Linear Integral Equations Theory and Technique
RAM P. KANWAL
Pennsylvania State University
University Park, Pennsylvania
ACADEMIC PRESS 1971 New York and London COPYRIGHT © 1971, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS.
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United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. Berkeley Square House, London W1X 6BA
LIBRARY OF CONGRESS CATALOG CARD NUMBER: 77-156268
AMS (MOS) 1970 Subject Classification: 45A05, 45B05, 45C05, 45D05, 45E05, 45E10, 45E99, 45F05, 35C15, 44A25, 44A35, 47G05
PRINTED IN THE UNITED STATES OF AMERICA IN LOVING MEMORY OF MY GRANDMOTHER PREFACE
Many physical problems which are usually solved by differential equation methods can be solved more effectively by integral equation methods. Indeed, the latter have been appearing in current literature with increasing frequency and have provided solutions to problems heretofore not solvable by standard methods of differential equations. Such problems abound in many applied fields, and the types of solutions explored here will be useful particularly in applied mathematics, theoretical mechanics, and mathematical physics. Each section of the book contains a selection of examples based on the technique of that section, thus making it possible to use the book as the text for a beginning graduate course. The latter part of the book will be equally useful to research workers who encounter boundary value problems. The level of mathematical knowledge required of the reader is no more than that taught in undergraduate applied mathematics courses. Although no attempt has been made to attain a high level of mathematical rigor, the regularity conditions for every theorem have been stated precisely. To keep the book to a manageable size, a few long proofs have been omitted. They are mostly those proofs which do not appear to be essential in the study of the subject for purposes of applications. We have omitted topics such as Wiener-Hopf technique and dual integral equations because most of the problems which can be solved by these methods can be solved more easily by the integral equation methods presented in Chapters 10 and 11. Furthermore, we have con centrated mainly on three-dimensional problems. Once the methods have been grasped, the student can readily solve the corresponding plane problems. Besides, the powerful tools of complex variables are available for the latter problems. The book has developed from courses of lectures on the subject given by the author over a period of years at Pennsylvania State University.
XI xii PREFACE
Since it is intended mainly as a textbook we have omitted references to the current research literature. The bibliography contains references to books which the author has consulted and to which the reader is referred occasionally in the text. Chapter 10 is based on author's joint article with Dr. D. L. Jain (SIAM Journal of Applied Mathematics, 20, 1971) while Chapter 11 is based on an article by the author (Journal of Mathematics and Mechanics, 19, 1970, 625-656). The permission of the publishers of these two journals is thankfully acknowledged. These articles contain references to the current research literature pertaining to these chapters. The most important of these references are the research works of Professor W. E. Williams and Professor B. Noble. Finally, I would like to express my gratitude to my students Mr. P. Gressis, Mr. A. S. Ibrahim, Dr. D. L. Jain, and Mr. B. K. Sachdeva for their assistance in the preparation of the textual material, to Mrs. Suzie Mostoller for her pertinent typing of the manuscript, and to the staff of Academic Press for their helpful cooperation. INTRODUCTION CHAPTER 1
1.1. DEFINITION
An integral equation is an equation in which an unknown function appears under one or more integral signs. Naturally, in such an equation there can occur other terms as well. For example, for a^s ^b9 a ^ / ^ b, the equations b /(*) = JK(s,t)g(t)dt, (1) a b g(s)=f(s) + JK(s,t)g(t)dt, (2) a b g(s) = jK(.s,t)lg(t)Y dt, (3) a where the function g(s) is the unknown function while all the other functions are known, are integral equations. These functions may be complex-valued functions of the real variables s and t. Integral equations occur naturally in many fields of mechanics and mathematical physics. They also arise as representation formulas for the solutions of differential equations. Indeed, a differential equation can be 1 2 1 /INTRODUCTION replaced by an integral equation which incorporates its boundary con ditions. As such, each solution of the integral equation automatically satisfies these boundary conditions. Integral equations also form one of the most useful tools in many branches of pure analysis, such as the theories of functional analysis and stochastic processes. One can also consider integral equations in which the unknown function is dependent not only on one variable but on several variables. Such, for example, is the equation
g(s)=f(s) + JK(s9t)g(t)dt9 (4) where s and / are ^-dimensional vectors and Q is a region of an n- dimensional space. Similarly, one can also consider systems of integral equations with several unknown functions. An integral equation is called linear if only linear operations are performed in it upon the unknown function. The equations (1) and (2) are linear, while (3) is nonlinear. In fact, the equations (1) and (2) can be written as Llg(s)]=f(s), (5) where L is the appropriate integral operator. Then, for any constants c1 and c2, we have
L[clgl(s) + c2g2(y)] = clL[_gl(j)] + c2L\_g2(s)] . (6) This is the general criterion for a linear operator. In this book, we shall deal only with linear integral equations. The most general type of linear integral equation is of the form
h(s)g(s) = f(s) + X j K(s, t)g(t) dt, (7) a where the upper limit may be either variable or fixed. The functions/, h, and AT are known functions, while g is to be determined; X is a nonzero, real or complex, parameter. The function K(s, t) is called the kernel. The following special cases of equation (7) are of main interest. (i) FREDHOLM INTEGRAL EQUATIONS. In all Fredholm integral equations, the upper limit of integration b, say, is fixed. (i) In the Fredholm integral equation of the first kind, h(s) = 0. 1.2. REGULARITY CONDITIONS 3
Thus, b f(s) + kJK(s,t)g(t)dt = 0. (8) a (ii) In the Fredholm integral equation of the second kind, h(s) = 1;
b g(s)=f(s) + lJK(s,t)g(t)dt. (9) a (iii) The homogeneous Fredholm integral equation of the second kind is a special case of (ii) above. In this case,/"C?) = 0;
b g(s) = XJ K(s,t)g(t)dt. (10) a (ii) VOLTERRA EQUATIONS. Volterra equations of the first, homo geneous, and second kinds are defined precisely as above except that b = s is the variable upper limit of integration. Equation (7) itself is called an integral equation of the third kind. (iii) SINGULAR INTEGRAL EQUATIONS. When one or both limits of integration become infinite or when the kernel becomes infinite at one or more points within the range of integration, the integral equation is called singular. For example, the integral equations
00 g(s)=f(s) + X | (cxp-\s-t\)g(t)dt (11) — oo and s f(s) = I [l/(j-0*]g(t) dt, 0 < a < 1 (12) 0 are singular integral equations.
1.2. REGULARITY CONDITIONS
We shall be mainly concerned with functions which are either con tinuous, or integrable or square-integrable. In the subsequent analysis, 4 1 /INTRODUCTION it will be pointed out what regularity conditions are expected of the functions involved. The notion of Lebesgue-integration is essential to modern mathematical analysis. When an integral sign is used, the Lebesgue integral is understood. Fortunately, if a function is Riemann- integrable, it is also Lebesgue-integrable. There are functions that are Lebesgue-integrable but not Riemann-integrable, but we shall not encounter them in this book. Incidentally, by a square-integrable function g (7), we mean that
b j"|0(O|2
(a) for each set of values of s91 in the square a^s^ba^t^b, bb 2 ^\K(s,t)\ dsdt < oo , (2) a a (b) for each value of s in a ^ s ^ b,
b j \K(s,t)\2 dt < oo , (3) a (c) for each value of Mn a ^ / ^ ft,
b j \K(s,t)\2ds < oo. (4) a
1.3. SPECIAL KINDS OF KERNELS
(i) SEPARABLE OR DEGENERATE KERNEL A kernel K(s9t) is called separable or degenerate if it can be expressed as the sum of a finite number of terms, each of which is the product of a function of s only and a function of / only, i.e.,
K(sj) = tai{s)bi{t). (1) i=i 1.5. CONVOLUTION INTEGRAL 5
The functions at(s) can be assumed to be linearly independent, otherwise the number of terms in relation (l) can be reduced (by linear inde pendence it is meant that, if clal+c2a2-\ hc„tf„ = 0, where c{ are arbitrary constants, then cv = c2 = ••• = cn = 0). (ii) SYMMETRIC KERNEL A complex-valued function K(s,t) is called symmetric (or Hermitian) if K(s, t) = K*(t,s), where the asterisk denotes the complex conjugate. For a real kernel, this coincides with definition K(s, t) =K(t,s).
1.4. EIGENVALUES AND EIGENFUNCTIONS
If we write the homogeneous Fredholm equation as b j K(s,t)g(i)dt = M(S), JI = 1M, a we have the classical eigenvalue or characteristic value problem; \i is the eigenvalue and g (t) is the corresponding eigenfunction or character istic function. Since the linear integral equations are studied in the form (1.1.10), it is X and not 1/A which is called the eigenvalue.
1.5. CONVOLUTION INTEGRAL
Many interesting problems of mechanics and physics lead to an integral equation in which the kernel K(s, t) is a function of the difference (s—i) only: K(s,t) = k(s-t)9 (1) where A: is a certain function of one variable. The integral equation s g(s)=f(s) + ljk(s-t)g(t)dt, (2) a and the corresponding Fredholm equation are called integral equations of the convolution type. 6 1/INTRODUCTION
The function defined by the integral s s jk(s-t)g«)dt = jk(t)g(s-t)dt (3) 0 0 is called the convolution or the Faltung of the two functions k and g. The integrals occurring in (3) are called the convolution integrals. The convolution defined by relation (3) is a special case of the standard convolution 00 00 1 k(s-t)g{t) dt = J k(t)g(s-t) dt. (4)
— 00 — 00 The integrals in (3) are obtained from those in (4) by taking k(t) = g (t) = 0, for / < 0 and / > s.
1.6. THE INNER OR SCALAR PRODUCT OF TWO FUNCTIONS
The inner or scalar product (0, \jj) of two complex ^-functions $ and ^ of a real variable s, a ^ s < b9 is defined as b {M) = \m**(t)dt. (i) a Two functions are called orthogonal if their inner product is zero, that is, cj) and ifr are orthogonal if ((/>, i/O = 0. The norm of a function (j)(t) is given by the relation b b Ul = [ J HtWii) dt]A = [ j |
1.7. NOTATION
For Fredholm integral equations, it will be assumed that the range of integrations is a to b, unless the contrary is explicitly stated. The quantities a and b will be omitted from the integral sign in the sequel. INTEGRAL EQUATIONS WITH CHAPTER 2 SEPARABLE KERNELS
2.1. REDUCTION TO A SYSTEM OF ALGEBRAIC EQUATIONS
In Chapter 1, we have defined a degenerate or a separable kernel K(s, i) as
K{s,t)=tai{s)bi{t), (1) /= 1 where the functions al(s)9...,an(s) and the functions bl(t)9 ...,&„ (0 are linearly independent. With such a kernel, the Fredholm integral equation of the second kind, g(s)=f(s) + XJK(s,t)g(t)dt (2) becomes g(s) = f(s) + X t ai(s) j MO0(0 dt . (3)
It emerges that the technique of solving this equation is essentially dependent on the choice of the complex parameter X and on the definition of
ct = fbt(t)g(t)dt. (4) 8 2.1. REDUCTION TO ALGEBRAIC EQUATIONS 9
The quantities c( are constants, although hitherto unknown. Substituting (4) in (3) gives n g(s) =f(s) + XY,ciai(s)9 (5) 1=1 and the problem reduces to finding the quantities ct. To this end, we put the value of g(s) as given by (5) in (3) and get
t at(s) {ct - f 6f(r)[/(0 + X t ckak(m dt} = 0 . (6) /=i J k=\
But the functions at(s) are linearly independent; therefore,
ct - f bt(t)W(t) + *i ckak(m dt = 0, i=l,...,« . (7) J k=i Using the simplified notation
J*i(0/»=/i, jbt(t)ak(t)dt = aik9 (8) where /j and aifc are known constants, equation (7) becomes
(9) that is, a system of n algebraic equations for the unknowns ct. The determinant D(X) of this system is
l—Xall —Xaxl "- —Xaln
-lalx \-la22 ••• -Xa2n D(X) = (10)
-Xanl -Xan2 ••• \-Xann which is a polynomial in X of degree at most n. Moreover, it is not identically zero, since, when X = 0, it reduces to unity. For all values of X for which D(X) # 0, the algebraic system (9), and thereby the integral equation (2), has a unique solution. On the other hand, for all values of X for which D(X) becomes equal to zero, the algerbaic system (9), and with it the integral equation (2), either is insoluble or has an infinite number of solutions. Setting X=\/n in 10 2 / INTEGRAL EQUATIONS WITH SEPARABLE KERNELS equation (9), we have the eigenvalue problem of matrix theory. The eigenvalues are given by the polynomial D(X) = 0. They are also the eigenvalues of our integral equation. Note that we have considered only the integral equation of the second kind, where alone this method is applicable. This method is illustrated with the following examples.
2.2. EXAMPLES
Example J. Solve the Fredholm integral equation of the second kind
g(s) = s + XJ(st2 + s2t)g(t)dt. (I) o The kernel K(s, t) = st2 + s2t is separable and we can set
2 ci = j t g(t)dt, c2 = j tg{t)dt . o o Equation (1) becomes 2 g(s) = s + Xcx s + Xc2 s , (2) which we substitute in (1) to obtain the algebraic equations
cx = i + Uct + \Xc2 , W c2 = $ + &i + Uc2 . The solution of these equations is readily obtained as 2 2 c, = (60 + >l)/(240-120A-A ) , c2 = 80/(240- 120A-2 ) . (4) From (2) and (4), we have the solution g(s) = [(240-60/1)s + 801y2]/(240-120/1 -X2) . (5) Example 2. Solve the integral equation
I g(s)=f(s) + k\(s + t)g(0dt (6) o and find the eigenvalues. 2.2. EXAMPLES 11
Here, al(s) = s, a2(s)= l, bx(i)= l, b2(t) = ty
1 I
au = j tdt = \ , al2 = j dt = 1 , o o
l I 2 a2l = j t dt = $, a22 = j tdt = i, o o I I
/, = //(/)*, f2 = jtf(t)dt. 0 0 Substituting these values in (2.1.9), we have the algebraic system
(\-±X)Cl - kc2 =fv , -±Act + (1 -±A)c2 =/2 . The determinant D(X) = 0 gives A2 + 12A-12 = 0. Thus, the eigen values are
lx =(-6 + 4^3), A2 = (-6-4^3). For these two values of A, the homogeneous equation has a nontrivial solution, while the integral equation (6) is, in general, not soluble. When X differs from these values, the solution of the above algebraic system is 2 ci = [-12/! + Wi - 12/2)]/a + 12A - 12) ,
2 c2 = [~ 12/2 - A(4/i - 6/2)]/(A + 12A - 12) . Using the relation (2.1.5), there results the solution
16{X-2)(s + t)-l2kst-4X g(s)=f(s) + X A2+ 122-12 f{t) dt ■ (?)
The function T(s, t;X),
V(s,t;X) = [6(A-2)(* + r) - 12Arf - 4A]/(22 + 12A - 12) , (8) is called the resolvent kernel. We have therefore succeeded in inverting the integral equation because the right-hand side of the above formula is a known quantity. 12 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS
Example 3. Invert the integral equation In 9 (s) = f(s) + A j (sin s cos t) g (t) dt . (9) o As in the previous examples, we set In c = f (cost)g(i)dt to obtain
g(s) = f(s) + Xcsms. (10) Multiply both sides of this equation by cos s and integrate from 0 to In. This gives In c = J (cos t)f(i)dt. (11) 0 From (10) and (11), we have the required formula: In
g (s) = f(s) + A f (sin s cos 0/(0 dt . (12) o Example 4. Find the resolvent kernel for the integral equation
I 9(s) = f(s) + lj(st + s2 t2)g(t) dt . (13)
For this equation, 2 2 fliCO = s , a2(s) = s , bY{t) = t, b2(t) = t ,
flu = f > 0i2 = 021 = 0 , tf22 = I >
2 / = jtf(t)dt9 f2= \t f(t)dt.
Therefore, the corresponding algebraic system is
Cid-iX)=fi9 c2(l-iX)=f2. (14)
Substituting the values of cx and c2 as obtained from (14) in (2.1.9) yields the solution 2.2. EXAMPLES 13
l 9W-/W + 1 jrVc^]™*-
g(s) = XJesetg(t)dt. (17) o Define I f c = j e g(t)dt9
so that (17) becomes g(s) = Xces. (18) Put this value of g(s) in (17) and get I Xces = Xes j e'llce^ dt = \k2 esc(e2-\) o or Xc{2-k(e2-\)} = 0. If c = 0 or X = 0, then we find that g = 0. Assume that neither c = 0 nor X = 0; then, we have the eigenvalue A = 2/(e2-l). Only for this value of X does there exist a nontrivial solution of the integral equation (17). This solution is found from (18) to be g(s) = [2c/(e2-l)]^. (19) Thus, to the eigenvalue 2/(e2 — 1) there corresponds the eigenfunction es. Example 6. Find the eigenvalues and eigenfunctions of the homo geneous integral equation 14 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS
2 g(s) = XJtst + (\lst)-]g(t)dt. (20)
I Comparing this with (2.1.3), we have
ax(s) = s , a2(s) = \/s , bx(t) = 19 b2(t) = \jt .
The formula (2.1.9) then yields the following homogeneous equations:
(\-U)c1 -Xc2 = 0, -Xcx +(\-$X)c2 = 0, (21) which have a nontrivial solution only if the determinant 1—1A -A Z)(A) = = 1 - (17/6)1 + (1/6)A2 -A 1-A vanishes. Therefore, the required eigenvalues are *i = i(17 + V265) ~ 16.6394 ; (22) 22 = 1(17-^265) ~ 0.3606.
The solution corresponding to v^ is c2 — —2.2732c,, while that corre sponding to X2 is c2 ~ 0.4399c/. The eigenfunctions of the given integral equation are found by substituting in (2.1.5): flf, (s) ~ 16.639c [> - 2.2732(1/*)] , t (23) 02(s) ~ 0.3606c/ |> + 0.4399 (1/*)] , where cx and c/ are two undetermined constants.
2.3. FREDHOLM ALTERNATIVE
In the previous sections, we have found that, if the kernel is separable, the problem of solving an integral equation of the second kind reduces to that of solving an algebraic system of equations. Unfortunately, integral equations with degenerate kernels do not occur frequently in practice. But since they are easily treated and, furthermore, the results derived for such equations lead to a better understanding of integral equations of more general types, it is worthwhile to study them. Last, 2.3. FREDHOLM ALTERNATIVE 15 but not least, any reasonably well-behaved kernel can be written as an infinite series of degenerate kernels. When an integral equation cannot be solved in closed form, then recourse has to be taken to approximate methods. But these approximate methods can be applied with confidence only if the existence of the solution is assured in advance. The Fredholm theorems explained in this chapter provide such an assurance. The basic theorems of the general theory of integral equations, which were first presented by Fredholm, correspond to the basic theorems of linear algebraic systems. Fredholm's classical theory shall be presented in Chapter 4 for general kernels. Here, we shall deal with degenerate kernels and borrow the results of linear algebra. In Section 2.1, we have found that the solution of the present problem rests on the investigation of the determinant (2.1.10) of the coefficients of the algebraic system (2.1.9). If D(X) ^ 0, then that system has only one solution, given by Cramer's rule
ct = (Dufx + D2if2 + ••• + Dnifn)ID(X) , / = 1,2, •.-,/! , (1) where Dhi denotes the cofactor of the (/z, /)th element of the determinant (2.1.10). Consequently, the integral equation (2.1.2) has the unique solution (2.1.5), which, in view of (1), becomes
n
g(s) = f(s) + X y —— at(s) , (2) while the corresponding homogeneous equation
g(s) = ij K(s,t)g(t)dt (3) has only the trivial solution g (s) = 0.
Substituting for/f from (2.1.8) in (2), we can write the solution g (s) as g(s)=f(s) + WDW]
x f { £ lDubx{t) + D2ib2(t) + - + DMty]ai(s)}f(t)dt. (4) Now consider the determinant of (n+ l)th order 16 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS
0 ax(s) a2(s) ••• an(s)
bl(t) 1— Aan —laX2 ••• — Xaln
D(s,t;X) = - b2{t) -ka2X \-Xa22 ••• -hi2n (5)
Kit) -XanX -Xan2 ••• \-lann By developing it by the elements of the first row and the corresponding minors by the elements of the first column, we find that the expression in the brackets in equation (4) is D(s, t;X). With the definition
r(s,t;X) = D(s9t;X)/D(X)9 (6) equation (4) takes the simple form g(s)=f(s) + ljr(s,t;X)f(t)dt. (7) The function F(s, t\X) is the resolvent (or reciprocal) kernel we have already encountered in Examples 2 and 4 in the previous section. We shall see in Chapter 4 that the formula (6) has many important con sequences. For the time being, we content ourselves with the observation that the only possible singular points of Y{s, t\k) in the A-plane are the roots of the equation D(X) = 0, i.e., the eigenvalues of the kernel
The above discussion leads to the following basic Fredholm theorem.
Fredholm Theorem. The inhomogeneous Fredholm integral equation (2.1.2) with a separable kernel has one and only one solution, given by formula (7). The resolvent kernel r(s, t\X) coincides with the quotient (6) of two polynomials. If D(X) = 0, then the inhomogeneous equation (2.1.2) has no solution in general, because an algebraic system with vanishing determinant can be solved only for some particular values of the quantities fv To discuss this case, we write the algebraic system (2.1.9) as (I-AA)c = f, (8) where I is the unit (or identity) matrix of order n and A is the matrix (au). Now, when D(X) = 0, we observe that for each nontrivial solution of the homogeneous algebraic system (I-/lA)c = 0 (9) 2.3. FREDHOLM ALTERNATIVE 17
there corresponds a nontrivial solution (an eigenfunction) of the homo geneous integral equation (3). Furthermore, if k coincides with a certain eigenvalue k0 for which the determinant D (k0) = 11 — k0A | has the rank /?, 1 ^p ^ n, then there are r = n—p linearly independent solutions of the algebraic system (9); r is called the index of the eigenvalue k0. The same holds for the homogeneous integral equation (3). Let us denote these r s an linearly independent solutions as g0i(s)9g02(s), •••,0orC )> d let us also assume that they have been normalized. Then, to each eigenvalue k0 of index r = n—p9 there corresponds a solution g0(s) of the homo geneous integral equation (3) of the form r 9o(s) = Z *k9ok(s) ,
where ak are arbitrary constants. Let m be the multiplicity of the eigenvalue k0, i.e., D(k) = 0 has m equal roots k0. Then, we infer from the theory of linear algebra that, by using the elementary transformations on the determinant |I — kA\, we shall have at most m + 1 identical rows and this maximum is achieved only if A is symmetric. This means that the rank p of D(k0) is greater than or equal to n — m. Thus, r = n — p ^ n — (n — m) = m ,
and the equality holds only when ati = ajt. Thus we have proved the theorem of Fredholm that, if k = k0 is a root of multiplicity m ^ 1 of the equation D{k) = 0, then the homo geneous integral equation (3) has r linearly independent solutions; r is the index of the eigenvalue such that 1 ^ r ^ m. The numbers r and m are also called the geometric multiplicity and algebraic multiplicity of k0, respectively. From the above result, it follows that the algebraic multiplicity of an eigenvalue must be greater than or equal to its geometric multiplicity. To study the case when the inhomogeneous Fredholm integral equation (2.1.2) has solutions even when D(k) = 0, we need to define and study the transpose of the equation (2.1.2). The integral equation1
^(s) = f(s) -f k j K(t,s)\Kt) dt (10)
1 We shall consider only real functions here even though the results are easily extended to complex functions. Outcome is the same in both the cases. 18 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS is called the transpose (or adjoint) of the equation (2.1.2). Observe that the relation between (2.1.2) and its transpose (10) is symmetric, since (2.1.2) is the transpose of (10). If the separable kernel K(s,t) has the expansion (2.1.1), then the kernel K(t, s) of the transposed equation has the expansion
K(t,s)= faMbiis). (11)
Proceeding as in Section 2.1, we end up with the algebraic system (I-2AT)c = f, (12) T where A stands for the transpose of A and where ct and f{ are now defined by the relations
ct = j at(t)g(t) dt, ft = j aMfit) dt . (13) The interesting feature of the system (12) is that the determinant D(X) is the same function as (2.1.10) except that there has been an inter change of rows and columns in view of the interchange in the functions ax and bt. Thus, the eigenvalues of the transposed integral equation are the same as those of the original equation. This means that the transposed equation (10) also possesses a unique solution whenever (2.1.2) does. As regards the eigenfunctions of the homogeneous system |I —AAT| c = 0, (14) we know from linear algebra that these are different from the correspond ing eigenfunctions of the system (9). The same applies to the eigen functions of the transposed integral equation. Since the index r of A0 is the same in both these systems, the number of linearly independent eigenfunctions is also r for the transposed system. Let us denote them by xl/o\ixl/o2i'">xlJQr anc* let us assume that they have been normalized.
Then, any solution \\J0{S) of the transposed homogeneous integral equation iKj) = X \ K(t,s)\Kt)dt (15)
corresponding to the eigenvalue X0 is of the form where /?,• are arbitrary constants. 2.3. FREDHOLM ALTERNATIVE 19
We prove in passing that eigenfunctions g(s) and \jj(s) corresponding to distinct eigenvalues Xx and X2, respectively, of the homogeneous integral equation (3) and its transpose (15) are orthogonal. In fact, we have
g(s) = Xlj K(s, t)g(0 dt, i/>(s) = X2 j K(t, s) xjj(t) dt.
an Multiplying both sides of the first equation by X2\l/(s) d those of the second equation by X{g(s), integrating, and then subtracting the resulting equations, we obtain b
(X2-Xl)jg(s)ifr(s)ds = 0. a an But Xx / X2, d the result follows. We are now ready to discuss the solution of the inhomogeneous Fredholm integral equation (2.1.2) for the case D(X) = 0. In fact, we can prove that the necessary and sufficient condition for this equation to have a solution for X = A0, a root of D(X) = 0, is that f(s) be orthogonal to the r eigenfunctions \\j0i of the transposed equation (15). The necessary part of the proof follows from the fact that, if equation (2.1.2) for X =X0 admits a certain solution g(s), then
jf(s)^oi(s) ds = | g(s)il/oi(s) ds
-X0jil/0i(s)dsJK(s,t)g(t)dt
= j g(s)^oi(s)ds
-X0j g(t) dt \ K(s, i) \^Qi{s) ds = 0 ,
because X0 and \\/0i (s) are eigenvalues and corresponding eigenfunctions of the transposed equation. To prove the sufficiency of this condition, we again appeal to linear algebra. In fact, the corresponding condition of orthogonality for the linear-algebraic system assures us that the inhomogeneous system (8) reduces to only n — r independent equations. This means that the rank of the matrix (I — XA) is exactly p = n — r, and therefore the system (8) or (2.1.9) is soluble. Substituting this solution in (2.1.5), we have the solution to our integral equation. Finally, the difference of any two solutions of (2.1.2) is a solution of 20 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS the homogeneous equation (3). Hence, the most general solution of the inhomogeneous integral equation (2.1.2) has the form
g(s) = G(s) + a^oifa) + oc2g02(s) + ••• +
Fredholm Alternative Theorem. Either the integral equation g(s)=f(s) + XJK(s,t)g(t)dt (17) with fixed X possesses one and only one solution g(s) for arbitrary ^-functions f(s) and K(s, t), in particular the solution g = 0 for /= 0; or the homogeneous equation g(s) = XJK(s,t)g(t)dt (18) possesses a finite number r of linearly independent solutions g0h i= 1,2, •••,/*. In the first case, the transposed inhomogeneous equation ifr(s)=f(s) + XJ K(t,s)il,(t)dt (19) also possesses a unique solution. In the second case, the transposed homogeneous equation
^(J) = XJK(t9s)^(t)dt (20) also has r linearly independent solutions \j/0h i= 1,2, •••,/•; the in- homogeneous integral equation (7) has a solution if and only if the given function f(s) satisfies the r conditions
(Mod = jfMotW ds = 0, i = 1,2, -,r . (21) In this case, the solution of (17) is determined only up to an additive linear combination X?=i ci9or The following examples illustrate the theorems of this section. 2.4. EXAMPLES 21
2.4. EXAMPLES
Example 1. Show that the integral equation In g{s) = /(*) +(1/TT) J [sin (* + /)] #(/)
5 = ai(s) = sins, d^C ) coss, £,(/) = cos/, b2(t) = sin/. Therefore, 2;r
axl = sin/cos/^// = 0 = a2 o
2TT 2 fl12 = COS t dt = 71 = fl2l 0
1 — ATT £>(!) = ~ = 1-A22^.7i22 . (2) -ht 1
The eigenvalues are Xl = \/n9 A2 = — 1/TT and equation (1) contains At =l/7r. Therefore, we have to examine the eigenfunctions of the trans posed equation (note that the kernel is symmetric) In g(s) = (\ln)jsin(s+t)g(t)dt. (3) o The algebraic system corresponding to (3) is
ci — Xnc2 = 0 , — Xncx + c2 = 0 , which gives c c i = i for Al = l/7i ; ci = — c2 for A2 = — 1/TZ .
Therefore, the eigenfunctions for At = 1/TT follow from the relation (2.1.5) and are given by 22 2/INTEGRAL EQUATIONS WITH SEPARABLE KERNELS
g(s) = c(sins + cos*?) . (4) Since In
(ssins + scoss) ds = —2n^09 while In I (sins + coss) ds = 0 , we have proved the result. Example 2. Solve the integral equation
I g(s)=f(s) + lj(l-3st)g(t)dt. (5) 0 The algebraic system (2.1.9) for this equation is
(l-X)Cl+i^2=fl9 -\kcl +(l+X)c2 = /2, (6) while l-X fA D{X) = = i(4-A2) . (7) -\k \ + X Therefore, the inhomogeneous equation (5) will have a unique solution if and only if A ^ ± 2. Then the homogeneous equation
g(s) = lj(\-3st)g(t)dt (8) has only the trivial solution. Let us now consider the case when X is equal to one of the eigenvalues and examine the eigenfunctions of the transposed homogeneous equation I g(s) = lj(\-3st)g(t)dt. (9) 0
For k = +2, the algebraic system (6) gives c{ = 3c2. Then, (2.1.5) gives the eigenfunction g(s) = c(l-s), (10) 2.5. AN APPROXIMATE METHOD 23 where c is an arbitrary constant. Similarly, for X = - 2, the correspond ing eigenfunction is g(s) = c(l-3s). (11)
It follows from the above analysis that the integral equation
g(s)=f(s) + 2J(l-3st)g(t)dt 0 will have a solution if f(s) satisfies the condition
I
j(l-s)f(s)ds = 09 0 while the integral equation
g(s) = f(s) -2| (l-3st)g(t)dt o will have a solution if the following holds:
j(l-3s)f(s)ds = 0.
