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Consider a Fuchsian type differential equation written as $$\frac{d^2 y}{dz^2} + \frac{p(z)}{(z - z_1)(z - z_2) \cdots (z - z_m)} \frac{dy}{dz} + \frac{q(z)}{(z - z_1)^2 (z - z_2)^2 \cdots (z - z_m)^2} y = 0, \quad m \geq 2,$$ where $z_1, z_2, \ldots, z_m$ are distinct regular singular points, and $p(z)$ and $q(z)$ are polynomials. How does one relate the solutions to this differential equation to the solutions to the hypergeometric differential equation $$z(1 - z) \frac{d^2 y}{dz^2} + [c - (a + b + 1)z] \frac{dy}{dz} - aby = 0$$ around each regular singular points?

The case $m = 2$ essentially reduces to Riemann's differential equation (if $z = \infty$ is a regular singular point) whose solutions can be written in terms of the hypergeometric functions. Can this be done in general, or for at least four or five regular singular points?

Update: From what I have gathered so far, all homogeneous linear differential equations of the second order having four regular singularities in the extended complex plane, can be transformed into Heun's differential equation whose solutions are related to the hypergeometric functions.

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    $\begingroup$ It cannot be done in general, even for Heun's functions. $\endgroup$ – Start wearing purple Jul 30 '17 at 11:28
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This very depends on degrees of $p(z)$ and $q(z)$ .

Assume $p(z)$ is on degree $k_1$ and $q(z)$ is on degree $k_2$ ,

The ODE can simplify to the form

$\dfrac{d^2y}{dz^2}+\left(\sum\limits_{n=0}^{k_1-m}P_nz^n+\sum\limits_{n=1}^m\dfrac{Q_n}{z-z_n}\right)\dfrac{dy}{dz}+\left(\sum\limits_{n=0}^{k_2-m}R_nz^n+\sum\limits_{n=1}^m\dfrac{S_n}{z-z_n}+\sum\limits_{n=1}^m\dfrac{T_n}{(z-z_n)^2}\right)y=0$

It is known that when $m=4$ and both $\sum\limits_{n=0}^{k_1-m}P_nz^n$ and $\sum\limits_{n=0}^{k_2-m}R_nz^n$ vanish,

some great work done by e.g. Solutions in terms of the hypergeometric functions can convert the ODE to the Heun ODE.

Note that either $\sum\limits_{n=0}^{k_1-m}P_nz^n$ or $\sum\limits_{n=0}^{k_2-m}R_nz^n$ presents often leads the nature of the ODE extremely complicated.

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