It's taken me ages, and I really am struggling to understand this, so I am not sure if this is correct but here goes...
[tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] [tex]\frac{\partial G(x,0)}{\partial y} = 0[/tex] for [tex]y \geq 0 \ - \infty < x < \infty[/tex]
Let [tex]r = |\underline{x} - \underline{x}_0 | = \sqrt{(x - x_0)^2 + (y - y_0)^2}[/tex]
be the distance between [tex]\underline{x}[/tex] and [tex]\underline{x}_0[/tex]
Then [tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] becomes
[tex]\bigtriangledown^2 G = \frac{1}{r} \frac{\partial}{\partial r} \Big( r \frac{\partial}{\partial r} \Big) = 0[/tex]
everywhere (although not at [tex]r = 0[/tex]) and subject to
[tex]\iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1[/tex]
which gives
[tex]G(r) = A \ln r + B[/tex]
[tex]A = \frac{1}{2 \pi}[/tex]
Choosing [tex]B = 0[/tex]
[tex]G = \frac{1}{2 \pi} \ln r = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|[/tex]
I don't think this is the method of images though...(Headbang)