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Let $D=\{z\in\mathbb{C}|z\le1\}\setminus{-1,1}$.Find an explicit function $u$ on $D$ such that u is continuous on $D$ and harmonic in the interior $D$ and $u(z)=1$ on $|z|=1,Im\ z>0$ and $u(z)=-1$ on $|z|=1,Im\ z<0$.

My first guess is to use the Poisson Kernel to represent such a harmonic function,but Poisson Kernel need the harmonic function continuous up to the whole boundary,there are two jumping points here is the formula still valid?

Even though we use the Poisson Kernel,it seems that the integral cannot produce an explicit function.

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  • $\begingroup$ Poisson kernel is a special case of the theory of harmonic measure. $\endgroup$
    – user43423
    May 19, 2014 at 13:20
  • $\begingroup$ Hint: If $\varphi$ is a holomorphic function, and $u$ harmonic, then $u\circ\varphi$ is harmonic. Try to find a $\varphi$ such that the boundary conditions for $u\circ\varphi$ are easy to satisfy. $\endgroup$
    – Daniel Fischer
    May 19, 2014 at 13:20
  • $\begingroup$ @Daniel Fischer Aha!Thank you for your hint,I can invoke a Mobius transformation map the domain onto right half plane minus the origin,then I can define a branch of logarithm,of which the imaginary part is harmonic and constant on the boundary--imaginary axis!Thanks so much! $\endgroup$
    – Daniel S.
    May 19, 2014 at 13:24

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Solved in comments... the function $\dfrac{1}{\pi}\arg z$ on the upper half-plane often appears in this context: it is harmonic and takes values $0$ and $1$ on complementary half-axes. The modification $\dfrac{2}{\pi}\arg z - 1$ takes values $-1$ and $1$. It remains to map the unit disk to halfplane so that the jump points go to $0,\infty$.

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