Partial Differential Equations: an introduction

PARTIAL DIFFERENTIAL EQUATIONS

AN INTRODUCTION

BERNARD EPSTEIN PI%OTZSSOR OF

GRADUATE SCHOOL OP UNIVIRS1TY

McGRAW-HILL BOOR COMPANY, INC. NEW

SAN FRANCISCO

TORONTO

1962 LONDON

PARTIAL DIFFERENTIAL EQUATIONS (.'ompany, Inc. Printed ® 1962 by the McGraw-hut in the United States of America. Mi rights reserved. This book, or parts thereof, may tot be reproduced in any form without permission of tbe publisher'. J4brary qf Congress Catalog card Number 61-17838 t9640 THE &SAPLZ PRESS COMPANY,

YOfl, PS.

PREFACE

It has been the purpose and hope of the author in writing this book to help fill a serious need for introductory texts on the graduate level in the field of partial differential equations. The vastness of the field and— even more 8igmficantly—the absence of a comprehensive basic theory have been responsible, we believe, for the comparative scarcity of introductory books dealing with this subject. However, the importance of this field is so tremendous that the difficulties and pitfalls awaiting any. one who seeks to write such a book should be looked upon as a provoca-. tive and stimulating challenge. We hope that we have achieved some measure of success in meeting this challenge. Any book dealing with a subject possessing substance and vitality is bound to reflect the particular interests and prejudices of the author. Even if the field is well organized and has been worked out with a considerable degree of completeness, the author's inclinations will be reflected in the manner in which the subject matter is presented.

When,

in addition, the subject is as extensive and incompletely developed as that here under consideration, they will also be reflected in the choice of material. We are well aware that many important topics are sented briefly or not at all. However, we are consoled by the thought that in the writing of a book of moderate size the omission of much signi.flcant material was inevitable; and by the hope that our presentation will be such as not only to interest the student in the topics presented here, but also to stimulate him to pursue some of them, as well as topics not touched upon here, in other books and in the research journals. Throughout this book the stress has been on existence theory rathcz than on the effective determination of solutions of specific classes of problems. it. is hoped that the presentation will complement usefully

Preface

any text which emphasizes the more "practical" or "applied" aspects of the subject. A word is in order concerning the intimate relationship between physics

and the theory of differential equations, both ordinary and partial. Physics has certainly been the richest source of problems in thia field, and physical reasoning has often been an invaluable guide to the correct formulation of purely mathematical problems and to the successful development of techniques for solving such problems. In this connection we would strongly urge every prospective student of differential equations (indeed, every prospective student of mathematics) to read, and deliberate on, the splendid preface to the Courant-Hilbert masterpiece, "Methods of Mathematical Physics." Although little is said in the following pages concerning the physical origins of many of the mathematical problems which are discussed, the student will find that his understanding of

these problems will be heightened by an awareness of their physical counterparts. The author has good reason to follow the tradition of acknowledging gratefully his wife's unselfish aid in the seemingly endless task of typing successive drafts of the manuscript. More important than the typing,

however, was her constant encouragement to carry the writing task through to its completion. it is hoped that her encouragement was directed to a worthwhile objective.

Bernard

TERMINOLOGY AND BASIC THEOREMS

For convenience we list here a few terms, notations, and theorems that will be used frequeT: I 1j.

A domain is an open

set (in the plane or in a higher-dimen-

space); sn equivalent definition is that a domain is an open set which cannot be expressed as the union of two disjoint nonsional

vacuous open sets.

The "Kronecker delta" assumes the values I indices i, j are equal or unequal.

0, according as the

denotes the closure of the set S.

A disc is (he set of all points (in the plane) satisfying an inequality of the form (x — Xo)2 + (y —
The symbols €, C, 'sJ, are used with their customary set-theoretic significance: a A means that a belongs to (is an element of) the set A, and A C 8 means that every element which belongs to A also belongs to

B (A is a subset of B). Note that A C B does not imply that A is a proper subset of B. .4 Li B and A B denote the union and interseetion, respectively, of the sets A, B. The distance between two sets is the minimum distance between a pair of points, one from each set. If both sets are closed and at least one is hounded, thin the minimum is actually attained.

Terminology and Basic Theorems The Heine-Borel theorem: Given a compact (i.e., bounded and closed) set S in a euclidean space and an open covering of B (i.e., a collection of open sets whose union contains S), then it is possible to extract from this covering a finite number of open sets which suffice to cover S.

A real-valued function which is defined and continuous on a compact set is uniformly continuous, is bounded above and below, and attains its maximum and minimum values. A function defined in a domain 1) is said to be of class C" if all partial derivatives of order up to and including the nth exist and are continuous throughout D. If thefunctionsf(x,y), g(x,y) are of class C" in a neighborhood of (xo,yo),

and if the Jacobian equations

—

does not vanish at that point, then the

f(x,y), = g(x,y) can be solved in a sufficiently small neighborhood of (xo,yo) for x and y in terms of E and say X y = and the functions #(E,t,) are also of class C'.

CONTE NTS

cwma 1. Some Preliminary Topic.

.

.

.

.

.

.

.

1

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1. Equicontinuous Families of Functions. 2. The Weieratrass Approximation Theorem. 3. The Fourier Integral 4. The Laplace Transform 5. Ordinary Differential Equations 6. Lebesgue Integration 7. Dini's Theorem CHAPTER 2. Partial Differential Eqtialion. of Fires Ordrr 1. Linear Equations in Two Independent Variables

1

4 9 13

17

25 27 28

.

28 33 36

2. Quasi4inear Equations 3. The General First-order Equation 3. The Cauchy Probiem 1. Classification of Equations with Lines.r Principal Parts. 2. Characteristics 3. Canonical Forms 4. The Canchy Problem for Byperbolic Equations. 5. The One-dimensional Wave Equation 6. The Ripwnann Function

7. Clsifloation of Second-order Equations in Three or More Independent Variables 8. The Wave Equation in Two and Three Dimensions. 9. The Legendre Transformation

42 • • •

•

42

.

44 .46 .4853 65

•

.5860 .

65

CHAPTER 4. The Fredhoim AUernative in Banach Spaces 1. Linear Spacea

2. Normed Linear Spaces 3. Banach Spaces 4. Linear Functionals and Linear Operators. 5. The Frodhoim Alternative vu

69 71 •

•

74.

. .

76

82

Contents ci,

5. The Fredhoim A

Hithert Spaces

•

1. Inner-product Spaces

•

2. Hubert Spaces... Jleruiitian and Complelely 5. The Fredhoirn Alternative 6. Integral Equations 7. Hermitian Kernels 8. illustrative Example.

•

111

118

.

121

127

.

'I'heor'j

VHAI'TER El. ElementB of

99

.104

Operat4'rb

4.

.

¶11)

95

Adjoint

Liitear

3.

90

•

I. Introduction 2. Laplace's Equation and Theory of Analytic Functions 3. Fundamental Solutions. 4. The Mean-value Theorem 5. The Maximum Principle 6. Formulation of the Diriehlet Problem. 7. Solution of the l)iricblet Prolilem for the I 8. The of the Mean-value Theorem 9. Convergence Theorems 10. Strengthened Form of the Maximum Principle 11. Single and Layers 12. Posson's Equation . .

130 130

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•

•

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.

131

.

.

133

135

6

.

CHAPTLR 7. The J)iric.hlet Problem

1. Subbarmonic Functions. The Method of Balayage 3. The Perron-Remak Method 4. The Method of Integral Equations 5. The Dirichiet Principle 6. The Method of Finite Differences 7. Conformal Mapping .

2.

.

8.

The Heat Eqtiation

.

.

The Initial-value Problem for the Infinite Rod 2. The Simplest Problem for the Semi-infinite Rod. 3. The Finite Rod 1.

CHA

1

9. Green's Fu,wtions and Separation of Variables.

1. The Vibrating String 2. The Green's Function of the Operator 3. The Green's Function of a Second-order 1)ifferential Operator. 4. Eigenfunction Expansions 5. A Generalized Wave Equation, 6. Extension of the 1)eflnition of Green's Functions

2

35 .

39

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•

SOLUTIONS TO SELEcTED EXERCISES SUGGESTIONS ron FURTHER STUDY INDEX

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•

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237

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211 243 253 2El7

1. SOME PRELIMINARY TOPICS Before entering on the subject of partial differential equations, it seems appropriate to devote a chapter to some concepts and theorems with which the reader is perhaps not yet acquainted. The reader may choose to study the topics covered in this chapter as the need for them arises later, rather than beginning by going through this chapter (or

those parts of it which he has not previously encountered) systematically.

1. Equicontinuous Families of Functions A basic difficulty that besets many mathematical investigations is the fact that there exists no simple extension to families of functions of the Bolzano-Weierstrass theorem, which asserts (in one of its several possible formulations) that from every bounded sequence of real numbers it is possible to extract a convergent subsequence. If, instead of a sequence defined on a of numbers, we consider a sequence of functions fixed interval, say 0 z 1, then, from the hypotheses that each function

is

continuous on this interval and that these functions are

uniformly bounded [i.e., there exists a positive number M independent of z and n such that <MJ, it does not follow that it is possible extract a subsequence convergent throughout the interval. (Cf. Exercise 1.) However, there is a certain more stringent condition of continuity, which is frequently found to be satisfied by sequences (or families) of functions encountered in problems of analysis, and which suffices, when taken together with the condition of uniform boundedness, to assure the existence of a uniformly convergent subsequence; this more stringent condition will now be explained. A. basic theorem of analysis asserts that a function f(x) continuous on

a compact (bounded and closed) set S is uniformly continuous on 8; 1

Equations

Partial

2

that is, given any positive number there exists a positive number S such that whenever the conditions xi 8, E 8, lxt — X2( <S are satisfied, holds. Given, any finite number of — f(x2) the inequality I

continuous functions defined on S and any positive number , it is possible to find a single number S which will suffice for each of these

functions, for one ctrn determine a suitable & for each of these functions and then choose the smallest of these numbers. However, this method fails for an infinite family of functions, for there may not exist a positive lower bound to the corresponding set of numbers S. (This occurs in the family of functions considered in Exercise 1.) We may describe this

situation as a lack of "over-all" uniformity, despite the uniform continuity of each individual function of the family. We are thus led to formulate the following more restrictive concept of uniform continuity, which refers to a family of fun ctions, not to a single function. DEFINITION 1.

A family of functions defined on a set S of zeal num-

be& is said to be "equicortinuous" if for every positive nUmber

e

there exists a positive number S such that, for every function f(x) of the family and every psr of numbers Zt, Xa contained in S and satisfying the inequality IXL — X21 <5, the inequality Jf(xt) — f(x,)j <e holds. (It should be emphasized, for clarity, that each function of an equicontinuous family is uniformly continuous. Also it may be noted that no restriction of boundedness or closure is imposed on 8.) A simple example of an equicontinuous family is furnished by any set of functions defined and continuous on a fixed interval (open or clcsed) and having, at all points of this interval, a first. derivative whose absolute value never exceeds some fixed number C; for then, by the theorem of mean value, we have, for any function f(x) of the family and any two numbers xl, the interval, the inequality If(zt) — f(x,)I

Jf'(E)(xt

—

c,)j Cjxi

— x21

(Xi < E

< x,)

(1)

so that, for any given the choice a = will suffice. (Note that in Exercise 1 the derivatives of the functions under consideration are not uniformly bounded.) We now state the following striking theorem, which accounts for the important role played in analysis by tho coflcept of equn,ontinuity.

Theorem 1. Ascoli Selection Theorem. Let? be an infinite, uniformly bounded, equicontinuous family of functions defined on a I For simplicity, we formulate the definition only for functions of one real variable; the extension to more general classes of functions is quite straightforward. the theorem of this section is stated in the one-dimensional case, the extension to functions of any (finite) number of variables being elear.

1. Some Preliminary Topics

Then from every sequence ff,,(z)J chosen from F it is possible to select a uniformly convergent subsequence. Proof. Select any countable dense subset Si of 8, such as the set of Let a all rational numbers in S, and enumerate them: ri,

finite closed interval 8: a x b.

sequence

be

selected from F. Then the sequence

satisfies the hypothesis of the Bolzaño-Weierstraas theorem, and so we of the original sequence which conmay select a subsequence By applying the preceding argument to verges at the point rt. which converges at and also at r1. (for we obtain a sequence any subsequence of a convergent sequence is also convergent, and has the same limit). Repeating this argument, we obtain further sequences each of which is a subsequence of the preceding . . In one. and such that the kth sequence converges at rt, , order to obtain a single sequence which will converge at all the points of &, we employ the "diagonalization procedure," originally employed by .

Cantor to demonstrate the uncountability of the set of real numbers. Consider the sequence formed by taking the first function of the first subsequence, the second function of the second subsequence, etc. This last sequence is convergent at each point of for it is evidently a subsequence of the first sequence

and, aside perhaps

from the first k — I terms, a subsequence of the sequence f,i(x) fc

2, 3

It now remains to show that the sequence

converges throughGiven any 0, choose

out 8, and that the convergence is uniform. i> 0 such that, for every function / in F, and a fortiori for every function of the sequence the inequality txi — xtf implies that

<e. Now we select a subset of such that each point of S differs by Ices than from at least one point of 8:. (This can be accomplished, for example, by dividing S into adjoining segments, each of length not exceeding and selecting one point of in each of Jf(x1) — f(x2))

these segments.) We next determine a positive integer N so large that,

for n,m> N, the inequality <e holds at each point — y of S2. Then for any pointxofSweehoosey S,suchtbatlx — y( < 8, and we obtain, for n,m> N, the chain of inequalities

+

+

—

f,.,,(z)I

<3e (2)

Since the index N has been chosen independently of x, (2) implies the uniform convergence of the sequence { and the proof is complete. Two brief remarks may be helpful in clarifying the significance of the two hypotheses (uniform boundedness and equicontinuity). First, the proof of the existence of a subsequence of the original sequence which

Partial Differential Equations converges at all points.of a preassigned countable dense set requires only the existence of a pointwise bound (not a uniform bound) on the family F', and

does not involve equicontinuity. Secondly, the proof given

above may be easily modified to may be left as a simple exercise.

the following corollary, which

Let F be an infinite family of functions defined on an open interval .S: a <x
bounded at some point a <
The important Montel selection theorem of the theory of analytic functions is closely related to (the two-dimensional version of) this corollary; the essential point of the proof of this theorem consists in showing that a family of analytic functions uniformly bounded in a domain is equicontinuous in every compact subset of the domain. EXERCISES 2,

1, 1. Consider the sequence of functions tsin nTzj on the interval 0 x 1, n nixj < 1. . . . ; these functions are uniformly bounded on this interval, for

Prove that there does not exist a subsequence which converges uniformly at each point of the interval. 1, n 1, 2. Consider the sequence of functions {xhIJ on the interval 0 2, . .. ; in contrast. to the preceding exercise, this sequence (and hence every subsequence) converges throughout the interval, but the limit function is discontinuous at the end point z — 1. Prove directly from the definition that this sequence of functions is not equicontinuous. (This fact also follows, of course, from the selection a <1, the obove functions theorem.) Show that in any smaller interval, 0 i in agreement with the fact that the limit function is continuous in are this smaller interval. 3. Prove that a. sequence of continuous functions which converges uniformly on a compact set forms an equicontinuous family.

4. Show that a family of functions equicontinuous on any bounded set S and bounded at one point of S is uniformly bounded on S.

2. The Weierstrass Approximation Theorem In many branches of there are theorems whose proofs have to be given, in two parts: First the theorem is proved subject to certain additional hypotheses, and then it is shown, by the use of suitable approximation techniques, that the additional hypotheses may be dropped. To cite only one example, we may mention the Riemann-Lebesgue lemma, of fundamental importance in the theory of Fourier series and integrals, which asserts that1 fpr any function f(x) which is (absolutely) integrable

1. Some Preliminary TopIcs dx approaches over an interval I, finite or infinite, the quantity f' zero as the real parameter X becomes infinite. It is a simple matter to prove this theorem under the additional assumptions that the interval I is finite and the function f(s) is continuously differentiable, for in this case an integration by parts yields the desired result immediately. One can then extend the proof to continuous functions by using the fact that

a continuous function can be suitably approximated by continuously differentiable functions; then, similarly, one uses the fact that any integrable function can be suitably approximated by continuous functions; finally, the restriction to a finite interval is easily dropped. One of the most important and striking approximation theorems is the following, which will be used subsequently a number of times.

Theorem 2. Weierstrass Approximation Theorem. Let f(xi,x,, . be defined and continuous on any compact set R. Given any positive e, there a polynomial P(z1,x2, . . such that the •

.

inequality

—

P(x1,x2,

.

.

<

.

holds at all points of R. Proof. For simplicity, we consider the case n = 2; the modifications required for any other value of n will be clear from the proof to be presented. (Of. Exercise 6,) First we make the additional assumption that the set R is a rectangle and that f vanishes at all boundary points of R. We extend the function f over the entire plane by assigning it the value zero at all points outside Clearly the extended function is uniformly continuous, not only in but also over the entire plane.

Consider the one-parameter family of functions: =

f[

f(E13E2)P(E1 —

— x2, t)

(t> 0)

dE2

(4)

where P(Ei — Si,

— x2,

t) (or, for brevity, = (at)' exp (—

—

+

— 52)2]

On account of the uniform continuity of I we can choose 8 > 0, independent of and such that whenever — f(xi,x2)1 (Ei — xi)2 + (E2 — x2) < Taking account of the fact that the right side of (4) has the value one when the function f is replaced by the constsnt function 1, we obtain the chain of inequalities

Partial Differential Equatlou.

If,(xi,z2)

f(xt,x2)f

—

=

ff (/(Ei,b) —

ff If(Ei,b) —

+ 2M Jj

P1 dE1

fj P,

dE,

dE,

P1

f2r

+ [Here M = max Ills while

and

f

dr dO =

+

denote the disc

xi)' + (E2 — x,)' < 8' and its complement, respectively.] We now choose t (independent of With this choice of t, and x,) such that 2M exp (— —

(5) yields the inequality If,(xz,x2) — f(xa,.r2)I

<

(This inequality holds everywhere, not or1ly in R.) Taylor expansion of the function in the form u

We now write the

u'

whenever p'/t. A and the area and diameter, respectively, of fl. (That N can be so chosen follows either from the elements of the theory of analytic functions or from Taylor's theorem with remainder.) Letting — xi)2 + — 14 = choosing

N so large that

p denotin3

P(xi,xs) = (it)—'

and

we obtain dEtdE,

—

— it

—

A

This is valid at all, points of 1?. Combining (6) and (8), we obtain —

P(xi.x,)t

+=

(8)

I. Some Preliminary Topics

By inspection, it is clear that P(xi,x,) is a polynomial, so that the proof of the theorem is complete, subject to the restrictions concerning R and f that were made at the beginning. To eliminate these restrictions we require the following lemma. LEBESOVE EXTENSION THEOREM.

If a function f(xi,z2,

is defined and continuous on a compact set R, it is possible to . . extend f continuously to any larger set. Momentarily accepting this lemma, we now choose a (closed) rectangle R' containing tbe compact set R in its interior, and extend f by defining it to be zero on the boundary of H'. Clearly 118 still continuous under this extension, and its domain of definition is still compact. We invoke the lemma to extend f continuously to all of R'. Now the preceding argument is applicable, so that f can be approximated within throughout R', and a fort.iori throughout H, by a polynomial. It remains to prove the lemma. Again we may restrict attention to •

the case of two independent variables. First, we consider the very simple particular case of extending a continuous ftmction defined on the boundary of a square to the interior. We merely assign to the center of the square the mean of the values at the four vertices, and then define the function along each line segment connecting the center to a boundary point by linear interpolation. It should be noted that this method of

extending the given function assigns to each interior point a value between the minimum and maximum values which are assigned on the boundary. Now let a continuous function f be defined on any compact set R, and let R be contained in the interior of a square S whose boundary we denote by F. We then extend f continuously to the compact set by defining / to vanish everywhere on F. To prove the lemma it will suffice to show that f can be extended continuously to all of 5, for if this can be done, we can accomplish the continuous extension of f to the entire plane, and hence to any specified set in the plane, by defining f to vanish everywhere outside F. Let G denote the set of points inside F which do not belong to H. Since 0 is open (and nonvacuous), it is possible to construct a network of equally spaced horizontal and vertical lines sufficiently fine that at least one square of this network lies, together with its boundary, entirely in G. Then the original network is refined •by adding horizontal and vertical lines midway between those originally

Those squares (if any) of the finer network which lie (together with their boundaries) in 0 but whose interiors are disjoint

constructed.

from the square or squares previously selected from the original network are now determined. By repeatedly refining the network and selecting

squares, we evidently break 0 down into a countable union of closed whose interiors are disjoint. Let V be defined as the set of all

Partial Differential Equations We points of G which appear as vertices of thia collection of points define f at each point Q of V as the minimum value of / at of RjF which are closest to Q. (Since these points constitute a compact set, this definition is meaningfuL) it is readily seen from the manner in which the squares were chosen that f has been defined at the vertex of each square, and perhaps also at a finite number of additional points on the boundary of each square. Let / be defined along the boundary of each square by linear interpolation between successive points of the

set V, and then letf be defined inside each square by the method described earlier in this paragraph. Then / has evidently been defined throughout

G and is continuous there, but it must still be shown that this function possesses the proper behavior near the boundary of (1. Let any boundary point of G, say T, be selected, and let a positive number be given. Then &( >0) can be so chosen that, for all points 7" of RuF whose dis— tance from T does not exceed &, the inequality lf(T) — f(7")I < holds. it is evident For any point P of V whose distance from T is less than closest to P all lie within a distance less than ö that those points of if, finally, P lies within a distance from T, so that — f(P)I less than

from T, it is readily seen that P lies inside or on the boundary

of a square such that all the points of V on the boundary of this square lie within a distance less than from 7'. From the maimer in which f was defined on the boundary and inside each square, it follows that 11(T) — f(P)I <e, and the proof is complete. EXERCISES 5. Carry out in detail the proof of the IUemann-Lebesgue lemma sketched at the beginning of this section, for any function absolutely integrable over the real axis (in the Riemann sense). 6. Modify the proof given in the text to apply equally well to any value of a. 7. Prove the Lebeague extension theorem in the one-dimensional case. (This case is decidedly simpler than in higher dimensions.) 8. Carry out in detail the proof of the Weieratraas theorem in one dimension which

is outlined here: By the preceding exercise, we may assume that the function I is defined on a closed interval, rather than on an arbitrary compact set. By uniform continuity, / can be approximated uniformly by a polygonal function (i.e., a function which is continuous and sectionally linear, with only a finite number of Thia polygonal function can be expressed as a finite sum of polygonal functions, each having only one "corner." Each such function, in turn, can be expressed as the sum of a linear function and a function of the form constant- Ix — It therefore suffices to prove that the function IxI can be uniformly approximated by polynomials on the interval —.1 x 1. To do this, consider the identity !xl — (1 — (1 — and the Taylor series of the function (1 — ti about the point u 0. (This proof is due to Lebeague.) 9. Prove the following extension of the Weieratrass theorem: If/fr) is of class C" on

the interval a x b (i.e., f")(z) exists and is continuous on the interval), then,

9

I. Some Preliminary Topics given any

> 0, there exists a polynomial P(z) such that Jf(h)(x) — P(b)(x)I <4,

k—0,1,2,...,n.

10. Let /(x) be continuous and strictly increasing in the interval a x b. Show that, in this case, one can impose the additional requirement that the approximating polynomial ebould also be increasing. Hint: Approximate / by a suitable function of class Ci, and use Exercise 9. 11. A family of functions (f, defined on a fixed domain D (in any number of dimensions) is said to be "concentrated" at Q, where Q is some point of 1), if the following 1 as t—. 0 conditions are satisfied; (a) fg 0 throughout I), (b) fDt'* 1, (c) (or some other limit) for each neighbiwhood of Q. Prove that If the funetion g is g(Q). (It is assumed that u is such gf, bounded in D and continuous at Q, then

that the integrals Jb gf, exiat. It is worth noting, however, that JD g need not exist.. For example, if the integrals are taken in the Lebesgue sense (ef. Sec. 6) it suffices that

g be bounded and measurable, even though

g may then fail to exist if the measure

of 1) is infinite.) The functions P, used in the present section constitute a sequence concentrated at (ZI,z,). n — 0, 1, 2, . . . ,be a sequence of polynomials orthonormal with 12. Let

respect to an intervai a

z

b [i.e.,

p,1(x)p..(x) dx — ö,,.j,

and let

be of

+1, the p., are, aside from constant factors, identical degree n. (For a — —1, b with the Legendre polynomials.) Let f(x) be continuous in the sfore-mentioued interval, and let the coeffioientaf, be defined as foUows:f.. Jbf(x)pu(x) dx. Prove that

th. series

/.,'

converges

and that its sum is equal to f/1(z) dx.

Ar in the case of Fourier series, there is no assurance that the series actually converges unless suitable additional conditions are imposed on f(s).

3. The Fourier Integral In the theory of Fourier series one investigates the relationship between a periodic function f(x) and its formal expansion in exponential functions,

1(x) —21a,exp

L

where Li is the period of f(x) and the coefficients (I. I z f(s) exp L

are given by

)

(11)

We assume that the reader is acquainted with the elements of this theory, and shall devote the present section to a brief discussion of the

Partial thiferential Equations Fourier Integral, which can be looke4 upon as the limiting case of (10) obtained by letting L become infinite. Given any finite interval of the real axis, the number of integers n for which n/L lies in this interval one would expect (10) to assume increases with L; in the limit as L —p the quantity nIL assumes all real value8, so that the a form in summation should be replaced by an integration. The following theorem renders this heuristic remark more precise. Theorem 3. Fourier Integral Theorem. Letf(x) be defined for all dx exist. Then the real x and let the integral' f's (12)

=

function g(X) so defined is continuous. For exists for all real X, and of the any value of x, say E, at which the derivative f'(x) exists, the by the formula2 function may be recovered from

= Proof.

The existence of the integral (12) for all real values of X fol-

lows from the absolute integrability of f(x) and the boundedness of 1.) The continuity of g(X) is (Since X and x are real, easily proved in the manner suggested in Exercise 14. To establish (13) we need the following lemma, known as the 1'ourier single-integral Under the same hypotheses as above, urn

1.

\

sin

I

—

dx =

0

(14)

Temporarily assuming this lemma, we establish (13) as

f

=

'In the general theory it to be made here it

[.1:.

dx]

A

essential to employ the Lebesgue integral, hut for the to interpret all integrals in the Riem:inn sense.

'As may be shown by a simple example (ef. Exercise 17), thc integral I

A

may fail to exist; the approach to infinity of the limits of integration indiis therefore essential. If l's does not exist but lim I does, this cated in limit is rcf;.,rred to as the value" of the divergeat integral $ Note that for finite A the two iterate4 integrals appearing in (1Z) arcs absolutely ronvergent and, hence, equal to each other: the passage to the limit A —* is performed after the of order of

11

1. Some Preliminary Topics Subtracting from (15) the equality E)dx

1(E)

(ef. Exercise 15), we obtain =

sin A(x

—

E)dx

We now obtain (13) by applying the above lemma to (16).

it remains to establish the lemma. The existence of the limit appearing it (14) is readily seen as follows. Let two numbers x1, x2 be chosen such that xi < E <x2. The integral 1(E) Jxz

sin A(x

dx

—

is well defined on account of the assumption that f'(E) exists, for this if defined as for implies that the function [f(.r) — 1(E))! (x — The integral x = is continuous at Jx: extsts,

—

X —

E)dx

since the integrand is dominated by the function Jf(x)f/(x2 —

E).

As for 1(E) sin A(x —

Jxs X

it is reduced by the substitution t

exists,

t

Thus, the integral

(Cf. Exercise 15.)

("1(x) Jams

— 1(E)

Z—E

dx

to the integral

A (x —

A(xi-O

which is known to exist.

E)

sin A(x —

E)

dx

and the same statement is, of course, true for the integral f(x) —

sin A(x

— E)

dx

j_..

Having established the convergence of the integral appearing in (14), we now consider its behavior for large A. Given e> 0, we impose on the quantities and x2, which were introduced previously, the additional conditions dx <e

f"

dx

Partial Differential Equations

The possibility of satisfying the two latter conditions is assured by the existence of f'° If(x)I dx. Splitting the integral appearing in (14) into the sum of the following five integrals, (zt j—.. x —

Il

= =

14 =

sin A (z —

dx

E)

E

dx

sin A (x —

f(x)

— 1(E)

sin A (x —

dx

Jx&

1(x) sin A(x —

("

J74 x fwa

Is = I Jzi

X

—

dx

E

'" sinA(x —

E)dx

<e, regardless of the value of A. immediately see that <e, Employing the substitution t = A (x — used previously, we see that for all sufficiently large values of A the inequalities lIst < €, us! < —+ 0 as hold. Finally, by the Riemann-Lebesgue lemma (cf. Sec. 2), we

A

for A sufficiently large.

so that 1131

Hence, the inequalities

<5e

(17)

bold for all sufficiently large values of A, and this assertion is equivalent to (14).

The function g(A) associated with f(x) according to (12) is termed the "Fourier transform" of 1(x), and will be denoted by the symbol EXERCISES 13. Derive Theorem 3 formally by performing suitable limiting operations in (10) and (11). f(z)I dx exists, that the function 14. Prove, under the assumption that f defined by (12) is continuous, and that g(X) -4 0 as X becomes infinite. Hini Divide dx over the two semithe real axis into three parts such that the integrals infinite intervals are small, and apply the Riemann-Lebeague lemma to the integral over the finite interval. 15. (a) Prove the trigonometric identity + coa U

+ cos 2u +

•

.+

—

(b) Prove that the f%,nction 1/u — 1/2 be equal to zero there.

sin (ii

+

(n — 1,2,

.

.

.

)

(18)

is continuous at u — 0 if deflnedto

1. 9e Preliminary Topics

13

S

(c) Use (a) and (b) together with the Riemann-Lebesgue lemma to show that'

i

ifk>O

0 Is—i

ifk
1

•

I

iiJ._.

£

ifk—0

(19)

be shoWn,) (In particular the existence of the integral 16. Modify the proof given in the text to show that (13) holds under the weaker hypotheses that the left- and right-hand limits

1(r)

—h) h>O

h>O

and the left- and right-hand derivatives

—1(r)

urn

-

—h

h—.O

h>O

A>O

all exist and thatf(E)

17. Determine a function f(x) values of

.

lun h—.O

+f(t)1. such that f

h

dz exists and (13) holds for all

yet such that for some value of E the integral f

to liin

dx (in contrast

fails to exist.

18. By explicitly evaluating the Fourier transform of the function employing Theorem 3, evaluate the integral

f"

and

then

-

cosax

19. Evaluate the integrals appearing in Exercises 15 and 18 by employing complex integration (residues). 20. Prove, under suitable hypotheses, that ixff(f), and, more gener&ly, (ix)Rff(f)

(20)

21. Prove, under suitable hypotheses on the functions /(x) and g(x), that (21)

11
where f s g, the "convolution" or "faltung" of / and g, is given by

f

4

f

4 —

g*f

(22)

4. The Laplace Transform Closely related to the Fourier transform is that named after Laplace, which will now be discussed briefly. Applications of both these transforms will be made in Chap. 8 in connection with the equation of heat conduction. The function defined by (19) is termed the "signum function," denoted age k.

PartIal Differential Equation.

14

Let 1(x), assumed defined for a- 0, be such that for some value of the complex parameter a the integral f(s)

fe" e_1zf(x) dx

= converges (not necessarily abolutely).' [To give several simple examples, the integral (23) exists for all values of 8 if f(x) e-", for any value of a but for no value of a if 1(x) = such that Re (8 — c) > 0 if f(x) = Then the function f(s) thus defined for those values of a for which the integral (23) exists is called the "Laplace transform" of f(x). A most fact about integral (23) is that the set of values of a for which it exists [that is, the domain of the function j(s)J is always of a very simple form, as described by the following theorem.

Theorem 4. For any given function f(x), the integral (23) converges for (1) no values of a, or (2) all values of a, or (3) all values of 8 whose real part exceeds a certain real number a [the "abscissa of convergence" of 1(x)), and, perhaps, for some or all values of a whose real part equals a. In cases 1 and 2 the function f(s) is analytic for all a and in the half plane Re a> a, respectively. Proof. it is readily seen that the first part of the theorem can be reformulated in the following more concise form: If Re Re so and if

the integral (23) converges for s = a0, it also converges for a

8i.

While this statement is trivial if "converges" is replaced by "converges absolutely" [for, since a- 0, the inequality must I

hold), it is conceivable that when the integral converges only conditionally for a = 8o it might fall to converge for a = (Cf. Exercise 22.) To

rule out this possibility, recourse is made to integration by parts, as follows. Let g(A) dx; by hypothesis, urn g(A) exists. =

f

A—i.

Then, through integration by parts,2 we obtain

f

dx

Since Re (si.

— So)

e_(a_t.)Ag(A) + (si.

— So)

fA e (.rt.)rg(x) dx

(24)

> 0 and urn g(A) exists, the quantity

A becomes infinite.

Since g(x) is continuous (being an integral) and approaches a finite limit as a- —' it is bounded in absolute value, say fg(a-)t <M. Therefore, dx is dominated by fe" 1

However,

it is necessary to

ICf. the expression

ing (25).I

that

If(s)! dx

for every finite A.

'The integration by parts is valid eves without the assumption that f(s) is con-. tmous; it suffices that f(s) be absolutely integrablo on every finite interval 0

z A.

(Cf. preceding footnote.)

1. Sonic PreUminay Toptes

dx, and is consequently itself the convergent integral f convergent. The right-hand side, and hence the left-hand side, of (24) has been shown to possess a finite limit asA becomes infinite. The first part of the theorem is thus proved. Turning to the assertion concerning analyticity, we begin by showing dx, which that, for any fixed (finite) value of A, the integral for convenience we denote by 174(3), is analytic for all values of s. For any complex numbers a and h (h # 0), we readily obtain F'A(8 + h) —

A

+

h

dx

=

h

/

/

f 41

( —

JO

x'

h hz8

+

f(x) dx (25)

— .

< 1 we immediately find that the right side of (25) is dominated For by the expression •

tA

A1

+

+

dx

! Jo

It follows that the left side of (25) must approach zero with h, and hence

that

exists and equah —

dx, a result that could have

Jo

been obtained formally by differentiation under the integral. Next we replace Si in (24) by s and subtract the resulting equation from (23). We thus obtain 174(8)

f(s) =

— (a — so)

f

dx

(26)

(It is understood, of course, that 8 is again subjected to the restriction 0.) This immediately yields the inequality

Re (s —

—

f(s)l Me—A 1I*(.-..,)

+ Me4 R (.—*.)

Is —

—

(27)

Given any e> 0, choose positive numbers & R, and A0 such that the inequalities

Re(s—s0)>.6

—


± I?) <

(28)

are all satisfied. From (27) and (28) we then conclude, that the analytic functions 174(8) converge uniformly to f(s), as A —' in the domain defined by the first two inequalities of (28); by a basic tbeorem on analytic functions, f(s) is itself analytic at. 8. Tlais completes the proof.

Equations

Partial

It may happen [and usually does for the functions f(z) most commonly encountered in specific problemsi that the function J(s) can be defined analytically in a region larger than the half plane of convergence. Thus, 1 the abscissa of convergence is given by a = 0, but the transif f(x) 1/8 can be continued analytically over the entire plane form 1(8) except for the origin. It may happen, furthermore, that J(s) becomes multiple-valued when continued analytically. (Cf. Sec. 8-2.) If we formally set 8 = A and define f(x) to be zero for negative values of x, (23) goes over into the Fourier transform (12). The connection between the two types of transform is rendered more precise by the following theorem.

Theorem 5. Laplace Inversion Formula. Let f(x) be such that the integral (23) is absolutely convergent' for some real value of s, say g, and letf(x) 0 for x <0. Then at any value of x, sayx at which f(x) is differentiable, aj+i.'

f where the integral is defined as a Cauchy principal integral (of. Sec. 3), 1

(29)

ds

taken along the vertical line Re s = in the complex plane, being any real number exceeding cr.t Proof. For any s on the path of integration we may write s + A, X reaL Then (23) assumes the form

f(s)

= .7

+ A)

(30)

dx

f°

Replacing f(x) in (12) by exf(x), we obtain, in place of (13), the inversion formula

=

A

f

A) (31b) reduces to (29), thus completing

the proof. EXERCISES 22. Construct a pair of continuous functions g(x), h(x) such

(b)h(x)/p(x) > 0 (for all x such that g(x)

f°

0), (c)

that (a) lh(x)I

g(x) dx converges, but (d)

Ii(z) dx diverges.

1 This assumption can be weakened, but it introduced in order to permit an immediate application of Theorem 3. t In particular, then, the integral (29) equals zero for any negative value of

17

1. Some Preliminary Topics

.23. Let f(s) exist for a value ci whose real part equals the abscissa of convergence. within a "wedge" ($tolt Prove that Urn f(s) exists and equals f(so) ifs approaches a,

region) defined by the inequalities —e srg (a— a.) 9, • being any pesitive [This theorem is 4oeely related to Abel's theorem, which constant less than converges to a, then the sum of the series states that if the series ci + a, + (LIZ

approaches a if z approaches the value one within a wedge inequalities --9 arg(1 — z) 9, 0 <9 <11/2.1

+ az' +

by the

•

24. Determine a function whose Laplace transform converges conditionally for all values of s. 25. Determine the proper analogues of Theorems 1 and 5 for the "bilateral Laplace f(s) dz, where f(s) is now defined for all real values of a. transform" f —.

This application of formula, which is the analogue of (20), plays a fundamental role in the the Laplace transform to the solution of linear differential equations, both ordinary and partial. 21, prove, under suitable hypotheses on the functions 27. In analogy with f(z) and g(x), the convolution formula 26.

on f(s), that f' —

Prove, under suitable

—

sf(s)

—

(32)

28. Let f(s) be periodic (for x 0). Prove that f(s), as defined by (23), exists at lenit for Re 8 > 0, hut not necessarily in a larger half plane. On the other hand, show that 7(a) can be analytically continued to the entire plane except perhaps for simple poles on tbe imaginary axis.

5. Ordinary Differential Equations In the customary introductory course in ordinary differential equations, methods are taught for solving various classes of such equations (linear

with constant coeflicients, Riccati, Clairaut, etc.)., but the question of existence and uniqueness is usually disregarded. This question may he formulated as follows: Under what conditions does a given differential equation possess solutions, and when solutions do exist, what conditions may be imposed in order to assure uniqueness? A large part of the theory of differential equations deals with this question, and the basic result of this part of the theory is given by the following theorem.'

Theorem 6. Cauchy-Pieard. Let f(x,y) be defined and continuous in a closed rectangle !z — a, Iu — YOI b, and let f(x,y) satisfy, in this rectangle, a Lipscbitz condition with respect to y; that is, a positive constant k is med to exist such that the inequality

f(x,yr)I throughout this rectangle. Let M = —

holds

— yiJ

max

If(x,u)j in the afore-

'A thorough mastery of this theorem and its proof is urged. The proof provides a comparatively simple, yet typical, illustration of the method of successive a powerful and frequently employed technique in analysis.

PartIji Differential EquatiOns

18

mentioned rectangle and let c = miii (a,b/M). Then there exists for c a function g(x) such that g(x) — yoj b, g(z.) = ho, and — x0! f(x,g(z)) i.e., g(x) is a solution of the difTerential equation

Finally, the solution is unique; i.e.. if h(s) also satisfies the differential equation (33b) in the afore-mentioned interval (x — c, and if h(so) = ho, then h(s) g(x). The existence portion of the theorem will be proved by actually constructing a solution. Let a sequence of functions n = 0, 1, 2, . . . , be defined for Ix — xol c as follows: Proof.

go(s) gn(z)

f = Yo +

(34c)

Yo

dE

(n 1)

(34b)

First, it must be shown that the definition of the functions gn(x) provided by (34b) is meaningful. Assume that g,(x) is defined and continuous for

c and satisfies in this interval the inequality b. — Then the right-hand side of (34b) is meaningful (and continuous) when fx —

n—

i is replaced by j, and we obtain for g1÷i(x) the estimate

Mc b — (35) Yol so that satisfies the conditions imposed above on g,(x). Since, in particular, go(s) satisfies these conditions, the induction is complete. —

Next it will be shown that the sequence of functions is uniformly convergent. Replacing n in (34b) by n ± 1 and subtracting, we obtain (x) —

dE

—

(36)

From (36) and the hypothesis that f(x,y) satisfies a LipsohItz condition, we readily obtain the inequality' —

while

for

k

(n 1)

—

n = 0 the appropriate inequality is the following: Jgi(x) —

kM(x —

So)

(37b)

For convenience, we restrk't to the interval Zo z S x, + C; the reasoning employed obviously appliea, wit trivial modifications, to the interval x, — C x xo as well.

19

1. Some

By an easy induction (cf. Exercise 29), we xeadily obtain from (37a) and (37b) the inequality — —

+ 1)!

and, a fortiori,

).s+1

(

— g,1(x)f

(ii

38

+

1)!

Therefore, the series —

is dominated by the series of constani terms (n + 1)! L, n—O

which converges for all values of M, k, and c, and so series (40) converges

uniformly (as well as absolutely). This is equivalent to the assertion that the sequence of partial sums SM(x)

(m> 0)

—

(41)

is uniformly convergent. The sum on the right' is readily seen to equal Therefore, the sequence of functions or gm+i(x) — — I/o. is uniformly convergent to a limit, say g(x), and from (34b) we obtain

g(x) = yo +

Urn

(42)

z.

It is now necessary to show that the limiting operation appearing in (42) can be performed under the integral sign; i.e., that the operations urn and are permutable. This may be done as follows: Since, as b for all n, the same inequality must hold shown earlier, — in the limit, namely, lg(x) — Yol

b

(43)

Since, furthermore, g(x) is the uniform limit of continuous functions g,,(z), it is itself continuous. Therefore, f(x,g(x)) is well defined and continuous, and the expression is meaningful. We may A series, flnito or "teJeacoping."

of the form Z(u,1+i — UR), such as (41), is said to be

Partial Differential Equations

20

therefore rewrite (42)

g(x) =

Yo

as

follows:

+ lim

+

—

(44)

Taking account once again of the Lipschitz condition, we find that the second integral in (44) is dominated by max

klz —

Jg(x) —

Qn(X) $

1x—x,I

hence must approach zero as n becomes infinite. Therefore, the last term in (44) vanishes, and we conclude that g(x) satisfies the integral identity (45) g(x) dE yo + and

This equation is eciuivalent to the statement made above concerning the permutability of the operations lirn and 7S appearing in (42). Since g(x) and f(x,y) are continuous functions of the indicated variables, it follows that f(z,g(x)) is a continuous function of z; the elementary rule for differentiation of an integral with respect to its upper limit is therefore applicable, and we obtain

Ef(x,g(x))

(46)

Furthermore, it is clear from (45) [or simply from the fact that gn(xo) Ye for all nj that g(x0) Thus the existence portion of the theorem has been proved. It is worth stressing that the fact that the function g(x) is differentiable does not follow by any means from the mere fact that it is

the uniform limit of the differentiable functions g,1(x); the differentiability of g(x) and the further fact that g(x) satisfies the identity (46) are established only by having recourse to (45). This example illustrates the usefulness of integral representations in establishing properties of differeriliabiitty.'

Uniquene8s is now easily established. Suppose that h(s) is also a solution of (33b) satisfying h(s0) = Ye. Then by integration we get

h(s) =

+

(45')

Subtracting (45') from (45) and exploiting the Lipschitz condition, we obtain $g(z)

—

h(x)$

k

—

dE

(47)

1 In a quite different connection, the reader may recall that the fact that a function of a complex variable having first derivatives throughout a domain baa derivatives of

aU orders

throughout the domain is proved with the aid of the Cauchy integral

representation for analytic functions.

21

1. Some Preliminary Topies

(As before, we restrict attention, for conveniences to values of x x0.) — h(x)I. Then, from (47), we obtain, sue= max Let cessively,

— h(x)f

— h(x)I

k

kI

(48a)

— Xo)

kM(E — Xo)

=

(48b)

and, by induction, (48c)

jg(x) — h(z)I

w 0.

— Letting n become infinite, we conclude from (4&) that This completes the proof of uniqueness. (Cf. Exercise 31.)

The hypothesis that f(x,y) satisfies a Lipschitz condition was exploited in establishing both the existence and the uniqueness parts of the above theorem. One might conjecture that;, by employing more powerful methods, one might prove this theorem without the Lipschitz condition—

that is, merely under the hypothesis that f(x,y) is continuous. It is of interest that the existence can be proved under this weaker hypothesis, but not the uniqueness. To demonstrate that uniqueness may now fail, consider the differential equation (49)

=

The function fyi'1 is easily seen to satisfy a Lipschitz condition in any

rectangle which (including its boundary) lies entirely in the upper (or lower) half plane, but not in any rectangle containing a segment of the x axis in its interior or on its boundary. (Cf. Exercise 32.) Through the point (0,0) there exist, among others (cf. Exercise 33), the following four solutions: y

0

y=

y=

(x,O)]2

y=

(x0)J*

Turning now to the question of existence under the weaker hypothesis

that f(x,y) is continuous, we approximate f(x,y) uniformly in R by a sequence of polynomials (Cf. Sec. 2.) Corresponding to each n we define quantities

M and a Lipschitz condition in R (cf. Exercise 32),

we can, by Theorem 6, associate with each n a (unique) function G1(x) such that Gn&0) = I/o

(JX — xof

c.)

(50a)

PartIal Differential Equation.

22

or, in integral form, Gft(x) = Yo +

f:

dE

(SOb)

are uniformly bounded in the intervals

Now, the functions

for, as is evident from (50b),

—

(51) + (max c,1)(max denotes the maximum of IP,,(x,y) I in the rectangle R. (Since where approaches the quantity M defined in the statement of Theorem 6, the quantity max M1 exists.) Furthermore, from (50a),

hiof +

(52)

are equicontinuous.t Hence Theorem I so that the functions of the sequence applies, and we can, therefore, select a which converges uniformly to a function 0(x). Confining attention to this subsequence, we let n become infinite and obtain, from (50b),

0(x) =

i's

+ lim

(53a)

x.

to f(x,y) and of Taking account of the uniform convergence of to 0(x), we easily show that (53a) may be rewritten in the form

0(x) =

L1(E,0(E)) and that G(x) satisfies in and from this it follows that 0(xo) = c the dilferential equation (33b). lx — Ye +

The above theorem is readily extended to systems of differential equa-

tions, a. indicated by the following corollary, whose proof, which is a routine modification of the proof of the theorem, is left to the reader as Exercise 34.

Let the

.

a,

.

be defined and

(i,j 1, 2, . n), and let each of the functions f, .atisfy a Lipschitz condition with respect to each of the variables y(i); i.e., there exists a constant k such that, for each i and each j, continuous in a region fx

—

—

.) —

.

.

.

.

.

.)J

(54)

Since the quantity r., varies in general, with n, the of definition of the is not necessarily fixed. However, the reasoning of Sec. I holds with only minor modifications, which the reader should supply.

23

I. Some Preliminary Topics

there Then for Ix — xol c, where c is some suitably chosen constant, and = exist uniquely determined functions g(')(x) such that .

(55)

So far only differential equations and systems of first order have been considered, but equations (or systems of m equations in m unknown functions) of higher order are readily handled. Consider, for example, the differential equation dTy —

f

dy dty

(56) ,

This single equation is readily seen to be equivalent to the following system of the form (55) [the so-called "canonical system" associated with (56)1:

where

r> 1.

dv

y(I) (a)

dx

(57) dx

= ,y(r))

An application of the above corollary shows that a solution of (56) exists and is uniquely determined by prescribing the values of and its first (r — 1) derivatives for a specified value of x, assuming, of course, that f eatisfies the neceseary continuity and Lipschitz conditions. The restriction in the statement of Theorem 6 (and similarly in the corollary) to a sufficiently small interval about the point Xo is essential. This is readily shown by the following simple example. Consider the

differential equation

Although the function f(x,y) = 1 + y2 is defined over the entire x,y plane, the unique solution passing through the origin, namely, the function tan x, becomes infinite as x approaches either of the values ± T/2. hi contrast to the essentially "local" character of Theorem 6 (which

must be kept in mind whenever differential equations or systems are solved) is the much more satisfactory situation which exists in the linear

Partial Differential Equations case.

If (33b) aseumes the form q(x)y + r(x)

(59)

where q(x) and r(x) are defined and continuous on a finite or infinite interval I, there exists a unique solution of (59) defined on the entire interval I and passing through any prescribed point (x0,y0), where x0 is any (inner) point of I. This follows immediately from the familiar rule for solving (59) with the indicated condition, namely,

[Jx

y = exp

q(E) dE]

{yo +

q(t) dt] dE}

exp [...

(60)

However, this fact can be easily established, without making of (tb), by suitably extending the proof of Theorem 6. The obvious analogue of the above statement also holds when the system (55) is linear, i.e., of the form (61)

In particular, it follows [of. (56)] that, given n + 1 continuous functions a0(x), aj(x), . . , a,,...1(x), b(x) defined on an interval I, n constants Ye, yi, . . . , y,,_,, and any (inner) point z0 of I, there exists a uthque function 1(x) defined on the entire interval I satisfying there the differential equation .

'I—I

dIsy_V

— Li

+

,,, )

i—0

and the conditions

(i = 0, 1,

.

.

.

,n

—

1)

(63)

EXERCISES 29. Prove (38) by induction.

30. Show that an attempt to prove (39) directly by induction, rather than as a

consequence or (38), fails. 31. Let g(z) and (.7(x) he solutions of (33b) defined in a common interval.

Under

the hypotheses of Theorem 6, prove that if the inequality g(z) <0(x) holds at one point of the interval, it holds at all points. 32. Prove that if f(x,y) is continuous in the (closed) rectangle R and possesses a bounded (not necessarily continuous) derivative with respect to y throughout the interior of R, then f(x,y) satisfies a Lipechitz condition in R. 38. Determine the totality of solutions of (49) by elementary quadrature, and confirm both the existence and uniqueness parts of Theorem 6.

1. Some Preliminaiy Topica

25.

34. Carry out in detail the proof of the corollary on page 22. 35. Prove that a function f(x,y), defined in a rectangle and satisfying a Upeehits

condition in each variable, is necessarily (in contrast to the fact that continuity in each variable separately does not assure continuity as a function of both variables).

6. Lehesgue Integration Tn Chap. 5 it will be necessary to employ Lebesgue's, rather than Riemana's, definition of the integral. For convenience we state here, with utmost brevity and without any proofs, the most significant facts about the Lebesgue integral that are needed for our purposes. For simplicity we discuss the case of one independent variable, but the extension to a greater number of variables offers no essential difficulty. A set E of real numbers is said to be a "null set," or set of measure zero, if, for every e> 0, it is possible to find a finite or denumerable col-

lection of intervals <x
Li


<X2,

"integral"

.

.

.

,

C2,


.

<

.

.

,

on intervals

(= b)

(a=) x0 <x <xi,

respectively,

a(x) dx we mean the sum

and by the

(If the

ci(xa — z_t). '-I

interval extends to — oc (or + cc) the constant Ci (or is required to be zem.J If a nondecreasing sequence of step functions is given [i.e., 8,,(x) for all x and for n = 1, 2, . . .1, then their integrals

evidently form a nondecreasing sequence of numbers which approach either a finite limit or + A nonnegative function f(x) is said to be "measurable" if there exists a nondecreasing sequence of step functions which converges almost everywhere on the specified interval to f(x). (By "almost everywhere" we mean everywhere except perhaps for a null set.) The limit of the sequence .(fb dx). is independent of the particular sequence of step functions which is employed, and therefore represents a property of the function 1(x). If this limit is finite, f(x) is said to be "suinmable," or integrable, and Lbfx dx is defined as the limit of the integrals of the step functions. In particular, if the interval a <x
PartIal Differential Equatlona

26

where f÷(x)

j

+ f(x)J and f(x) =

—

f(x)J.

f(x) is said to be

that

defined

integrable if f+(x) and f_(x) are both integrable, and f f(x) dx dx —

as the difference f

fbf_(x)

dx.

ENote

A compler-valued function

1(x) is said to be integrable if its real and imaginary parts are both integrable, and the integral f"f(x) dx is defined, of course, as [Re f(x)] dx + j

[Tm 1(x))

All the familiar properties of the Riemann integral carry over to tJ.ie Lebeague integral, and need not be stated here. Any function whic) is

absolutely integrable (ef. Exercise 36) in the Riemaun sense is integrable in the Lebesgue sense, and the integral fbf(x) dx has the same value in both senses. The essential difference in power between the two methods of integration lies in the fact that much stronger convergence theorems hold for the Lebesgue method. In particular, the

following theorem, which is not true for Riemann in

of analysis.

Theorem 7. Rlesz-Fischer. Given a sequence

of "quadratically integrable" functions—i.e., is measurable and I' is integrable—such that urn 0, there exists a quad— fm!t dx raticafly integrable function f(x) such that lim

j

—

(

=

We also state, without proof, two further theorems concerning Lebesgue

integration which will be employed in Chap. 5.

Theorem

8.

Fubini. Let f(x,y) be integrable over the rectangle R:

(Anyorallofthequantitiesa,b,c,dmaybe

infinite.) Then for almost all x the integral

fdf(xy) dy exists, and the function g(x) so defined is integrable; i.e., the iterated integral

ijdf(x.y) dy] dx L" exists. Similarly, the iterated integral

jd [fhf(xy) dx] d7,

1. Some Preliminary Topics also exists, and the values of both iterated integrals are the same as that of the integral

I Theorem 9. Let f(x) be integrable over the interval 1: a x 6, is a conand let c be any fixed number in 1. Then F(x) = tinuous function of x, and F'(x) exists and equals f(z) almost everywhere in I. EXERCISE dx doee not exist in the Lebesgue sense.

36. Show that the integral /

7. Dini's Theorem This is a simple but striking partial converse to the theorem that the sum of a uniformly convergent series of continuous functions is itself continuous.

Theorem 10. Let the nonnegative functions

each be continuous

on a bounded closed interval I, and let the series

converge

everywhere on I to a continuous function f(x). Then the convergence of the series is uniform. Proof.

Let the remainders

be denoted Rzjr(x). These II

functions are continuous, nonnegative,

monotone in N; i.e., for

each x of I, the inequalities

0 Rw.i(x) RN(X) hold. Given e> 0, we can associate with each numb&

(64)

of I an,

open interval 1(x5) containing x, and an index N, depending both on e and x0, such that for all x contained in both I and I(xo) the inequality RN(X) <E

(65)

holds. By the Heine-Borel theorem, a finite number of the intervals 1(x0) suffice to cover I. If we let N' denote the largest of the indices N

associated with this finite collection of intervals, it follows from (64) that the inequality


This is equivalent to the

2. PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

I. Linear Equations in Two Independent Variables Let three functions a(z,y), b(x,y), c(x,y) be defined in some domain I) By a "solution" of the equation

of the x,y plane.

(I)

is meant a function f(x,y) defined in all or part of I) such that (1) reduces to an identity in the variables x andy when z is replaced byf(x,y). (1) is termed a first-order linear partial differential equation; "first-order" because no derivatives of order higher than the first appear,

and "linear" because of the following identity, valid for arbitrary constants ai, and arbitrary (differentiable) functions z1, (2) + a,z,) = aiL(zi) + Our principal objective in this section is to demonstrate the nature of the conditions which sufilce to determine uniquely a solution of (1). It will be shown that a very simple and complete theory of this equation exists, the essential feature in its development being the reduction of (1), by means of a suitable change of independent variables, to an especially simple form which can be treated as an ordinary differential equation. Actually, it is possible to develop a more general theory which applies equally well to the analogue of (I) in any number of independent variables, namely, GLZgt

+

+

+ G,.ZJ.,

C

(3)

Furthermore, the general theory covers the case that the coefficient., a1, . . , Ga, c depend on z as well as on the independent variables 28

2. Partial Differential Equations of First Order

29

(3) 18 then termed "quasi-linear." However, in order to introduce the basic ideas as easily as possible, we confine attention ZI,

.

.

.

,

in the present section to (1), deferring the discussion of the genera) quasi-linear equation (3) to the following section.

By analogy with the theory of ordinary differential equations, as presented in Sec. 1-5, it appears plausible that, under suitable hypotheses on the coefficients a, b, c, a unique solution of (1) is determined by the

requirement that the surface z f(x,y) represerting the solution shall contain a specified curve C of the x,y,z space. More precisely, let a curve C be defined by a set of three equations,

x=

y= i(t),

z=

t

t2)

(4)

where the functions are assumed to be sufficiently differentiable for the purposes of the following discussion. Also, we assume that E'(t) and do not vanish simultaneously for any value of t; this assumption assures that C possesses, for each value of t, a well-defined tangent which, furthermore, not parallel to the z axis. (The latter

condition must be satisfied in order that a surface containing C may possess a nonvertical tangent plane; a vertical tangent plane would, of course, be inconsistent with finite values of z, and ;.) We also require that C have a one-to-one projection on the x,y plane; this is necessary in order to assure the single-valuedness of the solution. Now, assuming that a solution of (1) exists which contains C, then it should be possible to determine from (1) and (4) the values of and; at each point of C, for at any such point Za and must satisfy, in addition to (1), the equality = zE'(t) + ;,'(t) (5) By elementary algebra (Cramer's rule), (1) and (5) determine uniquely the values of and s, unless the quantities E'(t) and i'(Z) satisfy the "characteristic condition"2 a

= a,'(t)

—

=

0

(6)

When (6) holds, (1) and (5) are inconsistent unless the "compatibility condition"

a:b:c

(7)

is also satisfied. When (6) and (7) are both satisfied, (1) and (5) admit infinitely many - solutions.2 Furthermore, when (6) is not satisfied at i If a and b both vanish at a given point of C, no restriction is imposed by (6); we set aside this degenerate case. 'That is, when (1) and (5) are considered as algebraic (not differential) equations in the unknowns s and s, (at a given point of C).

Partial Differential Equations

30

a given point of C, all higher derivatives of z at this point are also (formally) uniquely determinable from (1) and (4) (assuming, of course, that possess derivatives of all orders), whereas, if p, the functions a, b, c, both (8) and (7) are satisfied at all points of C it is possible to derive from (1) and (4) consistent, hut not uniquely solvable, equations for the higher derivatives of z. (Cf. Exercise 1.) It may therefore be expected,

and we shall prove, that when (6) is never satisfied on C there exists a unique solution of (1) containing C, and that when both (6) and (7) are satisfied everywhere on C there exist infinitely many solutions of (1) containing C. Rewriting (6) in the form b dy —=— a dx

dx a\ /ior—=—I

\

d'j

(8)

1)/

we see that the characteristic condition determines at each point of 1) a unique "characteristic ground direction." A curve lying in D which possesses at each of its points this direction is termed a "characteristic ground curve." These curves are obtained by sok-ing the ordinary differential equation (8); referring to Theorem 1-6, we conclude that, under suitable restrictions on a and b, a unique characteristic ground curve passes through each point of 1). [Note that neither the function c(x,y) nor the third rectangular coordinate z is involved in determining these curves.) Clearly, (6) asserts that the (orthogonal) projection of C on the x,y plane possesses (at the projection of the point, of C under consideration) the characteristic ground direction, and (7) asserts that when this occurs, the curve C must possess at this point the "characteristic direction" determined by the direction numbers a, b, e. A curve (in ;y,z space) possessing at each point this direction will be termed a "characteristic" of equation (1). Now let a change of independent variables be introduced; we assume that the functions expressing the new variables in terms of the old, g(x,y)

y'

h(x,y)

(9a)

and the inverse functions

y=

(9b)

possess continuous first partial derivatives. into the equation + =

Then (1) can be converted

x

where

y=c

and the characterlstie ground direction at any point. (x',y') is the image of

the characteristic ground direction at the corresponding point (x,y). the latter assertion can be proved (ef. Exercise 2) by an ele-

P'iiaI Differential Equations of First Order

31

mentary computation, it is instructive to prove it without computation by the following argument: Given a noncharacteristic direction at (z,y), equations (1) and (5) uniquely determine and z1,, and these, in turn, uniquely determine z7' and z,,' at the corresponding point (x',y'); hence, the corresponding direction at (x',y') is noncharacteristic. Conversely, by symmetry, each noncharacteristie direction at (x',y') corresponds to a and hence the characteristic ground noncharacteristic direction at directions must also correspond. Now, if it is possible to introduce new variables in such a manner that the quantity appearing in (10) vanishes identically, then the theory of (10) [and hence that of (1)] becomes, as will be shown soon, an almost obvious modification of the theory presented in Sec. 1-5. According to (ii), it suffices to find a (noneonstaut) function h(x,y) such that

+

= 0

(If such a function 4(x,y) is determined, any function g(x,y) independent of h(x,y) may then be employed in (9a).J We leave the analytical discussion of (12) as Exercise 3, and instead employ here the following geometrical argument. In sorne sufficiently small neighborhood D' of any given point (x0,y0) of I) we construct the set of all characteristic ground curves. These will cover D'. Let h(x,y) bc defined so as to a constant value on each one of these curves, the of the constant being different for each pair of curves.' From the remarks made in the preceding paragraph concerning the correspondence between character-

istic ground directions under change of variables, it follows that the characteristic ground curves of (10) will be given by y' = constant. Taking account of (8), or, rather, its analogue dy' dx'

a We have therefore proved that (1) can be

see that $ must vanish. reduced, by a suitable change of variables, to the form we

= c(x,y)

(14)

Now consider the problem of obtaining a solution of (14) which has specified values along a ground curve r; i.e., the surface representing the solution is to contain a curve C whose projection is r. First, we assume

that F never possesses the characteristic ground direction; this means that C can be represented parametrically in the form (4) with t(t) 1. We consider any fixed value of y and interpret (14) as an ordinary 'For example, draw through (xo,yo) the orthogonal trajectory of the family of characteristic ground curves, and associate with each ground curve F the arc length along the trajectory between (Xo. y,) and the iMerecction it h F.

