The Heun equation is a generalization of the hypergeometric D.E. to the case(adsbygoogle = window.adsbygoogle || []).push({});

of four regular singular points. With the singular points at z=0,1,a,and inft,

the Heun equation takes the form,

z(z-1)(z-a)w''+(c_1*z^2+c_2*z+c_3)w'+(c_4*z+c_5)w=0

(a) In terms of k_1, k_2, k_3, k_4 and a, calculate c_1, c_2, c_3, c_4 and e,

such that the Riemann P-function for this Heun equation is

w=P( 0 a 1 inft )

( 0 0 0 k_1 z)

( 1-k_3 1-k_4 1-e k_2 )

(b) In the special case when k_4=0 and the general solution is free of

logarithms at z=a, show that the Heun equation reduces to a hypergeometric

equation. Give the general solution in terms of hypergeometric functions.

(c) Obtain a quadratic equation for the accessory parameter c_5 in the

special case when k_4=-1 and the general solution is free of logarithms at

z=a. (In this case, z=a is an apparent singularity of the differential equation)

I've done part (a) using indicial equations to find exponents, and then equate.

Part B: From part a, i got using indicial equations that k_4=0 implies that c_1*a^2+c_2*a+c_3=0, so c_1*z^2+c_2*z+c_3 = c_1*(z-a)*(z-a') for some number a'. Then in the equation, if c_4*z+c_5 has a as a factor, i.e. c_5=-a*c_4, then we can divide through by z-a and get a hypergeometric D.E. So i need to prove that if the general solution is free of logarithms, then c_5=-a*c_4; i'm not quite sure how to do it rigorously. I tried doing it by using a power series about z=a, and proved that we reach a contradiction if there are no terms of negative power (i.e. (z-a)^(-t)) in the Laurent series, but I'm not sure how to justify that.

for part (c ), i'm clueless.

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# Hypergeometric D.E.

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