TITLE: Solving Laplace's equation with fourier transform
QUESTION [6 upvotes]: I am trying to solve $$u_{xx}+u_{yy}=0$$ where $-\infty < x <\infty$ and $0<y<1$, subject to $$u(x,0) = H(x)e^{-x}$$ $$u(x,1) = 0$$ where $H(x)$ is the Heaviside step function.
After applying transform in $x$, I get that the transformed solution is $U(\omega,y) = A(\omega)e^{\omega y}+B(\omega)e^{-\omega y}$, and, the boundary conditions become $$U(\omega,0) = \frac{1}{1+i\omega}$$ $$U(\omega,1) = 0$$
After applying these, I get stuck with $$B(\omega) = \frac{e^\omega}{2(1+i\omega)\sinh(\omega)}$$ and $$A(\omega) = \frac{1}{1+i\omega}\left(1-\frac{2e^{\omega}}{\sinh(\omega)}\right)$$
However, plugging this into the solution above and trying to invert this leads me nowhere. I'm not quite sure how to proceed. Any help or guidance would be great.

REPLY [2 votes]: After enforcing the boundary conditions, we find that 
$$A(\omega)=-\frac{e^{-\omega}}{2(1+i\omega)\sinh(\omega)}$$
$$B(\omega)=\frac{e^{\omega}}{2(1+i\omega)\sinh(\omega)}$$
Then, applying the Fourier inverse transform, we obtain
$$u(x,y)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega x} \frac{\sinh(\omega(1-y))}{(1+i\omega)\sinh(\omega)}\,d\omega \tag 1$$
To evaluate the integral in $(1)$, we use contour integration.  We note that the integrand has poles at $\omega =i$ and  $\omega =in\pi$, $n\ne0$.  
Enclosing the contour in the upper-half plane when $x>0$, application of the residue theorem yields
$$u(x,y)=\frac{e^{-x}\sin(1-y)}{\sin(1)}+\sum_{n=1}^\infty \frac{(-1)^ne^{-n\pi x}\sin(n\pi (1-y))}{n\pi -1}$$
Enclosing the contour in the lower-half plane when $x<0$, application of the residue theorem yields
$$u(x,y)=-\sum_{n=1}^\infty \frac{(-1)^n e^{n\pi x}\sin(n\pi (1-y))}{n\pi +1}$$