Partial Differential Equations differential equation; that is, we solve (14) subject to the condition that

z=

1=

when x =

E(t) E = E(v)J.

We immediately obtain as the

solution

=

+

c(X,y) dX

(15)

is readily apparent from (14), the curve obtained from (15) for each axed value of y is a characteristic, and so the solution may be generated by the simple procedure of constructing the characteristic through each p'.)int of C.

On the other hand, if r meets a given horizontal line y ye at two distinct points (xi,yo) and (x2,y0), the values of z which are prescribed at these two points must satisfy the relation' z(x2,yo) — z(xi,yo)

c(x,yo) dx

= In particular, if F consists entirely of the line y =

(16)

yo, or a segment of this line, the value of z at any point determines the value everywhere on F, so that if the given curie C is not a characteristic there is no SOiUtiOfl, but if C is a characteristic there exist infinitely many solutions cont.aining C. Such solutions may be obtained by constructing the characteristic through each point of any curve C' which intersects C and whose ground curve never intersects any line y = constant more than once. The foregoing remarks are readily illustrated by the simplest of all possible examples, namely, the equation z2 = 0. If C is given by the equations x = E(t), y = t, z = the unique solution is (t1 t <x < byz Clearlythis giveninthestript, y surface is generated by passing through each point of C the characteristic, namely, the line to the x axis. However, if C is itself a char-

acteristic there are infinitely many cylindrical surfaces and, hence, infinitely many solutions containing C. Finally, if y, but not z, is constant along C, then the characteristics through C form part of a vertical plane, so that no solution exists. Returning to the general case of (1), and taking account of the fact that

characteristic ground curves and characteristics are preserved under change of (independent) variables, we conclude that (1) admits infinitely many solutions containing the curve C if C is a characteristic, no solution if C is not a characteristic hut possesses a characteristic ground curve, and one solution in the remaining case, namely, that the ground curve is noncharacteristic.

Finally, we comment briefly on the previously mentioned fact that from (1) and (4) it is possible to determine formally (i.e., even before tAssuming that c(x,y) ci. Exercise 4.

is

defined on the linc segment connecting the two points;

2. PartIal Differential Equations of First Order

33

establishing the existence of & solution), at any point P of C at which (6) is not satisfied, the values of all derivatives of the solution, not merely the values of the first derivatives and ç. In order to carry out this formal procedure it is necessary, of course, that the coefficients a, b, c possess which partial derivatives of all orders, and thai the functions define the curve C poacess derivatives of all orders with respect to the parametric variable t. Once the derivatives of the presumed solution z have been determined at P, a formal Taylor series solution Z

""n —0

can be written; here (x0,y0) represents the projection of the point P on the x,y plane, and the coefficients Can are, aside from numerical factors, equal to the values of the derivatives that have been determined. The

ipiestion then arises whether this series actually converges [in some and defines a solution of (I) containing the curve neighborhood of (?. might be expected, the answer can be shown to be affirmative if,

in addition to the afore-mentioned assumptions concerning a, b, c, it is assumed that these functions are analytic in their arguments; by this is meant, of course, that each of the functions a, b, c can be expanded in a Taylor series in some neighborhood of (x0,y4), and that each of the functions E(t), ,(t), can be expanded in a Taylor series about o, the value of t corresponding to the point P in the parametric representa-

tion (4) of C. This assertion constitutes (the simplest form of) the celebrated Cauehy-Kowalewski theorem. EXERCISES 1. Obtain formulae for s., z,, a,,,, and a,,, from (1) and (5), assuming that (6) is not satisfied. Also show that when (6) and (7) hold, there exist infinitely many solutions of (1) and (5) for the above five quantities. 2. Prove analytically the assertion following (ii). 3. Work out the analytical counterpart of the geometrital discussion following (12). 4. The illustrative example presented in the text shows that it is not possible to construct a solution in the entire half plane v 0 of the equation z, — 0 with the prescribed conditions: a — 1 — z2 on the right half of the parabola y — .r', z — on the left half of the same parabola. Rowever, show that if the portion y 1 of the y axze is deleted, it is possible to obtain a unique solution defined in the remain portion of the half plane. Explain why this does not conflict with the illustrative

example.

2. Qua.i-lineaz Equations For ease in geometrical visualization, we shall first consider the quasilinear equation in two independent variables; that is to say, we shall con-

Equationa

Partial

34

skIer the coefficients a, b, c in (1) to depend on z as well as on x andy. We assume these coefficients to be defined in some domain G of x,y,z space,

and to satisfy conditions adequate to permit the following arguments to be applied.1 As in the preceding section, we define a characteristic to be a curve whose tangent has at each point the direction numbers a, b, c. of G there exists a unique characteristic, Through each point which can be obtained by solving the system = a(x,y,z)

b(x,y,z)

= c(x,y,z)

with initial conditions x(O) y(O) = yo, z(O) = zo; the solution will give a parametric representation of the characteristic. In contrast to the linear case, where the characteristics through two points having the same z and y coordinates can be obtained from each other by vertical displace-

ment, the characteristics do not have any simple relationship in the quasi-linear case, and, in particular, the characteristic ground curves (i.e., the projections on the x,y plane of the characteristics) do not in general constitute a one-parameter family, so that a reduction to form (1.4) cannot be attained. Nevertheless, as will be seen, the characteristics and

characteristic ground curves play an important role in the case of the quasi-linear equation also. Let any solution of (1) be considered. Since the normal at any point of the surface representing the solution has direction numbers 23) z,,, — 1, (1) may be interpreted requiring that the normal shall be orthogonal

at each point of the surface to the characteristic through that point. This, in turn, is equivalent to requiring that the characteristic shall be tangent to the surface.

It follows that if a (continuously differentiable)

function f(x,y) has the property that the surface z = contains the characteristic through each of its points, then the given function satisfies (1). Suppose, in particular, that a curve (7 is given, and that by constructing the characteristic through each point of C there is obtained a smooth surface, no two of whose points have the same projection Oil the ;y plane. Then the surface evidently represents a function satisfying (1). Conversely, it will be shown that evcry solution of (I) can be obtained by such a construction. The essential step in proving the last assertion consists in showing that a surface S representing a solution f(x,?/) of (1) must contain the characteristic through each one of its points. Let a point (x0,y0,z0) be seJected on 5, and let a curve r be determined on S by means of the system

= a(xry, f(x,y))

b(x,y, f(x,y))

z=

(18)

'in particular, it wifl suffice if these functions satisfy the Lipschitz condition (cf. Sec. 1-5) and if a and b do not both vanish at any point of 0.

2. Partial Differential Equations of First Order where the first pair of equations is to be solved with the initial conditions x(O) = x0, y(O) = Ye. Then along this curve the identity (19)

a must be satisfied. Since, by (1), the curve 1' must coincide comparison of (18) and (19) with (17), that [This argument shows that if with the characteristic through point P in common, then they actually have two solutions of (1) possess a

in common the chara cteristie. through P. Hence, if the surfaces representmust be a ing two different solutions intersect along R curve, this Conversely, it will be evident from tho discussion in the characteristic.

following two paragraphs that any number of solutions can be con-

structeci which will intersect along• any prescribed characteristic, and also that any number of solutions can be constructed which will coincide on one side of a given characteristic hut not on the other side. These remarks may be summed up in the statement that the characteristics are "branching lines" of solutions.]

It therefore follows that if a curve C is drawn on S and meets each characteristic on S at exactly one point, the surface can be reconstructed by drawing through each point of C the (unique) characteristic. While such a curve might be tangent at one or more of its points to the characteristic, we rule out this case. It then follows that the curve C [parametrized as in (4)1 does not satisfy (6) at any point. [While this is obvious geometrically, it also follows analytically from the fact that if (6) were satisfied at any point of C, then (7) would also have to hold, and this would contradict the condition that has been imposed on C.) Conversely, if a curve C is given which does not satisfy (6) at any point,' and if tl)e characteristics are drawn through each point of C. it is geometrically evident that a surface is thus generated which represents a single-valued function satisfying (1), at least if attention is confined to a sufficiently small neigh horhood of C. (Cf. Exercise 5.) Analytically, this follows from the following simple argument. The surface may be considered parametrized in the form

x = x(s,t)

z = z(s,t) where 8 and t are the parameters appearing in (17) and (4), At any point of C the Jacobian reduces to —

(20)

y

—

is understood, of course, that C is required to satisfy the conditions imposed in

Sec. 1.

'Each characteristic has assigned to it the value of t corresponding to the point where the characteristic meets C, and each point on this characteristic also has a value of a associated with it by (17) and the Initial conditions.

Pavdal Differential Equations

which does not vanish, by assumption. By continuity, the Jacobian remains distinct from zero in some neighborhood of C, and therefore a and t, and hence also z, can be expressed as single-valued functions of x and yj. We sum up the above disëussion in the form of a theorem.

Theorem I. If C is a curve which does not satisfy (6) at any point, then there exists a unique solution of (1) containing C, which can be constructed by passing the characteristic through each point of C. If C is itself a characteristic, there exist infinitely many solutions of (1) containing C, each of which can be obtained by selecting an arbitrary curve C' which intersects C and never satisfies (6), and then repeating the above construction. If C satisfies (6) but not (7) at some point, no solution of

(1) exists which contains C.

It is now obvious that the same reasoning may be applied to the general quasi-linear equation (3) in n variables. We define the characteristics as the curves in (n + 1)-dimensional space satisfying the system dx1

...,

dx2

Then every solution z = f(xl,x2,

dx,,

dz

,x.) of (3) is generated by an . . (n — 1)-parameter family of characteristics, and conversely, given any (n — 1)-dimensional "surface" C in the (n + 1)-dimensional space of the variables Xa, . . . , x,,, z satisfying suitable restrictions analogous to .

those imposed in the above theorem (cf. Exercise 6), there exists a unique

solution of (3) containing C, and this solution may be constructed by passing the characteristic through each point of C. EXERCISES 5. Show analytically that if C satisfies (6) but not (7) at a certain point P, then the surface generated by the characteristics through C possesses a vertical tangent plane

at P

6. Determine the "suitable restrictions" mentioned in the last paragraph of the present section.

3. The General First-order Equation We consider the equation F(x,y,u,p,q)

0

(p

z3,q

where F is defined and of class C' in some domain I) of the ;y,z,p,q space. We assume that the equality (22) is satisfied at some point (zo,yo,zo,p5,q0)

2. Partial Differential Equations of First Order

31

of D and that at this point at least one of the inequalities (23)

holds, so that (22) can be solved in some neighborhood of this point for p or q in terms of the four remaining variables. It is then evident that (22) may be interpreted geometrically, in the x,y,z space, as imposing on the surface z(x,y) the condition of being tangent at each of its points to one of a one-parameter family of planes which pass through the point. More precisely, assuming (without loss of generality) that (22) can be solved for p in terms of x, y, z, and q, we permit q to vary, for each fixed set of values of x, y, z, over its range, and we consider all planes through (x,y,z) whose normals possess direction numbers (p, q, —1). If (22) assumes the linear or quasi-linear form (1) at any point (x,y,z), the associated family

of planes form a coaxial bundle, whose common line is known as the "Monge axis," but otherwise these planes envelop a conical surface with vertex at (x,y,z). This surface is known as the "Monge cone," and (22) may be interpreted as requiring that the surface z(x,y) shall be tangent at each of its points to the Monge cone associated with that point. The generators of the Monge cone associated with any point (x,y,z) can be determined by the following simple argument. Assuming, as above, that p is expressible in terms of x, y, z, and q, we choose distinct values of and then determine from (22) the corresponding q, say q and q + The planes whose normals possess direcvalues of say p and p +

tion numbers p, q, —1 and p + —1 intersect in a line q+ perpendicular to both normals arid hence possessing direction numbers 1, —

Letting

p—

approach zero and taking account of the relationship

[which is obtained by differentiating (22), considered as an identity in p and q for fixed values of x, y, and zJ, we obtain for the direction numbers of the generators the quantities F,, Fq, pF, + qFQ. It was shown in the two preceding sections that when (22) is linear or quasi-linear, each surface representing a solution of (22) can be generated by a one-parameter family of curves, the characteristics, and this simple observation served as the foundation of a very satisfactory theory. It will be shown here that a somewhat similar situation holds in the general case. However, the theory is slightly more difficult, the essential difference between the linear or quasi-linear case and the general case

being that in the latter we must deal at each point with a family of

U

Partial Differential Equations

directions, namely, those of the

of the Monge cone, instead of

with a single direction. Assume for the moment that a solution z(x,y) of (22) has been obtained, so that (22) reduces to an identity in x and y when z, p, and q.are replaced and z,(x,y), respectively. At each point by the functions z(x,y), of the surface determined by the function z(x,y) a unique direction is determined [cf. (23)1 by the condition (25) dx:dy = Thus, introducing an auxiliary variable s, we find that the pair of differential equations (26)

determines in a unique manner, by Theorem 1-6, a one-parameter family of curves x(s), 1= z(x(s), y(s))] which generate the surface Furthermore, the functions z(x,y), p(x,y), q(x,y) satisfy along z(x,y). each of these curves the differential equations dz

dp

dy

dx

dx

Taking account of (26) and the equality alternative forms

dy

qr, we obtain for (27) the

= p,F,, +

+

+

dx

dq

(28)

Now, eonsidering as an identity in X and y, as explained above, and differentiating with respect to each variable, we obtain the equations

+ pF ±

+

0

F, + qF +

+

= 0

(29)

We can therefore rewrite (28) in the following form: dz

= pF,

+

d'

=

—

+ pP,)

d

=

—

(F, + qP1)

Equations (26) and (30) constitute a system of five first-order ordinary differential equations in the five unknowns x, y, z, p, q. (Note that the

reason for replacing (28) by (30) is that the former set of equations involves the additinijal quantities pz, p,, qx, qy.] Referring again to Theorem 1-6, we conelude that, given a point. (xo,yo,zo) on the surface and the values at this point of and z, [which must, of course, be consistent

with (22)), the equations (26) and (30) determine a "strip" (x(s), y(s), z(s), p(s), q(a)) lying on the surface—i.e., the curve (x(8),y(s),z(8)) lies on the surface, and the normal to the surface at each point of this curve has the direction numberE p, q, —1.

2. Partial Differential Equations of First Order

39

Now, disregarding the manner in which equations (26) and (30) were obtained, we consider them as determining a system of curves in the x,y,z,p,q space. ISince s dses not appear on the right side of any of the differential equations, it is not necessary to consider solutions of (26) and (30) as determining curves in the six-dimensional x,y,z,p,q,s space.j Taking account of the identity dp dy dz dx dF F F + F d.q by the right sides of (26) and (30), we con-

and replacing

clude that.

vanishes, and hence that F remains constant,' along each

solution curve of the system of differential equations under consideration. Those solution curves on which the constant value of F reduces to zero— and such curves must exist, in accordance with the assumption made at

the beginning of the section-—-are termed "characteristic strips" of equation (22).

We consider, as in Sec. 1, a curve C defined by equations (4) and satisfying the same conditions as those stated following (4). We assume that functions P(1) and Q(t) are given such that the equalities

r(t) =

+ ,P(t) ,Q(€))

0

(32a) (32h)

arid the inequality —

FqE'(t)

0

hold for each value of t. [It should be 8tressed that wheti (22) assumes the linear or quasi-linear form (1), the above inequality asserts that the

projection of C does not possess the characteristic ground direction at any point; as explained in Sec. 1, it is theii unnecessary to prescribe P(1) and Q(t), for these quantities arc then uniquely determined by (32a) and For each value of I we construct the characteristic strip which for a = 0 reduces to i.e., we determine the functions x (s,t), y(s,t), z(s,t), p(s,t). q(8,t) which satisfy (26) and (30) and are such i1(t). z(0,t) = P(I), q(0,t) = Q(i). p(O,I)

that x(O,() =

Taking account of (26) and (32c), we observe that the Jacobian — x,y, does not vanish on the segment t1 t 1,, s = 0, and hence (by continuity) in some of the s,t plane containing this segment. It is therefore possible to express s I as functions of x and y, and from this it follows, in turn, that the equations .v = x(8,t), y = y(s,t), z = Given a Pystein of differential equationi, a function of the (independent and dependent) inbies which remains constant along each solution curve is termed an "integral" of th5 system.

Partial Differential Equatione define a surface 8, which can be expressed in the form

z=

(33)

(34) = that Thus, the surface S contains curve C. It is to be expected that when p(s,t) and q(s,t) are expressed as functions of x and y they reduce to z, and and that (22) reduces to an identity in x and y. First, it follows from (32b) and the constancy of F along each characteristic strip that (35) ,q(s,t)) = 0 F(x(s,t) ,y(s,I) ,z(s,O Therefore, if we can show that (36)

the proof that S furnishes a solution to (22) will be cos.nplete.

By

to the identities

±

ZX4 +

Zg

zytJg

(37)

and the inequality (32c), we observe that to prove (36) it will suffice to establish the identities

z,—pz—qy.=O

ze—px4—qyg=0

(38)

The first of these identities is an immediate consequence of (26) and the first equation of (30). Referring to the left side of the second identity as V, we obtain, by differentiation with respect to s (and reversing the order of differentiation several times), the equalityt ôz,

ox.

(39)

•

Replacing the quantities x., y., z, p., q. by the right. sides appearing in equations (26) and (30), we obtain, after an elementary computation, V

(Ffx, + Fry, +

+

The quantity in parentheses is equal to

+ F,q,) — OF

(40)

and it now follows from (35)

that (40) simplifies to

—FV Solving (41) with the initial condition = 0 fcf. (32a)j, we conelude that V(s,t) m 0, and the proof is complete. A very close analogy has thus been established between the linear or quasi-linear equation (1) and the general first-order equation (22) in the 'The symbol

is used several times in (39) in preference to the subscript notation,

for clarity of exposition.

2. Partial Differential Equations of }'irst Order

41

of oourse, to the noncharacteristic çaee (32c). The analogy characteristic case, in which (32c) is replaced by the equality —

F,E'(t)

0

The extension to the ease of three or more independent variables is also entirely routine. The reader should formulate a theorem analogous to (and including as a particular case) Theorem 1. EXERCISES 1. Illustrate the theory developed in the present section by determining the two solutions of the equation z,z, — 1 passing through the straight line z 2t, y — s—st. 8. At a given point P in the x.y,z space, let the characteristic strip corresponding to each generator of the Monge cone (which is assumed not to degenerate to a single line, as in the linear or quasi-linear case) be determined. Each characteristic strip determines, in an obvious manner, a. characteristic curve tangent at P to one of the generators of the Monge cone. Show that these curves form & surface z(z,y) containing P which, in general, a conical vertex at P but which satisfies (22) elsewhere. This surface is termed the "characteristic conoid" associated with the point P.

3. THE CAUCHY PROBLEM

1.

Classification of Equations with Linear Principal Parts

In this chapter we shall be concerned primarily with partial differential equatons of the form q = ui,) (1) — f(;y,u,p,q) = 0 + + (p where a, b, and c are given functions of the independent variables x andy, while f is a given function of the five indicated variables. A large fraction of the partial differential equations arising in mathematical physics are of this form. We list here three especially important examples: Wave equation (2a)

Heat equation (2b)

Laplace equation (2c)

Any equation of form (1) is termed a "second-order equation with linear principal part," the expression "second-order" referring, of course, to the fact that (1) contains derivatives of second order but not higher, while the "principal part" is the combination of terms containing second derivatives, namely,

L(u) = + cu,, + L(u) is "linear" in the that the equality = aiL(u,) + a,L(u2) + 42

(3)

(4)

3. The Cauehy Problem

holds for arbitrary oonstants al, a, and arbitrary (sufficiently differeütiBy a "solution" of (1) is meant a function of z able) functions is,, and y define4 in a domain D such that (1) reduces in D to an identity in x and y when is is replaced by this function.

Equations of form (1) fall quite naturally into a three-way classification which can, perhaps, he introduced most simply by considering the effect of performing a change of independent variables. Let a pair of functions g(z,y) and h(z,y) be given, and suppose that in some neighborhood of the these functions are of class C'. If the Jacobian gh, — point does not vanish at (xo,yo) (and hence, by continuity, near this point), we are, assured that, in some sufficiently small neighborhood of this point, the pair of equations = k(x,y) g(x,y) say can be solved for x and y as single-valued functions of and y = &(E,,). Furthermore, these new functions are of class C'

corresponding to (x0,y0), so that. any in a neighborhood of the point C2 function of x and y is also of class C2 when considered as a function of The princpal part of (1) may be expressed in terms of the new and variables as follows (cf. Exercise 1):

+

+ cti,,,, =

a = =

=

± +

+

+ CE'

+

+ +

involve only ti and its first and the quantities indicated in (6) by derivatives. Evidently (1) is transformed into a second-order equation and au explicit whose principal part is precisely + + calculation (ef. Exercise 2) shows that $2 — ay

(6' —

—

(8)

does not vanish, it follows that the disSince the Jacobian — criminants b' — ac and $2 — x-y are both positive, both zero, or both negative at any given pair of corresponding points. Thus, if equation (1) is classified at any given point according to the sign of its discriminant, its classification will not change under (sufficiently differentiable) change of We now make this classification by defining equation (1) to be "hyperbolic," "parabolic," or "elliptic" at. a given point, or throughout a given domain, according to whether the discriminant — ac is posit.ive, zero, or negative at the given point, or throughout. the given domain. We note that equations (2a), (2h), and (2c) are hyperbolic, parabolic, and elliptic, respectively, in any domain. It is of interest and importance variables.

Partial Differential Equations by a suitable (especially in the hyperbolic case) that (1) can be whose principal part coinchange of independent variables, into a form cides with that of one of the three equations appearing in (2). This fact will be established in See. 3. EXERCISES 1. Work out (6) and (7) in detail. 2. Prove (8).

2. Characteristics Reasoning by analogy with the considerations of Sec. 2-1, one is led to

expect that a unique solution of (1) is determined by the requirements that the surface u(x,y) shall contain a given curve C and be tangent at each point of C to a specified plane. These conditions ("Cauchy data") are i1(t), h1(t), formulated analytically as follows. Five functions, h2(t), are given, and a function u(z,y) is sought which satisfies (1) and the further equalities u1(E(t),,,(t)) = h2(t)

(9)

[We assume, as in Sec. 2-1, that and do not vanish simultaneously, and also that all functions encountered possess whatever differentiability properties may be needed.J'

We observe that the five given functions of t cannot be prescribed entirely independently, for the equality (10)

+ h2(t),i'(t)

must be satisfied. [Cf. Assuming that (10) holds and that a function u(z,y) satisfying (1) and (9) exists, we proceed to determine the values of its second derivatives, at points of C. Just (= as (10) is obtained by differentiating the first equation appearing in (9), differentiation of the remaining two equations yields =

+

+

=

and respectively, we obtain in (1) and (11), for each fixed value of t, a system of three linear equations in the unknowns u,,, Evaluating the Replacing z, y, u, p, and q in a, b, c, andf by

determinant of this system, we find that the second derivatives are uniquely determined at any point of C unless, for the corresponding value of t, the "characteristic condition" cE'(t)2

—

26E'(L)i,'(t)

+

0

3 Tb. Cauchy Problem satisfied.

When (12) holds, the system possesses either infinitely many

solutions or none, according to whether the "compatibility condition" 2b c a f (13) 14(i) <3 0 Rank E'(t) 14(t) 0 E'(t) i,'Q) is satisfied or not. It is not difficult to show that the derivatives of u of all orders beyond the second are also uniquely determined at any point of C at which (12) does not hold. (Cf. Exercise 3.) It is therefore to be expected (since the Taylor expansion of the function ts(x,y) about a given point is determined by the values of all the derivatives at that point) that, in order to guarantee existenbe and uniqueness of a solution, it is necessary and sufficient that the characteristic condition (12) should not be satisfied at any point 'of C. 'It will be seen later that this expectation is confirmed for hyperbolic eque.tions.

It is to be noted that (12) involves only functions a, b c, E(t), and ,,(i), not r(i), h1(t), h2(t), or /; geometrically, this means that two excep-

tional, or characteristic, directions are determined at each point of the ;y plane by (12), and that if the projection of C on the ;y plane (the "ground curve" of C) does not either of these directions at any point, the higher derivatives are uniquely determined by C and the prescribed values of the first derivatives. A ground curve possessing a characteristic direction at each of its points is termed a characteristic ground curve, and the curve C in x,y,u space is termed a characteristic if it possesses a characteristic ground curve. From (12) we find that the characteristic ground curves are obtained by solving the differential equatIons

=

(14)

where Xj(x,y) and X2(x,y) are the roots of the quadratic equation'

am—2bm+c=O

(15)

The assertion made in the preceding paragraph that condition (12) determines two characteristic directions at each point of the x,y plane is, of course, not strictly correct, for the quadratic equation (15) may possess

complex conjugate roots, or real coincident roots, rather than real distinct roots. Examining the discriminant of (15), we find that the number of (real) characteristic directions at a given point is zero, one, or two, according to whether equation (1) is elliptic, parabolic, or hyperbolic there. tIf a vanishes at any point must be

a trivial modi6cstlon Is neeesry; however, it

that coefficients a, b, and c do not vanish

Partial Differential Equations

46

EXERCiSES 3. Prove that if the ground curve of C does not possess a characteristic direction at Rflv point, then all derivatives of u are determined at each point of C by the Cauchy data (9). 4. Prove that under a (sufficiently differentiable) change of independent variables characteristic directions map into characteristic directions (and hence characteristic curves map into characteristic curves).

Canonical Forms

3.

In this section we consider the problem of reducing equation (1) to an especially simple, or "canonical." form by means of a suitable change of variables. First we consider the case that (1) is hyperbolic at a given point (xo,yo).

Then the quadratic equation (15) possesses distinct real roots in some neighborhood of this point. Let functions be determined which satisfy the homogeneous linear first-order equations (16)

and the inequalities (17)

in a neighborhood of (x0,y4,).

The existence of such functions is assured then — assumes the form (X2 — and hence does not vanish, so that. E and ij may be used as independent variables in place of x and y; furthermore, by taking account of (7), (1.5), and (16), we see that the quantities a and7 defined in (7) must vanish. Therefore, (1) now assumes a form in which the principal part consists of the single term Dividing1 by we conclude that (1) may be reduced to the form

by the theory developed in See. 2-1. The Jacobian

u,u.t,'u!) = 0

(iS) This is the canonical form of a hyperbolic equation, and will be used in establishing the basic theory of such equations. An alternative canonical —

form is obtained by performing the additional change of variables =

E

+

'i'=

E—

this leads to the form fcf. (2a)1 —

—

= 0

(19)

We note that the two systems of characteristics of (18) are given by the equations constant, i = constant, while those of (19) are given by the equations E' ± = constant. If 8 vanIshes at any point, then we see from (7), by interchanging the roles of the 2

two pairs of variables (x,y) and (i,,,), that a, b, and c all vanish, a degenerate which we exclude from consideration. (Cf. the preceding footnote.)

4,7

3. The Cauchy Pràbi.m

If (1) is parabolic throughout a neighborhood of (xe,yo) we may obtain, but the function 'n exactly the same way as above, the function — since the Jacobian would now be dependent on differentinow vanishes. Instead, we choose for E(x,y) any (sufficiently 0. Then; as before, the introducable) function such that sib, — to vanish in a neighborhood of (zo,yo). causes tion of the variables Thus, (1) 0, we conclude that fi also Since, by (8), ay — assumesa form in which the principal part reduces to auE€; division by a yields the canonical form Un

0

—

X2 become complex Finally, if (1) is elliptic at (xo,yo), the functions a formal repetition of the reasoning used in the hyperconjugates, bolic case would lead to (18), where and are complex conjugates. — ii), we would Introducing the real variables E' = + n),',' =

then obtain the form

+

=

—

The foregoing procedure can be

(21)

0

stifled with little difficulty if the defini.

tions of the functions a, b, c can be extended to complex values of the independent variables—more precisely, if there exist analytic functions of the complex variables z + ix', y + iy' which reduce to a(x,y), h(x,y), c(x,y) for x' = y' = 0. When such an extension is not possible, the proof that (1) can he transformed into form (21) is quite deep. To sum up, we list the canonical forms which have been introduced in this section.' Type of (1) Hyperbolic

condition —

ac

Canonical form

+

>0

0

or .

Parabolic Elliptic

ac h2 — ac —

—

+

0

<0

+ a1,1,

+

.

.

— Q —

0

EXERCISE 5. Reduce the following equations to canonical form: (0) (b) (c)

(1

+

(1 + + (1 +

— (1 + y')¼1,1, 0. — 2(1 + x')(l ± x')tu1,1, — 0.

For convenience we replace the letters (21) by x and y. 1

+ (1 + y')2u,, (or

and

0.

(or v,') appearing in (18) to

Partial Differential Equations

4. The Cauchy Problem for Hyperbolic Equations The "Cauchy problem" is the problem of obtaining a solution of (1) which satisfies conditions (9) along a given curve C. The heuristic argument presented in Sec. 2 suggests that, for any point (zo,yo) at which C has a noncharacteristic direction, there exists a neighborhood within which the Cauchy problem is solvable. Although the afore-mentioned argument seems to apply equally well to hyperbolic, parabolic, or elliptic equations (subject to the afore-mentioned restriction on the direction of C, which is always satisfied in the elliptic case), a satisfactory theorem

can be obtained only in the hyperbolic case. We shall later discuss briefly the difficulties that arise when a Cauchy problem is formulated for a nonhyperbolic equation. Taking account of the discusston in Sec. 3, we confine attention, when (1) is hyperbolic, to the case a c 0, 2b 1, so that (1) assumes the form Uapp f(x,y,u,p,q) is1,) (p — u, q (22) the two systems of characteristics are given in this case by x constant, y — constant, the curve C can be expressed in the form y where g'(x) is always positive (or always negative). The Cauchy data (9) can then be expressed in the form Since

u(x,g(z)) — h1(x)

u,(x,g(x))

?&,(x)

u,,(x,g(x))

hs(x)

(23)

and the "consistency condition" (10) assumes the form + h1(x)g'(x) (24) Now, without loss of generality, we may assume that functions hi(x), h1(x) vanish identically. This assertion is justified by introducing a new dependent variable, h,(x)

v

is — h1(x)

—

— g(x)Jhs(x)

(25)

Then equation (22) is converted into a simiiar equation, while the Cauchy data now vanish identically. (Cf. Exercise 6.) Therefore, it suffices to consider equation (22) in conjunction with the simplest possible form of Cauchy data, namely,

ónC

(26)

We now state the principal result of this

Theorem I. Let f(z,y,u,p,q) be a continuous function of the five Indicated variables defined in a neighborhood N of the point (xo,y,,O,O,O), tThe analogy, both is content and in method of proof, with Theorem 1-6 noted carefully.

Problem

3. The

where (xo,yo) lies on a smooth curve C whose tangent is never horizontal

or vertical. Let / satisfy in N a Lipschitz condition (cf. Sec. 1-5) with respect to each of the variables u, p, and q. Then, in a sufficiently small neighborhood of (xo,yo), there exists a unique

solution of (22) satisfying the Cauchy data (26). Proof.

Let 5> 0 be chosen so small

that the region S defined by the inequalities lx — xol 5, 1!! — Vol < tul ö,

a,

B

lies entirely inside N; let m denote the maximum of Ill in 8; let k1, ic,, k3 denote the Lipschitz conS

stants associated with u, p, q respectively; and let Ill max (m,ki,k2,lc,). Now choose

3

-

a positive number 1 so small that all four of the following

inequalities are satisfied: MI2 < S

ML < a

Ml(Z + 2) < 1 1 < 25 Let A and B be the two points where C intersects the square formed by the lines x x0 ± 1/2, y ± 1/2, and let R be the rectangle formed by the horizontal and vertical lines through A and B. (Cf. Fig. 3-1.) Then we define within R three sequences of functions (p.), (q,) as follows:

u.(z,y)

0

(n 0)

dE di

=

=

(27a) (27b) (27c)

(27b) D denotes the region bounded by C and the horizontal and vertical lines through the point (x,y). It is necessary, before proceeding further, to show that the definitions of the functions u,,, p., q,. are meaningful. We assume that for some In

value of n, say n

k, the functions

pn, q, are defined and continuous

in R, and satisfy there the inequalities Then the function

5,

is

thus the definitions of

ö,

I

8.

well defined in R, and

Pk+1(x,y), qk+L(x,y) are meaningful. Furthermore, from (27b), we obtain the inequality (28)

Similarly, by

differentiating

pk+1(x,y)

=

(27b) with respect to z, we

obtain d,7

(29)

•

Partial Differential

50

where the integration is performed over the vertical boundary of D. We immediately obtain from (29) the inequality (30a) Mt < o and, similarly, we obtain

Ml < S

(30b)

Inequalities (28), (30a), and (BOb), together with the fact that for n = 0 the hypotheses imposed above for n = k arc obviously satisfied, show that definitions (27b) and (27c) are meaningful for aU n. Replacing it in (27b) by .n + 1, subtracting, and taking account of the Lipechitz condition,we obtain, for any point in R, the inequality —

max

u.,+t(x,y)I MZ2SI,

+

+ IP'.44 —

— u,11

Similarly,

—

from (29) we obtain (32a)

Pn+:(Z,Y) —

and analogously we obtain (32b)

MZSII

—

Adding these three inequalities and taking the maximum of the lefthand side, we obtain Mi(l

a

+

2)

tiivial induction we

S,,

(34)

&'S0

and we conclude that each of thc three series

ft0

—

uft)

!

(1m+i — p,.)

—

(35)

p&0

hence, taking

is dominated by the convergent geometric series IL

—0

account of the telescoping character of the series (35) and the fact that 0, we conclude that the seqiwiices are tb (p, }, qo

uniformly convergent throughout R to continuous functions, which we denote by u, p. q, respectively. Passing to the limit under the integral sign in (27b) and (29) (as in Sec. 1-s). we conclude that u(x,y) =

and

p(x,y) =

dE

(36)

(37)

3. The Cauchy Problem

Differentiating (36) with respect to z and (37) with respect to y, we obtain f(x,y,u,p,q)

= p Similarly,

(38) (39)

q

From (38) and (39) we conclude that u1,

=

Furthermore, from (36). and (37) we conclude that and p (and similarly q) vanish if (x,y) is on C, for then D shrinks to the single point equation (22) and the Cauchy data (26). (x,y). Thus, u

The uniqueness portion of the theorem is established very much as in the proof of Theorem 1-6. Suppose that. in the afore-

was

meiitioned rectangle R a second solution, U(x,y), of the same Cauchy problem exists. Then U and P = are expressible in exactly the same forms as (36) uid (37). Subtracting (36) and (37), together with

an analogous equation for q, from the corresponding equations for U, P, and Q (.= Ui,), adding, and exploiting the condition, we obtain

± jq — Ml(t +2) max (ju

max (tu — UI + jp —

—

UI

+ !p — Pf

+

— QJ)

(41)

Since MZ(i + 2) <1, we conclude that each side of (41) must vanish, and hence the uniqueness is established. We conclude this section with a few remarks concerning various aspects of the Cauchy problem. 1. The proof of Theorem I shows that the value of the solution u of (22) at a point P does not depend on alt the Cauchy data (23), but only on that portion pertaining to the segment of C (the segment of determination) which is intercepted by the and vertical lines through P. Conversely, the Cauchy data at a point Q on C influences the values

of the solution only in a portion of the region in which the solution is defined; this portion (the region of effect of Q) is readily seen to consist of two of the four quadrants determined by the horizontal and vertical lines through Q, namely, those quadrants which do not contain the curve C. Analogous statements hold, of course, for hyperbolic equations in noncanonical form, with horizontal and vertical lines replaced by characteristics. 2. A slight ezteusion of the analysis presented in the proof of Theorem I shows that Funder very mild hypotheses on the function f(x,y,u,p,q)J the Cauchy problem for equation (22) is a "stable" problem, in that slight changes in the Cauchy data (23) result in alight. dhanges in

52

Partial Differential Equlitiona

the solution. This fact is of great physical significance, for a mathematical problem which is not stable cannot be expected to correspond accurately to a physical problem. 3. As indicated at the beginning of the present section, it is not possible to extend Theorem 1 in any straightforward manner to elliptic or pam.

bolic equations, despite the heuristic argument presented in Sec. 2. This is demonstrated conclusively by considering the Laplace equation (2c) with the following Cauchy data, preecribed on the z axis: u(x,O) = 0 =0 v1,(x,O) = g(x)

Anticipating from Chap. 6 (or from the theory of analytic functions of a complex variable) the fact that a function satisfying (2c)—even on only one side of the x axis—and the first condition of (42) must possess derivatives of all orders on the x axis, we see, in particular, that the function g(x) appearing in (42) must be indefinitely differentiable. Thus, for example, there can be no solution if g(z) for g"(x) fails to exist at x 0. Continuing further, we may note that even when g(x) is so chosen that the Cauchy problem ía solvable, the solution will fail to possess the property of stability discussed in aspect 2. The classical illustration of this fact is furnished by the following example: If g(x) — 0, the solution is given by u 0, but if g(x) n—1 sin nx (which can be made uniformly small by choosing the parameter n sufficiently large), the solution is gwen by u n2 sinh ny sin nx, which does not become small as n increases.

A counterpart of the foregoing remarks is furnished by the fact that the Dirichiet problem (cf. Sec. 6-6) is not, in general, solvable for a hyperbolic equation. Referring to the especially simple hyperbolic equation 0, we observe that any solution must be of the form /(z) + g(y), and from this fact it follows that if P1,P, and are pairs of opposite vertices of any rectangle with sides parallel to the axes, then u(P1) + u(P,) = u(Q1) + u(Q,). Thus, if values of ti are prescribed on the boundary of the rectangle which are not consistent with the above equality, no solution can be found. 4. In complete analogy with the concluding remarks of Sec. 1-5, we note that the restriction in Theorem 1 to a sufficiently small neighborhood Is unavoidable, but that a "global" or "in-the-large" theorem

holdi when equation (22)18 linear.

More precisely, consider the equation (43)

where a, b, c, and / are given functions of x and y defined in a rectangle R. Then, for any sufficiently smooth monotone curve C lying entirely in R and any set of Cauchy data prescribed on C, there exista a unique

53

3. The Cauchy Problem solution

of (43) defined throughout the rectangle determined by the

horizontal and vertical lines through the end points of C. EXERCISE 6. Prove that substitution (25) reduces the Cauchy problem consisting of equations (22) and (23) to another such problem in which the Cauchy data vanish identically.

5. The One-dimensional Wave Equation Some of the material presented in the preceding sections, in particular of the role played by the characteristics, can be illustrated the

by the especially simple example of equation (2a), which permits all the equations involved in the analysis to be written out quite explicitly. As explained in Sec. 3, equation (2a) is equivalent to the equation = 0, but we prefer form (2a) on account of its immediate physical significance; we shall, however, replace the variable y by t, to represent the time variable. First, we consider the problem with Cauchy data u(x,O)

u2(x,O) = f'(x)

f(x)

0

(44)

where f(x) is defined for all x. Then the solution of the Cauchy problem defined by (2a) and (44) is given by (cf. Exercise 7)

u=

(45) + t) + f(x — 1)] Thus, a solution exists and is uniquely determined for all values of x and t. in order that (45) should actually constitute a solution of (2a), it is necessary that f"(x) exist for all values of x. Suppose, however, that f"(x) fails to exist for a certain value of x; for the of dealing with a specific example, suppose that f(x) xt for positive x and

f(x) for negative x, so that f"(O) fails to exist. the following form' (ef. Exercise 9): (

x2+gt 21x

Then u assumes

x>1

ixIt

x<—t

It is immediately evident from (46) that. u, and 14 are continuous without exception, but that each of the second derivatives, u., fails to exist along the lines x — I = 0, x + I = 0, undergoing jump diecontinuities as point (x,t) crosses either of these lines. It is to be noted that lines are the characteristics through the point (0,0) at which a aWe confine attention, here and subsequently, to positive values of I (as is donem most physical but (46) holds, with trivial modifications, for 1 <0.

Partial Differential Equations

54

breakdown in the differentiability properties of the prescribed Cauchy data occurs, so that this discontinuity may be said to propagate along the characteristics. The precise nature of the nropagated discontinuity can be seen most. conveniently by introducing the change of variables so that (46) assumes the form (when account is x+t= x — t taken of the restriction of t to positive values) I

E>O,i?<0

n<E<0

and remain continuous across the From (46') it is seen that 0, while u1, is discontinuous across this line; similarly, line = 0, is remain continuous across the Jizce = 0, <0, while u,, and discontinuous across this line. Returning now to (46), we may assert that the second-order normal derivative is discontinuous across each point of the alore-mcntioned characteristics x ± t O,.while the second derivatives involving at. least one differentiation in a characteristic direction remain across a characteristic. Essentially the same phenomenon occurs for the most general hyperbolic (or parabolic) equation.

If the function /(x) appearing in (44) is defined only on some finite

segment of the x axle, say the interval 0 x 1, then the Cauchy data determine a solution only in the triangle bounded by the x axis and the lines x = t, x I — t. Physically, this corresponds to the fact that the vibrations of a string occupying in its rest positión the specified portion of the x axis are influenced by the constraints which are imposed on the ends of the string, the time at which the point P with abscissa x is first affected by the constraints being equal to the distance of P from the nearer end of the string. Thus, it may be said that the influence of the constraints on the ends (boundary conditions) is propagated along the string with unit velocity. (Similarly, in the case of the equation U,1 — the influence is propagated with velocity c.) Confining attention to the simplest boundary conditions, namely, u(O,t)

0

u(1,t)

0

(t

0)

(47)

readily confirm that the solution to the problem consisting of (2a), (44), and (47) ía still given by (47), provided that the function f(x), originally defined on the interval 0 z 1, is now defined for all we

values of z by extending its definition in the following manner:

1(x) = —f(—x) (Cf. Exercise 10.)

ffr + 2)

f(x)

(48)

3. The Cauchy Problem EXERCISES 7. Derive (45). 8. Show that when the third condition of (44) is replaced by u.(x,O) solution is given by is

—

[f(x +t) +f(x

g(z), the

u(E)dtj

—1)

Show that in this case, an well as in the case of (45), the displacement u(x,t) can be interpreted as the superposition of two disturbances moving with unit velocity to the right arid left, respectively, without change of shape. 9. Derive (46). 10. Confirm the solution given for the last problem discussed in the text. and show

that the displacement u(x,t) can be interpreted an the superposition of two dinturbances which move with unit velocity and are reflected, with change of sign, at the end points.

6. The Riemann Function For a linear hyperbolic equation (43), the solution of the Cauchy problem can be expressed in a form which brings out clearly the

of uniqueness and stability which are mentioned in Sec. 4.

Let the

left side of (43) be denoted by Lu, and let the "adjoint" operator L* of L be defined by the identity = — (av), — +cv It is easily confirmed (cf. Exercise 11) that the identity vLu — uL°v = (hvu), — aiw). + — + but,)1 holds for arbitrary functions u and v, and so, with the aid of the divergence M

JrX

FIg. 3-2

theorem (cf. Sec. 6-3), we obtain, referring to Fig. 3-2, ffD (vLu —

—

uL5v) dx dy —

i: + buy) dx + i: —

=

+f

+ auv) dy —

+ auv) dy

-

+ buy) dx

(51)

Partial Differential Equations

56

Employing integration by parts, we obtain

f.

dx

dy +

—

—

+

u(av —

v,,)

dy +

u(bv —

dx

(52)

Now suppose that it is possible to determine a function v, depending such that the following on the pair of points P [= (x,y)} and Q (= conditions are all satisfied: L*V

av —

(53a) (53b) (53c)

0

by —

=

0 0

on the line y

on the line x =

(53d)

(It is understood that the independent variables in the above equations of Q are to be are the coordinates x, y of P, while the coordinates Then, taking account of (51) and (52), we considered as parameters.) obtain for any function u satisfying (43) the integral representation

+

u(Q)

+

f,'

—

+ JfDfrdxdv + buy) dx —

—

+ auv) dy

This formula, which can be written down once the appropriate function v

is known and the Cauchy data are prescribed, demonstrates, quite independently of Theorem 1, the unique determination of the solution of the Cauchy problem by the Cauchy data, and it is easy to prove, on the other hand, from this formula the existence part of the theorem as well. (Cf. Exercise 13.) Furthermore, it is immediately evident that small changes in the Cauchy data result in small changes in the solution, so that the Cauchy problem is shown to be a stable one. The function v defined by equations (53a) to (53d) is known as the "Rieniann function" associated with the operator L, and is denoted as EL(P ;Q), or Ri.(x,y;E,i,). The question of existence of this function is readily settled. For each choice of the parameter point Q equations (53b) and (53c) become ordinary differential equations of first order,

so that in combination with (53d) they uniquely determine v along the characteristics through Q; in fact, v is given along these lines by the formulas v(x,i)

exp [f b(t,i) SC]

v(E,y)

exp [L' a(E,t) di]

We are thus led to a so-called "characteristic Cauchy problem," in

S. The Cauchy Problem

which a solution of a hyperbolic equation [(53a) in the present case) is sought whose values are prescribed on a pair of intersecting character(Cf. Exercise 16.) A minor modification of the proof of Theorem 1 shows that this problem is solvable, in essentially the same manner as

istics.

the noncharacteristic Cauchy problem. Thus the existence of the Rienmnn function is established.

In two simple cases the Riemanri function can be expressed in an Lu) = then it is seen that the Riemann

explicit form. If Lu function is given by

1

If L(u)( L'u) is given by

+ Cu, c being any const4nt, the Riemann func$ioa Jo(2[c(z

[Since

—

—

the Taylor expansion of the Bessel function

namely,

contains only even powers of £, the ambiguity in the square root appearing in (57) is of no importance.] EXERCISES U. Confirm (50), and show that L is completely determined by the requirement that the left aide of (50) ihall be expressible as a "divergence eapression" + (.

.

12.

Prove that any linear ecoond-order differential operator L, Lu —

+ 2bu1,, +

+

+ nit, + fit

has * unique adjoint L such that the requirement imposed in the preceding exercise is s&tisfled, and then show that

— L.

13. Prove that the function it defined by (54) satisfies (43) and the Csuehy data that are inserted on the right side. 14. Show that the Rieinann functions and RL. are related by the equality —

so that if

.el.f-adjofnt (i.e., L — L*), RL is symmetric in the pair of points (x,y),

15. Show that it is possible to determine from the values of a solution of (22) on a pair of intersecting characteristic, the values of it, and it, on both characteristics. 16. Show that a characteriatfe Cauchy problem can be reduced, by means of a substitution similar to (26), to one in which the unknown function i8 required to vaThah on both characteristics. Then show that the iteration procedure empkiy.d in the proof of Theorem 1. applies in the present cue.

Partial Diffetential Equatlona

7 ClassIfication of Second-order Equations in Three or More Independent Variables We consider briefly in this section a generalization of equation (1) , to the case of three or more independent variables: xi, X2, We begin by considering the effect of a (real) nonsingular homogeneous linear transformation (59)

= Z on an expression of the form

L(u) = jAt —1

are given constants. Restricting attention to the we may assume that case that u is of class C2, so that =

where the coefficients

= An elementary computation shows that under the substitution (59) the expression L(u) assumes the form L(u) =

(62) j,k—1

where

— iM' —1

Now, by the theory of quadratic forms, it is possible to select the coefiii ficients cij appearing in (59) in such a manner that aim 0 If and = 1, —1, or 0. While there is wide freedom in the choice of the it is of importance that for every possible choice the coefficients

afore-mentioned numbers 1, —1, and 0 appear the same number of times, which we denote by ni, n_1, n0, respectively. Thus, the numbers n1, n_i, n0 are invarianUy with the differential operator L appearing in (60), so that L(u) can be reduced to a uniquely determined canonical form, *i —4—n..t

L(u)

—

k—i

k—'i*+l

n or now classify the differential operator L as follows: If = n, L is termed elliptic; if n0> 0, L is termed parabolic; if n0 = 0 and 0
59

S. The C*uchy Problem

The same classification can be formulated in terms of the quadratic

If this form is positive- or negative-definite, L is

form

elliptic; if the form is degenerate (i.e., reducible by a suitable linear transformation to a form in fewer than n variables), L is parabolic; if the form is not degenerate arid not definite, L is hyperbolic. (Cf. Exercise 18.) Now we consider a differential equation of the form

±—1

—

f(x1,x2,

.

.

.

.

.

.

,p.)

0

(p

(65)

where the aJk may be given functions of the x's, all defined in some domain

D of the

.

.

.

It is now no longer true, in general,

,x,,) space.

that the right sides of (60) arid (62) are equal under the substitution (59).

However, the right sides of the afore-rnentioned equations will differ only by an expre.'ision involving the first derivatives of u; this will hold true even if the change of variables is no longer linear, as in (59), provided the quantities c,, now are taken as

so

that

'I

LI

tm—i

If (65) is now classified at each point P according to the classification assigned above to the "principal part" L(u) (the coefficients

being assigned their values at P), it follows from the above discussion that (65) will become, under a sufficiently differentiable change of variables, an equation of the same type (elliptic, parabolic, or hyperbolic) at the point P' corresponding to P. In contrast to the case of two independent variables (and of an arbitrary number of independent variables with constant coefficients it is not possible, in general, to effect a reduction of the principal part of (65) to a canonical form (64). This fact is suggested by the observation that (66), taken together with the symmetry condition (61) (which

implies the symmetry of the ujk) and the requirement that the new principal part shall be proportional to the right side of (64), constitutes a system of + 1) — I equations in the n unknown functions Ek(X1,X2, . . . Since + 1) — 1 exceeds n for n 3, it is to be that this system will be unsolvable. Although this counting argument does not constitute a proof, the impossibility of a reduction to canonical form can be exhibited by means of a specific example.

Partial Dlfferentl*A Equations

charFinally, we consider briefly the formulation of the concept acteristics and of the Cauchy problem for equations of the form (65). By the Cauchy problem we now mean the problem of deterniining a solution u of (65) defined in a domain containing a given (n — 1)-dimen-

sional surface 8, such that 'u and all its first partial derivatives assume specified values (Cauchy data) on & In analogy with Sec. 2, we shall' define a characteristic surface as one on which the Cauchy data fail to determine [together with (65)] the values of all the second derivatives. that a given surface S be a characteristic, let To obtain a or all of the plane us first consider the case that S consists of 0. From the given values1 of u, U,1, ui,, . , u. it is evidently possible to determine at each point of S the values of all the second derivatives except u,,,,, which, in turn, can then be uniquely determined from (65) unless a11 = 0, in which case either no value or infinitely many values can be determined. Now, turning to the general case of a surface defined by an equation of the form E(xi, X2, . . , = 0, we determine additional functions of the x's such that the Jacobian . , of

.

o(E,

. . .

,

. .

.

.

, X,iI.

does not vanish.

Changing variables in (65), we con-

by the argument presented above, that the given surface is a characteristic if and only if the coefficient au vanishes. Referring to (66), we thus conclude that the surface = 0 (or, more generally, E = constant) is a characteristic if and only if

dude,

'I

'V' S., — I

aO

EXERCISES 17. Work out the details leading to (62) and (63). 18. Prove the equivalence of the two definitions following (64).

8. The Wave Equation in Two and Three Dimensions In this section we shall consider briefly a specific Cauchy problem for the simplest of all hyperbolic equations in three and in four independent variables, namely, the wave equations U2,

tL1, +

+ + IL,, —

Ugg

= 0 =0

'Actually only u and u,, need be given; in analogy with (tO), the given values of , u, would have to agree with those obtained by differentiating the values of ti given on 2. SimIlarly, in the general case only u and its derivative in thó normal to the surface can be prescribed freely.

ti1,, u,, ...

3. The Cauchy Problem

[in keeping with the physical significance of the independent variables, we denote them as x, y, z, t (time) rather than as in the preceding section.J Since equation (67) assumes the forms (69a) =0 + — — (89b) 0 and + +

is not a characteristic of either (68a) or (68b). It is therefore plausible, although Theorem 1 does not carry over in any straightforward way to higher dimensions, that the Cauchy problem is solvable for Cauchy data prescribed on the afore-mentioned

it follows that the surface t =

0

surface.

Somewhat surprisingly, the most effective procedure for obtaining solutions of this pair of problems involves solving first the one in the greater number of variables, and then solving the other by treating it as a particular case of the problem that baa been solved. We therefore begin by considering (68b) in conjunction with the Cauchy data: u(x,y,z,0)

u.(x,y,z,O) =

It suffices to show how to obtain a solution in the particular case that 0; for if we momentarily denote by a solution of (68b) satisfying (70) with 0, then a simple calculation (cf. Exercise 19) shows that u u,(#) satisfies (68b) and (70) with 0. Thus, by superposition, we find that + u,(#) solves the given Cauchy problem. If depends only on the distance r from a fixed point Q, then a solution depending only on r and t can easily be obtained, for the assumption u = u(r,t) and the substitution v = ru reduce (68b) to the form

(r>O) and the conditions (70) (with

(71)

0) assume the form

v(r,0) = 0

r./,(r)

(72)

Equations (71) and (72) constitute a linear noncharacteristic Cauchy problem, and hence determine a unique solution in the region () < JtJ
r> 0. However, the original problem requires a solution defined for £> 0, r> 0, and this can be obtained by the following device. The definition of v in terms of u shows that v must vanish for r = 0; if the definition of

is extended to negative values of r as an even function •(—r)1, then (71) and (72) determine the solution v

which vanishes for t

1 —

('+t Jr—

(73)

0. Since the integrand in (73) is odd, the lower limit may be replaced by Ir — tj. Reverting to the variable u, we thus

PartIal Differential Equatlona

62

obtain the solution

(r > 0, 1 > 0)

u(x,y,z,t) = u(r,t) =

(74)

Ir—tt

If the function

prescribed in (70) is expressible as the sum of a (finite) number of functions . . . , each depending only on the then the solution is given by a sum , distance from points . . of the corresponding solutions each having the form (74), namely, .

'u(xy.zt) =

2r1

(r, fr.—tI

It may then be expected that the solution for an arbitrary function can be obtained by carrying out a suithble limiting process on (75). It is more convenient, however, to return to (74) and rewrite it in a manner which immediately suggests the solution for an arbitrary Referring to Fig. 3-3, in which S

denotes the sphere of radius I with center at P and Q denotes the center

Fig.

of symmetry of

we find by elementary geometrical considerations

(cf. Exercise 20) that (74) can be rewritten in the form' u(x,y,z,f) =

ff dS

(76)

Since (76) involves integration over a surface determined entirely by P [= (z,y,z)1 and I, it follows that equation (75) can also be written in form (76), and it is natural to expect that (76) furnishes the solution in the general case of a more or less arbitrary function This expectation will now be confirmed.

Noting that the right side of (76) can be rewritten in the form where 3(8) denotes the mean value of over the sphere 8, so that lim 3(S) = it follows that

u(P,t) =

0

u,(P,b) = li ti(P,t) = urn 3(8) =

(77)

The prescribed Cauchy data are therefore satisfied. It remains to show that (68b) is satisfied. (76) in the form

= If

For this purpose, it is convenient to rewrite

+ ait, x2 + a2t, x3 + a3t) dfl

(76') 1Although Q i.e shown outside S in Fig. 3-3 (corresponding to the inequality i
3 The Cauchy Problem denote direction cosnea of the where Zi = z, Z2 y, X3 z, radius vector connecting the center P to the point of integration, and dii dS). represents the element of solid angle subtended at P by dS (dii = obtain from (76') Performing the indicated differentiations, we

t'

+ and

2/f

4Tuu

+ u,) dii

t

+

ff tJf

dii

(79)

1.-i

using in (78) the fact we obtain = t, and in each integral the fact that

Next we rewrite these integrals by setting

that

(78)

dii

V.'

+

+

=

— Vu)

+Y

=

—

Ei#u,)] dS

(80)

The integrand is seen to be the (outward) normal component of the vector whose jth component is the quantity appearing in brackets. Applying the divergence theorem,1 we conclude that the right side of (80) is equal to the integral over the interior of S of the quantity

j'.l

—

—

and a simple computation (cf. Exercise 21) shows that this quantity vanishes identically.

Thus, u 8atisfies (68b), as was to be proved.'

Turning to equation (68a), one might attempt, as above, to obtain first a solution radially symmetric Cauchy data by reducing the problem to one in only two independent variables (t and the distance from the center of symmetry). However, it is easily shown that such an attempt must fail. Instead, as indicated above, we employ a very simple but effective method (Hadaniard's method of descent) which furnishes the solution to the present problem by considering it as a particular case of the problem which has been solved. We simply consider the given function as a function of the three variables z, y, z by employing the obvious definition çb(x,y,z) =

(82)

The integral of the outward normal component of a vector field over a closed surface equals the integral over the enclosed volume of the divergence; cf. Sec. 0-3. 2 We omit the proof of the fact that (70) furnishes the unique to the Cauchy problem under consideration.

Partial Differential Equation. The integral (76) is clearly independent in this case of z, for changing z

over a sphere S' obtained by translating S parallel to the z axis. The function u(xy,z,t) thus obtained may be merely involves averaging

considered as a function of the three variables z, y, 1, and since identically, (68b) reduces to (68a).

vanishes

Although the foregoing remarks may be considered as furnishing the solution to the problem under consideration, it is instructive to develop for

the solution an explicit formula, which will demonstrote a remarkable contrast between the wave equations in three variables (68a) and in four variables (68b). To obtain such a formula, we merely observe that, since 4) now depends only on x and y, the integral (76) can be converted into an integral over the disc A in the ;y plane with center at (z,y) and with radius t. A simple geometric consideration shows that the. element of surface area dS located at on S and its projection dA located at are related by the simple equation dS !asht dA where

is defined following (76'). —

Furthermore,

t[t2

— ('1

and so (76) assumes the form

u(;y,t) =

Jf

—

(84)

(The appearance of the factor l/2ir,ins tead of l/4ir, is explained by the fact that two surface elements on S project onto each element of A.) We now point out the striking contrast between (76) and (84) which has been referred to previously. If, for three dimensions (four independent variables), the function 4)(x,y,z) differs from 0 only in a bounded region R (so that the function u(x,y,z,t) may be considered as describing the propagation of a disturbance in a medium which was originally (t = 0) at rest except in R], then u remains zero at a point P located outside R until a time equal to the minimum distance from P to This follows from the observation that for 0
t>

again consists entirely of points which were originally undisturbed, so that the value of u at P returns permanently to the value zero. Thus, we may assert that the influence of an initial disturbance of limited extent is also limited in its duration at each point. In contrast to this, we note from (84) that, while a disturbance concentrated in a bounded region R at

______________________ 65

3. Tb. Caiseby Problem

time t = 0 does not exert an influence on a point P outside R until some subsequent time to, it is not true that the effect of the disturbance dis appears when I exceeds the value Li defined above. Instead, (84) shows that for I> the value of u at P may be expressed in the form x,y,I)

=

1"

27

JJR[t2_ (E—x) —(el—

In general this quantity will not vanish; in particular, if

d d '1

85

is positive

throughout E, u will become and remain positive at point P after the effect of the initial disturbance begins to be experienced. By taking account of the remarks which follow (70), we can see that the contrast described above between the behavior of the solutions in two and in three dimensions persists when the restriction that shall vanish identically is removed. The one-dimensional wave equation (2a) shows a behavior intermediate between that of (68a) and (68b). can be seen by referring to (45'), if the initial values of u and differ from zero only on a bounded portion of the x axis, the effect of the initial disappears at any given point after a finite time, but the influence of the initial velocity g(z) is felt permanently, for, at each point, u reaches (after a finite time, and keeps permanently thereafter) the value fg(x) dx, this integral being extended over the entire x axis, or, equivalently, ovar the portion of the x axis to which a nonzero velocity was initially imparted. For a greater number of variables, it is found that solutions of the wave equation in an even number of dimensions (odd number of independent

variables) exhibit a behavior similar to that found above in the twodimensional ease (68a), while in an odd number of dimensions the behavior of solutions is similar to that obtained in the three-dimensional case (68b). This striking differentiation in the nature of wave propagation in even- and odd-dimensional spaces is known as "Huygens' principle." EXERCISES 19. Prove the assertion which fellows (70). 25. Derive (78) from (74). 21. Prove that expression (81) to zero. 22. Solve (68b) with the Cauchy data 9.

z° + 2zy + 3z'.

The Legendre Transformation

In Sec. 3 it. was shown that equation (1) can be reduced to one of several simple (cahonical) forms by means of a suitable change of independent variables. In the present section we present a brief discussion of the

Legendre transformation, which often proves effective in the study of

Partial Differential

66

partial differential equations possessing a more complicated structure than (1). On the other hand, the Legendre transformation, in contrast to the transformations considered in Sec. 3, involves the (unknown) solution to the differential equation under consideration, and the resulting difficulties connected with reformulating the conditions which are imposed on the solution (aside from the differential equation itself) may outweigh the advantages furnished by the simplification of the differential equation.

