# Does the wave equation have unique solutions?

1. Nov 1, 2012

### aaaa202

Well title says it all pretty much. My question is if one set of boundary conditions uniquely specifies the solutions to the wave equation.
My speculation comes from the fact that my book introduces electromagnetic in a bit weird way I think. It shows how Maxwells equations lead to the wave equation but it then confines it interest to just plane waves of the simplest form since they solve the wave equation. But what about other kinds of waves that solve the wave equation? Are they not interesting? I am not only talking about other kinds as in spherical waves etc. but also just of waves which dont have the boring form cos(kr - ωt) but maybe just something like cos(2kr - ωt) or whatever you can make up. Because generally as I see it you can't just absorb the 2 into k since that would ruin the nice relation between ω and k.

2. Nov 1, 2012

### Simon Bridge

One complete set of boundary conditions will restrict the possible solutions to the wave equation.
It may be that there is only one solution but not always - sometimes, as with waves confined to a cavity, there is a set of solutions.
There are other kinds of waves and they are, indeed, very interesting.
However - while you are learning, it is well to stick with the simple ones.
Plane waves have a lot of utility and are used to study other kinds of waves.
Don't worry - you'll get to see how to handle other kinds of waves soon enough.

3. Nov 1, 2012

### bengunn

Not directly related, but these stages of learning reminds me of Electromagnetic Field theory class, dealing mostly with waves using Maxwell's eq. and then on to Quantum Mechanics where wave theory introduces different ideas. Just saying it is a progressive learning thing and the basics need to be remembered when moving on.

4. Nov 2, 2012

### voko

The wave equation, of course, has infinitely many possible solutions. This is plain to see in the fact of wild diversity of electromagnetic phenomena, which all satisfy the wave equation. The equation does not determine the shape of waves, it merely describes how they propagate. The shape is determined by boundary conditions and initial conditions, and the propagation aspect can be studied with the simple case of plane waves.

Mathematically, uniqueness is a property of the the so called "Cauchy's problem", of which the wave equation is but one ingredient. This is similar to ordinary differential equations.

The significance of plane waves emerges when you consider that the wave equation is thus the principle of superposition holds: complex wave can be decomposed into (infinite) sums of simpler waves, such as plane waves.

5. Nov 2, 2012

### aaaa202

okay but even with plane waves my book restricts itself to them having the following form:
y = cos(k(r-vt)+δ) = cos(kr - ωt + δ)
But for some reason it always demands the wave vector to have magnitude of ω/c such that you would have for instance cos(k(x+y+z)/√3 - ωt + δ)
I guess demanding this normalized form provides a simple form but isn't it omitting all the functions of the form:
cos(2kr - ωt + δ) or any other forms which scale r relative to ωt. What is the benefit of this weird normalization?

6. Nov 2, 2012

### voko

Substitute the book's solution into the wave equation and see what restrictions you get for the constants involved.

7. Nov 2, 2012

### aaaa202

ahh thank you. It had not occured to me that they had to be normalized to satisfy the wave equation.

8. Nov 2, 2012

### Simon Bridge

I don't think that's a normalization so much as a consequence of the definitions of the quantities involved. It seems like wondering why mass should always be "normalized" to F/a. (Actually it is exactly like being surprised that $c=\nu \lambda$.)

If you take a snapshot of a wave at some convenient time, it could be something like $y=\cos(kx)$ - maybe some constants inside the trig function and so on but this time ($t:\delta-\omega t = 0$) is convenient for what I want to show you ... basically the number inside any trig function has to be an angle (because of the definition of, in this case, "cosine"). So $k$ has to have units of inverse-length. $kx$ has to cycle through $2\pi$ when you traverse one wavelength (because of the definition of "wavelength") so... $k=2\pi/\lambda$ is pretty much forced just by the semantics.

Once you have that you can see that $$k=\frac{\omega}{c} \Rightarrow c=\frac{\omega}{k} \Leftrightarrow c = \nu\lambda$$

Extrapolate for 3D and an arbitrary plane wave.
(Note: you may be used to an $f$ instead of $\nu$ for frequency.)

9. Nov 2, 2012

### aaaa202

hmm.. okay Simon. I am not sure I understand what you mean. My question was not about k but rather: Why would a function cos(k(r-vt)) be more interestering than a function cos(k(ar-vt)). But as Voko pointed out r simply has to me normalized for it to be a solution to the wave equation, so I guess there's no more a problem.

10. Nov 2, 2012

### voko

There is another reason. All solutions of the wave equation must be functions of (r - vt) and (r + vt), which are interpreted as waves travelling forward and backward in time. Say F(r - vt) is such a solution. Then F(k(r - vt)) is also a solution, but if we recast this as F(kr - wt), then kv must be equal to w. In your case, F is taken to be the cosine, which is OK since any (even) function can be given as a sum of cosines. cos(kr - wt) is really a "pure tone" of any EM wave.

11. Nov 2, 2012

### Simon Bridge

Voko is taking a math-first approach while I am trying for a physics first.
I seem to have got confused by:
Where a is a dimensionless constant?
You would be scaling r right? So changing coordinate system to one where the tick-marks are further apart (changes units for length).

Otherwise $ka=2\pi/\lambda$ (for reasons given above) while $kv=\omega$ ... is that consistent with $v=\nu\lambda$?

