Genera Iized Funct io ns THEORY AND TECHNIQUE
This is Volume 171 in MATHEMATICS IN SCIENCE AND ENGINEERING A Series of Monographs and Textbooks Edited by RICHARD BELLMAN, University of Southern California The complete listing of books in this series is available from the Publisher upon request.
Generalized Functions THEORY AND TECHNIQUE
Ram P. Kanwal Department of Mathematics Pennsylvania State University University Park, Pennsylvania
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IDX
Library of Conpess Cataloging in Publication Data Kanwal , Ram P . Generalized functions: Theory and technique. (Mathematics in science and engineering) Bibliography: p. Includes index. 1 . Distributions, Theory o f (Functional analysis) I . Title. I I . Series. Qe324.K36 I983 51 5 . 7 ' 2 2 3 83-2617 I SBN 0-1 2-396560-8
PRINTED IN THE UNITED STATES OF AMERICA 83 84 85 86
9 8 7 6 5 4 3 2 1
To Vimla. Neeru, and Neeraj
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PREFACE
xiii
1. THE DIRAC DELTA FUNCTION AND DELTA SEQUENCES 1 . 1 The Heaviside Function 1 1.2 The Dirac Delta Function 4 1.3 The Delta Sequences 5 1.4 A Unit Dipole 14 1.5 The Heaviside Sequences 16 Exercises 17 2. THE SCHWRTZ-SOBOLEV THEORY OF DISTRIBUTIONS
20 2.1 Some Introductory Definitions 22 2.2 Test Functions 2.3 Linear Functionals and the Schwartz-Sobolev Theory of Distributions 25 28 2.4 Examples 33 2.5 Algebraic Operations on Distributions 36 2.6 Analytic Operations on Distributions 2.7 Examples 42 47 2.8 The Support and Singular Support of a Distribution Exercises 48 3. ADDITIONAL PROPERTIES OF DISTRIBUTIONS 3.1 Transformation Properties of the Delta Distribution 59 3.2 Convergence of Distributions 3.3 Delta Sequences with Parametric Dependence 60 vii
52
viii
CONTENTS
3.4 Fourier Series 65 3.5 Examples 68 3.6 The Delta Function as a Stieltjes Integral Exercises 72
71
4. DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS 4.1 Introduction 75 4.2 The Psuedofunction H(x)lY, n = I , 2, 3, . . . 4.3 Functions with Algebraic Singularity of Order m 4.4 Examples 86 Exercises 103
80 82
5. DISTRIBUTIONAL DERIVATIVES OF FUNCTIONS WITH JUMP DISCONTINUITIES 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9
I05 Distributional Derivatives in R , 109 R,, n a 2; Moving Surfaces of Discontinuity Surface Distributions 113 Various Other Representations 115 First-Order Distributional Derivatives 117 Second-Order Distributional Derivatives 122 Higher-Order Distributional Derivatives 125 The Two-Dimensional Case 129 Examples 131
6. TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
6.1 Preliminary Concepts 137 6.2 Distributions of Slow Growth (Tempered Distributions) 6.3 The Fourier Transform 141 6.4 Examples 148 Exercises 166
139
7. DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS 7.1 Definition of the Direct Product 169 7.2 The Direct Product of Tempered Distributions 176 7.3 The Fourier Transform of the Direct Product of Tempered Distributions 178 7.4 The Convolution 179 7.5 The Role of Convolution in the Regularization of the Distributions 183 7.6 Examples 185 7.7 The Fourier Transform of the Convolution 194 Exercises 196
CONTENTS
IX
8. THE LAPLACE TRANSFORM
8.1 A Brief Discussion of the Classical Results 199 8.2 The Laplace Transform of Distributions 200 8.3 The Laplace Transform of the Distributional Derivatives and Vice Versa 202 8.4 Examples 204 Exercises 210 9. APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS 9.1 Ordinary Differential Operators 21 1 9.2 Homogeneous Differential Equations 212 9.3 Inhomogeneous Differential Equations: The Integral of a Distribution 2 14 9.4 Examples 215 9.5 Fundamental Solutions and Green’s Functions 217 9.6 Second-Order Differential Equations with Constant Coefficients 218 9.7 Eigenvalue Problems 227 9.8 Second-Order Differential Equations with Variable Coefficients 23 1 9.9 Fourth-Order Differential Equations 235 9.10 Differential Equations of the nth Order 238 24 1 9.11 Ordinary Differential Equations with Singular Coefficients Exercises 242 10. APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
10.1 Introduction 244 10.2 Classical and Generalized Solutions 246 10.3 Fundamental Solutions 248 10.4 The Cauchy-Riemann Operator 250 10.5 The Transport Operator 25 1 10.6 The Laplace Operator 252 10.7 The Heat Operator 257 10.8 The Schrodinger Operator 259 10.9 The Helmholtz Operator 26 1 10.10 The Wave Operator 263 268 10.11 The Inhomogeneous Wave Equation 10.12 The Klein-Gordon Operator 276 Exercises 28 I 11. APPLICATIONS TO BOUNDARY VALUE PROBLEMS 11.1
Poisson’s Equation
1 1.2 Dumbbell-Shaped Bodies
287
11.3 Uniform Axial Distributions
290 294
CONTENTS
X
1 1.4 Linear Axial Distributions 298 299 11.5 Parabolic Axial Distributions, n = 5 11.6 The Fourth-Order Polynomial Distribution, n = 7; Spheroidal Cavities 30 1 I I . 7 The Polarization Tensor for a Spheroid 303 1 1.8 The Virtual Mass Tensor for a Spheroid 306 1 1.9 The Electric and Magnetic Polarizability Tensors 308 11.10 The Distributional Approach to Scattering Theory 310 11.1 1 Stokes Flow 3 17 1 1.12 Displacement-Type Boundary Value Problems in Elastostatics 1 1.13 The Extension to Elastodynamics 325 11.14 Distributions on Arbitrary Lines 332 1 1.15 Distributions on Plane Curves 334 1 1.16 Distributions on a Circular Disk 335
319
12. APPLICATIONS TO WAVE PROPAGATION
12.1 12.2 12.3 12.4 12.5
Introduction 337 The Wave Equation 338 First-Order Hyperbolic Systems Aerodynamic Sound Generation The Rankine-Hugoniot Conditions
340 343 345
13. FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES AT AN INTERFACE 13.1 13.2 13.3 13.4
Introduction 347 Distributional Field Equations of the First Order 348 Applications to Electrodynamics 352 Magnetohydrodynamic Waves in a Compressible Perfectly Conducting Fluid 355 13.5 Second-Order Differential Equations 357
14. LINEAR SYSTEMS
14.1 14.2 14.3 14.4 14.5 14.6 14.7
Operators 360 The Step Response 362 The Impulse Response 363 364 The Response to an Arbitrary Input Generalized Functions as Impulse Response Functions The Transfer Function 366 Discrete-Time Systems 369
365
15. MISCELLANEOUS TOPICS
15.1 The Cauchy Representation of Distributions 37 1 15.2 Distributional Weight Functions for Orthogonal Polynomials
374
xi
CONTENTS
15.3 Applications to Probability and Statistics 390 15.4 Applications of Generalized Functions in Economics 15.5 Distributional Solutions of Integral Equations 409 References 419 Additional Reading Index 423
42 1
399
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Preface
There has recently been a significant increase in the number of topics for which generalized functions have been found to be very effective tools. Familiarity with the basic concepts of this theory has become indispensable for students in applied mathematics, physics, and engineering, and it is becoming increasingly clear that methods based on generalized functions not only help us to solve unsolved problems but also enable us to recover known solutions in a very simple fashion. This books contains both the theory and applications of generalized functions, with a significant feature being the quantity and variety of applications of this theory. I have attempted to furnish a wealth of applications from various physical and mathematical fields of current interest and have tried to make the presentation direct yet informal. Definitions and theorems are stated precisely, but rigor is minimized in favor of comprehension of techniques. Many examples are presented to illustrate the concepts, definitions, and theorems. Except for a few research topics, the mathematical background expected from a student is available in undergraduate courses in advanced calculus, ordinary and partial differential equations, and boundary value problems. Accordingly, most of the material is easily accessible to senior undergraduate and graduate students in mathematical, physical, and engineering sciences. The chapters that are suitable for a one-semester course are furnished with sets of exercises. I hope that this book will encourage applied mathematicians, scientists, and engineers to make use of the powerful tools of generalized functions. My thanks are due to many former students and my colleagues whose reactions and comments helped me in the preparation of this text. In particular, I thank A. Alawneh, R. Ayoub, R. Estrada, D. L. Jain, A. Krall, S. Obaid, R. Rostamian, B. K. Sachdeva, and I. M. Sheffer. A special word of gratitude goes to S. Obaid who also checked the manuscript. I am also grateful to the staff of the Academic Press for their cooperation. xiii
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The Dirac Delta Function and Delta Sequences
1 .l.THE HEAVISIDE FUNCTION
The Heaviside function H ( x ) is defined to be equal to zero for every negative value of x and to unity for every positive value of x ; that is, H(x)
=
0,
x < 0,
1,
x > 0.
It has a jump discontinuity at x = 0 and is also called the unit step,firnction. Its value at x = 0 is usually taken to be ). Sometimes it is taken to be a constant c, 0 < c < 1, and then the function is written H,(x). If the jump in the Heaviside function is at a point x = a, then it is written H ( x - a). Observe that
W-x)
= 1 -
H(x),
H(a - x)
= 1 -
H ( x - a).
(2)
The functions H ( x ) , H ( x - a), and H ( a - x) are drawn in Fig. 1.1. We shall come across the Heaviside function with various arguments. For example, let us examine H(ax h). If a > 0, then this function is zero when ax h < 0 or x h/a < 0 and is unity when x b/a > 0; that is,
+
+
+
+
1
2
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
IY
IY
IY (C)
Fig. 1.1. (a) H ( x ) ; (b) H(x
-
a ) ; (c) H(a
- x),
H ( a x + b) = H ( x + b/a), a > 0. Similarly, when a < 0, we set a = - A , where A > 0, and find that H ( a x + b) = H( - x - b/a). Thus H(UX
+ b) = H ( x + b/a)H(a) + H( -x - b/a)H( -a).
(3) The step function permits the annihilation of a part of the graph of a function F(x). For instance, y = H ( x - a ) F ( x ) is zero before x = a and equal to F ( x ) after x = a. Similarly, the function H ( x - a)F(x - a) translates the graph of F(x), as shown in Fig. 1.2. The function H ( x ) will prove very useful in the study of the generalized functions, especially in the discussion of functions with jump discontinuities. For instance, let F ( x ) be a function that is continuous everywhere except for the point x = <, at which point it has a jump discontinuity:
Then it can be written F ( x ) = F , ( x ) H ( ( - X) It is shown in Fig. 1.3.
+ F,(x)H(x
-
5).
(5)
1.1.
3
THE HEAVlSlDE FUNCTION
Fig. 1.2. (a) F ( x ) ; (b) H ( x - a ) F ( x - a).
This concept can be extended to enable one to write a function that has jump discontinuities at several points. For instance, we can write the function
i
xz,
F(x) =
as
3, 0,
0 < x < 1, 1 < x < 2, x > 2,
F ( x ) = x’H(x)- H(x - 1)( - 3
+ x’)
-
3 H ( x - 2).
IY € Fig. 1.3. F ( x ) = F , ( x ) H ( { - x )
X
+ F , ( x ) H ( x - 0.
4
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
One can similarly write functions with infinite jump discontinuities. For example, the periodic function sin x is such that its positive and negative values alternate; sin x < 0
for (2n - 1)n < x < 2nn,
sin x > 0
for 2nn < x < (2n
+ 1)n.
Accordingly, H(sin x) = =
0, 1, m
1
n=-m
(2n - 1)n < x < 2nn, 2nn < x < (2n + 1)n, [ H ( x - 2nn) - H ( x - (2n + 1)n)l
(7)
The functions sin x and H(sin x) are shown in Fig. 1.4.
1.2. THE DlRAC DELTA FUNCTION
In physical problems one often encounters idealized concepts such as a force concentrated at a point 5 or an impulsive force that acts instantaneously. These forces are described by the Dirac delta function 6(x - 0, which has
5
1.3. THE DELTA SEQUENCES
several significant properties :
and 6(x - 5) dx
= 1.
(3)
Equation (3) is a special case of the general formula
(4) wheref(x) is a sufficiently smooth function (this will become clear as we go along). This formula is called the sijting property or the reproducing property of the delta function, and (3) is obtained from it by puttingf(x) = 1. Although scientists have used this function with success, the language of classical mathematics is inadequate to justify such a function. Indeed, properties (1) and (2) are contradictory, because if a function is zero everywhere except at one point, its integral is necessarily zero, without regard for the definition used for the integral. Fortunately, certain sequences of classical functions exist and have property (4).For instance, the well-known Dirichlet formula
satisfies the sifting property. This suggests that we may define the delta function as the limit of a sequence of suitable functions. The next section is devoted to this concept.
1.3. THE DELTA SEQUENCES
Here we consider various sequences whose limit is the delta function. We have already mentioned the sequence (5), which we now discuss in detail as our first example. Example 1 s,(x)
=
sin mx/nx,
m = 1,2,... .
6
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
Fig. 1.5. s,(x)
=
(sin m x ) / n x .
It is clear that for fixed m as 1x1 becomes large s,(x) becomes small (see Fig. 1.5). Recall the formula
JOrn
71
=
-.
dy
=
-.
dx
=
1.
dx
2
Changing x to - y , we obtain
J-, Y O
Adding (2) and (3) yields
sin y
sin x
71
(3)
2
The result of changing x to mx in the latter formula is
J-
00
d x = 1. 00
This takes care of one of the properties of the delta function. Let us now attend to the sifting property and examine
where f ( x ) is differentiable with f ’ ( x ) continuous and bounded. Observe that, for any b > a > 0, sin mx d x S, Srn(x) d x S, 7 b
=
1
=!
71
1;
y d y ,
(4)
7
1.3. THE DELTA SEQUENCES
and
Similarly, for a < h < 0,
Thus
sin mx J;, 7 dx]
=
f(O)[;;;
=
f ( 0 ) lim
sin mx
[m-m
dx
[-a
]
=
.f'(O),
where E is any positive number, no matter how small, and we have used relation (4)as well as the mean value theorem of integral calculus. This proves the sifting property. Example 2. A very important example of a delta sequence is sm(x)=
1 ~
7c
m
1
+ m2x2'
It is instructive to interpret ( 5 ) as a continuous charge distribution on a line (see Fig. 1.6). It is clear that for m $ 1 , sm(x)4 1 , except for a peak of m/n. The total charge rm(x)to the left of the point x is 1 1 - tanmx, 2 7 c
sm(ir) du = -
+
(6)
the cirmzrlative charge distribution (see Fig. 1.7). Since s,(u) du
=
1
(7)
for every m, it follows that the total charge on the line is always equal to unity.
8
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
Y
X
Fig. 1.6. s,(x)
=
m/[n(l
+ mZx2)].
~~
Fig. 1.7. r,(x)
=
+ (I/n)(tan-' mx).
9
1.3. THE DELTA SEQUENCES
Furthermore, lim s,(x)
=
rn- m
0,
x,
x < 0, x = 0, x > 0.
0,
lim r,(x)
=
m
rn-
x # 0, x=o;
{ 1. I,
(8b)
As m increases, the charge is pushed toward the origin. Thus limrn+m s,(x) describes the charge density due to a positive unit charge located at x = 0. It therefore resembles (we have still to prove the sifting property) the Dirac delta function and is not an ordinary function. The corresponding cumulative charge distribution, which arises from limrn+ r,(x), is the Heaviside function a1
W ) .
The sequence s,(x) will characterize the delta function if we can prove that it satisfies the sifting property
1m
lim
m-m
(9)
f(x)srn(x) dx = f ( 0 )
for a functionf(x) which is bounded, integrable, and continuous at x The proof follows by writing it as
J-
m
where
m
J-
m
f(x)srn(x> dx
2
+ J-
=
0.
ou
m
f(o)srn(x) dx
a1
g(x)srn(x)dx,
(10)
g(x) = .f’(x) - .f’(O).
In view of (7), (10) becomes
J-
m m
f(x)srn(x)
= f(0)
+
g(x)srn(x) dx.
JPam
Thus, the sifting property will be satisfied if we prove that lim ~-mmg(x)sm(x) dx
=
0.
m- m
We have, therefore, to establish that for any such that
1
J:mg(x)sm(x)
dx
< E,
E
> 0 there exists an index N m > N.
10
1.
THE DlRAC DELTA FUNCTION A N D DELTA SEQUENCES
For this purpose let A be a positive number (soon to be specified) that divides the interval - m to cc into three parts, so that
J-mz-
J-
-A
ys, d x =
gs,
dx
+ JAmg.srnd x + /YAgsrndx = I , + 12 + I , .
W
For the integral I , , let the maximum of Ig( in - A I x I A be denoted M ( A ) . Then
Since g(0) = 0 and g(x) is continuous at x = 0, we have limA-,, M ( A ) = 0. Consequently, for any E > 0, there exists a real number A sufficiently small that I I , I < 4 2 , and this holds independent of m. With the number A so chosen, it remains to show that I I I + I , I is small for sufficiently large m. Sincef’(x) is bounded and Ig(x)l < l,f(x)l If’(O)l, it follows that I g(x) I is bounded in - r;c < x < co,say, I g I < b. Then
+
/ I , + I , ~ ~ h [ ~ - ’ s m d x + ~ ~ s m d x--tan-’mA ] 2= b ( ~ 7L
-a2
With the number A fixed, limm+m(2/7c) tan-’ mA = 1. This means that we can find N such that
h[1
- (2/7c) tan-’ mA]
< ~/2,
m > N.
With this choice of N , we have I/:mg(X).i,(x)dx
IIIi
+ 1 2 + 1 3 1 I 1 1 1 + 1 2 1 + (131 < r .
and relation (1 1) follows. This completes the proof.
Example 3. Various books in physics leave the reader with the impression that 6(x) = 00 at x = 0. The following sequence illustrates that this is not always true:
+
-m,
s,(x)
(n
= 2m,
1x1 < 1/2m,
1/2m I1x1 Il/m, otherwise.
(12)
This sequence is shown in Fig. 1.8, which shows that lim, s,(x) = - 03, although it yields the unit positive charge for large m ; indeed, it is a delta sequence, as will now be proved.
11
1.3. THE DELTA SEQUENCES
Y
Fig. 1.8. A delta sequence whose limit tends to
- m.
The cumulative charge distribution rm(x)is sm(ii)du
=
0,
1 x < - -, m 1
=
1
and
< X I - - ,
1 2m'
1 2m
12
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
In particular, sm(x)dx
=
1.
(13)
Let f‘(x) be bounded, integrable, and continuous at x
J-
m
f(X)sm(x) dx
./‘(o)J-msm(x)dx + J-m m
W
=
=
j ( o )+
J;
ou
m
=
0. Then
g(x)sm(x) dx
g(x)sm(x) dx,
where g(x) is defined in (10). From the definition of sm(x)we have g(x)s,(x) dx
g(x) dx - m
= 2m
Since g(x) is continuous at x = 0 and g(0) c > 0 there exists an integer N such that
=
J:‘2m g(x)dx 1i2m
0, it follows that for any given
1x1 < 1/N.
Ig(x)l < 4 3 ,
Thus from (14) we find that, for all m > N ,
J-
g(x)sm(x) dx
J-
- 1/2m
m
2m
1lm
Ig(x)l dx
sf;’”
+m
liZm
I&)l dx
1 im
[ (- - + - i)
< ~ / 32m
+ 2-(;
41,
im
-
+ m
-+(L
:m)
= E,
and the sifting property follows. Example 4. Next we show that the Gaussian sequence from the theory of statistics, (m/n)1’2e-mx’, (15) defines a delta distribution. We prove that for a functionf(x) that is bounded, integrable, and continuous at x = 0 lim00 m-+
ST,
ePmx2f(x)dx
=
f’(0).
13
1.3. THE DELTA SEQUENCES
Indeed, proceeding as in the previous examples, we find that
Also
(18)
where A is a finite real number. In view of the boundedness of J(x), and we have
I&)l
=
I.f’(x>- .f’(O)l < M ,
-
J;r M
+O
:1
dy
m
as m + m .
Similarly, the second integral on the right side tends to zero as m + 00. Finally, we exploit the continuity off‘(x) at x = 0. Given E > 0, there exists a 6 == 0 such that for 1x1
I f ( x ) - f(0)l < E
-= 6.
If we choose A such that 2 A < 6, then we have
>
-+ E
e-s2
G ’-:J
(/y
Adm
as m - c o .
(19)
Combining (17)-(19), we observe that the sifting property (16) has been established. From these examples we find that a sequence of functions each of which has its maximum value at x = 0 and, as we move along the sequence, the maximum value increases while the graph of the function gets narrower, so
14
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
as to lead to the sifting property. The sequences of functions which lead to the delta function in this manner are called delta-convergent sequences. We, therefore, have the following definition : Definition. A sequence s,(x) is called a delta-convergent sequence if lim m-m
s , ( x ) f ( x ) dx
=
m
for all functionsf(x) sufficiently smooth in - co that for a delta-convergent sequence lim s,(x)
=
f(O),
-= x < co.Thus we can say
6(x).
m+ m
In these examples we have taken the unit charge located at x located at x = 5, then the preceding formulas become
J-
=
0. If it is
m
lim
m- m
mSm(X
- < > j ( X ) dx
=~(4;)
and lim s,(x - 5 ) = 6(x - 5). m-m
For example, lim
-
n
=
6(x - 4;).
The sifting property,
is interpreted as the action of the generalized function 6(x - 4;) o n , f ( x ) ; that is, when 6(x - 5) acts on a suitably smooth function . f ( x ) ,it sifts out the value,f({) at x = 5.
1.4. A UNIT DIPOLE
We have seen in previous examples how certain convergent sequences converge to a delta function and represent idealized concepts such as a unit
15
1.4. A UNIT DIPOLE
Fig. 1.9. t,(x)
=
2m3x/rr(l
+ m2x2)2.
charge. We shall now prove that the derivative with respect to x of a deltaconvergent sequence gives a sequence that represents a unit dipole. Let charges fm be located at x = fE, respectively. The product 2 m is ~ known as the dipole moment of the charge configuration. When we let E -,0 and m + co in such a way that 2 m = ~ 1, we get the dipole moment equal to 1, i.e., a unit dipole. Our contention is that it can be approximated by a continuous charge distribution that is the derivative of a corresponding distribution of a unit charge. For this purpose we shall take the sequence s,(x) = m/n(l + m 2 x 2 )and show that t,(x) = -
dsm dx
~
=
2m3x n(1 + m2x2)”
describes a unit dipole. A sketch of t,(x) for large values of rn is given in Fig. 1.9. Furthermore, t,(x) d x = s,(u) - s,(b) =
m
n(1
+ m2a2)
-
m
n(l
+ m2b2)’
(1)
If neither a nor b is zero, each term on the right side of (1) approaches zero as m + co,and the total charge in any such interval goes to zero as m + co. However, t,(x) dx + + 00 (like a positive point charge just to the right of the origin), and 1; t,(x) dx + - co (like a negative point charge just to the left of the origin), whereas the first moment about the origin is JTrn xt,(x) dx = 1. Consequently, for large m, c,(x) approaches a unit dipole located at x = 0 and is thus a dipole sequence.
fi
16
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
Finally, let us consider the action A,[f'] of r,(x) on a suitably smooth function .f'(x), A,[j']
=
m+
=
J-
m
lim 00
t,(x)f'(x) d x m
ds,
lim
m-m
= lim [ - ~ , f ' ] 0 3 ~ m+ m
=
J'(4 d x
+ lim
n+ m
(df'/dx)(O).
l-mm s , -dl' dx dx
(2)
In the next chapter we shall prove that a generalized function with property ( 2 ) is the derivative of the delta function.
1.5. THE HEAVISIDE SEQUENCES
Finally, let us mention that we can also define a Heaviside function on the same lines. Indeed, we have already come across such a sequence, namely, (1.3.6). Another example is h,(x)
=
L
mx
x > 1/2m, - 1/2m I x I 1/2m, x < -1f2m.
+ 3, 1
(1)
To prove that it is a Heaviside sequence, observe that
J- mhm(x)f'(x)d x
1-
- li2m
m
=
+ As n
---f
hm(x)J'(x)d x
1i 2 m
h,(x)f'(x) d x
+
m
J12m
f ( x )dx.
m we have
jm
lirn
m+ m
h,(x)f'(x) d x
=
lom
f ' ( x ) d x = ( H ( x ) , .f'(x)).
where the symbol ( 4 ( x ) , $(x)) stands for JEOO ~ ( x ) $ ( xd)x . Note that occasionally, when the range of integration is the entire space, we shall omit the limits on the integral sign.
17
EXERCISES
EXERCISES
Show the following:
+ x)H(t - x) = H(r2- x2)H(t).
1. H ( t - 1x1) = H(r
i
min(a, b) < x < max(a, b), x < min(a, b) and x > max(a, b).
2. H{(x - a)(x - b)} = 0, 1,
3. H(er - n) = H ( t
-
In
71)
=
i
0, 1,
t < In n, t > Inn.
4. 5. Identify the Heaviside functions H(cos x), H(sinh x), and H(cosh x).
6. Consider the sequences of functions
1x1 > 1/2m, -1/2m 5 x 5 0, (b) sm(x)= 4m2x + 2m, -4m2x 2m, O I x I 1/2m.
(0,
+
Sketch these functions and the corresponding cumulative distributions and prove that limm-,msm(x)= 6(x). 7. Show that the sequences (a) +me-"IXI and (b) (l/n)m/(emx delta sequences.
+ eVmx)are
8. Let sm(x)be a sequence of nonnegative functions such that
s,(x)dx (b) mlim + m Cs,(x)dx
=
1;
=
0, 1,
a, b
> 0, or a, b < 0,
a < 0 and
b > 0.
Show that sm(x)is a delta sequence. 9. By differentiating the sequence of Example 3 show that it yields a unit dipole and is thus a dipole sequence. 10. By differentiating the Heaviside sequence (1.5.1) show that sm(x)= dh,/dx is a Dirac delta sequence.
18
1.
THE DlRAC DELTA FUNCTION AND DELTA SEQUENCES
11. Show that the sequence
+
sx
< 0, O Ix < 1/m, all other values of x
{;2x m, s,(x) = m - m2x,
- l/m
is a delta sequence. 12. I f f ' ( x )is a nonnegative function satisfying jTODf ' ( x )dx = 1, show that {m.f(mx)}is a delta sequence. 13. Let s,(x) be a delta sequence consisting of even functions. Let g(x) be an integrable function having a jump discontinuity at 0. Show that OD
m-
OD
where g(0 +) stands for the limits of g(x) as x * 0 from the right and from the left; i.e., g(O+) = lime+,, g(0 k).
+
14. Show that the sequence O I x < l/m, - l / m < x < 0, (XI 2m
-m2,
is a dipole sequence. 15. Show that each member of the sequences
and
1 s,(x) = - {tan-' n(x 27L
+ t ) - tan-'
n(x - r)},
satisfies the wave equation d 2 u / a t 2 - a2u/ax2= 0. Deduce that s(x) = ){6(x + t ) + 6(x - t ) } are also solutions of this equation. 16. (a) Prove that
and
s(x) = * { H ( x + t ) - H ( x - t ) }
19
EXERCISES
is a delta sequence. Sketch s,(x) for n = 1, 2, 3, 4. (b) Show that the relation
J1
P,(x)
=
1
.f(x
+ t)s,(t)
dt,
0 I x I1,
yields a sequence of polynomials. (c) With the help of relations (a) and (b), prove Weierstrass’s approximation theorem: If a functionf(x) is continuous on the closed interval [a, b ] , then there exists a sequence of polynomials P,(x) such that limn+mP,(x) = f(x). Hints. (i) There is no loss of generality in taking the interval [0, 11and in assuming thatf(x) vanishes at x = 0 and x = 1. (ii) The required polynomials are the ones given in (b).
The Sc hwartz-So bolev Theory of Distributions
2.1. SOME INTRODUCTORY DEFINITIONS
Let R , be a real n-dimensional space in which we have a Cartesian system of coordinates such that a point P is denoted by x = (xl, x,, . . . ,x,) and the distance r of P from the origin is r = 1x1 = (x: + x: + ... + x:)”’. Let k be an n-tuple of nonnegative integers, k = (Al, k , , . . . , A,), the so-called mlrltiinciex of order n ; then we define
Ikl = k l + k2 + ... + k,, k ! = kl!k2!...k,!.
Xk
=
x:Ixk,2
...
.>,
and
where D j = c7/axj,j = 1, 2, . . . , n. For the one-dimensional case D k reduces to d/dx. Furthermore, if any component of k is zero, the differentiation with respect to the corresponding variable is omitted. For instance, in R,, with k = (3,0,4), we have D~ = d7/dx:
ax;
20
=
0;~;.
2.1.
21
SOME INTRODUCTORY DEFINITIONS
A differential operator L of order p is defined as L
ak(X)Dk,
= Iklsp
where ak(x) are given functions and the sum is taken over all multiindexes k of order n. For example, when n = 1, we have the ordinary differential operator L
=
u,(x) dP/dXP+ a,- ,(x) dP- ‘/d~’-
+ . . . + u,(x).
(3)
As another example, the second-order partial differential operator in R , is
L
=
a,,,(x)
a2/aX:
+ a , , l(x) a2/ax, ax, + aO,,(x) a2/ax:
+ a,.o(x) W x 1 + a,,,(x)
Wx,
+ ao,o(x).
(4)
We shall use the following notation for the integral (in the Lebesgue sense) of a functionf(x) = f(x,, x,, . . . , x,) over an n-dimensional region R :
1... s,
f(x,, x,, . . . , x,) dx, dx, . . . dx,
=
s,
. f ( x ) dx,
(5)
where dx = dx, dx, . . . dx,. In the sequel we shall encounter similar integrals over hypersurfaces of dimensions n - 1 in R,, such as a two-dimensional surface S in R , and a curve C in R,.
Definition. A function f ( x ) is locally integrable in R, i_f f R I f(x) I dx exists for every bounded region R in R,. A functionf(x-) isfocafl’fi integrable on a hypersurface in R, if jsI ,f(x)l dS exists for every bounded region S in R,-
,.
The class of locally integrable functions is rather wide. All piecewise continuous functions are locally integrable. Some functions that are infinite, such as l / r m ,m < n, are also locally integrable. However, 6(x) is not a locally integrable function for the following reason. Let I , , I , , . . . , I , be a sequence of nested intervals (i.e., each interval is included in the preceding one) whose length tends to zero; then lim,,,J,n f(x)dx = 0, but 6(x) gives a finite value. Another concept that plays a crucial role in the theory of distributions is the support of a function, defined as follows:
Definition. The support of a functionf’(x) is the closure of the set of all points x such that ,f(x) # 0. We shall denote the support offby supp ,f.For example, forf’(x) = sin x, x E R , , the support off(x) consists of the whole real line, even though sin x
22
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
Fig. 2.1. A function with compact support.
vanishes at x = nn. If supp .f is a bounded set, thenfis said to have compact support. For instance, the support of the function (Fig. 2.1)
I0, f ( x >=
x+l, 1-x,
--co
<Xl-l,
-1<x
is compact.
2.2. TEST FUNCTIONS
We have observed that an operational quantity such as 6(x) becomes meaningful if it is first multiplied by a sufficiently smooth auxiliary function and then integrated over the entire space. This point of view is also taken as the basis for the definition of an arbitrary generalized function. Accordingly, consider the space D consisting of real-valued functions 4(x) = 4(xl,x2,.. . ,xn), such that the following hold: 1. 4(x) is an infinitely differentiable function defined at every point of R,. This means that D k 4 exists for all multiindices k. Such a function is also called a C" function. 2. There exists a number A such that 4(x) vanishes for r > A . This means that 4(x) has compact support. Then 4(x) is called a test function.
The prototype of a test function belonging to D is
23
2.2. TEST FUNCTIONS Y
Fig. 2.2. The test function defined by Eq. (2.2.1) for n = 1
shown in Fig. 2.2 (where n = 1). Its support is clearly r I a. By taking a = 1 and multiplying 4 ( x ) = $(x, 1) by a suitable normalizing factor we can construct the function
satisfying the conditions
4 E D, J R n 4 ( x )d x
= 1,
supp 4
= A(
= r I l), x E R,.
(2)
so that c-'
=
J-
r s1
exp(- 1/1 - r 2 ) d x .
This function, in turn, gives us the test function M X ) =
C&4(X/d =
with the properties JR.
$&(x)d x
=
i,
c, exp( -
1,
c, = Jarexp(
E2
r < 8, r2
SUPP
-
-),
-)
4, E2
(3)
E,
= A( = r IE ) ,
dx*
(4)
24
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
We can also form a delta sequence out of this function. Indeed, the sequence
where c',
=
J
r s lim
exp( - L) m2r2 dx,
is a delta sequence. The proof follows by appealing to the generalization of Exercise 8 of Chapter 1. We shall show in Chapter 3 (see Example 2 of Section 3.2) that the sequence 4,(x), depending on a parameter E, as defined by ( 3 ) also approaches d(x) = d(x,, . . . , x,) as E + 0. The following properties of the test functions are evident.
(1) If 4, and 42 are in D, then so is c14, + c 2 4 2 , where c , and c2 are real numbers. Thus D is a linear space. (2) If 4 E D, then so is 0'4. (3) For a C" function ,f(x) and for a 4 E D, ,f4 E D. (4) If 4(x,, x 2 , . . . , x,) is an m-dimensional test function and $(xm+ x,, 2 , . . . , x,) is an (n - m)-dimensional test function, then 41) is an n-dimensional test function in the variables x,, x 2 , . . . , x,.
,,
Note that the definition of D does not demand that all its elements have the same support. Take, for example, the functions 4(x) as defined by (1) in R , , then 4(x) and +(x - 3 ) are both members of D although their supports are, respectively, ( - 1, 1) and (2,4). It is more convenient to work with the basic notions of convergence than to introduce an inner product into this space. The kind of convergence we need is defined as follows:
Definition. A sequence { 4 m }m, = 1, 2, . . . , where if the following two conditions are satisfied:
4mE D, converges to 4o
(1) All 4mas well as 4o vanish outside a common region. (2) D k 4 , + D k 4 , uniformly over R , as m + m for all multiindices k .
It is not difficult to show that 4o E D and hence that D is closed (or is complete) with respect to this definition of convergence. For the special case 4o = 0, the sequence {q5m} is called a null sequence.
2.3.
25
LINEAR FUNCTIONALS AND THE SCHWARTZ-SOBOLEV THEORY
where $ ( x , a) is defined by (I), is a null sequence. However, the sequence { ( l / m ) $ ( x / m ,a)} is not a convergent sequence, because the support of the function $(x/m, a) is the sphere with radius ma, which is different for different m. In addition to the space D of test functions, we shall use certain subspaces of D. For a region R in R,,the space DR contains those test functions whose support lies in R , that is, { $ : $ E D, SUPP$ c R } . (7) DR It is clearly a linear subspace of D. For example, D, and D, are two onedimensional subspaces of test functions 4 ( x ) and 4(v) and are contained in DxJ,which is the space of test functions $ ( x , y ) in R , . The convergence in DR is defined in the same manner as that in the space D.
2.3. LINEAR FUNCTIONALS AND THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS A linear functional t on the space D is an operation (or a rule) by which we
assign to every test function # ( x ) a real number-a ( t , 4), such that ((7
Cl4l
+ CZ42)
for arbitrary test functions that
$ 1and
= Cl(L $1)
functional-denoted
+ c , ( t , $2)
$, and real numbers c1 and
(r. 0 )
=
(1) c 2 . It
follows (2)
0,
and
where cjare arbitrary real numbers. The next concept is that of the continuity of the linear functionals. It is defined as follows:
7
Definition. A linear functional on D is continuous if and only if the sequence of numbers ( t , $ m ) converges to ( r , 4) when the sequence of test functions {4m}converges to the test function $ (in the sense of the convergence as defined in the previous section). Thus lim ( r , $ m ) = ( t , lim 4,). (4) m- m
m-+ w
We now have all the tools for defining the concept of distributions [l].
26
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
Definition. A continuous linear functional on the space D of test functions is called a distribution. Regular Distributions
The set of distributions that are most useful are those generated by locally integrable functions. Indeed, every locally integrable function f ( x ) generates a distribution through the formula
Linearity of this functional is obvious. To prove its continuity we observe that
I(L 4)l 5 max
If(x>ldx <
I+(X)lJ
XSSUPPO
0.
SUPP6
Thus, if the sequence {4,,,} converges to zero, then so does (1; 4,). Hence, it is continuous. Distributions defined by ( 5 ) are called regular. All other distributions are called singular. However, we may use formula (5) symbolically for a singular distribution also. Distributions can be defined by partial differential operators as well. If f ( x ) is locally integrable function, we can define a distribution as
(L 4)
=
J
R"
4 E D.
f ( X ) D k 4 ( 4dx,
(6)
Remark 1 . The constant c has three meanings in this book: (a) c is a number; (b) c is a constant point function; and (c) as a constant point function, c is locally integrable and generates the distribution (c,
4)
=
JRn
c4(x)dx = c
JR"
4 ( x ) dx.
(7)
It will be clear from the text what c stands for whenever it occurs. The zero distribution on D has the property (094) = 0,
4 E D.
(8)
Remark 2. The definition of a distribution can be extended to include complex-valued functions of a real variable. The arbitrary constants c1 and c2 occurring in the foregoing definition are then complex numbers. The space of test functions is then called D@).Accordingly, we have the following definition :
2.3.
LINEAR FUNCTIONALS AND THE SCHWARTZ-SOBOLEV
THEORY
27
Debition. A distribution t is a complex-valued functiona on D(‘) such that (1)
((9
Cl4l
+ c242)
( 2 ) 1imm-m ( t 3 4 m ) =
+
= c ‘ l ( t , 41) c2(t, ( t , W,,+m &),
42),
where &(x) are elements of D(‘). Remark 3. The distribution (f,4) will also be denoted f.
Space D’ The space of all distributions on D is denoted D’. The distributions t , and t 2 give rise to a new distribution t = c l t 1 + c , t 2 such that ( 9
4)
= (Cltl
+ CZf2,4)
=
Cl(tb4)
+ c2(t2,4).
(9)
It is easily verified that this satisfies the requirements of a distribution. Thus D‘ is itself a linear space. It is called the dual space of D and is a larger space than D.It forms a generalization of the class of locally integrable functions because it contains functions such as 6(x) (see Example 2 in Section 4)that are not locally integrable. For this reason distributions are also called generalized or symbolic functions. We shall use the terms “distribution” and “generalized function ” interchangeably. Another basic idea in the theory of distributions is embedded in the following theorem. Theorem. Two continuous functions that produce the same regular distributions are identical. Proof. The theorem follows from ( 5 ) as the only continuous functionf(x) for which (f,4) = 0, for all 4 E D, is f(x) = 0, x E R , . This is because if there is a point xo such thatf‘(x,) # 0, then there exists a spherical neighborhood of radius E about xo in whichf’(x) # 0 (say it is positive). Now the test function ~ J ( X- xO)/&]defined by (2.2.3) is positive in the sphere Ix - xoI < E and vanishes outside it, for this test function (f,4) is greater than zero, which contradicts the hypothesis. The same is true forf(xo) < 0. This proves the theorem. For locally integrable functions this theorem does not hold, because we may alter the values o f j ( x ) on a set of measure zero without altering the regular distribution. Accordingly, we stipulate that iff(x) and g(x) are locally integrable functions and are equal almost everywhere, then they generate the same distribution, and we have
28
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
2.
2.4. EXAMPLES
Example I. The Heaviside distribution in R , is (HR,
,
4)
=
J 4 ( x ) dx,
where H R ( x ) =
R
XER, 0, x $ R . 1,
(1)
For R ,( 1 ) becomes (H,
4)
=
JD4(X) 0
dx.
(2)
Since H ( x ) is a piecewise continuous function, this is a regular distribution. Example 2. The Dirac delta distribution in Rn is (6(x - 8 , 4 ( x ) > =
(3) for 5 is fixed point in R , . Linearity of this functional follows from the relation (67
Cl41
+ c242)
=
c141(5)
+ c242(5)
4 ( 0 3
= Cl(6741)
+ c2(6,42)9
where c 1 and c2 are arbitrary real constants. To prove continuity we observe that limm+m(6, 4 m )= limm+m&(<). However, if 4 m ( -+ ~ )0, then &(() -+ 0, and we have continuity. Thus the delta function is a distribution. We observed earlier that the delta function is not locally integrable. This distribution is therefore a singular distribution. Example 3. Generalized functions related to the delta function. Various related generalized functions are useful in electrical and electronic engineering problems. For example, an infinite sequence of impulses (i.e., an infinite row of delta functions) is described by
cm
III(x) =
S(X -
n).
n=-w
(4)
This is called the sampling or replicating function because it gives the information about a function , f ( x ) at x = n: u
W x ) f ' ( x )=
( W X ) ,
.f'(x)> =
1
n=-cc
f ( x ) 6 ( x - It),
as is clear in the Fig. 2.3. Other quantities of interest are the impulse pair functions, which are defined as (Fig. 2.4) II(x) = +s<x I,(x) = + S ( X
+ +) + fscx - i), + f) - f 6 ( x - f).
(5)
(6)
2.4.
29
EXAMPLES
Fig. 2.3. The sampling function.
Fig. 2.4. The impulse pair functions.
Example 4. The function l/x does not define a regular distribution, because f T m [ ~ ( x ) / x ]dx is not convergent for all the test function (e.g., 4(x) = c). However, we can employ the notion of the Cauchy principal value and define
30
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
where the * to the right of the integral sign indicates the principal value. This limit exists for the following reason. Since 4 ( x ) is differentiable at x = 0, there is a function $ ( x ) continuous at x = 0 such that 4 ( x >= 4(0)
+ x$(x>.
(8)
Let [ - A , A ] be the support of the test function 4 ( x ) ,so that for E > 0,
=
J
A > 1x1 > E
$(x)dx
A +
j - A * ( x ) dx,
as
E + 0.
(9)
The functional ( 9 ( l / x ) , 4) so defined is clearly linear. To prove its continuity we appeal to relation (9) and find that
(9($ 4)
= JYA$(x) dx I 2 A max I $ ( x ) I,
-A I x I A,
by the mean value theorem. Thus 9(l / x ) is a distribution. This distribution is also called a pseudofunction; we shall write it as Pf( I / x ) in the sequel. Many more pseudofunctions are defined and analyzed in Chapter 4. Example 5. From Example 4 it follows that the functions 6'(x)
=
f 6 ( x ) T (1/2ni)Pf(l/x)
(10) are also singular distributions on D and are called the Heisenberg distributions. We can further show that
To prove this relation we proceed as follows. Let +(x) be a test function, and let $ ( x ) be the continuous function satisfying (7) and where the support of 4 ( x ) is [ - A , A ] . Then for each E > 0,
x - is
=
2i4(O) tan- E
+
dx
+
A
x-iE
x*(x) dx
x $ ( x ) dx.
2.4.
31
EXAMPLES
Thus,
where we have used results (4) and (6). Similarly,
Relations (12) and (13) are called Plemelj formulas. Writing them as
we obtain the required formulas (1 1). Symbolically, we can write them as 6*(x) = T lim
e+o
1
7 2711 x ~
- 1 1 1 =+-2ni x f io’ f ie
(15)
so that 6(x) = 6+(x) - Pf
ii
(1) -
=
+ 6-(x)
= lim
1
6-(x) - 6+(x) = lim
E
n x2 + E2’
-
E-.O
~
1
x
n x2 + e 2 ’
~
E-.O
(16)
~
Note that by setting E = l/m we get the delta sequence of Example 2 in Section 1.2. Formulas (1 1) are sometimes written
l/(x I/(x
+ i0) = - i d ( x ) + Pf(l/x), - i0) = ind(x) + Pf(l/x)
and called the Sokhotski-Plemelj equations. Example 6. Let us find the solution of the equation (x
-5)O
-
5) = dx).
The homogeneous equation (X
- 5Mx -
5) = 0,
(18) (19)
32
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
clearly has the solution 6(x - 4;), because The generalized particular solution is
oo
( x - 5)6(x - 5)4(x) dx
= 0.
as can be readily verified. Accordingly, the solution of equation (20) is t(x -
5 ) = 6(x - 5 ) + g ( x ) Pf
-
(x
5).
This solution was first derived by Dirac and is very useful in transport theory. Example 7. Single-layer distribution. Let us discuss a distribution in R, that is zero outside a hypersurface S and that is not the zero distribution. Distributions of this kind are called surface distributions and have important applications. They are generalizations of the 6 function. Let o(x) be a locally integrable function defined on S. In the language of electrostatics, we have a charge density of strength a ( x ) dS, spread over S ; i.e., the surface element dS at 4; E S carries a concentrated source density of strength o(4;)dS. The corresponding volume source density in space is a({) dS, 6 ( x - 5). Accordingly, the total volume density is t ( ~= )
L
The action off on a test function
0(4;)6(~- 4;) dS,.
4 is
(24) Note that we can write this relation (S(x)S(S)), where 6(S) denotes the surface distribution of unit density S, because
From (24) it follows that we may call t the distribution corresponding to a surface layer of sources spread over S with density a(x). This layer is called a single (or simple) layer. Example 8. Many physical quantities can be expressed by distributions in a natural way. In this example we consider a material system formed by a
2.5.
33
ALGEBRAIC OPERATIONS ON DISTRIBUTIONS
closed interval [a, b] on which a mass is distributed with density p(x). The total mass M of the system is M =
1
p(x)dx.
Ax
The abscissa X of the mass center is the first moment X
=-
x ~ ( x dx. )
Next, we extend the function p ( x ) outside [a, b] by zero so that we can write (26) and (27) as the distributions M = ( 1 , p),
respectively. As a special case, let us put p M
=
X
= md(x -
X
(1, md(x - A ) ) = m,
(28)
= ( l / M ) ( X , p>,
A). Then (28) becomes
= ( l / m ) ( x , m6(x -
A))
=
mA/M
=
A.
Thus, the distribution md(x - A ) completely characterizes the material part of abscissa A and mass M . This concept can be readily generalized to include a system of material points as well as various other moments.
2.5. ALGEBRAIC OPERATIONS ON DISTRIBUTIONS
We have already defined taking the sum of two distributions and multiplying a distribution by a constant. We now define some other algebraic operations.
Linear Change of Variables Let (f, 4) be a regular distribution generated by a function , f ( x ) that is locally integrable in R,. Let x = A y - a, where A is an n x n matrix with det A # 0 and a is a constant vector, be a nonsingular linear transformation of the space R, onto itself. Then we have
-
1
1f ( x ) 4 [ A - '(x + a)] dx ldet A l
where A - ' is the inverse of the matrix A.
R,
34
2.
THE SCHWARTZ-SOBOLEV
THEORY OF DISTRIBUTIONS
Fortunately, the same definition applies to a singular distribution because [ A - '(x a ) ] is a valid operation on @(x). Accordingly, we have
+
(t(AY - a)?4 ( Y ) >
Remark 1 . yields
=
(t(xX
4CA - '(x + 4l/l det A I >.
For a simple translation, i.e., when A ( f ( Y - a),
=
(2)
I , the unit matrix, ( 2 )
m>= ( t ( X > , 4 ( x + 4 ) .
(3)
For example
(S(Y - a), 4 ( Y ) >
=
4(x
(6(x),
+ a>> = +(a).
(4)
S(y - a ) is sometimes denoted S,(y). A distribution t is said to be invariant with respect to translation over a distance a, or to be periodic with respect to a, if
( N Y - 4,4(Y)>=
(W 4(x + a>> =
( t 3
4).
(5)
For instance, the regular distribution (on D")),
1m
( o i l u x , +(XI>
=
eiwx4(x> dx,
a
which is generated by the locally integrable function eiwx,is invariant with respect to translation over the distance 2nn, n = 0, 1, 2,. . . . That is, ((,im(xf
2nn) 7
4(x)>
= (eiWX7#(XI>.
Remark 2. For a simple scale expansion, A = cl, a = 0, (2) becomes A distribution is called homogeneous of degree I if
(7)
t(cx) = cAt(x),
for c > 0. This means that the relation (t(CY),
4(Y)>= c - " ( w ,
4(x/c)> = c"t,
4 > 3
or (r(x), 4(x/c))
= ci.+"(t,
4),
(8)
holds. For the special case of the delta function, ( 6 ) becomes (4CY),
4(Y>> = ( 1 / l C n l K w 7
4(x/c)> = ( 1 / I C " l ) ( W > 4(x)>,
or 6(cx) = (1/1 c" 1)6(x).
(9)
35
2.5. ALGEBRAIC OPERATIONS ON DISTRIBUTIONS
Thus, 6(x) = 6(x1,. . . , x,) is a homogeneous distribution of degree -n. The reflection of a distribution in the origin means that we take x = -y, after which (2) yields (t(-Y),
4 ) = (t(x), 4(-x)>.
(10)
For delta distribution, (10) becomes
4- Y h
4(Y)> = (6(x),
4( -x)>
=
4(0) = (6(x), 4(x)>,
or 6(-x) = 6(x); that is, the delta function is an even function. A distribution that is invariant under reflection in the origin, so that (t(-Y), 4(Y)> =
d4-X))
(tc-4
=
4),
(1 1)
is called a centrally symmetric or even distribution. Similarly, a distribution with the property (f(-X)9 $(XI>
=
(t(x),
4(-x)>
is called a skew symmetric or a odd distribution.
=
- ( t 3
4L
’,
(12)
Remark 3. For a simple rotation, a = 0 and A’ = A - where A’ and A are, respectively, the transpose and the inverse of A ; we then have (t(AY), 4(Y)> = (t(x), 4(A’x)lldet A I >.
’
(13)
Product of a Distribution and a Function In general, it is difficult to define the product of two generalized functions. It may not be possible to do so even if they are locally integrable, as is clear from the examplef’(x) = g(x) = l/x’”. However, we can always assign a meaning to the product of a generalized function t(x) and an infinitely differentiable function $(x) by setting ($L because
4)
=
(h
$4),
(14)
$4 is an element of D.
Example 1. For a constant c, we know that (ct, 4 ) = c ( t , 4). With this definition of a product distribution we have (ct, 4 ) = ( t , c 4 ) = c(t, 4 ) for a function c. Thus both results are consistent. Example 2.
<+&4) = (6, $4)
= +(0)4(0) = J/(O)<6, 4 ) =
$(X)W) = $(0)6(x).
($(OM, 4 ) or (15)
36
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
It follows that in this special case it is sufficient for the function $ ( x ) to be continuous at the origin. More generally, $(X)W Example 3.
- 5) = $ ( 5 P ( X
-
5).
(16)
Let us show that xn Pf(l/x)
=
xn-l,
n a positive integer.
(17)
We have
2.6. ANALYTIC OPERATIONS O N DISTRIBUTIONS
In this section we present the definition of the operation of taking the derivative of a distribution. This definition should yield consistent results when we differentiate a distribution that is also a classical function. Therefore, let us first consider a regular distribution generated by a continuously differentiable function f ( x ) = f ' ( x l , x 2 , . . . , xn),
This integral representation has a real and not a merely symbolic meaning. When we integrate the integral
by parts, we get
where we have used the fact that 4 ( x ) has compact support. Since d 4 / a x j is also a test function, we write'(1) (aflax,,
4 ) = (1; - a 4 l a x j >
=
-(,f, a 4 / a x j > .
This helps us in defining the distributional derivative a t / a x j :
2.6.
37
ANALYTIC OPERATIONS ON DISTRIBUTIONS
This result is remarkable, for we have transformed the difficulty of differentiation of a generalized function to the differentiation of a test function that has derivatives of all orders. This is precisely what happened to the algebraic operations in the previous sections. Let us satisfy ourselves that the functional defined by (2) is a distribution. Linearity is obvious, and continuity follows from the fact that if {&} is null sequence, then so is the sequence {a4,/axj} (recall the definition of convergence). The partial derivatives of higher orders can be defined by repeating this process. This results in the general relation
where D k is defined by (2.1.1). In R , we have only ordinary derivatives, and relations (2) and (3) become and
Properties of the Generalized (or Distributional) derivatives ( 1 ) From (3) and ( 5 ) it follows that a generalized function is infinitely differentiable. (2) The equality t'(x) = 0 holds if and only if the distribution t(x) is a constant. From this relation it follows that if t;(x) = t;(x), then r , ( x ) and t 2 ( x ) differ by a constant function. (3) The generalized derivative agrees with the classical one whenever the latter exists. (4) Because
-
(f.
a ax, a x , ) (2 ax, (5). ax, =
it follows that
a, t dx, ax,
-
a2t
ax, ax,
- D't.
(b),
38
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
Thus, the result of differentiation does not depend on the order of differentiation. (5) The derivative of the product of a distribution t and a function f ( x ) E C is
a
-( , f t ) d xj
Indeed, for
=
-
a xj
t
+ f -.aatx j
6 E D,
which is the same as (8). The higher-order derivatives of the product can be defined in a similar fashion and we have Dk(,ft)=
n, t I 1 = k
(k!/m!n!)(D"f')(D"t).
A Differential Operator and Its Adjoint
With the differential operator as defined in Section 2.1 and the principles of differentiation as given in this section, we have
where the pth-order differential operator L*, given by
2.6.
39
ANALYTIC OPERATIONS ON DISTRIBUTIONS
is called the formal adjoint operator to L. If L = L*, the operator L is selfadjoint. For example, the Laplacian operator V2,
is self-adjoint, because (v't, 6) = ( t , V26>,
6 E D.
When L is an ordinary differential operator, L = ap(x)
,
dP dP- 1 + u p - __ dxp dXP~
+ . . . + a , dxd + a,, -
(1 1 )
(10) yields
Definition. A distribution E is said to be a,fundamenta/ solution for the differential operator L if LE = 6. (13) We end this section with two important theorems in which we use, among other concepts, Examples 1 and 2 of the Section 2.7.
Theorem 1. If f(x) is a classical function and f(x)
+ a,6(x
-
5) + .* .
+ a,d'"'(x
-
5) = 0
on the whole axis, - cc, < x < m, thenf(x) = 0 and a,
=
. . . = a,
(14) = 0.
Proof. We proceed by induction. The case n = 0 is obvious because 6(x - 5 ) is a generalized function andf(x) is a classical function. Suppose that the theorem holds for n - 1. Then it follows thatf(x) = 0 for x # 5, so (14) integrates into C'
+ u,H(x
-
5 ) + a,6(x - 5 ) + ... + u,S'"-"(X - 5 ) = 0.
However, we have assumed that the theorem holds for n - 1, so
+ a,H(x - g) = 0, a , = a2 = ... = a, = 0. From the relation c + a,H(x - 5 ) = 0, we find that a, = 0, and we have c
proved the theorem.
40
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
Let a functionf(x) be n times continuously differentiable; then
Theorem 2.
f(x)d(")(x)= ( - l)"f'")(O)d(x)+ ( - l)"-'nf"- '(O)d'(x)
+ (-
1)"-2
n(n
- 1)
2!
f'"-2'(0)d''(~)+ . . . + f(O)d'")(x). (15)
After a succession of similar integrations by parts, we obtain
1m
m
J-
m
{ ~ ( x ) @ ( x ) } ~ ' "dx ) ( x=) (- 1)"
Substitution of the formula
m
{ ~ ( x ) ~ ( x ) } ' " ' dx. ~(x)
in the preceding relation and the application of the sifting property yields m
{,f(x)P"(x)}4(x) dx
= (-
+
+ nf'"-
1)" f'"'(0)4(0)
n(n - 1) f'"-2'(0)f"'(O) 2!
')
(0)4'(0)
+ .. + /(O)d'"'(O)}. '
Because
J-
m
J-
m
W
4(X)d'k'(X) dx
= ( - l)k
+'k'(X)d(X) dx m
= (-
l)"'k'(O),
2.6.
41
ANALYTIC OPERATIONS ON DISTRIBUTIONS
relation ( 1 7) can be written
+ ( - 1)-
'n,f("- "(0)(6'(x),4 ( x ) )
+ . .. + ( -
which is equivalent to the symbolic formula (15). For the special cases n = 1,2, 3, ( 1 5 ) becomes f(XP'(X) =
I
l ) - " ( P ( x ) ,&x)> ,
-,f'(0)6(x) + f(0)6'(x),
,f(x)b"(x)= f"(0)6(x)- 2f'(O)d'(x)+ f(O)S"(x),
(18)
(19)
, f ( x ) 6 " ' (= ~ )- , f " ' ( 0 ) 6 ( ~+) 3,f"(0)6'(~) - 3,f(O)S"(x)+ .f(O)S"'(x), (20) respectively. For instance, it follows from (18) that sin x d'(x) = -(cos O)f'(O)S(x)
+ (sin 0)6'(x) = -6(x)
(21)
and cos x 6'(x) = - ( -sin 0)6(x)
If we introduce the function 6(x use the binomial coefficient
+ (cos 0)6'(x) = 6'(x).
(22)
5 ) instead of 6(x) in formula ( 1 5 ) and
(i)
n! = k ! ( n - k)!'
we have the general formula f(x)h'"'(x - t;)
n
=
(- 1)"
1( - l)&
k=O
f'"-k'(t;)6'')(~ - t;).
(23)
Corollary (f(x)H(x))'"'= f'"'(x)H(x)+ f ' n - "(0)6(x)+ f'"- Z'(O)S'(X)
+ . .. + f ( 0 ) 6 ' n -
"(x).
(24)
42
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
Proof. Observe that (,f(x)H(x))'"'= f'"'(x)H(x)
+ n p - "(x)H'(x)
+ . . . + n(rz - 1) . . r.!. ( n - I' + 1) f',-r'(x)H'r'(x) + . . . + .f(x)H'"'(x). Because H"'(x) = 8 r - 1 ) ( x ) ,we get (24) on using (15) and striking out the terms that cancel each other. When we put H ( x - 5 ) for H ( x ) in relation (24), we have
(f(x)H(x -
5))'"' = f'"'(x)H(x - 5 ) + f'""'(5)b(x - 5 ) +(,f'"-2'(5)h'(~ 5) + ... + ,f(<)G("-"(x - 5). (25)
2.7. EXAMPLES Example 1. Recall that the Heaviside distribution in R , is (H, 4) =
/
m
-m
H ( x ) 4 ( x ) dx
=
l O m 4 ( x )d x .
According to the foregoing definition of a derivative, we have
(H', 4 )
= -( H ,
4') =
-
/
m
0
4'(x)dx
=
4(0)
=
(6,4),
or
dH(x)/dx = 6(x);
(1)
i.e., the derivative of the Heaviside distribution is the Dirac distribution. Example 2. For the delta function we have, first, (as(x -
tyax,, 4 )
-
= -(d(X
51, a 4 / a x j ) =
-a4/axji,=,.
(2)
In R , , ( 2 ) becomes
(6'(x -
5 1 9
4 ( x ) > = -+YO.
(3)
Comparing (3) with (1.3.2), we find that - 6 ' ( x ) is the dipole distribution. This process can be continued to yield in R , ,
( D k d ( x - 0, 4)
=
( - l)lk'Dk41x = t ,
(4)
2.7.
43
EXAMPLES
and, in R , ,
(q',$) = (-1Y-
dx"
Second, for a function f(x) E C", we have
I
x=t
.
or
Note that this result is valid even when f is merely a C' function. Example 3.
Let us study the distribution (XI. 0
00
(IXL4)
= J-,IxI4(x)dx
= JOrnX4(X)dX - S_mx4(x)rx.
(7)
Then (IXI'? 4) = - < l x l ,
4')
= - J-mm
I X l 4 ' ( 4 dx
J-mxq5'(x)dx - J0"x ~ ' ( x )dx, 0
=
,
which, when integrated by parts, yields ( I x 1')
4)
J-m4(x) dx J-
m
0
= Jom4(x) dx -
=
sgn(x)4(x) dx,
(8)
where sgn x, which denotes the signum jirnction, is sgn x
=
x > 0,
(9)
44
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
2.
Thus (Ixl’, 4)
=
(sgn x, 4),
1x1‘ = sgn x.
or
(10)
Note that sgn x
=
H(x) - H ( -x).
(1 1)
Hence the second differentiation is readily obtained:
I x I”
=
6(x)
+ 6( - x) = 26(x) = sgn’ x.
(12)
These and their associated functions play a useful part in many fields of analysis. For example, we can evaluate
j- Ix I 1
I
=
f”’(x) dx,
by using the foregoing results:
11
I
= “XI
f’(x)]L1 -
= f‘(1) -
1
f ’ ( - 1) -
=
,f’(1)- f ’ ( - 1)
=
,f’(1) -
Ixl’f’(x) dx
J1
C l ~ l ‘ f ~+~ ~ IIxl”f(x) ~ l dx 1
+ 2 J1
-
Csgn(x>f(x)1’1
f T - 1) - f ( 1 ) - f ( -
1)
1
&x)f(x) dx
+ 2f(O).
In passing we observe from (1 1) that H(x) = 81
+ sgn(x)).
Moreover, if we define the reversed function
4”(s)
=
4
(13) (s) by
4(-s),
(14)
then we can write (1 1) as sgn(x)
=
H(x) - H v ( x ) .
(15)
Example 4. The Heisenberg distributions, explained in Example 5 of Section 2.4, can also be obtained by differentiating the regular distribution In(x iO), which is generated by the locally integrable function
+
ln(x
+ i0) = lim In(x + k). E+O
45
2.7. EXAMPLES
We may write ln(x and i lim arg(x c+o
+ is) = lnlx + it;( + i arg(x + is),
+ is) = i lim tan-
Thus lim ln(x
E-0
I: -
X
=
inH( -x).
+ is) = In 1x1 + inH( -x),
E+O
and we have
=
(PI($
- ina(,),.4J>,
f$ E D,
which is equivalent to relation (2.4.6)except for a constant factor. The result (2.4.7) can be obtained in a similar manner. Example 5. Double-layer distribution. Let S be a piecewise smooth double-side hypersurface in R , and let A be the unit normal to S. On S we define a locally integrable function T(X)that varies continuously on it. Consider the generalized function - t(<)(d/dn)h[(x- <)], E S. This is clearly a generalization of the dipole distribution -6(x). In the language of electrostatics, We have normally oriented dipoles spread over S with surface density ~ ( 5 ) Thus, . the surface element dS, at carries a dipole field whose volume density is
<
<
This gives the total volume density
whose action on a test function f$ is
46
2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
We can also write (18) as ( - (a/dn)(th(S)),
4), because
Accordingly, ( a / a n ) ( d ( S ) )defines a double layer (or a dipole layer) over the surface S . Example 6. Multipoles. Let us consider the case n = 3. Then the generalized function 6(x - xo) represents a uniform surface charge along the plane x = xo, while 6(x - xo)6(y - yo) represents a uniform line charge along the straight line x = xo, y = y o . Similarly, the function 6’(x - xo) represents a plane layer of dipoles with axes along the x axis whereas 6’(x - xo)S(y - yo) represents a straight line x = xo, y = yo of dipoles with axes along the x axis. Similarly, S’(x - xo)6(y - yo)6(z - zo) represents a dipole at the point (xo, y o , zo) with axis along the x axis. We can continue in this fashion and define multiples. For example, a quadrupole defined as the limit of positive and negative dipoles with axes along the x axis displaced an infinitesimal distance along that axis has a volume charge density S”(x - xo)6(y - yo)6(z - zo). Incidentally, all these charge densities have unit strength. Various vector quantities such as electric current density J can be represented by the generalized functions. For instance. 6(x - xo)By, where 2, is a unit vector in the y direction, represents a uniform surface current of charge on the plane x = xo moving in the y direction. The function 6(x - X O ) S ( Y - Yo)&
represents a line current on the line x = xo, y = y o moving in the z direction. Similarly, the function 6(x - xo)6(y - yo)6(z - zo)P, represents a point charge moving in the z direction. Order of a Distribution
From Example 1 we find that 6(x) is the derivative of a regular distribution H ( x ) . For this reason, the delta distribution is called a distribution of order 1. In general, a distribution is said to be of order n if it can be expressed as the nth derivative of a regular distribution but cannot be expressed as a derivative of lower order of a regular distribution. This means that the higher the order of a distribution, the more singular it is. A distribution of order zero is a regular distribution. To generalize and summarize these ideas: If t is a distribution of finite order, then there exists a locally integrable function f and a multiindex k such that t = Dkf.
2.8.
THE SUPPORT AND SINGULAR SUPPORT OF A DISTRIBUTION
47
2.8. THE SUPPORT AND SINGULAR SUPPORT OF A DISTRIBUTION
The reader must have realized by now that we cannot talk about local properties of distributions. We can only consider their action on the class of test functions. Nevertheless, we can say that the distribution t vanishes in the neighborhood N of a point x if (t, 4) = 0 there for all 4 E D. A distribution t vanishes in a region R of R , if it vanishes in a neighborhood of every part of R. Accordingly, the distributionsfand g are said to be equal in R if f - g = 0 for all x E R. A point x is called an essential point of a distribution t if there does not exist a neighborhood of x in which t vanishes. The closure of the set of all essential points is defined as support oft and denoted as supp t. Technically, we can describe the support as the least closed set outside of which t vanishes. Iffis an ordinary function, then its support as a distribution coincides with its support as a function. Furthermore, it follows that if supp t and supp 4 have no points in common, then ( t , 4) = 0. It is interesting to compare this with the zero distribution; if x # supp t , then there is a neighborhood of x such that t = 0 in that neighborhood, and t also vanishes in any region consisting of such neighborhoods. Example 1 . Supp H(x), where H ( x ) is the Heaviside function, is the halfaxis x > 0. Example 2. The support of 6 ( x ) and each of its derivatives is a single point x = 0. The converse is also true. A distribution t that has support of one point is a finite linear combinations of 6(x) and its derivatives (see Exercise 25). Example 3. If the distribution t = 0 for x so that supp Dkt c supp t.
E R,
The proof follows by observing that if Accordingly, ( D k t ,4 )
=
(- 1)lk'(t,
from which it follows that Dkt = 0 for x
E
then also Dkt = 0 for x E R,
4 E DR, then also D k 4 E DR.
D k 4 ) = 0,
R.
Example 4
supp a(x)t(x) c supp a n supp t . This may be proved by noting that X E supp a(x)t(x) implies that a(x)t(x) # 0 for all neighborhoods of x . This means that a(x) # 0 and
48
2. THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
t(x) # 0 for all neighborhoods of x. Accordingly, x E supp a and x E supp t ; that is, x E supp a n supp t, as required. Example 5. The distribution It l 2 = t : + t t + . . . + t,' has the entire space R , as its support, although I r l2 vanishes at t = 0. The reason is that it is the behavior of a distribution over the neighborhood of a point that determines whether the point is in the support oft.
Another useful concept is that of a singular support. This is defined as the closed set of points where t(x) is not a smooth function. It is the least closed set outside of which t is equal to a functionf(x) possessing derivatives of all orders. For example, both 6(x) and H(x) have as singular support the single point x = 0. Similarly, the distribution 1x1 also has the point x = 0 as singular support. Example 6. As a final example, let us consider the functional
11
By writing it as
(sin 2x)4(x) dx.
(sin 2x)4(x) dx
m
=
where sin 2x,
- 1 < x < 1, otherwise,
it becomes clear that the support of this functional is the closed interval [- 1, 11 while the singular support consists of the two points - 1 and 1.
EXERCISES 1. Let a sequence of test functions { 4 m ( ~ ) m } , = 1, 2, . . . converge in D to zero, and let +(x) be an infinitely differentiable function. Show that also converges to zero.
2. Compute the first four distributional derivatives of sin x, lsin X I , x cos x, and Ix cos X I . 3. Find the first distributional derivatives of (a) [l - H(x)] cos x, (b) {H(x + 2) H(x) - H(x - 2)}eP2".
+
49
EXERCISES
4. Find the second distributional derivatives of e-lxl and sin 1x1. 5. Show that f(x) = tanh (l/x),
(draw it !) has a jump discontinuity at x
=
x #0
0. Find its distributional derivative.
6. Find the second distributional derivative of the function
f(x) =
("
1, -1,
1x1 51, x > 1, x<-1.
7. Show that (a) ex6(x) = 6(x), (b) x ~ ' ( x )= -6(x), (c) (sin ax)b'(x) = -aS(x), (d) (d2/t/xZ)[H(x) sin a x ] = d ( x ) - a 2 H ( x ) sin ax.
8. Show that [.f'(r)]" = - [ j " ( t ) ] ' , where the symbol " is defined in Example 3 of Section 2.7. 9. Recall that J,(x) is a solution of the Bessel equation d2y dx2 ~
1 dy + -+ y(x) xdx
=
0.
Show that J,(x)H(x) is also a solution. 10. Consider the function
and prove that anH(x) ax,
ax, . . . ax,
=
S(X).
11. If a(x) < b(x), show that
{ H ( t - a(x)]H[b(x)
- t]f(x, t ) } d t .
12. (a) Show that m < n, m! (m - n ) !
6'" - "'(x),
m > n.
50
'2.
THE SCHWARTZ-SOBOLEV THEORY OF DISTRIBUTIONS
(b) Deduce that
13. Verify that 1
-
1
a(") =
- -P +l )
n+l
X
+ c6,
where c is a constant. 14. Show that the collections {( - l)m6(m)(x)}~=o and { ~ " / n ! } ~form =~ a biorthogonal set. That is,
((-
); {
l)m6(m'(x),
=
0
if n # m, if n = m.
1
15. Power distributions x i and XL are defined as <xi,
4)
1 m
=
0
sA4(s)ds,
(x?,
4)
J0
=
m
Isla4(s) ds,
respectively, where Re I > 0. This means that the distribution x< is to be identified with the function x:
=
? ;
; {
x > 0,
x I 0,
and a similar definition holds for x? . Show that
These functions are studied in detail in Chapter 4. 16. Show that
17. Show that the sequence of test functions {&(x)}, where r < l/m, r 2 llm,
and c;'
is a delta sequence.
=
J
r<m--'
erp(-
)
1m2r2 m2r2 d x ,
51
EXERCISES
18. Show that 2ad(at + x)b(at - x) = d(x)b(t). Hint. Examine the action of2a6(at x)6(at - x)and then use the transformation ofcoordinate system s = a t - x , y = at x.
+
+
19. Let the function [x] be defined as the mean of the greatest integer less than x and the greatest integer less than or equal to x. Show that (a) [XI’= III(x), (b) (d/dx){[x]H(x)} = III(x)H(x) - $ 6 ( ~ ) . 20. Show that II(x) = 6(2x2 - $). 21. Show that
22. Derive the expansion off(x)Dkd(x) in R , and R , on the same lines as Theorem 2 of Section 2. Then show that in R,,
f(X)DkS(X)
= ’( - 1)Ikl
c
psk
k! P ! ( k - p ) ! (Dk-Pf(O))(DPS(x)).
( - 1)’P’
23. Show that the singular support o f t is contained in the support oft. 24. Find the support and the singular support of the functionals s 2 m ds, (a) (b) Jh e - l i S + ( s )ds, (c) J t e - l i ( 1 - s Z ) d 4 S ) ds. Jl?
25. Show that every distribution t(x) that has compact support is of finite order [2]. Then deduce that every distribution r(x) whose support is the cihcn(x). point x = 0 has the form t(x) = 26. Show that lim In z y-0
where z
=
x
f
=
lnlxl
+ in(1 f H(x)),
+ i y and derive Eqs. (2.4.18) and (2.4.19).
27. Find the distributional solutions of the equations: (a) xmt4(x)= 6(x),
(b) (sin nx)tr(x) = 1.
Additional Properties of Distributions
3.1. TRANSFORMATION PROPERTIES OF THE DELTA DlSTR I B UTlO N
Some algebraic operations on the delta function were studied in the last chapter. In subsequent chapters we shall be required to transform this function to certain curvilinear coordinates. For this purpose we devote an entire section to this topic. Let us first study the meaning of the function S[f(x)] and prove the result
where x, runs through the simple zeros off’(x). Suppose that,f(x) has a simple zero only at x = x1 and thatf’(x,) > 0, so that the transformation is one-to-one in the neighborhood of xl. Then the substitution y = f(x) in the functional f T m d[f(x)]&x) dx gives
52
3.1.
TRANSFORMATION PROPERTIES OF THE DELTA DISTRIBUTION
53
because only a neighborhood of y = 0 contributes to the integral. For the casef(x,) < 0, the direction of integration is reversed, and the right side of (2) becomes -+(xl)/f‘(xl). Combining these two cases, we have sCf’(x>I = 6(x - x1)/ I f’(x1) I.
