How to shorten the derivation of the Laplacian in polar coordinates?

Question

Yesterday, I computed the formula for the Laplacian in polar coordinates. $ \newcommand{p}[2]{\frac {\partial #1} {\partial #2}} \newcommand{s}[2]{\p {^2 #1} {#2 ^2}} \newcommand{m}[3]{\p {^2 #1} {#2 \partial #3}} \newcommand{po}[0]{\p {}} \newcommand{so}[0]{\s {}} \newcommand{mo}[0]{\m {}} \newcommand{t}[0]{\theta} \newcommand{w}[1]{\left( #1 \right)} \newcommand{ct}[0]{\cos \t} \newcommand{st}[0]{\sin \t} \newcommand{f}[1]{\frac #1 r} \newcommand{g}[1]{\frac #1 {r^2}} \newcommand{jx}[2]{\ct #1 - \f \st #2} \newcommand{jy}[2]{\st #1 + \f \ct #2} \newcommand{px}[2]{\jx {\po r #1} {\po \t #2}} \newcommand{py}[2]{\jy {\po r #1} {\po \t #2}} \newcommand{pox}[0]{\px {} {}} \newcommand{poy}[0]{\py {} {}} \newcommand{one}[0]{\sin^2\t + \cos^2\t} \newcommand{zero}[0]{\st\ct - \st\ct} $ \begin{align*} x & = r \ct \\ y & = r \st \\ r^2 & = x^2 + y^2 \end{align*}

First, I used the chain rule to relate $\po r$ and $\po \t$ to $\po x$ and $\po y$:

\begin{align*} \po x = \p r x \po r + \p \t x \po \t = \pox \\ \po y = \p r y \po r + \p \t y \po \t = \poy \\ \end{align*}

Applying each of $\po x$ and $\po y$ to itself:

\begin{align*} \so x + \so y & = \po x {\w {\po x}} + \po y {\w {\po y}} \\ & = \px {\w \pox} {\w \pox} \\ & \qquad + \py {\w \poy} {\w \poy} \\ & = \ct \w {\ct \so r + \g \st \po \t - \f \st \mo r \t} \\ & \qquad+ \f \st \w {\st \po r - \ct \mo \t r + \f \ct \po \t + \f \st \so \t} \\ & \qquad + \st \w {\st \so r - \g \st \po \t + \f \ct \mo r \t} \\ & \qquad + \f \ct \w {\ct \po r + \st \mo \t r - \f \st \po \t + \f \ct \so \t} \\ & = \w \one \so r + \g \zero \po \t \\ & \qquad - \f \zero \mo r \t + \f \one \po r \\ & \qquad - \f \zero \mo \t r + \g \zero \po \t \\ & \qquad + \g \one \so \t \\ & = \so r + \f 1 \po r + \g 1 \so \t \\ \end{align*}

This was long and tedious. Is there some way to make the calculation shorter.

Answer 1

If you are familiar with the language of differential forms, the Laplacian is $$\star d \star d,$$ where $\star$ is the Hodge star operator and $d$ is exterior derivative.

In polar coordinates, the volume form is $r dr \wedge d\theta$.

The Hodge star operator sends the volume form to $1$ ($\star r dr \wedge d\theta = 1$) and acts on the $1$-form basis as follows:

$$\star dr = r d\theta,\\ \star r d\theta = -dr.$$

So for a function $f$ we have: \begin{align} \star d \star d f &=\star d \star (f_r dr + \frac{f_\theta}{r} rd\theta)\\ &=\star d (f_r r d\theta-\frac{f_\theta}{r} dr)\\ &=\star (f_{rr}r dr \wedge d\theta + f_r dr \wedge d\theta - \frac{f_{\theta\theta}}{r} d\theta \wedge dr)\\ &=\star (f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2}) r dr \wedge d\theta\\ &=f_{rr} + \frac{f_r}{r} + \frac{f_{\theta\theta}}{r^2} \end{align}

Answer 2

Apparently, Math.SE does not allow me to “delete for real” my previous wrong answer, so I will try to provide a non-wrong answer.

Let $\newcommand \t \theta F(x,y) = f(r,\t)$, where $(r,\t) = g(x,y)$. Of course, in our case, $g$ is not a globally well-defined function, but who cares? Differential calculus is all about working locally. By the chain rule, $$F_{xx} = (f \circ g)_{xx} = (\nabla f \cdot g_x)_x = (\nabla f)_x \cdot g_x + \nabla f \cdot g_{xx} = g_x^T \cdot Hf \cdot g_x + \nabla f \cdot g_{xx}$$

Of course, analogously, we have $$F_{yy} = g_y^T \cdot Hf \cdot g_y + \nabla f \cdot g_{yy}$$

Adding everything, the Laplacian expands to $$F_{xx} + F_{yy} = g_x^T \cdot Hf \cdot g_y + g_y^T \cdot Hf \cdot g_y + \nabla f \cdot (g_{xx} + g_{yy})$$

Call the sum of the first two terms $A$ and call the last term $B$.

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In our particular case, $r^2 = x^2 + y^2$ and $\tan \t = y/x$. Thus, $$r_x = x/r, \qquad r_y = y/r, \qquad \sec^2 \t \cdot \t_x = -y/x^2, \qquad \sec^2 \t \cdot \t_y = 1/x$$

Rearranging the last two expressions, we have $$\t_x = -y/r^2, \qquad \qquad \qquad \t_y = x/r^2$$

By the chain rule, $A$ reduces to $$f_{rr} (r_x^2 + r_y^2) + 2f_{r\t} (r_x \t_x + r_y \t_y) + f_{\t\t} (\t_x^2 + \t_y^2) = f_{rr} + \frac 1 {r^2} f_{\t\t}$$

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Differentiating once again, we have $$r_{xx} = 1/r, \qquad r_{yy} = 1/r, \qquad \t_{xx} = 2yr_x /r^3, \qquad \t_{yy} = -2xr_y / r^3$$

Rearranging the last two expressions, we have $$\t_{xx} = 2xy / r^4, \qquad \qquad \qquad \t_{yy} = -2xy / r^4$$

By the chain rule, $B$ reduces to $$f_r \cdot (r_{xx} + r_{yy}) + f_\t \cdot (\t_{xx} + \t_{yy}) = \frac 1r f_r$$

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Combining everything, we have

$$F_{xx} + F_{yy} = f_{rr} + \frac 1r f_r + \frac 1 {r^2} f_{\t\t}$$