Disambiguating the Sturm-Liouville form of a differential equation with two candidate eigenvalues

Question

Context

There are some differential equations (like the one presented below) where there appears to be more than one way to rewrite the equation in Sturm-Liouville form. I am posting this question to attempt to understand how to resolve the ambiguity.

• Please bare in mind that there are more than one conventions for where positive and negative markers need to be in order for a differential equation to be considered to be in Sturm-Liouville form. This should not impede resolving this matter.

Question

Rewrite the equation below in Sturm-Liouville form and identify what symbol (either $\ell$ or $m$) is the eigenvalue. \begin{align*} \rho^2\,\frac{d^2 P_{\ell m}}{d\rho^2} + \rho \,\frac{d P_{\ell m}}{d\rho } + \left(\ell^2\,\rho^2 - m^2 \right)P_{\ell m} &= 0. \end{align*}

My two answers

From [2], I know that if both sides of \begin{align*} \frac{d^2 P_{\ell m}}{d\rho^2} + \frac{1}{\rho} \,\frac{d P_{\ell m}}{d\rho } + \frac{\left(\ell^2\,\rho^2 - m^2 \right)}{\rho^2}\,P_{\ell m} &= 0. \end{align*} are multiplied by \begin{align*} p(x) &= \exp{\left( \int_{\rho_0}^\rho \frac{1}{z} \,dz \right)} = \exp{\left( \ln{\left(\frac{\rho}{\rho_0}\right)} \right)} = \frac{\rho}{\rho_0}, \end{align*} then the differential equation is put in the formally self-adjoint form \begin{align*} \frac{d }{d\rho }\left( \frac{\rho}{\rho_0}\frac{d P_{\ell m}}{d\rho }\right) + \frac{\rho}{\rho_0}\frac{\left(\ell^2\,\rho^2 - m^2 \right)}{\rho^2}\,P_{\ell m} &= 0. \end{align*}

Option 1.

\begin{align*} \frac{1}{\left[ \frac{1}{\rho_0\,\rho} \right]} \left( -\frac{d }{d\rho }\left[ \frac{\rho}{\rho_0}\frac{d}{d\rho }\right] + \left[-\frac{\rho}{\rho_0}\,\ell^2 \right] \right)P_{\ell m} &= -\left[m^2 \right]\,P_{\ell m} . \end{align*} This option works. This is because $ \left[\frac{\rho}{\rho_0}\right]$, $\left[ \frac{\rho}{\rho_0}\right]'$, $\left[-\frac{\rho}{\rho_0}\,\ell^2 \right]$, and $\left[ \frac{1}{\rho_0\,\rho} \right]$ are real and continuous, and $\left[ \frac{1}{\rho_0\,\rho} \right]>0$ and $ \left[\frac{\rho}{\rho_0}\right]>0$ on the interval on the interval $[a,b]$, where $b>a > 0$.

Option 2.

\begin{align*} \frac{1}{\left[ \frac{\rho}{\rho_0}\right]}\left(-\frac{d }{d\rho }\left[\frac{\rho}{\rho_0}\frac{d }{d\rho }\right] + \left[ \frac{ m^2 }{\rho_0\,\rho }\right] \right)P_{\ell m}(\rho) &= -\left[-\ell^2\right] P_{\ell m}(\rho) . \end{align*} This option works. This is because $ \left[\frac{\rho}{\rho_0}\right]$, $\left[ \frac{\rho}{\rho_0}\right]'$, $\left[ \frac{ m^2 }{\rho_0\,\rho }\right]$, and $\left[ \frac{\rho}{\rho_0} \right]$ are real and continuous, and $\left[ \frac{\rho}{\rho_0} \right]>0$ and $ \left[\frac{\rho}{\rho_0}\right]>0$ on the interval on the interval $[a,b]$, where $b>a > 0$.

My Question

Even though I know from [1] that Option 2 is considered the correct way to interpret the governing differential equation (so that the eigenvalue is $\pm\ell^2$), I do not understand how come Option 1 is incorrect as an interpretation. Can you please explain how to resolve this ambiguity?

Bibliography

[1] Arfken and Weber, 5th Edition.

[2] Zwillnger, "Handbook of Differential Equations", page 97, 1st Edition.

Answer 1

Consider the heat equation for $\psi(r,\theta,t)$ on the unit disk centered at the origin. $$ \frac{\partial\psi}{\partial t}=c\nabla^2\psi,\;\; 0\le r\le 1,\;0\le\theta\le 2\pi,\;\;t \ge 0. $$ Using polar coordinates $r,\theta$, the Laplacian may be written as $$ \nabla^2\psi = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2\psi}{\partial\theta^2}. $$ To perform the separation of variables, assume the following form of solution: $$ \psi(r,\theta,t)=R(r)\Theta(\theta)T(t). $$ Plug this into the heat equation and divide the result by $\psi$. Here's what you get: $$ \frac{1}{c}\frac{T'}{T}=\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)+\frac{1}{r^2\Theta}\frac{d^2\Theta}{d\theta^2} $$ You can separate out the $\theta$ components after multiplying both sides by $r^2$: $$ \frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2}= \frac{r^2}{c}\frac{T'}{T}-\frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right) $$ So there is a parameter $m$ such that $$ \frac{1}{\Theta}\frac{d^2\Theta}{d\theta^2}=-m^2, \;\; -m^2 = \frac{r^2}{c}\frac{T'}{T}-\frac{r}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right). $$ The parameter $m$ is determined at this point by periodicity in $\theta$ and $\Theta(\theta)=A\cos(m\theta)+B\sin(m\theta)$. Assuming periodicity, $m=0,1,2,3,\cdots$ (negative values of $m$ do not give anything new.) So you don't get to choose the values of $m$ when you're working on a circular disk such as this. The values of $m$ are dictated by the geometry and periodicity in $\theta$.

The final separation of variables is $$ \frac{T'}{cT}=\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{m^2}{r^2} \\ \frac{T'}{cT}=\nu,\;\; \nu=\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{m^2}{r^2} $$ The equation in $r$ may be written as $$ \frac{1}{r}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\left(\frac{m^2}{r^2}+\nu\right)R=0 \\ \frac{d}{dr}\left(r\frac{dR}{dr}\right)-\left(\frac{m^2}{r}+\nu r\right)R=0 \\ r\frac{d^2R}{dr^2}+\frac{dR}{dr}-\left(\frac{m^2}{r}+\nu r\right)R=0. $$ Or, if you don't want $r$ in the denominator, $$ r^2\frac{d^2R}{dr^2}+r\frac{dR}{dr}-\left(m^2+\nu r^2\right)R=0. $$ I have an extra negative for the one of the terms, or at least I think I do. Anyway, you get the idea: The parameter $m$ is determined by periodicity; it is not an arbitrary parameter.