Generalized Functions: Theory and Technique (Mathematics in Science and Engineering, 171)

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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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 = 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<xldx
=

(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&( =

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

=

=

(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.



=

-(,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( = -+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( 1, xas 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 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, = 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)

=

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 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?

=



=

<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 -

[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((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( regions, respectively. In the present case, a = 0, b = 1, so that G, = x(5 - 1) and G, = 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) 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( = 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( 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)

/'( 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

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.

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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 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