2.5. AN APPROXIMATE METHOD
The method of this chapter is useful in finding approximate solutions of certain integral equations. We illustrate it by the following example:
I g(s) = es-s-js(est-\)g(t)dt. (1) o Let us approximate the kernel by the sum of the first three terms in its Taylor series: st 2 3 2 4 3 K(t9 s) = s(e - 1) - s t + is t + is 1 , (2) that is, by a separable kernel. Then the integral equation takes the form
I g(s) = es - s - J" (s21 + ±s312 + is4 t3)g(t) dt. (3) 24 2 / INTEGRAL EQUATIONS WITH SEPARABLE KERNELS
Since the kernel is separable, we require the solution in the form s 2 3 4 g(s) = e - s + ct s + c2s + c3s . (4) Following the method of this chapter, we find that the constants cl9c2,c3 satisfy the following algebraic system:
(5/4)c1+(l/5)c2 + (l/6)c3 = -2/3,
(1/5)*! + (13/6)c2 + (l/7)c3 = (9/4) - e , (5)
(1/6) Cl + (1/7) c2 + (49/8) c3 = 2e- (29/5) , whose solution is
cx = -0.5010 , c2 = -0.1671 , c3 = -0.0422 . (6) Substituting these values in (4), we have the solution
g(s) = es - s - 0.5010s2 - 0.1671s3 - 0.0423/ . (7)
Now the exact solution of the integral equation (1) is
g(s) = 1 . (8)
Using the approximate solution for s = 0, s = 0.5, and s= 1.0, the value of g (s) from (7) is
0(0) = 1.0000 , g(0.5) = 1.0000 , g(\) = 1.0080 , (9) which agrees with (8) rather closely. In Chapter 7 (see Section 7.7), we shall prove that an arbitrary j^-kernel can be approximated in norm by a separable kernel.
EXERCISES
1. Consider the equation I g(s)=f(s) + XJK(s,t)g(t)dt 0 and show that for the kernels given in Table I the function D(X) has the given expression. EXERCISES 25
TABLE I
Case Kernel D(k)
(i) ±1 1 + A (ii) st 1 - (A/3) (Hi) s2 + t2 l-(2A/3)-(4A2/45)
2. Solve the integral equation In g{s) = f(s) + lj cos(s+t)g(t)dt
and find the condition that f(s) must satisfy in order that this equation has a solution when X is an eigenvalue. Obtain the general solution if f(s) = sins, considering all possible cases. 3. Show that the integral equation n g(s) = A | (sin s sin 2i)g (t) dt o has no eigenvalues. 4. Solve the integral equation
g(s) = l+X je^'-tgWdt,
— It considering separately all the exceptional cases. 5. In the integral equation
2 g(s) = s + j(sinsf)g(t)dt9 o replace sin st by the first two terms of its power-series development
C*03 sm st = st — + ••• and obtain an approximate solution. METHOD OF SUCCESSIVE CHAPTER 3 APPROXIMATIONS
3.1. ITERATIVE SCHEME
Ordinary first-order differential equations can be solved by the well- known Picard method of successive approximations. An iterative scheme based on the same principle is also available for linear integral equations of the second kind:
g(s)=f(s) + XJK(s,t)g(t)dt. (1)
In this chapter, we present this method. We assume the functions f(s) and K(s, t) to be i??2-functions as defined in Chapter 1. As a zero-order approximation to the desired function g(s), the solution g0(s), g0(s)=f(s), (2) is taken. This is substituted into the right side of equation (1) to give the first-order approximation 91 (s) = f(s) + A j K(s, t) 0o (0 dt . (3)
This function, when substituted into (1), yields the second approxi mation. This process is then repeated; the («+l)th approximation is 26 3.1. ITERATIVE SCHEME 27 obtained by substituting the nth approximation in the right side of (1). There results the recurrence relation
Qn+i(s) =f(s) + X j K(s,i)gn{t)dt. (4)
If gn(s) tends uniformly to a limit as «-> oo, then this limit is the required solution. To study such a limit, let us examine the iterative procedure (4) in detail. The first- and second-order approximations are
g1(s)=Xs) + lJK(s,t)f(t)dt (5) and g2(s)=f(s) + XJK(s,t)Xt)dt + X2 j K(s, i) [ j K(t, x)f{x) dx\ dt. (6) This formula can be simplified by setting
K2 (s, i) = j K(s, x) K(x, t) dx (7) and by changing the order of integration. The result is
2 02(*) = f(s) + *j K(s, 0/(0 dt + A j K2 {s, 0/(0 dt. (8) Similarly,
g3(s)=f(s) + lj K(s,t)f(t)dt 2 3 + A J K2 (s, 0/(0 dt + l j K3 (s, 0/(0 dt, (9) where K3 (s, i) = f K(s, x) K2 (x, t)dx . (10) By continuing this process, and denoting
KJs,0 = | K(s,x)Km_x{x,t)dx, (11) we get the («+ l)th approximate solution of integral equation (1) as
m 9n (*) = f(s) + i^ (Km (s, 0/(0 dt. (12)
We call the expression Km(s9 i) the mth iterate, where Kx (5, i) = K(s, t). Passing to the limit as n -► 00, we obtain the so-called Neumann series
m g(s) = limgn(s) = f(s) + I A f Km(s, 0/(0 dt. (13) 28 3 / METHOD OF SUCCESSIVE APPROXIMATIONS
It remains to determine conditions under which this convergence is achieved. For this purpose, we attend to the partial sum (12) and apply the Schwarz inequality (1.6.3) to the general term of this sum. This gives
2 2 2 | J *„(*, 0/(0 dt\ *£ (I \Km(s, 0| dt) j |/(0| dt. (14) Let D be the norm of/,
2 2 D = j \f(t)\ dt 9 (15)
2 and let Cm denote the upper bound of the integral
2 \ \Km{sj)\ dt.
Hence, the inequality (14) becomes 2 2 2 \JKm(s,t)f(t)dt\ 2 2 2 \Km(s,t)\ ^ j \Km^(s,x)\ dx j \K(x,t)\ dx , which, when integrated with respect to t, yields 2 2 2 j\Km(s,t)\ dt^B C m^, (17) where B2 = jj\K(x,t)\2dxdt. (18) The inequality (17) sets up the recurrence relation 2 2m 2 2 Cm ^ B ~ Cl . (19) From (16) and (19), we have the inequality 2 2 2 2m 2 | j Km(s, 0/(0 dt\ < C D B ~ . (20) Therefore, the general term of the partial sum (12) has a magnitude less m -1 than the quantity DC1 |A| ZT , and it follows that the infinite series (13) converges faster than the geometric series with common ratio \A\B. Hence, if \MB<\, (21) the uniform convergence of this series is assured. 3.1. ITERATIVE SCHEME 29 It will now be proved that, for given 2, equation (1) has a unique solution. Suppose the contrary, and let gx (s) and g2(s) be two solutions of equation (1): gl(s)=f(s) + AJK(s,t)gl(t)dt9 g2(s)=f(s) + XJK(s9t)g2(t)dt. By subtracting these equations and setting gi(s)-g2(s) = (j)(s), there results the homogeneous integral equation 00) = A j K(s9t)(j)(t)dt . Apply the Schwarz inequality to this equation and get \ 2 2 2 2 j \ In view of the inequality (21) and the nature of the function (f)(s) = g\(s) — g2(s)9 we readily conclude that 0) = 0, i.e., gx(s) = g2{s). What is the estimate of the error for neglecting terms after the nth term in the Neumann series (13)? This is found by writing this series as g(s) = As) + f r f Km(s, 0/(0 dt + R„(s) . (23) Then, it follows from the above analysis that n + x \Rn\ ^ DC, \l\ 57(1 -\X\ B). (24) Finally, we can evaluate the resolvent kernel, as defined in the previous chapter, in terms of the iterated kernels Km(s, i). Indeed, by changing the order of integration and summation in the Neumann series (13), we obtain 1 1 g(s) =/(*) + A f [ £ A "" Km(s, 0]/(0 dt. Comparing this with (2.3.7), g(s) = f{s) + k j TO, t;X)f(t) dt, (25) 30 3 / METHOD OF SUCCESSIVE APPROXIMATIONS we have 00 m r(s,t;X)= X l -*Km(s,t). (26) m=l From the above discussion, we can infer (see Section 3.5) that the series (26) is also convergent at least for \X\ B < 1. Hence, the resolvent kernel is an analytic function of A, regular at least inside the circle \X\ < B~l. From the uniqueness of the solution of (1), we can prove that the resolvent kernel T(sj\X) is unique. In fact, let equation (1) have, with X = A0, two resolvent kernels r^(s, t\X0) and T2{s, t\X0). In view of the uniqueness of the solution of (1), an arbitrary function f(s) satisfies the identity f(s) + X0 j Tx (s, t;X0Mt) dt = f{s) + X0 J T2(s, t;X0)f(t) dt . (27) Setting H'C?,t\X0) = Tx(s,t\X0)-T2(s,t\k0), we have, from (27), JV(J,/;A0)/(OA = 0, for an arbitrary function/(/). Let us take/(/) = 4/*(5*, t; 2), with fixed .s. This implies that 2 j |V(J,/U0)| A = 0, which means that ^(.y, t; A0) = 0, proving the uniqueness of the resolvent kernel. The above analysis can be summed up in the following basic theorem. Theorem. To each j£?2-kernel K(s, t), there corresponds a unique resolvent kernel r(s, t\X) which is an analytic function of A, regular l at least inside the circle \X\ 3.2. EXAMPLES Example 1. Solve the integral equation I g(s)=f(s) + ljes-'g(t)dt. (1) 0 Following the method of the previous section, we have s t Kl(s9t) = e - , s f K2(s,t) = ]>-*£>*-' dx = e - . o Proceeding in this way, we find that all the iterated kernels coincide with K(s, t). Using (3.1.26), we obtain the resolvent kernel as V(s,f,X) = K(s,t)(\+X + l2 + --) = eS-'Kl-X) . (2) Although the series (1+A + A2H—) converges only for \X\ < 1, the resolvent kernel is, in fact, an analytic function of X, regular in the whole plane except at the point 2=1, which is a simple pole of the kernel T. The solution g(s) then follows from (3.1.25): I g(s) = f(s) - [A/(2- 1)] J* e*"'/(0 dt . (3) 0 Example 2. Solve the Fredholm integral equation g(s)= l+lf(l-3st)g(t)dt (4) o and evaluate the resolvent kernel. Starting with g0(s) = 1, we have 32 3 / METHOD OF SUCCESSIVE APPROXIMATIONS gx(s) = l+XJ(\-3st)dt = 1 +2(l-iy), o 2 g2(s)= 1 + X j (l-3st)[l+W-it)]dt = 1 +A(l-i*) + iA , o 2 3 g(s) = 1 + 1(1 -$J) + ±A + iA (l -ij) + TVA4 + TV15(1 -fc) + - , or 2 4 g(s) = (l+il + 1VA +-)[l + A(1-MI . (5) The geometric series in (4) is convergent provided \X\ < 2. Then, 2 3W = [4 + 2A(2-3J)]/(4-A ), (6) and precisely the same remarks apply to the region of the validity of this solution as given in Example 1. To evaluate the resolvent kernel, we find the iterated kernels K{ (s, t) = l-3st, 1 K2(s,t) = f (l-3sx)(\-3xt)dx = 1 -i(s+t) + 3^ , o K3(s,t) = j (\-3sx)[\ -$(x + t)-3xf]dx o = ±(1-3JO = ±*I(M), Similarly, KA(s,t) = iK2(s9t) and Kn(s9t) = iKn_2(s,t). Hence, = (l+i^+^4+-)^l+^(l+^2+^4+-)^2 = [(l+A)-|A(j+0-3(l-A)^]/(l-iA2), (7) |A| < 2 . 3.2. EXAMPLES 33 Example 3. Solve the integral equation n g(s) =1+^J [sinCy + O]0(O dt . (8) o Let us first evaluate the iterated kernels in this example: Kx(s,t) = K(s,t) = sin (^+0 , n &2 (s> 0 = Esm (s + •*)] sm (x + 0 dx 0 = ^TT [sin s sin f + cosscosf] = %ncos(s—t) , ^3 CM) = i^r I [sin(s + xy]cos(x — t)dx o = \n (sin s cos x + sin x cos s) o x (cos x cos t + sin x sin i) dx — (?ri)2 [sins cost + cosssinf] = (^n)2s'm(s+t) . Proceeding in this manner, we obtain ^4 CM) = (i7c)3cosCy-0, ^sCM) = (irc)4sin(j+0, ^6 C*> 0 = (i^r)5 cos (5 - 0 , etc. Substituting these values in the formula (3.1.13) and integrating, there results the solution g(s) = 2A(cos*)[l + (±n)2X2 + (i7i)4A4 + ...] + 227r(sins)[l + (i7i)2A2 + (i7r)4A4 + •••] , (9) or 2 2 2 g(S)= 1 + [(2/1 COS* + A 7TSinj)/(l - ±A 7T )] • (10) Since £2 = f f sin2(5+0 dsdt = \%2 , 00 the interval of convergence of the series (9) lies between — yj2/n and y/2/n. 34 3 / METHOD OF SUCCESSIVE APPROXIMATIONS Example 4. Prove that the rath iterated kernel Km(s,t) satisfies the following relation: Km(s,t) = JKr(s,x)Km_r(x,t)dx, (11) where r is any positive integer less than n. By successive applications of (3.1.11), Km{sj) = f - f K(s,x1)K(xl9x2)'.'K(xm-ut)dxm-i-dxl . (12) Thus, Km(sj) is an (m— l)-fold integral. Similarly, Kr(s,x) and Km_r(x, t) are (r— 1)- and (m-r— l)-fold integrals. This means that j Kr(s,x)Km-r(x,t)dx is an (m— l)-fold integral, and the result follows. One can take the g0(s) approximation different from f(s), as we demonstrate by the following example. Example 5. Solve the inhomogeneous Fredholm integral equation of the second kind, g(s) = 2s + XJ(s+t)g(t)dt9 (13) o by the method of successive approximations to the third order. For this equation, we take g0(s) = 1. Then, g(s) = 2s + X j (s + t) dt = 2s + X(s + ±) , o I g2(s) = 2s + X j (s + t) {2t + A[f+ (1/2)]} dt o = 2s + A[> + (2/3)] + 22|> + (7/12)] , 2 g2(s) = 2s + X j (s+t) {2/ + A[f + (2/3)] + A [f+ (7/12)]} A o 2 3 = 25 + A[J + (2/3)] + X [(7/6) 5 + (2/3)] + A [(13/12> + (5/8)] . (14) 3.3. VOLTERRA INTEGRAL EQUATION 35 From Example 2 of Section 2.2, we find the exact solution to be g(s) = 112(2-X)s + 8A]/(12-2A-A2) , (15) and the comparison of (14) and (15) is left to the reader. 3.3. VOLTERRA INTEGRAL EQUATION The same iterative scheme is applicable to the Volterra integral equation of the second kind. In fact, the formulas corresponding to (3.1.13) and (3.1.25) are, respectively, Q(s) = As) + f W Km(s9 0/(0 dt, (1) s g (s) = f(s) + X j T (s, t; X)f{t) dt, (2) a where the iterated kernel Km(s, t) satisfies the recurrence formula s Km(s,t) = j K(s,x)Km_x(x,i)dx (3) with Kx (s, t) = K(s, t), as before. The resolvent kernel F(s,t;l) is given by the same formula as (3.1.26), and it is an entire function of X for any given (5,0 (see Exercise 8). We shall illustrate it by the following examples. 3.4. EXAMPLES Example 1. Find the Neumann series for the solution of the integral equation s g(s) = (\+s) + lj(s-t)g(t)dt. (1) 0 36 3 / METHOD OF SUCCESSIVE APPROXIMATIONS From the formula (3.3.3), we have K1(s9t) = (s-t)9 C (s-t)3 v K2(s,t) = (s-x)(x-t)dx = 3! ' ' t (s-x)(x-t)3 (s-t)5 K (s,t) = J 3 3! ^ = —' and so on. Thus, ,w.1+I+^+^+^+^+.... For A= 1, g(s) = es. Example 2. Solve the integral equation 9(s)=f{s) + x\e*- s x x f K2(s9t) = j e - e - dx = (j-f)^"', t s x , K3(s,t) = !(x-t)e'-'e - dx = ^T"^"'. (m-\)\ The resolvent kernel is r(s,t;X) = \ Z, (m-\)\ { o, m=' Hence, the solution is 3.5. RESULTS ABOUT THE RESOLVENT KERNEL 37 s g(s) = f(s) + X j e s g(s) = 1 + fstg(t)dt. (6) o For this example, Kt (s9 i) = K(s, i) = st9 s 2 K2(s9 i) = j sx tdx = (s*t- st*)/3 , t s 4 A 7 4 4 K3 (s, 0 = J l(sx) (x 1 - xt )IJ\ dx = (s 1 - 2s t + sf)/18 , t s 7 4 7 K4(s9 i) = j l(sx)(x 1 - 2x* t + ;cf )/18] dx t 10 1 4 1 10 = (s t-3s t + 3s t -st )/l629 and so on. Thus. s* s6 s9 s12 ,(,)-!+ _+_+_ +__+.... (7) 3.5. SOME RESULTS ABOUT THE RESOLVENT KERNEL The series for the resolvent kernel T(s9 t\X)9 T(s9t\X) = £ X*-*Km(s,t)9 (1) can be proved to be absolutely and uniformly convergent for all values of s and / in the circle \k\ < \/B. In addition to the assumptions of Section 3.1, we need the additional inequality 2 2 f \K(s91)\ ds < E , E = const . (2) 38 3 / METHOD OF SUCCESSIVE APPROXIMATIONS Recall that this is one of the conditions for the kernel K to be an j2?2-kernel. Applying the Schwarz inequality to the recurrence formula Km(s9t) = j Km_l(s9x)K(x9t)dx (3) yields 2 2 2 \Km(s9t)\ < (j\Km,l(s9x)\ dx)j\K(x9t)\ dx9 which, with the help of (3.1.19), becomes 1 \Km(s9t)\ < QZHr- . (4) Thus, the series (1) is dominated by the geometric series with the general m l m_1 term Cx E(k ~ £ ), and that completes the proof. Next, we prove that the resolvent kernel satisfies the integral equation T(s,t;X) = K(s9t) + lj K(s9x)T(x9 t\X)dx. (5) This follows by replacing Km(s9t) in the series (1) by the integral relation (3). Then, oo r_1 T{s9t\X) = Kl(s,t)+ X Km_l(s9x)K(x9t)dx 1 = K(s, O + ifr" f Km(s9x)K(x9 t) dx r °° x = K(s9 t) + X [ X r~ Km(s9 *)] K(x9 /) dx , J m= 1 and the integral equation (5) follows immediately. The change of order of integration and summation is legitimate in view of the uniform convergence of the series involved. Another interesting result for the resolvent kernel is that it satisfies the integrodifTerential equation dr(s9t;X)/dl = jr(s9x;X)V(x9t;X)dx . (6) In fact, r o o o r 1 1 1 r(s,x;X)T(x,t;X)dx = £ A "" Km(s9x) £ A"" Ktt(x9t) dx . * J m=\ n=\ On account of the absolute and uniform convergence of the series (1), we can multiply the series under the integral sign and integrate it term by term. Therefore, EXERCISES 39 oo oo m+ 2 T(s,x;X)r(x,t;X)dx= ^ £ X "~ Km+n(s,t) . (7) J =\n=l m Now, set m + n=p and change the order of summation; there results the relation m+ 2 x x x '- Km+m(s,t) = £ 'iVx&o ra = 1 AJ = 1 p = 2 n= 1 ) = £ (^-l)A-^(,)0 = ^^i . (8) p = 2 VA Combining (7) and (8), we have the result (6). EXERCISES 1. Solve the following Fredholm integral equations by the method of successive approximations: (i) g{s) = e>-\e + \ + \\g{t)dt. 0 TT/2 (ii) g (s) = (sin s) - is + ± J" j# (f) A . o 2. Consider the integral equation I g(s) = 1 +ljstg(t)dt. o (a) Make use of the relation |A| < B~x to show that the iterative procedure is valid for \X\ < 3. (b) Show that the iterative procedure leads formally to the solution g(s)=l+ sl(m + (A2/6) + (^3/18) + ...]. (c) Use the method of the previous chapter to obtain the exact solution g(s) = 1 +[3AJ/2(3-A)] , X ± 3 . 3. Solve the integral equation I g(s) = 1 +lj(s+t)g(t)dt9 40 3/METHOD OF SUCCESSIVE APPROXIMATIONS by the method of successive approximations and show that the estimate afforded by the relation \X\ < B~x is conservative in this case. 4. Find the resolvent kernel associated with the following kernels: (i) \s—t\, in the interval (0,1); (ii) exp-\s-t\, in the interval (0,1); (iii) cos(.s + /), in the interval (0,2n). 5. Solve the following Volterra integral equations by the method of this chapter: s (i) g(s)=\+f(s-t)g(t)dt, 0 s (ii) g(s) = 29 + 6* + J* (6s-6t + 5)g(t) dt. o 6. Find an approximate solution of the integral equation s f s g(s) = (sinhs) + j e - g(t)dt9 o by the method of iteration. 7. Obtain the radius of convergence of the Neumann series when the functions/(s) and the kernel K(s, i) are continuous in the interval (a, b). 8. Prove that the resolvent kernel for a Volterra integral equation of the second kind is an entire function of I for any given (s, t). CLASSICAL FREDHOLM CHAPTER 4 THEORY 4.1. THE METHOD OF SOLUTION OF FREDHOLM In the previous chapter, we have derived the solution of the Fredholm integral equation g(s)=f(s) + kJK(s9t)g(t)dt (1) as a uniformly convergent power series in the parameter A for |A| suitably small. Fredholm gave the solution of equation (1) in general form for all values of the parameter L His results are contained in three theorems which bear his name. We have already studied them in Chapter 2 for the special case when the kernel is separable. In this chapter, we shall study equation (1) when the function/(J) and the kernel K(s, t) are any integrable functions. Furthermore, the present method enables us to get explicit formulas for the solution in terms of certain determinants. The method used by Fredholm consists in viewing the integral equation (1) as the limiting case of a system of linear algebraic equations. This theory applies to two- or higher-dimensional integrals, although we shall confine our discussion to only one-dimensional integrals in the interval (a, b). Let us divide the interval {a, b) into n equal parts, s1 = tl=a, s2 = t2 = a + h, ... , sn = tn = a + (n-\)h , 41 42 4 / CLASSICAL FREDHOLM THEORY where h = (b — d)\n. Thereby, we have the approximate formula f K(s, t) g(t)dt~h£ K(s9 sj) g (sj) . (2) J 7=1 Equation (1) then takes the form g{s)~f(s) + Xh£ K(s9sj)g(sj)9 (3) which must hold for all values of s in the interval (a,b). In particular, this equation is satisfied at the n points of division sh i= 1,...,«. This leads to the system of equations 0fo) = f(sd + to Z K(si9Sj)g(Sj) , / = 1, ...,* . (4) 7=1 Writing /fe) = ft , 9(st) = gt , K(shsj) = Kij , (5) equation (4) yields an approximation for the integral equation (1) in terms of the system of n linear equations gt- Xh^K^gj = ft , i = !,...,/? , (6) in n unknown quantities gl9...,gn. The values of gt obtained by solving this algebraic system are approximate solutions of the integral equation (1) at the points sl9s29 ...,sn. We can plot these solutions gv as ordinates and by interpolation draw a curve g(s) which we may expect to be an approximation to the actual solution. With the help of this algebraic system, we can also determine approximations for the eigenvalues of the kernel. The resolvent determinant of the algebraic system (6) is \-XhKxl -XhKl2 -MKln -XhK2X \-XhK22 -XhK2n DH(X) = (7) -XhKnl -XhK„ \-XhKni The approximate eigenvalues are obtained by setting this determinant equal to zero. We illustrate it by the following example. 4.2. FREDHOLM'S FIRST THEOREM 43 Example. 71 g(s)- X fsin(j + 0^(0* = ° • By taking n = 3, we have h = 7i/3 and therefore sx = tx = 0 , t2 = 7T/3 , S3 = t3 = 27T/3 , and the values of Ku are readily calculated to give 0 0.866 0.866 (*y) = 0.866 0.866 0 0.866 0 -0.866 The homogeneous system corresponding to (6) will have a non- trivial solution if the determinant 1 -0.907^ -0.907/1 DH(X) = -0.907/1 (1-0.907 A) 0 = 0, -0.907>* 0 (1+0.907/1) or when 1-3(0.0907)2A2 = 0. The roots of this equation are X = ± 0.6365. This gives a rather close agreement with the exact values (see Example 3, Section 3.2), which are ±yj2/n = ± 0.6366. In general, the practical applications of this method are limited because one has to take a rather large n to get a reasonable approxi mation. 4.2. FREDHOLM'S FIRST THEOREM tne The solutions gi,g29---,9n °f system of equations (4.1.6) are obtained as ratios of certain determinants, with the determinant Dn(X) given by (4.1.7) as the denominator provided it does not vanish. Let us expand the determinant (4.1.7) in powers of the quantity ( — Mi). The constant term is obviously equal to unity. The term containing ( — Mi) in the first power is the sum of all the determinants containing only one 44 4 / CLASSICAL FREDHOLM THEORY column -XhK^, fi=\,...,n. Taking the contribution from all the columns v = 1,...,«, we find that the total contribution is -Xh £v=i Kvv. The factor containing the factor (- Xh) to the second power is the sum of all the determinants containing two columns with that factor. This results in the determinants of the form KPP Kpq {-W K K where (p,q) is an arbitrary pair of integers taken from the sequence 1,...,«, with/? < q. In the same way, it follows that the term containing the factor ( — Xh)3 is the sum of the determinants of the form K K K PP PQ Pl (-A/,)3 K K K qp qq qr Krn Kra Krr where (/?, q, r) is an arbitrary triplet of integers selected from the sequence 1,...,n9 withp 2 (-Xh){ Dn(X)= \-Xh^Kvv + -^f-^ Kan K p,q=i m K K K (-M)3 PP pq PI K K K +- 3! qp qq qf + Krn Kra Krr KP (-Xh)" K„ K„ K„ + ■ (1) n\ Pl,P2,-;Pn=l K„ K„ Kn where we now stipulate that the sums are taken over all permutations of pairs (p,q), triplets (p,q,r), etc. This convention explains the reason for dividing each term of the above series by the corresponding number of permutations. 4.2. FREDHOLM'S FIRST THEOREM 45 The analysis is simplified by introducing the following symbol for the determinant formed by the values of the kernel at all points (sh tj) *(*„/,) K(sltt2) ••• K(sutn) K(s2,tt) K{s2,t2) - K(s2,tn) = K (2) tl9 t2, -.-,tn K(s„, /,) K(s„ t2) K(s„, t„) the so-called Fredholm determinant. We observe that, if any pair of arguments in the upper or lower sequence is transposed, the value of the determinant changes sign because the transposition of two arguments in the upper sequence corresponds to the transposition of two rows of the determinant and the transposition of two arguments in the lower sequence corresponds to the transposition of two columns. In this notation, the series (1) takes the form {-Xhf SP9SC Dn(X) = \-Xh2^K{sp,sp) + 2! 2* P=\ (-Ah)3 + (3) + 3! p£tx Wsq9sJ If we now let n tend to infinity, then h will tend to zero, and each term of the sum (3) tends to some single, double, triple integral, etc. There results Fredholm's first series: 2 X SuSl D(X) = 1 - X ,s)ds + — K[ \dsxds2 + (4) Hilbert gave a rigorous proof of the fact that the sequence Z)„(X)-» D(X) in the limit, while the convergence of the series (4) for all values of X was proved by Fredholm on the basis that the kernel K(s, t) is a bounded and integrable function.1 Thus, D(X) is an entire function of the complex variable X. We are now ready to solve the Fredholm equation (4.1.1) and express 1 For proof, see Lovitt [10]. 46 4 / CLASSICAL FREDHOLM THEORY the solutions in the form of a quotient of two power series in the parameter X9 where the Fredholm function D(X) is to be the divisor. In this connection, recall the relations (2.3.6) and (2.3.7). Indeed, we seek solutions of the form g(s)=f(s) + xjr(s9t'9X)f(t)dt9 (5) and expect the resolvent kernel T(s, t;X) to be the quotient T(s9t;X) = D(s9t;X)/D(X) , (6) where D(s9 f9X)9 still to be determined, is the sum of certain functional series. Now, we have proved in Section 3.5 that the resolvent T(s9 t;X) itself satisfies a Fredholm integral equation of the second kind (3.5.5): T(s9t;X) = K(s9 t) + X J K(s9x)T(x9 t\X)dx . (7) From (6) and (7), it follows that D(s9 t; X) = K(s9 i) D(X) + X J K(s, x) D(x91; X) dx . (8) The form of the series (4) for D(X) suggests that we seek the solution of equation (8) in the form of a power series in the parameter X: ( C 9 D(s,t;X) = C0(s,t) + J —f- A0 • ( ) For this purpose, write the numerical series (4) as D{X) = i + 2(—rc»' (10> P=\ ^' where S Sp K( ^- \dSX.:dSp. (ll) \sl9s2,...,spJ The next step is to substitute the series for D(s, t;X) and D(X) from (9) and (10) in (8) and compare the coefficients of equal powers of X. The following relations result: C0(s9t) = K(s9t), (12) Cp(s9t) = cpK{s9t)-p j K(s9x)Cp.i(x9t)dx. (13) Our contention is that we can write the function Cp(s91) in terms of the Fredholm determinant (2) in the following way: 4.2. FREDHOLM'S FIRST THEOREM 47 ^"-J-K"::^::::^*'•■•*•■ (14) In fact, for p = 1, the relation (13) becomes C\ (s9 i) = cxK{s i)- \ K{s9x) C0(x91) dx = K(s9 i) f K(x9 x) dx - f #($, JC) K(X9 i) dx = K''K ) dx , (15) where we have used (11) and (12). To prove that (14) holds for general p9 we expand the determinant under the integral sign in the relation: K(s9t) K(s,Xl) K(s,xp) s x ...,x K{xut) K{xuxx) K(xuxp) K 9 i9 p * 9 X i, ..., X. K{xp9i) K(xp9xx) K(xp9xp) with respect to the elements of the given row, transposing in turn the first column one place to the right, integrating both sides, and using the definition of cp as in (11); the required result then follows by induction. From (9), (12), and (14) we derive Fredholm's second series: (-A)" S, X j, . . ., Aj D{s,t;k) = tf(M) + 2 K dxl ••• dxp . (16) t9Xi, ...9xt This series also converges for all values of the parameter L It is interesting to observe the similarity between the series (4) and (16). Having found both terms of the quotient (6), we have established the existence of a solution to the integral equation (4.1.1) for a bounded and integrable kernel K(s, t)9 provided, of course, that D(X) # 0. Since both terms of this quotient are entire functions of the parameter X9 it follows that the resolvent kernel Y(s9 t\X) is a meromorphic function of A, i.e., an analytic function whose singularities may only be the poles, which in the present case are zeros of the divisor D(X). Next, we prove that the solution in the form obtained by Fredholm is unique and is given by 0(s) = As) + I j T(s9 t;X)f(t) dt . (17) 48 4 / CLASSICAL FREDHOLM THEORY In this connection, we first observe that the integral equation (7) satisfied by r(s, t\X) is valid for all values of X for which D(X)^0. Indeed, (7) is known to hold for \X\ < B~l from the analysis of Chapter 3, and since both sides of this equation are now proved to be meromorphic, the above contention follows. To prove the uniqueness of the solution, let us suppose that g(s) is a solution of the equation (4.1.1) in the case D(X) ^ 0. Multiply both sides of (4.1.1) by Y{s9 t\X)9 integrate, and get T(s9x; X)g(x) dx = T(s9x'9X)f(x) dx + X j [J T(s9x'9X)K(x9t) dx] g{i) dt. (18) Substituting from (7) into left side of (18), this becomes | K(s,t)g(t)dt = JT(s9x;X)f(x)dx9 (19) which, when joined by (4.1.1), yields 0(s) = f(s) + XJ T(s9 t\X)f(t) dt, (20) but this form is unique. In particular, the solution of the homogeneous equation g(s) = XJK(s9t)g(t)dt (21) is identically zero. The above analysis leads to the following theorem. Fredholm's First Theorem. The inhomogeneous Fredholm equation g(s)=f(s) + XJK(s9t)g(t)dt, (22) where the functions f(s) and g (/) are integrable, has a unique solution g(s)=f(s) + XJr(s9t;X)f(t)dt9 (23) where the resolvent kernel T(s9 t\X)9 T(S9t;X) = D(s9t;X)/D(X), (24) with D(X) # 0, is a meromorphic function of the complex variable l9 being the ratio of two entire functions defined by the series u X D(s9t; X) = K(s,t) + K[';* '"' ')dxl.