For simplicity, we consider a differential equation of first or second order, but it will be apparent how the method extends to equations of higher order. Let the given differential equation be expressed in the form

= 0 If a solution u(x,y) exists in a neighborhood of a given point and if — — u.), or does not vanish in this neighborhood, then the pair of equations (87)

can be solved in a neighborhood (usually smaller than the afore-mentioned.one) of (z0,y0) for and y in terms of and q, so that (87) defines a valid transformation of independent variables. If we introduce a new dependent variable, in place of u(x,y) by the substitution (88)

we readily obtain from (86) an analogous equation which must be satisfied by w. Differentiating both sides of (88) with respect to E and taking account of (87), we obtain (89)

Similarly,

= y. Thus, if (86) does not contain any second derivatives, the new dependent variable w must satisfy the differential equation obtained from (86) by replacing x,i y, is, tsr, and is, with WE, EØE + ',, respectively.

If (86) actually contains second derivatives, we proceed as follows: Differentiating the first equation of (87) with respect to E and we obtain — w, E,

I=

+ +

= + = + Solving this pair of equations, we obtain (cf. Exercise 23)

=

ts_

t&cxX,,

=

,

—-

—

(91)

—

Similarly, WI, — Will

(92)

Thr Caue

6'

Problem

'these equations evidently citable us to convert (86) into a differenti& equation of second order for the dependent variable We repeat that the Legendre transformation is applicable only if does not vanish. The vanishing of this expression has v — geometrical interpretation, namely, that the surface u(r,y. very simple

is developable—i.e., a sufficiently small neighborhood of any point on the surface can be mapped onto a plane region without changing the length of any curve. Thus, when the Legendre transformation is to be employed,

separate consideration must be given to determining whether (86)

possesses developable solutions. (Cf. Exercises 24 and .25.) We conclude this section wit.h a very brief discussion of the application

of the Legendre transformation to a problem of great physical significance. The plane steady (i.e., time-independent) flow of a perfect fluid can be described by means of either of two functions: the "velocity or the "stream function" #(x,y). The rectangular potential" components U, V of the velocity vector at each point of the flow ar, related to and by the equations p denotes the density of the fluid. Differentiating the• first equation with respect to x and the second with respect to y and adding, we obtaiu

+ (P#w)N = 0

If p is constant throughout the fluid, then (94) reduces to the Laplace equation + •,, = 0. However, if the fluid is compressible, then the variability of p must be taken into account. We assume, that an "equathe density p and the pressure p is known, and that tion of state" ((12 + VI)½1 is satia"Bernoulli's law" relating p, p, and the speed fled, namely,

+f

constant

(The integral may be expressed in terms of either p or p with the aid of the equation of state.) Since p may now be expressed as a function of q, we obtain from (94) the more explicit say p /(q) + form

f + [Here

r

f'(q)

qj)#ZX + J

This equation is not linear, for the function

appears (through its first derivatives) in the coefficients of the second derivatives; however, since the derivatives of highest order appear

Partial Differential Equations linearly, (96) is said to be "quasi linear." If we now apply the Legendre transformation we immediately see that (96) goes over into a linear equation, namely, 0

(Here q =

(E'

(97)

Thus, instead of considering the flow in the

+

"physical" plane of the variables x, y, one may study it in the "hodograph" plane of the variables U= V= vrhe

discriminant of (97) works out to — I

From (95) we obtain

=

—pg,

cf/f,

or —1 —

and from this we find that the dis-

criminant can be rewritten in the form

of sound or "acoustic velocity," is equal to is positive.)

—

— 1),

where a, the velocity

•

(It is assumed that

u-.)

Thus, equation (97) is elliptic or hyperbolic according to

whether the "Mach number" M = q/a is below or above one, respectively. if the 83w is "subsonic" (31 <1) in a portion of the region under consideration and "supersonic" I) elsewhere, equation (97) must be treated in a region of the plane within which the type does not remain fixed

EXERCISES 23. Auming that the Legendre transformation is applicable (i.e., that — — u,1 does not vsüish), prove that — 1, so that the determinant of system (90) does not vanish. 24. Prove that the only nondevelopable solution of the equation ZU* +

is given by it

—

+ u,2

+ 14V'.

25. Prove that the Legendre transformation cannot be applied to the equation X1Lz + Vuw — U.

4. THE FREDHOLM ALTERNATIVE IN BANACH SPACES

The principal objective of the present chapter and the one following it is to develop the theory of linear integral equations (of Fredhoim type). There is a very close connection between integral and differential equations, for it is often possible to formulate an integral equation which is

equivalent to a given differential equation plus additional (initial or boundary) conditions. This has already been seen in Chap. 1, where the taken together, differential equation (l-33b) and the condition g(XQ) = are equivalent to the integral equation (1-45). Further examples of this equivalence will be presented in Chaps. 7. and 9. Instead of dealing explicitly with integral equations, we shall present a

more abstract development, in terms of "normed linear spaces," which includes, as will be shown, the theory of integral equations as a particular "concrete" interpretation. Thus, in addition to providing an exposition of the theory of integral equations, these two chapters may serve as an

elementary introduction to the important subject of abstract linear analysis.

1. Linear Spaces This section is devoted to a brief discussion of the most important purely algebraic concepts relating to linear spaces. 1. A "linear space" (or vector space) V is a set of objects (elements or vectors) in which two operations, + (addition) (multiplication), are defined, where these two operations satisfy the following requirements. 1. For every pair of elements)', g of V, the elementf + g is defined, and

f+g = g+f.

60

Partial Differential Equations

10

2. For all elements f, g, h, the equality I + (g + h) = (1+ g) + h

From this it follows, exactly as in the development of the realnumber system, that any finite aunif, + /s + • + of elements has holda.

an unambiguous meaning. 3. There exists a sero element o such that, for every element!,! + o — f. 4. For every sealar' a and every element I, the element a •f is defined. 5. For every element f, 0 ff = o and 1 f = f.

a •f+ a•g,

6. For alt scalars a, $ and elements!, g, (1+9) = (a$).f.

It is scarcely necessary to point out that the real- and complex-number systems constitute linear spaces. (Cl. Exercise 1.) Aside from these trivial examples, the most familiar linear spaces are furnished by classical vector in ouclidean space of two and three dimension& In order to help illustrate some of the subsequent d nitions, we introduce here two rather simple examples of linear spaces. 1. The set. V, of all sequences of scalars, {aj,as. . . ,a.j, where n is a fixed positive integpr, the operations + and being defined in the obvious manner: •

•

.

,a,,) A

+ rai,a,,

. . .

.

= {a,

.

.

+

÷

,a,,)

.

The set V, of all infinite sequences and being defined as above 2.

•

. . . ,

a. + $.,}

. . .

the operations +

A finite set of elements Ii, . . . , f,, is said to be (linearly) independent if the equality2 at/i + as/s + + implies that each of the scalars at, • , a. must vanish. An DEPTNTTTON 2.

*

.

infinite set of elements is said to be independent if every finite subset is indrpendent. A (nonvacuous) subset M of a linear space V is said to be a (linear) manifold if, for every pair of elements f, of M and every pait' of scalars a, $. the linear combination af + $9 also belongs to AL V is itself a manifold. and any manifold is itself a linear space.) T)EFINITION 3.

'The term "scalar" is to he interpreted corrn8tcn(h reel number or complex number, V being accordingly termed a "real" or "complex" linear space. More generally, the scalars may be chosen from an arbitrary field F, in which case V is

termed "linear space over F," but we shall not be concerned with this generalization. f Often the same symbol is used (or the zero acalar 0 and the scm element o, since they cannot be confused. The + is used here with two different meanings. signifying addition between scalars on the left and addition between veciors on the right. '}reuceforth we omit the symbol . the multiplication of an element by a aoalar.

4. The Fredhoim Alternative In Banach Spaces DEFINITIoN 4. A linear space is said to be "finite-dimensional" if such that every elethere exists a finite set of elements . . , ment f can be expressed as a linear combination .

f=-aifl+a2f2+ these elements.

(1)

If the a&rre-mentioned set of elements is independent,

it is termed a "basis" of the space. The Proof of the following Exercise 2.

but important theorem is left

Theorem 1. Every finite-dimensional linear space admit.s a basis. Any two of its bases contain the same number of elements, so that it is meaningful to define the "dimension" of the space as the nwnbcr of eIernents in any basiM. Given an n-dimensional linear space, a set of elements is a basis if and only if it consists of exactly n independent elements. The . , a,, appearing in ropresentation (I) of any given clement f are Um(lUe)y determined by I and the basis. .

.

V., and Referring to the defined above, we note tbat the elements I ,0, . . . ,0), 0, 1, . . ,0 }, . . . , 0,0, . . . ,l form a basis of V.,, so that this space is n-dimensional, while on the other hand, is evidently not finite-dimensional. As a simple example of a linear manifold (other the given space itself), we may take the of all elements (in either V,. or whose first is zero. entry, .

EXERCI SES

1. Prove that the real-number system forms a real linear space of dimen&on one, while the cnmplex-number syetem may be eonsidered as either a complex space of dimension one or a real space of dimension two. 2. Prove Theorem 1. 3. Prove that a set of elements . . . , a,,(k)l, I k n, forms a basis of V,, if and only if the not vanish. 4. Provc that every manifold M of a linear space V of dimension a is of dimension a, equality holding only if M — V. 5. Prove that the zero element o is unique. 6. Prove that if I,, fit . . . , are any finite set of elements belonging to a manifold, then every linear combination of them also belongs to the manifold. Hint: Use induction.

2. Normed Linear Spaces in this section we introduce the concept of a norm, which is a generaliza..

tion of the absolute value of a real or complex number and of length ift elementary vector analysis.

Partial Differential Equations

72

A normed linear space is a linear space in which there is associated with each element f a real number, termed the "norm" of / and dejioted ff11. The norm is required to satisfy the following conditions: (2a) 0 and 11/I! = 0 if and only if f = o DEFINITION 5.

For all elements f and g (triangle inequality) Ill + gil 11111 + For each scalar X and each element f

(2b)

(2c) = il/il We now present a number of simple examples of normed linear spaces. I. Consider the linear space Vft defined in Sec. 1, and let the norm of any element be defined as II

2.

Consider

.

.

+ (a21

.

+

+

.

again, but let the norm now be defined as follows:

f

.

.

max

.

.

.

.

Consider the class of all continuous complex-valued functions 1(x) defined on the unite closed interval a x b, addition of and multiplication by scalars being defined in the obvious manner. Let the norm be defined as follows: 3.

= max

(5)

We denote this uormed space as C(a,b). In each of these three examples it is quite trivial that (2b) is satisfied; we now present two examples in which the proof is not quite obvious.

That (2b) actually holds in these cases will follow from the Schwarz inequality, which is established in the following chapter. 4. Consider again the space V,,, as in examples 1 and 2, but with norm defined as follows:

+ + Ia2P + Consider the class of functions employed in example 3, but with II

5.

{aI,02,

.

.

II

norm defined as follows: Il/Il =

We denote this norined space as H(a,b). The norm makes it possible to introduce the concept of convergence, in close analogy with the real- and complex-number systems.

4. The Fredhoim Alternative in Banach Spaces

'13

A sequence of elements {f'.} is said to be "convergent 0; i.e., if for in norm," or to be a Cauchy sequence, if lim flJ, — whenever 0 there exists an integer N such that every — m and n both exceed N. DEFINITION 6.

is said to be "convergent" 7. A sequence of elements 0. Since the eleif there exists an element f such that urn — ment f, if it exists, is unique (cf. Exercise 8), f may properly be termed the "limit" of the given sequence. DEFINITION 8. An elementf is said to be a "limit point" of a set S of elements if there exists a sequence f,' of elements of 5, each distinct from f, which converges to f. DEFINITION 9. A set S of elemeuts is said to be "closed" if every limit point of S belongs to S.

Returning now to linear manifolds, we introduce the following definition. DEFINITIoN 10.

A closed linear manifold is termed a "subapace."

We conclude this section by noting that, as follows immediately from Definition 8, any normed space is closed, but that, on the other hand, as shown by Exercise 12, a linear manifold in this space may fail to be closed (although, of course, when considered as a normed space by itself, without regard to the larger space, it is indeed closed). Thus, Definition 10 is not devoid of content. EXERCISES 7. A metric space is defined as a nonvacuous set of elements such that, for every pair of elementa/, g, a "distance" p(f,g) i. defined which satisfies the following conditions: (a) p(f,f)—O,p(f,g)>Qiff$g. (6) p(f,g) — (c) p(f,h) p(/,q) + Prove that any normed linear space becomes a metric apace with the obvious definition of distance:p(f,g) — llf—gI!. $. Prove the assertion made in Definition 7. Hint: Use (2b).

9. Provethatfiaa limit point ofa set Sif and only if, foreach c > 0, thereexiatat 1, 2. (Why cannot "at least — Iii <e, Ic two" be replaced by "at least one"?) Prove that "at least two" can be replaced by least two elementsf1, I, of B such that fl!,

"infinitely many." 10. Prove that the norm is a continuous function; more precisely, if flu — <s then — IIcUl U. Show that any set consisting of a finite number of elements is closed. 12. Let V denote the subset of consisting of all bounded sequences, the norm . . .J fi being defined as sup !) Let K denote the subset of V consisting

Partial Differential Equation.

74

of those sequences containing only a finite number of nonsero entries. is a manifold, but not a subapace, of V.

Prove that M

3. Banach Spaces Among the five normed spaces introduced in Sec. 2, it will be noted that the space H(a,b) differs in an important respect from each of the. other four examples.

defined

Consider the sequence of functions

as follows:1 0

!+Lion(x_!) (n=1,2,3,.

.

.)

(8)

Each of theso functions clearly belongs to H(O,1), and the sequence is readily seen to be convergent in norm.

(This sequence is also convergent

in the ordinary sense for each value of a' under consideration, but this fact does not concern us here.) However, there does not exist a function f(x) of the space 11(0,1) such that

=0

It is perhaps worthwhile to prove this statement in detail.

Suppose

there were such a functionf(x). Then, from the relationship Jo' there would follow, a fortiori,

JcIfff2do where c is any positive constant less than

Now from (8) we see that, except perhaps for a finite number of values of n, 0 in the interval 0 z c, so that the condition

fClfl!dx

0

must be satisfied; since!, being a member of H(O,l), is continuous, thifl last equality implies thatf 0 for 0 a' c, and hence forO x <3d. Similarly, we conclude that f 1 for
For convenience we choose a = 0,

b

I.

7$ Spaces 4. The Fredhoim Alternative in contradicting the assumption that it belongs to be continuous at x

11(0,1).

Thns, it; has been shown that a Cauchy sequence in H(a,b) may fail to be convergent. This is in contrast, as stated at the beginning of this section, to the other norrned spaces defined in Sec. 2. In particular, the space C(a,b) consists of the same functions as II(a,b), hut the norm in C(a,b) is such that convergence in norm is synonymous with uniform convergence, so t.hat, by a fundamental theorem of analysis, a Cauchy sequence in C(a,b) is convergent. Similarly, the same assertion is easily proved for the other three normed spaces. (Cf. Exercise 16.) The afore-rnentioned contrast between H(a,b) and the other normed spaces serves to introduce the concept of completeness, which we now define.

A normed linear space V is said to be "complete" of elements of V there exists an eleif for every Cauchy sequence DEFINITION 11.

ment I of V which is the limit of the sequence. A complete normed linear a "Banach space." space is EXERCiSES belonging to H(a,b) such that 13. Construct a sequence of functions also (a) f,,(x) converges at every point, but not in norm, to a function 1(x) belongs to H(a,b). converges in norm, but not pointwise for any value of x, to a function (b) f(r) which also belongs to H(a,b). 14. Show that if a sequence of functions t belonging to H(a,b) converges in. norm to a (unction /(x), and pointwise to a function g(x) also belonging to H(a,b), then f(z) — g(x). 15. .CoiUintuUion of Exercise 14: Let a — 0, b — 1. Show that the sequence of functions converges in norm to the (unction f — 0, while it converges pointwlse to the function g(x) which vanishes identically except that go) 1. (Note that this does not. contradict the assertion of Exercise 14.) 16. Show that the spaces 1, 2, and 4 defined in Sec. 2 are complete.

17. Lot A denote the set of all functions f(x) defined on the inter.ral U r <1 such that f(x) yanishes everywhere with the possible exception of a countable number of points, the values of f(x) being subject to the condition that converges, the summation being extended over those values of x where f(x) 0. Let !!f I! be deftned as the above sum. Show that, with the obvious definition of + and ., .4 is a Banach space.

18. Let ft.

.

.

.

, f,, be

a fixed linearly independent set of elements of a normed

space. Show that there exists a positive constant c such that, for every choice of the scalars X1, . . . , the inequality HAifi + At/2 + . . . + c(fXi( + holds. H int: Investigate the minimum of IlXift + +. . .+ +

under the restriction Ixti + 1. + ± Use this result to prove that a flnite.dimensional normed space must be not only complete, but also "locally compact"—i.e., every bounded sequence 1/4 possesses a convergent subsequence. •

Partial Differential Equations 19. A normed space B is said to be "separable" if it possesees a denumerable dense such that for every element g of B and every aubaet—i.e., a sequence of elements Prove: — fJ <€. • > 0 there exists an element /' of the sequence such that (a) Every finite-dimensional Banach space is separable. (b) The space C(a,b), which is obviously not finite-dimensional, is separable. Hint.

Either construct a dense sequence of "polygonal" functions, or employ the Weierstrass approximation theorem to construct a dense sequence of polynomials. (c) The space 4 defined in Exercise 17 is not separable. 20. Prove that every finite-dimensional linear manifold of a Banach space is closed (i.e., ft is a aubepace).

21. Prove that every subapace S of a Banach apace B is complete, and hence Is

itself a Banach space. Furthermore, if B is separable, so is S.

4. Linear Functionals and Linear Operators Two types of functions are of especial importance in the study of Banach spaces. First, there are functions which associate with each elementf a scalar, which may be denoted as 1(f), and, second, there are functions which associate with each element f of the space an element, which may be denoted as Tf. Following common usage, functions of these two

types will be denoted as "functionals" and "operators," respectively. The only functionals and operators th&t we shall consider are those which are additive and continuous; we proceed to define these two terms. A functional 1 (or an operator T) is said to be additive if for every pair of elementsf, g and every pair of scalars a, the equality ai(f) + fil(g) Eor T(af + 1(af + = a7f + flTg) holds. ISetting a = = 0, we find that 1(o) = 0, T(o) 0.) DEFINITIoN 12.

DEFINITION 13. A functional I (or an operator T) is said to be "continuous at the element f" if for every e> 0 there exIsts a ô> 0 such that the inequality 111 — < implies the inequality (1(f) — <e (or l(Tf —
Closely connected with the concept of continuity is the concept of boundednees, which we now formulate. The connection will then be established in Theorem 2.

A functional 1 (or an operator T) is said to be "bounded" if there exists a constant C such that, for each element f, the inequality 11(/)1 (or holds. The bound, or norm, of I (or T) is the minimum value of C for which the above inequalities are valid for alif. (Cf. Exercise 22.) The norm is denoted by 111(1 (or DEFINITION 14.

4. The Fredhoim Alternative in Banach Spaces

We are now in position to prove the following simple but important theorem.

Theorem 2. An additive functional (or operator) is continuous, and hence linear, if and only if it is continuous at o, and also if and only if it is bounded. If I is a linear functional and the sequence {f,j converges tof, then l(f,,)) converges ti) 1(f); similarly for a linear operator. Proof. It clearly suffices to give the proof for a functional. If I is continuous at o, we can find a ö > 0 such that 11W — = 11(1)1 1 whenever < S; by additivity, for any / we obtain 11(1)1 ö' H/il; this shows that 1 is bounded. Given any element / and any 0, the < implies, again by additivity, that inequality hg — Il(o)

—

1(/)1

I1(g —

1)1

< S—'

—

Thus, I is continuous at I, and hence at all points of the space. Thus us a continuous functional, and, in particular, is continuous at o. We have thus shown that any one of the three conditions (continuity, continuity at o, boundedness) implies the other two, and the proof of the first part is therefore complete. The latter part of the theorem follows immediately from the inequality 11(f — 111(1 Hf —

Renceforth it will be understood that all functionals and operators under consideration are linear. Let us explain addition and multiplication of operators, and scalar multiplication of operators, by the following rather obvious definitions:

(T1 + 7't)f a T1 + T,f

(XT)f = X(Tf)

(T1T2)f

T1(T2f)

(9)

By elementary arguments (cf. Exercise 23) we obtain, from the first two of these definitions, 1! T1 —f T211

Ii T211

—I—

Furthermore, if the operators f sense that urn — TPIIh =

}

IIXTII

=

IXI

fl TI1

constitute a Cauchy sequence, in the

we can conclude that this sequence is convergent to an operator 7', in the sense that urn 7' — 0. For, given any element f, the sequence { Tj} is a Caucliy sequence (since —

flTWI — TalL

0,

il/li

0)

and hence, on account of the

of the space, convergent to an element which we define to be Tf. 7' is then readily shown to he additive, bounded? and the limit of the sequence 7') in the sense explained above. (Cf. Exercise 24.) Let R(B) denote the set of all operators defined on the Banach space B. Then the foregoing remarks, taken together with the obvious fact that the only operator having zero norm is the zero operator 0 defined in the obvicompleteness

Partial Differential Equatlone

78

one manner

Of=o

(11)

Furthermore, returning to (9), we see that R(B) admits an operation of multiplication between any two of its elements, and the following inequslity is readily proved (ef.

show that R(B) is itself a Banach space. Exercise 23):

(12)

A Banaeh space admitting, aside from the operations of addition and scalar multiplication, an operat)on of multiplication between its elements are now understood to denote any two satisfying (12) (where T1 and elements of the space. not necessarily operators on some Banach space) is known as a Banach algebra; thus, the set of all operators on any Banach space constitutes a Banach algebra.

A particular class of linear operators will prove to be of especial importance; these are the operators having the property of continuity," which we now proceed to formulate. DEFINITION 15. An operator T is said to be completely continuous (or
In a finite-dimensional Banach space this concept is of no significance. (Cf. Exercise 29.) On the other band, if B is any infinite-dimensional Banach space, the identity operator I, which is, of course, defined by the identity (13)

is not completely continuous. This follows from the fact, easily shown with the aid of Lemma 1 of Theorem 6, that such a space possesses a j sequence of unit elements {f,,) such that. The > —

operator 0 is, trivially, completely continuous, but a more interesting example is furnished by integral transforms. Let be any continuous function defined on the square 0 z, y 1, and let an operator K be defined

on the

g(x) =

space C(O,l) (cf. Sec. 2)

I(f =

Clearly g E C(01) and K the

trivial estimate

as follows:

K(x,y)f(y) dy

(f E

C(O,1)1

(14)

is additive, while its boundedness follows from if fif, which leads to the follow-

max IK(x,y)I .

ing inequality for if Kit: IIKII

max

IK(x,y)1

(15)

4. The Predhoim Alternative In Ianaeh Spaces

79

Now consider any bounded sequence ff4, and let M be an upper bound on the norms; given any e> 0, we can, by uniform continuity, choose 6> 0 such that whenever lxi — z2j <6 the inequality (:16)

IK(xi,y) — is satisfied.

From this it follows that the inequality —

<

31 dy

101

(17)

e

also holds. From (15) and (17) we observe that the functions of the sequence {Kf,IJ are uniformly bounded and equicontinuous; invoking Theorem 1-1, we conclude that the latter sequence contains a uniformly convergent subsequence---Le., a sequence convergent in the norm defined on the space C(0, 1). This argument establishes the complete continuity of K. The interval 0 x 1 employed in this example may, of course, be replaced by any other finite interval. (Cf. Exercise 37.)

As was shown earlier in this section, a sequence. of operators Tft j which converges in the sense that urn T. — = 0 also converges jJ

in the sense that there exists an operator T such that lim

—

= 0;

the analogous assertion for completely continuous operators is also true. While the proof of this fact is similar to that of Theorem 1-i, we present it in detail.

Theorem 3. Let the sequence of completely con tinuo'us operators converge to 'the operator T; then T is also completely Proof. Let {f4 be any bounded of elements, <2W. Since T1 is completely continuous, we can select a subsequence { fii such that the sequence I is convergent. From the sequence {fii we select a subsequence such that I converges. Continuing this procedure and then "diagonalizing," we obtain a single sequence such that for each n the sequence con verges. Given c> 0, we choose any m such that the inequality {

(IT— T,IIjJ

holds.

(18)

Keeping m fixed, we then select N so large that (19)

Partial Differential Equations

holds whenever j,k > N; but then, IIT(f1, — fkk)It

—

— T.,)(fj

— T11,)(j,, —

< liT —

+ T(f,.1 —

—

(IlfialI +

fkk)ll

ll.4k11)

+ +

— fws)Ii

2M + =

(20)

The complete continuity of T is thus established. We now turn to the important concept of the inverse of an operator. Suppose that T is an operator which maps the Banach space B onto itself in a one-to-one manner, so that for each element g of B there exists a unique element f such that

Tf=g

(21)

Then f may be considered as a (single-valued) function of g, and we Thus, (21) is equivalent to denote this functiona' relationship by

T'g = f

(22)

Clearly T', like T, is additive, and it can be shown that T1 is bounded, is and hence linear. However, we shall not prove this assertion. known as the "inverse" of T. It is evident that T is the inverse of T', and that both of the equalities TT—' =

I

T—'T

I

(23)

hold. Conversely, these equalities characterize the inverse operator, as is shown by the following theorem.

Theorem 4. Suppose that the operators S and T, defined on a Bausch space B, satisfy the equalities

ST=1

TS=I

(24)

Then S and T are the inverses of each other. Proof. Suppose that S carries elements f and g into the same element— that is, Sf = Sg. Multiplying both sides of this equality by T and taking S account of the second equality of (24), we conclude that f = g. carries distinct elements into distinct elements. Now, according to the first equality of (24), 8 maps the range of T (i.e., the set of all elements of B which are expressible in the form Tf, f E B) onto all of B; a fortiori, the we have proved that S provides range of S must coincide with B. a one-to-one mapping of B onto itself, and the second equality of (24) now shows that T is the inverse of S. As pointed out above, S must then be the inverse of T; this is also evident from the symmetry of (24). [It is noteworthy that both equalities of (24) are used in the proof. As shown by Exercise 35, neither one of them implies the other.J

4. The Fredhoim Alternative In Banach Spaces Suppose, in particular, that T can be expressed in the form I—

T

IC

ItKl!

<1

81

(25)

Since an obvious induction based on (12) shows that 1, 2, 3,

(n

.

.

.)

(26)

it follows that the series 4-

.+.

—I—

, and is dominated by the geometric series 1 + UKU + IIKII' + therefore 'onvergent. According to Exercise 27, it is meaningful to define an operator S by the "Neumaftu series"

is

(27)

From Exercise 28 it now follows that. both equalities (24) are satisfied. We have therefore proved the following simple result, which, on account of its importance, is stated as a theorem.

Theorem 5. Let the operator T be expressible in the form (25). Then T possesses an inverse, which is given by (27). EXERCISES 22. Show that the set of values of C which can be used üi Definition 14 is closed, so that it is proper to speak of the minimum value (rather than of the inflmum). Prove that for a linear functional 1 the none is given by 11111

— sup

11(/)1

Prove by a simple example that "sup" may not be replaced by "max." 23. Prove the aaeertions made in (30) and (12). 26. Prove the assertions made in the text concerning the operator T which was defined as the limit of the Cauchy sequence (7's end similarly for an operator 1'.

25. Prove that if 7' Ia the limit of the Cauchy sequence 7' J, then lim exists end equals 11Th; in particular, it follows that the quantities flT.1I are bounded. 26. Prove that if the series/i . . 'converges tof(in the sense that the partial sums converge to f), then the series Tfa + Tfi + .' 'converges to I'J. 27. Show that if is convergent, then c,, where the series c, + c, +

the series T1 + 7', + .

such that

lizn

.

.

is

convergent in the sense that there exist. sfl operator 7' —

117' —

0; furthermore, 11T1f b—i

28. Show that if urn

IIT

— lim lIST — ST.Ifl — 0.

T.,lI — 0

and if S Ia any operator, then urn ITS

—

Partial Differential Equations

82

29. Prove that every addithre operator defined on a finite-dimensional Banack

Use Exercise 18. space is bounded and ornpletely continuous. of completely continuouS operators is 30. Prove that any (finite) linear co2rbination also completely continuous. 31. Prove that if S is- any operator and T is completely continuous, then so are ST and TB. 32. Prove that if K1, K2, . . . , K are each completely continuous, then (I — is also completely (1 — K,,) may be expressed as I — T, where (1 — K,) continuous. 1 of Theorem 6, prove the assertion following (13), sad then 33. show that the set of all elements f such that (I — K)f — o, where K 'a completely continuous, is a finite-dimensional aubepace. 34. Prove that if operators and exist such that B,T TB, I, then & —

..

by Theorem 4, T' exists. 35. Show that neither equality of (24) implies the other. Hint: Consider V, as

hence,

defined in Exercise 12, and let the operators S and T be defined as follows: 'I'

.

.

.

S {ai,as,aa,

. .

.

(O,at,cu,

.

.

. .

.

I .

the first equality of (24) we call T a "right-inverse" of B, and Sa "left-inverse" of T.j 36. In contrast to the preceding exercise, show that either equality of (24) implies the other if B is finite-dimensional.

tlf S and T

37. Prove that. (15) may be replaced by the stronger inequality (1 Jo

max 0s

IR(x,y)i dy. Generalize both of these mequalthea from C(O,1) to C(n,h).

that I — K 38. Suppose that flK"fl <1 for some positive integer m. Then is invertible and that ite inverse is given by (27), even though K need not satisfy the condition imposed in (25). 39. Obtain (27) formally by writing the equation f — (I — K)'g in the form f g + Kf and replacing Jon the right aide of the latter equation by the entire right side. (This is an abstract formulation of the "method of successive approximation"—cf. Theorem 1-6.)

5. The Fredbolni Alternative The theory of linear integral equations shows a strong resemblance to

a much more elementary theory, namely, that of a finite number of simultaneous linear (algchraic) equations in the same number of Indeed, the first successful approach to the theory of integral equations, due to Fredhoim, was based on considering an integral equation as a limiting case of such an algebraic system. However, we shall develop in this section a theorem concerning linear equations in Banach spaces which includes as particular cases the basic theorems concerning both linear algebraic systema and linear integral equations. These two particular cases are stated as corollaries of Theorem 6. It is of interest unknowns.

that the proof of the latter is not based, as one might expect, on the first corollary, which is 'the simplest of all particular cases. Indeed, it

4. The Fredhoim Alternathe in Banach Spaces

8$

will be found instructive to recast the proof of the theorem into fOrms specifically pertaining be each corollary.

Theorem 6. Fredholm Alternative. Let K he a completely continuous operator defined on a Banach space B. If the equation

f—Kf=o admits only the trivial solution /

(28)

o, then the equation

f—Rf=ç,

(29)

has one and only one solution for each element g. Otherwise, the solutions of (28) form a finite-dimensional manifold, there exist elements g for which (29) is not solvable, and whenever (29) is solvable the solution is not unique, since the sum of a nontrivial solution of (28) and a of (29) constitutes a new solution of the latter equation. Proof. The proof will be out by establishing a number of lemmas, some of which are of interest in themselves.

if B, is a proper sulispace of B and a is any positive

L1MMA 1.

number less than one, there an element x in B of unit norm such that llx — till a for every element tj of B,. if B, is finite-dimensional, a may even be chosen to one. Proof. Let x, be any element. of B not belonging to B,, and let = mi

foryE

—

B,

If ö were zero, x,

would be a limit point of .B,. and hence an element of B, (since B,, being a subspace, is closed). Therefore, 6 iaa,.positivc, and, by its definition, there exists an element iio in B, such that —

yoll

<

For any element y of B, the element y, +

—

yolly

also belongs to B,,

and hence ö < I!xo

+ llxo —

—

yolly)11

=

IIxo —

— till

<

Itx — till

where X

= Hzo

yoll1(x, —

Yo)

> a, and since 1!zll = 1, the proof of the first part of the Thus, lix — lemma is complete. If B, is finite-dimensional, we select a sequence yi,

.

.

.

of

elements of B, such that

— xoiI

each of these elements in terms of a basis Ilk

+ X,(k)f1 +

.

+

and express of B,: f,.)

—+ 6,

. . . ,

(30)

Partial Differential Equations are bounded, for, by the triangle inequality,

The quintities

+ (Yb — Xo) H llzaII + 11YIC —

llybII

It

then follows from Exercise 18 that the X's appearing in (30) are

bounded. Hence, a subsequence of the sequence . . which each of the sequences Consider the element

is the limit of the sequence where each Bc,, and from the inequalities IIxo —. 911

11xo — LhII

+

Clearly 9 belongs to

1.

llyir — 91!

+

!lXo

we conclude, by letting k become infinite, that 11z0

and hence must equal S.

} may be chosen for is convergent.

{X,P)

+

+

9 = X1f1 + X2ft

,

—

—

cannot exceed

may now repeat the argument used

We

replaced by the previously, but with the inequality 11x0 — v.11 S. We then conclude that the unit vector equality — x satisfies

the inequality

= —

— 9)

—

yfl

I for each element y of .B0.

LEMMA 2. (k = 0, 1, 2, . .) denote the null space of (I — K)1 Let —i.e., the set of all elements f such that (I K)bf = o. (Note that .

N. consists exclusively of the element o.) Then: (1) Each N1 is a finitedimensional subspaoe of B. (2) N0 C NL C N2 C . (3) There exists an integer v such that N,, N1÷1 if and only if k Proof. Part 1 follows directly from Exercises 32 and 33. Part 2 follows from the fact that

(I—K)1+'f= so that, if f E N,,, To prove part 3, we proceed in two steps.

First, if N, =

(I— K)'f= o Replacing / by (I — K)g, we conclude that (I —

o

then

4. The Fredhoim Mternative In Banach Space. implies that

$5

K)'"g o By induction, N, =

(I —

so that Ni42

N,.

= N,41

N,41 implies

that

Secondly, suppose that for each index j 1, 2, 3 the spaces N, and N,41 are different. Then N, 18 a proper subspace of N,÷1, so that, by Lemma 1, we can choose unit elements zi in Ni, xi in 1 whenever y E N,_1. Now, on the N,, . . . such that — N,

N,4,., t

one hand, the sequence (Kxi,Kx,,

.

.

}

.

must admit a convergent

subsequence, while, on the other hand, the equality (31)

so that

shows that, for i > j,

—

Kx1fl

—

1.

Thus,

1

so that, in contradiction with an earlier statement, we conclude that the sequence fKxi,Kx2, .) does not admit a convergent subsequence. Thus the proof is complete. .

.

Let M,, (k

LEMMA 3.

0, 1, 2,

.

.

.) denote the range of (I —

—i.e., the set of all elements expressible in the form (I — K)kg, g E B.

(It is understood that M0 B.) Then for each k there exists a positive number such that, for every element z E the equation (I — K)it possesses a solution whose norm does not exceed okllzll. Proof. For k = 0 the lemma is trivial, with a0 = 1. Turning to the

case k = I, we first show that for each z E

z = (I or

—

the equation

K)x

x=z+Kx

(32)

possesses a solution whose norm is a minimum. For, if we select a of solutions of (32) whose norms approach the minimum, sequence { we can, since K is completely continuous, select a subsequence such that is convergent. Referring to (32), we conclude that the subsequence converges, and its limit satisfies (32) and has the minimum norm. (The solution of minimum norm is not necessarily unique.) Now assume the lemma were false. Then we could find a sequence of of M1 such that converges to o and the corresponding sequence of solutions of (I — K)x or }

(33)

having minimum norm would consist entirely of unit elements. As above, we could find a subsequence of lz.1} such that converges. Referring to (33), we see that the indicated subsequence of would

Partial Differential Equations

converge and that its limit, z, would satisfy the homogeneous equation

x=Kx

(34)

+ K(x — xe), we find that Itzn — xli 1, — x) = Now, since for no solution of (33) can have norm less than one. This inequality converges to x. This cannot hold, however, since a subsequence of

contradiction proves the lemma for k =

1.

For k> 1 we merely

note that since, according to Exercise 32, (1 — K)k can be written in the form I — Kk, where Kk is also completely continuous, the above argument applies equally well to this case. (1)

LEMMA 4.

space of B.

ill0 J M1 J M2 J

.

.

(3) There exists an integer p

(2) Each such that

M1 is a sub=

Mk+L

if and

only if k p. Proof. 1ff

then! = (I — K)kh, where h = (I — K)g. (I — Thus, any element of Mk+1 is also an element of Mk. This proves part 1. If fa = (I — K)Lgi and 12 = (I — K)kg2, then

aifi +

a2f,

(I — K)k(cgigi + a2g!)

This shows that Mk is a linear manifold. To show that Mfr is closed, we sciect any limit point z of and a sequence } of elements of Mz, converging to z. According to the preceding lemma, we can find a sequence = such that and (I — or (of. preceding proof)

=

+ Kkx4

(35)

Once again we conclude that a suitably selected subsequence of converges, and that the limit of this subsequence satisfies the equality = z+ or

(I—K)"x=Z

(36)

Thus. z E Mk, and so M,. is a subspace. This completes the proof of

part 2. To prove part 3, we first show that if Mk+1, then = Mk÷,, from which it follows (by induction) that Mb, n = 1, 2, 3 Consider any element! E Mk+1. Then f — (I — for some element q. and

so /

(I —

But (7 — K)kg can be expressed as (I — — K)k+1h = (1 — K)k+2h. Thus,

Mb+2.

It remains to prove that for some value of k the equality Mb = must hold. In the contrary case, Mfr+1 would be a proper subspace of Mb for each value of k, and then, according to Lemma 1, we could choose unit elements E M0 (= B), Xj E M1. . . . , such that — y belonging to M,÷1. (Note that, according to Exercise

4. The Fredholm Alternative in Banach Spaces

87

21, eath M1 is a Banach space, so that Lemma 1 applies.) We now refer to (31), where we assign to xi and; the meanings developed in the present proof. Expressing Xi and x1 as (I — K)Eg and (I — K)'h, respectively, and confining our attention to the case i <j, we obtain for A1 the from which it — ± g — (I — expression (I — K) and so it is not — Kx1tj > follows that E Mi-,.i. Therefore, Since K is possible to extract a convergent subsequence from { Kr completely continuous, we have obtained a contradiction, and so the proof is complete. LEMMA &

If g E

and k is any positive integer, the equation

= g has one and only one solution belonging to M1. g can be expressed as (I — K)s*+cy for some Since M,. = K)My belongs to and is a solution of the given y, and (I — equation. To establish uniqueness, suppose that there were two solutions,fi andf2, such that ft E Let h1 = Ii — f2. Then E (I — K)kh1 (I — — (I — K)kf = g2 —02 = (I — Proof.

By the existence part of the present lemma, which has already been proved, we can find elements since MM is a manifold.

Also, h1 E h2,

.

.

.

,

which beJong to MM and satisfy the equations (I —

ti

1, 2, 3.

Then (I —

o

h1

= (I

(I —

Thus, h,,+1 would belong to N(*+t)k but not to

—

K)kht =

o

This would contradict

Lemma 2, and the proof is complete. LEMMA 6.

'=

p.

Let f belong to NM+l. Then (I — = o, and so the equation (I K)g = o possesses the solution g = (I — K)Mf belonging to M,,. According to Lemma 5, g = o, and therefore f belongs to NM. v. Thus, NM = NM+1, and so p Suppose now that p.> v. Then 1, and so we can choose an element f in not belonging to MM. Then f may be expressed as (J! — Let h = (I — so that ii E By Lemma 5, the equation (1 — = h possesses a unique Proof.

solution x in M. Then and 80 g — z E NM. (I —

On the other hand,

x) = (I

—

—

(I

—

f—

(I

—

Partial Diferentla! Equations laince /

M,, while the equality (I —

(I —

K)'(i

—

K)'y

(I —

—

E 1141. Thus, for some element y, shows that (I — The assumption and so p v. N,...1. Hence, N,_1 that p> is therefore incorrect, and it follows that p and are equal.

valid g— x

Turning back now to the theorem, we observe that if v and p both vanish, M1 = B and N1 consists exclusively of the element o. Therefore, as f ranges over all of B, (I — K)f does the same, so that (29) is always solvable and (28) possesses only the trivia! solution. That the solution of (29) is unique follows from the observation that the differequal ence between two solutions would be a solution of (28), and to the element o. If, on the other band, v and p are both positive, is a proper subset of B, so that there exist elements g for which (29) is not solvable. In this case, N1 contains elements other than o, and so whenever (29) is solvable the solution is not unique, for to any solutioii we may add any element of W1. Furthermore, Lemma 3 shows that

when uniqueness holds the solution of (29) satisfies the inequality whereas, when uniqueness does not hold, it is still true 1111 that (29), if solvable,

a solution satisfying the above inequality.

possesses

It may be remarked that the full strength of Lemma 6 has not been used, for we have exploited only the fact that p and v are either both positive or both zero. Now, consider the space V,, defined in Sec. I, and let it be normed in any manner, such as in examples 1, 2, or 4 of Sec. 2. Consider the operator K defined as follows:

KIai,a2,

.

.

where

.

19;

.

.

.

=

(37)

(38)

the being n2 given scalars. Since V,, is finite-dimensional, K is trivially bounded and completely continuous, and so Theorem 6 assumes in this case the following form, if we set = —

The system (39)

admits a unique solution for each choice of the scalars the homogeneous system obtained by setting each admits only the trivial solution.

if and only if equal to zero

z The Fredhoim Alternative In Banach Spaces

89

This is, of course, the familiar basic theorem on systems of linear

algebraic equations, but its formulation differs from the customary one in that no reference is made to determinants, which may therefore be looked on simply as a computational aid for solving (39) by Cramer's rule.

Finally, referring to the integral operator K defined on C(O,1) by (14), and taking account of the obvious fact that the interval 0 x 1 may be replaced by any other finite (closed) interval, we obtain the following result, which constitutes the basic theorem of the Fredhoim theory of integral equations.

Let K(;y) be defined and continuous in the closed square a x, y b. Then the integral equation1 COROLLARY.

g(x)

f(x)

—

K(x,y)f(y) dy

(40)

possesses a unique solution for each continuous function g(x) if and only if the homogeneous equation obtained by setting g m 0 in (40) possesses only the trivial solution. EXERCISES 40. Show that the eecond corollary is, in some plausible sense, a generalization of the first. 41. It might appear that a more natural generalization of the first corollary than the second would be the following: The equation K(z,y)f(y) dy — g(z)

(40')

admits a continuous solution for each continuous function g if and only if the equation

K(z,y)f(y) dy —0 admits no continuous eolution other than / 0. However, show that this is not true. 42. Let K(x,y) be defined and continuous in the closed triangle 0 y a a, where a is any positive constant.

Let the operator K be defined on C(O,a) as follows:

Kf —

Prove that ffK'I1

K(x,y)f(y) dy

where M max IK(z,y)t. Therefore, aooording to

Exercise 38, the "Volterra integral equation" f(x) — g(x)

+

K(x,y)f(y) dy

pooocsco a unique eolution for each continuous function g(z).

'(40) is known as a "Fredhoim integral equation of the seoond kind." The function K(x,y) is known as the "kernel" of the equation.

5. THE FREDHOLM ALTERNATIVE IN HULBERT SPACES

In this chapter an important class of Banach spaces, known as In particular, it will be shown that the Fredholin alternative (Theorem 4-6) admits in these spaces a more precise formulation, especially when the (completely continuous) operHUbert spaces, will be studied.

ator K appearing in (4-29) possesses a certain additional property, namely, that of being hermitian. In this case a very complete theory of equations of the form (4-29) can be obtained, which can be applied to integral equations whose kernels satisfy very mild conditions.

1. Inner-product Spaces 1. An "inner-product space" is a linear space in which to every ordered pair of elements f, q there is associated a scalar, denoted (f,g) and termed the inner product of f and g, such that, for all elements j, g, h and all scalars A,

(Ia)

(Xf,g) = X(f,g)

unless f — (f,f) > 0 (f + g, h) = (f,h) + (g,h) in a real linear spa& f (g,f) (1,9)

(Ib)

(lc)

in a complex linear space

We shall find it convenient henceforth to restrict our attention to complex spaces, but there will be no difficulty in seeing what modificat Setting A — 0 and g — o in (la), we obtain (0,0)

O(/,o) — 0.

'In this case every inner product must be real, for by (lc) and the first line of (Id) we obtain 2(f,g) — (/ + g, / + — (f,f) — (g,g), and the right side of this iiquality is real by (ib).

90

91 5. The Fredholm Alternative in HUbert Spaces tions, if any, are needed for real spaces. Two very simple, but important, examples of inner-product spaces are now introduced: defined in Sec. 4-1, and let the inner 1. Consider the linear space product of any two elements {ai,a*, . . a .

snd b

be

.

.

.

given by (2)

(a,b) = k—i

The inner-product space obtained from V.. in this manner will be denoted U1, "unitary n-dimensional space."

2. Consider the class of functions appearing in examples 3 and 5 of Sec. 4-2, and let the inner product of any two elements be given by (3)

dx

(fe)

The resemblance between these two inner-product spaces and the normed spaces defined in examples 4 and 5 of Sec. 4-2 is quite apparent, and will be discussed later in this section. First we return to Definition

1 and observe that (lb) suggests that any inner-product space may be considered as a normed space by defining the norm of any element in the obvious manner (4) = Indeed, comparison of Definition 1 with Definition 4-5 shows that it is necessary only to establish the "triangle inequality" (4-2b). For this ti/il

purpose we need the following theorem, which is of major importance.

Theorem 1. Schwarz Inequality. For any elements f and g, I(f,g)1

il/il

(5)

Poll

The equality holds if and only if f and g are linearly dependent. Proof. 1ff = o, (5) holds trivially, for then (f,g) (Of,g) = O(f,g) = 0. o, let h = (so that IIhII = 1) and let a (g,h). Using If f

Definition 1 repeatedly, we obtain o

(g — ah, g — a/k)

(g,g) — a(h,g)

—

+ 1a12(h,h) 11911* — Rh,g)I'

(6)

and hence j(h,g)I

(7)

which is equivalent to (5). Equality holds in (7), and hence in (5),

PartIal DifferentIal

92

onlyig_ah'iuo,andthisisequivalenttotheusertiOflthatfafldg are linearly dependent. The triangle inequality now follows readily. Theorem 2. For any elements f and g,

(If + gil ill +

•

(8)

The equality holds if and only if f and g are positive multiples of each other.' Proof. Using Definition 1 and Theorem 1, we obtain Il/ + 91(2 •

= (1+ g, I + g)

11111' + 2 Re (f,g) + + ((gil' + 2((f((

11111'

(1(111 +

ilgil)'

(9)

Comparing the end expressions of (9), we obtain (8). Equality holds The latter equality implies that only if Re (f,g) = I(f,g)1 = 111(1 Xg, while the former equality then implies that Re = N; that is, f is real and positive. Thus, we have proved the fact suggested previously, which we now state as a theorem. Theorem 3. Any inner-product space becomes a normed linear space if the norm is defined by (4).

By comparing illustrative examples 1 and 2 of this section with examples 4 and 5, respectively, of Sec. 4-2, we find that we have justified

the assertion made there that the triangle inequality does indeed hold. The essential point in the argument is, of course, that it is possible to

introduce into each of the two spaces under consideration an inner product which yields a norm coinciding with that originally defined. The question as to whether this can always be done. That the answer Ia negative follows readily from a simple example based on the "parallelogram law," Ill + gil' + (If — g(jt = 2((fjj' + 21!g((' which holds in any inner-product space. (Cf. Exercise 2.) Now, consider the apace V1 with norm as in example 1 of Sec. 4-2. For convenience, we taken = 2 and letf and g denote theelements f 1,0) and (O,1}, respectively. Clearly il/il (If — gfl = 1, (11 + gil 2, so that (10) is violated. It is of interest that (10) is known to be sufficient, as well as necessary, for the existence of an inner product consistent with the specified norm. An important role is played by the concept of orthogonality, which we now define. 'We set aside, for convenience, the trivial case that either of the giveo element. lao.

Aiteruative in HUbert Spacee

5. The DKnNITI0N 2.

93

Two elements are said to be "orthogonal" if their

inner product vanishes.1 An "orthogonal set" of elements is a set any two of whose elements are orthogonal; if each element possesses unit norm, the set is said to be "orthonormaL" We note that any orthonormal set is linearly independent, for we

obtain from the relationship (11) + crj,, = 0 czifi + a2fi + by taking the inner product of each side with any one of the elements 0. the equality From the relationship (aifi + a2f21g) a1(fa,g) + at(f2,g), it follows that the set of all elements orthogonal to a given element is a linear manifold. With the aid of the Schwarz inequality we easily show that this manifold is a subspace, for if the sequence f3, f, and if each element of the sequence is orthogonal to g, then

+

=

(f,g)

0.

More generally, it is evident from the above

argument that if S denotes any (nonvacuous) set of elements, the set 5.1 (the orthogonal complement of S) of elements which are orthogonal to every element of S is a subspace. Theorem 4. Every finite-dimensional inner-product space possesses an orthonormal basis. and let e1 so Proof. Select an arbitrary ba8isft,f2, . . . 1. Then the scalar a is so chosen that Cf2 — ae1, e,) = 0; that # o, for otherwise the original We note that f2 — i.e., a = set of n elements would not be independent. We may therefore divide — aei by its norm, thus obtaining a unit element e2 orthogonal to ej. Similarly, we can choose scalars ft and in a unique manner such that — 7et. Since the latter vector both ei and e2 are orthogonal to f, — may divide it by its norm, thus obtaining a unit is distinct from o, we vector orthogonal to and ej. Continuing this procedure, we obtain an orthonormat set of elements e1, e2, . . . , eR which are seen, either

directly or by combining Theorem 4-1 with the remark immediately following Definition 2, to form a basis. it is clear that the "Gram-Schmidt" orthogonalization procedure used

in the above proof may be applied to replace any denumerable set of independent elements in an infinite-dimensional inner-product space by an equivalent orthonormal set. By "equivalent" is meant that any finite linear combination of elements of either set i.s also expressible as a finite linear combination of elements of the other. I

It follows from (ld) that the relationship of ort!iogonality is symmetric.

Partial Differential Equatione

94

Given an element f of an inner-product space and an orth.onormal set . we consider the problem of choosing scalars ai, , ez, . This problem — a,enll. — a2e2 — — aiet so as to minimize 1f is solved by the following simple argument. Let the "Fourier coefficients" f, (1 = 1, 2, . . . , ii) be defined as follows:

= (f,ej

(12)

Then an elementary calculation furnishes the following result: hi

a4 =

—

111112

1142

+

— a42

(13)

From (13) it follows that the left side is minimized by choosing each of

the coefficients a equal to f,, and that the minimum value of the left side is

—

lId

Since the latter quantity cannot be negative, we

obtain "Bessel's inequality," (14)

!iflP

with equality holding if and only if f is a linear combination of the elements It follows that the "Parseval equality"

Z

i—i

(15)

1f42

holds for every element f if and only if the elements form a basis of (span) the space, which would t.herefore have dimension n. Finally, we note that if there exists a denumerable set of elements e1, e2, ea, in the space, then (14) holds for each n, and hence, by a passage to the limit, we obtain (16)

1142

This inequality, which in particular assures the convergence of the series appearing on the left, is also known as Bessel's inequality. EXERCISES 1. Prove the Schwarz inequality by exploiting the fact that (J negative for all values of X. 2. Prove (10), and explain the term "parallelogram law."

—

f—

non-

3. Show that (13) may be interpreted as a generalization of the theorem that the shortest distance from a point to a fine is the perpendicular. 4. Prove that any n-dimensional Ihner-produot apace is isomorphic to U..

5. The Fredholin Alternative In ifilbert Sp.*ces 5. Given n elemens f1,

.

.

.

,

f. in ary inner-product space, prove that set

positive if the given elements are independent and zero otherwise. 6. Work out (13) in detail. is

2. Hubert Spaces Just as we concentrated in Chap. 4 on complete normed linear spaces,

here we shall concentrate on compkie inner-product spaces, or "iii!bert spaces." Clearly, any Hubert spnce is a Banach space, but not conversely. According to Exercise 4-18, any finile-dimensionat inner-product space,

is a J-lilbert in particular A less elementary example is furnished by the infinite-dimensionai generalization of by this we mean the set of infinite, sequences lal,a2, . of complex numbers .

such that

converges, addition of element.s and multiplication

by scalars being defined in the obvious manner, while the inner product is defined by the obvious generalization of (2), namely, (a,b)

= We denote this space as 12. Because of the infinite summation involved, it is necessary to show that (17) is meaningful. This is readily seen by taking account of the elementary inequality2

+ which shows that series (17) converges whenever a and b belong to 1,.

It is also necessary to show that the addition of elements is meaningful; this follows from summing over all values of k the inequality3 tak +

2lakj2 + 2

These considerations show that 12 is well defined as an inner-product space. The proof of completeness, which is somewhat more subtle, proceeds as follows. Let. the sequence4 . be convergent in norm (i.e., in the Cauchy sense), and let denote the kth component .

of

.

Since — 1

Sometimes the condition of 8eparability is imposed, and sometimes also the addi-

tional requirement of infinite rlirnensioruility.

0

= fat;- +

—

Ja* + c3ef' +

—

=

— 21a4 . -F

= lake [Cf. (10).)

+

—

'It is necessary to distiuguinh carefully between the sequence of scalars {ai,a,, }, which an element of 1,, and the sequence {a,,a,, . . 4 of elements of

PartIal Differential Equations

96

constitute a it follows that, for each fixed k, the scalars Cauchy sequence, and therefore converge' to a limit, which we denote as It will now be shown that the sequence {ai,a,, .), which we denote as a, belongs to and is the limit of the given sequence of elements {a4. From Exercise 4-10 we conclude that the quantities 11a11, constitute a convergent, and hence a bounded, . of I Ia2II, real numbers. Letting At denote an upper bound on these numbers,

and N any positive integer, temporarily fixed, we conclude that, for all n,

M' Letting first n and then N become infinite, we conclude that

M' Hence, a does in fact belong to 12.

Now let > 0 be Then, since convergent in norm, we can find an integer such that — <E whenever n, m p. Furthermore, letting denote the element whose first i. components agree with those of b and whose remaining components are all zero, we choose v so large that — <

the sequence ai,

is

.

Then, by applying the triangle inequality to the right side of the equality

a

—

= (a —

+

+ (a(') —

—

+ +

—

(18)

+

—

(19)

aM)

we obtain Ito —

—

II

+

—

—i—

IIOM

—

+

II

—

It follows from the manner in which the integers IL and v were chosen that the first, t.hird, fourth, and filth' terms on the right-hand side of (19) are each less than (for n p). Since —

is

given by the finite sum —

'It

should be stres8ed that we exploit here and e'sewhere the completeness of the complex-number syaten,. For the fifth term we use the trivial inequality — — a,")fI

5. The Frcdholm Alternative in HUbert Spaces

97

we may employ a termwise passage to the limit to conclude that, with for — an(P>(i approaches zero, and hence is less than increasing n, all sufficiently large n.

Therefore, for a sufficiently large, l!a

—

<

The completeness of 12 is thus established.

A second example of an infinite-dimensional Hubert space, which plays an important role later in this chapter, is obtained by suitably extending the inner-product space defined in example 2 of Sec. 1. As shown in See. 4-3, this space is not complete, but from Theorem 1-7 it is evident that completeness may be attained by dropping the requirement of continuity and admitting into the space all quadratically integrable functions (in the Lebesgue senscj. However, a new difficulty then arises, for it is evident that any "null funetion"—i.e., a function which vanishes almost everywhere—possesses zero norm, not only the identically vanishing function. This difficulty is overcome by "identifying" any two functions whose difference is a null function. By this we mean that we form "equivalence classes" of functions, two functions being in the same class if and only if their difference is a null function. (Of course, we use here the trivial fact that if f — g and g — h are null functions, then so is their sum f — h.) Thus, we form a Hubert space whose elements are certain classes of functions, not individual functions. Nevertheless, one may speak of "the element f(x)" instead of "the class

of all functions differing from f(x) by a null function." The inner product between any two elements is given, of course, by (3), where f(x) and g(x) are any functions chosen from the two elements. This Hubert space is denoted L2(a,b). Its infinite dimensionality is obvious, while its separability follows from the fact, immediately evident from

the definition of the Lebesgue integral, that the step functions with (complex) rational values and rational points of discontinuity form a denumerable dense set. Closely related to Lt(a,b) is the Hubert space Lt(a,b) of all functions f(x,y) quadratically integrable on the square o <x, y
ParLial Differential Equations

98

linear combination of elements of C; i.e., given > 0 and an element f of of s.nd 8caltJs the space, there exist elements #a , < a. — that — at, . . , .

-

.

We shall now show that every separable' space contains a finite or be . . denumerable basis. Let. the sequence of elements o, for otherwise we could dense in the space. We may assume that merely change the numbering of the elements. We now seek the first element- in this sequence which is a linear combination of its predecessors; if no such element. exists, the given sequence clearly constitutes a basis, but if there is such an elenicut, we delete it. from the sequence and then repeat the above procedure with the diminished sequence. The eleby this procedure are then readily seen to ments that are never form a basis. If we arc dealing with a separable inner-product, space, we may apply the Gram--Sehm!dt procedure described in the preceding section to the basis obtained by the foregoing met-hod, and in this manner we obtain an orthonzrmal basis, finite or dciumerablo, according to whether the space is finite- or ni te-dimensionaL Finally, suppose that we are dealing with a separable infinite-dimensional Hilbcrt space H. (It may be stressed that completeness does play any role in the two preceding Let the elements e1, .

e3,

.

.

.

constitute an orthonormal basis; let f be any element of 11;

and let the coeffwients f be defined by (12).

Letting

fCn)

we obtain, for m > n, the equality

and,

V

—

.Ij

.

1

by taking acount of (16), we conclude that the sequence

is convergent in norm. On account of the completeness of the space, the sequence converges to au element of If. We shall now show that g f. Letting h g '- f, We have, for any n, {

.

.

.

e,,) + — f, For m > n, the last of these inner products obviously vanishes, and so, with the ald of the Schwarz inequality, we obtain (h,ek)

(g

.

i Even without the

enStence of a bfiss by using be denunierable.

—

—

of separability, it is possible to -demonstrate the induction, buts basis of a nonseparable 8paCe

The Frectholm Alternative in Thibert Spaces

Thus, h is orthogonal to every Given e > 0, it is element of the orthonOrmal. basis €1, e2 possible, by the definition of a basis, to select a finite tiumber of scalars such that a!, . . . , £dt.ing m —'

we

obtain (h,e,.)

— a1e1 —

0.

<E

—

—

the other hand, by an argument employed in Sec. 1, the left side of equal to (h,ei), which this inequality is minimized by choosing each <e, and hence h = o. Thus, we conclude that 28 equal to zero. Therefore g f, and so we may write

01)

f=

(20)

This equation is to be interpreted as mcaning that f = urn Without going 5nto detail, we remark that the above reasoning can be modified to give the following result: Given any orthonormal basis •

.

.

and

any set of scalars

then the

f,,

.

.

.

such

that

converges,

converges, in the sense explained above, to an

element f which is related to the scalars

by (12), and the infinite-

dimensional analogue of (15), namely, If

= IL112

holds. EXERCISES

7. Prove that

is separable, and that the elements (1,0,0,

(0,0,1, . . .1, . . •. form an orthonormal basis. Prove the statements mnde in the last paragraph

•

. 4, (0,1,0,

.

.

of the present section. 9. Let H denote the set of all complex-valued functions defined on isterval 0 <x I, vanishing everywhere except on a finite or denumerable set (which may differ from function to function) and such that converges, the suinniation being extended over the points wheref 0. Prove that, if (f,g) then If is a nonseparable Hubert space. Determine an orthonormal basis of H.

3. Projections, Linear Functionals, Adjoint Operators Given any vector in three-dimensional euclidean space and any plane line), it is possible to the vector as the sum of two vectors, of t.hem aud othor porpendicular to the given plane

PartIal Differential Equations

100

line); furthermore, the decomposition is unique. Our first objective in the present section is to establish the corresponding fact (Theorem 5) in any Hubert space. DEFINITION 4.

A subset M of a linear space is said to be "convex" if,

for every pair of elements f, g in M, and every real number t between zero and one, the element If + (1 — t)g also is in M.

This is an obvious generalization of the definition of convexity for figures in two- and three-dimensional euclidean space.

A simple example

of convexity is furnished by the set of all elements of (of. Sec. 4-1) whose first component a specified value a; note that this set is a manifold if and only if a 0.

Theorem 5. Let M be any nonvacuous closed convex subset of a Hubert space H. Then there exists a unique element h of M having minimum norm; i.e., if g E M and g h, then > Proof.

Let the infimum of hf I! for all I E M be denoted by 8. Then we can select a sequence h,. of elements of M such that 11/i,. —, ö. From the equality —

=

+

(22)

+ h,,.)112

—

[cf. (10)] and the fact (which follows from the convexity of M and the definition of 6) that (h,4 + he,.) 5, we conclude that, given any 0, we can choose N so large that the inequality

+

— hmlh2

+

=

(23)

holds for all in and n exceeding N. Hence, the sequence is convergent in norm; since I! is complete, there exists an element h E H such that IIh — 0, and, since M is closed, 4 E A!. The continuity of the norm then assures that 5. It remains only to prove the uniqueness of 4. Suppose tbat there are two distinct elements 4 and in M each having norm 5. Then, by convexity, + also belongs to M, and from the equality

+

=

+

—

—

=

—

—

(24)

conclude that M contains an element of norm less than 8, contrary to the definition of 6. Thus the uniqueness of 4 is proved.

we

COROLLARY. Let M be any nonvacuous closed convex subset of a Hilbert space H, and let f be any element of H. Then there exists a unique eletnent 4 of M such that !lf hll
element of M distinct from 4.

