# Orthonormal functions for squared Mathieu functions

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.

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*}