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.


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?


[1] Arfken and Weber, 5th Edition.

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

  • 1
    $\begingroup$ When there are two parameters, one of this is usually determined by a previous eigenvalue problem involving one of the parameters, which means those values are not up for grabs, unlike the second parameter. A single eigenvalue problem is not going to determine two different parameters. $\endgroup$ Apr 27 at 5:22
  • $\begingroup$ What I have observed anecdotally is that for $n$ variables, there will generally be $n$ eigenvalue problems along with $n-1$ separation constants. What Disintegrating By Parts says is sensible, but I wonder if there is more to it. $\endgroup$ May 5 at 17:46

1 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.


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