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Spring 2015 - Problem 11

harmonic functions

Let $\Omega = \set{z \in \C \mid 0 < \abs{z} < 1}$ . Prove that for every bounded harmonic function $\func{u}{\Omega}{\R}$ there is a harmonic function $\func{v}{\Omega}{\R}$ obeying

\pder{u}{x} = \pder{v}{u} \quad\text{and}\quad \pder{u}{y} = -\pder{v}{x}.

Solution.

Let

h\p{re^{i\theta}} = \frac{1}{2\pi} \int_0^{2\pi} \Re\p{\frac{1 + \p{1/2}e^{it}}{1 - \p{1/2}e^{it}}} u\p{e^{it}} \,\diff{t}

be the solution to the Dirichlet problem on the disk $B\p{0, \frac{1}{2}}$ with boundary condition $\res{u}{\abs{z}=\frac{1}{2}}$ . Let $v\p{z} = u\p{z} - h\p{z}$ and observe that $v\p{z} = 0$ on $\abs{z} = \frac{1}{2}$ by construction.

Let $\epsilon > 0$ and notice that $v_\epsilon\p{z} = v\p{z} + \epsilon\log{2\abs{z}}$ is harmonic away from $0$ . Moreover, $v_\epsilon\p{z} = v\p{z} = 0$ on $\abs{z} = \frac{1}{2}$ by construction. Since $v$ is bounded (because $u$ is bounded), we also see that

v_\epsilon\p{z} \xrightarrow{z\to0} -\infty,

so there exists $r > 0$ such that $v_\epsilon\p{z} \leq 0$ when $0 < \abs{z} \leq r$ . Applying the maximum principle to $v_\epsilon$ on the annulus $\set{r < \abs{z} < \frac{1}{2}}$ , it follows that $v_\epsilon\p{z} \leq 0$ on the entire punctured disk.

Sending $\epsilon \to 0$ , we see that $u\p{z} \leq h\p{z}$ on the punctured disk also. Running the same argument with $v$ replaced with $h - u$ instead, we see that $h\p{z} \leq u\p{z}$ on the punctured disk as well, and so $u\p{z} = h\p{z}$ except at the origin. In other words, $u$ extends harmonically to the entire disk, which is simply connected, so $u$ admits a harmonic conjugate.