Consider the PDE,
$$\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\bigg(\sin{\theta}\frac{\partial u}{\partial\theta}\bigg)+\frac{1}{\sin^2{\theta}}\frac{\partial^2 u}{\partial\phi^2}=-\lambda u.$$ We can do separation of variables to obtain the following ODEs: $$\frac{1}{f}\bigg(\sin{\theta}\frac{d}{d\theta}\bigg(\sin{\theta}\frac{df}{d\theta}\bigg)\bigg)+\lambda\sin^2{\theta}=m$$ $$\frac{1}{g}\frac{d^2g}{d\phi^2}=-m$$
In this case, the latter equation determines the possible values of $m$, and the former the possible values of $\lambda$, which we can find after knowing the constraints on $m$ by solving the latter.
Now consider the following instead: $$\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\bigg(\sin{\theta}\frac{\partial u}{\partial\theta}\bigg)+\frac{1}{\sin^2{\theta}}\frac{\partial^2 u}{\partial\phi^2}=\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}.$$
We can again do separation of variables to obtain the following:
$$\frac{d^2 T}{dt^2}=-c^2\lambda T$$
$$\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\bigg(\sin{\theta}\frac{\partial Y}{\partial\theta}\bigg)+\frac{1}{\sin^2{\theta}}\frac{\partial^2 Y}{\partial\phi^2}=-\lambda Y$$
My question is this:
There are no boundary conditions on $t$, which represents time (as far as I can tell), so we cannot determine $\lambda$ using the $T$ ODE. Instead, we must solve the first PDE I considered and determine $m$ and $\lambda$ using those solutions instead as there are well-defined boundary conditions (i.e. $f(\theta)=f(\theta+\pi)$ and $g(\phi)=g(\phi+2\pi)$). So, how do we write the $T$ ODE in Sturm-Liouville form, if there is no parameter it determines? Is it just an extraneous equation with parameter $\lambda$?