While studying Functional Analysis, Sobolev Spaces and Partial Differential Equations (Brezis), I came across this theorem ($E$ is a Banach space):
In class we studied the problem $$ \left\{ \begin{array}{lcc} (p(x)y'(x))'+q(x)y(x)=r(x) \\ y(a)=y(b)=0 \end{array} \right. $$ With $q,r\in C([a,b])$ and $p\in C^1([a,b]), \ p(x)>0 \ \forall x\in [a,b]$. This was the Fredholm alternative we were given:
Consider the previous problem. Then there are two possibilities:
- If the homogeneus problem ($r=0$) only admits the trivial solution, then the non-homogeneus problem admits an unique solution.
- If the homogeneus problem admits non trivial solutions then the non-homogeneus problem admits solution (an infinite number) iff $$ \int_{a}^{b}r(x)\psi(x)dx=0 \ \forall \psi \ \text{solution of the homogeneus problem} $$
The proof we were presented in class has nothing to do with the theorem in Brezis' book, but I would like to relate them, as I think they are the same theorem. From the first theorem I get that the equation $u-Tu=f$ can
- Have a unique solution for every $f\in E$, by $(c)$
- If $u-Tu=0$ has $n$ independent solutions (by $a)$), then equation $u-Tu=f$ has solution iff $f\in\ker{(I-T^*)}^\perp$
In the context of my problem I guess that $E=\{y\in C^2([a,b]):y(a)=y(b)=0\}$, which is a closed subspace of a Banach space ($C^2$ functions with its complete norm) and therefore a Banach space itself. However, I don't know which compact operator $T$ I have to use in order to prove the theorem. I have thought of $Tu=(1-q)u-(pu')'$, which is linear but I don't think it's valid, as it does not always return a $C^2$ function. In addition, I would have to calculate $\ker{(I-T^*)}^\perp$ or something that gives me the last condition. Any help please?