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I have been asked to find a bounded function $u(z)$ that is harmonic on $\Omega= \{z\in\mathbb{C}|\,|z|>1\,,\Im(z)>0\}$ that satisfies \begin{equation} u(z)=\begin{cases} 1\quad z\in\mathbb{R},|z|>1\\0\quad |z|=1,\Im(z)>0. \end{cases} \end{equation} Now i have been able to find a conformal map from $\Omega$ to the upper half-plane $\mathbb{H}=\{z|\Im(z)>0\}$, namely $F(z)=z+\frac{1}{z}$. Then the problem reduces to finding a bounded harmonic function on the upper half-plane that satisfies \begin{equation} u(x,0)=\begin{cases} 1\quad x\in[-2,2] \\0\quad x\in\mathbb{R}\setminus[-2,2]. \end{cases} \end{equation}

My first question: is there another conformal map $\phi$ that maps $[-2,2]$ to the vertical line $\Re(z)=1$ and that maps $\mathbb{R}\setminus[-2,2]$ to another vertical line $\Re(z)=0$. Because in that case I can find the correct harmonic function: $u(z)=\Re(\phi(F(z)))$

My second question: I saw here, Bounded harmonic function is constant, that a bounded harmonic function is a constant but clearly a constant function does not satisfy the boundary conditions so is it even possible to find the correct harmonic function?

Thank you in advance!

88888888
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  • You are leaving out continuity assumptions on the boundary. – zhw. Jan 26 '21 at 21:22
  • Because of the discontinuity at $z = 1$, your harmonic function will not be bounded there. It has to have an essential singularity at $1$ to exist at all. – Paul Sinclair Jan 27 '21 at 03:14
  • Does the problem has a solution if the condition of boundedness is omitted? Or only bounded on $\Omega$ and not on whole $\bar{\Omega}$? – 88888888 Jan 27 '21 at 07:27

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