(2) When the functionf(x) has simple zeros at x l r x 2 , . . . , x,, we have the required relation (1). Iff(x) has higher-order zeros, no significance is attached to fiCf(x)l. The following alternative approach is instructive and helpful in solving specific problems [3]. To fix the ideas, let us again assume that,f(x) has a simplezeroonlyat x l (and thus,f(x,) = 0)but that,f’(x,) # 0, which weagain assume to be positive. Furthermore, let us first consider a finite interval a x1 < p. Thus J(x) increases monotonical from f ( a ) to f(p) as x goes from a to p, so that H[f(x)l = H(x - Xl), (3)
-=
where H is the Heaviside function. Differentiating (3), we obtain (d/dx)HCf(x)l
or
=
6(x - XI),
“(x)l = Cf’(X1)1-16(x - x1). For a monotonically decreasing function, “(X)l
(4)
= H(x1 - XI,
(5)
whose differential is 6[f(x)] = - [f’(x,)]- ‘6(x - xl). Combining this with (4) we rederive (2). The extension to more simple zeros and infinite interval is now trivial. Indeed, if there are n zeros off(x), then
as desired. Incidentally, it follows from (3) and (5) that W”)I
=
1
m = l 2
6(x - x,){sgnf(x,
+ E,)
-
sgnf(x, - &,)I,
(6)
where E, is chosen so small that the interval (x, - E,, x, + E,) contains no zero other than x,. We can go a step further and evaluate d’[f(x)], wheref(x) is a twice continuously differentiable function that increases monotonically from f ( a ) tof(p) as x increases from a to p. Iff(<) = 0, where a < < p, then
<
54
3.
= -
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
[L{””’}l]
dx f”(x) J ” ( x ) x = c
which proves (7). Alternatively, we can use the formula S [ f ( x ) ] = [ f ’ ( 5 ) ] - ’ 6 ( x - 5), and obtain
where in the last step we have used 9(x)6’(x - 5 )
=
s(5)6’(x - 5 ) - 9‘(5P(X -
0
7
from Section 2.6. Example 1
+ b) = ( l / l ~ 1 ) 6+( ~b/a). The proof follows by differentiating H(ax + b). 6(ax
Example 2 6[(~ U)(X - b)]
=
[l/(b - u ) ] [ ~ ( x U)
+ 6(x - b ) ] ,
b > a.
(9)
3.1.
TRANSFORMATION PROPERTIES OF THE DELTA DISTRIBUTION
55
The function f(x) has two simple zeros, x1 = a, x2 = b. Appealing to (l), we obtain 6 [ ( ~- U)(X- b)] =
6(x - a) la-bl
- b) 1 -+ 6(x Ib-al b-u -
as required. In particular, if b > 0 and a
= - b,
6(x2 - b2) = [ 6 ( ~- b)
[ 6 ( ~- a)
+ 6(x - b)],
then
+ 6(x + b)]/2b.
+ 3x) = @(x). In this case f'(x) = x3 + 3x, x2 + 3 = 0. j ( x ) = O for x = 0, Thus x = 0 is a simple zero andf"(x) = 3x2 + 3,f"(O) = 3, so that 6(x3 + 3x) = @(x - 0) = f6(x).
(10)
Example 3. 6(x3
Example 4. G(tanx) = 6(x), -4n < x < in. Since x in this interval, we have f'(x)
=
tan x,
.f"(x)
=
sec2 x,
=
(11)
0 is a simple zero
j"(0) = 1.
Thus &tan x)
=
6(x).
(12)
Example 5. ~ ( C Ox)S = 6(x - n/2), 0 < x < n. Here x zero andf"(x) = -sin x,J"(x/2) = - 1, so that G(C0S x)
=
6(x - n/2)/1 - 1 I
=
=
742 is a simple
6(x - n/2),
(13)
as required. - n2)dx: Example6. With the help of Example 2 let us evaluate f?Znenx6(~2 2n
enx6(x2- n2)dx
enx[G(x
=
n2 + n) + 6(x - n)] dx = -cosh . n
Example 7. Result (1) remains true even for an infinite set of simple zeros. For instance, b(sin x)
=
1 OD
"=-=,
c
6(x - nx) OD = 6(x - nn). I(c0snn)l n=-m.
This can also be obtained by differentiating the identity (1.1.7). Example 8. Let us evaluate G'(sinh 2x). Since sinh 2x appeal to (7):
=
0 at x = 0, we
56
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
Substituting in the values j”(0) = 2 cosh 0 = 2, f”’(0) derive the required value G’(sinh 2x) = 6’(x)/4.
=
4 sinh 0
=
0, we
Example 9. Let us evaluate the integral m
I =
(cos x
+ sin x)6’(x3 + x2 + x) dx.
(15)
Since,f’(x) = x3 + x 2 + x and,f(O) = 0, we have d’(X3
i
(6x + 2)lx=o + x 2 + x) = (3x2 + 2x1 + 1)21x=o 6’(x) + __ a(,,) (3x2 + 2x + l ) l x = o =
6‘(x) + 26(x).
(16)
Substituting (16) in (15), we get I
1
m
=
(cos x
J-m
=
-[-sinx
+ sin x)(6’(x) + 26(x)) dx + C O S X ] ~ =+~ ~ ( C O S Xsin^]^=^ = -1 + 2 = 1.
Curvilinear Coordinates
Let us now study the form of the delta function when we change from Cartesian to curvilinear coordinates. Since there is no basic difference in the analysis whether we study the general case of n-dimensional space or the special case n = 2, we first assume the latter for simplicity. We shall then state the result for the general case [4]. Accordingly, let us transform from Cartesian coordinates (x 1, x2) to curvilinear coordinates ( 5 1, t2)where x1
=
Ul(51, C2)9
x2
=
[12(51?5 2 x
(17)
and ill, 2 i 2 are single-valued continuously differentiable functions of their arguments. The transformation is assumed to be nonsingular, so the Jacobian (18) J = d(lIl, 2!2)P(tl, 5 2 ) . is nonzero. Suppose that under this transformation the point (PI, P2) in the 5 system corresponds to the point (a1,a2) in the x system (see Fig. 3.1). Then with the change of coordinates (17), the equation JJR$(X17 X 2 ) G l - a,)6(x2 - u2) dx1 dx2 = +(a13 a2)9 becomes
//Rml,52P[U1(51, 52) - a116“12(51r 52) - a2llJI d51 d52
=
+(B13 P 2 X (19)
3.1. TRANSFORMATION PROPERTIES OF THE DELTA DISTRIBUTION
I
57
XI
Fig. 3.1. Transformation to curvilinear coordinates.
where 4(x1,x2)E D. From (19) it follows that d[lll(tl? 52)
-
a136"r2(517
52)
-
a2llJI
= 451
- 81)8(52 - Bz),
or 6(Xl - a1)6(x, - a2) = S(51 - B l P ( 5 2 - B 2 ) / l J I .
(20)
This analysis is readily extended to n dimensions and yields the following theorem: Theorem. Let curvilinear coordinates g l , t2, . . . , 5, be obtained from Cartesian coordinates x l , x 2 , . . . , x, by the transformation law
. . , n. Furthermore, let the point P with coordinates (a1,.. . ,a,) in the x coordinate system have the coordinates (B1,. . . , p,) in the 5 coordinate system. Then
ti= ui(xl, . . . , x,),
i
= 1,.
where the Jacobian J of the transformation is nonzero. Example 10. The transformation from rectangular Cartesian coordinates (x, y) to plane polar coordinates (r, 8) is x
Thus, J
=
=
r cos 8,
r sin 8.
(22)
6(r - rl)6(8 - Ol)/r.
(23)
y
=
r, and (21) becomes
6(x - xl)S(y - y l )
=
The transformation from three-dimensional rectangular Cartesian coordinates (x, y, 2 ) to spherical polar coordinates (r, 6, p) is x
=
r sin 8 cosp,
y
=
r sin 8 sin cp,
z = r cos 8.
(24)
58
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
This yields J = r2 sin 8, and
6(x
-
xl)S(y - y1)6(z - zl)
= 6(r -
rl)6(8 - Ol)6(p - (pl)/r2sin 8.
(25)
In the event that J = 0, so that at some point ti= pi,i = 1, . . . , n, (9) fails to give us the required transformation. Such a point is called a singular point of the transformation. In the foregoing example of plane polar coordinates J = r = 0 at the origin (0,O).At this point, r = 0 and 8 may take on any value. A coordinate, such as 8 in this example, which is either many valued or has no determinate value at a singular point of the transformation, is called an ignorable point. In the case of spherical polar coordinates, J = rz sin 8 vanishes for all points on 8 = 0 (the z axis) and there 4 may take on any value. Thus 4 is an ignorable coordinate in this case. Also, J = 0 at the origin, where both 8 and 4 are ignorable. Let us pursue the case J = 0 and confine ourselves to n = 2. Suppose that (al,a 2 ) is a singular point of the transformation (17) while t2 is an ignorable coordinate. Thus. at the singular point, t1 = p1and t2 is indeterminate, so that the test function $(tl, depends only on t1= p1 there. Accordingly, we denote it 4(p1).Then the generalized function
c2)
6(Xl
- a1)6(x2 - U 2 )
also becomes a function of tl only, and we denote it r(tl).Then relation (19) should yield
This equation is satisfied if we set J 1 = J IJI d t 2 , where the integration is along the entire domain of t2,and take r(t1)= a(<, - pl)/J1.Thus
6(x
- a 1 m - a2) = 6(51 - Bl)/Jl.
(27)
For instance, in the case of the plane polar coordinates, the origin is the singular point and J = r. Thus J 1 = Ji"r d8 = 2nr, and (27) becomes
6(x)6(y)
= 6(r)/2nr.
(28)
In the case of spherical polar coordinates, cp is the ignorable coordinate. Thus J 1 = ji" r2 sin 8 dcp = 2nr2 sin 8, and (27) becomes 6(x)6(y)d(z - zl) = 6(r - rl)6(8)/2xr2sin 8.
(29)
3.2.
59
CONVERGENCE OF DISTRIBUTIONS
However, at the origin both 8 and cp are ignorable coordinates, so that .I1 =
Jeff
Jo2*r2sin 8 d8 dqo = 4nr2,
which yields h(x)d(y)h(z) = h(r)/4nr2.
3.2. CONVERGENCE OF DISTRIBUTIONS
A very important concept in the theory of distributions is that of convergence. A sequence of distributions t, E D', m = 1,2, . . .,is said to converge to a distribution r E D' if lim (r,
m+ a,
4)
= (t,
4)
for all
4 E D.
(1)
When the sequence of distributions consists of locally integrable functions f,(x), there is the following relationship between classical convergence and distributional convergence (also called weak convergence): If { f,(x)} converges uniformly to f ( x ) over each finite interval, then (f,, 4) -,( J 4). Indeed, lim (f,,
m+ a,
4)
J-
a,
= lim
m- w
J-,f a,
=
f,(x)$(x) dx a,
(XW(X)
dx9
4 E D,
where we have used the uniform convergence of the sequence { f,(x)}. An important result in connection with the convergence of the sequence of distributions {t,} is given in the following theorem.
Theorem 1. If the sequence {t,} of distributions converges to t, then the sequence {D't,,,} converges to {D't}. Proof
lim (D't,, m-a,
4)
= lirn (m-
w
= (D't,
l)~k~(r,, D k 4 ) = (- 1)lk1(t, 0'4)
4),
s, conoerges to t if the sequence of converges to t. We write C;=, ,s = t. In
Definition. A series of distributions partial sums {t,} =
{C,=ls,}
4ED.
60
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
view of Theorem 1, we have the important result that every convergent sequence or series of distributions can be differentiated term by term as ojien as required. We have seen above that if a sequence of locally integrable functions converges uniformly to a function j ; then it also converges tof in the sense of distributions. An important question in this connection arises regarding the sequence of functions that converge pointwise or fail to converge. Theorem 1 suggests the following procedure. Let { f m ( x ) }be the sequence in question. Suppose that an integer k can be found such that { j J x ) } is the kth derivative of a sequence {g,(x)} that converges uniformly to g(x). Then + ;} g"), where g'k) is the kth generalized it follows from the theorem that {I derivative of g . Let us extend the definition of distributional convergence to a sequence of distributions t,(x) depending on a real parameter a.That is, t,
+t
as a + a,,
if
lim
(fa,
4)
=
(t,
4),
ED.
(2)
1-20
It can be readily proved that t is a distribution. The differentiation theorem in this case gives lim ( D k f l r4) a-io
=
lirn ( - 1)~'~(tz, D k 4 ) = (- 1)lk1(t, 0'4)
=
(D't,
4).
,-lo
Inasmuch as the limits of the sequence { t m } and { t , } are themselves distributions, we have the following important result:
Theorem 2. The space D is complete with respect to weak convergence.
3.3. DELTA SEQUENCES WITH PARAMETRIC DEPENDENCE
Section 1.2 gave various delta convergent sequences. With the help of the parametric dependence, we can construct a sequence of functions that approach a delta function. We shall present this analysis in n dimensions, but we must first study the volume and surface area of the n-dimensional sphere of radius r. Let V,(r) and S,(r) denote, respectively, the volume and the surface area of the n-dimensional sphere of radius r. To obtain a formula for V,(r) we first relate it to the volume of an (n - 1)-dimensional sphere. Let z denote a coordinate on an axis through the origin. If A is a constant smaller than r , the
3.3.
61
DELTA SEQUENCES WITH PARAMETRIC DEPENDENCE
Fig. 3.2
intersection of the hypersurface z = A and the n-dimensional sphere is an (n - 1)-dimensional sphere of radius ( r 2 - z2)l" (see Fig. 3.2, drawn for n = 2 and 3). Then
v,- l[(r2
-
z2)'12] dz.
(1)
For example, V,(r) =
v2(r)= V&)
=
J'p JI;(r2
=
2r,
-
z2)1/2
n = I,
dz
- zz) dz =
=
Jo Sb 1
4r2
2nr3
(1 - z12)1/2 drr
n = 2,
=m2,
(1 - u2) du = 47cr3,
n
=
3.
Continuing in this manner, we have V,(r) = Cnr",where C , is defined by the recurrence formula
c,, 1 = 2 c ,
Jo (1 1
112)n/2
du.
(2)
This formula can be put in a simple form if we appeal to the beta function
as well as to the relation of P(p, q) to the gamma function T(p):
P(p, q ) = r(p)r(q)/r(p + q),
where T(p) =
JOrn
t P - 'e-' dt.
62
3. ADDITIONAL PROPERTIES OF DISTRIBUTIONS
Then it can be easily proved that
Thus, for an integer n, it follows that
It is interesting to observe that the volume of unit sphere first increases with n and then decreases. The surface area S,,(r) of the n-dimensional sphere of radius r may be derived by considering the volume of a thin spherical shell. To the first order of thickness Ar, this is S,(r) Ar, S,(r) Ar = V,(r) - V,(r - Ar)
nn/2
=
n/2 !
~
[r" - (r - Ar)"]
From these above relations it follows that
Suppose we are dealing with a functionf'(x) that depends only on r (say, g ( r ) ) ; then O0
JRnf'(x) dx = Jo g ( r )
dV,
dr dr =
g(r)S,(r) dr
=
S,(l)
g(r)rn-' dr.
(6)
The construction of a sequence of functions approaching the delta function is explained in the following theorem:
Theorem. Let j ( x ) = f'(x,, . . . ,x,) be a nonnegative locally integrable function satisfying the condition JR"f.(X) dx = 1,
(7)
. f ( x ) = f ( x ) = (I/a")f'(x/Cc) = (1/afl).f(.x1/a,. . ., x,/a),
(8 )
and let
3.3.
63
DELTA SEQUENCES WITH PARAMETRIC DEPENDENCE
where A is a suitably chosen constant. These properties can easily be proved for one-dimensional case, and their extension to R , is obvious. The theorem is proved if we show that
or, equivalently (in view of Property l),
lim 1-0
1.
la(X)C$(X)
- $(O)l dx
4 E D.
= 0,
(10)
Division of the region R , into the two parts, 1x1 < A and 1x1 > A, gives
+
I
i;xl,Afa(x)[4(x)
-
4(0)l dx .
Now the boundedness of 4(x) ensures the boundedness of [+(x) - 4(0)]. Let x ) 4(0)]. Then, using this bound be M . Similarly, let N ( A ) = m a x l X l s A [ ~ (the nonnegativity off’(x) and Property 1, we have
I h.
f&)C4(X) -
dx 5 “ A )
+M
i;
XI > A
fU(X> dx.
Since the function [+(x) - +(O)] is continuous and is zero at x = 0, N ( A ) -,0 as A -+ 0. Thus for an arbitrary E, we first choose A sufficiently small that N ( A ) < ~ / 2 Then . we appeal to Property 2 and choose c1 sufficiently small that ~ l x l , A f J x ) dx < E / ~ MThis . gives
which proves (10).
64
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
3.
Special Case. WhenJ'(x) is a function only of r is,f'(x) = g(r), (7) reduces to
J-oa'rn-
g(r) dr
=
(xf
+ ... +
that
1
= -
S"(1)'
where S,( 1) is the surface area of the unit n-dimensional sphere and we have used (6).For n = 3, ( 1 1) and (8) become, respectively, I0"r2g(r) dr
=1
4Tc and lim(l/m3)g(r/m) = 6(xl,. . . ,x,) = 6(x). 2-0
(13)
Exumple 1. Let us prove that
In case (a), g(r) = 1/27c(r2 + l)3/2.Because
.lom + r
2n(r2
1)3/2
- -1 - 2x3
condition (1 1) is satisfied and the result follows. In case (b), g(r)
=
l/n2(r2
+
and f"
r2
1
which satisfies (12). This proves the result. Example 2. Recall the test function $ E ( ~as ) defined in (2.2.3);
r 2
E,
where c, is the normalizing factor such that f 4&(x)dx = 1. By the foregoing analysis it follows that $&(x)-, 6(x) as E -, 0.
65
3.4. FOURIER SERIES
3.4. FOURIER SERIES
It is known that the Fourier series [ S ] m
1
m=-w
cmeimx,
(1)
converges uniformly if, for large m, I c, I I M / m k ,k 2 2, where M is a constant and k is an integer. It therefore converges distributionally. This series may diverge for other values of k , but by applying Theorem 1 of Section 3.2, we observe that (1) is a distributionally convergent series when k is any integer, because it can be obtained from the uniformly convergent series
+
by k 2 successive differentiations. On the other hand, we can expand a periodic and locally integrable functionf'(x) in a Fourier series. To prove this assertion, let us take the period to be 2n. Since the Fourier coefficients of.f(x) are given as
we have to prove that the Fourier series m
converges distributionally tof(x). This is proved by considering a partial sum spq of the series (l), 4
C
spq(x)=
m= - p
cmeimx.
Its scalar product with a test function 4(x) is 4
(spq,
4) = C
m= - p
Cm(eimx,
4)
(4)
or (spq?
4)
=
2~
C 4
CmJmimr
m= - p
where the d , are the Fourier coefficients of 4 and the bar labels the complex conjugate. We now appeal to the famous Parseval theorem of classical
66
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
Fourier analysis. It states that if {urn};= - is a complete set of orthonormal functions, and if the Fourier coefficients ofj'with respect to 11, are c, while those of 4 are d,, then cmdm= ( j ; 4 ) . In the present case, 11, = eimx/&; thus we find that as p , q + 00, ( s p q ,4 ) 3 ( j ; 4). This proves the result. Accordingly, for a locally integrable periodic function, the classical convergence and the distributional convergence of the Fourier series are the same thing. The Fourier series of every periodic distribution t converges to t in the distributional sense. Let us illustrate this by expanding f(x) = 6(x - y), f ( x ) =f'(x + I), 0 < y < I, in terms of the Fourier sine series on the half range (0, 1):
c:,
(5)
The partial sum sq is s4(x, y )
=
-
1' sin ("I") sin("?). ~
l m = l
Its scalar product with a test function 4(y) gives
which is the partial sum of the Fourier series of the smooth function accordingly converges to 4. Therefore,
4 and
lim (s4(x,YX 4(Y)> = lim cq(x) = 4(x) = (6(x - Y), 4(Y>>. 4-
4-00
00
This proves that the distributional limit of sequence s4(x, y) is the delta distribution. Note that because of the assumption of the periodicity the left side of ( 5 ) is really an infinite row of delta functions situated at the periodic points. Moreover, we expanded an even function in terms of the sine series in the half-range (0, 1). Consequently, we have the odd extension of this function in the full range. The periodic row of delta functions in the full range (- 1. / ) is 1 " 6(x - 2m1) = cmeinmx". m = - 00 21,=-, Q.
1
1
3.4.
67
FOURIER SERIES
The coefficients cmare given formally by I
Cm =
j-;(x)e-inmx'l dx
=
1,
so that the relation
holds true distributionally even though the series on the right side is divergent. Incidentally, for the special case of 1 = 7c, we have m m=-w
6(x - 2m7c)
=
-
27c
[
1
m
+ 2 1 cos mx m=l
(6)
If we restrict x to the interval (-n, n),then only one of the delta functions in (6) is nonzero, and we have the representation (7)
Thus
For a periodic function,f'(x) with period 2n, we have
which is the Fourier series ofJ'(x). This analysis can be readily extended to the distributional convergence of Fourier Bessel series (see Exercise 17 at the end of this chapter). We can also extend the notion of the distributional convergence to integrals, as illustrated in Examples 1 and 2 in Section 3.5. Regularization of divergent integrals is discussed in Chapter 4.
68
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
3.5. EXAMPLES Example 1. It is easily verified that the integral
exists for every value of x and that the function fU(x)
=
J;
dy
converges uniformly to.f’(x) in every bounded region of x as M + cc. Thus fj(x) converges in the sense of distributions as well, and we can differentiate it as many times as we desire. For instance, by taking the second derivative of (1) we find that re
-
J, cos yx dy
is the limit of the convergent distributions {d*fjl(x)/dx’>as M + co,although the integral (2) does not exist in the classical sense as tl+ m. Example 2. To establish the relation 6(x -
5) = (1/27c)
exp{i(x - 5)s) ds,
lWm
(3)
we set t,(x - 5 ) = (1/2n) s”I exp{i(x - 5)s) ds. Then
Because 4(5)has a compact support, the integral with respect to 5 is finite, and we can change the order of integration. Thus
Appealing to the classical theory of Fourier transform, we obtain from this formula limg-m ( l a , 4 ) = $(x) = (6(x - 0, 4(4)), which proves (3). In particular,
69
3.5. EXAMPLES
An alternative derivation of (4) is simple and interesting. Recall from advanced calculus that sin xs
x > 0, x < 0,
x, -x,
and
iL 7 rn cosxs
ds
= 0.
With the help of these formulas we have
sin xs
-ds
=
{
x. -x,
x > 0, x < 0.
Thus
{",
x
=
3 sgn x = H(x) - f.
Differentiating both sides of this relation with respect to x , we recover (4). Similarly, it can be shown that 6(x)
i
m
=
2x
- ixs ds.
-m
The addition of (4) and ( 5 ) yields m
x
Many more relations can be derived from these formulas. For instance, differention of (4) and ( 5 ) with respect to x yields a representation for 6'(x):
We shall have more to say about these ideas in Chapter 6. Example 3. Show that
70
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
To prove this result we appeal to the relations
Jorn
,.2ne-r2
and
Jornr2.+
1e-r2
dr
= n1/22-2n
dr
=-
(2n - l ) ! (n - l)! '
n! 2
(from calculus). Next we use (3.3.5) and obtain (2n - l ) !
=
2 2 y n - l)!nn/s2n+l(l).
Thus (9) becomes Jornr12n+
1)- 1
e - r 2 dr
1)/2
#n+
= SZn+
l(1)'
Similarly, n ! = 2n"' '/S2,,+2(1), and (10) takes the form nn+
Jornr(2n+
2 ) - 1,-
r2
dr =
~
s2,
1
+ 2(1) *
Combining (11) and (12), we find that JOrnrn-lp-r2
dr
=
~
W)'
is valid for all positive integers n. Finally, we set ga(r) = ( l/n"/2)( l / d ' ) e - r * / u z ,appeal to Theorem 1 of Section 3.3, and obtain lim g,(r)
=
6(xl, . . . , xn).
a+O
Example 4. The sequence { f m ( x ) } = {sin m x } is not a convergent sequence in the classical sense unless x is a multiple of n. However, if we take the sequence g,(x) = { -cos(mx)/m}, which converges to zero uniformly in -co < x < 00, then { g k ( x ) } = {f,(x)} converges to zero in the sense of distributions. Example 5.
In Section 1.3 we studied the sequence t,(x) = 2 m 3 x / n ( i
+ m2x2)2,
which describes the dipole distribution. This sequence converges pointwise but not uniformly to zero in - co < x < 00. However, it is the derivative of the uniformly convergent sequence -m/n(l + m2x2), which converges to
3.6.
71
THE DELTA FUNCTION AS A STIELTJES INTEGRAL
the delta sequence, as proved in Example 2 of Section 1.2. Accordingly, we can appeal to Theorem 1 of Section 3.2 and get t,(x)
-+
- 6'(x)
or
(r,, 4)
=
4'(0).
(14)
3.6. THE DELTA FUNCTION AS A STIELTJES INTEGRAL
Let us examine the Stieltjes integral
wherej(x) is a function continuous on the interval [a, b], 5 is some fixed number in the interval (a, b), and H,is the Heaviside function that takes the value c, 0 I c I1, at x = 5. Let us partition [a, b] into n parts such that a
=
x, < x1 < ... < x , - ~< x,
=
b,
and let the quantity Akg = g(xk) - g(Xk-l), where g(x) is some fixed, monotone increasing function. Consequently, we have two possibilities : (1)
5 is an interior point of some subinterval, say, xk- < 5 < xk, so that Akg
=
H,(xk - 5 ) - Hc(Xk-l -
Apg = 0,
5) = 1 - 0 = 1,
for every p # k.
Thus n
1 f(Xp) Apg =f(Xk),
p= 1
(2) that
5 is a
where xk-
1
I xk
I xk-
boundary point of two adjacent subintervals, say, g
(2) =
XI., so
Akg = H,(Xk - 5 ) - H"(Xk-1 -
5 ) = C - 0 = C, Ak+lg = Hc(Xk+l - 5) - Hc(xk - 5 ) = - c, while all other terms Apg vanish. Thus n
f(xp) Apg = Cj(Xk)+ (1 - c)f(xk+ 1) =f(xk+
p= 1
1)
+ c[.f(xk)
-f(xk+
1113 (3)
wherex,.,
I xk I4 I xk+l I xk+l.
72
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
3.
From (1)-(3) and the hypothesis, it follows that, in the limit,
Ja
which is precisely the required sifting property of the delta function.
EXERCISES
1. Show that if ad - bc # 0,6(ax + by)S(cx
+ dy) = 6(x)6(y)/lad - bcl.
2. Evaluate (a) Jh e2"6(2x - 1) dx, (b) (sinh 2x)6(5x + 2) dx, (c) [ F n e-'%(sin nx) dx, (d) J', (sin-'x)6(4x2 - 3) dx.
s?
3. Interpret the expressions 6(ex),6(ln x), and G(sinJx1). 4. Show that (a) 6'(x3 3x) = +S'(x), (b) 6"(3x3 X) = 6"(x) - 26(x), (c) 6(3x - +sin 2x) = +6(x). 5. Prove that for x > 0
+ +
a(a -
): { =
(l/a2)6(x - 1/a), 0,
a > 0, a I0.
6. Prove that (x2 - l)S(x2 - 1) = 0. 7. Evaluate (a) 6"Cf(x)I and (b) sgn"f(x)l. 8. Show that (a) lima+msin ux/nx = 6(x), (b) (1/7c)(sin2 ax/ax2) = 6(x), (c) Iima+o(1/2(&), exp(-x2/4a) = 6(x).
9. Recall the definitions of Pf(l/x), 6(x), S'(x), and F ( x ) from Chapter 2 and show that (a) 6*(x) = Tlim,,o (1/2ni)[l/(x f is)], (b) 6(x) = lim,+o (l/n)[&/(x2 &')I, (c) Pf(l/x) = [7cix/(xz + &')I. (d) 1nJxJ inH( -x) = ln(x i ~ ) .
+
+
+
73
EXERCISES
10. Consider the following sequence of locally integrable functions of the single variable x :
(“1x < 0,
L(x) =
me-a214x
x>o.
@x3‘2’
Show that limu+ofJx)
=
6(x).
11. Considering the function
f*(x) = J:2 cos(2nyx) dy = sin 2ncIx/nx, and show that Jg 2 cos(2nyx) d y
=
6(x).
12. Compute the Fourier series of the distributions 6(x - a), 6’(x - a) and H(x - a), - 1 < a , x < 1.
13. Show that the series m sin mx, converges to -6’(x). This result can be extended to show that any Fourier series with coefficients that are polynomials in m converges in the distributional sense to a polynomial expression in the derivatives of the delta function. 14. Compute the Fourier series of the following periodic functions:
- 00 < t < 00. f ( t ) = f ( t - 2), (b) f ( t ) = e‘, 0 5 t < 2n;f(t) = f ’ ( t - 2n), - 00 < t < 15. The Fourier series
f
+ :-I‘
00.
sin mx
m= 1
converges to x/2 when - n < x < n and to 0 when x = fn. (a) Explain why the convergence cannot be uniform on [ - n, 111. (b) We might hope that term-by-term differentiation would yield Cz= ( - l)”’+ cos mx = $ for - n < x < n, but this is not valid. Show that this series diverges for all rational numbers x. (c) Explain why ( - l)m+ l cos mx converges -in the distributional sense. To what distribution does it converge? 16. Find a sequence of locally integrable functions gm(x) converging Hint. Set gm(x)=f:)(x), where (in D’)to the generalized function C~(~’(X). fm(x) is a delta sequence of infinitely differentiable functions.
74
3.
ADDITIONAL PROPERTIES OF DISTRIBUTIONS
17. Given an arbitrary complete system of orthonormal functions {Qm(x)} in an interval (a, b), form the Fourier Bessel series with respect to {Om}of a functionf(x) that is integrable along with its square and whose support lies in (a, b), namely,
c c , , , ~ ~ ( x ) , cm m
=
m= 1
[f(x)am(x) nx, m
=
1,2, . . . .
Show that this series converges tof(x) in the sense of distributions.
18. Show that
0 < x < 2n.
,
m= 1
19. Prove that lirn,+" crJ,(aax) function.
=
6(ax - l), where J,(crax) is the Bessel
20. Prove that in the distributional sense 03
W
m= 1
m= 1
C sin mx = 4 cot(x/2),
21. Show that the series
m cos mx
=
-a
csc2(x/2).
c
1 " cos mx u(x) = 2 + ___ 2a m = l a2 + m2 satisfies the boundary value problem d2u/dx2 - a2ti = -n6(x), ~ ' ( n=) u'( - n) = 0. Use the fact that its solution is
11(x) =
n cosh a(n + x ) 2a sinh an '
- n < x < 0,
n cosh a(n - x) 2a sinh an '
O<x
and show that n -coth 2a
an
=
As a particular case, prove that
1 2a2
~
" 1 +1 m2 + a2' ~
m=l
(l/m2)
=
n2/6.
22. Show that a line source of uniform density CJ distributed along the positive z axis can be represented in the notation of Section 3.1 by the density function (2nr2 sin 0)- 'a6(0). State the corresponding result for a source distributed along the negative z axis.
Distributions Defined by Divergent Integ ra Is
4.1. INTRODUCTION
In the previous chapters we have defined various singular distributions. One of them is Pf(l/x), defined in Example 4 of Section 2.4. The function l/x is not integrable on any neighborhood of the origin. We succeeded in regularizing this function by defining the functional Pf(l/x) by the principal value of the singular integral defined by the quantity (4, l/x). The aim of this chapter is to extend this idea and to regularize various singular integrals and thereby define the corresponding distributions. Let us start with a simple example. The locally integrable function
generates the distribution
75
76
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
4.
The distributional derivative ofj’(x) is ( f ’ , 4)
=
-(I; 4’)
= -
J
m
0
4YX)lJx dx.
The function,f’(x)has the classical derivative (except at x
=
=
0)
-iH(X)/X3’2.
(3)
We should very much like to see that the ordinary derivative (3) produces a distribution identical to the distributional derivative (2). However, f”(x) as given by (3) is not a locally integrable function and therefore does not generate a regular distribution. We thus have the following embarrassing situation : Every locally integrable function is identifiable with a distribution and thus has a distributional derivative, but the classical derivative of a locally integrable function may not be identical with the distributional derivative. To remedy this situation, let us look at the distributional derivative (2) a little more carefully. Let us write it as
J%
( f ’ , 4) = -1im
dx
E-0
and integrate by parts. The result is
In view of the mean value theorem, we have 4(x)
=
440) + x+’(rx)
=
4(0)
+ x$(x),
where $(x) = 4’(tx). Thus (b(~)/$ = +(O)/J, 4(0)/& as f: -, 0. Then (4)becomes
0 < t < 1,
+ JqY(ts)
and
4(s)/&
(5)
-,
The function $(x) defined by (5) is the remainder in Taylor’s formula for the function 4(x). It therefore is a continuous function of x for all x # 0 and can be extended so that it is continuous at x = 0. If R is a real number sufficiently large that 4(x) = 0 for x > R , then from (5) it follows that
4.1.
77
INTRODUCTION
or
The first term on the right side of (7) is a real number, while the second term remains finite because + ( x ) is continuous at x = 0. Now we let R -, m so that the left side of (7) becomes the right side of (6). Hence, the right side of (6) is the finite limit of the difference of two terms each of which tends to infinity. This is the finite part, in the sense of Hadamard, of the divergent integral -
1 2 JOrn
g
dx
and is written
It can be easily shown that it is a linear continuous functional (Exercise 1); it is called the pseirtlofirnction. This is a regularization of the divergent integral. Finally, we combine ( 6 ) and (9) to write -
A2 J
00
4(x)Pf [ H ( x ) x - 3 / 2 1 d x
= -
J-00m4yx)H(x)x- l j 2 d x
-m
or (-+Pf(H(x)x-3/2), 4(x))
= -(H(x)x- li2,
4’(x)).
(10)
Thus we have the formula for the distributional derivative
This analysis can be generalized to include functions , f ( x )= H(x)x-‘z+”, Indeed, in this case due to ( 5 ) we have
0 < a < 1.
(12)
78
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
for any test function 4(x). The integral term on the right of the preceding relation converges absolutely as E + 0, while the second term tends to fcc, unless +(O) vanishes. Accordingly, the integral term is the required Hadamard finite part, and we have
Now
where the last term vanishes with E because ( 4 ( ~-) @(O))/E" = c - " 4 '(tf-:), and 1 - a > 0. Thus from (1 3) and (14) we have - a J-mm4(x) Pf(H(x)x-'"+
1')
dx
or (4(x), - a Pf(H(X)X-(JI+1 ) ) )
=
lrnm
=
( -+'(x))(H(X)x-")
rix,
H(x)x-"),
-(+'(x),
which means that d (H(x)x-") = - a
Pf(H(x)x-'"+
dX
9.
Let us now examine the function f'(x) = H(x)x-("+~),
0 < a < 1.
(16)
To take its inner product with a test function 4(x), we need the Taylor expansion for +(x), $(x)
so that
=
4(0) + x4'(0)
+ tx24"(rx),
0 < t < 1,
(17)
4.1.
79
INTRODUCTION
Here again the integral term converges as finite part is m
~ ( ~ ) H ( X ) X - dx ' " +=~lim )) r-0
E +
0, so that the Hadamard
9 dx
The action o f J ' ( x ) on 4 ( x ) as given by (18) defines the required pseudofunction P f ( H ( x ) x - ' a + 2 ' ) .To prove that d dX
+ 1) Pf(H(x)x-'"+2'),
- [ H ( x ) x - ' " + " ] = -(a
(19)
we integrate the middle integral of (18) by parts:
( 4 ( ~ P) ,f ( H ( x ) x - ' " + 2 ' ) ) = F P
As E =
+
J-
W
$ ( x ) H ( x ) ~ - ' " + dx ~' m
0, the second term on the right vanishes because $ ( E ) - $(O) - E ~ ' ( O )
ic24(te), while the first becomes
and (19) follows.
80
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
4.2. THE PSEUDOFUNCTION H ( x ) / x n n ,
=
1, 2, 3, . . .
Let us now examine the function
= H(x)/x", (1) where n is a positive integer. The interesting feature of this function is that upon regularizing it in the manner of Section 4. I there appear the delta function and its derivatives for I I 2 2. To illustrate this, we first consider the simple case n = 1 and therefore study the integral
lrnm y H(xFx)
Since x
=
rlx =
JOm
rlx.
0 is the point of singularity, we examine the integral
If we put $(x)
=
4(0) + +'(tx), 0 < t < 1, this relation becomes
Jrn39 dx X
39 X rix + i , I
= Jlm
[q
+ +.cl.,] (/X
from which it immediately follows that the finite part is
When we integrate by parts, (2) becomes
Since 4(e) - 4(0) = E$'(fc), I 4 ' ( t s ) I is bounded, and lime-o (F: In E ) = 0, we finally have the required value of the Pf(H(x)/x):
4.2.
81
THE PSEUDOFUNCTION H ( x ) / x ” ,n = 1, 2, 3,
Hence, Pf(H(x)/x) = (d/dx)(H(x) In J xI). For n = 2 we use the relation
+ X+’(O) +
4(x) = +(O)
+X2@”(fX),
0 < t < 1,
and proceed as before, so
Thus the finite part is
which, when integrated by parts, yields
This finally yields the required relation: Pf(H(x)/x2) = -(d/dx)Pf(H(x)/x)
- d’(X).
(3)
We can now continue this process to prove (see Exercise 2) Pf(-) H(x) Xm+ 1
= -
d Pf(-) H ( x ) &
mxm
(-1y P y x )
+-- m !
m ,
m2l.
(4)
82
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
4.3. FUNCTIONS WITH ALGEBRAIC SINGULARITY OF ORDER M
Fortunately, these ideas can be extended to a rather large class of functions. A functionj’(x)has an algebraic singularity at x = u if for a certain integer rn > 0 the functionf’(x)Ix - aim, that is, the function f’(X1,x2,. . . , x,)((x, - u1)2
+ (x2 -
u2)2
+ . . ’ + (x, - un)2)m’2, (1)
is locally integrable in the neighborhood of the point x = a. The smallest nonnegative integer rn for which (1) is locally integrable is called the order of the algebraic singularity ofj’(x). For instance, the order of the algebraic singularity of the function f’(x) defined by relation (4.1.3) is 1 because ( - 1/2x- 3’2)x = - 1/2xwhich is locally integrable. Let us, for simplicity, confine ourselves to the functions that have only one algebraic singularity, occurring, say, at the origin. Then the foregoing discussion leads us to the following definition for the regularization principle. Definition. The regularization of the divergent improper integral
is a distribution identifiable with the functionf‘(x) in every region that does not contain the origin. In the neighborhood of the origin it is the finite part, in the sense of Hadamard. The following theorem generalizes the concepts of Section 4.1 to the case of a function,f(x) that has an algebraic singularity of order rn at the origin: Theorem 1. Let the function f’(x) have an algebraic singularity of order rn at the origin. Then its regularization is
1
(rn - l ) !
XiXj.. . .,]H( where
E + 0.
1
-
y)]
dx,
(3)
4.3.
FUNCTIONS WITH ALGEBRAIC SINGULARITY OF ORDER m
83
Proc$ The expression in square brackets is the Taylor polynomial of degree m - 1. Hence,
where 0 is the usual symbol for the order of magnitude. Accordingly, the product f(x)$(x) is integrable in the neighborhood of the origin. Moreover, inside the sphere 1x1 < E, H(l - I x ~ / E ) = 1, while for 1x1 > E , it is equal to zero, so the expression in square brackets disappears, and the integral is convergent for all $(x) E D. If for a bounded region R not containing the origin we have a test function belonging to DR,then (3) is transformed into (2). Accordingly, (2) generates a distribution that is regular in R and is thus identifiable withJ'(x). We leave it to the reader (see Exercise 3) to prove that (3) defines a linear continuous functional. Several important questions remain to be answered. First, whit kinds of functions f ( x ) do not admit a regularization? Our contention is that if l.f(x)I > Am/lxlm
(4)
in the neighborhood of the origin for m = 0, 1, 2, . . . , then it admits no regularization (see Exercise 4). The second question relates to the uniqueness of the regularization. Since the positive constant E can be chosen arbitrarily in (3), we observe that the improper integral ( 1 ) has infinitely many regularizations. The following theorem clarifies the situation. A regularization. if it exists, is uniquely determined apart from a linear combination of the delta function and its derivatives concentrated at the origin [the point of the singularity of,f'(x)].
Theorem 2.
Proof: Let the distribution t be a regularization of (2). Then t + t o , where sum of the delta function and its derivatives concentrated at x = 0, is also a distribution. This distribution can be identified with .f'(x) in every region R that does not contain x = 0. For +(x) E D R , 4(x) and all its derivatives vanish at x = 0. This gives
t o is the
((t
+ to), 4 ) =
0
3
4 ) + ( t o , 4 ) = ( t 7 4).
which is a regular distribution generated byJ'(x) in view of the foregoing discussion.
84
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
4.
On the other hand, if the distributions t , and t 2 are two different regularizations of (2), then t 1 - t 2 is also adistribution.This means that for $(x) E D, we have
0,- t 2 7 4 )
=
(tl,4) - ( t 2 , 4 )
=
0,
because both t , and t 2 are identifiable withf(x) in R . Consequently, t l - t 2 = t o is a distribution concentrated at the origin and is, therefore, a linear combination of the delta function and its derivatives. This means that whenever we want to write the general form of the regularization of the divergent integral (2), we arbitrarily choose a small value of E and add the distribution t o (as defined here) to it. The third important question is that ifj(x) has a regularization and possesses the classical derivativef’(x) except at x = 0, then is the distributional derivative of a regularization off(x) a regularization off’(x)? The answer is in the affirmative. Let us prove this for a single variable; the generalization to higher dimensions is immediate. Let Dfdenote the distributional derivative off(x). Then for all 4 E D
r
c
+ . . ’ + (m - l ) !
i, --E
=
-
f(X)&(X) dx -
Im
f(X)@(X) dx
is a regularization o f j ( x ) where we have used (3) and Integrating by parts, we have (Of;
4)
=
c -f(x)4(X)lI;
-
C.f’(X)4(X)l? +
4 replaced by 4’.
J-,f’(XW(.X) -&
dx
4.3.
85
FUNCTIONS WITH ALGEBRAIC SINGULARITY OF ORDER m
J-; CE
+
1
r f”(x) 4(x) -
xm [4(0) + 4’(0)x + . . . + 7 4(:o) m.
dx
+ ... +
4(0) - +‘(O)C
+-
I)
/
m!
where a,,, a,, . . . , a, are constants. The right side of this relation gives the . . . am( - 1)”6‘”’, regularization off”(x) plus the distribution ao6 - a,6’ which is concentrated at x = 0. This proves our assertion. The fourth question is concerned with the function,f’(x),which has several singular parts. Indeed, if .f’(x) has finite or countably infinite number of singular parts such that they are not contained in a finite interval, then we can represent it as f(x) = ,fr(x), where .f,(x) possesses only a single singular point. The regularization off’(x) is obtained by summing the regularizations ofj;(x). Following the previous notation, we call the regularization of (2) a psuedofunction and denote it by
+
+
x, Pf
s
,f’(x)4(x)dx.
Thus in the case whenf(x) is not locally integrable, it is possible to define a distribution Pf(f’(x)) by
86
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
We can recover many results of the previous two sections with the help of the analysis of this section. For instance, the function (4.1.3) has an algebraic singularity o f order m = 1 at the origin. Accordingly, we write (Pf(-+.x-3’2),
4) = J
m
- 4(O)H(1 -
(-+x-”2){4(x)
Ixl/e)} rlx
-a.
dx
= J--& m( - + X - 3 / 2 ) 4 ( 4
(-+x-3’2){4(X)
+
- +(O)}
fix
3 ’ 2 ) 4 ( x ) dx.
-+x-
All the terms on the right side are convergent integrals. Processing them slightly we recover (4.1.9). Many more examples are presented in the next section.
4.4. EXAMPLES
Example 1. The locally integrable function f(x) = (In x ) , = =
;f
x > 0,
x < 0,
x,
H ( x ) In x
defines the distribution $JE
D.
Its distributional derivative is (f’. 4) =
(-.L 4’) =
-
J
m
0
4 ~ x In1 x rlx,
(3)
while the derivative of the function defined by (1) is H ( x ) / x as defined by (4.2.1) with ti = 1. However, we know that this function is not locally inte-
4.4.
87
EXAMPLES
grable. We can use the analysis of Section 4.3 and prove that the distributional derivative (3) coincides with P f ( H ( x ) / x ) . Now, from (3) we have ([(In x ) + ] ’ , 4)
=
-((ln x ) + , 4’)
=
-1im C-0
S,
U t
+’(XI
=
-
so’
# ( x ) In x d x
In x r/x
Because
(4)becomes
-
Jo .:[ 4 ( x ) ~
=
( P f ( Y ) ,
-
4(O)H(1 - x ) ] tlx
4).
Thus or ( d / d x ) [ H ( x )In x ]
=
Pf[H(x)/x],
in the sense of distributions. Similarly (see Exercise 5), [(In x ) - ] ’
=
P f (l / x ) - ,
where (In x ) -
=
In( - x ) ,
x > 0, x < 0,
88
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
and
Pf(l) x -
x > 0, x < 0.
{O?
=
-l/x,
Exumple 2. In Example 4 of Section 2.4 we defined the singular distribution Pf(l/x) as the principal value of l/x. Let us examine this distribution in the light of the present discussion. The function l/x has an algebraic singularity of order m = 1 at x = 0. Therefore, (4.3.3) gives a regularization of l/x:
and both terms of the right side of (9) are convergent and agree with the known result. Thus the regularization of the function coincides with the Cauchy principal value. For the locally integrable functionJ‘(x) = lnlxl, we have,J”(x) = I/x, for all x # 0, in the ordinary sense of the differentiation. Hence [ln Ix 11’is equal to a regularization of I/x in the distributional sense (the proof is similar to that in Example 1); that is, (ClnlxIl’3
4)
=
Also lnlxl
=
(In x),
(Pf(l/x),
=
(10)
+ (In x)-,
and the previous example yields Pf(l/x)
4).
Pf(l/x)+
+ Pf(l/x)-.
(11)
Example 3. The derivative of the function Pf( l/x) can also be evaluated by this method. Indeed, we proved that if a function,f’(x) has a regularization and possesses a classical derivative f’(x) at x = 0, then the distributional derivative of a regularization is a regularization of,f”(x). Accordingly,
We can continue this process and obtain the following expression for the mth derivative of the principal value distribution: (Pf(l/x))‘”’
=
( - l ) m m ! Pf(l/X“+’),
(13)
89
4.4. EXAMPLES
',
where Pf(l/x"+') is a regularization of l/xm+ namely,
Example 4. Recall the Heisenberg delta distributions (2.4.10): 6'(x)
=
#x)
T (l/hi)Pf(l/x).
We can combine the results obtained for the derivatives of the delta function and the principal value distributions. This leads us to the useful relations
Example .5(a). The distributions x i , x t , Ix J',Ix 1' sgn x. The function x:
x > 0, x < 0,
=7 ; ; {
is locally integrable for Re A > - 1 and defines the regular distribution (x: , 4 ) = X"(X) dx, whose distributional derivative is ((x:Y,
4)
= - <x:
7
4').
(17)
On the other hand, the classical derivative of (16) is Ax"',
x > 0, x < 0,
which is not locally integrable. However by using the method of Section 4.3 we can show that the distributional derivative (17) coincides with a regularization of
JomAx1-
'4(x) dx.
90
DISTRIBUTIONS DEFINED B Y DIVERGENT INTEGRALS
4.
which can be integrated by parts to yield
The integral on the right side of (20) converges both for x = 0 and x = a' ( - 1 < Re A < 0) and is a regularization of (19). It even coincides with (19) if we choose a test function 4(.u) such that 4(0) = 0. Now let -2 < Re A < - I . For this case, x: has an algebraic singularity of order m = 1 at the origin. Accordingly, from (4.3.3) we derive
=
J;x"4(x)
-
4(0)] tlx
+ Jmx"b(x)
tlx.
Since two regularizations differ from one another by a distribution centrated at the origin, we put I: = 1 and obtain the general form
4)
(x:,
J x"c#)(x) - 4(0)] d x + 1
=
0 .
La
x"q5(.u) tlx
4)
f x%$(x) dx 00
=
0
1
= fo I
=
fol
x"(x)
.2[4(x) - 4(0)] r l x
= fo X"4(X)
- 4(0)]
rlx
dx
+ Jlm4(x)xi tlx
+ J1x"(x) Cct
+
Jla
x"(x)
tlx
I
+ fo +(O)XA t/x
tlx
+ Ac#)(O) + 1' ~
or
Comparing (21) and (22). we note that ( t o ,
4)
=
(21)
4) in the following
Next, we select a particular value of to by writing (x", way : (x",
+ ( t o . 4).
ro con-
( 6 / ( A + 1). 4).
4.4.
91
EXAMPLES
Our contention is that (22) yields a particular regularization of x: in the strip -2 < Re A < - 1. Indeed, the first term on the right of (22a) is defined for Re A > -2, the second term for any A, and the third for A # - 1. Thus, (22) has yielded a regularization of x: in the entire half-plane Re A > - 2 except at A = - 1. Before we continue this process, let us give the following new interpretation to this discussion. The function
F(A) =
JOU'
is analytic in A in the half-plane Re A > respective to A given by
F'(A) =
rix,
x"(x)
-
1 since it has the derivative with
s:'
x' In x&x) dx.
The foregoing analysis shows that if we seek the analytic continuation of the function F(A) in the strip -2 < Re A < - 1, we obtain a regularization of the improper integral JoaxA4(x)d x
( - 2 < Re A < - I).
(24)
Furthermore, (22) shows that F(A) has a simple pole at A = - 1, where its residue is $(O) = (6, 4). Proceeding in this manner, we can continue the functional x: analytically to the domain Re 1 > - n - 1, A # - 1, -2. . . . , - n . The result is
4(x)
-
4(0) - x$'(O)
-
. . . - ~EL
(ti
-
I)!
The right side of this formula provides a regularization of the improper integral (24) for Re A > - n - l , A # - 1, -2, . . . - n , because x: has the algebraic singularity of nth order at x = 0. Equivalently, it provides the analytic continuation of the function F ( A ) in this domain, and we observe that it has simple poles at the points 1 = - 1, -2, . . . , -n. The residue at A = -I(I>O)is
.
92
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
or ( - 1)'-1
(1 - l)!
6"- '(x).
Because for 1 I 1I n, j;^xi+'-' dx = - 1/(1 + l), (25)can be converted to a simple form in the strip --n - 1 < ReA < --n;
4 ) = J'x"[+(x) - +(O)
(x:,
0
- XqY(0) -
... -
Xn-
(n
~
1
- t)!
f#P-yo)] dx.
(27) In addition, (1 8), which is valid for Re 1 > 0, can be continued analytically throughout the 1 plane (except at the points 0, - 1,. . .). Incidentally, this discussion explains the special behavior of the pseudofunction Pf(H(x)xA)when 1 is a negative integer, as we found in Section 4.2. Exrrmple 5(h). In a similar fashion we can define the distribution xt,which corresponds to the function
.u:
=
that is,
i".
x > 0, x (0;
(-x)A,
Since we can write
(xl @(x)) = (xi 3
3
d4- x)),
it follows from the foregoing analysis that x l can be defined, by analytic continuation, for all complex values of 1 with the exception of the points 1 = - I ( / = I, 2, . . .) where xh has simple poles. This analytic continuation is equal to a regularization of this function. For instance, the value of this functional in the strip - n - 1 < Re A < - n is given by
(x!
1
+(XI>
=
j-oaxf&x)
-
4(0) + XqY(0) - . . . -
(n
- l)!
4.4.
93
EXAMPLES
To find the residue at the poles we note that because we replace $(x) by 4( - x ) in the analysis of x: we have to replace the quantities 4(J’(O) by ( - l)’+’(O). Accordingly, we find from (26) that the residue of x! at the pole A = -1is
6“ - 1 yX) (1 - I ) ! . Example F(c).
distribution
From the distributions x: 1x1’ = x:
and x!
we can form a new
+xi,
(33)
which is even, because ( [XI’, &x)) = ( Jxl’, +( -x)). It follows from the discussion in Examples 5(a) and 5(b) that I x 1’ can be continued analytically in the entire A plane except at certain poles; its analytic continuation is a regularization of the improper integral j T r n I x ~ ’ ~ ( x ) dx, Re A < - 1. Furthermore, functions :x and x! have poles at A = -1 with residues (26) and (32), respectively. It follows that the poles at A = -21 (1 = 1,2,. . .) cancel each other, and lxlAhas poles just at A = - 1, - 3 , . . . , -21 - 1. The residue at A = -21 - 1 is 262‘(x)/21!.
(34)
At the points A = -21, 1x1’ is well defined and is written Y2’. We can readily obtain the explicit expression for I x 1’ from those for x: and x? . For example, we can derive the explicit expression for 1x1’ in the strip -2m - 1 < Re A < -2m + 1 by substituting 2m for 11 in (27) and (31) and adding the two relations. The result is
Furthermore,
= il.ul”
sgn x,
A
+ 0. - 1,. . . ,
(36)
both sides of which admit analytic continuation to negative even values of A.
94
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
Example 5(d). Similarly, we can form the distribution JxlAsgn x = x:
-
Re A < - 1.
xi,
(37)
It is an odd distribution because (IxlAsgn x, 4(-x)> =
-(lXlA
sgn x, 4(x)>.
Proceeding as in Remark 5(c), we find that, when we analytically continue 1x1' sgn x, the poles of x: and x t at A = -(21 + I), I = 0, 1 , 2 , . . . , cancel, and therefore this distribution is meaningful for these values of A. It has simple poles at A = -21, I = 1,2, . . . ,with residues
- 262'- 1(x)/(2l
-
(38)
l)!.
The explicit expression for 1x1 sgn x in the strip -2m - 2 < Re A < -2m follows on substituting n = 2m in (27) and (30) and subtracting the two. That is, ( 1x1' sgn x, $(x)) =
[
+(x) - +( -x) - 2 x+'(O) X2m- 1
+ ... + (2m - I)!
+(2m-
x3 +4"'(0) 3!
" ( O ) ] ] dx.
(39)
Also,
except at A = 0, - I , . . . . Both sides of (37) admit analytic continuation to the negative values of A. Example 5(e). The distribution X-", n = 1,2,. . . . Combining the results of Examples 5(c) and 5(d), we find that the distribution X-" is meaningful for all integral values of n. Indeed, putting A = - 2 m in (35) and A = -2m - 1 in (39), we obtain (X-2m, 6) = (IXl-2m, 4)
4.4.
95
EXAMPLES
=
[
Jomx-2m-1{4(x) - 4(-x) - 2 XqY(0) X2m- 1
+ * . . + (2m - l ) ! 4(2m-"(O)]]
+ x3 3! qY"(0) -
dx,
respectively. The distributions x-", n = 1,2,. . . ,are defined by (41) and (42). Relations given in Examples 2 and 3 agree with these results. For instance, from (36) and (40) it follows that (d/dx)(x-") = - nx-"- '. Example 5 ( f ) .
Multiplication by a Function. In the formula
m 1. - m + l (43) x x+ - x + the left and right sides have independent meanings. To prove that they define the same distribution, we observe that the both sides of (43) are analytic in A for Re A > - 1 and coincide for these values of A. Therefore, they coincide over their full region of analyticity, that is, for all A except A = - 1, - 2 , . . . . However, the right side is also analytic for A = - 1, -2, . . . , -m. This implies that the factor on the left side is also analytic at these values of A. In particular, it follows that 7
lim xmx: = xm-' + ,
a+-1
l = l , 2,..., m,
so that (43) is valid for these values of A if interpreted as an appropriate limit. By similar arguments we have xmxl = ( - l)"X?!++".
(44)
xmIxIA= Ixlm+"(sgnx ) ~ ,
(45)
xmIxIasgn x = lxlm+A(sgn x)'"+',
(46)
for all A except A = - m - 1, -m - 2 , . . . . This concept can be used to multiply these functions by a function f(x) that is infinitely differentiable and has an m-fold zero at x = 0; i.e.,f(x) = x"g(x), g(0) # 0. For example, f(x)x:
= g(x)x"xi = g(x)x",+".
(47)
Example 5(g). The distributions x $ , x.l , lxl', and lxl'sgn x may be normalized by dividing them by an appropriate gamma function, since the latter has same kind of singularities. Indeed, the distributions
96
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
have the property that functionals defined by them, such as
(x:/r(A + I), 4(x)),
(49)
are entire functions of the complex variable A. The proof is as follows. Since the residue of the gamma function r(A + 1) = 1 ; xaee-Xdx, at A = - 1 is (-l)'-'/(l - 1), we find from (26) and (31) that the first two generalized functions in (48) are well defined at those points where x:, x.l have poles. Indeed,
for 1 = 1,2,. . . . The residue of [(A functional equation
-
+ 1)/2] at a pole -21
2
2
A
-
1 can be determined from the
A+5
+ 1 A + 3 '(7)
Thus Res,= - 2 1 -
A
+1
=
2(-1)1 ~
I!
'
where Res stands for the residue. Combining (51) with (34), we find that
Similarly (see Exercise 6),
Example 5(h). The other interesting combinations of the distributions are (see Exercise 9)
x i and x:
(x
f io)d
=
lim E-.
fO
(x2
+ E2)W2eiaarg(x+ic) = x:
+ e f i n xLa- ,
(54)
4.4.
97
EXAMPLES
valid for Re 1 > 0.' These distributions are again defined by analytic continuation for other values of 1.By expanding (x: ,$) and (e*i"'xx.', $) (for algebraic details see [ 6 ] ) we can prove that the poles originating from both terms of the right side of (52) cancel. This leads to the important formula lim(x _+
iE)-"-l
=
(x
k iO)-"-'
=
X-(m+l)
&+O
which agrees with (2.4.18, 19) form Pyx)
= (-
=
+ in( -m!1)"-
P'(X), (55)
0. Combining (15) and (55), we have
2ni 1)" -(x & iO)-"-l m!
Example 6. The Distribution r' = (x: + x : + . . . + xi)"2. For Re 1 > - n, r' is locally integrable and thus defines the distribution
x $(XI, x 2 , . . . ,x,) dx, d x , . . . dx,.
(57)
Since formal differentiation yields d (r', $) d1
-
=
j r ' In r$(x) d x ,
and r' In r is locally integrable, r' represents an analytic function of 1 for Re 1 > - n. For real 1 5 - n, r' is not locally integrable, and we use analytic continuation to define it. This can be done because r' has an algebraic singularity at the origin. We can reduce the distribution r' to x: and then use the results of the previous example. It is convenient to use polar coordinates r, el, 0 2 , .. . , 8,- Then (57) becomes
,.
(r', $)
=
/omrAIJ:(x)
dS]r"-' dr,
where dS is the surface element on the surface S of the unit sphere in R,. By virtue of the mean value theorem of integral calculus, we can express the inner integral )x(!J
dS = S,(l)$(r,
e\O),
ei0),. . . , Oko)),
(59)
98
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
where S,( 1 ) denotes the surface area of the unit sphere (see Section 3.3), and 0i0',Oio), . . . , (I:? are certain fixed values of the polar angle (depending only on the test function 4). If we write
4(r, O\", OiO',. . . , Olp')
=
Q,(r),
we find that (58) and (59) become
Our contention is that Q,(r) has compact support and derivatives of all orders. Moreover, all its derivatives of odd orders vanish at r = 0. Then (61) is well defined because Q,(r) is a test function, and we can transfer the results of Example 5 to this case. This assertion is proved as follows. Since 4 ( x ) vanishes for sufficiently large r, so does its mean value Q4(r). Thus Q,(r) has compact support. For r > 0, the differentiability of Q,(r) follows from the definition (60) and the fact that 4 has derivatives of all orders. In order to prove the differentiability part of the assertion for r = 0, we use Taylor's theorem with the remainder term and expand 4 ( x ) through terms of order r2,. Then (62) yields
where R 2 , is the remainder term. Because each term in the integrand containing an odd number of factors x i is an odd function (except the remainder term), its integral vanishes in the course of integration. O n the other hand, the term containing an even number, 2m, say, of factors x i yields a term of the form urnr2,. Accordingly, we have
Q,(r)
=
4(0)+ u l r 2 + u2r4 + ... + u2,rZn1+ o(r2"I),
where o has its usual meaning in measuring the magnitude of a term. This shows that we can differentiate Q,(r) 2m times at r = 0 and that odd derivatives vanish. This completes the proof. Consequently, Q,(r) is an even function of r in D, and the integral (61) is a well-defined function. To transfer the results of Example 5, let us write (61) as (note that r > 0) (Y*,
4)
= S n ( 1 ) ( x $ , Q,(x)),
11 =
1 + I1
- 1.
(63)
4.4.
99
EXAMPLES
which is an analytic function of p for Re p < - 1 or Re A < -n. Its analytic continuation to the rest of the p plane follows from the discussion of Example 5. The simple poles of occur at the points p = A + n - 1 = -1, -2, - 3 , . . . o r II = - n , - n - 1,... .Thevalueofthe residueat thepolep = m, readily derived from (26), is S"(l)( - 1 y 1 ( P -
R,$(x))/(m - I ) !
'1,
(64)
Since the derivatives of odd orders of R,(x) vanish at x = 0, no poles exist for even numbers m. To sum up, we have the following result: The distribution (r', 4 ) can be defined in the whole complex A plane with the exception of the points A = -n - 21 (I = 0,1,2,. . .), where this functional has simple poles with residues (65)
S,( l)Ry(0)/21! At the point A we have
=
- n , i.e., for I = 0, this result can be simplified. From (62)
S,( 1)fl,(O)
=
4(0)
J fls S
=
S,(1 )4(0).
Then ( 6 5 ) reduces to Sl,(1)4(0); that is. Res,,
I-'= S,( 1)6(x).
The distribution .'.1 can be normalized by introducing
because both the numerator and the denominator have poles at A = -n, - n - 2, - n - 4. The residue of 2ri at A = - 17 is 2S,,( l)d(.~), and the residue of r[(A + n)/2] at A = -n can be found as follows. From the functional equation
rfT)=% 2 r (A + n + 2
),
we have the relation (as A = - n ) ,
Therefore, the residue of
r[(A + n)/2]
R-"
=
at A
= -n
6 ( ~ ,x, 2 , . . . , x,)
=
is 2r(1)
6(~).
=
2. Accordingly,
(69)
100
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
By repeated applications of the Laplace operator V2 R"", we obtain
=
r 1-"(ra- d/dr) to
(V2)'R"2' = 2'(1 + 2)(I + 4) . . . ( A + 21)RA.
(70) This relation is valid for I > 0, and hence also for Re I I0, by the principle of analytic continuation. From (69) and (70) it follows that R-Il-21 =
(- l)'(V2)%(X) 2'n(n + 2 ) (n + 21 - 2)'
1 = 1,2,... .
(71)
Moreover, by combining (68) and (70) we have (v2)Ir-n+21 =
S,,(1)2'-'(1 - I ) ! ( - n + 2)(-n 1 = 1,2, ...,
which for the special case I
+ 4 ) . . . ( - n + 21)6(x),
(72)
1 yields
=
vZr-n+2
=
-S,,(l)(n
-
2)6(x).
(73)
We shall discuss this result in more detail in Chapter 10.
Example 7. Decomposition of r' and 6(x) into Plane Waves. Let o = (ol, 0 2 ,... ,a,,denote ) a point on the surface S of the unit sphere in R, and let the scalar product of w with a point x in R, be denoted (0. x) = o l x l + . . . + w,x,,. We attempt to evaluate the integral
JsI (o. x ) 1' dS
=
LI
w x 1' d o = Y ( x , I),
(74)
which exists as a proper integral for Re A > 0 and as an improper integral for Re A > - 1. The function Y is spherically symmetric in x , for if we substitute Ax for x in (74),where A is the matrix describing simple rotation (A' = A where A' is the transpose of A), we obtain
',
Js
Js
Js
Accordingly, Y(x,A) is a function of r and I only-Y(r, A). Moreover, Y(r,A) is a homogeneous function of degree A. Indeed, substituting cx for x , c > 0, in (74),we have
lo-cxl'ddw
= c'
lo.xl'ddo
=
c'Y(r,I).
This means that Y is proportional to r':
Y(r,A) = C(A)r'.
(75)
4.4.
101
EXAMPLES
To determine C(A) we take x to be the unit vector x in (74) and (75). This gives C(A) = "(1, A ) =
cos 9,- 1 ,
dS
=
C = (0, 0,
. . ., 1)
s,
Iw,)' dS.
Now, in spherical coordinates g l , g 2 , .. . ,%,on=
=
(76)
1,
sin"-2
dS,-
d%,-
where 9,- is the angle between ii and o,and dS,- is the area element of the surface of the (n - 1)-dimensional unit sphere. Then (76) becomes C(A) =
=
s,
Iw,J' d S
=
lcos 9,-
2S,- 1(1) / o f f ' 2 c ~ , %
1'
9 d%.
1)
d%,- dS,-
(77)
Since
and
(77) yields the following value for C(A):
Combining (68), (74), (79, and (78), we have
which, as already proved, is an analytic function in the entire A plane. This equation represents the decomposition of R' into plane waves, a concept similar to the Fourier decomposition. In the next stage, let us examine the integral on the left side of (79). From the analysis of Example 5 we know that, for an even integer A = -21, the
102
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
functional Ix.ol'/[(A + 1)/2] has no singularities, whereas for odd II (A = -(21 + I ) ) , its value is ( - l ) 1 1 ! 6 2 ' ( o . x ) / 2 1 !On . the other hand the value of R a is 6 ( x ) for all integral values A = -n. Consequently, (79) gives us the plane wave decomposition of the delta function;
6'"- " ( ( ax.) ) (lo,
n odd, (80)
6(x) =
These plane wave expansions solve Radon's problem, i.e., the problem of representing a test function 4 at any point x in terms of averages of 4 and its derivatives on hyperplanes 0 .x = a constant. Example 8. In this example we consider a function f ( x ) that is homogeneous and continuously differentiable outside the origin. Recall that a homogeneous function f ( x ) , x = ( . y 1 , . . . , x,), of degree I satisfies the functional relation . f ' ( t x l , t x , , . . . , tx,) = t ' f ( x , , x 2 , . . . , X,).
In our discussion we take 1 = - m - 1 . The function i3fydxi defines the distribution
where
4 E D and d S
nx = XilIXl,
is the element of surface on the sphere 1x1 = I:. Since
vi
103
EXERCISES
As for the second integral on the right side of (82), we observe that it is independent of E because the expression f'(x)(xi/I x 1 ) is homogeneous of degree - (m - 1) while dS is homogeneous of degree m - 1. Accordingly, if we let R + 0 in the first integral and set E = 1 in the second integral on the right-hand side of (82), we find that relation (81) becomes
This means that
From relation (83) we can recover (73) (see Exercise 17).
EXERCISES
1. Show that the functional defined by (4.1.9) is linear and continuous.
2. Prove (4.2.4). 3. Show that (4.3.3)defines a linear continuous functional.
4. Let f(x) be a locally integrable function except in a neighborhood of the origin, where l , f ( x ) l > A m / l x l m 0, < x < x O ,rn = 0, I , 2 , . . . . Show that f ( x ) cannot be regularized. 5. Show that [(ln x ) - ] ' = Pf(l/x)-. 6. Establish the relations (4.4.53). 7. Show that
r(A) = Jomx-i
'ePX d x m
1 (-
1)"
"1
m!
dx
+ Jrxa-
ReA> - n -
-
dx
1, A # - 1 ,.... - n ,
m
-11 m=O
1e-x
- 1
< ReA < - n .
104
4.
DISTRIBUTIONS DEFINED BY DIVERGENT INTEGRALS
8. Show that the general solution of x m t ( x ) = 1, is
+
t ( x ) = .cm 9. Writing ( x + ic)A
c a s
c16"-"(x).
1=1
= e A ' n ( x + i c ), prove
(4.4.54).
10. Discuss Examples 6 and 7 of Section 4.4 for the special case n
=
3.
11. Establish (4.4.70). 12. Evaluate the finite part of the integral
and determine the distribution it defines. 13. Show that lim 1 - cos R X
=
Pf($.
R+w
14. For n
=
2, define the function Pf(l/r2) as V2(+ In2 r ) = Pf(l/r2),
and show that
15. Derive the relations of Section 4.2 with the help of the results in Example 5 of Section 4.4. 16. Show that if -1 < A < 0 , k
17. In (4.4.83) set f ( x ) (4.6.73). 18. Show that (m2
=
+ P k iO1-l
=
1,2, 3, ..., then
d g / d x , , where g ( x ) = J x J - ( ~, -to ~ ) derive
=
(mZ+ P I - '
where m is a nonzero real number and P
=
x:
ind(m2 + P ) ,
+ x: + x i
-
t2.
D ist r ibut io na I Derivatives of Functions w i t h Jump Discontinuities
In boundary and initial-value problems relating to the potential, scattering, and wave propagation theories, we encounter functions that are defined inside or outside some surface S if the surface is closed and on both sides of it if it is open. However, these functions or their first- or higher-order derivatives have jumps across S. Classical theory is based on solving such problems on both sides of the boundaries and then attempting to satisfy the boundary conditions or jump conditions across S, as the case may be. There are many problems, however, that cannot be solved by classical techniques. Our aim is to develop the vector analysis of functions with jump discontinuities across surfaces and boundaries. With the help of this analysis we can solve many unsolved problems in the potential, scattering, and wave propagation theories. Furthermore, problems whose solutions are already known can be solved by this method in a very simple manner [7]. To distinguish between the classical and distributional derivatives we shall put a bar over the latter whenever there is an ambiguity. 5.1. DISTRIBUTIONAL DERIVATIVES IN R,
Let F ( x ) be a function of a single variable x that has a jump discontinuity at of magnitude a l but that everywhere else has a continuous derivative.
x =
105
106
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
Let the derivative in the interval x < t 1and x > t , be denoted F'(x). This derivative is undefined at x = tl. With the help of generalized functions, however, the distributional derivative F'(x) is obtained by setting .f(x) = F(x) - u , m -
(1)
5117
where H is the Heaviside function. The function,f(x)is continuous at x = tl. Its derivative coincides with that of F(x) on both sides of tl. Accordingly, we differentiate both sides of (1) and obtain F'(x)
5,)
(2)
+ U 1 6 ( X - 5,).
(3)
a,6(x -
= P(x) -
or Fyx) = F'(x)
Equation ( 3 ) is easily generalized to a function F(x) that has jumps of magnitude a,, u 2 , .. . ,al at tl, t 2 , .. . , t l .The result is
Let us now consider a function F ( x ) that admits derivatives up to the that has a jump discontinuity of second order on both sides of the point strength a,, and whose derivative has a jump discontinuity of strength bl at this point. To obtain F"(x), we substitute F'(x) for F(x) in (2) and obtain
(,,
(F')'
=
(F')' - b , 6 ( ~- ( 1 )
F"
=
F"
=
1'"
-
a16'(x
- 51)
-
blfi(x -
51X
or
+ a16'(x - tl) + b6(x - 51).
(5)
This process can be continued for higher derivatives and for singularities at several points. Thus, a function F ( x ) that admits continuous derivatives up to the mth order in each of the intervals (ti-1, t j ) ,j = 1, 2, . . . , I, has mth order distributional derivative F'"),
Ficm)(x) = F"(x)
1
+ C [uj6'"i= 1
"(x
-
ti)
+ bj6(m-2)(x tj>+ . . . + {jS(X - tj)], -
where aj
and [
=
[F(x)l<,,
bj = [F'(x)]t;,~
. . .,
1'.= [ p -1 )(x)leJ?
, I
It, stands for the jump in the quantity across the point t j .
(6)
5.1.
107
DISTRIBUTIONAL DERIVATIVES IN R ,
Example 1. The generalized solution of the initial value problem
LY
=
[
Ao(x)
(in
'/I-1
+ A , ( x ) rix" ~
-
1
+ ... + A, Y
= .f'(x),
x 2 0,
is equal to the solution of the inhomogeneous equation
Since we are interested in the value of y ( x ) only for positive values of x, we may assume that y ( x ) is identically zero for negative values of x . The result then follows by using (6) for in = n,17 - I , . . . , I , so that
Exumple 2. Consider the function 1 f(x) = 2
x 2x'
- -
x E LO, 2x1,
(7)
which is a 2x-periodic function; i.e.,f(x + 2 x ) = f ( x ) (see Fig. 5.1). It has jump discontinuities at x = f 2 m n . The magnitude of the jump at each
Fig. 5.1. f(.u) is 2n-periodic.
