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Green's Function and Parallel Plate Waveguide

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1. The problem statement, all variables and given/known data
A parallel plate waveguide has perfectly conducting plates at y = 0 and y = b for 0 ≤ x < ∞ and -∞ < z < ∞. Inside that bound, the waveguide is filled with a dielectric with k as a propagation constant.

The Green's function to be satisfied is

[itex] \nabla^2G + k^2G = -\delta(x-x')\delta(y-y')[/itex]

where 0 ≤ x' < ∞ and 0 ≤ y' ≤ b. G = 0 at the walls and there is no dependence on z. Find the Green's function everywhere inside the waveguide.

2. Relevant equations
[itex] \nabla^2G + k^2G = -\delta(x-x')\delta(y-y')[/itex]

[itex]G = f(x)g(y)[/itex]

[itex] f(0) = 0, f(∞) = 0, g(0) = 0, g(b) = 0[/itex]
limit as [itex]x→x'^- f(x) =[/itex] limit as [itex]x→x'^+ f(x), [/itex]limit as [itex]x→x'^- f'(x) = [/itex]limit as [itex]x→x'^+ f'(x)[/itex]

3. The attempt at a solution
We first look for solutions for the homogenous equation:

[itex] \nabla^2G + k^2G = 0 [/itex]

Using separation of variables, it is possible to form an ODE from this PDE. If k^2 = k_x^2+k_y^2 then the PDE looks like

[itex] \frac{∂^2G}{∂x^2} + \frac{∂^2G}{∂y^2} + (k_x^2+k_y^2)G = 0 [/itex]

Let G = f(x)g(y) and the two ODE that form are the following:

[itex]f''(x)+k_x^2f(x) = 0[/itex]
[itex]g''(y)+k_y^2g(y) = 0[/itex]

Now some basic general solutions to f(x) are:


and actually, I believe the most general solution to f(x) would be a summation of one of the two solutions above, like the following:

\begin{align}\overset{∞}{\underset{n=1}{\sum}} A_ne^{ik_xx}+B_ne^{-ik_xx}\end{align}

g(y) has similar solutions to f(x) and here are the two basic general solutions:


I just need to choose 2 of the 4 basic equations here and work with the boundary conditions. Lets say I choose the following:

[itex]f(x) = Ae^{ik_xx}+Be^{-ik_xx}[/itex]
[itex]g(y) = Ccos(k_yy)-Dsin(k_yy)[/itex]

The boundary conditions that I see are as follows:

[itex] f(0) = 0, f(∞) = 0, g(0) = 0, g(b) = 0[/itex]
limit as [itex]x→x'^- f(x) =[/itex] limit as [itex]x→x'^+ f(x), [/itex]limit as [itex]x→x'^- f'(x) = [/itex]limit as [itex]x→x'^+ f'(x)[/itex]

This is what I have been able to determine so far. When I start to apply boundary conditions is when I feel like I don't know what I am doing. Lets apply the BC for g(y):

[itex]g(0) = 0 = Ccos(k_y*0)-Dsin(k_y*0)[/itex]
This tells us C = 0.
[itex]g(b) = 0 = -Dsin(k_y*b)[/itex]
Now for the right side to be 0, [itex]k_y*b=n\pi[/itex] where n are integers. This means that [itex]k_y=n\pi/b[/itex]. So if I did stuff right here, g(y) is the following:

[itex]g(y) = -Dsin(n\pi*y/b)[/itex]

Lets look at the BC for f(x):

[itex] f(0) = 0 = A + B[/itex]
So this says that A = -B which I don't have too much of a problem with. I think that this only applies for when 0 ≤ x ≤ x'. The f(x) for this region would then be:

[itex]f(x) = A(e^{+ik_xx}-e^{-ik_xx}) = i*2A*sin(k_xx)[/itex]

Where my problem arises is for the following BC when x' ≤ x < ∞:

[itex]f(∞) = 0 = Ae^{ik_xx} + Be^{-ik_xx}[/itex]

With this equation, I can't seem to figure out how to make this equation true without it being the trivial solution (ie, A=B=0). If A = 0 then we have:

[itex]f(∞) = 0 = Be^{-ik_xx}[/itex]
[itex]f(∞) = 0 = Bcos(k_xx)-iBsin(k_xx)[/itex]

but I still don't know how to handle that. I don't know what happens when a complex exponential goes to -∞. Another idea would be to work with the sine of the complex exponential and make it so it equals zero by using k_x. The problem here is that I can't do that without making k_x dependent on x which isn't allowed.

[itex] k_x = n\pi/x[/itex]

The only other thing to do would be to set k_x = 0 but that just gives the trivial solution. Perhaps the answer to my problem is just so obvious that I am oblivious to it but can someone point me in the right direction?

Boundary conditions

Hello hover

I wonder why do you assume that f(∞) must be zero. In a realistic scenario (I mean, within a finite width parallel-plate) this should be the case, otherwise the energy necessary to sustain such a field would be infinite. But if you assume the presence of two unbounded plates, the electromagnetic fields, thus the Green function, can never decay and it will oscillate all along the x-axis, since an infinite cross-section needs an infinite amount of energy, therefore, a non-vanishing electric field all along the x-axis.

I believe that the boundary conditions that you must apply should be more related to the physics of electromagnetic fields on perfect electric conductors, i.e., E_{parallel} = 0 = B_{perpendicular} on the boundaries.

If you relate the Green's functions to the electric and magnetic fields you will be able to find the right boundary conditions: E_z (y=0) = 0 = E_z (y=b) = E_x (y=0) = E_x (y=b) = 0

I hope that this hint is useful.
Hi hover,

I think the way you're trying to solve the PDE by separation of variables is wrong is this case.

Let's assume that the plates are located at +/- infinity, i.e. we are in 2D free space. What kind of field would we expect? A field traveling away from the point source in the radial direction. This already tells you that G=G(r)=G(sqrt(x^2-y^2)) and separation of variables in x/y direction is impossible. For such a case it would make a lot of sense to formulate the problem in cylindrical coordinates. Then the problem is separable again but then in r and phi. This results in a PDE in the form similar to a Bessel equation. The solution to this equation is often written as a Hankel function.

Now to complicate things, consider the two plates moving back into place. What would happen is that the cylindrical waves reaching the top or bottom plate will be reflected back and forth and start to travel into x +-infinity direction. A simple way of formulating the Green's function in this case would be to apply the mirroring principle. This would mean that the point source gets mirrored in the top and bottom plate infinitely many times. The result is an infinite sum over Hankel functions. Although this formulation is correct, it will converge rather slowly.

I also agree with esorella that the vanishing field requirement at x +/-infinity does not hold. The fields cannot escape the waveguide. What you could envision though, is that the medium inside the waveguide is not completely lossless. This would mean that kx becomes complex (Note that signs are always a matter of convention). Now for x-> +/- infinity the Hankel function H(Z) will behave as ~exp(+/-i*Z ). Depending on the direction, this corresponds to a decaying field going to zero as x->+/-infinity.

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