# Correct form of the Laplacian on a 1-D ellipse, and it's solutions

I wanted to derive the Laplacian operator for a 1-D ellipse, and it seemed to me that there are two equivalent approaches:

1) Start with 2-D elliptic coordinates $$x = a \cosh(\mu) \cos(\nu)$$ $$y = a \sinh(\mu) \sin(\nu)$$ Then, follow the usual working to get the full 2-D Laplacian $$\nabla^2 = \frac{2}{a^2(\cosh (2\mu)-\cos(2\nu))}(\frac{\partial^2}{\partial\mu^2}+\frac{\partial^2}{\partial\nu^2})$$ Finally, set $$\mu$$ to be a constant ($$b = \cosh 2\mu$$) leading to $$\nabla^2 = \frac{2}{a^2(b-\cos2\nu)}\frac{\partial^2}{\partial\nu^2}$$

2) Start with elliptic coordinates and set $$\mu$$ to be a constant from the get-go $$x = a \cos(\nu)$$ $$y = a \sqrt{1-\gamma^2} \sin(\nu)$$ where $$\gamma$$ is the eccentricity. The Laplacian can now be derived with the knowledge of the metric $$g = a^2(1-\gamma^2\cos(\nu)^2)$$ $$\nabla^2 = -\gamma^2 \frac{\cos \nu \sin \nu}{a^2 (1-\gamma^2 \cos(\nu)^2)^2} \frac{\partial}{\partial \nu}+ \frac{1}{a^2(1-\gamma^2 \cos (\nu)^2))}\frac{\partial^2}{\partial \nu^2}$$ which is obviously quite different from 1).

Thus, my first question is why 1) and 2) give different answers in elliptic coordinates. After all, one can work through an analogous derivation for a ring in spherical coordinates and both approaches will give the same result. I see how it comes down to the different metric, but I lack any physical understanding.

In the end, what I'm really interested in are the solutions of the Laplacian equation on a 1-D ellipse. And, while I found a lot of literature on 2-D elliptic planes (Mathieu's functions) and ellipsoidal harmonics, I couldn't find anything on my problem.

I would greatly appreciate any help you could provide, including pointing me to the correct background literature.