James V. Herod*
Page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.
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,
and
u = limp [[phi]]p
then
||u - [[phi]]m || < F(rm+1,1 - r) ||f||.
THE RESOLVENT.
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
Inductively,
and
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.
EXERCISE 1.5.
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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