-.dx,, (25) r l Aj, ..., X„ pD= 1 •> v 9 4.3. EXAMPLES 49 and {-xy X1 j, ... , X.p D(X) =1+2 K[" "'"' \dxx--dxpy (26) P=I Kxu ...,xpJ both of which converge for all values of X. In particular, the solution of the homogeneous equation g(s) = XJK(s9t)g(t)dt (27) is identically zero. 4.3. EXAMPLES Example 1. Evaluate the resolvent for the integral equation I g(s)=f(s) + lj(s + t)g(t)dt. (1) The solution to this example is obtained by writing (2) where Cp and cp are defined by the relations (4.2.11) and (4.2.13): c0 = 1 , C0(s,t) = K(s,t) = (s+t). (3) cp = j Cp-l(s,s)ds , (4) Cp = cp K(s91) - p J K(s, x)cp.l (x, t) dx. (5) o Thus, cx = \ 2s ds = 1 , o QCM) = (* + /)-J"(.y + x)(x + /) 2 C2 = J(j-J -i)&=-i, 50 4/CLASSICAL FREDHOLM THEORY 1 C2(s,t) = -Us+t)-2J(s + x)[±(x + t)-xt-$y]dx = 0 Since C2(x,i) vanishes, it follows from (5) that the subsequent co efficients Ck and ck also vanish. Therefore, r (J+0-R(J+0-J'-£U 6 CM;4) = l; - x, - TTTJT^(i2\\2) > ( ) which agrees with result (2.2.8) found by a different method. Example 2. Solve the integral equation I g (s) = s + X J* [rf+(«)*] 3 (0 A • (7) 0 In this case, c0 = 1, CO(J,0 = st + (st)*, 2 cx = j(s + s)ds = |, o 0 1 2 % c2 = J(ij + iJ-fj )&= 1/75, o C2(J,O = O, and therefore all the subsequent coefficients vanish. The value of the resolvent is r(M;*) = 1-IA + (1/150)A» • (8) The solution g(s) then follows by using the relation (4.2.23), , , 150i + A(60VJ-75^) + 2U25 'W ^-125A + 150 • (9) Example 3. As a final example, let us solve the integral equation 4.4. FREDHOLM'S SECOND THEOREM 51 g(s) = 1 +A j[sin(j+O]0(/)A, (10) o which we have already solved in Section 3.2 by a Neumann series. For the present kernel, c0 = 1 , C00,0 = sin(•? + /) , ct = ) s'mls ds = 0 . o n Cl(sit) = 0— sin(5 + x)sin(x + /) dx = — %n cos (s — t) , o 2 c2 = — \ ^ncosOds = —\n , o n , C2{s,t) = — i^sinCs+O + 2 ^7isin(5 + x)cos(x — /) rfx = 0. o Hence, the resolvent is , , sin(j + 0 + ±7rAcosCy-0 r(M;1r n) = "^W ' and the solution is 2 '(J) = 1 +2A(COSJ—r^w—) +A 7r(sinj) ' (11) which agrees with the solution (3.2.10). 4.4. FREDHOLM'S SECOND THEOREM Fredholm's first theorem does not hold when X is a root of the equation D{X) = 0. We have found in Chapter 2 that, for a separable kernel, the homogeneous equation g(s) = lJK(s9t)g(t)dt (1) has nontrivial solutions. It might be expected that same holds when the kernel is an arbitrary integrable function and we shall then have a 52 4/CLASSICAL FREDHOLM THEORY spectrum of eigenvalues and corresponding eigenfunctions. The second theorem of Fredholm is devoted to the study of this problem. We first prove that every zero of D(X) is a pole of the resolvent kernel (4.2.24); the order of this pole is at most equal to the order of the zero of D(X). In fact, differentiate the Fredholm's first series (4.2.26) and interchange the indices of the variables of integration to get D'(X) = - j D(s,s;X)ds. (2) From this relation, it follows that, if X0 is a zero of order k of D(X), then it is a zero of order k— 1 of D'(X) and consequently X0 may be a zero of order at most k— 1 of the entire function D(s, t;X). Thus, X0 is the pole of the quotient (4.2.24) of order at most k. In particular, if X0 is a simple zero of D(X), then D(Xo) = 0, D'(Xo)^0, and X0 is a simple pole of the resolvent kernel. Moreover, it follows from (2) that D(s, t;X)^0. For this particular case, we observe from equation (4.1.8) that, if D(X) = 0 and D{s, t\ X) * 0, then D(s, t;X), as a function of s, is a solution of the homogeneous equation (1). So is ocD(s9t;X), where a is an arbitrary constant. Let us now consider the general case when X is a zero of an arbitrary multiplicity m, that is, when ( (w D(X0) = 0 , ... , Z) '>(X0) = 0 , Z) > (X0) # 0 , (3) where the superscript r stands for the differential of order r, r = 1, ..., m—\. For this case, the analysis is simplified if one defines a determinant known as the Fredholm minor: tu ^2> •••> tn \ti,t2,...,tj £, where {st} and {/,}, /= 1,2,...,«, are two sequences of arbitrary variables. Just as do the Fredholm series (4.2.25) and (4.2.26), the series (4) also converges for all values of X and consequently is an entire function of X. Furthermore, by differentiating the series (4.2.26) n times and comparing it with the series (4), there follows the relation 4.4. FREDHOLM'S SECOND THEOREM 53 n d D(X) ...,sn n (-1)" D, A]dsl--dsn . (5) dX V^ 1' * * *» Sn From this relation, we conclude that, if X0 is a zero of multiplicity m of the function D(X), then the following holds for the Fredholm minor of order m for that value of XQ: D„ ^o h^ 0 . tu t2, ..-, tm Of course, there might exist minors of order lower than m which also do not identically vanish (compare the discussion in Section 2.3). Let us find the relation among the minors that corresponds to the resolvent formula (4.2.7). Expansion of the determinant under the integral sign in (4), Kisutj K(sut2) K(sl9tH) K(sux{) K(suxp) Kis^t,) K(sl9t2) K(sl9tn) K(s2,Xl) K(s29xp) (6) K^tJ K(sn9t2) K{sn9tn) K{sn9xx) K(sn9 xP) Kix^t,) K(xut2) K(xl9tn) K(xuxx) K{xp9tx) K(xp9t2) K(xp9tn) K(xp9x,) K{xp9xp) by elements of the first row and integrating p times with respect to xl9x29 ...,xp for p ^ 1, we have Su S, Xl Xp i..!K( ' " '' )dx1-dxp J J \ti,...,tn,Xu...,XpJ n S2 S J x f- \K( ' ■■■' -' -' »>*i,->*P\ dXldx2...dXp P ■f2("1)"+n~1 h=\ w I \ vt x J^2v) Sn,Xl9 X29 ..-, Xh9 ..., Xj J J \t\9 •••» ^#1— 1 » ^«>-^l> •••»:V/l-l»^/l+ 1> ""»Xl x ^/xx '"dxp (7) 54 4/CLASSICAL FREDHOLM THEORY Note that the symbols for the determinant K on the right side of (7) do not contain the variables s{ in the upper sequence and the variables th or xh in the lower sequence. Furthermore, it follows by transposing the variable sh in the upper sequence to the first place by means of h + n — 2 transpositions that all the components of the second sum on the right side are equal. Therefore, we can write (7) as 5"i, ..., Sn, Xj, ..., Xp ■K* 1 9 " '9 */p -M 9 • • •» Xp +i = x (-\y K(Suth) h=\ J J Vl> ••■>'/i-l>'/i + l> •••> '»>*!> ■■■iXp/ X> ^2» • • '9 $n9 X 1, ..., ^p— 1 K(sltx) tl9t2, ..., tn,Xi, ...9Xp_ j c x rf^ •- dxp_nA | S19 -"9^n 2 A, ^=t(-ir>A:(Jl,/4)fl._Y* '- 'M X X2 + X!K(S1,X)D( ' ' ■''"] A hfa. (9) Expansion by the elements of any other row leads to a similar identity, with x placed at the corresponding place. If we expand the determinant (6) with respect to the first column and proceed as above, we get the integral equation s 91 • • •9 n s h+l Sl9 ...9Sh_ Sl9h+ ,x ..., Sn A A = ^(-\) K(shJl)Dn_lh' + . (10) J \x,t29...,tnJ and a similar result would follow if we were to expand by any other 4.4. FREDHOLM'S SECOND THEOREM 55 column. The formulas (9) and (10) will play the role of the Fredholm series of the previous section. Note that the relations (9) and (10) hold for all values of X. With the help of (9), we can find the solution of the homogeneous equation (1) for the special case when X =k0 is an eigenvalue. To this end, let us suppose that A = X0 is a zero of multiplicity m of the function D(X). Then, as remarked earlier, the minor Dm does not identically vanish and even the minors D{, Z)2,..., Dm_, may not identically vanish. Let Dr be the first minor in the sequence Du D2,..., Dm_x that does not vanish identically. The number r lies between 1 and m and is the index of the eigenvalue A0 as defined in Section 2.3. Moreover, this means that Dr_ i = 0. But then the integral equation (9) implies that s,s , ...,s 9i(s) = D, 2 r (11) >!,. u is a solution of the homogeneous equation (1). Substituting s at different points of the upper sequence in the minor Dr, we obtain r nontrivial solutions gf/Cs), /=l,...,r, of the homogeneous equation. These solutions are usually written as $ 1» • • •» ^i - 1 ? $i $i + 1 > • • •» ^r A h tu... .', *i(s) = , /= 1,2 r. (12) s S S S S D. 1 j -"9 i-i9 i9 i+l9 '"9 r K ti9... >*, Observe that we have already established that the denominator is not zero. The solutions Of as given by (12) are linearly independent for the following reason. In the determinant (6) above, if we put two of the arguments si equal, this amounts to putting two rows equal, and consequently the determinant vanishes. Thus, in (12), we see that S9 iJj, ..., Sr oj, ..., Sr H(s,t;X) = D D. (13) r+i x, X0 t,tu...,t, tu...,tr In (10), take n to be equal to r, and add extra arguments s and / to obtain S, Si, ..., Sr Dr+] X0) = K{s,t)Dr t, t19 ..., tr tu...,tr s,slyS Sl...,sh_i,sh+l, ...,sr + y(-lfK(sh,y)Dr( ; '-' AQ S Si ,Sr + l0fK(x,t)Dr+1( ' '- A0 )dx . (14) J \x,tu...,t, In every minor Dr in the above equation, we transpose the variable s from the first place to the place between the variables sh_x and sh+l and divide both sides by the constant sl9 ...,sr Dr 0 * o. ti,...,tr to obtain //(■M;A)- K(s,t)- X0 H(s,x;X)K(x,t)dx = -£ fffo.0 **(*)• (15) If g(j) is any solution to (1), we multiply (15) by g(t) and integrate with respect to t, g(t) H(s,t;X) dt - ^ - [g(x)T{s,x;X) dx r -I **(*), (16) where we have used (1) in all terms but the first; we have also taken A0 j* K(sh, t)g{t)dt = g(sh). Cancelling the equal terms, we have 4.5. FREDHOLM'S THIRD THEOREM 57 g(s)= X g(sh)^>h(s). (17) /»=i This proves our assertion. Thus we have established the following result. Fredholm's Second Theorem. If A0 is a zero of multiplicity m of the function D(X), then the homogeneous equation g(s) = X0JK(s,t)g(t)dt (18) possesses at least one, and at most m, linearly independent solutions i = l,...,r ; 1 < r < m (19) not identically zero. Any other solution of this equation is a linear combination of these solutions. 4.5. FREDHOLM'S THIRD THEOREM In the analysis of Fredholm's first theorem, it has been shown that the inhomogeneous equation g(s)=f(s) + lJK(s,t)g(t)dt (1) possesses a unique solution provided Z)(A)/0. Fredholm's second theorem is concerned with the study of the homogeneous equation g(s) = lfK(s,t)g(t)dt9 when D(X) = 0. In this section, we investigate the possibility of (1) having a solution when Z>(A) = 0. The analysis of this section is not much different from the corresponding analysis for separable kernels as given in Section 2.3. In fact, the only difference is that we shall now give an explicit formula for the solution. Qualitatively, the discussion is the same. Recall that the transpose (or adjoint) of equation (1) is (under the same assumption as in Section 2.3) 58 4 / CLASSICAL FREDHOLM THEORY ^(s)=f(s) + AJK(t,s) iKs) = f(s) + X j \_D(Us\X)ID(X)-\f(t) dt, (4) provided X is not an eigenvalue. It is also clear that not only has the transposed kernel the same eigen values as the original kernel, but also the index r of each of the eigenvalues is equal. Moreover, corresponding to equation (4.4.12), the eigen- functions of the transposed equation for an eigenvalue for X0 are given as *o) <::. • > h - 1 > '» U + 1 > • *,(/)= y..-.-.-»-.-.^.-.-r / ^ (5) *o) f(s) + k0 j H(s, t\k)f{t) dt = As) + ^o J K{s91) xU(t) + ko\H(t9x\k)f{x)dx\dt or jf(t) dt [H(s9 t\k)- K(s9 t)-k0 j K(s, x) H(x, t;k)dx]=0. (8) Now, just as we obtained equation (4.4.15), we can obtain its "transpose," H(s, t;k)- K(s, t)-k0j K(s, x) H(x, t;k)dx = -£K(s,th)Vh(t). (9) h=l Substituting this in (8) and using the orthogonality condition, we have an identity, and thereby the assertion is proved. The difference of any two solutions of (1) is a solution of the homogeneous equation. Hence, the most general solution of (1) is g(s) = f(s) + X0 f H(s, t;X)f{t) dt + £ Ch®„(s) . (10) J h=\ The above analysis leads to the following theorem. Fredholm's Third Theorem. For an inhomogeneous equation g(s)=f{s) + k0\K{s9t)g(t)dt9 (11) to possess a solution in the case D(k0) = 0, it is necessary and sufficient that the given function /(s) be orthogonal to all the eigenfunctions 60 4 / CLASSICAL FREDHOLM THEORY 4>.(V)} /= 1,2, ...,v, of the transposed homogeneous equation corre sponding to the eigenvalue X0. The general solution has the form .w-/»+^M;::r:;MK:::r-:: Ac (12) £X£/?C/S£S 1. Use the method of this chapter and find the resolvent kernels for the following three integral equations: I (i) g(s)=f(s) + lj\s-t\g(t)dt, 0 1 (ii) g (s) = f(s) + X j (exp -\s-t\)g(t) dt , o In (iii) g (s) = f(s) + X j [cos(j + /)] g (t) dt . o 2. Solve the following homogeneous equations n (0 0(j) = ±J[sin(j+f)]0(OA. o I (ii) g(s) = V/(e2-\)-]J2e°e The theories of ordinary and partial differential equations are fruitful sources of integral equations. In the quest for the representation formula for the solution of a linear differential equation in such a manner so as to include the boundary condition or initial condition explicitly, one is always led to an integral equation. Once a boundary value or an initial value problem has been formulated in terms of an integral equation, it becomes possible to solve this problem easily. In this chapter, we shall consider only ordinary differential equations. The next chapter is devoted to partial differential equations. 5.1. INITIAL VALUE PROBLEMS There is a fundamental relationship between Volterra integral equations and ordinary differential equations with prescribed initial values. We begin our discussion by studying the simple initial value problem f + A(S)y' + B(s)y = F(s)9 (1) 61 62 5 / ORDINARY DIFFERENTIAL EQUATIONS y(a) = % > y'ia) = <7i > (2) where a prime implies differentiation with respect to s, and the functions A, B, and Fare defined and continuous in the closed interval a < s < /?. The result of integrating the differential equation (1) from a to ^ and using the initial values (2) is y'(s)-ql = -A(s)y(s)-j\.B{s1)-A'(sl)]y(sl)dsl a s + JF(sl)dsl + A(a)q0 . a Similarly, a second integration yields s s s2 y{s)-q0 = - j A(sl)y(sl)dsl - jj [B(sx)-A'{sxy]y(sx) dsxds2 a a a s s2 + jj F(Sl) dsx ds2 + [_A(a)q0 + q{\{s-a) . (3) a a With the help of the identity (see Appendix, Section A.l) s s2 s jj F(Sl) dst ds2 = j(s-t) F(t) dt , (4) a a a the two double integrals in (3) can be converted to single integrals. Hence, the relation (3) takes the form y(s) = q0 + \.A(a)q0 + q^(s-a) + j (s-t)F(t) dt a s - j {A(t) + (*-/)[/?(/) - A'(t)-]}y(t)dt. (5) a Now, set K(s,t) = - {A{t) + (s-t)lB(t) - A1 (ty\} (6) and s f{s) = j (s-t)F{t) dt + IA(a)q0 + q^(s-a) + q0 . (7) 5.1. INITIAL VALUE PROBLEMS 63 From relations (5)-(7), we have the Volterra integral equation of the second kind: s y(s)=f(s) + JK(s,t)y(t)dt. (8) a Conversely, any solution g(s) of the integral equation (8) is, as can be verified by two differentiations, a solution of the initial value problem d)-(2). Note that the crucial step is the use of the identity (4). Since we have proved the corresponding identity for an arbitrary integer n in the Appendix, Section A.l, it follows that the above process of converting an initial value problem to a Volterra integral equation is applicable to a linear ordinary differential equation of order n when there are n pre scribed initial conditions. An alternative approach is somewhat simpler for proving the above-mentioned equivalence for a general differential equation. Indeed, let us consider the linear differential equation of order n: dny dn~xy dy ~W + Al(s)~d^ + "■ + An-l(s)Js + An(s)y = F(S) ' (9) with the initial conditions y(a)=q0, /(«)=?„ ..., /'-"(a) =?„_,, (10) an where the functions AuA2,-..,An d F are defined and continuous in a < s ^ b. The reduction of the initial value problem (9)-(10) to the Volterra integral equation is accomplished by introducing an unknown function d"y/ds" = g(s). (11) From (10) and (11), it follows that d" dsr-1 r (12) j»-2 p (continued) -^ry = Us-i)g{i)dt + {s-a)qn-l +q„-2 , 64 5/ORDINARY DIFFERENTIAL EQUATIONS dy [(s-ty-2 (s-a)"-2 (s-a)"-3 a + ■■■ +(s-a)q2+ql , (12) a + ••• +(s-a)q1 + q0 . Now, if we multiply relations (11) and (12) by 1, Ax(s), A2(s), etc. and add, we find that the initial value problem defined by (9)—(10) is reduced to the Volterra integral equation of the second kind s g(s)=f(s) + JK(s,t)g(t)dt, (13) a where ^ (s-t)k~l (j (14) jf(J>o = -2^ %=i)r and f(s) = F(s) - qn-iAi(s)-l(s-a)qn-l+qH-2]A2(s) {[(j»fl)""V(i-l)!]^-i + -+(s-a)q1+q0} *An(s). (15) Conversely, if we solve the integral equation (13) and substitute the value obtained for g(s) in the last equation of the system (12), we derive the (unique) solution of the initial value problem (9)—(10). 5.2. BOUNDARY VALUE PROBLEMS Just as initial value problems in ordinary differential equations lead to Volterra-type integral equations, boundary value problems in 5.2. BOUNDARY VALUE PROBLEMS 65 ordinary differential equations lead to Fredholm-type integral equations. Let us illustrate this equivalence by the problem y(s) + A(s)/ + B(s)y = F(s)9 (1) y(a) = yo > y{b) = yt . (2) When we integrate equation (1) from a to s and use the boundary condition y(a) = y0, we get s y'(s) = C + j F(s) ds - A(s)y(s) + A(a)y0 a s + f[A'(s)-B(s)]y(s)ds, a where C is a constant of integration. A second integration similarly yields s s2 y{s) - y0 = [C + A(a)y0-](s-a) + jj F(st) s s s2 - j A(Si)y(Sl) dsY + jj lA'(Si) - Bis^yis,) ds, ds2. (3) a a a Using the identity (5.1.4), the relation (3) becomes s y(s)-y0 = [C + A(a)yo](s-a) + j (s-t)F(t)dt a s - j {A(t) - (s-t)lA'(t) - B(tmy(t) dt . (4) a The constant C can be evaluated by setting s = b in (4) and using the second boundary condition y(b) =yx: yi - yo = LC + A(a)y0-](b-a) + j (b-i)F(t)dt - j {A(t) - (b-t)[_A'(t) - B(ty]}y{t)dt, or 66 5/ORDINARY DIFFERENTIAL EQUATIONS C + A(a)y0 = l\l{b-ay]{(yl-y0)-\{b-t)F{t)dt + J* {A{t) - (b-tKA'(t) - B(ty]}y(t) dt) . (5) From (4) and (5), we have the relation s y{s) = yo + j (s-t)F(t) dt + l(s-a)l(b-a)l a xl{yx-yo)-\{b-t)F{t)df] s - \ {A(t) - (s-t)lA'(t) - B(ty\}y(t) dt a + j l(s-a)/(b-a)-] {A{i) - (b-t)tA'(t) - B(t)^}y(t) dt . (6) Equation (6) can be written as the Fredholm integral equation y(s)=f(s) + JK(s9t)y(t)dt, (7) provided we set s f(s)=y0 + f(s-t)F(t)dt a + l(s - a)/(b - a)] l(y, -y0)-j(b-1) F(t) A] (8) and ( [_(s-a)l(b-a)-] {A(t) - (b-t)[A'(t) - 5(0]} , s < t, K{s,t) = (9) A(t){L(s-a)Kb-ay] - 1} -lA'(t) - 5(0] x 10 - a) (b-s)l(b-a)] , s > t . For the special case when A and B are constants, a = 0, b = 1, and y(0) = y(\) = 0, the above kernel simplifies to ( Bs(\-t) + As , s < t, K(s,t) = x (10) [Bt(l-s) + As- A , s > t . Note that the kernel (10) is asymmetric and discontinuous at / = s, unless A = 0. We shall elaborate on this point in Section 5.4. 5.3. EXAMPLES 67 5.3. EXAMPLES Example 1. Reduce the initial value problem f(s) + i.y(s) = F(s)9 (l) j(0)=l, /(0) = 0, (2) to a Volterra integral equation. Comparing (l) and (2) with the notation of Section 5.1, we have A{s) = 0, B{s) = L Therefore, the relations (5.l.6)-(5.1.8) become K(s,t) = A(t-s), s f(s)=l+j(s-t)F(t)dt, (3) 0 and s s y(s) =1+J (s-t)F(t) dt + XJ (t-s)y(t) dt . 0 0 Example 2. Reduce the boundary value problem ?(s) + XP(s)y= Q(s), (4) y{d) = 0 , y(b) = 0 (5) to a Fredholm integral equation. Comparing (4) and (5) with the notation of Section 5.2, we have ,4=0, B = XP(s), F(s)=Q(s), y0 = 09 y^=0. Substitution of these values in the relations (5.2.8) and (5.2.9) yields s f(s) = \(s-t)Q (t) dt - Us - a)/(b - a)] \{b-i)Q (0 dt (6) a and (XP(t)[_(s -a)(b- t)l(b-a)-], s y(0) = 0 , ytf) = 0 . (9) Then, the relations (6) and (7) take the simple forms :f(s) = 0, and f {Xsjt)(£-i) , s < t, K(s,t)= (10) [(Xtlf)(f-s)9 s> t. Note that, although the kernels (7) and (10) are continuous at s = t, their derivatives are not continuous. For example, the derivative of the kernel (10) is (A[l-(///)], s The value of the jump of this derivative at s = Ms ~dK(s, t) 'dK(s9t)l ds \t + 0 ds Jr-o Similarly, the value of the jump of the derivative of the kernel (7) at s= t is L * l+o L * l-o Example 3. Transverse oscillations of a homogeneous elastic bar. Consider a homogeneous elastic bar with linear mass density d. Its axis coincides with the segment (0,0 of the s axis when the bar is in its state of rest. It is clamped at the end s = 0, free at the end s = /, and is forced to perform simple harmonic oscillations with period Injco. The problem, illustrated in Figure 5.1, is to find the deflection y(s) that is parallel to the y axis and satisfies the system of equations 5.3. EXAMPLES 69 Figure 5.1 2 A o d £-**'-■ k = IF GO y(0) = /(O) = 0 , (12) (13) where El is the bending rigidity of the bar. The differential equation (11) with the initial conditions (12) can be reduced to the solution of a Volterra integral equation if we stipulate that /(O) = C2 , /"(O) = C3 , (14) and subsequently determine the constants C2 and C3 with the help of (13). Indeed, when we compare the initial value problem embodied in (11), (12), and (14) with the system (5.1.9)-(5.1.15), we obtain the required integral equation S S g(s) = k*i -C2 + j;C: + k-■l"#■9(t)dt (\5) where g(s) = d*y/ds4 (16) 70 5/ORDINARY DIFFERENTIAL EQUATIONS The solution y(s) of the differential equation (11) is C(s-t)3 s3 s1 K }\s) = -^- g(t)dt + - C3 + - C2 . (17) We leave it to the reader to apply the conditions (13), determine the constants C2 and C3, and thereby complete the transformation of the system (11)-(13) into an integral equation. The kernel of the integral equation (15) is that of convolution type and this equation can be readily solved by Laplace transform methods as explained in Chapter 9 (see Exercise 11 at the end of that chapter). 5.4. DIRAC DELTA FUNCTION In physical problems, one often encounters idealized concepts such as a unit force acting at a point s = s0 only, or an impulsive force acting at time t = t0 only. These forces are described by the Dirac delta function 3(s — s0) or S(t —10) such that ( 0 , x ^ x0 , S(x-x0) = (1) [ 00 , X = X0 , where x stands for s in the case of a unit force and for / in the case of an impulsive force. Also, - f 0 if x0 is not in (a, b) , S(x-x )dx= (2) J 0 [ 1 if x0 is in (a,b) . This function is supposed to have the sifting property 00 I S (x - x0) (j) (x) dx = $ (x0) (3) — 00 for every continuous function (f){x). The Dirac delta function has been successfully used in the description of concentrated forces in solid and fluid mechanics, point masses in the 5.4. DIRAC DELTA FUNCTION 71 theory of gravitational potential, point charges in electrostatics, im pulsive forces in acoustics, and various similar phenomena in other branches of physics and mechanics. In spite of the fact that scientists have used this function with success, the language of classical math ematics is inadequate to justify such a function. It is usually visualized as a limit of piecewise continuous functions/(x) such as 0 , 0 ^ x < x0 — ^e , fix) p, \x-x0\ ^ is, (4) i:: x0 + ie < x < £ , or as a limit of a sequence of suitable functions such as ' k, 0 < |JC| < 1/Jfc, /*(*) = (5) 0 , for all other x , where k = 1,2,3,..., and 1 s'mkx A(x) = (6) Our aim in this and the next chapter is to determine integral repre sentation formulas for the solutions of linear ordinary and partial differential equations in such a manner so as to include the boundary conditions explicitly. To accomplish this task, we have to solve differential equations whose inhomogeneous term is a concentrated source. This is best done by introducing the theory of distributions, but that is not on our agenda. We shall, therefore, content ourselves with the above-mentioned properties of the delta function. Furthermore, we shall need the Heaviside function H(x)\ [0, x < 0 H(x) = (7) 1 , x > 0 and the relation dH{x)ldx = S(x). (8) 72 5 / ORDINARY DIFFERENTIAL EQUATIONS 5.5. GREEN'S FUNCTION APPROACH We shall consider the initial and boundary value problems of Sections 5.1 and 5.2 in a different context. Let L be the differential operator Lu(s) = iA{s)—2 + B(s)- + C(sy]u(s) , a Mv(s) =— lA(s)v(s)-\ ds2 2 - ^ lB(s)v(sJ] + C(s)v(s) , a < s < b . (2) as It follows by integration by parts that j (vLu- uMv) ds = [A{vu' - uv') + uv(B-A')fa. (3) This is known as Green's formula for the operator L. It is traditionally proved in the theory of ordinary differential equations that the relation A(s)f + B(s)yf + C(s)y = d>(s) (4) can be converted to the form IK] + q(s)y = F(s), (5) which is clearly self-adjoint. The function p(s) is again continuously differentiable and positive and q(s) and F(s) are continuous in a given interval (a,b). Green's formula (3) for this operator takes the simple form f (vLu - uLv) ds = [/? (s) (vu' - uv')fa . (6) The homogeneous second-order equation d ^--.