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.

  • $\begingroup$ You obtained one solution from Maple. The general solution will contain 2 arbitrary constants. They and $\lambda$ are determined from the boundary conditions. $\endgroup$ Commented Nov 7, 2021 at 13:06
  • $\begingroup$ @AlexandreEremenko just made the edit $\endgroup$
    – Student
    Commented Nov 7, 2021 at 13:08
  • $\begingroup$ You say you get a "division by zero error". But that's not always true. Find the conditions under which it's not true, then you have the eigenvalues. $\endgroup$ Commented Nov 7, 2021 at 14:25
  • $\begingroup$ The dlmf might be helpful especially chapter 14. $\endgroup$
    – username
    Commented Nov 7, 2021 at 20:29

2 Answers 2


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 $.


What Maple tells you is that you could try a function of the form $Y(\theta)=(\sin\theta)^\frac14 y(\cos\theta)$. And indeed, when you do, you find that $y$ satisfies a Legendre equation, leading to the solutions provided. Your boundary condition in $y$ is a bit complicated at $x=\pm1$. Literally it is $$ \lim_{x\to\pm1} \left(\frac14 \frac{x}{(1-x^2)^\frac34}y(x)-(1-x^2)^\frac54 D(y)(x) \right) =0. $$ Which explains your division by zero conundrum.

Your initial problem is a degenerate Sturm-Liouville problem, since your leading order coefficient vanishes at the end points : it is possible that additional constraints are required to finish.

The choice of functions suggested by Maple imposes $Y(0)=Y(\pi)=0=Y^\prime(0)=Y^\prime(\pi)$, if $y$ is non singular, which seems a bit much. Maybe a better idea is to compromise and simply choose $Y(\theta)=y(\cos\theta)$ which encodes $0=Y^\prime(0)=Y^\prime(\pi)$ but assumes nothing about the values of $Y$ at the endpoints. In that case, you find $$ (1-x^2)y^{\prime\prime}-\frac32 y^\prime +\lambda y=0. $$ You now have no condition on $y$, so, again, you need to come up with some constraints.


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