The Fredholm Alternative Theorems

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

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

CHAPTER II. ORDINARY DIFFERENTIAL EQUATIONS

SECTION 3 THE FREDHOLM ALTERNATIVE THEOREMS

Before developing a technique for solving these ordinary differential equations with boundary conditions, attention should be paid to the statement of the Fredholm Alternative Theorems in this setting. You may wish to compare it with the alternative theorems for integral equations and for matrices.

Suppose that L is an nth order differentiable operator with n boundary conditions. B1, B2, ...Bn. The problem is posed as follows: Given f, find u such that L(u) = f with Bp(u) = 0, p = 1, 2, ...n.

I. Exactly one of the following two alternatives holds:

(a)(First Alternative) if f is continuous then L(u) = f, Bp(u) = 0, p = 1, 2, ..., n, has one and only one solution..

(b)(Second Alternative) L(u) = 0, Bp(u) = 0, p = 1, 2, ...n, has a nontrivial solution.

II. (a) If L(u) = f, Bp(u) = 0, p = 1, 2, ...n, has exactly one solution then so does L*(u) = f, Bp*(u) = 0, p = 1, 2, ...n have exactly one solution.

(b) L(u) = 0, Bp(u) = 0, p = 1, 2, ... n, has the same number of linearly independent solutions as L*(u) = 0, Bp*(u) = 0, p = 1, 2, ...n.

III. Suppose the second alternative holds. Then L(u) = f, Bp(u) = 0, p = 1, 2, ...n has a solution if and only if < f, w > = 0 for each w that is a solution for
L*(w) = 0, Bp*(w) = 0, p = 1, 2, ...n.

EXERCISE 2.3.

1. Decide whether the following operators L are formally self adjoint, and whether they are self adjoint on M. Decide whether the equation L(y) = f on M is in the first or second alternative.

(a) L(y) = y'', M = {y: y(0) = y'(0) = 0 }.

(b) L(y) = y'', M= {y: y(0) = y(1) = 0 }.

(c) L(y) = y'' + 4 \pi 2y, M={y: y(0) = y(1), y'(0) = y'(1)}.

(d) L(y) = y'' + 3y' + 2y, M = {y: y(0) = y(1) = 0}.

2. Suppose that L(y)(x) = y''(x) + 4 \pi 2y(x). B1(y) = y(0) and B2(y) = y(1).

(a) Show that the problem L(y) = 0, B1(y)= B2(y) = 0 has sin(2 \pi x) as a non-trivial solution.

(b) What is the adjoint problem for {L, B1, B2}?

(c) What specific conditions must be satisfied by f in order that L(y) = f, y(0) = 0 = y(1) has a solution?

(d) Show that y''(x) + 4 \pi 2y(x) = 1, y(0) = 0 = y(1) has

[ 1 - cos(2 \pi x) ] /4 \pi 2 as a solution.

3. Show that y'' = x, y'(0) = 0 = y'(1) has no solution.

4. Show that y'' = sin(2 \pi x), y'(0) = 0 = y'(1) has a solution.