Laplacian of a radial function

Question

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a radial function, i.e. $f(x)=f(r)$ with $r:=\left\|x\right\|_2$. As stated at Wikipedia $$\Delta f=\frac{1}{r^{n-1}}\frac{d}{dr}(r^{n-1}f')$$ What's the most elegant way to prove this fact?

Answer 1

On the one hand,

$$\frac{1}{r^{n-1}}\frac{d}{dr}\left(r^{n-1}f'\right) = \frac{1}{r^{n-1}} \left( (n-1)r^{n-2} f' + r^{n-1} f'' \right) = \frac{n-1}{r}f' + f''$$

On the other, the gradient of $f$ is $\displaystyle \nabla f = \frac{\vec{r}}{r}f'$ and thus the Laplacian of $f$ is

$$\nabla^2 f = \left( \frac{n}{r} - \frac{\vec{r}\cdot\vec{r}}{r^3}\right) f' + \frac{\vec{r}\cdot\vec{r}}{r^2}f'' = \frac{n-1}{r}f' + f''$$

Therefore $\displaystyle \Delta f = \frac{1}{r^{n-1}}\frac{d}{dr}\left(r^{n-1}f'\right)$.

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I don't know the history of writing the Laplacian of a radial scalar field this way. But it does make certain calculations fast. For example, in 3D do the scalar fields $f_1(r) = A/r$ or $f_2(r) = B/r^2$ have zero Laplacian?

It also makes calculating the Fourier transform of such a Laplacian easier.

Answer 2

Here's an alternative, it uses some heavy machinery (if some points are unclear perhaps the comment at the end might help) but casts a little light on the symmetry of the situation.

Let's distinguish $f$ from $\phi$ with $f(x)=\phi(\|x\|)$, and from now on we fix $x$ and put $\|x\|=r$. $B_{r}$ is the ball centered in the origin with radius $r$

We have that \begin{equation*} s_{n}r^{n-1}\phi'(r)=\int_{\partial B_{r}} \nabla f . \vec u \end{equation*} where $\vec u$ is the normal vector to $\partial B_{r}$, and $s_{n}$ is the surface area of $\partial B_{1}$.

By divergence theorem \begin{equation*} \int_{\partial B_{r}} \nabla f . \vec u = \int_{B_{r}} \Delta f \end{equation*}

But \begin{equation*} \frac{\partial}{\partial r}\left(\int_{B_{r}} \Delta f\right) = \int_{\partial B_{r}}\Delta f \end{equation*}

So \begin{equation*} \frac{\partial}{\partial r}\left(s_{n}r^{n-1}\phi'(r)\right)=\int_{\partial B_{r}} \Delta f \end{equation*}

Since $\Delta f$ is also a radial function \begin{equation*} \frac{1}{s_{n}r^{n-1}} \int_{B_{r}} \Delta f= \Delta f(x) \end{equation*} which concludes our proof (the $s_{n}$ cancel out).

A first problem with this argument is that it makes use of the fact that $\nabla f(x) = \phi'(\|x\|)\frac{x}{\|x\|}$ and that $\nabla f$ is also a radial function. Proving this properly requires more or less as much calculations as computing directly the laplacian. But this properties can be easily seen when picturing a radial function. The gradient must be orthogonal to the plane tangent to $B_{r}$ in $x$ (because $f$ is constant on $B_{r}$), so it is colinear to $x$ and must have norm $\phi'(r)$. The laplacian can be expressed from the local mean of $f$, so it should by invariant by isometry by radiality of $f$ and hence only depend on $r$.

The advantage of this proof is that it gives a faint idea of why a $r^{n-1}$ goes in and out the expression of the laplacian: the radiality of $f$ makes it possible to compute the laplacian on the whole $\partial B_{r}$ at once and not only in one point, so the $r^{n-1}$ appears when taking means over the $\partial B_{r}$.

Answer 3

You can read the section1.3 in Here, they proved that in the polar coordinate (here $r_n$ just means the $r$ in n-dimension.) $$ \sum_{i=1}^n D_{x_i}^2=D_{r_n}^2+\frac{n-1}{r_n} D_{r_n}+\frac{1}{r_n^2} \Lambda_n $$ where $\Lambda_n$ is a second order differential operator involving only the angular variables $\phi_2, \phi_3, \ldots, \phi_n$, (i.e., it does not involve $r_n$ ). Its proof is based on induction, since for radial symmetric function, we don't need the angular variables, so the only thing left is $\Delta f=\frac{1}{r^{n-1}} \frac{d}{d r}\left(r^{n-1} f^{\prime}\right)$.

Answer 4

Since I can't comment yet I'll point out here that in the last equation of the second answer (the one right before "which concludes our proof") I believe there's a typo and the integral should be evaluated over the border $\partial B_r$ of the ball $B_r$, and not on the ball itself, otherwise the last equation would simplify to $\phi^{'}(r) = \Delta f(x)$, which makes no sense.