Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL EQUATIONS

we may write thenth-order approximation as

yn(x)=f(x)+

∑n

m=1

λm

∫b

a

Km(x, z)f(z)dz. (23.35)

The solution to the original integral equation is then given by y(x)=

limn→∞yn(x),provided the infinite series converges. Using (23.35), this solution

may be written as

y(x)=f(x)+λ

∫b

a

R(x, z;λ)f(z)dz, (23.36)

where theresolvent kernelR(x, z;λ) is given by

R(x, z;λ)=

∑∞

m=0

λmKm+1(x, z). (23.37)

Clearly, the resolvent kernel, and hence the series solution, will converge

providedλis sufficiently small. In fact, it may be shown that the series converges

in some domain of|λ|provided the original kernelK(x, z) is bounded in such a

way that

|λ|^2

∫b

a

dx

∫b

a

|K(x, z)|^2 dz < 1. (23.38)

Use the Neumann series method to solve the integral equation

y(x)=x+λ

∫ 1

0

xzy(z)dz. (23.39)

Following the method outlined above, we begin with the crude approximationy(x)≈
y 0 (x)=x. Substituting this under the integral sign in (23.39), we obtain the next approxi-
mation

y 1 (x)=x+λ

∫ 1

0

xzy 0 (z)dz=x+λ

∫ 1

0

xz^2 dz=x+

λx

3

,

Repeating the procedure once more, we obtain

y 2 (x)=x+λ

∫ 1

0

xzy 1 (z)dz

=x+λ

∫ 1

0

xz

(

z+

λz

3

)

dz =x+

(

λ

3

+

λ^2

9

)

x.

For this simple example, it is easy to see that by continuing this process the solution to
(23.39) is obtained as

y(x)=x+

[

λ

3

+

(

λ

3

) 2

+

(

λ

3

) 3

+···

]

x.

Clearly the expression in brackets is an infinite geometric series with first termλ/3and