108
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
5.
discontinuity is 1, so the value of the distributional derivativef'(x) follows from (4): m
1
f'(x) = - - + 6(x - 2mn). 27c m = - m
1
(8)
The Fourier series of the function f ( x ) is f(x)
= -
i -
27i
" 1 . 1 -ermx, m
m # 0,
m=-m
(9)
which, in view of the analysis of Section 3.4, can be differentiated term by term in D'any number of times. Thus 1 ,f'(x) = -
2n
Comparing (8) and
(lo),
" 1
eimx
m=-m
1
--
(10)
271'
we obtain
which agrees with (3.4.6). With x
=
2ny this result becomes
Using the relation 6{2n(y - m)} = (1/2n)d(y- m), in (1 1) and relabeling y and x, we obtain m= - m
m= 1
m=-m
We already encountered this row of deltas in Section 2.4. Relation (12) is a distributional formula in the sense that
c m
m= - m
(6(x - m), 4) =
c W
(ei2nmx, 4),
m=-w
4 ED.
Since every test function vanishes outside a finite interval, the summation on the left side of this relation is finite. By successive differentiations of (12), we derive many more interesting Fourier series. For instance, m
m
1
m= - m
6'(.u- rn)
=
-4n
m= 1
m
1
m=-m
1 m sin 2nmx, m
6"(x - m) =
-
8n2
1m2 cos 2nmx,
m= 1
5.2. R,,, n 2 2; MOVING SURFACES OF DISCONTINUITY
109
and so on. These formulas give us the effect of the poles, dipoles, quadrupoles, etc. distributed at the points x = f m on the x axis. Note that we could have started with the function g(x) whose kth derivative is the functionf(x) as defined in (7). For instance, the function 1 O0 1 . g(x) = - elmx, 27~m = - - m m
1
m+ 1
is such that g'(x) = f(x). The series given by (13) is the Fourier series for a periodic function consisting of parabolic arcs. Comparing (10) and (13), we find that
5.2. R,, n 2 2; MOVING SURFACES OF DISCONTINUITY
The analysis can be extended to higher-dimensional spaces. Our aim is to develop systematically the distributional partial derivative of a function F(x), x = (xl, . . . ,xn),in terms of the jumps [ F ] , [grad F ] , and [grad grad F ] as well as the surface distributions. Since these partial derivatives play an important part in the theory of wave propagation, we shall include the time variable t in our derivation. Accordingly, in this section we present the kinematics of a moving surface. The analysis is presented for the general case of n 2 2. The results for n = 2 can be read off from this analysis. All the same, we present the special case separately, with the help of complex variables, in Section 5.8. Let C ( t )be a moving surface in the p-dimensional space R,. Such a surface can be represented locally either in an implicit equation of the form u(x1,. . . , x,, t ) = 0,
(1)
or in terms of the curvilinear Gaussian coordinates u l , . . . , u p - on the surface: xi = Xi(V1, . . . , U p - 1 , t). (2) We shall assume that the surface C ( t ) is regular so that all functions in (1) and (2) have derivatives of all orders with respect to each of their arguments and that for all fixed values of t the corresponding Jacobian matrices of transformation have appropriate ranks; that is, v u # 0,
(3)
110
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
where V stands for the gradient, the vector of components d/axi, i and rank(dxi/auj) = p - 1.
= 1,. . . ,p ,
(4)
Furthermore, this wave front divides the space in two parts, which we shall call positive and negative. It should be remarked that X ( t ) could be considered as a submanifold of the ( p + 1)-dimensional spacetime. Although we shall sometimes use this interpretation, it is in general more convenient to distinguish the time f from the space coordinates. Our regularity assumptions imply that at every point of the surface there exists a well-defined tangent plane and a well-defined normal line. We shall denote by A (ni, i = 1 , . . . ,p ) the unit normal vector pointing in the positive direction such that
au/axi = I vuI ni.
(5)
Letfbe a function oft defined on the surface X(t) in some interval, and let 6f /St denote the derivative with respect to time as it would be computed by an observer moving with the surface. This 6 derivative has the following geometrical interpretation. Let Po be a point on the surface at time t = ro. Construct the normal line to the surface at P o . At time t = t o + At, At an infinitesimal, this normal meets the surface X(t + At) at the point P , = P , (to + At). Then the 6 derivative is defined as Sf (Po t ) 9
6t
=
lim
Af-0
f W 1 ) -
At
.f(Po)
This means that if G denotes the speed of the wave front X(t)G
=
lim AslAt, At-0
where As is the distance between Po and P,-then 6xi - . h i . AS AX. _ - Iim = Iim - 2 = Gni, 6t A1-0 At A,-.o At AS ~
and
or
au
-= at
-GIVul.
5.2.
R,, n 2 2; MOVING SURFACES OF DISCONTINUITY
111
Sometimes we shall interpret (10) by saying that - G is the component of the normal vector in the time direction. Let us note that the essential feature of the 6 derivative is that it is computed on the surface or, what is the same, that it remains constant. Let u l , . . . , up be a local system of coordinates with u , = u such that
aul au. ax, ax,
j = 2, . . . , p ,
-__c=o,
where we have used the summation convention here as well as in the sequel. Because
for a l l j and the vector Vu,, . . . ,Vu, form a basis for R , , it follows that
Accordingly, if F(x, t ) , x hood of X ( t ) , we have dF
=
(x,,. . . , x,), is a function defined in a neighbor-
1
Suppose that a functionfis defined only on Z ( t ) and that it is equal to the restriction of F to the surface. Then we have
or
K ---d F at
at
dF +G-. dn
Since (16) holds irrespective of the way F extendsf, we use it as our formal definition of 6 time derivative. Observe also that this definition does not depend on the local system we take. It is also convenient to introduce 6 derivatives with respect to the space variables. They are defined by i5f
-=-
sxi
dF
axi
-
dF n, -, dn
112
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
which is analogous to (16) and implies that
4f / 6 x i is tangent to Z(t). Indeed,
where 9,p =
dx' axi
-- =
dun au,
. .
x;x;,
are the components of the first fundamental form of the surface and where the Greek indices assume values from 1 to p - 1. The quantities
(20) will play an important role. Observe that pij is a symmetric surface tensor; that is, pij = 6ni/6xj,
p i j n j = 0.
pij = pji,
The symmetry is preserved even with respect to time, since pi, = 6n,/6t = 6( - C)/SX, = p , i .
(21) We shall denote by p!;) the entries of the rth power of the matrix p, and set p$')
=
6.. II - n 1. nI'.
(22)
so that 6ni
p ! ! )= p . . = -
pjf) = pikpkj,
3
IJ
6Xj
pi;)
= p{:)pkj.
In general p(,n) IJ
The trace of the matrix
= p!n-m) Ik
(m)
pkj .
(23)
is denoted or:
(24) and we shall usually write -2Q instead of w,, because in the three-dimensional case -c0,/2 is equal to the mean curvature R of Z. The 6 derivatives do not commute. A direct computation using (17) gives Q, = p $ ) ,
5.3.
113
SURFACE DISTRIBUTIONS
Multiplication of these equations by ni and summing on i gives the useful relations
5.3. SURFACE DISTRIBUTIONS
In this section we study certain distributions defined on the wave front
C ( t ) and their extensions to the whole space. The basic distribution concentrated on Z(t) is the Dirac delta function, whose action on a test function +(x, t) is given by
( 6 ( ~ )4 , )
= Jm -m
J
T(t)
4(x, t) ~ S Wd t .
(1)
Observe the special treatment of time in (1). The integration with respect to the space variables is surface integration while that with respect to time is ordinary integration. If t is treated as an ordinary variable, we shall obtain a different delta function @), which is related to 6(Z) by
+ G')-
"'&Z),
(A,
4E).
A^ is usually denoted A&C),
since
6(Z)
(1
(2) because going from 6 to 8 amounts to considering the unit normal on R p + which is obtained from n,, . . . , n p , - G upon multiplication by ( 1 + G')"~'. Given any distribution on C,there are several ways to extend its action to the whole space. For our purposes this extension is achieved as follows. If A is a distribution on C and +(x, t ) is a test function on R p + then A can act on 4 by the transposition, =
( A , 4) This extension
( A , 4)
=
J
=
~ ( x tW(x, , t) ~
(3)
xt~ ,
(4)
Z(l)
whenever A is a locally integrable function. In the present study we shall obtain the extension of a distribution A by multiplying it with 6(C) so that, using (2), we have A6(C) = (1
+ Gz)- '/'A.
(5)
114
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
Observe that 6 ( X ) and 6(C) have two different meanings, either that of distributions on R,, or that of extension operators, and we shall distinguish them clearly as we go along. Another basic distribution concentrated on C is the normal derivative of the delta function: 6'(C)
=
a-
-
axi
[G(X)]ni,
(6)
where the bar labels the distributional derivative as mentioned previously. Note that a'@) is not a normal derivative operator. Indeed, the normal derivative operator will be denoted d n 6 ( X )and its action given by
From their definitions we find that
=
(d"6W
+ 2Q6(Xc),4h
where R is the mean curvature of the surface X ( t ) and is given by
-2R
=
6ni/6xi,
as mentioned earlier. Hence
6'(X) = 2R6(X)
(8)
+ dn6(X).
(9) More generally, let Q(x,, x2, . . . ,x p ) be any distribution concentrated on the surface X ( t ) . We define its normal derivative as Q' = (a - Q/axi)ni,
(10)
and by iteration, the rth-order derivative as Q(r)
= ( Q @ -1))'.
Then the following result holds:
Lemma 1. If Q is any distribution concentrated on X, then 3Q
Q"' =
axil . . . axirnil . . . nir.
(1 1)
5.4.
115
VARIOUS OTHER REPRESENTATIONS
Proof. The result is true for r holds for r, so that
-
=
1 because of (10). Let us suppose that it
-
ax,
d'+lQ ni, . . . ni,nk axil . . . axi,
because (6/6x,)n, is zero for any distribution on C(t). Letting k lemma follows. Let us now assume that aQ/axi = Q'n.. 1 9
=
r
+ 1, the (13)
then
J2Q JQ' ni axi a x j a x j
6ni + Q'-.6Xj
(14)
By symmetry, then, it follows that
Multiplying both sides of (15) by nj and summing on j yields JQ'/axi = Q"ni.
(16)
We thus obtain by induction the following lemma:
Lemma2. If Q is a distribution concentrated on C that satisfies relation [9], namely, 8Q/axi = Q'ni, then for any r 2 0 $Q(r)/axi= Q(r+ ' ) n i .
(17)
5.4. VARIOUS OTHER REPRESENTATIONS A few singular distributions closely related to 6(C) have been considered by Gel'fand and Shilov [6], Jones [8], and De Jager [9]. Let us compare them and connect them with the results of Section 5.3.
116
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
Let us suppose for the moment that time is absent in our analysis. Let P ( x l , . . . ,x,) be a smooth function such that the surface P = 0, coincides with our regular surface C and such that V P # 0 on C. Then, according to Gel’fand and Shilov [ 6 ] , the distribution d(P) is defined as follows. Let ul, . . . , up be a coordinate system with u1 = P , and let 4 ( x ) be a test function. Then, performing a formal change of variables, we have
(d(P), 4)
=
j d ( P ) + ( x ) dx =
=I
u,=o
$(O,
u2,.
s
d(ul)$(ul, . . . , u,)J d u , . . . du,,
. . , u,)J(O,
112,.
. . , u p )dlf, . . . nu,,
(1)
where $ ( i l l , . . ., u p ) = 4(xl,.. ., x,) and J is the Jacobian of the transformation. Gel’fand and Shilov prove that the value of the integral in (1) is independent of the coordinate system and accordingly it defines a distribution concentrated on C. The relation between d(P) defined this way and d(C) of Section 5.3 is r
1
where dS is surface measure on X.Thus
d(P) = d(X)/IVPl.
(3)
With the help of (3) and the relation aP/dxi = lVP1tii, we can write many of the formulas with d(P) instead of d(X). For example,
qc) = ( a P / a x i ) q P ) . We shall use this representation occasionally in subsequent chapters. Note that while the distribution d(X) depends only on the surface C, the distribution d(P)depends on P , that is, on the way the surface C is represented. Let Q ( x ) be a nowhere vanishing smooth function; then Q ( x ) P ( x ) = 0 also represents C,while, according to (3), we have
& Q p ) = (l/Q)&P).
(4)
Another way of introducing the distribution d(P) is used by De Jager [9]:
(d(P), 4)
=
lim -
Similarly, its higher derivatives can be defined as
5.5.
FIRST-ORDER DISTRIBUTIONAL DERIVATIVES
117
which are denoted in the literature 6(k)(P).However, they are not the normal derivatives as defined in Section 5.3. Indeed,
and it follows from (1) and (5.5.8) (in the next section) that
a
-P ( C ) =
axi
6‘k
+
l)(X)fli.
(8)
A few other related concepts are also discussed in the literature. For instance, the distributions concentrated on manifolds of smaller dimensions, such as those on lines and curves, can be represented in terms of the delta function of several arguments. Similarly, the distributions on intersecting surfaces can be considered [S].
5.5. FIRST-ORDER DISTRIBUTIONAL DERIVATIVES
The main objective of this work is to study the behavior of functions that are singular on the surface X ( f ) . The word “singular,” of course, has many meanings, and hence our first aim is to clarify its definition. Let u l , . . . , up be a local system of coordinates with u 1 = u as described in Section 5.2. For a small E > 0, the equation
represents a moving surface, which we shall designate C, and which can be described by the same curvilinear coordinates u 2 , . . . , u p as on X ( t ) . Given a function F defined on one side of the surface, say, in u > 0, we can form by the relation
a sequence of regular distributions on C ( t ) that act on test functions whose support is contained in the open set where the local system is defined. If this sequence converges to a distribution A , we say that A is the boundary value of F in the open set. The distribution A is the boundary value of F if it is in every open set of the above form.
118
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
Fig. 5.2. Wave front Z(u.
I).
,
Definition. A function F defined on R p + will be called a regular singular function with respect to X ( t ) if (1) F has derivatives of all orders outside C(r), and (2) F and all its derivatives have boundary values from both sides of
W).
The set of regular singular functions is a vector space closed under multiplication by C" functions as well as under ordinary (classical) differentiation. We shall denote this vector space by E(X) or merely E when the surface is clear from the context. Given F E E, its jump is the distribution
[F]
=
F+ - F-,
(3)
where F + and F - are the boundary values of F from the positive and the negative sides, respectively, of X (see Fig. 5.2). Every F E E is a regular distribution on R,, and hence we can apply both the ordinary and generalized derivatives to it. As before we shall indicate generalized derivatives with a bar. With this notation we have Theorem 1. If F E E,
8F - aF _ - - - [F]GG(X). at
at
Proof. It is sufficient to prove these results locally. If U is an open set that does not meet X ( t ) , the results are true because b(X) = 0 there. Near the surface we take a local coordinate system u l , . . . , up defined on an open set
5.5.
FIRST-ORDER DISTRIBUTIONAL DERIVATIVES
U and observe that for a test function we have
119
4(x, t), with support contained in
U,
which proves (4). Relation ( 5 ) can be proved similarly. However, it is more interesting to realize that it is a particular case of the first result. Indeed, if we consider C ( t ) as a submanifold of R,, it follows from the foregoing result that
,,
8F
- --
axi
dF dxi
+ [F]rni&C),
where m,,. . . , rn,, rn, are the components of the unit normal vector to X ( t ) in R,, Clearly, rn, = -(I + G2)-”’G. mi = (1 + G 2 ) - ” 2 n i , (7)
,.
Thus, for i = t , (6) reduces to (5). It is also true for the most of the results that follow that no separate consideration o f t is required. From Theorem 1 we can draw some very basic and interesting conclusions. Theorem 2.
2
axi 6(C) = 6’(X)ni.
-
2
-
at
6(C)
=
-G6’(C).
(9)
Proof. Let F , the function defined in Theorem 1, be such that it is 1 on the positive side of X ( t ) and 0 on the negative side. Then Theorem 1 yields
-8F_
axi - ni6(C).
120
5.
DISTRIBUTIONAL DERIVATIVES WITH J U M P DISCONTINUITIES
When we take derivatives of both sides of this relation with respect to x j and recall that generalized derivatives commute, we obtain
2 2 ~-
a
6ni
axiaxj axj Cni6(C)3 = ax -
- j
6nj -
6Xi
6(C)
-
6(X)
2
+ ni axj 6(E) -
2
+ n j axi 6 ( ~ =) -.dx,;12Faxi -
However, 6ni/6xj is symmetric, and it follows that
a
-
(3
ni -6(C) = n . - 6(X).
ax
J
axi
Multiplying (1 1) by ni and summing over i, we obtain (8). Another basic result that follows from Theorem 1 is that if [ F ] = 0, we should have
[ a ~ ~ a=~ Bni, ,]
(12)
[a~/at= ] -BG,
where B = [dF/dn] = [aF/dxj]nj. Indeed, a2F
axi axj axi axj + -
[iJ -
niS(C) =
axjaxi
~
a2F
+ [g]njd(X) =
-ax,,
a2F
because the second-order generalized derivatives commute. Hence,
[ a ~ / a =x [~a ~ /~a ~x ~ n ~ , and (13) follows by multiplying both sides of the foregoing relation by n, and summing over i. When [ F ] does not vanish, the jump in the gradient of F contains not only a normal component but a tangential component as well. Consequently, we can write it in two parts: F(u1,. . . , lip, t )
F(u,,. . . , u p , r )
=
+ H(tr,)A(u2,. . .,u p ,t ) ,
(14)
where A = [ F ] and H is the Heaviside function. Observe that the partial derivative with respect to u 1 is a derivative in the normal direction and that therefore B Also
=
8F
[dF/dn] = [ d F / d n ] . 8 9
dA + H(u,)--, aui aui aui
-=
__
(15)
5.5.
121
F I RST- 0 R D E R D ISTR IBUTlONAL D E R IVATIVES
and hence by the chain rule we have in Cartesian coordinates [dF/ax,] Since [ F ]
=
=
[ d F / a x i ] + aA/axi.
(17)
0, we obtain [dF/dx,]
=
[ d F / d n ] n , = Bn,,
(18)
[dF/dx,]
=
Bn,
+ GA/Ox,.
(19)
and therefore There is an alternative way in which (19) can be deduced. Indeed, by differentiating A = [ F ] along the surface, we have
(20) where A , , = aA/du". When we multiply both sides of (20) by gUpx$,sum on the repeated indices, and use the relation g@x;x$
=
6''
-
ninj
(21)
3
where the 6" are the components of the Kronecker delta, we obtain A,,g@x$
=
[dF/dxi](6'j - n'nj),
or
[dF/ax']
=
Bn,
+ gUPA,,x~, (22)
which is the same as (19). Thereby we have proved the following theorem: Theorem 3. If F E E, then
where A
=
[ F ] and B
+ 6A/dxi,
[dF/dx,]
=
Bni
[dF/dt]
=
-BG
=
+ 6A/6t,
(23) (24)
[dF/dn].
Many other interesting results can be deduced from the relations of this section. For instance, from (4)we find that for a vector F we have
div F curl F
+ A . [F]S(C>, = curl F + A x [ F ] S ( Z ) . =
div F
(25)
(26)
As mentioned earlier, the formulas of this section can also be obtained by considering C ( x , t ) as a p-dimensional submanifold swept out in R , , I by the moving wave front. We briefly present the corresponding definitions. Let,f(x, t ) and F(x, t ) denote scalar and vector distributions, respectively.
122
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
Similarly, let $(x, t ) and 4(x, t ) denote scalar and vector test functions, respectively. Then we define the fcllowing relations:
(div F , $)
=
- ( F , grad $),
(32)
(4 F , 4)
=
( F , curl +),
(33)
= - ( f , all//at>,
(34)
aww,
(35)
(3” /at, $) and
( a F i a t , 4) = - ( F ,
where the dot denotes the scalar product. It is left as an exercise for the reader to derive formulas (4), (5), (25), and (26) with the help of these definitions.
5.6. SECOND-ORDER DISTRIBUTIONAL DERIVATIVES
In order to generalize the results of Section 5.5 to second-order derivatives, it is convenient to introduce some notation that will simplify our formulas. Let us first observe that the tensor formed by the second-order generalized derivatives is symmetric. The same is true for the tensor formed by secondorder ordinary derivatives of functions in E . This symmetric behavior of these quantities suggests that the appropriate framework for our study is the algebra of the symmetric tensors. Accordingly, we shall understand that the product of the two tensors will be their product as elements of the symmetric algebra and not the ordinary tensor product. In particular,
( A i ) ( B j )= $(AiBj
+ AjBi).
123
SECOND-ORDER DISTRIBUTIONAL DERIVATIVES
5.6.
For F E E, DF is the tensor formed by the first-order derivatives (i.e., the gradient), D2F is the tensor formed by the second-order derivatives, and so on. For quantities defined only on the surface, D A is the vector of first-order 6 derivatives, while P A 6A ( D 2 A ) i j= -- p . n . -. 6xi 6xj J k ’ 6xk
Furthermore, we can consider DF as a p-dimensional or (p + 1)-dimensional object, according to whether time is one of the variables. Previous remarks show that it really makes no difference so long as we consider - G to be the component of the normal vector in the time direction. In view of this hypothesis, the theorems of the previous section can be written
DF = DF
+ AAG(C),
(2)
D6(C) = 6’(C)A, [DF] Theorem 1. For F
D2F
=
E
=
BA
(3)
+ DA.
(4)
E,
D2F
+ (BA’ + 2DAA + ADA)G(C) + AA2fi’(C).
(5)
Proqf. Using ( 2 ) - ( 4 ) we have
D2F
D ( D F ) = D(DF + AA6(C)) = D ( D F ) + D(AAG(C)) = D2F + A @ [DF]G(C) + ( D A @ A + ADA)G(C) + AADG(C) = D2F + (BAA + A @ D A + D A @ A + ADA)G(C) + AA26’(C) = D2F + (BA’ + 2DAA + ADA)G(C) + AA26‘(C), =
where
A’
=
AA
=
ninj,
( A @ B)
=
AiBj,
A @ DA
+ D A @ A = 2DAA.
(6)
When t is one of the variables, ( 5 ) has the following components:
J2F axi axj axi axj
axi at
at2
-
=
Bninj
6A 6A +nj + ni + A axi dXj -
6A 6A + ( -BniG + ni - - G + A axiat 6t 6Xi -
at2
+ (BG2
-
2G
6t
6(C)
6(C)
+ Aninjd’(X),
S(C) - AniG6‘(C),
+ AG26’(X).
124
5.
DISTRIBUTIONAL DERIVATIVES WITH J U M P DISCONTINUITIES
In particular,
+ ( B - 2RA)6(C) + A6’(C) = V’’F + B6(C) + Atl,h(Z).
V2F = V2F
Theorem 2. If F
E
(8)
E, then
[D’F] = CA2 + 2DBfi
+ BDA + D’A,
(9)
where C = [d2F/dn2] = [d2F/iixid x j ] n i n j .
(10)
Proof. It is again convenient to use the notation F . for d F / d x i . Then from (5) we have
r+]
Multiplying both sides with ni and summing on i gives =
Cnj + 6B + d2A ni . 6Xj 6Xj6Xi ~
Thus if we use this relation to find [tlF,i/dn], plug it back in ( 1 I ) , and use (5.2.27), we obtain [F.ij]
6B Sn. h2A 6A - p . 17.--, + Cninj + _b6xnB,j + -6Xj ni + B - + 6xj 6xi bxj Ik
6Xk
which is nothing but the component form of the required relation (9). From (13) we read off the following results:
As a special case we have [V’F]
=
C - 2RB
+ h2A/6xiSxi.
(13)
125
5.7. H IG H ER - 0R D ER D ISTR I B UTlONAL D ERl VAT1VES
From (5.3.6) it follows that Q 2 yields the following results:
=
6(C) satisfies (5.3.13), and hence Lemma
Theorem 3. Let r be any nonnegative integer. Then if 0 5 s
(526(C)
oxiaxj
=
Fij6’(C) + 6”(C)ninj,
+ cs”(C),
V2S(C) = -2!26‘(C)
526( C)
axi at
r, we have
= /ti, 6’(C)
-
(21)
G H6”(C), ~
5.7. HIGHER-ORDER DISTRIBUTIONAL DERIVATIVES
The formulas for generalized derivatives of order higher than 2 become very complicated since they contain a rather excessive number of terms. In this section we present the formula for the third-order derivative of a regular singular function [lo]. We also obtain formulas for the biharmonic operator
VF
=
V ~ ( V ~=F F) I ~ Faxi / axj ~ ~axj. ~
Third-Order Deriuatioes. Let F be a regular singular function with respect to C. Then by using the formulas for the second order derivatives and computing a new derivative we obtain D3F
=
+ (CA3 + 3DBA2 + 3D2AA + 3BpA + 3DAp + 3ADp)6(C) + (BA3 + 3DAA2 + 3ApA)S’(C) + AA36”(C); (1)
D3F
126
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
the symmetric product is understood. For instance,
as
Similarly, we can write this formula in terms of the operators d, and (dJ2
+ ((C + 2QB + (4R2 - 0 2 ) A ) A 3 + 3(DB + 2QDA)A2 + 3(B + 2QA)Ap + 3 p D A + 3D2AA + 3 ( 3 0 p + 2p'2'A)A)6(C) + d , [ ( ( B + 4RA)A3 + 3DAA2 + 3AAp)6(C)] + (d,)2[AA3S(C)]. ( 2 )
D3F = D3F
If we take Z to be a closed surface that encloses a volume V, and F is a function that vanishes outside of I/ u C,then (2) is actually the abstract form of an integral relation. To illustrate the ideas, let us compute (a/laxk)(T2F). We multiply by a test function 4 and integrate, obtaining
where the surface quantities are given by
Bihurmonic Operutor In the study of the boundary value problems in elasticity one encounters the biharmonic equation V 4 F = 0. Indeed, many plane problems of elasticity, when studied with the help of analytic functions, reduce to the solution of the two-dimensional biharmonic equation. Similarly, the discussion of the theory of elastic plates and shells leads to the three-dimensional biharmonic equation. Although it is algebraically very involved to derive the distributional derivatives of fourth order when there are discontinuities and boundaries present in the field, it is relatively simple to evaluate the distributional biharmonic equation, as we shall explain.
5.7.
127
HIGHER-ORDER DISTRIBUTIONAL DERIVATIVES
We shall denote by A j + the jump [djF/dn,i]for j = 1 , 2, . . . . Observe that A l = A, A 2 = B, A 3 = C. The symbols a, p, and y will designate the following jumps, respectively:
Lemma.
If F E E , then
a
= A 3 - 2RA2
+ V’A,,
p
= A , - 2RA3
+ V 2 A 2- u 2 A 2 + n j V 2
y = A,
-
2RA,
(9)
(R,))
+ V 2 A 3- 2w2A3 + 2njV2
~
Proof: The first result follows by putting i = j in the jump relation (5.6.13). To prove (9), we evaluate [ F , i i j ] in the following two ways. First, [F,iij]= pnj
+ 6a/6xj,
(1 1)
which is obtained from (5.5.19) by substituting V 2 F for F. Second,
[F.iijl = [F,iji] = [dF.ij/dn]ni+ 6[F,ij]/6xi.
(12)
Now we multiply both sides of ( 1 1) and (12) by n j and equate the results, producing 6[F
..I
D = r 2 1 n . ‘n . + 3 6Xi nj =
+
A3ninj
[F,ijk]ninjnk
6A2 6A2 +n . + __ ni 6Xi ’ 6Xj
d2A,
+ A2pij + Gxifixj - p j k n j “6‘1j.x k =
A , - 2RA3 + V 2 A 2
d3A1 + 66pX .i .n j A 2 + nj 6x,2 6Xj’ ~
which proves (9) on our observing that (6puJ6x.)n.= - p . . p . . = U
1
J
1J
1J
-0
2.
which proves (10). We are now ready for the evaluation of V 4 F :
V4F = V 2 ( V 2 F ) , which with the help of (8) becomes
V4F = V 2 ( V 2 F+ ( B - 2QA)d(C) + Ah’(C)) =
V Z ( V 2 F+ ) ( B - 2!20!)6(C)+ d ’ ( C ) + V 2 ( B - 2RA)d(C)
+ ( B - 2RA)V2d(X)+ V2A6’(C)+ AV26’(X). Thus we obtain Theorem. I f F
04F
=
E
E, then
i
V4F + A 4
-
4QA3 + 2V2A2 + (4R’ - co2)AZ
hR 6 A , __ - 2(V2Q)A,j6(2) + /ljVZ(hA1/hxj)- 4RV2A1 - 4 :--h x j 6Xj
+ { A 3 4 R A l + 2V2A1 + 4 R Z A 1 ) 6 ’ ( C ) + (A2 - 4nA1)d”(x)+ Ad”’(C). -
(14)
Let us now find the jump [V4F].T o this end, wc replace F by V Z Fin ( 8 ) and obtain [V4F] = [ V 2 V 2 F ]= 7 - 2Qg
+ V’O!.
5.8.
129
THE TWO-DIMENSIONAL CASE
The use of (8)-(10) in this expression gives us the desired formula:
[V4F]
=
A , - 4RA4
+ 2V2A3 + (4R2 - 2tu2)A3 + 2njV2(6A2/6xj)
5.8. THE TWO-DIMENSIONAL CASE
Let us now consider the distribution 6(Z) concentrated on the curve C.We assume in our discussion that the time t is not involved [l 13. If the curve C is described by the equation Y = dx),
(1)
then 4Z) =
rr
-
S(X>l/L-1
+ (g’(xN2 1l / 2
7
(2)
and
We can also describe C in terms of its arclength s :
i = 1, 2, (4) so that x1 = x, x2 = y , and A = (n,. n2), the outward unit normal vector, is xi
=
xi(s),
n , = xi, n2 = -,xi, x; = k nJ.’ (5) where the prime denotes differentiation with respect to s and k is the curvature of C.In this case, (5.2.18) reduces to tjf/6Xi =
In particular,
.f’(s)x;(s).
(6)
130
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
and r = l , 2 , 3 ,...,
a,=(-k)*,
(9)
where these.quantities are as defined in Section 5.2. Now we proceed as in the lemma in Section 5.3 and define the distribution 6'G) = (d/dxi)[W)]ni,
(10)
which is the normal derivative of the delta function. The corresponding normal derivative operator d,,(C) is defined as
These two distributions are connected as follows:
so that
6'(C) = dn6(C)+ k6(C).
The distributional derivatives (4.5.4) and (4.6.5) have the same form except that the terms have the present interpretation. To find the jump discontinuities across C, we consider it embedded in the complex plane C and let s measure the arclength along C in the counterclockwise direction, i.e., z = x + iy = z(s), 0 I s I L. Then the unit vector is i(s) = z'(s), i'(s) = k(s)li(s),and from (6) it follows that
Let a function F be analytic on C\C
f = [ F ] = F , - F - be its jump across
(i.e., the complement of C), and let X. Since
F'(z) = aF/iix = - i aF/ay,
the jump formula (5.5.19) yields [F'] = [ d F / d x ]
=
[dF/tin]n,
+ 6f / 6 x
and [F'] = [ - i a F / a ~ j ]= - i([dF/dn]n,
+ #'/dy).
(1 4a)
(14b)
5.9.
131
EXAMPLES
When we subtract (14b) from (14a) and use (13) and - i7(s), we obtain
(3,namely,
A(as)
=
E]
= -if%),
When we set F
=
[ F ’ ] = Qs)f ’(s). U + iV and f = u + iu, (1 5) yields
[dU/dnl
(16)
= U‘(S),
[d V l d n ] = - u’(s). Next, we apply the jump relation (16) to F’(z), obtaining [F”] =
+ kif”(s)}.
(19)
? ( S ) { ? ( S ) ~ ’ ( S ) } ’= ( ? ( S ) ) ~ { ~ ’ ’ ’ ( S )
Similarly, from ( 1 5) it follows that
+
(20) = ieis
[d2F/dn2]= - { , f ” ( s ) kif’(s)}.
In the special case that C is a circle, we have z(s) = n(s) = eiS, t ( s ) while k(s) = 1, so the foregoing relations simplify. Let us now present the corresponding formulas for the biharmonic equation. For this purpose we observe that the repeated use of (6) gives V’f =
ffl,
+ k’,f njnkV2(d2f/dxjdxk) = 4k2,f” + 3kk’f n j V 2 ( e f/ a x j ) = 2kf “
I,
I.
If we substitute these expressions in (5.7.14) and recall that k V4F = V 4 F
+
=
20, we obtain
+ {A4 - 2kA3 + 2A‘; + k 2 A 2 - k‘A‘, - k ” A , } d ( Z ) - 2kA2 + 2 A ; + k2A1}d‘(C) + ( A 2 - 2kA,)d“(C) + Ad”’(,).
(A3
Similarly,
+
[V4F] = A , - 2kA4 2A[; + ( k 2 - 2k)A, A;’ (6k2 - 4k)A; + 5kk‘A;.
+
+
+ 2kA; - (3k3 + k)A,
5.9. EXAMPLES Example 1 . For the Laplacian operator v2
= a2/aX:
+ a2/ax; + ... + a 2 / a X ; ,
132
5. DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
we found in Section 2.6 that
( V 2 F , 4 ) = ( F , V24).
Substituting (5.6.8), namely, V2F = V2F + B&C) obtain (V2F
+ A d,,(C), in
(1)
( I ) , we
+ B&C) + A d,,(X), 4 ) = ( F , V24).
(2)
For the special case that F vanishes outside C,(2) reduces to the classical Green’s second identity:
where R is the region enclosed by C. Example 2. Consider the following distribution defined in R , :
where r
=
1x1 = (x:
+ x ~ ) l ” .Then from (1) we have
Applying Green’s theorem (3) to the region lying between a small circle C of radius E and a circle of sufficiently large radius, we have 1
In - V 2 4 d x , dx, =
where ds is the element of length along C. Since V2(ln l / r ) = 0 for r 2 lim
E+O
lim E-0
I=, t)
=
(In
l= f (In): 4
E,
and
0,
ds = - 27440) = - 27c(b, 4),
(5) reduces to ( V2 ln(l/r), 4 ) = -2n(6,
4)
or
V2(ln l / r ) = -2nb(r),
(6)
so that (1/2n)In( l / r ) is the fundamental solution of the two-dimensional Laplace operator -0,.
133
5.9. EXAMPLES
Similarly, it can be shown that
0 2 ( l / r n - 2 )= - ( n
2)Sn(I)h(r),
-
(7)
where Sn(l)is the surface of the sphere of unit radius in n-dimensional space. We have already proved (7) in Example 6 of Section 4.4 in a different context.
Example 3. Let us consider the distribution 6 ( t 2 - r2), where r2 = x: + x: + x: so that C ( t ) is given by r2 - x: - x: - x: = 0. Then 6(C) is concentrated on a sphere in R 3 , and we have
(W), 4 ( x i ,~
/mm s, dt
2 ~ , 3 t ) ;> =
4 ( x i , ~ 2 1 x 3t;) dS.
(8)
Now we introduce the coordinates
x 1 = r sin 0 cos cp
=
x 2 = r sin 0 sin cp
=yo2,
x 3 = r cos 0 u = t2 -
=
rwl,
rw3,
r2.
Thus dS and du
=
=
r2 sin 0 d0 d q dr = r2 dR dr,
2t dt, or
where dR alent to
Since on u
=
du/2(u +
=
dt,
sin 0 d!2 dcp is the element of the solid angle. Thus (8) is equiv-
=
0, r2
=
t 2 , i.e., r =
(m),4) = ;J
r, this relation becomes
r=l
=
4ltl
4(ltIUl,
Itlw2, l t l o 3 ) l t l
J 4(ItlQJl? R
Itlto2,
dQ
Itlw3)dQ.
Example 4. Consider the four-dimensional distribution 6(C) = 6(x2 - m 2 ) with x2 = t 2 - x: - x: - x: and m a real constant. The surface C is given by u = ( r 2 - x i - x i - x:) - m2 = 0. For a test function 4(r,x ) we have
(W), 4) = Jm -a,
1 z
4 ( t s . x l , x 2 , x 3 )dt dxl dx2 dx,.
(9)
134
5.
DISTRIBUTIONAL DERIVATIVES WITH JUMP DISCONTINUITIES
To evaluate this we introduce new variables, as in the previous example:
x 1 = rwl, so that dt
=
du/2(u
x 2 = rw2,
+ r2 + m2)1/2
x3
=
and
rw3,
u
= t2 -
dx, dx, dx,
=
r2 - m2, r2 dSZ dr.
Then (9) can be written as
(10)
Performing the SZ integration and writing
we find that (10) is equivalent to
The distributions 6 + ( x 2 - m’) and &(x2 - m2),which are concentrated only on the upper and lower sheets of the hyperboloid x2 - m2 = 0, are given by
135
5.9. EXAMPLES
and <S-(x2 - m2), $(XI>
=
respectively.
( r 2 + m2)-”2&r, + JOrn
-(r2
+ m2)li2)
(/I-,
Example 5 . By the methods outlined in this chapter we can determine the boundary conditions on the obstacle placed in a field. Let us illustrate the approach for the electromagnetic field. Maxwell’s equations in the sense of distributions are
curl E + aB/at G IH
-
=
0,
(1 la)
divB=O,
(1 lb)
8D/at
J,
( 1 Ic)
divD=p,
(1 Id)
=
where E is the electric field, D the displacement current, B the density of magnetic flux, H the magnetic field, J the current density, and p the charge density. The quantities p and J include the surface densities, i.e., P = Pv
+ PSKQ
J =Jv
+J s K n
(12)
where p v and J v are the volume densities, p s and J s are the surface densities, and S is the surface of the obstacle. Now we apply relations (5.5.4), (5.5.5), (5.5.25), and (5.5.26) to system (1 1) and get
+ A x [E]s(s) + aB/at, div B = div B + A . [B]S(S), GIH - %/at = curl H + A x [H]s(s) - aD/at, div D = div D + A . [D]S(S), GIE
+ &/at
= curl E
where we have used the fact that G
=
(1 3 4
(13b) (1 3c) (1 3 4
0. We substitute these values in system
(1 1) and equate the singular parts, yielding
A x [El
=
0,
( 14a)
which is the well-known set of electromagnetic boundary conditions on S.
136
5.
DISTRIBUTIONAL DERIVATIVES WITH J U M P DISCONTINUITIES
The methods of this chapter enable us to convert readily many boundary value problems to integral equations [12]. We end this chapter with an application of (5.5.26) to aerodynamics as given by Farassat [13]. It was in this paper that the notation of putting a bar over the distributional derivative was first introduced.
Example 6. In aerodynamics it is found that there is a thin layer of vorticity, the vortex sheet, behind the lifting surface. Across this surface there is a jump in the velocity field u. The rest of the flow is irrotational. Let C denote the lifting surface together with its vortex sheet. Then from (5.5.26) we have
-
curl u
=
curl u
+ ii
x [u]S(C) =
A
x [u]6(X),
(15)
where we have used the fact the flow is irrotational on both sides of C.The solution to (1 5) follows from vector analysis:
where r
=
Ix
- x'l, P = (x -
x')/lx
-
x'l.
Tempered Distributions and the Fourier Transforms
6.1. PRELIMINARY CONCEPTS
In attempting to define the Fourier transform of a distribution t ( x ) , we would like to use the formula (in R , ) 7(u) = F ( t ( x ) ) =
J-mm
eiuxt(x)d x .
(1)
However, eiuxis not a test function in D, so the action oft on eiuxis not defined. We could try Parseval’s formula from classical analysis,
Jm
Q(x)g(x) dx =
m
f(x)O(x) dx,
(2)
which connects the Fourier transform of two functions , f ( x ) and g(x). That is, we define
(f, 4)
=
6),
6
4 E D.
(3)
We again run into trouble because may not be a test function even though 4 is one. These difficulties are circumvented by enlarging the class of test functions and by introducing a new class of distributions. 137
138
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
Test Functions of Rapid Decay Definition. The space S of test,functionsofrapid decay contains the complexvalued functions $(x) = 4(xl, . . . , x,) having the following properties: 1. 4(x) is infinitely differentiable; i.e., $(x) E Cm(R,). 2. +(x), as well as its derivatives of all orders, vanishes at infinity faster than the reciprocal of any polynomial.
Property 2 may be expressed by the inequality
I XpDk4(X) I < cpk
(4)
9
where p = ( p l , p , , . . . , p,) and k = ( A l , k , , . . . ,k,) are n-tuples of nonnegative integers and Cpk is a constant depending on p , k , and 4(x)..(Recall that xp and Dk4(x) are short notations for the expressions xp
=
xp'xp,. . . . x-,
where I k I =
D k 4 = d'"4(x)/ax:l
axk,' . . . ax:,
ki.)
It is evident that S 2 D,because all test functions in D vanish identically outside a finite interval, whereas those in S merely decrease rapidly at infinity. For instance, the Gaussian function exp(-1xI2/2) belongs to S but not to D.The test functions in S form a linear space. Furthermore, if 4 E S , then so is xPDk4 for any n-tuples p and k .
Convergence in S A sequence of test functions {$,,,(x)} is said to converge to 40(x) if and only if the functions &(x) and all their derivatives converge to 4o and the corresponding derivatives of 4o uniformly with respect to x in every bounded region R of R , . This means that the numbers Cpk occurring in (4)can be chosen independently of x such that
IXp(Dk4rn-
Dk40)l< Cpk
(5)
for all values of m. It is not hard to show that b0 E S. Thus the space S is closed with respect to this convergence. When 40(x) = 0, the sequence {&(x)) is called the null sequence. Remark. The space D is dense in S. To prove this assertion, take an arbitrary C" function a(x) that equals 1 for 1x1 I 1 and that vanishes for 1x1 2 2. When 4(x) E S , the test functions 4,,,(~) = a(x/m>4(x),
m
=
1,2, . . . ,
are test functions belonging to D such that the sequence {&(x)} converges to +(x) in the sense of S .
6.2.
DISTRIBUTIONS OF SLOW GROWTH (TEMPERED DISTRIBUTIONS)
139
Functions of Slow Growth
A functionf(x) = f(x,, x2, . . . ,x,) in R , is of slow growth iff(x), together with all its derivatives, grows at infinity more slowly than some polynomial. This means that there exists constants C , m, and A such that lDk.f(x)l I CIXI",
1x1 > A.
(6)
6.2. DISTRIBUTIONS OF SLOW GROWTH (TE M PER ED DISTR I B UTI0 NS)
A linear continuous functional over the space S of test functions is called a distribution ofslow growth or tempered distribution. According to each 4 E S, there is assigned a complex number (t, 4) with the properties (1)
(t9
Cl4l
+ C Z 4 2 ) = c,(t, 41) + cz(t9 4 A 4") = 0, for every null sequence {4"(x)} E S.
(2) limm+m(t,
We shall denote by S' the set of all distributions of slow growth. It follows from the definitions of convergence in D and in S that a sequence {4"(x)} converging to the function 4(x) in the sense of D also converges to 4(x) in the sense of S . Accordingly, every linear continuous functional on S is also a linear continuous functional on D and, therefore, S' c D'. Fortunately, most of the distributions on D that we have discussed in the previous chapters are also distributions on S. Only those distributions on D that grow too rapidly at infinity cannot be extended to S . For instance, the locally integrable function exp(x2) E D' but is not a member of S' (as the reader can easily verify after reading the following analysis). In fact, just as the locally integrable function formed a special subset of D', the functions of slow growth play that role in S . The corresponding result is given by the following theorem:
Theorem. Every function f(x) of slow growth generates a distribution through the formula (1)
Proof. It is clearly a linear functional. To prove continuity, we should show that if {+"} is a null sequence in S , then (f, 4,) -+ 0 as m + co. Now, for each m,
140
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
where I 2 0 is an integer. When 1 is sufficiently large, ,f(x)/(l absolutely integrable, and we have
The right side approaches zero as sequence { 4,}, proving continuity.
4,
-+
0. Thus (J;
4),
+0
+ 1 ~ 1 ~ ) is'
for a null
Note that S' can contain certain locally integrable functions that do not have slow growth. Take, for example, the function [cos(ex)]' = -ex sin(ex); it is not a function of slow growth, but it is still a member of S', as can be seen from the formula ((cos ex)',
4) =
- [cos(ex)+'(x) J
dx,
4 E s.
As in the case of D', we define convergence in S' as a weak convergence. The exact definition is as follows:
Definition. The sequence {t,} of distributions belonging to S' converges to t E S' if, for every 4 E S , ( t , , 4 ) + (c, 4) as m + co. From this definition and because S' c D', it follow that a sequence of distributions t , converging in S' to a distribution t E S', converges also in D' to the distribution t . Let us sum up the foregoing results in the form of a theorem:
Theorem. D c S and S' c D'. Furthermore, convergence in D implies convergence in S , and weak convergence in S' implies weak convergence in D'. All the singular distributions that we have studied in the previous chapters are in S' as can easily be verified. Moreover, the operations that were defined for distribution in D' remain valid in S' because S' is a subspace of D'. However, the result of some operations on a tempered distribution may not be a tempered distribution. If an operation does produce a tempered distribution, the space S' is said to be closed under that operation. Those operations are as follows: (1) (2) (3) (4)
addition of distributions, multiplication of a distribution by a constant, the algebraic operations given in Section 2.5, and differentiation.
Note that in these operations we now allow the test functions to traverse S.
141
6.3. THE FOURIER TRANSFORM
An example of an operation under which S' is not closed is the multiplication of a distribution by a function that is infinitely smooth. Take, for instance, the distribution t(x),
c 6(x m
t(x) =
m= 1
(2)
- m),
in S'. But ex2t(x)is not in S', because, for 4(x) = e-"' E S , ($(x), ex't(x)) = 1 + 1 + . . . + 1 + . ., which does not converge. On the other hand, if we take 4 E D, then (+(x), e"'t(x)) = em'4(m) possesses only a finite number of nonzero terms and therefore converges.
cz,,
6.3. THE FOURIER TRANSFORM Fourier Transform of Test Functions Let us first consider the Fourier transform of the test functions
4 E S,
wherei1.x = ulx, + u2x2 + ... + unxnandu,,u2,..., ~ ~ a r e r e a l n u m b e r s . The inverse transform is
We shall build up the theory for R,. The extension to n-dimensional space is straightforward. We shall, however, mention the n-dimensional generalization of a result when it is necessary. Using definition (1) and the definition ofthe space S , we prove the following theorem :
4 is in S , then $(u) exists and is also in S. In view of the rapid decay of +(x) at I x I = 00, the integral
Theorem 1. If Proof.
I m
(ix)kei""q5(x)dx,
k
=
m
0, 1 , 2, . . . ,
converges absolutely. This integral is the result of differentiating k times, It therefore represents the kth under the integral sign, the expression for derivative of 4,
s-,
a,
dk$(u)/duk =
4.
(ix)kei""4(x) d x
=
[ ( i ~ ) ~ "4( u] ) .
142
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
From this relation it follows that
I dk6(u)/dukI I
Ix k 4I d x ,
J-mm
so the quantity on the left side is bounded for all u. Also
where we have used integration by parts. Since the term in square brackets is in S, the function [ ( d p / d x P ) ( i x ) k ~ ( x ) ] e is i U absolutely x integrable, and therefore I up dk&u)/duk I is bounded for all u. The numbers p and k being arbitrary, we find that &u) E S . Thus, the members of S are mapped by Fourier transform into members of S. The inverse transform operation F - has analogous properties. Moreover, by the inversion formula we have
or
This shows that every function in S is a Fourier transform of some function in S . Also, if = 0, then 4 = 0; i.e., the Fourier transform is unique. Thus, the Fourier transform is a linear mapping of S onto itself. This mapping is also continuous. The proof is as follows. Let brn-, 4 as m -, co,in S . Then from the foregoing analysis we find that
6
From this inequality it follows that dk dxk
up-
-, u p
dk 6, dxk ~
which was to be proved. The analogous results hold for the inverse Fourier transform. We can summarize these results in the form of a theorem:
143
6.3. THE FOURIER TRANSFORM
Theorem 2. The Fourier transform and its inverse are continuous, linear one-to-one mappings of S onto itself. In order to obtain the transform of a tempered distribution, we need some specific formulas for the Fourier transform of 4. Let us list and prove them. The transform of d k 4 / d x k is
wherein we have integrated by parts. Thus [d'4/dxk]
A
= ( - iu)'&u),
or [(i d / d ~ ) ~A4(11) ] = uk&u).
(4b) Let P ( 1 ) be an arbitrary polynomial with constant coefficients. Then (4a) and (4b) can be generalized to give [ ~ ( d / d x ) $ ] ( u ) = P( - iu)&u)
and
CP(i d / d x ) 4 1
0.4) = J Y U ) & U ) ,
(5a) (5b)
respectively. From the relation m
(ix)k4(x)e'"xdx =
duk
-m
4(x)eiUxdx,
we have the kth derivative of the Fourier transform, or Their generalizations are CP(iX)41 A ( u ) = JW/du)&u)
(7a)
and CP(x)41 A (u) = P( - i d/du)&u),
(7b)
respectively. A translation $ ( x ) to 4 ( x - a) in S leads to the multiplicative factor e'"", as can be readily seen by substituting x - a for x in (1): Indeed,
J00
m
4 ( x - a)e'"" d x = e'""
Jm
m
c$(y)e'"Y dy,
144
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
or [ + ( x - u ) ] ^ ( u ) = eiau&u). On the other hand, the translation of a Fourier transform is
+ a ) = [eiU"4]^ (u),
[43 " ( 1 1
as is readily verified by replacing u by 11 Similarly, the scale expansion yields
[ 4 ( U X ) l " (11)
=
(8b)
+ u in (I).
( I / I a I >6(11/a>.
In particular, the reflection taking 4 ( x ) to tion &u) of & - 11):
(9a)
4( - x ) has the effect of the reflec-
[4( - X ) l ( u ) = & - u).
(9b)
The ri-dimensional analogs of the foregoing properties of the Fourier transform are easily written. They are
( u ) = P( - iu,, - iu,,
JX,
( u ) = P ( u , , u , , . . . , u,)$(u).
[ ( i ~ ) ~" @ ( u ]) = Dk$(u), [ D k 4 ]^ ( 1 1 ) = ( - iD)k&il),
[ P ( i x l , i x , , . . . , i x n ) 4 ] " ( u )= P [ P ( x , , x , , . . . , x , ) ~ ] ^ ( u= ) P [&x - a ) ] " ( 1 1 )
[4] and
A
(u
. . . , - iu,>qij(u),
a = eiu."$(u),
+ a ) = [ e i a ' " 4 ]" ( u ) ,
(12a) (12b)
145
6.3. THE FOURIER TRANSFORM
where A is a nonsingular matrix, A’ is its transpose, and xk = x:lxy . . . xk. For a pure rotation of coordinates, A’ = A - ’ and det A = 1. In this case (1 7a) becomes (1 7b)
C$(Ax>l “ ( u ) = 6 ( A u ) .
Example. We have already observed that the Gaussian function $(x) = exp( - Ix 12/2) is a member of S . In order to compute its Fourier transform, let us take n = 1 and note that $(x) satisfies the differential equation $’(x) = -x$(x). Taking Fourier transform of both sides of this equation, we have from (4a) and (6b) - iu&u) =
i(d/du)&u),
or
( d / d u ) [ & ~ ) e ~ ”=~0. ]
This gives the solution & u ) = Ce-u2i2, where C is a constant. Since &u) j Z m exp( -x2/2) dx = it follows that C = and we have
&,
&,
=
(18)
$(u) = (2.)”24(u).
That is, the Gaussian function is its own inverse except fcr a multiplicative factor (this factor disappears by a slight change in the definition of the Fourier transform). This property also holds for I I > I , as the reader can readily verify.
Fourier Transform of Tempered Distributions In Section 6.1 we notcd the difficulties we ran into by using Parseval’s formula to define the Fourier transform of a distribution in D. When we traverse the space of test functions in S and use tempered distributions, these difficulties disappear. We then have the following definition:
Definition. The Fourier transform i(u) of a tempered distribution provided by Parseval’s formula :
0, 4)
= (t,
4>,
4 E s.
4
t(x)
is
(19)
The functional on right side of (19) is well defined because E S . It is clearly linear. To prove continuity, we observe from Theorem 1 that, whenever $ m + 0, then 4m+ 0 also. Thus, for all 4mE S , whenever 4m-+ 0 (t,$m)+~
as m + m.
From (3) and definition (19) we derive the result [ i ] ” = 274 -x). Therefore, every distribution in S’ is a Fourier transform of some member of S’. Furthermore, this relation implies
(b), m>= (t(x>, &x>>
= 2X
146
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
This relation is also frequently used as a definition of the Fourier transform instead of (19). Let a sequence { t , ( x ) } of distributions in S' converge weakly in S' to a distribution t E S'. Then the sequence {i,(u)} also converges in S', so that its distributional limit is Z(u). Indeed, lim (im(xX 4 ( x ) > = lim ( t m ( u X &u>> = > = ( i ( x ) , +(x)>, m-m
m-. m
as was to be proved. Summarizing, we have a theorem analogous to Theorem 2:
Theorem 3. The Fourier transform is a continuous linear mapping of S' onto itself. The same is true for the inverse Fourier transform, defined as (F-
%I, 4 ) = ( t , F - '(4)).
(20)
It is vital that definition (19) be consistent with the classical definition whenever the latter is applicable; hence we give the following theorem: Theorem 4. Definition (19) is consistent with the classical definition for the Fourier transform of an ordinary function .f(x). Proqf m m
m
m
where F ( x ) is the classical Fourier transform o f f ' ( x ) .This proves the theorem. From definition (19) it follows that (4)-(9) carry over to the tempered distributions by transposition. The corresponding formulas are
[dkr/dxk]^ ( u ) = ( - iuIki(u),
(21a)
[(i d / d ~ ) ~^ (t u] ) = uki(tr),
(21b)
[P(d/dx)t]^(u)
=
[P(i d/dx)t] ^(u)
[(ix),?] ^ ( u ) [xkt] ^ ( u )
=
=
P( -iu)i(u), = P(LI)I(U),
(d/du)ki(u),
( - i d/dLl)ki(u),
(224
6.3.
147
THE FOURIER TRANSFORM
[ t ( x - a ) ] " ( u ) = eiaui(u), [t]"(u
+ a ) = [e'""t]"(u),
Cr(ax)l " ( u ) = (1 / I a I. [t(-x)]"(u) =
@la),
i(-U).
Let us prove one of the above results, say, (21a):
atk]
4(u)> = =
(t'k'(UX
(-
&4>
l)k(t(U),
(dk/dUk)f$(U))
= (t(uX
C( - i U ) k 4 1 " (u)>
=
(-
(@I,
iU)k4(U>>
= (( - iu)kKu),4(u)>,
which is (21a). The n-dimensional analogs of the foregoing formulas follow from the corresponding relations (lo)-( 17). We list them here for completeness and future reference:
[ i ] " ( u ) = (2n)"t( - x ) , [ D k t ]" ( u ) = (- iu)"i(u), [(iD)'t] " ( i t ) = uk2(u),
[ (a:,
P -,
-,
a:,
[P(i
. . . ,L ) r ( x ) ] " ( u ) ax,
=
P( - iu,, - iu,, . . . , - iu,,)i(u),
&, &, i
.
. . . ,i
( u ) = P(u,, u 2 , . . , u,,)i(u),
(29a) (29b)
[(ix)'?] "(u) = Dk7(u), [ x k t ]^ ( u ) = ( - iD)k7(u),
[P(ixl, i x , , . . . , ix,,)t]"(u)= P [ P ( x , , x , ,..., x , ) t ] " ( i t ) = P
[r(x - a ) ] " ( u ) i(ir
a = eia'ui(ii),
+ a ) = [eia."t] (it),
[ t ( A x ) ] " ( u )= (det A I-'i((A')-'u).
(32) (33) (34a)
148
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
For a pure rotation in R , , (34a) reduces to [?(AX)]
^ ( u ) = 7(Au).
(34b)
Thus the Fourier transform of the rotation of a distribution is the rotation of its Fourier transform. We find as a special case that the Fourier transform of a spherically symmetric distribution is a spherically symmetric distribution. We can obtain the Fourier transform for the volume and surface distributions as well. For instance, for the single-layer distribution a(x) spread over a closed bounded hypersurface S E R , , (2.4.25) yields
Thus
i(u)
=
(35)
Ja(x)d"'.' dS. S
From the analysis of this and the previous sections it follows that the theory of distributions does not provide any quick methods for computing Fourier transforms. It does, of course, provide us with the Fourier transform of generalized functions, for which the classical theory is helpless. We present in the next section various interesting examples to that effect.
6.4. EXAMPLES
Example 1. The delta function
Thus 8(x) = 1. (b) According to (6.3.27) we have [l]
^ =
[8]
=
(2n)"6(-x) = (2n)"6(x),
6.4.
149
EXAMPLES
or F - ' [ ~ ( x ) ]= 1/(27r)".
(3)
For n = 1 this gives the well-known integral representation formula for the delta function, which we derived in Example 2 of Section 3.5 in a different manner,
(c) The application of (6.3.22a) and (6.3.29a) yields [ ~ ( d / d x ) ~ ( x()u]) = P( - iu)b(u) = P( - iu) and
In particular, [d(k)(X)] ^(u) =
(-
iU)k
and [D%(x)] ^ ( u ) = ( - iu)k. For k
=
2m and 2m
+ 1, (7) becomes [ i P ) ( x ) ]^ ( u )
= (- 1)muZm
and [#z"
+
"(x)] ( u ) A
= ( - 1)"
+
1i b i 2 m
+
respectively. From (8) we find that n
[V26(X)]^(U)= -
1 u:,
i= 1
where V z is the n-dimensional Laplacian. (d) When we appeal to formula (6.3.32) we obtain [ 6 ( x - a)] ^ ( I ! ) = cia". (e) The result of combining (6.3.33) with (2) is [ e i a . x^] ( t i )
or F-'[b(u
=
(2n)"6(u
+ a).
+ a)] = (1/2n)"eia'".
1,
150
6.
For n
= 1,
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
(1 3) becomes [eiax]^ ( u )
Because sin o x
=
=
2n6(u
(15)
(elwx- eFLWx)/2i we observe from (15) that
+ w ) - 6(u - w ) ] .
[sin w x ] ^ ( u ) = -irc[b(u Similarly,
+ a).
[cos w x ] ^ ( u ) = n[6(u + w ) + 6(u
-
o)],
[sinh w x ] ^(it) = - n[6(u + iw) - 6(u - i w ) ] ,
[cash w x ] ^ ( U ) = n[6(u + iw) + 6(u - iw)].
(16) (17) (18) (19)
The n-dimensional analogs of these results can also be easily written. For instance, [sin(w. x ) ] ^ ( 2 1 )
= - fi(2n)”[h(u
+ w ) - 6(ir - w ) ] .
(20)
Exumple 2. The Heaviside function, n = 1. Since H ’ ( x ) = S(x), we find from (6.3.21a) that [ H ’ ( x ) ] ^ ( u )= -iuH(u), or
(21)
1 = -iu[H(x)]^(u).
Now recall from Example 5 of Section 2.4 that the solution of the equation -iut(u) = 1 is t ( u ) = c6(u) + i Pf(l/u), where c is a constant. Hence, it follows from (21) that [ H ( x ) ] ^ ( u ) = cd(u)
+ i Pf(l/u).
(22)
Changing x to - x in this formula, we find
[ H ( - x ) ] ^ ( u ) = c6(u) - i Pf(l/u).
+ H ( -x)
To find c, we use the relation H ( x ) [H(X)l ^ (u)
or c
=
=
1. Then
+ CH( - x ) l ^ ( u ) = 2nd(u),
n. Thus
[H(x)]^ (u)
=
n6(u) + i Pf( 1 /u).
(23)
[H(-x)]^(u)
=
n6(u) - i Pf(l/u).
(24)
If we write (23) I T r n H(x)eixud x imaginary parts, we obtain
=
n6(u) + i Pf(l/u) and separate real and
spcos(ux) rix
=
n6(u),
sin(ux) d x
=
Pf(l/u).
Jorn
6.4.
151
EXAMPLES
We also find that the function H ( a - 1x1). where a is a constant, has a Fourier transform in the classical sense: eiuxdx = 2 sin(au)/u.
[ H ( a - Ixl)] " ( u ) =
(27)
Example 3. The Signum Function sgn x and x - ~ m, > 0. First recall that sgn x = H(x) - H ( -x). Therefore, from (23) and (24), we have [sgn(x)] ^(u) = H(x) - H( -XI
=
2i Pf( l/u).
(28)
Next, we use (6.3.27) to derive [(sgn x)"(u)] "(x)
=
or
271 sgn( -x),
[2i Pf( l/u)] ^(x) = 271 sgn( -x),
which relabeled yields [Pf( l/x)] ^ ( u )
=
in sgn u.
=
ineiausgn u.
Next, we use (6.3.32) and obtain [ I/(x - a)] (u)
(29b)
Since
we find from (6.3.21a) that [x-"]"(U) = ( -
1y- 1 [ - j u ] m - l [ k ] A
( m - l)!
=
imn ( m - l)!
sgn u.
(30a)
Then, with the help of (6.3.32), we obtain
which reduces to (29b) for m
=
1.
Example 4. Heisenberg's delta distributions The Fourier transforms of Heisenberg's delta functions can be obtained by combination of the Fourier transforms of other distributions. Since 6+(x) = @(x) - (1/2ni) Pf(l/x), we have [S'(x)]"(u)
+[S(x)]"(u) - (1/2ni)[Pf(l/x)]^(u) - r - (1/2ni)i71 sgn u -
=
= +(1
- sgn u)
=
H ( - u).
152
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
Similarly, [h-(x)]A(tl)
= H(l1).
(32)
Exumplr 5. [ P f ( l / ~ ~ ~ ) ] ~=( u-2y )
where 1' is Euler's constant,
and the distribution Pf(l/lxl) is defined as
For all 4
E
S,
and ( 3 3 ) follows.
-
2 In(u),
(33)
6.4.
153
EXAMPLES
Example 6. In (6.3.24b), namely, [ P ( x ) t ( x ) ] t = 1 and use (2); the result is
A
=
P( - i d/du)Z(u), we put
[ P ( x ) ] ( u ) = 2nP( - i d/du)b(u). Let P ( x ) = a,
+ U I X + . . . + u,x".
(34) (35)
Then (34) becomes [P(x)]
A
(I/) =
+ . . . + ( -iya"8'nj)(l/).
2n(u08 - ia,6'
(36)
In particular, i
= -2 n i 8 ( I / ) ,
[x']
A
( u ) = - 2nb"(u),
...,
[x"] ( I / ) = ( - i)"2nS'"'(u). (37) A
E s u m p l c 7. Poisson Summation Formula. In Section 3.4 we found that every locally integrable periodic function / (.u) can be expanded in the Fourier series
c
I
f (.u)
=
m= - x
clf,cJtmx,
which converges to / (s)in the distributional sense. Consequently, we can take the Fourier transform of right side of this equation term by term. Using ( 1 3), we obtain [/ ( x ) ]
cc
A
1
(I/) = If'
x
=
cm[P'n'x] A
= -a
1
nt=-ic
27Ccm(5(11+ m).
The same result holds for a singular distribution. In particular, let us consider the series
which we studied in Example 2 of Section 5.2. If we set .Y = X / A , where A is a , real number, in (39) and use the relation S ( X / A - m ) = IAlS(X - d ) we obtain
154
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
Multiplying both sides by an arbitrary test function $ ( X ) and integrating from --x to Y,. we have r
Setting A
=
ra
x
I
1 in (40), we obtain cc
1
Equations (40) and (41) are called Poissow's ~ i i i n n i ~ i t i o ,/imii,lns. ii They generally transform a slowly converging series to a rapidly converging one. As an example, consider the test function $(x) = e-"'. Then $ ( I , ) = (7r)1'Ze-"2'4, and (40) becomes
The series on the left side converges rapidly for large A, that on the right sidc for small A. Fortunately, (40) remains valid even for functions of a much wider class than the test functions in S . E.xanip/e 8. Consider the quadratic form
2 uijsixj
=
i. j = 1
( A s , x),
A
= (Llij),
(42)
which is rcal and positive definite. that is, ( A x , .x) 2
hl.xl2,
b > 0.
Then [exp(-(Ax, x))IA(u) = nfiiZ(detA ) - " 2 exp(-i(ir, A - ' u ) ) .
To prove this result we define a nonsingular real transformation s which reduces the quadratic form ( A x , .Y) to diagonal form such that ( A X ,X)
=
( A B y , By) = (B'ABJ'.4')
=
(43) =
By,
l~jl',
where B' is thc transpose of B. This means that B'AB is the identity matrix, and we have A - ' = BB',
det A(det B)'
=
I.
6.4.
155
EXAMPLES
Thus [exp( - ( A x , x))] ^ ( u ) =
s
exp[ -(Ax, .u)
s
+ i ( i r . x)] dx
exp[-(ABy, By)
+ i(u, B y ) ] dy
=
ldet BI
=
(det A ) -
=
(det A ) - ’ ”
=
d”’2(det A ) - ” ’ exp( - $ I B ’ i r l Z )
=
n””(det A ) -
I”
exp[ -$(if,
BB‘u)]
=
d”’2(det A ) -
‘Iz
exp[
A-
J
I”
cxp( - ly12) + i(B’ir, y)] dy.
+ i ( B ’ ~ r ) ~dyj y~]
exp[ -y;
-*(ti,
‘ir)],
as required. Ex~lriiplr9.
From (6.3.18) it follows that (c,-rlxlz)
A
= (./[)“/2
(,-
142/4f.
(44)
If we substitute t = -is, we encounter an ambiguity when n is odd, namely, which square root to take for (n/(- is))”l2. T o remedy that, let us think of t as a complex variable z . Since we d o not want e-zlx12 to grow too fast at infinity, we must keep Re z 2 0. Now, for real z the square root is positive, so if we require that -n/2 I arg z I n/2, then the square root is uniquely determined. That is, e(n/2)iS -I
S =
Accordingly,
With this choice, (44) becomes (eiSIXl2) ’
It remains to verify that
=
e-(n/2)i
s,
s,
s >0 s
156
for all
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
4 E S. To prove this, we again appeal to (44), which ensures that
and accomplish the substitution by analytic continuation. For this purpose we consider
for fixed 4 E S. For Re z > 0, the integrals converge and can be differentiated with respect to z , so they define analytic functions in Re z > 0. Now from (46) we know that @ and Y are equal if z is real and positive. Since an analytic function is determined by its values for real z , we have
@(z) = Y(z) for Re z > 0. Furthermore, both @ and Y are continuous up to the boundary z = is for s # 0, and the result follows by taking the limits of @ ( E is) and Y ( E is) as E + 0 from positive values of E. For s = 1 and n = 1, (45) reduces to
+
Example 10(a). Fourier Transform of x: x:
+
Inasmuch as
x ' f f ( x ) = lim (e-rxx'H(x)), r-0 + < 0,
=
we have, for - 1 < Re A
J-
a:
[xtlA(u) = =
x: eiux d x
=
lim
r-O+
m
som
X'ei(u+ir)x d x
lim Jomx'eisx Ax, +
(48)
r-0
where s = ii + if. Since t > 0, Im s > 0, and therefore 0 < arg s < n.Let us compute the integral on the right side of (48), by setting isx = - (, or x = -(/is, dx = -d 0, and the contour is the ray L shown in Fig. 6.1. Accordingly, roo
J0
Xleisx
dx
=
(i/s)"+'
r
J
L
e-55' d ( .
6.4.
157
EXAMPLES
Fig. 6.1. The 5 plane.
Now, for Re 5 > 0, e - < is exponentially damped. Hence, by Cauchy's theorem, JLe-c5A d5 = Jorn.i,~ d5 = r(n + I), and (48) gives [x:]^(u)
=
+ 1) =
lim (i/s)'+'r(A 1-o+
-
ein(A+ 1)/2r(A +
l)(U
+
lim [ e i n ( A + 1(U) /+ 2 ir)-A-lr(A + I)]
t+O+
i0-A-
(49)
13
but from (4.4.54) we have (u
+ io)-"-'
=
(l,+)-I.-l
+ ein(-d-
Combining the above two relations, we find that [ x : ] ^(u) = r(n + l)[eiff('+1)/2( U + ) - d - 1 + ,-in(A+ -
1.
l)(t,-)-i-
r(A + l ) e i n ( ~ + l ) s g n ( u ) / 2 ~ U ~ - ~ - 1 ,
( u - )- A - ' 1
l)/2
(50)
for - 1 < Re A < 0. Thereafter we use analytic continuation with respect to A.
=
lim JomxAe-iSx dx,
1-0-
s =u
+ it.
158
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
Proceeding as in Example 10(a), we have
Example II(a). x: In x + The result of differentiating (49), namely, [x:]
^(u) =
ieiAni2 r(n
with respect to 1 is
+ l ) ( u + iO)-l-l,
+ + + + + +
I)(U iO)-"-' [ x i In x+]^(u) = ie'""/[T'(n (in/2)r(jl l)(u iO)-"-' - r(n l)(u iO)-"-' ln(u
+
+ iO)].
6.4.
159
EXAMPLES
As a special case, we set A = 0, obtaining
+
iO)-' + (in/2)(u [ln x+]^(u) = i[l-'(l)(u - (u + i 0 - l ln(u + io)]. = i [ +xi u
+i0-I
1-
+ r y i ) - ln(u + io) + i0 u + i0
(55)
Example I l ( h ) . x? In x - . Similarly, when we differentiate relation (51), namely, [xt]"(u)
=
-ie"""'r(A
[x-a In x-]"(u)
=
-i[e-'"''r'(A
we get
+ l)(u - iO)-'-',
+ l)(u - iO)-'-' - (in/2)e-""''r(A + I)(U - iO)-"' - e-iar'2r(A + 1)(u - iO)-'- ' ln(u - io).
For the particular case A
=
0, this becomes
[ln x-]"(u) = -i[r'(l)(u - iO)-' - (in/2)(u - iO)-' - (u - i0)- ln(u - io)]
'
= i[
ln(u - i0) - -+xi + r'(1) u - i0 u -iO
Adding and subtracting (55) and (56), we obtain [ln Ixl]^(u)
=
i[r'(l) - i(u
+ in/Z](u + iO)-'
+ iO)-'
ln(u
I.
(56)
i[r'(l) - in/2](u - iO)-' i(u - iO)-' h ( u - i0) (57)
-
+ io) +
and [lnIxIsgnx]"(u)
=
i[r'(l) - i(u
+ in/2](u + iO)-' + i[r'(l)- in/2](u
-i0-I
+ i0)-' ln(u + i0) - i(u - i0)- ' ln(u - iO),
(58)
respectively. Example 12. ra
=
(x:
+ x i + . . . + x:)"~. ga(u) = [ra] "(u) =
1
Let dx,
(59)
where - n < Re A < 0. We shall first show that ga(u) is a homogeneous function of degree -1 - n, that is, g&u)
= t -A-"ga(u).
(60)
160
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
From (59), for r > 0 we have ga(ru) = [ r a e i ( r u . xd)x =
which upon setting
I
raei(""")d x ,
x j = t - ' y j , j = 1 , 2, . . . , n,
r
=
lyl
t-'(y),
becomes ga(tu) =
s
=
(y:
dx
=
t-"y,
+ . . . + y,2)"2,
lyl't-'-'ei(u.y) dy
=
r-a-
nga(tl)?
which is (60). Since the Fourier transform of a spherically symmetric (radial) function is also spherically symmetric, we should have g,(u) = cap-'.-",
p
= (u:
+ u: +
* * *
+ u,2)1/2.
(61)
4 E s.
(62)
To calculate the value of C, we appeal to the relation (f(x),4(-x)> For 4(x) = e - r z / z ,we have
=
C1/(27c)"l(f(X),&>>, n
c
n
(63)
(27c)" Jrae-r'/2
Integrals on both sides can be evaluated by transforming to spherical coordinates by writing d x = rn-' dr dw,
du = pn-' d p dR,
where dw and dR are the solid angles in the x and ti spaces, respectively. The quantities J dw and dR give the areas of the unit sphere in the x and u spaces, respectively . Dividing by the area of the unit sphere, the integrals on both sides of ( 6 3 ) can be replaced by one-dimensional integrals. These integrals and their values are
1
JOm e - p z / 2
p - a - 1 d p = 2-(""-
'I-( -A/2),
6.4.
161
EXAMPLES
and
and (63) yields
Consequently, from (59) and (61) we have
For other values of A we appeal to the analytic continuation. For instance, for A = 2 - n, this formula yields
which for n
=
3 reduces to F-'[l/p2]
=
1 47tr
-.
Example 13. For the case n = 2 we introduce the generalized function Pf(l/r2), r = (x: + x:)~'~, through
Our contention is that [Pf( 1/r2)]
where p
= IuI = (u:
+ u;)''~and
= - 27t
In p - 27tC,
(67)
162
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
and J, is the Bessel function of order zero. The proof is as follows
=
s1A
Ila
+ =
[+(u) f‘[exp(irp
o r
2n
so1
+ 2n
J ~ ( u so2’exp(irp ) cos d) dd du dr
[4(u)[J0(rp) - 11 du dr
Ilaf
s
= - 2n
cos d) - 13 dd d i i d r
0
s 4 ( u ) J o ( r p ) du dr
4(u)(C
+ In p ) du,
from which we get the required formula (67). Example 14. The function e-‘l”l, t > 0, is a rapidly decaying function but is not in S , as it is not differentiable at x = 0. In the simple case of n = 1, we can find the Fourier transform directly:
p(.)
m
=
J-,e-t~x~ei~x
=
[“‘“i“]O
dx =
+
0
dx
J-,p+iux
[“‘-t+iu)]m
-m
-t+iu
/omp-tx+i~x
1
-
~
t+iu
+
t
+ iu
1 -t
+ iu
The inverse Fourier transform yields du.
dx -
2t t2
+ u2‘
(68)
6.4.
163
EXAMPLES
To find the corresponding formula for n > 1, we attempt to write e-'lxl as an average of Gaussian functions e-s1x12,that is, in the form [14] eLrr = Jomg(t, ~ ) e - ~ " d s ,
r
=
1x1.
