eigenvaue of Sturm Liouville problem Let the limit probem
$$
\begin{cases}
(P(x) y')' + q(x) y' + \lambda r(x) y=0\\
\alpha_0 y(0)+ \alpha_1 y'(0)\\
\beta_0 y(l) + \beta_1 y'(l)
\end{cases}
$$
with
$\alpha_0^2 + \alpha_1^2 >0$ and $\beta_0^2 + \beta_1^2 >0$
How can we prove that the eigenvalues of this problem are simple, i.e. the eigenfunctions associated to eigenvalues of this problem are orthogonal? 
Thanks for the help.
 A: The eigenfunctions will not necessarily be orthogonal with respect to an unweighted space. For the sake of discussion, assume 


*

*$P$, $q$ and $r$ are real;

*$P$ is continuously differentiable on $[0,l]$ and strictly positive;

*$r$ is continuous and strictly positive;
Then this problem falls into a nice classical setting where you can multiply by a factor to put the equation into selfadjoint form. The final form is not unique, but any form will do. For example, starting with
$$
             Py''+(P'+q)y'+\lambda r y = 0,
$$
you can multiply by $\frac{1}{P}\exp\left\{\int \frac{P'+q}{P}dx\right\}$ and have
$$
               \frac{d}{dx}\left(a\frac{dy}{dx}\right)+\lambda w y = 0,
$$
where
$$
            a=\exp\left\{\int\frac{P'+q}{P}dx\right\},
\;\;\; w = \frac{r}{P}\exp\left\{\int\frac{P'+q}{P}dx\right\}.
$$
Then the eigenfunctions are orthogonal with respect to the weighted inner product
$$
          (f,g)_{L^2_w[0,l]} = \int_{0}^{l}f(t)\overline{g(t)}w(t)dt.
$$
The orthogonality follows from the Lagrange identity for the operator
$$
                 Lf = \frac{1}{w}\frac{d}{dx}\left(a\frac{dy}{dx}\right)
$$
which is
$$
        (Lf)g-f(Lg) = \frac{1}{w}\left[\frac{d}{dx}\left(a\frac{df}{dx}\right)g-f\frac{d}{dx}\left(a\frac{dg}{dx}\right)\right] \\
   = \frac{1}{w}\frac{d}{dx}\left[\left(a\frac{df}{dx}\right)g-af\left(\frac{dg}{dx}\right)\right]
$$
This identity gives you symmetry of the operator $L$:
\begin{align}
    (Lf,g)_w-(f,Lg)_w & =\int_{0}^{l}\{(Lf)\overline{g}-f(\overline{Lg})\}wdx \\
    & = a\{f'\overline{g}-f\overline{g}'\}|_{0}^{l} \\
    & = \left(a\left.\left|\begin{array}{cc} \overline{g} & f \\ \overline{g}' & f'\end{array}\right|\right)\right|_{x=0}^{l}
\end{align}
The right side vanishes if both $f$ and $g$ satisfy the same real endpoint conditions as you described. Therefore,
$$
               (Lf,g)_w = (f,Lg)_w
$$
for all twice continuously differentiable $f$, $g$ that satisfy the required endpoint conditions. Using the same argument you use for symmetric matrices then proves that (a) the eigenvalues are real and (b) the eigenfunctions corresponding to different eigenvalues are orthogonal in the $(\cdot,\cdot)_w$ weighted inner product.
