Prove a function is harmonic

Question

This problem is from Conformal Mapping by Zeev Nehari:

If $u(x,y)$ is harmonic and $r=(x^2+y^2)^{1/2}$, prove $u(xr^{-2}, yr^{-2})$ is harmonic. The hint is obvious: "Use polar coordinates."

I have used the Laplacian for polar coordinates, but keep ending up with

$$\frac{1}{r^2} \left( -\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2}=\text{?}0$$

I have been working on this problem for awhile now. Any help would be much appreciated.

Answer 1

Write $u = u(r, \theta)$. Then the transformation

$$(x, y)\mapsto \left(\frac x{r^2}, \frac y{r^2}\right)$$

correxponds to

$$(r, \theta)\mapsto \left( \frac 1r , \theta\right)$$

in polar coordinate. Write $g(r, \theta) = u(r^{-1}, \theta)$. Then

$$g_r = -\frac{1}{r^2} u_r, \ \ u_{rr} =\frac{2}{r^3} u_r + \frac{1}{r^4} u_{rr},\ \ g_{\theta\theta} = u_{\theta\theta}.$$

Then \begin{equation} \begin{split} \Delta g (r, \theta) &= \left( g_{rr} + \frac 1r g_r + \frac{1}{r^2} g_{\theta\theta}\right) (r, \theta) \\ &=\left( \frac{2}{r^3} u_r + \frac{1}{r^4}u_{rr} -\frac{1}{r^3} u_r +\frac{1}{r^2} u_{\theta\theta} \right)(r^{-1}, \theta) \\ &= \left( \frac{1}{r^4}u_{rr} +\frac{1}{r^3} u_r +\frac{1}{r^2} u_{\theta\theta} \right)(r^{-1}, \theta) \\ &= \frac{1}{r^4} \left( u_{rr} + \frac{1}{r^{-1}} u_r+ \frac{1}{(r^{-1})^2} u_{\theta\theta}\right)(r^{-1}, \theta) \\ &= \frac{1}{r^4} \Delta u(r^{-1}, \theta) = 0 \end{split} \end{equation}

as $u$ is harmonic.

Answer 2

You have some sign errors in the polar representation for the 2D Laplacian. The polar coordinate $ r=\sqrt{ x^2 + y^2}$.

The Laplacian can be written in polars (sneakily) as
$r^{-2} [r(ru_r)_r+ u_{\theta \theta}] = r^{-2} [ D_r (D_r u)+ u_{\theta\theta} ] $

where $D_r= r\frac{\partial }{\partial r}$ is an operator that just changes sign when you replace $r$ by its reciprocal $1/r$. Since $D_r$ occurs twice, the Laplacian is transformed to a multiple of itself under this change of variables. This implies that the harmonic functions stay harmonic under this change of variables. (The broader idea is that log-polar coordinates are ideally suited to studying harmonic functions in the plane because if $ w= \ln r$ then $dr/r= dw $ and $D_r = \frac{\partial}{\partial w}$. Thus the Laplacian becomes $ e^{- 2w} ( u_{ww}+ u_{\theta \theta})$.

Answer 3

Let's be careful with the hypotheses. I suppose you are assuming $u$ is harmonic on $\mathbb C$ and you want to show your transformed function is harmonic on $\mathbb C \setminus \{0\}.$

Recall that a harmonic function composed with an analytic function is harmonic. Thus $v(z)=u(1/z)$ is harmonic on $\mathbb C \setminus \{0\}.$ But conjugation in the variable $z$ also preserves harmonic functions. Thus $v(\bar z)= u(\overline {1/z})$ is harmonic on $\mathbb C \setminus \{0\}.$ Because

$$\overline {1/z} = x/(x^2+y^2)+iy/(x^2+y^2),$$

we have the result.