Solve Dirichlet Problem: Find u(x,y) Harmonic on Upper Half-Plane

[skrat]

Homework Statement

Find function $u(x,y)$ that is harmonic on the upper half-plane $0<Im(z)$. Note that
$u(x,0)=0$, $x<0$
$u(x,0)=-1$, $0<x<1$ and
$u(x,0)=1$, $x>1$.

Homework Equations

$u(z)=\frac{1}{2\pi i}\int _{I}\gamma '(s)u(\gamma (s))\frac{\alpha '(\gamma (s))}{\alpha (\gamma (s))}Re(\frac{\alpha (\gamma (s))+\alpha (z)}{\alpha (\gamma (s))-\alpha (z)})ds$

Where $\alpha$ is holomorphic function that maps $Im(z)>0$ into open unit disk, and $\gamma (s)$ parameterization of the edge.

The Attempt at a Solution

For this problem $\alpha (z)=\frac{1-iz}{1+iz}$ and it is also obvious that I will have to make at least 3 integrals.

Let's firstly take a look at $u(x,0)=0$, $x<0$:

This tells me that $u(z)=\frac{1}{2\pi i}\int _{I}\gamma '(s)u(\gamma (s))\frac{\alpha '(\gamma (s))}{\alpha (\gamma (s))}Re(\frac{\alpha (\gamma (s))+\alpha (z)}{\alpha (\gamma (s))-\alpha (z)})ds$ will be equal to $0$ for all $x<0$.

But the question here is: What about $y$ ? None of the conditions tell anything about $y$. What do I do here?

The same question is for $u(x,0)=-1$, $0<x<1$:

Here the integral for $x$ goes from $0$ to $1$. But again, $y$ can be anything from $0$ to $\infty$. ?

I guess my question here is: What should I do with $y$ to get $u(z)$?

 

[skrat]

Ok, I found out that my relevant equation in original post was a very stupid equation to begin with. I think it would wiser to use the Poisson integral formula for the upper half plane: $u(x,y)=\frac{y}{\pi }\int_{-\infty }^{\infty }\frac{u(t,0)}{y^2+(x-t)^2}dt$ where I am guessing $t$ stands for some kind of parameterization of real axis? http://www.diss.fu-berlin.de/diss/s...ionid=76E8A9CC9C7EE802A569873649AAF2F7?hosts= page 59.

Anyway, now with that in mind:

$u(x,y)=\int_{-\infty }^{0}...+\int_{0}^{1}...+\int_{1}^{\infty}$ where the first integral is obviously $0$.

$u(x,y)=-\frac{y}{\pi }\int_{0}^{1}\frac{1}{y^2+(x-t)^2}dt+\frac{y}{\pi }\int_{1}^{\infty }\frac{1}{y^2+(x-t)^2}dt$

$u(x,y)=\frac{y}{\pi }[\frac{1}{y}arctan(\frac{u}{y})\mid _{u=x}^{u=x-1}-\frac{1}{y}arctan(\frac{u}{y})\mid _{u=x-1}^{u=-\infty }]$

and finally

$u(x,y)=\frac{1}{\pi }[2arctan(\frac{x-1}{y})-arctan(\frac{x}{y})+\frac{\pi }{2}]$

This should be a bit better than my first attempt.