Laplacian in elliptical coordinates I'm trying to calculate the laplacian in elliptical coordinates, just with the chain rule (because I don't know other method for doing this), but I have found difficulties to find the right expression. So this is the coordinates system of the problem:
\begin{align*}
x = cosh(u)cos(v) \\y = sinh(u)sin(v)
\end{align*}
with $u>0$ and $ v \in [0,2\pi]$
So after I calculate the terms using chain rule I found this first expression for $\partial_u^2 $ 
\begin{align*}
\partial_u^2 (\frac{1}{x^2- cos^2v} + \frac{1}{y^2+ sin^2v}) 
\end{align*}
I have try various ways to simplify this, but any one of this haven't work to me. Is suppose to be found the following expression:
\begin{align*}
\partial_u^2 (\frac{1}{sinh^2u} + \frac{1}{sin^2v}) 
\end{align*}
Can you say if that expression, that I have found is right? if yes can you help me simplify this?
 A: Here's a trick, call ${\bf r} = x \hat{x} + y\hat{y}$, and calculate the vectors
\begin{eqnarray}
\frac{\partial {\bf r}}{\partial u} &=& \cos v \sinh u \hat{x} + \sin v\cosh u \hat{y} \\
\frac{\partial {\bf r}}{\partial v} &=& -\sin v \cosh u \hat{x} + \cos v\sinh u \hat{y} \tag{1}
\end{eqnarray}
Now calculate the magnitudes of each vector
\begin{eqnarray}
h_u &=& \left|\frac{\partial {\bf r}}{\partial u} \right| = [ \sin^2 v \cosh^2 u + \cos^2 v \sinh^2 u]^{1/2} \\
&=& [\sin^2 v \cosh^2 u + \cos^2 v(\cosh^2u - 1)]^{1/2} \\
&=& [\cosh^2 u - \cos^ v]^{1/2} = h_v \tag{2}
\end{eqnarray}
So you have
$$
h_u = h_v = [\cosh^2 u - \cos^2 v]^{1/2} \tag{3}
$$
The Laplacian in these coordinates is 
\begin{eqnarray}
\nabla^2 \phi &=& \frac{1}{h_u h_v} \frac{\partial }{\partial u}\left(\frac{h_u h_v}{h_u^2}\frac{\partial \phi}{\partial u}\right) + \frac{1}{h_u h_v} \frac{\partial }{\partial v}\left(\frac{h_u h_v}{h_v^2}\frac{\partial \phi}{\partial v}\right) \\
&=& \frac{1}{\cosh^2 u - \cos^2 v}\left(\frac{\partial^2 \phi}{\partial u^2} +\frac{\partial^2 \phi}{\partial v^2} \right)
\end{eqnarray}
