Solving Laplace's equation with fourier transform

Question

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.

Answer 1

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

$$u(x,y)=-\sum_{n=1}^\infty \frac{(-1)^n e^{n\pi x}\sin(n\pi (1-y))}{n\pi +1}$$

Answer 2

Separation of variables can work. Assuming boundedness of the separated solutions in the $x$ direction, $$ X_{\lambda}(x)=C(\lambda)e^{i\lambda x},\;\; Y_{\lambda}(y)= A(\lambda)\sinh(\lambda y)+B(\lambda)\cosh(\lambda y). $$ Because you want $Y_{\lambda}(0)=0$, then $Y_{\lambda}$ can be written as $$ Y_{\lambda}(y) = D(\lambda)\sinh(\lambda(1-y)) $$ "Summing" the separated solutions and combing constants gives an unknown coefficient function $c(\lambda)$ such that $$ u(x,y) = \int_{-\infty}^{\infty}c(\lambda)e^{i\lambda x}\sinh(\lambda(1-y))d\lambda. $$ The coefficient function $c(\lambda)$ is determined by the condition $u(x,0)=H(x)e^{-x}$: $$ H(x)e^{-x}=\int_{-\infty}^{\infty}c(\lambda)e^{i\lambda x}\sinh(\lambda)d\lambda $$ Applying the Fourier transform gives $c(\lambda)\sinh(\lambda)$: $$ \frac{1}{2\pi}\int_{0}^{\infty}e^{-x}e^{-i\lambda x}dx =c(\lambda)\sinh(\lambda) \\ \frac{1}{2\pi(1+i\lambda)} = c(\lambda)\sinh(\lambda) $$ Therefore $u(x,y)$ is $$ u(x,y)=\frac{1}{2\pi }\int_{-\infty}^{\infty}e^{i\lambda x}\frac{\sinh(\lambda(1-y))}{(1+i\lambda)\sinh(\lambda)}d\lambda. $$