Small Integral Kernels, II

Integral Equations and the Method of Green's Functions

James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, All rights reserved.

Page maintained by Evans M. Harrell, II,


There is an alternate, and independent, concept of K being small which leads to convergence of the iteration process in the norm of L2[0,1] This alternate hypothesis asks that

THEOREM If K satisfies the condition that

then limp [[phi]]p(x) exists and the convergence is in norm- meaning that if u = limp[[phi]]p then

limp || u(x) - [[phi]]p(x) || = 0.

INDICATION OF PROOF. The analysis of the nature of the convergence will go like this:

|| [[phi]]1 - [[phi]]0 || 2

is defined to be

As before,


u = limp [[phi]]p


||u - [[phi]]m || < F(rm+1,1 - r) ||f||.


Before addressing the final case - where

K does not have a separable kernel,

we generate "resolvents" for the integral equations.

Re-examining the iteration process:

[[phi]]0(x) = f(x),

[[phi]]1(x) = K[[phi]]0(x) + f(x)

[[phi]]2(x) = K(K([[phi]]0))x + K(f)(x) + f(x)



One writes [[phi]]0=f, [[phi]]1=Kf+f, [[phi]]2 = K[Kf+f] + f = K2f+Kf+f, .....

In fact, with

Hence, the kernel K2 associated with K2 is



We have, in this section, conditions which imply that

[[Sigma]]p=1 Kpf

converges and that its limit y satisfies y = Ky + f. Many authors call this series of operators the "resolvent" and denote

R = [[Sigma]]p=1 Kp.

Note that R is a function which operates on elements of L2[0,1]. One writes that y = Ky + f

has solution

y(x) = [( 1 + R ) f](x) = f(x) + I(0,1, ) R(x,t) f(t) dt.

Suggestive algebra can be made by identifying (1 + R ) as

(1 - K ) -1 = 1 + K( 1 - K ) -1, so that R = K ( 1 - K ) -1.

Please refer to the accompanying notebook for the solution by iteration of a typical integral equation, including error estimates.


1. Suppose that

Give a formula for

2. Compute

for each K in the previous exercise set. ans: 1/12 and 1/6.

3. Let

For this K, find y such that y(x) = Ky(x) + x. Note that

What is the significance of this observation?

ans: x +1/8

4. Let

For this K, find y such that y(x) = Ky(x) + x. Note that

What is the significance of this observation?

ans: exp(x) - 1

5. Suppose that

so that the kernel of K is cos(x+t) and the kernel of H is sin(x+t). What is the kernel of K[H]?


6. Find the kernel for the resolvent of the K whose kernel is K(x,t) = x t.

Ans: R(x,t) = K(x,t) + K2(x,t) + K3(x,t) + . = 3xt/2.

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