Let us begin by computing
From (69) and (70) it follows that
where we have performed the u integration first. Consequently,
-
2n7[(n-1)/2 Jom
tS(n- 1)/2e-s(tz+~2)
where p = IuI. To evaluate this integral we set u = s(t2 preceding relation becomes
ds,
+ p2) so that the
which agrees with (68) for n = 1. Because F - ' ( e - ' " ) = [1/(27c)"]F(e-'Q))(-x), we find from (71) that
In the next example we consider a general radial distribution. Example 15. A distribution is called radial if its value depends only on r = 1x1 = (x: + x: + . . . + Since, the Fourier integral is invariant with respect to a rotation of orthogonal axes, it follows that the Fourier
164
6.
TEMPERED DISTRIBUTIONS AND THE FOURIER TRANSFORMS
transform off(x) is also radial. We have already discussed radial functions in a few examples. In this example our aim is to show that [ 151
where p = 1 u 1 = (u: + u: + . . . + it:)'/2. To prove (73) we evaluate the Fourier transform
f(p)
=
J
f(r)eiu'x
r/x,
(74)
R"
using spherical polar coordinates and taking the polar axis along the u direction, so that u . x = pr cos el, where x j = r cos O j , j = 1, . . . , n. Then (74) becomes "m
cn
x sin"-
rn
r2n
,
0, sin"-3 8, . . . sin 0,- den- den- . . . dO, dr.
Because
the preceding relation reduces to
Finally, we use the identity
where real part of v is greater than -$ and find that (75) is equivalent to (73), as desired. Example 16. Equation (6.3.35), which we may write
i(u) =
JS
o(x)e""" d S , ,
6.4.
165
EXAMPLES
gives the Fourier transform of the single-layer density over a sphere S of radiusa.Since,a = 1 , u . x = alu)cosO,anddS,,= Sn-l(l)an-lsinn-20d0, where I9 is the angle between u and x, ( 7 6 ) becomes
qU)=
an-
1 s
n-1
(1)
Jeff
eialulcose
0 d0.
(77)
For n = 2, ( 7 7 ) is
i(u) = a ~ , ( 1 J) o f f e i ~ ~ ~do~ c=o ~$aSl(l) ~ ~ ~ f f ~ i u ~ u d~ 0c o= s 2aJ0(a R I u I ), (78) where we have used the integral representation formula for Jo(a I I I fact that S l ( l ) = 2. For n = 3, S2(1) = 27c, so ( 7 7 ) becomes
=
I) and the
case/ialul]E = 4x0 sin a l u l / l u l ,
- 27crra2[ei4~I
or, in the notation of the previous examples (i.e., 1 u I i(u)
= 47ca
(79)
= p),
sin ap/p.
Example 17. In Chapter 10 we shall use the Fourier transforms to obtain fundamental solutions of partial differential equations. Let us examine here the concepts involved in that process. For this purpose we consider the equation
LE(x)
=
G(x),
(80)
where L is a differential operator with constant coefficients. Applying Fourier transformation to both sides of this equation, we get P(u)E(u) = 1, where P ( u ) is some polynomial. The particular solution E,(u)
=
I/P(U),
(81)
E , of ( 8 1 ) is (82)
while the complementary solution Ec is obtained by solving the homogeneous equation P(u)Ec(u)= 0.
(83)
Accordingly, 8, is the surface distribution AG{P(u)}where A is an arbitrary constant. Adding this and (82), we derive the general solution:
B(u) =
1/P(U)
+ AG{P(u)}.
(84)
166
6.
TEMPERED DISTRIBUTIONS A N D THE FOURIER TRANSFORMS
Finally, we take the inverse Fourier transform and obtain the required fundamental solution:
+ AGIP(tl))l.
E ( x ) = F-'[l/P(u)
(85)
EXERCISES 1 . Show that iff(x) is a function of slow growth on the real line, then
Iim (f(x)e-&l"l,
=
&-.Of
( f , 4).
Thus in the distributional sense limE.+o+ ,f(x)e-'I"I = f ( x ) . 2. Let t , be a sequence of distributions on S such that t, distribution on S. Show that 2, -,i. 3. Prove that for m 2 0 (a) [x"H(x)] ^ ( u ) = (- i)"'nG"'(u) + Pf[m !/(- iu)"+ (b) [x"' sgn x ] ^ ( u ) = 2 PfCm!/(iu)"'+ '1.
4. Prove that [ 1 / 1 ~ 1 ~ (] u^ ) = 2n2/lul,n
-+
t , where t is a
'1 ;
= 3.
5. Show that
[ { x - (a
e+ ib)}-'"]^(u)= 2niH(-bu) sgn b ((-m iu)m- l)! m>0,
iu(a+ ib) 9
b#0.
Note that the Fourier transform of (x - a)-"' is not the limit of that of { x - (a + ib)}-" as b -+ O+ or 0- ; rather, it is half the sum of these two limits. 6. Find the Fourier transform of (a) (xz - 4)-', (b) [ x 2 ( 1 x z ) ] - ' , (c) x3(x2 4x 3)-1. Hint. Write them as partial fractions.
+ + +
7. Find the Fourier transform of (1
-
x)-~"H(~ x).
8. Find the Fourier transform of (sin x - x cos x ) / x 3 . Hint. Use relations (5.3.23a) and (5.4.13).
9. Find the Fourier transform of xm6(")(x), m and n being positive integers.
167
EXERCISES
10. Considering the function e-x2-21i0x prove that
11. Show that (a) ( F - 'g)'"' (b) F - '(g'"') 12. Prove that
= =
F - '(( - iu)"g), (ix)"F- '(9).
[(ln x+)']^(u) = +xi + ryi) - ln(u [(In x-)']^(u)
= +xi
+ io),
- ryi) - h(u - io).
13. Let k be a positive integer, and let f be a tempered distribution that satisfies the equation xkf(x) = 1, for all x. By using the Fourier transformation, find all possible solutions for f : 14. The generalized function t(x) is said to be even (odd) if
1m
t(x)+(x) dx
=
0,
for all odd (even) functions +(x) E S . Prove that (a) Pf(l/x) is odd; (b) 6(x) is even; c~'~'(x) is even (odd) if k is even (odd); (c) if t(x) is even (odd), then t'(x) is odd (even); (d) if t(x) is even (odd), then Z is even (odd). 15. Show that a tempered distribution is the finite-order derivative of a regular function which is O( Ix I") for some a. 16. Prove that
f ( x ) = (1
+x
y = c,
JOmta-
le-rb12dt,
x E R,, a > 0, and c, is a positive constant. Show that f ( u ) is a positive function. 17. Verify (6.4.69) using the calculus of residues.
18. Let g(t) be the even function of the single variable that for positive t is identifiable with the functionf(x) in R , ; that is,f(x) = g(r), r = 1x1. Since g(t) is even and in R , , we have O(u) = 2 Iomg(t)cos ut dt,
u E R,.
168
6.
TEMPERED DISTRIBUTIONS A N D THE FOURIER TRANSFORMS
Now use (6.4.73) and show that for n = 2k
+ 1 we have
where s = p 2 . For n = 3 ( k = l), this reduces it f ( u ) = -$'(p)/27tp; derive this formula distributionally. 19. Write (6.4.73) in terms of Jo(r&), the recurrence relation
s = p 2 , for n = 2k
Similarly, write (6.4.73) in terms of J - 1,2(r&) n = 2k + 1. 20. In (6.4.40) set A
=
1 and
1 + e-" 1 - e-"
4(x)
=
+ 2, by using
( l / 7 t r f i ) 1 ' 2 cos(r&)
for
= e-"lx1 to show that
= c a 2 +2a47t2m2' W
m=-m
2 1. Prove the following generalization of Poisson's summation formula (6.4.40):
Ibl
m
C
m=
a?
+(a
+ mb) = C
e-2mnio'b$(27tm/b).
m=-cp
-OD
22. Show that
+
(a) F-'[l/(p2 0 2 ) ]= (1/471)e-"'/r, (b) F-'[l/(p2 t- 4ioirl)] = e2"*1r-2"'/47tr, where p = I u 1, r = I x 1, and o is a constant. 23. Show that
[ 6 ( . ~ , ) 6 ( r- a)8]" (u)
where r
=
=
(27tia/U)(eI x l r ) J l ( U a ) ,
& G,8 is the unit azimuthal vector, el the unit vector along
x, axis, LI constant, U =
x
UI,
and J , a Bessel function.
Direct Products and Convolutions of Distributions
7.1. DEFINITION OF THE DIRECT PRODUCT Let R , and R , be Euclidean spaces of dimensions m and n respectively, and let x = (xl,. . . , x), and y = (y,, . . . , y,) denote the generic points in R , and R,, respectively. Then a point in the Cartesian product R,,, = R , x R, is (x, y) = (xl, . . . , x, y,, . . . , y,). Furthermore, let us denote by D,, D,,and Dm+,the spaces of test functions with compact support in R,, R , , and R , + , , respectively. When f(x) and g(y) are locally integrable functions in the spaces R , and R , , respectively, then the functionf(x)g(y) is also locally integrable function in R,+,. It defines the regular distribution:
or
169
170
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS Y
Fig. 7.1
for $(x, y) E D,,,. distributions s(x)
Let us denote by s(x) 0 r(y) the direct product of the and t ( y ) E 0; according to ( 1 ) :
E 0;
Y)>), 4(x?~ ) ~ D m + n r(3) and check whether the right side of this equation defines a linear continuous For this purpose, we prove the following lemma: functional over D,,,.
(4x1 0
4 ( x , Y ) > = (~(x),(t(y), $(x,
Lemma 1. The function $(x) = ( t ( y ) , 4(x, y)), where t E 0; and $(x, y) E D,,,, is a test function in D,, and (t(Y>?D:4(X9 Y ) ) (4) for all multiindices k , where Dk implies differentiation with respect to (xl, x 2 , . . . , x), only. Also, if the sequence {$I(x, y)} + +(x, y) in D,,, as 1 + 00, then the sequence $I(x) = {(~(y),41(x,y))} + $(x) in D , as I +.- co. Dk$(X) =
Proof. For every point x E R,, 4(x, y) is a test function in D, and as such is well defined in R,. To prove that it is continuous, we fix x and let a sequence { x l } + x as 1 -+ co. Then $(XI Y ) 44x7 Y X (5) Y E Rn because the supports of $ ( x l , y) are bounded in R, independently of I (see Fig. 7.1) and for all q : 7
XI,
JI)
+
9
+
Q$(x, JJ) as I
+
JJE Rn.
m,
Now we appeal to the continuity of the functional t ( y ) on D, and find from (5) that (t(YX 4 ( X l ? Y)> (t(YX = $(x) as xI + x.
$(XO =
+
This proves that $(x) is a continuous function.
$(X?
Y))
7.1.
171
DEFINITION OF THE DIRECT PRODUCT
To prove (4) we again fix a point x in R , and set hi where h is located at the ith place in the row. Then
=
( 4 0 , . . .,h, . . . ,0), as h + 0,
in D,. Also, the supports of f have for all q D'x"'(Y)
=
are bounded in R , independently of h, and we
)
( I / h ) [ D ; 4 ( x + hi, Y ) - Q 4 ( x , .Y)]
+ D:
~
as h + 0, y E R ,
)'
ax
Accordingly, we can use (6) as well as the continuity of t ( y ) and observe that
4(x
+ hi, Y ) - 4 ( x , Y ) h
as h 0. ( r , x")) = (r(y), a 4 ( x , y ) / 8 x i ) Thus, (4) is valid for k = (0, 0, . . . , 1, . . ., 0), where the 1 is located at ith =
-+
place. By repeated applications of the preceding steps, we derive (4) in its full generality. We have thereby proved that $ ( x ) E C'"(R,). It still remains to be proved that $(x) has a compact support. But this follows from the fact that $(x, y) = 0 for 1x1 > R (see Fig. 7.1); for these values of x , $ ( x ) = ( r ( y ) , 0) = 0. Thus, $ ( x ) is a test function. Finally, to prove the third part it suffices to show that if { 4 1 ( x y)} , is a null sequence then so is { $ l ( x ) } .Now, the supports of 4 ( ( x , y) are bounded in R,+, independently of I, so the supports of {t,hI(x)}are also bounded independently of 1. Accordingly, we have only to prove that as I
Dk$,(x) -+ 0
co,x E R,. (7) Suppose that (7) does not hold and we can find a number c0 > 0, a multiindex k o , and a sequence of points x I such that D k o [ $ I ( ~ l2 )] ~
0 ,
--*
I
=
1, 2,. . . .
(8)
Since the supports of Icll(x,y) are bounded in R , independently of I, it follows that the sequence { x l } is also bounded in R,. Appealing to the BolzanoWeierstrass theorem, we can choose a convergent subsequence x l j + xo as j co. Then, -+
D 9 4 1 j ( x l jy) ,
-+
0
as j
-+
co.
172
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
Consequently, the distribution r(y) satisfies the relation D k o $ l , ( ~ l ,= ) ( t ( ~ >D?411(xI,. . Y>> - (r(y), 0)
=
0
as j
+
w. (9)
We have the required contradiction, and therefore the lemma is proved. Returning to definition (3), and using Lemma 1, we find that $(x) = (t(y), 4(x, y)) ED, for all 4 E D,,,. Thus, the right side of (3), namely, (s, I)), is defined for any distributions s and t and is a functional over D,,,. The linearity of this functional follows from the linearity of the functionals s and t . To prove the continuity of this functional, let the sequence {41} 4 as I -+ cx, in D,,,. Then, in view of Lemma 1, -+
( ~ ( Y X 4I(x, Y ) )
+
( t ( ~ )4(x, , 1')) in
D m
as I
--t
r*.
Since the functional is continuous, we have
(W, (t(Y>, 4I(X, Y ) ) )
(s(x), (t(Y), +(X? Y)>>l
(10) as I -,m. This proves the continuity of the functional defined by the right side of (3) and s(x) 0 t ( y ) is a generalized function in DL+,. +
Some Properties of the Direct Product
Property 1. Commutativity. The direct product is commutative. Proof. Let the test function
f
I=1
4(x, y) E D,,,
~'I(X)'J'I(YX
have the form
41(x)E D m , $I(Y>E Dn.
( 1 1)
Then according to definition (3) we have the following expression for both (s 0 t , 4) and (t 0 s, 4 ) : D
Now, to prove that this result holds for any test function 4(x, y), we show that the test functions of the form (1 1) are dense in the space Dm+,.For this purpose, let us denote the space of the test function of the form (1 1) D, 0 D, and prove the following lemma:
Lemma 2.
D, 0 D, is dense in D , + ,.
The lemma will be proved if we can show that for any function +(x, y) E
D, +, there is a sequence of test functions {41(x,y)} of the form (1 1) converging to +(x, y). Suppose that 4(x, y) has the support R( Ix I I a, lyl I a); that is, it vanishes outside this block R . Then for a given
E =
1/1, we can construct,
7.1.
173
DEFINITION OF THE DIRECT PRODUCT
by virtue of Weierstrass’s theorem, polynomials P,(x, y ) that differ in the region R’( I x I I2 4 I y I I2a) from 4(x, y ) by less than E. The same is true for all derivatives of order k. Let e(x) be the function
Property 2. Continuity.
When the sequence { s r ] M X )
-, s
in D; as I
+
co,then
0 t ( Y ) ) -, s(x)t(y),
in Dk+,. Proof: According to Lemma 1, for $(x, y ) Dmfnr ~ $(x) = (r(y), +(x, y)) D,. Thus,
E
(six)
0r(J% 403 Y ) ) = (s,(x),
(t(YX 4 k Y)>> = ( S h $) (s, $> = <s, (t(YX 4 k Y ) ) = 0 0 1 , 4(x, y ) > as I -, 30.
+
The theory of the direct product of two distributions can be readily extended to the direct product of any finite number of distributions.
Property 3. Associativity For s E DL,t E Db,and u E DZ we have s(x) 0 Ct(Y) 0 4.711 =
CHX)
0t(Y)l 0 4 z ) .
(13)
r p is the dimension Proqf: Let 4(x, y , z ) be a test function in R m + n + pwhere of the z space. Then
Property 4. Support supp(s 0r ) = (supp s) x (supp t).
(14)
This means that the set of points in supp(s 0t ) consists of just those points (x, y ) in R,,,nin which the first coordinate x belongs to supp s and the second coordinate y belongs to supp t . Proof. We want to show that when a point xo lies outside supp s(x), then every point (xo, y o ) also lies outside supp(s(x) 0r(y)), no matter what value
174
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
yo takes. By the definition of the support of the distribution, given in Section 2.8, we find that under this premise there is some neighborhood R(xo) of xo such that (s(x), 4(x)) = 0 for every 4(x) E D , whose support is contained in this neighborhood. For a test function 4(x, y)eD,+, whose support lies in the block R(xo) x R(yo), (ie., 4(x, y) is zero whenever x is not in R(xo)), we find that the support of the test function + ( t ) = (t(y), $(x, y)) is contained in R(xo). Hence, (s(x) 0 t(yX
4 k Y)> = (s(x), W)>= 0.
Next we assume that yo is outside the support of t(y). By the same arguments we conclude that in this case also (xo, yo) is not in the support of s(x) 0 t(y), no matter what value xo takes. Thus, we have proved that supp(s 0 t ) c supp s x supp t.
(15)
O n the other hand, let x , ~ s u p p sand y , ~ s u p p t .Then according to Lemma 2 we can determine a test function 4(x)+(y) for any neighborhood R(xo, yo) of the point (xo, yo), where 4(x) E D , and +(y) E D,, such that the support of +(x)+(y) is contained in R(xo, yo). Thus, (s 0 4
4*>
=
G4, t*>
f 0,
which means that supp s x supp t c supp(s 0 t). Combining (1 5) and ( 16) we have (14). Property 5. Differentiation
which proves (17a). Similarly, it follows (see Exercise 1)
Property 6. Multiplication by a C" function
For a(x) E C" we have
a(x)Cs(x) 0 tW1 = Ca(x)s(x)I
0 t(y).
(18)
7.1.
175
DEFINITION OF THE DIRECT PRODUCT
which is equivalent to (18). Property 7. Translation (S
Q t)(x
+ h, y ) =
S(X
+ h ) 0 t(y).
and we have the required result. Example I. The direct product of the delta functions over R , with that over R , yields the delta function:
Example 2. Just as an ordinary function is said to be independent of y if it is of the formf(x) Q l(y), we define that a distribution is independent of y if it is of the form s(x) Q l(y). It acts according to the rule M X )
0 l(Y) , 4(x, Y ) > =
=
( I ( Y ) Q s(x), 4(x> Y)>.
+ED,+,.
176
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
In other words, we have the relation
Example 3. Let H ( x ) be the Heaviside function of n variables: H(x) =
1, 0
x , > o , X,>O, elsewhere.
..., x , > o ,
This is clearly the direct product H ( x , ) 0H ( x 2 )0 . . 0H(x,). By virtue o f (17b) and the relation dH(xi)/dxi= 6(xi),we have
anH ax, ax, . . . ax,
= 6 ( x 1 ,x 2 , .
. . ,XJ.
This means that the function H ( x ) = H ( x 1 ) H ( x 2.). . H(x,) defined in Exercise 10 of Chapter 2 coincides with this direct product.
Example 4. The direct product of 6 ( x l )and a locally integrable functionf‘ of the variables x 2 and x 3 E R , is (6(x,)f’(x2,X 3 h
4(Xl?
=
<6(Xl),< . m 2 3
=
J
x27 x 3 ) ) Xd7
44x19 x2. x 3 ) ) )
+(o, x2 x3).f(x2
9
9
~
3
dx2 ) dx3.
R2
Consequently, 6 ( x , ) f ( x 2 ,x 3 ) is the volume source density corresponding to a simple layer of sources spread on the plane x , = 0. In particular, 6(x,)1(x2,x,) corresponds to a simple layer of unit surface density on the plane x , = 0.
7.2. THE DIRECT PRODUCT OF TEMPERED DIST R IB UTI0NS
We use the notation and the definition of the previous section. Let s(x) E Sk, X E R,, and t ( y ) ~ S ;Y, E R,. Since S’ c D‘,the direct product s(x) 0f ( y ) E D,+,, x , Y E R,+, = R , x R,. Our aim is to prove that s(x) 0t ( y ) ~ S ’ ~ + , . In view of the definition of the functional s(x) 0t(y), namely,
( d x ) 0t(YX +(x, Y)>
=
<s(x), > > l
(1)
7.2.
THE DIRECT PRODUCT OF TEMPERED DISTRIBUTIONS
177
where 4 ( x , y) now traverses the space S , we should show that the right side of We proceed as in the previous section and first state the following lemma, whose proof is analogous to Lemma 1 of Section 7.1.
(1) is a linear continuous functional on S,,,.
Lemma. The function +(x) = ( t ( y ) , 4 ( x , y ) ) , where ~ES:,,~ES,,,,, is a test function in S,, and Dk+(X) =
( f ( Y X D M X , Y)>?
(2)
By virtue of this lemma, (1) defines a linear continuous functional on S,,,. Thus s(x) 0 t ( y ) E S ; + , . Most of the properties that hold for the direct product in D;+, hold also in S g , , . The proof is similar. We state them for the sake of completeness. Property 1. Commutativity
Property 2. Continuity If sI s(x) 0 r(y) in S g , , as 1 + m.
-+
s in SL as 1 -+ m, then sI(x)0 t ( y ) -+
Property 4. Support
supp(s 0 t ) = (supp s) x (supp t ) . Property 5. Differentiation
(6)
178
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
7.3. THE FOURIER TRANSFORM OF THE DIRECT PRODUCT OF TEMPERED DISTRIBUTIONS
Let s(x) E Sk and t ( y ) E Sn;then [s 0 t ] ^ = 3 0 2.
For 4(u, w ) E Srn+,,,
where F , means the Fourier transform of the function so that
4 for the argument y ,
F W F , [ 4 ] ( x ,y) = I e i w y[eiUx4(v,w ) do dw.
Thus
which proves (1). Note that in this process we have also proved that
Example. For the case n H(x, Y ) =
=
2, consider the function H ( x , y),
1,
0
x > 0, y > 0, for all other values of x , y.
which can be written H ( x , y) = H ( x ) 0 H(y). When we use (1) for the Fourier transform of the direct product, we find that CH(x, Y ) l ^ = CWx) 0 W y ) l ^ = R X ) 0 &Y) = (z6(u)+ i Pf(l/u)) 0 (zS(w) + i Pf(l/w)).
7.4.
179
THE CONVOLUTION
7.4. THE CONVOLUTION
The convolutionf
* g of two functionsf(x) and g(x), both in R , , is defined as
It is clear that
whenever the convolution exists. Let us assume that functionsf(x) and g(x) are locally integrable in R , . Then ,f * g is locally integrable in R , and hence defines a regular distribution ( f * g, 4) :
or
Equation (3) seems to reveal a property of the convolution that might be used to define the convolution of two distributions. That is, the convolution of two distributions s and t in D is (s
* f, 4)
=
(s 0 l, 4(x
+ Y)).
(4)
A small problem arises: The function $(x + y ) does not have compact support. (Its support is the infinite strip that lies between x + y = A and x + y = - A , where the constant A depends on the supports of s and t . ) In order to ensure that the formula works we have to make certain assumptions.
180
7.
DIRECT PRODUCT AND CONVOLUTIONS OF DISTRIBUTIONS
Y R
-R
Fig. 7.2
We have seen in Section 7.2 that supp(s 0 t ) = supp s x supp t . Accordingly, (4) will become meaningful if the intersection of the supp(s 0 t ) and supp 4(x y ) is bounded. Indeed, in that case, we replace 4(x y ) by a finite function @(x,y ) that is equal to +(x + y ) in this intersection and vanishes outside it. In the sequel, when we write 4(x y ) we mean such a function 4(x, y). The boundedness of the intersection of the supp(s 0 r ) and supp 4(x y ) can be achieved in the following two ways:
+
+
+
+
1. The support of one of the distributions is bounded. Let, for example, the support of t be bounded. In this case, the support of 4(x y ) is contained in a horizontal strip of a finite width [x, y : Ix yl I A, l y l < R ] (see Fig. 7.2). Thus, by virtue of the definition of the direct product we have
+
((s
* tX 4 ) = ((s 0 t), 4(x + Y ) )
= (dx),
+
(t(Y>, 4b + Y ) ) ) .
(5)
On the other hand, if the support of s is bounded, then the support of y ) is contained in a vertical strip of a finite width. Under either of these circumstances, the function $(x) = ( t ( y ) , $(x + y ) ) is a member of D,, as proved in Section 7.1.
4(x
+
Y
Fig. 7.3
7.4.
181
THE CONVOLUTION
2. Both s and t have supports that are bounded on the same side. For example, let s = 0 for x > R , , and let t = 0 for y > R , . In this case the supy ) is contained in a quarter-plane lying below some horizontal port of +(x line and to the left ofsome vertical line (see Fig. 7.3). Therefore, the right side of (5) is again well defined.
+
Properties of the Convolution of Distributions Property 1. Commutativity s*t=t*s.
This is an immediate consequence of the definition (5) and the commutativity of the direct products s 0t . Property 2. Associativity (s
* t ) * 11 = s * ( t * u )
(7)
if the supports of the two of these three distributions are bounded or if the supports of all three distributions are bounded on the same side.
The proof of this result is a straightforward extension of the corresponding proof given for the validity of definition (5). Property 3. Differentiation If the convolution s * t exists, then the convolutions (Ilks)* t and s * (D't) exist, and
(D"s) * t PrmJ j = 1,
Dk(s * t ) = s * (Dkt).
(8)
It is sufficient to prove that (8) holds for each first derivative a / a x j ,
..., n. For + G D,
(a/axj(s * t ) , 4 ) = ( - i)(.y = = =
or
=
* t, a 4 / a x j ) = ( -
i)(s
o t , ( a / a x j )+(x + Y ) )
(s, ((- 1)t- (a/axj)4(x + Y ) ) > (s, ( w a x , , 4 ( x + YD) (s at/axj, 4 ( x y ) ) = (s * d t / a x j , 4),
o
+
(a/axj)(s * t) = s * a t / a x j .
(9)
By virtue of the commutativity of the convolution, we interchange s and t in (9) and get (10)
Combining (9) and (lo), we have (8), as required.
182
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
If L is a differential operator with constant coefficients, we find from (8) that
(ts) * t
= L(s * t ) = s * (Lt).
(1 1)
These results imply that, in order to differentiate a convolution, it suffices to differentiate any one of the factors. Property 4. Continuity In certain case6 the convolution is a continuous operator. The following theorem embodies this result.
Theorem. Let the sequenceofdistributjons {sI}+ s a s l + 00, then {sf * t } + s * t under each of the following conditions: 1. All distributions sl are concentrated on the same bounded set. 2. The distribution t is concentrated on a bounded set. 3. The supports of the distributions s and t are bounded on the same side by a constant independent of 1.
Proof. In view of (9,we have
If condition 1 holds, we can replace (r, 4(x + y ) ) by a test function $(x) that vanishes outside the region on which all the distributions sl are concentrated. Then (SI
* t , 9 ) = ( S f 0 t, 4 ( x + Y)> = (s11 = (SI,
*) ---* (s, *) = (s
( t . 4)) * t?4 ) ,
(13)
and we have sf*t-+s*t,
as l + m ,
as required. In the second case, $(x) = ( t , 4(x + y ) ) is a test function, and we follow the steps leading to (13) to derive our formula. Finally, in case 3 we suppose that the support of the distributions s and t are bounded on the left. Then the support of the function $(x) = ( t , #(x + y ) ) is bounded on the right. The rest of the proof proceeds as in the other two cases.
As a corollary to this theorem, we have the continuity of the distribution s, depending on a parameter tl under each of the following conditions: (1) All the s are concentrated on the same bounded set. (2) The distribution t is concentrated on a bounded set. (3) The supports of the distributions s and t are bounded on the same side by a constant independent of u.
7.5.
ROLE OF CONVOLUTION IN REGULARIZATION OF DISTRIBUTIONS
183
As a special case of this corollary we find that if as&?a exists, then (a/aa)(s,
* t ) = as,/au * t ,
(14)
because the derivative asu/du is the limit of as u + uo. Convolution of Tempered Distribution
The foregoing analysis can be extended to distributions of slow growth. We have the same definition (4) for the convolution of two distributions. The restrictions are also similar. To establish various properties we thus appeal to the direct product of tempered distributions. In applications, it is the convolution t * 4 of a tempered distribution r and a test function 4 E S that plays an important part. 7.5. THE ROLE OF CONVOLUTION IN THE REGULARIZATION OF THE DISTRIBUTIONS
In Chapter 1 we defined the Dirac delta function. In Chapter 2 we defined the class of test functions, which helped us define not only this function but many more generalized functions. Now we study the convolution of a distribution with some test function in D.This operation converts the distribution into a function that is infinitely smooth. Since $ E D has compact support, this convolution exists. Moreover, this convolution satisfies an interesting equality. Specifically, we have the following theorem :
Theorem. s*
*
=
M Y ) , *(x - Y ) ) E C"(R,),
*
E
D.
(1)
Prooj; The infinite differentiability of the right side of (1) is established in the same way as in Lemma 1 of Section 7.1. To prove the equality we appeal to (7.4.5) and find that, for 4 E D, (s
*
$9
4)
=
M Y ) 0 *(z), 4(Y + 2 ) )
184
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
Note that the function +(x)+(x - y) belongs to D(R,,).We can, therefore, use (7.1.21) to obtain from (2) the result (s
* 4) +?
w - Y)> dx
=
Ic(x)(s(sx
=
((S(Y), +(x - Y)>, 4),
as desired. Let E D be the function defined in Section 2.2,
where c, is such that 4&(x)dx = 1. When we use this function in the definition (l), we have the regularization SAX) = s * 4,
=
(S(Y), 4&(X- Y)).
(3)
Exercise 12 of this chapter illustrates this process for the pseudofunction Pf(Wx)/x). Recall that in Example 2 of Section 3.3 we proved that +&(x)-+ 6(x) as E 0. Combining it with the continuity of the convolution s * 4&with respect to 4&,we find that -+
s,(x)
-+
s(x)
as
E
-+
0.
This means that each distribution is a weak limit of its own regularization. These considerations lead to the following very important result :
Theorem. Each distribution s is a weak limit of a test function. That is, the space D is dense in D’. This theorem is really proved by the remarks preceding its statement if all the s, are of bounded support. Otherwise, we introduce a sequence [,(x) of ~ vanish for Ix I 2 2 / ~Then . cutoff factors that are identically 1 for I x I 5 1 / and lim(ia(x>s,(x), 4(x)> = lim(s,,
&+O
E-0
id) = W s , , 4)
for all 4 E D. That is, the sequence [,(x)s,(x) theorem is proved.
&+
=
0
-+ s(x)
as E
-+
(s,
4),
0 in D’, and the
Since D c S c S’ c D’, it follows that S and S’, as well as D, are dense in
D’.
185
7.6. EXAMPLES
The Space E From the analysis of these sections the reader must have observed the important part the distributions of bounded support play in this theory. Accordingly, the subspace of D' all of whose members have bounded support is given a separate symbol. We shall denote it E'. It is the linear space of all distributions having bounded support. A sequence of distributions {t,} is said to converge in E' to a limit t if it converges in D' to t and if all the t , have their supports contained in one fixed bounded region R . Then clearly the limit distribution t is also in E' and has its support contained in R . We shall have more to say about this space in Section 15.1.
7.6. EXAMPLES
Example 1 .
(a) We can express the integral
as a convolution. Setting y = x - z, we find that
J-
m
. f ( x ) = J:mq5(y) d y
= -
JOm4(x- z ) dz
=
H(z)c$(x - z ) dz
=
H
* 4.
OD
(b) Let t be an arbitrary distribution. The convolution of 6 ( x ) and t(x) is
(6 * 6 4 )
=
( t ( x ) , ( 6 ( z ) ,4 ( x
or
+ 4))
=
( W , 4(x)>,
4 €4
6 * t = t.
Thus the delta function is an identity element in D' for the operation of convolution. (c) 6,
*t
=
sh, t,
(24
where 6, = 6(x - a ) and the symbol sh, shifts t by a translation a. The translated distribution is also denoted fx-, or t(x - a ) ; that is, r(x - a ) = t ( x ) * 6(x - a).
(2b)
186
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
- _.
. ._. . .
. -
.,
which is (2a). In particular
6,
* 6b = bo+b.
(d) Next, we combine relations (7.4.8) and (1) to find that aslax, * t
=
6 * atlax, = atlax,.
(4)
This result can be extended to a differential operator L of order p. Indeed, from (7.4. I I ) and (2) we have (L6)* t
=
6 * L(t) = L(t).
(5)
Thus, every linear differential operator with constant coefficients can be represented as a convolution. An interesting particular case of (5) is fym)
*t
=
p).
(6)
Example 2. From relation (6) we derive that 6’ * 1 = 0. Thus (1 * 6’) * H = 0. On the other hand 6’ * H = 6, and therefore 1 * (6’ * H ) = 1. This means that (1 * 6 ‘ ) * H # 1 * ( 6 ’ * H ) ,
so the convolution is not necessarily associative. Example 3. Relation (7.4.8) states that (Dks)* t = s * (Dkt).The mere existence of the convolutions Dks* t and s * Dkt,is not sufficient for the existence of the convolution s * t ; specifically, these convolutions may not be equal. For instance, H’* 1 = 6 * I = 1, while H * 1’ = H * 0 = 0. The trouble is that H * 1 does not exist, because neither the support of H ( x ) nor that of 1 is bounded. Example 4. One of the most important uses of convolution theory is to obtain the particular solution of a differential equation Lu
=.L
(7)
where Lu = L(D)u, is a linear differential operator. This is achieved by appealing to the fundamental solution E, given by LE
=
6.
Indeed, our contention is that u = f * E.
(9)
7.6.
187
EXAMPLES
This follows by applying the operator L to both sides of (9). The result is Lu = L ( f * E ) = f * L E = , f * G =,f, as desired. Let us illustrate this concept by considering Poisson’s equation in R , , V*U(X) = -p(x).
(10)
We find from (4.4.73) that the fundamental solution of the operator -V2 in R 3 is (1/4xlxl). Accordingly, the particular solution of (10) is
This is called the Newtonian or volume potential for mass density p and is written V3. Its generalization in R, is
1
1
1
V2(x) = p * - In - = 2x 1x1 271
1
lx - Yl
dY?
where S,( 1) is the surface area of the unit sphere. Example 5. Single-layer distribution We have already come across the concept of the single-layer density in previous chapters. With the help of the definition of convolution, we can study this concept in more detail. Let o(x) be a locally integrable function defined over a bounded piecewise smooth two-sided surface S. Then o(x)G(S) is the single layer over S with surface density o. The potential generated by this distribution is
vi0)= o(x)G(S) * 2x and is called the surface potential of the single layer with density o. It is expressed
where for n = 2, S is a curve.
188
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
Let us prove the case n 2 3. For this purpose, we use definition (7.4.5) and find that for all 4 E D
from which (14a) follows. We can obtain a more general result by repeating the steps in ( 1 5). Indeed, let,f(x) be a locally integrable function in R , and let oS(S) be a single layer spread over a bounded smooth surface S with density rs. Then, for 4 E D
or
Example 6. Double-layer distribution Let T ( X ) be a continuous function over S (as defined in Example 5 for a single layer density) and let -(d/dn)[z(x)b(S)] be the double layer over S with density T . Here n stands
189
7.6. EXAMPLES
for the unit normal to S. The potential generated by this distribution of singularities is
vb”
= -
d
-
dn
vil)= - a
-
dn
1 1 ( n - 2)~,(1)I x I ” - ~ ’
[r(x)G(S)]
*
[r(x)G(S)]
1 1 *In -,
2n
n 2 3,
(17a)
1x1
and is called the surface potential of the double layer. The function V;” is a locally integrable function in R” and can Le expressed by the formulas
As in the previous example, we shall prove the case n 2 3. Indeed, for 4 ED, we have
from which the required formula follows. Example 7. We have come across many generalized functions that vanish from some point to infinity such as x< = x a H ( x ) . Formulas for their
190
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
convolution take a simple form. For instance, the convolution off and g that vanish on the negative axis is
f * 9 = /-mx
f(y)g(x - Y ) dY
=
JOrnf(Y)dX - Y ) dY
x < 0,
because g(x - y) = 0 for y < x. Let us denote the distribution
x,- l/r(A), which we studied in Example 5 of Section 4.4, as CD, and show that it satisfies the relation
* CDp
@,
(20)
=
For Re A > 0, Re p > 0 we have
and =
xA+p-
+
1
/w +
Thus (20) will hold for Re A > 0, Re p > 0 if we can prove that
but this follows by setting t the gamma functions,
p ( ~p) , =J
= XT and
using the identity involving the beta and
1
0
T ~ '(1-
- Ty-1
dT
=
r(n)r(p)/r(n + p).
By the principle of analytic continuation, (20) holds for all other complex values of A and p. Example 8. Consider an integrable function f(x) such that f ( x ) = 0 for x < 0. Its convolution with @,(x) (n = 1, 2, . . .), mentioned in Example 7,
yields
191
7.6. EXAMPLES
It is known [16] that
r x
1
Combining (22) and (23), we have By virtue of (4.4.50),we have &"(X)
=
6'"'(x).
Thus I-.,f=
0 - " ( x ) * f ( x ) = 6'"'(x)* f ( x ) = (d"/dx").f(x).
(25)
This means that the convolution off with 0 , ( x ) gives an n-fold integration, that with O-,,(x) an n-fold differentiation. The foregoing relations are readily generalized to the case that 1 is an arbitrary real or complex number and, instead off; we have a distribution t that is concentrated in x 2 0. Relations (24) and (25) then become i,t
=
(x:- l/r(I)) *t
(26)
and I - L t = dat/dxa.
(27)
O,+,, * t = (0, * O,,) * t = 0,* (O,, * t),
(28)
Because we find by putting 1 = - p that differentiation and integration of the same order are inverse operations with respect to one another. Equation (28) further implies that
is true for all 1 and p. When 1 is not an integer, the quantities I , and I - are called the fractional integral and fractional derivative, respectively, of order 1.
,
Example 9. With the help of the previous two examples we can prove the following interesting results in the theory of ordinary differential equations and integral equations.
192
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
(a) The only solution of the ordinary differential equation das/dxa= t(x),
s, t
E
D',
(30)
is s(x) = @a
* t.
This follows on writing (30) as (Wa* s) of both sides by @a.
(31)
=
t and then taking the convolution
(b) By virtue of the notation of Example 8, we can write the Abel integral equation
as t(x) = @-,+
1
* s,
(33)
which can be inverted to give s(x) = 0-1 * t .
(34)
Two remarks are in order here. The condition a < 1, which guarantees the convergence of (32), has been omitted because of the previous discussion. Second, to guarantee the existence of the convolution (33), we assume that s(x) and t(x) vanish for x < 0. (c) Let us follow Schwartz [l] and define .@,Xx) = eaXOa(x).
(35)
For instance, a@-l(x) = eaX@-,(x) = eax6'(x) = 6'(x) - aS(x).
(36)
Because eaxs(x)* eaxt(x) = eax{s(x)* t(x)}, it follows that a @ h )
* a@;)
=
a@'n
+ &).
(37)
Using these results in case (a), we find that the only solution of the ordinary differential equation (d/dx - a)'ss(x) = t(x)
or a @ - a
* s(x)
= t(x)
(38)
is s(x) = a@a
* t(x).
(39)
193
7.6. EXAMPLES
Example 10. Convolutions arise naturally in many physical problems. We demonstrate this by considering the initial value problem azulat2 - a z U / a x 2 u(x, 0)
=
W),
=
0,
(40) (41)
(aulat)(x, 0) = $(x>.
This is the famous problem of a string given initial displacement 4 ( x ) and initial velocity $(x). The general solution of (40) is clearly u(x, t )
=.m+ t ) + g(x
-
(42)
t),
as can be verified by direct substitution. Then from (41) we get 4x7 0) = . f ( x )
+ g(x) =
(43)
and
au(x, oyat = fyx)
-
gyx)
(44)
= +(XI.
Integration of (44) with respect to x yields
where the constant of integration has been incorporated into the lower limit. Relations (43) and (45) enable us to solve f o r , f ( x )and g(x), so that f ( x ) = +@(x)
+f
Jux$(s) ds,
g ( x ) = f&>
-
f r $ ( s ) ds. U
When these expressions are substituted into (42), we obtain 4 x 9 t ) = +C&
+ t ) + 4 0 - t>l +
+
x+r
Jxpt
4%)
ds,
(46)
which is the well-known d’Alembert formula. Let us interpret this formula in the light of the present theory. For this purpose, let us first take the special case + ( x ) = 0. Then (46) becomes
We can express it as a distribution with the help of the generalized function E(x, t ) defined as
= +H(f
+ x ) H ( t - x ) = +H(C - Ixl),
(48)
194
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
so that (47) takes the form
1m
u(x, t ) =
m
E(x
-
S,
t)*(s) ds
=
E
* $.
(49)
In Chapter 10 we shall show that E(x, t ) is the fundamental solution of the wave equation (40). As a second special case, let us set $(x) = 0. Then (46) reduces to
+ t ) + 4(x - t ) ]
u(x, t ) = +[&x -1 -2
J-m m
[d(t
Now from (48) it follows that
dE(x
+ x - S) + s(t - x + s ) ] ~ ( s ) ds.
+ x - s)H(t x + s) + s(t - x + s)H(r + x - s)],
- s, t ) / d t = +[s(t
which for t > 0 becomes
aE(x
(50)
-
+ x - s) + s(t - x + s)].
- s, tyat = &s(t
(51)
Combining (50) and (51) we obtain aE(x - s, t )
u(x, t ) =
4(s)
ds
aE
=at
* 4,
t > 0.
(52)
The general formula now follows by adding (49) and (52):
u(x, t ) = E
*
*+
(aE/at) * 4.
7.7. THE FOURIER TRANSFORM OF THE CONVOLUTION
Let ,fand g be two locally integrable functions. Then the Fourier transform of the convolution,f * g is [ f * g] ^(u)
J-m,f* geiuxd x m
=
m
=
J-,eiUx
= m
m
dx J-,.f(x
g(y)ei"Y d y
-
1m
y)g(y)d y
mf(x - y)eiu(x-Y) d(x
- y)
7.7. THE FOURIER TRANSFORM OF THE CONVOLUTION
195
Thus, the Fourier transform of the convolution offand g is the product of their Fourier transforms. A similar result holds for inverse Fourier transforms. Relation (1) also holds for two distributions s and t under certain restrictions. For instance, if at least one of these distributions has compact support, then relation (1) holds. For further study of this subject the reader should see reference [17]. We illustrate this concept with a few elementary examples. Example 1. In view of (7.6.6) and (6.4.7) we have [~(~'(x)](u) = [cYk)* t] ^ ( u ) = ( - iu)kt(u), A
which agrees with (6.3.21a). Example 2. Using relation (7.6.2b) we obtain [t(x - y)] u
=
A
[t(x) * 6(x - y)]
A
=
Z(u)eiU",
which agrees with (6.3.25a). Example 3. Consider the relation 6 ' ( ~- a) * 6 " ( ~- b) =
J-
m m
6 ' ( ~- u - y)S"(y - b) d y = 6 " ' ( ~- u - b),
which follows as a special case from Exercise 10.Taking the Fourier transform of both sides and using (6.4.7) and (7.6.2b), we find that ( - i U p u a ( - iu)Zeiub = ( - iU)3eiu(a + b),
which is an identity. Example4. We end this chapter with a physical example in which the Fourier transform, the direct product, the convolution, and the Fourier transform of convolution are all used. In optics F(u, u) is the Fourier transform of a functionf(x, y); it gives the far-field behavior due to a source of light whose strength (or intensity) isf(x, y) per unit area in the x, y plane. Let us consider the special case when an infinitely narrow slit parallel to the y axis is a light source with unit intensity, so that/(x, y) = 6(x) 0 l(y). Accordingly, the far-field behavior is F(u, u) = l(u) 0 2118(v), which is a light line perpendicular to the slit. Next consider two slits with intensity ,f,(x, y) and fz(x, y). Then the farfield behavior of the slit with intensityf,(x, y) *fi(x, y) is F,(u, o)F,(u, u). By extending these concepts one can study the interference phenomena due to a system of slits [l5].
196
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
EXERCISES 1. Show that
DW;Cs(x) 0 t(Y)l = CD5;s(x)l0 CDftf(Y)l. 2. Show that the direct product is linear in the following sense: If c1 and B are arbitrary numbers and if sl, s 2 , t l , and f 2 are arbitrary distributions defined over R,, then s1
0 (at1
(as1
+
+
Ps2)
D 2 )
0 tl
+ B(s1 0 t 2 ) , 0 t l ) + B(s2 0 tl). 0 tl)
=4Sl = @I
3. Prove that D:(s(x) 0 l(y)) = 0, Ikl # 0.
4. Prove the equations (a)
fa
*.fp
=f a + p ,
.fa(x) =
jU(x) 5. If s(x, y) E &+,,
=
H(x)xa-
l
c1
> 0;
+ x2’
c1
> 0.
a
u2
~
eax,
~
show that D:D~Fx[s]
=
F,[(~X>”D~,S],
and F,[Df:D;s]
=
( - iz~)~D:F,[s],
where F , denotes Fourier transform in R,, the x space.
6. Establish the following distributional convolution formulas : (a) [xH(x) * exH(x)] = (ex - x - I)H(x), (b) (H(x) sin x) * (H(x) cos x) = tH(x)x sin x, (c) 6’(x) * Pf( l/x) = Pf(H( - x)/x2) - Pf(H(x)/x2).
7. Evaluate (a) e - Ixl * e - IxI, (b) e - u x z* (c) xe-axz* x e - ~ ~ ’ .
8. By replacing A by -A in Example 7 and proceeding as in Example 8 of Section 7.6, show that
197
EXERCISES
Since 1 + = H ( x ) , on setting u ,
= 1
this formula becomes
dAHH/dxa= x-'H//T(l - A). Also deduce that
9. (a) From relation (7.6.21) deduce the following:
+ + 1 is a negative
Discuss its validity. (For instance, when A p integer, the right-hand side is to be replaced.) (b) For A = 0, (a) reduces to
From this relation derive
JoX6(-'(y)(x - y)"H(x - y ) d y = (c) Similarly, for A = p Show that
Jom6(m)(y)d y 10. Show that
JJ-
m m
=
=
p!X"-m ( p - m)!.
0, we have JTm H ( y ) H ( x - y ) d y = x H ( x ) .
Jox""-"(y)6"- l ) ( x - y ) dy
6'"'(y)6'"'(x - y ) d y
=
=
6'"-
S("+")(X),
W
J-
6(a - y ) 6 ( x - a) d y = 6 ( x - y),
m
m
6'"'(a - y)6'"'(x - a ) d y = S("+")(X - y).
m
Note that first of these relations implies that 8")* 6'")
1 1 . Prove that, in the notation of Section 7.5,
=
6("+").
(4.
1)
198
7.
DIRECT PRODUCTS AND CONVOLUTIONS OF DISTRIBUTIONS
12. With the help of the analysis of Section 7.5, regularize the psuedofunctionf(x) = Pf(H(x)/x) = Pf(l/x+). Hint. Split R , into three intervals: (i) In the interval E < O,f(x) = 0. (ii) In - E < x < E, we can appeal to the analysis of Chapter 4 and take the Hadamard finite part of (.L 4). (iii) In x 2 E, the function is regular. Sketch the regularization for E = and compare with the graph of.f(x). 13. Show that (a) H(x) * Pf(H(x)/x)
(b) Pfr?)
=
* Pf(x)H(x)
H(x) In x, =
dX2 d 2 Jt In 5 ln(x - 5 ) d5.
14. Show that
x - 5)
n.
c,"=
15. Letf(x) be a continuous function in R , that is periodic with period 27~. Show thatf(u)= - m b,6,(u - n), and relate b, to the coefficients of the Fourier series of,f(x).
16. Show that
Pf( l/x) * Pf( l/x)
= - 7T26(x).
The Laplace Transform
8.1. A BRIEF DISCUSSION OF THE CLASSICAL RESULTS
The main applications of the Laplace transform are directed toward problems in which the time t is the independent variable. We shall therefore use this variable in this chapter. Let f ( t ) be a complex-valued function of the real variable f such that ,f(r)e-c'is absolutely integrable over 0 < t < co,where c is a real number. Then the Laplace transform of j ' ( r ) , r 2 0, is defined as 03
Z(s) = 6 P { f ( t ) } =
0
f(t)e-"' dr,
Re s
=- c,
(1)
where s = 0 + io.The Laplace transform as defined by (1) has the following basic properties. (1) Linearity Let the Laplace transforms of the functionsf(t) and g(t) beJ(s) and g(s), respectively, and let a and j3 be any constants. Then the Laplace transform of the function h ( t ) defined by h(t) = a f ( t ) pg(t) is R(s) = aJ(s) pi($. (2) The uniqueness theorem IfJ(s) = g(s) on some vertical line in their region of convergence, then j ( t ) = g(t). (3) The transform of the nth derivative If the functionf(t) is n times continuously differentiable, then
+
+
9 { , f ( " ) ( t ) }= [ f ( " ' ( t ) ] " = s"J(s) - s"-'f(O) 199
- s"-*f'(o)
- .*.-f("-')(O).
(2)
200
8.
THE LAPLACE TRANSFORM
(4) The convolution theorem Let f(s) and (j(s) be the Laplace transforms of the functionsf(?) and g(t). Then the Laplace transform of the convolution h(r) = .f(r)g(t - r ) dr, t 2 0, is
So
m mm.
(3)
=
(5) The inverse transform transform is
The formula for the inverse of the Laplace
*
.fW = 9l{J(s)l = 27ri
a + im a-im
f(s)es' ds,
(4)
where o is Re s. The usual method of evaluating this complex integral is by analytically continuingf(s) into the complex plane for Re s < c, converting the line integral into a contour integral, and then applying the residue theorem. (6) 9-{!(as
+ b ) } = (l/a) exp( -bt/a).f(t/a).
For the special case when a
=
1, b
2-' { J ( s
= -a,
(5)
( 5 ) becomes
- a)} = e"'f(t).
(6)
(7) If 9 - ' { f ( s ) } = f ( t ) when c 2 0, and if,f(t) is assigned arbitrary values for -c 5 t < 0, then g - l { e - c s J ( s ) }= f ( t - c ) ~ ( t c).
(7)
where H(r - c) is the Heaviside function. All these results are discussed in many other texts, to which the reader is referred.
8.2. THE LAPLACE TRANSFORM OF DISTRIBUTIONS
Recall that when we attempted to define the Fourier transform of a distribution by the classical formula, in Chapter 6, we got into difficulty, which we resolved by defining a new class of test functions. The same difficulty arises in the present situation. Indeed, if we formally extend the definition (8.1.1) to a distributionf(t) whose support is bounded on the left at 0, we have
8.2. THE LAPLACE TRANSFORM OF DISTRIBUTIONS
201
We examine this relation assuming that there exists a real number c such that e - " f ( t ) is a distribution belonging to S' (the class of tempered distributions). Then we can rewrite (1) f ( s ) = ( e - c ' f ( t ) , H(t)e-(s-c)'),
(2)
where H ( t ) is the Heaviside function. For Re s > c, the function H(t)e-(S-c)' is a test function in S, and the definition (2) makes sense. However, this is not the case, as is clear from the distributionf(t) = H ( t ) e " ; f ( r ) is a member of D',but there is no value of c for which H(t)e'z-c' E S. Accordingly, we define a new class of test functions and its dual [ 181.
Definition 1. The space K of testfunctions ofexponential decay is the space of the complex-valued functions +(t) satisfying the following properties: 1. +(t) is infinitely differentiable; i.e., +(t) E C"(R,). 2. + ( t ) and its derivatives of all orders vanish at infinity faster than the reciprocal of the exponential of order c ; that is, I ecrDk+(t)I c M , Vc, k.
Definition 2. A function f ( t ) is of exponential growth if and only if f ( t ) together with all its derivatives grows at infinity more slowly than the exponential function of order c ; i.e., there exist real constants c and M such that IDkf(t)I I Me". Definition 3. A linear continuous functional over the space K of test functions is called a distribution of exponential growth. This dual space of K is denoted K'. In view of these definitions we find that all the distributions belonging to K' have the Laplace transform based on the definition f(s) =
Jmf(t)e-"
dt
=
( f ( t ) , e-"),
0
(3)
for real s E (c, a).This follows from the relation
(4) the right side of which is finite for Re s > c, and from e-" E K . Since this definition agrees with (1) of the classical transform, most of the known formulas remain the same. For example, all the properties of Section 8.1 hold for a distribution. However, Property 3 will be examined carefully in the next section.
202
8.
THE LAPLACE TRANSFORM
Example 1 . The Laplace transform of the Heaviside function is [H(t)]- =
rw
J
0
1
c - ~ d' f = A. S
Example 2. The delta function and its derivatives (a) [6(t - a)]" (b)
[a'([
=
s::
- a)]- =
e "8(t - a) dt
J
=
e-sa;
-
e-"'6'(t - a ) dt = - {(d/dr)(e-s')}t=o= se-Sa. ( 7 )
0
Continuing this process, we obtain (c)
[6'"'(t
- a)]" = ( -
By,formally setting a (d) [6(t)]" (e)
=
=
l)"sne-Sa.
0 in (6) and (8), we obtain
1.
[6(")(r)] = ( -
1)"s".
We have emphasized the word "formally" because the integral JF e-"'s(t) dt is not defined. Indeed, J? e-s'6(t) dt = fZrn e-"H(t)G(t) dt. This means that 6(t) = H(O), which is usually taken to be $. However, definition (9) is consistent with the fact that 6 ( t ) is the derivative of the Heaviside function, as explained in the next section.
8.3. THE LAPLACE TRANSFORM OF THE DISTRIBUTIONAL DERIVATIVES AND VICE VERSA
By definition (8.2.3), we have
P { f ' ( r )=) <j'(f), e-").
(1)
In order to reconcile this result for distributions with Property 3 of Section 8. I , we should first evaluatef'(r) and then apply the right side of (1). For this purpose, let us discuss the function
8.3.
LAPLACE TRANSFORM OF DISTRIBUTIONAL DERIVATIVES
203
where a > 0 and gl(t) and gZ(t) are continuously differentiable functions. The classical derivative f ' ( t ) of (2) is .f'(t) =
g>(t)H(a - t )
+ gXt)H(t - a),
(3)
for all t # a, while the distributional derivative is
f'@) = f ' ( t ) + CfIW - a),
(4)
where we have used the notation, as in Chapter 5, [,f] = ( f ( u + ) - .f(a-)). The Laplace transform of (3) is Jorne-qf)(f) dt = JOag>(r)e-"'d t =
=
[e-"'gl(t)];
+ rgi(t)e-"
df
+ s J gl(t)e-"' d t + [e-S'g2(t) dt]: ra
0
+s
dt
s{ ~;gl(r)e-"' d t
+ r g 2 ( r ) e - " d t } - e-asC.fI - Sl(0)
= s m - .f(O) - C.fle-"", where we have written ,f(O) for gl(0). On the other hand,
y { f ' ( t ) }= ( { ' ( t ) , e-"') = ( f ' = Jomff(t)e-"'
dt
+
(5)
+ [f]d(t
[f]e-Os
- a), e-"')
= sT(s) -
,f(O),
(6)
where we have used (5). Thus Property 3 of Section 8.1 holds. Relation (5) makes sense even when we allow u to tend to zero, because in that case we have
J
s{f'(t)} =
g;(t)e-s' d t
0
=
s J g2(t)e-"' dt - gz(0) 0
consistent with (5). To examine the situation in a different way, we write (2) with a .f(t) = g , ( t ) H ( -t ) gz(t)H(t). Then for r > 0,
+
=
0:
204
8.
THE LAPLACE TRANSFORM
For (6) and (8) to agree we should have &t) = 1, as stipulated in (8.2.9). Finally, we check consistency by using (8) for , f ( t ) = H ( t ) : 9{17'(t)} =
s(l/s) - H ( O + )
+ [H(O+)
- H(0-)]
= 1,
where we have used (8.2.5). For functionsf(t) that have support only for positive values o f t , so that f ( 0 - ) = 0, (8) becomes
Y{f'(t)>=
m.
(9)
Continuing in this fashion, we find that LZ{f(n)(r)}
(10)
= 9f(S).
Let us now prove, by induction, the result
( d k / d s k ) [ f ( s )= l
(f(t),
(- t)ke-sr)7
(1 1)
which gives the formula for the derivative of the Laplace transform. It is true for k = 0, by the definition of the transform. Now let us assume (1 1) to be true for k replaced by k - 1 ; i.e.,
( d k - '/dsk-' ) { f ( s ) }
= ( f ( t ) , (-
t ) k - 'e-").
Then
as desired.
8.4. EXAMPLES
Example 1. Since H ( t ) In r is a regular distribution, we have
Y { H ( t )In t } = On our setting st
= 11, S
Som
In t e-"' dt.
this integral becomes
Jome-n(ln q - In s) dq
1
= -S (y
+ In s),
8.4.
205
EXAMPLES
where y
=
-SF e-" In q dq
=
0.5772.. . is Euler's constant. Thus
9 { H ( t ) In t } = [ H ( t ) In t]"
=
+ In s).
-(l/s)(y
(1)
Example 2. We can use the result of the previous example to evaluate the Laplace transform of the singular distribution PfCH(t)/t]. Because PfCH(t)/tl
=
(d/dt)CH(t)In
tl,
we obtain, from (1) and (8.3.9),
+
9{PfCH(t)/t]} = 9 { ( d / d t ) [ H ( t )In t]} = s [ ~ ( l / s ) ( y In s)]
-(y
+ Ins). (2)
Similarly, because [see (4.2.3)]
P f ( y
=
= -
;
pf(?)] - d'(t),
we find that y{PfCH(t)/t2]} = s(ln s
+ y) - s = s(ln s + y - 1).
(3)
Example 3. Let us find the Laplace transform of the function t i = H ( f ) r L , where A # - 1, -2, -3, . . . . Since t: E K', we have 9 { t : } = f? e-s'tAdt. Letting u = st for s > 0, it follows that
In particular, for 1 = 0 we recover formula (8.2.5). Example 4. The Laplace transform of a periodic function Let f ( t ) be a function that vanishes identically outside the finite interval (0, T). The periodic extension off(t) of period T is the function obtained by summing the translatesf(t - kT), for k = 0, k 1, +2, . . . (as shown in Fig. 8.1): m
.fT(t) =
f(t+T)
1
k=-m
f(t -
f(t)
w.
f(t-T)
Fig. 8.1. The periodic exension of,f(t).
(5)
206
8. THE LAPLACE TRANSFORM
Then we can show that PT
1
The proof follows on writing ( 5 ) as the convolution k=-m
k=-w
Now 6p {k=:
1 (t - k m
}
~ = ) LZ
Cs(t - k
{kyO
}
~ )
This summation will be valid if le-st1
=
le-(a+io)TI = e-oT
< 1,
which is true for all 0 > 0. Now we apply the convolution theorem (8.1.3) to (7), so that
and (6) follows. Example 5. In many physical applications we encounter the convolution equation
* x ( t ) = s(t).
(10) Applying the Laplace transform to this, we obtainf(s)T(s) = g(s), so n(s) = (f)-'g. By inversion we obtain x ( t ) = ( f * ) - ' * g, where Y { ( f * )= } (f(s))-', and we assume that (f(s))-' is the Laplace transform of a rightsided distribution. For example, when f(t)
f
=
2
akDks(t),
k=O
so that (10) is an ordinary differential equation, we have 2{(f*) '} -= l /
akSk. k=O
The method of partial fractions now helps in solving the problem.
8.4.
207
EXAMPLES
Example 6 . Let us find
9l{++
lI2
,
The factors of { 1/(1 + s2)}'+ 11' are { l/(s relation
+ i)}'+
112
= I/(s -
Lf{H(t)e"t'-'/r(v)}
v >
s>o,
-4.
and { l/(s - i)}"
A)v,
v > 0,
which is easily proved, shows that we can set f ( t ) = {H(t)ei'tv- lI2/(v
+ t)},
g(t) = {H(t)e-"t"- 'I2/r(v
and
so that
f(s) = (L) s-1 , v+
112
(13)
+ i)},
(A.) . v+
as) =
'". The
112
Next, we use the convolution theorem to obtain hv(t) = f ( t ) * g(t); and find that the required value is
H ( r ) C f i / W + 4)l(t/2)'Jv(t). Example 7. Let us solve the initial boundary value problem =
o < < a, t > 0,
a2u/atz - aZo/ax2
=
0,
u(x, 0)
=
0,
(14b)
( a u / a t X x ,0) = 0,
( 14c)
(1 4 4
(144
o(0, t ) = h(t).
We multiply both sides by e-s'and integrate from t = 0 t o t = co.Integrating the first term by parts twice and using the initial conditions and the fact that e-S' is exponentially small at t = 00 for Re s > 0, we obtain the boundary value problem -d2ij/dX2
+ s2ii = 0,
ij(0,s)
=
0<x <
h"(s),
00,
(15a) (1 5b)
where ii(x, s) and h(s) are the Laplace transforms of u(x, t ) and h(t), respectively. To make this two-point boundary value problem complete, we add the requirement that lim C(x, s ) = 0,
x- m
Re s > 0.
( 15c)
208
8. THE LAPLACE TRANSFORM
The solution of the system (15 ) is ~ ( xs),
Because
~--I{h(s= ) ) h(t),
(16)
R(s)e-‘,,
=
L F 1 { e - s x } = d(t - x ) ,
we find by inverting (16) and using the convolution theorem that t > x, h(t)b(t - x - T ) dz = {;!t - x ) , t < x, = h(t - x ) H ( t - x ) .
Example 8. Let us attempt to solve the initial-boundary-value problem
azU/ax2- a2u/at2 = 0, u(x, 0) u(0, t )
x > 0, t > 0,
(1 7 4
0,
(1 7b)
= ecr,
(17c)
=
u,(O, t ) = 0, (1 7 4 with the help of generalized functions and Laplace transform. We set (18)
v(x, t ) = H(x)H(t)u(x,t ) , &/ax
=
G(x)H(t)u(x,t )
+ H(x)H(t)u,(x, t )
= G(x)H(t)u(O,0 + H(x)H(Ou,(x, t), a2U/ax2 = G’(x)H(t)u(O,t ) G(x)H(t)u,(x, t ) H(x)H(t)u,,(x, t ) = G’(x)H(t)u(O,t ) G(x)H(t)u,(O, t ) H(x)H(t)u,,(x, t ) = G’(x)H(t)e“ H(x)H(t)u,,(x, t),
+
+ +
+ +
where we have used conditions (1 7c) and (17d). Similarly,
= H(x)G(t)ur(x,0)
+ H(x)H(t)urr(x, t).
Consequently, the system (17) is equivalent to the inhomogeneous equation
82v/ax2- a2U/at2 = ec’H(t)G’(x)- u ~ ( xo,) H ( x ) G ( ~ ) ,
(19)
which we solve by taking the Laplace transform of both sides with respect to x . Then
( d 2 / d t 2 - s2)l/(s,t )
=
-secrH(t)
+ A(s)G(t),
(20)
where V(s,t ) and A(s) are the Laplace transforms of u(x, t ) and u,(x, O), respectively.
8.4.
209
EXAMPLES
The solution of (20) is derived by setting (21)
V ( S ,t ) = G(s, t ) H ( t ) ,
so that d V ( S ,t ) = G(t)G(s,0) dt
-
t) + H ( t ) dG(s, ___ dt '
a2 V(s, t ) = G'(t)G(s, 0) + G ( t ) dG(s,O) + H ( t ) d2G(s,t )
-
dt2
~
dt
~
dt2
'
Then the inhomogeneous equation (20) is equivalent to the initial value problem d2G(s,t)/dt2 - s2G(s, t ) = -sect,
(22a)
G(s, 0) = 0,
(22b)
dG(s, O)/dt = A(s).
(22c)
The solution of this system is easily found to be
so from (21) we obtain V(s, t )
=
{-
Because the value of A(s) is unspecified, we can eliminate the undesirable term containing eS'.We set A(s) = s/(s - c), and (23) then reduces to ~ ( s t, ) = [s/(s2
- c2)](ecr- e-"')H(t).
(24)
Now we use the relations
2-[s/(s2
- c2)] = cosh cx
H ( x ) and 2'(e-"F(s)) = f ( x
- t)H(x - t);
with these, inversion of (24) yields e""-")H(t)H(x), e"'-"')H(t)H(x), Finally, appealing to (18), we obtain
t < x, t > x.
210
8. THE LAPLACE TRANSFORM
EXERCISES
1. Recall that CT = Re s and prove that (a) 9 { H ( t ) e * ' " ' } = l/(s f iw), CT > 0 ; (b) 2 { H ( t )cos wt} = s/(s2 w2), Q > 0 ; (c) Y { H ( t )sin wt} = w/(s2 + w2), CT > 0; (d) 9 { H ( t ) J o ( t ) }= (s' 1)-1'2, Q > 0.
+
+
2. Establish the following relations: (a) z{.f(t - r)H(t - z) = e-Srf(s), (b) 9 { e - " f f ( t ) }= f(s a), (c) 9 { f ( @ l = ( l / a ) f ( s / a ) . Find the appropriate value of Re s for which each holds.
+
3. Show that (a) 2 { d ( t p)} = (l/lcrl)e"'"; (b) 9{PfCH(t)sin t / t 2 ] } = (i/2){(s ia) ln(s ia) - (s - icr) ln(s - ia)} + cr(1 - y), where y is Euler's constant; (c) 2{Pf[H(t) cosh crt/t]} = ln(s2 - a') - y.
+
+
+
4. The square wave function f ( t ) is defined by -1,
O
and
f(t
+ 2T) = f ( T ) .
Show that its Laplace transform is f ( s ) = (l/s) tanh $sT.
Applications t o Ordinary Differential Equations
9.1. O R D I N A R Y DIFFERENTIAL O P E R A T O R S
In Section 2.6 we defined the differential operator L,
and its formal adjoint L*,
c n
L*$
=
m=O
( - 1)"
dm(am(X)~)/dXm,
(2)
where the coefficients a,(x) are infinitely differentiable functions, t is a distribution, and 4 is a test function. These operators are related by the equation
(Lt,
4) = ( I , L*4).
This means that the action of Lt on test function $,I = L*4.
(3)
4 is equivalent to the action oft on the 21 1
212
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
Our aim is to find the solution of the ordinary differential equation m
Lt
=
1 u,(x)
dmt/dxm= 7,
m=O
(4)
where z is an arbitrary known distribution. As defined in Section 2.6, the fundamental solution is the solution for T = 6(x). A distribution t is a solution of (4)if for every test function 4 we have ( L t , 4)) = (G 4 ) ,
(54
4).
(5b)
or, equivalently, ( t , L*4) = (z,
In searching for a solution t of differential equation (4)we may have the following situations: (1) The solution t is a sufficiently smooth function, so that the operations in (4) can be performed in the classical sense and the resulting equation is an identity. Then t is the classical solution. (2) The solution t is not sufficiently smooth, so that the operation in (4) cannot be performed, but it satisfies (5) as a distribution. It is then a weak solution. (3) The solution t is a singular distribution and satisfies (5). It is then a distributional solution.
All these solutions are called generalized solutions. We really gain no new insight into the solutions of the classical problems in ordinary differential equations by using the theory of distributions. However, the theory does enhance our knowledge if discontinuities are present in these equations or if we want to find the fundamental solutions. This will become clear in the next sections.
9.2. HOMOGENEOUS DIFFERENTIAL EQUATIONS
Let us start with the simplest differential equation, dtldx = 0.
We know that the only classical solution to this equation is t = c. We now prove the following theorem:
Theorem. The only generalized solution to (1) is t = c. To prove this theorem we need the following lemma:
9.2.
21 3
HOMOGENEOUS DIFFERENTIAL EQUATIONS
Lemma. Any test function $(x) can be represented as the derivative of another test function 4(x) if and only if J-mm$(x) dx = 0.
ProoJ
When $(x) = 4'(x), $(x)
J-
E D,
we have
m
m
$(x) d x = [$(x)]
=
0.
-m
m
fi
On the other hand, $(x) = $(u) du is an infinitely differentiable function, and $(x) and $'(x) vanish outside the same interval. Proof of the Theorem. Equation (1) implies that (t', 4 ) = - ( t , 4') = 0. Thus
(3)
( t , *) = 0
for every test function that is the derivative of another test function provided (1) is satisfied. Let us now take a fixed test function cjo(x) that is normalized such that
and write
Let
Then
J-
m m
J-
m
$(x) dx =
4 ( x ) dx
J-
m
-
J-
m
4 o ( x ) dx
4 ( u ) du = 0,
where we have used (4). Then, in view of the lemma, $(x) is the derivative of some test function, and from (3) it follows that ( t , $) = 0. Accordingly, (5) becomes
214
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
But ( t . &) is a constant, say, c, so the preceding relation becomes
4)
1 W
=
M u )
= (c,
4),
(7)
-03
or t = c. From this theorem we immediately have the following two corollaries.
Corollary 1. If two generalized functions s and t have the same derivative, then s = t + c. Corollary 2. Let T = ( t , , t,, . . . , t,) be an n-dimensional vector distribution. Then the solution of the differential equation
0 (8) is T = C, C = (cl, c 2 , .. . ,c,). These ideas can be extended to the nth-order ordinary differential equation dT/dx
Lt = ant'"' + a,-
l)
=
+ . . . + a l t ' + aot = 0,
(9)
where u j , j = 0, 1, . . . , n, are infinitely differentiable functions and a, # 0. Indeed, this equation can be transformed to a system of linear first-order ordinary differential equations, as will be explained in Section 9.10. The solution in this case has also the same form as the classical solution.
9.3. INHOMOGENEOUS DIFFERENTIAL EQUATIONS: THE INTEGRAL OF A DISTRIBUTION
The simplest inhomogeneous ordinary differential equation is
dt/dx
=
f;
(1)
wheref'is a continuous function. A distribution t is a solution of this equation if ("9
or (t,
4)
=
(L 4)-
(2a)
4')
=
(.L - 4 ) .
(2b)
for every 4 E D.To solve (2b) we appeal to the decomposition (9.2.5): m
m
9.4.
21 5
EXAMPLES
Then we have (3)
where $ ( x ) is defined in (9.2.6). Since $ ( x ) is the derivative of the function 41(x) =
Lm
4(u>h -
[Irn 40(4 J-
m
do
4(u) d2.4
(4)
we can use $ ( x ) for & ( x ) and 4 1 ( x )for +(x) in (2). Thus ( t , $) = (,f, -41), which means that ( t , tj) is known, and we can set ( t , tj) = ( t o , 4). Also, ( t , 40) is merely an arbitrary constant. Accordingly, we can write (3) as ( t , 4) = ( c , 4) + ( t o , 4), so that t =c
+ to,
(5)
which agrees with the classical solution; c is the solution of the homogeneous equation and t o is the particular solution. As a corollary, we find that for a vector equation (6)
dT/dx = F, where T
= (tl, t,,
. . . , t,) and F
=
(,f1,,f2,.
. . , ,f,), the solution is
T=C+To,
(7)
where C is a constant vector distribution and To is the particular solution which arises in a manner similar to (4). The generalized solution of the inhomogeneous ordinary differential equation ant'"'
+ a,-,t'"-
l)
+ . f . + aor
=
j,
(8)
where a,, . . . , a, are infinitely differentiable functions, t is a scalar distribution, a, # 0, and ,f is a continuous function (or the generalized solution of its equivalent system of the first order equations), is identical to the classical solution. We shall discuss in Section 9.10 the equation obtained from (8) by substituting 6 ( x ) for the right side.
9.4. EXAMPLES
Example I . To find the general solution of the equation
xm dtldx
=
0,
m 2 1,
21 6
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
we appeal to the relation dH/dx = 6(x) and use the derivatives of 6(x). Indeed, we assume that t(x) = C' so that
+
c2 H ( x )
+
c3
6(x)
+ cq6'(x) + . . . + c+,
+ c 36'(x) + - .. + c+,
r'(x) = c 2S ( X )
16,-
yx),
(2)
6"- ' ( x ) .
Thus (xrnt'(x), 4)
=
(C2Xrn6(X),4 ( x ) >+ (c3xrnb'(x),4(x)> . . . + (Cm+'xm6(m-1)(X), 4 ( x ) ) = 0,
+
and we have xmt'(x)= 0 as required. Hence, (2) is the general solution of (1). For m = 1 the solution reduces to r(x) = c1
+ c,H(x).
(3) The Heaviside function H(x),although an ordinary function, is not differentiable. Therefore (3) is a weak solution. For m 2 2, (2) is the distributional solution. Example 2. We show that the general solution of x dt/dx = 1 is t = + c,H(x) + 1nJxl.The complementary part c1 + c2H(x) follows from Example 1. To prove the rest, we differentiate t , obtaining
c1
dt/dx Thus
=
c26(x)+ Pf(l/x).
( x dt/dx, 4(x)>= ( ' 2 ( X W ,4 0 ) ) + ( x Pf(l/x),4(x>>
lim
=
s_,
&+O
m
as required.