101 5. The Fredhoim Alternative in Hubert Spaces Proof. Let M, déiiote the set of all elements of H of the form / — and convex, so that g E M. Then M, is, like M, nonvacuous, closed, the above theorem is applicable.

Theorem 6. Projection Theorem. Let M be any aubspace of a

Then every elementf of H can be expressed in a unique manner in the form

Hubert space II.

f= g+h

(25)

Proof. Let g be that element of M, whose existence is assured by the for all elements of M > above corollary, such that 11/ — — other than g, and let h = / — g. Then, for any element e in M and any real number 111 — (g

+ Xe)(j2

— = — (g + obtain from (26) the inequality

Since

(26)

11h112

— 2X

tjhlL2 + — 2

Re (h,e), we

Re (h,e)] > 0

(27)

Now, if Re (h,e) <0, X can be chosen negative and so close to zero that the second factor of the left side of (27) is positive, and thus a contradic-

tion with (27) would be obtained. Similarly, the possibility that Re (h,e) > 0 is ruled out, so that Re (h,e) is shown to be zero. T&king X as pure imaginary instead of real, we can show similarly that Im (h,e) 0. Thus, (h,e) = 0; i.e., h is orthogonal to every element of M, as was to be proved. It now remains only to establish the uniqueness of the decomposition (25). Let another such decomposition be given by (g

—

—(h

—

= Thus g —

and hence =

o,

h—

—

—

(g,h)

+

= 0 (28)

= o, and the uniqueness is established.

The element g which is uniquely associated with each element f of H by (25) is termed the (orthogonal) projection off on M; we denote this relationship as follows:

g=PMJ From (25) and the resulting equality

(29) 11/112 = 110112

+

we easily

oonclude that the operator is additive and bounded—hence linear— 1 except in the trivial case that M consists of the single with IlPaill element o; in this case, is the zero operator and 11P4 = 0. Furthermore, whenever f E M the decomposition (25) evidently assumes the form / = f + o, and so we conclude that PM2f = PMJ for every element f. Thus, Pu* = PM; an operator having this property is said

Partial Differential Equations

to be "idempotent." We also note that for every pair of elements fi, the equality (30) (PMfI,f3) = (fl,Puf2) holds; the proof follows trivially from (25). With the aid of the concept of projection which has just been introduced we can now give a strikingly simple description of all linear functionals in a Hubert space.

Theorem 7. Rleez Representation Theorem. Let 1 be any

linear functional defined on a Hubert space H. Then there exists a unique element e of H such that, for every element f, 1(f) = (f,e)

(31)

Conversely, foi each element e of H, (31) defines a linear functional.

If 1(f) 0, 1 can be defined by (31) with e o. In any other case, the set of elements f such that 1(f) = 0 is a proper suhspace Proof.

(That 2W is a manifold follows from the additivity of 1. while the. closure follows from the boundedness of 1.) Next we show that any two elements of 1W must be linearly dependent.' To show this1 we select two such elements, h arid h'. If either is i.he element. o, the asser211 of II.

tion is trivial. Otherwise, the quantities 1(h) and 1(h") must both be and so the element ft" ft — [l(h)/l(h')}h' is well different from defined and, like ft and h', orthogonal to 2W: On the other hand, 1(4")

0, and so h" E if. Therefore h"

1(4) —

the linear dependence of ft and h' is thus established.

fixed element of )W.l. othcr than

Then,

o;

Now, let ft0 be any

in view of the foregoing

discussion, each clement I of H can be expressed in the form

1=

From (32) we get.

(32)

(J,ho) =

and

+ ()Jto,h0)

1(f) = i(PMf)

Mh0,ho)

±

Eliminating X from (33) and (34), we obtain — —

l(ho)(f,he)

— —

(

"1

ThUs (31) holds with e

trivial, for if (f,e) — (f,e'), then, in particular, the equalities (c, e — e') 0 and (e', e — e") = 0 must hold. Sub= o, or tracting, we obtain (e — e — e') = 0, and hence e — The uniqueness of

'The reader may find it helpful to viaualize the following argument in threospace.

5. The Predhobn Alternative in Hubert Spaces

e = e'. Finally, it is easily established that each element e of H deter.. mines a linear functional. Additivity follows from the properties of the inner product, and boundedness from the Schwarz inequality: so that the norm of the functional is at most By flfll

settingf = e, we conclude that the norm cf the functional is precisely Now we turn to the concept of the "adjoint" of a given operator T.

For as

fixed element g we define a linear functional, which we denote follows:

19(f) = (Tf,g) The additivity of

an immediate consequence of the additivity of 7 and of the inner product (as a function of the first element); the boundedness follows from the Schwarz inequality and the bouudedness of 7, for

1,(f)

C=

C11f11

11 T11

1g11

By the preceding theorem, we can associate with g a uniquely determined

element 7*g such that

iq(f) = (f,Tg) From (36) and (38) we obtain

(38)

(f,Ttg) (39) The additivity of is immediately established, (Tf,g)

for all elements f. g. for, from the identities (I,

+ g2)) = (Tf, = (7f,

+ = (Tf,gj) + (Tf,92) (f,T*gi) + (f,T*g2) = (I, T'g1 + 71*gj) (40) = X(Tf,g) = X(f,T*g)

(f,XT*q)

(41)

we conclude, as in the proof of the uniqueness of e in representation (31), that + g,) T*gi + T*g2, T*Xg = The boundedness of 71* is established by the consideration. For any g H, (T*g,T*g) (T(T*g),g) = J(T(T*g),g)I ugh

If

o, we divide by

IlTIb

(42)

Iholl

and obtain 1171!

whereas if T*g =

(43)

the inequality (43) holds trivially. Hence 7* Tfl. We are therefore assured of the existence of an adjoint (T*) * of 7*; from the identities o

bounded, and

(Tf,g) = (f,7*g) =

(q,(7?*)*/)

= ((T*)*f,g)

'(44)

_____

Partial Differential Equations

104

we conclude that (T*)* = T, and hence IITII Summing up, we have the following theorem.

ll(T5)11

a uniquely determined Theorem 8. Every operator T = adjoint T5, and Finally, we state a number of elementary facts concerning adjoint Operators, leaving the proof8 as Exercise 11.

Theorem 9 (a) (c)

(T1 + T2)5 = (T1T2)5 =

T

(XT)5 = XT*

(b)

= (T5)1

(d)f EXERCISES

10. Prove that every nonvacuous subset M of a !inear space cohtains a unique hull"—i.e., a convex set such that M C M and M C N for every convex N containing M. Hint: Consider the class of all convex sets cotitainhig If, and show that convexity is preserved under intersection. 11. Prove Theorem 9. 12. Prove that the adjoint of a completely continuous operator is alsO oornpletely %iöhtinuous.

Hint: Use the identity (TTf,f)

fl

13. A sequence of operators {

is said to converge "strongly" to aft Operator T if, element f, the sequence convêrge8 to Show, by means of aS

tot'

example, that, in contraat to Theorem 4-3, strong cohvergente of a se4uence of completely continuous operators does not imply the complete of the limit. 14. Prove that, in a complex Hubert space, the identical vanishing of (Df,f) Implies

that

/

—

0.

Hint: Replace f in the expression (Tf,f) by f + g, /

— g,

/ + ig, and

—

space.

Prove that the assertion of the preceding exercise Is ziot trite in a real Hint: Consider the rotations of the plane about the ought.

16. Give an alternate proof of Theorem 6, in the case that It is separable, by constructing orthonormal bases for M and M'.

4 flermitian and Completely Continuous Operators t)EFINITI0N 5.

adjoint') if T =

An operator T is said to be "hermitian" (or selfT5.

By referring to (30), we see that any projection

is herniitian.

Theoretit 10

An operator T is hermitian if and only if (Tf,f) is real for every elentent I. Proof.

If T is hermitian, then (Tf,f) = (f,T*f) = (f,Tf)

and so (Tff) t Assertion (d) and (ii) holds.

is real. is

Conversely, if (Tf,f) is always real, then

to be understood as follows: If T! exiats, then (T*)_1 also

1 A distinction is made between these two terms in the theory of uabounded operators.

105

5. The Fredhoim Alternative In Hilbert Spaces (T*f,f), and hence ((T — T*)f,f) (Tf,f) — (Tf,f) — (f,T*f) from which we conclude, by Exercise 14, that T = T.

Theorem

11.

Let

N, =

N,

If T is hermitian, then

I(Tf,f)l.

sup Ill—i

0,

Proor Whether or not T is hermitian, we obtain, from the Schwarz

inequality,

L(Tf,f)l and hence

(45)

11111 IITII

N,

(46)

Now, if T is hermitian and X is any positive number, + X—1Tf), Xf + X—1Tf)

lIT/Ill

— —

+ )ttTfll2 ÷ llXf

(T(Xf — X—'Tf), Xf —

= The expression

+

+ X—'ll

(47)

Till' is minimized by choosing

=

II

TILl

1, we obtain

and, with this choice of X and with IlfIl

211 7Yl1 =

and hence (even if Ti

X—1Tf)1

X_l7hfII!)

—

N,tl TfII

(48)

o)

N,

11

(49)

Since (49) holds for any unit element f, we conclude that

11Th N,

(50)

Combining (46) and (50), we conclude that

TI1

= N,.

We now turn to a class of operators having an especially simple structure, the degenerate operators. DIIPINITI0N 6. An operator K is said to be "degenerate" if there exists a finite-dimensional subspace M such that Kf E M for every clement f.

Let • . . , •,r and neceesaxily all distinct), and let

.

.

,

be elements of H (not

= Cleatip K is a degenerate operator, and its adjoint is also degenerate, for

Partial Differential Equations

106

an elementary verification showø that K*f

N

(52)

Furthermore, K is completely continuous.

To prove this it suffices, on

account of Exercise 4-30, to 8how that any operator of the form the has this property; Given any bounded sequence Kf = is also bounded (by the corresponding sequence of scalars {

Schwarz inequality), and hence possesses a convergent subsequence j. Then the sequence Kf.k is convergent, and so the proof is complete.

Now we show that this example is in no way exceptional.

Theorem 12. An operator K is degenerate if and only if it can be expressed in the form (51), from which it follows that K is completely continuous. If K is degenerate, so is K*, which is given by (52). Proof. Let M be a finite-dimensional subepace such that Kf E M for every elementf, and let . • •N form an orthonormal basis of M. .

Then for every element f there exist N uniquely determined scalars li(f), 12(f), . . . , 1,,(f) such that

Kf=

N

may be determined by forming the inner product of Each coefficient In this way we obtain both sides of (53) with

6(f) = Letting

we

=

obtain

6(f) — (f,*i) Substituting into (53), we obtain (51). The remaining assertions of the theorem have been covered by the discussion following Definition 6. Next we prove an important approximation theorem concerning completely continuous Operators.

Theorem 13. Any completely continuous operator K defined on a separable' Hubert space can be expressed as the limit of degenerate operators. If K is hermitian, it can be expressed as the limit of degenerate hermitian operators. Proof. Let Let P1 denote . . . be an orthonormal basis in H. the orthogonal projection on the subepace spanned by #2,. . . , #1., 1The theorem is true for nonseparable epaces also, but the proof given here applies only to a separable space (.rnce a denitmerable basis is employed).

5. The Fredholm Alternative In HUbert

Spaces

for all elements!, it Eollows that Since KJ and let J converges to K,, is degenerate. In order to show that the sequence K, we shall prove that the contrary assumption leads to a contradiction. Suppose then that the above assertion is false; it follows that there exists a positive number t such that the inequality ilK — K,Jj>.

holds for infinitely many values of n. Restricting attention henceforth each of unit norm, sueb that to such values of n, we can find elements

KJUj) > is bounded, and so, by confining attenNow the sequence { (I — tion to a suitable subsequence, we may assume (because of the complete continuity of K) that the sequences Kf. } and K(I — are both convergent, to limits g and h, respectively. We now prove three asser-

tions: (1) h =

o; (2)

0; (3) Lf(I —

—'

—

—+

0.

We prove assertion 1 as follows:

(h,h) = urn (K(I —

h) =

Urn

(f,, (I

lirn

—

(Here we have used the fact that

[K(I — Pa)1* =

(I

= (I

—

—

and also that, for any fixed elementf (in this case Kh), (I

—

P.)f—+ o.)

—, ilhlt

0 (59)

The proof of assertion 2 proceeds as follows: IIK(I —

—

= ilK(I — Turning now to assertion 3, we write fl(I

= IRI 11(1 — P,)(Kf%

—

— g)

+ (I —

P.)g$}

+ il(I — — ill — + il(I — = + il(I — — Taking account of assertions 2 and 3, we obtain IRK —

=

+

11(1 —

11(1 —

— gil

—' 0

(60)

0

(61)

—

+

—

—*

Since (61) contradicts (57), we have proved that the sequence converges to K. If K is hermitian. so is PIIKPII Eby part (c) of Theorem 91. The proof of the theorem is thus complete.

We now turn to operators which are both hermitian and completely First we present an important definition.

continuous.

Partial Differential Equations

108

Given an operator K, a scalar p i&said to be a "characteristic value" of K if the equation 7.

Kf—uf possesses a nontrivial solution, which is termed a "characteristic vector" of K (corresponding to ii). t

Theorem 14

If K is hermitian and completely continuous, and not the zero operator, it possesses at least one nonzero characteristic value. Proof. According to Theorem 11, there exists a sequence {f,, of unit elements such that where = ± IIKII 0. Then —

Now,

so

lim sup

— pi)1n112

+ pt2

—

that

= urn sup

+

—

si' —

+

= 0 (64)

Since the left side of (64) is nonnegative, it must vanish, and this is possible only if

K is completely continuous, we can select a convergent subsequence from and it then follows from (65) that the subsequence is also convergent. Since 0, the subsequence must converge to an element of unit norm, and by referring to (65) we find (since f —, implies that —, Kf) that

f

(66)

With the aid of the above theorem, we shall now obtain a very precise representation of the operator K. If g is any element orthogonal to we find, taking account of (66), that

=

=

=0

(67)

Hence, K maps the consisting of all elements orthogonal to into itself. K may therefore be considered as an operator defined on this subspace, and it is evidently still hermitian and completely continuous. If K reduces to the zero operator on this subspace, then Kf is given, for any elementf of H, by the formula

Kf = t The German term "elgenvalue."

(68)

"elgenwert" and "eigenvector" are often ueed, asia the hybrid

5. The Fredhoim Alternative In Hilbert Spaces + g, from which we immedi-

[This may be seen by expressingf as ately conclude that = 0, so that

Kf = K does not reduce to the zero operator on the subapace, we may repeat the argument used in the proof of Theorem 14 to show that there exists a unit element orthogonal to and a nonzero real number M2 such that = Reasoning

as before, we find that either, in analogy with (68), the

operator K is expressible by the formula

+

KJ =

or we can continue the above procedure. If this procedure terminates after a finite number of steps, we obtain for K the formula

Kf =

N

k—I

If, on the other hand, the procedure does not terminate, we obtain a denumerable orthonormal system of elements real numbers P1,

.

.

.

such

(k =

= I

.

that 1,

2,

. .

.

.

and nonzero (72)

.)

Before obtaining the analogue of (71) in the latter case, we show that must decrease to zero. The inequalities Iui! IMSI ,L4

(73)

are assureff by the manner in which the elements } were determined, so } must converge to a nonnegative limit ö. Now,

that {

ilK.1

+

—

2o2

(j

k)

(74)

and so, if 8 were positive, the sequence could not possess any convergent subsequence. Since this would contradict the complete }

continuity of K, we conclude that 8

0.

We now proceed to show that the obvious generalization of (71),

namely,

Kf =

(75)

holds when the procedure described above does not terminate. belongeto

the subspace spanned by the elements bi,

.

.

, we

If / may

Partial Differént4al Equations

apply K term by term to the expansion

/=

(76)

This immediately yields (75). Next we note that, according to the were determined, arty element f manner in which the elements must satisfy the inequality orthogonal to 4'i, . . {

}

,

(77)

Hence, if f E M1 (i.e., if f is orthogonal to all the 4"s), (Kf.f) =

0.

Since, furthermore, the reasoning folLowing (67) shows that K may be interpreted as a hermitian operator defined on the Hubert space M1, we conclude from Theorem II that Kf o. Now, given any element f of II, we may, according to Theorem 6, express it in the form g + h, where According to the foregoing remarks, g E M, h

Kf = Kg =

=

(78)

Thus the validity of (75) is established for any element I of H. We sum up the results of the above discussion in the form of a theorem.

Theorem 15. If K is any hermitian completely continuous operator other than 0, it may be represented in the form (71) or (75). according to whether the procedure described above terminates or not. The elements 4'i, 4's, . . span H if and only if the equation Kf = o possesses only the trivial solution f = o. .

EXERCISES

17. Prove that an operator is a projection if and only if it is herthitian and idempotent. 18. Prove that every hermitian operator whoàe square is the identity operator I

may be expressed in the form! — 2P, where P is aj,rojection. 19. Prove that all eigenvalues of a herinitian operator are real, and that eigenvectors corresponding to diatinct eigenvalues are orthogonal.

20. Prove the following statements: (a) The sum of

hermitian operators is

hermitian. (b) A real scalar multiple of a operator is hermitian. (c) The inverse of a herniitisn operator (if it exists) is hermitian. (d) If S and T are hermi-

tian, then ST is hermitian if and only if S and T commute—i.e., ST TB. 21. Prove that if P and Q are projections, then: (a) PQ is a projection if and only if PQ QP, (b) P + is a projection if and only if PQ = 0, (c) P — Q is a projection if and only if PQ Q. 22. Prove that a hermitian operator T possesses an inverse if there exists a positive number c such that (Tf,f) for every element f. Hint: Use Theorems 11 and 4-4.

23. Prove that the product of two o itself degenerate.

of which at b"tst one is degenerate is

5. The Fredhohn Alternative In filbert Spacea 24. An operator K (other than 0) said to be "nilpotent" if K" = 0 for some integer n > 1. (a) Give a simple example of a nilpotent operator. (b) Prove that a bermitian operator cannot be nilpotent. 112 is always true, prove that tho equality 25. AJtbougb the IIKUk. Then prove that, more generally, IIXr holds if K is hermitian. IIK'iI

26. Closely related to hermitian operators are "normal" operators, which are

characterized by the relation KK* KK. Prove that K is normal if and only if. IIX*fH for every element f. IlKf II 27. Prove directly from (75) that IlK II is equal to 28. A "unitary" operator U is one satisfying the conditions U11 U'U Prove that U is unitary if and only if it possesses iuverse and is for every elementf. —i.e., 110/Il Prove also that in a finite-dimensional space, hut not in an infinite-dimensional apace, the norm-preserving property implies the existence of an inverse, and hence suffices to characterize unitary operators.

5. The Fredhoirn Alternative 'rho characteristic feature of the Fredholni alternative in a HUbert

apace is that a close relationship is established between the equation

f—Kf=g

(4-29)

and the "adjoint" homogeneous equation f—

it

is

understood throughout this section

continuous.

that K is

completely

Theorem 16. The Fredhoim Alternative in a Hubert Space.

(1) The solutions of (79) and of

f—Kf=u

(79')

form subapaces of the same finite (zero or positive) dimension D. (2) If = 0, the inverse (1 — K)—1 exists, so that (4-29) is solvable for every element g, and 11/li for some sufficiently large constant C. (3) If D> 0, (4-29) is solvable if and only if g is orthogonal to every solution of

(79), and then the solution is not unique, for to any particular solution may be added an arbitrary solution of (79'). Proof.

While part 2 is covered by Theorem 4-6, it. is preferable to give an independent proof of the entire theorem. First, we consider the case

that the HUbert. space in question is basis

We select the orthonormal

where

is the element whose kth entry is one while all the others are zero. Then, expressing the element .

.

.

,

f

.

.

.

Partial Differential Equations

in the form (80)

I...' we

find that

Kf =

ii

(81) 1

the n' scalars c,, are determined by the equalities

where

(82)

Similarly, with the aid of n' scalars we can represent K5q, for every in the form element g = {$i, $,, . , .

K*g =

(83) i.J —1

where

the scalars

are determined, in analogy with the

by the

equalities K *e.

(84)

=

Prom (81) and (83) we obtain

(Kf,g) =

(f,K*g) =

=

(85) ij—1

Since the sums appearing in (85) must be equal for all choices of the 2n fi,,, the n2 equalities ci,, or scalars al, a2, . . . , an, th, $2, . .

,

Cs1 — Cis

must hold. Equations (79'), (79), and (4-29) assume the following forms1 respectively,

=0 =

d45 =

Cj1

(87)

0

= (1) Systems and (88) have the same number of linearly independent solutions.

and the three parts of the theorem may be stated as follows.

5. The Fredhoim Alternative In Hilbert Spaces

113

(2) If systems (87) and (88) have only the trivial solution = a,, = 0 al = at = then system (89) has a unique solution for every choice of the scalars th, •

.

.

stant C.

,

C

$,, and

for some sufficiently large con-

(3) If (87) and (88) possess nontrivial solutions, then (89) is = 0 for every solution

solvable if and only if

.

•

.

of

Now, these three statements, which are an extended version of the first corollary of Theorem 4-6, constitute the basic theory of systems of linear (algebraic) equations, which we assume to be known.' (The boundedness condition of part 2 is assured by Exercise 4-29, but may also be proved from Cramer's rule by inspection.) Next we consider the case that K is a degenerate operator defined on any Hubert space H. Taking account of representation (51), we find that (4-29) assumes the form (88).

/=g+ so that any solution, if it exists, is expressible in the form

I = C' + Replacing f on both sides of (90) by the right side of (91), we find that (90) assumes the form

=

[(g,#a) +

J

(92)

Since the elements may be assumed independent,t we . . . , obtain from (92) the algebraic system =

—

(93)

which reduces to (89) if we let = and = Con— versely, if the scalars ai, at, . . , satisfy (93), the element f defined by (91) is readily seen to satisfy (90), and hence (4-29). In particular, .

It is of interest that in order to prove the present theorem, unlike Theorem 4-6, wo must first know the simplest particular case of the theorem. 'Otherwise (51) could be rewritten as a SUm of fewer terms. 1

PartIal Differential Equations

114

setting g

o, we

find that equation (79') assumes the form

=

a,

0

(94)

0

(95)

—

while equation (79) assumes the form' a, —

These equations reduce, of course, to (87) and (88), respectively, when are indeare defined as above. Now, since the the quantities ,cz11} of U,, corresponds to pendent, any manifold of elements (al,a,, . .

of the same dimension in H.

a manifold of elemeirts

Thus, t.he

solutions of (79') and (79) form manifolds of the same dimension, respec-

tively, us the solutions of (87) and (88). Since part 1 holds in U,,, it must therefore hold in H. When 1) = 0, (93) is uniquely solvable for each set of scalars and hence (4-29) is uniquely solvable for eaeh element g. Furthermore, letting C have the same significance as in the discussion of U,, and using the Schwarz inequality, we obtain the inequalities fa42

<

and hence, from (91),

(I

+

(97)

Thus, part 2, including the boundedness condition, is proved.

Turning to part 3, we merely have to note that (93), and hence (4-29), is solvable

if and only if a,,

. . . , a,, of (95). whenever / satisfies K is degenerate.

aM) 0 for every solution This is equivalent to asserting that (q,f) = (g,

0

Thus the proof is completed for the case that

Finally, turning to an arbitrary K, we express it as the sum of a degenerate operator K, and a "small" operator K,; more precisely, we 'Bare we use (52) and the independence of the elements #i, which . . . , by the following argument: If the . . , were depeiident, (52) could he rewritten es a sum of fewer terms, but then so could (51), by symmetry. (Thus, we have shown that the ranges of K and K have follows from that of the equal dimensions.)

.

5. The Fredbolrn Mteiiiative In HUbert Spaces select K1, in accordance with Theorem 13, such that

I Then (I — K2)' exists, by Theorem 4-5, and so (4-29) may be rewritten in the form (98) / — (I — K,)—'Ki/ = (I — K2)'g Since (I — K,)'X1 is degenerate (cf. Exercise 23), we may assert that and its I— have null spaces of — I — (I — equal dimension. Now, on the one hand, (79') is entirely equivalent to

[I

(I —. K2)—1K,)f

o

while, on the other hand, I — — may bc rewritten in the K*)(I — from which it follows that its null space has the same dimension as that of I — K*. Part I is thus proved. Part 2 follows, in its entirety, from applying the (bounded) operator [I — (I — to both sides of (98). Now, when .0 > 0 we know that (4-29), being equivalent with (98), is solvable if and only if (h,(I — K2)'g) 0 form (1 —

whenever [1 —

=

—

o

and this is equivalent to saying that ((I —

0

whenever

(I

K*)(I

=

—

o

Replacing (1 —

by f, we obtain part 3, thus completing the proof. We now consider a simple modification of (4-29), namely, f — XKf =

g

where A is a (scalar) parameter. Then, by replacing K and K* everywhere in the above proof by AK and respectively, we that, for each value of A, either equation (99) is uniquely solvable for each g, and the adjoint homogeneous equation

f_XK*f=o

(100)

admits only the trivial solution, or (99) fails to admit a solution for

certain choices of the element g, in which case (100) and the homogeneous form of (99). namely,

f—XKf==o (101) both admit nontrivial solutions, the subspaoe of solutions being of the (finite) dimension for both equations. In the latter case, A is

Partial Differential Equations

termed a "singular value" of K. Since zero obviously cannot be a singular value, we may divide (101) by A, and thus conclude that the

singular values are simply the reciprocals of the eigenvalues of K. (Cf. Definition 7.)
(102)

These elements must be independent, for otherwise we could determine a are independent while #21 unique index N such that . . . , may be expressed in the form (103)

to both sides of (103) and taking account of

Applying the operator (102), we obtain

(104)

By subtraction, we obtain

Ni Since

the first (N —

are independent, and since the X's are all

1)

. . . , CN—I would have to vanish, which is impossible in view of (103). Now that. the independence is established, to obtain we may apply the Gram-Schmidt process to the sequence { an orthonormal sequence g,1 }. From the fact that each gn is a linear combination of the first n we readily find, by taking account of (102), that is also a combination of the first n but that — Qn iS

distinct, the coefficients Ci, C2,

a

combination of the first (n —

1)

"s, which, in turn, may also be

expressed as a combination of the first (n — 1) g's. Therefore, for ii> m we find that X$Kg% — — is a combinati3n of the first (n — 1)9's, and hence —

= g, +

5. The Fredhoim Alteni*tlve In Bilbeit Space. for suitably chosen scalars orthonormality relation

a2,

.

.

.

,

a.,_,.

Taking account of the

öq, we conclude that

1

—

and hence that the sequence K(X,g.) J possesses no convergent subsequence. However, this result contradicts the complete continuity of do not exceed R. The proof is thus complete.

K, since the norms

We point out that the above result concerning the distribution of singular values is contained in (75) for hermitian K, and that (75) also makes possible a very simple proof of Theorem 16. By using this representation of the hermitian operator K in (99) and repeating the argument used in the portion of the proof of Theorem 16 dealing with a degenerate operator, we obtain the unique solution in the form

f = (I + XK,)g

(107)

where K),, the "resolvent kernel" of K, is defined for all nonsingular values of X by the formul&

=

1—

If, on the other hand, X coincides with one or more of the quantities ur the corresponding inner products must vanish. If this orthogonality condition is satisfied, however. (107) still furnishes a solution of (99) if those coefficients in (108) which assume the indeterminate form 0/0 are assigned completely arbitrary values. Thus Theorem 16 is proved. EXERCISES 29. Let the scalars

be

associated with the operator K defined on

Show that K is unitary if and only if the nt

C,?,,

—

as in (82).

&, are satisfied.

(A square matrix whose elements satisfy these conditions is therefore termed "unitary.") . 30. Let two different orthonorinal bases . . and be selected in U., and let the operator K be represented, in analogy with (82), in terms of each basis by the scalars and respectively. Show that where

—

2.7—1

'Since the factors

—

appearing in (108) approach zero, the convergence

of the series is assured by the convergence of graph of Sec. 2.J

(16) and the final parai—i

Partial Differential Equatloiis

118

6. Integral Equations Our first objective in the present section is to show that there is a natural eorrespondence between the elements of L2(a,b) on the one hand and some (but not all) of the completely continuous operators on L2(a,b).

(Henceforth we usually suppress reference to the interval a <x
.

.

.

,

=

This follows from the obvious equality'

= and the triangle inequality. function

If we associate with each function f in L2 the

h(x) = ff(y)K(x,y) dy we obtain, from the Schwarz inequality, Ih(x)12 llftltfIK(x,y)P dy, and hence, by integration, Il/Il

Thus the function K(z,y) determines, through (110), a linear operator on we denote this operator by K. (The use of the same letter for the function and the operator is, of course, intentional.) We note from (Ill)

that the norm of the operator K is at most equal to the norm of the

function K; it is easily shown, as in Exercise 31, that, in general, the Iwo norms are not equal. The operator K is completely continuous, for it maps L2 into the fithte-dimeusjonal subspace spanned by dii, . . , çb1. Conversely, according to Theorem 12, every degenerate operator K may be expressed in the form (110), where the kernel K(x,p) has the form (109) for suitable choices of the and #'s, and the adjoint operator K5 corresponds to the Icernel K*(x,y), where, as is to be expected,

K(x,y) — K(y,x)

(112)

Now, the step from (110) to (111) does not involve the assumption that K(x,y) has the form2 (109), and so it follows that every in L: determines, by (110), an operator on L,; the formula (112) for the kernel We use the symbols and Ill Ill for norms in and L,, respectively; the former symbol will also beued for norms of operatote on L1. A function having the form (100) is, in analogy with the ease of operators, termed "degenerate."

__________

5. The Fredhoim Alternative In HUbert Spaces

corresponding to the adjoint operator is immediately established for the general ease by the identity' dy dx

(Kf,g)

55f(y)K(y,x)g(x) dx dy

(f,K*g)

(113)

In order to show that the operator determined by the kernel K(x,y) is completely continuous, we use a routine approximation argument. According to the definition of the Lebesgue integral, it is possible, given any 0, to approximate K(x,y) by a step function S(x,y) such that <e, and it then follows from a remark following (11]) that the II IC — norm of the nperator K — S cannot exceed €. Since any step function is

degenerate (cf. Exercise 32) and determines a degenerate operator, it follows from Theorem 4-3 that K is completely continuous. It now follows that Theorem 16 is applicable to the integral equation f(x)

g(x) + Xff(y)K(x,y) dy

The Fredhoim alternative assumes the following form: If K(x,y) E L2, then either (114) has a unique L2 solutionf for each L2 function g, in which where C is a constant depending on X and K(x,y), or the homogeneous equation case !jf(j

f(x) = )jf(y)K(x,y) dy (115) possesses a finite number of independent solutions, in which case the homogeneous equation h(x)

dy

(116)

possesses the same number of independent solutions, and (114) is solvable if and only if dx 0 for every solution of (110). We stress that (114) must be interpreted in the L2 sense. 13y this we

mean that the integral appeartng in (114) may fail to exist for a certain null set of values of x, and that there may be an additional null set for which the two sides of (114), although both well defined, have different values. If f, g, and K are replaced by equivalent functions, (114) will continue to be satisfied in this generalized sense. However, if the limits a and b are finite and if the functions g(z) and K(x,y) are continuous in

the closed interval a x b and square a x, y b, respectively,

then (114) may be interpreted in the conventional sense of holding pointwise, without exception. This follows readily from the observation that, according to the Schwarz inequality, jff(y)[K(xi,y)

—

K(z2,y)J

max IK(xi,y) — K(x,,y) 4v

(b —

'We freely use here and elsewhere Theorem 18, the Fubini theorem.

(117)

Partial Differential Equations

120

Since the first factor on the right side of this inequality can be made small small [this follows from the uniform continuity of by making — K(r,y)J, the right side of (114) must be defined and continuous throughout the closed interval. Therefore, any solution of (114) in the L2 sense is equivalent to a certain continuous function which satisfies the integral equation pointwise. If the operators J and K correspond to the L, kernels J(x,y) and K(x,y), respectively,

then the operator JK corresponds to the kernel L(x,y),

where

L(z,y) = JJ(x,z)K(z,y) dz and L(x,y) also belongs to L2. The first assertion follows by imitating suitably the reasoning following (110), for, just as the equality h = Kf is expressed analytically in the form (110), so the equality k = Jh = JK/ is expressed in the form

k(x) = fh(z)J(x,z) dz = fJ(x,z)(ff(y)K(z,y) dyj dz and (118) follows by inverting the order of integration. The second assertion is proved by applying the Schwarz inequality to (118) to obtain JL(x,y)j2

dzj[fJK(z,y)J2 dzj

(120)

and

then integrating (with respect to x and y) to obtain < 00 U!.JtII In particular, if .1 = K, we find that the operator K! is represented by a kernel construtted from K(x.y) by an integration procedure. This kernel, known as the "first iterate" of K(x,y), is denoted by K(!)(x,y). Similarly, the operators are represented for i& = . by "higher iterates" K(l)(x,y), For the purposes of the next section it is necessary to show that different kernels determine different operators. By linearity, it suffices to show that the only kernels which determine the zero operator are the null functions. If the kernel K(x,y) "annihilates" every L2 function, then the first equality of (113) shows that, for every pair of L2 functions f and g, the equality dy dx = 0 (122) must hold. By choosing f and g as the characteristic functions' of arbitrary (finite) subintervals of a <x
.

The characteristic function of any subset of a given set is equal th one at each point of the gubaet and zero elsewhere.

_____ 121

3. Tb. Fredholin Alton3Iativø in HUbert Spaces

It is of great importance the study of some integral equations that defined by (118) may turn out to be an L function, the kernel though one or both of the kernels J(x,y), K(z,y) may not be. In particular, it may happen that a kernel not belonging to L1 may have an iterate which belongs to L,. A simple example of such a kernel is given by K(x,y)

y < 1)

—

(123)

in which case the first iterate is given by the L, kernel K(s)(x,y)

+

+ 2 log

+ (1 — y)9J

—

—

2

log (x

—

(124)

t4

(Cf. Exercise 35.) The importance of such a situation can be indicated by showing bow the Fredhoim theory may then be applied, by means of a

simple artifice, to the integral equation (114) despite the fact that the kernel is not in L2. If we replace f(y) on the right side of (114) by the entire right side of (114) (with suitable change of letters), we obtain dy (125) f(z) g(x) + \JK(x,y)g(y) dy + dy and K(2)(x,y) belong to L2 and L2, respectively, If g(x) + then the Fredhoim theory applies, so that, if X2 is not a singular value of K(,)(x,y), there will exist a unique L2 solution of (125), and (114) cannot have any L2 solution other than this one. [The converse problem, namely, whether a solution of (125) also satisfies (114), must be studied

in each case.] EXERCISES

31. Let a —

0,

&

I, .K(z,y) —

1

+

Prove that

—

1,

while

—

32. Prove that every step function of two variables is degenerate, as defined in the footnote on page 118. 33. The kernel K(z,y) is said to be hermitian (ef. Sec. 7) if K(x,y) — K(y,x); according to (112), a hermitian kernel determines a herinitian operator. Prove that all the iterates of a hermitian kernel are bermitian. (This is to be expected formally, since the powers of a hermitian operator are hermitian.)

34. Prove that (122) implies that K(x,y) is orthogonal to every step function; from this result prove that K(z,y) is orthogonal to itself, and hence is a null function. 3$. Work out the details leading from (123) to (124), and prove that K(,)(z,y) is an L1 kernel but that K(x,y) is not.

7. Herinitian Kerncls Let K(x,y) denote any hermitian L, function, and let K denote the corresponding hrrnitian operator on If the eigenvaluee' and 'We aume here that there are infinitely many eigenvalues and otherwise the results of this seition are trivially true.

Differential Equatione

122

of K are obtained in the manner described in Sec. .4 eigenveetors } j evidently form an following Theorem 14, then the functions (.(x) orthonormal set in L2, and the Fourier coefficients of K(x,y) with respect to this set of functions are given by

dy dx

ffK(x,y)41k(x)cbk(Y) dx dy

dx =

= From (16) it follows that the series

(126)

Q is convergent, so that the series (127)

Z If we let

converges in norm to a hermitian L2 function

denote

the operator on L2 determined by the kernel R(x,y), it is evident from (127) that, for any element f of L2, the expansion =

Pkd)k(X)

f

(128)

dy

[We emphasize that (128) holds in norm, not necessarily in the sense of pointwise convergence.] On the other hand, according to Theorem 15, the right side of (128) is also equal to Kf. Therefore, the determine the same operator on L2, and so, kernels K(x,y) and coincide almost according to a result stated in Sec. 6, K(x,y) and everywhere. Thus we have shown that the kernel K(x,y) is expressible in terms of its eigenvalues and eigenfunctions by the series expansion holds.

K(x,y)

(129)

and that, for any element f of L2, the element Kf admits the expansion

f f(y)K(x,y) dy =

f f(y)

dy

From (129) and the Parseval equality (21) it also follows that

Now, from (75) we obtain

for K's, n 2, the expansions (132)

5. The Fredhohn Alternative in Hubert Spaces Since the operator K' is determined by the iterated kernel K(,1)(x,y) and no other kernel, it follows by a repetition of the above argument that, in analogy with (129), (130), (131), (129')

=

f f(y)

(x,y) dy

11(v)

=

dy

(130') (131')

!11K'oUI2

We sum up these results in the form of a theorem.

Theorem 17. Expansion Theorem. Any hermitian L3 function K(x,y) and its iterates K(2)(z,y), K(Z)(z,y), . . . are expressible, in and eigenfunctions of the operator terms of the elgenvalues { K determined on L2 by K(x,y), by the expansions (129) and (129'). The functions Kf, K2J, K'f . . . are expressible, for any L2 function f(x), by the expansions (130) and (130'). The L2 norms of K(x,y) and its iterates are given by (131) and (131'). The right sides of (129), (129'), (130), (130') are convergent in norm, but not necessarily pointwise, to the functions appearing on the left.

It is to be expected that under suitable hypotheses the series appearing in (129), (129'), (130), (130') will converge pointwise to the functions appearing on the left sides. In particular, suppose that the numbers a and b are finite and that the kernel K(x,y) is continuous in the closed square a x, y b. In this case, as explained in the paragraph following (116), the eigenfunctions •*(X)) and the function Kf Ifor any L2 function f(x)J may be considered as continuous functions. With this understanding, we can prove that the right side of (130) converges uniformly to the left side by the following argument. For any fixed value of x, the Fourier coefficients of K(x,y) are given by

f so

dy =

that, by Bessel's inequality (16),

f

(1$)

From our hypotheses, the right aide of (133), and hence also the left side, is bounded for all x by some constant C.

By the Schwarz inequality

Partial Differential Equatlona

we now have, for any positive integers n and m, n
f f(y)#*(y)

dy

C

f

dy 12

(134)

By appealing once again to Bessel's inequality we see that the right side, and hence the left side, of (134) can be made smaller than any preassigned positive number, uniformly in x, by choosing n sufficiently large. Therefore, the right side of (130) converges uniformly, and sines each term is a continuous function of x, the sum of the series is a continuous function, which we denote by g(x). In order to show that g(x) coincides with the left side of (130), we note that both of these functions are the limits, in norm, of the partial sums of the series on the right side. Since a Cauchy sequence in any normed space has at most one limit element, it follows that g(x) — ff(y)K(x,y) dy is a tiull function. Since, furthermore, the

last expression is the difference between two continuous functions, it must vanish identically. Thus, we have proved that (130) holds true uniformly pointwse as well as in norm. The same argument holds good, obviously, for (130'). Turning to (129), it is not difficult to show by examples that it does not necessarily hold in the sense of pointwise

convergence under the assumptions stated at the beginning of the present paragraph. However, we can show that (129') holds for all the iterated kernels. For any fixed value of t we replace f(y) in (130) by K(y,t). The left side becomes K(!)(x,t) and the right side assumes the

Replacing t by y, we conclude that (129')

form

holds, with n =

2,

uniformly in x for each fixed value of y (and, by

symmetry, uniformly in y for each fixed value of x). To show that the convergence is uniform in both variables jointly, we argue as follows. Setting x = y, we find that the identity K(2,(x,x)

=

in the sense of convergence. Now, according to Dial's theorem, the convergence must be uniform. Therefore, given 0, we can find such that whenever
< holds for all x.

By the Schwarz inequality it then follows that 12

< €'

(136)

5. The Fredhoim

125

in Hubert Spaces

for all x and y, and this inequality asserts that the convorgencA is uniform jointly in x and and also that the convergence is absolute. The same reasoning obviously applies to the higher iterates. We state the results of this paragraph as a corollary of the above theorem.

If K(x,y) is hermitian and continuous in the closed square a x, y b, where a and b are finite, the expansions (129'), (130), COROI,IIARY.

and (130'), but not necessarily (129), hold pointwise and uniformly.

The above corollary is supplemented by the following remarkable theorem.

Theorem 18. Mercer. The above corollary holds for (129) if the eigenvalues of K(x,y) are all of one sign. Proof. Without loss of generality, we assume that the cigenvalues are all positive. From the proof of Theorem 15 we then know that, for every function f(x) of L2, the inequality

dx dy 0 must hold. If K(z,z) were negative for any value of x, we could, by continuity, determine numbers a and p such that Re K(x.y) would be By choosing J(x) as the negative throughout the square a <x, y characteristic function of the interval <x
<0 in contradiction with the preceding inequality. Thus, we have shown that a continuous hermitian kernel without negative eigenvalues cannot be negative along the diagonal x = y. Now, the kernel K(x,y) —

is also continuous, hermitian, and without negative eigenvalues, since all its elgenvalues are obviously obtained by suppressing the first n eigenvalues of K(x,y). Therefore, for any positive integer n, we have the inequality

K(x,x) k—i

and by a paesage to the limit we obtain (138)

Partial Differental Equations

126

Since K(x,x) is bounded, we may also assert that, for some constant C and all y, (139)


0 and any fixed value of x, choose N so large that whenever

Given

N
(140)

Then for all y the inequality

holds.

j2

k—n

[ k—n

< & (141)

[ k—i

Therefore the series (127) converges for all x and y, uniformly in y for fixed x, so that (127) defines a function K(x,y) continuous in y for each fixed x. Furthermore, taking account of the uniform convergence (with respect to y only), we may assert that for each function of L2 the equality will hold.

f holds. L2 and

f

dy

(142)

Taking account of (130), we conclude that, for each function / of for each value of x, the equality ff(y)L(x,y) dy

0

(143)

holds, where L(x,y) = R(x,y) — K(x,y). Since L(x,y) is continuous as a function of y, we may assert that it cannot ever be different from zero; the argument is the same as that employed earlier in this proof to show that K(x,x) 0. Therefore K(x,y) R(x,y), and we have shown that (129) holds in the sense of pointwise convergence. Finally, to show that the convergence is uniform jointly in r and y (not merely in y for each fixed value of x), we note that, since (129) is now known to hold pointwise, the equality

K(x,x) =

(144)

must hold for each value of x. We may therefore invoke Dini's theorem, exactly as in an earlier part of the present section, to show that the right side of (129) converges uniformly to K(x,y). The proof is therefore complete.

5. The Fredholni Alternative in Hubert Spaces At the beginning of Sec. 6 it was indicated that there exist completely continuous operators on L2 which cannot be expressed as L2 kernels. The results of the present section enable us to give an example of such an operator. There exists a converse to Theorem 15 which asserts that, given in any Hubert space 11 an orthonormal sequence of elements and a sequence of nonzero real numbers converging to zero, there exists a hermitian completely continuous operator K on H expressible in the form (75). (Cf. Exercise 36.) Now, let the quantities be so }

chosen that

diverges.

Then, taking H as L2 and taking account

of (131), we see that K cannot be represented by means of an L1 kerneL

in fact, the convergence of the series

is readily

from the

n—i

developments of the present section, to be necessary and sufficient for the existence of such a kernel. EXERCISE 36. Prove the theorem stated in the laat paragraph of the present section.

8. illustrative Example In this section we illustrate the foregoing theory by considering in some

detail a symmetric (real hermitian) kernel for which all the required computations can be carried out explicitly. Let the kernel K(x,y) be defined as follows in the square 0 X, y 1: K(x,y)

(145)

= {

It is obvious that K(x,y) is continuous and symmetric, so that it defines a hermitian and completely continuous operator K on L,(0,1). We ahall first show that all the eigenvalues of K are different from zero. if not, there would exist a function f of positive norm such that1 K(x,y)f(y)

0

(146)

Taking account of the definition of K(x,y), we rewrite (146) in the more

specific form

f

yf(y)

dy + x f'f(y) dy

x 101 yf(y) dy

(147)

left side of (148) must vanish everywhere, without the possible exception of a null set., according to the argument presented in the paragraph following (116).

Partial Differential Equations

128

We may now, by Theorem 1-9, assert that each of the two integrals on the left side of (147) defines a function which is differentiable for almost all z,

that the derivative may be found in the usual manner; thus, the equality

dy = must hold almost everywhere.

j1 yf(y) dy

(148)

Since the left side of (148) is a continuous

funetinn of x and the right side is a we conclude that (148) must hold true without exception. Setting x = 1, we conclude that f' yf(y) dy 0., and. hence that

i: f(y) dy

0

(149)

(149), we conclude that 1(x) 0 almost all x, and 0, contrary to the original assumption. We are now assured that the eigenfunetions of K(x,y) span L2(O, 1). Let bo any cigenvalne of the operator K; then, since p 0, the equation Kf =• pf may be written iiithe form

hence. that

f(x)

1:);' yf(y) dy + x ft f(y) dy

—

yf(y) dy]

x

(150)

As before, we conclude that (150) holds everywhcre; differentiating twice, we obtain

f"(z) = --p''f(x)

(151)

Now, from (150) we observe that. f(O) = f(1) 0. 1-fence, p must be such that admits a nontrivial solutIon vanishing at both ends of the interval 0 x 1. By routine methods, we find that this requirement restricts p to thc values p

p,,

=

(n = 1, 2, 3,

.

(152)

that the corresponding normalized solutions are given by sin nwx

(153)

Conversely, by direct substitution we find that f, (x) is indeed a solution. of (150). Thus, we have obtained the well-known theorem from the theory of Fourier series that. the functions (153) span L2(O,1). Since the eigenvahies arc all of erie sign, Mercer's theorem is and we obtain the identity 2 sin

(154)

5. The Fredhoim Alternative in Hubert Spaces

129

The equality (131) assumes in the present case the form (155) 1

Now let g(x) be any

and let X be any constant.

in

Then the integral equation dy

J(x) —

g(x)

possesses the unique solution [cf. (108)) pa

f(x) = g(z) + if

nrxJ__g(y) sin nirydy -—. —

— -—

distinct from all the quantities whereas, if X conclude that (156) is solvable if and only if is

0; when this condition is satisfied, (156) possesses infinitely many solutions,

each of the form

f(x) =

where

g(x)

+

+ csinkwx (158)

the constant c is completely arbitrary. By rewriting (156) in £he

form f(x) — g(x)

X

J°1

K(zy)f(y) dy

and appealing to Theorem 17, we see that the difference f(x) — g(x) possesses a uniformly convergent Fourier expansion, and hence that equations (157) and (158) in the sense of pointwise, and indeed uniform, convergence, not only in the

of convergence in norm.

37. Work out the details leading to (155). 38. Derive fro,n (154) the I

39. Work out

and obtain

analogue øf (155) for this kernel.

6. ELEMENTS OF POTENTIAL THEORY

1.

Introduction

The simplest of all elliptic equations, and, in many respects, the basis for the study of more general equations of elliptic type, is the so-called Laplace equation, which in n dimensions, with rectangular coordinates xz,

.

.

.

,

reads

u2, +

±

•

•+

0

(1)

A function u(x1,x2, . . . ,z,,) possessing continuous partial derivatives of second order in a domain G and satisfying (1) at all points of G will be said

to be "harmonic" there. The present chapter is devoted chiefly to a study of some of the most important properties of harmonic functions, while the following chapter contains a fairly detailed study of one particular problem arising in the theory of harmonic functions—the Dirichiet problem. Equation (1), particularly in two and three dimensions, has been one of the most carefully studied of all partial differential equations. In addi-

tion to its interest as a purely mathematical problem, it is of great importance in mathematical physics, arising in the study of such subjects as electrostatics, fluid dynamics, elasticity, and heat conduction. Indeed, much of the stimulus for the study of Laplace's equation and many of the most important concepts that have been built up around it have had their origin in physical problems. The theory of Laplace's equation and the closely related equation of Poisson (ef. Sec. 12) Is usually termed

"potential theory," on account of its applicability to various physical problems involving potential fields.

We remark that the left side of (1) is, in vectorial terminology, the divergence of the gradient of the scaler fteld u(xj,x,, . . . it is 180

131

6. Elements of Potential Theory

frequently denoted by the 8ymbols V Vti, Vtts, or Mi. We shall use the last of these three notations. EXERCISES

I. Prove that (2a)

(2b)

e(8m Os.).

—

in plane and spherical polar coordinates, respectively. rin (2b), 0 denotes the colatitude and the azimuth.J 2. Prove that (1) is unchanged under the transformation of coordlnate8

+

(i —

1, 2, . .

.

,n)

0) and are arbitrary. 3. Prove that (1) is unchanged under any orthogonal transformation of coordinates; i.e., a linear transformation where the constants c

—

(1 — 1, 2, .

,n)

for which the quslity —

i—1

always

2. Laplace's Equation and Theory of Analytic Functions In the two-dimensional case the theory of Laplace's equation is essentially the same as that of analytic functions of a complex variable. The connection between these two theories is established by the well-known theorem Mating that the real and imaginary parts of an analytic function f(z)

f(x + ly) = u(x,y) + iv(;y)

harmonic, and that, conversely, given any function u(x1y) harmonic in a domain (7, there exists a (not necessarily single-valued) function v(x,y), the harmonic conjugate of u, such that the function f(z) defined by (6) is analytic in 0. Thus, harmonic functions are the real parts of analytic functions, and the powerful techniques of analytic function theory become available for the study of harmonic functions. For one simple, but highly interesting, Illustration of the use of analytic function are

1Exeroiaes 2 and 8, taken together, aort that a function harmonic in a domain 0 remains harmonic if 0 is subjected to any deformation which can be composed of rigid reflections, and zitretchinp.

PartIal Differential Equationa

132

theory in the study of harmonic functions, we refer to Exercise 29, in which an alternative derivation of (26b) is obtained. (However, the method described in this exercise is, by its very nature, confined to the plane, but the proof given in the text of Sec. 7 is, with obvious changes, applicable in any number of dimensions.) As ar*other illustration, we refer to Exercise 30, where the Neurnanp problem for a plane domain is converted into a Dirichiet problem; no comparable device exists in higher dimensions. A striking fact about Laplace's equation in the plane is that it remains invariant under analytic (conformal) mapping. More precisely, if u is a

harmonic function of x an4 y, aad if a one-to-one analytic mapping .r + iy + is given, then y(E,i)) is also harmonic as a function of and;. This fact, which is basic to the use of conformal mapping in determining potential fields (cf. Sec. 7-7), is easily proved as follows. We determine the harmonic conjugate v of u and form the analytic function/(z), as in (6). SInce an analytió function of an analytic function is itself analytic, it follows that ti, the real part of an analytic function of + i;, is harmonic In and (An alternative proof is outlined in Exercise 4.) An important consequence of the above result is the following. If u is harmonic in a domain 0, and if the function U is defined as follows: (T(P*) where

u(P) P E 0

P and P' are inverse points (ef. Sec. 7) with respect to the unit

circle, then U is harmonic in the domain G consisting of all points inverse to those of 0. This is proved by noting that the inversion may be accomplished by following the mapping E + i; (x + with a reflection

in the real axis. Since each of these transformations

har-

monicity, so does their resultant, If G contains the origin, then we find it convenient to consider as including the "point at infinity." Indeed, a function 11 harmonic for all sufficiently large values of r2 = + is

defined to be "harmonic at infinity" if the function u obtained by the inversion is harmonic at the origin. (Cf. S.) For the sake of brevity, in this and the following chapter will be formulated in two dimensions, but in order to make clear that the theorems hold (with straightforward modifications) in higher dimensions as well, tbe proofs presented will not use techniques. In studying the theorems, the reader should think through the corresponding formulations and proofs in higher dimensions. EXERCISES 4. Prove the invariance of Laplaoe's equation under conformal mapping by con-

verting differentiations with respect to x end y into differentiations with respect to

6. Elements of Potential Theory (Of course, the Cauchy-tUemaun equations zs and exploited.

y,, c,

must be

5. Provc the invariance of Laplace's equation under inversion by performing the substitution P 1/r, — 8 in (2a).

3. Fundamental Solutions An important rote in the development of potential theory is played by of the Laplace equation which depend only on the distance r from some fixed point. It is readily proved (cf. Exercise 6) that in the plane the only such solutione arc of the form those

u—alogr+b

(7a)

whereas iu n( >2) dimensions the solutions have the form

u=

(7b)

where a awl b are arbitrary constants. These solutions are employed in establishing the most important properties of harmonic functions. This is accomplished by the use of the theorem and some of its immediate corollaries. Though these are presumably familiar, we review them here briefly for convenience.

Divergence Theorem. Let 6' be a bounded plane domain whose boundary r Consists of a finite number of disjoint smooth (or sectionally smooth) simple closed and let. the functions f and a be defined and continuous first denvatives in (2. Then

fJ0

+

11' dy —

dx dy

(The integration over the boundary is understood to be in the positive

sense with respect to the domain G.) Now, if u possesses continuous first derivatives and v continuous second derivatives in we ObtAiIL. by settingf g Creen's first identity,

j1 where

(u

+

.

Vu)

dx dy

I

denotes differentiation in the direction of the

normal.

If u also possesses continuous secoiid derivatives, then it is permissible to

]nterehange ii and ii in (9), and by subtraction we then obtain Green's second identity,

ff(u4v—vAu)clxdy=

Equations

Partial

134

Several

particular cases are worthy of note. If u

1, (9) reduces to

If JJQY = and if, furthermore, v is harmonic, (ha) simplifies to

(llb) If v is harmonic and u is set equal to v, (9) yields

fj(vx2 +

dxdy =

Finally, if u and u are both harmonic in 0, (10) yields

1/iu——v-—ids=0 8v

I

Jr \

(13)

On,

ôfl

Now, let Q be any point of G, let r denote the distance from Q, let v = log r, and let u be any function having continuous second derivatives in 0. it is not permissible to apply (10), for at the point Q the function v violates the hypotheses. However, (10) is applicable to the domain G obtaine4 by deleting from C a disc of radius with center at Q, which, together with its circumference r4, lies entirely in 0. There is obtained (since v is harmonic in C.)

—

Jf

log r

dx dy

f (u

log r — log r —

I

da

e

On1

(14)

From the elementary inequality —

we

2ir€u(Q)j

IfreEU

—

u(Q)1d81

—

u(Q)I

obtain the limiting relation

I

.—.o 2we jr.

uds

u(Q)

(15a)

0

(15b)

Similarly, from the inequality

= we obtain the limiting relation Urn log

lou ds

185

6. Elements of Potenti*I Theory From (14), (15a), and (15b) we huniediately conclude that

=

—

—

(16)

This formula, expressing the value of u at any point of G in terms of the

values of u and

in U, is of great

on r and of

and brings

This fuitction, and in higher dimensions the functions r2", are known as fundamental solutions of Laplace's equation. out the essential role played by the function log (1/r).

EXERCiSES 6. Prove, as asserted in the text, that all harmonic functions depending only on the distance from a fixed point are gwen by (7a) and (7b). 7. Prove that if G possesses a sufficiently smooth boundary r, v is harmonic in (7, and v is. constant on r, then v is constant. in 0. (Actually, the result holds for any bounded domain, regardless of the nature of its boundary, but the proof in the general case, which is given in Sec. 5, runs quite differently.) 8. Show that the obvious genera)izatitin of (16) holds in higher dimensions, with and the factor ,r on thc left side replaced by an appropriate log (I/r) replaced by constant r,,. I)etermine the values of c, for n 3, 4

4. The Mean-value Theorem If U is a disc of radius R with center at Q and if u is harmonic in O, then (16) simplifies to

u(Q)

1

ifau

if

(17)

right vanishes by (1 Ib), while the Furthermore, the first term on second term is simply the mean value of u over r. Thus, we have the interesting and important Theorem 1. Mean-value Theorem. If u is harmonic in a disc G, including the boundary F, then the value of u at the center is equal to the mean value of ti over the circumference; i.e., (18)

Now, suppose only that u is harmonic in the (open) disc G and conin (7. The above argument can then be applied to a slightly smaller concentric circle r'. Since u is uniformly continuous in (7, it follows that the mean value of u over r is equal to the limit obtained by Jetting V1 expand to F. Since the mean value over I" is always equal to

Partial Differential Equations

136

u(Q), it follows that the same must be true for the mean value over F. This result will be stated as a CouoLt.&aY. Theorem 1 holds under the weaker hypotheses that u is harmonic in G and eon tinuous in O.

If we multiply both sides of (18) by 2TR and integrate with respect to R, we obtain

u dx dy, or

=

(19)

u(Q)

Thus, we have obtained the following corollary, which may be termed the "areal" mean-value theorem.

If u is harmonic in a disc G arid continuous in (), the value of u at the center equals the mean value of u over G. EXERCISES 9. By differentiating equation, one is led formally to the result that all derivatives of a harmonic function are also harinonio. However, such a procedure that the dnrivntrveg of higher order exist and that the necessary changes in order of differentiation are valid. A'eepting these facts Cd. Sec. 7), prove the fol-

lowing: If u is harmonic in a disc G of radiu*s ft with center at Q and if 5 throughout G, then 4M/irR. By a alight refinement of the argument, show that + 10. Lot u be harmonic in a disc C radius 2? with center at Q. Let —K

Prove that

and, more generally, if P is any point of C,

jii(P)J . If

Consider the Integral

—'VI

Eu —

dx

5. The Maximum Principle Theorem 2. Firat Form of Maximum Principle. If u is har-

moriic in a bounded domain G and in then the maximum value of u Ia attained at one or more points of the boundary of (1. Proof. The existence of a point where the maximum is attained is assured by the compactness of G, and it is therefore required only to show

137

6. E)eme.its of Potential

that if the maximum is attained at a point Q of (1, it is also attained at a boundary point. Let R be the distance from Q to the boundary, and let F be the circle of radius R with center at Q. By combining the first

corollary of Theorem I with the fact that, for every point P of u(Q)

u(P), we obtain u(Q)

fu

f ds =

'u(Q')

(21)

from which it follows that equality must hold between the second and third terms. Since u is continuous on r, the equality can hold only if, at all points P of F, iz(P) = u(Q). Since, from the definition of F, it has at least one point ui common with the boundary of G, the theorem is proved.

A second proof, quite different from the above, will also be given. Suppose that. the maximum is not attained at any boundary point. Then it must. be attained Si. a point Q of G, and, for sufficiently small positive the function v(x,y) = u(x,y) + €(x is also larger at Q than at any boundary point, xo denoting the abscissa of Q. Therefore, the maximum of u is attained at some point Q' of 0. Since

it that at least one of the > 0, >0 must be satisfied; but this cannot occur at a maximum. If u is harmonic but not constant in a bounded domain 0 and continuous in (?, the maximum value of u cannot be attained at any point of 0. Proof. Suppose that the maximum value of u is attained at a Q of 0. The argument employed in the first proof of the above theorem shows that u must equal u(Q) at all points of any disc with center at Q which lies entirely in G.

Hence, the subset of (2 on which the maximum of u is attained is open. On the other hand, the subset of (2 where the maximum of u is not attained is also open (by continuity), and the latter subset is nonvacuous, since u is assumed not to be constant. Thus 0 would be representable as the disjoint union of two nonvacuous open sets; implies that G is not connected, and so the proof is complete. By applying the above theorem to —a, we obtain a minimum principle, which need not, be stated explicitly. Theorem 3 Given a bounded domain (2 and a continuous function / defined on the boundary, there exists at most one function harmonic in 0, continuous in O, and equal to / at. aft boundary points. Proof. The difference of two such functions would be harmonic in 0, continuous in O, and zero stall boundary points. By the maximum and minimum principles, the difference would then vanish throughout 0.

Partial Differential

138

It should be stressed that Theorem 3 does not assert the existence of a harmonic function with prescribed (continuous) boundary values, only its uniqueness. However, throughout much of linear analysis uniqueness goes hand in hand with existence—recall the Fredhoim alternative (Chaps. 4 and 5)—and so it is plausible to expect that the given functionf can be extended continuously to in such a manner as to be harmonic in G. We are thus led to the Dirichiet problem, which will be discussed at

some length in the remainder of the present chapter and the entire following one. EXERCISE 11. Show by an explicit computation that the function I— 1 —

—

——

2z + z' + y2

is harmonic throughout the unit disc. Sinee the numerator v8nishes at all pointa of the unit circle, the maximum and minimum principles might be thought to imply that u 0. Find the error in this reasoning.

