How to solve this 2nd order Ordinary Differential Equation

Question

I was reading this, and wasn't able to solve equation (2.34). The equation is:

$$\Big[\nu^2 + \frac{\rho^2 -1}{\rho^2} \partial_{\rho}(\rho^2 (\rho^2 -1)\partial_{\rho}) \Big]f(\rho) = 0,$$

where $\rho$'s range is $(1,\infty)$.

I tried solutions of the form $f(\rho) = \frac{g(\rho)}{\rho}$, and further $\rho = \cosh[x]$. Then in the asymptotic limit $x \to 0$, the solution goes like $$g(\cosh x) = \left(\coth {\frac{x}{2}}\right)^{i\nu} g_1(\cosh x) $$

The differential equation for $g_1$ becomes then $$\frac{d^2g_1}{dx^2} + [\coth x -2i\nu\, \text{cosech}\, x]\frac{dg_1}{dx}-2g_1=0$$

I don't know how to proceed from here. I tried out the solutions using Mathematica also, but that didn't help. How do I solve the same? Thanks.

Answer 1

Writing $f(\rho) = \frac{g(\rho)}{\rho}$ is a good idea, you then get $$ (1-\rho^2)^2 g'' -2 \rho (1-\rho^2) g' + (2(1-\rho^2) + \nu^2) g = 0. \tag{*} $$ This is a form of the (associated) Legendre equation, which has solutions given by the associated Legendre functions $P_1^{i \nu}(\rho)$, $Q_1^{i\nu}(\rho)$. In this case, these take a relatively simple form in $\rho$; the general solution to $(*)$ is given by $$ g(\rho) = c_1 G(\rho) + c_2 G(-\rho), $$ with $$ G(\rho) = (\rho - i \nu) \left(\frac{1+\rho}{1-\rho}\right)^{\frac{i\nu}{2}}. $$