Consider the following Sturm-Liouville problem,
$$(\sqrt{\sin \theta} Y')' + \lambda \sqrt{\sin \theta} Y =0$$
where $Y(\theta):[0,\pi] \to \mathbb{R}$ with boundary conditions $Y'(0)=Y'(\pi)=0.$


I used maple and got the following explicit solution,
$$Y(\theta) = \sin(\theta)^{1/4} \left(c\_1P^\mu\_{\nu}(\cos \theta) + c\_2Q^{\mu}\_{\nu}(\cos \theta)\right)$$
where $\mu=1/4$ and $\nu = \frac{\sqrt{16\lambda+1}}{4}-\frac{1}{2}.$ When I try to differentiate this expression and plug in the boundary conditions, I get division by zero error. What other ways can I use to compute the eigenvalue of this expression?


**Edit:**
I tried to convert the above ode into the following form $u''+Vu=0$
where
$$u(\theta) = \sin^{1/4}(\theta) Y(\theta)$$
and
$$V(\theta) = \lambda + \frac{1}{4}\csc^2(\theta)-\frac{1}{16}\cot^2(\theta).$$
I am wondering if some property of the potential $V$ can be exploited to find perhaps upper or lower bounds for the eigenvalues.



Asymptotic for eigenvalues for the following ode?
I asked Mathematica about the boundary behavior. First, the $P^{1/4}\_{\nu } $ solution: We have, for $\epsilon \searrow 0$,
$$
(\sin \epsilon )^{1/4} P^{1/4}\_{\nu } (\cos \epsilon ) = \frac{2^{1/4} }{\Gamma(3/4)} + O(\epsilon^{2} )
$$
and
\begin{eqnarray\*}
(\sin (\pi -\epsilon ))^{1/4} P^{1/4}\_{\nu } (\cos (\pi -\epsilon) ) &=& \frac{2^{3/4} \pi }{\Gamma(3/4)\Gamma(-\nu )\Gamma(1+\nu )} \\
& & -\frac{2^{1/4} \pi }{\Gamma(5/4)\Gamma(-1/4-\nu )\Gamma(3/4+\nu )} \sqrt{\epsilon } \\
& & + O(\epsilon^{2} )
\end{eqnarray\*}
So, the $P^{1/4}\_{\nu } $ solution automatically satisfies the boundary condition at $\theta =0$, whereas at $\theta =\pi $, we have to eliminate the term proportional to $\sqrt{\epsilon } $. That determines the eigenvalues: We need either $-1/4-\nu $ to be a negative integer or 0, or $3/4+\nu $ to be a negative integer or 0.
The specification of $\nu $ in the OP suggests the constraint $\nu \geq -1/2$; this excludes the second alternative, and therefore we obtain the spectrum $\nu = n-1/4$, $n=0,1,2,3,\ldots $.


The $Q^{1/4}\_{\nu } $ solution, on the other hand, exhibits behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$, with coefficient
$$
\frac{\pi^{2} }{2^{1/4} } \frac{\cos ((4\nu +1)\pi /8) \Gamma (-\nu /2 -1/8) \Gamma (\nu /2 +9/8) - \sin ((4\nu +1)\pi /8)\Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +5/8)}{\Gamma (5/4) \Gamma (-\nu /2 -1/8) \Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +3/8)\Gamma (\nu /2 +7/8)}
$$
A plot as a function of $\nu $ suggests that, for $\nu \geq -1/2$, this is positive and monotonically rising (I have not attempted to verify this analytically); the behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$ can therefore not be eliminated, nor can it be compensated by admixture of the $P^{1/4}\_{\nu } $ solution. Thus, there are no further solutions involving $Q^{1/4}\_{\nu } $.


In summary, the complete spectrum is given by $\nu = n-1/4$, $n=0,1,2,3,\ldots $, or, in terms of $\lambda $, $\lambda = n(n+1/2)$, $n=0,1,2,3,\ldots $.


