I'm trying to solve the following problem using the Fourier transform:
$$u_{xx}+u_{yy}=0$$ on the domain $\;0\lt y\lt b$ , $-\infty\lt x \lt \infty \;$ with the following conditions:
$$ u(x,0)= \begin{cases} 1 & \text{for }-a< x<a\\ 0 & \text{for}\; \; \; | x |>a\\ \end{cases} $$ $$u(x,b)=0, \ -\infty\lt x \lt \infty $$ So, here's what I tried: First, I took the Fourier transform of in the $x$ variable, and the problem now looks like this: $$ \frac{\partial^2 \hat{u}(\omega,y)}{\partial y^2}-{\omega}^2\hat{u}(\omega,y)=0 $$ With the conditions: $$\hat{u}(\omega,0)=\frac{2\sin{\omega a}}{\sqrt{2\pi}\omega}\ \ \ \text{and} \; \;\hat{u}(\omega,b)=0 $$ Note that I'm using the following definitions for the Fourier transform and its inverse: $$ {\mathscr{F}}[f(x)]= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{i\omega x}f(x)\,dx \;,\;\; {\mathscr{F}}^{-1}[\hat{f}(\omega)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-i\omega x}\hat{f}(\omega)\,d\omega$$ Then, I solved the second order equation for $\hat{u}(\omega,y)$, which has the solution: $$ \hat{u}(\omega,y)=Ae^{\omega y}+Be^{-\omega y} $$ Using the boundary conditions I found that: $$ A+B =\frac{2\sin{\omega a}}{\sqrt{2\pi}\omega} \\Ae^{2\omega b} + B=0 $$ Now, here's where I'm stuck. The problem says that I need to establish a "bounding condition". So, the condition I proposed was that $u(x,y)\rightarrow 0$ as $x\rightarrow \pm \infty $ . My question is: does this condition carry on to $\hat{u}$? i.e. Must $\hat{u}(\omega,y)$ vanish as $ \omega \rightarrow \pm \infty$? I don't really know what to do from here on or whether I'm on the right track. Any help would be appreciated!