Separable Integral Kernels

Integral Equations and the Method of Green's Functions James V. Herod*

Page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.

SECTION 3. SOLVING y = K(y) + f WHERE K HAS A SEPARABLE KERNEL.

We now analyze the equation y = Ky + f in three cases:

K has a separable kernel,

K has norm less than one,

and K is approximated by K's with separable kernels.

We suppose that there is an integer n and functions

such that, for each p , ap and bp are in L2[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!

EXERCISE 1.3:

I. With K, f, and an interval as given, solve the integral equation y = Ky + f.

(a) K(x,t) = 2x-t, f(x) = x2 on [1,2]. ans: y(x) = x2 - (75x - 61)/6.

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

(c) K(x,t) = 2x2 -3t, f(x) = x on [0,1]. ans: y(x) = x +(6x2-13)/28.

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

(e) K(x,t) = xt2+1, f(x) = x on [0,1] ans: y(x) = -3.

(f) K(x,t) = 1/2 + x t, f(x) = 3x2-1 on [-1,1]. ans: y(x) =3x2 + c

(g) K(x,t) = x t, f(x) = exp(x) on [0, ln(7)]. ans: y(x) = ex+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].

II. Show that if f is continuous and 1 + [[lambda]]/2 - [[lambda]]2/240 != 0,

then

y(x) = -[[lambda]] I(0,1, ) ( x2 t + x t2 ) y(t) dt + f(x)

has a solution.

III. (a) For what functions f does the equation

have a solution?

IV. Solve the integral equation y = Ky + f where

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