Method of Images - 1st Quadrant

Aug 2010
10
0
Solve

\(\displaystyle \bigtriangledown ^2 G = \delta (x - x_0)\)

in the first quadrant \(\displaystyle (x \geq 0, y \geq 0)\) with \(\displaystyle G = 0\) on the boundaries in 2 dimensions.

Now I know the solution to this problem for the half-plane\(\displaystyle (y \geq 0, - \infty < x < \infty)\) is determined by the fact that

\(\displaystyle G = \frac{1}{2 \pi} \ln |r - r_0|\)

where \(\displaystyle r = \sqrt{(x - x_0)^2 + (y - y_0)^2}\)

and \(\displaystyle r_0 = \sqrt{(x - x_0)^2 + (y + y_0)^2}\)

which leads to

\(\displaystyle G = \frac{1}{4 \pi} \ln \Big((x - x_0)^2 + (y - y_0)^2 \Big) + \frac{1}{4 \pi} \ln \Big( (x - x_0)^2 + (y + y_0)^2 \Big)\)

So to solve this problem, however, I think I need to add more image sources, so at a guess I would have...

\(\displaystyle r = \sqrt{(x - x_0)^2 + (y - y_0)^2}\)

\(\displaystyle r_1 = \sqrt{(x + x_0)^2 + (y - y_0)^2}\)

\(\displaystyle r_2 = \sqrt{(x - x_0)^2 + (y + y_0)^2}\)

\(\displaystyle r_3 = \sqrt{(x + x_0)^2 + (y + y_0)^2}\)

and then due to superposition, Green's function would be given by

\(\displaystyle G = \frac{1}{2 \pi} \ln |r - r_0 - r_1 - r_3|\)

or something to that effect. Am I on the right track with this?