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.


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 $$ G(\rho) = (\rho - i \nu) \left(\frac{1+\rho}{1-\rho}\right)^{\frac{i\nu}{2}}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.