat equation equation for a heterogeneous heterogeneous medium is 1.
p(x)e(x)
au at - V· (K(X)VU) = 0,
x E
n, t > to·
3. For this example, [V.F= [
10
180
F·n=1.
5. Using for scalar obtain 5. Using the the product product rule rule for scalar functions, functions, we we obtain 3
v· (vVu) = L i=1
a
a
aXi [v o:J
3 ov OU = '"" L...J ox; ox;
i=1
02U
'""-2 3
+ v L...J ox'
= Vv· VU+ v~U.
i=1'
Appendix C. Solutions odd-numbered exercises exercises Appendix C. Solutions to to odd-numbered
570
co
-7
2.5 x 10
= 1044
2 1.5
0.5 Oli;;:::------:::::I.
-0.5 -1
-1.5
-2 -2.5 ' - - - - - 0......- 2 - - - - 0..... 4 - - - - 0.....6 - - - - 0....8 - - - - - ' 0 x
Figure C.27. C.27. Solution (7.26) with with an oscillatory forcing forcing term Figure Solution to to (7.26) an oscillatory term (see (see ExerExerfrequency of forcing term term is is 1044, fourth natural natural frequency. cise 7.4.3). cise 7.4.3). The The frequency of the the forcing 1044, the the fourth frequency. However, does not not exhibit exhibit resonance resonance since since the the point point source source is is placed placed at at However, the the solution solution does fourth standing aa fixed point of fixed point of the the fourth standing wave. wave. 7. Suppose Suppose u, u,vv G Then 7. E C™(O). C~ (n). Then (Lmu,v)
=-
In v.6.u
= { Vu· Vv
in
=
In Vu·Vv
={
ian
=-
u
In
-1an
v : (Green's first identity)
av - { u.6.v (Green's first identity) an
in
u.6.v
— (u,Lmv). (u,Lmv). = The boundary terms terms vanish vanish because because the the product product that that forms the integrand integrand is is zero zero The boundary forms the over the the entire boundary. For For example, over entire boundary. example,
{ v
au
ian an since on F2. since v = 00 on on FI fl and and du/dn au/an = 00 on f z.
=0
Appendix C. Solutions Solutions to to odd-numbered Appendix C. odd-numbered exercises exercises
571 571
Section 8.2 1. 1.
(a)
(b) The The graphs graphs of of the error are given in C.28. (b) the error are given in Figure Figure C.28.
X10-3
o
0
0.2
2
Figure Figure C.28. C.28. The error in approximating approximating ff(x,y) ( x , y ) by thefirst4 the first 4 terms (top) and the the first first 25 terms terms (bottom) (bottom) of of the (See Exercise and the double double Fourier Fourier sine sine series. series. (See Exercise 8.2.1') 8.2.1.) 3. The solution solution is 3. The is given given by by 00
u(x)
=L
00
L
m=l n=l
~mn sin (m7rXl) sin (n7rX2),
I\mn
cmn are the coefficients where the Cmn coefficients from Exercise 1 and
572
5.
Appendix C. C. Solutions Solutions to odd-numbered odd-numbered exercises
(a) The IBVP is ThelBVPis pc
au at - KC!..U = 0.02, u(x,O) u(x, t)
= 5, = 0,
x
E
n, t > 0,
x E n, x E
an, t > O.
The domain n the rectangle $1 is the rectangle
{x E R2 : 0 < Xl < 50, 0 < X2 < 50} . (b) The solution is u(x, t)
~ L...J ~ amn(t) sm . (m7rXl) = L...J -w sm. (n7rX2) 50 ' m=l n=l
where a mn (t)
=
b
=
mn
Cmn Amn
(b mn _ Cmn \
)
KAmn
20 (-1
e -t
+
Cmn
\
K~mn
,
+ (_I)m) (-1 + (_I)n) , mn7r 2
2 (-1
=
+ (_1)m) (-1 + (-It) 25mn7r 2
=
(m 2
'
+ n 2 )7r2 2500
(c) The steady-state temperature temperature Us us (x) satisfies the BVP -KC!..U
u(x)
= 0.02, x E
= 0,
x E
n,
an.
The solution is Us (X)
~~ Cmn . (m7rXl) . (n7rX2) = L...J L...J KAmn sm -w sm 50 . m=l n=l
(d) The maximum difference between between the the temperature temperature after after 10 10minutes minutes and and the th steady-state steady-state temperature temperature isisabout about 11degree. degree. The Thedifference differenceisisgraphed graphedininFigure Figur C.29. C.29. 7. The minimum temperature seconds. temperature in the plate reaches 4 degrees Celsius after 825 825 seconds 9. The leading edge of of the the wave wave isis initially initially 2/5 2/5 units units from from the the boundary, boundary, and and the th wave travels at a speed of of 261..;2 261-\/2 units units per per second. second. Therefore, Therefore,the the wave wavereaches reaches the th boundary after after ..;2/1305 V^/1305 ~= 0.00108 0.00108 seconds. seconds. Figure Figure 8.6 8.6shows showsthe thewave waveabout aboutto t< reach the boundary after 1010~33 seconds. seconds. 11. The difficult task is is to compute compute the the Fourier Fourier coefficients coefficients of of the the initial initial displacement displacemen ill. If If we write ?jJ. write 00
?jJ(x)
00
= 2: 2: bmn sin (m7rX l) sin (n7rX2), m=l n=l
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
573
60
o
0
x,
Figure Figure C.29. C.29. The difference difference between the temperature temperature in the plate after after 10 minutes and the steady-state temperature. temperature. (See (See Exercise 8.2.5.) then we find then we find that that 0,
m=/=-n,
ain 2 (mrr/Z) lZ50m 2 rr2 ,
m=n.
We then have that 00
u(x,t)
=L
00
L amn(t)sin(mll'x )sin(mrxz), 1
m=l n=l
where amn(t) a m n (t) satisfies satisfies where
dZa
mn dt2 + Amn C
2
= 0, amn(O) = bmn , damn (0) = o. amn
dt
The result is
amn(t)
={
m=/=-n,
°b'mm cos (n--) Cy Ammt , m=n.
Thus the solution is 00
u(x, t)
=
L bmm cos (CVAmmt) sin (mll'Xl) sin (mll'xz). m=l
Four snapshots of uu are shown in Figure C.30.
Appendix Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
574
X10-4
-1 1
-1
1
X10-4
-1
-1
1
1
Figure C.30. Four snapshots of the solution to Exercise 8.2.11: t = = 00 Figure C.30. (top left), left), t = T/8 T/8 (top (top right), right), tt = T/2 T/1 (bottom (bottom left), left), t = 5T/8 5T/8 (bottom (bottom right). right). The (top The solution is periodic with period T. T. 13. 13. The The solution solution is is
=L 00
u(x)
. . L 4sin(m7r/2)sin(mr/2) + 7r sm (m7rxI) sm 00
(m 2
n 2)
2
m =l n=l
The The graph graph of of uu is is shown shown in in Figure Figure C.3l. C.31. 15. 15.
(a) IBVP is (a) The ThelBVPis pc
au at - K~U = 0.02,
x E 0, t
> 0,
u(x, 0) = 5, x E 0, u(x,t) =0, xEr}Ur 4 , t>O,
au (x, t) = an
0, x E r2 U r3, t
> O.
°
The domain domain O is is the the rectangle The rectangle
{x E R2
: 0 < Xl
< 50, 0 < X2 < 50} .
(n7rX2).
Appendix C. to odd-numbered Appendix C. Solutions Solutions to odd-numbered exercises exercises
575
0.5
o -0.5 1
o
0
Xl
Figure C.31. 8.2.13. C.31. The solution to Exercise 8.2.13. (b) The solution is (b) The solution is
u (x, t)
=~ ~ (t)·sm (2m 100 -1) JrX l) . (2n -1) JrX 2) ~ ~amn sm 100 ' m=l n=l
where amn(t)
= (b mn _
C~n
) e-I
80 = .,......----0-.,......----,-_o_ (2m - 1)(2n - 1)Jr 2
Cmn
8 = .,----,.,.----:-.,...,.....---;---;:2
mn
C~n
,
KAmn
bmn
A
+
~Amn
25(2m - 1)(2n - 1)Jr _ «2m - 1)2 + (2n - 1)2)Jr 2 10000 .
(c) the BVP BVP (c) The The steady-state steady-state temperature temperature us(x) us(x) satisfies satisfies the -,,~u
u(x)
au
= 0.02, x E n, = 0, x E fl U f4,
an (x) = 0,
x E f2 U fa.
The The solution solution is is Us
(
_
~~
x) - ~ ~ m=l n=l
Cmn
"Amn
•
sm
(2m-l)JrXl) . (2n-l)JrX2) 100 sm 100 .
Appendix exercises Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises
576 576
(d) tem(d) After After 10 10 minutes, minutes, the the temperature temperature uu is is not not very very close close to to the the steady-state steady-state temthe difference difference u(x, u(x, y, 600) 600) —u -u.(x, y) graphed (Note: perature; the (x, y) is graphed in Figure C.32. (Note: s The The temperature temperature variation variation in in this this problem problem may may be be outside outside the the range range in in which which the linear linear model model is is valid, valid, so so these these results results may may be be regarded regarded with with some some skeptiskeptithe cism.) cism.)
o -2
-4 -6 -8 -10 60 60
o
0
Figure C.32. C.32. The The difference difference between between the the temperature temperature in in the the plate plate after after 10 10 Figure minutes minutes and and the the steady-state steady-state temperature. temperature. (See (See Exercise Exercise 8.2.15.) 8.2.15.)
Section Section 8.3 1. are 1. The The next next three three coefficients coefficients in in the the series series for for g9 are C40 ~
-0.0209908,
C50 ~
3. 3. The The solution solution is is
0.0116362, 00
u(r, B)
=
C60 ~
L amoJO(SOm r ), m=l
where where
amo
Cmo = -2-' sOm
-0.00722147.
Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises Appendix
577
f(r, 9) of the Cmo cmo are the coefficients coefficients of /(r, 0) = 1 -— r, and the SOm som are the positive roots of Jo. A direct calculation shows that Jo. alO
a20 a30 a40
a50 a60
== 0.226324, == -0.0157747, == 0.00386078, == -0.00148156, == 0.000716858, == -0.000399554.
The solution (approximated by by six terms of the series) is graphed in Figure C.33. C.33.
0.25 0.2 0.15 0.1 0.05
o 1
- 1 -1
Xl
Figure C.33. C.33. The The approximation approximation to solution u, u, computed computed with six terms of of the (genemlized) (generalized) Fourier Fourier series (see Exercise Exercise 8.3.3). 8.3.3). 5. The The solution solution is 5. is
00
u(r,9)
=L
amoJO(SOm r ),
m=l
where amo
Cmo = -2-' sOm
the f(r, 9) Jo. the CmO Cmo are are the the coefficients coefficients of of /(r, 0) = r, and and the the 80m som are are the positive positive roots roots of of JQ. A direct calculation shows shows that alO
a20
== 0.141350, == -0.0371986,
Appendix C. Solutions to to odd-numbered odd-numbered exercises Appendix C. Solutions exercises
578
a30 a40 a50 a60
== == == ==
0.0106599, -0.00537260, 0.00283289, -0.00182923.
The solution solution (approximated (approximated by by six six terms terms of the series) series) is graphed in Figure C.34. The of the is graphed in Figure C.fr
0.15
0.1
0.05
o 1
- 1 -1
x,
Figure C.34. C.34. The approximation to to solution solution u, u, computed computed with with six six terms Figure The approximation terms of the (generalized) (generalized) Fourier Fourier series series (see Exercise 8.3.5). of the (see Exercise 8.3.5). 7. If If we write 00
¢(r,8) =
L cmoJO(OmOr ) m=l 00
+L
00
L (Cmn cos (n8) + dmn sin (n8)) In(Omn r ),
m=l n=l Cmn d mn are known, and where the c mn and d 00
u(r, 8, t)
=L
amo(t)JO(OmO r )
m=l 00
+L
00
L (amn(t) cos (n8) + bmn(t) sin (n8)) In(Omn r ),
m=l n=l
then then a mn (t) -- Cmne -t
bmn(t)
= dmne-t
m -- 1, 2, 3, ... , n -- 0 , 1 , 2, ... ,
m, n
= 1,2,3, ....
Appendix C. C. Solutions Solutions to odd-numbered odd-numberedexercises exercises
579
However, since since
m
= 1,2,3, ....
Therefore, 00
u(r, 0, t)
=L
aml(t) cos (O)Jl(amlr),
m=l
with a m i(£) given given above. above. After After 30 30 seconds, seconds, the temperature distribution distribution can can be be with aml(t) the temperature approximated accurately accurately using a single eigenfunction eigenfunction (corresponding to the largest largest approximated eigenvalue, AU), An), so we need only Cll ~
9.04433.
The temperature temperature distribution distribution after after 30 30 seconds is graphed graphed in Figure C.35.
10
-10 -10
Figure C.35. C.35. The The approximation approximation to to solution at tt = 30, 30, computed computed with with Figure solution uu at one term of the (generalized) (generalized) Fourier series (see (see Exercise 8.3.7). 8.3.7). nA2, the same area as a square drum of side 9. A circular drum of radius A has area 7l'A2, length "fir ^/TrA. The fundamental fundamental frequency frequency of of such such aa square square drum drum is A. The is length C
Comparing to Example 8.9, we we see see that the circular drum sounds a lower frequency frequency than a square drum of equal area.
Appendix C. Solutions Solutions to exercises Appendix C. to odd-numbered odd-numbered exercises
580
Section 8.4 1. The mesh mesh for for this this problem which the free nodes nodes are 1. The problem is is shown shown in in Figure Figure C.36, C.36, in in which the free are labeled. stiffness matrix matrix is The stiffness is labeled. The 4.4815 -1.1759 K ~ [ -1.1759 0.0000
-1.1759 4.9259 0.00000 -1.3426
-1.1759 0.0000 4.9259 -1.3426
0.0000 ] -1.3426 ' -1.3426 5.8148
the load load vector vector is is while the while F ~
0.11111 ] 0.11111 0.11111 . [ 0.11111
The resulting weights for the the finite finite element given by by The resulting weights for element approximation approximation are are given
u
= K-1F ~
0.048398] 0.044979 0.044979 . [ 0.039879
Figure C.36. mesh for for Exercise Exercise 8.4.1. Figure C.36. The The mesh 8.4-13. for this this problem is shown shown in Figure C.37, C.37, in free nodes are 3. The The mesh mesh for problem is in Figure in which which the the free nodes are labeled. stiffness matrix is 16 16 x 16 and and neither it nor the load be labeled. The The stiffness matrix is x 16 neither it nor the load vector vector will will be reproduced here. The The matrix is singular, as is is expected for aa Neumann problem. matrix is singular, as expected for Neumann problem. reproduced here.
581 581
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
The unique solution to KU KU = = F with its last component equal to zero is
u=
-0.071138 -0.047315 -0.012656 0.001458 -0.067934 -0.047349 -0.014145 0.001566 -0.065432 -0.046198 -0.015160 -0.000329 -0.062914 -0.046391 -0.016065 0.000000 16
12
4
o
0.2
0.6
0.4
0.8
Xl
Figure for Exercise Figure C.37. C.37. The The mesh mesh for Exercise 8.4.3. 8-4-3. 5.
(a) A direct calculation calculation shows that the the inhomogeneous Dirichlet problem
= f(x), u(x) = g(x),
-V'. (k(x)V'u)
x E
n,
x EOn,
has weak form
u E G + V, a(u,v)
= (f,v)
for all v E V
= eben),
Appendix C. Solutions to to odd-numbered odd-numbered exercises Appendix C. Solutions exercises
582
where G is is any any function function satisfying satisfying the the condition g(x) for for xx E an, and where condition G(x) = p(x) € dQ, and
G+V
= {G + v
: v E V} .
Substituting u = w, where where w E into the the weak weak form form yields Substituting — G + w, G V, into yields wE V, a(w,v)
= (f, v) -
a(G,v) for all v E V.
In the the Galerkin Galerkin method, method, we we replace replace V by by aa finite-dimensional finite-dimensional subspace In subspace Vn Vn and and solve solve wE Vn , a(w,v) = (f,v) -a(G,v) for all v E Vn . When using piecewise piecewise linear linear finite we can can satisfy the boundary boundary conconWhen using finite elements, elements, we satisfy the ditions approximately approximately by by taking taking G to to be be aa continuous continuous piecewise piecewise linear ditions linear function function at the the boundary boundary nodes nodes agree agree with with the the given given boundary boundary function whose values values at whose function g. For simplicity, we we take G to be zero at g. at the interior nodes. The resulting load then given given by by load vector vector is is then
Ii = (f, CPi) -
a(G, cp;), i
= 1,2,3, ... , n.
Since G is is zero zero on on interior nodes, the the quantity quantity a(G, CPi) is nonzero nonzero only only if if (free) Since interior nodes, &) is (free) node ii belongs to a triangle adjacent to the boundary. boundary. (b) The regular triangulation triangulation of the the unit square having 18 18 triangles has only four each one belongs to to triangles triangles adjacent adjacent to to the the boundboundinterior (free) nodes, and interior (free) nodes, and each one belongs that every every entry entry in in the the load load vector vector is modified (which (which is is not ary. This This means means that ary. is modified the typical typical case). Since I/ = = 0, 0, the the load load vector vector ff is is defined defined by by the case). Since
= -a(G,CPi), i = 1,2,3,4.
f; The load vector is f
==
0.22222222222222] 1.55555555555556 0.22222222222222 ' [ 1.55555555555556
while the solution solution to Ku Ku = = ff is 0.27777777777778] ...:... u -
7. 7.
0.61111111111111
[
0.27777777777778
.
0.61111111111111
(a) (a) Consider Consider the the BVP BVP -V· (k(x)Vu)
= I(x),
in 0"
(C.5)
au an (x) = h(x), x Eon.
E V = C 22(O) (TI) and applying Green's Multiplying the PDE by a test function function vv € Green's first first identity identity yields yields
1n
kVu . Vv
=
r + ionr
in
Iv
kvh for all v E V.
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
583
We will apply Galerkin's method with Vm Vm equal to the space of all continuous piecewise linear function relative to a given triangulation T. We We label the standard basis of Vm Vm as as i, 02, ... - - - ,, CPn, n, and and the as zi, Z 2 ,... . . . ,, z standard basis of CPl, CPz, the nodes nodes as Zl, Z2, Zn. (Every n . (Every node in the {CPl, CP2, ... the mesh is now now free.) Thus Vn Vn = — span span{0i,^2, • - - ,, CPn}. (j>n}Galerkin approximation Wn wn by We then define the Galerkin Wn
E Vn , a( Wn, CPi)
= (f, CPi) +
r
Jon
khcpi, i
= 1,2, ... ,m.
Writing Writing m
Wn
=L
(JjcPj
j=l
and and
mXm we obtain obtain the the system system Kw Kw = = F. The stiffness matrix K E GR R mxm js given given by we The stiffness matrix K is by
Kij
= a(cpj,CPi),
i,j
= 1,2, ... ,n,
and the load vector FER by F e Rmm by
Fi = (f, cp;)
+
r
Jon
khcpi, i = 1,2, ... , n.
The boundary integral in the the expression for Fi is zero unless Zi z» is a boundary node. the following calcula(b) The compatibility condition for (C.5) is determined by the tion: tion:
In f = - In V·
(kVu)
__ r k OU = - r kh. Jon
- Jon an (c) Now consider (C.5) (C.5) with k(x)
= 1,
f(x)
3
= XlXZ + 4'
hex)
3xi
= -5'
where n O is the unit square. We will produce the finite element solution using 16x16 Rl6Xl6 the regular grid with 18 18 triangles (16 (16 nodes). The stiffness stiffness matrix K E eR is singular, as should be expected, and there are infinitely many solutions. The
Appendix C. Solutions Solutions to to odd-numbered odd-numberedexercises exercises Appendix C.
