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We're given the spherical Laplacian,

$$\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\bigg(\sin{\theta}\frac{\partial u}{\partial\theta}\bigg)+\frac{1}{\sin^2{\theta}}\frac{\partial^2 u}{\partial\phi^2}=-\lambda u.$$ We can do separation of variables to obtain the following ODEs: $$\frac{1}{f}\bigg(\sin{\theta}\frac{d}{d\theta}\bigg(\sin{\theta}\frac{df}{d\theta}\bigg)\bigg)+\lambda\sin^2{\theta}=m^2$$ $$\frac{1}{g}\frac{d^2g}{d\phi^2}=-m^2$$

I'm not sure how to write the latter of these in Sturm-Liouville form, as there is no dependence on $\lambda$. Any help would be appreciated.

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  • $\begingroup$ Your $\lambda$ here is the $\pm m^{2}$ (I say $\pm$ because you have a typo, either the first ODE in $f$ should have RHS $-m^{2}$ or the second in $g$ should have RHS $m^{2}$). Using the notation on the Wikipedia page $$g'' = -m^{2} g \implies \frac{d}{dx} \left[ 1 \cdot \frac{dg}{dx} \right] + (0 - (-m^{2}) \cdot 1) g = 0$$ $\endgroup$ – Mattos Nov 29 '17 at 15:53
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To perform separation of variables on the first equation, assume a form of solution $u(\theta,\phi)= \Theta(\theta)\Phi(\phi)$ and put everything involving $\Theta$ on one side and everything involving $\Phi$ on the other. In this case, you start with $$ \frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta \frac{d\Theta}{d\theta}\right)\Phi+\frac{1}{\sin^2\theta}\Theta\frac{d^2\Phi}{d\phi^2}=-\lambda\Theta\Phi. $$ Now use the standard trick of dividing by the original function $\Theta\Phi$ to obtain $$ \frac{1}{\Theta}\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{1}{\sin^2\theta}\frac{\Phi''}{\Phi}=-\lambda. $$ Next multiply by $\sin^2\theta$ to remove the $\theta$ dependence from the $\Phi''/\Phi$ term: $$ \frac{1}{\Theta}\sin\theta\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{\Phi''}{\Phi}=-\lambda\sin^2\theta. $$ Now rearrange to put everything involving $\theta$ on the left, and everything involving $\phi$ on the right: $$ \frac{1}{\Theta}\sin\theta\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\lambda\sin^2\theta = -\frac{\Phi''}{\Phi} $$ The conclusion is that, for any separated solution $u=\Theta\Phi$, there must be a constant $\mu$ such that $$ \frac{1}{\Theta}\sin\theta\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\lambda\sin^2\theta = \mu,\;\; \mu=-\frac{\Phi''}{\Phi}. $$ The reason for this is that you can fix $\theta$ on the left, and then the right side cannot vary in $\phi$ if the previous equation is to hold. Likewise, you can fix $\phi$ in order to see the left side must be constant in $\theta$. That's why you can separate the two sides by injecting an unknown constant into the equation, but it must be the same constant for both sides if you are going to have a legitimate solution of the original equation in the form $u(\theta,\phi)=\Theta(\theta)\Phi(\phi)$. Constraints on one side or the other will end up determining the workable separation constants, assuming there are any workable ones. The constant $\mu$ will be determined by the equation and constraints in $\phi$. Then these $\mu$ will be substituted into the equation in $\theta$, which will determine the values of $\lambda$. Each $\mu$ will lead to a potentially different set of $\lambda$.

The Sturm-Liouville forms are $$ -\frac{d}{d\phi}\left(1\frac{d\Phi}{d\phi}\right)=\mu 1\Phi, \\ -\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{\mu}{\sin\theta}\Theta=\lambda\sin\theta\Theta. $$ The first equation determines the possible values of $\mu$. These are then substituted into the second equation to obtain a family of problems in $\theta$ where the possible values of $\lambda$ depend on the values of $\mu$ as well as the equation in $\theta$ and any constraints in $\theta$.

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