Find a bounded harmonic function \(u:\Delta\to\mathbb R\) such that \[ \lim_{r\to1}u(re^{i\theta})=1 \quad\text{for }0<\theta<\pi, \] and \[ \lim_{r\to1}u(re^{i\theta})=-1 \quad\text{for }\pi<\theta<2\pi. \]
Solution
Define \[ \Phi(z)=\frac{1+z}{1-z}. \] This maps the unit disk biholomorphically onto the right half-plane. On that half-plane choose the branch of the logarithm with argument in \((-\pi/2,\pi/2)\), and set \[ \boxed{ u(z)=\frac{2}{\pi}\operatorname{Im}\Log\!\left(\frac{1+z}{1-z}\right).} \]
1. The function \(u\) is harmonic and bounded on \(\Delta\).
For \(|z|<1\), \[ \operatorname{Re}\frac{1+z}{1-z} =\frac{1-|z|^2}{|1-z|^2}>0. \] Thus \(\Phi(\Delta)\) lies in the right half-plane. The chosen branch \(\Log\Phi(z)\) is therefore holomorphic on \(\Delta\), so its imaginary part is harmonic.
Moreover \[ -\frac{\pi}{2} <\operatorname{Im}\Log\Phi(z) <\frac{\pi}{2}, \] and hence \[ |u(z)|<1. \]
2. For \(0<\theta<\pi\),\[ \Phi(e^{i\theta})=i\cot(\theta/2) \]lies on the positive imaginary axis.
Using \[ 1+e^{i\theta}=2e^{i\theta/2}\cos(\theta/2), \qquad 1-e^{i\theta}=-2i e^{i\theta/2}\sin(\theta/2), \] we obtain \[ \Phi(e^{i\theta}) =i\cot(\theta/2). \] If \(0<\theta<\pi\), then \(0<\theta/2<\pi/2\), so \(\cot(\theta/2)>0\).
3. For \(\pi<\theta<2\pi\), the same boundary value\[ \Phi(e^{i\theta})=i\cot(\theta/2) \]lies on the negative imaginary axis.
Now \[ \frac{\pi}{2}<\frac{\theta}{2}<\pi, \] so \(\cot(\theta/2)<0\).
4. The radial limits of \(u\) are the required constants.
3.Consequently\[ \lim_{r\to1^-}u(re^{i\theta}) =\frac{2}{\pi}\cdot\frac{\pi}{2} =1 \qquad(0<\theta<\pi), \]while\[ \lim_{r\to1^-}u(re^{i\theta}) =\frac{2}{\pi}\cdot\left(-\frac{\pi}{2}\right) =-1 \qquad(\pi<\theta<2\pi). \]