= Solution
For sufficiently smooth functions satisfying the same <boundary conditions>, <integration by parts> gives the boundary form
$$
\begin{aligned}
\langle f,\mathcal Lg\rangle-\langle\mathcal Lf,g\rangle
&=\int_{-L}^L\left[-f^*(pg')'+(pf^{*\prime})'g\right]dx\\
&=\left[p\bigl(f^{*\prime}g-f^*g'\bigr)\right]_{-L}^{L}.
\end{aligned}
$$
The stated endpoint relation annihilates this expression. This proves the required <self-adjoint differential operator> identity; for the usual regular <Sturm-Liouville problem>, separated real <boundary conditions> give a self-adjoint realization on its domain.
In this problem the transformed <diffusive flux> is $J=-e^{-\lambda t}py'$. Thus the physical <boundary conditions> are
$$
\boxed{py'(\pm L)=0,}
$$
equivalently $y'(\pm L)=0$ when the endpoint values of $p$ are positive. In particular $pf'=pg'=0$ at the endpoints, which makes the boundary form vanish directly. The <inner product> used here is the ordinary integral; the weight $w$ enters the <generalized eigenvalue problem>, not this boundary-form calculation.
Back to article page
