I would like to know if there is a simple formula for the 1-D Laplacian $\Delta$ on an Ellipse
$$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1.$$ So far I've considered the following options
1) I first tried to write $\Delta$ in terms of the Arclenght of the ellipse, but since the Arclenght is given in terms of a difficult integral this was a dead end.
2) Polar coordinates: The ellipse can be parametrized by $$x = r\cos\theta, $$ $$y = r\sin\theta, $$
where $$r=\frac{ab}{\sqrt{b^{2}\cos^{2}\theta+a^{2}\sin^{2}\theta}}.$$
The Laplacian in polar coordinates is $$\Delta=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}$$
So the chain rule can be used to write $\Delta$ in terms only on $\theta$ and $\frac{\partial}{\partial\theta}$, but the result is a nigthmare.
3) The ellipse can also be parametrized by $$x = a\cos\theta, $$ $$y = b\sin\theta, $$
where $\theta$ here is not the polar angle. I have no Idea where to start in this case.
4) It also crossed my mind that $\Delta\sim\frac{\partial^{2}}{\partial\theta^{2}}$ like for a circle, but that seems too simple to be true.