Various forms of the Confluent Heun Equation

The Confluent Heun equation is expressed in various forms. It's non-symmetrical canonical form is: $$\frac{d^{2}y}{dx^{2}}+(\beta+\frac{\gamma}{x}+\frac{\delta}{x-1})\frac{dy}{dx}+\frac{\alpha\beta x-q}{x(x-1)} y =0$$

with a particular transformation it is possible to change it to a Schrodinger-like equation: $$\frac{d^{2}y}{dx^{2}}+(A+\frac{B}{x}+\frac{C}{x-1}+\frac{D}{x^{2}}+\frac{E}{(x-1)^{2}})y=0$$

What is the transformation in order to go from one form to another? and how are $A, B, C, D$ and $E$ related to $\alpha, \beta, \gamma, \delta$ and $q$?

• To eliminate the first derivative you may try an ansatz $y(x)=\alpha(x)z(x)$. Choose $\alpha(x)$ so that the terms containing the first derivative vanishs Aug 25, 2015 at 17:40

If $y(x)$ is a solution of the first equation, let $$z \left( x \right) =y \left( x \right) {{\rm e}^{\beta\,x/2}}{x}^{ \gamma/2} \left( x-1 \right) ^{\delta/2}$$
Then if I'm not mistaken, $z$ satisfies the second differential equation with
$$A = -\dfrac{\beta^2}{4},\ B = \dfrac{2q + (\delta-\beta)\gamma}{2},\ C = \dfrac{2\alpha\beta - (\beta+\gamma)\delta - 2 q}{2},\ D = \dfrac{(2-\gamma)\gamma}{4},\ E = \dfrac{(2-\delta)\delta}{4}$$