Which solutions to use for the following hypergeometric equation?

Question

I am currently trying to understand the solution to a hypergeometric equation given in a paper on scalar fields and rotating black holes by S. Detweiler (https://journals.aps.org/prd/abstract/10.1103/PhysRevD.22.2323).

The equation I have trouble with is (17):

$$z(z+1)\frac{d}{dz}[z(z+1)\frac{dR}{dz}]+[P^2-l(l+1)z(z+1)]R=0$$

When I try to solve it in mathematica I get associated Legendre polynomials:

$$R(z) = C[1]LegendreP[l, 2 I P, 1 + 2 z] + C[2] LegendreQ[l, 2 I P, 1 + 2 z]$$

I know these can be written in terms of hypergeometric functions but they don't seem to match up when I try.

In the paper, the author provides a solution of the following form:

$$R(z)=(\frac{z}{z+1})^{iP}G(-l,l+1;1-2iP;z+1)$$

where G is any solution to the hypergeometric equation.

From here he uses two independent hypergeometric functions $U_3$ and $U_4$ (which are in turn linear combinations of two other solutions $U_1$ and $U_5$):

$$U_3 = (-z)^lF(-l,-l-2iP;-2l;-z^{-1})$$ $$U_4 = (-z)^{-l-1}F(l+1,l+1-2iP;2l+1;-z^{-1})$$

I am unsure how, given his solution in terms of G, the author knew which independent hypergeometric functions to use in the following steps to give the correct solution.

Thank you in advance for any advice you can give me!

EDIT: As I was unable to find them in the paper, I have calculated what I think are the correct boundary conditions. The requirement is that at the horizon, $r \rightarrow r_+$, there is only an ingoing wave solution. It is useful to know that in the above equations:

$$ P = (am - 2Mr_+ \omega)/(r_+ - r_{-}) $$

$$ z = (r-r_+)/(r_+ - r_-)$$

then the boundary condition at $r \rightarrow r_+$ is

$$R \sim (r-r_+)^{-i\alpha}$$ where $$ \alpha = \frac{Mr_+\omega-ma/2}{\sqrt{M^2-a^2}} $$

Answer 1

The result in terms of Associated Legendre functions seems correct (See below).

The relationship between Associated Legendre functions and a Gauss hypergeometric function can be found in mathematical handbooks, for example in https://en.wikipedia.org/wiki/Associated_Legendre_polynomials (first equation in §.11).

This leads to a solution $\quad \left(\frac{z}{z+1}\right)^{i\,p}\:_2 F_1\left(-l\:,\:l+1\:;\:1-2ip\:;\:-z \right)$

To be compared to $\quad \left(\frac{z}{z+1}\right)^{i\,p}\:_2 F_1\left(-l\:,\:l+1\:;\:1-2ip\:;\:z \right).\quad$

I cannot say why the sign of $z$ differs.

Answer 2

$z(z+1)\dfrac{d}{dz}\left(z(z+1)\dfrac{dR}{dz}\right)+(P^2-l(l+1)z(z+1))R=0$

$z(z+1)\dfrac{d^2R}{dz^2}+(2z+1)\dfrac{dR}{dz}+\left(\dfrac{P^2}{z(z+1)}-l(l+1)\right)R=0$

$\dfrac{d^2R}{dz^2}+\dfrac{2z+1}{z(z+1)}\dfrac{dR}{dz}+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z(z+1)}-l(l+1)\right)R=0$

$\dfrac{d^2R}{dz^2}+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\dfrac{dR}{dz}+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}-l(l+1)\right)R=0$

Which relates to Riemann's differential equation.

Let $R=z^a(z+1)^bS$ ,

Then $\dfrac{dR}{dz}=z^a(z+1)^b\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S$

$\dfrac{d^2R}{dz^2}=z^a(z+1)^b\dfrac{d^2S}{dz^2}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S=z^a(z+1)^b\dfrac{d^2S}{dz^2}+2z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S$

$\therefore z^a(z+1)^b\dfrac{d^2S}{dz^2}+2z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\left(z^a(z+1)^b\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S\right)+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}-l(l+1)\right)z^a(z+1)^bS=0$

$\dfrac{d^2S}{dz^2}+\left(\dfrac{2a}{z}+\dfrac{2b}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S+\left(\left(\dfrac{1}{z}-\dfrac{1}{z+1}\right)\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}\right)-\dfrac{l(l+1)}{z(z+1)}\right)S=0$

$\dfrac{d^2S}{dz^2}+\left(\dfrac{2a+1}{z}+\dfrac{2b+1}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{a^2+P^2}{z^2}+\dfrac{2ab+a+b-P^2-l(l+1)}{z(z+1)}+\dfrac{b^2+P^2}{(z+1)^2}\right)S=0$

Choose $a=iP$ and $b=-iP$ , the ODE becomes

$\dfrac{d^2S}{dz^2}+\left(\dfrac{2iP+1}{z}-\dfrac{2iP-1}{z+1}\right)\dfrac{dS}{dz}-\dfrac{3P^2+l(l+1)}{z(z+1)}S=0$

Let $z=-t$ ,

Then $\dfrac{d^2S}{dt^2}-\left(\dfrac{2iP+1}{-t}-\dfrac{2iP-1}{-t+1}\right)\dfrac{dS}{dt}-\dfrac{3P^2+l(l+1)}{-t(-t+1)}S=0$

$\dfrac{d^2S}{dt^2}+\left(\dfrac{2iP+1}{t}-\dfrac{2iP-1}{t-1}\right)\dfrac{dS}{dt}-\dfrac{3P^2+l(l+1)}{t(t-1)}S=0$

Which relates to Gaussian hypergeometric equation.