# Fourier Transform on Infinite Strip Poisson Equation

Im trying to solve the following Poisson equation:

$$u_{xx} + u_{yy} = \exp(-x^2)\ \text{for}\ x \in (-\infty, \infty)\ \text{and}\ y \in (0,1)$$

$$u(x,0) = 0,\ u(x,1) = 0$$

$$u(x,y) \to 0\ \text{uniformly as}\ |x| \to \infty\ \text{(i.e. compact support).}$$

I want to solve this using the Fourier Transform. I've tried taking the Fourier Transform with respect to $x$ to get an ODE in the $y$ variable but I'm having trouble understanding since the equation is not homogeneous. I'm also getting stuck with $y$ being a finite interval instead of the half line. Any help and suggestions are appreciated!

You have the right idea...take FT in $x$ and get

$$\frac{\partial^2}{\partial y^2} \hat{u}(k,y) - k^2 \hat{u}(k,y) = \sqrt{\pi} e^{-k^2/4}$$

$$\hat{u}(k,0) = 0 \quad \hat{u}(k,1) = 0$$

This is a simple BV problem with a constant on the RHS (yes, the $k$ is a constant as far as the diff eq'n is concerned). The general solution is

$$\hat{u}(k,y) = A(k) e^{k y} + B(k) e^{-k y} + C(k)$$

You get $C$ by plugging back into the equation:

$$C(k) = -\frac{\sqrt{\pi}}{k^2} e^{-k^2/4}$$

You get $A$ and $B$ by applying the boundary conditions. There is some algebra involved which I leave to you, but I get

$$A(k) = \frac{\sqrt{\pi}}{k^2} e^{-k^2/4} e^{-k/2} \frac{\sinh{(k/2)}}{\sinh{k}}$$

$$B(k) = \frac{\sqrt{\pi}}{k^2} e^{-k^2/4} e^{k/2} \frac{\sinh{(k/2)}}{\sinh{k}}$$

You then get the solution from the inverse FT:

$$u(x,y) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{u}(k,y) \, e^{-i k x}$$

BONUS

I get

$$\hat{u}(k,y) = \frac{\sqrt{\pi}}{k^2} e^{-k^2/4} \left (2 \cosh{\left [ k \left (y-\frac12 \right) \right]} \frac{\sinh{(k/2)}}{\sinh{k}} - 1 \right )$$

The factor of $1/k^2$ looks worrisome. Do not worry: $\hat{u}(0,y)$ is finite.