Find a solution $u(x,y)$ of Laplace’s equation on the domain
$-\infty< x < \infty$ and $0 <y<$ $\infty$ for which $u(x,0)=x^{1/2}$ for $0<x< \infty$. What is $u(x,0)$ for $-\infty <x< 0$?
I am solving older qualifying on Applied, complex part. And we study the conformal mapping has application for Laplace equation. Usually intersection between two circles with constant boundaries condition that we send to a strip with conformal mapping and we can see that the solution does not depend of one of the variables so.. you get the solution has linear form $cx+d$ and then you use the boundaries condition and get the answer.
But here… The condition is not constant, and it is only on some part of the boundary.
So… My question is , if I send it to a strip too with a $\ln z$ I only have the boundary condition for $y=0$ that is $x^{1/2}$ and I need to find an answer for the region and for the boundary $y= \pi$. Or what could be another region I can send it? The idea is use the less PDE theory, so I do not want to send it to the circle and use Poisson. (Since this for people who has no PDE courses too)
Should be the other part of the boundary condition $u(x,0)= -(-x)^{1/2}$ for $-\infty <x< 0$ ?
Best Answer
Since the general solution of Laplace’s equation is $u(x,y)=F(x+iy)+G(x-iy)$ , for matching the only condition $u(x,0)=\sqrt{x}$ , you can obviously take $F(x)=\dfrac{\sqrt{x}}{2}+f(x)$ and $G(x)=\dfrac{\sqrt{x}}{2}-f(x)$ so that you can get $u(x,y)=\dfrac{\sqrt{x+iy}+\sqrt{x-iy}}{2}+f(x+iy)-f(x-iy)$
Note that this solution not only satisfy on $-\infty<x<\infty$ and $0<y<\infty$ but also satisfy on $x,y\in\mathbb{C}$ .