Harmonic function on an annulus

Question

I have stumbled across the following fact in complex analysis and I was trying to prove it, but didn't get anywhere:

Let $R=\{r<|z|<R\}\subset\mathbb{C}$ where $0<r<R<\infty$ be an annulus in the complex plane and $u$ a harmonic function on $R$. Then there exists a constant $C\in\mathbb{R}$ and a holomorphic function $f$ on $R$ such that $$u(z)=\mathrm{Re}f(z)+C\log|z|$$

The problem arises because $R$ is not simply connected (on simply connected domains this is clearly true for $C=0$ by the CR equations). I just don't see where that $\log$ should come from.

Can somebody help me and provide an easy proof?

Any potentially useful approaches/hints are welcome.

Just a thought: try slicing the annulus the same way you slice the plane to get a branch of $\log(z)$. Do it in two ways: 1) Remove positive real axis, 2) remove negative real axis. Each one of these annuli is simply connected, so gives two different ways of expressing $u(z)$ as real parts of two functions $f(z)$ and $g(z)$, one on each split annulus. Maybe look at the difference of these on each component of the intersection of the split annuli? — bryanj, Jul 03 '13 at 17:49

score 6 · Accepted Answer · answered Jul 03 '13 at 17:55

Let $g=u_x-iu_y$ and $\gamma(t)=l\cdot e^{it}$ for $t\in[0,2\pi]$ and $r<l<R$. Clearly $g$ is holomorphic on $R$. Now set $$C=\frac{1}{2\pi i}\int_\gamma g(z) dz$$ This can be seen to be a real number. Now note that if $\tilde{\gamma}$ is a closed path in $R$ with winding number $m$ around $0$, then $$\frac{1}{2\pi i}\int_\tilde{\gamma} \left(g(z)-\frac{C}{z}\right)\,dz=\frac{m}{2\pi i}\int_\gamma g(z) dz - \frac{C}{2\pi i}\int_\tilde{\gamma} \frac{dz}{z}=mC-mC=0 $$ where the first equality follows from Cauchy's theorem since $m\gamma-\tilde{\gamma}$ is nullhomotopic in $R$ and $g$ holomorphic. Hence the function $$f(z)=\int_{z_0}^z\left(g(z)-\frac{C}{z}\right)\, dz+u(z_0)$$ is well-defined and holomorphic on $R$, where $z_0\in R$ fixed and we integrate over an arbitrary path from $z_0$ to $z$ within $R$.

Now the Cauchy-Riemann equations imply $$u(z)=\mathrm{Re} f(z)+C\log|z|$$

@ Anonymous999 : Could you explain "$g$ is holomorphic on R" in detail? — user64494, Jul 03 '13 at 17:59
Just check the Cauchy-Riemann equations: $u$ is harmonic, i.e. $u_{xx}=-u_{yy}$ and Schwarz' theorem implies $u_{xy}=u_{yx}$. — Anonymous999, Jul 03 '13 at 18:02

score 2 · Answer 2 · answered Feb 27 '17 at 17:53

2

Here is a paper with more general resoult yet similar reasoning.

answered Feb 27 '17 at 17:53

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Harmonic function on an annulus

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