The Dirichlet Problem

The Dirichlet problem asks for a harmonic function on a domain with prescribed boundary values. Conformal mapping and the Poisson integral provide powerful tools for its solution.

Statement of the Dirichlet Problem

Definition8.2Dirichlet problem

Given a bounded domain $D \subset \mathbb{C}$ with boundary $\partial D$ and a continuous function $g: \partial D \to \mathbb{R}$ , the Dirichlet problem is to find $u: \overline{D} \to \mathbb{R}$ such that:

$u$ is harmonic on $D$ ( $\Delta u = 0$ ).
$u$ is continuous on $\overline{D}$ .
$u = g$ on $\partial D$ .

Theorem8.5Uniqueness of the Dirichlet problem

If the Dirichlet problem has a solution, it is unique. This follows from the maximum principle: if $u_1, u_2$ are two solutions, then $u_1 - u_2$ is harmonic with zero boundary values, so $u_1 - u_2 \equiv 0$ .

Solution on the Disk

Theorem8.6Poisson integral solution

The Dirichlet problem on the disk $D(0, R)$ is solved by the Poisson integral:

$u(re^{i\phi}) = \frac{1}{2\pi}\int_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr\cos(\theta - \phi) + r^2}\,g(Re^{i\theta})\,d\theta$

for $r < R$ . This function is harmonic in $D(0,R)$ , and $\lim_{r \to R^-} u(re^{i\phi}) = g(Re^{i\phi})$ at every point of continuity of $g$ .

ExampleDirichlet problem with piecewise constant data

Find $u$ harmonic on $|z| < 1$ with $u(e^{i\theta}) = 1$ for $0 < \theta < \pi$ and $u(e^{i\theta}) = 0$ for $\pi < \theta < 2\pi$ .

By the Poisson integral: $u(z) = \frac{1}{2\pi}\int_0^{\pi}\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta$ .

At $z = 0$ : $u(0) = 1/2$ (the average of boundary values). Along the real axis for $-1 < x < 1$ :

$u(x) = \frac{1}{\pi}\arctan\frac{1-x^2}{2x\cdot 0}$

which simplifies to $u(x) = \frac{1}{\pi}\arg\frac{1-x}{1+x} + \frac{1}{2}$ (expressible in terms of the argument function).

Solution via Conformal Mapping

Theorem8.7Conformal invariance of harmonicity

If $u$ is harmonic on a domain $D$ and $\varphi: D' \to D$ is holomorphic, then $u \circ \varphi$ is harmonic on $D'$ . Therefore, the Dirichlet problem on any simply connected domain $D$ can be solved by:

Finding the Riemann map $\varphi: D \to \mathbb{D}$ .
Solving the Dirichlet problem on $\mathbb{D}$ via the Poisson integral.
Composing: $u_D = u_{\mathbb{D}} \circ \varphi$ .

ExampleDirichlet problem on the upper half-plane

The Poisson integral for the upper half-plane $\mathbb{H} = \{z : \text{Im}(z) > 0\}$ is:

$u(x, y) = \frac{y}{\pi}\int_{-\infty}^{\infty}\frac{g(t)}{(t-x)^2 + y^2}\,dt.$

For $g(t) = 1$ on $[0, 1]$ and $0$ elsewhere:

$u(x,y) = \frac{1}{\pi}\left[\arctan\frac{1-x}{y} - \arctan\frac{-x}{y}\right] = \frac{1}{\pi}\arctan\frac{y}{x(x-1)+y^2}.$

Green's Function

Definition8.3Green's function

The Green's function $G(z, w)$ for a domain $D$ is the unique function satisfying:

$G(z, w) = -\frac{1}{2\pi}\ln|z - w| + h(z, w)$ where $h$ is harmonic in $z$ .
$G(z, w) = 0$ for $z \in \partial D$ .
$G(z, w) > 0$ for $z \in D$ , $z \neq w$ .

For the unit disk: $G(z, w) = -\frac{1}{2\pi}\ln\left|\frac{z - w}{1 - \bar{w}z}\right|$ .

RemarkDirichlet problem via Green's function

The solution to the Dirichlet problem can be expressed as $u(w) = -\oint_{\partial D} g(z)\frac{\partial G}{\partial n}(z,w)\,ds(z)$ where $\partial/\partial n$ is the outward normal derivative. For the disk, $-\partial G/\partial n$ reduces to the Poisson kernel.