If we are given $\Delta \Phi(x)=-\frac{\rho(x)}{\epsilon_0}, x \in V$ and $\Phi(x)=\omega(x), x \in \partial V$, we can find a solution to $\Phi(x)$ on $V$ using Green‘s function:
$\Phi(x)=\frac{1}{\epsilon_0}\int_V G(x,y)\rho(y)d^3y-\oint_{\partial V}\vec{n}\cdot \vec{\nabla_y}G(x,y)\omega(y) d^2y$
With $\Delta_yG(x,y)=-\delta(x-y), \quad x,y\in V, \\ G(x,y)=0, \quad \quad \quad \quad \quad y \in \partial V$
I hope this is correct thus far.
I think I understand the physical interpretation of the first integral. If $V=\mathbb{R}^3, \omega(x)=0$, then $G$ is the potential of a point charge at y, then we convolve this potential with $\rho$ to get the total potential of the charge distribution. If $V$ is bounded but $\omega$ is still $0$, we are not only convolving the potential of a point charge at $x$ with $\rho$ but also it‘s „image charge“ somewhere in the complement of $V$. Where this image charge is, i.e. finding G, has to be done case by case, depending on $V$.
But what if now $\omega(x) \neq 0$ then, to my understanding, G doesn‘t change. It is still (for a chosen $y$) the potential of these two charges. But now we have to consider the surface integral over $\partial V$. It looks to me as if we were subtracting the convolution of the normal component of the electric field caused by these two charges with the potential at the boundary, but why does that make sense? What even is potential times electric field?
