Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

version of 1 June 1996

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

y = **K**y + f = **K _{n}**y +

or y - **G**y = **K _{n}**y + f.

Use the resolvent for **G**:

(1-**G**)^{-1} = 1 + **RG**,

to get that

y = **K _{n}**y +

=[**K _{n}** +

Define z to be **RG**f + f, or, what is the same, solve the equation

z = **G**z + f.

We can solve this equation because **G** is small. Now, we seek y such that

y = (**K _{n}** +

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 = **K**y + 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 persons
knowledgeable about the techniques of
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 \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 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 = **K**y + f where

The engineer will want to know about approximations. Here are two appropriate questions:

(a) Suppose one hopes to solve y = **K**y + 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?

**Exercises XIII**

**XIII.1**. With K, f, and an interval as given, solve the integral equation y =
**K**y + f.

(a) K(x,t) = 2x-t, f(x) = x^{2} on [1,2]. ans: y(x) =
x^{2} - (75x - 61)/6.

(b) K(x,t) = x + 2xt, f(x) = x on [0,1]. ans: y(x) = -6x.

(c) K(x,t) = 2x^{2} -3t, f(x) = x on [0,1]. ans: y(x) = x
+(6x^{2}-13)/28.

(d) K(x,t) = t(t+x), f(x) = x on [0,1] ans: y(x) =(18+48x)/23.

(e) K(x,t) = xt^{2}+1, f(x) = x on [0,1] ans: y(x) = -3.

(f) K(x,t) = 1/2 + x t, f(x) = 3x^{2}-1 on [-1,1]. ans:
y(x) =3x^{2} + c

(g) K(x,t) = x t, f(x) = exp(x) on [0, ln(7)]. ans: y(x) =
e^{x}+ax where a is3(7ln(7)-6)/(3-(ln(7)^{3})

(h) K(x,t) = x - t, f(x) = x on [0,1]. ans: y(x) = (18x-4)/13

(i) K(x,t ) = sin([[pi]]x) cos([[pi]]t), f(x) = sinh(x) on [0,1].

**XIII.2**. Show that if f is continuous and 1 + [[lambda]]/2 -
[[lambda]]^{2}/240 != 0,

then

y(x) = -[[lambda]] I(0,1, ) ( x^{2} t + x t^{2} )
y(t) dt + f(x)

has a solution.

**XIII.3**. (a) For what functions f does the equation

have a solution?

**XIII.4**. Solve the integral equation y = **K**y + f where

and f(x) = x. (Hint: take the derivative of both sides.)

**XIII.5**. Let

(a) Show that if 0 <= x <= 1,

In fact,

(b) Toward solving y(x) = **K**[y](x) + x , compute [[phi]]0, [[phi]]1, and
[[phi]]2.

(c) Give a bound on the error between the solution y and [[phi]]2.

ans: |y - [[phi]]2| <= 1/4

(d) Solve y(x) = **K**y(x) + x in closed form for this K. (Reference

**XIII.7**. Suppose that

Give a formula for

**XIII.8**. Compute

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

**XIII.9**. Let

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

What is the significance of this observation?

ans: x +1/8

**XIII.10**. Let

For this K, find y such that y(x) = **K**y(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]?

**XIII.11**. Find the kernel for the resolvent of the **K**
whose kernel is K(x,t) =
x t.

Ans: R(x,t) = K(x,t) + K_{2}(x,t) + K_{3}(x,t) + . = 3xt/2.

**XIII.12**. 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 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)

**XIII.13**. Consider the problem

(a) Show that the associated **K** is small in both senses of this
section.

(b) Compute \phi_{2} where f(x) = 1.

(answer)

(c) Give an estimate for how much \phi_{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_{2} for **K**^{3}.

(answer)

(e) Compute the kernel for the resolvent of this problem.

(answer)

(f) What is the solution for y=**K**y+f in case f(x) = 1.

(answer)

**XIII.14**. Consider the problem

(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.

**XIII.15**. Suppose that

(a) Show that

(b) Solve the problem y = **K**[y] + 1.

(answer)

**XIII.16**. (a) Find a nontrivial solution for y = **K**[y] in L^{2}[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 x^{2 }meet this condition?

(answer)

Go to a test at this point of your studies.

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