Solve the Laplace equation ${\nabla}^{2} u = 0$, x > 0, y < 0. Problem
Solve the Laplace equation ${\nabla}^{2} u = 0$, $x > 0$, $y < 0$ with the conditions
$u(x, 0) = 0,\quad x > 0$
$u(0, y) = \begin{cases}
b, & -4 \le y \le -2 \\
0, & \text{$y \lt -4$ or $-2 \lt y \lt 0$}
\end{cases}$
$|u(x, y)| < M$
My current progress
I use separation of variables, let $u = XY$, and the Laplace equation becomes $\frac{X''}{X} = -\frac{Y''}{Y}$, and I let this equation = $-{\lambda}^2$ and use the boundary condition $u(x, 0) = 0, x > 0$.
The function $u$ becomes
$$u = \sin(\lambda y)[A_1e^{\lambda x} + B_1e^{-\lambda x}]$$
but I don't know how to use the rest of the condition.
Can anyone help me or give me some hints? Thanks!
 A: You are attempting to solve
$$
       \nabla^2 u(x,y)=0,\;\; x > 0, \; y < 0,\\
           u(x,0)=0,\;\; x > 0, \\
           u(0,y)=b\chi_{[-4,-2]}(y),\;\; y < 0, \\
        |u(x,y)| \le M,\;\;\; x > 0, y < 0.
$$
Here $b$ is a real constant. The constant $M$ is apparently unimportant, but must exist. Using separation of variables leads to a separation parameter $\lambda$ such that
$$
           \frac{X''}{X}=-\lambda^2 =-\frac{Y''}{Y}
$$
In order to have bounded separated solutions in the fourth quadrant where $x > 0, y< 0$, it is sufficient to assume separated solutions of the form
$$
             C(\lambda)e^{-\lambda x}\sin(\lambda y),\;\;\lambda > 0.
$$
The proposed separated solution becomes
$$
     u(x,y) = \int_0^{\infty}C(\lambda)e^{-\lambda x}\sin(\lambda y)d\lambda.
$$
The coefficient function $C(\lambda)$ is determined by
$$
           \int_0^{\infty}C(\lambda)\sin(\lambda y)d\lambda = b\chi_{[-4,-2]}(y),\;\; y < 0.
$$
This is a straightforward inverse Fourier sine transform problem.