[
=
J;:~(X)
dx
+
4 ( x ) dx]
4(X>dx= (1,4),
Example 3. We show that the general solution of x 2 dtldx C'
+ c,H(x) + c36(x) - + Pf(l/X2).
From Example 1, the complementary function is c1 The particular solution is obtained by noting that
=
P f ( l / x )is
+ c,H(x) + c36(x).
9.5.
21 7
FUNDAMENTAL SOLUTIONS AND GREEN'S FUNCTIONS
9.5. FUNDAMENTAL SOLUTIONS AND GREEN'S FUNCTIONS
Recall that the fundamental solution for the differential operator L is defined as LE(x) = 6(x), where L is defined in (9.1.1). Let us start with the simple ordinary differential equation
LE(x) = (a, d/dx
+ ao)E = 6(x).
Since there is no loss of generality in taking a, equation in its customary form,
LE
=
dEfdx
=
1, we discuss this differential
+ UE = 6 ( ~ ) ,
(1)
+
where we have set a. = a. If a = 0, the solution is clearly t ( x ) = H ( x ) const, where H ( x ) is the Heaviside function. Fortunately, upon setting E(x) = e-'"t(x), ( 1 ) takes the same form, namely,
dtldx
=
6(x)e"" = 6(x),
(2)
because for all
(h(x)e'", 4 ( x ) ) = 4(0) = ( 6 ( x ) , 4 ( x ) ) , Thus
E(x) = e-""t(x) = e-""(H(x) + C ) = H(x)e-""
4 E D.
+ Ce-"",
(3)
where C is a constant. Inasmuch as U = e-"" is the unique solution of the initial value problem
dU/dx we can write (3)
+ aU = 0,
UI,=,
=
1,
E(x) = H ( x ) U ( x ) + Ce-"".
(4) (5)
Just as in the analysis of Sections 9.2 and 9.3, these arguments extend to a system of first-order ordinary differential equations
LE
=
dE/dx
+ AE = 6 ( ~ ) 1
(6)
where the fundamental solution E is now an n x n matrix, A is a given n x n matrix (whose entries are infinitely differentiable functions), and I is the identity matrix. In this case the initial value problem that corresponds to (4) is
dU/dx
+ A U = 0,
UI,=o
=
I.
(7)
218
9. APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
Its solution is U = e - x A . To find the solution of (6) we let B denote an arbitrary n x n matrix whose entries are distributions. Then, applying Leibniz formula, L ( U B ) = ( L U ) B U B = UB',
+
we find that the solution of (6) is equal to UB, where B satisfies the differential equation dB/dx = U - ' i j ( x ) = exA6(x)= 16(x). (8) Its solution is B = H ( x ) l + C , where C now is a constant n x n matrix and where H ( x ) l is the diagonal matrix with all its diagonal entries equal to the Heaviside function. Since E = U B = C x A B ,we have E
= H(x)e-XA
+ e-XAC.
(9)
This solution helps us in solving the inhomogeneous equation for a distribution vector f, dtjdx + At = T, (10) where T is a distribution vector with a compact support. Indeed, the particular solution is the convolution f = E ( x ) * T,
(1 1)
while the complementary solution is the same as the classical one. The same is true for the single differential equation (1). Incidentally, (9) is called the right fundamental solution of the operator L. The left fundamental solution is the distribution that satisfies the equation dE/dx
+ E A = 61.
Its solution is E = H(x)e-XA
+ CeKXA.
(12)
(13)
The next natural step would be to consider the general nth-order ordinary differential operator. We shall, of course, attend to it, but first we want to discuss second-order ordinary differential equations, because they are the cornerstone of mathematical physics and theoretical mechanics.
9.6. SECOND-ORDER DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
The simplest equation of this type is d2E/dX2 = 6 ( x - 0.
9.6.
SECOND-ORDER EQUATIONS WITH CONSTANT COEFFICIENTS
21 9
Since H ’ ( x ) = 6(x), integration of (1) gives
5 ) + 40,
d&x, 5)/dx = H ( x -
(2)
where a(<) is an arbitrary function. Next, we integrate (2) and obtain E ( x , t) = J;,H(X
-
5 ) dx
+ xu(<) + B(t>
= (x - t W ( x
- 5)
+ x 4 t ) + B(5X
(3)
where /I(() is another arbitrary function. It is easily verified that (3) satisfies (1). Equation ( 3 ) shows that E ( x , 4 ) is a continuous and piecewise differentiable function. This solution helps us in solving the inhomogeneous equation d2t/dx2 = T ( x ) ,
where
T(X)
(4)
is a distribution with compact support. Indeed, t(X) = E
* T,
(5)
as is easily verified by operating on both sides of (5) with d2/dx2. Let us see if we can solve with the help of the above analysis an initial or boundary value problem in the classical theory of ordinary differential equations. Let us consider the boundary value problem d2U/dXZ = f ( x ) ,
0 I x I 1,
~ ( 0= ) ~ ( 1 )= 0,
(6)
where f ( x ) is an integrable function with compact support. Appealing to relations (3) and (9,we find that
The boundary conditions (6) then give
220
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
From (7) we find that
p(5) = (H( - 5). Then (8) yields
Thus the required solution is u(x) = -
-
-
where
The function G(x, 5) is called a Green’s function. It satisfies differential equation (1) and the same boundary conditions as does u(x), namely, G(0, 5) = G(1, 5) = 0. (12) We can go a step further and examine the case of inhomogeneous boundary values, that is, u(0) = a, u ( 1 ) = b. For this purpose we split the function u into two parts u1 and u2 such that u = u1 + u 2 . The function u1 satisfies the same differential equation and has the same boundary values as the function u in (6). Accordingly, its value, from (9), is u1 = JOX(X
-
O f ( 0 d5
1
- x Jo (1
- Of(<)d5.
The function u2 satisfies the boundary value problem d2u2/dx2 = 0,
~ 2 ( 0 )=
a,
~ 2 ( l )=
b.
We can look at the quantity a as the strength of the jump discontinuity at 0 and - b as the strength of the jump discontinuity at x = 1 (so that the functions rises at 0 and falls back at x = 1). Then we can appeal to (5.1.5) and obtain d2u2/dx2 = u ~ ’ ( x ) bd’(x - I),
uZ(0) = 0, uZ(1) = 0.
9.6.
SECOND-ORDER EQUATIONS WITH CONSTANT COEFFICIENTS
Its solution follows from (9): 11’
221
bx - a(x - 1). Thus
=
Let us now consider the initial value problem d2u/dX2
+ Q’U
=6(~),
~ ) , = o + = 0,
u‘),=O+
= 1.
(14)
Here u is defined for x 2 0. Accordingly, the solution is such that u is zero for x < 0 and satisfies the differential equation d2u/dx2+ a2u = 0 for x > 0. The solution that satisfies the conditions u(0) = 0, u‘(0) = 1, is sin(ax)/a. Thus the solution of the initial value problem (14) is u(x) = ( l / a ) H ( x ) sin(ax).
(15)
It is the fundamental solution E (also called the Causal solution) of the operator d2/dx2 + a’. Let us use this fundamental solution to construct the solution of the initial value problem d2u/dx2
+ a2u = f ( x ) ,
~ ( , = o + = uo,
Il’lx=o+
= Ul,
(16)
wherefis a continuous function for x 2 0. For this purpose we continue the functions u and f in the following way: u(x) = {!iX),
x < 0, x 2 0;
{>(x),
x < 0, x 2 0.
=
We then find from (5.1.3) and (5.1.5) that ij’(x) = u’
+ uo 6(x),
ij”(x) = u“
+ uo S’(X) + u,6(x).
Accordingly, the function u satisfies the differential equation dx2
+ a2u = g ( x ) + uo S’(X) + u , 6 ( x ) ,
* (g
+ u06’ + u,6) = E * g + uoE‘ + u , E
d2V ~
(17)
whose solution is u(x)
= =
E a
Joxj(y) sin a(x - y) d y
sin(ux) + uo cos(ax) + u 1 __ . U
(18)
222
9. APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
An Alternative Approach
We can obtain (3) and various other interesting results by an alternative approach. For this purpose let us first recall (2.7.12),
d2
--(("Ix rix2
-
51) = 6(x
-
5).
Accordingly, the solution (Green's function) to the boundary value problem
G(0,
5) = 0,
is G(x, 5 ) = + t X -
5I
G(1,
5 ) = 0,
(21)
+ xA(5) + B(5).
(22)
Applying the boundary conditions (21), we find that
B ( t ) = -45. (23) Substituting these values in (22), we recover (1 1). At this stage it is useful to introduce the symbols x < and x > . They stand for the values A(<) =
5
and x,
=
max(x,
-
4,
i
5 ) = 5,
X,
usxs5, t s x s b ,
as shown in Fig. 9.1. The corresponding quantities G, and G, stand for the values of G in the x < and x> regions, respectively. In the present case, a = 0, b = 1, so that G, = x(5 - 1) and G, = <(x - 1). The functions G, and G, satisfy differential equation (1) in the x < and x, regions respectively. The function G , satisfies the boundary condition G(0, () = 0 while G, satisfies the condition G(1, 5 ) = 0. At x = 5 these two solutions are equal. However, there is a jump in their derivatives; that is, [dG,/dx - dG,/dx],=5 = 1. (24)
-
<
I
a
xc
Fig. 9.1
x,
;
9.6.
SECOND-ORDER EQUATIONS WITH CONSTANT COEFFICIENTS
223
The preceding remarks are valid for a general Sturm-Liouville problem,
G(a, 5 ) = 0, G(b, 5 ) = 0, (26) where p and 4 are real-valued functions on a Ix Ih, p , p‘, 4 are continuous in this interval and p is positive. In this case, G(x, 5)Ix=<+ = G(x, 5>Ix=<-
and
dG(x3 5 ) dx
I
x=<+
W x ,5) dx
1
x=<-
--
(27) 1
At)’
(28)
where (28) follows from (25) by first integration. Example 1 . In quantum mechanics there is a very interesting phenomenon in which a particle passes through a potential barrier that classical mechanics predicts is impenetrable. Because this phenomenon implicitly involves motion, let us begin with the time-dependent Schrodinger wave equation: at
where $(x, t ) is the wave function, V is the potential, and E is a suitable parameter. To illustrate the phenomenon of tunneling, we make two very simple choices. First, we assume that the time dependence of $(x, t ) is purely oscillatory, so that $(x, t )
=
(30)
y(x)eiE‘.
Second, we choose V ( x ) = 6(x),the Dirac delta function. Then (29) reduces to 2y”
+ Ey = y(x)b(x)= y(O)d(x).
(31)
We solve this equation by the procedure outlined in this section. The solutions of (31) are y ( x ) = ac-ixfi/&+ beixfi/&,
and y ( x ) = ce-iX&&
+ deiXJE/C,
x < 0,
(32)
,0.
(33)
There are four constants a, b, c, d in these solutions. Two can be determined by physical considerations. Suppose we aim a monoenergetic unit-amplitude
224
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
incident beam of particles moving toward x = 0 from the left. Then a = 1. The term beiXJEiE in (32) is then a reflected wave, moving left, for x I 0. In the region x > 0 there will be only a transmitted wave, moving right. Accordingly, d = 0 in (33) and we have the following solutions:
The remaining two constants are determined by the following considerations:
1. y(x) is continuous at x = 0; 2. [Idy/dxlo+ - CdY/dXlO- = Y(O)/E2. We thereby obtain
b=
2646- 1 4.c2E 1 ’
+
c=
2&JE(2cJE 4g2E 1
+
+ i)
(36)
The quantities R = lb12 and T = (c12are called the rejection and transmission coeficients, respectively. Their values are R =
4~~E T=4c2 1
1
+
4c2E + 1 ’
’
(37)
In the language of quantum mechanics, R is the probability that an incident particle of energy E will be reflected, and T is the probability that the incident particle will be transmitted. Note that the total probability that a particle will be reflected or transmitted is unity between T + R = 1. When E + x ) , R -+ 0 and T + 1, and when E + 0, T + 0 and R + 1. For these two limits the classical and the quantum-mechanical predictions agree. Example2. Let us find the Green’s function that satisfies the SturmLiouville problem
G(0)
-
G’(0) = 0,
G(n) + G’(n) = 0.
(39)
To find the complete solution of this system, we first find the solutions in the regions x , (0 I x I 5 ) and x, ( 5 < x I 1) regions. In the x < region, the solution of the homogeneous part of (38) is G(x,
so that
t) = J s i n x + B cos x,
G’(x, t) =
cos x -
B sin x.
(40) (41)
9.6.
225
SECOND-ORDER EQUATIONS WITH CONSTANT COEFFICIENTS
In view of the boundary condition at x solution in the x < region is G(X,(1
=
0, we obtain
A’= B. Thus the
+ cos x) = ,,hA’[(l/,,h) sin x + (l/,,h> cos XI. = A sin(x + n/4), A =fix. (42) =
A”(sin x
In the x, region, we take the solution of the homogeneous part of equation (38) to be G(x, 5) =
z;sin x + d cos x,
(43)
G’(x, 5 )
c cos x - b sin x.
(44)
so that =
Then the boundary condition (39) yields 2; = G(X, 5) =
c sin(x -
-d.
Thus
c = J5C.
(45)
The condition of continuity is A sin(x
+ x/4)I,=r
=
C sin(x - 7~/4)1,=~,
or A
sin(( - 4 4 )
-
C
sin((
+ n/4) = B = const.
(46)
This leaves undetermined only B,which we find by applying the jump condition dG/dx(,=e;+ - dG/dxI,=e;-
or
+ n/4) cos((
BCsin(4 which yields B
=
-
=
1
7~14)- sin(( - n/4) cos(5
+ n/4)]
=
1,
1. Thus
G(x, 5 ) =
sin(< - n/4) sin(x + 71/4), sin(( + 4 4 ) sin(x - n/4),
{
x < (, x > <.
(47)
Example 3. Let us use the information of the previous example to solve the boundary value problem u”(x)
+ u(x) = .f(x), f(x) =
{
0 Ix 5 n,
fi(X) = 0, fz(X) = 2471,
u(0) - u’(0) = 0,
u(7r)
(48)
0 Ix < 4 2 , n/2 < x 5 71.
+ u’(n) = 0.
(49)
226
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
Thus, the jump discontinuity is at the point 5 = n/2. In view of the jump in the functionfand the Green's function at x = 742, we expect the solution u and its derivative u' to have jump discontinuities at 7-42. As in Chapter 5, let square brackets label the jump of a function at x = 4 2 . We thus set
[ul
=
$+) (;-)
[u7
= a,
-
..(I.) d(;-) -
=
=
p.
Now recall the values of the distributional derivatives, namely,
U' = u'
+ aS(x - n/2),
U" = u"
Thus (48) can be written as U" so that G(x, t)[f(t)]
+ ii = f(x) + PS(x - 4 2 ) + aS'(x
4
s'n '
-
n/2),
+ pS
( I) : ( I) ;), ( :)( i)+ 9)
G(x, t).f'(t)dt -
+ a8(x - n/2) + pS(X - 4 2 ) .
-
-
+ PG
1-
X,
-
-
CI - X, -
pG( X,
(x,
-a
where we have used the solution for G from (47). As in the previous example, we find the values of regions.
if
in the x, and x,
In the x < region,
Next, we substitute the required values from the previous example, namely,
+ 71/4), sin(x + x/4),
+ n/4)lni2= (&2) dG,/dt(,=n,2 = [cos(t - n/4)sin(x + 7~/4)]"/~ = (&2)
G,(x, r)lr=n,2 = [sin(t - n/4) sin(x
sin(x
and obtain u , = (l/$)[sin(x
In the x, region, '
n/4)(1
( !)( );
u,=-sinx+-
4
-
I--
-
4/n)
+ sin(x + 71/4)(p - a)].
( ;)
+pG,x,-
-a-
d;
(.,;).
(50)
(51)
227
9.7. EIGENVALUE PROBLEMS
Because G,(x, t )
=
sin(t
Jz.
+ n/4) sin(x - n/4), we have mG>(x,;)= a
G>(x,i)=ism(x-i), Thus (51) yields 11,
=
sin(x - 7c/4)[p
(1/fi)
+ a + 1 - 4/n].
(52)
Combining (50) and (52), we have u(x) =
~
+
sin(x - n/4)( 1 - 4/n) + sin(x n/4)(/? - a ] , sin(x - n/4)[/3 a 1 - 4/n],
+ +
0 Ix I42, n/2 < x I n.
9.7. EIGENVALUE PROBLEMS
Let us now give a brief description of the spectral theory of the SturmLiouville problem with general boundary conditions. For this purpose we find the Green’s function G(x, (, A) that satisfies the system [19,20] LG
+ ArG = (pG’)’ + qG + ArG = 6(x - (),
a I x I b, (la)
BUG = (cos a)G(a) - (sin a)G’(a) = 0,
(1b)
BbG = (cos p)G(b)
(1c)
+ (sin B)G’(b) = 0,
where the quantities p , q, a, and b have been defined in Section 9.6. The function r(x) is a continuous real-valued function that is positive in [a, b], A is the eigenvalue, and a and 1are given real numbers, 0 Ia < n,0 Ip < n. The signs in the boundary conditions (lb) and (lc) have been chosen so that the eigenvalues decrease as a or p increases. In the notation of Fig. 9.1, the system (1) can be split into two simpler systems.
In the x < region, G(x, <, A) is a solution of the homogeneous equation
+
LG ArG = 0, BOG = 0. (2) We can connect the solution of this system to the solution 41(x)of the initial value problem WAX)
+ Ar4Ax) = 0,
(34
4n(a) = sin a,
(3b)
cos a,
(3c)
&(a)
=
228
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
so that B , ~ , ( u )= 0.
(34
Thus G(x, 5, A) is a constant multiple of +a(x) in the x < region. In the x > region, by the same token let xA(x)be the unique solution of the initial value problem
so that B,Xa(b)
0.
Thus G(x, t, 1) is a constant multiple of xA(x)in the x, region. From this discussion it follows that
t, 4= A4,(X<)Xl(X>).
(5) where A is a constant. G is clearly continuous at x = 5. To find the constant, we appeal to the jump condition (9.6.28), which holds for (la) also. Thus G(x,
"a(5)x;(r)
-
4;(t)XAt)l =
l/P(t).
The quantity inside the brackets is the Wronskian W,($,, evaluated at x = 5. Thus AW5(4n9X J
=
xa) of
l/P(t).
and
xa (6)
Now we appeal to Abel's formula for the Wronskian, namely,
Y 4 a 9 xa)
(7) where o ( A ) is independent of x but may depend on 1. From (6) and (7) we find that A = - I/o(A), and we have from (5) = 4n(X)XXX) - xa(x)+Xx) = o(wP(Xx
4= 4Ax < ) X l ( X > 1/44.
(8) It follows from (7) that the zeros of w(A) coincide with the eigenvalues of the system (1). Indeed, let I = p be a zero of o ( A ) , so that w ( p ) = 0. From (7) it follows that the Wronskian of 4,(x) and ~ , ( x )is zero, and therefore these functions are linearly dependent. However, neither of these functions can vanish identically because of their initial values. Accordingly, &(x) = kx,(.w), where k is a nonzero constant and both these functions satisfy the two boundary conditions of the system (1). Thus, p is an eigenvalue of (l), while 4,(x) is the corresponding eigenfunction. Furthermore, we shall soon find [see (14)] that at an eigenvalue Green's function has a singularity and G(x,
t 3
229
9.7. EIGENVALUE PROBLEMS
therefore o ( A ) must vanish. Thus, the zeros of o ( A ) coincide with the eigenvalue of system (1). Let us label these zeros
A, < A 2 < . ' . A, < ...,
A,+ x ,
and set
4a,(x) = k,XA,(X).
(9)
To normalize these eigenfunctions we appeal to the residue of G at A which is 4Ln(X < )XA,(X > ) -
w'(A,)
Note that xn,(x<)xA,(x,) Similarly, C$~,(X ,)4,n(x >) residue of G at A = A,, is
kfl
XA,(X < )XL.(X
>
(.'(A,)
1 - 4a.(x < )4& kl W
f
l
=
A,,,
>) )
.
= xA,(x)xAn(5) in both the x, and x > regions. = and the required value of the
4An(x)~An(<),
k , xA,(X)xan(5)/~'(~.> = 4An(X)4Ln(5)/~,Q'(A,) = U f l ( X ) ~ f l ( 5 X where u,(x) are the orthonormalized eigenfunctions of the system (1) and U,(x) are their complex conjugates. From this relation it follows that
*
* [~fl/m'(Afl)I
(10) Next we expand Green's function in terms of the orthonormalized eigenfunctions u,(x), as obtained in the foregoing, so that u,(x)
=
4a"(X)/Ck,m'(An)l 112 =
1'2xa,(x).
where
To find gn we multiply (la) by U,(x) and integrate from a to b. We then use (12) and the sifting property. We thus obtain ( L G , u,) - Lg,,= En(<). Because ( L G , u,) = ( G , Lu,) = Angn, we have Sfl(5,
From ( 1 1) it then follows that
A,)
=
~,(S)MA,
-
4.
(13)
230
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
Two interesting results follow from this discussion: first, we find that the solution w(x, L) of the inhomogeneous system (pw’)’
+ qw + Lrw = f ( x ) ,
a I x I b,
Bow = B ~ w = 0,
(154 (15b)
is
Second, when we expand G(x)/r(x),we obtain
c W
=
n= 1
un(x)un(5).
We end this section with an example. Example. Let us consider the system (XU’Y
+h
u
lim xu’
= 0, =
0,
x-0
u(1) = 0. The only solution that satisfies (18a) and (18b) is Jo($x). The boundary condition (18c) implies that .lo(&) = 0, which gives the eigenvalues of system (18). The corresponding eigenfunctions are un(x) = J z J o ( f i x ) / J o ( J z ) .
Then from relation (1 7) it follows that
9.8.
23 1
SECOND-ORDER EQUATIONS WITH VARIABLE COEFFICIENTS
9.8. SECOND-ORDER DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS
Let us consider the differential equation
where 6(”)(x)is the nth derivative of 6(x). To fix the ideas, we shall take N so that we have only to solve L y = [u(x) d2/dx2
+ b(x) d/dx +
C(X)]Y
= POS(X -
=
5 ) + /J16’(~- 5).
1
(1)
Since this is a nonhomogeneous equation, its solution is expressed as Y(x) = Yh(X)
+ Yp(x),
(2)
where Yh(X) is the solution of the homogeneous equation L y = 0 and yp(x) is a particular solution. The part yh(x) is the same as we study in the classical analysis; we therefore attend only to the particular solution, which we assume to be
51,
(3) where H(x) is the Heaviside function and G(x) is an unknown function. Then we appeal to (2.6.25), obtaining Y&X)
Yb(X) =
= G(x)H(x -
5 ) + G(5)&x
G’(x)H(x -
Y;(x) = C”(X)H(X -
-
0,
(4a)
5 ) + G’(()~(x - 5 ) + G(<)~’(x- 5).
(4b)
Thus b(x)Y;(x) = b(x)G’(x)H(x - 5 ) u(x)Y;(x) = u(x)G”(x)H(x - 5 )
+ b(5)G(5)6(x
- 5)
+ 45)G‘(5)6(x
= a(x)G”(x)H(x
+ a(5)6‘(x
-
= a(x)G”(x)H(x
013
+ a(<)G(5)6’(x
-
-
+ u(x)G’(~)~(x- 5 ) + u(x)G(S)G’(X 0
-5)
+ G(5)C -a’(5)6(x
5) + Ca(W’(5) - a’(t)G(016(x 5).
-
5)
-
Substituting (5) in (l), we get
5 ) + C a ( W ’ ( 0 - a‘(t)G(5)16(x - 5 ) + 45)G(5)6’(x - 5 ) + b(x)G’(x)H(x - 5 ) + b(l)G(5)6(x - 5 ) + c(x)G(x)H(x - 5 ) = Bo&x - 0+ B16’(x - 0,
a(x)G”(x)H(x -
(5a)
3
-
5)
-5)
(5b)
232
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
9.
or (Jmx))H(x - 5) + Ca(OG’(0 - a’(5M5) + b(5)G(5)16(x + a ( O G ( m ’ ( x - 5) = po 6(x - 5) p , 6‘(x - 5).
5)
+ (6) Equating derivatives of 6 ( x )of equal orders on both sides of this equation, we have LG
=
u(x)G”(x)
+ b(x)G’(x) + c ( x ) G ( x ) = 0,
a(OG’(5)
+
-
a’(t))G(t)
=
PO?
a(OG(5) = P I .
(7a) (7b) (7c)
System (7) is the initial value problem for G(x) and can be solved by the classical method, thereby solving the original problem completely. = 0, we have the fundamental solution For the special case when E(x - 5):
E(x -
5) = E ( x , 5) = G ( x ) H ( x - 5).
(8)
Let us illustrate these concepts with the following examples. Example 1 . Let us find Green’s function for the harmonic oscillator, that is, d2y/dX2
+ k 2 y = 6(x - 5).
(9)
Comparing (9) with (I), we find that U ( X )= 1,
b(x) = 0,
po = 1,
Thus the system (7) reduces to (d2/dX2
C(X)
=
k2,
p1 = 0.
+ k 2 ) G ( x ) = 0,
The general solution of (10a) is G ( x ) = A sin(kx
+ B).
(11)
Then application of the ini,tial conditions (lob) and (1Oc) yields G(5) = A sin(k5 + B) = 0, so that B = - k t and Ak cos(k5
Thus A
=
+ B ) = A k cos 0 = 1.
l/k. Substituting these values in (1 I), we obtain G ( x ) = [(l/k) sin k(x - { ) I ,
9.8.
SECOND-ORDER EQUATIONS WITH VARIABLE COEFFICIENTS
233
and the general solution of (9) and the fundamental solution are y ( x ) = CI sin(kx
+ B) + [(l/k) sin k(x - 5 ) ] H ( x - 5 )
and E ( x - <) = [ ( l / k ) sin k(x - < ) ] H ( x -
0,
respectively. This fundamental solution agrees with (9.6.15). Example 2. Consider (1 - x2)-d 2Y - 2x dx2
Here a(x) = 1 - x2, b(x) becomes
=
-2x,
nr
-
=
6(x -
dx
C(X) =
0,
5).
= 1,
=
0. System (7)
d2
[(I - x2) dx2 - 2x
The general solution of (1 3a) is G ( x ) = A In
l + x ~
1-x
+ B.
Initial condition (13c) yields A In _ ‘ +_ ‘+B=O.
1-E
Since 1 I+x
G’(x) = A(-
-)
+ 1 -1 x
=
~
2A 1-x2’
from condition (13b) we find that ( 1 - t 2 ) 2 A / ( 1 (15) gives the value of B as
r2) = 1, so A = i.Then
Substituting these values of A and B in (14), we obtain G ( x ) = - In-2 1+<1-x
234
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
so the general solution of (12) is
The fundamental solution E(x, 5 ) is
Example 3. Let us extend the method of the previous examplcs to solve the differential equation d2y/dx
+ 3tly/dx + 24’ = 6“‘(x).
(18)
It is clear that the terms S(x) and 6’(x) will appear in the expression for y(x). However, in view of (2.6.15) we need only assume that the coefficients of S(x) and 6’(x) are constants. Accordingly, we assume
+ U & X ) + bS’(x).
Y ( X ) = G(x)H(x)
(19)
Differentiating twice we obtain $(x)
=
G’(x)H(x)+ G(O)S(x)+ aS’(x) + bS”(x),
+
+ G(O)d’(x)+ aS”(x) + bd”’(x).
Y”(x)= G”(x)H(x) G’(O)S(x)
Substitution of these in (18) yields the following system:
G”(x) + 3G’(x) + 2G(x) = 0,
+ 3G(O) + 2a = 0, G(0) + 3a + 2b = 0, a + 3b = 0,
G’(0)
b Solving this system, we obtain G(x) = -e-” from (19) the particular solution is
y(x) = (-e-x
=
1.
+ 8e-2”, a = -3,
+ 8 e - 2 ” ) H ( ~-) 36(x) + 6’(x).
b = 1, so
(20) Example4. Let us find the fundamental solution E ( x ) of the differential operator d2/dx2- a 2 with the help of the Fourier transform. Accordingly, we take the Fourier transform of the equation d2E/dx2 - a 2 E = 6(x) and obtain - (it2 + a2)8 = 1 or .&id) = - 1/(u2 + it2). Taking the inverse transform, we find that
9.9. FOURTH-ORDER DIFFERENTIAL EQUATIONS
235
9.9. FOURTH-ORDER DIFFERENTIAL EQUATIONS
Fourth-order ordinary differential equations arise in various steady state and vibration problems in applied mechanics. We present the Green’s function theory of such an equation in this section. Let us consider the differential equation [2t]
where we have stopped at the term 6"'(x) for the sake of simplicity; the following analysis can be easily extended to include P"(x - 5 ) for n 2 4. Again, the general solution y ( x ) is
+ Y,(XX
(2) where y h ( x ) is the solution of the homogeneous equation and y,(x) is the particular solution. Our aim is to find the particular solution, which we again assume to be Y,(X) = G ( x ) H ( x - t). (3) Differentiating this relation four times we obtain Y(X) =
YdX)
where we have used (2.6.25). Similarly, we obtain
236
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
The next step is to substitute these values in (I), so that (LG)H(x -
5 ) + Ia(OG”’(0 - CaYO - h(OIG”(5)
+ Ca”(0 - b ’ ( 0 + C(OlG’(<) - Ca”’(0 - b“(5) + (.YO - 45)1G(5)16(x - 5 ) + Ia(OG”(5) - CW5) - b(OIG’(5)
+ C3U’YO - W 5 ) + c(5)1G(5)1S’(x 5 ) -
+ (a(OG’(5) - C3a’(O - b(5)1G(O)~”(X- 5 ) + 45)G(5)S”’(x- 5 ) = p , S ( X - 5 ) + B I S ’ ( X - 5 ) + P 2 S ” ( X - 5 ) + /j3b”’(x- 5). (6) We equate the coefficients of generalized functions on both sides of (6) to obtain the initial value problem: LG
= a(x)G”(x)
+ b(x)G”’(x)+ c(x)G”(x)+ d(x)G‘(x)+ e(x)G(x)= 0,
a ( W ” ‘ ( 5 )- Ca’(5) - b(5)IG’YO - Ca”’(0 -
+ Ca”(0 - b’(0 + c(5)1G’(5)
b ” ( 0 + c ’ ( 0 - d(OIG(5) = P o .
(7a) (7b)
a(<)G”(<)- P ‘ ( 5 ) - b(5)]G’(5)+ [3a”(O - 2b’(<)+ C(5)IG(O= P i , (7C) a(W’(5)-
C3U’(O
-
(74
b(5)1G(5)= P 2 ,
4 5 ) G ( 5 ) = P3.
To find the fundamental solution of the operator L, we set
P1 = p 2 = p3 = 0.
Po
(7e) = 1 and
Example. Let us solve the differential equation
+
(d2/dx2)[E~,(1 m
~d2y/dx2] ) ~ = plS’(x - +/),
(8) which embodies the static problem of finding the deflection y of a linearly tapered beam that is free at x = 0, clamped at x = I, and subjected to a couple /J1 at its midlength (see Fig. 9.2). The quantity E is Young’s modulus, I = I,( 1 + mx3)is the moment of inertia of the cross section, and I. and m are constants. Written explicitly, (8) becomes Elo( 1
+ m ~ ) ~ y +’ ’ 6E10m( 1 + mx)2jr”’+ 6E10m2( 1 + mx)y”
(9)
plS’(x - +/). Comparing (9) with (l), we have =
+ m ~ ) ~ b(x) , = 6E10m(l + mx)2, c(x) = 6E10m2(l + mx), d ( x ) = e(x) = 0.
a(x) = EI,(1
To find the particular solution we start with (3) with 5 in this case becomes 6m G”‘ + 6m2 Giv + ___ G“ = 0. 1 + mx (1 mx)2
+
=
(10)
+/, so that (7a) (1 1)
9.9.
237
FOURTH-ORDER DIFFERENTIAL EQUATIONS
1-
Y
Fig. 9.2. The deflection of a tempered beam.
Its solution is G ( x ) = c1
+ c 2 x + c3(l + mx) + c4 ln(1 + mx),
(12)
where the cs are constants. To find these constants we substitute (12) in the initial conditions (7b)-(7e). After some algebraic work we obtain
c1 (‘2
= -
=
B1I
4mEl0(1
+
B1
2rnE1,(1
+ $mI)2’
B,
-
2m2EIo(l C3
=
+ +MI)’
B1
2m2E10’
~
(13) c4 =
0.
From (3), (12), and (13) we find that the particular solution of (9) is Y , ( x ) = G ( x ) H ( x - il)
The general solution can now be readily obtained by adding to (14) the solution to the homogeneous equation, namely, y h ( x ) = b,
+ b 2 x + b3(l + mx) + b4 In(1 + mx),
(15)
where the bs are constants. These constants are obtained with the help of the boundary conditions that the beam is free at x = 0 and clamped at x = I, that is, y(I) = 0,
y‘(l) = 0,
y”(0) = 0,
y”’(0) = 0.
(16)
238
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
We leave the algebraic details to the reader (Exercise 6). The solution is 2mE10 B1
Y(X) = -
(1
+ ml)2
-
1
2( 1
-
+
1
m(1
+ ml)
-
m(1
+ +ml)
+ m(1 + mx) H(x - il). 9.10. DIFFERENTIAL EQUATIONS OF n T H ORDER
Finally, we consider the fundamental solution of the nth-order differential operator (9.1.1)with a,, = 1 . This means that we have to solve the differential equation 2"-1E 2 E . . . + aO(x)E = &x). + a,,- 1(x) dx"
+
~
Using the standard transformation
we can write (1) as a first-order system, dU/dx
+ AU
=
0
...
F,
(3)
where 0 0
-1
0
-1
...
0 0
Let us now recall a few well-known results from the theory of ordinary differential equations. First, the general solution of the equation LU =
239
9.10. DIFFERENTIAL EQUATIONS OF nTH ORDER
rlU/t/x + A ( x ) U = 0, where A ( x ) is a given n x n matrix of differentiable functions, can be written U ( x ) = K ( x ) C , where C is a constant vector and K ( x ) is a fundamental matrix of the differential equation, namely, a solution that satisfies the condition det K ( x ) # 0. From the matrix K we form the Green's function (matrix) for the initial value problems associated with the operator L by setting G ( x , y ) = K(x)[K(y)]- '. Second, G ( x , y ) is independent of the particular fundamental matrix K that we use. Moreover, if A is a constant, then G ( x , y ) depends only on x - y ; i.e., G ( x , y ) = g ( x - y), where g ( x ) satisfies the initial value problem dg/dx
+ Ag = 0,
g(0) = I .
The Green's function G(x, y ) enables us to solve the inhomogeneous equation (3) where F is a vector whose components are continuous functions. Indeed, Jo
where C is a constant vector, is the solution of (3) and satisfies the initial condition U ( 0 ) = C . We are interested only in evaluating the fundamental solution E . Accordingly, in view of (2) we take the first component of (5) and obtain
=
G1
O)H(x) +
n
C1 ci G
i=
1
i(x, O),
(6)
where Gij and ci are the components of G and C respectively. Suppose that u , ( x ) , . . . , u,(x) are n linearly independent solutions of the homogeneous equation associated with (1). Then
is a fundamental matrix of (3). Thus
240
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
where the quantity Ani is the (n, i ) element of the inverse of K , i.e., A nl. = ( - 1 y + i u1
1
u1
: &-2) 1'
...
ui-1
. ..
uI"-2)
...
un
...
Ui+l
...
1
I+
u y 2 )
-1
Comparing (8) and (9) we obtain
Then from (6) we find that the fundamental solution is n
E(x)
=
Gln(x,O)H(x)
+ 1c ~ u ~ ( x ) . i= 1
In view of this discussion, (1 1) can be written as n
E(x)
=
g(x)H(x)
+ i1 ciui(x), = 1
where g satisfies the initial value problem
g(0) = g(0) =
. . . = g'"-2)(0)
=
0,
g'"-"(O) = 1.
If the polynomial p(x) = xn
+ an-lXn-l +
. * *
+ a , x + ao,
has roots A l , . . . , A, that are simple, then it is easy to show that n
g(x) =
1 eAlx/p'(Ai),
i= 1
and so n
n
9.1 1.
ORDINARY EQUATIONS WITH SINGULAR COEFFICIENTS
241
We can readily extend the analysis of this section to the case in which the right side of (1) contains terms such as 6’(x), 6”(x),. . . ,6 ( k ) ( on ~ ) the same lines as presented in the previous sections.
9.11. ORDINARY DIFFERENTIAL EQUATIONS WITH SINGULAR COEFFICIENTS
Linear homogeneous systems of differential equations with infinitely smooth coefficients have no solutions in the space of generalized functions other than the classical solutions. However, for equations with singularities in the coefficients, there may appear new solutions in generalized functions. For instance, y ( x ) = C[S”(x) + @ ( x ) ] , where C is a constant, is a solution of the Bessel equation x2y” + xy’ + ( x 2 - 9 ) y = 0, as can be easily verified by using Exercise 12(a) of Chapter 2, namely,
rn < n,
Indeed, by the help of this formula we can verify that, in general,
where the symbol [rn/2] stands for the greatest integer I4 2 , is a solution of the Bessel equation x’y”
+ xy’ + { x 2 - (rn + 1 ) ’ ) ~ = 0.
(2)
Let us now consider the confluent hypergeometric equation xy”
+ (b - x)y’
-
a y = 0.
When a and b are positive integers and b 2 a solution
+
da- 1
y ( x ) = cdx“-
This is also easily verified. A solution of the differential equation xy”
+ ay’ + b x y = 0,
(3) 1, it has the distributional
b-a-
1
(4)
242
9.
APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS
where b # 0 and a is a positive even integer, is Y ( X )= C(d2/dx2
+ h)‘”-2)’26(~).
(6)
Similarly, the distributional solution of the equation XY”
+ ( X + u + b)y’ + UY = 0,
(7)
where a and h are positive integers, is
These concepts can be extended to higher-order differential equations. For instance, the third-order differential equation xy”’ - ( x
+ p)y” - ( x
-
p - 1)y’ + ( x
-
I)y
=
0,
(9)
has a distributional solution when p is a negative odd integer, p I -3. This solution is Y(X) = C ( d 2 / d x 2 - 1 ) - ‘ P + 3 ) ’ 2 6 ( ~ ) .
(10)
Distributional solutions to ordinary differential equations may also be found by deviations of arguments, as in the system
where the A , ( x ) are matrices and l j are numbers lying between - 1 and I, while y is a column vector. Indeed, under certain assumptions it can be shown that equation (1 1) has a distributional solution concentrated at x = 0. Although the research in this direction is not sufficiently developed, one paper [22] contains a rigorous discussion of these formulas. It also contains references to previous work in the field.
EXERCISES
1. Solve Example 1 of Section 9.8 by the method of Section 9.6. 2. Show that the solution of
243
EXERCISES
for G(x,
5) bounded as x -,0 and G( 1, 5) = 0, is
3. Solve the boundary value problem of Example 3 of Section 9.6 for the special case 0, f(x) = {xz171,
0 I x I 7112, 7112 s x < 71
4. Solve the boundary value problem of Example 3 of Section 9.6 with
.f(x) replaced by f'(x) on the right side of (9.6.48).
5. Solve Exercise 4 for the special value of the function . f ( x ) given in Exercise 3.
6. Derive (9.9.17), completing the algebraic details. 7. The governing differential equation that describes the response of a viscoelastic incomplete ring to a suddenly applied force at its free end is of the form
where a, b, a,Po, pl, BZ are appropriate constants relating to the problem. Solve this differential equation with a = b = 01 = Po = = Bz = 1.
8. Show that the infinite series
formally satisfies the first-order ordinary differential equation X'
dyfdx - 2y = 0.
It is interesting to observe that x ential equation.
=
0 is an essential singularity for this differ-
I0 Applications t o Partial Differential Equations
10.1. INTRODUCTION
Recall from Chapter 2 that the differential operator L of order p in n independent variables x l , x2,. . . x,, is where the coefficients ak have partial derivatives of all orders. Its formal adjoint operator L* is defined as L*u =
1 (-
Iklsp
l)kDk(akv).
(2)
Here u and u are functions having derivatives of order p in R , . An interesting relation involving the operators L and L* is ULU- uL*v = div J(u, u), (3) where J is a vectorial bilinear form in u, u, and derivatives of u and u of order p - 1 or less. We prove this result for p = 2. The generalization to higher dimensions offers no difficulty. For p = 2 we have
(4) 244
245
10.1. INTRODUCTION
"1
where we assume that A i j = A j i , so that
axiaxj - ax, [(.a,,)
~ A i j aZu
-
~
Also
axi
-
["
axi
(UAij)
"1
axj
au a a axj = nxj ( ~ B j u-) u ax, (0B.).
uB.-
J
Thus
=
where J
=
uL*u
+ div J ( u , u),
(J1,J 2 , . . . J , ) and
and we have proved (3). Finally, we apply the divergence theorem and obtain JiuLu - uL*u) dx
=
J li
+
J dS,
(7)
S
where R is a bounded region in R , with boundary S whose exterior unit normal is denoted li = (n1, n 2 , . . . n,). Example 1 . For the Laplace operator
it follows that L*
=
L, so it is a self-adjoint operator. Then ( 5 ) becomes
( u V2u - id V2u)= div(u grad
id -
u grad
v),
so that (7) yields Green's second identity, JR(u V2u - u V2u)dx
=
Js( :: :3
u - - u - dS.
(9)
246
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Example 2. For the heat operator L
= apt
- v2,
we have L*
and U L U - uL*u
=
=
a
-(uu) at
Next, we set J =
-a/at
+1 a i= axi n
(
-
1
au
VU,-U---V,
- v2
[
u-
a0
axi
a0
...,-u-
ax,
ax,
- 0-
ax, a"
U-,
.
)
and let P, denote a unit vector in the t direction, so that we can write J as J
=
e,(uu)
+ u grad,
v - u grad, u,
where the subscript x indicates that the differential operator is with respect to x. Accordingly, JR[uLu - uL*u] dx dt
=
s,
[P,uu
+ u grad, v - u grad, u] . A dS.
(12)
Example 3. The wave operator L=
02
(13)
= a 2 1 a t 2 - v2,
where O 2 is called the DAlembertian operator, is clearly self-adjoint. In this case (7) becomes
= Js [PI(.
-u
a)
+ u grad,
I
u - u grad, u . fi dS.
(14)
10.2. CLASSICAL A N D GENERALIZED SOLUTIONS
The classification about the classical and the generalized solutions made in the previous chapter is valid in the present case as well. However, there is one important difference. In the case of the ordinary differential equation Lu = 0 with constant coefficients, every solution is the classical solution. The matter
10.2.
247
CLASSICAL AND GENERALIZED SOLUTIONS
is quite different for partial differential equations. The solutions in a similar situation may now include generalized functions. For instance, au/ax, = 0, in R 2 , has among its solutions the generalized function b(x,),
J-
W
(8(x2),
4(Xl,
x2)) =
4(x19 0 ) dx,. W
A function u(x) is called a classical solution of the partial differential equation
Lu = s,
(1)
where s(x) is a given smooth function in R,, if u(x) has continuous derivatives of order p and satisfies (1) identically. If s(x) is a distribution, then u(x) is said to be a generalized solution of (1) if (Lu, 4 )
=
(s,
4)-
(2)
that is, if u(x) satisfies (1) in the distributional sense. In view of relation (10.1.3) we have (3) (4L * 4 ) = (s, 4). Accordingly, we have only to verify that, for each 4 E D, the action of u on L*$ is the same as the action of s on 4. In case we are interested in the generalized solution of (1) in an open region R and not the entire R,, then we seek a distribution u that satisfies (3) for all 4 E D(R), that is, for all test functions with support in R. For the homogeneous part of (l), namely, Lu = 0,
(4)
we have the following result.
Theorem. (a) If u is a classical solution of (4)in R, then it is also a generalized solution in R. (b) Let u be a function with continuous derivatives of order p that is a generalized solution of (4) in R. Then u is a classical solution in R . Proof: (a) Let 4 be an arbitrary test function with support in the interior of R. Then, in view of (10.1.7), (u,
L*4) = JRuL*4 dx = J / L u dx
+
1
n. J dS =
Js
n . J dS.
However, 4 vanishes on S , so that J = 0, and we have ( u , L*$) = 0, so u is a generalized solution. (b) As in (a) we find that (u, L*4) = f R 4Lu dx. Since we have assumed that (u, L*4) = 0, we have f R 4Lu dx = 0. Now, if Lu # 0 in the
248
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
neighborhood of any point x o in R , say, it is positive there. Then we chose 4 such that 4 > 0 in a sphere with center at xo and zero elsewhere, so that SR 4 L u dx # 0, which is a contradiction. Of course, as pointed out earlier; we have generalized solutions that are not classical solutions.
10.3. FUNDAMENTAL SOLUTIONS
We shall mainly be interested in the equations wherein the coefficients are constants. The theory of partial differential equations stems from the intensive and extensive study of a few basic equations of mathematical physics, and the coefficients in all of these are constants. Such equations arise in the study of gravitation, electromagnetism, perfect fluids, elasticity, heat transfer, and quantum mechanics. Of great importance in the study of these equations are their fundamental solutions. Recall that a fundamental solution E(x) is a generalized function that satisfies the equation LE(x) = S(x). (1) This solution is not unique, because we can add to it any solution of the homogeneous equation. This understood, in the sequel we shall select the fundamental solution from among the particular solutions according to its behavior at infinity or other appropriate criteria. The next few sections are devoted to the fundamental solutions of various important partial differential equations of physics, mechanics, and engineering. In the study of these solutions the following interesting concept is helpful. It is called Hadamard’s method of descent. Given the solution of a partial differential equation in R,+ 1, we can find its solution in R , or in a still lower dimension. In doing so, we descend from the higher-dimensional problem to a lower-dimensional one. For instance, the solution of the initial value problem for the wave equation in two dimensions can be obtained from that in three dimensions. Specifically, let us consider a linear partial differential equation
in the space R , + of variables ( x , x,+ 1), where x
j = 1, . . . , n,fED’(R,),
=
( x l , . . . , x,), D is a / a x j ,
and Lq(D)are partial differential operators involving the variables x , , . . . x , .
249
10.3. FUNDAMENTAL SOLUTIONS
When we say that the generalized function g E D’(R,+ allows the continuation over functions of the form +(x)l(x,,+ ]) where E D(R,), we mean the following: Given an arbitrary sequence of functions {+,,,(x,,+])}, m = 1,2, . . . , belonging to D ( R , ) , where R , is the space with variables x,+ and converging to 1 in R , [is., 1(x,+ ])I, then there is the limit
+
lim (9, +(X)+m(xn+l)>= (g,+(X)l(xn+1)> = (go,+>,
m- m
+ € D ( R n ) . (4)
In view of the completeness of D‘, we find that go E D’(R,). Specifically, for g(x) such that g(x) = .f(x) 0 d(x,+ 1), the inhomogeneous term in (2), we have (go,+>
=
lim MX), +(x)+m(xn+1))
m-m
= lim (f(x) m-
=
m
0 d(xn + 1 1 9 +(x)+m(xn + I )>
lim (f(x>, d(xn + 1)+(x)+m(xn + 1 )>
m-m
Accordingly, the method of descent can be stated as follows: If the solution u E D’(R,+ of (1) allows the continuation (4),then the distribution uo E D’(R,) is the solution of the equation L O ( W U 0 = !(XI.
(6)
For instance, if the locally integrable function E(x, t) is the fundamental solution of the operator L(D, dldt), then the distribution
is the fundamental solution of the operator L o . Indeed, in view ofthe Lebesgue theorem on the passage of the limit under the integral sign, we have lim (E(x, t ) , 4(x)+m(t)> = lim
m-m
m- u
I
E(x, r ) + ( x ) + m ( t )
dx dt
250
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
where E , is defined in (7) and 4 E D. Moreover, this limit does not depend on the sequence {+Jr)}. Hence E o ( x ) is the fundamental solution of the operator Lo, as required.
10.4. THE CAUCHY-RIEMANN
OPERATOR
+
For two-dimensional space we can use the complex variable z = x iy. Then Z = x - iy, d z = d x + i d y , and the Cauchy-Riemann operator is
a
=
2( z.+) : i i
a
Accordingly, we proceed to solve the partial differential equation
to derive the value of the fundamental solution E(x, y ) . We take the Fourier transform of ( 2 ) with respect to y, dE(x, u)/dx
+ uE(x, u ) = 6 ( x ) ,
(3) where we have used the notation of Chapter 6. We have studied this equation in Section 9.5. All its fundamental solutions are given by (9.5.5), and in the present case the required solution is E ( x , 11)
=
[H(x)
+ C(u)]e-"x.
(4)
Now we encounter a difficulty. The function e-ux is not tempered when x > 0 and ti + - w or when x < 0 and u + w.We can remedy this by defining C(u) as follows: C(u) = - H ( u ) =
Then (4) becomes E(x, u ) =
oir
l7
- H ( - x ) e - ux, H(x)e- ux,
11 l'
'
< 0. O7
11 11
and we have a tempered distribution. Its inverse is
> 0, < 0,
10.5.
251
THE TRANSPORT OPERATOR
From ( I ) , (2), and (7) we find that l/nz is the fundamental solution of the Cauchy-Riemann operator 8/85.
10.5. THE TRANSPORT OPERATOR
We now study the transport equation 1 8E(x, t )
-~
at
+ (u.grad, E ) + ctE = 6 ( x , r),
1111
=
1,
where E(x, t ) is the velocity distribution function, u is the velocity, and v and ct are some suitable parameters. Applying the Fourier transform to this equation with respect to x , we obtain
Its solution is E(u, t )
=
vH(r) exp{[iu. u - ctlvt),
(3)
which is tempered. Accordingly, we can apply the inverse Fourier transform, obtaining E(x, t )
=
'
F; [&u, t)]
=
vH(r)e-""F' [exp((iu . u ) v t ) ] .
Because from (6.4.12), F-'[exp((iu. u ) v t ) ] = 6 ( x - vtu), we find that the fundamental solution of the transport operator is E(x, t ) = vH(t)e-*"'6(x - vtu).
(4)
Here we encounter our first opportunity to use the method of descent, explained in Section 10.3. We determine the fundamental solutions Eo(x, t ) of the steady state operator u . grad E o ( x )
We have ( ~ ( xt ), , 0 ( x ) l ( t > >= v
J
+ crEo(x) = 6 ( x ) .
W
e-*'*'(b(x - vru), 4(x)) di
0
=
=
v JOme-*.'+(vtu) dt
($
-
=
i), 0).
(5)
252
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Comparing this with (10.3.8), we obtain ~ ~ ( =x ( e~- a ')x l / ( x ( 2 ) d ( u- x / l x l ) .
(6)
10.6. THE LAPLACE OPERATOR
We have already derived the fundamental solution of the Laplace operator in different ways (in Chapters 4 and 6). Let us rederive and examine this solution starting with the n-dimensional Laplace operator
When .f(x) depends only on r = 1x1 = (x: + X $ + . . . + x,2)'I2, then the Laplacian of this function takes a rather simple form,
from which it follows that if r # 0, then r 2 - " is a solution of V 2 f = 0. Since the function r 2 - " is a locally integrable function in the neighborhood of x = 0, we can define a distribution r 2 - " through the relation
Since the differential operator V2 is self-adjoint, we have
=
=
= lim E+O
s
s
r2-"V2+ dx
r 2 - " H ( r - c)V24 dx.
(4)
When we apply Green's second identity to the region R
=
{ X ER , - (1x1 < E ) } ,
we obtain
where S is the surface of the sphere of radius F with the center at the origin. We have seen in Section 3.3 that the surface area of S is (27~"'~t?-')/r(n/2).
253
10.6. THE LAPLACE OPERATOR
Accordingly, the right side of (5) has the value
where little o is the usual symbol for the order of magnitude of a quantity. Letting E -+0, we find from (4) and (5) that
which agrees with (4.4.73), found by a different method. Recalling the definition of a fundamental solution for the differential operator given in (10.3.1),we find from ( 6 ) that if n > 2, E(r) =
7c-
"12r(n/2)r2-'1 2(n - 2) (n - 2)S,(l)rn-2'
(7)
is the fundamental solution of the Laplace operator -V2. In particular, (8)
E(r) = 1/4m
is the fundamental solution in R , . The case n = 2 requires special consideration. In this case we replace r2-, by In r. The result, which corresponds to (5), as explained in Example 2 of Section 5.5, is /Jn
I'
V 2 4dx
= -
which yields
s,[
;: t. ]
In r - - - 4 dS
-V2
-
In
Thus E(r)
=
1
I'
=
=
-27c4(0)
+ o(E),
d(x).
(1/27c) In r
(9)
(10)
is the fundamental solution for this case. Solutions (8) and (10) are the Newtonian and logarithmic potentials, respectively. Alternatively, we can derive the fundamental solution (7) by taking the Fourier transform of both sides of the equation - V2E(x)=
h(x),
x
=
(X1,
~
2
. ., . , x,) E R,.
The result, in the notation of Chapter 6, is
lu128(u) = 1
or
B(u) = l/lu12. Taking the inverse transform and using (6.4.65), we recover (7).
(1 1)
254
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Single- and Double-Layer Potentials
We now derive the formulas for the discontinuities across a surface element of the second-order partial derivatives of a harmonic function due to single- and double-layer potentials. These formulas follow readily from the analysis of Sections 5.5 and 5.6. Let F(x) be a singular function with respect to the surface S. The Laplacian of F takes the form
+ ( B - 2QA)6(S) + A6’(S) = V 2 F + ( B - 2QA)b(S) + 2AQ6(S) + dn6(S) = V 2 F + B6(S) + Adn6(S).
V 2 F = V2F
If F is harmonic in the complement of S , then
v 2 F = B6(S)+ Ad,6(S). When F vanishes at infinity, ( 1 3 ) yields by convolution
where E ( x ) is the fundamental solution already derived, namely, for n
=
2,
E(x) =
for n 2 3. If we let F E R , such that
V2F = 0
outside S ,
CF1 = 0, [dF/dn]
=
B.
Then F is the single-layer potential. From (14) we find that the solution is F(x) = -
s,
B(<)E(x - 5 ) dS,.
(19)
Furthermore, from the analysis of Sections 5.5 and 5.6 we find that [c?F/8xi] = Bni, 0 = [ V 2 F ] = C - 2QB,
(20)
255
10.6. THE LAPLACE OPERATOR
or
[d2F/dn2]= 2 R B ;
[&]
= Cn,nj
(21)
6B 6B 6n. +nj + n, + B 2 6Xi 6Xj 6Xj -
= (2nn,nj
ax:
=
+ $)B + 6B nj + 6 6B ni;
(,,,i + 3).+ 6Xi
6B 2 -n,, 6Xi
i not summed;
I f F E E satisfies the system outside X,
V 2 F = 0,
E] CFI
=
A,
=
0,
then (14) reduces to
and is the double-layer potential. The required jump relations are
0 = [V2F]= C
+ V2A,
or
[d2F/dn2]= - V 2 A ; 6A
6Xi 6Xj -
a2A dXi6Xj
pjkni
6xk
6A V2An,nj - pjkni-; 6xk
6A V2An? - p n. -, ik ’ 6xk
i not summed;
256
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Poisson's Integral Formula
We can obtain the well-known Poisson's integral formula from our analysis in a very simple manner. Our aim is to solve the Dirichlet problem V2u(r,8, cp)
=
0,
r < a,
(34)
u(a, 8, cp) = f ( 8 , cp),
(35) where (r, 8, cp) are spherical polar coordinates and r = a is the sphere on which the boundary value,f(O, cp) is prescribed. In order to use our formula, we must introduce the function that has a jump discontinuity across the sphere. This is achieved by defining the function C(r, 8, cp):
so that
-
in the entire space.
V2C = 0
Accordingly,
(37)
Substituting these values in (14)leads to the integral
x a' sin 0' d0' dcp'.
Since x
=
(r, 0, Q),
5
=
(r', O',
Ix -
where
Q'),
we have
51 = (r2 + rI2 - 2rr' cos O ) ' I 2 ,
cos 8 = cos CI cos 0' + sin CI sin 0' COS(Q - v').
Substituting this value in (39), we obtain
{'
u(r, 8, cp) = - x
LI
x a'
1
(r2 + r" - 2rr' cos @'I2
-
2(r - r cos 8) (r2 + r" - 2rr' cos 8)3'2
sin 0' do' dcp'
1 f ( 0 , cp) sin 0' d0' dq' - r 2 ) Jon Jo (r2 + u2 - 2ar cos 0)3/2' 4n which is Poisson's integral formula. =
- a(a2
(40)
10.7.
257
THE HEAT OPERATOR
10.7. THE HEAT OPERATOR
Let us now study the heat operator qat - K v ~ ,
where t is time and K is a positive constant. For the sake of simplicity let us take K = 1; subsequently we shall restore it. Accordingly, our aim is to solve the differential equation ~ E ( xt)/& , - V 2 E ( x ,t ) = d(x)h(t), t 2 0, x E R,. (2) By taking the Fourier transform of this equation with respect to x, we obtain
dE(u, t ) / d t +
IIII'E(U,t ) =
d(t).
(3)
Its solution is E(u, t ) = H ( t ) exp( - t I 11 12) .
(4) This function decays rapidly at a and is therefore a tempered distribution. Its inverse transform is
E(x, t)
=
exp(-ix.u - tluI2)dir
~
where we have used the relation
J-WexP(- tY2) dY W
=
J./t.
Let us observe in passing that for t 2 0 / E ( x , t ) dx
=
(2&)-"
Jexp( -
!$)
dx
For the heat equation (2) the concept of the causal fundamental solution C(x, t) is of interest. It is defined as the particular solution of (2) that vanishes identically for t < 0. That is,
aclat
-
vzc = s(x)s(t), c = 0,
t < 0.
(8)
258
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Accordingly, C ( x , t ) = (2fi)-"H(t) exp( - Ix I2/4t).
(9)
It is a C" function except at t = 0. Furthermore, it is invariant under the space rotations because it depends only on lxI2. This solution is called causal (or nonanticipatiue) because no output signal can ever precede the input signal that gives rise to it. Furthermore, the causal fundamental solution C ( x , t ) coincides for t > 0 with the solution of the initial value problem
aulat - vzU= 0, u(x, O+) Indeed, if we define C ( x , t)
=
=
t > 0,
(10)
6(x).
(1 1)
H(t)u(x, t), then V 2 C = H ( t ) V2u,and
aclat = ~
+ u(x, t ) dHldt + u(x, o +)qq + s(x)s(t), - V2u] + h(x)d(t) = b(x)d(t)asrequired.
( t au/at )
~ ( tau/at ) = ~ ( t au/at ) =
sothataC/at - V z C = H(t)[au/at Let us now restore the constant K in (2) and obtain the results corresponding to those derived above. For instance, (5) becomes E(x, t)
=
exp( - I x I2/4.-t).
(2+)-"
(12)
To verify this solution, we observe that for all 4 E D (aE/at -
K
V = E ,4 ) = - (E,
=
+ K v24)
a4/at
-1im E - 0 O E ( x , t)(g +KV24)dxdt.
(13)
Next, we integrate ( 1 3) by parts, using aE/at - K V 2 E = 0, t > 0, and obtain
=
=
[
lim s E ( x , E)$(x, E ) dx &+O
lim e-0
s
E(x, E ) ~ ( xE ), dx
+
//g s cu
-K
V2E
E
=
lim 6-0
E(x, E ) ~ ( 0) x , dx,
(14)
259
10.8. THE SCHRODINGER 0PERAT OR
because, in view of (7), -
4 ( ~O)],
dx
I J IM
I:
E(x, E ) d x
=
ME,
where M stands for the maximum of b(x, E ) - +(x, 0). Moreover,
I
I E ( x , t ) [ 4 ( x ) - 4(0)] d x IM ( 2 6 ) - " /exp( -
= MI$
~ o m y " r - yd' y
=
T)
1x1 d x
C$,
where S,(1) is the area of the n-dimensional sphere of radius unity and M' and C are appropriate constants. Thus it follows that when t + O + we have
( E , 4)
=
J E ( x , t ) 4 ( x ) d x = 4(0)
= 4(0) = (634).
I
E(x, t ) d x
+
I&,
t)C4(x) -
4(0)1 d x (15)
Combining (14) and (15), we have the required verification. Let us use the method of descent to obtain the fundamental solution for the Laplace operator for n 2 3 from the fundamental solution of the heat equation. From ( 5 ) we have
which is precisely the value of the fundamental solution for the Laplacian.
10.8. THE SCHRODINGER OPERATOR
We now solve the partial differential equation
260
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
This is the free Schrodinger equation. Additional terms are included if there is a potential or other physical interactions are present. There is also a constant coefficient to the V2 term, just as in the heat equation. Taking the Fourier transform of (1) with respect to x, we find 1 d&u, t )
_ _ _
dt
i
+
lUI28(U,
t ) = d(t),
whose solution is &u, t ) = iH(t) exp( - i t 1 u 1)'.
To find its inverse Fourier transform, we observe that B(u, t , E )
=
iH(r) exp(-(E
+ it)lu12),
E
> 0,
converges in the distributional sense to &u, t ) as E -, 0. Then the inverse Fourier transform of B(r, E ) converges to that of &t). Accordingly, we proceed as we did in the derivation of (10.7.5) and find that
is..
~ ( xt ,, E ) = (2n)-"
exp(-ix. u - ( E
The limit of this quantity as E E(x, t )
-, 0 such
that
+ it)lu12) du
E
1
> 0, t > 0, is
( ~ T C ) - " ~ - "exp(~ ~ J " Ix12/4it),
(4) where J is the Fresnel integral f Y m e-'Y2 d y = (1 - i)n/,,h = neCini4. Thus =
E(x, t ) = exp[ - i(n - 2)n/4](4nt)-"/z exp( - I x I2/4it),
(5)
and the corresponding causal fundamental solution C(x, t ) is C(x, t ) = H ( t ) exp[ - i(n - 2)n/4](4nt)-"/2 exp( - 1xI2/4it).
(6)
It is easier to solve the initial value problem for the wave function $(x, t ) ,
$(& 0) = .f'(x).
The Fourier transforms of these equations are
261
10.9. THE HELMHOLTZ OPERATOR
respectively. The solution of (9) and (10) is
$@,t ) = e - i r l u l z -f (u). Taking the inverse transform, we obtain
It is interesting to observe from (1 1) that I $(u, t ) I Thus
I
= /(u)
I is independent oft.
is also independent o f t . Accordingly, we find that if the wave function is normalized at t = 0, it remains normalized.
10.9. THE HELMHOLTZ OPERATOR
Our contention is that the function [23]
~ ( x= ) eiklx1/4nI x 1, where k E (0, co), satisfies the Helmholtz equation in R 3 ,
-(V2 Indeed, in view of the relations
and
we have
+ k2)E(x) = 6 ( ~ ) .
262
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
where we have used the summation convention. Thus
However,
a2
1
axj a x j l x l
so that (3) reduces to
-
1 v 2 1x1 = -4n6(x), -
-(v2 + k2)eik1x1/4nIxl= d(x).
(4)
Similarly, we can prove that the complex conjugate ik1~1/4~ 1 1 (5) also satisfies the Helmholtz equation. Setting k = irn in (2), we find that
E
=
-
-(V2 = S(x), so that the fundamental solution of the operator -(V2
E
=
(6) - m 2 ) is
e-mlxl/4nIsl.
(7)
To extend these results to the n-dimensional case, we appeal to (6.4.73),
which gives the Fourier transform of a radial function,(( consider the general differential equation [24]
[XI),
1x1 = r. We
-(Vi - l)’E(x) = 6(x), where I is a positive integer. We take its Fourier transform, obtaining
fr(lul) Next, we use the identity [25]
= -(-l)‘/(l
+
lu12)‘.
(9)
263
10.10. THE WAVE OPERATOR
where K is the modified Bessel function, a > 0, and - 1 < Re v < 2 Re p - i.Hence, from (8), (lo), and (1 l), and considering the inverse Fourier transform, we get the fundamental solution for the operator -(Vi - 1)' in S:
From this relation we readily find that the fundamental solution for the operator -(V; - m2)' is
Similarly, the fundamental solution of the operator -( ff setting m = -ik in (13):
+ k2)' follows by
Because K,( - iz) = $ineinv12Hy(1)(z),the preceding formula becomes
(14) When I
=
1, this reduces to
For n = 3, we recover (1) through the relation H\+i(z) = -i(+n~)-"~e~'.A direct derivation of the formula for n = 3 is outlined in Exercise 4.
10.10. THE WAVE OPERATOR
To find the fundamental solution for the wave operator we solve the wave equation 02E(X,
t ) = -(VZE(x, t ) ) = 6(x)d(f),
d2E(x, t ) at2
X E R,,
=
6(x, t )
t~ R 1 ,
264
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
where O 2 is the DAlembertian operator. We have taken the wave speed c to be unity and the source time and source point to be t = 0 and x = 0, respectively, to simplify the algebra. Subsequently, we shall reintroduce c and take the source time and point to be T and y, respectively. Let us understand some of the basic concepts about this operator by first considering the case n = 1, that is, a 2 E / d t 2 - a 2 E / a X 2 = d(x)d(t),
(2)
x , t~ R , .
This equation takes a simple form if we introduce the variables (s, y ) : s
=
t
-
x,
y =t
+ x,
or
x = f(y - s),
t = +(y
+ s).
(3)
Then a test function 4 ( x , y ) E D becomes &s, y ) such that
and
Because
it follows that the value E(s, y ) of the distribution E(x, t ) in (s, y ) coordinates is E(s, y ) = s), +(y - s)]. Next we take a new test function $(x, t ) and let 4 = (a’/at’ - a2/ax2)$. Then, in view of (2) we have
+
W, 0) = $(o, 0) = ( $ ( x ,
t), d ( x ) d ( t ) ) = ($,
0’~ =) ( E , 02$>
Accordingly, E(s, y ) satisfies the differential equation 4 a2E(s, y ) / a s a y
=
~(~)d(~).
(4)
Its solution is found by taking the appropriate linear combinations of
P(s,y)
=
iH(s)H(y),
E3’(s,y ) = - iH(- s ) H ( y ) ,
E‘”(s, y )
= - i H ( s ) H ( -y ) ,
E(4’(s,y )
=
i H ( -s)H(
-
y).
(5)
265
10.10. THE WAVE OPERATOR
t
0 Fig. 10.1. The support of the fundamental solution to the wave operator in R , .
Finally, we can go back to ( x , t ) coordinates and evaluate E(x, t ) by using the relation E(x, t )
=
2E(t
-
x, t
+ x).
(6)
Thus
+ x), E‘:’(x, t ) = $H( - t + x ) H ( t + x), E\”(x, t ) = + H ( f - x)H(r
E‘:’(x, t )
= -f H ( t -
E\4’(X, t ) = fH(- t
x)H( - t - x ) ,
+ x ) H ( -t
-
x), (7)
where the subscript 1 indicates that we are in R , . Because E\”(x, t )
=
iH(t
-
x)H(t
+ X ) = f H ( t ) H ( t 2 - x’)
= +H(t -
Ix~),
(8)
+
its support is the sector t x 2 0, t - x 2 0 in the ( x , r ) plane (see Fig. 10.1). This sector is called the forward light cone. Recall from Section 2.8 that the singular support of a distribution is the smallest set outside of which it is a C“ function. Now, the distribution E ( x , t ) is equal to $ in the interior of the forward light cone and zero outside it. Thus, the singular support of E , ( x , t ) is precisely the boundary of this cone, that is, the union of two rays x+t=o,
x-t=o,
120.
(9)
The other three distributions in (7) have similar interpretations. From a physical point of view, we find that an initial impulse at the origin creates a zone of displacement E ( x , t ) = $ whose edges propagate along the lines x f t = 0. These lines are called the characteristic lines.
266
10. APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Let us now return to the case R , [26]. Taking the Fourier transform of
(1) with respect to x, we obtain
d2E(U, t ) / d f 2+ lU128(U, t )
=
6(t),
whose solution, which follows from (9.6.15), is E(u, t ) = ~ ( tsin(lult)/lul. )
Accordingly, the solution of (10) with support in t 2 0 is
E+(u, t ) = ~
( tsin(lult)/lul, )
and that for t I 0 is E - ( u , t ) = - ~ ( - t ) sin(lult)/lul.
The inverse Fourier transform of (1 1) follows from (6.4.79): E(x, t )
=
’
H ( t ) F - sin( 111 I t)/ I u I
=
H(r)( 1/4nr)6(S),
where 6(S) is single layer on the spherical surface of radius t . Thus,
where we have used the relation 6(S) = 2R6(R2 - 1x1’) when the radius of the sphere S is R , or
where we have used the fact that 6(t 1x1 = r, we have
+ 1x1) = 0 for t < 0. Thus, setting
E(x, t ) = H(r)G(t - r)/4nr.
(15)
From this relation it follows that the support of the fundamental solution of the wave equation for n = 3 has as support the surface of the forward light cone with vertex at the source point, namely, t 2 - 1x1’ = 0, t 2 0, (see Fig. 10.2). The upper part is called the,forward sheet and extends to the future ( t > 0). The lower sheet is called the backward or retrograde sheet and extends to the past ( t < 0). Next, we derive the fundamental solution for the case n = 2, using the method of descent, that is, E2(x,, x2. t )
=
OD
+ x:) + x,} 4nC-x: + xi) + x,]
H(r)G[t - {(x:
J-OD
1
2 1/2
2 1/2
dx3
267
10.10. THE WAVE OPERATOR
Fig. 10.2. The support of the fvndamental solution t o the wave equation in R , .
We set x: = v2 - (x:
+ x:),
so that
dx, = u dv/x, = u d v / [ v 2 - (x:
+ x:)]~’~,
in (16). We then have
This, of course, can be derived directly by the Fourier transform technique (see Exercise 5). From (17) we can recover the corresponding value for n = 1 by a similar approach. Indeed,
which agrees with (8) for r > 0.
268
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Finally, we introduce the wave speed c and take the source point and the source time to be y and T , respectively, so that the differential equation becomes d2E(x, t)/dt2 - c2 V2E = 6(x - y)&t - T);
(19)
the corresponding values of E3, E 2 , and El are
10.11. THE INHOMOGENEOUS WAVE EQUATION
We now present two forms of the solutions of the inhomogeneous wave equation in three dimensions and some related results [13, 27, 281, U2+(x, t)
=
a2d4x3 - c2 V’$(X, f,
at2
t) =f(x, t).
In view of the fundamental solution (10.10.20), we obtain by convolution
where 9=
- t
+ R/c,
J x- yI
=
R.
To obtain thejirstform, we substitute
in (2) and use the sifting property, so that (2) becomes
269
10.11. THE INHOMOGENEOUS WAVE EQUATION
Fig. 10.3. Surface area element of the sphere g
When g = 0, we have pot ent ial,
T =t -
=
0.
R/c, so (4) yields the well-known retarded
The secondjorm follows from observing that dy =
dY,
dY2
d9
I adaY 3 I
’
where dg/dy3 is evaluated for T and ( y l , y2) fixed. Thus
where dR is the element of surface area of the sphere g = 0 with center at x and radius r = c(t - 5 ) (see Fig. 10.3) and we have used
while n3 is the third component of the unit normal h on the surface of the sphere g = 0. Combining (6)-(S), we have dy = c dR dg,
so that (2) becomes
(9)
270
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
The Two-Dimensional Case Now we use the method of descent and obtain the solution of the twodimensional inhomogeneous wave equation
0;4 where
(11)
= S ( x , , x2, T),
4 ( x l ,x 2 ,t). This method applied to (10) yields
where from the Fig. 10.4 we find that
cos 0
dy2 do = dy1 cos 8 ’
= C C 2 ( t - TI2 -
(x1
~
Since the sphere g
-
1
2 112
Y J 2 - (x2 - Y 2 )
C(t - 5)
=
where, as before, R 2
0 is made up of two hemispheres, ( 1 2) becomes
=
Ix - yI2 = ( x l - y,)’
+ ( x 2 - y,)’.
An Initial Value Problem for the Wave Equation We now use the information gathered above to solve the following initial value problem in three dimensions [27] : 0 2 4 ( x , t ) = f ( x ,t ) ,
t 2 0.
(14)
4 ( X ? 0) = 9o(x),
(15)
W x , o)lat = S l ( X ) .
(16)
Fig. 10.4
10.11.
27 1
THE INHOMOGENEOUS WAVE EQUATION
Let us solve for 4(x, t ) H ( t ) instead. Distributional differentiation yields n2CH(t)4(x701 = H(t)f(x, t)
1
1
+ c" g,(x)W) + c' go(x)S'(t).
(17)
This problem can be split into three parts 4 = 41 + 42 + 43,where these functions are the solutions of the following equations:
U'CH4,I = N(t)f(x, t ) ,
O"H421
1
=7 C gl(x)w),
The solutions of these equations are obtained from the previous discussion :
where we have used 6(R/c - t ) = cd(R - ct). Now we observe that dy = R 2 dr dw, where w is the solid angle seen from y = 0. Then the previous relation becomes
where we have used the relation c2t2d o = dS, which is the element of surface on the sphere r = ct with center at x and Mt[gl(y)] is the mean of g,(y) on this sphere. Similarly,
H(t)&(x, t ) =
& fJq 6(r
-
r
+ R/c)G(r)dr dy
272
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
x i Fig. 10.5. The intersection of the source region and a sphere of radius ct
because the integrand is identical to that used in the evaluation of Combining these results, we obtain
42.