6. Formulation of the Dirichiet Problem As indicated at the end of the preceding section, Theorem 3 suggests the following problem, which has been, for over a century, the subject of intensive study. DIRICRLET PROBLEM. Given a bounded domain G and a continuous function / defined on the boundary of G, to find a function u harmonic in

G and continuous in G such that, for all points Q of the boundary, u(Q)

The condition that G be bounded can be dropped. and the condition that f be continuous can be relaxed somewhat, but we shall concern ourselves here with the more restricted formulation, as given above. It is worthwhile to demonstrate, by means of a simple example, that, contrary to the suggestion made in the preceding section, it is possible for some domains G to choose the function f in such a manner that the Dirichiet problem is not solvable. Let G be the domain defined by the inequalities (22)

i.e., the unit disc with the origin removed (so that the origin constitutes a component of the boundary). Let f be defined on the boundary as followu:

1

0

on z2 + y!

1

/(0,0)

1

(23)

of Potenti*I Theory

6.

By symmetry and uniqueness, it is clear that if the problem has a solution, it must possess rotational symmetry; i.e., the solution must be a function of r (x2 + y2)½, independent of the polar angIe 0. According to Sec. 3, the solution must have the form (7a),

u=alogr+b The first condition of (23) requires b to vanish; on the other hand, any choice of a other than 0 would cause u to be uubounded near the origin, while the choice a 0 would cause u to vanish identically, so that in

either ease the second condition of (23) cannot be satisfied.

(Cf.

Exercise 12.)

A domain having the property that the Dirichiet problem can be solved for every (continuous) function f defined on the boundary is termed a "Dirichiet domain." It will be shown in the next section, by suitably exploiting the geometrical properties of the circle, that every disc is a Dirichiet domain. Then, in the next chapter, a wide variety of domains will be shown to be Dirichict domains; in the various proofs to be

presented, it will be seen that the fact that discs possess this property plays a fundamental role. Therefore, the content of the next section, and in particular the Poisson formula (27), is of basic importance in the theory of harmonic functions. EXERCISES 12. Solve the Dirichlet problem for an annulue bounded by concentric circles of radii R1
13. With the aid of the maximum principle and the solution to the preceding exercise, prove: if u is harmonic inside and on the circumference P of a disc 0, except p.rhape at the center of 0, if u has a finite upper bound M in the "punctured" disc 6, and if max u(P) m, then in M.

r

7. Solution of the Dirichiet Problem for the Diac In order to demonstrate many important properties of harmonic functions, and thus prepare the way for proving the solvability of the Dirichlet problem for very general domains, it is necessary to obtain, for of domain, an integral representation of a function which is harmonic inside and on the boundary of the domain, in terms of its values on the boundary. As might be expected, the moat convenient domain for this purpose is a disc. After such a representation is obtained, the solvability of every Dirichiet problem (with continuous

at least one

Partial Differential Equationa

boundary values) for the disc will follow almost immediately. results will be obtained in the present section.

These

Formula (16) provides, if the function a is harmonic (in which case the double integral drops out), an integral representation for u at any point of G in terms of the values of u and its normal derivative on the boundary. From this representation there follows one of the most important and

striking properties of harmonic functions, namely, that a harmonic function possesses derivatives of all orders, a fact certainly not directly obvious from Laplace's equation. (The proof is immediate. If a is harmonic in a domain, one covers any given point of the domain by a disc

which, together with its boundary, lies in the given domain, and the disc by means of (10). Since any number of differentiations may be curried out under the integral signs, it follows that a has derivatives of all orders in the neighborhood of the selected point, and hence throughout the given domain.J However, (16) suffers a certain defect if it is to be used as a tool for studying harmonic functions, for, according to Theorem 3, a is completely determined throughout 0 by

its values on the boundary, and therefore the values of the normal derivative on the boulithiry are themselves determined by the values of a on the boundary. This suggests that one might expect the existence of a formula expressing u at any point of G exclusively in terms of its boundary values, not involving the normal derivatives at all. As indicated above, such a formula does exist in the case that G is a disc and, in fact, for much more general domains, as will now be shown. The derivation of this formula is based, in the case of a disc, on a very simple geometrical fact. Let 0 be the center of a circle of radius R, let Q be any point (other than 0) inside the circle, and let Q* be the inversc point of Q (i.e., R1 and the points Q and Q* lie on the same

half line terminating at 0). Then, as the point S describes the circumference, the ratio does not vary; the constant value of this ratio is therefore equal to the value which it assumes when S coincides with T, the end point of the radius through Q (cf. Fig. 6-1), in which cuse it evidently equals (R2/p — R),'(R — p), or R/p, where p = It then follows from (la) that the function

0(PQ) =

R

2rpcos(8 — 4)

=

+

— 2rp cos (0 — 4)

+

(where r, 0 and p, 4 denote polar coordinates of P and Q, respectively) possesses the following properties: 1. For any fixed point Q in the disc, G(P,Q) is harmonic throughout the disc, except at (2. 2.

The function (.7(P,Q) + log

including Q.

is

harmonic throughout the disc,

6.

141

of Potentisi

0 as P approaches the boundary of the disc. 3. G(P,Q) The function G(P,Q) is known as the "Green's function" of the disc (with pole at Q). Now, by repeating the argument which led to (16), but

Figure 6-!

with the function log (1/r) replaced by G(P,Q), we obtain —

j

ds —

ff

G(P,Q)

u is assumed harmonic in the disc (including its bound. ary), (25) thmpliflee to aG(P,Q)

-ds

Taking account of (24) and the fact that oG(P,Q)1 8r

we find that (26a) can be written in the form dO

J°2'

where points are, for convenience, represented in exponential form and denotes the 'Poisson kernel" — I" — 2Rp cos (8 —

+ p9

In (26h) we have a formula providing the desired representation of a function harmonic in a disc in terms of its boundary values.

Partial Differential Equations

142

We now remark, as indicated above, that (26a) is applicable for a wide

class of domains. Suppose that for a domain G there exists a function G(P,Q) satisfying the following obvious modifications of the above conditions 1, 2, and 3: 1'. For any fixed point Q of G, G(P,Q) is harmonic throughout (1, except at Q. is harmonic throughout G, including 2'. The function G(P,Q) + log Q, and continuous in O. 0 if P lies on the boundary. 3'. G(P,Q) Such a function is again termed the Green's function of (.7. If it exists, if the boundary F of G is sufficiently smooth, and if G(P,Q) possesses then t.he argument leading to adequate differentiability properties in (25) applies equally well to the present case, and similarly we find that (26a) holds if u is harmonic in it should be stressed that the derivation of (26b) does in itself prove

that the Dirichiet problem is solvable for the disc with arbitrary contin uous boundary values; at this point (20b) only provides a method for determining the value of a harmonic function at an arbitrary point' of the disc from its values on the circumference. However, if a given continuous function g(O) (defined on the circumference) can be extended

continuously to the closed disc so as to be harmonic throughout the interior, it appears plausible that this can be accompLished by replacing u(Re1) in (26b) by g(8). Indeed, Exercise 18 serves to show that if the Dirichiet problem can be solved for the disc, the solution must be obtaina-. ble in the manner just described. Thus we are forced, so to speak, to study the result of replacing u(Re') by an arbitrary continuous functioti g(8). This leads to the principal result of the present chapter:

Theorem 4. Let g(O), 0 fying g(O) = g(2ir).

8

2w,

be any continuous function satis-

Then the function =

dO

(27)

is harmonic inside the circumference p = R, and for every value of 0 it satisfies the condition urn U(pe") = g(O). Proof. Since the denominator is positive for all 8 and whenever p <.R, the integral is meaningful. There is no difficulty in justifying differentiation under the integral sign with respect to the parameters Strictly epeaking, (26b) was derived on the assumption that Q is not the center of the diet,, but in (28ô) gives the same result as Theorem 1, so that it is valid thmughout the disc without exception. I

143

6. Elements of Potential Theory p and

and so it follows that

+

Ml

(28)

(Cf. Exercise 1.) By direct computation (ef. Exercise 19) one easily and it shows that P(p,4,,R,8) is a harmonic function of the point To prove that U has the proper behavior at follows that U is harmonic. the circumference it clearly suffices, by symmetry considerations, to prove

that Now, for all

urn U(pe") =

(29)

g(0)

and for all p

I=

(30)

This can be proved, of course, by explicitly carrying out the integration, but., in fact, it is a direct consequence of (26b), with u 1. Multiplying both sides of (30) by g(O) and subtracting from (27), we obtain —

g(0)

do

—

.

(31)

Givezi any 0 we can, on acoount of the of g(O), choose a positive & so small that for —S 0 & Then, since — g(0)I <

the Poisson kernel is nonnegative, we obtain from (3!) the inequality •

P(p,4,R,O) do + 2M

U(pc")

dO

(32)

where M max The first integral on the right is less than unity [by (30) and the fact that is never negative]. llence, (29) will

be established if we can prove that the second integral approaches zero (for fixed 6) as R. Now, for — 6/2 6/2 and —6 0 2ir —6, we have R2 — 2Rpcos(8 — +p2 > R — p cos (8/2)12 > R'[I — cos (8/2)J2, and hence P(p,4i,R,O)

Therefore, for — 8/2

(33)

co:(8/2)J2

8/2,

f2r_8 do <21rR2[1_

(2ir — 26)

—

— cos (6/2)12

Letting p R, we conclude that the left side of (34) approaches zero, and the proof Is complete.

Partial Differential Equations

14.4

Formula (27), by which the Dirichiet problem is solved for the disc, is known as the "Poisson formula." It may be remarked that in proving (29) the continuity of g(0) only 0 was uaed, not the continuity of g(0) for all 0. Thia suggests, as 0 indicated in Sec. 6, the possibility of formulating in a reasonable manner s Dirichiet problem with discontinuous boundary values. We shall not discuss this possibility here, but refer to Exercise 20, which treats one particularly important case of a discontinuous boundary function. An alternative form of (27) io obtained with the aid of the identity

cosn(8—

(35)

I

22.)

(Cf.

in e awl we

Since for any fixed p(
[being dominated by the series I + 2

},

may multiply by g(8) and integrate termwise, thus obtaining ti(pe")

where

=

±

ens ni/ +

f' g(0) ens nO rIO,

f7 g(8)

sin

nO .10.

The right

side of (36) is identical with the Fourier series of except for the appearance of the factors (p/I?)". It should be emphasized that for = I? the series (36) may fail to converge; that is to say, the convergence of the Fourier series of does not enter into the derivation of (36). An important application of (36) is made in Sec. 10. Analogues of exist in higher dimensions; in particular, in three dimensions one is led to expansions in spherical harmonics. p

EXERCISES 14. Prove that the Green's function of a (bounded) domain is unique; i.e., therc. exists at most one function G(P,Q) satisfying the condiLions 1', 2'. and 3'.

15. Prove that the Green'8 function must

positive throughout the domain

(except at Q, where it ia undefined). 16. Symmetry of the Green's functien: Let Qi and Q be two distinct points of a

domain 0. Assuming that the Green's function exists, and making appropriate assumptions about the boundary of U and the behavior of the Green's function near the boundary, prove that G(Q2,Qa) (The render acquainted with the

element. of electrostatics should be ahiR to provide interesting physical interpretations of the Green's function and of the symmetry property.) 17. Work out the details leading to (26b).

145

6. Elements of Potential Theory

18. Prove, that If u is continuous in a closed disc and harmonic in the interior (with no dijjerenhiabililv aunmptiona on Lb. circumference), formula (26b) is still valid. thisi: Replacetho disc by a 8lightly smaller disc and employ a continuity argument, as in the proof of the first corollary of Theorem 1. 19. Acide from a limiting argument, prove the harmonicity of the Poisson kernel by considering the manner in which it I obtained from the without Green's function. 20. Let p(8) be continuous except at. 8 8o, and foi this value let the limits front be defined by (27), analyse the behavior the left and right both exist. Letting of this function near the point Ret.. 21. Let u be harmonic and bounded in a deleted neighborhood of a point p Prove Use Exercise 13, that is may be defined at P so as to be harmonic tbcrv. together with the faet that any disc is a L)iriohlet domain. 22. Prove (35). flint: Note that the right, side may be rewritten in the form

flbis funetion provides 23. Derive the three-dimensional analogue of the solution to the Dirkhlet problem for the sphere, in comp)et.e analogy with (27).l 24. Prove that if the expansion (36) of a harmonic function is carried out twice, disc of radius R' in a R), the series first in a disc of radius R and obtained are identical. 25. Poisson formula for the half plane: Let 1(x) be continuous and bounded for all a. Prove that tbe Dirichlet problem for the half plane > 0 with these boundary values is solved by the formula u(x,y)

I.

1

1(t) dE

(37)

1)iscusa the uniqueness situation in this ease. ,:The diWcrence between the present case and that of the disc arises front the noecon%paetneae of the half plane.) 26. Derive (37) from (27) by a conformal mapping. 27. Derive the three-dimensioiztd analogue of 37).

28. Derive (26b) by exploiting the mean-value theorem in combination with a suitable conformal mapping of the disc onto itself.

29. Suppose that is is harmonic in a disc, including its circumference V. By extending is to an analytic fun'tion f(s) aiid expressing f(s) in terms of its values on F by the Cauchy integral formula, derive equation (2.16). Hint: Set up also a second respect to F. Osuchy integral with the point z replaced by its image 30. Let G be a plane domain hounded by closed curve 1', and let a continuotis function g(s) be defined on T. Ph5 problem of determining a function is harmonic in G, continuous in O, and satisfying •— an

g(s)

on F is known as the

Neumann problem. Prove t the Neiiniann problem is not solvable unless the condition -' 0 is satisfied. Wben this condition is satisfied, show that the function u is uniquely determined up to an arbitrary additive constant. [Note that it is not asserted thata solution necessarily exists even when the condition 1 g(s) da 0

Jr

Pattlál Differential

146

is aatisficd.J 8how that the Ncumann problem can be reformulated as a Dirichiet. problem for the hannovie con tzgste of ii. (If the domain G is multiply connected,

being bounded by aev&al closed curves, the above method becomes complicated by the znultiple-valuedneas of the harmonic conjugate, but this difficulty can be overcome.)

8. The Converse of the Mean-value Theorem Theorem 5. Let u be continuous in a domain I)

and let it possess the mean-value property; i.e., for every closed disc lying in D the mean value of is over the circumference is equal to the value of is at the center.

Then u is harmonic in 1). Proof. Select any disc G which, together with its boundary F, lies in D. As was proved in Sec. 7, there exists a function v continuous in harmonic in G, and coinciding with u on F. Then the function w = u — v is continuous in 0, vanishes identically on F, and clearly possesses the mean-value property in G. The argument employed in the proof of Theorem 2 shows that w must assume its maximum and minimum values r; hence w vanishes identically. Thus is coincides with the harmonic function v in C, and is therefore harmonic in C. Since every point of D can be covered by such a disc U, is is harmonic throughout I). It may be pointed out that, as is readily seen from the above proof, it suffices to assume that for each point P of 13 there exists a positive number p(P) such that the mean value of u over every circle with center

at P and radius less than p(P) Is equal to u(P).

On the other hand,

Exercise 31 shows that It does not suffice to assume merely that, for each point P, there exists some circle with center at P over which the mean value of is is equal to n(P). It is remarkable that no differentiability assumption is imposed on is; it turns out that the mean-value property (which is a property involving integration) suffices (together with continuity) to assure that is is harmonic,, and hence indefinitely differentiable. On the other hand, the fact that the disc is known to be a Diriehiet domain plays an essential role in the above prooL Jt is instructive to present an alternative proof which does not make use of this fact, We begin by noting that u possesses the

area! mean-value property, for the proof of the second corollary of

Theorem 1 uses only the fact that is possesses the "circumferential" mean-

value property (and is continuous). Giveti any point P of D, with

coordinates yo, let R1 denote the distance from P to the boundary, and let C denote a disc with center at P and with radius R
tRZu(P)

is dx dy

u dx dy

(38)

6. Elements of Potential

hold, it follows that TR'[u(Ph) — u(P)1 can be expressed as the difference between the integrals of u taken over the two hines formed by 0 and as shown in Fig. 6-2. Taking account of the unIform continuity of u in the fact that each June is of constant width any compact subset of 1) Q

Hgure 6-2

Q and below Q', it follows readily

It, except for tho small portions

that — u(?) —

where F denotes the ciretunference of 0.

Jr

Therefore, it follows that u,(P)

exists and is given by (40)

the same repreeentation holdi for all points sufficiently close to P with a circle of the same size as F, it follows from the (uniform) contiuuity of u that v, not only exists, but is continuous, in a neighborhood of P; since P is chosen arbitrarily, (and similarly u4.) exists and is continuous throughout 1). It is therefore legitimate to employ (8), with / u, g =0. We may therefore rewrite (40) in the form Since

ff

d.c dy

(41)

Thus, we have shown that the first derivatives of exist, aie continuous, and possess the (areal) mean-value, property. The same reasoning as that employed above now shows that the second derivatives also exist and are continuous throughout 0. (In fact, they also posscas the mean-value property, and by repeatedly applying the above argument we can show that u possesses derivatives of all orders. However, we 8halJ not use this fact here.) Thus, the quantity is defined and continuous throughout

Partial Diffçreutial Equations Suppose that at some point P of 0, which for convenience we take as the origin, Au 0. We express i& in the neighborhood of P in the form' D.

u(x,y)

+ u,,(O,O)y +

u(O,0) +

+

+

+ + y2)

(42)

Integrating over the disc G defined by the inequality x2 + y2
where

Jf u dx dy —

+

irR'u(O,O)

+ Jf0

+ y2) dx dy

(43)

By hypothesis, the left side vanishes. Therefore, + u,,,(O,0)] = _8Jf0n(x,y)(x2 + y2i dx dy

(44)

The absolute value of the right side of (44) is at most equal to max

8

(x2+y2)dxdy = 4iR4 max

(z.y)EG

(x.v)EG

If we choose"!? so small that 4tnI < we obtain a contradiction. proof is complete.

throughout G, + Thus, Au must vanish everywhere, and the

It should be remarked that there frequently arise in the study of elliptic equations generalizations of the area! mean-value property, which are used to establish, as was done in the second proof of the above theorem, that the functions under consideration possess adequate differentiability properties.

The above theorem leads immediately to an important and striking method by which the domain in which a harmonic function is defined can be enlarged. COROLLARY.

SCRWARZ

PRINCIPLE.

Let u be harmonic

in a domain D, part. of whose boundary consists of a segment of a straight line L, and let D lie entirely on one side of L. If u approaches zero at all points of the segment, then the definition of u may be extended across the be segment by reflection. More precisely, for each point P of D let

the image of P with respect to L, and let u(P) = —u(P). Then u is harmonic in the domain consisting of D, its reflection D*, and the segment. (Of course, u is assigned the value zero on the segment.) Proof. By symmetry and the fact that u possesses the property in D, while it is it follows that u also possesses the mean-value property in obvious that u also possesses the mean-value property at all points of the

149 6. Elements of Potential Theory segment. Since it is obviously continuous throughout the enlarged domain, it. follows from Theorem 5 that u is harmonic there.

EXERCISES

31. Let z, (—

a

+ ib) denote any root of the equation J,(z)

1, where J.(z)

denotes Beseel's function of first kind and zero order; and let —

cos

ax

(45)

Prove that the mean value of u over any circle of radius one is equal to the value of u

at the centeir. Hiat: Make use of the integral representation J.(x)

1

— 2, jo

coe

(zeos e)de

(46)

is not harmonic, thus showing the "one-circle mean-value in contrast to Theorem 5, that a function property" is not necezzarily harmonic. The existsnce of infinitely many roots of the above equation follows easily from some of the earliest results in the theory of entire

(Note that for any root Se other than zero,

functiona.j

32. In contrast to the above exercise, prove that a function which is continuone in 1), where D is a bounded Dirichiet domain, and possesses the "one-circle mean-value must be htirinontc in D.

9. Convergence Theorem a A basic difileulty in. many parts of analysis is that important properties may not be preserved tinder limiting operations. For example, the limit of a sequence of contiuuowi functions need not be continuous; some addi-

tional hypothesis is required, the one most frequently used being, of course, that of uniform convergence. Going one step further, the convergence of the sequence may be uniform, each function of the sequence may be continuously differentiable, and yet the limit function may be nowhere differentiable. A remarkable contrast to thIs state of affairs is furnished by the theory of harmonic functions. It will be shown in this section that seemingly mild hypotheses turn out to assure that the limit of harmonic functions is itself harmonic, and that termwise differentiation of a sequence of harmonic functions is permissible.

Theorem 6. Ilarnack. If the functions

are harmonic in a

bounded domain G, continuous in O, and uniformly convergent on the boundary of G, then the sequence { w $ is uniformly convergent through-

out 0, and the limit function w is harmonic in (1. If 1. denotes any differentiation

then the sequence IWR converges through-

out G to 8w, uniformly in every compact subset.

Partial Differential Equations

150

For any t>

the hypothesis of uniform convergence assures the existence of N(€) such that Proof.

0,

—

wm(Q)(

<e

at all points Q of the boundary of G.

(n,m > N)

(47)

By the maximum principle (Sec. 5),

inequality (47) holds at all points Q of 0 as well; this proves the first assertion of the theorem. Now choose any disc in G, and represent each of the given sequence at any point Q of by the Poisson function formula the integral being extended over the circumference of £ Since the sequence is uniformly convergent and the Poisson kernel is bounded (for each fixed point Q) on the circumference of A, the passage to the limit may be carried out under the integral sign. The limit func-.

tion w is thus as a Poisson integral and hence, by Sec. 7, is harmonic in each point of G can be covered by such a disc, w is harmonic throughout G. Finally, by applying the operator under the

integral sign in the Poisson formula, it can easily be shown that the sequence ow4 converges, uniformly in any smaller disc to (ef. Exercise 33). The Heine-Borel theorem then guarantees that the convergence is uniform in every compact subset of 0. It is perhaps instructive to give an alternative proof that w is harmonic. The uniform convergence of the sequence in G, together with the mean-value property of the harmonic functions shows that the funetion w is continuous in G and possesses the mean-value property. By Theorem 5, w harmonic in G.

Theorem 7. Harnack's Inequalities.

Let 24 be

negative in a disc of ra.dllus R with center at Q. the following inequalities hold: u(Q)


u(Q)

harmonic and non-

For any point P of

(r

(JP)

(48)

Proof. First, suppose that u is continuous in A, so that u(P) can be expressed by the Poisson integral extended over the circumference of By integrating the obvious inequalities (which arc not correct if g(O) <01 (fl2 — r2)g(O)

<

R

<

(R2 —

r)

(49)

and then taking account of Theorem I and the Poisson formula (26b), one obtains (48). If no assumption is made concerning the behavior of u on the boundary of one carries out the above argument in a disc of radius fl' (where

r<


approach R.

151

6. Elementa of Potential Theory

Theorem 8. Itarnack's Theorem of Monotone Convergence.

are harmonic throughout a domain G, and if at each If the functions is monotone-nondeereasing, then the point P of G the sequence all points of G or converges at. sequence lw,, either diverges (to + points of G. In the latter case the convergence is uniform in compact subset of 0, so that (by Theorem 6) the limit function is harmonic in 0. does converge at some point Proof. Suppose that the sequence Let R denote the diaance from Q to the boundary the domain 0. Q of r < R. Given €> 0, we that QP = of G, and let P be any point can, on account of the convergence of the (numerical) sequence Wa(Q) I, choose N so large that for n > m > N the following inequality holds: }

0 w,,(Q) —

<s

(50)

(The left-hand inequality holds on account of the monotonicity of the By the right-hand inequality er (48), we obtain

sequence

0

W,,(P) --

[w,,(Q)

wm(Q)1 <e

(51)

which shows that the sequence fw4 } is convergent at P. Thus, the sequence converges in the largest disc with center at Q which lies entirely in 0. and clearly the convergence is uniform in any smaller disc. Now, suppose that there exists a point Q' of G where the sequence diverges. Then the sequence must also diverge at any point P' distance from Q' is less than R'/2, R' denoting the distance from Q' to the boundary. for the argument employed in the preceding paragraph shows that convergence at P' would imply convergence at Q'. (The distance from P' to the boundary will excecd R'/2, and hence Q' is closer than any boundary point to P'.) Thus, we have shown that the subsets of 0 on which the sequence converges and diverges, respectively, are open. As in the proof of the corollary of Theorem 2, we conclude that the sequence converges either nowhere or everywhere. Finally, the assertion

concerning uniform convergence follows from the Heine-Borci theorem. EXERCISES which was 33 Justify in detail the termwise differentiation of the sequence passed over in the proof of Theorem 6. 34. Obtain an analogue of the inequalities (48) in three dimensiona. 35. Prove that a function harmonic in the whole plane and bounded either above or below reduces to a constant. (This theorem is essentially the same as Liouvilie'a theorem in the theory of analytic

Partial Differential Equation.

152

10. Strengthened Form of the Maximum Principle The following theorem provides an interesting and important extension of Theorem 2 and its corollary. Theorem 9. Let u be harmonic in a domain G, and let assume a

local maximum at some point Q of C; i.e., for some 8 > 0, u(P) u(Q) whenever <ô. Then u is oonstant throughout C. Proof. Without lose of generality, it may be assumed that u(Q) 0. Let Ai be the disc of radius ô with center at Q. By the afore-mentioned corollary, u —0 in A,. Let R be the distance from Q to the boundary of G, and let A, be the disc of radius R with center at Q. Then the Fourier series representation (36) of u for the disc A, must be valid in the larger disc as well (cf. Exercise 24). However, all the coefficients obtained with the smaller disc obviously vanish; hence u es 0 in A2. Now let Q' be any point of C other than Q. Connect Q to Q' by a simple arc C. Since C is compact, its distance from the boundary of G is a positive quantity, which we denote by e. Now, we select a finite number

of points P,, P2,

on C such that each of the distances is less than s. Then, by the construction, u vanishes identically in some neighborhood of P1, and therefore, by the .

.

.

, P1

above argument, at least in the disc of radius e with center at P1. Since P, lies in this disc, we may carry out the same argument again. Thus, in a finite number of steps, we conelude that u(Q') = 0, completing the proof. COROLLARY. UNIQUE CONTINUATION PROPERTY OF HARMONIC FUNCTIONS. If u and v are harmonic in a domain C and if u v throughout a

subdomain C', then u v throughout G. Proof. Apply Theorem 9 to the function u — any point of the subdomain C'.

v,

taking as the point Q

U. Single and Double Layera Equation (16), which is of great importance in potential theory, admits a striking physical interpretation which we shall now discuss briefly. Letting r denote, as usual, the distance from a fixed point Q in the plane, the function u m log (1/r) represents the potential created by placing a mass' m at Q; i.e., the force exerted or a unit (positive) mass at any point P (distinct from Q) is given by the negative gradient2 of u. [In 'Frequently one speaks of (electric) charge rather than mass, sinee both positive and negative value, of m are to be allowed. Bowever, we prefer to speak of mase. 'The gradient I. understood to be the "field gradient"; i.e., the vector whose

ponent. are the derivative, with respect to the coordinates of P. The "sosrgo gradient," whose components are the derivatives with respect to the coordinates orQ, is evidently the negative of the field gradient.

153

6. Elements of Potential Theory

the three-dimensional case, which is of more immediate physical sig-

nificance, log (1/r) is replaced by l,/r.] If the mass, instead of being concentrated at a single point, 18 continuously distributed over a domain D with density f, the potential is given by

log!dEdn

u(z,y)

r

[CE

— x)2

+ (y —

(52)

Similarly, if the mass is distributed along a (sufficiently smooth) curve F with linear density (mass per unit length) o•, the potential is given by (53)

f o(8) log r

u(x,y)

The function u(x,y) defined by (53) is known as a "single-layer" potential We now proceed to define a "double-layer" potential. Let masses —m and +m be placed at points Q and Q', respectively. Now let Q' approach Q along a fixed direction n, and let m increase according to the rule (54) = constant = ilf m

The potential at a point P is given, in the limit, by

u(P) = where

IL(X,y)

logU/r)

= M

(55)

denotes differentiation with respect to Q in the direction n;

this is, of course, the component of the source gradient in the direction n. The configuration of masses just described is termed a "dipole" of strength

U. Now let dipoles be distributed continuously with density i along the curve F, the direction of the dipole at each point being normal to r. In this manner we obtain a double layer, which gives rise to a potential

u(P) =

a

=

log

(l/r)

dsQ

(56)

By taking account of (52), (53), and (56), we obtain for (16) the follow-

ing physical interpretation: Every function u defined and sufficiently differentiable in a domain having a sufficiently smooth boundary can be expressed as the sum of three potentials; one produced by a mass of density — (2v)'

density

distributed over D, one produced by a single layer of

distributed on the boundary of D, and one produced by

a double layer of density u/2r, also distributed on the boundary of I). If, in particular, u is harmonic in. D, it can be expressed as the sum of a single-layer and a double-layer potential.

Partial Differential Equations

154

if the function f is selected more or less arbitrarily, equation (52) defines, as will be shown in the following section, a function in the entire plane which is harmonic in the complement of D and which satisfies at any point of I) the Poisson equation Conversely,

•

Similarly, if o and are chosen arbitrarily,' then (53) and (56) are easily seen to define functions harmonic everywhere in the plane except on F; this is easily proved by differentiating each of these equations under the integral sign. Of particular importance is the question, to which we now turn, of the behavior of these potentials on and near r. A simple example, in which all the calculations can easily be carried

out in closed form, will help to suggest. the general situation. Let a single layer of constant density (which may be taken, for convenience, 0. equal to unity) be deposited on the line segment —1 x 1, y The potential arising from this mass distribution is given by

For any point to

(1 I

1

u(x,y) =

(58)

4)—i (x,y) not lying on the line segment this integral works out

+

2

—

y

(arctan

—

aretan

x— 1) (59)

(The principal branch of the inverse tangent is understood, and for points on the x axis, but not on the segment containing the mass, the last term is understood to vanish.) For any point lying on the segment the integral (58) becomes improper, but convergent, and works out to u(z,O) =

(x

— 1)

log (1

x) — (x + 1) log (x + 1) + 2 [u(±1,0) =

2—2

log2J

(60)

This result could have been obtained formally by setting y 0 in (59), hut doing so is tantamount to assuming that the potential remains continuous at the points occupied by the mass distribution. It must be emphasized that (60) can actually be obtained by first setting y 0 in

integrand appearing in (58), and then evaluating the resulting improper integral. In this manner, the potential u(x,y) is shown to exist and to be continuous everywhere in the plane, even at points the

occupied by mass. 1Subject, of course, to mild question are rnean!ngful. •

conditions

which suffice to assure that the integrals in

6. Element. of Potential Theory Next let us consider the gradient of the potential u. For all points not lying on the segment we obtain from (59)

_12 X

= 3'2 arctan

2

1

0

for y =

0,

1) ((Sib)

From (60) we obtain (— I

log

<x < 1)

(62)

which coincides with (61a) when y is replaced by zero. Therefore, 18 continuous over the entire plane, except at the end points of the segment on which the mass is deposited. The behavior of is quite different; if we choose any sequence of points converging to a point (x,O) of the segment (the end points excepted) from above, we obtain

=

+0; —i <x <1)

T

but, in contrast to this, if the sequence of points converges to (x,O) from below, we obtain

—0, —1
v(x,y)

—

+2

(Y

ti

a

i—i

,1,—o

log

+

E(E —

(y

',)9

(64)

The integral can, of course, be computed explicitly, but it is worth noting

that, since

—g'(y), (64) can be written in the aiterna-

'7

tive form (valid for points not on the segment)

v(x,y) =

log

—

z)' + y9 dE

(65)

Therefore, v(z,y) must coincide, except for sign, with the right side of (6Th). Thus, v(x,y)

= 34 arctan

—

aretan

(68)

Comparison with (63) immediately shows that v, like u,, approaches limits from above and below which differ by 2w. For any point on the segment, the integral (64) exists and has the value zero.

Partial Digerentlal Equations

The components of the gradient are given by V1

ii ii

= 21.(x + 1)2 +

=

x—1

y2

fl(x— I)' + y' —

(x

1)2

+

x+l

(x + 1)2

.1

(67a)

07b

Again it should be emphasized that these formulas have been denved fo points not lying on the segment. If we set y 0 in (67a), we obtain 0, in agreement with the previously mentioned fact that (formally) v1 v vanishes at all inner points of the segment. More important, however, is the fact that, as the point (x,y), y 0, approaches any inner point of the segment, v,, approaches a limit, namely, 1/(x2 — 1), independent of the manner of approach (from above, from below, or "mixed"). We sum up the results as follows: For the single layer, the potential

exists and is continuous everywhere, the tangential component of the gradient exists on the segment (except at the end points) and is continuous, but the normal component changes by 2w as the segment is crossed. For the double layer, both components of the gradient vary continuously the segment is crossed (at an inner point), but the potential chsnges by 2w, and on the segment it has a value equal to the mean of the limits from the two sides.

These relationships were found by explicit computation. It could again be shown by explicit computation that the numerical values 2r of the two kinds of discontinuities that were encountered are independent of the length of the segment, but it is more instructive to see this, without any computation, as follows. Take any segment Si, and select a portion s, of it. The potential arising from a layer, either single or double, on can be divided into the sum of the potentials arising from s, and — respectively. As the segment is crossed at any (inner) point of the potential arising from — behaves regularly, so that any departure from regularity of the entire potential is reflected in the same behavior in the potential arising from i.e., any "singular" behavior of the potential can be ascribed to that part of the potential arising from any segment, however small, containing the point where the original segment is crossed. Thus the value 2w which was obtained is independent of the length of the segment. Finally, we remark that if the density of the layer is an arbitrary constant u, the discontinuity obtained is simply 2wo. The argument used in the Last paragrapb is obviously restricted neither to straight-line segments nor to layers of uniform density. Any potential

arising from a single- or double-layer distribution on a curve must be harmonic, and hence differentiable infinitely often, at all points off the curve, and any discontinuities in the potential or its derivatives when the curve is crossed can be ascribed entirely to the portion ti'e

157

6. Elements of Potential Theory

curve restricted to an arbitrarily small neighborhood of the crossing point. Furthermore, it appears plausible that the behavior of the potential in the

isimediate vicinity of a point on the layer is essentially unaltered by

approximating the layer by a straight-line segment with uniform density. 'fhis last idea is rendered more precise in the following theorem, whose proof is left to the reader.

Theorem 10. Let a single or double layer of continuous density obe distributed along a smooth arc F Lie., one which can be parametrized in terms of are length a by functions x(s), y(s) having continuous first derivatives). Then, in the vicinity of any inner point P of F, the potential exhibits the following behavior: Single Layer. Potential and tangential components of gradient are continuous; normal component changes by 2rcr(P) across r. Double Layer. Potential changes by 2ro(P) across F; value of potential at P is equal to mean of limits from both sides; tangential and normal components of gradient are both continuous. EXERCISES

36. Prove that the potential produced by a double layer of unifonn density Q distributed on any (sufilciently smooth) simple cloeed curve r is equal, at any point P, to 0, or 2scr, aceording as P lies outside, on, or inside r. (The density is con-

sidered positive if the positive masses face inward.)

37. Prove, as a simple corollary of the above exeicise, that the potential arising from a double layer of uniform density . distributed along an open curse r is equal to o-8, where 8 is the angle subtended at the "field point." by r. (I)iscuss carefully the ambiguity in measuring 8, and also the value of the potential when the field point lies on F.)

12. Poisson's Equation In this section we shall study some of the properties of potentials of the form (52). It is understood that the domain D is bounded, and the is both continuous and bounded; these restrictions can be density somewhat relaxed without much difficulty, but we shall not concern ourselves with this matter. The integral (52) obviously converges for all

points (x,y) outside b, and u(x,y) may be differentiated any number of times under the integral sign; in thia way it is established, in particular, that is is harmonic. When the point P lies in I), the integral becomes improper, and a more detailed analysis is required. The absolute convergence of the integral follows from the inequalities

fff

log!

log!

dip M I

<

Partial Differential EquatE!n.

158

denotes a disc with

and

[here, and also later, M = sup

P which contains D.1 The continuity of the potential can also be easily proved directly, but will instead be established later as a consequence of the differentiability properties of u(x,y). Formally, the first derivative is obtained, as in the case that (x,y) is by differentiating the right side of (52) under the integral sign. outside In this manner, we are led to the formula —

11

(69)

The first step in justifying (69) is to establish the convergence of the integral; this follows easily from the inequalities x)

I fDf

y)2

(E

<M

(70)

Now we select an arbitrary point, which without loss of generality we take as the origin. From (52) we obtain, for any h 0, the difference quotient 1r'[u(h,0) —

u(O,O)J

—

ff

(E— h)' +

log

d21

(71)

The correctness of (69) is therefore equivalent to the limiting relationship '7

Jim

+

Is—OJJD

jdEdn =

0

(72)

The above integral is dominated by — h)2

+ h' log

M

+

+

1)isrcgardmg the factor M and introducing polar coordinates cos 8, = hr sin 8), we are led to the integral P2. fHflhj "— cos 8 + 1 fht

I

iog

7.1

(73)

dE dij

dr d8

(E

hr (74)

f

(I? radius of Since h is to approach zero, we may split the above expression into two parts:

IhljjO

(3

jO

,2u JR/Psi

+I"t jOI

(75) 3

The first term vanishes as h 0, sInce the integrand does not involve h. To estimate the second term, we apply the extended mean-vtilue theorem

159

6. flejnente of Potentiid Theory to the function log (1 — a), obtaining (for any real value of a < 1)

(70)

Setting

where y denotes some number between zero and unfty. a r2 (2r cos 0 — 1)

2 cos 0 + r log

we obtain

for r 3 (so that (76) is

and noting that IaI

r2 — 2r cos 0 + 1.

ra2

1

=

I

(1

—

+

ra2 (1

(77)

Note that we have employed the afore-mentioned inequality for Jaj only in the denominator; in the numerator we now employ the stronger inequal-

ity jaI 7/3r, and thus obtain

r2—2rcos0-j-1

2 cos 0 + r log —

C

—-

(78)

7

where C is a certain constant whose value, is of no importance. the second term of (75) is dominated by 2v

R/Ihj C

I1LI f f

—

dr do

log

.1?

Thus, (79)

approaches with h, we have established and the validity of formula (69).

Since the last

both the existence of The continuity of the first derivatives can also be easily established. Since, as explained above, the potential is harmonic outside D, it suffices at an arbitrary point P of b; without. loss of to cstablibh the generality, we may assume that P is located at the origin. Then, from the equation —

=

ff

/)2

±

— k)'2

dE d7

—

(80)

and the analogous equation for u1,(h,k) —

—

i[u,Q&,k)

—

we obtain

—

—

<2W Jf = MIwl

dE (1,1

—

(Here h + ik.) Now, let A be divided into four parts: E+ the discs and each of radius with centers at (0,0) and (4,k), respectively; region A. consisting of the disc minus the

Differential Equations

160

the remainder of L

aiore-mentioned discs; and, finally, readily obtain the inequalities1 f2. Pr

I,

iift

==

jj (ftjj,

ft < If2r I — Jo I

rdrd6=2T

< lw!2i2 —

—

I

I

Then we

'r(r —

r dr do

2w log

It now follows that the left side of (81) approaches zero with

and,

hence, that u2 and v1, are continuous. It follows, of course, that, as asserted earlier, the potential itself is continuous everywhere. It is of interest to note, however, that this fact follows more directly from (69), for from this equation we easily obtain the inequality (cf. Exercise 38)

M where d denotes the maximum distance between P and any point of D. Hence Ua (and similarly is hounded in any bounded subset of the plane, and from this it follows2 that u is continuous everywhere. We now consider what can be said about at a point P of D. First we remark that the value of and even its existence, is determined wholly by the behavior of the density in the immediate vicinity of To see this, let D be divided into a disc D1 containing P and D2, the remainder of D, and let u be considered as the sum u1 + u2 of the potentials arising from the mass distributed in the two parts of D. Thou, since P is an exterior point of D2, it follows that vanishes at P, and hence exists if and only if does, in which case their values are equal. Since the disc D1 may be chosen arbitrarily small, the statement made at the beginning of this paragraph is proved. Now, let us consider the particular case of a uniform mass distribution of unit density over a disc of radius R, whose center may be taken at the origin. In this case, (52) assumes the form u(x,y) = —

'ff denotes ff

fJ E'+,'
denote., as above, the radius of

—

x)2

+ (ti — y)2]

d,7

over the indicated region. The equahty of

and

(84)

tin (82c)l

follows by symmetry.

'By the theorem of moan value, u(x + h, y + k) — v(z,y) hu1(x + 81h, ii) + ku,(x -I-h, y + e,k), wbere 0< 8,<1. Hence +h, y +lc) — u(x,i,)I is at

most equal to (fhj + (k() times an upper bound on both d*termbz.d by the points and (z + h, v + k).

and

in the rectangle

161

Thewy

of

Although the integratirn can be carried out explicitly by tedious but

elementary methods, we shall evaluate (84) very easily by using some of the ideas that have been introduced in this chapter. Since u is harmonic outside the disc and obviously rotationally symmetric, it follows that u k, be of the form (7a), namely, a logr + ii. must, for r(= (x2 + determine the values of a and b, we note that To

of the disc and all points (x,y) outside, and hence, for all points taking logarithms and integrating over the disc, we obtain

—irR'log(r+R)
(86)

Dividing by log r (we make the additional restriction r> 1 if necessary) we find that the end terms both approach —TR', and letting r which must therefore be the value of a. To determine b, we subtract a log r (= —irR2 log r) from each term of (86) and obtain log

I +

R\
R\

(87)

—

conclude that b = 0. This proves, incidentally, the remarkable fact that the potential at any point outside the disc is the same as would be obtained by concentrating all the mass at the origin. Next we consider the potential produced at any point inside the inner a mass uniformly distributed with density one boundary of an over the annulus. As before, we conclude that the potential must be of the form a log r + b, but in this case a must vanish, since the potential must be finite at the origin. Therefore, the potential is equal, everywhere inside the inner circle, to its value at the center, in which case the potential is given by Letting r —+

we

=

—

f' r log r dr dO = —rIr log r —

where R1 and R2 denote the radii of the annulus. Returning now to the

potential at an inner point P of the disc, we consider the latter divided

into a smaller disc, containing P on its boundary, and an annulus. Applying the formulas developed above for the disc and for the annulus, we obtain u(P) ....v.2 Irig r — Tfr2 log r = — r' + R' — ?Rt log R (89)

We conclude that Au exists (as expected) at each interior point of the disc and has the value — 2r; if the density has the constant value po, the vaLue of is equal to 2TPO. Thus, the Poisson equation (67) has been

Differential Equation5

162 •

shown to hold at a point P if the density is constant in some neighborhood of P, however small. (Of course, the fact that u is harmonic outside 1) corresponds simply to the case = 0.)

is not constant in a Turning now to the general ease where is neighborhood of P, one might expect that the hypothesis that continuous would suffice to permit a proof that (57) holds. In fact, this is not true, as will be shown later by an example. A more stringent version of continuity, known as Holder continuity, does suffice, however. A function g defined in a neighborhood of P is said to be "HOlder-con-

tinuous" at P if there exist positive numbers c and a such that, for all — g(P)J Assuming that points Q in the neighborhood, the density is HOlder-continuous at a point P of I), which we again may place at the origin, we may, in view of the foregoing discussion, confine to the case that D is a disc with center at P and radius R andf vanishes at 1'; for 1ff does not vanish at P, we cousider u as the sum of two potentials, one produced by the constant density f(P) and the other by the density f — f(P). Thus, we must prove that and u,, both exist at P, and that their sum vanishes. Now, according to (69), — u2(O,O)J

=

h—1

ff f(E.'i) [E dE

+

lID

di,

(h

0)

(90)

lormaily, therefore, we are ted to the formula (91a) = First, we must show that the integral appearing in (91a) is meaningful. of our assumption that the density vanishes and is HOldercontinuous at (0,0), the integral is dominated by f2v c

I

•Jo

I

Jo

ra__.rdrdO

which converges, since a > 0.

Subtracting the right sides of (90) and (Ola), we find that the validity of (91a) is equivalent to the limiting

relationship

=

0

(92)

Taking account of the Holder continuity of f at the origin, we conclude that it will suffice to prove the limiting relation Jim h lID

dE

=

0

(93)

6. Elements of Potential Theory Dividing 1) into four parts and using simple upper bounds OA the intein each of these parts, as was done previously in establishing the

cvttinwty of u3 and us,,

readily show that (93) does indeed hold.

we

The details are left as Exercise 40. 'I'hus, (9 Ia) has been established; similarly, (9Th)

Jf

= vanishes at the origin, and Adding (9ta) and (91b), we conclude that this is equivalent to proving, as explained above, that the potential (52) satisfies (57) at any point where the density is Holder-continuous. We now prove, as indicated previously, that it would not be possible to is merely continuous. establish (57) under the hypothesis that Consider the density =

+ ',' < R', where k < 1.

is continuous Clearly 01, but it is everywhere in the disc [with the understanding that 1(0,0) not Holder-continuous at the origin. [This may be seen by setting = 0 and letting approach zero. For any positive constant a, we then have —.* cc.] The right side of (90) assumes in this log = case the form (after transforming to polar coordinates) (2w (Ii cos 28 r cos 28 — 4 C08 8 d d8 Jo Jo logr

in the disc

Now (cf. Exercise 41) J0

— hcos 0) r' — 24r cos 8 + h2 -

dO — —1

—irr3h—4

rr'

0 r < Ihl r > flu

Hence (95) is equal to IM —rh' Jo II log •—dr+T I r ff14

dr

r log The first integral is easily shown to approach zero with h, but the second integral is equal to [log log (1/R) — log log (l/jh))J and, hence, becomes infinite. Thus, u,, does not exist at the origin in this example. We sum up the results of this section up to this point as follows.

Theorem 11. The potential (52) is defined and continuous everywhere and possesses continuous first partial derivatives, given by (69) and its analogue for tie,. The potential is harmonic at all points outside and satisfies the Poisson equation (57) at every point of D wheie the density is Holder-continuous.

PartIal Differential Equatlona

164

We now consider a kind of converse to the problem discussed above. Given a funetionf(x,y) defined in a bounded domain D, we seek a function v(x,y) satisfying equation (57) and vanishing everywhere on the boundary. The question of uniqueness is immediately settled by the observation that the difference of two solutions would be harmonic in D and identically zero on the boundary, hence zero everywhere in 0. The question of existence is more delicate. If we assume that f is bounded in I), then (52) furnishes a solution u of (57) and which is continuous in I); if we assume further that D is a Dirichiet domain, we may determine a function w harmonic in D and sgreeing with u on the boundary and then obtain the unique solution to the problem in the form v = u — w. Suppose, however, that f fails to be Holder-continuous at a single point P of D; as has been shown by an example in this section, the function v(x,y) may fail to be sufficiently differentiable to satisfy (57) at P. The question arises whether a function satisfying (57) everywhere in D, including P, might nevertheless exist. That the answer is negative is easily shown as follows: The difference a — u WOuld be continuous throughout D and harmonic everywhere in 1) except at P. However, this is an impossibility, according to Exercise 21. We therefore specifically assume that j is bounded and HOlder-continuous in D, and that D is a Dirichiet domain. Then the problem under consideration is solvable. Now, instead of setting up the integral (52) and then solving a suitable

Dirichiet problem, one can, at least formally, solve the problem under consideration by setting up the integral2 v(P)

= lID f(Q)G(P,Q)

(97)

dE

Since G(P,Q) is equal to log i/Pa plus a function H(P,Q) harmonic in P,

we might argue (by differentiation under the integral sign) that the integral

defines a harmonic function, and hence that J(Q) log

i/Pu

=

—2ivf

Secondly, we might argue that, as P approaches the boundary, v(P) —* ffD 1(Q) lim G(P,Q) dij

0

'The index a would not have to be constant. For convenience, we denote the points with rectangular coordinates z, y and i,by P and Q respectively.

$

of Potential Theory

6.

165

However, we must justify carrying out the above operations under the integral sign. Rather than undertake this task, we shall justify the formula (97) only in the specific case that D is a disc—without loss of generality, the unit disc z2 + y2 <1. In this case, (97) assumes the form v(P)

=

ff

f(Q) log

(98)

(Here we have employed the first form of the Green's function given in denotes, of course, the (24) and exploited the symmetry in P and

image of P in the unit circle.) Since the Green's function is never negative (ef. Exercise 16), we obtain from (95) the inequality

(fJ log OP di + fj log —

(99)

The first integral is simply T log OP; the second integral represents the

potential of a uniform mass distribution at P; while the last integral Since the represents the potential of the same distribution at potential is continuous over the entire plane, it follows that the last two approaches unity. integrals cancel each other out as OP (and hence Finally, log OP approaches zero, and hence approaches zero, as P approaches the circumference. It remaini to show that = throughout. Breaking up the integral (98) into three parts, as in (99), we observe that it is necessary to prove merely that f(Q) log OP dE

+

1(Q) log

is harmonic. The first integral is simply

di,, and hence is harmonic.1 The second integral represents a potential outside the unit disc, and hence is a harmonic log OP ff,, f(Q)

function of the point

As shown in Sec. 2, the property of barmonicity

is preserved under inversion, so that the second integral defines a harmonic function of P. Thus, we have shown that formula (97) furnishes

the solution to the problem under consideration, at least for a disc. Actually, a more delicate analysis of the behavior of the Green's function near the boundary shows that (97) may be employed for domains whose boundary satisfies very mild hypotheses. 1A separate consideration is neoessary when P

but this is a minor matter.

0, for then both integrals become

PaxUaI

166

Differential Equations

EXERCISES 38. Prove (83). 39. Prove that u, and are bounded in the entire plane, not merely in any bounded set. Hint: Obtain an inequality similar to (83) valid for point. outside 1). 40. Prove (93) in the manner indicated in the text. 41. Prove (96). Hint: Exploit the resemblance to the integral (27). 42. In contrast to the procedure used in the text, derive the potential of a di8c of uniform density by exploiting the Poisson equation. 43. Dekrniine the potential of a (three-dimensional) sphere of uniform density in two ways, analogous to those used in the text and in Exercise 42, respectively.

7. THE DIRICHLET PROBLEM

In this chapter we present a fairly extensive treatment of the Dirichiet problem, which was formulated in Sec. 6-6. In addition to the great significance of this problem, whether considered from either the mathematical or the physical viewpoint, the study of a large number of different approaches to this problem is justified by the fact that it may serve to

acquaint the reader with some of the most important techniques of analysis, which can be used in a wide variety of problems. Most of the methods to be discussed in this chapter are applicable in any number of dimensions. Although we shall formulate all the methods only in two dimensions, the reader should consider in each case how the generalization to higher dimensions, if possible, is carried out.

1. Subharmonic Functions Among the methods for studying the Dirichlet problom which will be presented in this chapter, there are two which use subharmonic functions. These methods will be presented in the next two sections, and the present

section is devoted to a brief development of those properties of subharmonic functions which will be needed for the presentation of the afore-mentioned two methods. A definition of subharmonicity will be used which is somewhat simpler than is needed in more comprehensive treatments of this topic, but it will be adequate for the present purposes. DEFiwrrIoN I. A function u is said to be "subharxnonic" in a D if it is continuous in D and if, for each point P of D, there exists a

positive number p(P) such that for all r, 0
is satisfied.

I2r

u(P + re') dO —2irJo 167

(1)

PartIal Differential Equations

168

The reason for the use of the term subharmonic will be readily apparent from Exercise 7, or from a comparison of (1) with Theorem 6-1. Clearly any function harmonic in a domain is subharmonic there, and a function subharmonic in a domain is also subharmonic in any subdomain. The properties of subharmonic functions that will be needed later are developed in the following sequence of theorems.

Theorem 1. If Ui, and Ci,

C2, . .

.

+

+

.

.

.

are each subharmonic in a domain D,

are nonnegative constants, then the functions + cu,1 and u = max (ut,u2, . ,u,.) are also sub-

,

.

harmonic in D. Proof. The first part is obvious.

sider the case n =

2.

.

As for the second part, we first con-

The continuity of u follows from the explicit

representation

u

max (u1,u,)

+ U2

+

(2)

fu1 —

At any point P of D we have either u(P) = ui(P) or u(P) both).

In the former case we have, for sufficiently small r,

u(P) —

f2' u1(P + re') do

I f2f u(P

u:(P) (or

+ re") dO

(3)

and similarly if u(P) u2(P). For n > 2, the desired result is obtained by an induction in which the fact that max (Ui,U2, . . . max (u%, max (uj,u:, . . .

isuseci

Theorem 2. First Form of Maximum Principle. If u is sub-

harmonic in D and assumes a local maximum at some point P of D, then u is constant in some neighborhood of P. (Cf. Exercise 6.) Proof. Since u assumes a local maximum at P, there exists > 0 such

u(Q) u(P) Then for all positive r
(4)

we obtain, by combining (1) and (5)

The equalities must therefore bold in (5), and then, as in the proof of Theorem 6-2, we conclude that u(Q) must equal u(P) for sufficiently small; this completes the proof. Theorem 3. Second Forni of Maximum Principle. If u is subharmonic in a bounded domain 1) and continuous in 1), then u attains its maximum on the boundary; if u also attains its maximum at a point of D, then u is constant throughout 2).

169

1. Tb. DMcblet Piiulikm

It clearly suffices to prove the second assertion. Let A be the set of points of D at which u assumes its maximum value. Since any point of A is, a fortiori, a local maximum (but not conversely, as shown 6), it follows from Theorem 2 that the set A is open. The set by I) — A, on which u is lees than its maximum value, is also open (by continuity). As in the proof of the corollary of Theorem 6-2, we conclude that either A or D — A is void. This statement is equivalent to the second assertion. J'røof.

Tbeorem4. LetubesubharmothcinadomainD,andletAbeany

disc auch that A C D. Define the function v in I) as follows: v conv harmonic in & (By Theorem 6-4, v is tinuous in D, v m u in I) — uniquely determined by these conditions) Then v is also subharmonic

mD, andu vthroughoutD. Proof. For ali points of D

A the subharmonicity of w follows

ForanypointPofLithe

inequality (1), with u replaced by v, becomes an equality for all sufficiently small values of r (by the mean-value theorem). It remains only to consider the circumference r of & Now, u — v u + (—v) is subby Theorem 1, and vanishes identically on F. By harmonic in at all points of A, and hence at all pointeof D. Theorem 3, u—v Therefore, for any point P of F and any sufficiently $m,Lll r, t'(P)

ti(P)

f2"

u(P + red) de

v(P + re'0) dO

(6)

Thia completes the proof.

The procedure described in Theorem 4 of altering a subharnionio function u by redefining it inside a disc A as the solution of the Diriehiet problem with the boundary values derived from u (i.e., by replacing u in the disc by the Poisson integral formed with the values of u on the circumference) will be used systematically in the following two sections.

We shall refer to this procedure as "harmonization," and denote the Thus, if u is subharmonic in D, so is u4,

function thus obtained as UA. and U UA throughout D.

EXERCISES 1. Prove

atafunonuiaharmonicinadomainDifsndoplyifbothuand

are aubbarmonier in D.

2. Prove, by explicitly evaluating the right gid. of (1) for all P and all positive — x' + p'Issubbarmonicintheeutlreplane. (Cf. E*ercise 11.) 3. Prove that the equate ci a nonnegative subbarmonic function is also subharmosiki. Hj,sL Apply the Sebwars inequality to (1). 4 by $ simple example, that the assertion of Exercise 3 may be false if

the word "nonnegative"

omitted.

Partial Differential Equationa

170

5. Prove that the square of a harmonic function is subbarmomc. (Note that, in contrast tOL Exercise 3, it is not necessary to assume in this ease that the given function is nonnegative.) 6. Show, by a simple example, that (in contrast to harmonic, functions) a subharmonic function can attain a local maximum (and hence, by Theorem 2, be locally constant) without being constant throughout the domain. (Note the contrast with Theorem 3, where the function is assumed to attain not merely a local maximum, but the maximum for the entire domain, at a point of the domain.) 7. Let u and v be aubharuionic and harmonic, respectively, in a bounded domain and suppose that ii v on the boundary. Prove that u v D and continuous in

throughout D, and that the equality holds either everywhere or nowhere in D.

8. A seemingly more restrictive definition is obtained by modifying Definition 1 as follows:" . . for each point P of D and each positive r less than the distance of P from the boundary of D, the inequality . . . ." Show that in reality the two definitions are equivalent. 9. Prove, by a simple example, that if the continuity requirement is dropped, Definition I is no longer equivalent to the modified definition given iii the preceding .

exorcise.

10. Formulate a definition of subharmonicity based on Exercise 7, and prove its equivalence with Definition 1.

11. Prove that a function of class C' (i.e., possessing continuous second partial derivatives) in a given domain L) is subharmonic i and only if the inequality

us. + ur, 0 holds throughout I), by applying Green's second identity over an annulus to the functions tL and log r, where r denotes the distance from the center of the annulus. 12. Prove the sssertion of Exercise 11 by using Taylor's formula with remainder. Hint: Consider the second proof of Theorem 6-6. 13. Prove the assertion of Exercise 3 for functions of class C' by applying condition (7). 14. Formulate an appropriate definition of subharmonieity for a function of one variable and the analogues of the theorems of this section. 15. Extension of Theorem 3: Let v be subharmonic in a bounded domain D, and let a function g be defined on the boundary as follows: g(Q)— limsupv(P)

(The values ±

must be allowed.)

(PED)

Show that, for all P

1),

v(P) sup p(Q) [It may be assumed that sup 9(Q) is finite, for the case sup g(Q) — + is trivial and the case sup 9(Q) — — cannot occur. The latter assertion should, of course, be proved.1

As In Theorem 3, show that equality holds in (9) either everywhere or

nowhere in D.

2. The Method of Dalayage As explained in Sec. 6-11, a (sufficiently differentiable) function can be

represented in a domain (with sufficiently smooth boundary) as the potential resulting from a suitably chosen distribution of mass over the

7. The Dirichiet Problem

domain and its boundary; in particular, if the function is harmonic the mass is distributed entirely on the boundary. In the method of balayage (sweeping), due to Poincaré, one begins with. a function having the prescribed boundary values, but not harmonic, so that its associated mass

distribution is deposited, at least in part, within the domain. This function undergoes a succession of modifications such that at each stage the modified function is harmonic in some disc; the mass is thus removed from this disc. In subsequent modifications of the original function, however, new mass distributions are deposited in the disc. In the limit a harmonic function is obtained, so that the entire process may be looked upon as a scheme for sweeping the original mass distribution onto the boundary. Let a bounded domain 11) be given, and let a continuous function f be defined on the boundary. We assume that there exists a function continuous in 1), subharrnonic in D, and coinciding with I on the boundary. We shall presently show that the process of balayage leads to a harmonic function which, under certain conditions, provides the solution to the Dirichiet problem. While the assumption made in the preceding paragraph, namely, thatf can be extended into D as a subbarmonic function, appears extremely restrictive, it is noteworthy that if the Dirichiet problem is solvable for every such boundary function, it is solvable for every continuous boundary function. This is easily seen as follows. First, according to Exercise 20, every Dirichiet. problem whose boundary values can be extended into D as a polynomial will be solvable. Then, given any continuous function f defined on the boundary, we can, to the Weierstrass approxi-

mation theorem, select a sequence of polynomials

which

{

For each of these polynomials, we determine the solution of the corresponding 1Mrichlet problem. According to Theorem these solutions will converge to a function which solves the given Dirichiet problem. We select a denumerahie sequence of discs . whose union is precisely the domain D. This may be accomplished, for example, by enumerating the "rational" points of D (those whose coordinates x, y are converge uniformly to f.

.

..

both rational) and constructing about each such point the disc with center at this point and radius equal to half the distance from the point to the boundary. A second enumeration of these discs is also introduced, in which each disc appears infinitely often. This uew enumeration can be described as follows:

=

A4 = A = = (10) = . . . It is seen that the first two discs are enumerated, then the first three, the first four, and so on. A1

A2

A3

Partial Dilerentlal Equations A sequence of functions (the "modifications" referred to above) is now defined as follows:

(ti> 1)

f

(11)

By induction it is readily seen that the functions f, are well defined,

aubharnionic in D (by Theorem 4), and continuous in .1), and that they a11 coincide on the boundary with the original function f. Furthermore (again by Theorem 4), these functions constitute a nondecreasing sequence, and (by Theorem 3) each f, is bounded (above) everywhere in 1) by the maximum of f. Therefore, the sequence tf..) converges throughout D to a function u whose value nowhere exceeds the niarimum

off.

Next it will be shown that the limit function u is harmonic throughout of the set that was used in the balayage, and 1). We choose any disc consider the subsequence of f, consisting of those functions which were = we choose obtained by "harmonizing" with respect to (e.g:, if the /8, fit, * .). Since the functions of this subsequence Harnack's constitute a monotone sequence of functions harmonic in since The function n is therefore harmonic in theorem (6-8) applies. •

this argument can be applied to any one of the discs, and since these discs exhaust D, it follows that u is harmonic throughout D. (We are, of course, exploiting the fact that every subsequence of a convergent sequence is itself convergent to the same limit.) Although it is not essential for the rest of the discussion, it is instructive to settle a question that may be raised in connection with the process of balayage which has been described. The function u which has been constructed seems to depend not only on the function f originally defined on the boundary, but also on the particular (subharmonic) extension of /

that is selected and on the choice of the set of discs A1, At,.... It is noteworthy that the dependence of u on these two choices is only apparent; a different extension off, or a different choice of the discs, or both, has no effect on ii. We can prove this statement as follows. Let two subbarmonic extensions of f, say f(1) and and two sets of discs, . . . and . . . , be given, and let balayage be performed twice, first with the extensionf") and the and then with the extension f(t) and the discs We thus obtain harmonic functions and respectively. For ti = 1, 2, 3, . . 1, 2 we have, for any boundary point Q, urn

Since f,C.)(P)

f(}(P)

J(Q)

we also have

lirninf u")(P) 1(Q)

7. The Dirleblet Problem Setting v

I in (13) and r

2 in (12) and subtracting, we obtain — uC!)(P)]

urn áup p-eQ

0

Applying the result of Exercise 15 (with v replaced by

obtain

(14) —

we

0

(15)

u(2)(P) 0

(16)

0

(17)

f,,<1)(P) —

Letting n become infinite, we obtain u(1)(P)

By 8ymlnetry, we can also obtain —

From (16) and (17) we conclude that

Thus, the assertion that u depends only on the boundary function f has been proved.

