# Asymptotic for eigenvalues for the following ode?

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.

• You obtained one solution from Maple. The general solution will contain 2 arbitrary constants. They and $\lambda$ are determined from the boundary conditions. Commented Nov 7, 2021 at 13:06
• @AlexandreEremenko just made the edit Commented Nov 7, 2021 at 13:08
• 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. Commented Nov 7, 2021 at 14:25
• The dlmf might be helpful especially chapter 14. Commented Nov 7, 2021 at 20:29

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.
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.