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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?

ArjenRobben
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  • The classical case is $-y''=\lambda y$ with $y(0)=y(2\pi)$ and $y'(0)=y'(2\pi)$. The eigenvalues are $n^{2}$. For $n=0$, only the constant function is an eigenfunction. For $n=1,2,3,\cdots$, both $\sin(nx)$ and $\cos(nx)$ are eigenfunctions. The eigenspaces for periodic problems can be of dimension 1 or 2 for regular problems. Separated endpoint conditions for regular problems lead to 1 dimensional eigenspaces. – Disintegrating By Parts Dec 07 '14 at 00:53
  • thank you, but are operators with periodic b.c.'s l.c.? Or is satisfying periodic b.c.'s anything else? Conversely, afaik l.c. and l.p. lead to one-dimensional eigenspaces. This should imply that periodic b.c.'s are neither. – ArjenRobben Dec 07 '14 at 01:00
  • @T.A.E. you could also post this as an answer I think. – ArjenRobben Dec 07 '14 at 01:28

1 Answers1

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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:

  • Separated conditions $$ \cos\alpha f(a)+\sin\alpha f'(a) = 0 \\ \cos\beta f(b) + \sin\beta f'(b) = 0. $$ The linear conditions are characterized by $\alpha,\beta \in [0,\pi)$.
  • Periodic conditions $$ f(a) - f(b) = 0 \\ f'(a) - f'(b) = 0. $$ There are variants of the periodic conditions, such as antiperiodic, etc, but the periodic conditions are the most common.

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 $\lambda\in\mathbb{C}$.

