James V. Herod*
Page maintained with additions by Evans M. Harrell, II, firstname.lastname@example.org.
In case K neither has a separable kernel nor is small, then the next resort is to approximate K with an operator which has a separable kernel.
then there are kernels Kn and G such that
(1) K = Kn + G,
(2) Kn has a separable kernel,
In the succeeding pages, we show how to compute Kn and G. However, we first illustrate that the problem is - in theory - solved if we have such a resolution of K into Kn and G. We seek y such that
y = Ky + f = Kny + Gy + f
or y - Gy = Kny + f.
Use the resolvent for G:
(1-G)-1 = 1 + RG,
to get that
y = Kny + RG(Kny + f) + f
=[Kn + RGKn]y + (RGf + f).
Define z to be RGf + f, or, what is the same, solve the equation
z = Gz + f.
We can solve this equation because G is small. Now, we seek y such that
y = (Kn + RGKn)y + z.
Re-writing this as an integral equation, we seek y such that
What is astonishing is that this last integral equation is separable! To see this, suppose
So, here is the conclusion. If K is Kn + G as in the above Theorem, in order to solve y = Ky + f, use the fact that
to form the resolvent for RG; then find z such that z = (1+R G)f. Finally, solve the separable equation y = (Kn + RGKn)y + z
We now must address the question of how to achieve the decomposition of K into Kn + G. The ideas are familiar to persons knowledgeable about the techniques of Fourier series. In summary of those ideas, recall that if p and q are integers, then
We seek Apq such that
In fact, by integrating both sides of this last equation after multiplying by sin(m \pi x) sin(n \pi y), we have
From the theory of Fourier series,
in the sense that
as n -> * . Let n be an integer such that
Define Kn and G by
and G = K - Kn.
Then these three requirements are met:
(1) K = Kn + G,
(2) Kn is separable,
Thus, we have an analysis of an integral equation y = Ky + f where
The engineer will want to know about approximations. Here are two appropriate questions:
(a) Suppose one hopes to solve y = Ky + f and that Kn is separable and approximates K. How well does the solution u for u = Kn u + f approximate y?
(b) Suppose K = Kn + G and
Gp approximates RG. How well does the solution u for
u = [Kn+ SKn] u + [1+S ]f
1. Consider the problem
(a) Explain how you know this problem is in the second alternative.
ans: y(x) = c is a non-trivial solution
to the non-homogeneous problem.
(b) Find linearly independent solutions for the equation y=K*(y).
(c) Let f1(x) = 3x - 1 and f2(x) = 3x2 - 1. For one of these there is a solution to the equation y = K(y) + f, for the other there is not. Which has a solution?
(d) For the f for which there is a solution, find two.
2. Consider the problem
(a) Show that the associated K is small in both senses of this section.
(b) Compute \phi2 where f(x) = 1.
(c) Give an estimate for how much \phi2 differs from the solution y of y=K(y)+f.
(d) Using the kernel k for K, compute the kernel k2 for K2 and k2 for K3.
(e) Compute the kernel for the resolvent of this problem.
(f) What is the solution for y=Ky+f in case f(x) = 1.
3. Consider the problem
(a) Compute the associated approximations \phi0, \phi1, \phi2, and \phi3.
(b) Give an estimate for how much \phi3 differs from the solution.
(c) Give the kernel for the resolvent of this problem.
(d) Using the resolvent, give the solution to this problem.
(e) Using the fact that the kernel of the problem separates, solve the equation.
4. Suppose that
(a) Show that
(b) Solve the problem y = K[y] + 1.
5. (a) Find a nontrivial solution for y = K[y] in L2[0,1] where
K(x,t) = 1 + cos(\pi x) cos(\pi t).
(b) Find a nontrivial solution for z = K*[z].
(c) What condition must hold on f in order that
y = K[y] + f
shall have a solution? Does f(x) = 3 x2 meet this condition?
Go to a test at this point of your studies.
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