Green's function in electrostatics

[amjad-sh]

Sorry it may seem that my question is a homework question but it is not since I have the solution of the problem.
It is about obtaining Green function and using it to calculate the potential in space, provided the boundary conditions are satisfied.
the questions are like below (It is a problem from Jackson's book):
Consider a potential problem in the half-space defined by z≥0 with Dirichlet boundary conditions on the plane z=0(and at infinity)
(a) write down the appropriate Green's function G(x,x') .
(b) If the potential at plane z=0 is specified to be Φ=V inside a circle of radius a centered at the origin, and Φ=0 outside the circle, find an integral expression of the potential at the point P specified in terms of cylindrical coordinates(ρ,φ,z).
(c) show that along the axis of the circle (ρ=0) the potential is given by

$V=(1- \frac {z}{\sqrt{a^2+z^2}})$

(d) Show that at large distances (ρ^2 +z^2 >>a^2) the potential can be expanded in a power series (ρ^2 +z^2)^-1 and that the leading terms are

$Φ=\frac{Va^2z}{2(ρ^2+z^2)^{3/2}}[1-\frac {3a^2}{4(ρ^2 +z^2)} +\frac {5(3ρ^2a^2+a^4)}{8(ρ^2 +z^2)^2}+...]$
verify that the results of V and D are consistent with each other in their common range of validity.I really tried to solve the first part(a) alone but I couldn't.
When there a boundary condition in the problem( such that potential or surface charge density are specified) the formal solution of potential represented by

$Φ(x)=\int_v ρ(x')G(x,x')d^3x' +(1/4π)\oint_s [G(x,x')\frac {\partial Φ}{\partial n'} -Φ(x')\frac {\partial G(x,x')}{\partial n'}]da'$
Where$G(x,x')=1/|x-x'| + F(x,x')$
In this problem the solution was like follow:
(a) The green's function
G(x,x')=1/|x-x'|+F(x,x')
Dirichlet problem Which specifies the boundary condition at the surface G(x,x')=0 and G(x,x')=0 for z<0. It is given that at z=0 and Z=∞ Φ= constant=V(Why so it is constant, the problem didn't mention this?)
If cylindrical coordinates is used, the green's function so derived should not have F(x,x')(Why so ?)
The green function is then$G(x,x')=\frac {1}{\sqrt {ρ^2 +ρ'^2-2ρρ'cosθ +(z-z')^2}}+\frac {1}{\sqrt {ρ^2 +ρ'^2-2ρρ'cosθ +(z-z')^2}}$
The second term is an contribution from the image.(did he mean that he used the method of images to obtain this result and if so where did he locate the charges?)I'm really still not practiced enough in green functions and don't know where to use them properly in solving electrostatics problems.If somebody besides helping me understanding the problem above can give me insights about them or send me a link that explain them smoothly.

 

[NFuller]

Dirichlet problem Which specifies the boundary condition at the surface G(x,x')=0 and G(x,x')=0 for z<0. It is given that at z=0 and Z=∞ Φ= constant=V(Why so it is constant, the problem didn't mention this?)

It actually did, just not in an obvious way. Take a look at the surface integral term in your expression for $\Phi(\mathbf{x})$. Notice that it is expressed in terms of both a Dirichlet $\Phi(\mathbf{x}^{\prime})$ and a Neumann $\frac{\partial\Phi}{\partial n^{\prime}}$ boundary condition on the surface $S$. Such a set of conditions over-specifies the problem and the general solution does not exist on a closed boundary. So one of these terms must be zero. The problem stated that the half space had Dirichlet boundary conditions at $z=0$ and $z\rightarrow\infty$. So we demand that $G=0$ at these boundaries to eliminate the Neumann boundary conditions in the expression for $\Phi(\mathbf{x})$.

 

[amjad-sh]

It actually did, just not in an obvious way. Take a look at the surface integral term in your expression for $\Phi(\mathbf{x})$. Notice that it is expressed in terms of both a Dirichlet $\Phi(\mathbf{x}^{\prime})$ and a Neumann $\frac{\partial\Phi}{\partial n^{\prime}}$ boundary condition on the surface $S$. Such a set of conditions over-specifies the problem and the general solution does not exist on a closed boundary. So one of these terms must be zero. The problem stated that the half space had Dirichlet boundary conditions at $z=0$ and $z\rightarrow\infty$. So we demand that $G=0$ at these boundaries to eliminate the Neumann boundary conditions in the expression for $\Phi(\mathbf{x})$.

