# Greens function in EM with boundary conditions confusion

So I thought I was understanding Green's functions, but now I am unsure. I'll start by explaining (briefly) what I think I know then ask the question.

Background

Greens are a way of solving inhomogeneous differential equations by first solving the response to a unit impulse. I am aware they are a tool that can sometimes be given a physical meaning. I fear those two realms may be crossing for me where they shouldn't.

$\hat{L} f(x) =s (x)$

$\hat{L} G(x) =\delta (x)$

Using these we can obtain the following general expression for the solution:

$f(x)=\int G(x-x')s(x')dx'$

Physically, in EM, the Greens function appears to be in the form of a potential due to a single point charge. I.e. we can construct our solution for an arbitrary charge distribution by 'summing' over everything to create our known charge distribution.

$\nabla^2\Phi=-\frac{\rho}{\epsilon}$

$\nabla^2G(x-x')=-\frac{\delta(x-x')}{\epsilon}$

$\Phi(x)=\int G(x-x')\rho(x')dx'$

Where

$G(x-x')=\frac{1}{|x-x'|}$

When we start to talk about boundary conditions there is an ambiguity in the definition of our Greens function such that we can have a form as follows:

$G(x-x')=\frac{1}{|x-x'|}+F(x-x')$ so that $\nabla^2F(x-x')=0$

I guess this is where it starts to get blurred for me.

Question:

Jackson 2.7 a) Consider the half plane z>0 with drichlet boundary conditions on the z=0 plane. We wish to write down the greens function for this situation.

Many solutions I have found jump right to the greens function of:

$G(x,x')=\frac{1}{|x-x'|}-\frac{1}{|x-x''|}$

But I do not understand how they got here - they say it is obvious- but given what I understand I must be missing something.

How is this the greens function if there is no charge? They appear to allude to the existence of a charge and an image charge. It seems like we are trying to solve the laplace equation which is homogeneous (so I don't understand the use of greens here). I am guessing the boundary conditions are the critical thing I am not understanding fully.

Feel free to correct any of my misinterpretations and mistakes in general.

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Good questions; I'm sure a lot of people are confused on this stuff (as I was the first time I used Jackson).

Essentially your confusion boils down to being careful to consider the following fact:

The Green's function for a particular boundary value problem depends on the boundary conditions.

In particular, let's say you have a Dirichlet boundary value problem. Then, as Jackson shows on page 39, the appropriate Green's function for such a boundary value problem must (a) satisfy Poisson's equation with a delta function source in that region and (b) vanish on the boundary (see eq. 1.43) of that region. If you can find a function that has these two properties in the region you are considering, then you have found the Green's function for the Dirichlet problem.

So if we consider the half space ($z>0$) with dirichlet boundary conditions at $z=0$, then we are looking for a function that satisfies Laplace's equation in the upper half space with unit source and which vanishes on the boundary, which in this case is $z=0$ plus the "boundary at infinity."

You can check yourself that the function $$G(\mathbf x, \mathbf x') = \frac{1}{|\mathbf x - \mathbf x'|} - \frac{1}{|\mathbf x + \mathbf x'|}$$ has these properties. The intuition for this, and the reason why most people say you can just immediately write this down, is that the first term clearly satisfies Poisson's equation with unit source in the upper half space where $\mathbf x'$ is being taken to have $z>0$, and the second term corresponds to having a unit source in the lower half space with opposite sign. Our intuition about potentials of points charges indicates that this will cause that combination to vanish at $z=0$. Also, the Laplacian of the second term vanishes in the upper half space, so it doesn't affect the fact that in the upper half space, this function satisfies Poisson's equation with unit source at $\mathbf x'$.

I hope that was clear? I can definitely try to clean it up or expand on this. I know from personal experience that it's a confusing topic!

Green's functions are associated with a set of two data (1) A region (2) boundary conditions on that region. The function $1/|\mathbf x-\mathbf x'|$ is the Green's function for (1) All of space with (2) Dirichlet boundary conditions. This is because it (a) satisfies Poisson's equation with unit source in that region and (b) vanishes at the boundary of that region (which in this case is at infinity). In general, for any region $R$, for Dirichlet boundary conditions, as long as we simply find a function $G(\mathbf x,\mathbf x')$ that, (a) satisfies Poisson's equation with a unit source placed in that region, and that (b) vanishes on the boundary of the region, then we have found the Green's function for that Dirichlet problem (by the definition of a Green's function).

