Dirichlet problem for upper half plane

Question

The Dirichlet problem I read is as follows:

If $f$ is an integrable function, find a function $u$ such that for $x \in \mathbb{R}, y>0$ \begin{align} u_{xx} + u_{yy} & =0 \\ \lim_{y \to 0^+} u(x,y) &= f(x) \text{ almost everywhere} \end{align}

Does the method listed in this Find the solution of the Dirichlet problem in the half-plane y>0. also work if $u$ and $u_x$ are not required to vanish as $|x| \to \infty$ and $u$ is bounded? Since if we discard these conditions we can not exclude the case $\lambda>0$.

Answer 1

The final Poisson form of the solution will give you a non-unique solution: $$ u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-x')^2+y^2}f(x')dx'. $$ The non-uniqueness is easily verified because $v(x,y)=y$ is a solution of the Laplace equation that vanishes identically on the real line. So $u(x,y)+Cy$ is another solution, for any constant $C$.

Answer 2

In short, consider the Green function $$G_y(x) = \frac{y}{\pi (x^2+y^2)} = \frac{d}{dx}\frac{\log(x-iy)-\log(x+iy)}{2i\pi}$$ Then $G$ is harmonic for $y >0$ as well as $G_. \ast f$.

Integrating by parts and using that $$\lim_{y \to 0^+}\log(x\pm iy)- \log|x|=\pm i\pi 1_{x < 0}$$ the convergence being locally uniform away from $x=0$, you'll obtain $$\lim_{y \to 0^+} G_y \ast f(x) = \lim_{y \to 0^+}\int_{-\infty}^\infty G_y(t) f(x-t)dt = \int_{-\infty}^\infty 1_{t < 0} f'(x-t)dt = f(x)$$ at least if $f,f' \in L^1$