584
unique solution uu with with last last component component equal equal to to zero zero is unique solution is 0.3622 0.2984 0.1344 -0.0843 0.3728 0.3302 0.2081 0.0255 0.3803 0.3447 0.2334 0.0591 0.3665 0.3257 0.1878 0.0000
u=
9.1 Section 9.1 1. The The complex complex Fourier Fourier series series of of f/ is 1. is 00
n=-oo
where Co CQ = = 2/3 2/3 and and where Cn
=-
2(-1t
n
n 2 7r 2 '
= ±1, ±2, ....
approximating f/ by partial Fourier series are shown in Figure C.38. C.38. The errors in approximating by a partial 3. The The complex complex Fourier Fourier coefficients coefficients of of f/ are are 3.
=
Cn
(-It+ 1 sin (1) n7r -1
.
The magnitudes magnitudes of the error in approximating partial Fourier series are shown approximating f/ by a partial in Figure C.39. C.39. 5. 5. Let Let
00
g(x) = ao + L
(an cos
(n;x) + b sin (n;x)) , n
n=l
where
ao
=
;£ Ii g(x)dx,
!:
-i
l g(x) b = l Ii g(x)
an = n
-l
(n;x) dx, n= 1,2,3, ... , sin (n;x) dx, n= 1,2,3, ... , cos
Appendix C. Solutions Solutions to to odd-numbered Appendix C. odd-numbered exercises exercises
585 585
0.04r-------r-----r------r------,
-0.02
----.&....----. . .0.5. . .-------1 0
-0.04'-----.......... -1 -0.5
0.04r-----""T'"--------Il(~----..,..----.....,
0.02
Orvr-~--------------------------~~~vJ~ -0.02 -0. 04 '--_ _ _........._ _ _- - L_ _ _ _.........._ _ _....J -1 -0.5 0 0.5 x_ 0.04,......-----.r-------l.t-------,.-----.., 0.02
o~------------------------------------~ -0.02
..
-0.04'-------·.....----...I·~---- ·-----~ -1 -0.5 0.5 o
x
Figure C.38. The error in approximating approximating ff(x) - x22 by (x) = = 1— by its partial complex Fourier series with N = 10 10 (top), N = 20 20 (middle), (middle), and N = 40 (bottom). complex 40 (bottom). (See Exercise 9.1.1.) (see (6.25)), let (see (6.25)), and and similarly similarly let
hex) = Po + f: (Pn cos (n;x) + qn sin (n;x)) . n=l
The complex complex Fourier Fourier coefficient coefficient of! given by by The of / is is given Cn
=~ U
Ii !(x)e-
i1rn
;r;/i
dx.
-i
For > 0, For n n > 0, we we have have Cn
= 2~
J: (g(x)
1(1 Ii
="2 £
_£
+ ih(x))
(n;x) - isin (n;x)) dx .1 g(x)sm (n7rx) g(x) cos (n7rx) -g- dx - l£ -g- dx (cos
Ii . -i
Appendix C. Solutions Appendix C. Solutions to to odd-numbered odd-numbered exercises exercises
586
0.5
o """-
-1
o
-0.5
0.5
y
0.5
o~ -1
-0.5
o
0.5
o
0.5
x
0.5
0-
-1
-0.5
-
x
lx Figure C.39. approximating f(x] =e by Figure C.39. The The magnitude magnitude of of the the error error in in approximating f(x) = eix by its 10 (top), (top), N its partial partial complex complex Fourier Fourier series series with with N N = = 10 N = 20 20 (middle), (middle), and and N N = = 40 40 (bottom). (See (See Exercise (bottom). Exercise 9.1.3.) 9.1.3.)
+i
I: h(x)
E
cos
I:
(n;x) dx + E h(x) sin (n;x) dX)
= ~ (an + qn + i(pn -
bn )).
Similarly, if Similarly, if n > 0, 0, then then
Finally, Finally,
co
= ao + ipo·
7. 7. We We have have
n=O
n=O
e 21rij
1
_
- e 21ri j/N
-
l'
Appendix C. C. Solutions Solutions to odd-numbered exercises
587
the sum of a finite geometric series: using the formula for the
l6 27ry Moreover, since ei8 is a 27r-periodic function of 8, = 9, we we see see that ee27rij = 1 and hence that that N-l
L
= O.
(e7rijn/N)2
n=O
On the other hand, by Euler's formula,
~ + ~.. L...J (;7rjn/N)2 e =~ L...J ( cos (jn7r) N sm (jn7r))2 N n=O
n=O N-l
=L
(cos
(j~7r)
_ sin
e~7r) -
y:
2
2
(j~7r)
+ 2icos
(j~7r) sin (j;))
n=O
=
y:
cos
2
n=O
2
e~7r)
n=O
N-l
+2i
sin
L
cos
(j~7r) sin (j~7r).
n=O
Since this expression equals zero, we we must have
y:
cos
2
e~7r) -
y:
n=O
sin
2
e~7r) = 0,
n=O N-l
L cos (j~7r) sin (j~7r) = O. n=O
we set out to prove. prove. The first equation The second equation is one of the results we can be be written, using the identity sin sin22 (8) (9) = = 1I -— cos cos22 (8), (0), as written, using the trigonometric trigonometric identity can
y:
cos
2
e~7r) -
n=O
£ (1 -
cos
2
e~7r)) = 0,
n=O
which simplifies to N-l
2
or or
L
cos
2
e~7r) = N,
n=O
N ~COS2 e~7r) -"2' n=O
which is the desired result. The result
~ sin e~7r) = ~ 2
n=O
then follows.
Appendix C. Solutions to to odd-numbered Appendix C. Solutions odd-numbered exercises exercises
588
Section 9.2 1. 1. The The graph graph of of {|F {IFnI} is shown shown in in Figure Figure C.4D. C.40. n|} is
0.5.------......------r-------r--......- -.. . .
0.4
0.3 LL
c::
0.2
0.1
~O
5
10 n
20
15
magnitude of {Fn} from Figure C.40. The magnitude of the sequence {Fn} from Exercise Exercise 9.2.1. 9.2.1. 3. As shown in text, 3. As shown in the the text, en
where where
5 {F {Fn}~~-16 n}* =_16
== Fn, n = -16, -15, ... ,15,
is the the DFT DFT of {/j}jl_ is of {Ii }J~-16' 16, fj
= f(xj),
= -1 + :6' j = -16, -15, ... ,15
Xj
f( -1) = = /(!))• f(l)). Therefore, we we can estimate {f(Xj)} by taking the inverse (note that /(—I) {f(xj}} by DFT (using the inverse FFT) of of {en}. results are are shown shown in in Figure Figure C.41. DFT (using the inverse FFT) {cn}. The The results C.41. 5. We have
"A
N-l
~ n=O
N-l N-l
e2rrijn/N _ n -
..!. " " -2rrimn/N 2rrijn/N N ~ ~ame e n=O m=O N-l N-l
_ 1 ""
-
N~~ame n=O m=O
2rri(j-m)n/N
N-l {N-l } _ ..!. " "2rri(j-m)n/N N~am ~e
-
n=O
m=O
If jj — ra, then If = m, then N-l
L
n=O
N-l
e 2rri (j-m)n/N
=
2: 1 = N, n=O
.
589
Appendix Appendix C. Solutions to odd-numbered exercises 0.5.....-----~----"""T""----~-------.
-0.5~----~----~~----~----~
-1
o
-0.5
0.5
x
Figure C.41. The The function = x(l x(l -— xx22)) (the (the curve) curve) and and the the estiestiFigure C.4l. function f(x) f(x) = mates of f(xj) f(xj) (the circles) computed computed from from the Fourier coefficients coefficients of of ff mates of (the circles) the complex complex Fourier (see Exercise 9.2.3). (see Exercise 9.2.3). while if ^ m, is aa finite geometric series: series: while if jj =/= m, then then this this sum sum is finite geometric N-l
N-l
Le
21ri (j-m)n/N
= L (e 2rri (j-m)/N)n
n=O
n=O
(e 21l"iu-m)/N)N _ e21l"i(j-m)/N e 27Ti (j-m)
1
1
-
e 2rri (j-m)/N
1
-
1
=0.
Therefore, Therefore, N-l ~
am L.J e
2rri(j-m)n/N_
{
m=j, m =/=j,
-
n=O
and so so we obtain and we obtain N-l
..!. N-l
{N-l
m=O
n=O
~ A e 21rijn/N _ ~ L.J n - N L.J am n=O
1 -Naj N
= =aj, as desired.
~
L.J e
27Ti(j-m)n/N
}
Appendix C. C. Solutions Solutions to odd-numberedexercises exercises Appendix to odd-numbered
590
7. Below we we show the exact exact Fourier Fourier sine I, the the coefficients coefficients estimated by 7. Below show the sine coefficients coefficients of of /, estimated by the DST, and and the error. Since Since both and the estimated a are zero for n the DST, the relative relative error. both a an and the estimated an are zero for n n even, we show only only an an for for nn odd. odd. even, we show
n
estimated
an
1
an
relative error
1
6.2121 . 10- 6
1
2.5801.10-
2.5801 .10-
3
9.5560 . 10- 3
9.5511 . 10- 3
5.1571 . 10- 4
5
2.0641 . 10- 3
2.0555 . 10- 3
4.1824 . 10- 3
7
7.5222 . 10-
4
4
1.7340 . 10- 2
9
3.5393 . 10- 4
3.3531 . 10- 4
5.2608 . 10- 2
11
1.9385 . 10- 4
1.6778 . 10- 4
1.3448.10- 1
13
1.1744.10-
4
8.0874.10- 5
3.1135.10- 1
15
7.6448 . 10- 5
2.4279 . 10- 5
6.8241.10- 1
7.3918 . 10-
9. We We have 9. have N-1
(imr) N = 4"" f N-1 N-1
. 2 "~ Fn sm
~~ m
n=l
(mn7r) . (in7r) N
. -;;;- sm sm
n=l m=l N-1
N-1
m=l
n=l
= 4 Lim L sin (m;7r) sin e~7r). By the the hint, hint, By
~ . (mn7r) . (in7r) ~ sm -;;;- sm N n=l
is either N N/2 or 0, 0, depending on whether whether m ra = = ij or not. It It then then follows immediately is either /2 or depending on or not. follows immediately that that
~
. (in7r) N N = 4"2fJ = 2NfJ,
2 ~ Fn sm n=l
which wanted to which is is what what we we wanted to prove. prove.