(18)
Example 1. Let us use (18) to find 4(x, t ) for ( x1 > a that satisfies the system 0 2 4 =
0,
(19)
The solution is spherically symmetric, so the observer can be placed on one of the axes. We have 4 = 0 for t < ( R - a)/c and t > ( R a)/c, whereas 4 # 0 for ( R - a)/c I t ( R + a)/c. Because R 2 a, the intersection of the sphere of radius ct and the source region can be approximated by a circle (see Fig. 10.5) whose area is n[a2 - ( R - ~ t ) ~so] ,
+
M,"4(x, 0)l
1
= __
4ne2t2 Now we apply (18), obtaining
{ "
1 a 4 ( R , t ) = 4C2 at
0
-(;
a'
n[a2 - ( R - ~ t ) =~ ]
-ct)2
]
for
R-u ~
c
otherwise.
-
(R
- ct)'
4e2t2
R+a
Its-
c
' (21)
10.11.
THE INHOMOGENEOUS WAVE EQUATION
273
Moving Sources
Example 2. A moving point source n = 3 In radiation problems in acoustics and electromagnetism we encounter moving sources. Consider a point source moving with velocity u through an infinite medium that is at rest, so that the volume source density is [29]
(22) where y = (yl, y,, y 3 ) and T is a scalar. Accordingly, we have to solve the inhomogeneous wave equation dy,
= q o ( 7 ) X ~- ~ 5 ) ~
a24/nr2 - c 2 v24 = q o ( r ) ~ ( y
(23)
In Section 10.10 we observed that the solution of the equation ( a 2 / a t 2- C Z V Z ) E ( X , t ; y, z) = S(x - y)6(t - z),
(24)
yl/c)/4nlx - yl.
(25)
is E(x, t ; y , T)
=
b(t - z - Ix
-
Thus by convolution we have
Next we use (3.1.1), namely,
where the zi are the simple zeros off(z). In the integral of (26) we have
so that df -
-
V2t
-u.x
+ 1,
dr CJX- UTI and we have to find the roots of the equation Ix -
Utl/C
+ z - r = 0.
(27)
This is a quadratic in 7 and has at most two roots T~ and 7 2 . Accordingly, (26) takes the form
274
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Let us denote t1 and r2 as,'t (vector)
respectively, and introduce the Mach number M
and the separation vectors
(29)
= V/C,
R ' = x - u T ,*
in (28). We then have
where cos0*
M.R*/lMIIR'I,
=
(32)
is the cosine of the angle between the vectors R* and M. Note that (27) takes the simple form t - R*/c.
(33) Example 3. A moving line source (n = 2) Let us now take the radiation field from an infinitely long line source moving perpendicular to its own axis with uniform velocity u through a fluid at rest, so that its strength can be expressed [30] q(T)6(y - ut), where t is time and y = (yl, y,, y 3 ) . Then the velocity potential 4(y, t) for this field satisfies the wave equation 't
a24/aT2
-
=
v24 = 4(T)6(y - 0s).
c2
(34)
In Section 10.10 we found that the solution of the equation a2/at2
-
C2 v2qx,
y ; t,
=
qX- r)s(t -
(35)
is given by (10.10.21), that is, E(x, y ; t , t) =
H(c(t -
t) - Ix -
y()
2 7 K [ C 2 ( f - T)2 - IX -
J'12]1'2'
Comparing (34) and (35), we find that the value of the velocity potential 4(x, t ) is given by the convolution integral
For the spherical case of a source moving along the x 1 axis with a constant speed V such that V < c, we take q(r) = qoe- i"r, where qo is a constant. Then (37) reduces to
10.11 .
275
THE INHOMOGENEOUS WAVE EQUATION
Now, introduce a new variable 52
=
[‘+
( 1 - M2)2 - Mt)’ (1 - M ’ ) ( X ~ / C > ’
=
+
(X,/C
where M
5: (x,M/c)- t 1 -M2
V / c is the Mach number. Then (38) takes the form
This formula takes an elegant form if we use the integral representation of Hankel function H f ) ( x ) :
Then (40) becomes
x
Hb‘)
o [ ( x ~- V t ) 2 + ( 1 - M 2 ) ~ : ] ” 2 ~ (l M2)
Example 4. Moving surface sources equation is [13]
In this case the inhomogeneous wave
02 4 ( x , t ) = 4 ( x 7 t ) W ) = 4(x, t ) 191’Ih(.f),
(43)
where .f = f ( x , t ) . The surface .f(x, t ) = 0 can expand and move. Again by convolution, we find from (10.10.20) that
t - R/c) for a function t j and the subscript ret where [t)(y; x , t],,, = I&), stands for retarded time. Let F ( y ; x , t ) = [,f(y, T ) ] ~ , , . Then if the surface C is described by F ( y ; x , t ) = 0, with x and t fixed, we find that
=
d y , tiy, dF - d y , d y , ciF 1 --dF dC la~/ay,l I ~ F / ~ ~ , I / I V-F I V F .I
(45)
276
10.. APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
Because F
= [.f(y,
T ) ] =~, f~( y ~ , T - R/c), we have
where M, = - (gf/at)/(c I V f I) = u,/c is the Mach number based on the local normal velocity u, = -(df/(?r)/ I V f I of the surfacef' = 0, and R i = (xi - yi)/R. Thus
=
+
[ IVf 12( 1
+ M,Z - 2M, cos @Ire,= [ IVf I2A2Irel,
(46)
where A2 = 1 M,Z - 2M, cos 0 and the quantity 0 is the angle between the normal tof' = 0 and the radial direction x - y. Combining (44)-(46), we finally obtain
10.12. THE KLEIN-GORDON
Let p 02
= ( t 2 - x: - x i -
= a 2 / a t 2 - (?2/aX:
-
OPERATOR
. . . - xi)112.Then ,
for the d'Alembert operator
. . - d2/dx;, we find that
;,
OZu(t, x l , . . . , x,) = +fP (P.
$)?
for a function u(x, t ) . The proof is straightforward and is left as an exercise (see Exercise 6). This formula is similar to (10.6.2) for the Laplacian operator, whose fundamental solution we found to be -(l/(H - 2)S,(l)T2), which depends only on r. It is therefore natural to look for the fundamental solution of the d'Alembert operator that is a function of p only. For this purpose we redefine p as follows: P = PO, x1, * . t2
=
{0
-
. 9
x,)
- ... - xi,
t >r
= (x:
otherwise.
+ x: + . . . + x,2)''*,
(2)
277
10.12. THE KLEIN-GORDON OPERATOR
(cf. Example 6 of Section 4.4). Let us first define the distribution pa, where A is a complex number. This presents no problem in the half-plane Re 2 > -2, since pi is locally integrable for these values of A. Indeed,
(PA,4)
=
J
p w t , x) d x dL
1>0
(3)
where 4 is a test function, defines an analytic generalized function for Re A > -2. Using analytic continuation, one can extend ( 3 ) to a meromorphic function in the whole complex plane. This is achieved with the help of (1). Indeed, because 02pI =
A(A
+
11
-
l)p"-2
is valid for Re A > -2, it follows that for all A and (02p2.4) =
A(A
(4)
4 E D we have
+ I2 - l ) ( p " - 2 , 4 ) .
(5)
By iteration of (5) we obtain more generally that
(6) which gives the values of p' in the strip -2k - 2 < A I - 2k in terms of the values of p' in Re A > - 2. From this relation it also follows that the singular points of pA are A = - 2 k , k = 1, 2, 3 , . . . , and A = -17 - 2k - 1, k = 0, I, 2,3,. . . . Thus for n even all the singular points are simple poles, and for n odd the points - 2 , -4, . . . , -!I + 1 are simple while the points - 1 1 - I, - n - 3, . . . are double poles. We now normalize the function p' by putting
thereby removing the singularities of p* and producing an entire generalized function. To prove this, we observe that the points 2 = -2, -4, . . . for A > --n - 1 are simple poles of p i ; then the residues at these points follow from (6):
278
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
10.
where H = H ( t - ( x : Accordingly,
+ . . . + x : ) " ~ ) = po
lim 4(A/2
x
A+
- 2k
is the Heaviside function.
+ I ) . .. (A/2 + k - l ) [ ( A + 2)pA]
+ k + 1)
( - I ) k 0 2_k_ H
X
2 k - ' ( k - l)!(n - 2 ) . . . ( n - 2 k )
Next suppose that n is even so that A = n - 1 is a simple pole of pi. In order to compute the residue at this point, we set x i = y o i , i = 1 , . . . ,n ; then ( p a , 4) =
I Ia
( t 2 - r2)"'4(t, r l , . . . , r n ) r n - l dt dr dQ,
t>r
-
f ( t 2 - r2)'/'rn- ' $ ( t , r ) dr dt =
0
where
t""(D(t,
A) dt,
0
Thus A=
Res ( p a , 4) = -n-
1
lim (D(0,A)= &O, 0) I- -n-
1
(1 - r2)"'r"-' d r
(10)
279
10.12. THE KLEIN-GORDON OPERATOR
Subsequently,
=
6(x).
(14)
With the help of (6), the analysis can be generalized to give
so that Z - 2 k = OZk6. (16) Finally, for odd n it follows from a similar calculation that the coefficient c'!'~ of the expansion of ( p A , 4) about --n - 1 - 2k ( k = 0, 1,2, . . .) is
so that (16) is valid for this case as well. Note that from ( 6 ) it follows that Furthermore, we can convolute 2, and 2, since their supports lie in the forward light cone. We then have
z, * z,
=
ZP+".
The distributions 2, are called the Riesz distributions. Example 1 . Let us consider the set D'+(T) formed by all the distributions with support in the forward light cone r. The convolutions of the members of D'+(T)are also in D'+(I-). Moreover, it can be shown that this convolution algebra has no zero divisors. Hence 2, has the unique inverse 2-, in D'+(T). Consider now the differential equation
my= g,
(18)
where the distributions,fand g vanish for t < 0. Then z - 2 k
*f= ( 0 2 k 6 )* f = 6 * O 2 y = 9,
so that f = Z2, * g is the unique distributional solution. Because the initial values can be added to the differential equation, it follows that such an initial value problem has a unique solution.
280
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
The Distribution ,,,ZZk Our aim in this section is to find the solution of the Klein-Gordon operator
U2+ m 2 . For this purpose we let m2 be any complex number and define
(19)
+
mZ-2k = [(02 m2)6]*k, (20) where *k in the exponent means that we have k-fold convolution. Since these distributions are in D'+(T),they have unique inverses m Z 2 k such that
=
p=o
P (%+@
)(-
1)Pm2PZ2p+2k.
From this discussion it follows that the iterated Klein-Gordon equation
(0' + m2)"'= g (23) has a unique solution for the distributionsfand y vanishing for t < 0, and the same is true for the initial value problem. Indeed, the solution is
.f = m Z 2 k * 9.
(24)
Finally, we observe that on substituting (7) in (22) we obtain
where J k - ( n +
l)i2
is the Bessel function of order k - ( n
+ 1)/2.
Example2. Let us find the fundamental solution E(x, t ) of the KleinGordon operator, i.e., the solution of the equation
(02-t m2)E(x,t ) = 6(x, r ) (26) in R4, where E ( x , r ) = 0 for t i0. From the foregoing analysis we know that the solution is E ( x , t) = m Z 2so , from (25) we have
281
EXERCISES
Because 2, comes
=
(1/47c) Res,,
-
p’ = (1/27c)d(t - ( x :
1 d(t - ( x : + x : + x 27c Example 3. Let us solve the equation E(x, t )
=
-
+ x : + x:)”’),
y-)m Jlo 47c P .
(0’ + m : ) ( U 2 + m:)E(x, t ) = d(x, t ) in R , , where E(x, t ) vanishes for r < 0.
(27) be(28)
(29)
We can write (29) as m , Z - 2* m z 2 - 2 E ( x ,t ) = 6 , so that
-
47c
(30)
P
Many results of Sections 10.10 and 10.11 can be deduced from the present formulas and are left as exercises for the reader.
EXERCISES 1. Derive (10.10.16) with the help of the Fourier transform.
2. Use the method of Section 10.6 and obtain the Green’s function for the half-space problem, that is, solve (a) the Dirichlet problem V’u = , f ( x ) , x 3 > 0, u ( x , , x 2 , 0) = g ( x , , x 2 ) ; (b) the Neumann problem V’u =,f(x), x3 > 0, ( w a x , ) ( x , , x 2 ; 0 ) = & I , x2). 3. Derive the Poisson integral formula for a circle by following the steps given in Section 10.6 in the derivation of the corresponding formula for a sphere.
4. Find the fundamental solution of the Helmholtz equation in three dimensions, -(V’
+ k 2 ) E ( x ) = &x),
by the following steps: (a) Take the Fourier transformof both sides of this equation and obtain - ( k 2 - lu12)8(lul) = 1. (b) Show that the solution of the latter equation is E(u) = 6 ( k 2 - I u 1)’
- Pf
1 (k’ - 1 ~ 1 ’ ) ‘
282
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
10.
(c) Expand the terms of this equation. (d) Use the relation - i ( x - a) = ~ > ~ ~ . ‘ i ( u ) . (e) Use the fact that the Fourier transform of the step function H ( x ) is I?(+x)
=
nrfi(x)
+ i Pf(l/x),
as well as Exercise 18 of Chapter 6 .
5. Derive the two-dimensional fundamental solution E(x, r ) given by
(10.10.1 7) directly by the Fourier transform method.
6. Derive (10.12.1). 7. Show that the fundamental solution of the dissipative wave equation
(
7 2 - a2 -
is
E(x, x’, t , t ’ )
a
nt
=
-
”’)
--
c2 a t 2
E ( x , x’, t , t ’ ) = d ( x - x’)S(t - t’)
c exp[ --fa2c2(t 4 x 1 ~- x ’ I
r’)]
x J 1 [ - f u 2 c [ l x- X’I - 2 ( t - t 1 )2 ] 1 / 2 3
x H[(c(t - t’) -
IX
-
x’I]
1
H ( t - t‘).
8. Show that the value of the fundamental solution E ( x , t ) with support in the region t 2 0 for the n-dimensional wave operator d2/(?t2 - V,Z = -D,Z is
when n is odd and E(x, t )
=
1
2nlrn/Z (:t:- - vf)n’2[t2-
~
-
Ix12)H(t)
when n is even. 9. By taking the Fourier transforms with respect to x , of the fundamental solutions as given in the Exercise 8, replacing u, (the Fourier transform of x,) by k and then changing n - 1 to 11, show that the fundamental solution of the operator -(Of + k 2 ) with support in the region t 2 0 is E(.u.t)= [2“(7-~)”’~(n/2)!k]-’(U,2 + k 2 ) ” ’ 2 [ H ( t ) H ( t 2- Ix12)sin(k(t2- I
x~~)~’~)],
283
EXERCISES
when
17
is even and
E(s. t )
=[2"+ld"-1'2((17
x
when
II
+ l)/2))!/;]-l(u: +
X2)ln+l'2
[ H ( t ) H ( r Z - Ix12)(r2- I.xlz)' Z]Jl(k(t2 - I . X ~ ~ ) 2I ) ,
is odd.
10. Prove that the fundamental solution ofthe biharmonic operator in the three-dimensional case with support in t 2 0,
u4= (d*/(V - v2)(s2/ot2 - a VZ),
is
1 1. Let P ( D ) = P(d/dr, d/dx d/dx,, . . . , ii/dx,,,) be the homogeneous hyperbolic differential operator of order i i i . Find the distributional differential equation that is equivalent to the Cauchy problem P(D)tI
t 20
= ,/;
aktI/iitk= gk(x1,. . . , xn),
;/ = 0, 1.
. . .,m
- 1,
t = 0.
12. Show that the fundamental solution E(x, r ) of the Stokes equations VE(X,2)
=
Vp
0,
= ~1
V2E(x, 2)
where x is a given vector, is E ( x , a ) = a/r
+ 87~pa6(~),
+ ( a . x>.x/r3,I' = 1x1.
13. For a homogeneous, linearly elastic, isotropic medium, the displacement equations of equilibrium arc p[( 1 - 2 v -
v v . u(x) + V2II(X)] + f(x) = 0,
where I I = ( I ! , , 1 i 2 , i f 3 )denotes the displacerncnt field, p is the shear modulus, itis the Poisson ratio of the material, and,f'isthe body force per unit of volume. Show that for,f'(x) = 16scp(1 - v)6(x)a,where a is a given vector, the solution is tI(X x ) =
(3 - 4v)a/r
+ ( a . x)x/I',
r = 1x1
14. The equations of motion for steady state elastodynamics are 1 v .I I l'[nl
v
I
+ VZII + n7211
+./'= 0,
284
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
10.
where m is a suitable parameter and the other quantities are as defined in Example 13. Show that for .f'(x) = 411pLS(x)c( the solution u is u(x, u ) =
r[
- imr
u
+ m21 v ( a .
- imrr
- imr
-
r
where r = Ix I and 'T = ( 1 - 2v)/2( I - v). Derive the result of Exercise 13 from this value of u(x) as m + 0. 15. Show that the Green's tensor that satisfies the vector wave equation
is
+ 3I c2tH(t)6(x). -
16. With the help of Example 15, derive the formal solution of the electromagnetic wave equation, 1 a2E aJ -~ +Vx(VxE)=---, at
c2 a t 2
where E is the electric field and J is the current density. This is achieved by first writing the left side of this equation in distributional derivatives. The initial conditions and boundary conditions thereby appear as sources on the left side of the equation. Convolving the sum of these sources with Gij then yields the required solution. The result is simplified when the initial and boundary conditions are specified explicitly. 17. Show that the solution of the heat equation when initially a mass M is distributed uniformly on the surface S of a sphere of radius r o , i.e., the solution of the initial value problem u, - V2u = 0, x E R , , u(x, 0) = Md(S)/S,(l)r:is
where (', =
for n odd,
( n - 2)(n - 2 ) !
for n even.
285
EXERCISES
Deduce that the value of u(0, t ) is the same as for the case when the initial value u(x, 0) is replaced by Md(x - a), where la1 = ro. Thus an observer at the origin cannot tell the difference between the effect of a mass M at a single point at a distance ro or the mass M distributed uniformly over a sphere of radius ro. Hints. (i) In view of the equivalence of the initial value problem (10.7.10) and (10.7.1 1) with the inhomogeneous equation (10.7.2), show by convolution and the sifting property of 6(S) that
(ii) Cut the surface S into infinitely small rings of thickness dh situated at a distance h from the origin so that Ix - < I 2 is constant on each ring. Use the fact that the area of this ring is d S = S,- ](l)rO(ri - h2)‘”-3)/2dh. Then deduce that
where s
=
h/ro.
18. Consider Oseen’s flow equations div v
=
0,
dv/dx
+ grad p - V2u/40 = td(S),
where u = (vl, u 2 , u 3 ) is the velocity vector; x1 = x, x2 = y, x3 = z ; t = f i + g j + hk, i J , k being unit vectors along the x, y, z directions, respectively, p is the (scalar) pressure; 40 is the Reynolds number; and th(S) describes the action of the body S on the fluid motion. Take the Fourier transform of these equations and deduce that
p 2 - 4aiu1
where p 2 identities
=
u:
{ -iu(-iu.[t6(S)IA)
+ p2[t6(S)]^},
+ u: + u : . Determine their inverse transforms by using the
and the convolution theorem.
286
10.
APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS
19. Using the notation and the method of Exercise 18, show that the solution of the Oseen’s equations
div u = a6(x, y, z),
where a, b are constants and d1 is the unit vector in the x direction, is
Derive the value of the pressure p .
20. In the previous two exercises we have considered the nondimensionalized and steady Oseen’s equations. Let us now consider the unsteady Oseen’s equations in their physical dimensions, that is, div v = 0,
aqdt
+ uau/aX + i/p grad p - v V’V = bS(x)6(y)6(z)6(t)d2,
where U and b are constants, v is the kinematic viscosity, and d, is the unit vector in the y direction. Find the values of p and u with the help of following two hints. (i). Take the divergence of the second equation and show that p satisfies the Poisson’s equation V’p = pbS(x)6’(y)6(z)d(t). (ii) The solution of the homogeneous equation
af
-
at
+ u-axif - VV*f = 0,
is
where R’ = ( x - Ut)’ function
+ y’ + z2
and erfc is the complementary error
Applications t o Boundary Value Problems
In this chapter we present solutions to boundary value problems, arising in various disciplines of mathematical physics, for axially symmetric bodies, including dumbbells, elongated rods, and prolate bodies, of which spheres and spheroids are special cases. The method rests on exploring the fundamental solutions of partial differential equations, as presented in the previous chapters, and then taking a suitable axial distribution of the Dirac delta function and its derivatives on a segment of the axis of symmetry of the body. This idea is extended to include distribution of these functions on arbitrary straight lines, curves, and disks [31-391.
11 . I . POISSON'S EQUATION
Let us consider Poisson's equation in n dimensions,
where u ( x ) is the generalized n-dimensional potential, F ( x ) is some forcing function, and x = (xl, x 2 , . . . , x,) is the position vector. In Section 10.4 we
288
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
found that for the special case F(x) tion E(x):
=
6(x) we obtain the fundamental solu-
r = 1x1. (2) u(x) = E(x) = l / F 2 , This value of u(x) gives the generalized electrostatic potential due to an n-dimensional sphere of radius I . For a distribution F ( x ) of delta functions over a volume V, the corresponding value of u(x) is
where * denotes the convolution and the variable of integration is x’. Our aim is to use ( 3 ) to obtain the generalized electrostatic potential of a large class of axially symmetric bodies for which a sphere, a dumbbell, and a spheroid are special cases. For this purpose, we prescribe an axial distribution of sources along the axis of symmetry, which we take to be the x1 axis. Accordingly, F(x)
=
.f’( x1)6(x2)6(x,)...6(xn)
(-el
2
x1
I
c2),
(4)
wheref(x,) is the strength of the distribution and c 1 and c 2 are two positive constants. For bodies with fore-and-aft symmetry we have c 1 = c2 = c. In terms of cylindrical polar coordinates, the shape of an axially symmetric body can be written p
=
po(x),
--a
I x Ia,
po(+u)
=
0,
(5)
where we have let x denote x 1 and where p 2 = x: + x i + . . . + x i . In view of the symmetry, po(x) = po(-x). The geometry of this configuration is explained in Fig. 1 1.1. In the next stage, we substitute F given by (4)in ( 3 ) and obtain
where we have used the sifting property of the delta function. If the potential u is prescribed on the surface S to be g(x), then (6) becomes
This is a Fredholm integral equation of the first kind for,f(x). Using this technique we can solve the potential problems for a wide class of axially symmetric configurations and in various fields of mechanics. Furthermore, we can solve both the direct and inverse problems. For the direct problem, the body profile p = po(x), as well as g(x), is prescribed, and we evaluate the
11.1,
289
POISSON'S EQUATION x2
b
Fig. 11.1. Line distribution of delta functions along the .vl axis in c, 5 .vl 5 c2 o f a n axially symmetric body.
distribution f(x) and the parameters c I and c 2 from (7). For the inverse problem we are given ,/'(x), g(x), and the parameters c I and c2: then (7) provides a body profile. The value g(x) = 1 is of special interest bccause it leads to the evaluation of the n-dimensional capacity of S. This is defined as C(n) = (11 - 2 )
where r l x
= dx'
rlx' . . . tlx"
=
s,,
F(.Y) ( I s ,
(8)
t1V. This rclation, in view of (4), becomes (2
C(n) =
(I1
- 2)
(9)
/'(<) ti<.
Let us now consider an axial distribution of dipoles along the axis of symmetry such that F ( x ) = h(x)h'(V)h(x,). . . h(xn),
1'1
I s I
('2.
(10)
where x 1 = x, x2 = y , and h ( x ) is the strength of the distribution. Thus, the dipoles are directed along the y axis. Substituting (10) in ( 3 ) , wc obtain
290
11.
Let u(x, p )
= yl(x)
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
on the surface S. Then ( 1 1) becomes
which, like (7), is a Fredholm integral equation of the first kind for h(x), provided that the equation of the surface S , p = p,(x), is known. Consequently, if h(x), l(x), e l , and c2 are prescribed, then (12) yields us the body profile. In the sequel we take c 1 = c2 = c.
11.2. DUMBBELL-SHAPED BODIES
Let us take two isolated delta distributions of equal strength such that
+ c) + A,h(x
,f(x) = A,h(x
- c),
where A , is a constant. This gives a dumbbell-shaped body, as shown in Fig. 11.2. We substitute in ( 1 1.1.7), and find that for g(x) = 1 we have
1
=
Ao(l/R;-2
+ I/R;-2),
(1)
where Rl
= [(x
+ c)' + p2I1I2,
R2 =
+
(2)
[(x - c ) ~ p2]*I2.
To obtain the relationship between the geometric parameters c/a and d/a (see Fig. 11.2), we evaluate ( I ) at the terminal points x = f a , p = 0 and at the dumbbell neck x = 0, p = +d. The result is 1 A,
--
1
(a
1
+ cy-2 + (a - cy-2'
1
A, - (c2 -
2
(3)
+ d2)'"-2)/2'
Elimination of A, yields the desired relation:
Furthermore, from ( I ) and (3) we derive the equation for the n-dimensional dumbbell :
1
[(x
+ c)2 + p2]'"-2'/2
1
+
[(x - c)2
+
p2]'"-2"2 - (u
1 + (a + cy-2 1
Cy-2'
(5)
11.2.
291
DUMBBELL-SHAPED BODIES
P
-0
1
,
A0
x
-C
Fig. 11.2. Dumbbell-shaped bodies that can be represented by two delta functions
The capacity of this configuration follows from (1 1.1.9): C(n) = 201
-
2 ) A , = 2(n - 2)
1
(a
+ c ) ' - ~+ (a
-
cy-2
I-'
Electrostatics
This corresponds to the case n respectively,
C(3)
=
3. In this case ( 1 1.1.6) and (4)-(6) become,
= (a2 - c2>/a=
(c2
+ d2)1/2.
(10)
Thus, the capacity of a dumbbell in three dimensions is equal to that of a sphere of radius (c2 + d2)1/2. Axially symmetric configurations are presented in Fig. 11.3 for different values of d/u.
292
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
I
-1.0
I
-0.8
I
-0.0
I
I
-6.4
-0.2
V 0
I
0.2
I
1
1
0.4
0.6
0.0
1
1.0
x/o
Fig. 11.3. A class of axially symmetric bodies generated by a pair of delta functions. The numbers along the p/a axis designate values of d/a.
Perfect Fluids
The relation +(n - 2)
=
UpnP3(n- 3)-’[1 - ~ ( n ) ]
(1 1)
relates the stream function +(n - 2) for an ( n - 2)-dimensional body of revolution moving with a uniform speed U along its axis of symmetry in a perfect fluid (or, equivalently, the fluid streams past the body with speed U ) with potential u(n)in a space of n dimensions. This potential function assumes the value unity on the surface of the body and vanishes at infinity. Thus, for the flow of a perfect fluid in three dimensions, we need the value of 4 5 ) . Now, from relations (1 1.1.6), (1 1.2.3), and (1 1.2.4), for n = 5 it follows that
C(5) =
Then (1 1) yields
6(a’ - c ’ ) ~ a(a’
+ 3c2) = 3(c’ + d*)3”.
11.2.
293
DUMBBELL-SHAPED BODIES
This gives the value of the stream function for a wide class of dumbbellshaped bodies. Cavities in Elastic Medium
The solution of a torsion problem for a shaft with a cavity can be obtained from the solution of the electrostatic potential for a seven-dimensional body, with g( x ) = 1, provided the cavity and the body have the same meridian profile. This isoperimetric equality is (15) 11/(5) = ap4c1 - u(7)~, where 11/ now stands for the elastic stream function. The strain energy E is E
=
(2G/n)[(2~~/15)C(7) - V(5)],
(16)
where C(7) is the capacity, V(5) is the volume of the body, and G is the shear modulus. The present method is suitable for solving some of these problems in a very simple fashion. For instance, for a spherical cavity we takef‘(x) = a56(x). Then (1 1.1.6) and (1 1.1.9) yield u(7)
= a5/(X2
+ p2)5’2,
C(7)
= 5a5.
Substituting these values in (1 5) and (16) and using (3.3.3) for V ( n )for n we obtain 11/(5) = ip4[1 - a5/(x2
+
p2)5’23,
E
= ~ ~ 7 1 0 5 .
=
5,
(17)
Similarly, for a dumbbell-shaped cavity,
C(7)
=
5(a2 - c 2 ) 5
a(a4
+ 10a2c2 + 5c4)’
Accordingly,
and E
=
nG
4(a2 - c 2 ) 5
3a(a4
+ 1oa2c2+ 5c4)
(19)
294
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.3. UNIFORM AXIAL DISTRIBUTIONS
Capacity (n = 3)
Let the uniform distribution be
f (x) = A, a const. Then for g(x) = 1, (1 1.1.7) yields 1
-
A
=
In
R2
- (X
R1 - (X
- C)
+ c)’
where R , and R2 are defined in (1 1.2.2). This equation is equivalent to (p2
(p’
+ (x - C)2)1” - (x - c ) + (x + c)2)1’2 - (x + c ) =
where A is a constant, or X2
[c2(4A3
+ 8A2 + 4A)]/4A( I - A)’
+
[c2(4A3
P’
+ 81’ + 4A)]/( 1 -
= ’’
which is the equation of a spheroid. The equation of a prolate spheroid is x2/u2
+ p2/b2 = 1,
b2 = ( I
-
e2)u2, c = ue,
(3)
where e is the eccentricity. Accordingly (2) becomes 1 l + e -=InA l - e
=
2 tanh- e.
(4)
The value of the potential function u(x, p), from ( 1 1.1.6), is u ( x , p ) = (2 tanh-’ e)-’ In
R2
- (X - C)
R,
-
-
(x
+ c>’
(5)
and applying (1 1.1.9), we obtain the value of the capacity: ue(tanh-’ e)-’. (6) An interesting limiting c a w i s that of a slender body, which is obtained by setting b/u 6 1 . In this case (6) reduces to C
where
E =
b/u
=
=
(1 - e2)lj2 is called the slenderness parameter. Equation
11.3.
295
UNIFORM AXIAL DISTRIBUTIONS
(5) can be put in familiar form by using prolate-spheroidal coordinates
t, 4 4 :
x
=
ctu,
+ iz = c[(t2
y
e
l)(I -
-
l/to. (7)
=
Then (5) takes the form
where Qo(<) is the Legendre function of the second kind. The corresponding results for an oblate spheroid can be derived by replacing c by ic and e by ie(1 - e 2 ) -112 in (5) and (6) and by replacing t by i t in (8). The corresponding formulas are C(3)
=
c/sin-' e
=
be/sin-' e,
(9)
and u(x, p )
=
'
Qo(it)/Qo(iSo) = cot - t/cot -
' to.
(10)
+ 1 or to-+ 0, the oblate spheroid reduces to a thin circular disk of radius b and (9) and (10) reduce to
As r
C(3)
=
2b/~,
U(X,p )
=
2/7~cot - ' 5.
The results for a sphere follows upon setting e
=
0.
Perfect Fluids (n = 5 )
In this case, if we set,f(x) = A and g(x) = 1 in (1 1.1.7), we obtain
where R and R 2 are given by ( 1 1.2.2).For the semimajor axis (x and the semiminor axis (x = 0, p = b), (1 1) yields the relations 2ac(a2 - c2)-2 = 1/A,
2cb-'(b2
+ c')''~ = 1/A.
=
a, p
=
0)
(12)
Hence (a2 - c ~ = )ab2(b2 ~ + c~)'/~,
(13)
which relates the parameters b/a and c/a. The shape function p = po(x; a; c) is now known from (1 I)-( 13). These bodies are prolate shaped. The value of the potential 4 5 ) follows from relations ( 1 1.1.6): (14)
296
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Next, we use (11.2.1 1) to obtain the stream function for the motion of these configurations in a perfect fluid,
where A is given by (12). The capacity C(5) is C(5) = 6Ac =
3(a2 - c 2 ) 2 U
=
3b2(b2 + c2)’j2.
(16)
Two limiting cases are of special interest.
(I)
c + 0, A
+ cx) such
that c A
+
iu3.
In this limit we recover the results for a sphere. (2) b/a 6 1 leads to an elongated rod. In this case (13)-(16) reduce to C b 7 b 4 - - - 5 - 6 (;)3 o(;) ,
+
U
(20) respectively. The body shape in this limit, from ( I I), is b
x2(3c2 - x2) 4(c2 - x2)4
+
(c’ - x2) >> b2.
(21)
Cavities in an Elastic Medium ( n = 7)
The relation corresponding to (2) and (1 1) in this case is
while those corresponding to (12) are 2ac(a2 + c2) - 1 (a2 - c2)4 A’
2 4 2 2 + 3b2) - -1 3b4(c2 + b2)3/2- A .
(23)
11.3.
297
UNIFORM AXIAL DISTRIBUTIONS
Hence a(a2 (a2
+ c2)
-
2c2 + 3b2
-
- 3h4((.2 +
c2)4
(24)
h2)3/2'
which relates the parameters b/a and c/a. Furthermore, (22)-(24) provide the body shape p = p o ( x ; a ; c). The quantities u(7) and C(7) are u(7) =
["
+
3b4(c2 h2)3'2 _ +_c _ x_ ~ _ _ 2c(2c2 + 3b2)p4 R ,
C(7)
=
5(a2 - ~ ~ ) ~ / +a cz) ( n =~ 15h4(c2 + h2)3'2/2c2+ 3h2.
(26)
We can then appeal to relations ( 1 1.2.15) and (1 1.2.16) to obtain
[
$ ( 5 ) = ip4 1 x ~
3h4(cZ+ h 2 ) 3 / 2 2c(2c2 + 3h2)p4
+ c - .Y -
C'
R,
-
1 (x + c)3 3 R:
-
~
2c2 + 3h2
p-.
where we have used V ( 5 ) = in2 p:(x) dx. As c + 0, A + x' such that c A + 1/2a5, we recover the results for a spherical cavity in Section 11.2. For a cavity with the shape of an elongated rod, that is, for h/u 6 I , we have
C(7) =
[:' (a)" +
(c2 - x2) 2 h Z ,
+ O(&)'], E
=
[;
G7cab4 c ~
4
O(;)y,
-
8 - 3(!)4(;
+ i c y+ o + 2c4(a2 3(u2 -
(361
.
( 9 3
A class of configurations is sketched in Fig. 11.4 for various values of h/u.
298
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
x /a
Fig. 11.4. A class of axially symmetric cavities of uniform distribution of delta functions. The numbers along the p/a axis designate the values of h/a.
11.4. LINEAR AXIAL DISTRIBUTIONS
In this section we study the polarization potential, so that g(x) = -Eox, where E o is the uniform electrostatic field in which a conductor is lying. Furthermore, we take (1)
J(x) = Bx.
Then (1 1.1.7) yields -Eox
=
[
B xln
r2
- (x -
'I} - R ,
R, -(x+c)
+ R.].
Proceeding in the same manrier as we did following (1 1.3.2), we find that the right side of (2) is a multiple of x only if the shape of the body is a spheroid. For a prolate spheroid, as defined by (1 1.3.3), we find from (2) that B
=
[
E , 2e - In
(t ':)I-'
__
11.5.
299
PARABOLIC AXIAL DISTRIBUTIONS, n = 5
The potential 4 3 ) is obtained from (1 1.1.6): u(x, P ) =
2r - In[(l
Eo
+ t.)/(l
-
r)]
[ {”’ .Y
In
-
(x
-
‘)}-R1
R1 - ( x + c )
+ R,].
(3)
The polarization potential P ( 3 ) is
In terms of the prolate spheroidal coordinates defined by ( 1 1.3.7), (3) and (4) take the forms
4 5 , V ) = -cEo5VQi(O/Qi(5o) and
p(3)
-%5;c35i(5;
-
1)”’CQ~(0/Q1(5o)l7
respectively.
11.5. PARABOLIC AXIAL DISTRIBUTIONS, n = 5
Let us now take f(x) = A
Setting y(x)
=
+ B.u2,
Is1 5
(1)
C.
1 and substituting this value of,/’(.v)in ( I 1.1.7) yields X+(’
xz x + c
x+c +
p’
(T-
-)
x -c RZ
This relation is again valid only for a spheroid. For a prolato spheroid as defined by ( I 1.3.3), we find that
300
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which determines the values of both A and B. Appealing to (11.1.6), we obtain
This in turn enables us to use ( 1 1.2.11) to obtain the stream function for perfect fluids,
'+I,
In prolate spheroidal coordinates (1 1.3.7), (4) and (5) become u(x, p ) =
(5)
- In c2 A [ t225 - 1 5-1
-
~
respectively, furthermore, from (1) and (1 1.1.9) we obtain C(5) =
4a3e3 2e/( 1 - e2) - In[( 1
+ e)/( 1 - e)]
'
The values of these quantities for an oblate spheroid can be obtained by using the limiting process, as explained for the n = 3 case in Section 11.3. They are
. [ 2Ai
5
u(x, p ) = 7 - cotc 5 +1
$(3)
=
+
' 51.
fUcZ(1 t2)(1 - 1 2 ) - iUA(1 x [-cot-' 5 5/(1 593,
+
+
+ 52)(1 - $)
C(5) = -2b3e3[e(1 - e2)1'2 - sin-' e l- ' ,
'
'.
where in this case A = +b2e2i[e(1 - e2)1'2 - sin- e l - As e -+ 1 we derive the corresponding results for a thin circular disk of radius b : 4x9
P)
=
(2/.)C5/(t2
+ 1)
-
'
cot- 51.
$(3) = + U P ( 1 + 52)( 1 - q 2 ) - ( U P / n ) (1 x C-ccot-l t 5/(1 t2)3,
+
+
+ t 2 ) ( 1 - $)
C(5) = 4b3/n. Similarly, if we set e
=
0, we obtain the corresponding formulas for a sphere.
11.6.
FOURTH-ORDER POLYNOMIAL DISTRIBUTION
301
11.6. THE FOURTH-ORDER POLYNOMIAL DISTRIBUTION, n = 7; SPHEROIDAL CAVITIES
Let us take
With this value of,f'(x)and y(x)
=
1, (1 1.1.7) yields
where
n = o , 1 , 2, . . . , m = - l , l , 3 , 5 , . . . . These satisfy the recurrence relation
Thus
302
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
Ill., = In
R , - (X - C) R1 - (X c)'
+
By evaluating these quantities on the surface of the prolate spheroid (1 1.3.3) and with a slight algebraic manipulation, it follows that c4BS., - 2c2B5.,
l + e _ _ _2e + In ____ + €5, = 3(1 4e3 . - e2), 1 e2 I-e -
(8)
Comparing (2) and (8) we obtain A,
=
A4c4, A , = -2c2A,,
A,
=
[3(1
y:2)2
-__ 2e_ 1 - e2
+ In Eel. 1-e
Next, we make use of (1 1.1.6) and (1 1.1.9) and obtain
4 7 ) = A4[c4B5. 0 - 2c2B5,2 C(7)
=
+ B5,
(10)
41,
yu5e5A4.
( 1 1)
We substitute these values in (1 1.2.15) and ( 1 1.2.16) and derive
11/(5) = aP4[1 E
=
- A4(cL4Bs.o-
2c'Bs.z
(16Gnu5/45)[4e5A, - 3(1
-
+ BS.,)],
e2),],
where we have used V ( 5 ) = (8n2a5/15)( 1 - e2),,
The corresponding relations for an oblate spheroid are C(7) = -%b5es[(2e3 + e)(l - e 2 ) ' I 2 - sin-' E = -(16.rrG/45)[8h5e5~(2e3+ e)(l + 3b5(1 - e 2 ) ' I 2 ] .
el-'
e2)l', -
sin-' e}-'
When e -+ 1, these relations reduce to their corresponding values for a penny-shaped crack of radius b, C(7)
=
16b5/3n,
E
=
EGh5.
11.7.
303
THE POLARIZATION TENSOR FOR A SPHEROID
11.7. THE POLARIZATION TENSOR FOR A SPHEROID
The polarization potentials ui,i = 1, 2, 3, created by uniform electric fields in the x i (x = x I , y = x 2 , z = x3) directions, respectively, are obtained by solving the boundary value problems V2Ui =
-47cF,(x), on S ,
ui = - x i
+
ui
=
o ( l / r 2 ) as r
--+
x.
(1)
where r2 = x2 y’ + z 2 and S is the surface of the spheroid, as defined in ( I 1.3.3). The boundary value problem (4) for i = 1 can be solved by taking the line distribution of delta functions along the axis of symmetry (x axis), that is, F , ( x ) = .1;(x)h(y)h(z).Then, following the analysis of Section 11.1, we find that
Letting (x, p ) approach S and using the boundary condition, we have
which is a Fredholm integral equation of the first kind. This equation has been solved in Section 11.4; the solution is
Then (2) becomes ul(x, p )
=
[
2e - In
~
‘3”.
I-e +
hiR2 - (x R1 - (X
+ ‘I}
-
C)
-
R,
+ R2],
(5)
where R and R 2 are defined by (1 1.2.2). In prolate spheroidal coordinates, (5) takes the form ~ 1 ( 5 9
YI)= -doYIQi(O/Qi(to).
(6)
Solutions of the boundary value problems (1) for i = 2 , 3 are obtained by taking the line distributions of dipoles, that is, F 2 ( x ) = i2(x)h’(y)b(z)and F , ( x ) = j;(x)h(y)d’(z)along the axis of symmetry. The values of ui(x, p), i = 2, 3, follow by appeal to ( 1 1.1.1 1). Indeed,
304
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
In view of the boundary conditions, we find that (7) reduces to the Fredholm integral equation
The solution of (8) is achieved by taking
Then (8) yields .u+c
x - c
Rz - (X - C ) Rl
- (X
+
C)
11
,
1x1 I n .
(10)
This relation is valid for the prolate spheroid ( 1 1.3.3) if
Combining (7), (9), and ( I 1) we obtain the values of
Iti,
i
=
2, 3,
In terms of prolate spheroidal coordinates, these relations take the forms ~ z ( 5 Vt ,
CP)=
-C(COS
~p)C(sIi-
-
V ~ ) ] ” ~ Q ~ ( O / Q (13) ~(~~)~
and 143(5.
YI.cp) = -+in
q>C(5;- 1)(1 - Y I ~ ) I ” ~ Q ! ( ~ ) / Q : ((14) ~~).
The formula for the components Qij
= -
s
ui
Qij
of the polarization tensor is
du . rlS dn
From the symmetry of the spheroid it is clear that Qij = 0,
i # .j.
(16)
11.7.
305
THE POLARIZATION TENSOR FOR A SPHEROID
Next, we use the values of ui, i
=
( 1 5 ) and obtain
In terms of the eccentricity c Qii
=
1, 2, 3, as given by (6), (13), and (14) in
I/C0, these take the forms
=
+
4ntr3(1 - e2)[ln---1 e 3 I - r
Q 2 2 = Q33 =
-
x
4ntr3(1 - e 2 ) 2e--- 2 e 3 3 I - e2 2t?
[I
1
= -
4nh3(1 3
+ "I-' .
In
LEI I-e
In (20) I-r e2 in (19) and (20), we derive the corresponding ~~~~
-
Replacing e by ie( I - e 2 ) values for an oblate spheroid: QI
fi
2e ][ln - 2 ~ ] l ' , (19) 1-r2 I-e
("
[sin-' e
-
e2)Il2] sin- r x [r( I
- e2)Ij2 -
sin-
I
03- I .
(22)
-+ 1, the oblate spheroid reduces to a thin circular disk of radius h, and (21) and (22) become
As
Q11
=
0,
Q22
= Q33 =
16b3/3,
respectively. The limiting case of a sphere follows by letting r have
(23) -+
0. We then
(24) Q I 1 = Q 2 2 = Q 3 3 = 8xu3/3, as is well known. The limiting case of a slender body is obtained by taking h/u = ( 1 - L ) ~6 )1. ~Then ~ ~(19) and (20) reduce, respectively, to
306
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
In this section we consider the potentials ui(x), i boundary value problems
=
1, 2, 3, that satisfy the
V2ui = -4nFi(x),
i = 1, 2, 3.
i l . Vui = - A . i i on S,
ui = O( l/r2)
(1)
at
E,
(2)
where A is the unit vector normal to the surface S and the i i ,i = 1,2,3, are the unit vectors in the x, y, z, directions, respectively. Then, the u i , i = 1,2,3, are the velocity potentials of a body with surface S and moving with unit velocity in a perfect fluid in the x, y, z directions, respectively. For a prolate spheroid, fi is
A
=
( I :+,
( a 2 - e2x2)-'I2 - x,tl
CI
-
yi2
+
(3)
We solve these boundary value problems by the same distribution of singularities as in Section 11.7. That is, we set F,(x) = ,fl(x)S(y)G(z),F2(x) = .f2(X)b'(Y)8(Z), and F3W = .f2(X)&Y)WZ). The integral representation formulas for u l , u 2 , u3 are given by (11.7.2) and (1 1.7.7). Applying the boundary conditions (2) to these formulas, we obtain the integral equations
The solution of (4) is obtained by settingfl(x) =
A[a2B3. 1 - xB3.
=
Ax, 1x1 I c, so that
21,
where the quantities B,,,n are defined by (11.6.3). On the surface of the spheroid (1 1.3.3) this relation becomes x
=
2e
A [ -1 - e
-
In I+e] 1 - e X,
307
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
which yields the value of A . Thus,
Equations ( 5 ) are solved by setting
,fi(x) = A.
-
A2x2,
i
=
(7)
2, 3, 1x1 I c.
Substituting (7) in (9,we obtain, after a slight algebraic manipulation,
Hence,
Next, we substitute these values of ,fi(x) in the integral representation formulas ( I 1.7.2) and ( 1 1.7.7) and obtain
U,(.X,
p) =
-
u3(x,
p) =
-
2e(l - 2eZ) -
[
2e(l - 2 2 ) 1-e2 ~~
The virtual mass tensor
In Ice]-1y(czB3.0 1-e
-
1-e
B,.,),
(10)
Z(C~B, . ~B,.,).
(11)
In%]-', I-e
(13)
-
w.jis defined as
Substituting (9)-( 11) in (12), we obtain
w,,
=
W,,
=
4nu (1 - e2) 3 ~
W,,
=
4TU3 3 x
and
2e
--
I +e In-1-e
[
wj
=
0
I-e
I'
2e(I - 2eZ) 1 - LJ2 for i # ,j.
-'
308
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
By letting e -,0 in these relations, we recover the corresponding values for a sphere of radius a : W,,
=
W,,
=
W,, = 2na3/3,
which are well known. For a slender body, these relations reduce to
w,,= w,,=
~
w.j
The value of for an oblate spheroid follows from ( I 3)-( 15) on replacing e by ie(1 - e2)-'I2: W,,
=
W2,
= W33 =
(4nb3/3)[(1- e2)'12sin-' e - e][e(l
- e2)'12 -
(4nb3/3)(I - e')[e(I - e2)lI2 - sinx [(I - e2)lI2sin-' e - e - e 3 ] - ' ,
' el
sin-'
el-',
(19) (20)
and Rj = 0 for i # j . In the limiting case of a thin circular disk of radius b, (19) and (20) reduce to W1, = $b3,
W,,
=
W,, = 0.
(21)
11.9. THE ELECTRIC A N D MAGNETIC POLARlZABlLlTY TENSORS
The electric and magnetic polarizability tensors P i j and M i j are defined in terms of the tensor Qij and as follows:
wj
where Iij is the identity tensor and V is the volume of the body. Accordingly, we can use the results of Sections 11.7 and 11.8 to find the values of P i j and Mij.
11.9.
THE ELECTRIC AND MAGNETIC POLARlZABlLlTY TENSORS
For a prolate spheroid, 1+e 8na3e3 PI,= pn- 2e] 3 I-e
-1
(3)
,
~
309
In
I f e l ' , (4) I-e
~ , , 8na3e3 = ~ [ ~2e - l n ! + e , ] I-e - 1
(5)
(6)
1 - e2
P.. = M . . = 0,
i # j.
For an oblate spheroid,
P II
=
-(4nh3e3/3)(1 - e2)"'[(1
P,,
=
P 3 3 = -(8nb3e3/3)[e(1 - e2)li2- sin-' el-',
-
e2)'l2 sin-' e - e l - ' ,
(7)
(8)
M I , = -(4nh3e3/3)[e(l - e2)'/, - sin-' e l - ' , M,,
=
M,,
-(8nb3e3/3)(1 - e2)'/,[(1 - e 2 ) ' / 2 s i n - L e - e - e 3 ] - I ,
=
P,,
=
P2? =
MI,
=
M,,
P33 =
=
M,,
P . . = M.. = 0,
For a slender body, 4na (a2 - h2)(ln: 3
P,,
=
P,,
= P 3 3 = __
~
2nab2 (41n:
-
l)-'[l
Mll =
2na(a23- 3b2) (a'p
M,,
M,,
2na -(2a2 3
-
- jI 3b2)
In
4nu3,
=
ha3,
i # j.
,I):(.
(1 1) (12) (13)
+
- l ) ( l n T 2a
3
=
(10)
P I.J . = M IJ. . = 0, i # ,j.
For a sphere,
=
(9)
;)-'[
+
-
1
+
"')-'[I
+)'I,
+ (I(:)(15) ,],
310
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
For a thin circular disk,
PI,
P22
0,
=
M I , = $b3,
=
P33
=
yb3,
M22 = M 3 3 = 0.
(18) (19)
P I.J . = MlJ. . = 0,i # j . The values of component M, of the magnetic polarizability tensor and component PZ2of the electric polarizability tensor for various dumbbelland prolate-shaped profiles can be found using the isoperimetric equalities
,
M11(3) + V(3) = (2743)C(5), PA3)
+ V(3) = 2CMI1(3)+ W I ,
(20) (21)
where C(5) for these profiles is given in Sections 11.2 and 11.3. Indeed, using (1 1.2.6) with n = 5 and (1 1.3.16),we derive the following results. For dumbbell-shaped bodies, defined by ( 1 1.2.5),
+ d 2 ) 3 / 2- J : o p i ( ~ )dx],
(22)
where p = p,(x), is the equation of the profile and V(3) = ny-, pi(x) dx. Finally, for prolate-shaped bodies, defined by (1 1.3.1 l),
[
M, ,(3) = n 2b2(b2 + c2)l/' -
[:
p i ( x ) dx],
a
4b2(b2 + c2)II2 - /:ap;(~) dx].
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
In this section we derive an approximation for the scattered field by distributing delta functions along the axis of symmetry. We discuss only acoustic scattering and start with the inhomogeneous Helmholtz equation
(V'
+
k2)U"X)
=
-4nF(x).
(1)
11.10. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
For the special case F ( x )
=
31 1
6(x), (1) has the solution (see Section 10.9)
(2)
q X ) = eiklxl l l x l .
Let the scatterer S be defined as
Then for F ( x )
= ,f(x)h(y)6(z)
we find that
where R = [ ( x - < ) 2 + p 2 ] 1 / 2and p2 = y 2 + z2. We can now discuss the scattering of an acoustic wave by a soft spheroid (the Dirichlet problem) and that by a rigid spheroid (the Neumann problem).
The Dirichlet Problem
In the case of a soft spheroid the boundary condition is
where we have taken the incidence to be along the axis of symmetry. From
(4)and ( 5 ) we derive
J-c c
eikR
- f ( t )d t =
R
-eikx,
(x,p ) E S.
(6)
Now for small values of k we expand the functions eikx,eikR/R,andf(x) as follows : (ik)" n=O
R
and
n=O
(7)
312
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Substituting these expressions in (6), we obtain the following simple set of integral equations:
where (x, p ) E S. Proceeding as in earlier sections, we find that the solutions of (lo)-( 14) are
+ e)/(I - e)}]-' = a - ' , = -2ueaP2 + ( a - 2e)-'x,
.fo(x) = [In{(] .f;(x)
a(3 - e 2 ) - 2e X2, 2a[(3 - e2)a - 6el
-
,f4(~)
+
+
2u2e3[-6e 2e3 a(3 - 2e2)] 2e)'[ - 30e 8e3 3(5 - 3e2)a]
+ ae [- 10e + 4e3 + ( 5 - 3e2)aI - _ x2 + x3, 3a2 2(a - 2e)[-30e + 8e3 + 3(5 - 3e2)a] = Do + Dlx + D2x2+ D4x4, (a
-
+
11 .lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
31 3
where
- 2e}{2e + 2e3 - (1 + ( 4 3 - e 2 )32a{(3 e2)a - 6e}
- e’)’.}
- 2e2(3 - 2e2) 9a2
-
1 + 192a {10e - 6e3 + 3(1 ~
+ 2e
-
(1 - e2)a D, 2u2
-
e2)2a)
+ $ { - 6 e + 10e3 + 3(1
D,
=
2u3e3/9(a - 2 c ~ ) ~ ,
D2
=
-2a2[(3 - e2)a - 6el-l
-
i . ’ ) ’ a } ~ ~ ] (21a) .
(21b)
e(2e - (1 - r2)a){2ea(4- e 2 ) - 12r2 - m’} 2a3{(3 - r 2 ) @- 6 e ) ~~
+ (5
a{
-
6eZ + e4)a - 2e 32a
+ 6e3
2e4 3a2
--
1
+ 26e3 + 3(5 - e2)(1 - e2)a}D4 , D4 = (1/8a)[6ea(42 36e2 + 6e4) (3 - e2)(35 - 3 0 2 + 3e4)a2 84e2 + 44e4][210e 1 10e3 - 3(35 - 30e3 + 3e4)a]-’ x [(3 - e2)a - 6 ~ 1 ’. -
-
30e
-
-
-
(21c)
-
(214
We have obtained the value o f f ’ ( x )up to 0(k4). To find the far-field behavior of the scattered field u ( x ) we appeal to (4). In terms of spherical polar coordinates, x = r cos 0, y = r sin 8, we have for r > > 1, R
= J(x
-
o2+ p 2
2:
I / R = I / J ( x - t>2+ /T”
r -
< cos O,
= l/r,
and eikR/~
(eikr/,.)e-ik
Substituting this value in (4) we obtain iis(r,
0)
N
(eikr/r)jl(8),
(22a) (22b)
314
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
where j l ( U ) is the far-field amplitude, given by .jr(8)=
J
e-ikCCOSef(5) dt.
~
(24)
-C
We next substitute in (24) the value of,f’(x)we have obtained. Then (23) yields the far-field value of the scattered fields:
+--3a
u3e3
-
-
2u3e3 cos u 3(cr - 2e)
ik3{7 2u4e3
(- 8’
+? !
cr2
3r
3
D2
u3e3 + __ cos2 8 3cr
2u4e4
5e) cos’ 8 9 3cr2
-
2a5e5
+ ___ 5 D4
}
2a5e5[30e2 - 14e4 - 5(25 - 14e’)ecr + 3(5 - 3e2)a2] 15(cr - 2e)’[ -30e 8e3 3(5 - 3e2)a]
+
+
COS
40e - 2cr(10 + 3e2) + 3ecr2(3 - eZ) a5e4jcos2 t) +{F+ 30a2[(3 - e2)a - 6e] use5 cos3 o + uses C O S ~8 . 15(ln(l + e)/(l - e ) 2e) 60a 4a5e6
~
-
The value of the scattering cross-section u , is given by u1 = 2n: Joljl(0)12 sin 5
16n:a2[ ( : ) 2
+ 2)4{:r
1
o (10 -
(6)’
+ 3 c(5) r - +e2}
1
40e2 - 2ecr(10 + 3e2) + 3e2cr2(3- e2) 30a2((3 - e2)a - 6e)
(94[-q2 + 4 ():
+3cr
-
%]}I.
O
315
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
Neumann Problem for a Prolate Spheroid
In this case the boundary condition is
. VL,i
fj .VUS(X) = - f j
= -fj
.v p ,
(27)
where Applying (27) to (4), we obtain
I-, ',ikR
c
1 - ikR)(a2 - x ( ) , f ( ( )d t , (x, p ) E S. (29) R3 Next we use the same expansions as in the case of Dirichlet problem except that now $..p -
-(
2 ,fn(x)(iky+l. m
.f(x) =
(30)
n=O
Proceeding as in the previous case, wc obtain the following system of simple integral equations:
where (x. p ) E S. The values of,fi(x),i = 0, I , 2,3, are derived by a method similar to that of earlier sections. The result is
(Iz
.f;(-x) = - - ( a
2P
+ [I-
2r
=
2[4e(l -
2e)[2r
+ 2e(3 - 2r2) - 3(1 - OZ)a]-l(l
I-'
+ 4e - 3a x z , 6e~+ (3 - 2)a][22e + 9e2a -~22e/( I ~ - r 2 ) + 26r - 3(5 - ez)a][2e/( 1 - I?)
- ez
uz[.f2(x)
-
-
2)'
- eZ)] -
_
( 5 - @')a - 6e - 4r/( 1 - C J ~ ) 2(2~/(1- r z ) - a)[4e/(l - P') + 26c - 3(5 - (.')a] ~
~-
_
.Y33
(37)
31 6
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
and .f;(x) = Go
+ Glx + G2x’ + G4x4,
(38)
where G0
+ (a
G,
I - e’ 2e a - e)(( 2 I - e’
2e
-
=
(-
2e)G2
-
2e
1 - e’
+ 9a2[6e
-
+ 4e - 3a
4e3 - 3(1 - e 2 ) a 2 ] G 4 }
+ 4e - 3a) -’a’{? 1 [a’(1 2e- e’) + a - 4 e
2e
+ 4e - 3a 3e’)a
-
I-’
-
[8e3 - 30e
+ye3
66e
-
~
+ 3(5 - eZ)a]G4
1 - e2
+ (5 - 3e’)a +
’),
+ 4e - 3a]-
2e
and a is defined by (20). Putting this value off’(x)in (4) and using (22)-(24) withj, replaced by j 2 , we obtain the far-field behavior: u’(x)
N
-
eikr [ p 3 ( i k’ r
r(
e’)
-
- k’
2aeG0 + %a3e3G2+ 3a5e5G4
_ _a5e5 _ 30
+ 6(50
a)-’[ I
2e
~-
1 - e’
1392e’
-
48e’ 16e’)a’
-
1+
+ 5 a 5 e1 5- (e 4
ye2+ { + ’} [- + 26e
808
~
-
4e3] cos’ 8
a)-’
(3083 O}].
-
I-’
3(5 - e2)a
512 1-e
use2 2e cos 0 - __ 30 1 - e
x [5(l - e2)a - 1Oe -
2a3e3
I-’
4e - 3u
(39)
11.11.
31 7
STOKES FLOW
The scattering cross-section 0 , is given by 6,
5
47t -((ku)4u2 3 -kio2[ -
2e
(1 - e 2 ) 2
1
2e(I
-
u4
-
c2
c2)
e2(l - e 2 ) [5(l - e2)a - 10e 90
2e + 4e + 4e3][I- e2
-
I-'
+ 2 6 -~ 3(5 - e2)ci
+ 6(50
-
+
~6e')o2]
I-'
3ci
&?(A - a)-']}.
I-e
11.11. STOKES FLOW
Let us now study the flow of a viscous fluid in which a solitary body or a large number of bodies of microscopic scale are moving. They may be carried about passively by the flow, like solid particles in sedimentation, or moving actively, as in the locomotion of microorganisms. The linear differential equations (the inertia forces have a negligible effect on the flow) that describe their motion are the Stokes equations
v . u(x) = 0, V p ( x ) = pV2u(x)
(1)
+ F(x),
(2)
where u is the velocity vector, p is the pressure, p is the constant viscosity, and F is the external forcing function. We shall study only rotary motion. For this purpose we assume F to be solenoidal, that is, V . F = 0, so that (3)
F ( x ) = 47tpv x Q(x),
where Q is the vector potential. The far-field behavior of u and p is u
-+
0, p
--t
p m , a constant,
as r
=
1x1 -+ co.
Taking the divergence of both sides of (2), we find that V2p = p V 2 ( V . U )
+ V . F = 0,
(4)
318
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which means that p is a harmonic function having a constant value pm at infinity. Accordingly, p = pa, in the entire space, and the differential equation (2) reduces to V2l1 = - ( l / p ) F ( x ) = -471v x 51(x). (5) The fundamental solution of (5) is obtained by taking 51(x) = $(x) where
y is a constant vector. It is readily verified that
(6)
is then the solution of (5). This solution is called a rotlet or couplet. Physically, this solution provides the velocity generated by a sphere of radius a rotating about the y axis with angular velocity oo= 1 y !/a3. For a volume distribution 51(x) that is a superposition of delta functions the solution is
Following the method of this chapter, let us take an axial distribution of rotlets of the form
(8) wheref(x) is the line distribution and 2 is the unit vector along the x axis. Thus 51 = &j(x)6(y)6(z)
(-c1 I x I CZ),
For a body with fore-and-aft symmetry, this equation yields
where p2 = y 2 + z2. Thus
In terms of cylindrical polar coordinates (p, 8, z), (1 1) is equivalent to u = (0, 0, u~),and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
31 9
The boundary condition is u d x , P ) = W,P, on P = pO(x>, (13) where coo is the angular velocity of the body. From (12) and (13) we get the Fredholm integral equation,
This equation is a special case of (1 1.1.7). Accordingly, we can obtain the solution of the present problem by takingj'(5) given in Sections 11.2-11.6. For various examples in fluid mechanics solved by this method, the reader is referred to the work of Chwang and Wu [35]. The analysis for the rectilinear motion of bodies in Stokes flow can be obtained as an appropriate limit of their displacement field in elastostatics as given in the next section.
11.12. DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS IN ELASTOSTATICS
For a homogeneous, linearly elastic, and isotropic medium, the displacement equations of equilibrium are [32]
1+
1
vI v . un + v2u " [
.f' = 0,
where u = ( u l , u 2 , u 3 ) denotes the displacement field, p the shear modulus, v the Poisson ratio of the material, and,fthe body force per unit of volume in a three-dimensional Euclidean space. In Exercise 13 of Chapter 10 we found that by taking f to be the Kelvin force ,fk
=
16np(1
-
v)b(x)a,
(2)
where x is the position vector of the field point and t~ is a constant vector that characterizes the direction and magnitude of the force, we can obtain the fundamental solution Uk(x; a ) =
(3 - 4v)a r
+-( a rx ) x , '
r
=
1x1.
(3)
This solution expresses the displacement field corresponding to a concentrated force located at the origin of the coordinates. Let us note in passing that U k ( x ;a ) = O(l/r)
as r
-+
m or r + O ,
(44
320
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
and, for any constant A , (4b)
Uk(x;Aa) = ALlk(x; a).
The net force F experienced by the medium bounded by a control surface S enclosing the singular point is given by F
=
s,
fi. T dS =
Jv
V . T dV =
i
-fk
dV = - 1 6 ~ ~ / ~ (V1) U ,
(5)
where ii is the unit outward normal at S, T is the stress tensor, and V is the control volume enclosed by S . The net moment M is obviously zero. In the absence of boundaries, a derivative of (3) of any order in an arbitrary fixcd direction is also a solution of (l), the corresponding forcing function being the derivative offk of the same order and in the same arbitrary direction. These derivatives are readily obtained by considering the displacement field corresponding to a concentrated force located at a point p # 0. The formal multipole expansion (similar to that in classical potential theory) in a Taylor series about x yields
Uk(x - p ; X )
=
U k ( x ;a) - ( p . V ) U k ( x a) ; + + ( / ~ * V ) ' U ' ( Xa)+ ; ... .
(6)
The interpretation of various terms on the right of (6) is obvious: The first term has already been explained in (3), the second term represents a Kelvin dipole characterized by the vectors a and /3, the third term represents a Kelvin quadrupole, etc. We write a Kelvin dipole
+ 3(a . x)(r5p . x ) x
(74
the corresponding forcing function being
j k d= - I ~ T C / L ( I - v ) ( p . V ) [ G ( x ) ~ r=] -16np(l
-
v ) [ ~ . V G ( X ) ] ~(7b) C,
where the superscript kd stands for Kelvin dipole. For fixed x , a Kelvin dipole (7a) is, in general, not symmetric with respect to an interchange of a and p. It is more convenient to handle a Kelvin dipole in terms of its symmetric and antisymmetric parts. The antisymmetric part is itself a physical entity of special significance:
3 [ U k d ( x a, ; p) - U k d ( x /), ; a ) ] = -2(1 - v)(a x
P)
x x/r3.
(8a)
This singularity is called a center ofrotation, and we denote it Ur.Next, set y = -2(1 - v)(a x
8).
(8b)
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
321
Then (8a), in view of (8b), can be rewritten
where the superscript r indicates the center of rotation. The corresponding forcing function is ,fr =
f [ , f k d ( xa, ; p) - ,fkd(x;p, a)]
=
4 7 V~ x~ [ h ( x ) ~ ] .
(8d)
The displacement field (8c) has a vector potential y/r and no scalar potential. The net force F and the torque M experienced by a control volume V containing the singular point and bounded by a surface S are given by
and
= JV(fr
x X) dV = -8xy,
respectively. The symmetric part of a Kelvin dipole yields another fundamental singularity, called a strrsslet. Its displacement field and the forcing function are
+ 3(a . x)(r5
x)x
and ,fk"
= -8
7 ~I ~-( v ) [ p . VG(X)C~ + ~ 1 VS(x)fl], .
(lob)
respectively. Physically, a stresslet represents a self-equilibrating system and hence contributes no net force or moment to the medium. It is of the nature of two double forces with moments in the plane determined by the vectors cy and p, the moment of one double force being equal in magnitude but opposite in direction to that of the other double force.
322
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Yet another type of singularity is useful in the present investigation. It is called a center qfdilatarion, and it is characterized by a scalar. In this case the displacement field and the forcing function are
respectively. Thus, the displacement field is derivable from a scalar potential l/r and, like a stresslet, is a self-equilibrating singularity. A dipole formed by two centers of dilatation yields U dd (x;a)
=
-V(a.x/r3)
=
-(a.V)(-l/r)
=
-a/r3
+ 3(a.x)x/rS,
(13)
whereas a quadrupole formed by two dipoles or by four centers of dilatation leads Ud4(x; a, p) =
-(p
*
V)Lldd(x; a)
=
( f l . V)(a. V)Ud(X) = ( a . V)(fl. V)Ud(X)
= Ud4(x; /J, a ) = - V(fl . V)(a . V)( l/r).
(l4a)
This shows that the quadrupole Ud4 is symmetric with respect to an interchange of a'and p. In particular,
where C,, C,,, 2z are the base vectors in the x, y, z directions, respectively. These singular solutions are useful in constructing solutions to several boundary value problems involving different modes of displacement of rigid prolate and oblate spheroids, including their limiting configurationsthe sphere, slender body, and thin circular disk. We now proceed to demonstrate this by an example. Various other examples are given in reference [32]. Translation of a Prolate Spheroid
Consider a rigid prolate spheroid, as defined by (1 1.3.3), embedded in an unbounded elastic medium. It is given the displacement U
=
Uxex
+ U,e,.
(15)
There is no loss of generality in assuming the displacement to be in the form of (15) because the y and z axes can be so chosen that the z component of the specified displacement is zero. The displacement field in the elastic medium must (a) satisfy the homogeneous equations of equilibrium (1) (with.f = 0), (b) vanish at infinity, and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
323
(c) meet the boundary condition (15) on the surface S of the spheroid. We attempt to construct the appropriate solution by employing a line distribution of Kelvin solutions (3) and dipoles formed by centers of dilatation (13) between the foci x = - c and x = c ; i.e., [a,Uk(x -
5 ; b x ) + a2 U k ( X - 5 ; by)] (15
J --c
where 4 = 52,. The first integral in (16) represents a line distribution of Kelvin solutions of constant strength a , and a2 oriented in the x and y directions, respectively. The second integral represents a line distribution of dipoles formed by centers of dilatation,each of parabolic density and pointing in the x and y directions. It is thus obvious that (16) satisfies the homogeneous (withf = 0) equations (1) of equilibrium in the elastic medium and vanishes as ( xl -+ x).To satisfy the boundary condition (2) on the surface S of the spheroid, we use the integrated form of (16), namely,
where b,, = ( J + ~+ _-bZ)/pis the unit radial vector in the y , z plane, R , and R 2 are defined in (1 1.2.2), and the quantities B,,,n are given by (1 1.6.3). The particular values B , . o , B , . , , I?,.,, B 3 . and B5.0have already been evaluated in (1 1.6.7); the remaining particular values needed are BL.1
=
R,
-
Rl
+ xBI,o,
B5.1 =
&l/R;
-
l/R:)
+ xBS.0.
(18)
Next we evaluate these quantities on the surface of the spheroid S. We find from (17) that the displacement on S is
+
b2xyb, u2yp?, b2(a2- e2x2) ’
324
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where l + e L = ln-= 1-tJ
2tanhK'e.
Accordingly, the boundary condition (1 5) on the surface S is satisfied if we choose
+ 1 + 3e2 - 4ve2)L]-',
a, = -2e28,/(1 - e2) = U,e2[-2e
a2 = -2e2p2/(1
-
(20a)
e2) = 2U,e2[2e - (1 - 7e2 + 8ve2)L]-'.
(20b)
This choice of a,, p,, a 2 , and b2 in (17) furnishes the required solution. Next, we compute the net force F experienced by the spheroid by superposition of (9, F
= - 16np(l -
v)
s1,
(a,C,
+ a28,) d(
=
F,C,
+ F,&,,
(21a)
where the longitudinal and transverse components of the force are
+ (1 + 3e2 - 4ve2)L]-',
F,
=
-32np(l - v)U,ae3[-2e
F,
=
-64np(I - v)Uyae3[2e - (1
-
7e2
(21b)
+ 8ve2)L]-',
(21c)
respectively. By the limiting processes v + f and 2pvV , u/( 1 - 2v) -+ - p (where p is the pressure), the elastostatic equations (1) reduce to the equations of motion for Stokes flow as given in Section 11.11. In this case u denotes the velocity field. By setting v = 3 in (21b) and (21c), we obtain the corresponding drag formulas when a rigid spheroid moves in a viscous fluid with velocity U . In the limiting case of a sphere ( b + a, or e -+ 0), the displacement field (17) and the net force (21) reduce respectively to 3a
u ( x ) = 10 - 12v
and
F r 1(3 - 4v)U
+
( U .x)x r3
a3 u.x (22) 10 - 12v ( 7) 3
+
F,
=
-24np(l - v)aU,/(5 - 6v),
(234
F,
=
- 24np( 1 - v)a U,/(5
(23b)
-
6v).
These results can be summarized as follows: An embedded rigid sphere of radius a is given a displacement U . The field in the elastic medium is the same as would be produced by (i) a concentrated force - 24np( 1 - v)aU/(5 - 6v) and (ii) a dipole of strength -a3U(10 - 12v) formed by centers of dilatation, both located at the center of the sphere, namely,
325
11.13. THE EXTENSION TO ELASTODYNAMICS
For the other extreme case, a very slender spheroid,
F,
=
-
32r~p(1 - v)uU, (6 - 8 v ) ln(2/~)+ 1 [ I
+ O(E)1.
When the elastic medium is under torsion, the displacement field is quadratic even without any inclusion. The problem is therefore rather complicated when a rigid inclusion is embedded in it. However, by this method this problem can be solved effectively, as explained in reference [36].
11.13. THE EXTENSION TO ELASTODYNAMICS
The concepts in the previous section can be extended to cases of steady state dynamic singularities [37-391. The singularities can then be applied to obtain the solutions for the rectilinear oscillations of rigid inclusions in an infinite homogeneous isotropic medium. The dynamical equations of elasticity are
( A + p ) grad div ZI + pV2u - p
aZu
at
+ ,f = 0,
where u is the displacement vector, L and p are Lam& constants of the elastic medium, p is the density of the medium, andf is the body force per unit volume. Let us qssume that f = foeiw', u = uoei'". Substituting these values in (1) and dropping the zero subscript, we have
( A + p ) grad(div u)
+ p V2u + p o 2 u + .f = 0,
or 1
grad div u
1
+ V2u + m2u + f
=
0,
(2) (3)
where m2 = p o 2 / p , and v is the Poisson ratio. The fundamental solution Udkcorresponding to a force .fdk(X) = 4np6(x)u,
(4)
where x = ( x l , x 2 , x3)is the field point, u is a constant vector, and the superscript d and k stands for dynamical Kelvin, located at the origin of the coordinate system is (see Exercise 14 of Chapter 10) - imr
r
(5)
326
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
where r = 1x1 and T~ = (1 - 2v)/2(1 - v). This solution will be called the dynamical Kelvin solution. The net dynamical force F experienced by a control surface S enclosing the singular point is given by F =
I
Jv Jv
n.Tds= -
= 47rpa
+ m2p
V . T d V = /vfdkdV+m2pJv~dk(x;a)dV
Udk(x,a) d V ,
where A is the unit outward normal to S , T is the stress tensor, and V is the control volume enclosed by S. The net moment M is clearly zero. In free space a derivative of ( 5 ) of any order in arbitrary direction is also a solution of (3). The corresponding forcing function is the derivative off dk of the same order in the same direction. These derivatives can be obtained easily by expanding Ddk(x,p; a), where p is a given vector, in a Taylor series about x . This expansion is Udk(x,B; a)
N
udk(x,a) - (j?.V ) u d k ( x a) ;
+ $ p . V ) 2 U d k ( a)~ ;+ . . . .
