I'm currently trying to solve Laplace's equation outside of a prolate spheroid. The scalar field has to vanish far from the spheroid. Because of the geometry I thought it might be convenient to use prolate spheroidal coordinates (I'm using the first definition from that page, since it's the one I've been able to found more information on). I'm trying to solve: $$\nabla^2 f = 0 $$ $$ f \to \infty \quad \text{as} \quad \mu \to\infty$$
Additionally, my problem is axisymmetric, so it doesn't depend on $\phi$. There is another boundary condition at the surface of the spheroid, but it is not relevant for my question.
I thought that it might be convenient to use prolate spheroidal coordinates in this situation, but I've run into a problem when trying to satisfy the condition that the field vanish at infinity. The most general solution to Laplace's equation in these coordinates is:
$$f = \sum_{n= 0}^\infty (A_nP_n(\cos{\nu})+B_nQ_n(\cos{\nu}))\times(C_nP_n(\cosh{\mu})+D_nQ_n(\cosh{\mu}))$$
Here the $P_n$ are Legendre polynomials and the $Q_n$ are Legendre functions of the second kind.
However, I can't seem to make it satisfy my condition at infinity. This is because, as $\mu$ tends to $\infty$, so does $\cosh{\mu}$, and both the $P_n$ and $Q_n$ diverge at infinity! Additionally, the $Q_n$ become imaginary, which is not expected for my case.
How can you use this coordinate system to solve external problems? Or is it simply ill suited for this type of problems?
I could try to use spherical coordinates, but then account for the shape might be more trouble than it's worth.
When $x>1$, an alternative definition of $Q_n(x)$ is used where the $\ln\frac{1+x}{1-x}$ is replaced with $\ln\frac{x+1}{x-1}$ so that the functions are real.
Actually $Q_n(x)$ tends to zero as $x\rightarrow\infty$; there is a series representation of $Q_n$ in terms of negative powers of $x$, which can be found on page 58 of "the theory of spherical and ellipsoidal harmonics" by Hobson. So the combination $Q_n(\cosh\mu)P_n(\cos\nu)$ will tend to zero, far from the origin.
(Also an axisymmetric problem will still have $\nu$ dependence, but wont have $\phi$ dependence).