I'm looking for some verification on my work. This is a follow-up to this question where I am looking to solve Laplace's equation in the infinite strip $(x,y)\in(0,2\pi)\times(0,\infty)$ with slightly modified BCs:
$$\Delta u=0,$$ $$u\text{ periodic in $x$ with period }2\pi,$$ $$u(x,0)=f(x).$$
Through separation of variables, we have the solution
$$u=\sum_{k=-\infty}^\infty A_ke^{ikx}e^{-\vert k\vert y},$$ $$A_k=\frac{1}{2\pi}\int_0^{2\pi}e^{-ikx}f(x)\,dx.$$
By exchanging the sum and the integral, one can obtain the integral representation
$$u=\int_0^{2\pi}K_y(x-t)f(t)\,dt,$$ $$K_y(x)=\frac{1}{2\pi}\frac{e^{2y}-1}{1+e^{2y}-2e^y\cos(x)}.$$
Indeed, $\Delta K_y(x)=0$ and $\lim_{y\to0}K_y(x)=\delta(x)$ so it is a valid kernel. However, the desired behavior of the solution is $\lim_{y\to\infty}u=0$, but
$$\lim_{y\to\infty}u=\int_0^{2\pi}\lim_{y\to\infty}K_y(x-t)f(t)\,dt=\frac{1}{2\pi}\int_0^{2\pi}f(t)\,dt=\overline{f}.$$
In general, there is no restriction on the average $\overline{f}$, so the limit is not $0$ as desired. I can rectify the limit by omitting the eigenvalue $k=0$ in the separation-of-variables solution, but then $\lim_{y\to0}K_y(x)\not=\delta(x)$, so $K_y$ is not a valid kernel.
My thoughts are to generate a second harmonic function satisfying
$$\Delta v=0,$$ $$v\text{ periodic in $x$ with period }2\pi,$$ $$v(x,0)=0,$$ $$\lim_{y\to\infty}v=-\overline{f}.$$
It's clear the solution to this is the kernel itself:
$$v=-2\pi\overline{f}K_y(x).$$
So my desired solution is
$$u+v=\int_0^{2\pi}K_y(x-t)f(t)\,dt-2\pi\overline{f}K_y(x).$$
Is this correct? I suspect the singularities at $(0,0)$ and $(2\pi,0)$ in $v$ might be an issue. I appreciate any feedback!
