I have an equation similar to the Lamé differential equation in the Jacobi form, defined as

$$\frac{d^2y}{dx^2} + (a\,\mathrm{sn}(x)^2+b)y(x)=0$$

where the function $\mathrm{sn}(x)$ is one of the Jacobi elliptic functions. My equation takes the form

$$\frac{d^2y}{dx^2} = \left[k^2-\alpha_1\left(\frac{1-\gamma_1\mathrm{sn}^2(\gamma_2 x,\beta)}{r-\gamma_1\mathrm{sn}^2(\gamma_2 x,\beta)}\right)-\alpha_2\left(\frac{r-\gamma_1\mathrm{sn}^2(\gamma_2 x,\beta)}{1-\gamma_1\mathrm{sn}^2(\gamma_2 x,\beta)}\right)^2\right]y(x)$$

where $k,r,\alpha_1,\alpha_2,\beta,\gamma_1,\gamma_2$ are all real constants. I recognize this could also be viewed as Schrodinger's equation in one dimension with a tricky non-periodic potential, so I've looked at methods to solve that too.

For background on the Lamé differential equation, check Chapter XV of Higher Transcendental Functions (http://apps.nrbook.com/bateman/Vol3.pdf) and the last chapter in A Course of Modern Analysis (https://archive.org/details/courseofmodernan00whit).

The goal is to find an exact closed-form solution if possible. I believe there is some transformation I might be able to use to put the equation in a nicer form, but I can't quite find it.

EDIT: When I use the transformation $z=1-\gamma_1\mathrm{sn}^2(x)$ I get something that looks like Heun's differential equation, but with slightly different forms for the polynomials. Are there any known generalizations?

  • $\begingroup$ for what stands $sn$? $\endgroup$ – Dr. Sonnhard Graubner Apr 3 '17 at 15:04
  • $\begingroup$ It is one of the Jacobi elliptic functions, see mathworld.wolfram.com/JacobiEllipticFunctions.html $\endgroup$ – William J. Cunningham Apr 3 '17 at 15:05
  • $\begingroup$ the usual transformation for the lame equation is $z=am(x)$, where $am(x)$ is the Jacobi amplitude $\endgroup$ – Kiryl Pesotski Apr 3 '17 at 15:59
  • $\begingroup$ When I use that transformation I get a differential equation for $y(z)$ with a $dy/dz$ term, and it doesn't seem easier to solve. $\endgroup$ – William J. Cunningham Apr 3 '17 at 16:34
  • $\begingroup$ But it becomes algebraic, then you can work with usual methods. $\endgroup$ – Kiryl Pesotski Apr 4 '17 at 9:40

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