Solve

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

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

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

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

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

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

which leads to

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...

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

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

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

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

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

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?