Linear Methods of Applied Mathematics
Evans M. Harrell II and James V. Herod*
version of 11 May 2000
Here is a Mathematica notebook with calculations for this chapter, and here are some Maple worksheets with similar calculations:
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.
Theorem. If
then there are kernels K_{n} and G such that
(1) K = K_{n} + G,
(2) K_{n} has a separable kernel,
and
In the succeeding pages, we show how to compute K_{n} and G. However, we first illustrate that the problem is - in theory - solved if we have such a resolution of K into K_{n} and G. We seek y such that
y = Ky + f = K_{n}y + Gy + f
or y - Gy = K_{n}y + f.
Use the resolvent for G:
(1-G)^{-1} = 1 + RG,
to get that
y = K_{n}y + RG(K_{n}y + f) + f
=[K_{n} + RGK_{n}]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 = (K_{n} + RGK_{n})y + z.
Re-writing this as an integral equation, we seek y such that
where
What is astonishing is that this last integral equation is separable! To see this, suppose
Then
So, here is the conclusion. If K is K_{n} + 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 = (K_{n} + RGK_{n})y + z
QED
We now must address the question of how to achieve the decomposition of K into K_{n} + G. The ideas are familiar to us from chapter IV on Fourier series. In summary of those ideas, recall that if p and q are integers, then
We seek A_{pq} such that
In fact, by integrating both sides of this last equation after multiplying by sin(m x) sin(n y), we have
From the theory of Fourier series,
in the sense that
as n -> * . Let n be an integer such that
Define K_{n} and G by
and G = K - K_{n}.
Then these three requirements are met:
(1) K = K_{n} + G,
(2) K_{n} is separable,
and
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 K_{n} is separable and approximates K. How well does the solution u for u = K_{n} u + f approximate y?
(b) Suppose K = K_{n} + G and
G^{p} approximates RG. How well does the solution u for
u = [K_{n}+ SK_{n}] u + [1+S ]f
approximate y?
(a) K(x,t) = 2x-t, f(x) = x^{2} on [1,2].
then
y(x) = - I(0,1, ) ( x^{2} t + x t^{2} ) y(t) dt + f(x)
has a solution.
and f(x) = x. (Hint: take the derivative of both sides.)
(Answer)
(a) Show that if 0 <= x <= 1,
In fact,
(b) Toward solving y(x) = K[y](x) + x , compute _{0}, _{1}, and 2.
(Answer)
(c) Give a bound on the error between the solution y and _{2}.
(Answer)
Give a formula for
for each K in the previous exercise set.
For this K, find y such that y(x) = Ky(x) + x. Note that
What is the significance of this observation?
For this K, find y such that y(x) = Ky(x) + x. Note that
What is the significance of this observation?
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]?
(Answer)
Ans: R(x,t) = K(x,t) + K_{2}(x,t) + K_{3}(x,t) + . = 3xt/2.
(a) Explain how you know this problem is in the second alternative.
(c) Let f^{1}(x) = 3x - 1 and f^{2}(x) = 3x^{2} - 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?
(Answer)
(d) For the f for which there is a solution, find two.
(Answer)
(a) Show that the associated K is small in both senses of this section.
(b) Compute _{2} where f(x) = 1.
(Answer)
(c) Give an estimate for how much _{2} differs from the solution y of y=K(y)+f.
(Answer)
(d) Using the kernel K for K, compute the kernel K_{2} for K^{2} and K_{3} for K^{3}.
(Answer)
(e) Compute the kernel for the resolvent of this problem.
(Answer)
(f) What is the solution for y=Ky+f in case f(x) = 1.
(Answer)
(a) Compute the associated approximations \phi_{0}, \phi_{1}, \phi_{2}, and \phi_{3}.
(Answer)
(b) Give an estimate for how much \phi_{3} differs from the solution.
(c) Give the kernel for the resolvent of this problem.
(Answer)
(d) Using the resolvent, give the solution to this problem.
(Answer)
(e) Using the fact that the kernel of the problem separates, solve the equation.
(a) Show that
(b) Solve the problem y = K[y] + 1.
(Answer)
K(x,t) = 1 + cos( x) cos( 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 x^{2 }meet this condition?
(Answer)
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]?