Is this a regular Sturm-liouville problem Given
$$C_{t} + VC{x}=C_{xx}$$
with 
$$0<x<L and t>0$$
with BC: $$C(0,t) = 0 , C_{x}(L,t)= 0 $$
Then the associated Eigenvalue problem is 
$$X" - vX'=- \lambda X$$
With BC 
$$X(0) = X'(L) = 0$$
I'm asked if this is a regular eigenvalue problem. I don't think it is based on my workings but a confirmation would be helpful

The eigenvalue problem $Lf=\lambda f$, where
$$
           Lf=-\frac{d^{2}}{dx^{2}}f +V(x)\frac{d}{dx}f,
                 f(0)=f(L)=0
$$
can be put into Sturm-Liouville form in several different ways. For example,
$$
         -\frac{d}{dx}e^{-\int Vdx}\frac{d}{dx}f = \lambda e^{-\int Vdx}f
$$
In this context, the eigenfunctions will be orthogonal on $[0,L]$ with respect to the weight function $w(x)=e^{-\int Vdx}$, which multiplies $\lambda$ on the right. To see why, assume $f$ is an eigenfunction with eigenvalue $\lambda$ and $g$ is an eigenfunction with eigenvalue $\mu$, then
\begin{align}
        (\lambda-\mu)e^{-\int Vdx}(fg) & = (Lf)g-f(Lg) \\
             & =-\frac{d}{dx}\left(e^{-\int Vdx}(f'g-fg')\right).
\end{align}
Therefore,
$$
    \left.(\lambda-\mu)\int_{0}^{1}f(x)g(x)e^{-\int Vdx}dx=-e^{-\int Vdx}(f'g-fg')\right|_{0}^{2\pi} = 0.
$$
