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To take the simplest example,

$$\Phi(\vec{r}) = \begin{cases} V & r\leq a \\ 0 & r\geq a \end{cases}$$

For a plane conducting surface that extends all the way. Choosing an appropriate $G(\vec{r},\vec{r'})$

$$\Phi(\vec{r}) =\frac{1}{4\pi\epsilon_0} \int_V G(\vec{r},\vec{r'}) \rho(\vec{r'})d^3 r' -\frac{1}{4\pi}\oint_S \Phi(\vec{r'}) \frac{\partial G(\vec{r},\vec{r'})}{\partial n'} da' $$

And I can find all the stuff by making a rigged guess of a Green's function (from having solved the simpler 0 potential problem).

My question is really elementary though. The $\rho(\vec{r'})$ that appears in the problems only takes into account the initial charges present but not any induced charges on the plane. I cannot understand that convincingly. There are some issues floating in my mind:

  1. Induced charges appear as an effect of the inital charge distributions present, and are not part of the initial value problem. However, this suggests there is some kind of time evolution. Wrong reasoning, move on.

  2. Induced charges are an unknown for some known $\rho(\vec{r'})$. Our effort is to investigate the effect on the system. So we have the time when the effect due to the $\rho(\vec{r'})$ is not present, that is the instant we assume to solve the problem. Whatever we get however appears to apply to later time when effects have distributed themselves and accounts for the induced charges present.

All this sounds fudgy when I use the Green's function method to solve BVPs. The more elementary usage of method of images seemed highly intuitive and a natural consequences of the uniqueness theorem. We take an image charge, and that accounts for all induced charges. Can someone tell me how to reason in the correct way and why the two are equivalent. How do I intuitively think of the Green's function so that I naturally arrive at coherent conclusions like in the simplistic approach.

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I remember that during the mathematical physics course at the university, the Green functions method seemed to me exactly like the method of images.

The reason is that you have to solve the equation for Green function: $$\Delta \phi(\vec{r}-\vec{r}') = -2\pi \delta(\vec{r}-\vec{r}') $$ With corresponding boundary conditions. And to account for the boundary conditions you have to do virtually the same method-of-images-stuff.

Here is some reference I found, that supports my memory:

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