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Spring 2010 - Problem 2

harmonic functions

Let $p_1, p_2, \ldots, p_n$ be distinct points in the complex plane $\C$ and let $U$ be the domain

U = \C \setminus \set{p_1, \ldots, p_n}.

Let $A$ be the vector space of real harmonic functions on $U$ and let $B \subseteq A$ be the subspace of real parts of complex analytic functions on $U$ . Find the dimension of the quotient vector space $A/B$ , give a basis for this quotient space, and prove that it is a basis.

Solution.

We claim that if $u_k\p{z} = \log\abs{z - p_k}$ , then $\set{u_1, \ldots, u_n}$ is a basis for $A/B$ . In particular, $\dim\p{A/B} = n$ .

Write $p_k = a_k + b_ki$ so that $u_k\p{x, y} = \frac{1}{2}\log\p{\p{x - a_k}^2 + \p{y - b_k}^2}$ . Thus,

\frac{\partial^2 u_k}{\partial x^2} = \frac{-\p{x - a_k}^2 + \p{y - b_k}^2}{\abs{z - p_k}^4} \quad\text{and}\quad \frac{\partial^2 u_k}{\partial x^2} = \frac{\p{x - a_k}^2 - \p{y - b_k}^2}{\abs{z - p_k}^4}.

Hence, $\Delta u_k = 0$ , which means $u_k \in A$ for all $1 \leq k \leq n$ .

Let $\gamma_k$ be a small circle oriented counter-clockwise centered around $p_k$ with radius $r_k$ small enough so that the $\gamma_k$ are pairwise disjoint. Then

\int_{\gamma_k} u_k\p{z} \,\diff{z} = \int_0^{2\pi} \log\,\abs{r_k} \,\diff\theta = 2\pi\log\,\abs{r_k}.

This tells us that near $p_k$ , $u_k$ is not the real part of a holomorphic function. Indeed, if that were not the case, then Cauchy's theorem implies that this integral would be independent of the radius $r_k$ .

Next, if we define the conjugate differential of $u \in A$ to be ${}^*\diff{u} = -u_y \,\diff{x} + u_x \,\diff{y}$ , then

{}^*\diff{u_k} = -\frac{y - b_k}{\abs{z - p_k}^2} \,\diff{x} + \frac{x - a_k}{\abs{z - p_k}^2} \,\diff{y}.

Also, notice that $u_k = \Re\p{\log\p{z - p_k}}$ . So, if $k \neq \ell$ , $\log\p{z - p_k}$ is holomorphic on the ball $B\p{p_\ell, r_k}$ and so $v = \Im\p{\log\p{z - p_k}}$ satisfies $\diff{v} = {}^*\diff{u_k}$ , so $\int_{\gamma_\ell} {}^*\diff{u_k} = 0$ . Otherwise, if $k = \ell$ , then we can parametrize $\gamma_k$ via $z = p_k + r_ke^{i\theta}$ , which gives

\begin{aligned} \int_{\gamma_k} {}^*\diff{u_k} &= \int_0^{2\pi} -\frac{r_k\sin\theta}{r_k^2} \p{-r_k\sin\theta} + \frac{r_k\cos\theta}{r_k^2} r_k\cos\theta \,\diff\theta \\ &= \int_0^{2\pi} \,\diff\theta \\ &= 2\pi. \end{aligned}

This implies that the $u_k$ are linearly independent:

\begin{aligned} c_1u_1 + \cdots + c_nu_n = 0 &\implies \int_{\gamma_k} c_1 {}^*\diff{u_1} + \cdots + c_n {}^*\diff{u_1} = 0 \\ &\implies 2\pi c_k = 0 \\ &\implies c_k = 0 \end{aligned}

for any $k$ . It remains to show that the $u_k$ form a basis. Let $u \in A$ and set $c_k = \int_{\gamma_k} {}^*\diff{u}$ . Consider

\tilde{u} = u - \frac{1}{2\pi} \sum_{k=1}^n c_ku_k.

$\tilde{u}$ is harmonic, and its conjugate differential satisfies

\int_{\gamma_\ell} {}^*\diff{\tilde{u}} = \int_{\gamma_\ell} {}^*\diff{u} - \frac{1}{2\pi} \sum_{k=1}^n c_k \int_{\gamma_\ell} {}^*\diff{u_k} = c_k - c_k = 0

for all $1 \leq \ell \leq n$ . Set $f\p{z} = \tilde{u}_x - i\tilde{u}_y$ so that $f \,\diff{z} = \diff{\tilde{u}} + i{}^*\diff{\tilde{u}}$ . Notice that $f$ is holomorphic since it satisfies the Cauchy-Riemann equations. Hence, if $\gamma$ is any simple closed curve in $U$ , let $m_k$ be the winding number of $\gamma$ around $p_k$ and $\sigma = \gamma - \sum_{k=1}^n m_k\gamma_k$ , which gives

\int_\sigma f\p{z} \,\diff{z} = 0

by the residue theorem: $f$ can only have poles at $p_1, \ldots, p_n$ , and we cancel out the residues in the sum via the $\gamma_k$ . Since $\diff{u}$ is exact and $\gamma$ is closed,

0 = \int_\sigma f\p{z} \,\diff{z} = \int_\sigma \p{\diff{\tilde{u}} + i{}^*\diff{\tilde{u}}} = i\int_\sigma {}^*\diff{\tilde{u}}.

Thus,

\int_\gamma {}^*\diff{\tilde{u}} = \sum_{k=1}^n m_k \int_{\gamma_k} {}^*\diff{\tilde{u}} = 0.

In other words, ${}^*\diff{\tilde{u}}$ is conservative, and so

v\p{z} = \int_{\gamma_{z_0\to z}} {}^*\diff{\tilde{u}}

is a well-defined harmonic conjugate to $\tilde{u}$ . This means that $\tilde{u}$ is the real part of a holomorphic function on $U$ , so $u - \frac{1}{2\pi} \sum_{k=1}^n c_ku_k \in B$ , which shows that $\set{u_1, \ldots, u_n}$ spans $A/B$ , so it is a basis for $A/B$ .