Prove $\Delta(|x|^m)= m(m+n-2)|x|^{m-2}$ pointwise in $\mathbb{R}^n\backslash\{0\}$ for all $ m \in \mathbb{R}$

Question

Exactly what it says in the title. I'm trying to prove that the identity:

$\Delta(|x|^m)= m(m+n-2)|x|^{m-2}$ pointwise in $\mathbb{R}^n \backslash0 $ for all $ m \in \mathbb{R}$

holds (at least for $n\geq 3$). I've tried computing it directly, but I get something like

$\Delta(|x|^m)= m |x|^{m-2} [\sum_{i=1}^n (\frac{m}{2} -1) x_i |x|^2 +1]$

instead. (I'm actually trying to prove several identities like this one, but once I figure out what I'm doing wrong here I figure that I can get the rest sorted).

Bonus question: How does one come up with something like this? According to my book, these are the "key observation" to finding fundamental solutions to the biharmonic operator (once one finds a version that holds in $S'(\mathbb{R}^n)$, etc.).

Answer 1

Note that $$\frac{\partial}{\partial x_i}\left(|x|^2\right)=\frac{\partial}{\partial x_i}\left(\sum_{i=1}^nx_i^2\right)=2x_i$$ and thus, for any $k\in\mathbb{R}$, $$\frac{\partial}{\partial x_i}\left(|x|^k\right)=\frac{\partial}{\partial x_i}\left((|x|^2)^{k/2}\right)=\frac{k}{2}\cdot(|x|^2)^{k/2-1}\cdot2x_i=kx_i|x|^{k-2}.\tag{$*$}$$ So, for any $m\in\mathbb{R}$, $$\frac{\partial^2}{\partial x_i\partial x_i}\left(|x|^m\right)\overset{(*)\text{ for }k=m}{=}\frac{\partial}{\partial x_i}\left(mx_i|x|^{m-2}\right)\overset{(*)\text{ for }k=m-2}{=}m|x|^{m-2}+mx_i(m-2)x_i|x|^{m-4}$$ which implies \begin{align} \Delta\left(|x|^m\right)&=\sum_{i=1}^n\left(m|x|^{m-2}+m(m-2)x_i^2|x|^{m-4}\right)=mn|x|^{m-2}+m(m-2)|x|^2|x|^{m-4}\\&=m(n+m-2)|x|^{m-2}. \end{align}

Answer 2

Observe when $x\neq 0$ we have \begin{align} \Delta(|x|^m) =&\ \nabla\cdot\nabla |x|^m = \nabla\cdot\left(m|x|^{m-2}x\right)\\ =&\ m(\nabla |x|^{m-2})\cdot x+m|x|^{m-2}\nabla\cdot x \\ =&\ m(m-2) |x|^{m-4}x\cdot x + mn |x|^{m-2}\\ =&\ (m(m-2)+mn)|x|^{m-2}\\ =&\ m(m+n-2)|x|^{m-2}. \end{align}