It's taken me ages, and I really am struggling to understand this, so Im not sure if this is correct but here goes....
$\displaystyle \bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)$ $\displaystyle \frac{\partial G(x,0)}{\partial y} = 0$ for $\displaystyle y \geq 0 \ - \infty < x < \infty$
Let $\displaystyle r = |\underline{x} - \underline{x}_0 | = \sqrt{(x - x_0)^2 + (y - y_0)^2}$
be the distance between $\displaystyle \underline{x}$ and $\displaystyle \underline{x}_0$
Then $\displaystyle \bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)$ becomes
$\displaystyle \bigtriangledown^2 G = \frac{1}{r} \frac{\partial}{\partial r} \Big( r \frac{\partial}{\partial r} \Big) = 0$
everywhere (although not at $\displaystyle r = 0$) and subject to
$\displaystyle \iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1$
which gives
$\displaystyle G(r) = A \ln r + B$
$\displaystyle A = \frac{1}{2 \pi}$
Choosing $\displaystyle B = 0$
$\displaystyle G = \frac{1}{2 \pi} \ln r = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|$
I don't think this is the method of images though......
$\displaystyle
\displaystyle { G=\frac{1}{2 \pi} \; ( \; ln(\sqrt{(x-x_0)^2+(y-y_0)^2}+ln(\sqrt{(x-x_0)^2+(y+y_0)^2} \; )
}
$
$\displaystyle
\displaystyle { \frac{ \partial G}{\partial y}=\frac{1}{4 \pi} \; ( \; \frac{ 2(y-y_0)}{(x-x_0)^2+(y-y_0)^2}+\frac{ 2(y+y_0)}{(x-x_0)^2+(y+y_0)^2 } \; )
}
$
$\displaystyle
\displaystyle { \frac{ \partial G}{\partial y}|_{y=0}=0
}
$
I looked this link about method of images
The method of images.