Find an harmonic function $h(z)$ in the upper half plane with the following properties:

  1. $h$ is bounded in $\mathbb{C}_+$ and is continuous in $\{z | \Im(z) \geq 0 \}\setminus \{0 \}$

  2. $h(x)=1$ if $x>0$ and $h(x)=-1$ if $x<0$

Also, prove that there is exactly one harmonic function with these properties.

This time I really have no clue. Any hints?

  • $\begingroup$ Can you see the existence of $h$? For uniqueness, do you know Lindelöf's maximum principle in the linked page? $\endgroup$ – 23rd Jul 30 '13 at 19:15
  • $\begingroup$ @Landscape: I don't see the existence, and also as I understand the question - I need to actually find such a function. Unfortunately I don't know this theorem. $\endgroup$ – catch22 Jul 30 '13 at 19:19
  • $\begingroup$ No theorem is needed in constructing $h$(just consider something like the argument function); the theorem I cited is only used for proving uniqueness of $h$, as your question stated "exactly one". $\endgroup$ – 23rd Jul 30 '13 at 19:23

If we consider $f(z)=f(x,y)$, then this problem is equivalent to the 1st boundary problem of Laplas' equation:

$$\begin{cases}\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x^2}=0 \\ |f(x,y)|\le C \text{ for all } x \in \mathbb{R} \text{ and for } y>0\\ f(x,0)=\begin{cases}1,x>0\\ -1,x<0 \end{cases} \end{cases}$$

It has very simple solution with the help of fundamental operator:

$$f(x,y)=\frac{y}{\pi}\int\limits_{-\infty}^{+\infty}\frac{f_{0}(t)}{\left(x-t\right)^2+y^2}dt= -\frac{y}{\pi}\int\limits_{-\infty}^{0}\frac{1}{\left(x-t\right)^2+y^2}dt+\frac{y}{\pi}\int\limits_{0}^{+\infty}\frac{1}{\left(x-t\right)^2+y^2}dt$$

If you want to represent this function as a complex value function, you can calculate this integrals. The result is

$$f(x,y)=\operatorname{csgn}(\bar{y})\frac{\pi}{y} $$

-- this is that stuff which the program has calculated. In complex it is:

$$ f(z)=\operatorname{csgn}(\overline{\operatorname{Im} z})\frac{\pi}{\operatorname{Im} z}$$

According to the uniqueness of the solution, this solution is unique.

  • $\begingroup$ Thanks Paul. I took this question from a previous' semester exam and I have no idea what you're talking about :) I guess the curriculum was different that semester... $\endgroup$ – catch22 Jul 30 '13 at 19:31

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