Section 9.3 1. The The partial sine series, series, with 50 terms, is shown shown in in Figure C.42. It It appears appears 1. partial Fourier Fourier sine with 50 terms, is Figure C.42. that the to the the function F satisfying satisfying that the series series converges converges to function F F(x) = x - 2k, -2k - 1
< x < 2k + 1,
k = 0, ±1, ±2, ....
The period of this this function is 2. The period of function is 2. 3. 3.
(a) The extension fIOdd continuous on on [-£, [—I, £] t] only only if if /(O) = 0. Otherwise (a) The odd odd extension is continuous I(a) = a. Otherwise, 0dd is fodd has aa jump discontinuity at = 0. IOdd has jump discontinuity at xx = o. (b) even extension is continuous continuous on £] for for every (b) The The even extension lev fevenen is on [-£, [—•£,•£] every continuous continuous I/ [0, f\ -t -»- R. R. [a,£]
:
591 591
Appendix to odd-numbered Appendix C. C. Solutions Solutions to odd-numbered exercises exercises
1.5r-----.----....-----.----....----.........-----, 1
-1.5~---~------~------~----~-----~----~
-3
-2
-1
0
2
3
x
Figure partial Fourier Fourier sine sine series series (50 terms) of of ff(x) Figure C.42. C.42. The The partial (50 terms) (x) = = x. x. 5. The quarter-wave sine series of f/ : [0, [0, f]i] -+ —>• RR isis the the full full Fourier Fourier series series of ofthe the function function / : [—2i, 2i] -+ -» R R obtained by first =£ t (to [-2£,2f] obtained by first reflecting reflecting the the graph of f/ across the line x = (to obtain a function defined on [0, and then taking the [0, 2£]) 21]) and the odd extension of the the result. That is, is, /is is defined by by
1:
1
f(x), f(x) = f(2£ - x), { -f(-x), -f(x + 2£),
°:-: x:-::: x:-::: £, 2£, f<
-£:-::: x < 0, -2£:-::: x <-£.
Since this function is odd, its full Fourier series has only sine terms (all of the cosine coefficients the even sine terms also coefficients are zero), and because of the other symmetry, the drop out.
Section 9.4 1. 1. The Fourier series of f/ converges pointwise to the function F defined by x = ±(2k - 1)71", k = 1,2,3, ... , (2k - 1)71" < X < (2k + 1)71", k = 0, ±1, ±2, .... 3. There are many such functions f, /, but they all have one or more discontinuities. An example is x, 0:-::: x:-::: ~, f(x) = { x+ 1, ~<x:-:::l.
Appendix C. Solutions to to odd-numbered Appendix C. Solutions odd-numberedexercises exercises
592
5. If If {fN} {/N} converges [a, b], 6], then then it it obviously obviously converges (after 5. converges uniformly uniformly to to /f on on [a, converges pointwise pointwise (after all, (E [a, 6], between ( x ) converges to the maximum maximum difference, difference, over over x E [a,b], between /N(X) fN(X) and and ff(x) converges to all, the zero, so each converge to uniform zero, so each individual individual difference difference must must converge to zero). zero). To To prove prove that that uniform mean-square convergence, define convergence implies mean-square define MN
= max{lf(x) -
fN(X)1 : a:::; x :::; b}.
Then Then
lb
Jlb
If(x) - fN(X)12 dx :::;
M'j. dx
= MNVb-a. --+ 00 as N N —> --+ oo, 00, which gives the the desired result. By hypothesis, MN —> £] —>• --+ R is continuous. Then even, periodic extension of f/ 7. Suppose /f : [0, [0,^] Then ffeven, even, the even, to R, is defined by feven(x)
={
f(x - 2k£), f( -x + 2k£),
2k£ :::; x < (2k + 1)£, k (2k - 1)£ :::; x < 2k£, k
= 0, ±1, ±2, ... ,
= 0, ±1, ±2, ... .
Obviously, continuous on every interval interval (2k£, (2ki, (2k (Ik + 1)£) 1)1) and and ((2k ((2k — Obviously, then, then, ffeven is continuous on every even is 1)1,2k£.), except possibly and (2k -1)£, — l)i, k = — 0, 0, ±1, 1)£, 2k£), that that is, is, except possibly at at the the points points Ikt 2k£ and ±1, ±2, ±2, .... We have1 1H7_ lim
x~2W-
feven(x)
= =
lim
x~2W-
f(-x+2k£)
lim f(x)
x~o+
= f(O) and and lim
x~2kl+
feven(x}
=
lim
x~2kl+
f(x - 2k£)
= lim f(x) x~o+
= f(O). Since = /(O) is continuous continuous at at xx = 2n£. Since ffeven(2k£) f(O) by by definition, definition, this this shows shows that that ffeven 2/';,£. even(2kf) even is A similar calculation shows A similar calculation shows that that lim
x~(2k-l)l-
feven(x) =
lim
x~(2k-l)l+
and hence that feven is continuous at x
feven(x) = feven((2k - 1)£) = feR),
= (2k -
I)£,
9. Given Given f/ : [0, [0,£]f\ --+ -»• R, R, define! define /: [-21,21] --+ -»RR by by 9. : [-2£,2£] f(x), lex) = /(2£ - x), { -fe-x), -f(x + 2£),
0:::; x:::; £, £ < x:::; 2£, -£:::; x < 0, -2£:::; x < -f.
Appendix to odd-numbered Appendix C. C. Solutions Solutions to odd-numbered exercises exercises
593
Then F :: R R —>• --+ R R to to be of f/ to to R. Assuming I/ Then define define F be the the periodic periodic extension extension of R. Assuming is smooth, the quarter-wave sine sine series series of of I/ converges converges to if F F is is piecewise piecewise smooth, the quarter-wave to F(x) F(x) if is continuous at x, continuous at x, and and to to 1
"2 [F(x-) + F(x+)] if F discontinuity at at x. is continuous, continuous, then then its its quarter-wave quarter-wave sine sine if F has has aa jump jump discontinuity x. If I/ is series converges to I/ except at xx = O. 0. If If I/ is is continuous /(O) = = 0, 0, then then except possibly possibly at continuous and and 1(0) series converges to the quarter-wave sine sine series series of converges to at every € [0, [0,.e]. i}. the quarter-wave of f/ converges to f/ at every xx E
Section 9.5 1. The The function to satisfy satisfy h(—1) = h(l), so its its Fourier coefficients decay decay 1. function h(x) hex) fails fails to h(-l) = h(l), so Fourier coefficients like l/n 1/n and and its its Fourier Fourier series series is is the slowest to to converge. the slowest converge. The The function function I/ satisfies satisfies like /(—I) = /(I), df/dx has has aa discontinuity at rex = = a 0 (and (and the the derivative of fIper also f( -1) = f(l), but but dl/dx discontinuity at derivative of per also has discontinuities discontinuities at at xx = ±1). ±1). Therefore, Therefore, the has the Fourier Fourier coefficients coefficients of of f/ decay decay like like 1/n2. Finally, its first first derivative continuous, but its second second derivative derivative 1/n2. Finally, ggper and its derivative are are continuous, but its per and has aa jump at xx = ±1, ±1, so so its coefficients decay like 1/n 1/n33.. The has jump discontinuity discontinuity at its Fourier Fourier coefficients decay like The Fourier series of g is the fastest to converge. Fourier series of 9 is the fastest to converge. 3. Figure shows the the /, its partial series with with 21, 21, 41, 41, 81, 81, and and 161 161 terms, terms, 3. Figure C.43 C.43 shows I, its partial Fourier Fourier series and the line near xx = is indeed and the line yy = = 1.09. 1.09. Zooming Zooming in in near =a 0 shows shows that that the the overshoot overshoot is indeed about 9%; 9%; see see Figure about Figure C.44. C.44. 1.5~--------~---------r--------~--------~
>- 0.5
21 terms - _. 41 terms 81 terms 161 terms
-0.5
o x
0.5
Figure C.43. C.43. The The function its partial series Figure function f from from Exercise Exercise 9.5.3, 9.5.3, its partial Fourier Fourier series terms, and the line line yy = 1.09. with and 161 with 21, 21, 41, 41, 81, 81, and 161 terms, and the = 1.09.
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numberedexercises exercises
594
1.1051--..-----r----...----~====::::c:::===;_,
21 terms 41 terms 81 terms 161 terms
1.1
1.095
>.
1 .09 - - - - - - - - - - - - - I,-:- - - - - - - -I ...\ - - - - - - - - - - - ii' , 1.085
I
,
I
i i
I
-
,,
1.08 I
I
i
o
0.02
0.06
0.04
x
0.08
0.1
Figure C.44. C.43. Figure C.44. Zooming in on the overshoot in Figure C.43.
Section 9.6 1. 1.
(a) (a) The The infinite infinite series series
1
00
Ln
2
n=l
87 2 converges to a finite value. Therefore, {lin} and so, by by Theorem 9.36, value.87 {1/n} E G f£2 2 the L2(0, 1). the sine sine series series converges converges to aa function function in in L (0,1).
(b) Figure C.45 shows the sum of the first 100 100 terms of the sine series. The graph suggests suggests that that the the limit limit /f is is of of the the form form ff(x) ( x ) = m(l m(l -— x). The The Fourier Fourier sine sine series are series of of such such an an f/ are Cn
=2
10t
m(l - x) sin (mTx)
= 2m, n = 1,2,3, .... n~
Therefore, in order order that cn = = lin, 1/n, we must have have m = Tr/2. Therefore, Therefore, in that en we must m = ~/2. Therefore,
f(x) 87
87A A
~
= 2(1 -
x).
standard the theory that standard result result in in the theory of of infinite infinite series series is is that 00
L:k
n=l
converges converges if if kA; is is greater greater than 1. 1. This This can can be proved, proved, for for example, example, by comparison comparison with with the the improper improper integral integral
f
1
oo dx X
k'
Appendix C. C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises Appendix
595
2.-------~------_,--------T_------_r------_.
