Prove a particular property of Laplacian operator 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!!
 A: 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.