It now remains to decide whether the function u assumes the prescribed values on the boundary, or, more precisely, whether

lim u(P) = 1(Q)

p-eQ

In order to settle this question, it is necessary to introduce a new and Important concept, that of a barrier. DEFINFrI0N 2.

Let D be a bounded domain and Q a boundary point of

D. A function w defined in I) is said to be a "barrier for D at Q" if it satisfies the following conditions: w continuous in 1) cu harmonic in D

(20a) (20b)

w(Q)=O w(P)>Q PEI)—Q

(20c) (20d)

Before explaining the role played by the barrier in the study of the Dirich let problem, we give one simple but highly important example of a barrier. Suppose that there exists a disc A such that A and 1) are disjoint, but A and have a single point Q in common. Then the function w(P) = log

where 0 denotes the center of

is

Op

readily seen to be a barrier at Q.

Boundary points which admit a barrier of this type were termed "regular"

Partial Differential Equations

174

by Poincaré. (However, we shall assign a different meaning to this term in Definition 3.) More delicate conditions guaranteeing the existence of a

barrier are known, but need not concern us here. On the other hand, there do exist domains some of whose boundary points do not admit a barrier. barrier.

For example, an isolated boundary point cannot admit a (Cf. Exercises 17 and 18.)

A boundary point Q of a domain D is said to be regular if it admits a barrier. DEFINITION 3.

We are now in position to state the following important result.

Theorem 5. The function u defined by balayage satisfies condition (19) at every regular boundary point of D. Proof. We shall prove the inequalities urn in! u(P) f(Q) P-.Q

lirn sup u(P) 1(Q) P—Q

(22)

.which, taken together, are equivalent to (19). To prove the first inequality, we recall that

u(P) f(P)t for all points P of D, and hence

urn in! u(P) liin unf 1(P)

(24)

However, since I is continuous in 1), the right side of (24) is equal to 1(Q), and so the first inequality of (22) is established. (Note that this half of the proof does not depend on the existence of a barrier.)

NowletwbeabarrieratQ. Givene>O,chooeeadiscAwithcenter at Q such that for all boundary points Q' of .D lying — f(Q)I within A; also, we require that A should not include all of I). Let miii w(P); by the definition of a barrier, wo must be positive.

let M

max

Also,

Now we define the harmonic function W as follows: W(P) = 1(Q)

++

—

f(Q)1

(25)

Everywhere in 1), W 1(Q) + and, in particular, for all boundary points Q' lying in A it follows, from the definition of A, that W(Q')> f(Q'). On the other hand, by the definition of we see that for all f We denote the (aubbarznonic) extension of the boundarl function f by the same letter.

7. The Dh,Ieblet

boundary points Q" not lying in W(Q") 1(Q)

+ + (M — f(Q)1 = M + > f(Q")

Thus f — W <0 everywhere on the boundary, and since the function is

subharmonie in D it must, by Theorem 3, be negative throughout D. By Theorem 4, this inequality is preserved under any sequence of har.. monizations of f. Therefore, for any point P of D and for a = 1, 2, 3,

. .

.

,

must hold.

Letting P

inequality

<W(P)

flence,

v(P) = urn Q

W(P)

(27)

in (27) and taking account of (25). we obtain tim snp u(P)

1(Q) + e. Since is arbitrary, we conclude th2t the SeCOILd inequalitY of (22) holds, and the proof is complete. The following theorem is an easy corollary of Theorem 5. ft states, without making reference to balayage, the results that have been established by the use of that method.

Theorem 6. A neeessary and sufficient condition that a domain be a Dirichiet domain is that its boundary consist entirely of regular points. Proof.

The sufficiency follows from Theorem 5 together with the

approximation argument presented earlier in the present section. As for the necessity, let be any boundary point of a Dirichiet domain. Then,

in particular, the Diriehiet problem with boundary values f(Q) = possesses a solution which clearly constitutes a barrier at i.e., Qo is a regular boundary point. EXERCISES 16. Explain why it Would not suffice, in the selection of the sequence of discs A,, . . . , to take the "rational" points (or some other denuinerable dense set) and construct about each of thcse points aotnc disc.

17. Let I) be the punctured unit disc, 0 <x' + " <1, and let / be de6ned as followr /

I on the circumference, / — 0 at the origin. Determine the function

which bslayage assigns to this Diriehiet problem. (Cf. Sec. 6-8.)

II. Prove the assertion made in the text that an isolated boundary point of any domain cannot be regular. Hin*: Consider a 1)iricblet problem similar to that posed in the preceding exercise. 19. Prove that a bounded domain cannot possess a boundary consisting entirely or irregular points; in fact, the boundary must contain a nondenumerable set of regular points 20. Prove that every polynomial nan be expressed In a given hounded domain as the difference of two subharmonic polynomials. Hint: Add and subtract a suitable multiple of x1, and use the result of Exercise 11.

Partial Differential

3. The PerronuRemak Method The treatment of the Dirichiet problem to be presented in this section is

closely related to the method which was presented in the preceding

Both methods use subharmonic functions and assign to every Dirichiet problem a harmonic function which coincides with the solution whenever the latter exists. It is of interest to note that both methods assign the same harmonic function to any Dirichiet problem; this fact will be proved at the end of this section. The outstanding differences section.

between the two methods are that balayage is constructive while the method of the present section is not, and that the latter method, unlike the former, does not require any extension of the boundary function to the domain itself. The latter method is thus independent of the Lebesgue extension theorem and the Weierstrasa approximation theorem, and may

therefore be applied without additional complications to a domain situated on a Riemann surface. It may also be added that the latter method can be readily adapted to the treatment of Dirichlet problems with discontinuous boundary values. However, we shall not pursue these two remarks further. Given a continuous function f defined on the boundary of a bounded domain D, we define to be the family of all functions w continuous in 1), subharmonic in D, and satisfying at all boundary points the inequality

wf The class S, is nonvoid, for it certainly contains the constant function w minf. For any point P of D and any function w of the family SI, the inequality

w(P) max f

(29)

holds, by virtue of Theorem 3 and the inequality (28).

Therefore, the

following definition is meaningful:

u(P) =

sup w(P)

wEE,

(P E D)

(30)

and the function u also satisfies the inequality (29). It is by no means obvious that u is harmonic, or even continuous, in D, hut this will now be

shown.

Let be any disc such that A C D, and let P be any point of £ From the definition (30) of u it follows that there exists a sequence of functions Ui, u,, . . . of the family such that u,(P) —+ u(P)

177

7. The Dircblet Problem

From Theorem I, it follows that the functions max

.

(32)

.

.

are also members of S,, and, by their very definition, the U0 satisfy the inequalities

(33)

throughout D. Therefore the sequence (U. converges monotonely replaced by u0(P), (31) holds with throughout D, and, since It follows from Theorem 4 and the minimum principle for har. . . obtained monic functions that the sequence of functions by harmonizing the U,. with respect to A also constitutes a monotone sequence of members of 8,. By a repetition of the previous argument, it follows that U,,4(P) —* u(P)

By Harnack's theorem of monotone convergence, the limit of the sequence which we denote as U4, is harmonic throughout A. If we can u throughout A, the harmonicity of u will be established show that is a in A, and hence at all pointh of D. Since each of the functions member of it follows that u everywhere in D (even though u might not itself be a member of 8,). Thus it will suffice to prove that

u

throughout A

(35)

Suppose, on the contrary, that at some point Q of A the following inequality holds: < u(Q) Now a sequence of members of S, is selected which plays the same role at Q as the sequence did at F; i.e., u(Q)

However,

instead of imitating the definition of the functions U0, we

employ a different definition, namely, V0 = max

(Ul,V1,U2,V2,

.

.

.

It is evident from this definition that the sequence

f

V0

converges

monotonely, and that both of the following relations V0(P)

u(P)

V0(Q) —+ u(Q)

Next, we replace the sequence (V. by f

As before, we see that the functions V0A converge monotonely to a function which is bar-

Partial Differential Equations

178

inonic throughout £

Furthermore, the following three relations hold:

Uk

throughout I) = U4(P) = u(P) = u(Q) > From (39a) and (396) we conclude that the function and nonnegative in A, but vanishes at the point Q of A. principle for harmonic functions, we conclude that

(39b) (39c) UA is harmonic By the minimum

—

throughout A

0

U6

—

(39a)

(40)

We therefore have shown that

which contradicts (39c).

U6

throughout

u

and hence, as was to be proved, u is harmonic throughout D.

It remains to analyze the behavior of u near the boundary. The analysis is similar to that given in the preceding section. as might be expected, and the result is the same, namely:

Theorem 7. If the boundary point Q is regular, then urn u(P)

(42)

f(Q)

Proof. The proof of the second part of (22) applies without change up to and including the words: "Thus .f — W <0 everywhere on the

boundary." The rest of the paragraph does not apply, because f has not extended to .D, but we can now argue as follows. For any function w of the family S,, the function w — W is continuous in 1), subharmonic in D, and negative on the boundary. By Theorem 3, w — W is negative throughout D, and hence, by the definition of u, we have, for all P E I), the inequality

u(P) W(P)

(43)

Therefore,

urn sup u(P) lim sup W(P) = lim W(P) = P-iQ On account of the arbitrariness of

we

1(Q)

+e

obtain

lirn sup u(P) f(Q) P-'Q

Turning to the first part of (22), we consider the harmonic function V(P) =

—

—

EM

+ 1(Q))

(44)

7. The Dfriclilet Problem

Proceeding as in the previous section, we find that V
PED,

V(P) n(P)

(47)

From this we immediately obtain lim V(P) =f(Q) — urn infu(P) liming V(P) = P-Q P-SQ

Again exploiting the arbitrariness of e, we obtain

urn mi v(P) 1(Q) P—sQ

Combining (45) and (49) we obtain (42), and the proof is complete. We now prove that the Perron-Remak method leads to the same function as that obtained by balayage. (Of course, this statement is trivially true if D is a Dirichiet domain, for in this case both methods furnish the unique solution of the Dirichiet problem.) Let f be defined and continuous on the boundary, and let it, as in the previous section, possess a denote subharmonic extension to D, which we also denote by f. Let the function obtained by balayage and Uz the function obtained by the Perron-Remak method. Then, since the function I (defined in b) belongs to Sj, it follows that all the functions of the sequence f,, also belong to 8,. Thus, f, zsa throughout D, and hence u1 u2. Now at some point P of D. According to the suppose that u1(P) definition of u2, there would exist a function w belonging to 8, such that <w(P). By Theorem 1, the function max (f,w) represents a subharmonic extension of the function f originally defined on the boundary. Then the function u, which balayage would furnish with this new extension off would exceed Ui at P. However, it was shown in Sec. 2 that the

function furnished by balaya.ge is independent of the particular subharmonic extension of f which is employed. Hence, we conclude that Ui

u2, as asserted. EXERCISE

21. Solve Exercise 17 by studying the behavior of 1, 2, 3 (x' ÷ yS)ll*,

sequence of functions

4. The Method of Integral Equations In this section we employ the results of Sec. 6-11 to show that, for a domain with sufficiently smooth boundary, the Dirichiet problem can be formulated as an integral equation for an unknown function defined on the boundary. Let the domain D be the interior of a curve r possessing

Partial Differential Equations

480

continuous curvature; i.e., r can be parametrized in terms of arc length by functions x(a), y(s) possessing continuous second derivatives. Let the continuous function f be defined on r. We make the assumption, to be justified below, that the Dirichiet problem with boundary values f possesses a solution u which is expressible as the potential of a double layer with continuous density i distributed on F. To justify the above assumption we shall prove that: (1) the function i must satisfy a certain integral equation; (2) this integral equation possesses a unique continuous solution; (3) the potential produced by the double layer whose density satisfies the afore-meutioned integral equation

does, in fact, constitute a solution to the given Dirichiet problem.

Starting with the representation (6-56) of a double-layer potential, namely,

j(Q) ôlog(lJr) da

u(P) =

(r

P on F. According to Theorem 6-10, —

rr(P) = urn ti(P') P'-4P

(P' E 1))

From (50), (51), and the fact that the right side of (51) is supposed to equalf(P), we find that the density of the double layer must satisfy the integral equation

?(P) =

—1(P)

where K(P,Q)

+ f r(Q)K(P,Q) d8Q

This completes the proof of part 1.

Next, we note that K(P,Q) is jointly continuous in P and Q; this follows from a straightforward computation, which is left as Exercise 22. Therefore, in order to establish the unique solvability of (52) for every continuous f, it suffices, according to the Fredhoim alternative, to show that the corresponding homogeneous equation

,(P)

dsQ

Jr

(53)

admits no nontrivial continuous solution. be any continuous solution of (53), and let v be defined as the potential arising from the double layer of density r: a log (1/r) v(P)

j

(14

(54)

= If the point F' approaches any boundary point P from within D, the relation

v(P') —, v(P)

—

(55)

7. The Dfrichlet Problem must hold (by Theorem 6-10), while by (53) the right side of (55) must vanish. Hence, the function v is harmonic in 1) and approaches zero at the boundary, so that, by the maximum and minimum principles, v 0 in 1). The interior normal derivative of v therefore vanishes everywhere

on F, and by appealing once again to Theorem 6-10 we see that the exterior normal derivative also vanishes everywhere on F.

Now let be its boundary. By (6-12),

disc containing D, and let

ff

D

dz dy

(Va2 +

—

f

t'

da

be a

+jv

The first integral on the right vanishes on account of the vanishing of on F. and

Now, as the radius R

of

F1 increases, the center remaining fixed, v

its gradient approach zero at least as rapidly as c/R

and

c/R',

where c is a sufficiently large constant. This may be seen as follows. Introducing rectangular coordinates x, y and ,j for P and Q, respectively, we write (54) in the form respectively,

v(x,y) = where r' = (x — Ri, — y)

+

r

jr

By the Schwarz inequality,

(y

di,J2 — x) + and we now conclude from (54') that —

— x)

— y)']

+ di,')

lv(x,y)t This, in turn, yields the further inequality Iv(x,y) I

<

(mm

fr

I da

(59)

which implies the assertion following (56) concerning the behavior of

v.

By differentiating the right side of (54) under the integral sign with respect to x (or y), we obtain, very much as in the above argument, an upper bound on (or Iv,I). Jn thia manner it is shown that the gradient of v has the behavior asserted after (56); the details of the computation

are left as Exercise 24. It now follows that

approaches

v

zero at

f

least as rapidly as c2/F?3, and hence that the right side of (56) approaches

zero as R becomes infinite. Thus, we have shown that + dy vanishes when the integration is extended over the entire exterior of F. Since v2 + is nonnegative and continuous, and must vamsh identically outside r. This shows that v must be constant outside r, and from its behavior "at infinity" it follows, that v vanishes dx

Partial Differential Equations

182

identically outside r. Thus, v is continuous across 1', and by Theorem 6-10 we conclude that 0, as was to be shown. Thus part 2 is proved. Turning now to part 3, we note that the function u defined by (50) is harmonic throughout 7) and that, by Theorem 6-10, it satisfies the limiting condition

lim u(P')

(P' E D, P E r)

u(P) —

By comparing (52) and (60), we conclude that urn u(P') = f(P) The proof is thus completed. We illustrate the method of integral equations by showing that when applied to the unit disc, it leads to the Poisson formula. Letting 8 denote the polar angles of P and Q, respectively, we obtain

K(P,Q) =

1

0 log (ljr)

ti

]

1 — 2p cos (8 —

+ p2

(62)

This works out to K(P,Q)

—

Substituting into (52), we obtain for the density the integral equation f(9\ P2w 1 r(O) = — (64) — — I V 2VJ0 Hence, i(8) c — f(6)/T, and the constant c is determined by substitu-

tion into (64):

[2T[f(#)l

(65)

I From (65) we obtain 1

(2w

(68)

and thus —

r(o) =

f(•)

(67)

J°

Substituting from (67) into (50) and performing a few elementary inanipulations, we obtain for any point P of the disc the formula1 1 — r cos (8 — — u(P) = TJo — 2r cos (8— + r2 —

1

Jo 1 In (68) the

1

2r CoS (a — 8)

+ r'

d

r denotes the radius vector of F, not the length PQ

'. The Dirichiet Problem By elementary computation it can be shown that [2r 1 —rcos (a — 8) Jo

1

2r cos (a —

8)

+

da =

(69)

(68) therefore simplifies to

ii(P)

2rcos(O—

(70)

in agreement with equation A few remarks may be made eoncerning the scope of the method of integral equations. In addition to providing an existence proof, this method provides, in (50) and (52), a pair of equations which are, for some

domains, amenable to accurate computational techniques. A second advantage is that the basic ideas which were used in this case can be gen-

erHlized to provide existence proofs for boundary-value problems for certain classes of elliptic equations. On the other hand, a serious disadvantage of this method is that it is tied very strongly to the smoothness properties of the boundary. If the boundary consists of a single closed curve formed by two smooth arcs which form an angle at one or both of their common end points, the kernel of equation (52) becomes discontinuous. In cases such as this the method of integral equations can be modified, but requires, for example, a proof that a finite number of iterations leads to a continuous kernel, and that the solution of the integral equation obtained by iterating (52) this number of times is also a solution of (52). However, if the boundary is not "almost" smooth, in some sense, the method of integral equations becomes completely itiapplicable. EXERCISES

22. Prove that the kernel K(P,Q) is continuous (jointly) for all P and Q on r. (Recall that r is assumed to possess continuous curvature.) 23. Prove that the Neumann problem (Exercise 6-30) can be formulated as an integral equation similar to (52), with a layer matead of a double layer.

24. Prove, with the aid of (54'), the assertion made in the text concerning the behavior of the gradient of v "at infinity." 25. Work out the details leading to (68), (69), and (70).

5. The Dirichiet Principle The Dirichiet problem is readily converted, by a simple argument, into

a problem of the calculus of variations. Though the approach thus introduced into the study of the Dirichiet problem appeared at first to offer an elementary method of establishing the existence of a solution, it was subsequently realized that the argument employed was defective. However, alter a lapse of several decades it was shown that the varia-

Partial Differential Equations

184

tional approach could, in fact, be used to study the Dirichiet problem and, more generally, boundary-value problems involving a broad class of linear elliptic equations.

The argument leading from the Dirichiet problem to the "Dirichiet principle," as Riemann named the variational principle to be presented here, proceeds as follows.' Given a domain G and a function f defined on the boundary, associate with each sufficiently differentiable function u defined in and assuming the prescribed boundary values the number D(u) This quantity is termed the Dirichiet integral of the function u.

Since

D(u) is, as the integral of the sum of squares, nonnegative, there is a finite (in fact, nonnegative) lower bound on D(u). Let v denote a function for which this lower bound is attained. Then, for arty function w which vanishes on the boundary and for any constant s, the inequality (72) + ew) D(v) satisfies the prescribed boundary condition. must hold, since v + Writing out (72) in the form D(v) + + e2D(w) D(v)

D(v

where

D(v,w)

JIG

+

dx dy

(74)

must vanish. Suppose, for example, that D(v,w) were positive. Then we could choose negative and so close to zero that 2D(v,w) + would also be positive. Then 2tD(v,w) + would be negative, contradicting (73). Similarly, the possibility that D(vw) might be negativ' is ruled out. Hence, taking account of (6-9), with u replaced by w, we obtain

we easily show that

D(v,w) =

—

w Lw dx dy = 0

since the vanishing of w causes the boundary integral to vanish. Now, suppose that is different from zero at any point P of G. We select a disc which contains P and is so small that Lw maintains the same sign throughout the disc, and then construct a function w which is positive in the (open) disc and zero outside. (Such a function is easily constructed; ef. Exercise 26.) Then (75) is obviously violated. We conclude that v is harmonic throughout G, and hence that the function minimizing (71) subject to the given boundary condition also provides the solution to the Dirichiet problem with the same boundary condition. The uniqueness The reader acquainted with the calculus of variations will recognize tbat we are simply deriving here the equation associated with the integral (71).

1. The fMriehlet Problem of v, incidentally, would follow either from Theorem 6-3 or from the follow-

ing argument. Suppose that there were two ftrnctions which minimize (71), say v and 0. Letting w = U — v and taking account of the fact that D(v,w) vanishes, we obtain D(v) + D(w) D(D) D(w) = 0 and hence Therefore w reduces to a constant, which must be zero, since v and 0 coincide on the boundary. There are a number of defects in the above arguments. First, we note that the use of (6-9) tacitly assumes that v possesses second derivatives, that tbe boundary of G is sufficiently smooth, and that the behavior

near the boundary is "reasonable." Also, the argument involving the disc tacitly assumes the continuity of Aside from these objections, however, there is one which is absolutely basic, namely, that the fact that the integral (71) possesses a lower bound does not guarantee the existence of a function for which this lower bound is actually attained. Related to this basic objection is another one, namely, that it is possible to pre.scribe continuous boundary values such that, for every function continuously differentiable in 0, continuous in G, and possessing the prescribed boundary values, the integral (71) is divergent. (Cf. Exercise 27.) We devote the remainder of this section to showing how it is possible,

by suitably modifying the formulation of the Dirichiet problem, to approach this problem along the lines suggested by the foregoing incorrect arguments.

Until otherwise indicated, we make no assumption concerning the domain 0 except that it is bounded. We associate with 0 two normed linear spaces, as follows.

II denotes the space of real-valued functions continuous and quadratically integrable in 0; llullg is defined as u2 dx DEFINITION 4.

DEFINITION 5.

1) denotes the

of Ii consisting of functions

whose first derivatives also belong to H;

is defined as1

(u1' + u,2) dx We remark that II and D each become (real) inner-product spaces with the following definitions of inner products: H(u,v)

JIG uv dx dy

J)(u,v)

=

+

dx dy

(76)

1 A elight complication ariees from the fict that all constant functiona, not only the have zero norm in 1). This cauces no difficulty in the following deve'opments. zero function,

Partial Differential Equations

186

The proof that either inner product is meaningful when the functions and ii belong to the appropriate space is entirely elementary and may be omitted. We next define two linear manifolds of D. DEJTTNInON 6. By 1) we denote the subset of D consisting of those functions which vanish in some boundary strip;' that is, for each such function u there exists a positive number such that u vanishes at all points of G whose distance from the boundary is less than e. By 1) we

denote the subset of D consisting of those functions which are expressible as the limit in both norms of a sequence of functions of D; that is, u E 1) if there exists a sequence u,1 of functions of I) such that —+ 0, — —

u4,, —, 0.

Theorem 8. 1) is relatively closed in D; that is, if there exists a sequence (u,,j of functions of I) such that

0, llu — UJD —' 0,

— UnILH

thenuEb. Proof.

For each u,1 we can, according to Definition 6, find an element of 1) such that flu, —

<—

—

By the triangle inequality, —

VIIIIH <

Hence llu —

— UnIIE

0,

+

— t)sItR

and, similarly,

<

—

0.

—

+ By Definition 6,

We now proceed to reformulate the Dirichlet problem. Actually, we find it convenient to define two problems, which will subsequently be shown to have the same unique Then it will be shown that, under certain circumstances, this solution also solves the Dirichiet problem as originally formulated in Sec. 6-6. GENERALIZED Dm1CHLET PROBLEM.

a harmonic function u E D such that u ITARIAT1ONAL DIRIORLET PROBLEM.

that u

— g

1) and

Given a function g EE D, th find —

g

E I).

Given g E D, to find u E D such

is as small as possible.

It should be stressed that the condition u — g EE is indeed a boundary condition, in the sense that only the behavior of u and g near the boundary of the domain G is involved; for, if u, and U2 coincide in some boundary Such are said to be compact support," the support of u being the closure with respect to G of the set. on which tz doen not vanish.

187

7. The Dirichiet Problem strip, and similarly for g1 and U2, then, from the equality = (ui — + (U2 — u,) + (gi — g:) —

it follows that u, — Qi belongs to I) if and only if u1 — does. (Of are required to belong to D.) It may course, the functions also be noted that it is not assumed that g can be extended continuously from 0 to C; such an assumption will have to be made only in order to show that the common solution of the two problems formulated above possesses the boundary behavior required in the original formulation of the Dirich let problem. Before proceeding with the details, we indicate briefly the general out-

From the class of "competing" functions (i.e., the functions of D which satisfy the "boundary condition") we select a sequence {4,I} such approaches the minimum value. Then it is shown that that line.

is a Cauchy sequence with respect to both the H and the D norms. Although this does not suffice to guarantee convergence (either pointwise or in norm) of the sequence, it will be shown that a rather simple averaging process may be applied to the functions { } to define a function u, which then shown to be harmonic throughout G. Then it is shown that u is, in fact, the limit in the D norm of the sequence { }, from which it follows that u — g 1) and that IIUIID is actually equal to the minimum possible value. Thus, u provides the solution to both of the problems formulated above. Finally, as indicated above, it will be shown that, if the domain G and the "boundary function" g satisfy certain conditions, u also solves the Dirichiet problem as originally formulated.

Several simple theorems will be proved before the function

is

constructed.

Theorem 9. If

E D and is harmonic throughout G, and if

then

=

0

E

1),

(77)

If i), then is equal to JIG' dx dy, + where C' is a smoothly bounded domain obtained from G by discarding a suitable portion of the boundary strip in which vanishes. Since (6-9) is then applicable (with u and v replaced by and 4, respectively), and since vanishes on the boundary of C', we immediately obtain (77). If 1), we select a sequence of functions belonging to 15 and converging to in the D norm. Then, using the Schwarz inequality, we obtain Proof.

{

+

—

=

— 114'

so that (77) holds in

case also.

—,

0

Partial Differential Equations of 1), the inequality'

Theorem 10. For every function

H(#)

(78)

hoLds, y denoting a constant determined by 0. Proof.

=

4a2.


is now continuously differentiable in the entire square), at any dE, and point (x,y) of the square, the equality •(x,y) hence, by the Schwarz inequality, (since

(fx

4i2(x,y)

j2 dE)

(fz

2a

11a

Integrating with respect to y, we obtain

42(x,y) dy 2a

fa fa

dy

Finally, integrating with respect to x, we obtain the desired result, for

dx dp

=

= of functions belonging to 1) and .1), we select a sequence { If converging to in both norms. Then, from what has already been

holds for n =

shown, the inequality

1,

2

Passing to the limit and recalling that the norms of a convergent sequence converge to the norm of the limit element, we obtain (78).

Theorem 11. The "generalized Dirichiet problem" has at most one solution. Proof. Suppose there were two solutions, u and v.

in (77) could be replaced by u — — v)

0.

v,

Then both .p and

and we would conclude that

Then (78) would imply that H(u —

v)

=

0, or is

v.

Theorem 12. The "variational Dirichiet problem" has at most one

solution. Proof. Suppose that both u and v provide solutions; then, for all values of the parameter s,

+ sD(u — v) D(u) (82) since v — u E 1). Thus, the quadratic function of e appearing in (82) attains its minimum value for two distinct values of namely, 0 and 1, but this is possible only if D(u — p) = 0. As in the preceding proof, we L)(u + e(v — u))

D(u) + .2eD(u, v —

is)

conclude that is and

are abbreviations for

and IN#,4), respectively.

7. The Dirichiet Problem

Theorem 13. The solution of the generalized Dirichiet problem, if it exists, also solves the variational Diriehiet problem.

Let u solve the former problem and let w be any function Then, by Theorem g E O.

Proof.

satisfying the boundary condition; i.e., w — 0, D(u, to — u) = 0, and therefore

1)(w) = D(u) + D(w — u)

D(u)

showing that u solves the latter problem as well.

Since D(#) cannot be negative, it is meaningful to define the quantity d as follows:

d=infD(qS) Although it is not obvious that there exists a function for which the lower bound is actually attained (this is, of course, the essential difficulty of the entire analysis), there must in any case exist a "minimizing sequence" of competing functions such that Henceforth 4.,) denotes a fixed minimizing sequence.

Theorem 14. Let bounded D norms, say

be

any sequence of elements of 1) with Tfhen

2W.

0.

For any value of the parameter e,

Proof.

+

boundary condition, and therefore D(#.,

+

—d

=

+

—

satisfies the

0

+

(86)

The discriminant of the quadratic function of E must then be noapositive, and so — d] —+

—

0

(87)

Theorem 15. is a Cauchy sequence with respect to both norms. Proof. From (85) it follows that the quantities are bounded, say by M/2. Hence J1#u

+ H4¼.b M

— #WIIID

and, since — E 1), we can, by the preceding theorem, given choose N so large that


—

whenever

> 0,

n' > N. In particular, if n and m also exceed N, we obtain —

and hence —

=


—

—

—

<2ö

Partial Differential Equations Thus the Cauchy convergence is established for the D norm, and Theorem 10 now enables us to draw the same conclusion for the H norm.

This theorem suggests an interesting proof of Theorem 12. If there were two solutions of the variational problem, say u and v, then u, v, u, would be a minimizing sequence and hence, by Theorem 15, a • . Cauchy sequence. Then D(u — v) = 0, and therefore, as in the proof of v. Theorem 11, u We now proceed to construct the function which solves both of the .

afore-mentioned problems. Let G' denote any closed subset of G, and let p denote the distance between G' and the boundary of G. Then, for any

positive number R < p, the following definition is meaningful at all points P0 of (1':

dxdy

=

(88)

denotes the disc of radius R with center at P0. where inequality we obtain LUn.R(Po) — U,,.ji(Po)J2

(ff P dx

dy)

(ff

By the Schwarz

dx dy)

—

(89)

—' 0

—

throughout (1' is thus The uniform convergence of the sequence { is continuous (this follows from the assured; since each function it follows that the limit function, which we denote as continuity of UR, is also continuous. Now, suppose that the same construction is carried out with R replaced by a smaller radius, say R'. (The corresponding function UR' could then be defined in a set larger than 0', but this need not concern us.) With any point P0 of G' we associate the function 2 log

r

+ r2

2 log j + (1

R2)

R'

r

("

r;P)

From the continuity of and of its first derivatives (cf. Exercise 28) 1) (in fact, and the fact that it vanishes for r R, it follows that E b), and from Theorem 14, taking each we conclude that as 0. Now, by (6-9),t =

—

f Although the second derivatives of fail to exist on the circles r — R' and r the applicability of (6-9) follows readily by dividing (1 into the regions r


r
on the afore-mentioned circles.

191

7. The Dirichiet Problem

1/R2) for r <.R' and —4/R2 for R'
Since

D(4J)

dx dy

Jf

Thus, the assertion that

—' 0

—

is

ff to the following one: (92)

dx dy

(93) (P0 E G') This, in turn, is equivalent to the assertion that the function (wR2)' (Ia is independent of fl, so that we may denote it simply as a. Since R may be chosen arbitrarily small, we may consider a to be defined througha is defined at each point P of by averaging over an out G. arbitrary disc (independent of n) with center at P and then letting n —+ 0

—

Theorem 16.

The function a defined above is harmonic through-

out 0.

Select any point P of 0 and a disc of radius R with center at P such that A C 0. Given a > 0, it is possible to choose > 0 such
<.ö, and also such that it is possible to express A as the inequality the union of a finite number of nonoverlapping discs whose radii are eaeh less than & together with a remaining set S of area not exceeding a.

The first part of the above statement follows from tbe fact that a is uniformly continuous on any closed of 0, and in particular 1, while the second follows from a simple geometrical argument.' Let the small discs be denoted by rN;

.

.

and their centers by

a . max

arid (fJ5 we

.

their radii by

, .

.

Since

,

dx dy) (ff

dx

.

a dx dy

dx dy)

obtain the inequality (u

—

dx

dy

-

max Iu(Q)l

+

max

(94)

an arbitrarily large of the area of can be exhausted by constructing fine, equally spayed mesh and retaining those squares of the mesh lying entirely within it suflicee to prove that an arbitrarily largs fraction of the arcs of a given square can be exhausted by nonoverlspping discs. If the inscribed disc is removed from the given square, less than one-fourth of the original area A remains. The remainder is then exhausted to a sufficiently great extent by a set of squares, as above, and then the inscribed disc of each of hese new is removed. Now the remaining area amounts to than A/41. If this procedure is repeated a suiRcient number of times, the original square is exheusted to the desired extent. The botmdedueaa of the quantities ii is proved as follows. Since — 0, Theorem 10 E 1) and — that — —' 0. Since the norms of the olevients of a sequence arc convergent, thc dcsired result is established. 1

Since

a

H

Partial Differential Equations in

Next, expressing u —

and taking were chosen and of the

+ [u(Pi) —

as Eu —

account of the manner in which the discs and Ut,. we obtain

definitions of

ff (u —

4k")

Summing (95) over the discs

we

dx dy

(u —

+

.

dx dy

e

(95)

—

I

obtain TR2 +

IUr4(Pi)

(96)

—

Combiniug (94) with (96), and taking account of the arbitrariness of dx dy —p 0, or conclude that ffA (u —

irRtu(P) = ffudxdy

we

(97)

can be carried out for all points P of G and Since the above all values of R less than the distance between P and the boundary of G, we may invoke Theorem 6-5 to conclude that u is harmonic throughout as was to be proved. (Note that we are not exploiting here the fact Uiat every dise is a Dirichiet domain, for the second proof of Theorem 6-5 does not use this fact.) We remark that this proof may be looked upon as a rigorous version of the following heuristic argument: Since u (recall that was constructed as a generalized limit of the sequence the generalized limiting operation consists of averaging the functions before letting n. oo), it appears plausible that the result of applying the averaging operation to u is to yield u over again. Thus u should possess the areal mean-value property, and it may therefore be expected to be harmonic. Theorem 17. Let f be continuous in G and let G' be any subdomain (or, more generally, any open subset) of G such that ?7' C G. Then dx dy —4 0. (u — Proof. Given e > 0, we can exhaust the area of G' to within at most with a finite number of nonoverlapping discs which are so small that 11(Q) — f(Q')I
ff

—

dxdy

—

+

ff

(14 —

—

1(P)]

dy + ff5 (u

dx dy

7. The Diriehjet Probtein we proceed to analyze

term on the right. First,

Jf (u —

dx dy

0

This follows from (97) and the definition of u. The first sum therefore approaches zero. We estimate the second sum by noting that it is dominated by — #14dxdy

and, a fortiori, by which cannot exceed, by the Schwarz inequality, (area of G)h] [JIG' Iuj dx dy + Since the sequence is bounded (cf. preceding footnote), we have shown that the second term is dominated by where C is independent of and n. The last term on the right side of (98) is dominated, similarly,

by

max

dx dy +

(area of

and, a fortiori, by max

max PEG'

+

max

Combining the above results, we conclude that —

C1 and C2 are independent, of €. Since is arbitrarily small, the left side of (99) must vanish, and the theorem is proved. It shoidd be stressed that the condition C G is essential in the proof, for otherwise u and/ might unhqunded in G'.

Theorem 18.

flu

—

to u in the H norm. Proof. First we must prove

0; that is, the sequence {4,,J converges

u E H. Let G' be chosen as in Theorem 17, and let N be chosen such that — for all

n > N. Then we write (u —

dx dy

(u —

dx dy

—

+

(u

—

—

dx dy

(100)

and note that the first term on the right vanishes as n oo, by the preceding theorem. We apply the Schwarz inequality to the second

PartIal Differential Equations

194

term and conclude that it is dominated by dx

(u —

—

and hence by (u

dx

—

The latter expression is dominated, according to the triangle inequality, by

[(JJ u2 dx

+ I1#nIIH]

Taking accouiit of the boundedness of

and the arbitrariness of €,

we conclude that (u

—

dx dy —, 0

Since the norm is known to be a continuous function in a normed space (we moment*uily associate with (P a space analogous to if), it follows that

ff,u2dxdy= urn Jf

iimJf

(101)

G' may exhaust (7 to within an arbitrarily small remainder, we conelude that u2 dx dy (so that u E H) and, in fact, is not greater than lini > 0, we choose N as before, and then choose the open subset G' of (7 such that (in addition to the earlier condition C 0) Now, given

both of the following conditions are satisfied: ffo—e' u' dx dy

ff0—C' #N2 dx dy

For convenience we introduce the notation

and (fJ00, w' dx

dx —

= tIu — + (ltullFi"

+

and Itwt!H" for (JIG' tot

respectively. —

+

+

+

(102)

Since

—

—

< llu

—

+ (3d)'

(for n > N), it follows from Theorem 17 and the arbitrariness of that 0, as was to be proved.

llu

Then

ff4(u_

195

7. The Dirichiet Problem

Lot / be any function which is continuously differentiable in Then since ff f(u — 4,,')fJ dx dy A and vanishes on the boundary of Proof.

vanishes,' we may write

ff (u —

dx dy =

(u

—

—

dx dy

(103)

From Theorem 17 (withf replaced byfA), we conclude that the right side,

zero as n

and hence the left side, of (103)

Given

> 0, choose f so as to satisfy, in addition to the afore-mentioned conditions, the following: (1) (2)

(Here R denotes the radius of and r the distance from the center of £) it is obvious that such a function exists; for example, we may define f in sin2

(r — R)

fL (u —

dx dy

Then we write

fL (u —

4,,.)

dx dy

—

= fL (t4 — 4,n)g(1 — f) dx dy (104) where A denotes the annulus ii — e
is a Cauchy sequence in the D norm implies that { } is a Cauchy sequence in the H norm. Once Theorem 19 baa been established, it is clear that the line of reasoning which culminated in Theorem 18 may he repeated with u and 4,,' replaced everywhere by and We thus conclude that u,, and similarly v,,, belongs to H, and that 0

11uz —

—

0

Now (105) immediately implies that lu —

0

(106)

Thus we have shown that u is the limit of the sequence f4,,, in both 1Replacef and gin (6-8) by (u — on the boundary of &

.jf and 0, respectively, and recall that/vanishes

Partial Differential Equations

196

norms.

Therefore,

=

Urn

d,

and, furthermore, since

both norms, it follows from Theorem 8 that u — g E 1). Thus we have proved the following theorem, which is the principal result of the present section. —

g) converges to { u

—

Theorem 20. The "generalized" and "variational" Dirichiet prob-

lerns are both solvable, and the unique solution of each is furnished by the in sequence { function u which was constructed from the the manner described following Theorem 15.

Finally, we return to the original formulation of the Dirichlet problem, and establish the following theorem.

Theorem 21. Sappose that C has the property that there exists a positive constant fl such that, for every positive constant fl' less than R and for every boundary point Q, there exists a boundary point Q' whose distance from Q is exactly R'. Then, if the function g (belonging to I)) can be extended continuously from G to (7, the function u —

g

approaches

zero at the boundary, so that u provides the solution to the Dirichiet problem as originally formulated. Proof. Let P be any point of C whose distance, 2h, from the boundary is less than 2R/3, let be the disc of radius h with center at P, let 9 be a = 2h (at least one such point must exist), boundary point such that and let S be the set consisting of all points of G whose distance from Q is less than 3h. Then is open, contains and each circle Ck of radius k less than 3h with center at Q lies partially, but not entirely, in S. We

denote the portion of lying in ,S as rk; consists of one or more (perhaps infinitely many) circular ares whose end point& are boundary points of C. We shall show later that, for any function ing inequality holds:

(HA dx dy)2

belonging to i), the follow-

ff (4's' +

dx dy

Accepting . (107) mbmentarily, noting that ff3 dx dy is + increased it S is replaced by the set of all points of 0 whose distance from

the boundary is less than 3h, and that the new integral thus obtained must approach zero with h (this is merely a restatement that4' E 1)), we conclude that dx dy approaches zero as P approaches the boundary. Replacing 4' by u — g (recalling that u — g E 1)), we conFe may consist of all of except for a single point; consider, for example, the domain obtained by removing from a disc one of its radii.

7. The Dirleblet Problem

elude that, as P approaches the boundary,

ff

ti dx dy — g(P)

ff

—

[g — g(P)J dx dy —, 0

(108)

By the (area!) mean-value theorem, the first term on the left. equals u(P). — which is dominated The last term is dominated by max -

P'EA

in turn by max

(P E 0, P' E G, PP'
—

(It is at this point that we use the fact that g can be extended confor this guarantees the continuity of g in 0, and tinuoualy to hence the existence of the above maximum.) Since this maximum must

approach zero with h (again by uniform continuity), we conclude that is — g approaches zero at the boundary, as was to be proved. It remains to establish (107). Jt suffices to confine attention to the case that 4 E 1), for the more general case that 4 1) may then be treated by approximating suitably by functions in as in the proof of Theorem .10. Now, by the Schwarz inequality, the left side of (107) is

dominated by (IL P dx dy) (IL dx dy.

dx dy), and hence by iA'

4'

It will therefore suffice to prove that

4' dx dy

+

dx dy

Let polar coordinates r, 0 be introduced with origin at Q; then • is

expressible at. any point T of S in the form 4(T)

f+ado

where the integration is performed along an arc of 1',. whose two end points arc T and a boundary point of 0. (Here r = and we are using the fact that E D and also the fact that 1', meets the boundary.) By applying the Schwarz inequality and lengthening the path of integration to all of

we obtain

4'(T) (Jr, i' do) (Jr.. 4•2 do) 2ir Jr.

do

Since (111) holds for all points T on r.., we obtain, by integrating both sides,

Multiplying both sides of (112) by r dr and integrating from 0 to 3h, we obtain

ff42dxdy

(2w)2ff84,2dxdy

Partial Differential Equations

198

The right side of (113) is increased when the integrand is multiplied by (3h/r)2, since r <3h throughout S. Thus,

dx dy

(r1#.)' dx dy

36T2h2

Finally, noting that (r'#.)2 (r3#e)2 + #,2 =

+

we

find that

(114) implies (109), and the proof is completed.

We conclude this section by pointing out two simple but highly important consequences of the above theorem. First, since every polynomial p(x,y) belongs to D (for any bounded domain 0), and since, as pointed out in Sec. 2, a bounded domain is a Dirichiet domain if the Diriehiet problem can be solved whenever the boundary values are expressible as a polynomial, it follows that 0 is a Dirichiet domain if it satisfies the condition imposed in Theorem 21. Secondly, any domain of finite connectivity whose boundary contains no isolated points is readily seen to satisfy the afore—mentioned condition, and so is a Dirichiet domain. (This is consistent with the example given in Sec. 6-6 of a domain which is not a 1)iriehlct domain.) in particular, we obtain the remarkable result that every bounded simply connected domain is a Dirichiet domain. EXERCI SES

26. Let denote any disc. Define a function which possesses continuous derivatives of all orders throughout the plane, vanishes here outside A, and is positiv€ throughout 27. lladamard's e.cam pie: Let be defined For x2 + y3 by the series

u(re') —

Prove that for this function, which, according to (6-36), solves the Dirichlet problem

with the continuous boundary values g(e)

—

the Dirleblet Integral (71) is divergent.

cos n!O

28. Prove that the fuaction defined in (90) is completely determined, except for a constant factor, by the following requirements: (a) should depend only on P0P; (b) rahould possess continuous first derivatives everywhere; (c) r should vanish for PoP > fl; (d) should be constant in the disc P0P
7. The Dirlehiet Problem

199

that the boundary of G and the behavior of u and v near the boundary such as to permit free use of the "transformation theorems" of See. 6.3. 30. Prove that D(u) is invariant under (one-to-one) conformal mapping. (It is not assumed that it is harmonic.)

6. The Method of Finite Differenc& As iii the case of ordinary differential equations, partial differential equations are often approximated by replacing the derivatives with corresponding difference quotients. The idea underlying such a procedure is, of course, the intuitively plausible one that, if the increments of the independent arc chosen sufficiently small, an algebraic system of equations will be obtained whose solutions will be close, in some sense, to solutions of the given differential equation. An extensive literature has grown up about this idea. Not only has it been possible to

analyze, for many classes of differential equations, the relationship between the solutions of the differential equations and those of the approximating systems of algebraic equations, but in many cases finitedifference methods have been successfully employed in eRtablishing the existence of solutions. A comparat.ively simple, but highly instruc.. tive and important, example of the use of such methods will be presented in this section. For any positive number h we denote by Lh the set of all points in the plane whose rectangular coordinates arc integral multiples of h. Two points of Lh are termed "neighbors" if their &)sCIssaS coincide and their ordinates differ by h, or vice versa. Thus, each point P = (nh,mh) of L5 has four neighbors, which we denote PN, PR, J)s, J)W, as follows:'

(nh, (m + l)h)

P8 =

(n/i, (m — 1)/i)

= ((n + l)h, rn/i) = ((ii — l)h, mh)

115 )

A point P is termed an "inner" point of a subset 8 of L5 if P and its four

neighbors all belong to 8; a point of S which is not an inner point is termed a "boundary" point. Let a (real-valued) function is be defined on a subset S of LA. We define the difference quotient at each point P of S as follows:

ifPZE,S

•

(116a)

This section consists of a somewhat amplified presentation of a portion of an important paper by R. Courant, K. 0. Friedrichs, and H. Lewy appearing in Mathsmaliache Annaien, vol. 100, 1928.

We employ this notation in order to suggest the four directions of the compass.

Partial Differential Equations Three more (first-order) difference quotients are defined analogously:

u,(P) =

[u(P)

—

v(P")]

if

ES

(116b)

jf

(hOc)

u1,(P)=O —

if P3 E S

(116d)

if P5

We can then define difference quotients of higher or'der, in particular We (U,)p, defined respectively as are equal at any inner point P of S, and that readily find that and their common value is equal to h—' + ing this expression with the analogous one for

—

2u(P)).

Combin-

(= u,,,), we obtain for

the "discrete Laplacian" the formula

=

+

=

[u(P") + it(P') + u(P3) + u(PW) — 4ti(P)]

Thus, a "harmonic" function defined on S is one whose value at any inner point is equal to the arithmetic mean oi its values at .the four neighboring points. (Cf. Theorem 6-1.) In analogy with Theorem 6-2, we obtain the following.

Theorem 22. Principle of Maximum and Minimum. A function u harmonic on a finite subset S of LA attains its maximum and minimum values at boundary points of & Proof. As in the proof of Theorem 6-2, it suffices to confine attention to the maximum, and to show that if the maximum is attained at an inner point, it is also attained at a boundary point. If the maximum

value, M, is attained at an inner point P, it follows immediately from (117) that u has this same value at all four neighbors of P. By repeating this argument a finite number of times, we ultimately find a boundary point at which u has the value U.

We now show that the discrete analogue of the Dirichiet problem is always uniquely solvable (in contrast with Theorem 6-3, which settles only the uniqueness question), and that, furthermore, the problem may be formulated as the discrete analogue of the variational Dirichiet problem which was formulated in the preceding section. Theorem 23. Let a function f be defined on the boundary of a finite subset S of LA. Then there exists a unique function u harmonic at all

201

7. The DMcblet Problem inner points of S and coinciding with f at all boundary points. other function satisfying the "boundary condition," then

+ S

Proof.

>0

+

—

If a is any (118)

S

The uniqueness is established by applying the preceding

theoremtothe difference of two presumed solutions. As for existence, we note that by equating the right side of (117) to zero for each inner point P we obtain a system of linear equations equal in number to the number. of unknowns. By "Cramer's rule" we conclude from the uniqueness that a solution must exist. (Cf. Sec. 4-5.) on To prove (118), we note that, for any two functions v and w S, the following analogue of (6-9) holds:L + v,w,,)

+

—

(119)

denotes summation over the inner points of S and

where

denotes

an expression in which the values of w at the boundary points appear linearly, and hence vanishes if w vanishes at all boundary points. Identifying v and w with u and a — u, respectively, and taking account of the equalities óu

0 at inner and boundary points, respectively,

0, a — u

we find that the left side of (118) is equal to

(a

—

+ (a —

u),2,

S

which is positive unless a — is is constant throughout S;t taking account of the assumed equality of u and on the boundary, we find that equality holds in (118) only if is a.

We now associate with a given Dirichiet problem a sequence of algebraic problems of the type described above. Given a bounded domain 'Letting and denote the subsete of S consisting of thcae points whose "eastern" and "western" neighbors, respectively, also belong to S, we note that —

raw, Lcf. (1 16a)1, and hence —

w(P))

—

h'

v1(PW)w(P)

Adding the analogous formula for

—

v.i(P)w(P) + boundary term

v,w,,, we obtain (119).

t More precisely, Ii — u would be constant on each "component" of 3, this term denoting, in analogy with its customary use, a maximal subset of 3 such that any two of Its points P, Q can be joined by a sequence P, P,, F,, . . . , Pu, Q of point. which belong to the subset and where every suce.essve pair of points in the sequence

are neighbors.

Partial Differential Equations

G, let a continuous function / be defined on the boundary. By the

Lebesgue extension theorem (ef. Sec. 1-2), we can extend (in an unlimited

any function thus obtained is number of ways) f continuously to be the portion of denoted as f. For each positive integer n, let G whose coordinates lying in 0, so that (quotients of integers in which the denominator are both dyadic rationals (depending on P) of the jq a power of 2) lies in all hut a finite number Note that the set of all such points, which we denote as B, is sets dense in G. For each n. we solve the "algebraic Dirichiet problem" with boundary values given by the values of / at the boundary points of (1w; the functions thus obtained will be denoted as u,,. It will converges, now be shown that, as might be expected, the sequence under suitable hypotheses, at all points of B, and that the limit function can be extended to all of 0 as a harmonte function. The proof of these assertions will be presented in the form of a series of lemmas. LEMMA 1.

The sums h2

u2(P) are bounded; here h

and we

write u for u,,. Proof. Let 0 be covered by a closed square R of edge a, and let M

denote the maximum of jf! in 0. Then, since the boundary values of u lie between — M and M, the same assertion i8 true at inner points of G., by cannot exceed the number Theorem 22. Since the number of points in

of mesh points in R, which is at most (1 + a/h)2, it follows that the sum is dominated by M2(a + h)2, which approaches M2a, and hence

remains bounded as n increases, as was to be proved.

LEMMA 2.

sums

Let

be any (open) square such that ?7 C 0. Then the remain bounded as n increases. ± + u,2 +

Proof. We assume, for convenience, that the vertices of G* possess coordinates which are all dyadic rationals. (This is permissible be replaced by a larger square possessing this property, and the sums

will be dominated by the corresponding sums for the larger square.) We then select an (open) square G** such that 0 and C such that the vertices of Q**, like those of (1*, all have dyadic rational coordinates. Then, for all sufficiently h&rge n, can be expressed as the union of and a sequence ni "shells," each of Si, . . . , which lies inside the next; lies on the boundary of lies on the and boundary of a square slightly smaller than G**. [The value of N is easily seen to be equal to 2"-'(a — b) — 1, where a and b denote the edges of G** and G*, respectively.J From the elementary equalities

7. The Dirlàhlet Problem h'[u42(P) +

±

+

+ u,'(P) + u9'(P)) = u2(P1%') + ii2(P8) — + u(P8) + u(P1) + + u'(P5) + u'(P') + u'(P") — 4u'(P)f

obtain, by summing over all points P of G and So, the identity

ii'

+ tQ) =

(it.' + US' +

—

u'

—

8.

(120)

0,1

where 5" denotes summation over the vertices of both So and

S1.

Sim-

llarl3r, we can write analogues of (120) with the summation on the left where k may take any . . . , and the shells So, extended over Since the left side of (120) and each of , N — 1. . . . of the values 1, 2,

the analogous equations dominate the quantity

we

obtain the N

inequalities

z

(121)

—

—

8,

j—tj

S1..,

and, a fortiori,

(j =

u'

—

1,

2,

.

.

,

2,

.

.

N)

(122)

Sj_i

*

Summing the first m inequalities of (122), we obtain mX

—

u

u2

(in =

1,

.

.

,

N)

(123)

We now sum the inequalities (123) over all values of in and obtain

+ 1) ,,

rn—I

S.

it'

(124)

Observing that the right aide of (124) is increased if u' is summed over aU of and taking account of the formula given earlier for N, we obtain the further inequality

8(a — b)-'(a From the boundedness of h' •

— b —

> it2

(125)

ti' (Lemma 1) we now conclude that (1..

ft

remains bounded as n increases.

Let G* be any (open) square such that C G, and let w (= w.) denote any specified difference quotient of it (= u,,). Then the sums h' to' remain bounded as n increases. 3.

es

1 The last equality follows from the fact that

4ia(P) - it(P') + v(P') + v(P') + ti(PW)

Partial Differential EquatIons

u,, If w is any one of the first-order difference quotients, Suppose the present lemma is contained in the preceding one.

Proof.

u,,

next that to is a second-order difference quotient, say to u21. According are bounded, where H to the preceding lemma, the quantities ht

like u, is har-

denotes a square such that C' C H, fl C G. Since

monic, we may again apply the preceding lemma, with G replaced by H

+ ud +

We conclude that the sums ht

and u by

w1, remain bounded.

and a fortiori the sums

+

Similarly, if to is

a difference quotient of higher order, we introduce a sufficient number of nested squares, ali containing G and contained in G, and repeat the above argument a sufficient number of times.

Let Q* denote any subset of G such that C' C 0, and let or any specified difference quotient of denote either u (=

LEMMA 4.

to (=

Then to is uniformly bounded and equicoiitinuous in G*; i.e., there exists a

<M holds at

quantity M independent of n such that the inequality

every point of G, and, for any > 0, there exists

independent of n such that lw(P) — w(Q)l <e for any pair of points P, Q contained in satisfying the inequality <& Proof. First we consider the particular case that G* is an open square, of edge a, with vertices possessing dyadic rational coordinates. Let P

Figure 7-1

and Q denote two points possessing equal ordinates and belonging to G for all sufficiently large n. Referring to Fig. 7-1, we find by a simple computation, whose details are left as Exercise 31, that PQ'

'8'

w(P) — w(Q) + w(R) —

and hence lw(P) —

Summing (127) over all a/h

is' — 1

+

11

tw.

(127)

possible choices of the line RB, we

7. The Dirichiet Problem obtain —

—

ZIWZVI

(128)

where both sums are extended over the portion of G contained between

the vertical lines through P and Q (including the former but not the latter). Applying the Schwarz inequality' to each sum, and noting that each sum is extended over (a/h — 1) Pa/h points, we obtain + (a — (129) Iw(P) — Extending both summations over all of 3, we conclude that fw(P) — w(Q)I

and taking account of Lemma (130)

where C is independent of n. Thus the "horizontal equicontinuity" of the functions w,1 is proved; "vertical equicontinuity" is established in the same manner, and it then follows immediately that (130) holds (with a new value of C) for arbitrary positions of P and Q in Thus equicontinuity is proved. The boundedness of the functions w,, is now established by the following simple argument. Selecting P and Q, respectively, as points of G where assumes its maximum and minimum values and we find, by referring to (130), that the differences M,, — are bounded. If

becomes arbitrarily large (positive or negative) for some subsequence I w,4, the same must then be true of mM, and, hence, of the sum h2

contrary to Lemma 3. a.'. Having established the lemma in the particular case that is a square, we dispose of the general case by covering with a finite number of squares, each contained in G; the possibility of doing this is assured by the Heine-Borel theorem. Now, by the AsCOIL selection theorem, we can select a subsequence of the sequence (u, which converges everywhere2 in 0, uniformly in every closed subset. From this subsequence we can extract a further subsequence on which the difference quotients also converge throughout G. Repeating this procedure a denumerable number of times and "diagonalising" as in Sec. 1-1, we obtain a single subsequence such that the functions and each of their difference quotients converge throughout 0, uniformly in every closed subset. (It will be shown later it

— . 1)' — (number of terms) . 2Each function u Is defined only on a finite set of point., which varies with n, but is evident that the definition of v may be extended to all of G without destroying

the boundednesa and equicontinuity.

Partial Differential Equations

206

that, in fact, the selection of subsequences is unnecessary.) We now confine attention to this subsequence. is harmonic in G, LEMMA 5. The limit function U of the functions and each sequence of difference quotients converges to the corresponding derivative of U, uniformly in every closed subset. Proof. Let P and Q be two points of G having dyadic rational coordinates and lying on the same horizontal segment, as indicated in Fig. 7-2. Then, from the equality (131)

li

—

and the uniform convergence of the sequences by passage to the limit, U(Q) — U(P) where

= denotes the limit of the sequence

and

we obtain, (132)

By continuity, we observe that the restriction that P and Q have dyadic rational coordinates $—)( P

h K

Figure 7-2

Q'Q

can be dropped, so that (132) holds for any horizontal line segment PQ lying in G. Fixing P and differentiating with respect to the abscissa of we obtain, as was to be expected. that exists and equals U throughout G.

Similarly, U possesses partial derivatives of all orders, which are

the limits of the corresponding sequence of difference quotients. In 92(j 02U particular, the quantity = exists and is given by + U

lim

+

(133)

+ vanishes at each point of B, it follows that must also vanish everywhere on B. Since is continuous in G and B is dense in G, it follows that U is harmonic. Since

Thus far, the only assumptions that have been made are that the domain G is bounded and that the boundary function f is continuous. In order to show that the function U obtained above does actually constitute, in some sense, a solution of the given Dirichiet problem, we must make some stronger assumptions. First, we require that the extension of the boundary functionf to G which was employed in defining the sequence tu,, belongs to the space D. (Cf. Definition 5.) (Aceording to a remark made in Sec. 5, this requirement does actually represent a

207 7. The Dirichiet Problem restriction on f.) Secondly, we assume that G is bounded by a single smooth convex curve 1'. Actually. the latter assumption can be relaxed very considerably, and is introduced simply in order to avoid geometrical

complications.

The functions U and U —

LEMMA 6.

/ belong to the space I).

3,

'-S

Figure 7-3

Figure 7—1

Taking account of the definition of the functions sets G, we obtain from Theorem 23 the inequality Proof.

h2

+

(fe! +

and the (134)

Letting n increase and taking account of Lemma 5, we obtain

ff

+

dx dy D(f)

(135)

where G' is any subdomsin such that. U' C G. Letting G' exhaust G, we conclude that D(u) exists land does not exceed .D(f)J. Since D is a linear space, U — also belongs to D. LEMMA 7. Let S. denote the " boupdary strip" of G; i.e., the subset of C consisting of all points whose distance. from the boundary is less than Then, as e approaches zero, the inequality

f

holds for some constant Proof.

Let F' denote a portion of Il such that the tangent at each

point of r' makes an angle less than 300 with the vertical, and let horizontal lines be drawn through the end points of r', as indicated in Fig. 7-3. It will suffice to prove that. (136) holds with S replaced by the portion

of 2, intercepted by the afore-mentioned lines, since, aecording to the restriction imposed on C, the strip S, can be expressed as the (overlapping) union of a finite number of portions to whith the same argument can be applied. Rderring to Fig. 7-4, let P0 be any boundary point of and let P1 be any point of (1,, lying on the same horizontal aa P.. Letting

Differential Equations

= .u,.

f, and noting that v1.(Po) vanishes, we obtain

—

v1.(P,) =

(137)

hZv1.,3

where the summation is extended over the mesh points from P0 to P,w, inclusive. Applying the Schwarz inequality as in the proof of Lemma 4, and noting that the number of points involved in the above summation cannot exceed Ce/li, where C is a constant determined by G, we obtain (138)

Letting P1 range over all mesh points of the horizontal segment through Po and summing, we obtain

<

+

(139)

where all summations are extcnded over the entire segment. We now sum (139) over all the horizontal "mesh segments" in and obtain

+ Now, from

(u13

—

f)'

— f1-)2

and the analogous inequality for

+

+

we obtain

+

+

=

+ 2ff'

+

Extending the summation on the right side of (141) to all of C,, and then taking account of (134), we obtain

4h'C22

(f3' + fe,')

(142)

f)2 dx dy 4CY D(f)

(143)

0.

By uniform convergence,' we find that (U

thus completing the proof. 8.

The inequality (136) may be replaced by the stronger asser-

tion that

,-Of [1 (U —f)2dxdy = Proof.

0

For ease in exposition, we first prove the lemma for the case

that 0 is the unit disc, and then we indicate briefly how the argument 'Since uniform convergence is only on a compact subset of 0, we employ the familiar device of replacing 8 on the left aide of (142) by a slightly narrower strip lettingn -. .o, and then letting exhaust

209

7. The Dirichiet Problem

extends to more general domains. Introducing polar coordinates, JR denoting U — f as v, and writing v(R,8) obtain, by the Schwarz inequality, [v(R,8) —

(JR!

p

2(1

—

r) J

Integrating with respect to 8, f2V

[v(R,O) — v(r,9)]2

tip, we

— v(r,O)

we

v,2p dp

r < R < 1)

(145)

obtain (letting 1 — r = e)

dO 2t

v,2p 4,, 48

(146)

where D(v) denotes the Dirichiet integral

+

= Jf(v,,t +

dx

dp do

Letting r (and hence R) approach unity, we do (Cf. Sec. 4-2.) If this limit,

extended over the strip S. f°2w conclude that urn

which we denote as C, were positive, we would have (for p sufficiently close to unity) vt(p,O)p do>

and hence, by integration from 1 — e to

(147)

1,

ff&vdxdy >

(148)

which contradicts (136). Therefore, C must vanish, and so, fixing r and letting R approach unity in (146), we obtain

f

2w

v2(r,O)

do < 2(1 — r)Dt....,(v)

(149)

Replacing r by p, multiplying by p dp, and integrating, we obtain

vt dx dy 2

(1 — p)Di...,(v)p dp

(150)

Since the factors 1 — p, Di..,(v), and p are dominated respectively by e, D,(v), and unity, we obtain

ff&v2dxdy which obviously implies (144). case that (7 is the unit disc.

eD(v) dp = 2e'D(v)

(151)

Thus, the proof is complete for the

'Of course, the fact that v E 1), which is aesured by Lemma 0, is essential here.

