Finding a harmonic function on half-disk that is equal to $1$ on the semi-circle and $0$ on the diameter

Question

I first showed that the mapping

$$z + \frac{1}{z}$$

sends the upper semi-disk, $\{|z|<1, \operatorname{Im} z >0\}$, along with the real line from $-1$ to $1$, to the whole of the real line in the w-plane.

Specifically, it sends the upper semi-circular arc to the real interval $[-2,2]$, and sends $[-1,1]$ to the rest of the real line in the $w$-plane.

Now I want to find a harmonic function taking values $T=1$ on the arc, and $T=0$ along $[-1,1]$.

The arc mapping to an interval $[-2,2]$ makes it problematic to use the harmonic $\operatorname{Arg}(z)$ function.

I have tried scaling and translating the function but still not successful.

Answer 1

The arc mapping to an interval $[-2,2]$ makes it problematic to use the harmonic $\arg(z)$ function

But only a little. You can use $$\frac{1}{\pi}(\arg(w-2) -\arg(w+2))$$ which evaluates to $1$ on $(-2,2)$ and to $0$ outside of $[-2,2]$. There is no limit at the points of discontinuity, but this is to be expected.