How to solve 2D Laplace Equation over an infinite rectangular strip (bounded on two edges), with Dirichlet boundary conditions

Question

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?

Answer 1

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.

Answer 2

The solution under the boundary conditions $(T)_{y=0}=f(x)$ and $(T)_{x=0}=g(y)$ with the constraint $f(0)=g(0)$, is found by assuming $T$ to be of the form \begin{equation*} T=\int_0^{\infty} A(\lambda)\sin(\lambda x)e^{-\lambda y}\,d\lambda+\int_0^{\infty} B(\lambda)\sin(\lambda y)e^{-\lambda x}\,d\lambda. \end{equation*} Imposing the boundary conditions at $y=0$ and $x=0$, we get \begin{align*} A(\lambda)&=\frac{2}{\pi}\int_0^{\infty} f(\hat{x})\sin(\lambda\hat{x})\,d\hat{x}, \\ B(\lambda)&=\frac{2}{\pi}\int_0^{\infty} g(\hat{y})\sin(\lambda\hat{y})\,d\hat{y}. \end{align*} Substituting these expression into the original solution form, and carrying out the integration with respect to $\lambda$, we get \begin{align*} T&=\frac{y}{\pi}\int_0^{\infty}f(\hat{x})\left[\frac{1}{(x-\hat{x})^2+y^2}-\frac{1}{(x+\hat{x})^2+y^2}\right]\,d\hat{x} \\ &\quad +\frac{x}{\pi}\int_0^{\infty}g(\hat{y})\left[\frac{1}{x^2+(y-\hat{y})^2}-\frac{1}{x^2+(y+\hat{y})^2}\right]\,d\hat{y}. \end{align*} In order to verify that the boundary conditions are satisfied, one should first compute the integrals and then substitute $x=0$ or $y=0$.