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Laplace's Equation in 3 dimensions is given by $$\nabla^2f=\frac{ \partial^2f}{\partial x^2}+\frac{ \partial^2f}{\partial z^2}+\frac{ \partial^2f}{\partial y^2}=0$$ and is a very important PDE in many areas of science. One of the usual ways to solve it is by seperation of variables. We let $f(x,y,z)=X(x)Y(y)Z(z)$ and then the PDE reduces to 3 independent ODE's of the form $X''(x)+k^2 x=0$ with $k$ a constant.

This method works for a surprising number of coordinate systems. I can use cylindrical coordinates, spherical coordinates, bi-spherical coordinates and more. In fact in 2 dimensions, I can take $\mathbb{R}^2$ to be $\mathbb{C}$, and then any analytic function $f(z)$ maps the Cartesian coordinates of $\mathbb{R}^2$ onto a set of coordinates that is separable in Laplace's equation. This follows from analytic functions also being harmonic functions. For example $f(z)=z^2$ maps the cartesian coordinates to the parabolic coordinates.

Parabolic Coordinates

So how about in 3 dimensions? Is there a way to describe all coordinate systems such that the Laplacian separates in this way?I have no idea how to start. I think that Confocal Ellipsoidal Coordinates will work but I'm not sure how to verify it. I'm not familIar with differential geometry, so any help appreciated.

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  • $\begingroup$ The very powerful and elegant tool in $\mathbb{R}^2$, conformal mapping, does not work in $\mathbb{R}^3$ unfortunately (at least not that I know of). $\endgroup$ – Shuhao Cao Oct 28 '13 at 15:59
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    $\begingroup$ Willatzen - "Separable Boundary Value Problems in Physics" $\endgroup$ – bolbteppa Jun 10 '14 at 1:10
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The Helmholtz equation is separable only in ellipsoidal coordinates (and degenerations like polar coordinates, and cartesian of course). For Laplace, there are a couple more; see the MathWorld article about Laplace's equation.

A good book source on this subject is Chapter 5 of Morse & Feshbach, part I.

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