Linear Methods of Applied Mathematics
Evans M. Harrell II and James V. Herod*
In this chapter we shall learn how to solve integral equations in three situations:
First we suppose that there are an integer n and functions
such that, for each p , a_{p} and b_{p} are in L^{2}[0,1]. Then K has a separable kernel if its kernel is given by
With the supposition the K is separable, it is not hard to find y such that y = Ky + f, for this equation can be re-written as
or, using the notation of inner products,
One can guess that, if the sequence
of functions on [0,1] is a linearly independent sequence, then y will have this special form:
there is a sequence {cp} of numbers such that
In fact, supposing there is such a sequence, we determine what it should be.
Suppose
Substitute this in the equation to be solved:
and we see that
This now reduces to a matrx problem:
Define K and f to be the matrix and vector so defined that the last equation is rewritten as
c = K c + f.
We now employ ideas from linear algebra. The equation c = K c + f has exactly one solution provided
det( 1 - K ) != 0.
The Fredholm Alternative Theorems for matrices address these ideas. (Review the alternative theorems for matrices.) If the sequence
is found then we have a formula for y(x).
EXAMPLE: In Exercises 1.2, it should have been established that if
K(x,t) = 1 + sin([[pi]]x) cos([[pi]]t),
then
K*(x,t) = 1 + sin([[pi]]t) cos([[pi]]x).
Also,
y = Ky has solution y(x) = 1
and
y = K*y has solution y(x) = [[pi]] + 2 cos([[pi]]x).
It is the promise of the Fredholm Alternative theorems that
y = Ky + f
has a solution provided that
Let us try to solve y = Ky + f and watch to see where the requirement that f should be perpendicular to the function [[pi]] +2 cos([[pi]]t) appears.
To solve y = Ky + f is to solve
We guess that the solution is of the form y(x) = a + b sin([[pi]]x) + f(x) and substitute this for y:
From this, we get the algebraic equations
Hence, in our guess for y, we find that a can be anything and that b must be
and also must be
The naive pupil might think this means there are two (possibily contradictory) requirements on b. The third of the Fredholm Alternative theorems assures the student that there is only one requirement!
Take [[phi]]_{0}(x) to be f(x) and [[phi]]_{1} to be defined by
It is reasonable to ask: does this generated sequence converge to a limit and in what sense does it converge? The answer to both questions can be found under appropriate hypothesis on K.
THEOREM If K satisfies the condition that
then limp [[phi]]_{p}(x) exists and the convergence is uniform on [0,1] - in the sense that if u = lim_{p}[[phi]]_{p} then
lim_{p} max_{x} | u(x) - [[phi]]_{p}(x) | = 0.
SUGGESTION FOR PROOF: Note that
Furthermore, if p is a positive integer, the distance between successive iterates can be computed:
Inductively, this does not exceed
Thus, if
and n > m then
Hence, the sequence {[[phi]]_{p}} of functions converges uniformly on [0,1] to a limit function and this limit provides a solution to the equation
COROLLARY. If
and
u = lim_{p} [[phi]]_{p}
then
THEOREM If K satisfies the condition that
then lim_{p} [[phi]]p(x) exists and the convergence is in norm- meaning that if u = lim_{p}[[phi]]p then
lim_{p} || 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 = lim_{p} [[phi]]p
then
||u - [[phi]]m || < F(r^{m+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 = K^{2}f+Kf+f, .....
In fact, with
Hence, the kernel K_{2} associated with K^{2} is
Inductively,
and
We have, in this section, conditions which imply that
[[Sigma]]p=1 K^{p}f
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 K^{p}.
Note that R is a function which operates on elements of L^{2}[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 Mathematica notebook for the solution by iteration of a typical integral equation, including error estimates.
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