I am trying to get rid of the following homogeneous ode.
\begin{equation} \begin{split} u''(z)+\frac{1}{2} \left(\frac{1}{z}+\frac{1}{z-1}+\frac{1}{z-1}\right) u'(z)+\frac{2\left(A+B\right)- \left(2 B\right)z }{4 z (z-1) (z-1)}u(z)=0 \end{split} \end{equation}
The form of this equation is very particular and reminds me the algebraic representation of the Lamé equation, as shown in http://dlmf.nist.gov/29.2.E2, which is :
\begin{equation} \begin{split} \frac{{\mathrm{d}}^{2}w}{{\mathrm{d}\xi}^{2}}+\frac{1}{2}\left(\frac{1}{\xi}+% \frac{1}{\xi-1}+\frac{1}{\xi-k^{-2}}\right)\frac{\mathrm{d}w}{\mathrm{d}\xi}+% \frac{hk^{-2}-\nu_1(\nu_1+1)\xi}{4\xi(\xi-1)(\xi-k^{-2})}w=0 \end{split} \end{equation}
For $0<k<1$ and $\xi={\operatorname{sn}}^{2}\left(z,k\right)$ then the Lamé equation can be expressed in this form:
\begin{equation} \begin{split} { \frac{{\mathrm{d}}^{2}w}{{\mathrm{d}\zeta}^{2}}+\left[h-\nu(\nu+1) k^2 sn^{2}(\zeta,k)\right] w(\zeta)=0 } \end{split} \end{equation}
Thus,when comparing my equation and Lame's equation the only way to match is the case in which $k=1$, but that leaves $k$ out of the parameters.
If I ignored the fact that $k$ can't have such value, I could proceed and see how it looks like:
\begin{equation} \begin{split} { \frac{{\mathrm{d}}^{2}u}{{\mathrm{d}z}^{2}}+\left[2(A+B)-2B sn^{2}(z,1)\right] u(z)=0 } \end{split} \end{equation}
Actually, the Jacobi elliptic function $sn(z,1)$ is reducible into $tanh(z)$
And finally I end up with this equation.
\begin{equation} \begin{split} { \frac{{\mathrm{d}}^{2}u}{{\mathrm{d}z}^{2}}+\left[2(A+B)-2B \tanh^{2}(z)\right] u(z)=0 } \end{split} \end{equation}
This equation seems to be solvable in terms of Associated Legendre Polynomials. And some questions have arisen to me:
- My original equation can be considered a legitimate Lamé differential equation?
- If so, it will be correct to suggest that Lamé differential equation for the parameter value k=1 degenerates into some form of Associated Legendre differential equation?
I would really appreaciate any comments on this, and I apologize for any mistake I could have done while typing this thread.