Solving Laplace's Equation on $(-\infty, 0)\times S^1$ I'm interested in solving
$$\begin{cases}
\Delta u(x,\theta) = 0 & x < 0 \\
u(0,\theta) = f(\theta)
\end{cases}$$
for some sufficiently nice prescribed $f$, and $u$ periodic in $\theta$ with period $2\pi$. Here $x$ and $\theta$ are both real. I've tried considering an ansatz of the form $u(x,\theta) = a(x)\sin(\theta)$. Plugging this into the PDE gives $a''(x) - a(x) = 0$, so $a(x) = e^x$. But then the boundary condition implies that $a(0) = f(\theta)/\sin(\theta)$, and we are in trouble unless $f(\theta) \equiv 0$. I also tried an ansatz of the form $u(x,\theta) = a(x)f(\theta)$, but plugging this into the PDE gives $\frac{a''}{a} = -\frac{f''}{f}$, which I'm not sure how to handle.
I also considered using Green's function for the halfspace. If we did not require $u$ to be periodic in $\theta$, then we would be able to write down the solution
$$u(x,\theta) = -\frac{x}{\pi}\int \frac{f(\eta)}{x^2 + (\theta - \eta)^2}\,d\eta,$$
but it's not clear to me how to modify this to make it periodic.
Any hints on how to proceed?
 A: Hint:
Use separation of variables
$$u(x,\theta) = X(x)\Theta(\theta),$$
as you have noticed, then note that in the resulting
$$\frac{X''}{X} = -\frac{\Theta''}{\Theta} = \lambda,$$
both sides must be the same constant, giving you the two ODEs
$$
\begin{align}
\Theta'' &= -\lambda \Theta,\qquad\Theta(0) = \Theta(2\pi) \\
X'' &= \lambda X.
\end{align}
$$
When we solve the first ODE, we should get
$$
\Theta(\theta) = A\cos(\sqrt\lambda\theta)+B\sin(\sqrt\lambda\theta),
$$
assuming $\lambda>0$.
From $\Theta(0)=\Theta(2\pi)$, we get that $\lambda$ has to be a positive perfect square, $n^2$. We cannot determine anything more about $n$ yet, apart from it being an integer. However, note that changing the sign of $n$ does not affect the solution, so WLOG, we can assume $n > 0$.
The other cases ($\lambda<0$ and $\lambda=0$) are straightforward;  you will find that $\lambda<0$ is incompatible with $\Theta(0)=\Theta(2\pi)$, and that $\lambda=0$ fits in nicely with our existing solutions.
$$
\begin{align}
\Theta_n(\theta) &= A_n \cos(n\theta) + B_n \sin(n\theta)&n>0\\
\Theta_0(\theta) &= A_0 &n=0
\end{align}$$
We can then solve the second ODE:
$$
\begin{align}
X_n(x) &= C_ne^{nx} + D_ne^{-nx}& n>0,\\
X_0(x) &= C_0 + D_0x &n=0.
\end{align}$$
Taking linear combinations, we get a bunch of solutions to the PDE (ignoring the initial conditions):
$$
\begin{align}
u(x,\theta) &= \sum_{n=0}^\infty \Theta_n(\theta) X_n(x) \\
&= C_0 + D_0 x + \sum_{n=1}^\infty (A_n \cos(n\theta) + B_n \sin(n\theta))(C_ne^{nx} + D_ne^{-nx})
\end{align}
$$
Normally, there is also a boundary condition of the form $\lim_{x\to-\infty}u(x,\theta)=0$ or $u_x(0,\theta) = g(\theta)$. If we had the first condition, we could set all the $D_n=0$.
At this point, so long as $f(\theta)$ has a Fourier series (which we know because it is sufficiently nice), we can satisfy the initial conditions.
