3

I need to determine Greens function $G(x,x_0), x \mbox{ and }x_0 \in \mathbb{R}^2$ inside a semi-circle $(0<r<a, 0<\theta<\pi)$ with $\nabla G = \delta(x - x_0)$ and $G = 0$ on the boundary.

I know that the formula of the Green function in a circle is : $$\frac{1}{4\pi} \ln(a^2 \frac{r^2 + r_0^2 - 2rr_0cos\phi}{r^2r_0^2 + a^4 - 2rr_0a^2 cos\phi})$$. $\phi$ is the angle between $x$ and $x_0, r = |x|, r_0 = |x_0|$

I also know that I need to make a (positive or negative?) image such that $\nabla G = \delta(x-x_0) \pm \delta(x - x^*_0)$, but that are the only ideas I have, I have no clue how to find that image and proceed.

user54297
  • 1,036

1 Answers1

8

Presumably, $x_0$ lies inside of the upper semi-circle. Let $\bar{x}_0$ be its reflection in the horizontal axis. That is, if $ {x_0}$ has polar coordinates $(r_0, \theta_0)$, then $\bar{x}_0$ has polar coordinates $(r_0,-\theta_0)$.

Green's function for the upper semi-circle with pole at $x_0$ is $G(x,x_0)-G(x,\bar x_0)$. Why? Because it has the right kindof singularity at $x_0$, because it solves the Laplace's equation elsewhere in the domain, and because it is equal to $0$ on the boundary.

By the way, calculations in $\mathbb R^2$ are often made easier by complex notation. For example, $$\frac{1}{4\pi} \ln\left(a^2 \frac{r^2 + r_0^2 - 2rr_0\cos\phi}{r^2r_0^2 + a^4 - 2rr_0a^2 \cos\phi}\right) = \frac{1}{2\pi} \ln \frac{a\,|z-z_0|}{|a^2-\bar z_0 z|}$$ where $z=re^{i\theta}$ and $z_0=r_0e^{i\theta_0}$.

ˈjuː.zɚ79365
  • 3,004