We set G(x,x')=0 since we are using dirichlet boundary conditions.So $Φ(x)=\int_v ρ(x')G(x,x')d^3x' -1/4π\oint_S Φ(x')\frac {\partial G(x,x')}{\partial n'}$
here $Φ(x')$ is the potential at the boundary region. We will consider $ρ(x')=0$ Shall we? in this case $Φ(x')$=constant and not equal to zero since if it is equal to zero at the boundary $Φ(x)$ will be zero every where.

 

[NFuller]

We set G(x,x')=0 since we are using dirichlet boundary conditions.So $Φ(x)=\int_v ρ(x')G(x,x')d^3x' -1/4π\oint_S Φ(x')\frac {\partial G(x,x')}{\partial n'}$
here $Φ(x')$ is the potential at the boundary region. We will consider $ρ(x')=0$ Shall we? in this case $Φ(x')$=constant and not equal to zero since if it is equal to zero at the boundary $Φ(x)$ will be zero every where.

You're on the right track. $\rho$ does not have to be zero though. You already correctly noticed that setting $\rho$ to zero and the boundary conditions to zero gives a potential that is zero everywhere, but this is exactly what one would expect the potential to be with zero charge and zero boundary conditions.

So with the boundary conditions given, how would you construct the Green function in cylindrical coordinates?

 

[amjad-sh]

You're on the right track. $\rho$ does not have to be zero though. You already correctly noticed that setting $\rho$ to zero and the boundary conditions to zero gives a potential that is zero everywhere, but this is exactly what one would expect the potential to be with zero charge and zero boundary conditions.

So with the boundary conditions given, how would you construct the Green function in cylindrical coordinates?

Ok. Actually I do not know where to start. We have the green function is expressed by $G(x,x')=\frac {1}{|x-x'|} +F(x,x')$
If we used cylindrical coordinates $G(ρ,θ,z,ρ',θ',z')=\frac {1}{\sqrt {ρ^2+ρ'^2-2ρρ'cosθ+(z-z')^2}} +F(ρ,θ,z,ρ',θ',z')$
I tried to get something from the relation

Φ(x)=∫vρ(x′)G(x,x′)d3x′−1/4π∮SΦ(x′)∂G(x,x′)∂n′

but I couldn't( is it the key of the solution?)
I usually use the method of images to obtain the potential and then I divide the potential by q to get the green function. It is the the only way until now I use to get the green function.

 

[NFuller]

The method of images is a good idea because we know $G(z=0,x^{\prime})=0$. This is analogous to the case of an infinite grounded conducting plane where the potential would be zero at $z=0$. Thus we choose $F(x,x^{\prime})$ to satisfy the boundary condition at $z=0$.
$$F(x,x^{\prime}) = \frac{-1}{|x-(x^{\prime}-2z^{\prime}\hat{z})|}$$
If we measure $\theta$ to be the angle between $x$ and $x^{\prime}$, then
$$F(x,x^{\prime}) = \frac{-1}{\sqrt{\rho^{2}+\rho^{\prime\;2}-2\rho\rho^{\prime}\text{cos}(\theta)+(z+z^{\prime})^{2}}}$$

 

[amjad-sh]

The method of images is a good idea because we know $G(z=0,x^{\prime})=0$. This is analogous to the case of an infinite grounded conducting plane where the potential would be zero at $z=0$. Thus we choose $F(x,x^{\prime})$ to satisfy the boundary condition at $z=0$.
$$F(x,x^{\prime}) = \frac{-1}{|x-(x^{\prime}-2z^{\prime}\hat{z})|}$$
If we measure $\theta$ to be the angle between $x$ and $x^{\prime}$, then
$$F(x,x^{\prime}) = \frac{-1}{\sqrt{\rho^{2}+\rho^{\prime\;2}-2\rho\rho^{\prime}\text{cos}(\theta)+(z+z^{\prime})^{2}}}$$

G(x,x′)=1√ρ2+ρ′2−2ρρ′cosθ+(z−z′)2+1√ρ2+ρ′2−2ρρ′cosθ+(z−z′)2G(x,x′)=1ρ2+ρ′2−2ρρ′cosθ+(z−z′)2+1ρ2+ρ′2−2ρρ′cosθ+(z−z′)2G(x,x')=\frac {1}{\sqrt {ρ^2 +ρ'^2-2ρρ'cosθ +(z-z')^2}}+\frac {1}{\sqrt {ρ^2 +ρ'^2-2ρρ'cosθ +(z-z')^2}}

Very nice. I think I copied the solution wrongly.