Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL EQUATIONS

sides of (23.51) withyj, we find

∑

i

ai〈yj|yi〉=λ

∑

i

ai

λi

〈yj|yi〉+λ〈yj|Kf〉. (23.52)

Since the eigenfunctions are orthonormal andKis an Hermitian operator,

we have that both〈yj|yi〉=δijand〈yj|Kf〉=〈Kyj|f〉=λ−j^1 〈yj|f〉. Thus the

coefficientsajare given by

aj=

λλ−j^1 〈yj|f〉

1 −λλ−j^1

=

λ〈yj|f〉

λj−λ

, (23.53)

and the solution is

y=f+

∑

i

aiyi=f+λ

∑

i

〈yi|f〉

λi−λ

yi. (23.54)

This also shows, incidentally, that a formal representation for the resolvent kernel

is

R(x, z;λ)=

∑

i

yi(x)yi∗(z)

λi−λ

. (23.55)

Iffcanbe expressed as a linear superposition of theyi,i.e.f=

∑
ibiyi,then
bi=〈yi|f〉and the solution can be written more briefly as

y=

∑

i

bi

1 −λλ−i^1

yi. (23.56)

We see from (23.54) that the inhomogeneous equation (23.50) has a unique

solution providedλ=λi,i.e.whenλis not equal to one of the eigenvalues of

the corresponding homogeneous equation. However, ifλdoes equal one of the

eigenvaluesλjthen, in general, the coefficientsajbecome singular and no (finite)

solution exists.

Returning to (23.53), we notice that even ifλ=λja non-singular solution to

the integral equation is still possible provided that the functionfis orthogonal

to every eigenfunction corresponding to the eigenvalueλj,i.e.

〈yj|f〉=

∫b

a

y∗j(x)f(x)dx=0.

The following worked example illustrates the case in whichfcan be expressed in

terms of theyi. One in which it cannot is considered in exercise 23.14.

Use Schmidt–Hilbert theory to solve the integral equation

y(x) = sin(x+α)+λ

∫π

0

sin(x+z)y(z)dz. (23.57)

It is clear that the kernelK(x, z)=sin(x+z) is real and symmetric inxandzand is