Spectrum of this ODE

I noticed something interesting studying this Sturm-Liouville Problem:

$$\frac{d}{dx}\left(\sqrt{(1-x^{2})}\frac{df}{dx} \right)+\frac{\left(n \alpha x+\alpha^2 x^{2} + \lambda\right)f}{\sqrt{(1-x^{2})}}= 0,$$ where $\alpha \in \mathbb{R}$ with periodic boundary conditions on $[-1,1]$. (Lambda is the eigenvalue) From basic St.-Liouville theory we know that the spectrum will be discrete cause the Sturm-Liouville operator is self-adjoint and the problem is regular.

Furthermore, we will have that $-\infty <\lambda_0 < \lambda_1 \le \lambda_2 < \lambda_3 \le .. \ .$

Now I noticed by extensive numerical calculations that for odd $n$ we will have that $\infty < \lambda_0 < \lambda_1 < \lambda_2 <...< \lambda_n,$ but after $\lambda_n$ all eigenvalues occur pairwise $\lambda_{n+1}=\lambda_{n+2}$. Is there a way to explicitely show this?

Edit: Due to two good questions, I want to add some information to the question: The eigenfunctions seem to a double eigenvalue seem to be even/odd functions respectively. I assume $\alpha >0$, but maybe it is worth studying $\alpha=0$ first (is similar to Legendre's ODE).

Edit2: I think I could reduce my problem to a Linear Algebra problem. Due to the very different kind of question this observation includes I asked a new question about this: https://mathoverflow.net/questions/177814/explain-eigenvalue-structure-of-these-sequences-of-matrices .

• what about the pairwise eigenfunctions, do they have any remarkable "pairwise" behaviour (odd/even, or whatever...)? Aug 4, 2014 at 19:56
• Something is strange: parameter $n$ enters your problem only in combination $n\alpha$. Are you saying that your equalities hold for odd $n$ and all $\alpha$? Aug 4, 2014 at 21:36
– user54300
Aug 4, 2014 at 21:49
• How could this problem possibly have even/odd eigenfunctions? Doesn't the term $n\alpha x f(x)$ have the wrong parity? Aug 4, 2014 at 22:27
• Right; that's why your equation is regular. What I'm trying to say is that in Legendre's equation, you have $p=(1-x^2)$, so $1/p\notin L^1$, and this equation is singular. Aug 4, 2014 at 23:18

Let's rewrite your equation as a Schrödinger equation, as follows: Introduce the new variable $t\in (-\pi/2, \pi/2)$ by $x=\sin t$. Then if $f$ solves your boundary value problem (with periodic boundary conditions), then $y(t)=f(x(t))=f(\sin t)$ satisfies $$-y'' +V(t)y = \lambda y , \quad\quad y(-\pi/2)=y(\pi/2),\quad y'(-\pi/2)=y'(\pi/2) ,$$ with the potential $$V(t) = -n\alpha\sin t -\alpha^2\sin^2 t = \frac{\alpha^2}{2}\cos 2t - n\alpha\sin t-\frac{\alpha^2}{2}.$$ (We can do such a transformation for any SL equation; see for example here.)
The additional change of variable $t=2s-\pi/2$ (so $0<s<\pi/2$) identifies $V$ as a Whittaker-Hill potential (up to an irrelevant shift): $$V(s) = -\frac{\alpha^2}{2}\cos 4s -n\alpha\cos 2s - \frac{\alpha^2}{2}$$
The difference of two consecutive periodic or anti-periodic eigenvalues is a gap of the associated problem on $\mathbb R$ with $\pi/2$-periodic potential. The Whittakker-Hill potentials are indeed known to have the property you are interested in for suitable parameters, see here.
PS: This last part came as quite a surprise to me; you can see from the edit history that I first conjectured the opposite. It's perhaps interesting in this context to mention that the Mathieu potential $W = g\cos 4s$ (corresponding to $n=0$ above) has all its gaps open.