Let $r:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $r(x)=\|x\|,$ where $\large\|x\|=(x_1^2+\dots+x_n^2)^{\frac{1}{2}}.$ Compute $\nabla^2r,$ the Laplacian of $r.$
I perform the following computations,
$$\large\partial_{x_i}r(x)=\frac{1}{2} \cdot \frac{2x_i}{(x_1^2+\dots+x_n^2)^{\frac{1}{2}}}=\frac{x_i}{r(x)},$$
so that we get
$$\large\partial_{x_i}\partial_{x_i}r(x)=\frac{r(x)-\frac{x_i^2}{r(x)}}{r(x)^2},$$
and putting these together yields
\begin{align*}\large\nabla^2r(x)
&=\large\partial_{x_1}^2r(x)+\dots+\partial_{x_n}^2r(x) \\
&=\large\frac{r(x)-x_1^2\cdot r(x)^{-1}}{r(x)^2}+\dots+\frac{r(x)-x_n^2\cdot r(x)^{-1}}{r(x)^2}\\
&=\large\frac{n\cdot r(x)-r(x)^{-1}\cdot r(x)^2}{r(x)^2}\\
&=\frac{n-1}{r(x)}.\\
\end{align*}
Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere
Best Answer
If $u(x)=v(|x|)$ then $$\Delta u=\frac{n-1}{|x|}v'(|x|)+v''(|x|).$$ See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf