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I know how to transform a general second order differential equation of the form

$$\frac{d^2w}{dz^2} + \left(\frac{A}{z-\xi}+\frac{B}{z-\eta}\right)\frac{dw}{dz} + \frac{1}{z-\xi}\frac{1}{z-\eta}\left(\frac{D}{z-\xi}+\frac{E}{z-\eta}\right)w=0$$

into a hypergeometric equation. But now I have to solve

$$\frac{d^2w}{dz^2} + \left(\frac{A}{z-\xi}+\frac{B}{z-\eta}\right)\frac{dw}{dz} + \frac{1}{z-\xi}\frac{1}{z-\eta}\left(\frac{D}{z-\xi}+\frac{E}{z-\eta}+C\right)w=0$$

in terms of hypergeometric functions and I can't seem to find the change of variables adequate to obtain an equation like the first one. I'm sure I'm missing something very simple here.

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  • $\begingroup$ I don't think it is reducible to hypergeometric equation, since $z=\infty$ seems to be an irregular singular point of rank $1$. Instead, I would expect it to be reducible to confluent Heun equation. $\endgroup$ Commented Oct 7, 2015 at 14:19

1 Answer 1

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Let $w=(z-\xi)^a(z-\eta)^by$ ,

Then $\dfrac{dw}{dz}=(z-\xi)^a(z-\eta)^b\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)y$

$\dfrac{d^2w}{dz^2}=(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y=(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+2(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y$

$\therefore(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+2(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y+(z-\xi)^a(z-\eta)^b\left(\dfrac{A}{z-\xi}+\dfrac{B}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{A}{z-\xi}+\dfrac{B}{z-\eta}\right)\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)y+\dfrac{1}{(z-\xi)(z-\eta)}\left(\dfrac{D}{z-\xi}+\dfrac{E}{z-\eta}+C\right)(z-\xi)^a(z-\eta)^by=0$

$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y+\left(\dfrac{Aa}{(z-\xi)^2}+\dfrac{Ba+Ab}{(z-\xi)(z-\eta)}+\dfrac{Bb}{(z-\eta)^2}\right)y+\left(\left(\dfrac{1}{(\xi-\eta)(z-\xi)}-\dfrac{1}{(\xi-\eta)(z-\eta)}\right)\left(\dfrac{D}{z-\xi}+\dfrac{E}{z-\eta}\right)+\dfrac{C}{(z-\xi)(z-\eta)}\right)y=0$

$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{a(a-1)+Aa}{(z-\xi)^2}+\dfrac{2ab+Ba+Ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)+Bb}{(z-\eta)^2}\right)y+\left(\dfrac{D}{(\xi-\eta)(z-\xi)^2}+\dfrac{(\xi-\eta)C-D+E}{(\xi-\eta)(z-\xi)(z-\eta)}-\dfrac{E}{(\xi-\eta)(z-\eta)^2}\right)y=0$

$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{(\xi-\eta)a(a-1)+(\xi-\eta)Aa+D}{(\xi-\eta)(z-\xi)^2}+\dfrac{2(\xi-\eta)ab+(\xi-\eta)Ba+(\xi-\eta)Ab+(\xi-\eta)C-D+E}{(\xi-\eta)(z-\xi)(z-\eta)}+\dfrac{(\xi-\eta)b(b-1)+(\xi-\eta)Bb-E}{(\xi-\eta)(z-\eta)^2}\right)y=0$

Choose $(\xi-\eta)a(a-1)+(\xi-\eta)Aa+D=0$ and $(\xi-\eta)b(b-1)+(\xi-\eta)Bb-E=0$ , the terms of coefficient $\dfrac{1}{(z-\xi)^2}$ and $\dfrac{1}{(z-\eta)^2}$ can eliminate, and the ODE can reduces to Gaussian hypergeometric equation.

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