# Green's Function in the wave equation

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## Homework Statement

The Green function for the three dimensional wave equation is defined by,

$$\left ( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right ) G(\vec r, t) = \delta(\vec r) \delta(t)$$

The solution is,

$$G(\vec r, t) = -\frac{1}{4 \pi r} \delta\left ( t - \frac{r}{c} \right )$$

Here r = |r|. Use this Green function to find f(r, t) for all r and t, where f is a solution to the inhomogenous wave equation:

$$\left ( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right ) f(\vec r,t) = Ae^{- \alpha r^2} \delta (t)$$

## The Attempt at a Solution

My professor did not go over this nearly at all. I need a significant amount of help getting started.

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fzero
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The general theory of Green functions states that, given a differential operator $$\cal{D}_{\mathbf{x}}$$, the solutions $$\phi(\mathbf{x})$$ to the inhomogenous equation

$${\cal D}_{\mathbf{x}} \phi(\mathbf{x}) = j(\mathbf{x})$$

are given by

$$\phi(\mathbf{x}) = \int d\mathbf{x}' ~ G( \mathbf{x},\mathbf{x}') j(\mathbf{x}'),$$

where the Green function $$G( \mathbf{x},\mathbf{x}')$$ is the solution to the equation

$${\cal D}_{\mathbf{x}} G( \mathbf{x},\mathbf{x}') = \delta(\mathbf{x}-\mathbf{x}').$$

This is covered in just about any math methods text, as well as most EM and QM texts.

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Yes, I found some information in both Jackson's EM and a math methods book. However, the wave equations they solve are in different forms. I can't find any information anywhere about there being an exponential on the right hand side.

Also, this page derives the solution that is given to me in the problem,

http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node75.html

But what can I do with the exponential on the RHS?

The geometry of the problem offers the fact that, $r^2 = r_0^2 -(ct)^2 - 2rct cos \theta$

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fzero
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Everything on the RHS of the inhomogeneous equation is the current j(x). I haven't checked how tricky the integration is to do though.

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Ok, so I have to evaluate the integral,

$$f(\vec r, t) = \int \int G(r', t') j(r') d^3r' dt' = -\frac{A}{4 \pi}\int \int \frac{e^{- \alpha r'^2}}{r'} \delta\left ( t' - \frac{r'}{c} \right ) \delta (t') d^3r' dt'$$

(Why is the convention in several disciplines to put the dr term in front? I've noticed this a lot recently)

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fzero
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There is a time integration. In my notation $$\mathbf{x} = (t,\vec{x})$$. You can also see the time integration in the very last equation on that Duke U page that you linked. The appropriate form of the Green function is (11.63) on that page. Whoever wrote this problem up didn't bother to use a notation where it's clear that the Green function depends on a pair of coordinates: G(x,t,x',t'). The expressions that you wrote need to be modified accordingly.

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My professor is lazy and always just takes x'=t'=x0=t0=0. In fact, that page I linked is what he handed out to us, and was all the info I had until I looked in textbooks from other classes. At any rate, it is crucial that i finish this problem tonight, what process should I follow to evaluate the integral?

Should I find the fourier transforms of the delta functions to remove time dependence?

fzero
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My professor is lazy and always just takes x'=t'=x0=t0=0. In fact, that page I linked is what he handed out to us, and was all the info I had until I looked in textbooks from other classes. At any rate, it is crucial that i finish this problem tonight, what process should I follow to evaluate the integral?

Should I find the fourier transforms of the delta functions to remove time dependence?
No, I realized that you have two delta functions, one in the Green function and one in the current. Those completely fix the r' and t' integrals, so you'll just have a trivial angular integral to do. You should just be able to take G(r-r', t-t'), but you can double-check your expression with the formulas on that page.

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The angular integrals just give a factor of 4pi and an $r'^2$. The first delta function blows up at r' = ct', but the second at t'=0, so do I just evaluate the function for r'=ct'? Something doesn't feel right about evaluating the first delta function with r'=ct' since it has a t' term in it as well, and we are also integrating with respect to that.

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fzero
$$\delta\left( t -t' - \frac{|\vec{x}-\vec{x}'|}{c}\right).$$
I probably mispoke before when I said that this would fix the r' integration, since $$|\vec{x}-\vec{x}'|\neq r -r'$$. However it might still work out that way because of the spherical symmetry of the current. In any case, the t' integration should be fixed by the $$\delta(t')$$ factor.