+"-;('s)- ° <7> 5.5. GREEN'S FUNCTION APPROACH 73 has exactly two linearly independent solutions u(s) and v(s) which are twice continuously differentiate in the interval a < s Initial Value Problems Let us first consider the initial value problem s('s)+«--««• <8> yW - 0 , /(a) - 0 . (9) To formulate this problem into an integral equation, we consider the function 5 S w(s) = u(s) j v(t)F(t) dt -v(s)j u(t) F(t) dt, (10) a a where u and v are solutions of the homogeneous equation (7) as men tioned above. The relation (10), when differentiated, gives s s w'(s) = u'(s) f v(t)F{i) dt - v'(s) \ u(t)F(t) dt a a + u(s)v (s)F(s) - u(s)v(s)F(s) s s = u'(s) j v(t) F(t) dt - v'(s) j u(t) F(t) dt. a a Hence, w(a) = w'{a) = 0, and a _d_[ (s)dvl ds |_ dsj (equation continued) 74 5 / ORDINARY DIFFERENTIAL EQUATIONS s (0 F(t) dt 4- p(s) [u'(s) v(s) - v'(s) u(sy] F(s) = - q(s) w(s) + p(s) [u'(s) v(s) - v'(s) u(sft F(s) , (11) where we have used the fact that u(s) and v(s) satisfy the equation (7). In addition [dropping the argument (s) for /?, w, v]9 d d d — {p(uf v — v' u)} = — {pu') v —- (pv') u + pu' v' — pv' u' = 0 , ds ds ds also because u and f satisfy (7). This means that p(s)\u'(s)v{s) - v'{s)u(sj\ = ^ , (12) where A is a constant. The negative of the expression in the brackets in the above relation is called the Wronskian W(u, v;s) of the solutions u and v: W(u,v;s) = u(s)v'(s) - v(s)u'(s) . (13) From the relations (11) and (12), it follows that the function w as given by (10) satisfies the system d ( dw\ Z{p^)^ = AF(s), (14) w(a) = 0 , w'(a) = 0 . (15) Dividing (14) by the constant A and comparing it with (5), we derive the required relation y(s) as s y(s) = j R(s,t)F(t)dt, (16) a where R(s9t) = (\/A){u(s)v(t) - v(s)u(t)} . (17) Note that R(s,t) = -R(t,s). It is easily verified that, for a fixed value of /, the function R(s,t) is completely characterized as the solution of the initial value problem d LR = -\p(s)— r / /*i+ q(s)R = S(s-t), 5.5. GREEN'S FUNCTION APPROACH 75 dR 1 »i « (18) *l" = °- Is This function describes the influence on the value of y at s due to a concentrated disturbance at /. It is called the influence function. The function G{s;t), fo, s < t , G(s\t) (19) R(s,f)9 s > t is called the causal Green's function. Example. Consider the initial value problem y" +y = F(s) , 0 < s < 1 , y(0) = y'(0) = 0 (20) The influence function R(s, t) is the solution of the system d2R dR + R = 5(s-i), Rl., = 0, = 1 (21) ds2 ~ds The required value of R, clearly, is R(s, t) = sin(5 — /), and the integral representation formula for the initial value problem (20) is y(s) = jsm(s-t)F(t) dt (22) When the values of y(a) and y'(a) are prescribed to be other than zero, then we simply add a suitable solution cxu + c2v of (7) to the integral equation (16) and evaluate the constants c{ and c2 by the prescribed conditions. For example, y" + y = F(s) , y(0) = 1 , / (0) = - 1 (23) has the solution s y(s) = f [sin(s-t)'] F(t) dt + cx (sin J) + c2(coss) . (24) o With the help of the prescribed conditions, we find that c{ = — 1 and c2=l. 76 5 / ORDINARY DIFFERENTIAL EQUATIONS Boundary Value Problems Let us now consider the boundary value problems and start with the simplest one, a s b (25) I(^)+W=BF(J)- * * - y(a) = 0 , y(b) = 0 . (26) We attempt to write its general solution as an integral equation of the form s y(s) = j R(s, t)F{t) dt + cx u(s) + c2 v(s) , (27) a where u(s) and v(s) are the solutions of the homogeneous equation (7). When we substitute the conditions (26) in (27), we obtain c1u(q) + c2v(a) = 0 , cx u(b) + c2 v(b) = - j R(b, i)F(t) dt, which will determine a unique pair of constants cx and c2 provided the following holds for the determinant D: D = u(a)v(b) - v(a)u(b) * 0 ; (28) for the time being, we assume this to be true. Therefore, ci = Ma)ID]JR(b9t)F(t)dt s b = Ma)ID] j R(b, t) F{t) dt + \v(a)ID\ j R(b, t)F(t) dt , (29) a s c2 = -lu(a)/DlJR(b,t)F(t)dt s b = - [u(a)ID] j R{b, t) F{t) dt - [«(a)/Z)] J R(b, t) F(t) dt. (30) a s Putting these values of cx and c2 in the relation (27), we have the solution y{s) as 5.5. GREEN'S FUNCTION APPROACH 77 y(s) = j{R(s,t) + (\ID)lv(a)u(s)-u(a)v(s)-]R(b,t)}F(t)dt + | (\ID)tv(a)u(s) - u(a)v(s)]R(b,t)F(t) dt. (31) Using (12) and (17) and doing some algebraic manipulation, we find that R(s, t) + {[v(a) u{s) - u(a) v(s)^/D}R(b, t) = (\/AD)[u(a)v(t) - v(a)u(t)~] [u(s)v(b) - v(s)u(b)] . (32) Finally, we define the function G(s;t): (\/AD)[u(s)v(a) - v(s)u(a)-\ [u(t)v(b) - v(t)u(b)'] , s < t, G(s;t) = (33) (\/AD)lu(t)v(a) - v(t)u(a)-] [u(s)v{b) - v{s)u(by\ , s > t. Then the solution y(s) as given by (29) takes the elegant form y(s) = -JG(s;t)F(t)dt. (34) The function G(s;t) is called the Green's function. It is clearly symmetric: G(s;t) = G(t;s). (35) Furthermore, it satisfies, for all t, the following auxiliary problem: LG -IK] + q(s)G= -8(s-t), (36) G\s=a = G\s=b = 0 , (37) GL=r + 0 - Gls = (-o = 0 > (38) dG dG (39) ds i+o ds 'Pit) ' where by C7|s=r + 0 we mean the limit of G(s;t) as s approaches t from the right, and there are similar meanings for the other expressions. Thus, 78 5/ORDINARY DIFFERENTIAL EQUATIONS the condition (38) implies that the Green's function is continuous at s = t. Similarly, the condition (39) states that dG/ds has a jump dis continuity of magnitude — {\/p(t)} at s = t. The conditions (38) and (39) are called the matching conditions. It is instructive to note that the relation (39) is a consequence of the relations (35) and (36). Indeed, the value of the jump in the derivative of G(s;t) can be obtained by integrating (36) over small interval (t — s,s) and by recalling that the indefinite integral of S(s — t) is the Heaviside function H(s — t). The result is dG(t-E\s) q(x)G(x;t) dx = p(t-s)—V, - H(s-t) ,»*p + ds When s traverses the source point t, then on the right side the Heaviside function has a unit jump discontinuity. Since other terms are continuous functions of s, it follows that dG/ds has, at t9 a jump discontinuity as given by (39). Example. Consider the boundary value problem y" = F(s) , j,(0) = y(S) = 0 . (40) Comparing this system with the relations (25), (26), and (36)-(39), we readily evaluate the Green's function: (Cs/W-0, s or min(s, /), and or max (5, /), the relation (41) takes the compact form 5.5. GREEN'S FUNCTION APPROACH 79 G(J;0 = (1/OI><('-J>)], 0 <*,/*£'• (42) It follows from the properties (36)—(39) of the Green's function G(s;t) that dG(s;a)jdt satisfies the system of equations d ' d (dG(s;a) ,JG(s;a) a < s < b , Js P(s)ds\-^r dG(a;a) _ _\_ dG{b\a) 0. (43) dt ~plfl)'p(a) dt Similarly, dG(s;b)/dt satisfies the system d d \dG(s;b) ,JG{s;b) a < s < b ds\_ ds\ dt dt dG(a;b) dG(b;b) 1 (44) dt o, dt pipy Hence, the boundary value problem (py'Y + qy = F, y(d) = a, y(b) = fi (45) has the solution y(s) = - G(s;t)F(t)dt b (46) + <*P(a)—dtr. PP( )' dt as is easily verified. Finally, we present the integral-equation formulation for the boundary value problem with more general and inhomogeneous end conditions: (pyj + qy = F(s) , -ft.yia) + vxy(a) = a , fi2y'(b) + v2y(b) = fi . (47) When we proceed to solve this system in the same way as we did the system (l)-(2), the Green's function for the present case can also be derived provided the determinant D = l-fi{ u'(a) + vx u{aj\ {JJL2 v'(b) + v2 v(bj] - ["Mi v'(a) + v, v(a)-] \ji2u\b) 4- v2 w(6)] ^ 0 . 80 5 / ORDINARY DIFFERENTIAL EQUATIONS Indeed, G(s;t) possesses the following properties: LG -IK= 4 \p(s)^\ 1+ q(s)G = -S(s-t) dG\ -/*1 — \ +v, G = M + v2G\s=b ds \, = 0 l- d~s s = b J|s=r + 0 — "|»=t-0 = 0 , dG 1 *l -Xl =--- 9 (48) s=t+o ds and the condition of symmetry. With the help of the Green's function, the boundary value problem (47) has the unique solution y(s) = -\G(s;t)-\G(s;i F(t) dt + — <*G(s;a) + — /K7(J; b) , (49) Hi Hi provided/ij and/i2 do not vanish. If \ix = 0, then the factor {\\[i^)G(s\ a) is replaced by (\/vl)dG(s;a)/dt. By the same token, if ju2 = 0, we replace (l//i2)GCy;£) by -(\lv2)dG(s;b)ldt. In view of the relation Z) # 0, we cannot have both /^ and vx or both //2 and v2 equal to zero. When a and ft are zero, the relation (49) reduces to (34). The Sturm-Liouville problem consists in solving a differential equation of the form (py'Y + qy + *ry = F(s) (50) involving a parameter k and subject to a pair of homogeneous boundary conditions -/!,/(«) + v^(a) = 0 , ii2y'{b) + v2^(6) = 0 . (51) The values of X for which this problem has a nontrivial solution are called the eigenvalues. The corresponding solutions are the eigen- functions. In case/?(a) =p(b), the boundary conditions (51) are replaced by the periodic boundary conditions y(a)=y(b)9 y'(a)=y'(b). From formula (49), it follows that the solution of equation (50) subject to the conditions (51) is 5.5. GREEN'S FUNCTION APPROACH 81 y(s) = X j r(t)G(s;t)y(t)dt - j G(s;t)F(t)dt, (52) which is a Fredholm integral equation of the second kind. In this equation, G{s; i)r{i) is not symmetric unless the function r is a constant. However, by setting [r(s)l*y{s) = Y(s) , under the assumption that r(s) is nonnegative over the interval (a,b), the equation (52) takes the form yMlrW* = i. G(s; t) [/■(*)]* [/•(0]«4 \r(V\*y(t) dt G(s;t)lr(s)y>lr(t)y< dt. or Y(s) = X G(s;t)Y(t)dt (53) where G(s; i) = G(s;t)[r(s)']¥'[r(t)yA is a symmetric kernel. The above discussion on boundary value problems is based on the assumption that D = u(a)v{b) — v(a)u(b) does not vanish. If it vanishes, then the homogeneous equations ct u(a) + c2 v(a) = 0 , Cj u(b) + c2 v(b) = 0 have a nontrivial solution (cuc2), and the function w(s) = ct u(s) + c2v(s) satisfies the completely homogeneous system (pw')' + qw = 0 , w(a) = w(b) 0. (54) Therefore, if y is a solution of (25), (45), or (47), then so is y + cw for any constant c. This means these systems do not have a unique solution. This is not all. There is an additional consistency condition which must be satisfied for these systems to have a solution. Take, for example, the system (45). Multiply the differential equation (45) by w and integrate from a to b and get j w(s)F(s) ds = j w(s) [(/?/)' + qy] ds = [wp/ - w'pyfa + J y[(pw')' + qw)'] ds = p(a)w'(a)a-p(b)w'(b)p. (55) 82 5 / ORDINARY DIFFERENTIAL EQUATIONS Therefore, if (45) is to have a solution, then the given function F(x) must satisfy the consistency condition (55). For a = /? = 0, we get the consistency condition J w(s)F(s)ds = 0, (56) for the system (25). Thus, if D is zero, then we either have no solution or many solutions; but never just one. 5.6. EXAMPLES Example 1. Reduce the boundary value problem y" + ly = 0, (1) y(0) = 0, y'(l) + v2y(l) = 0, (2) to a Fredholm integral equation. From the properties (48), and (48)2, we must have ( A,(t)s , s < t, G(s;t) = [A2(t)l\+v2(l-s)-], s>t. The consequence of the symmetry of the Green's function is Ax = C[l+v2(l-0], A2 = Ct, where C is a constant independent of t. The jump condition (48)4 yields C/(-v2)-C[l+v2(l-/)] = -l or C= 1/(1 +v2). Thus, the Green's function is completely determined: f[l+V2(l-/)]5/(1+V2), S 1 +V2(1-J) * y(s) = A 0>(0 * 1 + v2 o l + ^-■J—— If[l+v [1 + v2((l-l -ty]y{t)i dt + -?- . (4) 1 ++ Vv22 JJ 1 + v2 Example 2. Reduce the Bessel equation ^2^ + ^ + (^2-l)j = 0 (5) with end conditions j(0) = 0, ^(1) = 0, (6) to a Fredholm integral equation. The differential equation (5) can be written as (sy'Y + [{-lls) + As]y = 0. (7) Comparing (7) with (5.5.50), we obtain p(s) = s , q(s) = - \/s , r(s) = s , F(s) = 0 . To find the Green's function, we observe that the two linearly independent solutions of the equation d2y dy s 5 = ~rds2f +S+ Tds ~y ° are s and \/s. Therefore, from (5.5.36)-(5.5.39), it follows that G(s;t)={ (8) Substituting this value of G in the relation (5.5.52), we get the required Fredholm integral equation. Example 3. Reduce the following boundary value problem to a Fredholm integral equation: f + *sy=l, y(0) = 0 , y(0 = 1 . (9) 84 5/ORDINARY DIFERENTIAL EQUATIONS The Green's function is the same as given by (5.5.41): f(j/O(^-0, s<*> G(s;t) = (10) \{tlt)Vs), s> t. The expression for the integral equation then follows from (5.5.46): y(s) = - J G(s;t) dt + X j G(s; t)ty(t) dt - dG{s\t)jdt . (11) o o The function dG{s\£)jdt satisfies the system of equations (5.5.44), which for the present case become whose solution is dG(s;f)/dt = -s\t . (13) Substituting (12) and (13) in (11), we have s l y(s) = (s//) - j WW-s) dt-\ WW-i) dt 0 s + k" JG(s'9t)ty(t)dt "o0 or y(s) = (s/2/)[2-f2 + st] + I f G(s;t)ty(t)dt . (14) 5.7. GREEN'S FUNCTION FOR /VTH-ORDER ORDINARY DIFFERENTIAL EQUATION The boundary value problem consisting of a differential equation of order n and prescribed end conditions can also be transformed to a Fredholm integral equation in the manner of Section 5.5. For instance, take the boundary value problem 5.7. GREEN'S FUNCTION FOR /VTH ORDER 85 d2y dy d? \X(S)M + ds \pis) Is] + '<*» ~ Xr(s)y = F(s) (1) where 1 is a parameter and the boundary conditions are (xy")' + py' = 0 , or y prescribed , (2) xy" = 0, or / prescribed . The Green's function G(s\t) for the problem (l)-(2) is such that it satisfies the differential equation d2 2 d S <^ y\^ \ ,Jy 5(s-t), (3) ds ds2 ds ds together with the prescribed homogeneous boundary conditions. In addition, (7, dGjds, and d2G/ds2 are continuous at s = t. The value of the jump in the third derivative of G is d3G d3G 1 (4) s = t + 0 IF s = t-0 x(s) Finally, G is symmetric. In terms of the Green's function with the above five properties, the boundary value problem (l)-(2) reduces to the integral equation y(s) = X j G(s, t)y(t) dt - j G(s, t)F(t) dt (5) Example. Consider the boundary value problem d4y + Xy = -/(s) , (6) ds4 y(0) = 0 = /(0) , y(l) = 0 = /(l) . (7) The homogeneous equation d4y\ds4 = 0 , has the four linearly independent solutions 1, s, s2, s3. Therefore, we take the value of G(s; t) to be 2 3 { A0{t) + A{(t)s + A2{t)s + A3{t)s , s < t, G(s;t) = 2 3 (8) { B0{t) + £,(/)* + B2(t)s + B3(t)s , s > t. 86 5 / ORDINARY DIFFERENTIAL EQUATIONS The boundary conditions at the end points give A0(t) = 0, Ax{t) = 0, B2 = -3B0-2BX, B3 = 2B0 + B{. Thus, the relation (8) becomes 2 3 f A2(t)s + A3s , s < t, G(s;t)={ 2 \(\-s) [B0(t)(\+2s) + Blsl, s> t. The remaining constants are determined by applying the matching conditions at s = t, which result in the simultaneous equations 2 3 2 3 2 t A2 + t A3- (\-3t + 2t )B0 - t(\-t) Bx = 0 , 2 2 2tA2 + 3t A3 + 6t(\-t)B0-(\-4t + 3t )Bl = 0, 2A2 + 6M3 + 6(1 -2f)*o + 2(2-3f)£i = 0 , 6A3- \2B0-6Bl = 1 , whose solution is 2 2 A2(t) = -±t{\-t) , A3(t) = £(l-0 (2/+l) , 3 2 B0(t) = it , B,(t) = -±t . Hence, 2 2 {ts (l-t) (2st + s-3t)9 s < t, G{s\t) = (9) [it2(\-s)2(2st + t-3s), s> t . The required Fredholm integral equation then follows by substituting this value in the relation (5). 5.8. MODIFIED GREEN'S FUNCTION We observed at the end of the Section 5.5 that, if the homogeneous equation Ly = 0, where L is the self-adjoint operator as defined in equation (5.5.5), with prescribed end conditions, has a nontrivial solution w(s), then the corresponding inhomogeneous equation either has no solution or many solutions, depending on the consistency 5.8. MODIFIED GREEN'S FUNCTION 87 condition. This means that the Green's function, as defined in Section 5.5, does not exist, because f S(s-t)w(s)ds + 0, a< t < b . (1) A method of constructing the Green's function for this type of problem is now presented. Such a function is called the modified Green's function and we shall denote it by GM(s;t). We start by choosing a normalized solution of the completely homogeneous system so that J" w2(s) ds = 1. The modified Green's function is to satisfy the following properties. (a) GM satisfies the differential equation LGM(s;t) = d(s-t)-w(s)w(t). (2) This amounts to introducing an additional source density so that the consistency condition is satisfied. Indeed, f [ (b) GM satisfies the prescribed homogeneous end conditions. (c) GM is continuous at s = t. (d) GM satisfies dG„ dGM ds ds -,15 • <3) Thus, the construction is similar to that for the ordinary Green's function except that the modified Green's function is not uniquely determined; we can add cw(s) to it without violating any of the above four properties. It is often convenient to choose a particular modified Green's function that is a symmetric function of s and /. This is accomplished by defining two functions (/MCS;^) and GM(s;t2) which satisfy the equations LGM(s;tx) = Sis-t,) - w(tx)w(s) , (4) LGM(s;t2) = S(s-t2)- w(t2)w(s) , (5) along with prescribed (same for both) homogeneous end conditions. 88 5 / ORDINARY DIFFERENTIAL EQUATIONS Multiply (4) by GM(s; t2) and (5) by C7M(5-; Z^, subtract, and integrate from a to b. Finally, use the Green's formula (5.5.6) and get cM(f i; h) - cM(>2; >i) + w(*,) j cM(*; h) H>(J) A -W(/2)JCM(J; 0^)^ = 0. (6) Now, if we impose the condition: (e) GM satisfies the property $GM(s;t)w{s)ds = Q, (7) then it follows from (6) that GM will be symmetric. Thereby, the Green's function is uniquely defined. Finally, we can reduce the inhomogeneous equation Ly = F(s) (8) with prescribed homogeneous end conditions into an integral equation when the consistency condition J F(s) w(s)ds = 0, is satisfied. Indeed, by following the usual procedure, one gets J (GMLy-yLGM) ds = j GM(*l t)F(s) ds - y(t) + j y(s)w(s)w(t)ds . (9) But the left side is zero because of the Green's formula (5.5.6) and the end conditions. Thus, we are left with the relation y(t) = j GM(s; t)F(s) ds + Cw(t) , (10) where C = $y(s)w(s)ds, is a constant, although as yet undetermined. By interchanging s and t in (10), we have y(s) = \GM(t',s)F(t)dt + Cw(s). (11) For a symmetric Green's function, the above result becomes y(s) = JGM(s;f)F(t) dt + Cw(s) . (12) When w(s) = 0, this reduces to the relation (5.5.34) [note that G in (5.5.34) is negative of the <7M in (12)]. 5.9. EXAMPLES 89 5.9. EXAMPLES Example 1. Transform the boundary value problem F s -(~idsi + h) = ( ) > /(°) = /(O = o, o^s^s (13) into an integral equation. The self-adjoint operator -d2y/ds2 = 0, 0 < s < £, with the end conditions y(0)=/(£) = 0 has the nontrivial solution y = const. We take it to be 1//I/2 because Jo (\/t)ds= 1. We therefore have to solve the system 2 d GM(s;t) 1 -^r = -^- = 0- (14) The solution of the system (14) for s < t that satisfies the end condition at 5 = 0 is A + (s2/2£). Similarly, the solution that satisfies the end condition at s = £ is B-s + (s2/2£). Hence, I A + (s2/2/) , s < t, GM(s;t) = ( £ - s + (s2/20 , 5 > f. The condition of continuity at s = t implies that B = A + /, while the jump condition on dGM/ds is automatically satisfied. Thus, ( A + (*2/20 , s The constant ^t is determined byt the symmetry condition (7): i( A+9}ds+\(A+t-s+si)ds = ° or 2 2 l// t t Thereby, the symmetric Green's function is completely determined: 90 5 / ORDINARY DIFFERENTIAL EQUATIONS t, s < t , GM( ;0 = + ( 2 + 2) (15) * 3 i * ' S , s > t . The expression for the integral equation follows from the formula (5.8.12), £ £ t y{s) = -A j GM(s;t)y(t) dt + JGM(s;t) F(s) ds + (I/O j y(s) ds. (16) 0 0 0 Example 2. Transform the boundary value problem T{(1-J2)?1 + ^ = °. -K-0, yiS) finite (17) ds ds into an integral equation. The operator — (dlds)[{\—s2)dylds] is a self-adjoint Legendre operator. The function w(s) = 1/^/2 satisfies this operator as well as the boundary conditions. Hence, we have to solve the equation i\o-*%i-*-o-, (18) For s # /, we have "-^-idGK " or CM = -i{log[(l+J)(l-*)]} + i/t{log[(l+j)/(l-5)]} + fi. (19) Thus, (iz(-i)log(l +5) - (i^+i)log(l -s) + B , s < t, GM(s;t) (20) (±C-±)log(l + s) - (ic + i)log(l -s)+D, s > t. EXERCISES 91 Since GM(s;t) has to be finite at -l and + l, we must take A = \, C = — ^, and the relation (20) reduces to GM(s;t)=\ (21) { -ilog(l+^)+ D , s > t . The continuity of GM at s = t implies that B - D = -±l0g(l +/) + ±log(l -0 . (22) The jump condition at s = t is automatically satisfied. Finally, the symmetry of GM yields I / I 0 = J GM(s; t) ds = j [B - ±log(l - J)] ds + jlD- ±log(l + *)] ds -l -l / or B+D = 2(log2) - 1 - ±[log(l +0 + log(l -0] • (23) From (22) and (23), we obtain the values of B and D as B = (log2) - i - ilog(l + 0 , Z) = (log2) - i - ilog(l -t) . (24) Putting these values in (21), we have (±log[(l-j)(l+0], *<*, GM(,;0 = dog2)-i- (25) (ilog[(l+j)(l-0], *> t. The required integral equation now follows by using the formula (5.8.12). EXERCISES Transform the following boundary value problems to integral equations: 1. y" + y = 09 y(0) = 09 /(0) = 1. 2. y + xy=\, y(0) = y(l) = 0 . 3. y+y = s, y(0)=/(i) = 0. 4. y+y = s, y(0)= 1, /(1) = 0. 92 5 / ORDINARY DIFFERENTIAL EQUATIONS 5. / + (y'/2s) + Xs*y = 0 , y(p) = y(l) = 0 . 6. f-y=f(s), y(0)=y(\) = 0. 7. Reduce the differential equation f + {P(s) - t?)y = 0 , rj a known positive constant, with the end conditions y(0) = 0, (y'/y) = -rj at s = s09 into a Fredholm integral equation. 8. Convert the initial value problem 2 y + Xs y = 09 y'(0) = 0 = ^(1) into a Volterra integral equation. 9. Find the boundary value problem that is equivalent to the integral equation I y(x) = lj{l-\x-t\)y(t)dt. 10. (a) Show that the Green's function for the Bessel operator of order n Ly = (dlds)(xdy/ds) - (n2/s)y , n # 0 with the end conditions *(0) = y(\) = 0 is [snltn{\-t2n), s < t, G(s; i) = \ n n 2n {t /s (\-s )9 s> t. (b) Use the result of part (a) to reduce the problem s2y" + sy' + (ks2-n2)y = 0 , ^(0) = y(l) = 0 to an integral equation. 11. Reduce the boundary value problem EXERCISES 93 y(0) = 0 = j(l) , /'(O) = 0 = y'"(l) to a Fredholm integral equation. 12. Extend the theory of Section 5.8 to the case when the completely homogeneous system has two linearly independent solutions wx (s) and w2(s). 13. Find the modified Green's function for the systems y"-Xy = 0, y(0)=y(l), /(0)=/(l) 14. Transform the boundary value problem f + y=f(s), y(0) = y(n) = o9 into an integral equation. Hint: The self-adjoint operator — (y"+y) = 0 has the nontrivial solution (2/7i)1/2sins and it satisfies the boundary conditions. Now proceed as in the examples in Section 5.9. APPLICATIONS TO PARTIAL CHAPTER 6 DIFFERENTIAL EQUATIONS 6.1. INTRODUCTION The applications of integral equations are not restricted to ordinary differential equations. In fact, the most important applications of integral equations arise in finding the solutions of boundary value problems in the theory of partial differential equations of the second order. The boundary value problems for equations of elliptic type can be reduced to Fredholm integral equations, while the study of parabolic and hyperbolic differential equations leads to Volterra integral equations. We shall confine our attention to the linear partial differential equations of the elliptic type, specifically, to the Laplace, Poisson, and Helmholtz equations, wherein lie the most interesting and important achievements of the theory of integral equations. Three types of boundary conditions arise in the study of elliptic partial differential equations. The first type is the Dirichlet condition. In this case, we prescribe the value of the solution on the boundary. The second type is the Neumann condition. In this case, we prescribe the normal derivative of the solution on the boundary. When we prescribe the Dirichlet conditions on some parts and the Neumann conditions on other parts of the boundary, we have a mixed boundary value problem. 94 6.1. INTRODUCTION 95 The divergence theorem and the two Green's identities will be used repeatedly in this chapter. They are as follows: Divergence theorem: j divAdV = J An dS ; CD R S Green's first identity: 2 [ dv j* uV vdVvd V = -|j ((gra d u - grad v) dV + u—dS ; (2) dn Green's second identity: 2 2 dv du\ _ J-{uV v-vV u)dV = (3) where A is a continuously differentiable vector field and the functions u and v have partial derivatives of the second order which are continuous in the bounded region R; S is the boundary of R9 while n stands for the unit normal outward to S. The surface S is a smooth (or regular) surface as defined by Kellog [9]. The differential operator V2 is the Laplacian, which in Cartesian coordinates xux2,x3 has the form 2 V = : 2 2 (4) dx{ dx dx ' Any solution u of the equation V2z/ = 0 is called a harmonic function. To keep the chapter to a manageable size, we shall confine ourselves mainly to three-dimensional problems. The results can be readily extended to two-dimensional problems once the technique is grasped. Boldface letters such as x shall signify the triplet (x1,x2,x3). The quantities R{ and Re will stand for the regions interior and exterior to S, respectively. Furthermore, we shall not follow the common practice of writing dSx or dS$ to signify that integration is with respect to the variable x or £. We shall merely write dS and it shall be clear from the context as to what the variable of integration is. We shall be interested in giving the integral-equation formulation mainly to the differential equations V2w = 0, the Laplace equation; 96 6 / PARTIAL DIFFERENTIAL EQUATIONS V2 u = — 4np, the Poisson equation; and (V2 + k2) u = 0, the Helmholtz (or the steady-state wave) equation. Here, p(x) is a given function of position and A: is a given number. 6.2. INTEGRAL REPRESENTATION FORMULAS FOR THE SOLUTIONS OF THE LAPLACE AND POISSON EQUATIONS Our starting point is the fundamental solution (or free-space solution) £(x;£) which satisfies -V2E = S(x-Q (1) and vanishes at infinity. This function can be interpreted as the electro static potential at an arbitrary field point x due to a unit charge at the source point £. Such a potential is given as £(x;$)= l/47rr= 1/4TT|X-$| . (2) For the two-dimensional case, the corresponding formula is (l/27r)log(l/r) = (l/27r)log(l/|x-§),wherex = (x1,^2)and§ = «1,{2). The fundamental solution can be used to obtain the solution of the Poisson equation -V2w = 4np . (3) Indeed, multiply (1) by W(JC), (3) by £(x; £), subtract, integrate over the region Rh and use Green's second identity as well as the sifting property of the delta function. The result is (after relabeling x and %) w(x) = l-%dS. (4) r An dn\r) 4n r on Suppose that from some previous considerations we know the values of p, w, and du/dn which appear in the formula (4): u\s = T , du/dn\s = o ; (5) then this formula becomes u(P) = + W®^. (6) ^ if«»s(>-s i 6.2. SOLUTIONS OF LAPLACE AND POISSON EQUATIONS 97 where P is the field point x and Q is a point ^ on S. The formulas (4) and (6) lead to many interesting properties of the harmonic functions. For details, the reader should refer to the appropriate texts [4, 9, 19]. We shall merely note the properties of the three integrals which occur on the right side of (6). The Newtonian, Single-Layer, and Double-Layer Potentials The integral \R (p/r) dV is called the volume potential or Newtonian G r potential with volume density p. Similarly, the integral \s( l ) dS is the simple or single-layer potential with charge (or source) density For the Newtonian potential u = \R(plr) dV, we have the following: 2 (1) V w = 0, for points Pin Re. (2) For points P within Ri9 the integral is improper but it converges and admits two differentials under the integral sign if the function p is sufficiently smooth; the result is V2u = —4np(P). The single-layer potential u = Js(o-/r) dS has the following properties: (1) V2w = 0, outside S. (2) The integral becomes improper at the surface S but converges uniformly if S is regular. Moreover, this integral remains continuous as we pass through S. (3) Consider the derivative of u taken in the direction of a line normal to the surface S in the outward direction from S. Then, C y =-2M/>)+U0 -^^ (7) and du cosg-x,n) = 2no(P) + (8) dn C{Q) |x-^ dS' s 98 6 / PARTIAL DIFFERENTIAL EQUATIONS where P+ and P_ signify that we approach S from Rx and Re, respec tively, and where both x and % are on S. From (7) and (8), we obtain the jump of the normal derivative of u across S: 1 fdu du (9) 4n\dn ~d~n Similarly, the double-layer potential u — \sx{djdn)(\jr) dS, has the following properties: (1) V2w = 0 outside of S. (2) The integral becomes improper at the surface but it converges if the surface 5 is regular. (3) The integral undergoes a discontinuity when passing through 5 such that , f cos(x — E, n) , u\P+ = 2TTT(/>)+ T(0 "_^2 dS (10) and cos(x-^,n) u\p_ = -2TLT(P) + T(0—j Trr-dS (11) in the notation of the relations (7) and (8). Hence, the jump of u across S is T = (1/4JT)[M|J,+ -II|P.]