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Spring 2017 - Problem 10

harmonic functions, residue theorem

Let $a_1, \ldots, a_n$ be $n \geq 1$ distinct points in $\C$ and let $\Omega = \C \setminus \set{a_1, \ldots, a_n}$ . Let $H\p{\Omega}$ be the vector space of real-valued harmonic functions on $\Omega$ and let $R\p{\Omega} \subseteq H\p{\Omega}$ be the space of real parts of analytic functions on $\Omega$ . Prove the quotient space $H\p{\Omega}/R\p{\Omega}$ has dimension $n$ , find a basis for this space, and prove it is a basis.

Solution.

We claim that $\set{u_1, \ldots, u_n}$ , where $u_k\p{z} = \frac{1}{2\pi}\log\,\abs{z - a_k}$ , is a basis for $H\p{\Omega}/R\p{\Omega}$ . Pick $\set{r_1, \ldots, r_n}$ small enough so that $\cl{B\p{a_k, r_k}} \subseteq \Omega$ and are pairwise disjoint.

First, we need to show that they are linearly independent. For each $u_k$ , consider its conjugate differential ${}^*\diff{u_k} = -\p{u_k}_y \,\diff{x} + \p{u_k}_x \,\diff{y}$ . Now suppose we have coefficients $c_1, \ldots, c_n$ such that $\sum_{k=1}^n c_k u_k = 0$ . Taking conjugate differentials, we see

\sum_{k=1}^n c_k {}^*\diff{u_k} = 0 \implies 0 = \int_{\partial B\p{a_j,r_j}} \sum_{k=1}^n c_k {}^*\diff{u_k} = \sum_{k=1}^n c_k \delta_{kj} = c_j

for all $j$ by direct calculation, so all the coefficients are zero, which means that the $u_k$ are linearly independent.

Let $\func{u}{\Omega}{\R}$ be harmonic and define $f = u_x - iu_y$ , which is holomorphic on $\Omega$ since $u$ is $C^2$ and satisfies the Cauchy-Riemann equations. If $f$ had an antiderivative $F = U + iV$ , then by the Cauchy-Riemann equations again,

u_x - iu_y = f = F' = U_x - iV_y \implies u = U + C,

where $C$ is some constant, which would mean that $u = \Re{f}$ . But this may not be the case since $\Omega$ is simply connected. To ensure that a holomorphic $g$ has an antiderivative, it suffices to have $\int_\gamma g\p{z} \,\diff{z} = 0$ for any closed $\gamma$ in $\Omega$ .

For each $k$ , let $c_k = \Res{f}{a_k}$ . Then by the residue theorem,

\int_\gamma f\p{z} - \sum_{k=1}^n \frac{c_k}{z - a_k} \,\diff{z} = 0

for any closed curve $\gamma$ in $\Omega$ . Thus, $g\p{z} = f\p{z} - \sum_{k=1}^n \frac{c_k}{z - a_k}$ is holomorphic in $\Omega$ with integral zero around any closed curve, so it has a primitive $G$ on $\Omega$ , by Morera's theorem.

If we set $\tilde{u}\p{z} = u\p{z} + \sum_{k=1}^n c_k\log\,\abs{z - a_k}$ , then

\begin{aligned} \tilde{u}_x\p{z} - i\tilde{u}_y\p{z} &= u_x\p{z} - iu_y\p{z} + \sum_{k=1}^n c_k\p{\frac{\Re\p{z - a_k}}{\abs{z - a_k}^2} - \frac{i\Im\p{z - a_k}}{\abs{z - a_k}^2}} \\ &= f\p{z} + \sum_{k=1}^n c_k\frac{\conj{z - a_k}}{\abs{z - a_k}^2} \\ &= f\p{z} + \sum_{k=1}^n \frac{c_k}{z - a_k}. \end{aligned}

Hence, by our argument above, we see that $\tilde{u}$ is the real part of a harmonic function $G$ on all of $\Omega$ , and so

u\p{z} + \sum_{k=1}^n c_k\log\,\abs{z - a_k} = \Re{G\p{z}} \in R\p{\Omega}.

In particular, $c_k$ is real, so the $u_k$ span and this completes the proof.