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Suppose $n \ge 2$ an integer and consider finding the first eigenvalue of $$ -\partial_\theta \left( \omega(\theta) \psi'(\theta) \right) = \mu_1 \omega(\theta) \psi(\theta)$$ for $ 0<\theta<\frac{\pi}{4}$ with $ \psi'(0)=\psi'( \frac{\pi}{4})=0$ and we want $ \psi'>0$. Of course the first eigenvalue is zero but we want the first with a nonconstant eigenfunction.

In the case of $(0, \frac{\pi}{2})$ one can explicitly compute the first eigenpair and the first eigenvalue is (i believe) $4n$. Additionally using Sturm-Liouville theory (and reflecting across $ \frac{\pi}{4}$) one can see this eigenvalue we are looking is maybe the third eigenvalue on $(0, \frac{\pi}{2}$).

Question. So I am interested in whether there is some trivial formula for the eigenvalue I am looking for (and I just can't see it or..).

thanks for all replies.

EDIT. We are taking $\omega(\theta):=\cos^{n-1}(\theta) \sin^{n-1}(\theta)$. Also I am taking $ \mu_1$ to be the first non trivial eigenvalue. I guess I should really just say $ \mu_2$.

EDIT 2 (Really a long comment). thanks for the comments. Let me explain where this is coming from. In another problem I had $ \omega(\theta) = \cos^{m-1}(\theta) \sin^{n-1}(\theta)$ and I was working on $ (0, \frac{\pi}{2})$ with the Neumann BC. Here $m,n \ge 1$ integers. By playing around with the problem I realized there was an explicit (and easy) first nontrival eigenpair. Now I am taking $m=n$ and reducing it to $( 0 , \frac{\pi}{4})$ and I don't see some nice easy formula. I can see by extending across $ \theta = \frac{\pi}{4}$ this is really maybe the third or so eigenvalue of the full problem on $(0,\frac{\pi}{2})$.

Edit 3. (sorry for all the edits). In the above example with $N=m+n$ on $(0, \frac{\pi}{2})$ I believe one has

$$ \mu_1=2N, \quad \psi_1(\theta)= \frac{m-n}{N} - \cos(2 \theta).$$

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    $\begingroup$ Maybe it's obvious for the other viewers but it isn't for me: what is your operator? Is it $\psi\mapsto-(\omega\psi')'/\omega$? Or does it act on $\omega$? Is $n$ the dimension? What is $\omega$, a smooth function? You're asking for the eigenvalue of an equality, is $\mu_1$ a notation for your eigenvalue? For me, your question is very difficult to read, and requires a few more precisely stated “Let ... be ...”. $\endgroup$
    – Pierre PC
    Commented Jan 30, 2022 at 10:17
  • $\begingroup$ I knew I wrote a sloppy question but I didn't realize it was this bad. Let me go edit it a bit. $\endgroup$
    – Math604
    Commented Jan 30, 2022 at 18:02
  • $\begingroup$ @PierrePC: Yes, that would be the standard operator associated with this SL equation. It acts on $L^2(\omega\, dx)$. $\endgroup$
    – Christian Remling
    Commented Jan 30, 2022 at 18:59
  • $\begingroup$ So normally one would insist that $\omega > 0$ ... $\endgroup$
    – Christian Remling
    Commented Jan 30, 2022 at 19:00
  • $\begingroup$ I was convinced there was no simple little formula but I made a mistake. After looking for solutions of the form $ \psi(\theta)= A - \cos(4 \theta)$ I believe there is a simple solution. So I guess I should close the question. Thanks for all the comments. $\endgroup$
    – Math604
    Commented Feb 2, 2022 at 17:25

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