. (12) (4) The normal derivative remains continuous as S is crossed. The reader who is interested in the proof should look up the elegant proof given by Stakgold [19]. Interior and Exterior Dirichlet Problems For the solution of a boundary value problem for an elliptic equation, we cannot prescribe it and du/dn arbitrarily on S. Therefore, equation (6) does not allow us to construct a solution for equation (3) such that u shall itself have arbitrary values on S and also arbitrary values for its 6.2. SOLUTIONS OF LAPLACE AND POISSON EQUATIONS 99 normal derivative there. As such, there are two kinds of boundary value problems for elliptic equations. For one kind, we have the value of the solution prescribed on S—the so-called Dirichlet problem. For the second kind, the value of the normal derivative of the solution is prescribed on S—the Neumann problem. We discuss the Dirichlet problem first. To fix the ideas, let us discuss the Dirichlet problem for the region exterior to the unit sphere. In order to get a unique solution, it is necessary to impose some sort of boundary condition at infinity along with the boundary value on the surface S of the sphere. Indeed, the functions w,(x)= 1 and w2(x)= \jr are both harmonic in the region Re and assume the same value 1 on S. But if we require that the solution vanishes at infinity, then u2 is the required solution. As a matter of fact, it is an important result in the potential theory [9, 19] that, when one solves the Dirichlet problem for the exterior of the unit sphere (by expansions in spherical harmonics) such that the potential vanishes at infinity, then one finds that the behavior of the solution is = 0[\). (13) From these considerations and from the value of the fundamental solution, it is traditionally proved that du dE\ lim r-+oo S on the surface of the sphere of radius r. We can now define and analyze the exterior and interior Dirichlet problems for an arbitrary surface S as follows. Definition. The exterior Dirichlet problem is the boundary value problem 2 V we = 0, xei?e; uc\s=f, <".'!)• £ -o," °5) where /(x) is a given continuous function on S. 100 6 / PARTIAL DIFFERENTIAL EQUATIONS Definition. The interior Dirichlet problem is the boundary value problem 2 V Wi = 0, xeRi9 ui\s=f. (16) Suppose that we are required to find the solution of the interior Dirichlet problem. We assume that such a solution u is the potential of a double layer with density T (which is as yet unknown): fT(5)cos(x-5,n) ^ «i(x) = -2 dSo . (17) s For Wj to satisfy the boundary condition (16)2 from within S, we appeal to the relation (11) and get the Fredholm integral equation of the second kind for x(P)\ T(P) = -(1/2TT)/(P) + J K(P, 0r(0 dS , (18) s where the kernel K(P9 Q) is 2 K(P9Q) = [cos(x-^n)]/27r|x-^| , (19) and P ( = x) and Q ( = £) are both on S. We solve the integral equation (18) for T, substitute this solution in (17), and obtain the required solution of the boundary value problem (16). In exactly the same way, the Dirichlet problem for an external domain bounded internally by S can be reduced to the solution of a Fredholm integral equation of the second kind. We can present an integral-equation formulation of the exterior and interior boundary value problems (15) and (16) in a composite medium when/(x) is the same function in both these problems. Recall that the fundamental solution £(x; £) satisfies the relation -V2E = S(x-JQ , for all x and £ . (20) Multiply (16) by E, (20) by wb add, integrate and apply Green's second identity. This results in 1\ * d") (o, ieR,, s 6.2. SOLUTIONS OF LAPLACE AND POISSON EQUATIONS 101 where n is the outward normal to R{ on S. The corresponding result for the exterior region is obtained by multiplying (15) by £, (20) by we, adding, integrating over the region bounded internally by S and externally by a sphere Sn and applying Green's second identity. The contribution from Sr vanishes as /■-► oo in view of the boundary condition at infinity, and we finally have 8ue dE\ , 0, $eR;, i( (22) where we have used the fact that the outward normal to Re on S is in the — n direction. The next step is to add (21) and (22) and observe that both u-, and we take the same value / as we approach the surface. Thus, we obtain dS = \ (23) M£-£) «.£), 2,eRe, and x e S. Let us make use of the relations (2) and (9), and relabel x and £; thereby, we end up with the relation *®ds=l«M. «*.. (24) I X-SI Ue(x), X6i?e, that is, a single-layer potential with unknown charge density a. Finally, using the boundary condition "i|s = We|s =/, in (24), we obtain the Fredholm integral equation of the first kind / Interior and Exterior Neumann Problems In this case we are required to find the solution of Laplace or Poisson equation when the normal derivative is prescribed. 102 6 / PARTIAL DIFFERENTIAL EQUATIONS Definition. The exterior Neumann problem is the boundary value problem = /, « U=0. (26) on e Definition. The interior Neumann problem is the boundary value problem 2 V ^ = 0, xeRi9 -p- /. (27) on For a Neumann problem, the prescribed function /(x) satisfies the consistency condition $M)dS = 09 $eS, (28) s which follows by integrating the identity 2 JQ7 uddV=09 and using the divergence theorem. The exterior and interior Neumann problems can be reduced to integral equations in a manner similar to the one explained for the corresponding Dirichlet problem. Indeed, we seek a solution of the interior Neumann problem in the form of the potential of a simple layer «i = j IXC)/'] dS , (29) which is a harmonic function in Rr It will be a solution of (27) if the density a is so chosen that du-, = f(P), PeS, (30) ~d~n Appealing to the relation (8), we have du f(P) = = 2na(P) + COS n) dn ff(0 V ^- (3.) 6.2. SOLUTIONS OF LAPLACE AND POISSON EQUATIONS 103 Thus, G(P) is a solution of the Fredholm integral equation of the second kind , x cos (£ — x, n) °(Q) \ 'dS. (32) In r2 The solution of the exterior Neumann problem also leads to a similar integral equation. Furthermore, we can give the integral-equation formulation of the problems (26) and (27) in a composite medium when /is the same function in both these problems. Proceeding as we did for the corresponding Dirichlet problem, we obtain a Fredholm integral equation of the first kind. Instead of a single-layer potential, we now get a double-layer potential. The details are left to the reader. Let us observe in passing that the solution of (27) is not unique, since an arbitrary constant can be added to a solution and the resulting function will satisfy (27). 6.3. EXAMPLES Example 1. Electrostatic potential due to a thin circular disk. Let us take 5 to be a circular disk of radius a on which the potential V is prescribed. Let us choose cylindrical polar coordinates (p,cp,z) such that the origin is on the center of the disk with the z axis normal to the plane of the disk. Thus, the disk occupies the region z = 0, 0 ^ p ^ a, for all cp. There is no loss of generality in taking the potential V on the disk as f(n)(p) cos ncp, where n is an arbitrary integer, because we can use the Fourier superposition principle. The charge density a will also then have the form a{n)(p) cos ncp. From (6.2.25), we have fn\p)cosncp = J [affi)/|x-$|] dS , (l) disk where x = (p, cp,0) and ^ = (t9cpuO). Or a 2n , , t But, by setting cpx-(p = ^, we find that In 2n-u coswcpj dq>l cosn((p + ij/) d\\f 2 2 2 2 [p + t - 2ptcos((p-(pi)^ J [P + t - 2ptcos(cp-(Pl)^ 2n — 0 2/r 2TT- T f f f cosml/ dij/ 1 = (cos^)|J+j+ j _____j -p 0 2TT In f cos /n/f ^ = (COSH(/>) (3) ■ 2p/cosi/T|1/2 ' where in the first integral we have put i/s' = \j/ + 2n. From (2) and (3), there follows the equation a 2n ta(n) (t) cos mj/ dtp dt where (50r is the Kronecker delta, and use the orthogonality of the cosine function. The result is the Fredholm integral equation of the first kind fi*(p) = fto<*\t)K0(t9p)dt9 (6) o where the kernel K0 (/, p) is 00 K0(t,p) = 2n j Jn(pp)J„(pt) dp (7) 6.3. EXAMPLES 105 For an annular disk of inner radius b and outer radius a, the formula that corresponds to (6) is a f"(p) = jta"(t)K0(t,p)dt. (8) b Example 2. Solve the integral equation (6.2.25) when S is a unit sphere and/= sin 6cos cp\ that is, solve the integral equation In . f f (sin flt) a{Ouipx)dBl simOcoscpn " = \dcpA — . (9) o o Here, we use the expansion formula ^I»,,.Zn'w"'^ <■« m where Yn (0,(p) are the spherical harmonics and 2n n a Nm,n = dtp (sin 0) | y.-co. 9»>l rf» 0 0 4n (H+H)! (11) ~ 2/f+l («-|m|)! Furthermore, we set *(0i,«»i) = I Z x a (0, q>) = (3/4TT) Pt (cos 0) cos 6.4. GREEN'S FUNCTION APPROACH The Green's function is an auxiliary function which plays the same crucial role in the integral-equation formulation of partial differential equations as it plays in the case of ordinary differential equations. This function depends on the form of the differential equation, the boundary condition, and the region. For instance, the Green's function G(x;^) for the Laplace equation in an open, bounded region R in three-dimensional space with boundary S is the solution of the boundary value problem 2 -V C = (5(x-$), G|s = 0, (1) where x and % are in R. In the language of electrostatics, the function G is the electrostatic potential due to a unit charge at % when the surface S is a grounded metallic shell. As such, G is the sum of the potential of the unit source at £ in free space and the potential due to the charge induced on S: C(x;« = (l/47c|x-5|) + «(x;©, (2) where y is a harmonic function which satisfies the boundary value problem 2 V v = 0, xeR, v\s = -E. (3) Let us show that the Green's function is symmetric. When <7(x;£) and G(x,rj) are the Green's functions for the region R corresponding to the sources at % and //, we have the relations 2 -V G(x;« = 5(x-$), G\s = 0, (4) 2 -V C(x;/7) =S(x-rj), G\s = 0 . (5) The result of the routine steps of multiplying (4) by G(x;rj), (5) by G(x; £), subtracting, integrating, and applying Green's second identity is f r^;©^ - G(*:*>^1ds = Note also that the fundamental solution E(\;^) is the free-space Green's function. Solution of the Dirichlet Problem We are now ready to give an integral-equation formulation to the boundary value problem 2 -V w(x) = 47ip(x), xeR, u\s=f, (7) in terms of the Green's function. For this purpose, multiply (1) by u and (7) by (7, subtract, integrate, use Green's second identity, and get Mft) = 4TT f C(x;Qp(x) dV- J7(x)[3G(x;$)/3/i] dS . (8) R S By interchanging x and % and using the symmetry of the Green's function, we find the representation formula u(x) = 4nj (7(x;$)p($) dV - J [3C(x;©/3/i]/(© dS . (9) R S For the particular case p = 0, the formula (9) becomes «(x) = - Jyi(§)[3C(x;§)/3/i] JS . (10) S When /= 1 on 5", then the solution u of the Laplace equation is clearly u = 1 for the interior Dirichlet problem. Thus, (10) yields an interesting relation, -J[flC(x;©]/5/i]rf5,= 1 , xeJR. (11) s Example. Poisson integral formula. The Green's function for the Laplace equation, when the surface S is a sphere, can be found by various methods. The easiest method is to express it as source and image point combination. Let the radius of the sphere be a (see Figure 6.1). 108 6 / PARTIAL DIFFERENTIAL EQUATIONS Figure 6.1 For any point P ( = x) with radial distance a within the sphere, we have an inverse point P' ( = x') on the same radial line at a radial distance /? outside the sphere, such that a/] = a2. If Q ( = £) is any point on 5, then the triangles OQP and OQP' are easily seen to be similar. Therefore, r'jr = a jet, or l/r = alar' . (12) Examining the relations (1) and (2), we readily find the value of the Green's function to be (13) 47i V r a r Having found the Green's function, we can solve the interior Dirichlet problem for the sphere: V2w = 0, r To use the formula (10) we need the value of dG/dn. This is obtained if we observe from the figure that a2 = a2 + r2 — 2arcos(\ — £,n) ; (15) 2 2 2 p = a + r' - 2ar'cos(x'-£,n) 6.4. GREEN'S FUNCTION APPROACH 109 Thus, dG 1 / 1 dr a 1 dr'\ _ 1 /cos(x-£,n) tf cos(x'-£,n)\ 2 dn An\ r2 dn (xr'2dnj An\ r2 "a V ) 1 (a2-a2-r2 af$2-a2-r,2t\ a2-oi2 (16) An \ lar a 2ar'3 J Anar where we have used the relations r'\r = a/a and a/? = a2. Substituting (16) in (10), we finally have n In 2 „2 \ CC f(0' Neumann Problem By defining the Green's function (7(x; £) by the boundary value problem 2 -V G(x;^) = S(x-JQ , dG/dn\s = 0 , (18) we can extend the above analysis to the Neumann problem. Indeed, the integral equation that corresponds to (10) for the Neumann problem -V2w = 0, xeR, du/dn=f (19) is u(x)=JGM)M)dS. (20) s Of course, the function/must satisfy the consistency condition (6.2.28). Finally, let us consider the interior and exterior Dirichlet problems for a body S enclosed within a surface I: 2 V Wi=0, xeRi9 u{\s=f, (21) 2 V We = 0, xeRe, Ke |s = A "e|i = 0. (22) The Green's function (7(x;%) satisfies the auxiliary problem (we absorb the factor An in G): -V2G = AnS(x-^) , Gl = 0 . (23) 110 6 / PARTIAL DIFFERENTIAL EQUATIONS We now follow the same steps as we did in deriving the formula (6.2.25). Thus, from the relations (21)—(23), we obtain /(x) = JG(x;$)a{%)dS , (24) s which reduces to (6.2.25) for an unbounded medium. 6.5. EXAMPLES Example 1. Electrostatic potential problem of a conducting disk bounded by two parallel planes. This problem is an extension of the problem considered in Example l of Section 6.3. We follow that notation and assume that the parallel planes are z = b and z = —c (b,c>0). The boundary value problem becomes 2 V V(p,(p9z) = 0 in D, (1) V(p9 q>9 0) = /<"> (p) cos nq> , 0 < p < a , (2) V(p, 0 © 0 0 0 © © I 1 1 ^-H 1__| 1 1 1 -4c-2b-z, -2c-2b+z, -2c-z, z=-c 0 z, z = b 2b-z, 2b+2c + z< 4b+2c-z, Figure 6.2 6.5. EXAMPLES 111 & = [t, %„' =[t, |x-$l „4llx-^l „f±,l*-$„+| „£,>-$. I (7) The next step is to use the identity 00 1/lx-SI = Jy0(/w)(exp-p|z-z11)*, (8) 0 2 2 V2 where w = \_p +1 — 2pt cos (cp — cp X)~\ . Then, the relation (7) takes the form r° r00 C(x;§) = (l/|x-^|) + j 70(/,m) [£ (exp -/? |z - 2n(b + c) - zx\) o l — 00 4- X (exp-p|z- 2H(6 + C)-Z1|) -l oo - X (exp-/?|z - 2«(Z? + c) + 2c + zx\) o -00 - X (exp -p \z - 2n(b + c) + 2c + zj) 00 G(x;^)= ]^T\+ \J°{pw) (c+Zl)p ( Zl)p e- [s\nhp(z-b)-] -c- "- [sinh/?(z + c)] J x • u 00 MP*>) = Z (2-50r)[cosr((p-(pl)']Jr(pp)Jr(pt) , (ll) r=0 in (10) is 1 (r) G(x;Q = , - + Y (2-50r)[cosr((p-(p1)^ (pJ,z,z1), (12) |x-q| r=o where 6>-p(c+z,)[sinh/;(z-/?)]-e-p(fc-z,)[sinh/7(z + c)] C?(r)(p,/,z,z ) = I sinh p(b + c) xJr(pp)Jr(pt)dp . (13) To derive the integral equation, we multiply (1) by G and (4) by V, follow the routine procedure, and get 1 dV dG\ K(p, In {n) (n) f (p)cosncp = I j ta (t)G(p,t, in) where a (t)cosn(pl = (\l4n)(dV/dzl+-dVldzl_). Settings l—q> = i/^,and following the steps which led from (6.3.2) to (6.3.4), we obtain In f cos™/^_ 'W(n\ = to r\p) J |x-5|,-„-c + J to*"' (0 [27rG(n) (p, /, 0,0)] * , (16) where we have substituted the value of G as in (12). This can be written as 6.5. EXAMPLES 113 /"(p) = JV"> (?)*:,(?, p) When 6-+ oo, c-> oo, we recover the formula (6.3.4). For an annular disk of inner radius b and outer radius a, the formula which corresponds to (17) is a ( / ">(p) = \ta^(t)Kx{t,p)dt (19) b with the same kernel as in (18). Example 2. Electrostatic potential problem of an axially symmetric conductor placed symmetrically inside a cylinder of radius b. We again take cylindrical polar coordinates (p, cp,z) with origin at the center of the conductor and z axis along its axis of symmetry (which is also the axis of the cylinder). For simplicity, we take V on the surface S of the conductor to be unity. Then, from the relation (6.4.24), we have the Fredholm integral equation of the first kind 1 = J"G(x;$M$)£/S, x,^e5, (20) s where G satisfies the system 2 -V G = 4nd(x-Z>) , G = 0 on p = b. (21) In terms of cylindrical polar coordinates, the differential equation for G becomes 2 2 1 / dG\ 1 d G d G _ 4T5T r(P^:)+n^- + 1T + +TT = -- (P-05(^)5(z-z ), (22) p V W V C1(x;^) = C(x;§)-(l/|x-§|) (23) 114 6 / PARTIAL DIFFERENTIAL EQUATIONS is finite in the limit as x-*£. We can calculate the solution of (22) by taking the Fourier series expansion ( ) C(x;§) = f (2-^0r)(cosnA)^ '' (p,/,z,z1) , (24) r=l where lit g(r) = (1/2TT) J"G(X;$)(COS/^)#. Multiply the differential equation (22) by (I/2n) cos r\j/ and integrate with respect to ij/ from 0 to 2n. The result is Next, we take the Fourier transform of equation (25) by setting 00 7Vr)) = (2JI)-* J>Vr> — 00 00 <7(r> = (2s)-'/2 | e""" 7Vr)) . (26) — 00 The system (21) becomes 2 2 2 2 P £~2 T(g") + pj T{g") - (p p + r )T^>) = --^ ^i <5(p-Pl), 7V'>) = 0 , p = 6 . (27) PV71 This boundary value problem can be easily solved by the method and notation of Chapter 5 and the solution so obtained is then inverted to yield 00 z = (l/jt) J><»- >{KrOv>K(w><) - lK,(Pb)irr(pby] Upp)ir(pt)} dp, (28) where Ir and Kr are modified Bessel functions. Finally, from (24) and (28), we find the value of G: 6.5. EXAMPLES 115 ,p<2 -z) G(x;$) = (l/jt) t (2-50,)(cosiV) f e ' x {Kr(pP>)Ir(pP<) - lKr(pb)IIr(pb)-]Ir(pp)Ir(pt)} dp. (29) When b-* oo, G = l/|x —£|, and we have from (29), l/|x-$| = (I/TT) X (2-(50r)(cosnA) [e^-^KApp^IAppJdp. r = 0 (30) Combining (29) and (30), we have (,) G(x£) = (\l\x-%\)+ f (2-50r)(cos/V)C (p,r,r,r1), (31) where 00 G^(p,t,z,Zl) = -(Ijn) j e-«*-*>>Ir{pp)Ir(pt)lKr{pb)IIr(pb)-\dp — 00 00 = -(I/TT) [J e-'t'-^UppWptKKripWUpby] dp 0 -0 + / e-'«*-">Ir(pp)Ir(pt) lK,(pb)IIr{pby] dp] . (32) — oo Changing p to —p in the second integral and observing that r rr(-z) = (-\yrr(z), K(-z) = {-\) Kr(z), and hence Kr(—pb)/Ir(—pb) = Kr(pb)/Ir(pb), we have from (32) Ir(pp)Ir(pt)Kr(pb) &*(p9t9z9zx) = -(2/TT) cos\_p(z — zx)~]dp. (33) So far, we have not used the fact that the conductor is axially symmetric. For an axially symmetric body, the Green's function is independent of cp. That leaves only one term in the series (31): (0 C(x^) = (l/|x-^|) + G >(p,?,z,z1). (34) 116 6 / PARTIAL DIFFERENTIAL EQUATIONS Equation (34) is of the form (23). Substituting (34) in (20), we have the required integral equation. When /?->oo, we recover equation (6.3.4) for/(">(p)=l. 6.6. THE HELMHOLTZ EQUATION The discussion of the previous two sections can be easily extended to the case of the Helmholtz equation (V2 + /l)w = 0. (i) The free-space Green's function or the fundamental solution E(x;^) is the solution of the spherically symmetric differential equation 2 -V E-XE = 8(x-$)9 (2) and which vanishes at infinity. Such a solution in three dimensions is Fflt.*\ _ exp(/|x-^lVI) _ expjiryfj) 47c|x —§| 4nr (i) When X is a complex number, then yfl is selected to be that root of X that has a positive imaginary part so that E vanishes exponentially at infinity. (ii) When X is real and positive, that is, X = a>2, w real, the solution 47r | x — ^ I 4nr is selected such that v/I = co>0. This represents an outgoing wave if we adjoin the factor e~l0)t. (iii) When X is real and negative, we again choose yfl in (3) to be the square root of X that has a positive imaginary part for the same reason as in (i). For the particular case X = — A:2, where k is real and positive, the formula (3) becomes E(x;JQ = e~kr\Anr . (5) The solutions which correspond to (3), (4), and (5) in two dimensions are 6.6. THE HELMHOLTZ EQUATION 117 (i/4)ffn|x-$|VA), OnH^(ca\x-S,\), (ll2n)K0(k\x-$\), { X) respectively. Here, the functions H 0 and K0 are the Hankel and modified Bessel functions, respectively. The integral representation formula for the solution of the inhomo- geneous equation (V2-k2)u = -4np (6) is obtained from the relations (2) and (6) by using Green's identity and is readily found to be 1 f d (e-kr\ — -dS.(l) M(X) = I>JL:V+I('4(-V-£ r on The interpretation of these integrals as volume, single-layer, and double- layer potentials is the same as for the corresponding formulas in Section 6.2. The properties of these potentials are also similar. For instance, the formulas that correspond to (6.2.7) and (6.2.8) are du = +27I(T(/>) + a{Q) dS (8) dn Pi dn\~> ' s where e~kr u = °(Q)[_ ]dS. (9) r s Similarly, the formulas that correspond to (6.2.10d ) and (6.2.11) are ± 2m{P) + f T(0 L (e-krlr) dS , (10) s dn~n where d «= \^Q)^(e-krlr)dS. (11) J on The rest of the notation is the same as in Section 6.2. The integral representation of solutions of the exterior and interior Dirichlet and Neumann problems is achieved in an analogous manner, as shall become evident from the examples of physical interest which are presented in the next section. We end this section by mentioning the 118 6/ PARTIAL DIFFERENTIAL EQUATIONS so-called Sommerfeld radiation condition. A three-dimensional solution of the Helmholtz equation (V2 + k2)u = 0 is said to satisfy the radiation condition if lim/Y-^ -iku\ = 0, (12) as /•-► oo. Physically, this condition implies that there are no incoming waves from infinity. In two dimensions, the corresponding condition is \im^(^-iku\==0, (13) as /--►oo. The free-space Green's function satisfies the radiation condition. 6.7. EXAMPLES Example 1. Acoustic diffraction of a plane wave by a perfectly soft disk. We follow the coordinate system and notation of Example 1 in Section 6.3. Furthermore, we assume the time dependence of the form e-io>t for ^ wave functions involved in the problem and omit this factor in the sequel. The time-independent part of the velocity potential u is u(p9 KS(P 2 2 2 V2 exp/A:{p + t - 2p/[cos((^-(^1)] + (z-Zi) } 2 2 2! . (6) 4n{p + t - 2pf[cos( Multiply (2) by E, (5) by ws, subtract, integrate, use Green's second identity, and obtain dus dE\ ws(x) = ETn-U*Tn )dS> (7) 5++5- where S+ and S" are the upper and lower surfaces of the disk, respec tively. On S±, we have djdn = +djdz, respectively. Thus, (7) can be written as a In t, dEl us(p,q>,z) = t dcpx at i ^i_|Zl = o + + 0 0 a 2TT /(T(?,<^1,0)£|ZI = 0#1 A , (8) where we have used the fact that us = —u{ on both sides of the disk and where duA a(t,cpu0) = (^ (9) o+ vz |z! = o- When we apply the boundary condition (3) in (8), we get a In 2 2 2 f ta(t,(pu0)expik[p + t - 2ptcos{cp-cp^ 2 2 1/2 4n [p + t - 2p/cos((/?-(/?1)] o o (10) x dcpldt. 120 6 / PARTIAL DIFFERENTIAL EQUATIONS In view of the Fourier superposition principle, we can assume {n) that «i(p, cp,z) and v(p,(p) are of the form ut (p, z) cos ncp and 2o{n)(p)cosncp, respectively. Then, proceeding as in the steps that led from (6.3.2) to (6.3.4), we obtain from (10) the integral equation a Uiin)(p,0) = jtaMiOKtfrp)*, (11) where the kernel Kx (t, p) is In expikjp2 + t2 - 2p cos ^ Ki(up)== ——= y— —ry—cos ny ay/. (12) h (p2 + t2 - 2p/cosi/0/2 The integral equation (11) is the required Fredholm integral equation of the first kind which embodies the solution of the boundary value problem (2)-(4). For an annular disk with inner radius b and outer radius a, the corresponding integral equation is a u^faO) = f to^WK^p)* . (13) b Example 2. Torsional oscillations of an elastic half-space. In terms of cylindrical polar coordinates, the axially symmetric boundary value problem d2v I dv v d2v . „ - co2 da2 ^^ -! + -- - +— + k2v = 0, k2 = , (14) opz p dp pz2 dz2 z p v = tip, z = 0, O^p^l, (15) dv/dz = 0, z = 0, p > 1 , (16) embodies the torsional oscillations of a homogeneous and isotropic half-space which occupies the region z ^ 0. A disk of radius a is attached to it and is forced to execute torsional oscillations with period 2n/a). All lengths are made dimensionless with a as the standard length, so that the disk occupies the region z = 0, 0 ^ p ^ 1. The quantities d and p are respectively the density and shear modulus of the elastic material, while k is a dimensionless parameter. 6.7. EXAMPLES 121 It is easily verified from (14) that the function w(p, cp, z), w(p, where, as before, x = (p,cp,z) and § = (t9 Qpcoscp = t 2 2 V2 f exp/A-[p + t — 2ptcos((p — cpl)'] cos Set (pl—(p = ij/, proceed as we did in the relations (6.3.2)-(6.3.4), and obtain Qp = \ t(j){t)Kx(t,p)dt, 0 < p < l , (25) where In exp ik(p2 + t2- 2ptcos \jj)x/l K cos \j/ d\\i (26) ^P) = Yn (p2 + t2-2ptcos\l/)i/2 Finally, we use the identities [21] exp ik(p2 + t2 — 2ptcos \\J)X/2 pJo LP(P2 + t2~ 2pt cos i/01/2] dp, (27) (p2 + t2-2pcos\l/)t/2 where 2 2 f -i(k -P )*9 k>P y = 2 2 (P -k )*, P>k and 2 2 Jotp(p + t -2ptcosiljy^ = X (2-S0r)(cosnl/)Jr(pp)Jr(pt) (28) r=0 in (26), use the orthogonality of cosine functions, and derive 00 *i (', p) = j (plrUi iPP)Ji (P» dp ■ (29) 0 When co-»0, we get the corresponding integral equation for the steady rotation of the elastic half-space, and (29) becomes Kd*,P) = jj1(pp)Jl(pt)dp . (30) For the case of an annular disk with inner radius b and outer radius a, the integral equation that corresponds to (25) is of the form 6.7. EXAMPLES 123 a Qp = j t(j)(t)Kl(t,p)dt , b V-T1=0, T^O as x->oo, (35) where I = (5l7, the Kronecker delta. It follows by direct verification that the system (34)-(35) has the representation formulas 2 2 Ti = (1/8TI)[IV - gradgrad^] , Pl = -(l/87i)gradV (/>, (36) V40 = V2V20 = -87r<5(x-£) • (37) The appropriate solution of the biharmonic equation (37) is (j) = r = |x-S|.Thus, 2 Tx = (l/87r)[lV |x-^| -gradgrad|x-^|] , (38) 2 Pl = -(l/87r)gradV |x-^| . (39) The required integral equation formula now follows by taking the scalar product of (32) by Tx and of (34) by q and using the usual steps of 124 6 / PARTIAL DIFFERENTIAL EQUATIONS subtracting and integrating. In the integral so obtained, there occur terms T* V/? and q*Vp and they can be processed by using the identities V-(qPi) = q*Vpt , V.