1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $|x|)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.