- 0 . 5 L . . - - - - - - - -.......- - -........----'----~ o 0.2 0.4 0.6 0.8
x
C.45. The partial sine series, with 100 terms, from Exercise Exercise 9.6.1. 9.6.1. Figure C.45. terms, from The infinite series 3. The
~ 00
(
1)2
fo
1
=~~ 00
series, a standard example of a divergent series. Therefore, {1/ fo} 0(j. is the harmonic series, {l/^/n} 2 1) function. tp2, , and so so the series does not converge to an an L2(0, L 2 (0,1) function. 5.
2 (a) The function ff(x) == xxkk belongs to L L2(0, 1) if and only if the integral (x) = (0,1)
is finite. finite. Provided k #^ —1/2, -1/2, we we have
1
1 X2k
dx == lim
o
r-+O+
lim
11
x 2k dx
r
1 - r 2k + 1 1 + 2k
r-+O+ 1
=={
1+2k' 00,
For kA; = -1/2, For == -1/2,
{1
10
x 2k dx ==
t
10
11
dx == lim dx x r-+O+ r X == lim -In (r)
==
r-+O+ 00.
Appendix to odd-numbered Appendix C. C. Solutions Solutions to odd-numbered exercises exercises
596
2 Thus we we see EL L2(0, 1) if and only if k > -1/2. see that f/ e (0,1) -1/2. 4 11 4 X- // ,, as computed by the (b) The first few few Fourier sine coefficients coefficients of ff(x) (x) = = a:" DST, are approximately
1.5816, 0.25353, 0.63033, 0.63033, 0.17859, 0.17859, 0.41060, 0.14157, .... 0.41060, 0.14157,.... (c) partial Fourier sine (c) The The graphs graphs of of /,j, the the partial sine series, series, with with 63 63 terms, terms, and the differdifference in Figure the two, two, are are shown shown in Figure C.46. C.46. ence between between the
6
--
4
y=x- 1/4 sine series (63 terms)
~
2
• 0.2
0.4
0.6
0.8
1
x 0.1n-------~~------~~------~~-------,---------n
o
-0.1U.------~--------~------~--------~------~
o
0.2
0.4
0.6
0.8
1
x
1 4 Figure C.46. C.46. The together with with its its partial partial Fourier / ,, together The function ff(x) ( x ) = xX-1//4 Fourier sine series (first (first 63 terms) (top), and the difference (bottom). See difference between the two (bottom). Exercise 9.6.3. 9.6.3. Exercise
7. Since Vn ~ v, there exists a positive integer N such that vn —>•
n? N ~
10
Ilv - vnll < 2'
Then, if if n, m m? > N, TV, we Then, we have have 10
10
IIvn - vmll ::; IIvn - vII + IIv - vmll < 2 + 2 = 10. Therefore, } is Cauchy. Therefore, {v {vnn} Cauchy.
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
597
Section 9.7 1. integration by parts (Green's identity): Sup1. The proof is a direct calculation, using integration pose £ eben). Cf)(fi). Then Then u, vv E pose u,
(KDU, v)
=
-10
=
1
=
10
n
= -
=
V'. (k(x)V'u) v
k(x)V'u . V'v -
{ k(x)v ~u
Jon
k(x)V'u· V'v (since v
vn
= 0 on an)
r V'. (k(x)V'v) u + Jon r k(x)u vn ~v
Jn
-10
V'. (k(x)V'v) u (since u = 0 on
an)
= (U,KDV). 3. If If t is is very very large, then we can approximate approximate u(x, -u(x, t) t] by the first term in in its its gener3. large, then we can by the first term generalized Fourier Fourier series. series. Using same notation notation as as before for the the eigenvalues eigenvalues and and the same before for alized Using the eigenfunctions of the the negative Laplacian on n, fi, we have
constant C1 c\ is the first generalized Fourier coefficient temperature The constant coefficient of the initial temperature 5: 5: In 5'1f;1 C1
= In'1f;i·
The approximation is is valid provided >' AI1 is is aa simple simple eigenvalue; is, there The approximation valid provided eigenvalue; that that is, there is is only only one linearly independent independent eigenvector corresponding to >'1. AI. Then Then all of the other terms in the generalized Fourier series decay to zero much more rapidly than does the term. the first first term.
Section 10.1 Section 10.1 1. 1. We must check that the one-point rule gives the exact values for the integral
when /(x) 1, /(x) or /(x) we have have when I(x) = = 1, I(x) = = xi, Xl, or I(x) = = x
and and
Appendix C. Solutions exercises Appendix C. Solutions to to odd-numbered odd-numbered exercises
598 With With /(x) f(x) = = #i, Xl, we we have have
1
f =
111-"'1 1 0
TR
= On the other hand,
Xl
dx 2 dx1
0
1 1
o
~f
2
(Xl - xI)
1
1
1
= -2 - -3 = -. 6
dX1
G, D= ~. ~ =~,
so the rule is exact in this case as well. By symmetry, the rule must hold for f(x) = /(x) = X2, which shows that the rule has degree of precision at least 1. 1. To show degree of precision is not greater than 1, we note that, for for /(x) f(x) = = xi, that the degree 1, we xl,
while 3. We have
On the other hand, the mapping from TR TR to T is 1
Xl
= 1 + Zl + 2"Z2,
X2
= Z2·
The Jacobian is The Jacobian is
J=[~
t],
and its determinant is 1. Therefore, with and 1. Therefore,
we have
11l-z1 ( + 1 1 1 f
T
=
9
TR
=
1
0
Zl
1
+ 2" Z2 ) z~ dZ2 dz 1.
0
This last last integral integral also also equals equals 1/8. This 1/8.
5. reader should consult Exercise (and its its solution) solution) for for an an explanation of 5. (The (The reader should consult Exercise 8.4.5 8.4.5 (and explanation of inhomogeneous Dirichlet condition is handled.) Let G be the piecewise linear how an inhomogeneous function whose whose nodal value is zero everywhere everywhere except at those boundary nodes lying the boundary boundary which which are are not not free. free. At At the the nonfree nonfree boundary boundary nodes, nodes, the the value value of of on the on
599 599
Appendix C. C. Solutions Solutions to odd-numbered exercises
G is is the assigned Dirichlet Dirichlet data. data. To To solve solve the inhomogeneous inhomogeneous Dirichlet Dirichlet problem, problem, we merely need to modify the load vector by replacing
by by
Fi
=
l
JrPi - a(G,CPi).
fa) is nonzero nonzero only if the free node nfi r\fi belongs to a triangle that The quantity a(G, CPi) also contains a (nonfree) boundary node. The point here is simply that this can easily be determined from the information in the data structure. As we we loop over we can determine, for for each, whether it contains both a free node the triangles, we node and a nonfree node. node. If it does, we we modify the load vector accordingly. accordingly.
Section 10.2 Section 10.2 1. A = U requires n -— 1 steps, and step ii uses 1. Computing Computing A = L LU
(n - i)(l
+ 2(n -
i))
= 2(n -
i)2
+ (n -
i)
arithmetic operations (this is determined by simply counting the operations in the the pseudo-code pseudo-code on page 473). The total number of operations required required is n-1
i=l
n-1
n-1
j=l
j=l
2n 3 n2 n --- 2 6 3 O(2n33/3) operation count given in the text. This agrees with the O(2n -1 The computation of L L-lb b requires n -— 1 steps, with 2(n -— i) arithmetic operations nfir stfvn. Thfi tnta.l is per step. The total is n-l
n-l
2 ~)n -1) i=l
= 2"Lj = n 2 -
n.
j=l
The final step of back substitution requires n steps, with 2( n -— i) + 2(n + 1 operations per step, for a total of of n
n-l
L {2(n - i) + 1} = 2 L j + n = n
2
•
j=l
i=l
2n2 -— n, n, which also agrees with the The total for the two triangular solves is 2n2 the count given in in the the text. text. given 3. With 3. With
cp(x)
1 = -x· Ax 2
b . x,
Appendix C. Appendix C. Solutions Solutions to to odd-numbered odd-numbered exercises exercises
600 we have have we
1
= "2 (x + y). A(x + y) 1
h· (x + y)
1
= "2x . Ax + y . Ax + "2 Y . Ay - b . x - b . Y =
+ (Ax -
b) . y
1
+ "2 Y . Ay.
The fact A is The fact that that A is symmetric symmetric allows allows the the simplification simplification 1
1
"2x . Ay + "2 Y . Ax
= y. Ax,
which was used above. above. We thus see see that that
+0
(1IyI\2) ,
with with V'0 all x, and only if = 0.
(b) (x,y) (x, y) == (y,x) (y, x) for all x,y. x, y. (b) for all (c) + ,8y, /3y, z) = a(x, + /?(y, for all x, y, ft. (c) (ax (ax + z) = a(x, z) z) + ,8(y, z) z) for all x, y, zz and and all all a, a,,8. These for These properties properties hold hold for
= X· Ay.
(X,y)A
The third would be for any any matrix matrix A, A, the the second second property The third property property would be true true for property obviously obviously A be be symmetric, first property A be be positive positive requires requires that that A symmetric, and and the the first property requires requires that that A definite. definite. 7. 7.
(0) (0) (a) With x xeD) = (X(D)) (a) With = (4,2), (4,2), the the negative negative gradient gradient is is -V'
The a = and so so The minimizer minimizer is is a = 5/14, 5/14, and X(l)
= (23, 23) . 7 14
(b) (23/7,23/14), the (b) At At x(1) x(1) = (23/7,23/14), the negative negative gradient gradient is is (-3/14,3/7), (-3/14,3/7), so so we we must must minimize minimize ,/, (X(l) _ r7,/, ( (1))) =.E...- a 2_ ~ a _ 529. '/' av,/, x 196 196 28
The minimizer is is aa = 5/6, 5/6, and and so The minimizer so x
(2)
= (87 2) 28"
The The desired desired solution solution is is (3,2). (3, 2).
