Bounded solution to $\Delta u = f$ with certain boundary conditions tend to 0?

Question

Let $T := \{(x, y) \in \mathbb{R}^{2}: x \geq 0, y \geq 0\}$ and suppose $f$ is a continuous function which vanishes when $x^{2} + y^{2} > R$ for some $R$. Suppose $u$ solves $$\Delta u = f \text{ in } T$$ with $u(x, 0) = u_{x}(0, y) = 0$. If $u$ is also bounded, must $u \rightarrow 0$ as $x^{2} + y^{2} \rightarrow \infty$?

Answer 1

Yes. One way to build an argument is this:

1. Reflect $u$ across the $y$-axis, to expand the domain to upper half-plane.
2. Let $\Omega = \{(x,y) : y>0,\ x^2+y^2>R\}$; the function $u$ is harmonic in $\Omega$.
3. Let $\Gamma = \{(x,y) : y\ge 0,\ x^2+y^2 = R\}$; this is the part of $\partial\Omega$ where $u$ is nonzero.
4. Observe that the harmonic measure of $\Gamma$ with respect to a point $z\in\Omega$ tends to zero as $|z|\to\infty$. This can be verified by mapping $\Omega$ to upper halfplane, so that $\Gamma$ is mapped to a line segment; the harmonic measure of a line segment on the boundary of a halfplane is known explicitly.
5. By the maximum principle (using the boundedness of $u$), we have $$|u|\le (\max_\Gamma |u|)\, \omega(z,\Gamma,\Omega)$$ where $\omega$ is the aforementioned harmonic measure.

Also relevant: Phragmén–Lindelöf principle.