When we are trying to find the Green's function for the Dirichlet problem on the upper half space, we first imagine putting a point charge in the upper half space so that condition (a) is satisfied, this leaves us with the function $1/|\mathbf x-\mathbf x'|$ where $\mathbf x'$ is a point in the upper half space. Then, we notice that although this function is an appropriate solution to Poisson's equation, it does not vanish for $x$ on the boundary, so this can't be the Green's function for this Dirichlet problem. We need to do something to this function which does not spoil the fact that it satisfies the unit source Poisson equation in the upper half space but such that the resulting function additionally satisfies the appropriate boundary condition.

So we ask ourselves "what can we do to this function so that (a) remains satisfied, but so that (b) is also satisfied in the upper half space. Well we notice that if we add any function that satisfies Laplace's equation (Laplacian equaling zero) in the upper half space to $1/|\mathbf x-\mathbf x'|$, then the resulting function will still satisfy (a).

Now what sorts of functions satisfy Laplace's equation in the upper half space?

The answer is any charge distribution whose charge density is only nonzero outside of the upper half space will create a potential that satisfies Laplace's equation in the upper half space.

So if we can find a charge distribution that when placed in the lower half space produces a potential that when added to $1/|\mathbf x-\mathbf x'|$ causes their sum to vanish on the boundary, then their sum will satisfy the properties required of a Green's function. This is where we notice that an "image" point charge will do exactly this!

All we are doing with this point charges is an intuitive way of finding a function that satisfies the appropriate mathematical properties (a) and (b) that a Green's function for a Dirichlet problem must satisfy.

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I think what you said is perfectly clear - I am just taking some time to digest it. I am understanding more about how the boundary conditions come into play. Take E.q. 1.42 for example. If we specify dirichlet conditions such that 1.43 must be satisfied, why don't both surface integrals go to zero? (i.e. if $G_D=0$ then why doesn't $\frac{dG_n}{dn'}=0$ also?) I have no problem accepting your intuition statement except for the fact I don't see how charges get involved at all. Is this where I am applying physical meaning to the greens function where there is none? –  Fire Feb 28 '13 at 4:54
Ok cool. Notice that if a function vanishes on a surface, then it doesn't mean that it's normal derivative vanishes on that surface in general (take by analogy any real-valued function of a real variable that passes through zero but who's slope is nonzero at that point). The intuition relating to point charges comes from the fact that $1/|\mathbf x-\mathbf x'|$ is, up to normalization, the electric potential of a point charge at position $\mathbf x'$. Imagine multiplying the Green's function expression in the response by $\frac{q}{4\pi\epsilon_0}$, –  joshphysics Feb 28 '13 at 6:42
(contd.) then notice that it becomes the sum of the electric potential for a point charge $q$ at $x'$ plus a point charge $-q$ at $-\mathbf x'$ (the so called image charge). If you think about this physical situation, then it's pretty immediately clear that the potential along $z=0$ vanishes for this configuration of charges, so we don't need to do a calculation to be confident that the expression for the Green's function we wrote down vanishes at $z=0$ as it should for Dirichlet boundary conditions. Let me know if that makes the intuition more clear. –  joshphysics Feb 28 '13 at 6:42
Well from the mathematics its clear but for the problem I don't understand why we are adding charges that aren't there so to speak. $1/|x-x'|$ is the greens function of a point charge, and sure adding two together symmetrically will construct the desired boundary condition in this case but doesn't this have ramifications elsewhere? For example, is the greens function for the same problem with a REAL charge at $z=+x$ the same? If not, I don't understand the difference. Abstractly, mathematically I am fine. Thinking physically as above is where I am confused. –  Fire Feb 28 '13 at 15:25
(hopefully I am being clear where my issue is, I don't want to be repeating the same thing over and over) –  Fire Feb 28 '13 at 15:26

Just wanted to add a final point to clear this up for you. The potential of a point charge and the Green's function for your problem are the same, up to the normalization constant. In a comment, you say this is by coincidence; It is not, it's physical!

Take your equation $\nabla^2 \Phi = -\frac{\rho}{\epsilon}$, and suppose you wanted the charge density in that equation to be that of a point charge. What would you put in for the charge density of a point charge? Well, it only exists at one point in space, and the way we express densities that exist at only one point is to use delta functions. So for a point charge, $\rho\sim q\,\delta(|{\bf x-x'}|)$. The exact proportionality constants don't matter so much as the fact that there is a delta function in the charge density. Now you can see that a point charge's charge density inserted into the equation get's you the differential equation for the Green's function, up to the constants like $\pi$ and $\epsilon$.

To reiterate, point charges have densities which are proportional to their charge and a delta function, so talking about Green's functions and responses to point charges is equivalent.

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I said coincidence since this was the confusing point for me - when it is physical, and when it is not. It is very intuitive for me to say: Lets build up a charge distribution from dirac deltas. Then we jump from the greens to the convolution of G and $\rho$. Thanks for the clarification though! –  Fire Mar 4 '13 at 18:26