(/>T,) = iy V/> , (40) where we have used the results V*q = V«Tj =0. The final result is q(x) = -|[(B-'-)"T'-'-(Sf-")]<'5- (4I) s From equation (34) and the divergence theorem, it follows that, if q is constant on S, then equation (41) reduces to q(x) = -Jf-T,rfS, (42) s where dq { = dn~Pn (43) Finally, using the boundary condition (33),, we have the required integral equation ei = -Jf-TirfS. (44) Example 4. Steady Oseenflow. The dimensionless Oseen equations are ^dq/dx, = -Vp + V2q , V-q = 0 , (45) q = e! on S ; q->0 asx->oo, (46) where 0t = ua/v is the Reynolds number, v is the coefficient of kinematic viscosity, and a and u are the same quantities as defined in the previous example. The Green's tensor T and Green's vector p for this problem satisfy the system 2 StffTjdxx = -Vp + V T + I(5(x-S), (47) V-T = 0, T->0 as x->oo. (48) The corresponding representation formulas are T = (1 /in) [I V2 cj) - grad grad 0] , (49) 6.7. EXAMPLES 125 p = -(l/87r)grad[V2<£ - £50/5*,] . (50) 2 2 V (V - 0tdldxx)4> = -8TT<5(X-$) . (51) To solve equation (51) for V2(l/|x-^|) = -4TE5(X-$) (52) in it and obtain 2 2 2 V (V - 0td\dxx)<\> = 2V (l/|x-£|) . (53) Thus, if (j) satisfies V20 - Md^jdx, = 2/|x-$| , (54) then (51) is satisfied. On the other hand, we can also write (51) as 2 2 (V -^d/dxx)V (l) = -8TT<5(X-£) . (55) By setting V2(j) = il/ (56) and using the identity 2 2 (x, l) a{x 2 (V -ff )[e-* -* £| = e- *-^\y -2odldxx-]\l/ , (57) we can write (55) as (V2-a2)[^-CT(x,-^)iA] = -%ne-a{x'-^d{x-Z) (58) with o = 201. Now, observe that, by the nature of the Dirac delta aiXl 4l) function, the factor e~ ~ influences the equation only at xx =£l9 where its value is unity. Thus, (58) yields *(x,$) = V2tf> = j-i_exp|-Hr|x-^|-|^| (*,-(?,) (59) From (54) and (59), we obtain 8 s=\x-$\-(al\a\)(xl-Zl), (61) we have 126 6/ PARTIAL DIFFERENTIAL EQUATIONS dxx ds dxx we find that dcfr/ds =(\-e-^s)/\a\s or \a\s 4>=0/| q(x) = - --,n,.q *(T-q)n, dS, (64) s HZ-~ where nx is the xx component of the outward normal. When we use the boundary condition q=ex on S, equation (47), and the divergence theorem, we obtain e, = -J"T-f<£S, (65) s where f is defined by (43). Example 5. Heat conduction. The boundary value problem V2u-k2u = -p(x), xe R,, — = 0 (66) s embodies the solution of the heat conduction problem of an infinite expanse of material containing a cavity S on which the temperature gradient is zero. Our aim is to give the system (66) an integral-equation EXERCISES 127 formulation. For this purpose, we assume that u can be represented as the sum of a volume potential and a single-layer potential 471 u(x) = I pft)£(x;$) dV+ j * Rt S where £(x;^) = [exp( — k\x — §|)]/|x — ^| and a is an unknown source density. The next step is to take the normal derivative of both sides of equation (67), let x approach S, and use relation (6.6.8). The result is dE(x;%) P ft) —z dV - 2nc (x) + otf-^dS, (68) on which is a Fredholm integral equation of the second kind in o(x). The reader who is familiar with the theory of heat conduction is advised to formulate the corresponding problem of the composite medium into a Fredholm integral equation of the first kind. EXERCISES 1. Show that when S is the surface of a unit sphere (a) the solution of the integral equation 1 pzcoscp = a(puz{) |x — £| cos is a = (5/12)/V (cos0); (b) the solution of the integral equation 2 (\/2)pz cos(p = (3/8TT) \ PSicosOJlx-QcoscpidS s + JffO^zOlx-Sl^cos^ dS s > 1 1 is(T = (l/47r)[(3/2)/ 1 (cos0) + (7/15)P3 (cos^)]. In the above relations, (p,cp,z) are cylindrical polar coordinates. Hints: (i) Use the formula 128 6 / PARTIAL DIFFERENTIAL EQUATIONS where x = r<\r> , r< = min^,^), r> = max(r,r,) , cosy = cos0cos9X + sin0sin0, cos(cp — (px), while (r,(p,6) and (ru = a2-pl f Aa,6)d6 HP'm In J a1 - lap [cos (9 - 4. Discuss the single-layer and double-layer potentials in two- dimensional potential theory by starting with the formula £(x;0 = (l/2;r)log(l/|x-5|) instead of (6.2.2). 5. With the help of the results obtained in the previous exercise, prove that the solution of the interior Dirichlet problem in two dimensions can be written as *(X) = J[(C0S*)/|x-$|]T($) ws and dujdz are continuous across z = 0, a < p < oo , and ws satisfies the radiation condition, reduces to the integral equation F —i(p,^,0)= ta(i) \—A . ' , ' dq>xdt, 0 where a(/) = (l/47r)( wsLi = o+~wslz = o-)- This boundary value problem embodies the solution of the diffraction of a plane wave by a perfectly rigid, circular thin disk. 7. By following the method and notation of Example 3 in Section 6.7, prove that the integral representation formula for the velocity vector when the fluid is bounded by a vessel I is q(X) = -\[(^-pa}T-q{fn-nP)]dS- Substituting the boundary condition q=e{ on S gives the Fredholm integral equation 130 6 / PARTIAL DIFFERENTIAL EQUATIONS ex = -ff'TdS. s Here, T satisfies the boundary value problem V2T-gradp = I<5(x-£), V-T = 0, T = 0 on I. 8. The dimensionless equations of elastostatics are (A + /i)grad0 + /*V2u = 0, 6 = divu, where u (ut. i= 1,2,3) is the displacement vector, and X and pi are Lame constants of the medium. The above equations have been made dimensionless by introducing a characteristic geometric length a. Consider the uniform translation of a light, rigid body B with surface Sl9 which is embedded in an infinite homogeneous and isotropic medium. Prove that the corresponding Green's tensor T{ and Green's dilation vector 9X are given by the formulas _ 1 p + 3p Su X + ii (x.-Qjxj-jj)! "J 87r [_A+2^ |x-^| X+2/i |x-li|3 J' " 4nA + 2/xL|x-§|3J" Also prove that the integral representation formula is n^ = -\htti+(X+'i)en\'T s -u-L —- + (A + Ai)91n lidS. Using the boundary conditions u = (d0/a) e on S ; u -► 0 at oo ; where d0 is the magnitude of the translation and e is the unit direction along the translation of B, show that the Fredholm integral equation that embodies the solution of the above problem is (d0/a)e=-jf-TldS9 where f = p(daldri) + (X+ii)On. EXERCISES 131 Extend the above results to a bounded medium. 9. Proceed as in Example 1 of Section 6.3 and Example 1 of Section 6.7 and give the integral-equation formulation of these problems for a spherical cap. 10. The Schrodinger equation V2^(x) - (2m/h2)V(x)il/(x) + k2il/(x) = 0 with boundary condition that il/(x)exp( — iEt/h) represents an incident plane wave, with wave vector k0, as well as the condition that we have an outgoing wave as x-> oo, describes the quantum mechanical theory 2 2 2 of scattering by a potential V(x). Here, k = k0 = 2mE/h , and other quantities have their usual meaning. Use the method of Section 6.6 and prove that we can transform this scattering problem into the integral equation iA(x) = (exp/k0-x)-—2 ^^vmdQdv. Find its solution by the method of successive approximations (the approximation obtained in this way is called the Born approximation). SYMMETRIC KERNELS CHAPTER 7 7.1. INTRODUCTION A kernel K(s, t) is symmetric (or complex symmetric or Hermitian) if K(s,t) = K*(t9s), (1) where the asterisk denotes the complex conjugate. In the case of a real kernel, the symmetry reduces to the equality K(s,t) = K(t9s). (2) We have seen in the previous two chapters that the integral equations with symmetric kernels are of frequent occurrence in the formulation of physically motivated problems. We claim that if a kernel is symmetric then all its iterated kernels are also symmetric. Indeed, K2(s,t) = j K(s,x)K(x,t)dx = j K*(t,x)K*(x,s)dx = K2*(t,s). Again, if Kn(s,t) is symmetric, then the recursion relation gives Kn+1 (s, t) = j K(s9 x) KH(x, i) dx = j K*(t9x)K*(x,s)dx = Kn+l*(t,s) . (3) 132 7.1. INTRODUCTION 133 The proof of our claim follows by induction. Note that the trace K(s,s) of a symmetric kernel is always real because K(s,s) = K*(s,s). Similarly, the traces of all iterates are also real. Complex Hilbert Space We present here a brief review of the properties of the complex Hilbert space Furthermore, let us define the inner product by the bilinear form (1.6.1): {ct>^) = \Ht)r(t)dt. (5) Then, H is a linear and complex Hilbert space <£2{a,b) (or j£?2). The norm ||c/>|| as defined by (1.6.2), WI = ({|*WI2«a)* (6) is called the norm that generates the natural metric Also recall the definition of an j£?2-kernel as given by (1.2.2)—(1.2.4). Another concept that is fundamental in the theory of Hilbert spaces 134 7 /SYMMETRIC KERNELS is the concept of completeness. A metric space is called complete if every Cauchy sequence of functions in this space is a convergent sequence. If a metric space is not complete, then there is a simple way to add elements to this space to make it complete. A Hilbert space is an inner- product linear space that is complete in its natural metric. The complete ness of JS?2-spaces plays an important role in the theory of linear operators such as the Fredholm operator K, K(t> = JK(s9t)4>(t)dt. (10) The operator adjoint to K is K*il/ = j K*(t9s)\l/(t)dt . (11) The operators (10) and (11) are connected by the interesting relationship (K^M-faK**), (12) which is proved as follows: (Kfr i/0 = j V is) [ j K(s9t) (f> (t) di\ ds = j 0(/)[J K(s9t)ilf*(s)ds']dt = \Hs)i\K{t9s)r(t)dt-]ds = f = (&***). For a symmetric kernel, this result becomes (KM) = (4>9m9 (13) i.e., a symmetric operator is self-adjoint. Note that permutation of factors in a scalar product is equivalent to taking the complex conjugate, i.e., (0, Kef)) = {K(f)9 0)*. Combining this with (13), we find that, for a symmetric kernel, the inner product (Kcj), (j)) is always real; the converse is also true and forms Exercise 1 at the end of this chapter. 7.1. INTRODUCTION 135 An Orthonormal System of Functions Systems of orthogonal functions play an important role in the theory of integral equations and their applications. A finite or an infinite set {cj)k} is said to be an orthogonal set if ( (0, i*j9 I 1 » ' = J> Individual functions (j) for which |||| = 1 are said to be normalized. Given a finite or an infinite (denumerable) independent set of functions {1/^,1/^, ...,i/^,...}, we can construct an orthonormal set {!, (j>2, ...,$*,...} by the well-known Gram-Schmidt procedure as follows. Let cj)l = *Ai/||iAi||. T° obtain 2, we define w2(s) = ^2(j)-(^2»0i)0i • The function w2 is clearly orthogonal to <\>x\ thus, (j>2 is obtained by setting (j)2 = w2/|| w2||. Continuing this process, we have k-i M^ = ^k(s) - £ 0A*> <£/),• > We have, thereby, obtained an equally numerous set of orthonormal functions. In case we are given a set of orthogonal functions, we can convert it into an orthonormal set simply by dividing each function by its norm. Starting from an arbitrary orthonormal system, it is possible to construct the theory of Fourier series, analogous to the theory of trigono metric series. Suppose we want to find the best approximation of an arbitrary function ij/(x) in terms of a linear combination of an orthonormal set (i,2, •••>«)• By the best approximation, we mean that we choose the coefficients al5a2, ...,a„ such as to minimize or 2 || i// — X/=i <*i0i II > > what is equivalent, to minimize ||^ — ]T/=i at- r*|| . Now, for any ai, ...,an, we have 136 7/SYMMETRIC KERNELS 2 2 2 2 ll*-I«i*J = lhH + IIOM()-««l -IIGM,)l . (14) i= 1 i= 1 i= 1 It is obvious that the minimum is achieved by choosing a,- = (\j/, 0,) = av (say). The numbers at are called the Fourier coefficients of the function ij/(s) relative to the orthonormal system {(/>,}. In that case, the relation (14) reduces to n*-i«i*iii2 = iMi2-i> /=i which, for the infinite set {0J, leads to the Bessel inequality 2 i H2 < m • d7) /= i Suppose we are given an infinite orthonormal system {,0s)} in j£?2> and a sequence of constants {a,}, then the convergence of the series 2 Z*=i \ock \ is evidently a necessary condition for the existence of an j£?2-function f(s) whose Fourier coefficients with respect to the system Riesz-Fischer Theorem. If {,C?)} is a given orthonormal system of functions in j£?2 and {aj is a given sequence of complex numbers 2 such that the series £*=i |afc | converges, then there exists a unique function f(s) for which a, are the Fourier coefficients with respect to the orthonormal system {(f)t} and to which the Fourier series converges in the mean, that is, ILfa)-I«i*ill-0 as /i-oo. /=1 If an orthonormal system of functions ,- can be found in j£?2-space such that every other element of this space can be represented linearly in terms of this system, then it is called an orthonormal basis. The concepts of an orthonormal basis and a complete system of ortho- 7.1. INTRODUCTION 137 normal functions are equivalent. Indeed, if any of the following criteria are met, the orthonormal set { (a) For every function \j/ in j^2, ^ = £0Mi)& = 5>^i- (18) i i (b) For every function i// in S£2, IWI2 = EIGM,)I2. 09) /= 1 This is called ParsevaVs identity. (c) The only function ijj in j£?2 for which all the Fourier coefficients vanish is the trivial function (zero function). (d) There exists no function ifr in <£2 such that {ij/9 ( 0 , j*k9 jr(t)4>j(t) Ml = llr(t)Mt)4>*(t)dt\*. (22) The space of functions for which \<\>\r < oo is a Hilbert space and all the above results hold. Some examples of the complete orthogonal and orthonormal systems are listed below. V2 iks (a) The system (j)k(s) = (2n)~ e is orthonormal, where k is any integer - oo < k < oo. 138 7/SYMMETRIC KERNELS (b) The Legendre polynomials n 2 i Kd (s -\Y P0(s) = l , PH(s) = — , n = 1,2,. 2 n! as are orthogonal in the interval (-1,1). Indeed, f 0, j*k, (23) -J, l 2/(2*+1), j = k. (c) Let ak>fl denote the positive zeros of the Bessel function J„(s), k= 1,2,...; n> —1. The system of functions Jn{cnkfns) is orthogonal with weight r(s) = s in the interval (0,1): J (0, y ^ k, LsJB(ai/illj)yfl(aMj)ds= ] (24) o 14.+ 1 W' y = *. A Complete Two-Dimensional Orthonormal Set over the Rectangle a^s^b9c^t^d Let {0,(V)} be a complete orthonormal set over a ^ s ^ b, and let {(//,(/)} be a complete orthonormal set over c ^ t ^ d. Then, we claim y , that the set 01 (5)\\Ji(t), 1(^)^2(0» •••>^2(' ) Ai(0» •• is a complete two-dimensional orthonormal set over the rectangle a^s^b9c^t^d. The fact that the sequence of two-dimensional functions { 7.2. FUNDAMENTAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONS FOR SYMMETRIC KERNELS We have discussed the eigenvalues and eigenfunctions for integral equations in the previous chapters. In Chapters 2 and 4, we found that 7.2. EIGENVALUES AND EIGENFUNCTIONS 139 the eigenvalues of an integral equation are the zeros of certain determi nants. It may happen that no such zeros exist, so that the kernel has no eigenvalues. There are many kernels for which there are no eigenvalues. Consider, for example, the homogeneous equation (cf. Exercise 3, Chapter 2) In In g(s) = X I (sinscos t)g(t) dt = A(sin^) \ g(t)cost dt . (1) o o Its solution must clearly be of the form g{s) = A sin s. Substituting this in (1) yields In Asms = A(sins) A (sm t cos t) dt = 0 . o Thus, the kernel K(s91) = sin s cos t (0 ^ s ^ In, 0 ^ t ^ In) has no eigenvalues. For a symmetric kernel that is nonnull (i.e., that is not identically zero), the above phenomenon cannot occur. Indeed, a symmetric kernel possesses at least one eigenvalue. The proof is briefly discussed in Section 7.8. An eigenvalue is simple if there is only one corresponding eigen- function, otherwise the eigenvalues are degenerate. The spectrum of the kernel K is the set of all its eigenvalues. In this terminology, the above assumption states that the spectrum of a symmetric kernel is never empty. The following are some important properties of the symmetric integral equations X j K(s, t) Q (0 dt = f(s), or XKg = f; K(s, t) = K* (f, s). (2) 1. The eigenvalues of a symmetric kernel are real. Let X and (f)(s) be an eigenvalue and a corresponding eigenfunction of the kernel K(s, t). This means cj)(s)-XK(t)(s) = 0 . (3) Multiply (3) by (f)*(s) and integrate with respect to s from a to b and derive the relation 2 ||0(J)|| -A(^,0) = O or i= ||4>(*)||2/W,4>). (4) 140 7/SYMMETRIC KERNELS Since both the numerator and denominator are real, we have the required result. 2. The eigenfunctions of a symmetric kernel, corresponding to different eigenvalues, are orthogonal. If (/>, and <\>2 are eigenfunctions correspond ing, respectively, to the eigenvalues Xx and X2, we have fa-^Kfa =0, <\>2-X2K<$>2 = 0. (5) Since X2 is real, the second equation in (5) may be written as (\>2*-X2K* (\>2* = 0. Then, by suitable multiplication and integration it follows that h *2 [ f f - ^ ^{s)K*{s9t)^(t)dtds\ = (^-^Mi,^)- (6) But K(s, 0 = K*(t,s), and the left side vanishes identically, and because kx # X2, the right side becomes ((/>l9 (\>2) = 0. 3. The multiplicity of any nonzero eigenvalue is finite for every 2 symmetric kernel for which \\\K(s9t)\ dsdt is finite. Let the functions > De tne (/)1A(>y), 02ACO> •••> a{ = j K(s9t)uix(t) dt = X~'uik{s) be the series associated with the kernel K(s,t) for a fixed s. Applying Bessel's inequality to this series, we have f lAr&oi^^ia-yiM*)!2, J i which, when integrated with respect to s, yields JJl^^oi2*^^!^-1)2. (7) 7.2. EIGENVALUES AND EIGENFUNCTIONS 141 The right side of (7) is m(X~1)2, where m is the multiplicity of L Since the left side is finite, it follows that m is finite. 4. The sequence of eigenfunctions of a symmetric kernel can be made orthonormal. Suppose that, corresponding to a certain eigenvalue, there are m linearly independent eigenfunctions. In view of the linearity of the integral operator, every linear combination of these functions is also an eigenfunction. Thus, by the Gram-Schmidt procedure, we can get equally numerous eigenfunctions which are orthonormal. On the other hand, for different eigenvalues, the corresponding eigenfunctions are orthogonal and can be readily normalized. Combining these two facts, we have the proof of the above property. 5. The eigenvalues of a symmetric j£?2~kernel form a finite or an infinite sequence {Xn} with no finite limit point. If we include each eigen value in the sequence a number of times equal to its multiplicity, then ttt„-l)2< ff \K{s,t)\2 dsdt . (8) Let {uk(s)} be the orthonormal eigenfunctions corresponding to different (nonzero) eigenvalues Xt. Then, proceeding as in the proof of the property 3 and applying the Bessel inequality, we have 1 2 2 Ear ) < jj\K(s9t)\ dsdt< oo. Hence, if there exists an enumerable infinity of Xh then we must have 1 2 En=i (V ) < °°- It follows that lim(l/2l)^0 and oo is the only limit point of the eigenvalues. 6. The set of eigenvalues of the second iterated kernel coincide with the set of squares of the eigenvalues of the given kernel. Note that the symmetry of the kernel shall not be assumed to prove this result. Let X be an eigenvalue of K with corresponding eigenfunction (j)(s), that is, (I—lK)(j) = 0, where / is the identity operator. When we operate on both sides of this equation with the operator (I+XK), we obtain (I-X2K2) 2 (j>(s)-X JK2(s,t) 2 which proves that X is an eigenvalue of the kernel K2(s, t). 142 7 /SYMMETRIC KERNELS 2 Conversely, let fi = X be an eigenvalue of the kernel K2(s9t), with (I-XK){I+UC)4> = 0. (10) If X is an eigenvalue of AT, then the above property is proved. If not, let us set (I+kK)4> = 4>'{s) in (10) and obtain (I - XK) $'(s) = 0. Since we have assumed that X is not an eigenvalue of K, it follows that , (f) (s) = 09 or equivalently (1 +XK)(j) = 0. Thus, —X is an eigenvalue of the kernel K and the above property is proved. We can extend the above result to the nth iterate. The set of eigenvalues of the kernel Kn(s, t) coincide with the set of nth powers of the eigen values of the kernel K(s, /). 7. If At is the smallest eigenvalue of the kernel K, then \I\X,\ =max[|(^,0)|/||0||] = max[| jj K(s,t)(j>(t)ct>*(s)dtds\/\\cl>n 0D or equivalently, 1/|^ | =max(*M), || 0|| = 1. (12) This maximum value is attained when (j)(s) is an eigenfunction of the symmetric J^'kernel corresponding to the smallest eigenvalue. For proof, see Section 7.8. 7.3. EXPANSION IN EIGENFUNCTIONS AND BILINEAR FORM We now discuss the results concerning the expansion of a symmetric kernel and of functions represented in a certain sense by the kernel, in terms of its eigenfunctions and the eigenvalues. Recall that we meet a similar situation when we deal with a Hermitian matrix. For instance, if A is a Hermitian matrix, then there is a unitary matrix U such that U~l AU is diagonal. This means that, by transforming to an ortho- normal basis of the vector space consisting of the eigenfunctions of A, the matrix representing the operator A becomes diagonal. Let K(s, t) be a nonnull, symmetric kernel which has a finite or an 7.3. EXPANSION IN EIGENFUNCTIONS AND BILINEAR FORM 143 infinite number of eigenvalues (always real and nonzero). We order them in the sequence Al9 A29...,/>„,... (0 in such a way that each eigenvalue is repeated as many times as its multiplicity. We further agree to denumerate these eigenvalues in the order that corresponds to their absolute values, i.e., 0< |AJ ^ |A2| ^ ••• ^ K| ^ \Xn+l\ < .... Let fW^iW W4- (2) be the sequence of eigenfunctions corresponding to the eigenvalues given by the sequence (1) and arranged in such a way that they are no longer repeated and are linearly independent in each group corresponding to the same eigenvalue. Thus, to each eigenvalue Xk in (1) there corre sponds just one eigenfunction (j)k(s) in (2). According to the property 4 of the previous section, we assume that they have been orthonormalized. Now, we have assumed that a symmetric j£?2-kernel has at least one eigenvalue, say 2.{. Then ^(s) is the corresponding eigenfunction. It follows that the "truncated" symmetric kernel Ar(2>(j,0 = ^,0-Wi(j)*i*(0Mi] we is nonnull and it will also have at least one eigenvalue k2 ( choose the smallest if there are more) with corresponding normalized eigenfunction s even 2(s). The function (j)l(s)^ (\>i{ ) if ^i = ^i-> since (2 K \s,t)<\>x(i)dt = K(s,i) = 0. Similarly, the third truncated kernel Ki3)<3)/(s,t)P A =- VW<-Ki2 \s,t)-- ^W^2*W 2 = K(s,t)-^ &(*)&*(') 144 7/SYMMETRIC KERNELS gives the third eigenvalue A3 and eigenfunction (j)3(s). Continuing in this way, we end up with two possibilities: either this process terminates after n steps, that is, K(n +l)(s,t) = 0, and the kernel K(s, t) is a degenerate kernel, K(s, t) = > , (3) or the process can be continued indefinitely and there are an infinite number of eigenvalues and eigenfunctions. Note that we have denoted the least eigenvalue and the corresponding in) eigenfunction of K (s,t) as Xn and <£„, which are the «th eigenvalue and the «th eigenfunction in the sequences (1) and (2). This will be justified by Theorem 2 below. We shall examine in the next section whether the bilinear form (3) is valid for the case when the kernel K(s, t) has infinite eigenvalues and eigenfunctions. The following two theorems, however, follow readily. Theorem 1. Let the sequence {(\>k{s)} be all the eigenfunctions of a symmetric i?2-kernel K(s,t), with {Afc} as the corresponding eigen values. Then, the series A Xn n= 1 " converges and its sum is bounded by C\2, which is an upper bound of the integral f\K2(s,t)\dt. Proof. This result is an immediate consequence of BesseFs inequality. Indeed, the Fourier coefficients an of the function K(s, t), with fixed s, with respect to the orthonormal system {(/>„*(s)} are an = j K(s9t)(j)n(t) dt = Thus, applying BessePs inequality, we have y^#% o^oi^^c,2. (4) 7.3. EXPANSION IN EIGENFUNCTIONS AND BILINEAR FORM 145 Theorem 2. Let the sequence (f)n(s) be all the eigenfunctions of a symmetric kernel K(s, t), with {Xn} as the corresponding eigenvalues. Then, the truncated kernel K^»(s,t) = K(s,t)-2 has the eigenvalues Xn+U A„+2,..., to which correspond the eigen (n+1) functions <}>n+\{s), (t>(s) -XJ Kin + 1)(s, 00(0 dt = 0 (6) is equivalent to (f>{s)-X |tf(M)0(O* + *y ^(0,0J^ = O. (7) If, on the left side of this equation we set X = X} and (j)(s) = (j)j(s) j ^ n+\, then, in view of the orthogonality condition, we have cj>(s) - Atf0(j) + A V ^ (0,0J = 0 . (9) Taking the scalar product of (9) with (j)j(s),j ^ n, we obtain (<(>, 4>j) -W, 0,) + (AM.) (, ^) = o, (l 0) where we have used the orthonormality of the 0y. But (K(f), $/) = (0, A^7) 1 = Xj~ ((j), cj)j). Hence, (10) becomes ( (j)(s) - X J K(s, 0 0(0 A = 0 , (12) which means that 2 and K(s,~, At) = V> In Chapter 2, we found that every degenerate kernel has only a finite number of eigenvalues. Combining these two results, we have the following theorem. Theorem 3. A necessary and sufficient condition for a symmetric j£?2-kernel to be degenerate is that it have a finite number of eigenvalues. 7.4. HILBERT-SCHMIDT THEOREM AND SOME IMMEDIATE CONSEQUENCES The pivotal result in the theory of symmetric integral equations is embodied in the following theorem. Hilbert-Schmidt Theorem. If f(s) can be written in the form f(s) = JK(s9t)h(t)dt9 (1) where K(s, i) is a symmetric if2-kernel and h(i) is an j£?2-function, then f{s) can be expanded in an absolutely and uniformly convergent Fourier series with respect to the orthonormal system of eigenfunctions (7.3.2) of the kernel K: 7.4. HILBERT-SCHMIDT THEOREM AND CONSEQUENCES 147 The Fourier coefficients of the function f(s) are related to the Fourier coefficients hn of the function h(s) by the relations fn = hjkny hn = (K(\>n), (2) where Xn are the eigenvalues (7.3.1) of the kernel K. Proof. The Fourier coefficients of the function f(s) with respect to the orthonormal system {(f)n(s)} are where we have used the self-adjoint property of the operator as well as the relation kn Kcj)n = „. Thus, the Fourier series for f(s) is /w~i/„^) = f 7^). (3) The remainder term for this series can be estimated as follows: k = n+l k k = n+l k=n+l n + p oo i 2 / \ k = n+l k=l K From the relation (7.3.4), we find that the above series is bounded. Also, 2 because h(s) is an j£?2-function, the series £*=i hk is convergent and 2 the partial sum X*=«+i hk can be made arbitrarily small. Therefore, the series (3) converges absolutely and uniformly. It remains to be shown that the series (3) converges to f(s) in the mean. To this end, let us denote its partial sum as n i m iw = yr ^) (5) and estimate the value of \\f(s) — il/n(s)\\. Now, /(S)-^) = tt-2f^) (6) m=1 m (continued) 148 7/SYMMETRIC KERNELS +1 = ^_2^-^m(,) = ^" »/I, (6) where K(n+1) is the truncated kernel as defined in the previous section. From (6), we obtain 2 +1 2 n + ||/(5)-iAn(5)|| = ||*<" >/z|| = (K^"KK' "h) (n+1) in + 1) in+1) = (h9K K h) = (h,K2 h) , (7) where we have used the self-adjointness of the kernel K(n+1) and also (n+1) {n+1) (n+1) the relation K K = K2 . If we use property 6 of Section 7.2 and Theorem 2 of Section 7.3, we find that the least eigenvalue of the (n + 1) 2 kernel K2 is equal to A„+1 . Furthermore, according to property 7 of Section 7.2, we have 2 +1 1M„+1 = max[(/i, *2<" >A)/(M)] , (8) where we have omitted the modulus sign from the scalar product {n+l) (/z, K2 h), as it is a positive quantity. Combining (7) and (8), we have 2 ( +1) 2 !