0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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 $.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.