(This is the $n$-dimensional analogue of the 3D case: Show that $H(x) := |x|^{-1} u(x/|x|^2) $ is harmonic if $u$ is harmonic)
Suppose that $u$ is a harmonic function on $\mathbb{R}^n$.
Prove that the function $\displaystyle H(x):=\frac{1}{|x|^{n-2}}u\left(\frac{x}{|x|^2}\right)$ is harmonic on $\mathbb{R}^n\backslash\{0\}$.
I tried to compute $\Delta H$ directly by following the brute force method in the linked question, but it gets tedious very soon. Therefore, I am wondering if there is a more elegant way?
I am thinking to use the converse of mean value property. So far I have worked out (probably) that
- Under the mapping $f: x\mapsto \frac{x}{|x|^2}$,
a circle with radius $r$ centered at $x_0$ would be mapped to a circle with radius $R=\frac{2r}{|x_0|^2-r^2}$ centered at $y_0=\frac{x_0}{|x_0|^2-r^2}$. - $f = f^{-1}$
- Its Jacobian is $|Jf|=|x|^{2n}$.
I am stuck after doing change of variables in the integral $\displaystyle\frac{1}{|B_R(y_0)|}\int_{B_R(y_0)} H(y)dy$ , since the terms do not magically cancel out as wished, and expressing $|B_R(y_0)|$ in terms of the corresponding $|B_r(x_0)|$ also yields a mess.
I really appreciate any help. Other methods (or more efficient brute force) are also greatly welcomed.
I'm putting this in as an answer as it's too long for a comment.
This will definitely be computation-heavy, but here are some formulas that could help. Note that the bolded nabla is understood to be the general tensor gradient, also known as the covariant derivative.
The gradient has a chain rule for $\varphi:\mathbb{R}^n\to\mathbb{R}$, $\mathrm{v}:\mathbb{R}^n\to\mathbb{R}^n$ $$\nabla(\varphi\circ \mathrm{v})=(\nabla\varphi\circ\mathrm{v})~\boldsymbol\nabla\mathrm{v}$$ The divergence has one as well, for $\mathrm{w}:\mathbb{R}^n\to\mathbb{R}^n$: $$\nabla\boldsymbol{\cdot}(\mathrm{v}\circ\mathrm{w})=\operatorname{tr}\big((\boldsymbol\nabla\mathrm{v}\circ\mathrm{w})\boldsymbol\nabla\mathrm w\big)$$