Solving Laplace Equation with Fourier Integral

Question

I have a problem with using the Fourier integral for solving the Laplace equation.

Problem. Consider the Laplace equation under Dirichlet type boundary conditions in the upper half-plane \begin{equation} \left\{ \begin{gathered} u_{xx}+u_{yy}=0,\quad-\infty<x<\infty,\ y>0\\ u(x,0)=\begin{cases} 1,&|x|<\pi\\ 0,&|x|\geq\pi \end{cases}\\ u\ \text{is bounded as}\ y\to\infty\\ u\ \text{and}\ u_{x}\ \text{vanish as}\ x\to\pm\infty. \end{gathered} \right.\notag \end{equation} We will use the Fourier integral concept to find the solution. ■

I can use Fourier transform and show that the solution is \begin{equation} u(x,y):=\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{\omega}\sin(\pi\omega)\cos(\omega{}x)\mathrm{e}^{-\omega{}y}\mathrm{d}\omega.\notag \end{equation} However, I cannot solve this problem by applying Fourier integral technique. I could also solve some heat equations but I could not succeed with this Laplace equation. Let me show you where I stuck.

Solution. Suppose that $u(x,y):=\varphi(x)\psi(y)$. Substituting this into the PDE gives us \begin{equation} \varphi^{\prime\prime}+\lambda\varphi=0 \quad\text{and}\quad \ddot{\psi}-\lambda\psi=0,\notag \end{equation} where $\lambda$ is the so-called separation constant. Then, we have \begin{equation} \varphi_{\lambda}(x):= \begin{cases} a(\lambda)\mathrm{e}^{-\omega{}x}+b(\lambda)\mathrm{e}^{\omega{}x},&\lambda<0\\ a(\lambda)+b(\lambda)x,&\lambda=0\\ a(\lambda)\cos(\omega{}x)+a(\lambda)\sin(\omega{}x),&\lambda>0 \end{cases}\notag \end{equation} and \begin{equation} \psi_{\lambda}(y):= \begin{cases} c(\lambda)\cos(\omega{}y)+d(\lambda)\sin(\omega{}y),&\lambda<0\\ c(\lambda)+d(\lambda)y,&\lambda=0\\ c(\lambda)\mathrm{e}^{-\omega{}y}+d(\lambda)\mathrm{e}^{\omega{}y},&\lambda>0, \end{cases}\notag \end{equation} where $\omega>0$ satisfies $\omega^{2}=|\lambda|$. Now, consider the discussion on the separation constant $\lambda$.

1. Let $\lambda<0$.
   • $\varphi_{\lambda}(x)$ and $\varphi_{\lambda}^{\prime}(x)$ do not vanish as $x\to\pm\infty$ if $a(\lambda)\neq0$ or $b(\lambda)\neq0$.
   • $\psi_{\lambda}(y)$ is bounded as $y\to\infty$.
2. Let $\lambda=0$.
   • $\varphi_{\lambda}(x)$ and $\varphi_{\lambda}^{\prime}(x)$ do not vanish as $x\to\pm\infty$ if $a(\lambda)\neq0$ or $b(\lambda)\neq0$.
   • $\psi_{\lambda}(y)$ is not bounded as $y\to\infty$ if $d(\lambda)\neq0$.
3. Let $\lambda>0$.
   • $\varphi_{\lambda}(x)$ and $\varphi_{\lambda}^{\prime}(x)$ do not vanish $x\to\pm\infty$ if $a(\lambda)\neq0$ or $b(\lambda)\neq0$.
   • $\psi_{\lambda}(y)$ is not bounded as $y\to\infty$ if $d(\lambda)\neq0$.

Then, we have \begin{equation} \varphi_{\lambda}(x):= \begin{cases} 0,&\lambda<0\\ 0,&\lambda=0\\ 0,&\lambda>0 \end{cases}\notag \end{equation} and \begin{equation} \psi_{\lambda}(y):= \begin{cases} c(\lambda)\cos(\omega{}y)+d(\lambda)\sin(\omega{}y),&\lambda<0\\ c(\lambda),&\lambda=0\\ c(\lambda)\mathrm{e}^{-\omega{}y},&\lambda>0. \end{cases}\notag \end{equation} Therefore, $u_{\omega}(x,y):=0$ for any $\lambda$... ■

At this point, depending on the ranges of $\lambda$, I cannot figure out that $\lambda$ must be positive, i.e., the solution should be in the form \begin{equation} u_{\omega}(x,y):=\{A(\omega)\cos(\omega{}x)+A(\omega)\cos(\omega{}x)\}\mathrm{e}^{-\omega{}y} \quad\text{for}\ \omega>0.\notag \end{equation} Should I consider a boundary condition of the form $\varphi_{\lambda}(\pm\ell)=0$ for $\ell\geq\pi$.

Answer 1

The separation of variables problem is $$ -\frac{X''}{X} = \lambda = \frac{Y''}{Y} \\ X''+\lambda X = 0 ,\;\;\; Y''-\lambda Y=0. $$ In order for the solution $X$ to be bounded for all $x$ it necessary to to have $\lambda$ to be real and positive. So, set $\lambda = \omega^2$. The solutions $Y$ must similarly be bounded. So, $$ X(x) = Ae^{i\omega x},\;\;\; Y(y)=Be^{-|\omega|y},\;\;\; -\infty < \omega < \infty. $$ A general Fourier integral solution is $$ u(x,y) = \int_{-\infty}^{\infty}c(\omega)e^{i\omega x}e^{-|w|y}d\omega, $$ and $c(\omega)$ is a coefficient function determined by the boundary condition: $$ \chi_{[-\pi,\pi]}(x) = u(x,0) =\int_{-\infty}^{\infty}c(\omega)e^{i\omega x}d\omega. $$ It leads to the same place, provided by you use $e^{-|\omega|y}$.