5
$\begingroup$

Find the Green function for the domain $Ω = {x = (x_1, x_2, x_3) ∈ \mathbb{R^3} : x_3 ∈ (0, L)}$. The operator in questions is the Laplacian and the Green's function is defined as:

$G(x,y) = ϕ(x,y) - ϕ^*(x,y)$

where $ϕ$ is the fundamental solution of the Laplace equation and $ϕ^*$ is the corrector term. So far I have considered the Green's function for the half space to start this problem, i.e.:

$G(M,M_0) = \frac{1}{4\pi|M-M_0|} - \frac{1}{4\pi|M-M_1|}$

Where $M = (x_1,x_2,x_3)$, $M_0 = (x_{10},x_{20},x_{30})$ and $M_1 = (x_{10},x_{20},-x_{30})$

I am having problems because in this problem x3 is bounded between 0 and L

$\endgroup$
  • 4
    $\begingroup$ You need to specify a lot more than that. What is the operator you are working with? Furthermore you should provide the definition of a greens function and any attempts you have made to use it to answer your question. $\endgroup$ – Spencer Oct 13 '13 at 6:00
  • 1
    $\begingroup$ Sorry, I forgot to add in the details. I have tried starting from the half space Green function case but then I am not really sure where to go from there $\endgroup$ – johnsteck Oct 13 '13 at 6:38
  • $\begingroup$ No problem at all, glad that you provided the details. Are you familiar with the method of images in electrostatics? It can provide you with intuition on how to construct the appropriate correction term. If I remember this greens function ends up being an infinite sum of corrections. $\endgroup$ – Spencer Oct 13 '13 at 6:43
  • $\begingroup$ Unfortunately, I am not familiar with the method you mention. Do you know where I could find more about it? So far in class we only did the Green's function for the half space and for a ball, and this question was assigned as homework. So I don't know a whole lot about the Green's function :( $\endgroup$ – johnsteck Oct 13 '13 at 6:58
  • 2
    $\begingroup$ Take the Green function $G'$ for half-space, and set $G(M,M_0)=\sum_{n\in\mathbb Z} G'(M,M_n)$ where $M_n=M_0+(0,0,2nL)$. $\endgroup$ – user8268 Oct 13 '13 at 9:47
3
$\begingroup$

The function $\frac{1}{4\pi |M-M_0|}$ satisfies the PDE and has the kind of singularity we want, so all is left is to enforce the boundary condition (which is the Dirichlet condition $G=0$). In a half-space this is done using the fact that an odd function of one variable is necessarily $0$ at the origin. So, we create an odd function (with respect to $x_3$) by taking $$\frac{1}{4\pi |M-M_0|}-\frac{1}{4\pi |M-M_0'|} \tag1$$ where prime means reflection in the $x_3=0$ plane. Now we'd like to apply reflection to (1), this time about the plane $x_3=L$. This creates two more terms and unfortunately destroys the symmetry about $x_3=0$. We can restore the symmetry about $x_3=0$ by reflecting there again... This process leads to an infinite series: see xkcd555.

To make the long story short, observe the following: an odd function of $x_3$ which is periodic with period $2L$ will necessarily be zero at both $x_3=0$ and $x_3=L$. (Why?) So, all we need to do is to create a periodic function out of (1). Like this: $$\sum_{n\in\mathbb Z}\left(\frac{1}{4\pi |M+2Ln e_3-M_0|}-\frac{1}{4\pi |M+2Ln e_3-M_0'|} \right)\tag2$$ where $e_3$ is the basis vector for $x_3$ axis. (This is what user8268 suggested in a comment.)

$\endgroup$
2
$\begingroup$

When talking about Greens functions we have to specify a few things. First of all we must specify the operator for which we are trying to find a Green's function, in our case that is the Laplacian operator $\nabla^2 = \partial_{xx}+\partial_{yy}+\partial_{zz} $. Furthermore every Green's function has some region over which it is defined, we will call this region $S$.

A Dirichlet Green's function in a given region is defined by the following properties,

  • $\nabla^2 G_D(\vec{x},\vec{x}') = \delta(\vec{x}-\vec{x}')$
  • $G_D(\vec{x},\vec{x}')\mid_S = 0$
  • $G_D(\vec{x},\vec{x}') = G_D(\vec{x}',\vec{x})$

The Dirichlet Green's function for all space in $\mathbb{R}^3$ is $$\phi=\frac{1}{\vert \vec{x}-\vec{x} '\vert}$$

Careful application of $\nabla^2$ to this function will result in a delta function and the function goes to zero at $\infty$ as required.

To lend some intuition $\phi$ can be thought of as the electrostatic potential at $\vec{x}$ due to a point charge at $\vec{x}'$. This motivates the idea of the method of images. For the simplest case we consider the greens function in the region $ \lbrace z \geq 0 \mid (x,y,z) \in \mathbb{R^3} \rbrace$.

To construct a Green's function for this region we first start by including $\phi$ in order to get the delta function and then we add solutions of Laplace's equation to enforce the boundary condition.

$$G_D(\vec{x},\vec{x}') = \phi + F \qquad (\nabla^2 F=0)$$

The difficulty is normally in finding the appropriate $F$. We want $G_D$ to disappear in the $xy$-plane, thinking of $\phi$ as placing a point charge at $\vec{x}'$ in $\lbrace z \gt 0 \rbrace$ we choose $F$ to be the potential of a point charge which is the reflection of the first on the other side of the plane with the opposite sign.

$$F = -\frac{1}{\vert \vec{x}-\vec{x}''\vert} \qquad \langle x'',y'',z''\rangle=\langle x',y',-z' \rangle $$

We can see that by taking the reflection of the $\vec{x}'$ about the $xy$-plane vector we just found the appropriate correction term to kill $G_D$ at $z=0$ without breaking the condition that $\nabla^2 G_D = \delta(\vec{x}-\vec{x}')$ in $z\geq 0$.

A physicist would want to say that we placed an image charge at $\vec{x}''$. If that language is not comfortable to you then feel free to just think of the vector $\vec{x}'$ being reflected about the $xy$-plane.

That was the simplest example which I think you have already seen. Now we have two planes as bounding surfaces with a spacing of $L$ between them. We can use the same approach but it will be a bit more complicated. The issue is that each image charge will make it necessary to place a new image leading to an infinite series of corrections for $F$.

As a simple illustration lets think of how it would look if we just considered the case where $\vec{x}'$ is halfway between the planes. That is $z'=L/2$. Thinking of each plane as a mirror we would expect to see a reflection on each side at $z=-L/2$ and at $z=3L/2$. If you've ever been in room with mirrors on opposing walls you're probably familiar with the effect of the reflections going on forever making an infinite number of copies of the room. This is going to happen here as well. The bottom reflection (z=-L/2) below is going to have a new reflection above at (z=5L/2) and vice versa. These new reflections will have further reflections each time trying to ensure that the potential on the planes is zero. No finite number of images can accomplish this but an infinite number can (the convergence is really slow).

Something important to keep in mind is that you never put a reflection in the original region of interest because this would break the first condition for $G_D$. Hope that helps.

You can find more on this in the following books:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.