= 0,0, the (c) With With xeD) x(0) = the CG CG algorithm algorithm produces produces (1) X(l) == x = (2.67456,2.34024), (2.67456, 2.34024),
with X(2) x^ = (3,2), (3, 2), the exact exact solution.
601 601
Appendix C. odd-numbered exercises Appendix C. Solutions Solutions to to odd-numbered exercises 9. If yy = x - x(O), x = y + x(O) =x x (0) , then x =y x (0) and and so so Ax = b
=> Ay + Ax(O) = b => Ay = b -
Ax(O).
We can therefore replace b by b -— Ax(O), Ax^ 0) , apply the CG algorithm to estimate y, and add x(O) x^ to get get the estimate estimate of x. x.
Section 10.3 10.3 Section 1. F(x) = 0). Then 1. Let the vector-valued function F be defined by F(x) = (f(x), (/(x),0).
'\7. (¢F) = cf/fVl. F
+ '\7¢. F = ¢ !f + !¢ f, vXl
VXl
so, by the divergence theorem,
inr¢~f + inr ~¢ f= inr'\7'(¢F)=lan ¢F.n=lan ¢fnl. VXl
VXl
If 0 ¢E G C8"([2), C'o0(fi), then the boundary integral vanishes, and we we obtain
r ¢~ - _ inr o¢ f
in
OXI -
OXI
•
The derivation for 0/ OX2 is exactly analogous. d/dx-2 3. A direct computation shows that
Ilf - 911L2 = ~ while
IIf -
911Hl
for all integers m,n,
= ~Jl + m 2 Jr2 + n 2 Jr2 for all m,n.
Thus the L2 L2 error is independent of m, n, n, while while the HI Hl error becomes arbitrarily oo. -+ 00. large as m, n —>
Section 10.4 Section 10.4 1. 1.
(a) Let u, Vh, u, vERn v € Rn contain the nodal values of piecewise linear functions 'Uh, Uh,Vh, respectively, so that n
'Uh
=
L
n
'Ui¢i, Vh
=
L
i=1
i=1
It follows that
n
n
i=1 j=1
=u·Mv.
Vi¢i.
602
Appendix C. Solutions Solutions to Appendix C. to odd-numbered odd-numbered exercises exercises (b) in the eigenvectors for problem (b) As As shown shown in the text, text, if if u u and and v v are are generalized generalized eigenvectors for the the problem T Ku = AMu, then then L LTTu u and and L LTv orthogonal in in the the Euclidean Euclidean norm. norm. But But Ku v are are orthogonal then then o = L T U • LTV = U . LLTV = U . M v = (Uh' Vh). 2 Thus and Vh orthogonal in in the the L L2 norm. norm. Thus Uh UH and VH are are orthogonal
3. 3.
(a) same method 10.5, we find that smallest eigenthe same method as as in in Example Example 10.5, we find that the the smallest eigen(a) Using Using the the pentagon pentagon is is approximately approximately 18.9. estimates obtained obtained on on five five value on the 18.9. (The (The estimates value on successively finer 19.5574, 19.0818, 19.0818, 18.9601, 18.9601, 18.9294. 18.9294. The The successively finer meshes meshes were were 21.2794, 21.2794, 19.5574, coarsest mesh is is shown in Figure Figure C.47.) C.47.) coarsest mesh shown in (b) It would would be be reasonable reasonable to to hypothesize hypothesize that that (b) It A(n) I
-t Al ,
where Ai Al = ~ 18.2 is the the smallest eigenvalue on on the the disk of area area 1. (The value value of of where 18.2 is smallest eigenvalue disk of 1. (The Ai found near of Section Al was was found near the the end end of Section 8.3.) 8.3.)
(c) Repeating calculation again of area area 1, 1, we find that (c) Repeating the the calculation again for for aa regular regular decagon decagon of we find that the smallest smallest eigevalue eigevalue is is approximately approximately 18.3. values obtained obtained on the 18.3. (The (The values on four four successively meshes were were 19.0328, 18.2755.) successively finer finer meshes 19.0328, 18.5055, 18.5055, 18.3268, 18.3268, 18.2755.) 0.8.------..----.......-----..-----, 0.6 0.4 0.2 xN
0 -0.2 -0.4 -0.6 _0.8l...-----L------'-----"-----~
-0.5
0
x
0.5
1
Figure The coarsest Figure C.47. C.4T. The coarsest mesh mesh from from Exercise Exercise 10-4-3. 10.4.3.
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Index Index aluminum, 15, 26, 163, 163, 217, 217, 276 aluminum, 15,26, 276 angle between two angle between two vectors, vectors, 55 55 area 329 area integral, integral, 329 automatic step control, automatic step control, 111 111
boundary conditions, 134 boundary conditions, 134 Dirichlet, Dirichlet, 13, 13, 34, 34, 144, 144, 168, 168, 169, 169, 197, 211, 197, 211, 280, 292, 332, 335, 339 essential, 266, 313, 388 essential, 266, 313, 388 for aa vibrating vibrating string, string, 30 30 for for the diffusion equation, equation, 20 for the diffusion 20 for the hanging bar, for the hanging bar, 24 24 for the heat heat equation, equation, 13 for the 13 homogeneous, 14, 14, 168 homogeneous, 168 inhomogeneous, 164, 169, 197, inhomogeneous, 14, 14,164,169,197, 222, 243, 273, 273, 298, 222, 243, 298, 391 391 inhomogeneous Dirichlet, 321 321 inhomogeneous Dirichlet, mixed, 13, 13,150,161, 301, 313, 313, mixed, 150, 161, 283, 283, 301, 333, 348, 388 333,348,388 natural, 266, 313, 313, 388 388 natural, 266, Neumann, 13, 27, 34, Neumann, 13, 27, 34, 152, 152, 231, 231, 273, 275, 282, 332, 335, 349, 273,275,282,332,335,349, 388 periodic, 153, 153, 246 periodic, 246 Robin, 143 Robin, 143 time dependent, 223 223 time dependent, boundary 14, 144 144 boundary value value problem, problem, 14, on aa rectangular rectangular domain, domain, 345 on 345 BVP, 14 14 BVP,
back substitution, 474 back substitution, 474 backward Euler method, 118, 263, 310 backward Euler method, 118,263,310 banded matrix, matrix, 386, 475 banded 386, 475 bar bar aluminum, 217, 276 276 aluminum, 217, chromium, 278 chromium, 278 245 circular, circular, 245 copper, 228, 243, 244, 264, 265, 278 elastic, 21, 21, 173 elastic, 173 heat equation and, and, 333 heat equation 333 heat flow flow in, in, 99 heat heterogeneous, heterogeneous, 276 276 iron, 214, 214, 217, 222, 228, 230, 236, 244, 264, 281, 281, 323 steel, 302 steel, 302 basis, basis, 50 50 orthogonal, 57 orthogonal, 55, 55, 57 orthonormal, 65 orthonormal, 57, 57, 65 Bessel function Bessel function of order order n, of n, 366 366 properties, 366 366 properties, Bessel's equation Bessel's equation of order order n, n, 364 364 of Bessel's inequality, 428 best approximation, 61, 61, 147,156,181, 147,156,181, best approximation, 182, 343 343 182, 491 in the energy norm, 491 in the energy norm, bilinear form, 180, 180, 270 bilinear form, 270 symmetric, 181 181 symmetric, body force, 23 body force, 23
C[a, b], 33, 33, 58 C[a,_b], 58 336 eC 22(ft), (n),336
ek[a, b], 33 Ck[a,b], 33 00 Co (ft), 489 489 e~(n), C£(n), 336, eben), 336, 377 377 e1 (n) , 338 338 C&(fi), carbon monoxide, monoxide, 220 carbon 220 Cauchy sequence, sequence, 451 451 Cauchy Celsius, 19 19 Celsius, chain rule, rule, 360 chain 360
607
608 change of of variables variables change in multiple integral, 471 in aa multiple integral, 471 characteristic characteristic polynomial polynomial of aa second-order second-order ODE, ODE, 82 of 82 of 69 of aa square square matrix, matrix, 69 characteristic characteristic root root of aa second-order second-order ODE, ODE, 82 of 82 Cholesky factorization, factorization, 474, 474, 496 Cholesky 496 chromium, 278 chromium, 278 co-domain, 32 co-domain, 32 coefficients coefficients discontinuous, 264 discontinuous, 264 compatibility condition, condition, 240, 240, 243, 243, 250, compatibility 250, 276,350 276, 350 for aa singular singular linear linear system, system, 45 for 45 complete complete normed vector vector space, space, 452 normed 452 complete 449 complete orthogonal orthogonal sequence, sequence, 449 completeness completeness of HI, 490 ^,490 2 of L L2, 453 , complex numbers, numbers, 394 complex 394 concentration concentration time evolution evolution of, of, 221 time 221 concentration gradient, gradient, 17 concentration 17 conditioning conditioning of matrix, 481, 486 of aa matrix, 481, 486 conjugate 485 conjugate directions, directions, 485 conjugate conjugate gradients gradients algorithm, 485 algorithm, 485 convergence, 486 convergence, 486 conservation conservation of of energy, energy, 173 173 continuously continuously differentiable differentiable function, function, 33 33 convergence convergence mean-square, 149, 419 mean-square, 149, 419 pointwise, 419 419 pointwise, uniform, uniform, 420 420 convolution, 424 convolution, 424 periodic, 425 425 periodic, copper, 278 copper, 228, 228, 243, 243, 244, 244, 265, 265, 278 d'Alembert's solution solution to the wave equad'Alembert's to the wave equation, 288 tion, 288 data data structure structure for describing describing aa triangulation, triangulation, 467 for 467 DCT, 414 DCT, 414
