# Poisson equation on semi-infinite strip

Function $$u(x,y)$$ satisfy the equation: $$\Delta u = e^{-2y}\sin x$$

in the semi-infinite strip: $$00$$ and the boundary condition: $$u(0,y) = u(\pi,y),\text{ }u(x,0)=\sin(3x),\text{ } \lim_{y\to\infty} u =0$$

Find $$u_y(x,0)$$

I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $$u_y(x,0)$$, so is it possible to find $$u_y$$ without solving the equation?

Let $$v=u-\frac{1}{3}e^{-2y}\sin(x)$$, where $$u$$ is a solution of the stated problem. Then $$v$$ is a solution of $$\Delta v = \Delta u - e^{-2y}\sin(x)= 0,$$ with conditions $$v(0,y)=v(\pi,y) \\ v(x,0)=u(x,0)-\frac{1}{3}\sin(x)=\sin(3x)-\frac{1}{3}\sin(x)$$ The solutions $$v$$ is $$v(x,y)=\sin(3x)e^{-3y}-\frac{1}{3}\sin(x)e^{-y}$$ So,
\begin{align} u(x,y)&=v(x,y)+\frac{1}{3}e^{-2y}\sin(x) \\ &=\sin(3x)e^{-3y}-\frac{1}{3}\sin(x)e^{-y}+\frac{1}{3}\sin(x)e^{-2y}. \end{align} And, $$u_y(x,0)=-3\sin(3x)+\frac{1}{3}\sin(x)-\frac{2}{3}\sin(x) \\ =-3\sin(3x)-\frac{1}{3}\sin(x).$$