Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1997,2000 by Evans M. Harrell II and James V. Herod. All rights reserved.

version of 14 March 2000

Here is a Mathematica notebook with calculations for this chapter, and here are some Maple worksheets with similar calculations

XIV. Solving Y = KY + f

In this chapter we shall learn how to solve integral equations in three situations:

• K has a separable kernel,
• K has norm less than one,
• K is approximated by K's with separable kernels.

Definition XIV.1 Suppose that there are an integer n and functions

such that, for each p , a_p and b_p are in L²[0,1]. Then K has a separable kernel if its kernel is given by

                             (14.1)

Another term for operators K of this type is finite-rank, and we shall see that they can be considered as matrices of finite rank.

With the supposition that K is separable, it is not hard to find y such that y = Ky + f, for this equation can be re-written as

                          (14.2)

or, using the notation of inner products,

                          (14.3)

We can see that if the sequence

of functions on [0,1] is a linearly independent sequence, then y will have the following special form:

there is a sequence {c_p} of numbers such that

Why is this? In (14.1), all the definite integrals over t are just numbers. Even though we do not know their values yet, we can call them c_p and procede to determine their values with a bit of algebraic labor.

Suppose

Substitute this in the equation to be solved:

and we see that

This now reduces to a matrix 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.

is found, we have a formula for y(x).

Example XIV.2: In the exercises of chapter XIII, it should have been established that if

       K(x,t) = 1 + sin(x) cos(t),

then

       K*(x,t) = 1 + sin(t) cos(x).

Also,

        y = Ky has solution y(x) = 1

and

       y = K*y has solution y(x) = + 2 cos(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 +2 cos(t) appears.

To solve y = Ky + f is to solve

As usual we see that the solution must be of the form y(x) = a + b sin(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 b must also 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 ₀(x) to be f(x) and 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 XIV.3. If K satisfies the condition that

                          (14.4)

then lim_{p p}(x) exists and the convergence is uniform on [0,1] - in the sense that if u = lim_pp then

       lim_p max_x | u(x) - _p(x) | = 0.

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 {_p} of functions converges uniformly on [0,1] to a limit function and this limit provides a solution to the equation

Corollary XIV.4. If

and

       u = lim_{p p}

then

Sometimes it is convenient to express the iteration as an infinite series, called the Neumann series, i.e., the sum of _n = _n-_n-1. We reason this way in the next example.

Model Problem XIV.5. Consider the integral equation

where g(x) is given. We wish to solve for u(x), and we try the method of iteration.

We begin with the guess ₀ = g(x), and calculate the next couple of iterates:

This integral can be simplified by reversing the order of integration. Setting the limits takes a moment of reflection, and may be helped by the following diagram:

The relationship of the variables is 0 < t < t₁ < x. If the first (inside) integral is in the variable t, then it runs from 0 to t₁, and then the second integral in the variable t₁ runs from 0 to x. If we reverse the order, the first integral, in the variable t₁, runs from t to x, and the second integral runs from 0 to x. We find that ₂(x) is:

If we now calculate the further iterates, we find inductively that

The special feature about this example is that the iterates can be summed up, and when we recall that we get: We leave it as an

It is a miracle when the series for K sums in closed form like this, but that is not important in applications, since the convergence of the Neumann series implies that we can calculate the answer to any desired accuracy.

                          (14.5)

Theorem XIV.6. If K satisfies the Hilbert- Schmidt condition (14.4), then lim_{p p}(x) exists and the convergence is in the r.m.s. sense, that is:

       lim_p || u(x) - _p(x) || = 0.

INDICATION OF PROOF. The analysis of the nature of the convergence will go like this:

||₁ - ₂|| ²

As a consequence, the sequence _n is Cauchy convergent:

Let's state the conclusion in a careful way:

Corollary XIV.7. If ,

Then

u = lim_{p p}

resolvent of an operator K is the inverse operator ( - K)^-1. This is same as the solution operator for the equation y = Ky + f Often the number is taken as 1, and unless stated otherwise we shall always do this.

Before addressing the final case - where

Re-examining the iteration process:

    ₀(x) = f(x),

    ₁(x) = K₀(x) + f(x)

    ₂(x) = K(K(₀))x + K(f)(x) + f(x)

One writes ₀=f, ₁=Kf+f, 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

_p=1 K^pf

(1 - K)^-1 = 1 + R, where
R = _p=1 K^p.                           (14.6)

Continue with this chapter

Link to