I missed voko using the word "normalize". I am used to "normalizing r" to mean replacing it with a unit vector with the same direction. In general, kr≠1. So can you see why I am having trouble?
I am concerned that you see the relationship as some kind of mathematical convenience or convention.

12. Nov 2, 2012

### aaaa202

well I do see it as a mathematical convenience. That if you scale r and not scale vt you don't get solutions to the wave equations any more - is that not how i should see it?

Just to clear any confusion (because I really find the notation confusing). We are only looking at waves of the form:
cos(k*r-wt)
and k is a vector of magnitude k=v/w and r is the vector r=(x,y,z). So if you had a wave travelling in the direction (1,2,3) you would have k(vector) = k*(1,2,3)/sqrt(14). You would never look at waves of for instance the form cos(k*2r wt) or any other weird function since that would scale r and wt unequally which isn't allowed by the wave equation solutions.

But yeh I am probably wrong. Overall I find the notion of an electromagnetic wave very confusing. Another point which irritates me: How do you know in which direction the fields E and B oscillate. My book usually says E is in the x-direction and then the direction of B is given. But how do we know which direction E oscillates?
Question arose because I had an exercise with an electromagnetic direction travelling in the direction (1,1,1) and it said the E-field was polarized parallel to the xz-axis. But as far as I can see there are to ways of achieving a field oscillating perpendicular to direction and propagation and parallel to xz-plane. That is (-1,0,1) or (1,0,-1). Which of these directions would the field oscillate?

Last edited: Nov 2, 2012
13. Nov 2, 2012

### Simon Bridge

No - if you scale r and not v then, physically speaking, you are measuring distance in two different units and then using them in the same relation[*] (eg. does it make sense to have distances in mm and velocity is m/s in the same equation? Does that describe something that happens in Nature?). It's scaling one side of an equation and not the other - then insisting that the equality still holds.

Math is a language to physics - and algebra, the grammar - when you use a language you need to be consistent in your use of the terms or you are talking nonsense. Just because you can write down particular words one after the other does not mean the result is a sentence.... same with math.

I tend to get a bit of a bee in my bonnet about this: it is about the relationship between Nature and math. In physics, our math is supposed to say something about what happens in Nature. Nature has Rules - not "just anything" can happen. The math has rules to make sure that it is describing consistent relationships.
However - some people find this a somewhat subtle (maybe trivial) point so I just have to accept that it takes some people longer than others.

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[*] though - it may be that you have introduced "a" because you did an experiment in which speed was measured by one device in (say) m/s while distance was measured by another in mm. In which case, a=0.001 is needed to make the units match. I don't think that's what you mean though.

14. Nov 2, 2012

### Simon Bridge

... because those are simple to learn, and are important for signal analysis later. You are only starting out - nobody wants to give you the really hard stuff too soon. Don't worry - the other waves will come and you'll gain the skills to use the wave equation (with a bunch of other tools) to handle any physical wave allowed in Nature.

Remember - the wave equation is not some handful of symbols thrown together on a whim - it is constructed from an understanding of physical principles found in Nature. If it is not in Nature it is not in the wave equation either (unless something went wrong :) ) and if some proposed wave function is not a solution to the wave equation then we do not expect to find it in Nature either.

You may need a particular form because it has to satisfy the wave equation but the reason it needs to satisfy the wave equation is because it has to satisfy Nature.

A healthy attitude for a scientist :) Don't worry, we are almost always wrong - you get used to it.
I'd like to answer that in two parts.
1. the x direction (and the others) is completely arbitrary - it is a fiction created by mathematicians to help label locations in space. Nature does not care which is which.
Usually we allow our equipment to decide on directions - or we use some feature of the symmetry we see in the phenomenon under investigation. In this case, there is a kind-of convention that the direction of propagation is always called z (due to cylindrical symmetry - makes the use of cylindrical coordinates easier). If we align the x-axis ($\phi = 0$) with $\vec{E}$ then $\vec{B}$ lines up with the y-axis. So your text-book is not trying to show you the relationship between the fields with some axis - it is trying to show you the relationship between the fields and the direction of propagation.
2. we know about the E field by observing the effect it has on charges. When you put a long thin wire (an antenna) in the path of the wave, and hook it to a galvinometer, then the current measured will oscillate with the wave with a strength that depends on E and the orientation of the wire wrt E. When the wire is lined up with E, then the signal strength is a maximum. Pick one direction along the wire and call it +x.
I want you to reread that passage. Can you see how your words are inconsistent? You talk about an x-z axis in one place and an x-z plane in another part for example. You need to be more careful.

As for your choices - the difference between them will be a matter of phase.
If the phase is important then set (x,y,z)=(0,0,0) and see. If you don't have that information then you have a decision to make.

Personally, in the absence of any other data, I'd line the time axis up[*] to make the phase as simple as possible (preferably 0, with initial dE/dt pointing in the +x direction relative to the direction of propagation). i.e. formally you let the wave define it's own coordinate system (x',y',z') with z' in the direction of propagation and x' in the direction of $\vec{E}$. Then all you need to do is describe the x' direction in terms of x,y, and z.

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[*] axis are arbitrary - the t=0 moment is just when I start my stopwatch and I can do that any time I like.