Solve the Laplace equation ${\nabla}^{2} u = 0$, x > 0, y < 0.

Question

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!

Answer 1

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.