Let $u$ be a harmonic function in $\mathbb{R}^3$ and let $a > 0$.
Show that the function $v$ defined in spherical coordinates by $v(r,\theta,\psi )=\frac{a}{r}u(\frac{a^2}{r},\theta,\psi)$ is harmonic in $\mathbb{R}^3 \setminus \{ {0}\}$ .
I tried solving it but I'm getting nowhere. Could someone help, please?
In spherical coordinates, the Laplacian of $v$ is given by
$$\Delta v = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial v}{\partial r}\right) + \frac{1}{r^2\sin^2 \psi}\frac{\partial^2 v}{\partial \theta^2} + \frac{1}{r^2\sin\psi}\frac{\partial}{\partial \psi}\left(\sin\psi \frac{\partial v}{\partial \psi}\right).$$
Let $\rho = \frac{a^2}{r}$, so that $v(r,\theta,\psi) = \frac{\rho}{a}u(\rho,\theta,\psi)$. Let $\tilde{\Delta}$ denote that Laplacian with respect to spherical coordinates $(\rho,\theta,\psi)$. I claim
$$\Delta v = \frac{\rho^5}{a^5}\tilde{\Delta}u.$$
In particular, $v$ is harmonic in $\Bbb R^3\setminus\{0\}$ if $u$ is harmonic in $\Bbb R^3$. The claim is proven by repeated applications of the chain and product rules.
\begin{align}\frac{\partial}{\partial r}\left(r^2\frac{\partial v}{\partial r}\right) &= \frac{\partial \rho}{\partial r}\frac{\partial}{\partial \rho}\left(r^2\frac{\partial \rho}{\partial r}\frac{\partial v}{\partial \rho} \right)\\ &= -\frac{a^2}{r^2}\frac{\partial}{\partial \rho}\left(-a^2\frac{\partial v}{\partial \rho}\right)\\ &=\frac{a^4}{r^2}\frac{\partial}{\partial \rho}\left(\frac{1}{a}u+\frac{\rho}{a}\frac{\partial u}{\partial \rho}\right)\\ &=\rho^2\left(\frac{1}{a}\frac{\partial u}{\partial \rho} + \frac{1}{a}\frac{\partial u}{\partial \rho} + \frac{\rho}{a}\frac{\partial^2 u}{\partial \rho^2}\right)\\ &= \frac{\rho}{a}\left(2\rho\frac{\partial u}{\partial \rho} + \rho^2\frac{\partial^2 u}{\partial \rho^2}\right)\\ &= \frac{\rho}{a}\frac{\partial}{\partial \rho}\left(\rho^2\frac{\partial u}{\partial \rho}\right), \end{align}
$$\frac{1}{r^2\sin^2 \psi}\frac{\partial^2 v}{\partial \theta^2} = \frac{\rho^2}{a^4\sin \psi} \frac{\rho}{a}\frac{\partial^2 u}{\partial \theta^2} = \frac{\rho^3}{a^5\sin \psi} \frac{\partial^2 u}{\partial \theta^2},$$ and
$$\frac{1}{r^2\sin \psi}\frac{\partial}{\partial \psi}\left(\sin\psi \frac{\partial v}{\partial \psi}\right) = \frac{\rho^2}{a^4\sin \psi}\frac{\rho}{a}\frac{\partial}{\partial \psi}\left(\sin \psi \frac{\partial u}{\partial \psi}\right) = \frac{\rho^3}{a^5\sin \psi} \frac{\partial}{\partial \psi}\left(\sin \psi \frac{\partial u}{\partial \psi}\right).$$
So then
\begin{align} \Delta v &= \frac{\rho^2}{a^4} \frac{\rho}{a}\frac{\partial}{\partial \rho}\left(\rho^2 \frac{\partial u}{\partial \rho}\right) + \frac{\rho^3}{a^5\sin^2 \psi} \frac{\partial^2 u}{\partial \theta^2} + \frac{\rho^3}{a^5\sin \psi}\frac{\partial}{\partial \psi}\left(\sin \psi \frac{\partial u}{\partial \psi}\right)\\ &=\frac{\rho^5}{a^5}\left\{\frac{1}{\rho^2}\frac{\partial}{\partial \rho}\left(\rho^2\frac{\partial u}{\partial \rho}\right) + \frac{1}{\rho^2\sin^2\psi}\frac{\partial^2 u}{\partial \theta^2} + \frac{1}{\rho^2\sin \psi} \frac{\partial}{\partial \psi}\left(\sin \psi \frac{\partial u}{\partial \psi}\right)\right\}\\ &=\frac{\rho^5}{a^5}\tilde{\Delta}u, \end{align}
as desired.