Sturm-Liouville theory.

Question

I'm trying to prove a theorem concerning inhomogeneous Sturm-Liouville system. Let's consider a system $$ \begin{cases} (p(x)y(x)')'+(\lambda r(x) - q(x))y(x) = f(x), \\ y(a)=A, y(b) = B. \end{cases} $$ From differntial equations we know that if $\lambda$ is not an eigenvalue for $L \equiv - \frac1r \frac{d}{d x} \left( p(x) \frac {d}{d x} \, \right) + \frac{q(x)}{r(x)}$, then the system has a unique solution. But I want to proof that if $L y_n = \lambda_n y_n$, then the following 2 statements are equivalent: $$ \exists y \, - \, \text{solution} \Leftrightarrow \int\limits_a^b f(x) y_n(x) d x = A p(a) y_n'(a) - B p(b) y_n'(b). $$

Answer 1

You have to assume $\lambda_n=\lambda$ is real, that the first integral to the right of the equivalence should be $\int_a^bf(x)y_n(x)r(x)\,dx$ and that $y_n(a)=y_n(b)=0$. Then all you have to do is integration by parts twice, like this: $$ \int_a^bfy_nr=\int_a^b(-(py')'+(q-\lambda r)y)y_n= (-py'y_n+ypy_n')|_a^b+\int_a^by(-(py_n')'+(q-\lambda r)y_n) =py'(a)y_n(a)-py'(b)y_n(b). $$