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I'm looking at solutions for the wave equation

$$\frac{\partial^2 z}{\partial t^2}=c^2 \nabla^2z,$$

in a finite 2D domain. Say that I have periodic boundary conditions on the left and right edges and free boundary conditions ($\nabla z \cdot \hat{n}=0$, where $\hat{n}=0$ is normal to the edge) on the other two edges (call them top and bottom). Example domain below.

If the top and bottom edges are smooth and at $y=0,1$ and the left and right edges are at $x=0,10$ then

$$z(x,y,t)=\cos(k_x x)\cos(k_y y)\sin(\omega t),\; \sin(k_x x)\cos(k_y y)\cos(\omega t)$$

are solutions as are

$$z(x,y,t)=cos(k_x x \pm\omega t)cos(k_yy ),$$

where $k_x =\frac{\pi n}{1}$, $k_y =\frac{\pi m}{10}$, and $\omega = c \sqrt{k_x^2+k_y^2}$ for positive integers $n,m$. (Hopefully got my math right there.) The top solutions are standing waves and the bottom solutions are traveling waves. For my purposes, the distinction between the two types of solutions is that all motion is in phase for a standing wave but not for a traveling wave; e.g. at $t=0$, all parts of a standing wave take their maximum or minimum value but that is not so for a traveling wave, where different parts of the wave take their maximum and minimum values at different times. (Let me know if the distinction is unclear, otherwise the real question won't make sense.) The two standing wave solutions at a given $\omega$ can be written in terms of the two traveling wave solutions of the same $\omega$ and vice versa.

Now, say that I have some irregular top and bottom edge. My intuition says that there will always be standing-wave-like solutions of the form $z(x,y,t)=A(x,y)cos(\omega t)$, where $A(x,y)$ is a (real) position-dependent amplitude. (In the above example, $A(x,y)=\sin(k_x x)\cos(k_y y)$, but $A(x,y)$ won't always have such a simple form.)

When I find solutions to the wave equation numerically, I do find standing-wave-like solutions. For example, below is a depiction of a standing-wave-like solution on an irregular domain. The color represents $A(x,y)^2$ (blue smaller values, red larger values)

Standing Waves

However, I've had trouble finding solutions with a position-dependent phase (a traveling-wave-like solution); i.e. a solution like $z(x,y,t)=B(x,y)cos(C(x,y)-\omega t)$. If I find numerical solutions for a smooth wire, I get the standing wave solutions I expect, and I can combine them into traveling wave solutions. However, with the irregular wire shown, each mode seems to have a different frequency, so there's no way to combine the standing waves of the same frequency to get a traveling wave of the same frequency.

Are there traveling-wave-like solutions that I'm not finding? (I could be doing something wrong with the numerics.) When will traveling-wave-like solutions exist? (Is there some symmetry condition that irregular wire has broken?) Is my intuition correct that standing-wave-like solutions will always exist?

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If power enters in one place and exits in another, then a traveling wave component must exist (otherwise the net Poynting vector across any plane would be zero). If there is no net flow of energy there can be no traveling wave in a steady state solution.

Note that it is possible, for a given shape of waveguide, that some frequencies are perfectly reflected (think stub matching circuit - except there you try to eliminate the reflection; but the principle is the same). For those frequencies only a standing wave solution would appear.

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  • $\begingroup$ That's very insightful. Two more questions: 1) Say I hook up leads to this structure so that energy can flow through it (e.g. I could get rid of the periodic boundary conditions and fire a wave packet in from the left and see what comes out on the right). Why does the modified domain now have traveling wave solutions? From getting rid of the periodic BCs? From adding the leads? 2) Do you have any thoughts on what domain geometry allows for traveling wave solutions? I'm guessing it requires some symmetry, but I'm not sure what. $\endgroup$ – lnmaurer Jan 9 '16 at 18:14
  • $\begingroup$ Really it means you have non-zero boundary conditions. That can only happen when there is a physical gap in your waveguide (input or output). That's all you should need. For specific boundary shape there may be some frequencies for which only a standing wave solution exists but that would be the exception not the rule. $\endgroup$ – Floris Jan 9 '16 at 18:17
  • $\begingroup$ Got it. I think we're using "gap" and "leads" to mean the same thing. Let me rephrase my second question: "Do you have any thoughts on what domain geometry allows for traveling wave solutions if there are no leads/gaps?" When I crunch the numbers on the above structure (which has no leads/gaps), I only get standing waves, so I'm curious which closed structures allow for traveling wave solutions. $\endgroup$ – lnmaurer Jan 9 '16 at 20:53
  • $\begingroup$ In principle you can have a traveling wave solution if you have a "packet" (pulse) bouncing around the cavity. However, a waveguide has dispersion, so after a time only a steady state solution survives. And that cannot involve energy flow and thus, no traveling waves. Note - this is my intuition, not a deep knowledge based on experimental data, speaking. $\endgroup$ – Floris Jan 9 '16 at 20:56

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