(7)
The interpretation of various terms on the right-hand side of (7) is as follows: The first term is the dynamical Kelvin solution just discussed, the second term represents the dynamical Kelvin dipole characterized by the vectors a and p, the third term gives the corresponding quadruple, and so on. The dynamical Kelvin dipole can be written u d k d ( x a, ; p)
=
- ( ~ . V ) U ~ ' ( Xa);
=
-a(/?.V)- r
- imr
1
-m2
[T
v(p.V ) ( a .V )
-1.
- imrr
- imr
-
r
(8)
while the corresponding forcing function is f
dkd
-
- 4 7rpL(p* V)G(x)a = -4lrpcp * VG(x)]r,
(9)
where the superscript dkd stands for dynamical Kelvin dipole. From (8) it follows that a dynamical Kelvin dipole is not symmetric with respect to a and p. Its symmetric and antisymmetric parts yield other important fundamental solutions. The antisymmetric part is + [ u d k d ( xa,; p) - U d k d ( xp, ; a)]
= +[ a( p. V)e-imr/r- /?(a.V)e-imr/r] =
-V x +(a x p)eWimr/r.
We shall call it the dynamical center of rotation and denote it Udr.If we set y = -12(a x p), the foregoing relation becomes Udr(x,y)
=
(V x y)e-'"'/r.
(10)
11.13.
327
THE EXTENSION TO ELASTODYNAMICS
The corresponding forcing function is
Since Ud' has only a vector potential ye-'"'/r and no scalar potential, the net force F vanishes, while the torque M experienced by the control volume V containing the singular point and bounded by S is
M
=
=
Js
x x ( f i . T ) d S =-
Jv,fdr
J"
(V.T) x xdV
x x dl/ + rn2p J " ~ d ' ( x ;x,
0)
x x dl/
The symmetric part of the dynamical Kelvin dipole gives another fundamental solution, which will be called the dynamical stresslet (dks). The corresponding displacement field and the forcing function respectively are
and
This fundamental solution represents a self-equilibrating system and accordingly contributes neither a net force nor a moment to the medium. Another useful fundamental solution in the present investigation is the potential dipole, which is characterized by a scalar function and is also useful in discussing vibrations in Stokes flow. The displacement field due to this potential is
We now demonstrate that these fundamental solutions are very effective in solving boundary value problems in elastodynamics. To fix the ideas we consider the case of a spheroid.
328
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Translation of a Prolate Spheroid Let the rigid spheroid S, x2/a2 + p2/b2 = 1,
p2 = y 2
+ z2,
c2 = ( a 2 - h2) = a2e2,
(16)
where e (0 Ie I1) is the eccentricity and 2c is the focal length, be embedded in an isotropic and homogeneous elastic medium. It is excited by a periodic force with period 2n/o acting in the direction of its axis of symmetry, so the displacement of the points on S is
u = u2,,
(17)
where 2, is the unit vector along the x axis and U is a constant. Our aim is to find the displacement field in the elastic medium so that (3) and the boundary condition (17) as well as the far-field radiation condition are satisfied. For this purpose we apply the technique of distributing the appropriate singularities. In the present situation we have the following line distributions between the foci x = - c and x = c: U(X)
=
s1,
F1(<)Udk(X- <,2,) dx
+ [ I C F 2 ( < ) ( c 2- t2)Udd(x- <,ex) d5,
x E S,
(18)
<
where 4 stands for 2,. The first integral in (18) represents the line distribution of dynamical Kelvin solutions (5) of variable strength F,(<), while the second integral is the line distribution of potential dipole (15) with strength (c2 - t2)F2(<).Clearly (18) satisfies the differential equation and vanishes as 1x1 + m. To find F,(<) and F,(<) we apply the boundary condition (17) on the surface S of the spheroid. We obtain Ug,
=
fcF1(C)Udk(x - t, 8,) d t
SIC
F2(<)(c2 - l2)Udd(x-
+
t, 2,)
d<,
x E S.
(19)
It is obvious that we cannot easily find a closed form solution to equation (19). Accordingly, we use a perturbation technique for small values of the parameter in. For this purpose, we expand all the functions in (19) in powers of in:
11.13.
329
THE EXTENSION TO ELASTODYNAMICS
where Uk(X,
a) =
(3
-
4v)a
r
+-(a .r3x)x ’
u, (x, a ) = dd
a
- -
r3
+ 3(a.x)x r5
________
are the Kelvin solution and potential dipole, respectively, for the elastostatic field, as derived in the previous section. Substituting the expansions (20) in (19) and equating equal powers of m, we get the following system of equations: 1
rc
330
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
The solutions of (21) and (22) are available in Section 1 1.12, and we thus have 2r2
1 1
4(1 - v ) '1'l 2
e2
2e2
-
-
1
.
2
f21 = Ue2[-2e+(1 +3r2-4ve2)L]-',
'
f2 =
2i(2
+ ?)
3
(24)
.fl ~4~
+
where L = In[( I e)/( 1 - e)]. Next we substitute (24) and (25) in (23) and obtain
where $0
+ T ~ +) (1 - e 2 ) L ] - u2J21(-2e + L ) + $iue(2 + = afll[e(l + r4) - ( I - e ' ) ~ ] - .f21[4f? - (2 - e2)L], 1 2 '
= zu
.lll[e(I
t3).f1 2,
$2
=
and C, = (yC, (26) we set
-[%I
+ zCJp
-
+ ,f21(-2e + L)I,
~~).f11e
(274 (27b) (27c)
is the unit radial vector in the y , z plane. To solve
Substitution of (28) in (26) and simplification yields
11.13.
THE EXTENSION TO ELASTODYNAMICS
where
4o = - 1/2(1 - v)e + L, 4l = a'[e - { ( I - e2)(3 - 2v)/4(1 4 2
=
-2u2[6e
Al
=
[(14 - 1 2 ~ -) (7 ~
A2
=
-36e
0,
=
-(2e
02 =
-
(3
-
-
6v
-
+ 2(9 - 3e2)L, + L)/2(1 -
+
2e2)L],
-
v)}L],
3e2 + 2ve2)L]/4(1 - v ) ,
\I),
1 6 ~ 8e/(l - r 2 ) - 12L.
Equation (29) is satisfied if we choose 1
4( 1
-
(A0
V)
+ c2A2) = -
This is a linear system of equations with the following solution:
The net force F experienced by the spheroid can be computed by superposition of (6);
J
-5
Jv
332
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where the volume integral is taken over the spheroid. Substituting the values of .fl ‘,f1 ,f, 3 , and II in this formula, we have =
PO[{l
+
12npuu2,
+
+ A,e’ + f 2 1 [ 2- 4(1 -(e-
where Po
=
1 1+
’ + 4e - ~ v e ) (urn)’ ~ ]
32npa(l
-
v)Ue3[-2e
zX
-
v)e2
O(~rn)~,
(39)
+ ( 1 + 3e2 - 4ve’)L]-’.
Various other rotary and rectilinear oscillation problems can be solved by this method and are presented in references [38, 391.
11.14. DISTRIBUTIONS O N ARBITRARY LINES
The concepts of the distribution of the delta function and its derivatives can be extended to include any straight line in the plane. Indeed, in the theory of bending and buckling of elastic plates one has to apply concentrated loads on lines inside the plate, the so-called line loads. The classical approach is to take a strip of finite width, so that the problem is first solved for the case of loads on an area; then the width is made to approach zero. The problem can also be solved easily with the help of the theory of distributions. As an illustration, let us find the bending of a rectangular plate 0 I x I a, 0 Iy I b of small thickness h when a periodic concentrated load of uniform strength P is applied on its diagonal line x 2 = (b/u)x, (see Fig. 11.5). This load p ( x , , x 2 , t ) can therefore be written [34] p ( x l , x 2 , t ) = PeiwrS(x2- bx,/u)(l
+ b2/a2)li2,
where the factor ( 1 + b2/a2)’/’has been added for dimensional reasons. The equation of motion for the plate in the foregoing situation is 9 V4v
+ h d2v/dt’
=
P e ilDi 6 ( x 2 - b x , / a ) ( l
+ b’/uZ)”’,
(1)
(2)
where u denotes the deflection of the plate in the x 3 direction, 9 = Eh3/12(l - v’) is the flexural rigidity (also called bending stiffness), E is Young’s modulus, v is Poisson’s ratio, p is the density, and the differential operator V4 is v4 = a4/ax:
+ 2 a 4 / a x : ax: + a4/ax;.
11.14.
333
DISTRIBUTIONS O N ARBITRARY LINES
Fig. 11.5. x 2
= (h/a)x,.
Let us assume that the plate has simply supported edges so that the boundary conditions are u
=
d2v/8x:
=
0,
x1 = 0 and
x 1 = a;
u
=
J2v/dx:
=
0,
x2 = 0 and
x2
=
(3)
b.
The solution of the boundary value problem (2) and (3) is given by the double series
where A,, is the amplitude, +,
is the phase, and
Vmn(xl,x 2 ) = sin(mnx,/u) sin(nnx2/b), are eigenfunctions for this boundary. The corresponding eigenvalues A,,, are
A,,
= x2[mZ/o2
+ n2/b2](9/ph)1/2.
Substituting (4) in (2), we obtain the following values for the amplitude and phase, respectively:
where
Amn =
(Frnn/Ain){[l
+,
tan- '[2[rnndm,,(&
=
+ 41in((0/1rnJ2)-
- ((~/Arnn)~I~ -
to2)- '3,
1'2,
334
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
a
0 Fig. 11.6. x 2 = cx,
+ d.
XI
and l,, is the modal damping factor. Using the sifting property of the delta function, we have F,, = 2P(u2
+ b2)'l2phub(l + b2/~2)1/2&,,n,
where,a, is the Kronecker delta. We have thereby obtained all the values of the required solution (4) to the deflection v of the plate. By the same token the load can be distributed on any straight line x2 = cxI d that intersects the plate (see Fig. 11.6). In this case the corresponding load is Peiorb[x2- (cxl + d)](l + c2)'I2.
+
11.15. DISTRIBUTIONS ON PLANE CURVES
Let us now distribute the generalized functions on a plane curve x2 = f(xl) that intersects the elastic plate as in Fig. 11.7. Accordingly, the value of the load is d x , , x2, t ) = PeiW'G[x2 - f(x1)lCl
+ (f'(x1)) 21
112
.
We have already discussed the delta function 6[x2 - f(xl)] in Section 5.8. As an example, let us take the curve to be a parabola, x2 = b(xl/u)2. In this case the differential equation for the deflection v of the plate is
Proceeding as in Section 11.14 [see (1 1.14.4)], we again obtain the value of
11.16.
DISTRIBUTIONS ON A CIRCULAR DISK
335
Fig. 11.7. x 2 = f ( x , ) .
the deflection ti(xl, x2, t ) , where the quantities Fmn are now
The next natural extension is to distribute the generalized functions on laminae and surfaces and to solve the corresponding boundary value problems for configurations such as shells. This is easily accomplished with the help of the surface distributions explained in Chapter 5. We content ourselves with discussing in the next section the distribution on a circular disk.
11.16. DISTRIBUTIONS ON A CIRCULAR DISK
The ideas of the previous sections can be extended by distributing the singularities on a circular disk. We take F(x) = ,f(p)h(x),
0 I 0 I 2 n , 0 I p I b,
(1)
where (x, p, 0) are cylindrical polar coordinates andf(p) is defined over the disk 0 p I b, 0 5 8 27c. Then (1 1.1.3) reduces to
336
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
If we set cc
=
8,
is easy to see that
- 8, it
Jo2n
+ p2 + 5’
[x’ =
do, - 2 5 p cos(8 - 8 1 ) ] 1 / 2 dcc
Jo2’
[x’
+ p2 + 5 2 - 2 5 p cos cc]””
Thus (2) takes the form
Next, we use the expansion [x’
+ p 2 + t 2 - 25p cos a ] - ” 2
in (3), where So, is the Kronecker delta and we use the orthogonality of the cosine functions. The result is u(x, p ) = 271
Suppose that u(x, p ) becomes
J; J ~ . I ( t ) t J o ( P P ) J o ( p 6 ) e -
(4)
l x l p d 5 dP.
= A(x, p )
on the surface S , p
= po(x), so
that (4)
where the kernel K ( x , 5, p ) is
Equation ( 5 ) is a Fredholm integral equation ofthe first kind forf(5) provided the equation p = p o ( x ) of the surface S is known. The inverse problem pertains to finding the equation of the surface when A(x, p),f(5), and b are prescribed. For the special case ,f(5) = S(5 - b), (4) reduces to u(x, p ) = 2 n b
J;
J o ( p p ) J o ( p b ) e - l x l pd p ,
which is the electrostatic potential due to the point sources on a ring.
(7)
12 Applications t o Wave Propagation
12.1. INTRODUCTION In Chapter 10 we have discussed various properties of the homogeneous and inhomogeneous wave equations in two and three dimensions. We derived their fundamental solutions and studied moving point, line, and surface sources. In Chapter 5 we considered various kinematic and geometrical aspects of the wave propagation in the context of surface distributions. In this chapter we consider some applications of these results and study partial differential equations whose solutions can be defined as regular singular functions. In problems of these kinds, singular surfaces can play two essentially different roles: one relates to the propagation of wave fronts, and the other, regular singular functions arise in the study of boundary value problems. In the latter case, the surface C. is given along with some data on it, and the solution is to be found in only some region of the space. A powerful method to attack these problems is to embed them in the whole space. This is achieved by extending the solution to the other side of the surface in some suitable fashion, as we did in deriving the Poisson integral formula in Chapter 10. We then obtain a regular singular function that satisfies the equation in the complement of X. 337
338
12.
APPLICATIONS TO WAVE PROPAGATION
If a generalized solution of an equation is a vector of regular singular functions with respect to C, then we say that C is a singular wave of that equation. More precisely, if F E E ( C ) is such that L F is continuous across C for all differential operators L of order less than k , but there exists an operator of order k such that L F is discontinuous, then it is said that C is a singular wave of order k . When k = 0 it is called a shock wave, and for k = 1 it is called an acceleration wave. In general, F describes the behavior of a mechanical system with different states in the two sides of C. The problem is to examine how F behaves at the wavefront. In this situation, F satisfies the equation in the whole space, even across C,and the discontinuities arise naturally [7].
12.2. THE WAVE EQUATION
One of the simplest cases in which singular waves exist is provided by the wave equation
We assume that C ( t ) is a sonic discontinuity; that is, A , the jump in F , cannot vanish. From the results of Chapter 5 we obtain 6t
+ (4
-
$+j’(C)
=
0.
For this equation to hold, both the regular part and the coefficients of 6(C) and S’(C)must vanish. Since A # 0, the coefficient of 6’(C) shows that G = fc. We choose the value G = c; this amounts to choosing a particular orientation. The fact that G is a constant implies that the propagation must be parallel, that is to say, that the wave fronts form a family of parallel surfaces. Indeed, if we let s be the arc length as measured along the orthogonal trajectories of C(t), then 6s/6t = c, dx,/ds = ni, and dni 1 6ni 1 6 ( - G ) d2xi - - - - - _ _ - -~ c 6t c 6xi ds2 ds
=
0.
12.2.
339
THE WAVE EQUATlON
This proves that the orthogonal trajectories of the wave fronts are straight lines, which is another way of saying that the propagation is parallel. Accordingly. if the wave front is known at a time t = t o , then we can construct it for all subsequent times. Next, we equate the coefficients of d(Z) to zero and obtain
+ 2c dA = 0, 6r
-2QA
’-A-- QA.
or
--
(3)
dS
We can integrate this equation to obtain / /.s
where CI is the distribution on the surface that corresponds to s three-dimensional case, Q(s)
=
(a,- sKo)/(l
-
2sQ0 + ? K O ) ,
= so. In
the (5)
where Q, and K O are the mean and the Gaussian curvatures, respectively, corresponding to s = so. Thus (4) becomes A
=
~ ( 1- 2sQ0
+s ~ K ~ ) - ~ ’ ~ .
(6)
To derive the jump in the normal derivative of F we appeal to (5.6.9) and obtain O=[02F]=G-2QB+V2A-or
Let T be a tube of rays that cut the surface Z ( t o ) and Z(t,) at elements of area S1 and S2 and R be the volume enclosed thereby. Then the divergence theorem applied to the vector field AZA yields (8)
Because div(A2A) =
d ~
ax,
(A’n,)
=
dA 2A -ni ax;
+A
where we have used (3), from (8) we derive
6n; 6Xi
340
12. APPLICATIONS TO WAVE PROPAGATION
Finally, let us consider the case of higher-order waves. Suppose that F and all its derivatives of order less than k are continuous but L F is discontinuous, where L is the partial derivative of the form
Since L F
=
LF, we have
and, accordingly, the foregoing analysis can be applied to the functions L F . The conclusions about the surface X ( t ) remain the same, as do the results for the quantity A , except that we now designate A = [ L F ] .
12.3. F I R S T - O R D E R HYPERBOLIC S Y S T E M S
Let us now consider the existence of the regular singular solutions of a system of quasilinear partial differential equations of the form
where aijpand b, are given infinitely differentiable functions of x , , . . . ,x,, t , and F , , . . . , F,, while Latin indices run from 1 to q and Greek indices from 1 to p . Furthermore, we shall employ the notation Aijkp = aUijp/dFk,
Bik
=
abi/aFk.
(2)
Our aim is to examine the existence of first-order singular waves. If we denote the jump of dFJdn by A, and recall the analysis of Section 5.5, we have
since F is supposed to be continuous. Next, we take the derivative of (1) with respect to x, and obtain
a2F
--
a
+ [-AiGn, + a i j p A j n , n p ] 6 ( X ) . ( 5 )
12.3.
341
FIRST-ORDER HYPERBOLIC SYSTEMS
Both the regular and the singular part of this equation must vanish. Multiplication of the singular part by ny yields -AiG
+ a i j a n a A j= 0,
(6)
which says that G is an eigenvalue and Ai is an eigenvector of the matrix aijana.For a singular wave to exist, this matrix should have at least one real eigenvalue. In this case the system is called hyperbolic. If G is an eigenvalue with multiplicity rn I q, then (6) enables us to write the strength Ai as a linear combination of m of them, say, A,, . . . , Am:
Ai = Ai(A,, A,, . .. , Am),
i
=
(7)
1,. . . , q.
It follows that the decay and growth of these discontinuities can be studied merely by determining the propagation of A,, . . . , Am. For this purpose we write the jump conditions for the regular part of (5):
t BijAjn7,
(8)
where Tj = [ d 2 F j / d n 2 ] .Multiplying (8) by n,,, we obtain 0
=
-Tic
[,,,
6A. + u i j s T j n p+ 2 + A~~~~
6t
dFj]
ny
~-
a x , axa
+ uijP6 A . + BijAj. 6xa
(9)
In order to eliminate the Tis from (9), we take rn linearly independent right , we designate Ay),r = 1, . . . , rn, and eigenvalues of the matrix a i j a n p which multiply (9) by each of them. This leads to 0
=
-
AT){ti + A,. ~
lJkP
ax,. ax,,
P F k aFj] ~~
,
r
= 1,.
. ., rn. (10)
These rn equations determine the propagation of A,, . . . , Am, and (7) will provide us with the rest. Let us also observe that we have
342
12.
APPLICATIONS TO WAVE PROPAGATION
and, accordingly, if F is known in the positive part, so will be the coefficients of this expression. A particular but quite important case is that in which the positive part is at rest, that is, for which F = 0 in the positive part; then -n,AjAk is the only nonzero term left in (1 1). Let us illustrate these results with the help of the equations of inviscid fluid dynamics, namely,
where i, B = 1, 2, 3; Fi are the components of velocity; R = In p, p is the density; a 2 ( R ) = p’(p); and p is the pressure. We assume that a sonic discontinuity is propagating into a region at rest and at constant pressure. In this case (6) leads to
+ F,n,Ai + u2rni = 0, - G r + F p n p r + Apnp= 0,
- GA,
where Ai
=
(14) (15)
[ d F J d n ] , r = [ d R / d n ] . From these equations we obtain G = Fpnp f
=
(16)
fa,
since F , = 0 on the surface. We take the positive sign to fix the ideas. Another relation that follows is (17)
Ai = urni,
and so only r needs to be determined. Furthermore, since the velocity G is a constant, it follows from the analysis of Section 12.2 that propagation is parallel. Next, we attend to (9), which in the present case reduces to - TiG
6A . + a2Fni + 2 + AiApnp+ a’ St
6r 6Xi
-
+ cr2izi = 0,
where r‘ = [ d 2 R / d n 2 ] and c = a a 2 / d R in the positive part. Multiplying (18) by ni and (19) by a, adding the resulting equations, and using (17), we derive the transport equation dr/ds - rQ
+ (1 + c)r2 = 0,
which determines the propagation of r.
(20)
12.4.
343
AERODYNAMIC SOUND GENERATION
12.4. AERODYNAMIC S O U N D GENERATION
We can effectively extend the results of Section 12.3 by including the dissipative terms in (12.3.12). Such a system has been studied by FfowcsWilliams and Hawkings [40]. To facilitate the comparison, we write the mass and momentum equations in their usual notation, namely,
3~ -+c7t
3(~ui) ?Xi
= 0,
where p is the density, the ui are the components of the velocity vector, and g i j is the symmetric stress tensor. Let X ( t ) be a singular wave of order zero. Then equating the singular part of (1) with zero gives CP(Ufl
-
G)1
=
0.
(3)
Let Q denote the common value of p(u, - G) on X; then for any function F discontinuous across C we have CFp(ufl -
GI1 = QCFI.
(4)
Similarly, equating the singular part of (2) with zero yields
QCuil = Coijlni. The particular values of the stress tensor lead to interesting results.
(5)
344
12.
APPLICATIONS TO WAVE PROPAGATION
The following type of problem associated with (1) and ( 2 ) is also of great importance. Let us suppose that a body of arbitrary shape and surface X(t) is surrounded by a continuum medium. Then the normal velocity u, of a point on the surface of the body is equal to G , the velocity of the fluid. This means that we have to solve the partial differential equations (1) and ( 2 ) subject to the boundary condition
u,
=
uini = G
on X ( t ) .
This boundary value problem can be transformed to an integral equation by means of the following embedding procedure. Let us extend the functions appearing in this boundary value problem to the inside of the body in such a way that the equations are satisfied there. A simple way to do this is to assume that all the flow parameters vanish inside the body. In that case the jumps across X(t) reduce to the boundary values from the fluid side. Because [ F 1 F 2 ] = [F1][F2], we obtain the following system of equations for the extended functions:
a-’ + - -a((pui) - ( - p G + po,)B(Z) at dxi
=
0.
(7) Taking the time derivative of (6) and the space derivative of (7) and subtracting the resulting relations, we obtain a2p at2 -
a2(puiuj
~
- Oij)
axi a x j
a + axi (qjnjbm). -
It is convenient to subtract po from p so that /j = p - po vanishes at infinity. Clearly, (8) holds if p is changed to p on the left side. Subtracting the term c2V2/j from both sides of (8), we obtain
where T j = p u i u j - oij - c2/jSijis the Lighthill stress tensor. It is interesting to observe that we can have extra source terms in (9) if we extend the parameters in a different fashion. For instance, if the density is assumed to be a constant, say, A, inside the body, then (6) is replaced by ap/at
+ a(pu,)/ax, = AG~(z),
(10)
12.5.
345
THE RANKINE-HUGONIOT CONDITIONS
while (7) remains unchanged. Then (9) becomes
The extra term on the right side of this equation will, however, not change the solution in the fluid because ij depends on A only inside the body. By giving other values to the parameters inside the body, we can get various other equations, with source terms that affect the solution only inside the body. Finally, the convolution with the fundamental solution of the wave equation gives the integral equation
X E(T
- t , 4’
- X)
dJJdT.
(13)
Assigning the value po to A and taking derivatives outside of the integral sign, we obtain the Ffowcs-Williams-Hawkings integral equation [40].
12.5. THE RANKINE-HUGONIOT
CONDITIONS
When the hyperbolic system (12.3.1) is of the form aFi/dt
+ a M i j / a x j= 0,
(1)
where M i j is a suitable tensor, we can obtain the jump conditions for a shock front by applying (5.5.4) and (5.5.5). Then (1) becomes
+ a M i j / d x j + [ M i j ] n j S ( C )= 0,
aFi/dt - [Fi]GS(C) or dFi/dt
+ d M i j / d ~ +j { [ M i j l t ~-j G[Fi]}6(C)
= 0.
The singular part immediately yields the jump conditions : [ M i j ] n j= G[Fi]. Now we show that the system of equations (12.3.12) and (12.3.13) can be readily put in the form of (l), which leads to the well-known RankineHugoniot conditions across a shock wave in fluid mechanics. For this purpose
346
APPLICATIONS TO WAVE PROPAGATION
12.
we use the conventional notation of Section 12.4, so that the equations for the conservation of mass, momentum, and energy are aP
at
a
-(POj)
dt
-a( l u . u .
a +axj
(POj)
a + axj
-( P V i U j
+ pel +
a t 2 1 1
= 0,
(3)
+ P 6 i j ) = 0,
(4)
ax
respectively, where all the terms have already been defined except for e, which is the internal energy. This system can be put in the form of (1) if we set
M''= lJ
i
+ +
PO2
PO1
PVllJ2
+P
pv2Ol
PVlV2 PO202
+p
(7)
P O 3 u1
P O 3 O2
Pv3 O3
PVlZ
P U 2 Z
PO3
+P
z
where Z = +viui e p / p . Substituting ( 6 ) and (7) in (2), setting v = [ p ] , which gives the strength of the shock front, and simplifying, we obtain V
[PI = l+v PI(G - u l n I 2 9
where h
=
e
V
CU~I = _ _ (G - O l n ) n , , l + v
+ p/p. These are the required jumps.
Functions That Have Infinite Singularities at an Interface
13.1. INTRODUCTION
In Chapter 5 we discussed the theory of distributional derivatives with jump discontinuities across points, curves, and surfaces. In the process we obtained the values of the jumps of various derivatives across these interfaces. These results do not in general apply if the fields have infinite singularities at a moving and a deforming surface. This happens in the theories of electromagnetism and magnetohydrodynamics, where surface densities can be generated by the sudden creation of electromagnetic multipoles in free space. Take, for instance, the Maxwell equations,
curl E = -aB/dt, -
div B
=
0,
curl H - 3D/dt = J , div D = p , where E is the electric field, D the displacement current, B the density of magnetic flux, H the magnetic field, J the current density, and p the charge density. 347
348
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
The corresponding equations for the electromagnetic potentials
A are
E
4 and
-
+ grad 4 = - -,3atA 1
34
- _ _ = c 2 at
-
div A,
(7)
where the speed c is a constant. Now, if the fields E , B,D,and H have only the jump discontinuities across an interface X,then from (3) and (4) it follows that the current density J and the change density p are of the order of the Dirac delta function (O(6)). Similarly, from ( 5 ) and (6) it follows that if the potentials 4 and A are O(6), then E and B are O(6’).Then from (1)-(4) we find that the densities J and p are O(6”). The same relationship is valid for then higher-order infinite discontinuities. For example, if E , B,D,Hare O(hcn)) J , p are O ( P +’). Similar situations arise in the study of vortex sheets, atmospheric disturbances, and magnetohydrodynamic flows. Our aim in this chapter is to use and extend the results of Chapter 5 to the present context. We follow the notation of that chapter and confine ourselves to R 3 so that x = (xl,x2, x3) denotes the field point and t denotes time. Some of the results of this chapter have been derived by Costen [41] by the classical technique of expressing the partial differential equations in their integral forms and then taking the usual limiting processes across the surface of discontinuity. We follow the method of generalized functions [42].
13.2. DISTRIBUTIONAL FIELD EQUATIONS O F THE FIRST ORDER
Let us discuss the first-order equations
where 8 $, ,I,and m are tensorial functions of x and t . The functions 8 and $ have jump as well as delta function singularities on a moving and deforming interface X. Thus, for a field F equal to 8 or $ we have F(x, t> = F d x , t )
+ Fa(x, t ) W > ,
(3)
13.2.
349
DISTRIBUTIONAL FIELD EQUATIONS OF THE FIRST ORDER
where the function F , has the representation F , = F + H A [see (5.5.14)], where A = [ F ] and H is the Heaviside function. In order to derive the distributional derivatives of these functions, we need some results from Chapter 5, which we collect here for ready reference:
SF 6t
dF dn
dF
-=-++---, c?t
-6 -F - -_ d F dxi
axi
I?,
nF
-;
ail
these are (5.2.15) and (5.2.16), respectively. Equations (5.6.18) for b and 3//axi yield
3
-
at
6(C) = - G6’(C) = - G(2Qd(C))
a .
+ d,6(C),
axi b ( C ) = ni6’(C)= ni(2Q6(C)) + d,6(C).
-
=
a/& (54 (5b)
Other relations that we need from Chapter 5 are
and
which are (5.5.5) and (5.5.4), respectively. With the help of (3)-(7) it follows that for a differentiable functionj‘(x, t) defined on C ( t ) we have
3
-
at
6f 6t
(,fd(C)) = - 6(C) - Gfd’(C)
and
where we have used (4).
350
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
With the help of (3) and (6)-(9) we find that the right sides of ( I ) and (2) become
and -
grad $
=
grad
$H
+grad($,G(X)) + 2AR$, +
+ dnCfi$aW>l,
(1 1)
where A = n i . If we neglect the last term in (1 0) and (1 1 ), we recover Costen’s results [41]. Next, we consider the first-order equations -
j = curl p
(12)
and
5
=
divq,
where the vector p and scaiar q also have the representation (3). To put the right side of these equations in the form of ( I l), we can start with the (5.5.25) and (5.5.26): -
+ A x [F]S(X), div F = div F + A . [F]S(X)
curl F = curl F
(14)
-
(15)
or find directly from ( I 1) that
curl p
= curl p H
+ dn[fi and
+ [A
x
b] + ~ R ( Ax
pa) + [curl - A
x paW)l
div q = div q H + [A. [q] + 20(ri. 4,) + dnCfi .q a W ) l ,
respectively.
x a/an]p,]h(~)
(16)
+ (div
-
fia/an)q6]6(x) (17)
13.2.
351
DISTRIBUTIONAL FIELD EQUATIONS OF THE FIRST ORDER
The coefficients of 6(C) in (lo), ( 1 I), (16), and ( 1 7) give the classical surface densities of A, m, j , and q in (l), (2), (12), and (1 3), respectively. Let us write them as follows:
A,
=
G[8]
+ 2RG0, - (a/& + G a/aH)8,,
(18)
+ 2RA$, + (grad - A J/dn)$,, ,ja = A x [ p ] + 2R(A x pa) + (curl - A x i?/dn)p,, q, = A . [q] + 2R(A. (1,) + (div - A . d/dn)q,. m, =
(19)
ti[$]
(20) (21)
These densities have interesting physical interpretations. The term G[8] in (18) represents the rate of increase of density due to the snow-plow action. The term 2RG0, accounts for an expanding (receding) bulge on C that tends to cause local dilution (concentration) of 0, as it appears to the observer moving with the surface Z. The term (J/dt + G 6/6n)0, = 68,/6r represents the time rate of change of 8, as it appears to the observer moving with the surface C. In (19) the terms A[$] and A2R$,are the normal contributions from the discontinuity in $ and from the curvature of C, respectively. The term grad -i? d/an is the tangential contribution. The terms in (30) and (21) have similar interpretations. The quantities d,[ - GH,G(Z)], d,[A$,G(C)], d,[A x pa6(C)], and d,[A . qa6(C)] in (lo), (1 I), (l6), and (1 7), respectively, yield both the tangential and normal contributions on the moving surface. Let us check the previous results for the expanding sphere r - ct = 0, where c is a constant. In spherical polar coordinates ( r , (ol, w 2 ) we may take 0 to be
+
0 = . f ( r , w,, (u2, t ) g(r, wl, w 2 ,r)H(r - ct) = 0, 0,6(r - ct).
+
+ h(r, tolr w 2 . t)s(r - ct)
Thus 0,
[8] = g,
=
h,
G
A
= c.
=
r,
R
=
-
l/r,
and (1) becomes
- -_ ?f- -&HI( r - ct) at
at
+ cl,[ch6(r - ~ t ) ] . Accordingly, [A]
= -ag/dt
and
+ [cg
-
2ch
-
(g +
c
;)h]d(r
-
ct)
352
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
which agrees with (1 8). The other densities can be checked in the same way. In Sections 13.3-13.5 we present applications of the formulas of this section. However, we neglect terms in the derivative operator d , from our analysis in order to compare the results with classical theory.
13.3. APPLICATIONS TO ELECTRODYNAMICS
Let us apply the results of Section 13.2 to (1 3. I S ) - ( 13.1.7) to the equations of electrodynamics :
E
8A +grad 4 = - -, at
a4-- - -1 _ div A . c2
at
(3)
Substituting the values of the distributional derivatives (1 3.2. lo), (1 3.2.1 l), (13.2.16), and (13.2.17) for functions with infinite singularities in (1)-(3) and using the fact that these equations are satisfied in their classical form on both sides of C,we obtain
Ed + ii[4]
+ 2Mi4, + (grad - A d/dn)4, = G[A] + 2RGAd - (a/& + G a/an)A,, B, = A x [A] + ~ R ( xA Pa) + (curl -ii x a/an)pa, G[4] + 2QG4, - (a/& + G d/an)4, = [A] + 2n(fi. A,) + c2(div - A . a/an)A,,
(4) (5) (6)
where 4, A are O(6) and E , B are O(6’). Next, we apply the results of Section 13.2 to the Maxwell equations (1 3.1.1)-( 13.1.4), namely,
-
curl E
=
-
div B
-aB/at, =
0,
GFl H - 8D/dt = J , div D = p.
353
13.3. APPLICATIONS TO ELECTRODYNAMICS
They yield the jump relations
A
+ 2R(n x E,) + (curl - A x d/Sn)E, G [ B ] + 2RGB, - (a/& + G d/dn)B,,
x [El =
[A. B ]
(1 1)
+ 2R(A. B,) + (div - A . d/Sn)B, = 0,
(12)
J , - fi x [ H I - 2R(A x H , ) - (curl - A x d/Sn)H, = G [ D ] + 2RGD - (a/& + G d/dn)D,, pa = i i . [ D ]
+ 2R(A. D,)+ (div
- A . d/dn)D,,
(13) (14)
where E , B, D , H are O(6) and J , p are O(6’) while J6 and pa denote the surface current and charge densities, respectively. Finally, we apply the analysis of Section 13.2 to the equation for conservation of charge, and get
div J
(15)
= -dp/dt,
+ 2R(A . J,) + (div - A . d/dn)J, = G [ p ] + 2RGp, - (a/& + G d/dn)p,,
A. [J]
(16)
where J , p are O(6). Note that in applications of ( I 3.2.18)-( 13.2.2I ) the order of singularities must be compatible. For instance, jump conditions (4) and (5) are not applicable simultaneously with (1 1) and (12) because the singularities in E and B are of different order unless 4, = A , = 0. Similarly, jump conditions (13) and (14) are not compatible with (16) unless H , = D, = 0. Static Limits
Let us consider the case of a fixed interface between two nondispersive media with zero net charge but across which 4 is discontinuous so that
C4I
=7/h,
where E,, is the permittivity of the vacuum and jump conditions (4)-(6) reduce to B,
E, = - A [ 4 ] , Accordingly, we may set E6 =
=
A
7
x [A],
-AT/E~,
D =E ~ E , Da
=
-AT,
pa = A = B = B, = 0.
is a given function. Then
A. [ A ]
=
0.
354
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
+
Then jump condition (14) becomes A . (E" E ) - 2R.r - div A T A . aA/dn = 0. With the identities div A = -2R and A . dA/an = 0, the preceding relation becomes A . [El
=
0.
(17)
Similarly, ( 1 I ) yields x
[El
2R(A x A) + curl(TA) - A x @/&)(AT) = (grad T ) x A + T curl A - T A x aA/an - A x A &/an = (grad T) x n + T(cur1 A - A x dA/dn). =
Because curl A = A x at?/&, the preceding relation reduces to A x [El
-(l/c0)A x grad
=
T.
(19)
Equations (17) and (19) agree with those given by Stratton [ 4 3 ] . Next let us consider a fixed interface devoid of surface current but across which A is discontinuous owing to a surface density of magnetization M such that A . [ A ] = 0,
A x [A]
=
POM, - poA(A.M,),
where p o is the permeability of the vacuum. Accordingly, we may set
ya),
B,
=
Po M - Po fi(fi.
H
=
B/pO - M ,
Ha
=
B,/PO - M , = - A ( f i . Ma),
J,
=
4
=
D
=
D, = 0,
where M is the volumetric magnetization. Then (12) reduces to [A. B]
+ 2poA.( M , - A(A.
M,)
+ po(div - A .
or
Now we use the identities (18) and
a / d n ) ( M , - A(A. M,)) = 0,
13.4.
MAGNETOHYDRODYNAMIC WAVES I N CONDUCTING FLUID
355
so that the preceding relation becomes
[A. B]
=
-po[div M , - (ii . M , ) div A - (A. hM,/hn)]
=
-p,,A.curl(A
(22)
x Ma),
Similarly, ( 1 3) becomes -A
x [B/po]
+ 2R{A x
(A(A. M , ) }
=
(curl - n x d/dn)A(A. Ma),
or
=
(grad(A. M , ) ) x A - ;A(
(A. M , ) }
+ (A. M , ) curl ii - ii x
{(A. M , )
};,
which, in view of ( 1 8), yields
A
x [ B ] = - p 0 A x grad(A. Mb).
(23)
In this derivation we have deleted the jumps in the volumetric magnetization M . Equations (22) and (23) agree with those given by Stratton [43].
13.4. MAGNETOHYDRODYNAMIC WAVES IN A C O M PR ESSl BLE PERFECTLY CON DUCT1N G FLUID
The system of partial differential equations that describe the motion of a compressible fluid in a magnetic field is
+ div(pu) = 0, p(du/dt + u . grad u) = -grad p + (1/4rc)(curl B ) x (3p/c3t
dp/dt
+ u . grad p = a2(dp/dr + u . grad p ) ,
(1)
B.
(2) (3)
where we have neglected the dissipation due to finite conductivity, viscosity, and heat conduction. The quantity p is the density, u the velocity vector, p the pressure, a the speed of sound (a2 is the derivative of p with respect to p at constant entropy), and B the density of the magnetic flux. Since the fluid is
356
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
perfectly conducting, the equation for the rate of change of B is dB/&
curl(v x B).
=
(4)
Let the departures p, v , p , a, and B from the uniform values p,, 0, p , , a,, and B,(where B , is a constant vector), respectively, be regarded as so small that their squares and products are negligible, so that (1)-(4) simplify to the linear system aP
-
at
av
--
at
-
=
1 -grad
-p,div
p
Po
+
u,
(5)
1
curl B )
47%
x B,,
dp = a 2, a-,P at
aB
-=
at
(7)
at
curl(v x B,) = B , . grad u - B , div v.
These equations can be analyzed with the help of the techniques of this chapter when the magnetohydrodynamic wave fronts carry delta densities. In that case we write the system ( 5 ) - ( 8 ) in distributional derivative form and use the formulas developed in Section 13.2. We thereby obtain
+ G a/an)p, + 2RA. [v,] + (div -A.
~ [ p +l ~ R G P ,- (a/ar =
G[v]
p o { A . [u]
a/an)u,},
+ 2QGv, - ( a / & + G a/an)v, = (l/p,){A[pl + 2RAp, + (grad -A a/dn)p,} - 1/4np0{A x [B] + 2R(A x B,) + (curl -A
(9)
x d/&)B,} x B,,
(10)
+
+
G C ~ I 2 ~ ~ -p( a,p t G a/an)p, = a;{c[pl + ~ S Z G ~ (, a p t G a/an)p,>, G[B]
+
+ 2RGB, - (a/& + G a/an)B, - B o . {A[v] + 2RAv, + (grad - A a/an)v,} + Bo{A. [v] + 2R(A. u,) + (div -A. a/an)v,}.
(1 1)
=
(12)
When the field B , is normal or tangential to the wave front, these formulas simplify. The interpretations of the various terms are similar to those given in Section 13.2.
13.5.
SECOND-ORDER DIFFERENTIAL EQUATIONS
357
13.5. SECOND-ORDER DIFFERENTIAL EQUATIONS
So far we have considered the system of partial differential equations (1 3.1.1)-( 13.1.7) independently of each other. However, if we combine them suitably we get various partial differential equations of second order. For example, (13.1.7) and (13.1.5), or
yield partial differential equations of second order either by taking the gradient of both sides of ( I ) and then substituting from (2) or by taking the time derivatives of both sides of (2) and then substituting from ( I ) . The same is true of various other combinations of equations of that system. Indeed, this system is of the form 1' = -3r]/i3t,
(3)
c = curl u, f: = div 21.
They are a specialized form of (13.2.1), (13.2.2), (13.2.12), and (13.2.13). By differentiating and combining (3)-(7) in appropriate pairs as mentioned, we get the second-order identities
grad y -
= -ag/i?t,
cur1.f = -%/at,
(8) (9)
-
div f' = - ;?r:/dt,
gradt; = curl c
+ v2v.
If r] and u are of O ( H ) ,then 7, g,!; c, E are of O(S),so that the corresponding delta densities are Y6
=
GCrll,
(12)
358
13.
FUNCTIONS THAT HAVE INFINITE SlNGULARlTlES
Next, wedevelop(8)-(11)in thesame manner as(l3.2.1),(13.2.2),(13.3.12), and (13.2.13) and obtain the jump conditions
+ 2RAy, + (grad - A d/dn)y, = G[g] + 2RGg, + G d/dn)ga, A x [ f ] + 2R(A x fa) + (curl -A x d/dn)f, = G[c] + 2RGc, - (a/& + G d/dn)c,, ri . [ f ] + 2R(A .fa) + (div -A. d/an)f,, = G[E] + ~ R G E, ( d / d t + G d/dn)c,, ACE] + 2RAs, + (grad -A a/&)&, = A x [c] + 2R(A x c,) + (curl - A x d/dn)c, + [Au/dn]. A[?]
(17)
(18) (19) (20)
In the next stage we substitute the identities (12)-(16) in (17)-(20) and use (5.2.21), namely, (d/dt
+ G d/dn)A = -(grad
as well as the relation D[u] curl. We thereby obtain
(a/& (d/dt
+ G d/dn)(A x
=
-A d/dn)G,
[Du], where D stands for
a/&,
grad, div, or
+ G d/an)[q] = - [ y ] + GA . [g], [ u ] ) = G [ c ] - A x [ , f ] - (curl - A
(21) x d/dn)G[u],
(22)
(a/& + G a/dn)(A. [ u ] ) = G [ E ] - A . [f] - (div -A.
d/dn)G[u], (23)
[du/dn] = ACE] - A x [c] + 2R[v] + (grad -A d/dn)A . [u] - (curl - A x d/dn)A x [u].
(24)
We can use these identities to derive the corresponding formulas for the electromagnetic field. For instance, let us consider the electromagnetic equations
J
-
H
=
(25)
-aD/at,
(26)
p=divD. When we set u = D,
.f = J - curl H,
E =
p,
13.5.
SECOND-ORDER DIFFERENTIAL EQUATIONS
359
then (25) and (26) correspond to (5) and (7), respectively. Accordingly, (23) yields
(a/& + G d/dn)(A . [ D ] ) = G [ p ] - A . [ J
-
curl H ] - (div - A . d/dn)G[D],
which generalizes (13.3.16) in the sense that it holds for discontinuities of arbitrary order. Indeed, this relation remains valid even when all the brackets are removed.
Linear Systems
14.1. OPERATORS
Recall that we define a function as a rule that maps (transforms) numbers into numbers, whereas a functional maps functions into numbers. An operator maps functions into functions. Let a class of functions be given, all defined for a variable, say, time t , - x, < t < a.Then an operator (transformation) T assigns a member of this class (inputs, excitations, or signals) to members of a second class of functions (outputs or responses). We shall use the symbol x(t) for an input and y(r) for the corresponding output. A system is a mathematical model of a physical device and is represented by an operator T. Then y ( t ) is called the response of x ( t ) due to the given system. A system is often represented by a box as shown in Fig. 14.1. Inputs, such as a force applied to a mass, a voltage applied to an electrical circuit, or a heat source applied to a vessel filled with a liquid, are given; outputs are to be calculated or measured. The outputs corresponding to the
x(t)
-
-
output
input
Fig. 14.1. A system.
360
y(t)
14.1.
361
OPERATORS
inputs we have mentioned might be the velocity of the mass, the voltage across a resistor, and the rate at which the liquid flows through a pipe. If the operator T satisfies the identity T[c,x,
+ c2xJ
= c']
+
T[xl]
('2
T[XJ
=
"IJ'1
+
('2J'2,
for all inputs x 1 and x2 and for all constants c', and c 2 , then T is called a linear operator. A system is called linear if and only if the operator which represents it is a linear operator. This means that a linear system obeys the superposition principle, which requires that the following two tests be satisfied. by
(1)
Multiplying the input by any constant c must multiply the output
c'.
(2) The response to several inputs applied simultaneously must be the sum of the individual responses to each input applied separately.
It follows that if x(r) = 0 for all t , then so is y(r). Thus, there can be no output without an input. Accordingly, there is no stored energy. A linear system is therefore also called a relaxed system. Let us write down a few examples of linear systems: reflection, x ( t ) -+ v ( t ) = x( - t ) ; (2) a constant gain amplifier, x(r) v ( t ) = cx(r), where c is constant; (3) the difference, x(t) y(r) = x(t + 1) - x ( r ) ; (4) an ideal time delay of duration a, x(t) -+ y ( t ) = x(r - N ) = xa(t), where 11 is a positive number: (5) the multiplier, x ( t ) y ( t ) = f ( t ) x ( t ) .where f ( t ) is a given fixed function; (6) the differential, x(t) -+ y ( t ) = f'(D)x(r),where / ( D ) is a linear differential operator, (1)
-+
-+
-+
with as that are functions o f t only;
r-
(7) the integrator x ( t ) y ( t ) = (8) the integrodifferential system -+
co
x(r) t l r ;
where a and b are constants; (9) the Fourier transform, x ( r ) -+ y ( t ) = JYm x(r)eirrdr; (10) the Laplace transform, x(t) -P y ( t ) = J?x(r)e-" dr.
362
14.
LINEAR SYSTEMS
So far we have implicitly assumed that all the functions and constants involved in the discussion are real. However, the analysis holds for complex quantities as well. In that case we write x(t) = x,(t) + ix2(t) and y ( t ) = yl(t) iyz(t),so that xl(t) -, y , ( t ) and xZ(t)-, y z ( t ) . Of course, in a physical system we expect all the input and output signals to be real valued. An operator is called stationary if for each fixed real number a we have
+
nxa1
=
(7-Cxl)a =
-
4.
The corresponding system is calledJixed or time invariant. Accordingly, for a time-invariant system, the only effect of delaying (or advancing) an input signal is to produce a corresponding delay (advance) of the output signals. We shall be mainly concerned with time-invariant linear systems. A system will be associated with a specific class of functions that are admissible inputs in the sense that the corresponding output functions are well defined. For instance, for a system defined by a differential operator we expect ~ ( tto) be differentiable. An operator is called continuous if, whenever a sequence of input signals converges to zero, then the corresponding sequence of outputs signals also converges to zero.
14.2. THE STEP RESPONSE
Let U ( t )be the response to the Heaviside function H ( t ) , as shown in Fig. 14.2. The function U ( t ) is called the step response. Example 1. Let us find the step response for the system
(a,D
+ a l ) y ( t ) = x(r),
a , # 0, x(t)
=
0 for t < 0.
(1)
This means that we have to solve the differential equation (uOD
+ a l ) U ( t ) = H(t),
a, #
0, U ( t ) = 0, t < 0.
(2)
Its solution is U ( t ) = (l/u,)(l - e-al"ao)H(t).
(3)
Example 2. Consider the system
[Dz+ (a + b)D
where x(t) = 0 for t < 0.
+ ab]y(t)
= ~(t),
a # 0, ab # 0,
(4)
14.3.
363
THE IMPULSE RESPONSE
Fig. 14.2. The step response function U ( t ) .
The step response for this system is given by [D'
+ (a + b)D + ab]U(t) = H ( t ) .
Its solution is
14.3. THE IMPULSE RESPONSE
From the analysis of Section 14.2 we find that the response to rectangular pulse of height A and width a, as shown in Fig. 14.3, is y(t) = A[U(t
+ +a)
-
U(t - +a)].
Fig. 14.3. Response to a rectangular pulse of height A and width a.
(1)
364
14.
If U ( t ) is a continuously differential function we write y(t) =
and let a + 0, A
--+ 30
[
U(t
ALI
+ )a) - U(t -
+U)
U
such that Aa
=
LINEAR SYSTEMS
1
1. We then have
(2)
y ( t ) = dU(t)/dt.
This is called a unit impulse response and is denoted by h(t). We know from Chapter 1 that the rectangular pulse (1) yields d(t) in the limit we have considered; accordingly, the response to the delta function is h(t). Example 1 . In Example 1 of Section 14.2 we found that the step response to the system (14.2.1) is (14.2.3). Hence, the impulse response of the present system is
(3)
h(t) = (e-a"i""/uo)H(t).
This is precisely the solution of the system
@OD
+ ally = so),
(4)
as is easily verified by setting y ( t ) = G ( t ) H ( t ) and following the method explained in Section 9.8. Example 2. Let us now find the impulse response of the system of Example 2 of Section 14.2, namely, [D2
+ (a + b)D + ub]y(t) = x(t).
(5)
Differentiating the step response (14.2.6), we obtain e-bi
h(t) =
- e-ai
a-b
H(f)?
which is precisely the solution of the system [D2
+ (a + b)D + ub]h(t)= d(t).
(7)
14.4. THE RESPONSE TO A N ARBITRARY INPUT
Having found the response function corresponding to the delta function, we can convolve to get the response to an arbitrary input. Thus y(t) = X(t) * h(t) =
I W
W
x(T)h(t - T ) dT =
m
X ( t - T ) ~ ( T dT. )
(1)
14.5.
GENERALIZED FUNCTIONS AS IMPULSE RESPONSE FUNCTIONS
.
365
U
U
Fig. 14.4. The Qutput z ( t ) due to the impulse responses h , and h2 connected in cascade.
Example 1. In Example 1 of Section 14.3we found that the impulse response to the system (14.2.1) is given by (14.3.3). Thus the output y ( t ) for an arbitrary x ( t ) is
Example 2. Proceeding as in Example 1, we find that the output y ( t ) for the system (14.3.5), given impulse response (14.3.6), is
By using the associative property of the convolution, we derive the impulse response h3 due to the impulse responses h , and hz connected in cascade, as in Fig. 14.4: h 3 ( t ) = hi
* hz
=
h,(t
-
~ ) h Z ( td) ~ .
(4)
14.5. GENERALIZED FUNCTIONS AS IMPULSE RESPONSE F U NCTlO NS
From the sifting property, x ( t ) = JZm x(t - T)~(T) dt, we infer that h(t) is itself that impulse response function of the time-invariant linear system having the property that it leaves every input signal x ( t ) unchanged. Similarly, from
J-
00
x(r - a )
=
x(t -
T)h(T
-
a ) rlt
m
we find that the impulse response function associated with an ideal time delay system of duration a is h(t - a). Next we appeal to the differential property of the delta function,
366
14.
LINEAR SYSTEMS
and infer that the generalized function dcn)(t)is that impulse response function characterising the time-invariant linear system transforming every n-times continuously differentiable input x ( t ) into the nth derivative x'")(t). Since the generalized functions are impulse responses for various systems, they can be convolved with other impulse responses, leading to very interesting cascade connections, such as 1. h * 6, = 6, * h. 2. s,(f) * a,([) = j_"oua,([ - T ) b b ( T ) d z = 6,+b(f), and 3. 6'"'(t) * 6'"'(t) = J?m 6'"'(t - T)6'")(T) dT = 6'm+")(t),
where m 2 0, n 2 0.
14.6. THE TRANSFER FUNCTION
The response y ( t ) for an exponential input x(t), x(t) = e"',
s
=0
+ io,
plays a key role in the analysis of linear systems. It is usually denoted H,(t). For a time-invariant linear system it can be deduced from the following two considerations :
( I ) From the time invariance we observe that for any given fixed value of T the response to the input x(r + z) is H,(t + T) and for x(t) = esr we have x(t + T) =
esU+r)
= esr est - eSTx(t).
(2) In view of the linearity ofthe system, we know that ifx(t) -+ H,(t), then x(t
Specifically, for t
=
+ z) + eSrHs(t).
(1)
0, ( I ) becomes X(T)-+ eSrHs(0)= eSrH(s),
where we have set H,(O) = H(s). Thus, the response of a time-invariant linear system to the input x(t) = es' is y ( t ) = H(s)e".
(3)
The quantity H ( s ) depends only on the parameter s = 0 + iw and on the particular system concerned. It is called the transfer function of the system and is defined in some suitable region of the complex plane ( 0 , ~ ) .
367
14.6. THE TRANSFER FUNCTION
The response to the input x(t) integral, is
= esf, obtained
rm I
from the convolution
m
es'h(t
- T)
dT =
m
fZ"'f-r)h(T) dt,
(4)
where h(t) is the impulse response of the system. Comparing (3) and (4), we obtain
or
J-me-"h(T) m
H(S)
=
dT.
(5)
This relation yields the following two results: (1)
Ifa
=
0, so that s
=
iw, we have
where h ( u ) is the Fourier transform of h(t). In this case the transfer function is called the.freyuency response junction. ( 2 ) For a system whose impulse response function h(t) vanishes for all negative values oft, we obtain from (5) H(S)
= /ome-srh(T)d S = y { h ( t ) ) ,
(7)
where 2'stands for the Laplace transform. This means that if the input x(t) vanishes for all t < 0, then so does the output y(t). Accordingly, no output signal can ever precede its input signal; that is, the signal is causal (or nonunticiputiue). Equation (7), which gives the transform function H ( s ) as the Laplace transform of the impulse response, facilitates its evaluation. Taking the Laplace transform of y ( t ) = STrn h(t - T)X(T) lit, we have y { J o > ) = H(s)Y{x(r)).
We illustrate these concepts by the help of the following example.
(8)
368
14.
LINEAR SYSTEMS
Example. The zero-state response Let the system be in the form of a general nth order input-output ordinary differential equation: d"Y %dt" + a,-, =
d"- y dt"~
d"x bmdt"
+ b,-
+ 1
* * *
+ a , dY - + aoy dt
d"- 'x dtm-
+ . . + box,
-
y(0) = 0, y'(0) = 0, . . . , y"-'(o) = 0, (9b) x'(0) = 0, . . . , xm- ' ( 0 ) = 0. (9c) x(0) = 0, Since, the initial conditions are zero, the stored energy is zero. This is called a zero-state response. Taking the Laplace transform of both sides of (9), we find (a,Sn
+ a,-
=
1.4-
(b,s"
+ . . . + a1s + ao)j(s)
+ b,-
+ . . . + b,)Z(s),
where we have used the notation of Chapter 8, so from (8) we find that
+ + + bo + a,-,s"-' + . . . + a,s +
b,Sm j j ( s ) = (a,Sn
bm-lSrn-'
Thus H(s) =
+ + a,-
b,Sm a,s"
* * .
LIO)
n(s) = H(s)Z(s).
+ ... + bo + . . . + als + a,'
b,,-l~"-' 1s"-
By factoring the two polynomials in (10) we can write
H(s) = K
( s - z l ) ( s - z 2 ) . . . ( s - z,) (s - PlMS - P 2 ) . . * (s - Pn)'
where K = b,/a,. The quantities z,, z 2 , . . . , z , are the zeros of H ( s ) , and p , , p 2 , . . . , pn are the poles. From this relation it is apparent that H(s) can be completely specified by its zeros, poles, and the multiplying factor K , which is assumed to be nonzero. The zeros and poles may be complex numbers and can be represented graphically in a complex plane, which is called the complex,frequency plane. Let us assume that the Laplace transform of the input x ( t ) can be written as a rational function of s, Z(S) = N(s)/D(s),
(12)
where N ( s ) and D(s) are polynomials. Then from (8), (1 l), and (12) it follows that
14.7.
369
DISCRETE-TIME SYSTEMS
By factoring N ( s ) and D ( s ) we can obtain all the zeros and poles ofj(s) from this expression. This facilitates the inversion of y(s), and we thereby obtain the required solution y(t). This analysis can be readily extended to the case of the nonzero initial conditions.
14.7. DISCRETE-TIME S Y S T E M S A discrete-time system contains a sampler (a switch) that closes every T sec (Fig. 14.5). For this reason it is also called sampler data system. The output of an ideal sampler is a string of impulses starting at t = 0 spaced at intervals of T sec of amplitude (power) x(nT), so that a r
y(t) =
1 x(nT)6(t - nT).
n=O
We take the Laplace transform of both sides of this relation, obtaining
This series is different from the one we encountered in Example 4 of Section 8.4, where we discussed the periodic functions. We handle this series by introducing the Z transform, which is defined as follows. Let the function f ( t ) be known to us only through its sample values at t = 0, 1,2,. . . ,n, namely
.f"b .f(l), .f(a
. . '
9
f(n).
Then the polynomial F(z) = Z[.f] = f ( O ) +.f(l)z-' +f'(2)z-2
+ ... + f ' ( n ) z - "
defines the Z transform. This definition can be extended to infinite series. To make use of this transform we set
(3)
z = PSI,
x(t)
*y(t)
Fig. 14.5. A discrete-time system.
370
14.
LINEAR SYSTEMS
x(i)
Fig. 14.6. A linear discrete-time system with impulse response h(t).
so that right side of (2) becomes m
z[x(t)] = X(z) =
1x ( n T ) z - " ,
(4)
n=O
which is the Z transform of x(t). Let us first consider the system of Fig. 14.6, a linear system with impulse response h(t). Then c(t) =
Joay(r)h(f -
T) d r =
J
m o o
0
1x(nT)h(r - nT)h(t - r ) dr
n=O
m
=
1 x(nT)h(t - nT).
n=O
Because the Z transform of c(t) is C ( z ) = Z [ c ( t ) ] = c2=o c(nT)z-", we find from (5) that m
W
x(rnT)h((n - m ) T ) =
f [x(mT)
m=O
where k
=
m
W
1
x(rnT)z-"C h(kT)z-&, &=O
& =- m
n - rn, and we have used that h ( k T ) = 0 for k < 0. Then, C(Z) = X ( z ) H ( z ) .
(6)
This relation is very useful in determining the 2 transform function for various systems connected in cascade.
15 Miscellaneous Topics
15.1. THE CAUCHY REPRESENTATION OF DISTRIBUTIONS
We have come across generalized analytic functions in Chapter 4, where we regularized the divergent integrals. We also encountered them in the study of the Riesz distribution in Section 10.12. The interplay between theories of distributions and analytic functions has been studied in detail, and various books are devoted entirely to this subject [44-461. The classical theory in which one studies the behavior of the harmonic and analytic functions at the boundaries can be extended with the help of the distributions. This is achieved by introducing various indicators that measure the growth of a harmonic function near the boundary. This involves [l 13 finding several relations among different indicators that enable us to characterize harmonic and analytic functions that have distributional boundary values. Since this topic involves concepts that have not been introduced in this book, we shall not include it in this chapter. We confine ourselves to the Cauchy representation of distributions.
The Spaces E and E Let E be the space of test functions $(x) that are continuous and have continuous derivatives of all orders. These test functions, unlike those in D,are 371
372
15.
MISCELLANEOUS TOPICS
not required to have compact support on the real axis. However, we need the following definition of convergence for functions $(x) E E: Definition. A sequence {$,,,(x)} of test functions in E is said to conuerge to zero if and only if for every fixed integer p the sequence {DP$,}converges uniformly to zero on every compact subset of the real axis. A sequence of functions {$,,,(x)} in E conoerges to anotherfirnction $(x) if the sequence {&(x) - $(x)} converges to zero in this sense. It is clear that under this definition of convergence, the space E is complete.
From this definition it is clear that the functions $ E D are also in E but that the converse is not true. Thus E 13 D. Indeed, it can be shown that D is dense in E. Let E' be the space of all linear functionals t on E that are continuous with respect to the present definition; that is,
(
)
lim (t, $,,,) = t, lim $,, . m+ m
m-m
Thus, if t is a distribution on E, then the sequence {(t, $,,,)} converges to zero if the sequence {$I~(X)} does. Thus E' is the dual space of E. An interesting feature of the distributions t E E' is that they have compact support, for otherwise we can always find a function $(x) E Cmwith sufficient growth at infinity to make the integral ( t , 4) divergent. Since D c E, we have E' c D'. From this discussion we observe that the test functions of compact support lead to the distributions of arbitrary support, whereas the test functions of arbitrary support yield the distributions of compact support. We encountered the space E' of distributions in Section 7.5 and found that they play an important part in the regularization process. The Cauchy Representation
Let z denote a point located in the upper or lower half-plane but not on the x axis. Then the function [271i(x - z ) ] -
is continuous and has continuous derivatives of all orders for all values of x and, hence, is a function of the space E. Accordingly, for a distribution t(x) E E , the operation
15.1.
THE CAUCHY REPRESENTATION OF DISTRIBUTIONS
373
generates a function t,(z) of the complex variable z for all values of z outside the real axis. This function is defined to be the Cairchy representation yj'the distribution t(x). Our contention is that t,(z) is an analytic function of z in the complement of the support of r(x). To prove this, we examine d dz
(3)
Because lim
-
I,. -E
1
-
1
-- (x -
z)2'
and t is continuous on E, we find from (3) that
so that the complex derivative of t,(z) exists for all z with Im z # 0 and hence is analytic in that region. Equation (4) also proves that
dt,(z)/dz
=
ti(z),
where t' is the distributional derivative oft. Example I. Let t ( x ) be a locally integrable function that vanishes outside some finite interval on the real axis and is the restriction to this interval of a function t ( z ) that vanishes for large values of z. Then the regular distribution t , ( t , 4 ) = {Tm t(x)+(x) d x , is a distribution of bounded support. The Cauchy representation of this distribution takes the form of the ordinary Cauchy integral
Example 2. In the case t ( x ) = 6 ( x ) we have
The Cauchy representation t,(z) of a distribution r on the space E has far-field behavior t,(z) = O( l/z)as z -+ 00. We observe from Example 2 that 6,(z) = 0(1l-) L as z+ a.
374
15.
MISCELLANEOUS TOPICS
The Fourier Transform of Tempered Distributions with Bounded Support
In Section 6.1 we attempted to define the Fourier transform of a distribution by the formula (5)
t(u) = ( t , eiux),
and got into trouble because the function eiuxis not in D.Then we tried the definition (i, 4) = (f, $), 4 E D,and found that $ may not be in D.This necessitated the introduction of the space S. However, if we define the Fourier transform of a distribution t of a bounded support, then (5) is clearly valid so long as we confine ourselves to the class of test function in S . For example, the delta function is a tempered distribution with bounded support, and thus & x ) = (6, dux)= 1, The hypothesis of bounded support in the representation ( 5 ) can be relaxed if we move from the real axis into the complex plane and consider the test functions eiAX, where L is a complex variable. Indeed, if t ( x ) is a tempered distribution that vanishes for negative values of x , then we define ?(A) by
i(L) = ( t , e’”) (6) for all values of L in the upper half of the complex plane. It follows by writing L = u + iu, u > 0, that eiAX = eiUXe-”Xis in S . Moreover, i ( L ) as defined by ( 6 ) and considered as a function of L is analytic in the upper half of the complex plane. Similarly, if the tempered distribution t vanishes for positive values of x , then ?(A) in ( 6 )exists for all values of L in the lower half of the complex plane, where it defines an analytic function.
15.2. DISTRIBUTIONAL WEIGHT FUNCTIONS FOR ORTHOGONAL POLYNOMIALS
Let P,(x) be a polynomial of degree n such that l P m ( x ) P n ( x ) w ( x dx ) = 0,
rn # n.
(1)
Then we have an orthogonal polynomial sequence with respect to the weight function w ( x ) on the interval (a, b). The weight function satisfies certain conditions. For instance, we should have l w ( x ) d x > 0.
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
375
Because Pn(x)is a polynomial, ( 1 ) leads us to integrals such as
[x"w(x)dx, b
11, =
which are called the moments. We stipulate that ,u, < cc for all n. The moments play an important role in the theory of orthogonal polynomials. Indeed, every polynomial can be expressed in terms of its moments [47]. From (2) it is clear that if we know the weight function of an orthogonal polynomial sequence, then we can calculate the moments. Morton and Krall [48] have shown that the theory of generalized functions helps in solving the inverse problem; that is, given the moments, we can determine the corresponding weight functions. For this purpose, we consider the concept of a linear functional w such that (2) can be written ,un = ( w , x")
for all n
=
0, 1,2, .. . .
(3)
This is clearly a linear functional. Let $ ( x ) be a real analytic function whose Taylor series converges to $ for all x . Then
Since $(')(O) can be represented in terms of the Dirac delta function and its derivatives by $'")(O) = ( - l)n(ti("), $), it seems reasonable to expect that
Accordingly, we can write
c (-1)nm
w(x) =
n=O
,una(n)(X)
n!
(4)
in the sense of distributions. When the moments { p i } ~are o those associated with the various classical orthogonal polynomials-the Legendre polynomials, the Laguerre polynomials, or the Hermite polynomials-the weight functions w yield the same results as the classical weight functions concerning orthogonality and norms. However, when the moments { , u i } ~ oare those associated with the Jacobi polynomials or the generalized Laguerre polynomials, then w remains a suitable distributional weight function.
376
15.
MISCELLANEOUS TOPICS
In the rest of the discussion we assume that the moments pi are given. Furthermore, A,,
i
=
/l"
'... ..
7
n = 0 , 1 , 2 ,...,
#O,
:
(5)
P2n
for some arbitrary but fixed constants c and M. Let us note that the collections {( - l)m6cm)(x)}m_o and { x " / n ! } ~ =form , a biorthogonal set [see Exercise 14, Chapter 21; that is,
); {
((
- l)msm(x),
=
n.
0 1
if n # m, if n = m.
Some Orthogonal Polynomials and Their Properties
1. The Legendre polynomials,
where [n/2] is the standard notation for the greatest integer I4 2 , have diferential equation weightfitnction
( 1 - x2)y" - 2xy'
+ n(n + l ) y = 0,
w(x) = 1,
Rodrigues's formula
orthogonality relation P,(x)P,(x) dx a recurrence relation and moments p2,
=
2/(2n
=
m # n, m = n,
+ l), (n + l)P,,+l ( x ) - (2n + l)xP,(x) + nP,+ l ) , p2,+
=
2/(2n
0.
l ( x ) = 0,
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
2. The Hermite polynomials,
have differential eqirution weightjirnction
y" - 2xy'
+ 2ny = 0,
w ( x ) = e-"', -co
<x <
CXJ.
generating junction
Rorlrigues's jormula
orthogonality relation
a recurrence relation H,(x) moments
- 2 x H f l -l ( x )
+ 2(n - 1 ) H n - 2 ( x )= 0, and
p2,, = J.n(2n)!/4"n!, p2,,+ = 0.
3. The Laguerre polynomials, Osx<m
n a positive integer, have
+ ( a + 1 - x)y' + ny = 0, w ( x ) = x"e-"/r(a + I ) ,
differential eqiration xy" weightjitnction
generating fitnction e-xl"l
( 1 - t)a+l Rorlrigues's jormula orthogonulity relation
c L;'(x)t", m
-I)
=
fl=O
Lf)(x) = ( n !)- lx-nexDn[xn+a e-x 1 7
a#-n,
377
378
15.
MISCELLANEOUS TOPICS
a reciirrenceformula n L f ) ( x )- (2n + a - 1 - x)LIpll(x) + ( n + a - l ) L f l z ( x )= 0, and moments p,, = T(n
+ a + l)/T(a + 1).
For a = 0, all these formulas reduce to those for the classical Laguerre polynomials. 4. The Jacobi polynomials,
diflerential equation ( 1 - x2)y”
+ [ p - a - (a + p + 2)x]y’ + n(n + a + D + l ) y = 0,
weight jirnction
generating jirnction
where R = ( 1 - 2xt
+ t2)’1z,
orthogonality relation P~9’)(x)P~.’)(x)(1 - xY(l
f 0,
+ x)’
dx m # n,
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
37 9
a recurrence relation 2n(n
+ o! + B)(2n + a + p - 2)P;.D)(x)
(2n + tl + /3 - 1 ) + a + p)(2n + a + p - 2)x + a2 - p’]P?:{’(x) + 2(n + - l ) ( n + p - 1)(2n + tl + p)P::<)(x) = 0. -
x (2n
o!
For different values of o! and p we derive special cases of Jacobi polynomials. For instance, for a = p = 0 we recover the Legendre polynomials. Similarly, the values a = p = T $ yield the Chebyshev polynomials of the first and the second kind. The moments about 0 are
so that Po = 1,
where (
)j
is Pochhammers’s symbol.
The Spaces P and P The spaces D, S , and E and their duals D’, S‘, and E’ that we have encountered in the previous analysis do not suit our purpose in this section. Our function space should include polynomials, and at the same time the dual space should include such functionals as those generated by the exponential functions, that is, without compact support. This necessitates the introduction of a new space P that includes polynomials such that D c S c P c E. The dual spaces then satisfy D’2 S‘ 3 P’ I> E’. Definition 1. The space P is the linear vector space that consists of all complex-valued infinitely differentiable functions +(x) satisfying, for all a > 0 and q 2 0,
-
lim 1x1
m
e-4.xl
+
(x) = 0.
(4)
Thus all the polynomials with complex coefficients are in P.
380
15.
MISCELLANEOUS TOPICS
Definition 2. A sequence { c $ ~ } in P is said to converge to zero in the sense of P provided that for each CI > 0 and 9 2 0 the sequence {e-"1"1C$~'(x)}
converges uniformly to zero. Definition 3. We let P' denote the space of linear functionals on P . From these definitions it follows that the connotation slow growth is appropriate for P, rapid decay for P'. Indeed, we have the following analogies: D, E':
compact support;
D', E :
no restriction on growth;
P, S':
slow growth;
P, S:
rapid decay.
The Spaces Z,,, and Z One of our aims is to find a linear space on which w is continuous. For instance, in the case of the Laguerre moments p,, = n ! , and for +(x) = e-"', which is in P , we have 1)(~"'(0)= ( - 1)"
2m 1
-L
m!
and
(w,I)>
=
1 (m
m=O
1)"
(2m)! -x,
which unfortunately diverges. Furthermore, the action of w on a test function means that
*
should be not only infinitely differentiable but also This implies that analytic. Accordingly, an obvious space to consider is Z , the space of Fourier transforms of elements in D that are analytic; but this space is also slightly too large. Thus we consider a slightly smaller subspace Z M ,which is defined as follows: Definition 4. For E > 0, Z,, is the subspace of all I) E Z such that F - ' ( $ ) is defined on the interval [ - ( M + e)-', ( M + &)-'I, where F - ' stands for the inverse Fourier transform.
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
38 1
We note that if (pnI i cM"n!, n = 0, 1, . . . , then lpnl l$"')(O)~/n! exists for all $ E Z , . To see this, let $ E Z , have inverse Fourier transform 4 E D,then $(x) =
fy
e i x ' 4 ( t )d t ,
m
which implies
Thus
c m
n=O
lpnll$(")(O)l/n! I c
f
n=O
f
(M+E)-I
(L)n
M
+E
In addition we can see that if Ip.1 < cM"n!, n
c-
IW)l dr,
-(M+E)-'
=
0, 1 , . . . , then
m
w(x) =
( - 1)"pn6'"'(x)/n!
n=O
is a continuous linear functional on Z , in the sense of Z . This is easy to show since (w, $) is well defined for $ E Z,. Suppose that $ j 5 0; then c # ~= ~ F - ' t j j 4 0. Hence
Since the integral can be represented by ( K [ - ( , + & ) K is the characteristic function of the interval [ - ( M namely, 1
K = {0
if t E [ - ( M otherwise,
+ E)-',
(M
Il(t), 4 j ) ,where
+ E ) - ', ( M + 8:)- '1,
+ &)-'I,
and since 4 j % 0, the integral approaches 0 and so does (w,$ j ) . Thus w is continuous. Moments of Extension
Our aim is to extend the weight function w from Z,, to a larger space, with the specific aim of extending w to act on P . For the moment, however, let us assume that a continuous extension w p of w is possible. Since, X" F$Z , for
382 any n
15.
= 0,
1, . . . , even though (w, x")
=
MISCELLANEOUS TOPICS
p, is defined, it is not obvious that
(Wp,Xfl) =
(7)
pn.
This can, however, be shown to be true by appealing to the delta sequence of functions 6rn(t) =
A,exp[-(l/m2 10.