Partial Differential Equations

210

For more general domains we replace the circles r constant and the radii 8 constant by the inner boundary curves of the stripe S, and their orthogonal trajectories, respectively. There is then no difficulty in imitiating the argument presented above, provided that, as assumed earlier, the domain is smoothly bounded. The function v (= U — /) belongs to 1). (Cf. Definition 6.) Proof. We present only the proof for the case that G is the unit disc, since, as in the preceding proof, the extension to more general domains is be defined (for n 1, 2, quite apparent. Let the functions Qn and LEMMA 9.

3,

. .

.) as follows: 1

n

I

nJ

1.—!
0

2n

—

(152)

We shall show that v is the limit in both the H and D norms (of. Definitions 4 and 5) of the sequence Since the functions of this sequence are readily seen to belong to .0, the lemma will then be proved. As for the H norm, we merely observe that H(v —

vga)

=

ff8

v2 dx dy - 0

(153)

As for the 1) norm, we observe that

=

D(v — vga)

Adding the quantity

+ — vJi.,,)2

dx dy

to the integrand,

+ (vhM.p —

we obtain D(z' — Vgn)

2

+

he,,,) + h*2(V.g'

+

dx

From the inequalities (cf. Exercise 32) we

h'n,x +htn.t—
then obtain D(v

—

2n2ir2 Jf

flu

h2<— v2

dx dy + 2D11,(v)

and, by Lemma 8, the right-hand side of (157) approaches acro as n Increases. The proof is thus complete. We now observe that the argument employed in Theorem 21 applies to

211

7. The Dirichiet Problem

the problem under consideration here, the functions u and g appearing in eec. 5 corresponding to the functions U and f, respectively. Thus, we

conclude that the function U exhibits the desired behavior near the boundary of the domain. Furthermore, we can assert that the selection procedure described following Lemma 4 to secure a convergent 8ubsequence of mesh fun ctiofla was not actually necessary; for if the entire did not converge everywhere, it would be possible (on sequence account of the boundedness of the mesh functions at each fixed point of B) to choose two subsequences converging to different values at some point, and then, by applying the afore-mentioned selection procedure, we could both of which would provide and obtain distinct functions,

solutions to the given Dirichiet problem, contrary to the fact that at most. one solution can exist. Therefore, the method described here serves to furnish the solution as well as to prove its existence. Finally, we remark that the assumption that the l)irichlet integral J)(f) is finite (or even that I possesses any differentiability properties) can he eliminated by an approximation argument similar to that employed in Sec. 2. We sum up the results of this section in the form of a theorem. function f be defined on the boundary Theorem 24. Let a If the of a bounded domain 0, and let f be extended continuously to boundary possesses suitable smoothness properties, then the sequence

converges to a function U which provides the solution to the Dirichiet problem, &nd each sequence of difference quotients converges, of 0, to the corresponding derivative uniformly in every comp8ct of U. EXERCISES 31. Prove (12t1). 32. Prove the. first equality of I 56). 33. Work out in detail the approximation argument mentioued near the end of section.

7. Conformal Mapping Perhaps the most striking single application of the fact that every bounded plane simply connected domain is a Dirichiet domain is to be found in the following theorem, which is justly credited to Riemann, despite the fact thai his proof invalid. Theorem 25. Riemann Mapping Theorem. Any two boundedt simply connected domains can be conformally mapped onto each other. By employing a suitable transformation, we can remove the requirement of boundednesa; it is actually only consists of the entire plane.

to exclude t}ie

ease that the domain

Partial Differential Equations

212

Proof.

It suffices to consider the case when one of the domains is the

unit disc, for the general case then follows by mapping one of the given domains onto the disc and then mapping the disc onto the other given domain. Given the bounded simply connected domain G, we select any point Po of G and define on the boundary the continuous function f(P) log IcP. Since (118 a Dirichiet domain, there exists a function u and coinciding on the boundary with f. harmonic in 0, continuous in It should be emphasized that ti(P) is not given by the function log PP In fact, it is for P (1, for this latter function is not harmonic at is the Green's function log readily seen that the function u(P) — cf. Sec. 6-7 and Exercise 6-14. Let v (J(P,P0) of C with pole at

denote a harmonic conjugate of u—that is, a real function such that u + iv is analytic throughout G. (Cf. Sec. 6-2.) The function v is determined to within an arbitrary additive constant, and its singlevaluedness is assured by the monodromy theorem. Theii the function (z0 and z = u + iv — log (a — a0) is analytic in G except at denote the complex numbers corresponding to Po and P, respectively.) The function (15S) (z /(z) is therefore ana'ytic (and single-valued) in 0, has a simple zero at a0, and it follows from does not vanish anywhere else. Since Exercise 6-15 that < 1 throughout G, while from the behavior of approaches unity there. G(P,P0) near the boundary it follows that Given any complex number C of modulus less than unity, we construct a smooth simple closed curve F which surrounds P0 and is so close to the boundary of Gthat lf(z)I > everywhere on F.f By the principle of the argument, it follows that the equation 1(z) = C has the same number of solutions inside r as the equation f(z) = 0, namely, one. It follows that maps 0 in a one-to-one manner onto the unit disc, and the proof is complete. The freedom in choosing

and the existence of an arbitrary additive constant in v show that an arbitrary point of G may be chosen to map into the origin, and that the argument of the derivative at the selected point may be prescribed.. INote that —v(Po).] Going over to the general case, where neither domain is nece&cuily a disc, we. conclude that any pair of points, together with a direction at each point, may be made to correspond. It is easily proved, finally, that these correspondences uniquely determine the mapping. While the above proof is due to Riemann, his argument was defective, for, as explained at the beginning of Sees the solvability of the Dirichiet problem, which plays an eSsential role in the proof, had not been rigorously established at that f We accept without proof the

plausible fat that such a curve exists;

7. The Diriehiet Problem

213

The above theorem makes no assertion concerning a correspondence of

boundary points, and in general it is impossible to extend the mapping eontinuously from the domains to their boundaries. However, in the simplest case, which is highly important in practical applications, as well as in the further development of the theory of conformal mapping, the existence of a boundary correspondence is guaranteed by the following theorem, which we state without proof.

Theorem 26. Caratheodory's Extension of the Riomann Mapping Theorem. If two domains are each bounded by a simple closed curve, then any conformal mapping of the domains onto each other can be contmuoiisly extended so as to provide a one-to-one correspondence of the boundary points.

Suppose now that a domain G boutukd by a simple closed curve is mapped conformally onto the unit disc. Then, in view of the above theorem and the fact that harnionicity is preserved under mapping (cf. Sec. 6-2), it follows that every Dirichiet problem for U can be converted into a Dirichiet problem for the disc. By solving the latter problem with the aid of the formula (6-27) and then mapping from the disc back to C, we can obtain the solution to the original problem. Turning now to doub'y connected domains, we begin by showing that the most obvious generalization of Theorem 25 is not true; that two doubly connected domains cannot, in general, be mapped coriformally onto each other. This will be proved by showing that two annuli eamiol; be mapped conformally onto each other unless they are similar. To show this, we suppose that the analytic function w f(z) provides a one-to-one mapping of the annulus < < R2 onto the annulus < < R. Then the function log If(z)I must be harmonic in the former annulus and must approach the constant values log and log as ki approaches R1 and .R2, or R2 and R1, respectively. Thus we have a Dirichiet problem in which both the domain and the boundary values possess rotational symmetry, and so, by uniqueness, we know that (159)

and hence

{ = constant

(160)

Since an analytic function of constant modulus must reduce to a constant. we conclude that constant . f(z) (161)

Now, f(z) can be single-valued only if a is an integer, and can furnish a one-to-one mapping only if this integer equals ±1. Thus, the mapping

Partial Differential Equations

214

must assume one of the forms = constant z

=

constant —

from which we immediately conclude, in either case, that R1/R1 =

as was to be proved.

The above example serves to suggest the correct generalization of Theorem 25, which we now state and prove.

Theorem 27. With each doubly connected domain G there is associated a unique number a> 1 such that G can be mapped one to one and whose outer and inner radii are in the ratio conformally onto an cr:]. Thus, two doubly connected domains can be mapped onto each other if and only if they have the same "modulus," 2w/log a. Proof. We may assume that G is bounded (of. footnote on page 211) and that neither of the two boundary components reduces to a single point, for otherwise the present theorem follows trivially from Theorem 25.t By Theorem 21, there exists a harmonic function u approaching the values one and zero, respectively, at the outer and inner boundary components. Accepting for the present the fact—which will be proved later—that the harmonic conjugate v of u is multiple-valued, changing by a positive constant c each time the Inner boundary component is surrounded in the positive sense, wc see that the function f(z)

(163)

is analytic and single-valued in G.

Furthermore, I <exp (2w/c), and (f(z)t approaches the indicated bounds as z approaches the inner arid outer boundary components, respectively. Since the argument. of f(z) (= 2wv/c) increases by 2w when the inner component is surrounded in the positive sense, it follows by applying the principle of the argument (in a manner entirely analogous to its use in t.hc proof of Theorem 25) that f(z) assumes each complex value C satisfying the inequalities 1
21.5

7. The Dirichiet Problem

beyond the circle, so that the analytic function u + iv may also be

extended beyond the circle. On the circle, the outward normal derivative of u must be nonnegative, since, by the maximum principle, u is less than one everywhere in G. According to the Cauchy-Riemaun equations,

the tangential derivative of v in the i itive sense must also be nonnegative.

we

If the latter derivative were zero everywhere on the

would conclude (recalling that the tangential derivative of u is zero is constant

everywhere on the circle, since d

there) that the derivative

± w)

also vanishes everywhere on the, eb'cle, and this would imply, by an elementary theorem concerning auuiytic functions, that. u ± iv is Conactually increase when the stant, which is certainly false. Thus, ." other once in the positive sense. Shwe outer boundary is

positive, circuit about the inner boundary component must give the same change in v, the desired result is established in the case nuder consideration. Turning tA) the case (If an arbitrary (bounded) doubly connected domain, it will suffice to map it onto a domain of the type that has just been considered. This may be accomplished if we can tied a simply precisely of connected domain C containing G whose boundary the outer boundary component of (I, for we can then map onto the tmit disc and confine attention to the part of the unit disc which is the image of G. Such a domain is easily found-—one draws a simple closed curve lying in 0 which surrounds the inner boundary component, and then adds to 0 the interior of this curve.

Theorem 27 admits interesting generalizations to domains of higher For example, if 0 is a triply connected domain (none of whose boundary components consists of a single point) we define, in each analogy with the above procedure, harmonic functions u1 and assuming the value unity on one of the boundary components and zero on the other two components. Their respective harmonic conjugates Vi and are determined, and then it is shown that real con.stants and C2 can be chosen such that the function connectivity.

f(z)

exp

+ iv1) ± 62(U2 + iV2)i

(164)

is single-valued in 0 and furnishes a one-to-one mapping. From the constancy of and a2 on each boundary component it follows that approaches constant values on each component, and it is then shown, by applying the principle of the argument, that the image o 0 consists of an annuhis from which a portion of a concentric circle has been removed. More generally, a domain of connectivity n (3) can he mapped conformally onto an annulus from which a — 2 concentric I

Partial Differential Equations

216

circular arcs have been removed. Again, the essential idea is to eOneach of which assumes the . . . , struct harmonic functions Ui, value unity on one of the boundary components and zero on the others. The respective conjugate functions vi, v,, . . . , v,1—1 are constructed, and such that the are cT, . . . , then suitable constants

function exp

Ck(Uk +

}

the mapping described

accomplishes

above.

By similar methods, it is possible to show that a given multiply connected domain can be mapped onto other types of domains ("canonical domains") possessing a simple geometrical structure. The formulation of a suitable Dirichiet problem may be used in each case as the starting point of the proof of the existence of the desired mapping. In particular, which are known as the "harmonic the functions Ui, U,, . . , measures" of the boundary components on which they assume the value unity, play an essential role in determining the mappings onto many of the known classes of canonical domains. .

EXERCISES . . . , u,. denote the harmonic measures of the boundary components 34. Let UI, of a domain 0 (of connectivity n). Prove that ui + U, + + u, — 1, but that no nontrivial linear relationship can exist among any smaller number of these functions. denote the change in ug, the harmonic conjugate of 85. Let resulting from a circuit around the jth boundary component. Prove that, with a suitiible convention concerning the sense of the circuit, the equality pq holds, and hence

that 36.

Prove that

37.

Prove that

except Xi — 38.

p,,

ii

0, and that

is negative except when i

is positive for all (real) values of —

. . . ,

t

—

Hint: Consider 1) (Z

— 0.

—

Prove that fur each set of numbers

only one set of numbers

c,,

.

.

. ,

.

.

.

,

such that the function

there exists one and increases by

when the boundary component is surrounded. Hint: Prove directly from Exercae 37 (or from the general theory of quadratic forms) that det(p%,)

(i,j — 1,2,

.

.

. ,n

—1)

does not vanish.

$9. Prove that the modulus of a doubly connected domain, as de6ned in the

statement of Theorem 27, is equal to the Diichiet integral of the harmonic measure of either boundary component.

8. THE HEAT EQUATiON In this chapter we shall consider a few problems dealing with the simplest of all parabolic equations, namely, the one-dimensional beat equation

(I) Under suitable physical assumptions and choice of units, this equation

governs the distribution of temperature on a homogeneous thin rod occupying part or all of the x axis, the variable t denoting, of course, the time.

1. The Initial-value Problem for the infinite Rod Let a rod occupying the entire z axis be given, and suppose that at a given instant, which may be taken as t 0, the temperature distribution

is described by the function f(x). On physical grounds, it appears plausible that there should exist one and only one solution of (1) satisfying the afore-mentioned initial condition, and this conjecture is supported by the heuristic reasoning employed in Sec. 3-2. We shall now see how the existence of such a solution can be rigorously established. The basic idea

to be employed is that the transform (with respect to x) of the solution (assuming that the latter exists and does indeed possess a trans.. form) can be determined quite explicitly, under reasonable assumptions on f(x).

By applying the inversion formula to the transform, we shall be

led to an apparent solution, and finally it will be verified that the apparent solution is actually a solution of (1) satisfying the prescribed initial condition. We temporarily assume, therefore, that a function u(z,t) exists for all values of x and all positive values of t, satisfying equation (1) and the 217

PartIal Differential Equations

218

prescribed initial condition, the latter being understood in the following sense:

lim u(x,t) = f(x) t—..

(2)

+0

and integrate with respect to x; applying (1-20) and taking account of (1), we readily find which is clearly a function of A and t, satisfies the that the function very simple differential equation For each positive value of t, we multiply u(x,t) by

=

(3)

depends The symbol of partial differentiation is used in (3), since on x as well as 1. This equation, which constitutes a notable simplification over (1), can be solved explicitly by separation of variables. We thus obtain (4)

ff(u)

where c, the "constant" of integration, may depend on A. c, we make the plausible assumption that u)

Jim

To determine (5)

Taking account of (2) and (4), we then obtain

c—ff(f) and hence

=

(7)

or, taking account of Exercise 4, (8)

Finally, by invoking the "convolution theorem" (1-21), we obtain, as the presumed solution of our problem, the function -

u(x,t) =

j:.

dE

is the "heat kernel" exp [-- — x)2/4t1. Now, disregarding the questionable manner in which was derived,

where

let us study this formula on its own merits. First we observe, by an explicit computation, that f(s) 1 (of. 1 yields the result u(x,t) Exercise 1), which satisfies both the differential equation (1) and initial condition (2). It is to be noted that even this simple case the function does not possess a Fourier transform, so that the procedure leading to (9) was entirety formal; nevertheless, this procedure has indeed

led to a solution of (1) satisfying (2) as well. Turning now to more

219

8. The Heat Equation

general choices of f(s), it is readily seen that the integral (9) converges for all x and all £ > 0 if f(s) is continuous and bounded for all and under these assumptions it is readily shown, by justifying differentiation under the integral sign (ef. Exercise 2), that u(r,C) satisfies (1). It remains to establish (2). For any x we subtract from (9) the equation f(x)

(10)

=

1]. [which follows from the foregoing remark concerning the case f(E) We obtain (11) [f(s) — f(x))K(E,z,t) dE u(x,1) — f(x) =

f.

e Given any e > 0, we can choose o > 0 so small that If(s) — we have the 8, while for all other values of whenever — 2M, where M denotes an upper bound on f(z) — inequality which we have assumed to exist. Taking account of the fact that

K(E,x,t) is always positive, we obtain from (11) —

f(z)I

f'

2M

dw

(12)

is employed in obtaining the last — x)/2 The first quantity of the last pair is equal to while the second

IThe substitution w = integral.)

+0. Taking account of the that (2) is satisfied. arbitrariness of e, we was not It should be noted that the explicit form of the kernel actually exploited in establishing (2)—only the fact that this kernel constitutes, as t —+ +0, a sequence of functions "concentrated" at the point x in the sense of Exercise I-Il. A brief remark concerning uniqueness is in order; by linearity, the

quantity clearly approaches zero as t

question of whether a solution of (1) satisfying (2) is unique is equivalent to the question of whether the only solution of (1) satisfying (2) in the 0 is given by u(x,t) 0. The answer to this is negative, as case f(x) is shown by the example n(x,i)

For any fixed value of x other than zero, the quantity exp (—x'/4t), and hence u(x,t), approaches zero as t does, while 0. However, a more profound analysis' shows that the only u(0.t) bounded solution (in fact, the only solution having either an upper or a (Cf. Exercise 3.)

lower bound) of (1) with f(x) m 0 is the trivial one u — 0.

[Note in (13)

IThe intereeted reader may consult the paper by D. V. Widder in Transactions of the American Mothe,naiicat Society, vol. 55, 1944.

Partial Differential Equations

220

that u is unbounded as z and t approach zero under the constraint c, c denoting

any positive constant.J EXERCISES

L

Prove that, when 1(E)

1 in (9), v(x,t)

I.

2. Justify the assertion made in the text that the function u(x,1) defined by (9) satisfies (1) under the assumption that 1(E) is continuous and bounded; more generally,

assume, aside from continuity (or even measurability), only that 1(E) ii "email" compared to exp in the sense that there exist two constants A and B (A > 0,

0
JN

K(E,O,t) cos E dE

Then deduce from (1) and (2) that gi(t) must satisfy the differential equation 1p'(t) + g — 0 and the additional condition g(0) — 1. Thus, the integral

f oxp

cos E

and, more generally, the integral

(—at'))

exp(—aE')cosxtdt

[

a denoting a positive constant.

5. Prove that if f to

If(x)I dx exists, then

dx also exists and is equal

/(x) dx for all t (>0). Give a physical interpretation of this result. 6. Prove that if 1(E) Ia bounded and has a mean value in the sense that Urn

exists, then u(x,t) baa, for each t (>0), the same mean value. Also prove that, for each x, lirn v(x,t) exiata and is equal to the afore-mentioned mean value.

7.Letf(x)*—lforx<0,f(z).lforz>O. Provethat —(—1 \T,

I

JO

and confirm the assertions of the preceding exercise. 8. Let f(x) be any bounded continuous function defined for: > 0; extend the definition of 1(x) to all values of x by imposing the condition f( —x) — —1(x) [so that, in particular, /(0) — 01. Show that formula (9) yields a solution of (1) satisfying (2) even at x — 0, although 1(x) will, in general, be discontinuous there.

8. The Heat EquatIon

221

2. The Simplest Problem for the Semi-infinite Rod In the remainder of this chapter we shall indicate, by presenting two examples in considerable detail, how the Laplace transform can be used to solve a wide variety of problems of heat conduction. In the present section we consider a very simple problem, which is well suited to serve as an introduction to the use of the Laplace transform. Let us consider a semi-infinite rod whose end is maintained permanently at a fixed temperature, and suppose that at a given instant the temperature at all points of the rod (except the end) is uniform, but different from

that at which the end is maintained. We seek to determine the distribution of temperature along the rod at any subsequent instant. By choosing the units of time, length, and temperature suitably, we may. formulate the problem as follows:

Urn u(x,t) =

0

(x > 0, 1 > 0) (t > 0) Urn u(x,l) =

1

(x > 0)

(14) (15)

We note that this problem can be solved by the method of the preceding section, through the device explained in Exercise 8; in this way we are

led to the problem of Exercise 7, and thus conclude that the problem under consideration here possesses the solution u(x,t)

fxf2g)'4

exp

(—} jo I

(16)

However,

we shall temporarily disregard this method of getting the solution, and instead investigate the problem directly by Laplace that the problem under consideration has a bounded solution tt(z,t) such that all the subsequent operations are permissible, we proceed as follows. For each fixed x > 0 we take the Laplace transform of both sides, obtaining' dl

Moving the differential operator

di

=

(17)

before the integral sign on the left

side of (17) and integrating by parts on the right, we obtain for the transform U(x,s) of u(x,t) the equation

(18) 'We might alao attempt to transform (14) with respect to x inetead oft; this would lead, In place of (18), to a differential equation of first. rather than second order, but having the disadvantage of involving the unknown quantity urn r—.O

Partial Differential Equations

222

Although this is, like (14), a partial differential equation (since U, like u,

of two variables), it has the advantage of involving difis a ferentiation only with respect to one of the variables, so that it may be solved as an ordinary differential equation. For the general solution of (18) we obtain

+ B exp

+ A exp

U(x,s)

(19)

where the "constants" A and B may depend on s. Before proceeding further with the solution of the problem, we clarify

the ambiguity in (19) arising from the presence of the radical is a two-valued function of the complex variable s, it may be rendered single-valued by restricting s to a simply connected domaiii not containing the origin and then assigning a definite value to the radical at any one point of the domain.' In particular, if a is confined to the half Although

plane Re a > 0. as is the case here,2 we may, and shall, interpret unambiguously as that value which is positive, rather than negative, for positive values of a.

Returning to (19), we shall now determine A. Since u(x,t) was assumed bounded, say

<M, we have, for any positive value of 8,

<M Jo1" e1t di =

S

(20)

If A were different from zero, U(x,s) would be unbounded as a function of x.

Thus A =

0..

Next., performing in the equation U(x,s) =

limiting operation under the integral sign, we obtain, taking account of the first condition of (15), the relation the

(21)

Thus we obtain the explicit formula U(x,s) =

—

exp

(22)

This is a simple illustration of the monodromy theorem of the theory of analytic functions. 2

From (19) it follows that.

integraL

a half plane tontaining the origin in

v(z,t) exp (—et) dl cannot converge in interior, for U(x,a), considered 08 a Laplace

transform, must be single-valued. Of course. thia does not contradict the fact that when analytically continued beyond the ball plane of convergence, becomes a multiple-valued function.

8. The Heat Equation 0 was established only for real values of 8, (22) is also subject to this restriction. However, a similar argument could have been used for nonreal values of s, or, more directly, we may argue that since both sides of (22) are analytic in a half plane and identical on a curve in that region, they are identical throughout the half plane I In order to obtain u(x,t), we apply the inversion formula, obtaining (for positive values of z and t) [Strictly speaking, siisce A

ts(x,t)

hrn

1

fir+$v1i

j

1

exp

1

da

(23)

[— —

a being any positive constant, and the path of integration being the vertical line segment connecting — ir to a + ii. Since we know

Figure 8-1

already (cf. Sec. 1-4) that the inverse transform of s—' is given by the function f(t) I (for t > 0), it remains to invert r' exp At this point, we exploit the fact that the latter function is analytic beyond the half plane of convergence. More precisely, it is analytic in the entire complex plane except the origin, so that if we cut the plane along the negative half of the real axis obtain a function single-valued and analytic in the remainder of the plane. By the Cauchy integral theorem, the vertical path of integration indicated in (23) may be replaced by any other curve which connects the end points a- ± ir and avoids crossing the cut. In particular, it will be found convenient to employ the path of Integration shown in Fig. 8-1.

Partial Differential Equations which

and j First, we shall determine upper bounds o& each approach zero as r —' and suffice to show that

Intro-

ducing polar coordinates on Ui, we obtain —j

+ tR exp (iO)] do

exp

exp

(24)

and hence, by replacing the integrand by its absolute value,

f' exp (—

fCi

Now, as 0 varies between a and cos

f exp

iR

cos 9 tRcos a

any

cos

cos is decreasing on the interval 0 in this interval, the inequality

exp

and hence

do

exp Since

(25)

+ LR C08 0) dO

cos

(26)

i-, we have, for

0

+ (i.

cos

dO

—

(27)

Given 0, we choose such thai — <€ exp (—ta), so that the second term on the right side is less than €, while, for all sufficiently large R, the flrst term also is less than €. Thus <2€ for R sufficiently I

large, and we have therefore shown that

urn

=

0.

Similarly,

limJCi =0. Turning to C, and C4, we note that on these lines is equal to and respectively. Hence, taking account of the directions of

integration, we get (by setting p =

+ fCi = — 2i

JR

—s

=

SN)

'H

— 4i I J

Thus, Jim R-.

+

exists and equals —4i

(Note that the separate integrals

as r

and

v—' sin

xv tr-" dv

r4 sin xv exp (—tv') dv.

do not approach finite limits

0.)

By f we mean fr' exp (—$4z + €1) da .Jccg the indicated curve; a — nrg

— aretan r/q R

+

+

8. The Heat Equation Finally, for C, we obtain

f [r' +

=

ds

(29)

C.

where #(s,x,t) 18, for each fixed (positive) x and I, bounded as 8 approaches

Hence, the integral of the second term of the integrand approacheF = 2ri. zero as r does, and we conclude that tim zero.

We have thUs established the result

st)ds =

'—u-

1

—

V

0

v—1sinxve-"4dv

(30)

Taking account of (23), we obtain as the presumed solution of our problem

j

ts(x,t)

sin xv e" dv

We could now analyze (31) and show directly that it provides a solution to our problem. Instead of doing this, we shall reconcile (3.1) with (16); this will suffice to establish the desired result, for we know that (16) does, in fact, solve our problem. Let us differentiate both sides of (31) with respect to x; as in the preceding section, there is no difficulLy in justifying differentiation under the integral sign, and we obtain

j Tj0 —

c" cos xv dv

Vj_. e" cos xv dv = —

(32)

(The second integral is obtained by observing that the integrand is an

even function of v, while the evaluation of the integral may be performed either by complex integration or by the method outlined in Exercise 4.) Observing that u(0,t) = 0, we obtain by integration

u(z,t) = and the change of variable

e"14' dy

=

(33)

now yields (16).

EXERCISE 9. Replace the second condition of (15) by the following: urn u(x,t) •—. +0

1(x), z > 0.

Gezieralise the method employed in the present section to thia case, and then reconcile the result with (9). (Cf. Exercise 8.)

Partial Differential Equations

226

3. The Finite Rod Let a rod of finite length be ghen, and suppose that at a given instant the temperature distribution is known. Also, suppose that thereafter the ends are maintained at prescribed temperatures, which may vary with time. On physical grounds it appears plausible that the subsequent temperature distribution along the rod is uniquely determined. We now proceed to investigate this question in some detail. The problem under consideration may be formulated as follows:

(O<xO) u(i,t)

u(O,t) = f1(t)

f2(t)

n(x,O)

(34) (35)

g(x)

and f2(t) are continuous for I 0, that g(z) is = Then it may be expected that a solution exists which is

We shall assume that

continuous on the closed interval 0 X 1, and that = g(l).

continuous in the closed region t 0, 0 x 1 of the x,t plane, and that this solution is uniquely determined. The uniqueness question is seUled, as iii the ease of the Dirichiet problem for Laplace's equation (of. Theorem 6-3), by establishing a suitable maximum principle for solutions of the heat equation (1).

Theorem 1. Maximum Principle. Let vcz,t) be continuous in the closed rectangle a x b, T1 t T2 and satisfy (1) throughout the interior. Then the maximum of is assumed at least at one point on the vertical sides (x = a, z = b) or on the bottom (t = of the rectangle. Proof. If the assertion is false, the zna.'dmum is assumed at a point

(xo,to) such that a <xo <

b,

T1 < t0 T2.

Let v

u+

.—

Xo)2,

where e is chosen positive and so small that v, like u, does not assume its maximum on the vertical sides or the bottom. Then v must also assume its mitximum elsewhere in the rectangle, say at If is an interior point of the rectangle, the equality = 0 and the inequality < 0 would hold at this point. However, these conditions would be inconsistent with the identity — Vt

=

— u,

+

2e (>0)

(36)

which holds by virtue of (1) and the definition of v. 'The remaining possibility, that (xj,t1) might be on the top side of the rectangle, cannot be ruled out by the same argument, for no differentiability assumptions are made on the boundary. However, this difficulty is circumvented by the device of replacing T, by a slightly smaller value, say where by "slightly smaller" is meant that u still does not assume its maximum (for the new rectangle) on the vertical sides or bottom. Then the maximum is

The Heat Equation

surely attained on the top of the new rectangle (not in the interior, by the preceding argument). Since u and v are now differentiable along the 0 at the point where the maximum is top, we may still assert that 0 is now replaced by Vt 0. attained, while the previous equality However, this pair of inequalities is again inconsistent with (36). Thus the proof is complete.

If we replace u by —u, we get a repetition of Theorem I with "maximum" replaced by "minimum." We now immcdiately obtain the following corollary, which plays exactly the same role here as does Theorem 6-3 in the theory of harmonic functions.

Given a rectangle and a conCOROLLAaY. UNIQUENESS tinuous function defined on the vortical sides and bottom, there exists at most one function continuous in the (closed) rectangle, the equation (1) in the interior, and coinciding with the given function where the latter is defined. Proof. If there were two solutions, their difference would vanish on

the vertical sides and on the bottom; by the maximum and minimum principles, the difference would then vanish throughout the rectangle.

Having settled the uniqueness question, we turn to existence. Although the general set of boundary and initial conditions (35) can be handled together, it is more convenient to split the problem into several simpler ones, and then treat each one separately. Consider the heat equation in combination with each of the following sets of boundary and conditions, where h(x) douotes the linear function coinciding with g(x) at x = o, x 1. u(O,t) = 0 u(O,t) = g(0) = 0 u(0,t) fi(t)

=

u(z,O)

0

u(l,t) = g(l) — g(O)

u(1,t) = f,(t) n(1,t) = 0

—

g(1)

g(x) — h(x) (37a) h(x) u(x,O) (37b) u(x,O) = 0 (37c) u(x,0) 0 (37d)

if solutions can be found for each of these problems, their sum clearly constitutes a solution to the problem posed at the beginning of this section. The first of these four problems will be solved by reducing it to one of the type considered in Sec. I, the solution to the second can be written down at sight, and solutions to the third and fourth will be obtained by the use of Laplace transform. We now proceed with the details. o

Let f(x) be defined for all real x as follows: f(x) = g(z) — h(x), x 1, f(—x) —f(x), f(x + 2) = f(x). Clearly f(x) is bounded

and continuous for all x, and so we may assert, ;M in SeC. 1, that the

Partial Differential Equatiohs

(unction ui(x,t)

j:. f(E)K(E,z,t)

(38)

dE

is a bounded solution of (1) and that urn ui(x,t) f(x). From the sym= 0, so that metry properties of 1(E), it is evident that u1(O,t) ui(x,t), when restricted to the strip 0 x 1, t 0, solves the first of the four problems. As for the second problem, it suffices to note that u2(x,t) = h(z)

constitutes a solution.

Turning to the third problem, and assuming that a solution u3(x,t) exists, that it admits for each value of x a Laplace transform U(x,s) with respect to 1, and that the formal rules of the transform calculus are applicable, we find that the transform satisfies the differential equation

= with "boundary conditions" U(O,s)

0

U(l,s)

4(s)

= f°•

where, for convenience, we have set ft(L) — g(1) obtain U(x,s) =

#(t).

We readily (42)

By invoking the Laplace inversion formula, we are led formaUy to the presumed solution us(x,t)

1

f

e"(J(x,s)

da

(43)

However, this formula may very well .be meaningless, for f(s) may fail to exist, as in the case #(t) exp (t'). This objection may be overcome by performing another formal operation. Assuming for the moment (we

shall justify this assumption later) that there exists a function w(x,t),

defined for 0 x < 1, t 0, whose transform is equal to sinh sinh s½, then the convolution theorem (Exercise 1-27) suggests, because of the form of (42), that u3(x,t) is the convolution of with w(z,t). Thus, instead of (43), we are led to consider the formula ua(x,t)

t — r) dT

(44)

We note that (44), In addition to involving no assumption concerning the

existence of the transform

is consistent with the uniqueness theorem

229

8. The Heat Equation

established earlier in this section, for the value of us at any instant in La (> 0) depends, according to (44), only on Uie defithtion of t

the interval 0 I to, not on its behavior for I >

We now turn to the existence of the function w(z,t) appearing in (44). For the present, the only properties of the function sinh xs½/sinh 54 thaI we shaU use are that, for each fixed value of x, 0 x < 1, it is analytic in > 0, the half plane Re s > 0 and that, on any fixed vertical line Re $ r), given of Tm a (= its modulus is, for large (positive or negative) values by + o(1)1

(45)

I

where 0(1) denotes a function of i which approaches zero as r becomes infinite. It follows that, for all values of x in the afore-mentioned range and for all values of I, it is meaningful to define a. function w(x,t) as follows: w(x,t)

J

nh sinh

(46)

Also, w(x,t) is continuous in both variables, vanishes identically on the line x 0 and on the segment 0 x < 1, t = 0. (The behavior of w(x,t) as x —' 1 is not immediately apparent from (46), but will be Furmined later after an alternative definition of w(x,t) is thermore, any number of differentiations with respect to x and t under the integral sign in (46) are easily justified because of the rapid approach of the integrand to zero for large r, as indicated by (45). Since a trivial computation shows that the integrand of (46) satisfies the heat equation, it then follows that the same is true of w(x,t). In order to investigate the behavior of w(x,t) near the line x = 1, we find it helpful to evaluate the right side of (46) by the method of residues. We note that the integrand is single-valued, despite the appearance of the radical st. This follows from the fact that the Maclaurin expansion of the function sinh z contains only odd powers of a, so that both sinh and sinh may be expressed as the product of and a series containing only powers of s. On the other hand, the presence of infinitely many poles necessitates a certain amount of caution. Noting that the singularities are all simple poles, situated at S (ii = 1, 2, 3, . . we select the parabolas

Re a

—

(n

—

+ (2n—

Tt

(47)

and define the closed contour C,, as the curve consisting of the portion of the vertical line employed in (46) lying inside the parabola (47), together with the portion of the parabola to the left of the line. (Cf. Fig. 8-2,

Partial Differential Equations

where for the sake of clarity we have selected a particular value of n, denote the vertical and parabolic namely, n 4.) If we let I, and respectively, we obtain by the method of residues the portions of following result:

ds =

ea

2Tk(— l)*+1 sin

kTx e"

(48)

—' 0. (Cf. Exercise 12.) Hence letting n —+ we readily find that the left side of (48) approaches w(x,t), and we obtain the series eKpansion

w(x,t)

=

2i-k(—- i)k+L sin kwx

(49)

For t o, t0 any positive constant, the series (49) is dominated by the convergent series of constants

2rk exp (— ktir2to), so that (49) con-

verges, for all positive £ and all x, to a continuous function. Since each term of (49) vanishes for z 1, we 4'ouclude that w(z,t), which was

Pigure 8-2

originally defined in the region t> 0, 0 x < 1, becomes continuous in the region i > 0, 0 x 1 if defined to vanish on the line t > 0, x = 1. Now, returning to (44), we immediately see that ua(x,t) is well defined,

continuous, and a solution of the heat equation for 0 <x < 1, t > 0, and continuous for 0 x < 1, £ 0, if defined to vanish for z 0 and also for t 0. It we formally set x = 1 in (44), might that Us vanishes, but this reasoning is inv-alid, on account of the highly singular behavior of w(x,t) near the point x = 1, t = 0. In order to overcome this difficulty, we consider the particular case n any positive integer. In this case #(t) does indeed possess a transform, given by and we can show without difficulty that (44) is entirely equivalent to (43).

231

8. The Heat Equation

that u3(x,t) possesses the Now, from (43) we can very easily 1, namely, thst ua(x,t) approaches to" proper behavior near the line x approaches the point (l,t0), to 0. By superposition, as the point (x,t) furnishes the solution to the problem whenever we conclude that (44) Now, for arbitrary continuous is a polynomial vanishing at t = 0. 4410 vanishing at t = 0 and any T> 0 we can, by the Weierstraas

in the interval 0 t T by a approximation theorem, approximate each of which also vanishes at t = 0. sequence of polynomials {

(x,t). Hence, by construction, the functions u3(")(x,t) converge uniformly on the

For each

we obtain by (44) a corresponding solution,

vertical sides and bottom of the rectangle 0 x 1, 0 £ T, from which we conclude by the maximum principle that they converge unifornily in the interior as well. The limit function, which we term clearly satisfies the prescribed conditions (37c). It remains only to show that u2(x,t) satisfies the beat equation in the interior. However, from the fact.. that w(x,t) is continuous for 0 x < 1 and all values of t, it readily follows that thc function us(z,t) coincides with the function defined

by (44), and there is no difficulty in justifying differentiation under the does satisfy (1). integral sign, from which it follows directly that Thus, a solution of (1) satisfying (37c) has been obtained. Replacing x by I — x and f2(t) — g(1) by f1(t) — g(O), we obtain a solution of (1) satisfying (37d). Thus, a complete solution has been obtained for the problem posed at the beginning of this section. EXERCI S ES

10. Prove that the function w(c,tj Ueftned by (46) vanishea in the region 1 <0,

0 z <1.

11. Show that w(x,t) may be interpreted as a limiting case of the temperniure distribution r'aulting when there is applied to a rod initially at temperature zero a specified amount of beat at the right end over a very Mhort time interval, after which this end is again maintained at zero while the other end is maintained permanently at zero temperature.

12. Prove the assertion made following (48) that

—'

0.

13. Consider the "backward heat equation," = — ti,. with the conditions 1,1) 0, t. Show that this problem Jeada to a function U(x,s) which cannot be a Laphtce transform. flint: ()bbcrve the location of the

u(O,t) — 0, u(x,0)

oingularitiee of (J(x,s).

9. GREEN'S FUNCTiONS AND SEPARATION OF VARIABLES

Among the most. usefuL and frequently employed methods of solving certain types of problems in partial differential equations is the one known "separation of variables." The reader is probably acquainted with the manner in which prol)lerns dealing with vibrating strings and circular membranes lead, by the use of the afore-mentioned method, to expansions

in series of trigonometric functions and Bessel functions, respectively; similarly, the Dirichlet problem for a sphere leads to an expansion of the solution in a series of Legendre functions (spherical harmonics). We shall be only incidentally concerned with the particular functions arising in the study of such expansions. Rather, our primary interest will be in developing the theory underlying such expansions.

1. The Vibrating String with a brief review (from a It is convenient to begin our viewpoint different from that of Sec. 3-5) of the familiar prob1em of the vibrating string. Under suitable hypotheses and with suitable choice of physical units, the vibrations of a uniform string which is held fixed at both ends in the rest position are described by the one-dimensional wave equation (1)

and the "boundary conditions" u(O,t)

u(1,t) 232

0

(2)

9. Green's Functions and Separation of Variables

233

At a given instant, which without loss of generality may be taken as 1 0, let the shape and velocity of the string be given by the functions f(x) and g(x), respectively. Then the study of the subsequeut motion of the string reduces to the determination of a function u(x,t) defined and continuous 0 and satlsfyJng equations (1), (2), and the "initial for 0 x 1, 1 conditions" (3) = g(x) u(x.,0) = f(x)

Disregarding (3) temporarily, we seek a function which satisfies (1) and (2) and expressible as the product of functions each depending on only one of the independent variables; i.e., we seek a solution of (1) having the form

u = X(x)T(t)

(4)

Substituting from (4) into (1), we obtain TX"(x)

XT"(t)

which is equivalent to

Since the left and right sides of (6) are independent of I and x, respectively, they must both reduce to constant, which we term —). Thus, X(x)

and T(t) satisfy the (ordinary) differential equations (7a) (7b)

In order that (2) be satisfied, it is necessary that

1(0) = X(1) =

0

(8)

Trivially, the function X 0 satisfies (8) as well as the equation (7a), regardless of t.he value of the "separation constant" X; however, a nontrivial function satisfying these two equations exists only if )t is one of the quantities n 1,2, 3, . . . , and for any one of these values of X the corresponding solution must be of the form

1(x) constant sin iwx (9) The proof of these facts is left to the reader as Exercise 1. The correspondmg function T(1) must then be of the form

T(t) = constant cos nfl + constant . sin n,1

(10)

and (4) therefore assumes the form u(x,t) sin flirZ (constant 008 twt + constant sin nrt) (11) Since (1) and (2) are both linear and homogeneous, it follows that any finite combination ot functions of form (11) will also satisfy the above pair

Partial Differential Equations

234

It now appears plausible to attempt to form an infinite combination of all the particular solutions which have been obtained of equations.

initial conditions (3) as well. which will satisfy not only (1) and (2) Temporarily disregarding questions relating to convergence, we assume a solution of the form it =

cos ni-t + b,

sin

nirt)

(12)

and obtain from (3) the conditions f(x)

and

(13a)

a,, sin

=

(13b)

sin

We recognize that these last two equations represent "half-range" trigonometric expansions, and so we obtain for the coefficients the formulas ti 2 1 f(x) sin b,. = — ( g(x) sin dx dx (14) Jo fl'jo Now, if the functions f(x), g(x) are such that the coefficients a,,, b,, approach zero sufficiently rapidly with increasing n, series (12) will indeed converge to a function continuous for 0 x 1, t 0 and will permit term-by-term differentiation, so that (1) and (2) will be satisfied. (For example, suppose that f(z) and g(x), when defined for all x as odd functions of period 2, possess continuous derivatives of fourth and third orders respectively. Then, by an elementary argument,1 and b,, cannot exceed Cn—4 for some sufficiently large constant C, and then there is no difficulty in justifying the term-by-term differentiations of (12) needed to show that (1) is satisfied.] Furthermore, from the general theory of trigonometric series, expansions (.13a) and (13b) will actually converge to

/(x) and g(x), respectively, so that the initial conditions (3) will be satisfied.

It should be stressed that the essential feature of the above method is that the family of functions obtained by determining all nontrivial solutions of (7a) which satisfy (8) is "large" enough to permit any "reasonable" function to be expressed in the form of a linear combination of these particular functions. Now, while this fact is known, as indicated above, from the theory of trigonometric series, which may be developed 1

Rewriting a1, in the form

we C

—

obtain

a,

max

(ni'r)'

f(x) sin

dx

f"(x) sin

dx,

similarly for

b%.

and integrating by parts four times, and

hence

tap! Cn', whas

and Separation of Variables

9. Green's

quite independently of equations (7a) and (8), the question arises whether the functions obtained by replacing this pair of equations by somewhat

different equations would still possess the afore-mentioned property. As suggested by the introductory remarks to this chapter, a satisfactory theory concerning such systems of functions exists, and we shall begin by demonstrating in the next section, quite independently of the theory of trigonometric series, that expansions of the form (13a) for any "reasonable" function are indeed possible. EXERCISES 3. Prove the assertions made in the sentence following (8).

2. Prove, without employing the explicit form of the general solution of (7a), that those values of X (if any) for which (7a) admits a nontrivial solution atisfying (8) must be positive. H(nt: Multiply both sides of (7a) by X, then add [X'(x)]t to each side. 3. Obtain (12) from (1), (2), (3) by Laplace transform.

2. The Green's Function of the Operator Let f(x) be any function of C2 defined on the interval 0 X 1 and vanishing at both end points. Then f(x) is unambiguously determined by its second derivative,f"(x); for if g"(x) f"(x), it follows that g(x) — f(x) must be a linear function, and hence, if f and g each vanish at both end points, their difference must vanish identically. Now, we shall derive an integral formula expressing 1(x) in terms of its second derivative, which, for convenience, we temporarily denote by —h(x). Jntegrat.ing the equation

f"(x) =

twice, we obtain

f(x) =

—h(x)

(15)

—

(16)

where c1 and C2 are arbitrary constants. The conditionsf(0) yield for these constants the following formulas: c1—

flfth(E)dEdt

C20

f(1)

0

(17)

If we interchange the order of integration in (16) and (17), we obtain —

Jo

IzJzh(E)dtdE+flflh()&d

(18)

Upon performing the inner integration in each case, we obtain f(x)

=

—

x)

+

h.(E)(x — Ex) dE

(19)

Partasl Differential Equations

236

or, finally,

= where

f'

J(l—x)E was studied in Sec. 5-8 in order to illustrate Now, the funcUon the expansion theorem for a continuous symmetric kernel, and it was shown that its eigenfunctions are precisely the functions given by (9), so that any lunction expressible in form (20) [with some continuous h(x.)J can also be expanded in a uniformly convergent series of the functions (9). As in Sec. 5-8, we stress here the method of proof, rather than the strength of the result, for it. is well known that the requirement thatf(x) be of class Ce is far more than is actually needed for this result to hold. While the existence of a representation of form (20) of f(x) is essential in the above considerations, t.he fact that 1(x) and 4(x) are related in the precise maniicr defined by (15) is of no importance. For example, let

three continuous functions, p(x), q(x), r(x), be defined on the interval a x b, and suppose that there exists a continuous symmetric function such that every twice continuously differentiable function f(x) vanieJiing at the ends of the interval can be expressed in the form f(x)

fb

h(E)K(x,E) dE

(22)

where

Lf pf"(x) + qf'(x) + ,1(x) —4(x) (23) Then, exactly as in the above example [where p 1, q r 0, a = 0, b = 1, K(x,E) the function /(x) will be expandible in a forinly convergent series of eigenfunctions of It will be shown in the following section that, under certain conditions on the functions p, q, and r, such a function exists. This function will be termed the

"Green's function" of the differential operator p particular, the function

the operator

+q

+ r; in

considered above is the Green's function of

(for the interval 0 x I). EXERCISES

4. Prove that the function G(x,E) may be characterized as follows:

(a) (7(0,E)

0. G(1,t) (b) G{z,E) is continuous in x for each fixed

(c) (d)

— 0, x

urn

—

thu

—1.

5. Prove that (20) is entirely equivalent to (1 tionaf(O) —f(1) 0.

together with the boundary condi-

9. Green's Functions end Separation of Variables

3. The Green's Function of a Second-order Differential Operator In this section we shall consider the problem of representing a function f(x) in the form (22), where h(x) is related to 1(i) according to (23). A.q in See. 2, it is understood that f(.c) is of 'chss C2 in the closed interval x x3, and vanishes at both ends. I: Xt We now impose on the functions p(x), q(z), r(x) appearing in (23) the following restrictions: 1. p(x), q(x), and r(x) are continuous in I, 2. p > 0 throughout 1. 3. p'(x) q(x), so that

p.f"(x) + aJ'(x) = Epf'(x)l' 4. The only solutAon of the equation

Pu" + qu' + ru

(24)

0

vanishing at both ends of I is the trivial one, u 0. (This condition is satisfied, for example, in the caae considered in the preceding section.) Now, letf1(x) denote the (unique) solution of (25) which vanishes at x1 and whose first derivative there is equal to unity. Similarly, let fs(x) denote the solution satisfying the same conditions at Then. the so-called "Wronskhtn determinant" W(f1,f2)

11

,r

(26)

f1.t! —

never vanishes in the Therefore, given any point. inside 1, there exist uniquely determined quantities c and c2( such that cJi(E) —

0

—

= Indeed, denoting the constant (ef. footnote) value of obtain = Cf2(E) = Now, consider the function defined as follows: K( E) — —

Cfs(E)f1(x)

t

X' E

z E X

X2

by C, we (28)

29

1(pWY a.F,(p1)' —f(--rf,) —ft(—th) 0. fence, pW constant., and therefore W vanishes everywhere or nowhere. To rule out the former possibility, we note that W retiuces at to .-f,(z,), so that if W vanished therc, f,(z) would be a nontrivial solution of (25) vanishing at bcth ends of I, contradicting condition 4.

Partial Differential Equations

238

K(x,E) is evidently continuous in the closed square x1 x, E symmetric in its arguments; i.e., K(x,E) = x3

x, and (30)

Let u(x) and v(x) be any two functions of class C' defined on an interval x x4 contained in the afore-mentioned interval I. Then,

denoting for brevity (pu')' + ru by L(u), we obtain

L' [uL(r) — vL(u)] dx

L' Lu(pv')' —

dx

(upv' — vpu')' dx = (upv'

vpu')

—

Now let E be any inner point of I and let us apply (31) twice, first to the interval x1 x < and then to the interval E x x,, replacing v(x) each time by K(x,E). [Note that in each of these two intervals K(x,E) is of class Ct, although in the entire interval 1 it is not. even of class C'.) We obtain, since L(v) = 0, —

and

L(u)K(x,E) dx = [upK2(x,E)

—

dx =

—

—

K(x,E)pu')

J'°

(32) (33)

Adding these two equations and observing, with the aid of a simple computatio;t based on (29) and the definitions of C and of the functions

andf,(x), that

K,(x,E) 1

—

1

=

—

J

+ u(s)

(34)

(35)

We note that (35) furnishes the solution to the differential equation L(f) = (p1')' + rf = —h(x) (36) with prescribed values of f(xj) and f(x,), namely, h(x)K(x,E) dx + •(37) = I:: In paiticular, if f(x) is required to vanish at both ends of 1, (37) reduces 1(E)

to (22). We conclude this section by giving a definition of the Green's function K(x,E) of the operator L which, in contrast to (29), emphasizes essential properties of this function, rather than the specific manner in which it is constructed. K(x,E) is now defined as follows: For each inner point E of

I, it is a continuous function of x, vanishing at both end points, and

9. Green's Functions and Separation of Variables is satisfies, for all values of £ except E, the equation L(K) = 0; nature of the discontinuity being given by (34). discontinuous at the

EXERCISES 6. If condition 3 immediately preceding (24) is omitted, (31) no longer hold.. Show that, in ite place, the following identity holds:

f

IuL(v)

dx —

—

+ piw' — (pu)'v)

where M, the "adjoint" of the operator L, is given by M(u) — pu" + (2p' — q)u' + (p" — q' + r)u

7. Prove that the adjoint of the operator M defined above is I,. 8. Prove that M is uniquely determined by the requirement that, for arbitrary (sufficiently differentiable) functions a and v, the left side of (38) shall be equal to a "boundary term"—i.e., an expression involving only the values of u and v and their first derivatives at the ends of the interval. 9. Relate the present use of the term "adjoint" to its use hi Chap. 5. 10. Let the operator L satisfy conditions 1 and 2 but not 3. Prove that there exists a positive function p(x) defined on I such that the operator p(x)L is seif-adjoint.

11. Prove that, if r(x) 0 throughout I, then K(x,E) > 0 throughout the open square xi <x, E

4. Eigenfunction Expansions Let the differential operator L satisfy conditions 1 through 4 imposed in

the preceding section, and let p(x) be a given function continuous and positive in the interval I. Then the "boundary-value problem"

L(u) + Xpu = 0 u(x1) = = 0 admits, for each value of the parameter A, the trivial solution u

/ However, as shown in Sec. 1

L =

p

1, x1 = 0, $2

0.

there

may exist certain particular values 'of A for which (40) admits a nontrivial solution. A comprehensive theory concerning the existence of these special values of A (eigenvalues) and the properties of the corresponding nontrivial solutions of (40) (eigenfunctions) is most readily developed by converting (40) into an integral equation. According to the results of the preceding section, any solution of (40) satisfies the equation

u(E) =

A

K(x,E)p(x)u(x) dx

(41)

(22)), where is the Oreen's function of L (for the interval x1 z x,). Conversely, if u satisfies (41), it also satisfies (40) (i.e., the boundary conditions as well as the differential equation). If p(x) 1, [ef.

we would conclude that the problem of determining eigenvalues and

Partial Diflerentia) Equations

240

eigenfunctions for (40) is equivalent to that of determining the eigenvalues

By a 8Jmple procedure, we shall show that and elgenfunctions of even when p(x) is not 1, it is possible to interpret the eigenvalues and eigcnfunctions of (40) as being, aside from a simple factor, the eigenvahtes and eigenfunctions of a symmetric kernel closely related to X(x,E). Let

= Then (41) assumes the form

=

X

f'

(43)

dx

is clearly continuous and symmetric (and not 0), we know that. it possesses a finite or denumerable set of eigenvalues X2, X3, . . and eigenfunctions Now, letting f(x) Since

.

denote any from (22) that

of class C2 vanishing at both ends of I, we conclude

=

dx

By the expansion theorem of Chap. 5, formly convergent series,

(44)

is expressible as a uni-

= the coefficients cft being given by the familiar formula

lIt is understood that the

are taken as an orthonormal set, = we obtain in

1sctting = place of (45) the entirely equivalent expansion

f(E) = while (46) may be replaced by

(47)

(48) = It may be noted that we have not indicated whether the summations in <45) and (47) are finite or infinite, but this matter is easily settled. If there were only a finite number of eigenvalues and eigenfunctions, the set of functions. of class C2 vanishing at both ends of I would constitute a finite-dimensional subspace of Lt(x1,x2) (ef. Chap. 5). In fact, how-

ever,

afore-mentioned functions constitute a dense subset of 1i2(x,,x2),

and so it follows that the functions 1z,,(E) actually constitute an orthonormal basis of (Cf. Exercises 13 and 14.)

9. Green's Functions and Separation of VarIables

241

We have, therefore, shown that the boundary-value problem (40) of for a denumerable sequence admits nontrivial solutions values of the parameter A. Any sufficiently smooth function vanishing at both ends of I admits the uniformly convergent expansion (47), the are coefficients c,, being given by (48), provided that the functions }

normalized as indicated following (46). Although the foregoing treatment employed the existence of a Green's

function, which, in turn, was based on the assumption that (40) admits only the trivial solution for A = 0 (i.e., that A 0 is not an eigenva.lue), the final result involves neither the Green's function nor the eigenva.Lues.

It therefore appears plausible that the proof might be modified so as to eliminate this assumption. This is indeed so, and is easily proved as follOws: The eigenfuuctions of (40) are not changed if L is replaced by L + Cp, and the cigenvalues are merely reduced by c, where c is any constant. Let c be so chosen that L + ep possesées a Green's function

(cf. Exercise 18); then the above argument is applicable, and we obtain the fact that the functions form an orthonormal basis in L2(xi,x2). EXERCISES 12. Prove that the Green's function K(x, cannot be degenerate. 13. Prove, as asserted following (48), that the set of functions of claea, C' vanishing a dense subset. of Hira: Approximate an arbitrary member of J',(zi,x,) in norm by a continuous function; then approximate the latter by a smooth function satisfying the end conditions.

at both ends of I

14. Prove that the functions Un(E), like il,,(t), constitute a basis for will not, in general, constitute an orfiurnonnul baeis.l 15. Prove that, if r 0, all the sigenvalues of (40) are positive. 16. Let and p be any distinct riunihers, X and let u and v be nontrivial solu-

Of course, the functions

tions of the equations L%l + — 0. Lv + Lpe — 0, respectively. Prove that between any two zeros of u there must exist at least one zero of v. Then use this result to prove that (40) can possess only a finite number of negative eigenvahiea. so that the elgenvalnes may be enumerated in ascending order, <)2 < • 17. Prove t.hnt the eigenfunction itt corresponding to the lowest elgeuvalue (of. preceding exercise) does not vanish in the interior of the interval 1. Hint: First make the additional assumption r(x) 0 and use Exercise 1; then show that this assumption may be dropped.

18. Prove, as asserted at the end of the present section, that the constant c can be so chosen that zero is not an eigenvaluc of the operstor L + cp. Hint: Exploit the fact that is separable, while the real-number system is uncountable.

5. A Generalized Wave Equation We shall now show bow the method of separation of variables can be used to solve boundary-value problems involving equations more dif-

ficult than (1). Let a(x), b(x), c(x) be defined and continuous on a bounded closed interval I: x1 r x,, and let a and 8 be two given

Partial Differential Equation

242

constants.' Let us consider the equation

+ cu =

+

(49)

ati,g +

together with the "boundary conditions" u(x,,t)

and "initial

u(x2,t)

(1 0)

0

conditions"2

u(x,O) = f(x)

u1(x,0) = g(x)

(49) contains (1) as a particular case, we may describe the former as a "generalized wave equation." Proceeding as in Sec. 1, we seek a function satisfying (49) and (50), obtaining for X(x) and T(t) the ordinary differential equations: Since

aX"

+ cX + XX = aT" + p7" + XT =

+ bX'

0 0

a "separation constant." Now, assuming that a is positive throughout I, we. multiply (52) by a suitable positive function p(x) such that the operator where

as in (7), denotes

d

d2

is ae]f-adjoint (of. Exercise 10). section, the cigenfunctions that the formal expansions

results of the preceding of (52) form a basis in Lz(zi,x2), so

Then, by the }

g(x) =

f(x)

(54)

gi,4n(x)

will actusily converge to f(x) and g(x) if, as we henceforth assume, the coefficients approach zero with sufficient rapidity as a —* oo. Let the functions (1) and be defined as the solutions of

aT" + j3T' 4

= 0

(55)

satisfying the conditions

= respectively.

0

1

0

1

(56)

Then the series u=

+

(57)

At least one of these is assumed different from zero, for otherpise the variable Actually, a and could be allowed to depend on t withoit causing any essential change in the following considerations. 'If a — 0, the second condition in (51) must be dropped. I

C

disappears from (49).

243

9. Green's Functions and Separation of Variables

Now, in made on the coefficients appearing addition to the assumption previously In the expansions (54), we assume that they approach zero with sufficient rapidity to assure the convergence of (57) for all £ > 0, and also to permit termwise differentiation of this series as many times as may be required. Then (57) will also satisfy (49) and the second initial condition of (51), and thila will provide a solution to the problem under consideration. It may be noted that, though the functions #(x)) and the coefficients only on the left are independent of a and $ and aide of (49) and on the data (51)J, the convergence of the series (57) is

certainly satisfy (50) and the first of the initial conditions (51).

= 1, a 1, b 0, influenced by a and 0. Thus, when x3 0, 0, there is a marked contrast between the cases a = 1, = 0 and c 0. Tn the former, we have, the problem considered in a —1,

Sec. 1, where we saw that very mild hypotheses on the functions f(z) and g(x) suffice to guarantce the existence of a solution. In the latter, on the other hand, we have the Laplace equation, and the functions and

are given by cosh

sinh nrt, respectively.

and

Since these functions become large, either for fixed n and large t or vice versa, much more stringent conditions must be imposed to assure the convergence of the series (57), even for some small range of the vanable t. This difference reflects the fact, pointed out in Chap. 3, that the Cauchy problem ia meaningful for hyperbolic, but not for elliptic, equations. EXERCISE

19. Consider the system of equations (40),. (50), (51) with a

4, b C 0, and for positive and the behavior of the functions (A physical interpretation of as a frictional for negative values of the constant —

1.

coefficient ii helpful.)

6. Extension of the Definition of Green's Functions In this section we indicate briefly how the definition of the Green's the scope of the method of separation of function can be broadened, variables correspondingly enlarged.

1. In the previous sections of this chapter the only boundary conditions that have been considered are the simplest possible ones, namely, that tbe functions under consideration vanish at both ends of the given interval. While the linearity of these conditi9ns (i.e., the fact that c1u1 + c,u2 satisfy them whenever and u, do) is essential in the of separation of variables, there is considerable flexibility in the nature

of the conditions that can be handled. For example, suppose that in Sec. I the conditions (2) are changed to the following: u(0,t)

u3(1,t)

0

PartIal Differential Equations

244

The method of separation of variables leads in this case to the functions sin (n — , which are the eigenfunctions of the symmetric kernel

lx [or, more concisely, mm (x,E)], just as in the previous case we were led to the functions {sin nwx}, which are the eigeitfunctions of the

symmetric kernel

defined by (21). Furthermore, just as (20) furnishes the solution to (15) with conditions f(0) = 0, the solution to (15) with conditions f(0) = f'(I) = 0 is given by (20')

= Jo'

Also, is uniquely determined by the conditions listed in Exercise 4 except for an obvious change in part (a). Therefore, the whole discussion of Sec. 1 carries over with straightforward modifications to the present problem. We term in analogy with G(x,E), the Green's function of the d2 operator for the boundary coziditions 1(0) = f'(l) = 0. More gen-

erally, given a differential operator L, as defined by (23), and any pair of linear boundary conditions, we can associate with L, by an obvious modification of the procedure carried out in Sec. 3. a Green's function for the given boundary conditions; as before, it is necessary and sufficient that the only solution of the equation = 0 which satisfies the boundary conditions is the trivial one u 0. Actually, the assump'ion that L is seif-adjoint is not essential insofar as the existence of is concerned; however, as might be expected, the se]1-adjointness of L is essential for the symmetry of K(x,E), without which a satisfactory theory of expansion in series of cigetifunctions cannot be obtained. One might expect that the self-adjointness of L is sufficient to assure the symmetry- of but this is not correct, as is indicated by the followd2 ing example: Let L and let the boundary conditions be given by a—i'

f(0) + f'(l) = f(1) + 2f'(O) =

0.

