Laplace equation can implies some result

Question

Prove that $$ \Delta u(x)=\Delta u(x_1,\ldots,x_n)=0 $$ also implies that $$ \Delta (|x|^{2-n}u(x/|x|^2))=0 $$ for $x/|x|^2$ in the domain of definition of $u$. I have no idea how to start with this question. I try to calculate and simply the result of $\Delta (|x|^{2-n}u(x/|x|^2))$ to get $0$. But I didn't get some good as we want. Hope some can get me some tips or the proof let me to reference and learn.

Answer 1

We start by recalling the Laplacian in $\mathbb{R}^n$ (up to convention on sign) is: $$ \Delta_{(r,\vec\theta)}=\underbrace{r^{1-n}\partial_r r^{n-1}\partial_r}_{=\partial_{rr}+(n-1)r^{-1}\partial_r=\text{radial part}}+\underbrace{r^{-2}\Delta_{S^{n-1}}}_{\text{spherical part}} $$ in spherical polars. The inversion $x\mapsto\frac{x}{\lvert x\rvert^2}$ is $(r,\vec\theta)\mapsto(1/r,\vec\theta)$.

So as remarked in my comments above, it suffices (apply to $u(1/r,\vec\theta)$) to establish $$ \Delta_{(r,\vec\theta)} r^{2-n}\operatorname{id} = \rho^{n+2}\Delta_{(\rho,\vec\theta)} $$ where $\rho=1/r$. The spherical part is obvious, so we just need to consider the radial part. By the chain rule, $$ \frac{\partial}{\partial r}=\frac{\mathrm{d}\rho}{\mathrm{d}r}\frac{\partial}{\partial\rho}=-\rho^2\frac{\partial}{\partial\rho} $$ or equivalently $$ r\partial_r=-\rho\partial_\rho. $$ So the radial part of LHS is \begin{align*} &r^{1-n}\partial_r r^{n-1}\partial_r r^{2-n}\operatorname{id}\\ &=r^{-n}[r\partial_r r\partial_r+(2-n)r\partial_r]\operatorname{id}\\ &=\rho^n[\rho\partial_\rho \rho\partial_\rho + (n-2)\rho\partial_\rho]\operatorname{id}\\ &=\rho^{n+2}[\partial_{\rho\rho}+(n-1)\rho^{-1}\partial_\rho]\operatorname{id} \end{align*} which agrees with the radial part of RHS $\rho^{n+2}\Delta_{(\rho,\vec\theta)}$. QED.