Problem: Let $u(z)$ be a bounded harmonic function in $\mathbb{D} = \{ |z| < 1 \}$ such that we have the limits $$\lim\limits_{r\rightarrow 1^-}u(re^{i\theta}) = \begin{cases}1 & \text{if } 0 < \theta < \pi\\ 0 & \text{if } \pi < \theta < 2\pi\end{cases}$$ Compute $u(\frac{1}{2})$.
Attempt: I consider applying Poisson formula to get that $$u(\frac{1}{2}) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1-|\frac{1}{2}|^2}{|e^{i\theta} - \frac{1}{2}|^2}u(e^{i\theta}) d\theta = \frac{3}{8\pi}\left[ \int_0^\pi \frac{1}{|e^{i\theta} - \frac{1}{2}|^2}u(e^{i\theta})d\theta + \int_\pi^{2\pi} \frac{1}{|e^{i\theta} - \frac{1}{2}|^2}u(e^{i\theta})d\theta \right] = \frac{3}{8\pi} \int_0^\pi \frac{1}{|e^{i\theta} - \frac{1}{2}|^2}d\theta $$ since $u(e^{i\theta})$ is given for intervals $0< \theta < \pi$ and $\pi < \theta < 2\pi$.
Now, we can compute the RHS by just working out the integral by hand. First we write out some ingredients so that everybody is on the same page:
Observe that $$|e^{i\theta} - \frac{1}{2}|^2 = |\cos \theta + i \sin\theta - \frac{1}{2}|^2 = (\cos \theta - \frac{1}{2})^2 + \sin^2 \theta = \frac{5}{4} - \cos\theta$$
We also have that: $$\int_\pi^{2\pi} \frac{1}{5-4\cos \theta} d\theta = \int_\pi^{3\pi/2} \frac{1}{5-4\cos \theta} d\theta + \int_{3\pi/2}^{2\pi} \frac{1}{5-4\cos \theta} d\theta = \int_{-\pi}^{-\pi/2} \frac{1}{5-4\cos \theta} d\theta + \int_{-\pi/2}^{0} \frac{1}{5-4\cos \theta} d\theta = \int_0^{\pi} \frac{1}{5-4\cos\theta} d\theta$$
Lastly, $$-2z^2 + 5z - 2 = -2(z-2)(z-\frac{1}{2})$$
Now, using the substitution of $z=e^{i\theta}$ and the three ingredients above, we get the following computation using the Residue theorem:
$$u(\frac{1}{2}) = \frac{3}{8\pi} \int_0^\pi \frac{1}{|e^{i\theta} - \frac{1}{2}|^2}d\theta = \frac{3}{4\pi} \int_0^{2\pi} \frac{1}{5-4\cos\theta} d\theta = \frac{3}{4\pi} \int_0^{2\pi} \frac{e^{i\theta}}{-2e^{2 i\theta} + 5e^{i \theta} - 2}d\theta = \frac{3}{4\pi i}\int_{|z|=1} \frac{1}{-2z^2 + 5z - 2}dz = \frac{3}{4\pi i} 2\pi i \lim\limits_{z\rightarrow \frac{1}{2}}\frac{1}{-2(z-2)} = \frac{3}{2} \cdot \frac{1}{-2(-\frac{3}{2})} = \frac{1}{2}$$
Question: Is my work correct? I feel a bit sketchy with the Poisson formula as I have never used it before. If so, is my method the standard method? Or is there a better way of computing this value?
Thank you.
The computations in the OP look correct and probably are the standard method but one can actually compute $u$ in this case (when it is called the harmonic measure of the upper semicircle wr circle) as $u(z)=\frac{2\theta(z)-\pi}{2\pi}$ where $\theta(z)$ is the angle subtended by the upper semicircle at $z$. For $z=1/2$ the angle is obviously $\pi$ so the result is indeed $1/2$.
Another way is to notice that if $v$ is the (bounded) harmonic function where you switch the boundary values, then $u+v=1$ and clearly by symmetry ($z \to \bar z$ which leaves $(-1,1)$ invariant) $u=v$ on $(-1,1)$ so $u(r)=v(r)=1/2, -1<r<1$