Index delta 318 delta function, function, 126, 126, 205, 205, 318 sifting property, 126 sifting property, 126 delta 126 delta sequence, sequence, 126 dense 186 dense matrix, matrix, 186 density, density, 10, 10, 330 330 descent 479 descent direction, direction, 479 determinant, 69 determinant, 69 DFT, 404 DFT, 404 differential equation, equation, 1 differential 1 autonomous, 109 autonomous, 109 constant coefficient, coefficient, 4, 4, 82, 82, 91 91 constant converting to to aa first-order system, converting first-order system, 79 79 first-order linear, linear, 87 first-order 87 homogeneous, 3, 3, 202 homogeneous, 202 inhomogeneous, 3, 96 inhomogeneous, 96 linear, 2, 2, 91 91 linear, linear system, system, 260 linear 260 nonconstant nonconstant coefficient, coefficient, 44 nonlinear, 33 nonlinear, order, 22 order, ordinary, 1, 1, 79, 79, 82, 82, 213, 213, 260 ordinary, 260 partial, 1 partial, 1 right-hand side, side, 44 right-hand scalar, 55 scalar, system, 5, 5, 91, 91, 108 108 system, differential differential operator operator symmetric, 137 symmetric, 137 differentiating an an integral, integral, 11 11 differentiating diffusion, 16 16 diffusion, diffusion diffusion coefficient, coefficient, 18 18 diffusion equation, equation, 18 18 diffusion inhomogeneous, 220 inhomogeneous, 220 dimension, 50 dimension, 50 Dirac delta delta function, function, 126, 126, 205, 205, 318 Dirac 318 Dirichlet 13 Dirichlet condition, condition, 13 Dirichlet 423 Dirichlet kernel, kernel, 423 discrete 414 discrete cosine cosine transform, transform, 414 discrete Fourier Fourier transform, discrete transform, 404 404 discrete sine sine transform, transform, 410 discrete 410 and 411 and the the FFT, FFT, 411 disk disk Fourier series series on on a, a, 359, 359, 372 372 Fourier displacement 21 displacement function, function, 21 divergence divergence operator, 329 operator, 329
609
Index Index 330 theorem, 330 domain domain 31 of aa function, function, 31 of domain of dependence, 289 domain domain of of influence, influence, 290 290 domain dot product, 45, 55, 71 dot DST,410 DST, 410 eigenfunction, 68, 68, 145, 145, 494 494 eigenfunction, complex-valued, 394 of of aa differential differential operator, operator, 139 139 eigenpairs on on aa disk, disk, 369 369 145 under Dirichlet conditions, 145 under mixed mixed boundary boundary conditions, conditions, under 151, 153 153 151, 233 under Neumann Neumann conditions, conditions, 233 under under periodic periodic boundary boundary condicondiunder tions, 247 247 tions, eigenspace, 70 eigenspace, eigenvalue, 68, 116, 116, 494 of aa differential differential operator, operator, 139 139 of 139 of a symmetric matrix, 139 eigenvalue problem general, 455 generalized, 495 eigenvector, 68 eigenvector, 68 of a symmetric matrix, 139 139 of a symmetric matrix, elastic 21 bar, 21 bar, elliptic elliptic regularity, 492 energy energy inner inner product, product, 182,379 182, 379 energy 182 energy norm, norm, 182 Euclidean Euclidean n-space complex, 395 Euler's Euler's formula, formula, 83, 83, 394 Euler's Euler's method, method, 117, 117, 261, 261, 310 backward, 310 backward, 118, 118, 263, 263, 310 improved, 104 even even extension, extension, 418 418 even even function, function, 415 415 and the the full full Fourier Fourier series, series, 417 and 417 existence, 238 existence, 238 38 of of solutions solutions to to aa linear linear system, system, 38 explicit method, 118
extension extension even, even, 418 418 odd, odd, 417 417 periodic, 425 periodic, 425 fast fast Fourier Fourier transform, transform, 393, 393, 407 407 FFT, FFT, 393, 393, 407 407 Fick's Fick's law, law, 17 17 fill-in, 477 fill-in, 477 finite finite element element method, method, 77, 77, 167, 167, 173 173 for for eigenvalue eigenvalue problems, problems, 495 495 for for the the heat heat equation, equation, 256 256 for for the the wave wave equation, equation, 306 306 Fourier Fourier coefficients, coefficients, 149 149 complex, complex, 397 397 rate rate of of decay, decay, 436 436 sine, sine, 149 149 Fourier 77, 149, 149, 155 155 Fourier series, series, 77, method method of, of, 166 166 complex, 401 complex, 397, 397, 401 double double sine sine series, series, 343 343 for for the the wave wave equation, equation, 291 291 full, full, 249, 249, 252, 252, 398 398 in in three three dimensions, dimensions, 355 355 on aa rectangular rectangular domain, on domain, 339 339 pointwise convergence, 429 pointwise convergence, 429 sine, 149, 149, 211 211 sine, Fourier's law, 331 331 Fourier's law, Fourier's of heat heat conduction, conduction, 12 12 Fourier's law law of Fredholm Fredholm alternative, alternative, 45, 45, 241 241 free free nodes, nodes, 379 379 function, 31, 35 35 function, 31, function space, 33 33 function space, fundamental frequency, 497 fundamental frequency, 497 of aa circular circular drum, drum, 375 of 375 of aa vibrating vibrating string, string, 294 of 294 fundamental theorem theorem of of calculus, calculus, 11, 11, fundamental 23, 101, 101, 327, 327, 330, 330, 338 338 23, Galerkin method, method, 173, 173, 180, 180, 193, 193, 258, 258, Galerkin 269, 275, 307, 378 269,275,307,378 Gaussian elimination, elimination, 43, 43, 77, 77, 196 Gaussian 196 for aa banded banded matrix, matrix, 476 for 476 operation count, count, 53,474,486 53, 474, 486 operation pseudo-code, 474 pseudo-code, 474
610 610
general general solution solution of 286 of the the wave wave equation, equation, 286 generalized 495 generalized eigenvalue eigenvalue problem, problem, 495 generalized 205 generalized function, function, 126, 126, 205 Gibbs's phenomenon, 148, Gibbs's phenomenon, 148, 344, 344, 435, 435, 443 gold,250 gold, 250 183,457 Gram-Schmidt procedure, 74, Gram-Schmidt procedure, 74,183, 457 27 gravitational gravitational constant, constant, 27 gravity, 27 gravity, 27 Green's 377 Green's first first identity, identity, 337, 337, 339, 339, 377 Green's 123 Green's function, function, 77, 77, 123 causal,124 causal, 124 for 203 for aa BVP, BVP, 203 for 127 for aa PDE, PDE, 127 for the heat 279 for the heat equation, equation, 279 grid, 188 grid, 101, 101, 188 irregular, 111 irregular, 111 guitar, 296, 318 guitar, 296, 318 HI, 490 Hl,m HJ,490 #<J,490 heat 19 heat capacity, capacity, 10, 10, 19 heat 275 heat equation, equation, 9, 9, 12, 12, 245, 245, 275 and finite elements, 257 and finite elements, 257 in 333 in aa bar, bar, 333 in 332 in three three dimensions, dimensions, 332 in 334 in two two dimensions, dimensions, 334 inhomogeneous, inhomogeneous, 13 13 weak form, 257 weak form, 257 heat flow, 99 heat flow, steady-state, 14 steady-state, 14 heat flux, 330 heat flux, 11, 11, 12, 12, 330 homogeneous homogeneous versus 13 versus heterogeneous, heterogeneous, 13 Hooke's 22 Hooke's law, law, 22 Hookean, 22 Hookean,22
IBVP, 14 IBVP, 14 implicit method, 118 118 implicit method, initial 246, 258 258 initial condition, condition, 14, 14, 80, 80, 246, for the wave wave equation, 24 for the equation, 24 initial initial value value problem problem for 285 for the the wave wave equation, equation, 285 initial-boundary initial-boundary value value problem, problem, 14,274 14, 274 for wave equation, 291 for the the wave equation, 291
Index Index
inner inner product, product, 56, 56, 59, 59, 181 181 complex, 395 complex, 395 2 complex L2, complex L , 396 energy, 182 energy, 182 inner product space, 56 inner product space, 56 integral integral area, 329 area, 329 iterated, 329 iterated, 329 line, 329 line, 329 surface, 329 surface, 329 volume, 328 volume, 328 integrating an ODE, 102 102 integrating an ODE, integrating 87 integrating factor, factor, 87 integration by parts, integration by parts, 138, 138, 178, 178, 307, 307, 336,489 336, 489 interpolant interpolant 492 piecewise piecewise linear, linear, 492 inverse inverse of 207 of aa differential differential operator, operator, 207 inverse 404 inverse discrete discrete Fourier Fourier transform, transform, 404 iron, 214, 217, iron, 19, 19, 214, 217, 222, 222, 228, 228, 230, 230, 236, 236, 244, 459 244, 260, 260, 264, 264, 281, 281, 323, 323, 459 iterated integral, 329 iterated integral, 329 iterative iterative method method for 478 for solving solving aa linear linear system, system, 478 Jacobian 471 Jacobian matrix, matrix, 471 jump discontinuity, 427 jump discontinuity, 427
Kelvin, 19 19 Kelvin, kinetic energy, 174 kinetic energy, 174 L L22 ,,445 445 L22 inner 60 L inner product, product, 59, 59, 60 Laplace's equation, 336, 336, 352 Laplace's equation, 352 Laplacian, 332 Laplacian, 332 in polar coordinates, 362 in polar coordinates, 362 lead,255 lead, 255 Lebesgue Lebesgue integral, 446 integral, 446 measure, 446 measure, 446 line 329 line integral, integral, 329 line 480 line search, search, 480 50 linear linear combination, combination, 50 linear 309 linear interpolant, interpolant, 309
611 611
Index
linear operator, 31, 132 132 linear operator, 31, definition, 35 definition, 35 linear operator equation, 36 linear operator equation, 31, 31, 36 linear linear system system 131 algebraic, algebraic, 31, 31, 131 linearly 52 linearly independent, independent, 52 load vector, load vector, 183 183 of, 467 467 computation of, computation for aa two-dimensional two-dimensional problem, problem, 378 378 for Lotka-Volterra predator-prey predator-prey model, model, Lotka-Volterra 108 108 LU factorization, factorization, 474 474 LU Maple, 147 Maple, 147 mapping, 31 31 mapping, mass matrix, matrix, 258, 258, 265, 265, 275, 275, 308, 308, 495 495 mass Mathematica, 147 147 Mathematica, MATLAB, 147 147 MATLAB, matrix, 31 matrix, 31 386,475 banded, 386, 475 dense, 186 dense, 186 Gram, 62 62 Gram, identity, 46 identity, 46 46 inverse, inverse, 46 invertible, 46 invertible, 46 mass, 308, 495 mass, 308, 495 53 nonsingular, 46, nonsingular, 46, 53 singular, 270 singular, 270 sparse, sparse, 173, 173, 186, 186, 189, 189, 386, 473 473 stiffness, stiffness, 308, 308, 495 495 272 symmetric, 71, 72, 72, 139, symmetric, 71, 139, 272 tridiagonal, 272 tridiagonal, 188, 188, 192, 192, 272 mean value value theorem, theorem, 431 431 mean mean-square, 60 60 mean-square, mean-square convergence, mean-square convergence, 149, 149, 419 of Fourier series, of Fourier series, 448 448 measurable measurable function, function, 446 set, 446 set, 446 mesh, 188 188 mesh, method of method of lines lines for the the heat heat equation, equation, 260 260 for middle C, 305 middle C, 295, 295, 305 modulus modulus 22 of of elasticity, elasticity, 22 Young's, 22
