Laplace's equation periodic in one dimension, from boundary values

Question

I'm trying to solve Laplace's equation in a domain that is semin infinite in one ordinate and periodic in the other.

That is, we consider a pair of functions $x(\xi,\nu),y(\xi,\nu)$ such that we have the information $$f(\xi) = x(\xi,0)+i y(\xi,0),$$ where $f(\xi)$ is the boundary value of the analytic function $F(\xi,\nu)$. The domain is such that $\xi \in (0,2\pi)$ while $\nu \in (-\infty,0)$.

I would like to find $x(\xi,\nu)$ and $y(\xi,\nu)$, that is, the analytic extension of these functions to the semi infinite strip.

Any tips, or suggestions, would be greatly appreciated.

Thanks

Answer 1

Generally, the approach depends on what the given function is. Some functions are easier to integrate against the Poisson kernel than others. Also, some functions can be related to a holomorphic function in an algebraic way.

To use the Poisson kernel:

1. Subtract the mean value from the given boundary function $x$. You can add it to $x$ later.
2. Extend $x$ by periodicity, i.e., make it a $2\pi$-periodic function on $\mathbb R$.
3. Convolve with the Poisson kernel $$x(\xi,\nu) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{-\nu}{(\xi-t)^2+\nu^2} x(t)\,dt $$ and the conjugate Poisson kernel $$y(\xi,\nu) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{ t-\xi }{(\xi-t)^2+\nu^2} x(t)\,dt $$

The integrals should converge (the second only conditionally, and that only thanks to removing the mean value of $x$). Whether or not you can actually evaluate them is anyone's guess.