Sturm Liouville differential equation and hypergeometric functions

Question

I'm trying to understand how to solve this differential equation:

$ [z^2(1-z)\dfrac{d^2}{dz} - z^2 \dfrac{d}{dz} - \lambda] f(z) = 0 $

I know the solution is related to the hypergeometric function $_2F^1$, but as I recall from many sources: this functions satisfies another differential equation:

$ [z(1-z)\dfrac{d^2}{dz} + (c - (ab + 1)z) \dfrac{d}{dz} - ab] f(z) = 0 $

with $a,b,c \in \mathcal{R} $

I've tried to transform it in this form:

$ \dfrac{d}{dz} [(z-1)f'(z)] + \dfrac{\lambda}{z^2} f(z) = 0 $

or use quadratic transformation or other properties of $_2F^1$, but I failed. Any suggestions?

Thanks in advance for the attention.

Answer 1

Let $\alpha$ be a root of $\alpha^2-\alpha-\lambda=0$. The change of the dependent variable $f(z)=z^\alpha w(z)$ reduces to the hypergeometric equation in $w$: $$z(1-z)w''+(2\alpha(1-z)-z)w'-\alpha^2 w=0.$$

Answer 2

The transformation $x = (z-2)/z$ takes your differential equation to $$ (x^2-1) f'' + 2 x f' - \lambda f = 0$$

which is a Gegenbauer differential equation. Its solutions can be written using Legendre P and Q functions:

$$f \left( x \right) =c_1 \,{\it LegendreP} \left( {\frac {1}{2} \sqrt {1+4\,\lambda}}-{\frac{1}{2}},x \right) +c_2 \,{\it LegendreQ} \left( {\frac {1}{2}\sqrt {1+4\,\lambda}}-{\frac{1}{2}},x \right) $$