Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL EQUATIONS

These two simultaneous linear equations may be straightforwardly solved forc 1 andc 2 to
give

c 1 =

24 +λ

72 − 48 λ−λ^2

and c 2 =

18

72 − 48 λ−λ^2

,

so that the solution to (23.12) is

y(x)=

(72− 24 λ)x+18λ

72 − 48 λ−λ^2

.

In the above example, we see that (23.12) has a (finite) unique solution provided

thatλis not equal to either root of the quadratic in the denominator ofy(x).

The roots of this quadratic are in fact theeigenvaluesof the corresponding

homogeneous equation, as mentioned in the previous section. In general, if the

separable kernel containsnterms, as in (23.8), there will bensuch eigenvalues,

although they may not all be different.

Kernels consisting of trigonometric (or hyperbolic) functions of sums or differ-

ences ofxandzare also often separable.

Find the eigenvalues and corresponding eigenfunctions of the homogeneous Fredholm

equation

y(x)=λ

∫π

0

sin(x+z)y(z)dz. (23.13)

The kernel of this integral equation can be written in separated form as

K(x, z)=sin(x+z)=sinxcosz+cosxsinz,

so, comparing with (23.8), we haveφ 1 (x)=sinx,φ 2 (x)=cosx,ψ 1 (z)=coszand
ψ 2 (z)=sinz.
Thus, from (23.11), the solution to (23.13) has the form

y(x)=λ(c 1 sinx+c 2 cosx),

where the constantsc 1 andc 2 are given by

c 1 =λ

∫π

0

cosz(c 1 sinz+c 2 cosz)dz =

λπ

2

c 2 , (23.14)

c 2 =λ

∫π

0

sinz(c 1 sinz+c 2 cosz)dz =

λπ

2

c 1. (23.15)

Combining these two equations we findc 1 =(λπ/2)^2 c 1 , and, assuming thatc 1 =0,this
givesλ=± 2 /π, the two eigenvalues of the integral equation (23.13).
By substituting each of the eigenvalues back into (23.14) and (23.15), we find that
the eigenfunctions corresponding to the eigenvaluesλ 1 =2/πandλ 2 =− 2 /πare given
respectively by

y 1 (x)=A(sinx+cosx)andy 2 (x)=B(sinx−cosx), (23.16)

whereAandBare arbitrary constants.