/(*)- *M\ = (A,*2 " A) < (M)M„+1 . Since A„+1->oo, we find that || f(s)-\l/n(s)\\ ->0 as n->oo. Finally, we use the relation II/-1H < ll/-^« + ll^-^ll , (9) where i// is the limit of the series with partial sum t/^, to prove that /= ij/. The first term on the right side of (9) tends to zero, as proved above. To prove that the second term also tends to zero, we observe that, since the series (3) converges uniformly, we have, for an arbitrarily small and positive e, \ll/n(S)-\l/(s)\ < 6, l/2 when n is sufficiently large. Hence, ||^n(s) — \j/(s)\\ < e(b — a) and the result follows. Remark. Note that we assumed neither the convergence of the Fourier series h(s) nor the completeness of the orthonormal system. We have merely used the fact that h is an j£?2-function. An immediate consequence of the Hilbert-Schmidt theorem is the bilinear form of the type (7.3.3). Indeed, by definition, 7.4. HILBERT-SCHMIDT THEOREM AND CONSEQUENCES 149 Km(s9t) = j K(s,x)Km_l(x,t)dx, m = 2,3,..., (10) which is of the form (1) with h(s) = Km_l(s, t); t fixed. The Fourier coefficient ak(t) of Km(s9 t) with respect to the system of eigenfunctions { m ak(t) = JKm(s,t)cl>k*(s)ds = lk- 4>k*(t) . It follows from the above theorem that all the iterated kernels Km(s, /), m ^ 2, of a symmetric j£?2-kernel can be represented by the absolutely and uniformly convergent series Km{s,t)= f V"&(*)&*«• (11) Jfe=l By setting s = t in (11) and integrating from a to b9 we obtain m £ V = [Km(s9s)ds = Am9 (12) where Am is the trace of the iterated kernel Km. Next, we apply the Riesz-Fischer theorem and find from (12) with m = 2 that the series A K k= 1 " converges in the mean to a symmetric J2?2-kernel K(s91)9 which, con sidered as a Fredholm kernel, has precisely the sequence of numbers {Xk} as eigenvalues. Definite Kernels and Mercer's Theorem A symmetric J^-kernel K is said to be nonnegative-definite if (Kcfr, (/>) ^ 0 for every if2-function 0; furthermore, K is positive-definite if, in addition, (Kef), ) = 0 implies is null. The definitions of nonpositive- definite and negative-definite symmetric kernels follows in an obvious manner. A symmetric kernel that does not fall into any of these four categories is called indefinite. 150 7/SYMMETRIC KERNELS The following theorem is an immediate consequence of the Hilbert- Schmidt theorem. Theorem. A nonnull, symmetric J^-kernel K is nonnegative if and only if all its eigenvalues are positive; it is positive-definite if and only if the above condition is satisfied and, in addition, some (and therefore every) full orthonormal system of eigenfunctions of K is complete. Proof, (a) From the Hilbert-Schmidt theorem, we have { f{s) = Kci> = ^ ^-4>n{s). (14) n=\ n The result of taking the inner product of (14) with cj) is ^T 1(0, Hence, if all Xn > 0, we have (A^>, $) ^ 0 for all 0. In addition, if any Xn l is negative, then (A^>„, „) = kn~ < 0. Thereby, the first part of the theorem is proved. (b) Let K be nonnegative-definite. From (15), it follows that (Kcj),^) = 0 if and only if (0, (/>„) = 0 for all n. Therefore, K will be positive- definite if and only if the vanishing of (0, $„) for all n implies Mercer's Theorem. If a nonnull, symmetric J^-kernel is quasi- definite (that is, when all but a finite number of eigenvalues are of one sign) and continuous, then the series IV1 (16) n= 1 is convergent and 7.5. SOLUTION OF A SYMMETRIC INTEGRAL EQUATION 151 ^ n=\ " the series being uniformly and absolutely convergent. Note that the continuity of the kernel is an absolutely essential condition for the theorem to be true. 7.5. SOLUTION OF A SYMMETRIC INTEGRAL EQUATION Let us use the Hilbert-Schmidt theorem to find an explicit solution of the inhomogeneous Fredholm integral equation of the second kind g(s)=f(s) + lj K(s,t)g(t)dt (1) with a symmetric i?2-kernel. It is assumed that k is not an eigenvalue and that all the eigenvalues and eigenfunctions of the kernel K(s91) are known and arranged as in (7.3.1) and (7.3.2), respectively. The first thing we observe is that the function g (s) — f(s) has an integral repre sentation of the form (7.4.1). As such, we can use the Hilbert-Schmidt theorem and write g(s)-f(s) = 2>* **(')• (2) where k= 1 with ck = j [.9 (s) - f{s)] fa* (s) ds = gk- fk (3) 9k = j Q(s) ck = XgJXk . (5) Since k is not an eigenvalue, from (3) and (5) we have ck = [A/(A4 - A)] h, gk = [At/(4 - m A • (6) Substituting the value of ck from (6) into (2), we derive the solution of the integral equation (1) in terms of an absolutely and uniformly con vergent series: 152 7/SYMMETRIC KERNELS ,(s)=f(s) + X^-^-4>Ak(s) (7) 4-1 or fit) dt. (8) *=i (A -K) Thus, the resolvent kernel F(s, t;X) can be expressed by the series T(Stt,k).fm*m (9) from which it follows that, the singular points of the resolvent kernel T corresponding to a symmetric i?2-kernel are simple poles and every pole is an eigenvalue of the kernel. The above discussion is based on the assumption that X is not an eigenvalue. If it is an eigenvalue, then it necessarily occurs in the sequence {Xk} and perhaps is repeated several times. Let X = Xm = Xm+ ! = •••= Xm.. For the indices k, different from m9m+l9...9m'9 the coefficients ck and gk in (6) are well-defined. However, if k is equal to one of these numbers, then/fc = 0. This means that the integral equation (1) is soluble if, and only if, the function f(s) is orthogonal to the eigenfunctions (f)m9(j)m+i, ..., f(s) = JK(s9t)g(t)dt9 (10) where the kernel K(s91) is a symmetric J£?2-kernel. We again assume that the sequence of eigenvalues {Xk} and corresponding eigenfunctions { fk = 9k/K or gk = Xkfk. (11) Because of the Riesz-Fischer theorem, there are only two possibilities: either (a) the infinite series IA2V (12) k=l 7.6. EXAMPLES 153 diverges and equation (10) has no solution, or (b) the series (12) con verges and there is a unique j£?2-function g(s) which is the solution of equation (10). This solution can be evaluated by taking the limit in the mean *,(*) = lim £ A^k(s). (13) n-* 7.6. EXAMPLES Example 1. Solve the symmetric integral equation 1 g{s) = (s+1)2 + j(st+s2t2)g(t)dt. (1) The eigenvalues and eigenfunctions for this symmetric kernel can be found by the method of Chapter 2 (see Example 4 in Section 2.2): (2) *2 = *. 4>2(*) = [0/72)71°] *2- /, = f(t2 + 2t+l)(ij6)tdt = ij6, -l (3) I 2 2 f2 = J (t + 2t+ l)(WIO)f * = (8/15)710 . Thus, '"-^(M'^G^y^ or g(s) = (25/9) 5P22 + 6^+1. (4) Note that, in equation (1), X = 1, which is not an eigenvalue. Example 2. Solve the symmetric integral equation 154 7/SYMMETRIC KERNELS g(s) = s2 + 1 + f J (st + s2t2)g(s) ds . (5) Here, A = Al=%, so we shall have the indeterminate form 0/0 in one of the coefficients. Fortunately, the function (s2 + l) is orthogonal to the eigenfunction (i^/6)^, which corresponds to the eigenvalue f. Following the procedure of Section 7.5, we obtain I 2 2 /i = 0 , f2 = j (t + l)(yV0)t dt = (8/15) VIO . (6) -1 Thus, the required solution is , x 3 (8/15) /TO (\ ,-\ 2 or g(s) = 5s2 + cy+ 1 , (7) where c is an arbitrary constant. Example 3. Solve the symmetric integral equation g(s)=f(s) + XJk(s)k(t)g(t)dt. (8) If we write j k{s)k{i)k{i) dt = (J [fc(0]2 *) *(*), we observe that h = i/jmtfdt (9) is an eigenvalue. The corresponding normalized eigenfunction is 2 The coefficient/! has the value /i = (/ [*(0]2 *}-* J7(o*« * • (i i) Thus, for A ^ Xu the solution is 7.6. EXAMPLES 155 g(s) = Wiltti-W g(s) = (ik(s) jf(s)k(s) dsj{\-X J lk(s)¥ ds\) +f(s) . (12) On the other hand, if 2 X = Ax = l/flk(sy] ds9 then/(V) must be orthogonal to i(V), and in that case the solution is g(s) = f(s) + ck(s), c an arbitrary constant. (13) Example 4. Solve the symmetric Fredholm integral equation of the first kind JK(s,t)g(t)dt=f(s) (14) o where K(s,t)=\ (15) [ (\-s)t, s > t. Recall that in Example 2 of Section 5.3 we proved that the boundary value problem dj_y +Xy = 0, y(0) = y(l) = 0 (16) ds 2 is equivalent to the homogeneous equation 1 g(s) = XJK(s9t)g(t)dt. (17) o The eigenvalues of the system (16) are 2 2 2 i1 = 7r , 22 = (2TT) , A3 = (3TT) , ..., and the corresponding normalized eigenfunctions are ^sniTLS, ^2sin27LS, y/2 sin 3ns, .... (18) Therefore, 156 7 / SYMMETRIC KERNELS /* = y/2J (sin knt)f(t) dt, (19) o and the integral equation (14) has a solution of class S£2 if and only if the infinite series k=l k=\ converges. Example 5. Solve Poisson's integral equation In „~. - r i g((x)d(x r UK f(0) 2TT J' 1- 2p[cos(0-a)]+p2 ' o 0 ^ 0 < 2TT ; 0 < p < 1 . (20) Here, the symmetric kernel K(9, a) can be expanded to give K(0,*) = [(1-P2)/2TT]{1 -2p[cos(0-a)] + P2}"1 = (1/2*) + (1/TT) £ pkcos[*(0-a)] . (21) k=\ It is a matter of simple verification that, by using the expansion (21), one gets ffi K(9, a) da = 1, or In j K(0,OL)(2n)-l/2 d In cos cos K(6, a) . ntxdx = pn . «0, w = 1,2,3,... , sin sin o we have ^2*-i =^2fc = P"k ; 02k-IW = H"** cos fey ; (22) 02it('s) = ft 1/2 sin fes , fc = 1,2,3,... . 7.7. APPROXIMATION OF A GENERAL J£?2-KERNEL 157 We can now readily evaluate the coefficients/fc in the series (7.5.12), and it emerges that the integral equation has an J*?2-solution if and only if the infinite series A p2" ' /!= 1 r where 2n In an = (\/n) j f(0)cosn6d6 , bn = (1 jn) j f(6)sinnO dO (23) o o converge. 7.7. APPROXIMATION OF A GENERAL J2VKERNEL (NOT NECESSARILY SYMMETRIC) BY A SEPARABLE KERNEL In Section 2.5, we approximated an analytic kernel s(est— 1) by a separable kernel. In this section, we show that we can approximate every ^-kernel in the mean by a separable kernel. The proof rests on the availability of a two-dimensional complete orthonormal set as discussed at the end of Section 7.1. Let K(s, i) be an if2-kernel and let (i/^C?)} be an arbitrary, complete, orthonormal set over a ^ s ^ b. Then, the set {i//t(s) ij/j* (t)} is a complete orthonormal set over the square a ^s,t ^ b. The Fourier expansion of the kernel K(s, t) in this set is K{sj)= f KtjMsWj*®, 0) where the Ktj are the Fourier coefficients Ku = jj KfrtWisWjO) dsdt . (2) Parseval's identity gives \\\K{s,t)\2dsdt= t l*yl2- (3) J J i,j= 1 Now, if we define a separable kernel k(s, i) as 158 7/SYMMETRIC KERNELS k(s9t) = £ tf^WlO, we find after a simple calculation that 2 2 \\\K(s9t)-k(s9t)\ dsdt = £ [Kyi . (4) J J U=n+ 1 But the sum in (4) can be made as small as we desire by choosing a sufficiently large n because the series (3) is convergent by hypothesis. This proves our assertion. 7.8. THE OPERATOR METHOD IN THE THEORY OF INTEGRAL EQUATIONS In this section, we shall show briefly how we can treat a Fredholm integral equation from the standpoint of present-day functional analysis. We have already analyzed the properties of a function space in Section 7.1. We now note that the transformation or operator K9 K *(0i +4>i) = *0i + Hi , *(«*) = «*# • The operator K is called bounded if \\K(j)\\ < M||0|| for an J^2-kernel K(s9t), an j£?2-function j,(s) = K 2 2 2 2 |*M*)| = \$K(s9t) |hH = ||Jty|| < H\\ [jj\K(s,t)\2dsdt]*, which implies that m^[jj\K(s,t)\2dsdt]^, (3) as desired. A rather important concept in the theory of linear operators is the concept of complete continuity. An operator is described as completely continuous if it transforms any bounded set into a compact set (a set S of elements (/> is called compact if a subsequence having a limit can be extracted from any sequence of elements of S). Obviously, a completely continuous operator is continuous (and hence bounded), but the converse is not true. Furthermore, any bounded operator K whose range is finite- dimensional is completely continuous because it transforms a bounded set in S£2{a9 b) into a bounded finite-dimensional set which is necessarily compact. Many of the integral operators that arise in applications are completely continuous. For instance, a separable kernel K{s, t), K(s,t) = £at(s)bl{t), /= 1 where a-Xs) and 6,(7) are ^-functions, is completely continuous, as can be proved as follows. Indeed, for each J^2-function Kg = \\t a^bMgit)] dt = £ c.a^s) , that is, the range of K is a finite-dimensional subspace of <£2{a,b). In addition, 160 7/SYMMETRIC KERNELS l|Affll = ||Ec,fl((*)||< Ek-lkll i= 1 /= l < I M f l*.(OI |flf(OI dt. (4) 1=1 J Applying the Schwarz inequality in (4), we have \\Kg|| ^ Af ||0|| , (5) where A/ = Z/=i \Wi\\ II^11- This means that K is a bounded operator with finite-dimensional range and hence is completely continuous. We can use this result to prove that an j£?2-kernel K(s, t) is completely continuous. We need only the theorem that, if K can be approximated in norm by a completely continuous operator, then A is completely continuous. If we assume this theorem, then our contention is proved because an j£?2-kernel can always be approximated by separable kernels, as shown in the previous section. Next, we prove the interesting result that the norms of K and of its adjoint ** are equal. To this end, we appeal to the relation (7.1.12): (*0,tff) = (0,**0), (6) which holds for each pair of j£?2-functions , i//. Substituting \j/ = Kcj) and applying the Schwarz inequality, we obtain (K4>,K4>) = {4>,K*K4>) ^ ||0|| ||tf*tf0|| , where we have used the fact that (Kcj), Kcj)) is nonnegative real number. Hence, \\K which implies (I/WII2 = vet,*) < wm u\\ < I^II ur. (?> From (3) and (7), we derive an upper bound |1M| < [jJK(s,t)2dsdty>. (8) When the j£?2-kernel is also symmetric, we can use the following result from the theory of operators: If K is a symmetric and completely continuous operator, at least one of the numbers \K\ or — \\K\\ is the reciprocal of an eigenvalue of K and no other eigenvalue of K has smaller absolute value. By recalling the definition of \K\ and the fact that a symmetric j£?2-kernel generates a completely continuous operator, we have proved property 7 of Section 7.2. In the process, we have also proved the existence of an eigenvalue. Suppose we have found the first eigenvalue kx and corresponding eigenfunction 7.9. RAYLEIGH-RITZ METHOD FOR FINDING THE FIRST EIGENVALUE Let us take a real, nonnegative, and symmetric j£?2-kernel K. We have found that the smallest eigenvalue X{ is characterized by the extremal (or variational) principle 162 7/SYMMETRIC KERNELS 1/Aj = max (A^, 0) , = 1 . (0 The Rayleigh-Ritz method rests on selecting a special class of trial functions of the form (j> = £,-=i a^Cs), where {ij/i(x)} is a suitably chosen set of linearly independent functions and {a,} are real numbers. The relation (1) implies that, to obtain a close approximation for A,, we must maximize the function (K^(t>) = (K\£ atUs)], £ «,*,(*)) = £ K^M , (2) v L/=l Ji=l 7 /,* subject to c a a |i(af^)|| = Z a i * = ! > (3) where Kik = Win) = OAP^W ; cifc = 0A<,W = W^) s (4) are known quantities. Thereby, we have transformed the extremal problem (1) into an extremal problem in the advanced calculus of several variables al9 ...,a„. We use the method of Lagrange multipliers and set ® = Z(*fcai"ak-ffCJkaJa*)> (5) where a is an undetermined coefficient. The extremal values of a; are determined from the equations SO/5af = 0; Z *»*a*-°" Z c/ka* = °> ' = J'--'" • (6) This linear and homogeneous system of equations in ctu. ,a„ will have a nontrivial solution if and only if the determinant ^11-^11 ^12-^12 ••' K\n- K C n\-° nl Knl-OCnl ••• Knn-GCm Also note that, by multiplying (6) by af and summing on /', one obtains The determinant (7), when expanded, yields an Azth-degree polynomial 7.9. RAYLEIGH-RITZ METHOD FOR FIRST EIGENVALUE 163 in a which can be shown to have n real, nonnegative but not necessarily distinct roots. Let at be the maximum of these roots. Then, from the above discussion, we infer that al ^ \jXx. This usually gives a good approximation to Xx. When we solve the equations (6) for the vector a (al9 ...,a„) with a = a1 and evaluate 0, i = l,2,...,« (8) k=l and *n- ^12 *i. = 0, (9) ^»1 ^«2 Km-a respectively. We illustrate this method by two examples. Example 1. Find the first eigenvalue of the integral equation g(s)-XJ K(s,t)g(t)dt = 0, (10) where (M2-0, s < t, K(s,t) = (11) s > t, which can be shown to be a positive kernel. This integral equation can be proved to be equivalent to a simple ordinary differential equation (see Example 2 in Section 5.3. The exact value of the smallest eigenvalue is 4.115.) To apply the Rayleigh-Ritz procedure, we take two trial functions: ^AnC?) = y/2sinnns , n = 1,2 , (12) which are orthonormal in the interval (0,1). Proceeding as above, we have 2 2 2 KX1 = 2/TT , Kl2 = K21 = - 1/2TT , K22 = l/ln . (13) 164 7 / SYMMETRIC KERNELS The relation (9) gives for this special case 2 (2/TT2) - a - l/2;r = 0 . (14) -1/27C2 (l/27T2)-(7 2 2 The largest root of (14) is <7 = 2.15/TT . Thus, Xx~ n /2A5 = 4.59. By including more functions in the sequence (12), we can improve the approximation progressively. Example 2. Find an approximation for the smallest eigenvalue of the positive-definite symmetric kernel s < t (15) s > t in the basic interval (0,1). To use the Rayleigh-Ritz method, we take two trial functions lM*)= 1, *2(') = (2*-l), (16) which are orthogonal but not orthonormal [they are Legendre poly nomials P0(2s — 1), P^ls— 1)]. Thus, c C\\ = 1 , \i = c2l = 0, c22 = 1/3 , *n = 1/3 , K12 = K21 = 1/12, K22 = 1/30 . Substituting (17) in (7) and evaluating the determinant, we have 2 ,/2 <7 -(13/30)<7 + (l/80) = 0. The largest root is ox = (1/60)[13 + (124) ]. Thus, the smallest eigenvalue is X{ = 1 /ax ~ 2.4859, which compares favorably with the exact value 2.4674. EXERCISES 1. Show that, if (K(j>9 0) is real for all 0, then A^is a symmetric Fredholm operator. 2. Determine the iterated kernels for the symmetric kernel 00 K(s,t) = YJ k~l sinknssinknt . EXERCISES 165 3. Show that the kernel K(s, t),0< s9t^\9 -i)9 s < /, K(s,t) = -s)9 s > t9 has the bilinear form 00 sinA:7LSsin/:7c/ K(s,t) = 4. Use the result in Exercise 3 and show that 00 1 2 V = ~6 ' n— i 5. Consider the eigenvalue problem -1 Differentiate under the integral sign to obtain the corresponding differential equation and boundary conditions. Show that the kernel of this integral equation is positive. 6. Determine the eigenvalues and eigenfunctions of the symmetric kernel K(s9 i) = min(s, /) in the basic interval 0 ^ s, t ^ 1. 7. Use the Hilbert-Schmidt theorem to solve the symmetric integral equations as given in Examples 1-3 and 6 in Section 2.2. 2 3 8. Apply the Gram-Schmidt process to orthogonalize \9s,s ,s in the interval — 1 ^ s ^ 1. Use this result to find the eigenvalues and eigen functions of the symmetric kernel K(s, t) = l+st + s2t2 + s3t3. 9. Consider the kernel K(s91) = log[l — cos(^ — f)], 0 ^ s9 t ^ In. Show that (a) it is a symmetric j£?2-kernel; (b) the following holds: f(s K{s9t) = -(log 2) + 21og[l-e -'>] ^ >0 cosCOS ns nS cos COS nt nt ^ \r^ sin ft?si n nl = -(log2) - 2 > 2 ^ n n= 1 and (c) its eigenvalues are XQ = — l/(27rlog2), Xn = —n/2n9 n = 1,2,. 166 7 / SYMMETRIC KERNELS with eigenfunctions 0(s) = C, $n(s) = A cos ns+B sin ns, where A, By and C are constants. 10. By combining Sections 5.5 and 7.4, show that the eigenfunctions of any self-adjoint differential system of the second order form a complete set. 11. Prove that, for a square-integrable function, the Fourier transform preserves norms. SINGULAR INTEGRAL CHAPTER 8 EQUATIONS 8.1. THE ABEL INTEGRAL EQUATION An integral equation is called singular if either the range of integration is infinite or the kernel has singularities within the range of integration. Such equations occur rather frequently in mathematical physics and possess very unusual properties. For instance, one of the simplest singular integral equations is the Abel integral equation s M = jl0(t)Ks-tr]dt, 0 u f(s)ds ds g{t)dt (2) I (us)1-' (u-sy-'j (s-ty The double integration on the right side of the above equation is so written that first it is to be integrated in the t direction from 0 to s and then the resulting single integral is to be integrated in the s direction from 0 to u. The region of integration therefore is the triangle lying below the diagonal s = t. We change the order of integration so that we first integrate from s = t to s = u and afterwards in the / direction from t = 0 to / = u. Equation (2) then becomes U f(s)ds ds U g(i)dt (3) s)1 sy-*(s-ty (0,0) Figure 8.1 To evaluate the integral u C ds jiu-sy-'is-ty 8.1. THE ABEL INTEGRAL EQUATION 169 one sets y = (u—s)/(u — t), and obtains u 1 j(u-sy-1(s-ty*ds = jy'-'il-yydy = n/sinan , / o where we have used the value of the Eulerian beta function B(a, 1 —a) = 7r/sin can. Substituting this result in (3), we have f As)ds sin arc g(t)dt, (us)1-' which, when differentiated with respect to u, and then changing u to t, gives the required solution: sin arc d .»-==£[//«(.-*-*] (4) The integral equation (1) is a special case of the singular integral equation [18] = f 9(t) dt As) J LHs) - Km , 0 < a < 1 , (5) where h(t) is a strictly monotonically increasing and differentiable function in (a,b), and h'(i) # 0 in this interval. To solve this, we con sider the integral s h'(u)f(u)du Jilh(s)-h(u)V-" and substitute for/(w) from (5). This gives g(i)h'(u)dtdu J. lh(u)-h(t)Tlh(s)-h(u)y-" which, by change of the order of integration, becomes 170 8/SINGULAR INTEGRAL EQUATIONS s s h'(u) du The inner integral is easily proved to be equal to the beta function B(a, 1 —a). We have thus proved that a a and by differentiating both sides of (6), we obtain the solution sin arc d [ h'(u)f(u)du '(') = —*J[A<0-AM]1-- (?) Similarly, the integral equation b g{t)dt A*) = j{u,""Z.™. 0 8.2. EXAMPLES Example 1. Solve the integral equation 5 = Cg(t)dt J(*-0'/2' (l) Comparing this with integral equation (8.1.1), we find that/(V) = s, a = 1/2. Substituting these values in (8.1.4), there results the solution: 0 -;|[-f<-+W-tf'I (2) n dt |_3 J ft Example 2. Solve the integral equation s g(t)dt 0^a Comparing (8.1.5) and (3), we see that a = 1/2, and h(t) = 1 —cos/, a strictly monotonically increasing function in (0, ri). Substituting this value for h{u) in (8.1.7), we have the required solution \_d_ (sin\u)f(u)du 1 g(t) = -u , a < t < b . (4) ndt (cos u - cosO/2J 172 8/SINGULAR INTEGRAL EQUATIONS Similarly, the integral equation b g(t)dt f(s) = 0^a (6) n dt LJ (cosr - cosw)/2J Example 3. Solve the integral equations ( t (a) f(s) =[^2 ^2 y, 0 < a < 1 ; a < s < b , (7) and D tu\ «\ f S(t)dt < a < 1 ; a < s < b . (8) From (8.1.5) and (7), we find that h(i) = t2, which is a strictly mono- tonic function. The solution, therefore, follows from (8.1.7): 2sina7r d ufju) du 0(0 = 2 2 a < t < b . (9) n dt (t -u y-* Similarly, the solution of the integral equation (8) is b 2 sin canan d d C uf(u) du 9(0 = —^ j-r— , a < t < b . (10) n dt) (u2-t2)1 * n at t The results (9) and (10) remain valid when a tends to 0 and b tends to + oo. Hence, the solution of the integral equation ' 00) dt 2 2 0 < a < 1 , (ID /w = (s -t )*' 8.3. CAUCHY PRINCIPLE VALUE FOR INTEGRALS 173 IS t 2 sin arc d f uf(u)du g{t) = (12) ^rdt)w^r*'o Similarly, the solution of the integral equation m-\$$r- 0<" 8.3. CAUCHY PRINCIPAL VALUE FOR INTEGRALS The theory of Riemann integration is based on the assumption that the range of integration is finite and that the integrand is bounded. For the integration of an unbounded function or an infinite range of integration, the concept of improper integrals is introduced. Consider a function f(s), defined in the interval a ^ s ^ b, which is unbounded in the neighborhood of a point c,a < c < b, but is integrable in each of the intervals (a, c — e) and (c + rj, b) where e and rj are arbitrary, small positive numbers. Then, the limit b c—e b jf(s) ds = lim [ J f(s) ds + jf(s) ds] , (1) if it exists, is called the improper integral of the function f(s) in the range (a, b). Here, it is implied that s and rj tend to zero independently. But it may happen that the limit (1) does not exist when e and rj tend to zero independently of each other, but it exists if e and rj are related. The classic example is the function/^) = \/(s — c), a < c < b\ the limit (1) in this case is 174 8/SINGULAR INTEGRAL EQUATIONS c-e b ds ds b — c e = log + log- + s — c c — a rj s — c C + tJ If e and rj tend to zero independently of each other, then the quantity \og(s/rj) will vary arbitrarily. However, if e and rj are related, then the above limit exists. In the special case e = //, this limit is b ds i ~bC = log (2) I s—c c — a and is called the Cauchy principal value or Cauchy principal integral. The same definition applies to a general function f(s). The Cauchy principal value of a function f(s) that becomes infinite at an interior point x = c of the range of integration (a, b) is the limit c — e b lim(J + \)f{s)ds, (3) £_+0 a c+e where 0 < 8 ^ min(c — a,b — c) . b b Such a limit is usually denoted as P \ af(s) ds or \* af(s) ds. We shall use the latter symbol in the sequel. A similar definition for the Cauchy principal value is given for integrals with an infinite range of integration. For instance, the limit (f(s)ds= lim lf(s)ds J J „ A->oo A may not exist when A and B tend to infinity independently of each other, but the limit exists when A = B. This limit, A liim (f(s)ds, (4) J A^co-►0 0 _AA is called the Cauchy principal value. The limits (3) and (4) are also called singular integrals. If a function f(s) satisfies certain regularity conditions, then the above-mentioned singular integrals exist. One such concept of regularity 8.3. CAUCHY PRINCIPLE VALUE FOR INTEGRALS 175 is the Holder condition. A function f(s) is said to satisfy the Holder condition if there exist constants k and a, 0 \Asi)-As2)\ < k\Sl-s2\'. (5) Such a function is also said to be Holder continuous. The special case a = 1 is often called the Lipschitz condition. It is not hard to prove that, if f(s) is Holder continuous, then the singular integral *b \U(i)l(t-sy]dt (6) a exists. Indeed, the relation (6) can be split as b b dt As) + MO-/W*. (7) t-s t-s The first integral has the principal value as proved by the relation (2). In the second integral, the integrand is such that < k\t-sY t-s Therefore, this integral exists as an improper integral for a < 1 and as a Riemann integral for a = 1. The function fx(s) defined by the singular integral (6), * has the following property, which we state without proof. If f(s) is Holder continuous with exponent a, a < 1, then J\ (s) is also Holder continuous in every closed interval (al,bl), where a < a± < x ^ bx < b. When/(X) is Holder continuous with a= 1, then/i (a) is Holder con tinuous with exponent /?, which is an arbitrary positive number less than unity. The Holder condition can be extended to functions of more than one variable. For example, the kernel K(s91) is Holder continuous with 176 8 /SINGULAR INTEGRAL EQUATIONS respect to both variables if there exist constants k and a, 0 < a ^ 1, such that \K(sutl)-K(s29t2)\ Figure 8.2 small circle of radius s with center at c. Let LE denote the part of the contour outside this circle. If a complex-valued function/(z) is integrable along L£, however small the positive number e, then the limit lim f(z) dz , if it exists, is called the Cauchy principal value and is denoted as j f(z)dz or L P j f(z)dz. L We shall be interested in the contour integrals of the Cauchy type, 8.4. CAUCHY-TYPE INTEGRALS 177 J"L/W/(T-*)]A, (9) L in the sequel. It is known in the theory of functions of a complex variable that, if f{z) satisfies the Holder condition l/(*i)-/(*2)l where zl9z2 is any pair of points on L, while k and a are constants such that 0 < a ^ 1, then the integral (9) exists for all points z on the curve L, except perhaps its end points. The function fx (z) defined by the integral (9) is also Holder continuous, with similar properties as given for the case of the corresponding real functions. The definition (10) can be extended to complex-valued functions of more than one variable as was done for the real-valued functions above. Incidentally, the function/(T) occurring in the integral (9) is called the density of the Cauchy integral. 