Using appropriate conformal maps, solve the Dirichlet problem (for Laplace's equation) for the following region and boundary condition: $U=\{\text{Im}(z)>0\cup \text{Im}(z)=0\}$, with boundary conditions $f(x,0)=0$ when $\mod(x)>1$ and $f(x,0)=1$ when $\mod(x)<1$

I have not been able to make much progress: by trial and error I have found a few functions satisfying the boundary conditions, but they do not satisfy Laplace's equation. any help will be greatly appreciated!

  • $\begingroup$ Does mod(x) mean absolute value of x? $\endgroup$ – Jo Wehler Jan 15 '15 at 6:24
  • $\begingroup$ yes. thanks a lot for the help! $\endgroup$ – user144361 Jan 19 '15 at 18:19

The answer to your question is given by the Poisson integral for the upper half plane $\mathbb H$. The solution is

$$P_f(x+iy) = \frac {1}{\pi} \int_{- \infty}^{\infty} \frac{y}{(x-\xi)^2 + y^2} f(\xi)d\xi$$

for $x + i y \in \mathbb H$. The integral evaluates as

$$P_f(x+iy) = \frac {y}{\pi} \int_{- 1}^{1} \frac{1}{\xi^2 -2\xi x + x^2 + y^2}d\xi = \frac {1}{\pi}[arc\ tg\frac {1-x}{y} - arc\ tg\frac {-1-x}{y}].$$

We have $P_f(x+iy)=f(x+iy)$ if $y=0, |x| \neq 1$ (I assume $mod(x)$ means $|x|)$.


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