Then an elementary computation

furnishes for the Green's function the formula K (x,E)

(59

it is readily seen (by interchanging and x in either of the above expressions and comparing the result with the other expression) that is not symmetric. 2. We now indicate how Green's functions can be associated with differential operators of order higher than the second. We shall conand

245

9. Green's Functions and Separation of Vaxtables

sider only the simplest of all fourth-order differential operators, rntxnely,

together with the linear boundary conditions f(O)

J'(O) =

J(l)

f'(l) =

0

Thea, by an elementary computation entirely analogous to that performed in Sec. 2, we find that the (unique) solution of the equation (60) satisfying the afore-mentioned boundary conditions is given once again

by (20), where G t

—

.v)2(3x

—

— 2Ex)

E x I

The symmetry of G(z,E) (which, of course, is termed the Green's function

of the operator

for the given boundary conditions) assures us, exactly

as is Sec. 2, that its cigenfunetions span L2(0, I). Thus, in complete. analogy with the manner in which separation of variables was uaed successfully to solve the system consisting of equations (1), (2), and (3), we may now solve the systhm consisting of the following equations (which play the same role in the theory of vibrating beams as that played by the former system in the theory of vibrating strings):

+ u(O,t)

u(x,0)

u?(0,t)

0

E

u(1,t)

= g(x)

f(x)

0

(62) (63) (64)

The eigcnfunctions and elgenvalues of the Green's function are obtained

in the present case as follows: We write out the general solution of the equation X" — =0 in the form1

+ h cos X = a sin + c sinh + d cosh When the boundary conditions are imposed on (66), we find, by an

elementary computation, that the coefficients a, b, c, d must all vanish unless A satisfies the transcendental equation cos cosh = 1 (67) This equation is readily seen to have infinitely many real roots, for the

left side vanishes when A

(2n —

and

exceeds unity when

'For x — 0, formula (66) breaks down, but in this case the gencrel solution of (65) is given by an arbitrary cubic pclynomial, and it is easily shown that no such (nontrivial) function can satisfy the boundary conditions.

Partial Differential EqualionB A = (2nA)4, n = 1, 2, 3. [Of course, the existence of infinitely many eigenvalues is also assured by the fact that the cigenfunctions must span

3. Finally, we consider briefly the problem of associating a Green's function with a partial differential operator and of using it to justify the

application of separation of variables to the solution of partial differential

equations in three or more variables. Only a single example will be considered, but this should suffice to indicate how more difficult problems

may be treated. It will be seen that, though the same basic ideas are employed as in the previous sections of the present chapter, a certain difficulty arises from the fact that the Green's function encountered here is unbounded. We consider the two-dimensional wave equation

together with the boundary condition

fortO

onx2+y2=l

(69)

and the initial conditions u(x,y,0)

f(x,y)

ug(x,y,0) = g(x,y)

+ y2 < 1)

(70)

This system, an obvious analogue of the system (1), (2), (3). describes the vibration of. an elastic circular membrane clamped in rest position along

the circumference, the displacement and velocity of each point of the membrane being specified at t = 0. We seek nontrhial solutions of (68) wbióh satisfy (69) and are of the form u = r(x,y) T(1)

By substituting (71) into (flS), we are immediately led to the pair of equations

+

=

0

(72a) (?2b)

T" + AT 0 Proceeding formally, we temporarily assume that (72a) admits, for a denumerable set of values of A, nontrivial solutions which vanish on the unit circle [so that (71) will satisfy (69)], and that these

functions constitute a basis for the hubert space of functions quadraticaily integrable 0 the unit disc D. g(x,y) are expanded in series,

Then the functions

f(x,y) = g(x,y) —.

and (73a)

b,,v.(;y)

(73b)

247 9. Green's Functions and Separation of Variables and the 8olution of The given problem is presumably represented by the

series cos

u

+

sjn

(74)

It is now necessary to justify the statements made in the preceding paragraph. First, by comparing (72a) with (6-57) and taking account of the discussion at the end of Sec. 6-12, we conclude that any function

which satisfies (72a) in D and vanishes on the circumference must satisfy the homogeneous integral equation

v(P) =

ff v(Q)G(P,Q)

(75)

where (i(P,Q), the Green's function of D, is defined by (6-24). (The step from (72a) to (75) is justified by the Holder continuity of v throughout .1), which is certainly assured by the fact that v satisfies (72a).J Next we shall show, conversely, that any (quadratically integrable) solution of (75) satisfies (72a) and vanishes on the circumference of D. We shall use the

following two facts, proofs of which are deferred to the following paragraph: (1) fl,, ffD G2(P,Q) dE di1 dx dy is finite, so that G(P,Q) and each of its iterates may be considered as a completely continuous operator the first iterate of G(P,Q), is con(of. Sec. 5-6); (2) in tinuous (jointly) in P and Q as these points vary throughout the closed From (75) and the first of the above assertions, it follows that v disc also satisfies the integral equation v(P)

/X\2

ft

JJ = It now follows from (76) and the second of the above assertions that v is coatinuous throughout D; returning again to (75), we conclude, by the argument presented in Sec. 6-12, that v vanishes on the boundary of D and satisfies (72a) throughout 1). In this manner we have established the

complete equivalence of the integral equation (75) with the partial ferential equation (72a), together with the condition that v shall vanish on

the circumference. Now it was shown in Sec. 6-12 that every function sufficiently differentiable in D and vanishing on the circumference is representable in the form (6-97).

Since such funetions are dense in L2(D)

(of. Exercise 20), it follows from the theory of hermitian completely continuous operators presented in Sec. 5-4 that the eigenfunctions

of

the symmetric kernel (2w)1G(P,Q) constitute an orthogonal basis of L,(D). Hence, if the sequences { a,. {b. of coefficients appearing in (73a) and (73b) approach zero with sufficient rapidity, these series will

Equations

Partial actually converge throughout 1) to f and g, respectively. may be obtained by the usual formulas, namely,'

= =

= JJD

The coefficients

dx dy

= JJD g(X,y)Vn(X,y) dx dy Thus, as in the simpler problems considered earlier in this chapter, series (74) will furnish, under suitable restrictions on the functions f(x,y) and g(x,y), the solution to the given problem. It should be pointed out that the eigen values { X,, are all positive, so that (74) actually involves trigonometric, rather than hyperbolic, functions of t. The proof runs as follows: If the nontrivial function u satisfies (72a) and vanishes every(Q,Vn)

where on the circumference, then it must have a positive maximum or negative minimum (or both), and, in either case, the extremum would be assumed in D, not on the circumference. At a positive maximum we find from (72a) that is opposite in sign to X; sinco v and must both be nonpositivc at an interior maximum, it follows that X is nonnegative. A similar argument applies for a negative minimum. The possibility that vanishes is ruled out by the maximum and minimum principles for harmonic functions. the only tenable conclusion is that X must be. positive. It remains to prove the two statements that were temporarily assumed in the preceding paragraph. As for statement 1, we exploit the inequality

G'(P,Q) 3(logt

+ log2

+ log'

which follows by applying the Schwarz inequality2 to (6-24) (with R 1). Introducing polar coordinates r, 0 and p, for P and Q, respectively, we obtain = J2IFJl log2 d, (log p)tprdp dr do (79) JJD ff,, .17

This integral obviously exists.

To show that

liD (liD log'

dx dy

exists, we replace the. region D in the inner integral by the disc of radius 2 with center at P. Since the latter disc contains D, we see that the integral under consideration is dominated by

lb (IL log2

dx dy

Since the inner integral exiSts and is independent of P, we conclude that 'The functions v,1

assumed normalized, so that

t(j •a+ I •b + I •c)' (I' +

+ 13)(a' +1,' +c2).

—

and Separation of Variables

9. Greeu'a

the dominated integral exists. JJD

249

Finally, we consider

(lID Iog2

dy) dE

In the inner integral we replace the region D by the circular sector S with vertex at Q which circumscribes D. (Cf. Fig. 9-i.) Introducing polar

Figure 9-1

coordinates with origin at Q*, we obtain ff3 log2

dx

+ "i

dy =

(log r)tr dr do

sin a = p] From the inequalities a sin a (cf. Exercise 22) and 1 + i/p 2/p we find that the integral appearing in (80) is dominated =

[p

by f 2/p which works out to

/

(logr)2rdr

2 2 log — — 2 log— + 1

The fourfold integral under consideration is

f

(2

log

dominated by

+ i) dp

which is readily seen to be finite. Thus the proof of statement I is complete.

Turning to statement 2, we write G(2)(P,Q)

JJD G(P,W)G(W,Q) (is dl

where a and I denote the rectangular coordinates of W.

and W be represented, for convenience, by complex

(81)

Letting P, Q,

re',

Partial Dlfterentisi Equation. respectively, we rewrite (81) in the form

j

O(,)(P,Q)

r7.eI(f:t)

j1 log

v dr da (82)

log

Without loss of generality (since P and Q play symmetric roles), we may

assume that r p.

r, r r

o

We split the range of r into three intervals: 1. In the first interval, G(P,W) and p, p i

G(W,Q) may be expanded into the following series:1 G(P,W) G(W,Q) (83b)

—log r +

—

r")

p+

—

p*) cos n(a —

—

log

n(a — 8)

(83a)

is Still valid in the second interval, but (83a) must be replaced by G(P,W)

—

logi +

cos

—

— 8)

In the third interval, (83c) is still valid, but now (83b) must be replaced by G(W,Q)

—

logT

+

— T") coB

n(a

Splitting the disc I) into three parts corresponding to the afore-mentioned

intervals, and taking account of the orthogonality properties of the trigonometric functions app&uing in the above expansions,' we find that (81) works out to the sum of thft three following expressions:

=

r

log r log p +

n'(2n +

X (ri' —

—

p")

cosn(8

—

'The following expansions are immediate consequences of the identity —R valid for 1*1 <1.

tSr

positive mt.geri a, 'a.

I

(84a)

____

9. Green's Functions and Separation of VarIables

ff

= ir {{vhlogi.

ii. —

+

c'

r*+2 1'

17$ —

X

ff

251

—

(84b)

p") cos n(8 —

=ilr[2J'llogtltir+ n"'

p

x JpI

1 —

(84c)

Thus, G(,)(P,Q) is expressed as the sum of three infinite series, each term

of which depends continuously on the variables r, 0,

If each of these

terms can be dominated by a constant in such a manner that all three of

the dominating series converge, the continuity of G(,)(P,Q) will be established. In (84a), the term r' log r log p is dominated by (r log r)', p < 1, and this dominating expression never exceeds e' in the since r

interval 0 r 1. tr'(2n + Since the series

—

Each term in the first summation is dominated by' r2")(l — p$fl), and a fortiori by the quantity n'.

is convergent, the continuity of (84a) is estab-

Similar crude estimates suffice to establish the continuity of (84b) and (84c), and in this manner the continuity of G(,)(P,Q) in f) is proved.

Lished.

EXERCISES

20. Prove that the functions of class C' in the unit disc I), continuous in 1) and vanishing on the circumference, constitute a dense subset of L,(D). and v,,1 associated with distinct 21. The ortbogonality of the functions follows from representation (75). An alternative approach would values i..,. and

be to integrate the identity v,

nv,,

—

—

over D and apply

Green'8 second identity and the fact that v,, and v,, vanish on the boundary. What is

the difficulty associated with this approach?

22. Prove that the inequalities 2/, sin a/a

1 hold in the interval 0 < a w/2. 23. Prove that ff0 ff0 G'(P,Q) dx dy di, is finite for any bounded domain D, where G(P,Q) denotes the Green's function of tbis domain. Hint: Cover D by a disc and compare the Green's functions of the two domains. 'Here we employ the inequalities 0

r

p

1.

SOLUTIONS TO SELECTED EXERCISES

CHAPTER

1

(sin n,rz — sin mwz)' dx — 1 holds for any two distinct posiI. The equality tive integers n and in. Hence the inequality lain nrx — sin mrxl <1 cannot hold for all z, and so no uniformly convergent subsequence can exist. (A more delicate

argument shows that not even a pointwise convergent sequence can exist.) 7. Let I denote the smallest closed interval, a x b, containing the compact set R. The get I — R, being open, consists of a finite or countable union of open intervals

whose end points belong to R, so that f(x) is defined at each of these end points. Extend /(z) linearly into each of these open intervals. Then 11x) is evidently defined and continuous in I. Finally, let f(x) — f(a) for z b. hu been extended continuously (and without altering the maximum and Then minimum) +o the entire real axis, and a fortiori to every set containing ft. 8. The Taylor series of (1 —

i414

about u — 0 is of the form I

where —

c. > 0. If

diverged (or even converged to a value exceeding one), we could Il—I

find an index N such that

c> 1, and then we could find (by continuity) a value.

of ti below one ouch that

1.

This would imply that (1 — u)½ can aanme

negative values in the interval tul <1.

lOf course, we are exploiting the fact that

the Taylor series converges to (I —

in this interval I

Thus,

converges,

and so. given e > 0, we can find N such that (1 — t,)½ is approximated within by I—

Setting u —

1

—

within e by the polynomIal 1 —

1, we conclude that

a', c.(

is approximated

Partial Differential Equations

254 N b

12.

dx is muunused by choosing c0—/,, and

For fixed N, f (i —

N

the mmimuni value of the above integral is found to be L P dx

(Cf. —

Sec. 5-2.) Since the minimum is nonnegative,

conclude that

we

dx; letting N —'

Given • >0, we can cheese N and a .poly..

1r.

nomial p(x) of degree N such that if — p1 <

throughout the interval; hence

<st(b—a). Sincethe p,1areof degreen, pean be expressedas Taking account of the minimizing property of the coeffioientef. explained

above, we conclude that

<e1(b

dx —

—

a).

Letting N —

and

aeiount of the inequality obtained previously, we find that 0 / j! dx — — a).

—

Since • is arbitrary, It follows that

fb11

dx.

17. Let f(x) — egn x. It follows from Theorem 3 and Exercise 16 that the formula (13) may be applied for all values of E. In particular, for — 0, (13) assumes the form —2A

On

the other hand, f's

dx does not exist, so that the use of a symmetric

interval of integration is essential.

20. Assuming that f(x) approaches sero at ±

and that Integration by parts is

Justified, —

21.

i:,. + *.

dx —

Assuming that all integrals are meaningful and that interchange of order of

integration is permissible,

j:.

•

dx —

—

f' f

[f's

f(E)g(x

—

dx] dE —

g(S)e4*dt]

dE —

23. Since s may be replaced by s — So in the following argument, we may confine attention to the cases. — 0. Employing the notation used in the proof of Theorem

Solution. to

Selected Exercises

in the fo.m

we exjire, for. (— Be.) >0, the difference J(.) — — ftO)l dx

a

and that !g(x)r <M for some M. Thus, for any A >0, — dx <2MAI.I + f 02M dx + max < f4 xA sothat Let .beconflnedtothewedgelarg.I Given. > 0, choose 4 so large that max Ig(x) — 7(0)1 <34. oosO,

Note that 7(0) — If(s)

—

5

s/ui

sec 8.

x

and then confine 8 to the intersection of the afore-raentioned wedge and the disc -. <./4MA. Then we obtain the inequality <34. + 34. —.. 28. Let L denote the period of f(x). For u (— Re 8) > 0, the integral (23) is

f e" If(x)I dz•

dominated by

the latter series converges, the integral (23) exists at least in the half plane. > 0, while the example f(x) — 1 shows that the ahecissa of convergence may equal zero.

For. > 0, the convergence of the above series justifies rewriting 7(a) in the form

fL

or (1 —

dx

e"f(x) dx. According to the proof

of Theorem 4, the latter integral defines an entire function of a, so that f(s) may be extended analytically to the entire plane except that simple poles will appear at the zeros of (1 — s—'.), namely, the points 2,wi/L (ti — 0, ±1, ±2, . . .), unless the

valucoftheintegralhappenstobezero. (For example, iff(x) -ainx, the tranaform is equal to (I + to be poles of J(s).1

eo

that only two of the zeros of (1 —

e") turn out

31. U the opposite inequality g(x') > G(x') held at some other point of the interval, would exist a third point a" mAch that g(x") - G(x"). By the uniquenese portion of the theorem, g and (7 would coincide. 32. By the theorem of mean value, if(z,gi) — f(x,yi)1 If(x,ya)i . for — acme value of between 11 and y,. Hence, If(x,vz) — f(x,yi)I kIy1 — if k is chosen as max + lf(x,ip) — <(f(x,y) — ftx,',)l + + Jz — This inequality obviously extends to a larger number of independent vanables. $5.

—

—

—

kQy —

Uf(x,y) — i,1

CHAPTER 3 7. Prom (5) we find that the relation 2., + 2r, 5 must hold everywhere on the given line, and since sas, — 1, it follows that 2z, + 2/a, — 5, so that either — 2 or — 34. If the former solution Ia chosen, the characteristic strip associated with dp

dq

We thus obtain a — 2t +34., y — 2t + 2s, a — 5t + 2., p —2, q —34. Solving the first two equations of the latter system for, and tand substituting into the third, we

Partial Differential Equations obthin z

2x +

z

2Y.

Similarly, if we choose za

along the given line, we obtain

CHAPTER 3 78. The substitution r — z + 4, s x — 4 converte (2a) into the form u,. 0, o(x + 4) + b(.c — 4). From this general solution, we u '= a(r) + b(s) and From the equations a(z) + — b'(z). oltain u(x,O) — a(x) + b(z), v4(x,O) + g(x)1 and hence g(x), we obtain cil(x) — b'(z) a'(x) — 5(x) f(s), + cJ, where c denotes an arbitrary constant. From

[ix + f

—

this we obtain b?z) .=

+ 4) + b(x

—

—4)

—

0, this simplifies to (45). + t) +f(x — 1) ± g(E) dE]. When 12. Applying integrstion by parts twice to the second derivatives and once to the first derivatives, we find that JfvLu d.r dy + dx dy -4 boundary integral.

+

+

—

(dv),

—

14. Let Q end Q' be Ilxed point., and let u(P) Rz.'(P;Q), v(P) R1AF;Q'). Apply (5L) to the rectangle formed by the horizontal and vertical lines through Q and Q'. Since Lu = L*v — 0, the right side of (51) must vanish. When account is taken of equations (53a) to (53d) and the corresponding conditions for u, the vanishing to u(Q') — v(Q), or If L•(Q;Q)

of the right side of (5J) is found to be RL(Q;Q').

15. On the vertical characteristic the assigned values of u determine q, and so (22) uiay be considered as an ordinary differential

g(y,p). Since the value

of the characteristics is determined by the assigned values of at the u on the horizontal characteristic, a unique solution p(y) is determined. Similarly, q can be foutid at each point of the horizoatal [so that 16. Let the data he of the form v(x.,,,.) v(zo,y) Let v = u -— -. + (xe). Then v must vanish on tho two characteristics and muM satisfy an equation similar to (22). The iterative procedure of

employed in the proof of Theorem 1 applies without any change to the present problem. the point (r,y) (cf. Fig. 3-1) approaches the horizontal characteristic, the segment

PM does not shrink to a point, and so, in contrast to the noncharacteristic problem. q need uot vanish on the horizontal characteristic; similarly, p need not vanish on the vertical characteristic. 20. Referring to Fig. we note that — 4' + 0 — 2rt 008 0, and so ri sin 8 dO.

given by plying by equal..

'Ihe area dS of t.he zone determined by colatitudes 0 and e + 40 ii (4irt)' Multitdn U do. and so dS — 2irir'E or '4 and integrating, we conclude that the right sides of (74) and (76) are

22. tc

x'4 + 2xyf + 3z4' + 41/3. The Legeudre transformation reduces the given equation to the form — Thus z — x — 2.,, V 4' + :4 + 24, y + 3.fr —

24.

+ ny')

—

+

25. The Legendre transformation, if applied formally, would yield 0,. a 0, y — 0, so that x and could not vary. Thus, all solutions of tho given equatinn must be developable. This may he seen without considering the Legondre — u,' 0 an easy consequence of the transformation, for the equation given equation.

Sol*atione to Selected Exercise. CHAPTER

4

of basis: Lot fi, .6, . . , fm be a set such that representation (1) is e, discard it, then eliminate the first element (if any) whib isexpressible as a linear combination of the preceding element,s. Repeat this procedure as long as possible. The elcment8 that are not eliminated vouEtitute a basis. of number of elsmenls iim basis: Let the dimension be momentarily redefined as the znuzUezt possible nuniber of elemeut.s in a basis. For a I uniqueness is trivial; apply induction on a. Next., suppose a given independent elements.6,.6, . . . , fail to a basis. and t.hen apply the elimination procedure deseribeirl a basis d'i, #:, . . . above to the (ordered) set None of the f's are I,, . . . , . . . , eliminated by this procedure; hence at least one must be retained. This would furnish a bssin consisting of at leut a + 1 elenienta, contradicting uniq'zeuese. The "only if" assertion from uniqueness. form a ba'à, the equality If A, . . . , or Z(s1 — implies, by independence, aj — 0. Thus the coefficients are uniquely determined. 8. If U —f'.ll—-' 0, —f4 0, then jlf — gfl — —f) — g)Jf Hf —Mi + g —14—+ 0. Hence l;/ — — 0,f —' g. 10. Ufit 11(1 — v) + gfl il/ — gil + hence — Hf — g!l. Interchanging f and g, we get — gft. Thus, ugH — — Hf — oil. 12. M is obviously a manifold. Consider the sequence 11.1, where f. — 13, . . , 1/n, 0, 0, 0, • . 4. Then converges t' the element II, 3f, . . 'I, which does uot. belong to M. 13. (a) For convenience let a 0, b .' Let. f,(x) st3* sin ax for 0 < a/n, 0 for a/n
.

If

•

triangular graph of fixed height within the subinterval. Let the aublntetvale

repeatedly move

the interval as their lengths approach aero. iS. By homogeneIty of the norm, it anfflcea to prove that has a positive lower hound under the restriction 1. Suppose the contrary; then there exist a (k 1, 2 a) suet, that a, 1. .

Since each sequence is bounded in absolute value, there cxi*ts a subsequence of I on which each f).a"'l converges, say to Then 1, while (by continuity of the norm) 0, or ZX,,f, — a. The last equality contradicts the independence of the f's. 81. If is bounded, then a suitable subsequence of f Tf,, is convergent; applying B to each term of the subsequence, we conclude that {871I4 is also convergent.

Therefore, RT is completely continuoua, The sequence 1814 is bounded; hence a suitable subsequence of is convergent. Therefore, TB i8 completely Cant inuous.

32. First, consider a — 2. Then the product may be written as I — K, where is completely continuous by the two preceding exerdees. For a > 2, a trivial induction may be used. 34. & — 8,1 — 8,(TS,) — (8,7)8, — iS, — 38. Let flKI! a and — B <1. Then series (27) is dominated by the con. vergent series I + a + + + + + $c + •. . + + + $'a + •+ + . . • . The proof presented in the text that 1 — K is invertible if
K = K, + K, —

•

Partial Differential Equation. 41. Leta —0, b K(x,y)f(y) dy —

w,

n'wnnxâiny.

K(x,y)

nil., sin nx, where vfu

By uniform convergence, sin fly dy.

—2

By ele-

K(x,y)f(y) dy 0 if and only if I — 0. On the other hand, (4(Y) is not solvable if g(O) ,' 0, since each term in the above series vanishes at x — 0. 42. Similar to convergence proof employed in Theorem 1-6.

mentary theory of Fourier seriea,

CHAPTER 5 the Schwarz inequality is certainly satisfied. If (f,g) is real an9 L If (f,g) — > 0 for all real values of — 2X(f,g) + positive, we obtain Ill — — If (f,g) ia not zero and not positive, we replace I In the and so I(f,g)I' — 1 and (4,g) > 0. above argument by qr, where 4. Let ., be any orthonormal basis in the given space and let each . . . ,

eleinentc4#i be mapped into the element 5. The "hermitian form"

ai,

.

. . ,

aol of

(f.,f,Yaö, can be written as

).,fi

and so is

always real-valued and nonnegative. If the elements sic independent, this form aemmes positive values except when 0, and if the elements — are not independent, the form assomas the value zero for nontrivial choices of the as well. According to the theory of herinitian forms, the determinant of the coefficient. must be positive and zero in these two cases, respectively.. 7. The set of all elements with only a finite number of nonzero components, each of which has rational real and imaginary part., is easily seen to be dense in and denumerable. The second part is obvious. jf is bounded, say <M, tben (T'II. is also bounded, and so { I contains a convergent subsequence. Confining attention to the subsequenceof IIT'(f,, (7'T(f., —f,.),f., lIfe <2M. IITT0I., — TT0f.,tI 0. Thus IT'f,d isaCauchy sequence; IITT°(f., since H is complete, T0f, is convergent, and so T° is comp1etely continuous. 13. Let be an orthonormal basis in a separable Hilbert space, and let T.J — For each f, T..f—+ f, so that

converges strongly to the identity

b—i

operator 1; each T., is completely continuous, but I is not. (Note that 1 — T,,fl 1, so that the sequence { T,. does not converge in norm to I.) 14. Prom the hint, we easily conclude that (Tf,g) — 0 for every pair of elements 1,9. Choose f and then let. g — Tf. We obtain (Tf,Tf) —0, and so Tf — o and T —0. 17. The necessity is proved by (30) and the remarks preceding (30). To prove sufficiency, let T be idempotent and herinitian, and let M be the subepace spanned h,

the range of the smallest subspace containing all element. of the form 7'f. 1ff EML, then (TJ,Tf) — (T'fJ) — (Tf,f) —0 (sincefE M1 and Tf EM), and so Tf — o.

1ff is in the range of 1', then / —

7'9 for

some element g, and ao/ —

$59

Solutions to Selected Exercises

the

1— Tf holds for all elements in M. (That T(Tg) — Tf. By Taking account of the projection range of T ía precisely M, not a proper subset.)

theorem and of the linearity of 1', we conclude that 7' PM. 19. If Tf pf,/ o, we obtain p (Tf,f)/(f,f). Since (Tf,f) is real, by Theorem (T/,g) — 10, ,i is real. If Tf — pf, Tg — ).g, then (p — XXf,g) (pf,g) — 0. X implies that (f,g) — p (I Tq) — (Tf,g) (Tf,g) — 0. Thus, 20. (a) and (b) are obvious. (c) For any element f, Is and so — since T is hermitian. Thus Thus T5' — TB. (d) ($T) — By Theorem 10, 7'-i is hermitian. real for each f. (ST)S ST if and only if ST — TS.

QP, then PQ is hermitian by the preceding exercise, and also

21. (a) If P9

P9, so that PQ is ideinpotent. By Exercise 17, PQ is a PPQQ (PQ)' — PQPQ projection. Conversely, if P9 is a projection it must be hermitlan, and 80 PQ —QP by the preceding exercise. (b) If PQ 0, then QP also equals 0, by part (a) (since o is a projection). Then (P + 9)' P' + Q' + PQ + QP — P + Q + 0 +0 — P + Q. Thus P + Q is (deinpotent, and it is hermitisa by preceding exercise. Thus P +9 is a projection by Exercise 17. Conversely, if P + Q is a projection, P + Q

and so PQ+QP—0.

Multiplying on the right by P, we get P9? + QPP = PQP + 9? — 0. Multiplying on the left by P we get PPQ + PQP — P9 + PQP 0. Subtracting, we get 0, PQ and QP each equal 0. Conversely, PQ 0 PQ — QP; but since P9 + QP impliesthatQP — Obypart(a). Tbua(P + Q)' P' +9' + P9 ÷ QP P + Q, and so P +9 ii idempotent. Since it is hermitian by the preceding exercise, it is a projectionbyExerciael7. (c) P—Qisaprojeotionlfandonlyifl —(P—Q),or — + is a projection. By part (b), (1 — P)Q — 0, or PQ Q, is necessary and eufficient. (b) If K is hermitian and . . .} — jot, 0, 0, . . 4. 24. (a) In 1, let Kb,, ai, Kf # o, then 0 <(Kf,Kf) — (K'f,f), and so K"f o. By repeating this argument, we see that K'f, Khf, K"/, . . . must all be distinct from o, and this could not

happen if K were nilpotent. IIKU'.

sup (Kf,Kf)

sup I(K'f,f)I

25. By Theorem 11,

ti/H—i

H/li—i

Similarly,

and, In general,

IIK'II' —

—

< jIKII', let m be any power of 2 exceeding k. Then IIK'i — = IIKII", which contradicts the previous result. (Note

{Kjj". If

<

that this includes part (b) of the preceding

(Kf,Kf) — (K"Kf,f)

26. If K is normal, IIK'fll'. Converse'y, if (KK*f,f). Thus, (Tf,f)

and so KK —

sup IKfll'

if/il—I

KK.

—

0, where T — KKS

—

(KKI,f) — (Ktf,K/)

(K'f,K'j), or (KKf,f)

flKffl, then (Kf,Kf)

KK.

By Theorem 11, IVTII — 0,

31. Let f(x) E L. possess the Fourier expansion 1(x)

fl' and Kf — fo +

so that tIKf!b' =

Thusfor fixed "$0.1 to aero, and in this case lIKfIl

foP

Thus

y we may write

xt(1 —tXl —v)dt+f'x31(1 —t)'d&

—

—

—

—

maximized by choosing each f, except Jo equal IIKH — 1.

is given by an elementary computation. 38. Set x — y — in (154).

39. For 0 x

+

then I/il'

K(.)(x,y) —

The equality I!IKIII

JZP(l

— x)(1

—

y)dt

+ f1' —vX2v—z'v'). —

Partial Differential Equations For a > y we merely interchange a and yin this expression.

n' —

Hence

1/9,450.

dx dy

we obtain

By elementary cornputa-

,t—z

CHAPTER 6 3.

Summing

—

An orthogonal transformation is cbaracterized

a.&aih)

'S'

over k, we obtain

ii—' k—i i.,. Therefore, the triple summation reduce. to

by the equalities b—i

or s—I

'.1—i

6. When f(r), r

f"

—

0,

is substituted for is, (1) ummtr's the form f'

=

or i"ir f (a — l)/r —

Integrating once we obtain f'

0.

+ con-

2, (71') for n > 2. stant r'". integrating again, we obtain (7a) for 7. Taking account of (1 lb) and of the constancy of v on F, we Sod that the right

side, and hence the left side, of (12) vanishes. Sinee the intsgrsnd the left side is 0, continuous and nonnegative, it must 'vanish throughout. C. Therefore, u, . and hence v is constant throughout f is dy jby (8)). Hence, max 9. irk'u,(Q) .a t&,dx dy

M 4R, or

Establishing a new rectangular coordinate system (x',y') by rotating the axes so that the x' axis has the direct.ion of the gradient

Jr

of

is,

4M/irR.

+

we obtain

(The fact that the

4M/TR.

Laplace equation is invariant under rotation plays an essential role.) tO. (20a) foUows directly by observing that the integral given in the hint is nonIt can negatwe and works out (using the second corollary) to K —is dx dy to obtain he proved by applying the Schwarz inequality to rR'u(Q)

dii)

—

.

follows by applying this result to the disc G' with center at P internally tangent K. to observing that ffg,uldxdY 12. The circular symmetry shows that is must have the form (7a), and the given data, yield for a and lithe values (20b)

a—

C,—Ci log

— log

log

-- log R1

Solutions tt. Selected Exercise.

26J

inner circumference u —. c1. Thus, as Letting Rj 0, we obtain a —, 0, b shrinks to a point its influence disappears. 13. P be any point of the punctured disc, let F' be any circle concentric with F and not containing or enclosing I', and let v be defined in the annulus as the function with values M on F' and m on F. By the maximum and minimum priueiples, and taking account of 0, and hence fu(P)! s(P). Letting F' v±

Exercise 12, we conclude that lu(P)i < in. hence M m; on the other hand.

M 2 m, by the definition of these quantitieè. 15. Apply the maximum principle to the region obtained by deleting from the domain a small disc containing Q. 16. Apply (10) to the region obtained by deleting small discs with centers at and Q2, i'espeetively, taking u and v as the two Green's functions. The dostred result. is then obtained by shrinking the discs. 24. Subtract the two etpanaions and employ the fact that the sum of a convergent. power series vanishes Identically only if all coefficients vanish. 29. Por simplicity let. P be the unit circle. (Otherwise minor modifications are needed.) Then, for any z (except 0) mule F,

f

I

21-IJV

Subtracting, setting r

es', z

and performing some elementary simplifica-

tiona, we obtain f(pei#) _

Taking the real part of both sides and noting that P is real, we obtain (26b). 30. The necessary condition is simply a repetition of (1 Ib). Aasume that u eolves the given Neumann problem. Then the harmonic conjugate v satisfies (by CauchyRiemann equations)

—

a(s),

which determines r on F to within an arbitrary addi.

constant. (Note that the condithin f g(s) cfs — 0 assures the single-valuedness of o in the case that r is a single curve.) Thus we obtain a I)irichlct problem for v. If v is found, its harmonic conjugate, with sign changed, solves the Neumann problem. 32. Let u be the given function, and U the harmonic function coinciding on thc boundary with u. Then n — U vanishes on the boundary and also possesses the "one-circle mean-value property" This suffices to permit thc conclusion that u — U

vanishes throughout 0. 35. If u 0, let R —+ u — e (or c — u).

in (48).

If u > c (or u

c), apply the same argument to

41. For 0 < r
J7[r(i

+00S40).— h(ooe 0 + eos 30)JP(r,0,h,6) do

Now, the Dirichiet problems with boundary (I + "os 40) and (cos + cos 30) on the circle of radlue h are solved by 1 + (r/h') and (nh) cos + (n/h)' cos respectively, which reduce for 0 to -+ (r/h)'l and (n/h + (n/h)'). [Cf. (36)j Thus, the left side of (96) equals r(h' -. (rf I + (r/h)9 — hfr/h ± (r/h)'J},

o—

PartIal Differential Equations

262

Similar argument. apply when 0 r < —h and whene > account of rotational 42. Choosing unit density in the disc r
symmetry, we find that the potential u(r) eatisfies the equation (ru')' + 2,r —0 in but the disc. Integrating twice, we obtain is — —J4wr' + log r + must since is remain, finite at the origin. The potential at the origin is readily computed, and this gives the value of Thus (89) Ia obtamed. For the exterior of the disc we exprees is in the form (7a), and evaluate a and b by matching the "interior" and "exterior" formulas for is and is' on the circumference. Thi& leads to the result previously obtained for is outside the disc, namely, is log r.

CHAPTER 7

3,4,5. The inequality (1) may be squared if ii(P) 0, and the equality (6-18) may be squared in any ease, to yield, with the aid of the Schwarz inequality, u2(P)

If u(P) <0 and equality does not hold in (1), the squaring of both aides of (1) and the subsequent derivation of the inequality u'(P)

2i

0

may lead to an incorrect result. Example: The function x' + y' — I is subharmonio everywhere, but its square is superharmonic in the disc x' + y' this may be seen either by direct computation or by applying the result of Eaercise 11. 6. Consider the function max (1, x' + y'). 17. By rotational symmetry, is — a log r + b, but a must vanish since is is bounded. Hence is reduces to a constant. According to Theorem 5, is demonstrates "proper" behavior near each regular boundary point, and so is i. 19. ChoOse any exterior point 0. If there is a unique boundary point Q closest to 0, the function (21)1. a barrier, and so regular. If there are two or more such boundary points, choose any one of them, say Q, and then select a point 0' between 0 and Q. Then (21), with 0 replaced by 0', furnishes a barrier, and so Q is regular in this case aloe. 21. By Exercise II, ea'h of the given functions is subhar,uonic. Since (z' + p')"" —. 1 for each point (x,y) of the punctured disc, u 1. However, is i by the maximum principle. Hence, is — i. 29. Let v+ be the solution of the Pirichiet problem with boundary values max (O,v), and let v be the solution with boundary values max (0, —a). Then is — ifr + v— — v. (These equalities are correct on the boundary, and then, by Theorem 6-3, they remain correct in the domain.) Then D(v) — D(v) — — a) — D(v+ + vj — —4D(v4,vj, which by (64) to

_4Jt r-d, At bodndary points where a

0, v

vanishes, an ci hence so do.e a-

At point.

where v<0, v'—O; once v40 everywhere inside (by minimwn pstndple),

263

Sohitlons to Selected Exercises +

0 everywhere on the boundary, and there-

fore D(v) — D(ti) 0. [A slight extension of the argument shows that the strict inbquality D(v) — D(u) > 0 holds whenever v assumes both positive and negative values on the boundary.) — w(P1), the indicated summation 31. Since h'w.,(P) — w(PNZ) + w(P) — covers each inner point of the rectangle PQRiS four times, twice with eaeh sign, while

each boundary point which is not a vertex is counted twice, once with each sign. Each vertex Ia counted only once, and examination of signs yields (126).

r ii, — J ôn

35. By (6.9) we get D(u4,u3) — I

Since iij vanishes on all boundary

thi.

I except F1 and equals one on that component, D(u,,u,) Jr, By the Cauchy-Rieznann equations, this equals the increase in resulting from a circuit of Fj (or any curve in G deformable into r,). 37. By Exercise 35, the indicated Dirichiet integral is equal to the given quadratic form, while by Exercise 34 the integral is positive unless all vanish. 39. Since by Exercise 30 the Diriohiet integral is conformally invariant, it suffices to prove the statement for an annulus of radii 1 and a. The harmonic measure of the outer circumference is given by u log f/log a, so that u' + (r log a)', 2w Fa (r log a)'r dr dl INu) 2r/log a. —

CRAPTER 8 1. The substitution 'p = (t p. 4.

reduces (9) in this case to the form u(x,t)

J —.

The substitution

p.

008 a J —

e, —

—z

reduces (9) in this case to the form u(x,t) —

(cci ,j)K(ap,O,t) dip = g(t) cos x.

Substituting the latter expression into (1)

yields g'(l) + g(t) — 0, while the initial condition yields g(O)

e' cci x, or f' exp (j-!.) obtain

we

5.

p. J

J—

—

f

exp

u(x,t)dz

J—. f(

(—ax')

cos Xx dx

Setting; —

—

5(exp

Thus ti(x,t) —

1. 3X1z

and t —

exp (—).'/4a).

f' [f1(E)K(E,x,t) df] dx

j:. f(E)[f:. K(E,x,t)dz]

[By Exercise I and the symmetry of the kernel, I' K(

i—'

dx

Physically, this result asserts that the total amount of heat is conserved. may be assumed tb of /(E) is zero, for otlierwiBewemay simply of the mean value. Letting dq F(E) and employing — 1.1

iategration.by parts, we write (9) in the form u(x,t) —

F( —

dE.

Then J

J—ut FWIK(e,v,t) — K(

->

—

—

I'

/ —

K( E,, —y,t)] dE

(1(E) — P( —E)IK(E,y,t)

Ow.n

em find positive zmmbers A, M such that — — bi <ejEj for > A-and 11W <M Thelsat integral is then dominated

Equations

Partial dE, and hence by

K(t,y,t) dt + e

by M

(M

+

dE

— y to the form

Thie integral is reduced by the substitution

f which i5 dominated by

Thus,

urn (2yY' f

d,E, and this integral works out

dx

Jim sup

and hence

dx exists and equals zero.

J" F(

the formula

dE, we obtain

—

E,O,t)

<(16i-L'ti'4 [j0

dE + f4

axp

(ME + 4') exp (—1)

exp (—if) dE] < clu'ie that

+ ,,I)K(s,O,e)

EM + e($yI +

to M + 'flyl +

exp (—

(M +

dE —

+

and taking account of the arbitrariness of e, we conLetting I approaches sero. Similarly, for any other fixed value of x,

x,, Urn

0.

it is merely neeessary to replace x in the above argument by

dE exists and equals zero. as in to observe that Jim (2yY1 f V.-'.. the case x,, for 0; this is an consequence of the houndednc.ss of x

--

(2y)—'

—•

—

and the latter expression is dominated by ixo/y! max tJ(E)!. IThe x, corresponding to ShOWS that the boundedness condition on 1(E) cannot be dropped.j 12. By adding jr (r Im s) to both sides of (47), we find that the relationship s — — + i(n — 34)sI' is satisfied on ± K2n —. From this we immediately obtain + i(n Isinh

—

> exp [(2n — that the integrand is dominated by — (I — xX2n — Since

ainh [(2n —

and Iainh seMI <exp Lx(2n — so f,,(r) — 2exp (—(a — + (2n — — (i + 2i(2n — di- and fri is bounded by c,. — (2n

— 1)wfr + (a — a traveries V,,1 the inequality
as

2irf exp

—

The numerator and denominator each become infinite with n; treating a as a continuous variable and employing L'iiôpital's rule, we find that the fraction approaches zero with increasing n.

Solutions to Selected Exercises sin 8J'4. This function has 13. Proceeding formally, we obtain U(x,e) — sin and so cannot be a Laplace transform. 2, . 1, (it . .), at the points

Taking account of the reversal of sign, we may assert that it is, in general. not

defined for all negcstwe values of t and possible to find a temperature distribution This result satisfying (1) such that u(x,O) coincides with a Bpecifled function the physical phenomenon of Irreversibility. is the mathematical counterpart of

CHAPTER 9

2. Multiplying both sides of (7a) by X and adding (X')', we obtain 2(XX')' + (I')'. Integrating over the interval and taking account of (8), we obtain X' dx

j11

(X')' dx. Since both integrals must be positive, )is also positive.

q• If the inner product (f,g) of two functions

defined aa f

(38) assumes the form (u,Ls) — (Mu,v) for vanishing at both ends. 10. We seek a positive solution of the equation (,ip)'

— pq;

dx, then functions u,v

such a function ls given

by p'(x) exp 11. Recalling the definition of fi(x) given in the text, we first prove that f(x) be the smallest never vanishes in the interval 1. Assuming the contrary, Jet is <Xi, value of x at which f(z) vanishes. Then, in the subinterval — ,f,, and the positive, and so fz increases monotonely from zero; since cannot decrease latter quantity in noonegative in the subinterval, the quantity

value at x,, naInel3', p(xi). This inonsistent. with the vanishing of It follows that increases monotonely throughout I, and so, in part ice JXr, ft(t,) > 0. This proves that the Green's function K(x,E) exists, and also that it either increases decreases monotonely in the interval it x f; by the same type of argument, K(x,t) shows the opposite monotone behavior on the interval E x Z3. and no K(x, E) IS either entirely positive or entirely negative throughout the interior of ) and attains an extremuni at 'l4he possibility of a minimum at is ruled out by (34), and we conclude that K(x,E) is positive for z, <x
I:' —

—

To show that v must vanish at least once in theopen interval <x it evidently sufficee to assume that is in positive there, and then rule out the possibility that v is also

positive throughout this interval. Under the afore-mentioned assumption, u'(xs) > 0, ts'(x4) <0. If a were poeitive in the interval, vpu' would have to be nonnegative and nonpositive, respectively, at arid end so each side of the above equation would be nonpositive. Recalling that pus, and hence v itself, (*flflOtbe positive throughout the interval. Now, if there were infinitely many negative eigenvalueu, it would Immediately

Partial Differential Equations follow from the above result that each eigenfunction must possess infinitely many zeros. By a simple argument based on Rolls's theorem, there would exist for each eigenfnnction a point in I at which the function and its first derivative both vanish, and this would imply the identical vanishing of the function. 17. Assuming momentarily that r(x) 0, we know from Exercise 11 that K(x,E) and the same is true for exists and is positive in the open square xi <x, the lowest eigenvalue, ía positive, and From Exercise 15 we know that hence lowest in absolute value, and so, according to the theory developed in Sec. 5-4, (normalized so that puz1 dx — l)ia charthe corresponding elgenfunction acterized by the fact thai, among all functions f(x) so normalized, it maximizes assumed both positive and negative values, the IIJt(x,t)f(x)f(t) dx dEl. If inequality

dx dEl >

If I

dx dEl

would hold

for

Thus we may assume that Ui(z) 0 throughout I. If vi vanished at any interior point of I, i4 would also have to vanish there, and this would imply vanishes only at the ends of I. Rence, the identical vanishing of The assumption '(x) 0 may be eliminated by the device of rewliting (40) in the form (pu')' + (r — cp)u + (X + c)pu 0, c being a positive constant

1(x)

tut(z)I.

large that r —

cp

0.

1 (as in Sec. 1). Then

— and (55) The roots of the associated quadratic — If are the roots are complex conjugates with ± Thus, any solution is expressible as the product of a sinusoidal real part — function and the exponential function exp (— For fl >0 this 4aetor provides

19. For convenience let Xi — 0,

assumes the form T" ±

+

—

0.

a damping effect on the sinusoidal function, while for ft <0 it provides an exponentinily increasing factor. If 4n'ir', the foregoing discussion must be modified, but this inequality can hold or only a finite number of values of a. 20. Similar to Exercise 13. 21. The differentiability properties of the cigenfunctions on the boundary would have to be investigated in order to justify the use of the Green's 23. The cancellation of logarithmic singularities at Q and the fact that P,Q), the Green's function of A, is positive on the boundary of D enable us te assert (by the minimum principle) that > G(P,Q) >0 for any pair of points P, Q in D. Thus ffD ffDGt is dominated by ffff and the latter is known to be finite.

SUGGESTIONS FOR FURTHER STUDY

B. B. Baker and E. T. Copeon, "The Mathematical Theory of Ruygheni' Principle," Oxford University Press, New York, 1939. Company, New Chelsea 2. 8. Banach, "Théorle dee York, 1955. (Reprint.) 3. H. Batenhan, "Partial Differential Equatione," Dover Publications, New York, 1944. (Reprint.) 4. 8. Bergman, "The Kernel Function and Conformal Mapping," American Mathematical Society, New York, 1950. 5. S. Bergman and M. Schiffer, "Kernel Functions and Elliptic Differential Equations in Mathematical Physics," Academic Press, Inc., New York, 1953. 1.

6. C. Caratheodóry, "Variationerechnung und partielle Differentlaigleichungen erater Ordnung," Teubuer Verlagsgeaellachaft, mhil, Stuttgart, 1935. 7. fl. 8. Carslaw, "Introduction to the Mathematical Theory of the tonduction of heat in Solids," Dover New York, 1945. (Reprint.) 8. H. 8. C.rslaw and I. C. Jaeger, "Operational Methods in Applied Mathematics," Oxford University Press, New York, 1941. 9. H. V. Churchill, "Fourier Series and Boundary Value Problem.," McGraw-Hill Book Company, Inc., New York, 1941. 10. E. A. Coddington and N. Levineon, "Theory of Ordinary Differential Equations," McGraw-Hill Book Company, Inc., New York, 1955.

11. R. Courant and D. Hubert, "Methoden der mathematiachen Physik," vole. I and II, Spsmger-Verlsg, Berlin, 1931 and 1937. (A slightly revised version of vol. I in Englich baa been published by Interscience Publiabera, Inc., New York.) 12. G. Doetech, "Theorie und Anwendungen der Laplace-transformation," Dover Publications, New York, 1943. (Reprint.)

13. G. F. D. Duff, 'Partial Differential Equations," University of Toronto Press, Toronto, 1956.

14. N. Dunford and.7. Schwartz, "Linear Operators," Interecience Publishers, Inc., New York, 1958.

15. B. Friedman, "Principles and Techniques of Applied Mathematics," John Wiley & Sons, Inc., New York, 1956.

16..7. Hadamard, "Lectures in Cauchy's Problem in Linear Partial Differential Equations," Dover Publications, New York, 1952. (Reprint.) 267

Partial Differential Equations 17. B. L. moe, "Ordinary Differential Equations," Dover Publications, New York, i944. (Reprint.) 18. 0. D. Kellogg, "Pountiatious of Potential Theory," Dover Publications, New York, 1953. (Reprint.) i9. W. V. Lovitt, "Linear Integral Equations," Dover Publications, New York, 1960. (Reprint.) 20. C. Miranda, "Equasioni sUe Derivate parsiale di Tipo ellitico," Springer-Verlag, Berlin, 1955.

21. Z. Nehari, "Conformal Mapping," McGraw-Hill Book Company, Inc., New York, 1952.

22. J. v. Netunaun, "Matheinatieche Crundlagen der Quant.enmechanik," Dover Puhheations, New York, 1943. (Reprint.) 23. 1. (1. Petrovaky, "Lectures on Partial Differential Equations," Lnt.erscience Publishers, Inc., New York, 1954. "Lectures on the Theory of Equations," Cray lock ?ress, 24. I. G. Rochester, N.Y., 1957.

25. F.

and B. Ss.-Nagy, "Functional Analysia," Frederick tTngar

Co., New York, 195.5.

26. 1. N. Sneddon, "Elements of Partial Differential Equations," McGraw-Hill Book Company, Inc., New York, 1957. 27. A. .7. \V. Sommerfeld, "Partial Differential Eqi&ationa in Physics," Academic Press, Inc., New York, 1949. 28. M. H. Stone. "Linear Trariaformationa in Hubert Space," American Mathematical Society, New York, 1932. 29. 1. Tamarkin and W. Feller, "Partial Differential Equations," Brown Unweraity, Providence, Ri., 1941. (Mimeographed.)

30. B. C. Titehmarsh, "The Theory of Function.," Oxford University Press, New York, 1939.

31. B. C. Titchmarsb. "Introduction to the Theory of Fourier Integrals," Oxford University Press, New York, 1937.

32. B. C. Titchmareh, "Elgenfunctiort Expansions Associated with Second..order Diffàrential Equations," Oxford UniversIty Press, New York, 1940. General references: [31, [111, [13], 123), [26), [271, [291

for particular topics: Chapter 1: Sec. 3: 19], (11, voL 1, chap. 2], [31] See. 4f8J, (12] Sec. 5: 110), [1'7j Sec. 6: (251, [301

Chapter 2: [6j, vol. U, chap. 21, 126, chap. 2) Chapter 3: (II ((or See. 8), (13. vol. Ii. chaps. 3, 5, 81, 1161 Chapters 4 and 5: [2], [11, vol. 1, chap. 3], [14), 1191, 122], (24], (25], (28] Chapters 6 and 7: [4] (for See. 7-7), [64, (11, vol. Ii, chaps. 4, 71, [181, 1201, [211, (26, chap. 4] Chapter 8: [F,], (7], [8], (91. [12], (26. chap. 6) Chapter 9: [91, (11, vol. 1, chaps. 5, 6l, [15), [.32]

INDEX

Abel's theorem, 17 Abscissa of convergence of Laplace trannform, 14 Acoustic velocity, 68 Additivity, 76 Adjoint homogeneous equation, 111 Ad joint operator, of differential operator, 55, 57, 239 in filbert apace, 103, 104 self-, 104 Algebraic equations, 88, 80, 113 Analytic functions, 145 (See also Conformal mapping) Areal mean-value theorem, 136 Ascoli sel.'ction theorem, 2, 3

"Backward" heat equation, 231 Balayage, 170—175

Banaeh algebra, 78 Banach space, 75 Barriers 173, 174 Bseia, 71, 97—99

Bernoulli's law, 67 Bessel function, 57, 149 Bessel inequality, 94 Bolzano-Weierstraas theorem, I Boundary point, regular, 374 Bou.ndedneaa, 76

Branching lines, characteristics as, 35 C", functions

of clan, z

Calculus of variations, 184

Canonical domains, 216 Canonical formA. 46, 47 Canonical system of differential equations, 23 Caratheodory's extension of Riemsnn mapping theorem, 213 Cauchy data, 44 Cauchy principal value, 10, 16 Canchy problem, characteristic, 56. 57

for hyperbolic equation., 48-52,80 linear, 52, 53, 55—57

for Laplace equation, 52 stability of, 51, 52 for wave equation, 53—55, 60-65 Cauchy sequence, 73 Cauchy-Kowaleweki theorem, 3B Cauchy-Picard theorem, 17 Csuchy-Riemann equations, 133 Characteristic Cauchy problem, 56, 57 Characteristic condition, 29,44 Charscteriatic conoid, 41 Characteristic direction, 30, 45 Characteristic ground curve, 30, 45 Characteristic ground direction, 30 Characteristic strip, 89 Characteristic surface, 60 Characteristic value, 108 Characteristic vector, 108 Characteristics, 30, 34, 36, 45, 60 as branching lines, 35 Circle, ix Classification of second-order equations, 43, 58, 59 Closed set, 73 269

Partial Differential Equations

270

Compactness, x local, 75 Compatibility condition, 29, 45 Complete oontinuity, 78, 106-110

Divergence, 130 Divergence expression, 67 Divergence theorem, 55, 63, 133 Domain, ix

Completeness, 75

Double layer, 154—157, 180—182

"Concentrated" family of functions, 9,

Doubly connected domains, conformal

219 Conformal mapping, 132, 211—216

Dyadic rational, 202

mapping of, 213—215

of multiply connected domains, 216, 216

Conjugate, harmonic, 131 Convergence, of Laplace transform, abscissa of, 14 in norm, 73 of operators, 77 Strong, 104 Convex hull, 104 Ponvexity, 100 Convolution, 13. Content, R1 199*. Covering (of a set), x Cramer's rule, 89 Degenerate function, 118*., 121 Degenerate operator, 105—107 Descent, Hadamard's method of, 83 Developable surface, 67 Disgonalization procedure, 3 Difference quotients, 199, 200 Dimension, 71 Dial theorem, 27 Dipole, 153 Dirichiet domain, 139, 175, 198 Diriohlet integral, 184 conformal invariance of, 199 Dfriebiet principle, 183—198

Dirichiet problem, 138, 139, 167 for disc, generalized, 186 for hyperbolic equation, 52 varIational, 186 Diac,ix Diziohiet problem for, 139—144 punctured, 139 Disoontinuities, propagation of, 54 (See also Branching lines) Discrete Laplacian, 200 43

Distance, in metric space, 73 between sets, ix

Eigenfunction expansions, 239-241 Elgenvalue, 108*., 240, 241 Eigenvector, 108*. Elliptic equation, 43, 58, 59 Equation of state, 67 Equicontinuity, 2 Euler-Lagrange equation, 184 Expansion theorem, 122-426 FaUung, 13 Finite differences, method of, 199-211 First-order equations, 17—22, 28-41 Fourier coefficient, 94 Fourier integral theorem, 10-12 Fourier series, 9, 128, 144, 233, 234 Fourier transform, 12 of derivative, 13 (See also Heat equation) Fredholni alternative, 83—88, 111—116 Fredhoim integral equation, 89, 118—121

Friedrichs, K. 0., 199*. Fubini theorem, 28 Function harmonic at infinity, 132 Functional, 76 linear, 76, 102 Fundamental solution, 135 Gradient, 130, 152 Gram-Schmidt procedure, 93 Green's function, 140—142, 144, 104, 165, 235—239, 243—251

Green's identities, 133 Hadamard's example, 198 Hadamard's method of descent, 63 Harmonic conjugate, 131 Harmonic functions, 130 convergence theorem. for, 149-151 maximum prineiple for, 136, 137, 152

Index

271

Harmonic functions, unique continuation property of, 152 uniqueness theorem for, 137, 138 differentiability of, 140 (See ales Dirichiet problem) Harmonic meseore, 216 Harmonization, 169 Harnack inequalities, 160 Harnack theorems, 149-151 Heat equation, 42, 217-231 "backward," 231 solutions of, prmciple for, 226

Heat kernel, 218 Heine-Borel theorem, x Hermitian kernel, 121-127 Hermitian operator, 104, 105, 108-110 Hubert 8pace, 95 Hodograph plane, 68 HSlder continuity, 162 Uuygens principle, 65 Hyperbolic equation, 43, 68, 59 Dirichiet problem for, 52

Idempotent operator, 101, 102 Identity operator, 78 Independence, linear, 70, 93 Infinity, function harmonic at, 132 point at, 132 Inner product, 90 Inner-product space, 90 Integrable function, 25, 26 Integral of a system, 89n. Integral Oquations, 89, 118-121 solution by, of Dirichiet problem, 179183

of Neumann problem, 183 Volterra, 89 Integral transform, 78 Inverse operator, 80 Inversion, 132, 133, 140 Iterated kernel, 120,121,249-251

Jacobian (determinant), x Kernel, 89n. heat, 218 hermitlan, 121—127

Kernel, iterated, 120, 121, 249-251 Poisson, 141, 144 reeolvent, 117 Kronecker delta, ix it, 95-67 L1, 97

L, kernel as operator on, 118-129 L,, 97 (See also Kernel) Laplace equation, 42, 130 and analytic functions, 131, 132 Cauchy problem for, 52 in polar coordinates, 131 (See also Harmonic functions; PoIsson equation) Laplace inversion formula, 16 Laplace transform, 14—16

bilateral, 17 of derivative, 17 Laplacian, discrete, 200 Lebesgue extension theorem, 7,8 Lebesgue integratIon, 26-27 Left inverse, 82 Legendre polynomials, 9 Legendre transformation, 65-68 Lewy, H., Limit, 73 of sequence of operators, 77 Limit point, 73 Linear functional, 76, 102 Linear independence, 70, 93 Linear manifold, 70 Linear ordinary differential equatIons, 23,24 Linear partial differential equations, of first order, 28—33 of second order, 52, 53, 55—67

Linear principal part, 42, 58, 59 second-order equations with, 42—44,59 Linear space, 69 finite-dimensional, 71 normed, 72

linearIty, 76 Liouville theorem, 151 condition, 17, 22, 24, 25

Mach number, 68 Manifold, linear, 70

Partial Dafferential Equations

272 distributions, 152—163, 170, 171

Matrix, unitary, 117 Maximum principle, for harmonic (uncttOflf*, 136, 137, 152

for mesh functions, 200 for eclu tions of heat equation, 226 for euhharmonic functiona, 168, 169 Mean value, 146, 149, 220 Mean-.value property, 146, 149 Mean-value theorem, 136 steal, 136

Operator, linear, seIf-adjoint, 104 Unitary, 111 zero, 78 (Eke also Adjoint operator; Convergence) Ordinary differential equations, 17—24 Orthogonal transformation, 131 Orthogonality, 93 Orthogonalization, 93 Ortbonornial basis, 93, 98, 99 Orthonormal sequence of polynomIals, 9

converse of, 146—148

Measurable function, 25 Mercer theorem, 126, 126 Metric space, 73 Minimizing sequence, 189 Minimum principle, 137. 200 Modulus of doubly connected domain, 214, 216 Monge axis, 37 Monge cone, 37 Monotone convergence, theorem of, 151 Montel selection theorem, 4 Multiply connected domains, conformal mapping of, 216, 216

Neumann problem, 145, 146, 183 Neumann series, 81

Nilpotont operator, ill Norm, 72, 76 convergence in, 73 Normal operator, 111 Normed linear space, 72 Null function, 97 Null set, 25 Null spsoe, 84

One-circle mean-value property, 149 Operator, linear, 76 ndjeint of, 103, 104 completely continuous, 78, 106—110 degenerate, 105-407 hermitien, 104, 105, 108—] 10

idempotent, 101, 102 identity, 78 inverse, 80 flhlpotent, 111 normal, 111 projection, 101, 110

Paraboli. equation, 43, 58 Parallelogram law. 92 Parsoval equality, 94, 99, 122, 123, 129 Perron-Remak method, 176-179 Physical plane, 68 Poinoaré method of balsyage, 170-175 Point at infinity, 132 Poisson equation, 154, 160—165 Poisson formula, 142, 144, 145, 182, 183 Poisson kernel, 141, 144 Polygonal function, 8 Polynomials, approximation by, 5-8 orthonorrnsl sequence of, 9 Potential, 152-163 67

Potential theory, 130 Pressure-denrity relation (equation of state), 67 Principal p5rt, 42, 59 ultrahyperbolic, 68 Principal value, Cauchy, 10, 16 Projection, 101, 110 Projection theorem, 101 Quadratically integrable function, 20 Quasi4inesr 28, 29, 33-36, 68 Range, 85 Refle.ctic.n principle, 148, 149 Jfcgion of effect, 51 Regular boundary point, 174 Resolvent kernel, 117 Riemann function, 56, 57 Eiernann mapping theorem, 211, 212 Caratheodorys of, 213 Remarin.Lebesgue lemma, 4, 5

Index Rieea representthon theorem, 102 Rises-Fischer theorem, 26

Supersonic flow, 68 Surface, developable, 67

(See also L,)

Right inverse, 82 Scbwars inequality, 91 &b,arz reflection principle, 148, 149 Setond-order equations, olaaeifleation of, 43, 58, 5') $egmeut of determination, 51 Seif-adjoint operator, 104 Separability, 76 Separation of variables, 2,2—234

Separation constant. 242 Signum function, 13n. Single layer, 154—157

Singular value, 116 Sound, velocity of. 68 Spheric*l harmoities, 44 Stability of problem, 51, 52 State, equation of, 67 Step function, 25, 121 Stolz region, 17 Stream function, €7 String, vibrating, 232—234 Strip, characteristic, 39 Strong convergence, 104 Subbarmonic functions, 167—170 maximum principe for, 168, 169 Subsouic flow, 68 Subepace, 73 Successive approximation, 17*., 82 Summable function, 25

Telescoping series, 19*. Trisugle inequality, 72, 92 Type, 43, 58, 59 91

principal part, 58 Uniform continuity, x, 1, 2 Unique continuation property, 152 of solution, 137, 138, 219, 227

Unitary matrix, 117 unitary operator, 111 Unitary space, 91 Vector space, 69 (See also Linear space) Velocity of sound. 68 Velocity potential, 67 Volterra integral equation, 89 Wave equation, 42,53—55, 60—65, 232—234 generalized, 241—243

Weierstraas approximatiolL theorem, 5—7

Lebesgte'b proof of, 8 Widder, I). V., 219n. Wrouskian determinant, 237 Zero element, 70 Zero operator, 78