Sturm–Liouville equation and the Eigenvalue general Problem (PDE)

Question

As I am studying for my Partial Differential Equations exam, I came across Sturm–Liouville equation where it says that it's solutions $y(x)$ are the eigenfunctions of the general problem $Ly=λy$. I do understand that the equation $py''+p'y'+(λw-υ)y=0$ is a special case of the eigenvalue problem $Ly=λy$, where $L=a(x)\frac{d^{2}}{dx^{2}}+b(x)\frac{d}{dx}+c(x)$, as it demands that $b=a'$, but I have some issues here. Maybe it's to early already as I haven't checked any exercises yet, but I'll ask anyway. Why is that particular form (Liouvilles' form) of the general eigenvalue problem so special, and why should I care about turning other equations (like Bessels') into that form? Can't I solve the general problem properly without it being in Liouvilles' form? Thanks in advance, I haven't got the time unfortunately to get into more details by myself so you are saving me a lot of time.

Answer 1

You want an operator that is symmetric. Suppose $$ Lf = a\frac{d^2}{dx^2}f+b\frac{d}{dx}f+cf $$ You want $a$ and $b$ to be such that $$ \int_{a}^{b}(Lf)gdx = \int_{a}^{b}f(Lg)dx+\mbox{evaluation terms}. $$ Symmetry of operators plays a critical role in Sturm-Liouville problems. It turns out that $L$ needs to have the form $$ Lf = \frac{d}{dx}\left(a\frac{df}{dx}\right)+cf $$ When you integrate by parts, you can see how you get symmetry out of that. And you can work out that this form is necessary. That means $b=a'$.

The reason that symmetry is so important is the same reason that symmetry is important for a matrix. if you view this in terms of an inner product \begin{align} (Lf,g) &= \int_{a}^{b}(Lf)gdx \\ &= ( \frac{d}{dx}(a\frac{df}{dx}),g)+(cf,g) \\ &= \mbox{evaluation terms} - (a\frac{df}{dx},\frac{dg}{dx})+(f,cg) \\ & = \mbox{evaluation terms}+(f,\frac{d}{dx}(a\frac{dg}{dx}))+(f,cg) \\ & =\mbox{evaluation terms}+(f,Lg) \end{align} The full Sturm-Liouville problem requires endpoint conditions that make the evaluation terms vanish, so that you end up with full symmetry.