Solving Potential of a Charge Outside a Sphere with Green Functions

[Mounice]

I was wondering if there is a way to deduce the solution of the potential of a charge outside a sphere given by the image method, though Green functions. Because of a Dirichlet condition (G_D(R,r')=0), I know that a solution can be written as G_D=Go+L, where ∇²L=0. But in order to approach this problem I proposed the Green solution Go, as the one associated to the poisson solution. So, by applying the Dirichlet condition I can get the boundary condition for the problem in L(r,r').
That's how I get
$$G_D(\vec{R},\vec{r})=\dfrac{1}{4\pi \epsilon_o}\left [\dfrac{1}{|\vec{R}-\vec{r'}|}+L(\vec{R},\vec{r})\right]=0$$
So,
$$L(\vec{R},\vec{r})=-\dfrac{1}{4\pi \epsilon_o}\dfrac{1}{|\vec{R}-\vec{r'}|}$$
But, because that expression is constant with respect to the variable r, I denoted it as Vo.
So I know that the solution for ∇²L=0 for the outside of a sphere with constant voltage Vo, is given by
$$L(\vec{r},\vec{r'})=\dfrac{V_oR}{r}=-\dfrac{1}{4\pi \epsilon_o}\dfrac{R}{r|\vec{R}-\vec{r'}|}$$
Such that,
$$G_D(\vec{r},\vec{r})=\dfrac{1}{4\pi \epsilon_o}\left [\dfrac{1}{|\vec{r}-\vec{r'}|}-\dfrac{R}{r|\vec{R}-\vec{r'}|}\right]$$
and it satisfies the boundary condition,
$$G_D(\vec{R},\vec{r})=\dfrac{1}{4\pi \epsilon_o}\left [\dfrac{1}{|\vec{R}-\vec{r'}|}-\dfrac{R}{R|\vec{R}-\vec{r'}|}\right]=0$$
But I know that this solution is incorrect because the typical solution to this problem is a image charge, such that
$$G_D(\vec{r},\vec{r})=\dfrac{1}{4\pi \epsilon_o}\left [\dfrac{1}{|\vec{r}-\vec{r'}|}-\dfrac{R}{r'|\vec{r}-\dfrac{R^2}{r'^2}\vec{r'}|}\right]$$
Can someone tell me what's the error in the solution that I am proposing or how am I supposed to arrive to the correct solution by Green functions?

 

[vanhees71]

I cannot understand your arguments, but usually you derive the Green's function of the Laplace operator for the Dirichlet problem of the sphere using the image-charge method, because it's the easiest way to do so. Another way is to use the multipole expansion, which finally of course leads to the same result.