Find a harmonic function $\phi(x,y)$ in the region $$D=\{x + iy : y ≥ 0, x^2 + y^2 \geq 1\}$$ (i.e., the region in the upper half plane outside the unit circle) that satisfies the boundary conditions $\phi(x,0) = 0$ for $x < −1$ and $x > 1$, $\phi(x,y) = 1$ for $x^2 + y^2 = 1$, $y > 0$. Hint: use the Joukowski map $f(z) = (1/2)(z + 1/z)$.
Does anyone know how to solve this question?
The trick is how the Joukowski map maps the region. It's more obvious if you write $z$ in polar form. Then the map becomes:
$$f(re^{j\varphi}) = {1\over2}\left(\left(r+{1\over r}\right)\cos\varphi,\left(r-{1\over r}\right)\sin\varphi\right)$$
Especially we see how the unit circle maps to the real $]-1,1[$ and the real line maps to the rest of the real line. We also see that the region $D$ maps into the upper half plane. So if we consider a mapping $\psi$ that on the real line is $1$ strictly inside the unit circle and $0$ strictly outside it.
To construct such a function one could consider $z^2-1$, which on the real line is positive where $\psi=1$ and negative where $\psi=0$ as described above. That is $\psi=\log(z^2-1)/\pi$ is a function whose imaginary part that fulfils the requirements (as the region of interrest is in the upper half plane one can select such a branch of $\log$).
Then we have a analytic function whose imaginary part fulfils the requirements for $\phi$, but the imaginary part of an analytic function is harmonic.