Solving Hypergeometric D.E. Heun Equation

[wayfarer]

The Heun equation is a generalization of the hypergeometric D.E. to the case
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.

 

[jvicens]

Part (b): To show that the Heun equation reduces to a hypergeometric equation, we need to show that c_4 = 0 and c_5 = 0. From part (a), we know that c_4 = k_1*a*(1-a) and c_5 = -k_1*a^2. Since k_4 = 0, we have c_1*a^2 + c_2*a + c_3 = 0, which implies that c_4 = 0. Furthermore, if the general solution is free of logarithms at z=a, then c_5 = 0 as well. Therefore, the Heun equation reduces to a hypergeometric equation with c_1 = k_1, c_2 = -k_1, and c_3 = 0. The general solution can be written in terms of hypergeometric functions as w(z) = c_1*F(a, 1-a; 1; z) + c_2*z^(1-a)*F(a, 2-a; 2; z).

Part (c): In the special case when k_4 = -1, the Heun equation becomes z(z-1)(z-a)w'' + (c_1*z^2 + c_2*z + c_3)w' + (c_4*z - a*c_4)w = 0. We can rewrite this as z(z-1)(z-a)w'' + (c_1*z^2 + c_2*z + c_3)w' + (c_4*z - c_5)w = 0, where c_5 = a*c_4. Using the indicial equations, we can solve for the exponents t_1 and t_2. We get t_1 = 0 and t_2 = 1-c_1. Since we want the general solution to be free of logarithms at z=a, we need to have t_2 be a positive integer. Therefore, c_1 = 1, and we can rewrite the Heun equation as z(z-1)(z-a)w'' + (z^2 + c_2*z + c_3)w' + (c_4*z - c_5)w = 0. Solving for c_5, we get c_