I have a conducting plate on $x$-$y$ plane.
So I have a boundary condition at $z=0$ $\Phi=0$ but,
for $z>0$ I have a point charge at z=a which is expected to create a potential.
$$
\nabla^2\Phi=\frac{\rho}{\varepsilon_0}
$$
I need a Green function which can be assigned as :
$$
G(r,r')=\frac{1}{|\sqrt{(x-x')^2+(y-y')^2+(z-a)^2}|}
$$
But this Green function does not satisfy Laplace $\nabla^2G=0$ at $z = 0$. Right? So I need an additional function to also satisfy Laplace such as :
$$
G(r,r')=\frac{1}{|r-r'|}+F(r,r')
$$
here is my question starts : What is proper function for F and how am I supposed to assign it ? How should I use the Laplace and Poisson properly?
I have chosen this unique and common problem to understand the Green function solutions but mainly my aim is to understand the use of Laplace and Poisson with boundary conditions.
Best Answer
I have to say that I don't fully understand your question, and there still seem to be inconsistent factors of $4\pi$. But I want to get at the heart of the Green function for your boundary conditions, and then you can always ask another question about how to handle it for other boundary conditions.
$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-a)^2}}$$ is not your Green function, and neither is
$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$, but I think
$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}-\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}$$
is the function you are looking for. It tells you the potential at $r$ due to a charge at $r'$, and when $z=0$ it gives $G=0$. Now for a charge distribution $\rho$ you can integrate to get $\int G(r,r')\rho(r')dx'dy'dz'$
The point of a Green function is that if you can find the solution at $r$ due to a single unit charge at an arbitrary point $r'$ that meets your boundary conditions, and call that function $G(r,r')$ then the work you did to get $G$ now allows you to solve for any charge distribution $\rho$ by doing an integral to get $V(r)=\int G(r,r')\rho(r')dx'dy'dz'$.
One Green function $G(r,r')$ that allows you to get $V(r)$ for many different $\rho$ is the whole point.