Consider the following singular Sturm–Liouville problem on $[0,R]$ $$ {d\over dx}\left(x^4{dy\over dx}\right)+Cx^4y=-\lambda x^4 y $$ for some constant $C$ and with the usual Robin boundary condition $\alpha y(R)+\beta y'(R)=0$ at $x=R$. This is a singular Sturm–Liouville problem as $x^4$ vanishes at $0$. (Or even more generally we can replace $x^4$ with, say, $x^d$ for $d\in\mathbb N$.)
Does the usual conclusion of (regular) Sturm–Liouville theory hold in this case? In particular, for example, does a smallest eigenvalue $\lambda=\lambda_0$ exist, and such that the corresponding eigenfunction has no zeros in $(0,R)$?
I've spent days looking for the answer on internet and textbooks, but many of them are just elementary textbook/webpage that talked only about the regular Sturm–Liouville problem; and the ones that do talked about the singular Sturm–Liouville problem are all really advanced esoteric non-accessible bits and pieces filled with jargons that seems to be only aimed at experts.
From what I can gather, apparently singular endpoints could be classified into limit-circle and limit-point type (possibly with more sub-types to do with oscillatory or not). But I could not, for the life of me, find out a clear criterion of how exactly I can classify particular examples of singular Sturm–Liouville problem into these types. And furthermore, even if I could, I'm still stuck since I also could not find clear statements about how being each of these types of singularity changes the conclusion of Sturm–Liouville theory.
For example, does the usual conclusion of (regular) Sturm–Liouville theory holds when one endpoint is of limit-point type? Any idea where I can find clear accessible statements or descriptions of singular Sturm–Liouville theory that does not require me studying 100 hours just to understand how I can apply it to particular examples of singular Sturm–Liouville problems? Or at the very least, what's the answer to the particular singular Sturm–Liouville problem I stated at the top here? Thanks a lot!
The problem you're considering can be cast in terms of the following operator defined on the weighted Hilbert space $L^2_{x^4}[0,R]$ with inner product $\langle f,g\rangle_{L^2_{x^4}}=\int_0^R f(x)g(x)x^4dx$: $$ Lf = -\frac{1}{x^4}\frac{d}{dx}\left(x^4\frac{df}{dx}\right)-Cf. $$ If you consider set $\lambda=-C$ and consider $Lf=\lambda f$, then the classical solutions are $$ \frac{d}{dx}\left(x^4\frac{df}{dx}\right)=0 \\ x^4\frac{df}{dx}=A \\ \frac{df}{dx} = Ax^{-4} \\ f = -\frac{1}{4}A x^{-3}+B $$ Only one of the two terms on the right is in $L^2_{x^4}[0,R]$. So $L$ is in the limit point case near $x=0$, which means that there are no non-zero evaluation terms at $x=0$ for $$ \langle Lf,g\rangle_{L^2_{x^4}}-\langle f,Lg\rangle_{L^2_{x^4}}. $$ And that is equivalent to not being able to impose any condition at $x=0$. On the other hand, $L$ is regular at $x=R$, which means that a condition must be imposed at $x=R$ in order for $L$ to be self-adjoint. Any standard condition at $x=R$ of the following form will work: $$ \cos(\theta) f(R)+\sin(\theta) f'(R)=0. $$