I am working with Laplace-Eigenvalues of ellipses and in this context I started working with Mathieu functions. Now I have reached a point where I can no longer go any further.

I am currently looking for functions that are orthogonal to the square of Mathieu functions, i.e., I am looking for a differentiable and $2 \pi$ periodic function $g$ such that, $$ \langle g(\cdot),ce^{2}_{m}(\cdot,q) \rangle = \int_{0}^{2 \pi}g(x)ce^{2}_{m}(x,q)dx=0$$ My first question is wether there must be a non-trivial solution $g$ such that $\langle g(\cdot),ce^{2}_{m}(\cdot,q) \rangle=0$ for all arbitrary but fixed $m \in \mathbb{N}$ and $q>0$.

I tried to use the explicit Fourier series representations of $g$ and $ce_m$ but it did not go anywhere. The representation I used can be found here on Wikipedia.

Since the Mathieu functions themselves are orthogonal I hope there might be some simplifying structure for "triplets" i.e.,
$$\int_{0}^{2 \pi}ce_{k}(x,q)ce_{m}(x,q)ce_{n}(x,q)dx=\int_{0}^{2 \pi}ce_{k}(x,q)ce^{2}_{m}(x,q)dx.$$

Unfortunately, I could not find any material (on triplets or squares of Mathieu functions) or derive any property myself. I am happy for any help, thanks a lot in advance.


1 Answer 1


I recently managed to solve the problem. The solution is actually quite intuitive and constructive. I thought I would share the answer here as well.

Any function $g$ with $g(\nu+\pi)=-g(\nu)$ ulfills the condition $$ \langle g, ce_m^2(\cdot, q_{m,n}^c)\rangle = 0. $$ The main idea of the proof is that the square of the Mathieu functions $ce^2_m$ are $\pi$ periodic. However, the scalar product is defined over the full interval $[0,2\pi]$. Thus, any point-symmetric function around $\pi$ vanishes the whole integral. The identity $g(\nu)=\nu$ would fulfill this requirement. However, the identity is not $2\pi$ periodic and thus not permissible. However, for example, $g(\nu)=\cos(\nu)$ also fulfills this condition.

Let $m\in\mathbb{N}$ and $q \in \mathbb{R}^{+}$. We first show that ${\text{ce}}_{m}^2(\nu+\pi,q)={\text{ce}}_{m}^2(\nu,q)$. This follows immediately from the $\cos$ representation of $ce_{m}$. Indeed we know that we can write $ce_{m}$ as an infinite sum, in particular \begin{align*} {\text{ce}}_{2m}(x,q)&=\sum _{r=0}^{\infty }A_{2r}^{(2m)}(q)\cos(2rx),\\{\text{ce}}_{2m+1}(x,q)&=\sum _{r=0}^{\infty }A_{2r+1}^{(2m+1)}(q)\cos \left[(2r+1)x\right] . \end{align*} Then clearly $ce_{m}$ is $\pi$ periodic since, \begin{align*} {\text{ce}}_{2m}(x+\pi,q)&=\sum _{r=0}^{\infty }A_{2r}^{(2m)}(q)\cos(2r(x+\pi))\\ &=\sum _{r=0}^{\infty }A_{2r}^{(2m)}(q)\cos(2rx) \\ &= {\text{ce}}_{2m}(x,q). \end{align*} Using the same steps we get the analogous result for the odd case $2m+1$. We now use this property to calculate the scalar product, \begin{align*} \langle g, ce_m^2(\cdot, q_{m,n}^c)\rangle &= \int_0^{2\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu\\ &= \int_0^{\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu + \int_{\pi}^{2\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu\\ &= \int_0^{\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu + \int_0^{\pi}g(\nu+\pi)ce_m^2(\nu+\pi, q_{m,n}^c)d\nu\\ &= \int_0^{\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu - \int_0^{\pi}g(\nu)ce_m^2(\nu, q_{m,n}^c)d\nu \\ &= 0. \end{align*}


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