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?


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}

  • $\begingroup$ Wow this is great, thank you! $\endgroup$ – Matheus Fachini Dec 15 '18 at 0:32
  • $\begingroup$ But why the laplacian must be divide by the magnitude? $\endgroup$ – Matheus Fachini Dec 15 '18 at 0:36

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