Prove a particular property of Laplacian operator

Question

I can't prove that Laplacian $\Delta(u(x))=\Delta(u(x_1,\ldots, x_n))=0$ also implies $$ \Delta\left(|x|^{2-n}u\left(\frac{x}{|x|^2}\right)\right)=0 $$ for $\frac{x}{|x|^2}$ in the domain of definition of $u$. Help me!!

Answer 1

Hint: Introduce $\mathbf{y}=\frac{\mathbf{x}}{|x|^{2}}$. Then \begin{equation*} (\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{\partial \mathbf{y}})u(\mathbf{y})=0 \end{equation*} Now express $\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{ \partial \mathbf{y}}$ in terms of $\mathbf{x}$ and $\frac{\partial }{ \partial \mathbf{x}}$.

Edit

It may help to use hyperspherical coordinates, see the contribution by Kienzler in What is the Laplace operator's representation in 3-sphere-coordinates?

Unless my calculation is incorrect it seems thar the desired property does not follow.

Notation: $\mathbf{x}$ is a vector from $\mathbb{R}^{n}$, $x$ its absolute value, $\Delta _{\mathbf{x}}=\frac{\partial }{\partial \mathbf{x}}\cdot \frac{\partial }{\partial \mathbf{x}}=\partial _{\mathbf{x}}\cdot \partial _{ \mathbf{x}}=\nabla _{\mathbf{x}}\cdot \nabla _{\mathbf{x}}$. We note that \begin{equation*} \Delta _{\mathbf{x}}=x^{1-n}\partial _{x}(x^{d-1}\partial _{x})+\frac{1}{ x^{2}}L^{2} \end{equation*} where $L^{2}$ only contains derivatives wrt. angles.

The function $u(\mathbf{x})$ satisfies \begin{equation*} \partial _{\mathbf{x}}\cdot \partial _{\mathbf{x}}u(\mathbf{x})=0 \end{equation*} Now the question is whether or not it is also true that \begin{equation*} \partial _{\mathbf{x}}\cdot \partial _{\mathbf{x}}\left( x^{2-n}u(\frac{% \mathbf{x}}{x^{2}})\right) =0 \end{equation*} Note that the part with angular derivatives gives no problems \begin{equation*} x^{-2}L^{2}\left( x^{2-n}u(\frac{\mathbf{x}}{x^{2}})\right) =x^{2-n}x^{-2}L^{2}u(\frac{\mathbf{x}}{x^{2}}) \end{equation*} We have \begin{equation*} \frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{\partial \mathbf{% y}}u(\mathbf{y})=0 \end{equation*} so, setting \begin{equation*} \mathbf{y}=\frac{\mathbf{x}}{x^{2}} \end{equation*} we have to express $\frac{\partial }{\partial \mathbf{y}}\cdot \frac{ \partial }{\partial \mathbf{y}}$ in terms of the $\mathbf{x}$-derivatives and we only need to consider the radial part \begin{equation*} y^{1-n}\partial _{y}y^{n-1}\partial _{y}=\partial _{y}^{2}+\frac{n-1}{y}% \partial _{y} \end{equation*} Note that \begin{equation*} \mathbf{y}=\frac{\mathbf{x}}{x^{2}}\Rightarrow y=\frac{1}{x},\;x=\frac{1}{y} ,\;\mathbf{x}=\frac{\mathbf{y}}{y^{2}},\;\partial _{y}=\frac{\partial x}{ \partial y}\partial _{x}=-\frac{1}{y^{2}}\partial _{x}=-x^{2}\partial _{x} \end{equation*} so \begin{equation*} X=y^{1-n}\partial _{y}y^{n-1}\partial _{y}=x^{n-1}(-x^{2}\partial _{x})x^{1-n}(-x^{2}\partial _{x})=x^{n+1}\partial _{x}x^{3-n}\partial _{x}=x^{4}\partial _{x}^{2}+(3-n)x^{3}\partial _{x} \end{equation*} Now

\begin{eqnarray*} Xx^{a} &=&x^{n+1}\partial _{x}x^{3-n}\partial _{x}x^{a}=x^{n+1}\partial _{x}x^{a+3-n}x^{-a}\partial _{x}x^{a} \\ &=&x^{n+1}x^{a+3-n}x^{-(a+3-n)}\partial _{x}x^{a+3-n}(\partial _{x}+\frac{a}{% x}) \\ &=&x^{a+4}\{\partial _{x}+(a+3-n)\frac{1}{x}\}(\partial _{x}+\frac{a}{x}) \\ &=&x^{a+4}\{\partial _{x}^{2}+(a+3-n)\frac{1}{x}\partial _{x}+\frac{a}{x}% \partial _{x}-\frac{a}{x^{2}}+a(a+3-n)\frac{1}{x^{2}}\} \\ &=&x^{a+4}\{\partial _{x}^{2}+(2a+3-n)\frac{1}{x}\partial _{x}+a(a+2-n)\frac{ 1}{x^{2}}\} \end{eqnarray*} This should be equal to $x^{a}X$ which requires \begin{eqnarray*} 2a+3-n &=&3-n\Rightarrow a=0 \\ a(a+2-n) &=&0 \end{eqnarray*} so $a=0$ is the only solution.