# How to find eigenvalues and eigenfunctions of simple-looking differential operator

What process can be used to solve for the eigenvalues and eigenfunctions of the following differential operator? $$H=A\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+\cos(x-y)$$

Where the solution is periodic for $x$ and $y$ on $(0,2\pi)$ and $A$ is an arbitrary constant. The eigenvalues of a 2D Laplacian operator with such periodic boundary conditions are $\lambda=m^2+n^2$ with eigenfunctions $$F(x,y)=\sum_{m,n=-\infty}^{\infty}a_{m,n}\cos(mx+ny)+b_{m,n}\sin(mx+ny)$$ but I am not sure if this will help me find a solution.

• @Winther I may be wrong but I can't see the solution $f = C\cos(v) + g$ working out. How do you manage the $\cos^2(v)$ term in the expression $A\nabla^2 f + f\cos(v)$? – Dylan Jul 4 '18 at 15:29
• @Dylan No it was me that's wrong. I forgot that one also multiplies $\cos(v)$ by $f$ in the eigenvalue problem so this does not work. Thanks, I'll remove the comment. – Winther Jul 4 '18 at 15:35

Substitute $2u=x+y$, $2v=x-y$, then

$$H(u,v) = \frac{A}{2}\left(\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2}\right) + \cos(2v)$$

Let $f(u,v) = U(u)V(v)$, then the problem becomes

$$\frac{A}{2} \left(\frac{U''}{U} + \frac{V''}{V}\right) + \cos(2v) = \lambda$$

which we can separate to get

\begin{align} U'' + \mu U &= 0 \\ V'' + \left(\mu - \frac{2\lambda}{A} + \frac{2}{A}\cos(2v)\right)V &= 0 \end{align}

You can look for solutions that are $\pi$-periodic in $u,v$, which will then be $2\pi$-periodic in $x,y$. The $V$ components are Mathieu functions, of which periodic solutions do exist for specific values of $\lambda$.