02 - Fredholm Alternative (Linear ODE Version)

Fredholm Alternative Theorem (Linear ODE Version)

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}Ly& =0& & (H)\\ Ly& =f\ne 0& & (N)\end{array}$$

for $0<x<a$ with linear homogenous boundary conditions of form

$$\begin{array}{rl}{B}_{1}y& ={\alpha }_{1}y(a)+{\alpha }_2{y}^{\prime }(a)+{\beta }_{1}y(b)+{\beta }_2{y}^{\prime }(b)=0\\ B_{2}y& ={\alpha }_{3}y(a)+{\alpha }_4{y}^{\prime }(a)+{\beta }_{3}y(b)+{\beta }_4{y}^{\prime }(b)=0\end{array}$$

where $({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2)$ and $({\alpha }_3,{\alpha }_4,{\beta }_3,{\beta }_4)$ are linearly independent

Define homogeneous adjoint equation

$$L^{\ast }w=0$$

and corresponding adjoint boundary conditions ($B{C}^{\ast }$) then

Exactly one of the following possibilities occur

1. Homogeneous Problem ($H+BC$) only has the zero solution
   Solution of $(N+BC)$ is unique

2. Homogeneous Problem ($H+BC$) admit non-trivial solution and so does $(H^{\ast }+B{C}^{\ast })$
   Hence there are two sub-possibilities

   2a) If $⟨f,w⟩=0$ for all $\mathbf{w}$ satisfying $(H^{\ast }+B{C}^{\ast })$ then $(N+BC)$ has a non-unique solution

   2b) Otherwise $(N+BC)$ has no solution