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\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
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\begin{document}

{\bf Question}

By considering the function $A\log(z)+B$, find a harmonic function
in the upper half plane $Im(z)>0$ which takes the prescribed
values

$$\phi(x,0^+)=\left\{\begin{array}{l} 1,x>0\\ 0,x<0
\end{array}\right.$$

\medskip

{\bf Answer}

Let $Phi=A\log z+B$, where $\Phi=Phi(z),\ z=x+iy$. Assume $A$ and
$B$ are real.

Then

$Re(\Phi)=A\log(x^2+y^2)^{\frac{1}{2}}$

$Im(\Phi)=A\arg(z)+B=\arctan\left(\ds\frac{y}{x}\right)\times A+B$
fro \un{$y>0$}

$\Phi$ is analytic except at $z=0$ (and along a cut from there
which can be taken along negative imaginary axis). Thus $Re(\Phi)$
and $Im(\Phi)$ are harmonic in $Im(z)>0$.

If $\phi(x,o^+)=\left\{\begin{array}{l} 1, x>0\\ 0, x<0
\end{array} \right.$

the obvious choice is $Im(\Phi)$.

Why?

Well

\begin{eqnarray*} Im(\Phi) & = & A\theta+B\\
\theta & = & \arctan\left(\ds\frac{y}{x}\right) \end{eqnarray*}
(1)

So for $\begin{array}{l} x>0,\ y=0,\ \theta=0\\ x<0,\ y=0,\
\theta=\pi \end{array}$.

Thus

$Im(\Phi)=\left\{\begin{array} {ll} B & x>0,\ y=0\\ A\pi+B & x<0,\
y=0 \end{array} \right.$ (2)

Compare (1) and (2) to see $B=1,\ A=\ds\frac{-1}{\pi}$.

Thus harmonic function is:

$$\un{\phi(x,y)=1-\ds\frac{1}{\pi}\arctan\left(\ds\frac{y}{x}\right)}$$

This is \un{unique}.


\end{document}
