<s>[INST] If $u(r, \theta)$ is a solution of Laplace’s equation show that $u(\frac{1}{r}, \theta)$ is also a solution. 
Suppose that $u(r, θ)$ is a solution of Laplace’s equation. Show that
  $u(\frac{1}{r}, θ)$ is also a solution.

So far, I know that if $u$ satisfies Laplace's equation, then
$$\Delta u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta}$$ since $u$ is given in polar coordinates. I let $w = u(\frac{1}{r}, \theta)$. I want $w$ to also satisfy Laplace's Equation. Thus $\Delta w = w_{rr} + \frac{1}{r}w_r + \frac{1}{r^2}w_{\theta \theta}$
I computed the partial derivatives of $w$ and plugged them into the above equation but can't seem to get the right result.
Thanks for the help!
 [/INST] Consider a function $v(r, \theta) = u\left(q, \theta\right),$ with $q = \frac{1}{r}.$
Then, using the chain rule:


*

*$v_{r} = q_r u_q = -\displaystyle\frac{1}{r^2} u_q$

*$v_{rr} = \displaystyle\frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq}$ 

*$v_{\theta\theta} = u_{\theta\theta}$
Then:
$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta} 
= \frac{2}{r^3} u_q + \frac{1}{r^4} u_{qq} -\displaystyle\frac{1}{r^3} u_q + \frac{1}{r^2}u_{\theta\theta}  = \frac{1}{r^4}\left(u_{qq} + ru_q + r^2 u_{\theta\theta}\right).$$
Therefore, since $r = \frac{1}{q}$, then:
$$v_{rr} + \frac{1}{r}v_r + \frac{1}{r^2}v_{\theta \theta} 
= q^4\left(u_{qq} + \frac{1}{q}u_q + \frac{1}{q^2} u_{\theta\theta}\right) .$$
</s>