Disintegrating By Parts
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  • I think I should quote from Teschl's book: If the operator is l.p. at one endpoint, it is clear that all eigenvalues are simples, sincere there is at most one solution of $(L-\lambda)u=0$ which is square integrable near this endpoint. If $L$ is l.c., this also follows since the fact that two solutions of $(L-\lambda)u=0$ satisfy the same boundary conditions implies that their Wronskian vanishes. – ArjenRobben Dec 07 '14 at 12:00
  • Since you did not mention, how periodic bc.'s are related to l.c. or l.p. I don't see how this( from Teschl's book) agrees with your answer. Could you elaborate on this? – ArjenRobben Dec 07 '14 at 12:01
  • @ArjenRobben : I added more for you about the limit point and limit circle cases. Look at the end the post. – Disintegrating By Parts Dec 07 '14 at 12:30
  • If I understand you correctly, then the argument that l.c. operators have one-dimensional eigenspaces in general is wrong? I mean the operator $-f''$ on $[0,2\pi]$ should be regular and therefore l.c., but the eigenspaces are, as you pointed out, two-dimensional. – ArjenRobben Dec 07 '14 at 12:49
  • the argument I quoted is from this book on page 192 (or 204 in the pdf file) (at the bottom of the page)http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf – ArjenRobben Dec 07 '14 at 12:51
  • sorry, forgot to ping you – ArjenRobben Dec 07 '14 at 13:09
  • @ArjenRobben : You'll notice that he states regular problems are in the limit circle case at both endpoints. And you'll notice that my periodic example is such a case. So, there is some way in which he is not considering periodic conditions for that theorem, or he is wrong. The example I gave could not be more well-known or classical. And that suggests to me he has somehow excluded periodic conditions for that theorem. Teschl would undoubtedly know that example. He's an expert in the field. – Disintegrating By Parts Dec 07 '14 at 13:13
  • do you think I could write him an email and ask about this? – ArjenRobben Dec 07 '14 at 13:28
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    @ArjenRobben : I have not looked thoroughly through his book to know if he has excluded periodic somehow. But I do know the theorem is false for the periodic case unless "simple" doesn't mean 1-d eigenspaces, or $\tau$ is defined in such a way as to tacitly exclude periodic. So it might be worth writing him to clarify this point. Remind him of the example of $-\frac{d^{2}}{dx^{2}}+V$ where $V=0$ and $f(0)=f(2\pi)$, $f'(0)=f'(2\pi)$, and mention the fact that the eigenspaces are two dimensional for all eigenvalues except $0$. He'll know how to resolve any apparent conflict with this example. – Disintegrating By Parts Dec 07 '14 at 13:36
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    @ArjenRobben : By the way, here is where his argument breaks down: "If $\tau$ is l.c., this also follows since the fact that two solutions of $(\tau-\lambda)u=0$ satisfy the same boundary condition implies that their Wronskian vanish." That's the error in the logic for periodic which is not an error for separated conditions. Both $\sin(nx)$ and $\cos(nx)$ satisfy the two periodic conditions, but their Wronskian does not vanish. That's his last statement on the page. – Disintegrating By Parts Dec 07 '14 at 13:44
  • but actually, I have troubles to understand this argument at all: He says, well assume that you have two functions satisfying the same boundary conditions, then the wronskian vanishes(okay, for periodic bc.'s we just saw, that this does not have to be the case). But from that, he apparently also concludes linear dependence. I mean, how does a vanishing wronskian imply linear dependence? – ArjenRobben Dec 07 '14 at 13:47
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    @ArjenRobben : The Wronksian theorem is a general theorem of ODEs. Find a decent book on ODEs, and you'll find the Wronskian theorem for arbitrary orders of regular ODEs. Here's the idea: If the Wronskian vanishes at a point 'a', then the data arrays of values of $f$ and its derivatives are linearly dependent. Then the linear combination of these data vectors adding to 0 gives the same linear combination of solutions satisfying $f(a)=f'(a)=f''(a)=\cdots=f^{(n-1)}(a)=0$ which, by classical theorems, means $f$ is identically $0$. Works for all regular ODEs of order $n$. – Disintegrating By Parts Dec 07 '14 at 13:53
  • @ArjenRobben : For order $2$, the Wronskian $W(f,g)(x)$ is the determinant of of the matrix with columns $[f(x),f'(x)]$ and $[g(x),g'(x)]$. If $W(f,g)(a)=0$, then $\alpha [f(a),f'(a)]+\beta [g(a),g'(a)]=[0,0]$ for some $\alpha,\beta$ which are not both $0$. Thus $h=\alpha f+\beta g$ is a solution of the ODE with $h(a)=h'(a)=0$. By uniqueness of classical solutions, $h=0$. That's the idea. – Disintegrating By Parts Dec 07 '14 at 13:57
  • so for general functions a vanishing Wronskian does not imply that solutions are linear dependent, but if they are both solutions to the same ODE, this holds?-thanks, did not know this. – ArjenRobben Dec 07 '14 at 13:58
  • @ArjenRobben : Yes, for regular ODEs, the vanishing of the Wronskian for solutions of the ODE forces linear dependence. So the Wronskian vanishes identically or not at all for such solutions. In fact, you can derive an ODE for the Wronskian $W(f,g)(x)$ by differentiating in $x$ and using the fact that $f$, $g$ are solutions of the ODE. Try it, and you'll be amazed how simple the ODE for $W(f,g)(x)$ becomes. – Disintegrating By Parts Dec 07 '14 at 14:02
  • but regular implies l.c. not conversely, right?-But here we use that the Wronskian for an l.c. operator vanishes and want to see that the functions are linearly dependent. – ArjenRobben Dec 07 '14 at 14:06
  • @ArjenRobben : Yes, regular implies limit circle, but there are cases of limit circle for singular. Example: $Lf=-\frac{d}{dx}(1-x^{2})\frac{d}{dx}f = -(1-x^{2})f''+2xf'$ is Legendre's operator, which is limit circle at both endpoints of $[-1,1]$ even though it is singular at both endpoints. I posted an analysis of this problem because I saw so much confusion about it. http://math.stackexchange.com/questions/886775/selfadjoint-restrictions-of-legendre-operator-fracddx1-x2-fracddx – Disintegrating By Parts Dec 07 '14 at 14:12
  • also please look at the preceding Theorem 9.10. He says that if $\tau$ is l.c. at both endpoints, then any self-adjoint extension has purely discrete spectrum and the eigenfunctions (which are simple(!)) form an ONB. I mean, apparently $\tau$ is the maximal operator, so which operator does he want to extend there? – ArjenRobben Dec 07 '14 at 14:18
  • @ArjenRobben : Yes, either you would have selfadjoint extensions of the minimal operator, or selfadjoint restrictions of the maximal. I posed the problem for Legendre in terms of selfadjoint restrictions of the maximal operator. And I did mention how periodic types of conditions are possible for Legendre, too, even though the conditions at the endpoints are not the regular conditions of evaluation of a function and its derivative. Singular gives strange, asymptotic endpoint conditions that are not generally equivalent to simple evaluations. Whatever comes out of the Wronskian is what you get. – Disintegrating By Parts Dec 07 '14 at 14:23
  • thank you, I will ask him now and will inform you about his answer. – ArjenRobben Dec 07 '14 at 14:26
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    @ArjenRobben : Correction to earlier comment: Regular ODE on a finite interval [a,b] implies limit circle at $a$ and $b$. Infinite intervals are different. – Disintegrating By Parts Dec 07 '14 at 14:36
  • you were right, we need separated boundary conditions, thanks. so this actually was an error. – ArjenRobben Dec 07 '14 at 15:42
  • @ArjenRobben : Thank you. I'm glad that you pursued the issue and were able to get an answer. Someone else expressed that belief to me recently, and I wonder if that's where they got it. Interesting. – Disintegrating By Parts Dec 07 '14 at 18:18