multiplicity of of eigenvalues, eigenvalues, 69 69 natural frequency natural frequency of aa string, 293, 318 318 of string, 293, Neumann Neumann condition, condition, 13 13 26 Newton, 26 Newton, Newton's law law of cooling Newton's of cooling and boundary conditions, 20 and boundary conditions, 20 and equation, 20 and the the heat heat equation, 20 Newton's second law 23, 29 29 Newton's second law of of motion, motion, 23, nodal values, 190 190 nodal values, nodes, 188 nodes, 188 nondimensionalization, 217 nondimensionalization, 217 59 norm, norm, 59 energy, energy, 182 182 normal derivative, derivative, 332 normal equations, equations, 62 normal normal modes modes of aa vibrating vibrating string, 293 of string, 293 null 134 null space, space, 134 of aa matrix, 42 of matrix, 42 trivial, 42 42 trivial,
odd extension, odd extension, 417 417 odd function, function, 415 odd 415 and full Fourier Fourier series, and the the full series, 417 417 ODE, ODE, 1 operator, 35 operator, 31, 31, 35 derivative, 36 differential, 36, 132 differential, 36, 132 linear, 132 linear, 132 matrix, 131 matrix, 35, 35, 131 nonlinear, 49 nonlinear, 35, 35, 49 wave, 285 wave, 285 order order of differential equation, equation, 22 of aa differential of numerical method, 103 of aa numerical method, 103 orthogonal basis, basis, 183 183 orthogonal orthogonality orthogonality of of functions, functions, 60 60 56 of of vectors, vectors, 56 partial derivative partial derivative weak, 490 490 weak, partial pivoting, 473
612 Pascal, 26 26 Pascal, PDE, PDE, 11 periodic convolution, 425 periodic extension, extension, 425 425 periodic periodic periodic function, function, 415 415 piano, 298 piecewise piece wise continuous, continuous, 427 piecewise linear approximation, 492 approximation, 492 basis function, 190, 190, 257, 257, 269 basis function, 269 function, 188, 188, 189 interpolant, interpolant, 259 piecewise polynomials, 173 piecewise smooth, smooth, 427 plate iron, iron, 459 459 pointwise convergence, 419 of complex Fourier series, 429 of aa complex Fourier series, 429 of aa Fourier series, 433 433 of Fourier cosine cosine series, of aa Fourier Fourier sine 433 of sine series, series, 433 of aa full series, 432 432 of full Fourier Fourier series, Poisson's equation, 335 Poisson's equation, 335 on aa disk, 373 on disk, 373 polar coordinates, 360 potential energy, 173 potential energy, 173 elastic, elastic, 173, 173, 174 174 external, 175 external, 175 gravitational, 175 minimal, 177 power series solution power series solution of equation, 364 of Bessel's Bessel's equation, 364 preconditioned conjugate conjugate gradients, gradients, 486 486 preconditioned pressure, 25 prime factorization and the FFT, FFT, 407 407 and the principle of virtual work, 177 product rule, 337 programming finite finite elements, 386 projection theorem, 182, 397 Pythagorean Pythagorean theorem, theorem, 55, 55, 428 428
quadrature, 102 quadrature, 102 choosing choosing aa rule, rule, 470 470 general formula, formula, 469 one-point Gauss rule, 469 one-point rule on aa triangle, 470 470 one-point
Index
over an an arbitrary arbitrary triangle, over triangle, 471 471 three-point rule rule for triangle, 470 470 three-point for aa triangle, quarter-wave cosine series, series, 153 quarter-wave sine functions, functions, 162 quarter-wave sine 162 151 quarter-wave sine quarter-wave sine series, series, 151 range range of aa function, function, 31 of 31 of linear operator, operator, 38 38 of aa linear resonance, resonance, 295, 295, 305, 305, 318, 318, 319 319 Riemann integral, integral, 446 446 Riemann Riesz Riesz representation representation theorem, theorem, 490 490 RK4, 107 RK4, 107 Runge-Kutta method, 106 Runge-Kutta-Fehlberg Runge-Kutta-Fehlberg method, method, 111 111
scalar, 32, 395 scalar, 32, 395 semidiscretization in space, space, 260 semidiscretization in 260 separation 225, 340, 340, 348 separation of of variables, variables, 225, 348 in polar coordinates, 362 in polar coordinates, 362 shifting 164, 197, 197, 222, shifting the the data, data, 164, 222, 299, 299, 321 321 sifting sifting property, property, 205 205 silver, silver, 255 255 Simpson's rule, Simpson's rule, 106 106 sink sink heat, heat, 12 12 snapshot snapshot temperature, 215 temperature, Sobolev space, 490 source heat, 12 span, span, 52 52 sparse matrix, 186, 386, sparse matrix, 186, 386, 473 473 sparsity sparsity pattern, pattern, 386, 386, 477 477 specific heat, 10, 330 temperature, 19 and temperature, spectral method, 156 156 spectral for aa BVP, for BVP, 144 144 for aa system system of of ODEs, ODEs, 96 for 96 for solving system, 167 for solving aa linear linear system, 167 speed wave, 288 of a wave, wave, 433 square wave, stability, 118 stability, of of aa time-stepping scheme, 310
613 613
Index stability 116 stability region, region, 116 stainless steel, 26 standing standing wave, wave, 294 steady-state steady-state and Neumann conditions, 237 temperature, 225 temperature, 225 steel, 302 steel, 302 steepest steepest descent algorithm, 481 algorithm, 481 direction, 481 direction, 481 stiff 260 stiff differential differential equation, equation, 117, 117, 260 stiff stiff system of ODEs, 117 stiffness stiffness 22 of of aa bar, bar, 22 stiffness stiffness matrix, 183, 191, 258, 265, 275, 275, 308, 308, 495 495 algorithm 466 algorithm for for computing, computing, 466 computation computation of, of, 465 465 for a two-dimensional problem, 378 null 271 null space, space, 271 singular, 275 singular, 270, 270, 275 strain, 22 strain, 22 string string elastic, 27, 131, 131, 160 sagging, 30 sagging, 30 strong strong form form of a BVP, 177, 267 subspace, 34 of Rn, R n , 39 superposition, 287 superposition, 287 principle of, 47, 85 support support of 382 of aa function, function, 192, 192, 382 surface integral, 329 symmetry and eigenvalues, 72 of 137 of aa differential differential operator, operator, 137 71 of of aa matrix, 71 of the the Laplacian, Laplacian, 337 of 337 temperature temperature gradient, gradient, 12 test function, function, 178, 267, 306 306 test 178, 267, thermal conductivity, 330 thermal conductivity, 12, 12, 19, 19, 330 101 time step, 101 time-stepping, 101 101 time-stepping, trace theorem, 490
transient transient behavior, 117, 118 transpose and product, 71 71 and the the dot dot product, of 45 of aa matrix, matrix, 45 trapezoidal trapezoidal rule, rule, 402 triangulation, 382 triangulation, 379, 379, 382 data 467 data structure structure for, for, 467 description 462 description of, of, 462 tridiagonal 192 tridiagonal matrix, matrix, 188, 188, 192 trigonometric interpolation, interpolation, 406 uniform convergence, convergence, 420 of a complex Fourier series, 439 of 441 of aa Fourier Fourier cosine cosine series, series, 441 of of aa Fourier Fourier sine sine series, series, 441 441 of of aa full full Fourier Fourier series, series, 441 441 of of continuous continuous functions, functions, 443 443 of cosine cosine series, series, 441 441 of uniqueness, 238 of to aa linear 42 of solutions solutions to linear system, system, 42 unit impulse, impulse, 126 unit normal vector, 328 unit normal vector, 328 unit point source, 126 variation of variation of parameters, parameters, 98 98 variational form of of a BVP, 173 vector, 31, 32 function 132 function as, as, 132 vector field, 329 vector 132 vector space, space, 132 complex, 395 definition, definition, 32 32 vibrating string, 292 vibrating string, 292 virtual work, 177 volume integral, 328
wave wave left-moving, left-moving, 288 288 right-moving, 288 right-moving, 288 wave wave equation, equation, 23, 23, 173 173 acoustic, 334 acoustic, 334 d'Alembert's solution, 288 general 286 general solution, solution, 286 homogeneous, 293 in 285 in an an infinite infinite domain, domain, 285
614 614
in three dimensions, 334 in two two dimensions, dimensions, 334 in 334 inhomogeneous, 296 on 374 on aa disk, disk, 374 on a square, 346 318 with a point source, 318 wave 285 wave operator, operator, 285 wave 296 wave speed, speed, 288, 288, 296 weak 490 weak derivative, derivative, 490 weak form weak form of a BVP, 173, 177, 266, 267 ofaBVP, 177,266, of of aa BVP in in two two or or three dimendimensions, sions, 378 of the the wave equation, 307 Young's modulus, 22
Index
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