I was wondering about this:
I know that if a 1-d Sturm-Liouville operator is limit circle or limit point then the eigenvalues are simple ( so no degenerated spectrum). But in the case of periodic boundary conditions it is actually possible that two eigenvalues agree, that's why people talk about spectral gaps, I think. Now if I have a regular Sturm-Liouville problem, then my operator is l.c. at both end-points, right?-So my spectrum should be simple. Thus, I have troubles to understand how the fact that two eigenvalues may agree for periodic boundary conditions fits into this l.c. and l.p. picture?
Best Answer
For a regular Sturm-Liouville eigenvalue problem on a finite interval $[a,b]$, say $$ Lf = \left[-\frac{d}{dx}p\frac{d}{dx}+q\right]f = \lambda f, $$ there are two types of standard endpoint conditions:
By "regular", one generally means that (a) $[a,b]$ is a finite interval (b) $p$ is real, continuously differentiable and strictly positive on $[a,b]$, and (c) $q \in L^{1}[a,b]$ a real function.
For regular problems with separated or periodic conditions, the eigenvalues are real and isolated because they're the zeros of an entire function $\omega(\lambda)$ which is a Wronskian type of determinant involving classical eigenfunction solutions.
For separated conditions, the eigenspaces are one-dimensional.
For periodic conditions, the eigenspaces can be one-dimensional or two-dimensional. For example, consider $$ Lf = -f'' = \lambda y,\\ f(0)=f(2\pi),\\ f'(0)=f'(2\pi). $$ The eigenvalues are $n^{2}$ for $n=0,1,2,3,\cdots$. The eigenspace for $n=0$ is one dimensional and spanned by the constant function $1$. The eigenspaces for $n > 0$ are two-dimensional and spanned by $\sin(nx)$, $\cos(nx)$. The eigenspaces for general regular periodic problems can be a mix of one- and two-dimensional spaces. The dimension of the eigenspace with eigenvalue $\mu$ is the order of zero of the aforementioned $\omega$ at $\mu$.
Singular problems are very different. A problem can be singular because (a) the interval [a,b] is infinite (b) the function $p$ vanishes at $a$ or at $b$ (or at both $a$ and $b$) or (c) $q$ is not absolutely integrable on the full interval. Spectral band gaps have to do with periodic potentials $q$ on $(-\infty,\infty)$; the spectrum is typcially continuous spectrum that comes in bands (i.e., intervals.)
Limit-Point/Limit-Circle: Ignoring any endpoint conditions for the moment, suppose $L$ is regular on $[a,c]$ for all $a < c < b$, meaning that $p$ and $q$ are as stated. The classical eigenfunction solutions of $Lf=\lambda f$ (no mention of endpoint behaviors or endpoint conditions) are in $L^{2}[a,c]$ for all $c$, but may not be in $L^{2}$ on $[a,b)$. The Weyl alternative theorem states that there are two possible cases which can occur, and these are named limit point and limit circle (these are historical names used to describe Weyl's original method of proof.)
The limit circle case is where all classical solutions are in $L^{2}[a,b)$ for some $\lambda$. If all classical solutions are in $L^{2}[a,b)$ for some $\lambda$, then the same is true of all $\lambda\in\mathbb{C}$.
The limit point case is the alternative. In this case, for all non-real complex $\lambda$, there is always at least one classical eigenfunction solution of $Lf=\lambda f$ which is in $L^{2}[a,b)$. For real $\lambda$ there may be no classical solutions of $Lf=\lambda f$ which are in $L^{2}[a,b)$ (this is case for $L=-\frac{d^{2}}{dx^{2}}$ on $[0,\infty)$, for example.) But there is always one for $\Im\lambda \ne 0$.
In the limit point case, you can never have more than a one dimensional eigenspace for any $\lambda$, regardless of what conditions you impose, when you consider $L$ as a operator on $L^{2}[a,b)$, which is where Hilbert space operators are defined. The limit point case is equivalent to not being able to impose any kind of endpoint condition at $b$. The limit circle case is equivalent to there existing two possible endpoint conditions at $b$ which are limiting expressions of some kind.
The limit point and limit circle cases only apply when one endpoint is regular. If $L$ is singular at $a$ and at $b$, then there do not have to exist any eigenfunctions in $L^{2}(a,b)$ for any $\lambda$. That's the case, for example, when you consider $L=-\frac{d^{2}}{dx^{2}}$ on $(-\infty,\infty)$.
Note: All regular operators on $[a,b]$ are in the limit circle because all classical solutions of $Lf=\lambda f$ are in $L^{2}[a,b]$ for all $\lambda\in\mathbb{C}$.