Spring 2021 - Problem 7

Solution.

First, we show uniqueness. If $u, v$ are two such harmonic functions, then $u - v = 0$ on $\partial\Omega$. Since $u - v$ is bounded, we may apply the Phragmèn-Lindelöf method to see that the maximum principle still applies, so $u - v \leq 0$ on $\Omega$. Applying the same argument to $v - u$, we see that $u - v = 0$ on $\Omega$, so we have uniqueness.

For existence, first consider $\phi\p{z} = z^2$ and $\psi\p{z} = \frac{z - i}{z + i}$. Observe that $\phi$ maps $\Omega$ conformally to the upper half-plane $\H$, the negative real-axis to itself, and the positive imaginary axis to the negative real axis. Similarly, $\psi$ maps $\H$ conformally to $\D$, maps the negative real axis to the arc $\p{0, \pi}$, and maps the positive real axis to the arc $\p{\pi, 2\pi}$. Thus, $\psi \circ \phi$ maps $\Omega$ conformally to $\D$, sending the positive real axis to the arc $\p{\pi, 2\pi}$ and the positive imaginary axis to $\p{0, \pi}$. On $S^1$, define

which is certainly in $L^1\p{S^1}$. Hence, its Poisson integral $v$ defines a harmonic function on $\D$ such that $v\p{z} \to f\p{e^{i\theta}}$ as $z \to e^{i\theta}$ at points of continuity of $f$. In particular, $v\p{z} \to 1$ on $0 < \theta < \pi$ and $v\p{z} \to 0$ on $\pi < \theta < 2\pi$ as $z \to e^{i\theta}$. Thus,

has the desired properties, since $\psi \circ \phi$ is continuous up to the boundary.