Sturm–Liouville theory - show that the linear operator is self-adjoint.

The goal of this post is to document my derivation that the linear operator for the Sturm Liouville theory is actually self-adjoint

The differential equation is

$\frac{d}{dx}(p(x) \frac{d}{dx}y) + (\lambda w(x) + q(x))y = 0$

Rearranging, we can write this as an eigenvalue problem as follow:

$\frac{d}{dx}(p \frac{d}{dx}y) + qy = -\lambda w y$

$\frac{1}{w}(\frac{d}{dx}(p \frac{d}{dx}y) + qy) = -\lambda y$

Define the operator $L = \frac{1}{w}(\frac{d}{dx}(p \frac{d}{dx}) + q)$, it is trivial to verify the operator is linear.

Consider the function space with inner product defined as $<a, b> = \int a b w dx$, we wanted to verify the operator $L$ is self-adjoint in this space, in other words, $<As, t> = <s, At>$.

$\begin{eqnarray*} <As, t> &=& \int \frac{1}{w}(\frac{d}{dx}(p \frac{d}{dx}s) + qs) t w dx \\ &=& \int (\frac{d}{dx}(p \frac{d}{dx}s) + qs) t dx \\ &=& \int \frac{d}{dx}(p \frac{d}{dx}s)t dx + \int qst dx \end{eqnarray*}$.

Similarly we can write the other side as

$\begin{eqnarray*} <s, At> &=& \int \frac{1}{w}(\frac{d}{dx}(p \frac{d}{dx}t) + qt) s w dx \\ &=& \int (\frac{d}{dx}(p \frac{d}{dx}t) + qt) s dx \\ &=& \int \frac{d}{dx}(p \frac{d}{dx}t)s dx + \int qts dx \end{eqnarray*}$.

So all we needed to prove is $\int (ps')'t dx = \int (pt')'s dx$, we do that using integration by part:

$\begin{eqnarray*} & & \int (ps')'t dx \\ &=& \int t d(ps') dx \\ &=& ps't - \int ps' dt \\ &=& ps't - \int ps't' dx \\ &=& ps't - \int pt' ds \\ &=& ps't -pt's + \int s d(pt') \\ &=& ps't -pt's + \int (pt')' s dx \\ \end{eqnarray*}$.

So we need to show $ps't - pt's = 0$. This should come from the boundary conditions, which is not specified here. The boundary of a Sturm Liouville problem should satisfy this. For example, $p(x) = 0$ for the Legendre's problem at the boundary.