Suppose $1>e>0, E>0$ are constants, and $f(x,y)$ and $g(x,y)$ satisfies the PDE:
\begin{align} \sqrt{\frac{1+e}{1-e}}\frac{\partial f}{\partial y}+\frac{\partial g}{\partial x}=\frac{ieE}{\sqrt{1-e^2}}\frac{f}{1-e} \\ \frac{\partial f}{\partial x}-\sqrt{\frac{1-e}{1+e}}\frac{\partial g}{\partial y}=\frac{ieE}{\sqrt{1-e^2}}\frac{g}{1+e} \end{align}
where $i=\sqrt{-1}$. One can verify that after eliminating $f$ or $g$, $g$ and $f$ satisfy the same Helmoholz equation.
Change to the elliptical coordinates: $$ y=\cosh \mu \cos\theta,\quad x=\sinh\mu\sin\theta $$
and separate variables $f=u(\mu)v(\theta)$, one can get the Mathieu function for $v(\theta)$, and modified Mathieu function for $u(\mu)$. Therefore, the general solution for $f$ is: $$ f=\sum_{m=0}\alpha_m \operatorname{ce}_m(q,\theta)\operatorname{Mc}_m(q,\mu)+\sum_{m=1}\beta_m \operatorname{se}_m(q,\theta)\operatorname{Ms}_m(q,\mu) $$
where $q=\frac{1}{4}(\frac{eE}{1-e^2})^2$.
$\operatorname{ce}_m$ and $\operatorname{se_m}$ are even and odd Mathieu function, chosen because we require $v(\theta+2\pi)=v(\theta)$. $\operatorname{Mc}_m$ and $\operatorname{Ms}_m$ are even and odd radial Mathieu functions, respectively.
Since $g$ satisfies the same equation, it also has the solution of the form: $$ g=\sum_{m=0}\gamma_m \operatorname{ce}_m(q,\theta)\operatorname{Mc}_m(q,\mu)+\sum_{m=1}\delta_m \operatorname{se}_m(q,\theta)\operatorname{Ms}_m(q,\mu) $$
Question is: what is the relationship between $\alpha_m,\beta_m,\gamma_m,\delta_m$?
In principal, the relationship between these coefficients can be determined by substitute the solutions to either one of the original pde equations. However, I can't find any resources which express $\operatorname{ce'}_m(\theta)$ or $\operatorname{se'}_m(\theta)$ in terms of $\operatorname{ce}_m(\theta)$ and $\operatorname{se}_m(\theta)$, therefore I don't know how to compare the coefficients.
One may need the partial derivatives of the following: \begin{align} \frac{\partial\mu}{\partial y}&=\frac{2\sinh\mu\cos\theta}{\cosh 2\mu-\cos 2\theta} \\ \frac{\partial\mu}{\partial x}&=\frac{2\cosh\mu\sin\theta}{\cosh 2\mu-\cos 2\theta} \\ \frac{\partial\theta}{\partial y}&=-\frac{\partial\mu}{\partial x},\quad \frac{\partial\theta}{\partial x} = \frac{\partial\mu}{\partial y} \end{align}