8.4. THE CAUCHY-TYPE INTEGRALS The integral equation 2m j x — z where L is a regular curve, is called a Cauchy-type integral. We shall first study the case when L is a closed contour. For the discussion of the integral equation (1), we need a result from the theory of complex- valued functions, which we state without proof. Let g{x) be a Holder-continuous function of a point on a regular closed contour L and let a point z tend, in an arbitrary manner, from inside or outside the contour L, to the point t on this contour; then the integral (1) tends to the limit [3, 15] i^d, (2) /*M-J»M + 5S x — t 178 8 /SINGULAR INTEGRAL EQUATIONS or /-<0 = -5*(0 + ^ — dx , (3) x — t respectively. The formulas (2) and (3) are known as Plemelj formulas. It is interesting to compare them with the formulas (6.2.10) and (6.2.11). Incidentally, we follow the standard convention of counterclockwise traversal of the closed contour L. This means that the first boundary value/* (t) relates to the value of the Cauchy integral inside the region bounded by L, while the second boundary value/"(f) relates to the value in the outside region. Let ,l(0 i f'JL)A, (4) 2ni x — t M (5 > -*i)i=i*2ni J x — t > L be two singular integrals, where g and gx are Holder-continuous functions and L is a closed contour. Can we compound these two integrals and obtain an iterated integral connecting the functions g2 and gl The answer is in the affirmative. To prove this assertion, consider the Cauchy-type integrals 2ni j T — Z and L A(z) = ^-.\ — dx. (7) 1 fg.(T) 2ni J T —z Using the Plemelj formula (2) and the integrals (6) and (7), we obtain the limiting values * /+(0 = ^(0 + ^-.f —A, (8) 2 271/J T — t L 8.4. CAUCHY-TYPE INTEGRALS 179 + W + (9) /i (0=2 01 2ni J x — t Comparing the relations (4) and (8) on one hand and the relations (5) and (9) on the other hand, we obtain + gl(t)=f (t)-$g(t), (10) + 92«)=fl (t)-$9l(t). 01) Substituting (10) in (7) yields /+(T) /i(z) = — f rfr ' dx (12) T — Z 2ni J T — Z Ani The value of the first integral in equation (12) is/(z) because its density /+ (T) is the limiting value of the function/(z), which is regular inside L, and therefore we can use the Cauchy integral formula. The second integral is one-half of the integral in (6). Hence, ft (z) = \f(z), which implies that /i + (0 = i/+(0- 03) From (10), (11), and (13), it follows that 9i(t) = i/+(0 - ±[/+(0-i<7(0] = iff(0 • (14) Finally, from (4), (5), and (14), we have the required iterated integral equation: 1 f dx, | (15) T —T, 4 L L the so-called Poincare-Bertrand transformation formula. It is interesting to note that, in the formula (15), it is not permissible to change the order of integration. Indeed, if we change the order of integration, then the left side of (15) gives 1 C dxx g{T)dx (16) {Inif 180 8/SINGULAR INTEGRAL EQUATIONS But where we have used the Plemelj formula (2), which gives for the present case L L Thus, the relation (16) is equal to zero and not \g(t) as in (15). 8.5. SOLUTION OF THE CAUCHY-TYPE SINGULAR INTEGRAL EQUATION (j) CLOSED CONTOUR. The problem is to solve the integral equation of the second kind ag(t)=f(t)- -. ;dT, (1) Til x — t where a and b are given complex constants, g (T) is a Holder-continuous function, and L is a regular contour. A fortunate aspect of this integral equation is that it can be solved simultaneously for the cases a # 0, and a = 0, so that the solution of the integral equation of the first kind follows as a limiting case. To solve (1), we write it in the operator form Lg = ag(t) + -\^dx=f(t), (2) ni ' - * and define an "adjoint" operator 8.5. SOLUTION OF THE CAUCHY-TYPE 181 M(j) = a(j)(t) . dx . (3) 71/ x-t From (2) and (3), it follows that L ft (4) 71/ Using the Poincare-Bertrand formula (8.5.15) and after a slight simplifi cation, equation (4) becomes 9(0 = 2 lM 2 2 ■■[-.*■ (5) a -b (a -b )ni i) x-t where it is assumed that a2 — b2^ 0. Substituting (5) back in the integral equation (1), it is found that the function g{t) indeed satisfies the original integral equation. The solution of the Cauchy-type integral equation of the first kind, At) (6) 71/ J X — t follows by setting a = 0 in (5): g(t) (7) bniJ x — t For 6=1, equations (6) and (7) take the form V(t) dx = 9(t) (8) raj t-/ ni x-t 182 8 /SINGULAR INTEGRAL EQUATIONS which displays the reciprocity of these relations. (ii) UNCLOSED CONTOURS AND THE RIEMANN-HILBERT PROBLEM. The analysis of the case (i) is based on the application of the Poincare-Bertrand formula. When the contour L is not closed, then this formula is not applicable and new methods have to be devised to solve the integral equation (1). However, the Plemelj formulas (8.4.2) and (8.4.3) are valid for an arc also when we define the plus and minus directions as follows. Supplement the arc L with another arc L' so as to form a closed contour L + L''. Then, the interior and exterior of this closed contour stand for the plus and minus directions. Thus, we have + 9(r) , / w = ^w + 2^. ax , (9) x — t fffW f-{t) = -\g{i) + ^-. dx , (10) 2 2ni for an arc L. These formulas can also be written as 0(O=/+(O-/~(O, (11) 1 (12) ni x — t Now, suppose that a function w(t) is prescribed on an arc L and that it satisfies the Holder condition on L. It is required to find a function W(z) analytic for all points z on L such that it satisfies the boundary (or jump) condition W+{i)-W~{i) = w(t), teL. (13) The formula (11) obviously helps us in evaluating such a function W(z). The problem posed in (13) is a special case of the so-called Riemann- Hilbert problem, which requires the determination of a function W{z) analytic for all points z not lying on L such that, for / on L, W+{t)-Z(t)W-(i) = w(t), (14) 8.5. SOLUTION OF THE CAUCHY-TYPE 183 where w(t) and Z(t) are given complex-valued functions. It follows by substituting the formulas (11) and (12) in the integral equation ag(t) = F(t)-- &, (15) 711 x — t that the solution of this integral equation is reduced to solving the Riemann-Hilbert problem (a + b)f+(t) - (a-b)f~(t) = F(t) . (16) We shall content ourselves with merely writing down the solution of (1). For details on the Riemann-Hilbert problem the reader is referred elsewhere [3, 13, 15]. Let L be a regular unclosed curve; then the solution of the singular integral equation (1) is a b (t'-u. 8 M ® = tt -&=?&{—p x f{T)—t + (t-ay-»(t-pr> (17) L where a and /? are the beginning and end points of the contour L and the number m is defined as 1 , a + b 2ni a — b The quantity c is an arbitrary constant and is suitably chosen so that g (t) is bounded at a or at p. In particular, the solution of the integral equation of the first kind (we can put b = 1 without any loss of generality), — 8.6. THE HILBERT KERNEL A kernel of the form K(s,t) = cotl(t-s)/2-], where s and t are real variables, is called the Hilbert kernel and is closely connected with the Cauchy kernel. In fact, the integral equation g(s) =f(s) - X j F(s, t) {comt-s)l2]}g(t) dt, (1) o where f(s) and F(s, t) are given continuous functions of period 2n9 is equivalent to the Cauchy-type integral equation 9 (0 = M) - A / LGiC, T)/(T - 0] 9 (x) dx , (2) L where C and T are complex variables and the contour L is the circum ference of the unit disk with center at the point z = 0. Let C and T denote the points of the boundary L corresponding to the arguments s and /, respectively: is lt C = e , T = e , so that dx . Therefore, ft-s\ , 2dx dx T + (dz cot —- )dt = --- = —-, (4) and equation (1) takes the form (2). 8.6. THE HUBERT KERNEL 185 The Hilbert kernel (1) is also related to the Poisson kernel in the integral representation formula for a harmonic function U(r,s): In 1 f 1-r2 (5) U(r9s) = — 2 —- «(/) A , 2n J l+r — 2rcos(t—s) inside the disk r < 1. The function w(/) = £/(l, 0 is the prescribed value of the harmonic function on the circumference L of the disk. Set 1 z = re ', T = e in the relation (5) and get x + z dx U(r,s) = Re u{t) (6) 2ni x — z x Now, let V(r9 s) be the function that is harmonic conjugate to U(r, s): 1 , x + z dx U(r,s) + iV(r,s) = u{i) , (7) 2ni x — z x such that V(r, s) vanishes at the center of the disk: V(r,s)\r=0 = 0. (8) Then, the function V(r,s) is uniquely defined. When r-y 1, so that z tends to a point C, of the circumference L from within the disk, we can apply the Plemelj formula (8.4.2) to the analytic function (7). Therefore from (7), (8.4.2), and (4), we obtain _1_ v(s) = - «(0cot(^)rff, (9) 2K where v(s) = V(\,s)is the limiting value of the harmonic function on L. The formula (9) thus connects the limiting values of the conjugate harmonic functions U(r9s) and V(r,s) on the circumference. We shall need the iterated formula formed by compounding the integrals with the Hilbert kernel: 186 8/SINGULAR INTEGRAL EQUATIONS *2n 9i(s) = Yn\ *WcotfcA dt, (10) and S g2(s) = I gi(t)cot( -^)dt. (11) If U(r, s), Ul (r, s), and U2 (>*, s) are the functions that are harmonic inside the disk r < 1, and whose values on the circumference r = 1 are equal respectively to g(s)9 gi(s), and g2(s), then from (9) it follows that Ul(r,s) is a harmonic function conjugate to U(r,s) and U2(r9s) is conjugate to U{(r9s). The Cauchy-Riemann equations then yield the relations dU __ dU2 SU _ dU2 dr dr ' d.s ds or J72 (r, J) = - U(r, s) + C , C = const. (12) But the constant C can be determined from the condition (8): 2;r 2n C= U(r,s)\r=0 = (ll2n)jU(l9t)dt = (\/2n) j g(t) dt, (13) o o where we have used the mean-value property of the harmonic function. Thereby, (12) becomes In U2 (r, s) = - U(r, s) + (1/2TT) jg(t)dt. (14) o Now, put r = 1 in the above equation, use the relations U(\,s) = g(s) and U2(l,s) = g2(s), and get In g2(s) = -g(s) + (l/2n)jg(t)dt. (15) 0 From (10), (11), and (15), we finally obtain the required iterated integral equation: 8.7. SOLUTION OF THE HILBERT-TYPE 187 cot| — \da g(t)cot[t-^-)dt In. o 2ii = -g(s) + — g{t)dt, (16) 2n which is called the Hilbert formula. 8.7. SOLUTION OF THE HILBERT-TYPE SINGULAR INTEGRAL EQUATION We can solve the integral equation of the second kind ,2* t-s ag(s) = f(s) - — | flf(r)cot( — )dt, (1) where a and b are complex constants, in the same manner as we solved the corresponding Cauchy-type integral equation (8.5.1). As in Section 8.5, we define the operators L and M as Lg = ag{s) + flf(0cot(^)A, (2) In Mcj) = acf)(s) — (3) 2n ^(Ocot(^)A- Then, MLg = a\a \aag(s) + — flf(0cot (T>] (equation continued) 188 8/SINGULAR INTEGRAL EQUATIONS .2* b_ dt 2% x \ag(t) + 3((7)cot(— \da = F(s), (4) In where F(s) = Mf = af(s) -- f(t) ot\'-^)dt 2n C Using the Hilbert formula (8.6.16) and simplifying, we obtain the relation 2/r (a2 + b2)g(s) - (b2l2n) j g(t) dt = F(s) . (5) The integral equation (5) has the simple degenerate kernel K(s,t) = 1, and can be readily solved by the method of Chapter 2. The result is In a .. , * t 2* ^ [ff(t)dt, (6) + 2na(a2 + b2)) J provided a2 + b2 / 0. For the particular case a = 0, the formula (6) is not applicable. Therefore, the solution of the Hilbert-type integral equation of the first kind **-i]0(/)cot(^)A , (7) cannot be deduced from that of the second kind. But the integral equation (7) can be solved by other methods. For instance, the method of Section 8.1 is applicable here. Indeed, let us consider equation (7) with the constant b incorporated in g (i): 8.7. SOLUTION OF THE HILBERT-TYPE 189 1 /(*) = 0(Ocot[^W (8) 2n In this equation, change s to t and / to F(s) = /(/)cot( l-f\dt. (10) In The integral equation (9) also has the simple kernel K(s, t) = 1 and therefore can be solved by the method of Chapter 2 if we set 2n (\l2n)jg(s)ds = c Then, equation (9) becomes g{s)-c = F(s), (11) which, when integrated with respect to s from 0 to 2n, gives In j F(s) ds = 0 . (12) From the relation (10), it follows that equation (12) holds for all values of the function/(s). Therefore, the constant c is an arbitrary constant and the solution of the integral equation (8) is l g(s) = - /(/)cot( -^)dt + c . (13) ~2K 190 8/SINGULAR INTEGRAL EQUATIONS Finally, when we substitute (13) in (8), we find that the function/(a) given by the relation (13) satisfies the original integral equation if, and only if, 2/t jf(s)ds = 0 (14) Hence, the condition (14) is necessary and sufficient for the Hilbert-type singular integral equation of the first kind to have a solution. The second method is to use the results of Section 8.6, where we have connected the Hilbert kernel with the Cauchy kernel. For this purpose, we write g(elt) = g(t), etc., and assume that g{t) and f(t) are periodic functions with period In. Further, we replace f(t) by f(i)ji. Then, the formulas (8.5.8) with the help of the transformation (8.6.3) yield the reciprocal relations *2w 2/r g{t)dt=f{s) (15) and ,2» In (s) (16) With the help of this pair of equations, the solution of the integral equation (8) can be easily deduced. It follows from the pair (15)-(16) that, for periodic functions f(t) and g(t)9 if the condition ffi f(i) dt = 0 is satisfied, then we also have ffig(t)dt = 0. We shall have more to say about the integral relations (15) and (16) in the next chapter. 8.8. EXAMPLES Example 1. Prove that the integral equation EXERCISES 191 -ia(w-v) F(v) = wf(w)- dw (1) w — v has the solution % -ia(v-w) dv. (2) «*~M£) M*» V — W The integral equation (1) arises in the discussion of various problems of mathematical physics. The solution follows by comparing (1) with (8.5.18) and (2) with (8.5.19), and by setting a = 0, fi = 1. Example 2. Solve the integral equation v* l f g(t)COt ['-y)dt. (3) > (an cos ns + bn sin ns) = — Observe that the function/'(s) = XT (ancos ns + bnsinns) is a periodic function with period In. Moreover, the condition jo*/(f) dt = 0 is satisfied. Therefore, from the reciprocal pair (8.7.15) and (8.7.16), it follows that cot g(s) = -— ^(ancosnt + bns'mnt) (^" \dt = /> (ansinns — bncosns) , (4) n= 1 where we have left the actual integration as an exercise for the reader. EXERCISES Solve the following integral equations. s g{i)dt 1. a + bs + cs2 + ds3 = 1 < s < 2, (cost — coss)' 192 8/SINGULAR INTEGRAL EQUATIONS where a, b, c, d are real constants. 2 2 ^33 = f g(t)dtw 2. a + bs + cs + ds = I * " vf/ , 1 < j < 2 . J (coss — cos/)1/2 ' _ f g (/) 2 4 2 f_£(0* w _ 4. s = JoI , \2-' 2V'i)*/ ', 2 < s < 4 5. as + fo2 = f£(0* o 6. Substitute the solution (8.5.5) in (8.5.1) and verify that this solution satisfies the given integral equation. 7. Prove that the solution of the integral equation (8.7.5) is (8.7.6). 8. Find the solution of the integral equation g(s) = (sins) - — g(t)cot(^j dt. o 9. Solve the integral equation s 2 -fi(s), 0 lOL-2t\ /• /(j) = 2,+ 2 ^W(a)" JI ' v-' r^w *, o < a < i, o is EXERCISES 193 9(t) = F^)^J"2"+2,+1('2~"2ra/(")'/"- 0 11. Prove that the solution of the integral equation oo 2s2" 2_ 2\a-1 »-2«-2f+1 (t2_s 2y-iP t {i)dt, 0 < a < 1 , As) T(a)J is ,23+2,-1 rf + 0(0 = - -2l l/-„2_A-(u2-t2y"Au)du.« T(l-a) A, 12. Solve the integral equation r f g(t)dt \ Ms), O^s^a, (t*-s*)1* - cWA -Ms), a 9.1. INTRODUCTION The integral transform methods are of great value in the treatment of integral equations, especially the singular integral equations. Suppose that a relationship of the form g(s) = jj r(s9x)K(x,t)g(t) dtdx (1) is known to be valid and that this double integral can be evaluated as an iterated integral. This means that the solution of the integral equation of the first kind, f(s) = JK(s,t)g(t)dt, (2) is g(s) = jr(s,t)f(t)dt. (3) Conversely, the relation (2) can be considered as the solution of the integral equation (3). It is conventional to refer to one of these functions as the transform of the second and to the second as an inverse transform of the first. 194 9.1. INTRODUCTION 195 The most celebrated example of the double integral (1) is the Fourier integral 00 00 g(s) = (l/2n) J* jeisxe-ix'g(t)dtdx, (4) — oo — oo which results in the reciprocal relations 00 As) = (2ny* je-"'g(t)dt (5) — 00 and 00 g{s) = (2n)-* je*At)dt. (6) — 00 The function f(s) is known as the Fourier transform r[#] of g (t) and g(s) as the inverse transform r_1[/] of f(s), and vice versa. The function f(s) exists if g (t) is absolutely integrable, and it is square- integrable if g(t) is square-integrable, as can be readily verified using BessePs inequality. In the sequel, we shall assume that the functions involved in the integral equations as well as their transforms satisfy the appropriate regularity conditions, so that the required operations are valid. As a second example, consider the double integral 00 00 g(s) = (2/TT) [ f (sin sx sin xt)g(t)dtdx . (7) o o This leads to the sine transform and its inverse, 00 f(s) = (2ln)*j(smst)g(t)dt (8) 0 and 00 g(s) = (2/n)* f (sin st)f(t)dt, (9) 0 respectively. For ease of notation, we shall also call the transform of/ as F and that of g as G, etc., for all the transforms. It will be clear in the context as to what transforms we are implying. 196 9 / INTEGRAL TRANSFORM METHODS 9.2. FOURIER TRANSFORM The Fourier transform T[/], 00 Tin = Hs) = (2n)-tt I At)e-is< dt, (l) — oo is a linear transformation: 00 Tlaf+bg-] = (2«)-a J [a/(0 + ^(0]e"w * ,A s v h = a{2n)- jf(t)e-' = arm + wfe]. (2) As such, we can use many properties of the linear operators. Further more, in Chapter 7 (see Exercise 11), we found that under Fourier transformation a square-integrable function preserves its norm. Hence, for such a function, we have l|7T/]|| = \\F\\ = H/ll • (3) Let us note some of the important properties of the Fourier transforms. They can be found in every standard book on the subject (see, for example, [17]) and, in fact, can be proved very easily by the mere use of the definitions above. These properties are: (i) TIAt-a)-] = e-™TU(ty\ , (4) where a is a constant. (ii) TU{aty] = {\l\a\)TUi.m^sla ■ (5) (iii) TU'ity] = isTUity] , (6) where the prime denotes differentiation with respect to the argument. Similarly, nf\ty] = (isfnnm, m where by fk(t) we mean the &th derivative off. 9.3. LAPLACE TRANSFORM 197 (iv) If t h{t) = jf(x)dx, (8) a then r[A(0] = (i/«) n/0]. (9) From (6) and (9), we see that the differentiation has the effect of multi plying the transform by is, whereas integration has the effect of dividing the transform by is. (v) The convolution integral 00 00 h{t) = (2m)-% j/(t-x)g(x) dx = (2n)-y> j g(t-x)f(x) dx (10) — oo — oo gives T\h{ty\ = T[f\T[ja\ > (11) or H(s) = F(s)G(s). (12) 9.3. LAPLACE TRANSFORM The Laplace transform L [/] of a function f(s) is defined as 00 Lm = F(p) = jf(s)e-"sds. (1) 0 The inverse L_1[F] is y+/oo, r'[f|= f(s) = (l/2*i) | F(p)e"dp . (2) y— /oo This transformation is also linear because Llaf+bg-] = aL\f] + bL[g\ , for two constants a and b. 198 9/INTEGRAL TRANSFORM METHODS The following are some of the basic properties of the Laplace transform: (i) F(p-a) = L[_e°sf{sy] , (3) (ii) LU{asy] = (lla)LW(s)-]^p/a, (4) (iiia) L[/']=jpL[/]-/(0), (5) k (b) £.[/*] = P Lin -/~7(0) -/-V'(O) f'Ho), (6) (c) dF(p)ldp = -L\_sf(s)2 , (7) where fk means kth derivative with respect to the argument. (iv) If S h(s) = J* f{x) dx , (8) 0 then L[A] = H(p) = {\lp)Lm ■ (9) (v) For the convolution integral s s h(s) = jf(x)g(s-x) dx = J g(x)f(s-x) dx , (10) 0 0 we have LIK] = L[/]L[sf] , (11) or H(p) = F(p)G(p), (12) which is the same as (9.2.12). 9.4. APPLICATIONS TO VOLTERRA INTEGRAL EQUATIONS WITH CONVOLUTION-TYPE KERNELS The basic information given here about the Fourier and Laplace transforms is sufficient to demonstrate their application to the solution 9.4. VOLTERRA INTEGRAL EQUATIONS 199 of integral equations. We shall apply only the Laplace transform in this section, although the Fourier transform can also be applied just as effectively. Let us first consider the Volterra-type integral equation of the first kind, s f(s) = jk(s-t)g(t)dt, (1) 0 where k(s — t) depends only on the difference (s — i). Applying the Laplace transform to both sides of this equation, we obtain F(p) = K(p)G(p) or G(p) = F(p)/K(p) . (2) The solution follows by inversion. The present method is also applicable to the Volterra integral equation of the second kind with a convolution-type kernel s g(s)=f(s) + jk(s-t)g(t)dt. (3) 0 On applying Laplace transformation to both sides and using the convolution formula, we have G(p) = F(p) + K(p)G(p) or G(p) = F(p)/ll-K(p)l, (4) and inversion yields the solution. We can also find the resolvent kernel of the integral equation (3) by integral transform methods. For this purpose, we first show that, if the original kernel k(s9t) is a difference kernel, then so is the resolvent kernel. Since the resolvent kernel r(s, t) is a sum of the iterated kernels, all that we have to prove is that they all depend on the difference (s — i). Indeed, s s—t k2(s9t) = f k(s-x)k{x-i)dx = f k{s-t-o)k(o) da , (5) / o 200 9 / INTEGRAL TRANSFORM METHODS where we have set a = x— t. This process can obviously be continued and our assertion is proved. Hence, the solution of the integral equation (3) is S g(s)=f(s) + jr(s-t)f(t)dt. (6) 0 Application of the Laplace transform to both sides of (6) gives G(p) = F(j>) + n(p)F(j>)9 (7) where Q(p)=L[T(s-m. (8) From (4) and (7), we have F(P)IU ~K(p)-] = F(p) [1 + Q(i>)] , (9) Sl(p) = K(p)/\_l-K(p)-]. (10) By inversion, we recover F(s — t). We illustrate the above ideas with numerous examples. Throughout these examples, we have left the evaluation of the Laplace transform and its inverse to the reader. There are numerous monographs that contain the required formulas [3, 12, 17]. 9.5. EXAMPLES Example 1. Solve the Abel integral equation s f(s) = flg(t)l(s-tyidt, 0 K(p)=p*-'T{\-o). (3) 9.5. EXAMPLES 201 From (2) and (3), it follows that p s G(/0 = ^^[J(.-0"-7(0^]. (5) 0 By virtue of the property (9.3.5), we finally have fM-^jfV/r'AO*. (6) n as J o which agrees with the relation (8.1.4) obtained in the previous chapter by a different method. Example 2. Solve the integral equation s s = je°-'g(t)dt. (7) 0 Taking the Laplace transform of both sides, we obtain l/p2 = K(p)G(p), (8) where K(p) is the Laplace transform of k(s) = es: 00 K(p) = fe>e-"ds=ll(p-l). (9) 0 The result of combining (7), (8), and (9) is G(j>) = (p-l)lp2 = (l/p)-(llp2), 202 9 / INTEGRAL TRANSFORM METHODS whose inverse is g(s)= l-s. (10) Example 3. Solve the integral equation s sins = [jQ(s-i)g(i)dt. (11) o Here, the function k(s)=J0(s), whose Laplace transform is known to be 1/(1+/?2)1/2. Also, the Laplace transform of sins is 1/(1 +p2). Therefore, when we take the Laplace transform of equation (11), there results the relation G{p)= \l(\+p2f\ which by inversion yields the solution g(s) = J0(s) . (12) Incidently, by substituting (12) back in (11), we get the interesting result s [ JQ{s-i)J0{i)dt = sins . (13) o Example 4. Recall that we solved integral equations of the type s f(s) = jk(s2-t2)g(t)dt, s>0, (14) o in the previous chapter for some special cases of the kernel k(s2 — t2). With the help of the Laplace transform, we can solve equation (14) for a general convolution kernel. For this purpose, the first step is to set s = u*, t = a* , gi(a) = \p-*q{p*) , Mu) = f{u*). (15) Then, the integral equation (14) takes the form u Mu) = j k(u-o)gi{o) da, u > 0 . (16) o 9.5. EXAMPLES 203 Taking the Laplace transform of both sides of (16), we get Gl (P) = Fl (p)/K(p) = pFx (p)lpK(p) . (17) By defining llpK{p) = H(p), (18) the relation (17) becomes Gl(p)=pH(p)Fl(p). (19) Using the relations (9.3.5) and (9.3.12), we have u Gi(/0 = L\JU [ h(u-a)Ma) dA (20) or u *i(«) = £ [*(«- ")/iw*. (2i) 0 where h{s) stands for the inverse of the function H(p). Finally, from (15) and (21), we have the required solution s 2 2 g(s) = 2Js!tf(t)h(s -t )dt. (22) 0 Let us solve (14) for some special cases: (a) k(t) = t~a, 0 < a < 1. This means that we have to solve the integral equation s As) = jlg(t)l(s2-t2)^dt. (23) 0 The solution follows from (22) if we can evaluate the function h(s) from the relation H{p) = l/PK(p) . (24) But 204 9 / INTEGRAL TRANSFORM METHODS K{p) = j r*e-ptdt = r(l-a)/?a~1 Therefore, sin arc a-i h(s) = L ■T—— s and 2 sin arc d f tf(t) dt g(s) = 2 2 (25) rc ds (s -t y-*' which agrees with the relation (8.2.9). (b) k(t) = t ,/z cos(/fr1/2), where ft is a constant. In this case, K(p) = nV2p-t/2exp(-p2/4p). Therefore, h(s) = L-'ln-^p-KexpiP2^)] = n'^'^cosh^). (26) This means that the solution of the integral equation _ Ccos[p(ss2-t22-t)1 ^ f(s) "J {s2-t2f -g(t)dt, s>09 (27) is 2 d_ cosh[j?(s2-*2)'/2] g(s) = tRt) dt (28) nds (s2-t2)* 6 Note that the relations (27) and (28) remain valid for 0< s< oo. Example 5. Solve the inhomogeneous integral equation s g(s)=l-j(s-t)g(t)dt. (29) Since k(s) = s, K(p)=l/p2, (30) 9.5. EXAMPLES 205