How to derive Laplacian in new coordinate system?

Question

I’m working with the Laplacian operator, in terms of using it in different coordinate systems. I’m managed to derive it in the polar and cylindrical basis, but I’m struggling with applying that to a new coordinate system, in this case I’m using x=0.5(u^2-v^2), y=uv, and z=z. I’m guessing I’ll have to find the partial derivative of u and v wrt x and y, but I’m struggling to put this into practice to actually get the new Laplacian.

Answer 1

$ \newcommand\PD[2]{\frac{\partial#1}{\partial#2}} \renewcommand\vec\mathbf $

The operator $\nabla$ can be expressed in any basis $\{\vec f_i\}$ with coordinates $\{f^i\}$ as $$ \nabla = \sum_i \vec f^i\PD{}{f^i}, $$ where $\vec f^i$ is the unique reciprocal basis satisfying $\vec f_i\cdot\vec f^j = \delta_i^j$. It also follows that $\vec f^i = \nabla f^i$.

In your case $x, y, z$ are the coordinates of the standard basis $\vec e_1, \vec e_2, \vec e_3$ (which is it's own reciprocal, i.e. $\vec e^i = \vec e_i$), and $u, v, z$ are the set of coordinates we're interested in. To compute $\nabla$ for the new coordinates, we need to determine $\vec f^u = \nabla u, \vec f^z = \nabla z$. Of course, we already know $\vec f^z = \vec e^3 = \vec e_3$. For the others, we could solve for $u, v$ in terms of $x, y$, but perhaps more simply we can take your equations $$ x = \frac12(u^2 - v^2),\quad y = uv $$ and apply $\nabla$ using the product rule to get $$ \vec e_1 = \frac12(2u\vec f^u - 2v\vec f^v) = u\vec f^u - v\vec f^v,\quad \vec e_2 = \vec f^uv + u\vec f^v, $$ which we readily solve for $\vec f^u, \vec f^v$ as $$ \vec f^u = \frac{u\vec e_1 + v\vec e_2}{u^2 + v^2},\quad \vec f^v = \frac{-v\vec e_1 + u\vec e_2}{u^2 + v^2}. $$

The Laplacian is $$ \nabla^2 = \nabla\cdot\nabla = \sum_{i,j} \vec f^i\cdot\vec f^j\PD{^2}{f^i\partial f^j} = \sum_i\vec f^i\cdot\vec f^i\PD{^2}{(f^i)^2} + 2\sum_{i<j}\vec f^i\cdot\vec f^j\PD{^2}{f^i\partial f^j}, $$ so computing $$\begin{aligned} \vec f^u\cdot\vec f^u &= \frac{u^2 + v^2}{(u^2 + v^2)^2} = \frac1{u^2 + v^2}, \\ \vec f^v\cdot\vec f^v &= \frac{v^2 + u^2}{(u^2 + v^2)^2} = \frac1{u^2 + v^2}, \\ \vec f^z\cdot\vec f^z &= e_3\cdot e_3 = 1, \\ \vec f^u\cdot\vec f^v &= \frac{uv - uv}{(u^2 + v^2)^2} = 0, \\ \vec f^u\cdot\vec f^z &= 0, \\ \vec f^v\cdot\vec f^z &= 0, \end{aligned}$$ we finally get the Laplacian as $$ \nabla^2 = \frac1{u^2+v^2}\left(\PD{^2}{u^2} + \PD{^2}{v^2}\right) + \PD{^2}{z^2}. $$