- l / m < t < l/m, otherwise,
- t2)-'],
where A, is chosen so that J'iY,, 6,(t) = 1. We have discussed this sequence in Chapter 2. These functions are infinitely differentiable. Furthermore, for m 2 M E , the support of6, is within the interval [ - ( M E)-', (M &)-'I and 6,(t) + d ( t ) as m -, co. Our aim is achieved by setting
+
+
$AX) = F C ~ m ( t > = l J;';/Xd,cr,
+
nt.
This sequence of functions is in Z , for m 2 M now takes little algebraic work to show that [48]
+ E,
by their definition. It
P
(i) $ m --t 1, (ii) limm+a,(w,$), = p o , and (iii) limm+m(w, xfl$,) = p,,,
so (7) holds. The Fourier Transform
Our major tool in the extension process is the inverse Fourier transform. We shall thus need an additional space of test functions in order to carry out conveniently the required calculations.
Definition 5. For E > 0, let DMME be the subspace of all 4 E D such that the support of 4 is contained in the interval [ - ( M &)-I, (M E)-'].
+
+
With this definition, the image of D,, under the Fourier transform is Z M c and F-'(Z,&) = L I M E . Since, for 4 E D or D M M EE, fD or D M E , the Fourier transform offis defined through (6.3.19), namely (f,4) = (1;6). If
J-
a,
+(x) =
and f
= g,
a,
4(t>eix' dt = $(x>
the inverse Fourier transform of g E Z' or Z$, is (F-'g,
4)
= (9, F-'4) = (9.
$)?
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
383
where $ E Z or Z M Eand ,
is in D or D M e .Let us recall the following results from Exercise 1 1 of Chapter 6: 1. (F-ly)'"' = F - ' ( ( - i x ) " y ) , = (it)"F- lg,
2. F - '(y('")
and, in particular [cf. (6.4.37)], 3. F -
'(P) = (it)"/2n.
Recall that the weight function has the value
so that it follows from (8) that
Because, by assumption, Ipnl < M"n!,F-'w represents an analytic function for It I < 1 / M . Furthermore, F - ' w is a regular functional, and we have (F-'w,
4)
S- " m
=
F - ' ( w ( f ) ) $ ( f )dt,
4 E D.
From this discussion we conclude that w has extensions wL that are distributions on Z and extensions ws that are distributions on S [48]. The Extension to P and E
In order to extend further the weight functional w , we need the following lemma, which we state without proof:
Lemma. Letf'(z) be analytic in the region IIm(z)I < so, with lf(z)l I ho(t). I.f"(z)l I h , ( t ) for all z = t + is and I s \ < so. Furthermore, we assume that lim h,(t) = 0 as It I + cc and that J T m h , ( t ) d t < co.Thenj'(t) has a classical Fourier transform g ( x ) and there exist constants G and r such that [48] Jg(x)l < Ge-rIXl/lxl.
(9)
384
15.
MISCELLANEOUS TOPICS
This lemma helps us in proving the following theorem: Theorem 1. Let f ( z ) be the analytic continuation of F-'(w), where w, as given by (4),is a weight distribution on Z M e Assuming . that z = s it,f(z) satisfies the following:
+
=-
1. .f'(z) is analytic in the strip IIm zI = Is1 < so for some so 0. 2. When Is1 < so, l,/'(z)l I ho(t) and If'(r)l I hl(t), where lim ho(t) = 0 as It1 + co and JZCOhl(t) dr < co.
Then w,(x), the classical Fourier transform of f ( t ) , is a continuous linear functional on P and is an extension of w. Proof. By the foregoing lemma, f ( t ) has a Fourier transform
J-
m
g(x) =
CO
f(r)eix'dt,
which satisfies (9) for all x. Because for $ E P
xs(-ao,
rn)
J-CO
it follows that (9, $) not only exists but is a continuous linear functional. Thus wp(x) = g(x) is the desired extension. As a corollary, it follows that if wp has compact support, then wp can be extended to a unique continuous linear functional wE on E . Regularization
In the previous analysis we have shown how w can be extended when the analytic continuation f'of F-'w has a classical Fourier transform. Let us now study the problem of extending w when a classical Fourier transform does not exist. The procedure of regularization studied in Chapter 4 helps us achieve this goal. Let us consider the regularization of h(x) that has a singularity only at x = 0. We assume that h(x) is integrable over every bounded region E not containing 0 either in its interior or on its boundary. A regularization of h(x) is a continuous linear functional that coincides with h(x) except at x = 0. That is, for every test function $(x) that vanishes in the neighborhood of 0, the functional has the same value as f Z r n h(x)$(x) dx. Let us restrict our discussion to those functions which can be written h(x) = pi(x)qi(x),
1
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
385
where the p i ( x ) are infinitely differentiable and each qi is one of the functions x: , x.l , or x - ~ . Let us assume that f ( z ) is the analytic continuation of F - l w and that .f""'(z),z = t is, is analytic with I j'("'(z)I I Izo(t), lf'("+I)(t)l I hl(r) when 1st < so, where lim ho(t) = 0 as I t ( + ;c, and !Trn h l ( t ) d t < m. Then the
+
following theorem gives the regularized extension.
Theorem 2. Let g ( x ) denote the classical Fourier transformf"")(t). Assume that y ( x ) / ( - i x ) m has a regularization h ( x ) . Then there exist constants ck, k = 0 , . . . ,m - 1, such that wp(x) = h ( x )
+
c
m- 1
Ck6'k'(X),
k=O
is a continuous linear extension of w to P . Proof'. The constraints on f'guarantee that JYm y ( x ) + ( x ) d x is continuous on P . Accordingly, the regularization of g ( x ) / ( - i ~ will) be~ continuous and linear on P . Recall that w p = .rf is an extension of w . Now we show that there exist constants Ck, k = 0 , . . . , m - 1, such that
which is a linear continuous functional because k ( x ) is so. The proof is as follows. Let y ( t ) = F - 'h(t).Then
(y'"", F 4 )
=
(Fy'"),
4(t)) = (( -ix)"h, 4).
Since ( - i ~ )is ~infinitely differentiable, ( - i x ) m h ( x ) = ( - i x ) " R [ y ( x ) / ( - ix)"] = R [ ( - ix)"y(x)/( - ix)"] = g (x),
where R stands for the regularization as defined in Chapter 4. Thus
(y'"', 4)
=
pn),
that $ m ) = f""). In order to evaluate ('k we appeal to the fact that every distribution in D has antiderivatives of mth order and conclude that SO
386
MISCELLANEOUS TOPICS
15.
Taking Fourier transforms, we obtain the required relation (10). Because ( w p , Xk ) = p k , k = 0 , 1 , 2,... and (6'"(x), x k ) = ( - l)kk!6 k 1 ,
we find that Pk =
(Wp,
Xk) = (h, Xk)
+
m- 1
C(6"'(X), X k ) = ( h , X k ) k=O
+ (-
l)kk!Ck.
Thus ck = (( - l)k/k!)[Pk -
(h, xk)].
As a corollary to Theorem 2, we observe that if this extension w p has compact support, then w, can be extended to a unique continuous linear functional w E on E .
Pn(x) =
An-
~
1
Po Pl
P1
Pn-I
Pn
:
1
112
... ...
x
... ...
Pn Pn+ 1
P2n-1
Xn
O n the assumption that the distribution w has been extended to w,, which is continuous on P , we have the following theorem:
Theorem 3. The sequence (Pn(x)},"=,is mutually orthogonal with respect to w,; that is, (w,,P,P,)
=
0
if m # n.
Proof. For k < n, we have
(w,, XkPn(X>)=
1 n-1
Po Pl
P1 P2
... ...
~
Pn-I
Pn
Pk
Pk+l
Pn
Pn+ 1
:
P2n- 1 " '
Pk+n
= 0,
15.2.
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
387
because the last row is identical with one above. Accordingly, for m < n, if m
Pm(x) =
1
CkXk,
k=O
then m
( w p , P m P n )=
1 ck(wp, Xkpn)= 0.
k=O
As a corollary we observe that ( w p , P i ) = ( w p , x n P n )= A,/(n - 1). Finally, we present the precise connection between w p and the classical weight functions. 1. The Legendre polynomials. Legendre polynomials are P2n
=
As already mentioned, the moments for the
2/(2n + 1x
P2n+ 1
= 0.
Thus from (4)and (8) we have
This is a Fourier series representation for f(t) =
(eif - ~?-~')/27cit = sin t/m.
This function has a classical Fourier transform :
2. The Hermite polynomials.
In this case the moments are
p 2 n= 4 3 2 n ) !/pn!,
pzn+ = 0.
Accordingly,
This is a power series representation for f ( z ) so = 1,
ho(t) = e-f2'4,
=
e-2214/2&. Thus
h,(t) = $ l t ~ e - ~ " ~
and the conditions for the extension of w to w p are satisfied. 3. The Laguerre polynomials. The moments for the Laguerre polynomials are p,, = n ! . Hence, 00
w(x) =
1( -
n=O
l > " ~ ( " ) ( x ) and
F-'w
=
1* -.(-it>" 2Tc
n=O
388
15.
MISCELLANEOUS TOPICS
',
This Fourier series representation converges when 1 t I < 1 to (1/27c)( 1 + it)so the analytic continuation of F - ' w isf'(z) = (1/2n)(l + iz)-'. To derive the extension w p , we see that 1
so = 7,
hl(r)
Thus liml+.m ho(t) = 0 and JTrn h , ( t ) dt < P. Cauchy's residue theorem then yields
03,
= 2741
1
+ 4?)'
so that w p = f extends w to
> 0, x I0.
w p = {x ;y
We have
w p = f ( x ) = e-XZ,
--03
< x < 00.
Generalized Laguerre polynomials. The generalized Laguerre equation is
xLi'") - ( x -.a
-
1)L;'")
+ nL?) = 0.
If a = 0, this becomes the equation for the polynomials already discussed. We now give the corresponding analysis for the general Laguerre polynomials LIP). When a # - 1, -2, . . . ,the moments of the polynomials L?) are p,, = ( n
+ a + l)/(a + 1).
(12)
These moments can be calculated by appealing to the recurrence relation p,,= (n a)p,,- and setting p o = 1, so they do not depend on the existence of a weight function. Furthermore, when a is a negative integer -no, all moments p,, = 0 for n > n o . In this case (ll), for P,, defines polynomials only up to degree n o . We omit this degenerate case. Furthermore, let us observe that the moments are not all positive when a < - 1. If - j - 1 < a < -j, the first j moments alternate in sign. The remaining moments retain the same sign as p j - 1. Appealing to (4), we find that the formula for the weight function for the generalized Laguerre polynomial { L:)},, = is
+
2 (-1y m
w(x) = Also
n=O
r ( n + a + 1) d(")(x) T(a + 1) n! '
For
389
DISTRIBUTIONAL WEIGHT FUNCTIONS FOR POLYNOMIALS
15.2.
tl
>
- 1, the
Fourier transform shows
so that
+ a + 1) r(tl + 1)
n!(n
rn = n. 9
It is possible to find a suitable weight function when tl < - 1 and is not a negative integer. Although F - ' W cannot be directly inverted, a suitable derivative can be. Let - j - 1 < a < - j , and replace 1 + it by z. Then F - ' w p ( z ) = z-"-'/2n.
When z
=
0, F - ' w p
(F-
=
0. Similarly, when 0 Irn < j ,
Wp)(m)(Z) =
( - l)"(a
--a-m-
+ 1) . . . ( a + rn)
2n
1
'
and ( F - ' W ~ ) ( ~=) 0. ~ Finally, ~=~ ( F - IWP)("(Z)
= (-
l)'(a
+ 1) . . . (a + j )
is the first derivative to become infinite at z classical Fourier transform, which is
=
z-a~
277
0; it is also the first to have a
Accordingly, (F-
1WP)(')(t)
=
-
( - 1)' ( - 1)'
j- I
dX.
390
15.
MISCELLANEOUS TOPICS
The result of integrating this relation j times from 0 to z is
Expanding the last term in powers of - i t and interchanging the summation indices, we have
The fact that ( - i t ) " ) = (e-itx)(') when evaluated at x = 0 suggests that the regularization of w is
which is the required linear functional wp. For moment p, given by (1).
II/ = x", this reduces to the
For a discussion of the Jacobi polynomials and another class, the Bessel polynomials, the reader is referred to Morton and Krall [48].
15.3. APPLICATIONS TO PROBABILITY AND STATISTICS
In order to present the applications of generalized functions to the theories of probability and statistics, let us start with very simple concepts. The outcome of an experiment is described by a real number x, called the random variable. For example, in tossing a coin we can assign the value x = 0 to heads; the outcome tails will then have the value x = 1. The probability distribution F ( t ) is defined as the probability that the outcome is a number x that lies in the interval (-00, t). Since certainty corresponds to a probability of 1, we have 0I F(t) I 1
for
-00
00.
15.3.
APPLICATIONS TO PROBABILITY A N D STATISTICS
391
Furthermore, there is a better chance of having the random variable in 00, t 2 ) than in ( - co, t l ) when t , < t 2 . It follows that F(r) is monotonic:
(-
F(r,) s F(t2)
for r l < t 2 .
Also lim,+mF ( t ) = 1, because it is certain that the random variable x will occur in ( - 00, co). The probability density function .f'(t) is defined as f(t) =
dF(t)/dr
(1)
and has the following properties: 1. f ( r ) 2 0 for all t in ( - a,00). 2. By setting F( - co) = 0 and F ( c o ) = 1, we find that
Lrnm.
f'(t)dt =
1.
This latter is a very important property of the density function. A probability distribution is called discrete or continuous according to whether t takes on discrete or continuous values.
Example 1 . Let us examine the tossing of a coin. We assign the value x = 0 if we obtain heads and the value x = 1 if tails. To evaluate the probability distribution F ( t ) we make the following observations. (a) Since we have only two possibilities, namely, x = 0 and x = 1, the probability that t < 0 yields the distribution F ( t ) = 0. (b) For a fair coin, on a single toss there is a probability of 4of obtaining heads, with an equal probability for tails. Hence for 0 I t I 1, F ( t ) = i. (c) For the case t > 1 we appeal to the definition of F(t), which is defined as the probability that the outcome is less than t . Since both possibilities (x = 0 and x = 1) are less than t in this case, we have F ( t ) = 1. Summing up these observations we find that F(t) = =
i
0, +,
0 I t I 1,
1,
t > l
f[H(t)
t < 0,
+ H ( t - l)],
(3)
where H(r) is the Heaviside function. From (1) and (3) it follows that in this case the probability density function is f(t) =
4[6(t)
+ 6(r - l)],
(4)
392
15.
MISCELLANEOUS TOPICS
where 6 ( t ) is the Dirac delta function. Property (2) is clearly satisfied by this density function. The foregoing concepts can be generalized as follows. If the random variable x takes the values a l , a z , . . . , a, with the probabilities pl, p 2 , . . . ,pn, respectively, such that P k = 1, then the generalized function
z=
n
is the probability density of x . Example 2. The binomial distribution has the density f ( t )=
i
k=O
($kqfl-k6(r
-
0 5 p 5 1, q
k),
=
1 - p.
Then we find that the probability distribution function F ( t ) is
Example 3. The Poisson distribution has the probability density
so that
Example 4. The Gaussian distribution is defined as dt,
where p is the mean and (T is the variance. The corresponding density function is
The Characteristic Function
Given a probability density function f(t), the characteristic function ~ ( s )is defined as
J-_ei‘”d F ( t ) J-mei‘sf(f)dt = f ( t ) . a
~ ( s )=
W
=
(9)
393
15.3. APPLICATIONS TO PROBABILITY A N D STATISTICS
SinceJ'(x) is the derivative of a bounded function F ( x ) , the Fourier transform exists in the generalized sense. Conversely, given a characteristic function ~(s), we have j'(r)=
2n
S
m
-cc,
X(s)eYifSds
=
(10)
F- '(I),
where F - ' stands for the inverse Fourier transform. Example 5. Let us take ~ ( s )= eiAS. Then from (10) we have
S"
eiAse f(r> = 2n -"
- its
ds = -
I'"
eis(x-
271 -"
where we have used (3.5.3).
A) d s =
6(x
- A),
Example 6. For the Gaussian distribution, for which . f ( t ) is given by (8), namely,
the characteristic function is x(s) = f(s)
1
J-"
=0J2n
eitse-
(f
dt = exp(ips - )azs2).
-r ~ i 2 a 2
For the special case of p taken to be zero, this relation becomes ~ ( s )= exp( -+02s2).
(11)
The Cauchy Representation for the Probability Density
Let us recall from Section 15.1 that the Cauchy representation of a distribution t in E' is t,(z) = 1/2ni(f, l/r - z ) . Because d(t) E E', the Cauchy representation of the probability density
is
For instance, for the binomial distribution studied in Example 2 we have
394
15.
MISCELLANEOUS TOPICS
Thus in this case the Cauchy representation is
The Cauchy representation exists for any probability density; the proof follows on showing that iff(t) is monotonic, then it is measurable. Since the number of discontinuities of a monotonic function is at most countable,f(t) is continuous almost everywhere and hence measurable almost everywhere. Furthermore, iff(t) is measurable on disjoint intervals A and B, then it is measurable on their union. Accordingly,f(t) is measurable on R '. Next, we show that since 0 IF ( t ) I1 and l/(t - 2)' is in L', therefore F(t)/(t - ')2 E L'. This follows on observing that
Accordingly, the functional ( F ( t ) , l/(t - z)~)is defined. From this relation we infer that 1/2ni(f(t), l/t - z ) is defined, which proves the assertion. Probability Fields
Let R be the set of elementary events and B be a class of the subsets of R such that (1) the family B contains the empty set 0and the total set R; ( 2 ) if A E B, then c A E B,for c is a real number; and (3) if the sets A,, A 2 , . . . ,A, belong to B,then their sum belongs to B. The probability measure P on B has the following properties: ( 1 ) P ( Q ) = 1. (2) If the sets A,, for i # j, then
. . . ,A , , . . .are mutually disjoint, that is, ifAi n A j = 0 P
U An
(n=1
)
m
=
1P(An)*
n=l
A random variable is a mapping x from the set R into the set of real numbers B;i.e.,
t such that the set of all points w E R where x(w) < t belongs to x- '(( - 03, t ) ) E
B
for any real number t. The system (a,B, P) is called the probabilityjeld.
15.3.
395
APPLICATIONS TO PROBABILITY AND STATISTICS
As mentioned in the introduction to this section, we associate with the random variable x the probability distributionfirnction F ( t ) such that
F(t) = P(x-'(-oo, t ) ) = P ( x ( 0 ) < t).
(12)
Our aim is to show that the probability distribution functions can be connected with the theory of distributions. (We must keep in mind that the terms probability distributions and distributions refer to different entities.) In this connection we note that any probability distribution function defines a distribution by the formula
(f,4 ) = Jm 40) dF(t),
4 E D3
-m
(13)
where .f and F are the probability density and distribution, respectively. Indeed, the density j as defined by ( 1 3) is valid not only for a 9 E D but also for all continuous functions (ie., #J E C) that have compact support. The fact that (13) defines a distribution is clear from the following considerations:
(1)
It is linear because
(49 C l f l + C 2 f 2 )
= J-mm#J(t)d(clFl(t)+ c a F 2 0 ) ) .
(2) Continuity follows from
I
I
l O ( r ) dF(t) I SUP IM)I
Jm
-m
I sup IM)I 1,
W) f E
[a, bl,
which implies that if the functions #J,, E C have all their supports included in the same interval [a, b] and if they uniformly converge to zero in this interval, then (,f,#J,,) + 0 in this interval. Accordingly, (13) defines a linear continuous functional, as desired. On the other hand, the probability distribution F(t), being a locally integrable function, defines a distribution (F, 4 )
= Jm
#J(t)F(t)dt,
#J
E
-00
Furthermore, from (2) and (3) it follows that (F',
4)
=
- (@,F )
=
-
1
m
-m
m
#J'(t)F(t)dt
D.
(14)
396
MISCELLANEOUS TOPICS
15.
so.f'(t)is the distributional derivative of F ( t ) . We have already found this to be true in Examples 1-3. In the light of the foregoing discussion, we can define the classical quantities in the following way: (1) The expectation value is
E(x) =
(t,f') =
(2) The variance is 0 2 =
Jam
t dF(t) =
((t - E ( X ) ) ~ , J= )
J
J*
(15)
x ( o )dP(0).
W
(t -
E ( X ) ) ~d F ( t )
-OD
(3) The mth moment is (tm,
1') =
1W
a,
t" d F ( t ) =
For n = 1, we recover (15). (4) The central mth moment is
s,
m
( ( t - E ( x ) ) " , f ) = [-,(t
-
E(x))" d F ( t ) =
[x(w)l" d ~ ( o ) .
J*
-
(~(0) E(x))"
(17)
dP(w). (18)
Equation (16) follows from (18) on taking m = 2, and the central moment of first order is zero. In this notation we can define the characteristic function ~ ( s )by (9) or by (19)
~(s)= E(eifs).
Also, when we evaluate the mean and variance of the Gaussian distribution j ( t ) as given in Example 6 we find that p = E ( x ) is the mean and 0 is the variance, as defined there. These concepts can be extended to a finite system of random variables xl, x 2 , . . . ,x , . This system may be considered as a mapping from the set R into the n-dimensional space R,. Such a mapping is called an n-dimensional random variable. The probability distribution is now F@,, t 2 , . . . , t,) = P ( X l ( 0 ) < t , , x 2 ( 0 ) < t 2 , . . . , X " ( 0 ) < t,) = P ( 0 ) .
The moments are given by the formula
s,.
t;'t;'
. . . t? df'
=
. . . ( ~ ~ (d0P)( 0')" .
15.3.
APPLICATIONS TO PROBABILITY AND STATISTICS
397
The Generalized Stochastic Process To study the stochastic processes in the context of the generalized functions, we need to set up a Hilbert space. Recall that a Hilbert space is an inner product space that is complete in its natural metric, namely, d(x, y) = IIx - yll. For this purpose we observe that the set of all the random variables that have finite expectation value and finite variance form a vector space with the usual operations of addition and multiplication. The inner product (in Hermitian form) in this space is [49, 501, (x, Y ) = J*x(w)J(w)d P ( o ) = E ( x j ) .
(20)
This form clearly satisfies the relation
so llxll = (x, x ) ” ~ gives the norm. This yields the Hilbert space H = L2(R, B, P ) . We take the subspace in which the expectation value is zero, and it is clear that this is also a Hilbert space. The value of the norm in this space is IIx112
= Jox(w)G)
dP
=
s,
x2(w) tlP =
I
(x - E(x))2 d P = 0 2 ,
because E ( x ) = 0. Thus the variance o is the norm in the space H . Definition 1. A stochastic process is a real or complex function x(t, w ) of two variables t and w, where t runs over a set of real numbers and w runs over the set R of a probability field (a,B, P ) . For any t , the mapping x : w -+ x(t, (0) = x,(w) is a random variable. In applications we assume that a variable runs over a finite or infinite interval on the real line whereas a random variable has a finite expectation and variance such that x -+ x,(o) is continuous on L2(R, B, P ) . The scalar product is given by (20). Definition 2. A generalized stochastic process is any distribution with values in H , that is, any continuous linear mapping of the space D into H . Our contention is that if x : t -+ x ( t ) is a continuous function defined on the real line with values in H , then it defines a linear and continuous mapping from D into H through the formula m
(22)
398
15.
MISCELLANEOUS TOPICS
The proof follows by first observing that this functional is well defined, because 4 has a compact support and therefore the above integral is finite. It is clearly linear. Continuity is ensured because
where [a, b] is the support of 4 and we have used that &r) and x ( t ) are continuous. Now, when 4,(t) -,0 in D,so does max I $,(t) 1, and we conclude that l I ( 4 n 9 x>ll 0. --+
Definition 3. The functional is called the correlation of the generalized stochastic process T. Definitioa 4. The characteristic functional (x, 4) of the generalized stochastic process is defined as E(eiT(”’),that is, the mean of the random variable eiT(@: ( x , 4) = E(eiT(@)) = E(ei<”*6)).
(23)
This generalizes ( 1 5 ) Example 7. If x = x(t) is a continuous function, then we can take a generalized function for $(t) such as ss(t - T), so that ,qei<X.
6)) =
E(eisx)= ~ ( s ) .
Since, then the distributional derivative of x is defined as (x(”),4)
=
(- l)”(x, #”)),we find that the characteristic functional for the nth derivative
is
E(ei<(n).6))= , c ( ~ ( - 1 ) n < x . 6 “ 9 ) = ( x , ( -
1)”4(“)).
Example 8 . A generalized Gaussian process is a process whose characteristic functional is a natural extension of the Gaussian characteristic function (1 1). That is,
4>*)1 for the generalized functions with mean E ( ( x , 4)) equal to zero. ,qeO+’.6<) =
exp[-9E((x,
15.4.
APPLICATIONS OF GENERALIZED FUNCTIONS IN ECONOMICS
399
The Stationary Processes We have already defined the concept of stationarity in Section 14.1, where we discussed time-invariant linear systems. In the present context it has a similar meaning. Let T be a generalized stochastic process as already defined. This process is called stationary if for any real number a we have the following:
T(4(t + a ) ) = T ( 4 ) ; (2) K ( 4 ( t + a)?$0 + 4) = K ( d ( t ) , $(t>). It can be shown [50] that any stationary process is of the form (1)
T ( 4 )= =
ST, (JImeisf4(t) 1 sib, (rm )4(t)dt sp, &) d M ( s ) =
dt d M ( s )
m
eisfd M ( s )
where M is an arbitrary measure. Thus x ( t )
=
=
x ( t ) 4 ( t ) dt,
JZrn cis'dM(s).
15.4. APPLICATIONS OF GENERALIZED FUNCTIONS I N ECONOMICS
Several studies of the optimal path of capital accumulation of a firm facing costs of adjustment have been made using the techniques of control theory. The mathematical analysis in these models is interesting in itself since economic reasons suggest that the investment schedule is not necessarily a function of time but rather a general Radon measure. The cost of adjustment, however, is a nonlinear functional of the investment schedule, and extension to the generalized schedules is not straightforward. There has been some controversy as to which is the more adequate form of the cost of adjustment function. Arguments for both a convex and a concave cost function and the corresponding analyses have been made. To learn the basic concepts and definitions of economic and mathematical analysis, the reader is referred to the bibliography given in reference [ 5 11, from which this material is taken. In this study we obtain the weak lower semicontinuous extension of the cost of adjustment functional. Weak lower semicontinuity accords well with the minimal properties that economics suggests that a cost should have. Weak continuity does not hold unless the cost of adjustment function is linear, since the extension should be linear with respect to the Dirac delta function. In deriving this extension functional, it will be shown that the cost
400
15.
MISCELLANEOUS TOPICS
of adjustment function, whatever its form, must be replaced by a convex one. We first give the rationale for using generalized functions in economic models. We then devote discussion to the basic model, where the discontinuous character of the cost of adjustment functional is portrayed. The next two topics of interest will be the derivation of the extension functional and some of its properties. We conclude this section with some comments on the uses of this model, in particular, when we should expect jumps in the capital stock and why. The Use of Generalized Functions
Most models of the dynamical behavior of an economic system usually assume (implicitly) that the variables of the system are functions of time. This is a very reasonable assumption for certain types of variables, namely, those that can be observed at almost each instant of time, such as the price of a certain commodity or the prevailing interest rate. For other variables, however, such an interpretation is problematic. Consider, for instance, the output produced at a certain factory (measured in physical units). Given a time interval I = [ t l , t J , we can evaluate the output Y ( I ) produced during that period, but the output Y ( t ) at time t = t , is not clearly defined. Continuing with this example, if the price of each unit of output at any time is given, the dollar value of output during any interval can be evaluated, or, alternatively, the discounted or present value of such an output can be evaluated. This example shows that certain types of economic variables are not ordinary functions of time but rather distributions: They give a definite number only when multiplied with a function (as prices or discounted prices) and integrated. Distributions also arise when certain analytic operations, such as the derivative, are applied to observable functions. In our particular problem the investment schedule gives the additions to the capital stock. The size of the capital stock can be observed at almost all times, but it can suffer jumps when additions are made in a very short period. A Dirac delta function placed at the instant of the jump is the best description of the investment. The Basic Model
The basic dynamic model for the investment decisions of a firm postulates that the investment schedule I ( t ) for t 2 t o is chosen at time t = t o in such a way as to maximize the present value of the future stream of profits, V ( t - to){R(t,K ( t ) ) - r(I(t))}dt,
(1)
15.4.
APPLICATIONS OF GENERALIZED FUNCTIONS IN ECONOMICS
40 1
where K ( t ) is the capital stock at time t , which, depreciation disregarded, is formed according to
R(t, K ) is the expected quasi-rent to be obtained from a capital stock of size K at time t ; r(1) is the cost of adjustment; and V ( t ) is the discount factor. We assume that R(t, K ) is smooth enough. We also assume that for each value o f t it is a strictly concave function of K in the interval [0, co) with a unique maximum at K * ( t ) and with two zeros, at K = 0 and K = K , ( t ) . Thediscount factor V ( t )is chosen by the firm and is assumed to be continuous, decreasing, and having compact support; the last assumption is made to explain the way the firm handles the uncertainty of R(t, K ) for large t . The cost of the adjustment function r(1) is positive if I # 0 and zero of I = 0; it increases for I positive and decreases for I negative. Smoothness for I # 0 is assumed, but discontinuities of r or any of its derivatives at I = 0 are allowed. Finally, we assume the asymptotic relations
lim r'(1) = b,
lim r'(1) = a,
I+-a,
1-03
(3)
where (4)
O
Let us observe that, for a given discount factor V(r),the cost of adjustment is a nonlinear functional of the investment schedule I ( t ) given by
This integral functional is well defined if l ( t ) is a locally integrable function of time. When l ( t ) is a delta function 16(t - t l ) , direct substitution in (5) is not possible, and thus a limit process is needed to evaluate this cost. Let nl,
I t It , otherwise. tl
+ l/n,
Then it is natural to set
Thus the cost of a discrete change in the capital stock can be computed as a limit of the cost of continuous changes. One would expect the reverse also to be possible-to recover the cost of a continuous change as a limit of the costs of a sequence of discrete approximations. This is not possible, however,
402
15.
MISCELLANEOUS TOPICS
since the cost functional cannot be continuous under the weak topology of signed measures unless r(I) is linear for I > 0 and I < 0. (The weak topology fn(f) = d(t - t , ) . ) For example, let g(t) be a is the one for which positive continuous function on [ t o , co), and let the support of V be contained in the interval [0, 23. Let us then consider the discrete approximations g n ( f )=
We then have
/(to k= 1
+ Xz)d(t
- to
g,, = g weakly in [ t o , t o
lim C(V, gn) = a
n-r
m
-
Xt).
+ 23, but
6."
V(t - to)g(t) dt,
(9)
which, in general, differs from (5) as long as r is nonlinear. The Minimum Cost Function
In order to derive a cost functional with some interesting analytic properties, it is necessary to introduce a new cost function. Let us assume that no discounting is made (V = 1). The function r(I),as implied by (9,gives the cost of augmenting the capital stock I units in one unit of time using a constant rate of investment. The cost of augmenting the capital stock I units in one unit of time, however, will in general be smaller than r(I). Let us define the cost function r*(I, t) giving the cost of investing I units of capital in z units of time by r ( f ( t ) )dt : f' E L'[O, t],
A change of variables in definition (10) shows that r*(I, z) is homogeneous of the first degree. If we denote r*(I) the unit cost function r*(I, 1). We then have
Lemma 1. The function r*(I, z) satisfies r*(I, t) = tr*(I/z). (1 1) We can therefore limit our attention to the function r*(I). This function can be constructed from r(Z) by a very simple method, as our next lemma shows. Lemma 2. The function r*(I) is the lower convex envelope of the function r ( I ) ; that is, it satisfies the following three properties: (a) r*(I) is convex; (b) r*(I) Ir ( I ) ; (c) If h(l) is a convex function that satisfies h(I) Ir(I), then h(l) I r*(I).
15.4.
403
APPLICATIONS OF GENERALIZED FUNCTIONS IN ECONOMICS
Proof'. Direct use of the definition of r*(l, T ) shows that if then r*(Till
+
+ T 2 1 2 , 1) 5 r*(TlIl, = Tlr*(ll) + T 2 r * ( I 2 ) .
T212) =
r*(Tili
51)
T~
+ z 2 = 1,
+ r*(T212,
T2)
Condition (b) is clear. For (c), let h be a convex function satisfying the stated conditions. Let I E R . Then for any constants tl,.. . , T,, u l , . . . , a, such that
Cqa, = I, i= I
we have n
h(1) i
i= 1
n
)
tih(ai)_< C ~ ~ r (=q ) r C a i g i ( t ) dt, Jo'
i= 1
(i:l
where the function gi(t) is given by i
i- 1
10
(14)
otherwise.
Now we can take the infimum of (14) over all constants ui, ti that satisfy (12) and (13). This infimum is equivalent to the infimum off:, r(.f'(t))dt over all piecewise constant functions whose integral from 0 to 1 equals I , which (according to (10)) reduces to r * ( l ) . Example. A particularly appealing type of basic cost function r(1) is one that is convex for I > 0 but whose right limit at I = 0 is a strictly positive constant (indicative of fixed costs). Let co = inf{r(l)/l : I
> O}.
(16)
Then if co < a, there exists a largest number I* > 0 such that r ( l * ) = co I*, and the cost function is given by
If co = a, then r * ( I ) = a1 for all I . In either case, r*(l, T ) has the interesting property that if the investment is small relative to the time interval, then the cost will not depend on the time interval.
404
15.
MISCELLANEOUS TOPICS
Extension of the Cost Functional
We shall use the notation
lim f "
=
j'(w),
n-m
to indicate thatf" converges weakly to,f, that is, for every continuous function 4 defined on [to, t l ] we have lim
[q5(l)fn(t)
n+W
(19)
dt = [ 4 ( t ) j ( t ) dt.
Economic considerations suggest that the cost of adjustment functional should have a minimal property of the following type: If limn+mj" = p ( w ) , then (20) lim inf C( V , f n ) 2 C( V, p), n-m
where C is the extension of the cost functional. It is thus natural to define the cost of adjustment of a generalized investment schedule represented by a Radon measure p as
1
inf C( V, 1"): lim J, = p ( w ) . n-m
(21)
When definition (21) is applied to an ordinary function, the original value
C( V, p ) is not necessarily recovered. Before we study generalized investments
we shall obtain a formula for C(V, p ) when p is a piecewise continuous function. We start with the convex case. Lemma 3. Let r(1) be convex. Let 4 be a continuous positive function with compact support on [O, a).Then i f f is piecewise continuous on [0, a), we have
C(4, f ' ) = C(4, , f ) = Proof: The inequality C I C is clear. Since 4 has compact support, we can assume that its support is contained in [O, 11. Let fn be a sequence of functions that converge weakly to f : Let J j = [ ( j - l)/k, j/k] for j = 1, . . .,k and let c # ~ ~.,. . , 4 k be a set of positive continuous functions on [O, 13 that satisfy the following conditions: n
C4i(t)=
s,
1
SUPP
i=1
4j(f)df =
4j C_
where ck is a positive constant.
(23)
I,,
(24)
+(t)
[ ( j - l)/k -
Ekr
j/k
+ ELI,
(25)
15.4. APPLICATIONS OF GENERALIZED FUNCTIONS IN ECONOMICS
Let E > 0. There exists N
=
405
N ( d 1 , .. . , b , k ) such that if n 2 N, then
Since r is convex, one has
for any pair of piecewise continuous functions $, 9 with $ 2 0. Hence k
*I
rl
for some constants sj E J j , T: and I: we obtain
As k
-+
00, if
the ck
-+
E
supp bj ( j
=
1, . . . , n). Therefore, for any k
0, the right-hand side of (28) approaches Jolb,(r)[r(.t(r))-
E l (it,
and if we then let I: -+ 0, the result follows. We can now derive a formula for C'(b,,j ' ) for a general function r(1). Lemma 4. Let b, be a continuous positive function with compact support on [0, a). Then if,f'is piecewise continuous on [0, E), we have
Proqf'. Let us suppose as in the previous lemma that supp b, c [0, 11 and let J j = [ ( j - l ) / k , , j / k ] as before. We shall suppose that f is positive. Let
406
15.
MISCELLANEOUS TOPICS
& ( f ) be a piecewise constant function that satisfies the following conditions f o r j = 1, ..., k :
Let g k be given by constants a,,,, on intervals Jm, (m = 1, . . . ,m(j)) whose union over all rn is Jj. Thus
+
We first show that limk+mg k = f ( w ) . Let E C[O, 11 be positive. Let E > 0 and let 6 > 0 be such that I+(x) - II/(y)I IE for Ix - yl I6. Then if k 2 1/6, we have I
I-1
for some constants T
1J I
fl
0
~ E ,
Jm, and T~ E J j . Thus I
+(t)(gk(t) - f(t)) dtJIE
I-1
J0 f(t)
dt-
A similar argument shows that C(4, g k ) converges to Jh 4 ( t ) r * ( f ( t ) )d t since, because of (33), C(4, g k ) is almost a Riemann sum of this integral. Therefore,
C(4,f ) 5 Jrn4(t)r*(f(t)) 0 dt.
(34)
For the reverse inequality, let C(r*, 4, f ) be the integral functional formed with the function r* and let C(r*, 4, f ) be the corresponding extension. Then C(r*, 4, f ) = C(r*, 4, f ) since r* is convex. On the other hand, since r Ir* we get C(r*, 4, f ) = C(r*, 4, f ) Iq 4 , f),
which combined with (34) gives the desired result.
15.4.
APPLICATIONS OF GENERALIZED FUNCTIONS I N ECONOMICS
407
We now turn to generalized investments. We study a single jump first.
Lemma 5. Let 4 be a continuous, positive function with compact support on [O, co). Letf'be piecewise continuous on [0, 00) and let T 2 0, c 2 0. Then
+
Proof. Let f'" be a sequence of functions converging weakly to f ( t ) 4 1 , ... , 4kbe chosen as in Lemma 3, except that T E supp 4i for only one i and 4 i ( ~=) 4(z). Then proceeding in a similar fashion we obtain
cd(t - T). Let
+ f(Ti) -
UE
1
.
(36)
As k + co and E -,0, the first term approaches C(4, f'), the second approaches . zero, and the third, by L'Hospital's rule, approaches U C ~ ( T ) Thus
C(4, f'
+ ch(t -
7)
2
C(4, f ' )
+ uc4(z).
The reverse inequality is obtained by taking a sequence g. + f' with C(4, g.) + C(4,f') and then adding a term g; that is n on the interval [T, T + l/n] and zero everywhere else, since gn + cgk -, f' + ch(t - T ) and C ( 4 9" + csh) C(4,f')+ ac4(z). We thus obtain +
Theorem 1. Let on [0, co). Let
4 be a continuous, positive function with compact support
where u j , /Ij 2 0, where zi # on [0, 00). Then
T > for all
m
rm
' ( $ 9
1') = J
0
i, j, and wherefo(t) is locally integrable
4 ( t ) r * ( f ( t ) )dt
+ j1 ( a ~ j 4 ( 7 j+ > b B j+( T >) >* = 1
(38)
The rationale for using the functional C instead of C is that, given an investment schedule f ( r ) , one can obtain an arbitrarily close schedule g(t)
408
15.
MISCELLANEOUS TOPICS
whose cost is as close to C( V , f) as desired. In considering the maximization of (I), the use of C would be justified only if the revenue part of the profits is weakly continuous with respect to investment. We thus provide the following theorem : Theorem 2. Let 4(t) be a continuous, positive function with compact support in [0, 00). Suppose aR/aK is bounded on each finite time interval. Then the revenue functional,
Y ( 4 , f , K O )= / o m 4 0 R (t , K O + /:f(9,
ds) Lit,
(39)
ns
(40)
is weakly continuous on f for every constant K O . Proof. Iff,
-, J'(w),
then
F,(t) = KO
+ fj&)
ds -+ KO
+
J:
f(S)
strongly as measures. Let A be a bound for dR/aK for t in the support of 4. We then have
I Y ( 4 ,f , KO) - Y ( 4 ,f", K0)l =
I:s
4(t)(R(r,F(tN - R(r, Ffl(t))
I A s u p { ) & t ) ) : t20). Further Comments
The foregoing analysis applies to a very broad class of functions R , r, and V. It is clear that unless more information is available only very general aspects of the solution can be obtained. The behavior of r(Z) for large I and the relative size of costs with respect to revenues are particularly important. Different types of investments, however, seem to behave quite differently, and thus a wide range of functions R, r, and V are found in practice. The model we have presented is then the basic framework upon which more detailed case studies should be made. It is very difficult to prescribe conditions that guarantee the existence or nonexistence of generalized solutions given this level of generality. A few remarks on the likelihood of such solutions can be made. If the function r(1) is linear for I > 0, then upward adjustments in the capital stock will probably be made in jumps, not continuously; the reason is, clearly, that r*(Z, T) = a1 ( I 2 0), and thus the benefits of making a given adjustment quickly can be derived without having to pay extra costs.
15.5.
DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
409
+
A cost function like v ( l ) - u l - ln(l l), on the other hand, would preclude jumps in almost all cases, since adjusting very fast is very expensive.
15.5. DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
Our aim in this section is to solve the integral equations that involve generalized functions. We shall limit our study to the convolution-type integral equations because generalized functions are very suitable for them. Indeed, from the sifting property we find that the solution of the integral equation - t).f'(t)d t =
k(x)
isf(r) = 6(t). As another example, it is easily verified that the solution of the integral equation Joxsin(x -
+ +
t ) X ( t ) dt =
1
+ x,
+
is f ( t ) = 1 t 6 ( f ) 6'(t). Another advantage is that by taking the Laplace or Fourier transform of the convolution type integral equation we get a simple algebraic equation to solve for the transformed functions [ 161. By setting h(x) = 6(x) k(x), we can write the Volterra integral equation
+
f(x)
+ JOXk(X - t ) f ' ( t )dt = g ( x ) ,
x 2 0,
(1)
as h*f=g,
(2)
where we have extended the functionsfand g for x < 0. Incidentally, by this substitution the distinction between the integral equations of the first and second kind has disappeared. In order to solve this equation it suffices to find the distribution h , such that
h * h , =6.
(3)
Then the solution of (2) is
h, * g . (4) We found in Chapter 7 that 6(x) plays the role of the unit element in the convolution algebra. Accordingly, we observe from (3) that h , is the inverse .f'
=
41 0
15. MISCELLANEOUS TOPICS
of h = 6 + k . Taking the inverse of 6 of 1 + 1,so we set
h,
=
6
+
+ k, however, is like taking the inverse
c (-l)"(k*)", 03
n= 1
where (k*)" is the convolution product of k by itself taken n times. When the kernel is suitably smooth, the series (5) can be shown to converge and thus can be substituted in (4)to yield the solution
where
c (-l)"(k*)". m
k , ( t - x)
=
n= 1
We now discuss a few important singular integral equations. Cauchy-Type Integral Equations
Recall that the function l/x defines the distribution
for a test function &x). Its Fourier transform is [l/x]" = -ni sgn 11. Let us use this information to solve the famous singular Cauchy-type integral equation
where the * means that the integral is the Cauchy principal value and where the functions l a n d g are tempered distributions. Taking the Fourier transform of both sides of this equation and using the convolution formula [(l/x) *j]" = [l/x]"
f=
-n$(u) sgn ZI,
we obtain
(1 - sgn u ) f ( u )
= $(ti).
The result of multiplying both sides of (8) by 1 + sgn u is
(1' - 1)f
=
10 + (sgn GI)$,
15.5.
DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
41 1
so that
where we have assumed that ,I2 # 1. Finally, we take the inverse Fourier transform of (9) and obtain ‘[sgn
81
which is the required solution. Carleman-Type Integral Equations
Our aim now is to solve the equation
in the context of generalized functions. For this purpose, we define the finite Hilbert transform [52] X : E’(R) + D’(R),
1 X ( j ‘ )= - IT y ( i ) * j ;
where E‘ and D’are the distribution spaces and 9 stands for the principal value while R is the real line. The finite Hilbert transform for distributions is q-1,’):
E’C-1, 11
-+
D’C-19 113
c % - i , ~ ) ( f )
=
dS(f)l
where p is the restriction from D’(R) to D’[ - 1, 13 and E’[-l,l]
=
D’[ - 1, 13 =
{tEE’(R):suppt G [-1, { t E D’( - 1,
131,
1) : t can be extended to D’(R)J
Solving the integral equation (1 1) in the context of generalized functions is equivalent to the following problem: Given g E D’( - 1, l), findf E E’( - 1,l) such that 4x>f(x) -
H x ) ~ ’i , -i ) ( f )
= 61
on ( - 1, 1).
To solve this we appeal to the Cauchy representation of distributions on E’( - 1, I), as explained in Section 15.1, which we now adapt to the present situation.
41 2
MISCELLANEOUS TOPICS
15.
,
Definition 1. A - is the set of analytic functions F ( z ) defined on C\[ - 1, 11, where C is the complex plane and \ stands for the complement, such that (a) F ( z ) = O(l/lzl) as i -+ 00; (b) I F(x iy)I 5 MI y I-", 0 < I y I < 1, for some constant M .
+
The set A
-,has the following properties:
(1) If F
E
A - ,, then the limits
F,
lim F(x
=
iy) = F(x f i0)
y-0+
exist in D'(R). (2) I f F E A - , , t h e n [ F ] = F + - F - ~ E ' [ - l , l l . (3) If F E E'[ - 1, 11, the Cauchy representation is
,
Here we have F E A - and [F] E'[-I, 11 and A _ , .
= f'.Thusf' c1 F
is an isomorphism between
(4) .Zf'= i ( F + + F - ) . ( 5 ) F , = if' - $iX(f'), F - = -if' - $ X ' ( f ' ) . ( 6 ) The distribution f' satisfies the Carleman equation if and only if F E A - I satisfies the Hilbert problem
(a
-
ib)F+ - ( a
+ ib)F- = g
on(-1, 1).
(12)
From this discussion it follows that we are required to solve the following problem: c , F + - c 2F F, - F-
=g =
0
on ( - 1, l), on ( -
(13)
00, - 1)
u (1, m),
(14)
where we have imposed no condition at x = k l , while c , = u - ib and c2 = a + ib. We need to put some restrictions on the coefficients c , and c2.
Definition 2. The quantities a and b are called proper if e l c2 = a + ib satisfy the following conditions:
=
a - ib and
(a) cj(x) ( j = 1,2) is C" and nonzero on ( - 1, 1). (b) There exist integers a j , flj such that lim x-.
-
dk dxk
~
[(l - x)Lllcj(x)]
and
lim x--l+O
dk
-
dxk
[(I
+ X)~JC~(X)],
15.5.
41 3
DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
exist for k that
=
0, 1, 2,. . . , J
=
1, 2,. . . and are nonzero for k = 0. This means
Tj(X) =
is in E ’ [ - 1,
(1 - XYJ(1
11 and t j ( x ) # 0 for
+ .U)”Cj(.X)
-1 I x 5 1.
Now let
and
Then (1) M i ’ ( z ) is analytic on C\[- 1, 13. (2) I M ? ’ ( z ) l = 0 ( 1 ) a s z + W . ( 3 ) If F E A - , , then M ’ I F E A - , , and ( M , F ) , R\{ - 1, 11.
=
(M,),F+
on
(4) MI+ MI -
~
=
on (-1, I), on(-cc, - l ) u ( l , x ) .
c,(x),
{l.
( 5 ) If 9 ED’(R), then M : ’ g (a distribution in D’R\{-1, extended to a distribution in D’(R).
1 ) ) can be
O u r problem thus reduces to ( M l + / M l - ) F +- (M2+/M2-)F-
=
$1,
(15)
where y’, is an extension of 9 to R that is zero for ( xI > 1, or
or
H+
-
H-
=
h
,
on R\{
-
1, l},
(17)
where H = M F / M 2 and h is an extension of ( M , - g 1 ) / M 2+ to D’(R). The solution of (17) is
H ( z ) = H,(z)
+ K(z),
(18)
414
15.
MISCELLANEOUS TOPICS
where
and K+ -K-
Thus [ K ]
=
onR\{-1,
0
l}.
k has support on { - 1, l}, so that
=
N
k
=
C iij6'"(x
-
j=O
Accordingly,
N
K(z) =
where ( - l)'j!2niaj
=
j=O
1) + hj6"'(x
a,(x - l ) - j - '
Lij andj!2nibj =
+ 1).
+ bj(x + l ) - j - ' ,
-hj.
F+
=
(21)
[ F ] . For this purpose we have from
- 9i#'(u2)} + K + ) = exp{+u, exp{+u, - +i#'(ul)}
M2 + -(Ho+ MI -
h=-M - lg M2 +
=
(20)
Then from (16) we find that
F ( z ) = C M , ( z ) / M , ( z ) l ( H o ( z )+ K ( z ) ) .
Let us recall that our aim is to findf' the foregoing
(19)
=
exp{ -+(ul
+ u 2 ) + $(ul
-
u 2 ) } g = e-"g,
which defines 11. Writing the similar expression for F - and doing some algebraic work, we obtain [52]
Abel-Type Integral Equations
We found in Example 9 of Section 7.6 that the solution of Abel's integral equation
15.5.
DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
is,f'(x) = y(x) *
l(x) where QA(x)= x;-'/l-(A).
lo
1
-
41 5
Thus
q'(t) rtt
(X -
t)l-l'
For 0 < c( < 1, the integral on the left side of (23) is convergent, and therefore this integral equation can be solved by classical means [16]. We have solved it by the distributional approach for all values of a. We can solve various integral equations related to Abel's equation by the help of generalized functions. We present two examples. Example I. First we solve the integral equation
where lim fI(.s) # lim .f;(s). s-a
S-+R
The formal solution of (25) is [16] y(t) =
Now set
11
=
2 d dt
- -
IT
J,
f'(u) (t2 - i
dll ~ ~ ) ' ! ~ '
t sin 0, so that (t2
-
and (27) becomes
2 d g(t) = - IT rtt
L r y 2
Jo n'Z
f(t
=
t cos 0
sin 0) d U
du = t cos 0 do,
j"(t sin 0) sin 0 do.
=
(28)
To make use of the generalized functions we write .f'(s> = f l ( s X 1 -
H ( s - a)]
+ .f';(s)H(s- .I,
which when differentiated becomes .f"(s) - .f'\(s)[l - H ( s - a)]
+ ,f';(s)H(s-
+ 6(s - a)C.f;(s) - .fl(s)l.
(I)
(29)
41 6
15.
Furthermore, we let t sin a g(t)
=
=
MISCELLANEOUS TOPICS
a. Substituting these values in (28) we obtain
sin 0)[1 - H ( t sin 0 - t sin a)]
Ji'2{,>(t
+ f ; ( t sin 8)H(r sin 0 - t sin x ) + 6(t sin 0 t sin a ) [ f 2 ( t sin 0) - , f l ( t sin 0)]} sin U (10. -
(30)
Now, in view of the (3.1.1), namely,
where the x m are the simple zeros of ,f'(x), we find that
Ji'26(r sin 0 =
-
t sin a ) [ j ; ( r sin
e) -fl(r
sin 0 ) l sin 0 (10
[.f;(r sin a ) - j , ( t sin a)](sin a)/(t cos a )
Substituting this value in (30) and using the definition of the Heaviside function, we obtain the required solution: sin 0) sin 0 d0
.f",(t
+
j'>(tsin 0 ) sin 0 (16'
i
0 I s I u, a < s < K,
Exumple 2. y(t) tlr
a,
(t2
-
?)"2
= j'(s) =
.f;(s), .f;(s),
(32)
with condition (26). Its formal solution is [16] g(r) = - - -
In this case we set
II =
(33)
t cosh 0, tlir = t sinh 0 do, so that (31) takes the form
1
r m
g(t) =
-
n o
f"(t
cosh 0) cosh 0 dU.
(34)
15.5.
DISTRIBUTIONAL SOLUTIONS OF INTEGRAL EQUATIONS
Writing the function j ( s ) as in (29), setting t cosh in the first example, we obtain g(f) =
- 2 { [ i f ’ , ( [ cosh 0) cosh 0 d0 IT
+
01
= a,
41 7
and proceeding as
1
f ; ( t cosh 0) cosh 0 d0
+ C.f;(a) - f;(a)la/rJ=2}. For the unified analysis for the distributional solutions of various singular integral equations, the reader is referred to Estrada and Kanwal [53].
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References
1. Schwartz, L., “Theorie des Distributions.” Hermann, Paris, 1966.
2. Rubinstein, Z., “A Course in Ordinaryand Partial Differential Equations.” Academic Press, New York, 1969. 3. Hoskins, R. F., “Generalised Functions.” Horwood, Chichester and New York, 1979. 4. Roach, G. F., “Green’s Functions.” Van Nostrand-Reinhold, New York, 1970. 5. Lighthill. M. J., “An Introduction to Fourier Analysis and Generalised Functions.” Cambridge Univ. Press, London, 1958. 6. Gel’fand, I. M., and Shilov, G. E., “Generalized Functions,” Vol.-l, Academic Press, New York, 1964. 7. Estrada, R.. and Kanwal, R. P., J. Inst. Math. Appl. 26, 39-63 (1980). 8. Jones, D. S., “Generalised Functions.” McGraw-Hill, New York, 1966. 9. DeJager, E. M.,“Mathematics Applied to Physics” (E. Roubine, ed.). Springer-Verlag, New York, 1970. 10. Estrada, R., and Kanwal, R. P., J. Math. Phys. Sci., 14, 285-293 (1980). 11. Estrada, R., and Kanwal, R. P., J . Math. Anal. Appl. 89, 262-289 (1982).
12. Jain, D. L., and Kanwal, R. P., J. Integ. Eqns. 3,279-299 (1981): 4,31-53, 113-143 (1982). 13. Farassat, F., J. Sound Vib. 55, 165-193 (1977). 14. Strichartz, R. S., Fourier Transforms and Distribution Theory, A Survival Kit, Lecture Notes, Mathematics Department, Cornell University, 1977. 15. Arsac, J., “Fourier Transforms and the Theory of Distributions.” Prentice Hall, Englewood
Cliffs, New Jersey, 1966.
419
420
REFERENCES
16. Kanwal, R. P., “Linear lntegral Equations, Theory and Technique.” Academic Press, New York, 1971. 17. Zemanian, A. H., “Distributional Theory and Transform Analysis.” McGraw-Hill, New York, 1965. 18. Liverman, T. P. G., “Generalized Functions and DirectlOperational Methods,” Vol. 1. Prentice-Hall, Englewood Cliffs, New Jersey, 1964. 19. Stakgold, I., “Boundary Value Problems of Mathematical Physics,” Vol. 1-2. Macmillan, New York, 1967, 1968. 20. Stakgold, I., “Green’s Function and Boundary Value Problems.” Wiley, New York, 1979. 21. Pan, H. H.,and Hohenstein, R.M.. Quart. Appl. Math. 39, 131-136(1981). 22. Wiener, J., J. Math. Anal. Appl. 88, 170-182 (1982). 23. Vladimirov, V. S., “Equations of Mathematical Physics.” Dekker, New York, 1971. 24. Maria, N. L., Elementary solutionsof partialdifferential operators with constant coefficients. Ph.D. Thesis, University of California, Berkeley, 1968. 25. Watson, G. N., “ A Treatise on the Theory of Bessel Functions.” Cambridge Univ. Press, London and New York, 1966. 26. Duff, G. F., and Naylor, D., “Differential Equations of Applied Mathematics.” Wiley, New York, 1966. 27. Farassat, F., “Class Notes of APSC-215, Analytic Methods in Engineering.” The George Washington University, 1979. 28. Farassat, F., AIAA J. 19, 1122-1 130 (1981). 29. Goldstein, M. R.,“ Aerocoustics.” McGraw-Hill, New York, 1976. 30. Tanaka, K., and Ishi, S., J . Sound Vib. 77, 397-401 (1981). 31. Alawneh, A. D., and Kanwal, R.P., SIAM REV. 19,437471 (1977). 32. Kanwal, R. P., and Sharma, D. L., J . Elasticity 6, 57-65 (1976). 33. Kanwal, R. P., Proceedings of the Second Symposium on Innovative Numerical Analysis in Applied Engineering Sciences, University Press of Virginia, pp. 531-542 (1980). 34. Seodel, W., and Power, D., J . Sound Vib. 65,29-35 (1979). 35. Chwang, A. T., and Wu, T. Y., J. Fluid Mech. 63, 607-622 (1974). 36. Datta, S., and Kanwal, R.P., J. Elusticity 10,435-442 (1980). 37. Alawneh, A. D., and Kanwal, R.P., Internat. J. Math. Math. Sci. 4, 795-803 (1982). 38. Datta, S., and Kanwal, R. P., Quart. Appl. Math. 37, 86-91 (1979). 39. Datta, S., and Kanwal, R. P., Urilitas Math. 16, 111-122 (1979).
40. Ffowcs-Williams, J. E., and Hawkings, D. L., Philos. Trans. Roy. SOC.London Ser. 264, 321-342 (1969).
41. Costen, R. C., J . Math. Phys. 22, 1377-1385 (1981). 42. Kanwal, R.P. A note on jump conditions for fields that have infinite integrable singularities at an interface. Unpublished report, 1981. 43. Stratton, J. A., “Electromagnetic Theory,” pp. 188-192; 247-250, McGraw-Hill, New York, 1941.
ADDITIONAL READING
42 1
44. Beltrami, E. J., and Wohlers, M. R., “Distributions and the Boundary Values of Analytic Functions.” Academic Press, New York, 1963. 45. Bremermann, H., “Distributions, Complex Variables and Fourier Transforms.” AddisonWesley, Reading, Massachusetts, 1965. 46. Roos, B. W., “Analytic Functions and Distributions in Physics and Engineering.” Wiley, New York, 1969. 47. Chihara, T. S., “An Introduction to Orthogonal Polynomials.” Gordon & Breach, New York, 1979. 48. Morton, R. D., and Krall, A. M., S I A M J . Math. Anal. 9,604-626 (1978). 49. Cristescu. R., and Marinescu, G., “Applications of the Theory of Distributions.” Editura Academici. Bucharest (also Wiley, New York), 1973. 50. Gel’fand, I. M., and Vilenkin, M. Ya., “Generalized Functions,” Vol. 4. Academic Press, New York, 1964. 51. Estrada, R., and Kanwal, R. P., Nonlinear Anal. Theory Meth. Appl. 5, 1379-1387 (1981).
52. Orton, M.. Appl. Anal. 9, 219-231 (1979).
53. Estrada. R. and Kanwal, R. P.. J . Inte,q. Eqns. to appear.
ADDITIONAL R EADlNG F. Constantinescue, ‘*Distributionsand Their Applications in Physics. Pergamon, New York, 1980. A. Erde’lyi, “Operational Calculus and Generalized Functions.” Holt, Rinehart and Winston, New York, 1962. S. Fenyo’, and T. Frey, ”Modern Mathematical Methods in Technology.” Vol. I . North-
Holland, Amsterdam, 1969.
A. Friedman, “Generalized Functions and Partial Differential Equations.” Prentice-Hall, Englewood, New Jersey. 1963. B. Friedman, ”Lecture on Application-Oriented Mathematics.” Holden-Day. San Francisco, 1969. L. Hiirmander. “Linear Partial Differential Operators.” Springer-Verlag. New York. 1963. E. Marx. and D. D. Maystre, J. Math. Phys. 23, 1047-1056 (1982). R. D. Milne, “Applied Functional Analysis, An Introductory Treatment.” Pitman. London. 1980.
E. E. Rosinger, “Nonlinear Partial Differential Equations, Sequential and Weak Solutions.” North-Holland, New York, 1980. G . E. Shilov, “Generalized Functions and Partial Differential Equations.” Gordon & Breach, New York. 1968. G . Temple,
I’
100 Years of Mathematics.” Springer-Verlag, Berlin and New York, 1981.
F. Treves, “Basic Linear Partial Differential Equations.” Academic Press, New York, 1975.
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Index
A
Cartesian product, 169 Cauchy principal value, 29 ff Cauchy problem, 283 Cauchy representation, 37 1 of a distribution, 371, 412 for a probability density, 393 Cauchy-Riemann operator, 250 Causal signal, 367 Causal solution, 221, 258 Cavities in elastic medium, 293, 296, 297, 30 1 Center of dilatation, 322, 323, 326 Center of rotation, 320, 321, 326 Central moment, 396 Characteristic function, 392, 393 Characteristic functional, 398 Characteristic lines, 265 Chebyshev polynomial, 379 Compact support, 22 ff Complementary error function, 286 Confluent hypergeometric equation, 24 I Concave function, 399 Convergence of distributions, 59, 60 Convex function, 399 Convolution, 169 ff definition, 179 Fourier transform of, 194 Laplace transform of, 200 properties, 181 Correlation, 398 Cost function, 402, 403 Cost functional, 402
Abel integral equation, 192, 414, 415 Abel’s formula, 228 Acoustic scattering, 310 Adjoint operator, 38, 39, 21 1, 244, 245 Algebriac operations on distributions, 33 Algebraic singularity, 82-86 Analytic operations on distributions, 36 Axially symmetric bodies, 287, 288 ff
B Bending of a rectangular plate, 332-334 Bessel equation, 49, 241 Bessel function, 49, 74 ff generalized solutions of, 241 Bessel polynomial, 390 Behavior of analytic functions at the boundaries, 371 Behavior of harmonic functions at the bondaries, 371 Biharmonic operator, 125-129, 131, 283 Biorthogonal set, 50, 376 Boundary value problem, 136, 208, 219 ff C
Capacity, 289 of a dumbbell-shaped body, 291 of an elongated rod, 296 of an oblate spheroid, 295 of a prolate spheroid, 294 of a slender body, 294 of a sphere, 296 Carleman-type integral equation, 41 1 423
424
INDEX
Couplet, 318 Cumulative charge distribution, 7 ff Curvature of a curve, 129 ff Gaussian, 339 mean, 112 ff Curvilinear coordinates, 56, 57, 109 ff
D d’Alembert formula, 193 d’Alembert’s operator, 246, 276 Delta convergent sequence, 14 Delta function, 4 ff decomposition into plane waves, 100, 102 integral representation, 68, 149 as a Stieltjes integral, 71, 72 Delta sequence, 5 ff with parametric dependence, 60 Differential operator, 21, 38 ff adjoint, 38, 39, 21 1, 244, 245 Dipole, 14 ff Dipole sequence, 15, 17, 18, 70 Dirac delta function, see Delta function Direct product, 169 ff Dirichlet formula, 5 Dirichlet problem, 31 1 Disk (circular) capacity of, 295 distribution on, 335, 336 elastostatic displacement of, 322 electric polarizability tensor for, 310 magnetic polarizability tensor for, 310 polarization tensor for, 305 stream function for, 300 virtual mass tensor for, 308 Distributional convergence, 60 ff Distributional derivative, 36 ff Distributional weight function, 374 ff Distributions algebraic operations on, 33 analytic operations on, 37 on arbitrary lines, 332 of bounded support, 185, 372, 373, 374 Cauchy representation of, 371, 372, 373, 41 I centrally symmetric, 35 convergence of, 59, 60,140 essential point of, 47 even, 35, 167 of exponential growth, 201 Fourier transform of, 137 ff
homogeneous, 34 integral of, 214 invariant, 34 Laplace transform of, 200 ff odd, 35, 167 order of, 46 periodic, 34 on plane curves, 129, 334 product with a function, 35 regular, 26 ff Riesz, 279, 371 singular, 26 ff singular support of, 47, 48 skew symmetric, 35 of slow growth, see tempered support of, 47 tempered, 139 ff transformation properties of, 52 Divergent integrals regularization of, 67, 82 ff Double layer distribution, 45, 188 Double layer potential, 189, 254, 255 Dual space, 27 ff Dumbbell-shaped bodies, 290 capacity of, 291 electric polarizability tensor for, 3 10 magnetic polarizability tensor for, 310 stream function for, 293 strain energy for, 293
E Eigenvalue problem, 227 Elastodynamics, 325 Elastostatics, 319 Electric polarizability tensor, 308-310 Electromagnetic boundary conditions, 135 Electromagnetic potentials, 348 Electromagnetic wave equation, 284 Elongated rod capacity of, 296, 297 strain energy for, 297 Embedding procedure, 344 Even distributions, 35, 167 Expectation value, 396
F First fundamental form, I12 Fourier Bessel series, 74 Fourier series, 65-67, 73, 108 ff Fourier transform, 68, 137 ff of convolutions, 194, 195 of direct product, 178
INDEX of radial functions, 160, 163, 164, 262 of tempered distributions, 145 ff of test functions, 141-144 Fourth-order polynomial distribution, 301 Fractional derivative, 191 Fractional integral, 191 Fredholm integral equation, 288, 290, 303, 304, 312, 315, 319, 336 Frequency response function, 367 Fresnel integral, 260 Functional, 25 ff linear continuous, 25 ff Functions of compact support, 22 ff Functions of slow growth, 139 ff Fundamental matrix, 217, 239 Fundamental solution, 39, 115, 217 ff of biharmonic operator, 283 of Cauchy-Riemann operator, 250 of dissipative wave equation, 282 of equations for elastodynamics, 283, 284, 325 of equations for elastostatics, 283, 319 of first-order ordinary differential equation, 217 of fourth-order ordinary differential equation, 236 of heat operator, 257 of Helmholtz operator, 261, 262 of Klein-Gordon operator, 280 of Laplace operator, 132, 133, 252, 253, 254, 259 of nth order ordinary differential equation, 238 of Oseen‘s equations, 285, 286 of Schrodinger operator, 260 of second order ordinary differential equation, 2 18-2 19 with variable coefficients, 232 of Stokes equations, 283 of vector wave operator, 284 of wave operator, 263, 264 ff
G Gaussian coordinates, 109 Gaussian distribution, 392, 393 Gaussian sequence, 12 Generalized analytic function, 37 1 Generalized derivative, see Distributional derivative Generalized electrostatic potential, 288
425 Generalized function, 2, 14, 27 ff, see also Distributions action of, 14 ff even, 167 odd, 167 Generalized Gaussian process, 398 Generalized solution of ordinary differential equation, 2 I2 ff of partial differential equation, 247 ff Generalized stochastic process, 397 Green’s function, 217, 220 ff Green’s matrix, 239 Green’s tensor, 284
H Hadamard finite part, 77 ff Hadamard’s method of descent, 248 ff Harmonic oscillator, 232 Heat operator, 246, 257 Heaviside function, 1 ff Heaviside sequence, 16, 17 Heisenberg distributions, 30, 44, 89, 151 Helmholtz operator, 261 Hermite polynomial, 375, 377, 387 Hilbert problem, 412 Hilbert space, 397 Hilbert transform, 41 I Homogeneous distribution, 34 Hyperbolic differential operator, 283 Hyperbolic system, 340, 341, 345
I Ideal sampler, 369 Impulse pair functions, 28, 29 Impulse response, 363-367, 370 Infinite singularity, 347 ff Inhomogeneous wave equation, 268 Input, 360 ff Initial value problem, 107, 153, 207, 217 ff Integral equation Abel’s type, 192, 414, 415 Carleman type, 4 I I Cauchy type, 410 Fredholm type, 288, 290. 303, 304, 312, 315, 319, 336 Volterra, 409 Integral of a distribution, 214 Integral representation of delta function, 68, 149 Invariant distribution, 34 Inverse Fourier transform, 142 ff Investment schedule, 400
INDEX
J
Jacobi polynomial, 378, 379, 390 Jump discontinuity, 2-4, 105 ff
K Kelvin dipole, 320, 321, 326, 327 Kelvin force, 319 Klein-Gordon operator, 276, 280
L Lagueme polynomial, 375, 377, 387, 388 Laplace operator, 100, 131, 132, 149, 245, 252, 259, 276 Laplace transform, 199 ff of convolution, 200 of distribution, 200 ff inverse, 200 ff of periodic function, 205 Legendre polynomial, 375, 376, 379, 387 Light cone, 265 ff Linear axial distributions, 298 Linear functional, 25 ff Linear operator, 361 ff Linear system, 361 ff Locally integrable function, 21 ff Logarithmic potential, 253
M
Mach number, 274-276 Magnetic polarizability tensor, 308-310 Magnetohydrodynamic waves, 355 Maxwell’s equations, 135, 347, 352 Moments of orthogonal polynomials, 375 ff Moving sources line, 274 point, 273 surface, 275 Multiindex, 20 ff Multipole, 46
N n-dimensional sphere surface area, 60, 62 ff volume 60-62 n-dimensional wave operator, 282 Neumann problem, 31 I , 315 Newtonian potential, 187, 253 Normal derivative operator, 114 ff Null sequence, 24 ff 0 Oblate spheroid capacity of, 295 electric polarizability tensor for, 309
magnetic polarizability tensor for, 309 polarization tensor for, 305 stream function for, 300 virtual mass tensor for, 308 Odd distribution, 35, 167 Operator, 360 ff continuous, 362 linear, 361 stationary, 362 Order of a distribution, 46 Orthogonal polynomials, see Polynomials Oseen’s equations, 285, 286 output, 360 ff
P
Parabolic axial distributions, 299 Parseval’s formula, 137 Penny-shaped crack capacity of, 302 strain energy for, 302 Periodic distributions, 34 Plemelj formulas, 31 Poisson distribution, 392 Poisson’s equation, 187, 287 Poisson’s integral formula, 256, 281, 337 Poisson’s summation formula, 153, 154, 168 Polarizability tensor electric, 308-310 magnetic, 308-310 Polarization potential, 298, 299 Polarization tensor, 303-305 Polynomials Bessel, 390 Chebyshev, 379 Hermite, 375, 377, 387, 388 Jacobi, 378, 379, 390 Laguerre, 375, 377, 379, 387 Legendre, 375, 376, 379, 387 Potential barrier, 223 Probability distribution, 390 ff binomial, 392 continuous, 391 density, 391 ff discrete, 391 field, 394 Gaussian, 392, 393 Poisson, 392 Product of a distribution, with a function, 35 Prolate-shaped bodies, 295 polarizability tensors for, 310 polarization potential of, 295
INDEX
427
Prolate spheroid acoustic scattering by, 31 1, 315 capacity of, 294 elastodynamic field for, 328 elastostatic field for, 322 electric polarizability tensor for, 309 magnetic polarizability tensor for, 309 polarization potential for, 299 polarization tensor for, 303-305 stream function for, 300 virtual mass tensor for, 306, 307 Psuedofunction, 30, 77 ff
R Radial distribution (spherically symmetric), 148, 160, 163, 262 Radon measure, 399, 404 Radon’s problem, 102 Random variable, 390 ff Rankine Hugoniot conditions, 345, 346 Rapidly decaying functions, 138 ff Regular distribution, 26 ff Regular singular function, 118 ff Regularization of distributions, 183, 184, 198 384, 385 Regularization of divergent integrals, 82 ff Replicating function, 28 Reproducing property, 5 Retarded potential, 269 Revenue functional, 408 Riesz distribution, 279, 371 Rotlet, 318
S Sampler data system, 369 Sampling function, 28, 29 Scattering theory, 310 ff Schrdinger operator, 259 Schrdinger wave equation, 223 Schwartz-Sobolev theory, 20, 25 Self-adjoint operator, 245, 246 Sifting property, 5ff Signum function, 43 ff Single-layer distribution, 32, 148, 165, 187, I88 Single-layer potential, 187, 254 Singular distribution, 26 ff Singular surface, 337, 338 Slender body capacity of, 294 electric polarizability tensor for, 309 magnetic polarizability tensor for, 309
polarization tensor for, 305 virtual mass tensor for, 308 Slenderness parameter, 294 Sokhotski-Plemelj equations, 3 I Sound generation, aerodynamic, 343 Square wave function, 210 Stationary process, 399 Step function, 1 Step response, 362, 363 Stieltjes integral, 71 Stochastic process, 397 Stokes flow, 317, 324 Stream function, 292, 293 Strength of the shock front, 346 Stresslet, 321, 327 Sturm-Liouville problem, 223, 224, 227 Superposition principle, 361 Support compact, 22 ff of a distribution, 47 of a function, 21 ff singular, 47, 48 Surface distribution, 32, 113 ff Symbolic function, 27 System, 360 ff discrete time, 369, 370 linear, 361 ff relaxed, 361 time invariant, 362, 366
T Tempered beam, deflection of, 236, 237 Tempered distributions, 137, 139 ff Test functions, 22 ff of compact support, 22 of exponential decay, 201 of rapid decay, 138 Time-invariant system, 362, 366 Transfer function, 366, 367 Transformation properties of distributions, 52-58 Transport operator, 25 I
U Uniform axial distributions, 294 Unit dipole, 14 ff Unit step function, 1
V Variance, 392 Virtual mass tensor, 306 Volterra integral equation, 409
428
INDEX
Volume potential, 187 see also Newtonian potential Vortex sheet, 136
W Wave equation, 18, 194 ff Wave operator, 246 ff Weak convergence, 59, 404 Weak limit, 184
Weak solution, 212 Weak topology, 402 Weierstrass’s approximation theorem, 19 Wronskian, 228
2 Z-transform, 369, 370 Zero state response, 368
