How to solve 2D Laplace Equation over an infinite rectangular strip (bounded on two edges), with Dirichlet boundary conditions Is it possible to solve Laplace equation $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}=0$, over an infinite rectangular strip defined by $0 < x < \infty$ and  $0 < y < \infty  $, with the boundary conditions prescribed on two of the edges? 
We have Dirichlet boundary conditions specified along two edges - $u(0,y) =f(y)$ and $u(x,0)=0$. In addition to this, we have the condition that the function decays to zero at infinity. i.e. $u(\infty,y)=u(x, \infty)=0$. 
Is there a conformal transformation with which the values at infinity can be prescribed for such problems?
 A: The standard method is to solve two poblems and add the solutions; one problem is where $f=0$ and $g$ is general and the other is where $g=0$ and $f$ is general.
I'll start with the problem where $g=0$ and $u(x,0)=f(x)$ is general. For this problem, $u(0,y)=0$. The separation of variables solution is
$$
                    -\frac{X''}{X}=\lambda = \frac{Y''}{Y}.
$$
We want solutions $X$ that remain bounded on $[0,\infty)$ and which vanish at $x=0$, which forces $\lambda > 0$; otherwise you get $\sinh$ terms that are not bounded for all $x > 0$. So $\lambda=\mu^{2}$ where $\mu$ is real. The separated solutions are
$$
                        X_{\mu}Y_{\mu} = A(\mu)\sin(\mu x)e^{-\mu y}.
$$
The general solution of this subproblem is then
$$
      F(x,y) = \int_{0}^{\infty}A(\mu)e^{-\mu y}\sin(\mu x)d\mu.
$$
In order to satisfy $F(x,0)=f(x)$,
$$
       f(x) = \int_{0}^{\infty}A(\mu)\sin(\mu x)d\mu \\
       \implies A(\mu) = \frac{2}{\pi}\int_{0}^{\infty}f(x)\sin(\mu x)dx
$$
The final solution of this subproblem is
$$
        F(x,y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(x')\sin(\mu x')dx'\right)e^{-\mu y}\sin(\mu x)d\mu
$$
Similarly, the solution for the subproblem where $f=0$ and $g$ is general is
$$
       G(x,y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(y')\sin(\mu y')dy'\right)e^{-\mu x}\sin(\mu y)d\mu
$$
The final solution is
$$
                u(x,y) = F(x,y)+G(x,y).
$$
I'll leave the limit properties for you.
Note about Uniqueness: These are solutions of the problem where $f=g=0$:
$$
             \sin(\mu x)\sinh(\mu y),\;\; \sinh(\mu x)\sin(\mu y).
$$
Adding linear combinations of these to $u(x,y)$ as described above gives another solution of the equation, but the required limiting properties at $\infty$ are not satisfied.
