How to find the Green's function Consider a domain $$D= \{(x,y) : x>0, y>0 \}$$ Let $\textbf x = (x,y)$ and $\xi =(\xi_x , \xi_y)$. Find the Green's function, $G(\textbf x , \xi)$ such that $$\nabla ^2 G=\delta (\textbf x - \xi), \, \, \, \, \, \, \textbf x \in D$$ subject to $$\frac{\partial G}{\partial x}(0,y,\xi)=0, \, \, \, \, \, \text{for} \, \, \, \, y>0$$
and $$G(x,0,\xi)=0, \, \, \, \, \, \text{for} \, \, \, \, x>0$$

Using method of images:
We have the source $\xi=(\xi_{x},\xi_{y})$ and images sources: 
$\xi_1=(-\xi_{x},\xi_{y})$, $\xi_2=(\xi_{x},-\xi_{y})$, $\xi_3=(-\xi_{x},-\xi_{y})$
So we have $\nabla ^2 G=\delta ( \textbf x - \xi )\pm \delta ( \textbf x - \xi_1 ) \pm \delta ( \textbf x - \xi_2 )\pm \delta ( \textbf x - \xi_3 )$
Can someone please clearly explain how you determine if the plus/minus signs are in fact plus or minus please.
I know that it is to do with the boundary conditions but don't understand how.
 A: The correct extension is
$$\nabla^2G = \delta(\mathbf{x}-\xi) + \delta(\mathbf{x}-\xi_1) - \delta(\mathbf{x}-\xi_2) - \delta(\mathbf{x}-\xi_3)$$
To see why, interpret $G$ as electric potential.  Then, the delta functions correspond to point charges, and we have the conditions that (i) the potential is symmetric across $x=0$, and (ii) the potential is zero along $y=0$.  The first condition implies that the sign on the delta function at $\xi_1$ should be the same as the sign on the delta function at $\xi$, while the second condition implies that the sign on delta function at $\xi_2$ should be opposite the sign at $\xi$.  Then, since $\xi_3$ is the reflection of $\xi_2$ over $x=0$, the delta function there should have the same sign as $\xi_2$ (alternatively, since it is the reflection of $\xi_1$ over $y=0$, it should have the opposite sign from the delta function at $\xi_1$).
Solving for $G$ yields
$$G = \ln\frac{1}{|\mathbf{x} - \xi|} + \ln\frac{1}{|\mathbf{x} - \xi_1|} - \ln\frac{1}{|\mathbf{x} - \xi_2|} - \ln\frac{1}{|\mathbf{x} - \xi_3|}$$
and we can verify that $G$ satisfies the required boundary conditions.
