Dirichlet problem on the unit disk using Poisson’s formula I’ve been trying to do the following exercise, of course without success, because I’m struggling with the integral. First things first, here's my exercise:
Writing the unit disk as $D \subset \mathbb{R}^2$, we define $g \in C(\partial D)$ by $g(x,y)=$ $\begin{cases}
       y \quad \text{if } y \geq 0\\
       0 \quad \text{otherwise}
     \end{cases}$.
If $u$ is a solution of the boundary problem
$\begin{cases}
       \Delta u = 0 \quad \text{in D}\\
       u=g \quad \text{on $\partial D$}
     \end{cases}$, compute $u(x,0)$ for $x\in (-1,1)$.
My intuition was to use Poisson's formula $u(x)= \frac{r^2-\lvert x \rvert ^2}{nr \omega_n} \int_{\partial D} \frac{g(y)}{\lvert x-y \rvert ^n} dS(y)$ and try to use the polar coordinates, but I'm instantly stuck because of the denominator. I'd really appreciate a hand with this!
 A: You can use the Poisson formula for the unit disk on the form
\begin{equation*}
  u(r\cos\theta,r\sin\theta) = \frac{1-r^2}{2\pi} \int_0^{2 \pi} \frac{h(\phi)}{1 - 2r \cos(\theta - \phi) + r^2} \, d\phi
 \end{equation*}
where $h(\phi)$ is the function giving the boundary data, in this case (since $y=\sin\phi$ on the unit circle)
\begin{equation*}
  h(\phi) =
  \begin{cases}
    \sin \phi, & 0 \le \phi \le \pi
    ,\\
    0, & \pi < \phi < 2\pi
    ,
  \end{cases}
\end{equation*}
so that
\begin{equation*}
  u(r\cos\theta,r\sin\theta) = \frac{1-r^2}{2\pi} \int_0^{\pi} \frac{\sin\phi}{1 - 2r \cos(\theta - \phi) + r^2} \, d\phi
  .
\end{equation*}
Since the exercise asks for the values of $u$ on the $x$ axis, take $\theta=0$ and $r=x$ (with $0 \le x < 1$ to begin with):
\begin{equation*}
  \begin{split}
    u(x,0)
    &
    = \frac{1-x^2}{2\pi} \int_0^{\pi} \frac{\sin\phi}{1 - 2x \cos\phi + x^2} \, d\phi
    \\ &
    = \frac{1-x^2}{2\pi} \biggl[ \frac{\ln|1 - 2x \cos\phi + x^2|}{2x} \biggr]_0^\pi
    \\ &
    = \frac{1-x^2}{4\pi x} \Bigl(  \ln|1 + 2x + x^2| - \ln|1 - 2x + x^2| \Bigr)
    \\ &
    = \frac{1-x^2}{4\pi x} \, \ln \frac{|1+x|^2}{|1-x|^2}
    \\ &
    = \frac{1-x^2}{2\pi x} \, \ln \frac{1+x}{1-x}
    .
  \end{split}
\end{equation*}
Since this turned out to be an even function, $u(-x,0)=-u(x,0)$, it is correct also for $-1 < x < 0$.
(The solution must be even with respect to $x$, since the boundary values are.)
So that's the answer to your exercise, if you add in the boundary values $u(\pm 1,0)=0$ “by hand”
(the expression above is undefined for $x=\pm 1$, but if you compute the limits as $x \to \pm 1$, you get zero).
As a bonus, note that with a little complex analysis we can compute the full solution for $u$ on the whole unit disk:
\begin{equation*}
  u(x,y) = \frac{y}{2} + \operatorname{Re} f(x+iy)
  ,\qquad
  f(z) = \frac{1-z^2}{2\pi z} \, \operatorname{Log} \frac{1+z}{1-z}
  ,
\end{equation*}
which after some calculations gives the following little expression:
\begin{equation*}
  \begin{split}
    u(x,y)
    & = \frac{y}{2}
    + \frac{x (1-x^2-y^2)}{4\pi (x^2+y^2)} \ln \frac{(1+x)^2 + y^2}{(1-x)^2 + y^2}
    \\&
    + \frac{y (1+x^2+y^2)}{2\pi (x^2+y^2)} \arctan \frac{2y}{1-x^2-y^2}
    ,
    \qquad
    x^2+y^2 < 1
    .
  \end{split}
\end{equation*}

