# An harmonic radial function in $\mathbb{R}^2$

I'm taking multivariable-calculus, and I got the following question:

A function $f$ in n variables is called harmonic if $\sum_{i = 1}^{n}{\frac{\partial ^2 f}{\partial x_{i}^2}} = 0$. Is there a non-constant, radial harmonic function in $\mathbb{R}^2$?

I found an almost-identical question here (PDF file) (number three), and I'm guessing that the answer is no.

One explanation I could think of, is that if there was such function, it would contradict the mean value property at the origin. However, we didn't learn about the mean value property (or about harmonic functions in general), so I'm not sure if I'm correct and either way I can't use it. I feel like there is something very simple I'm missing. Ideas?

Thanks!

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If you change your equation to polar coordinates you'll see that there are radial solutions. Those are considered elementary solutions. – Beni Bogosel Apr 10 '12 at 13:46

If $u$ is a radial harmonic function on the unit ball of $\mathbb R^2$, we can write $u(x,y)=g(x^2+y^2)$, where $g\colon\mathbb R_{\geq 0}\to\mathbb R$. We have $\partial_x u(x,y)=2xg'(x^2+y^2)$ and $\partial_{xx}u(x,y)=4x^2g''(x^2+y^2)+2g'(x^2+y^2)$ so $$\Delta u(x,y)=4(x^2+y^2)g''(x^2+y^2)+4g'(x^2+y^2).$$ We denote $h:=g'$, then for $r>0$ we have $rh'(r)+h(r)=0$, so $(rh(r))'=0$ and $rh(r)$ is constant. If we want $h$ non identically $0$ in order to have $g$ non constant, we have to take $h(r)=Cr^{-1}$ for $C\neq 0$, and $r\in\mathbb R_{>0}$. But $h$ cannot be defined at $0$, and so $g$ cannot be defined at $0$, so necessary $h$ is identically vanishing, $g$ is constant and $u$ is constant.