James V. Herod*

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

**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 n^{th} order differentiable operator with n
boundary conditions. B_{1}, B_{2}, ...B_{n}. The problem is posed as follows:
Given
f, find u such that L(u) = f with B_{p}(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, B_{p}(u) = 0, p = 1, 2,
..., n, has one and only one solution..

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

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

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

III. Suppose the second alternative holds. Then L(u) = f, B_{p}(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, B_{p}*(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 }.

(Answer)

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

(Answer)

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

(Answer)

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

(Answer)

2. Suppose that L(y)(x) = y''(x) + 4 \pi ^{2}y(x). B_{1}(y) = y(0) and
B_{2}(y) = y(1).

(a) Show that the problem L(y) = 0, B_{1}(y)= B_{2}(y) = 0 has sin(2 \pi x) as a
non-trivial solution.

(b) What is the adjoint problem for {L, B_{1}, B_{2}}?

(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 ^{2}y(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.

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