It is possible to show that functions of the form $f_{1}(kx-\omega t)$, $g_{1}(kx+\omega t)$, $f_{2}(\omega t-kx)$ and $g_{2}(\omega t+kx)$ are all solutions of the wave equation $\dfrac{\partial^{2}y}{\partial x^{2}}=\dfrac{1}{v^{2}}\dfrac{\partial^{2}y}{\partial t^{2}}$ for $v=\frac{\omega}{k}$, although we may write a general solution to the wave equation as a superposition of either the first two functions, or the latter two functions. There is, as far as I can tell, no real difference between the two choices.

Suppose we have a semi-infinite string extending from $x=-\infty$ to $x=0$, with a mass $m$ terminating the string at $x=0$. If the oscillations in the string are small, then we may take the tension $T$ in the string to be constant, in which case we have the boundary condition: $m\dfrac{\partial^{2}y}{\partial t^{2}}\vert_{x=0}=T\dfrac{\partial y}{\partial x}\rvert_{x=0}$. If we want to find the relative amplitudes of the incident and the transmitted waves, then it is typical to substitute a function of the form $y(x,t)=Ae^{i(\omega t -kx)}+A^{\prime}e^{i(\omega t+kx)}$ into the boundary condition and then solve for $\frac{A^{\prime}}{A}$. Although this is perfectly doable, what I don't understand is why instead substituting $y(x,t)=Ae^{i(kx-\omega t)}+A^{\prime}e^{i(kx+\omega t)}$ into the boundary condition doesn't produce an equation involving $A$ and $A^{\prime}$ that doesn't involve $t$, as the other choice of function did. If there isn't a difference in how they describe a wave propagating along a string, why should it make a difference to analysing wave reflection at a boundary?


It's a generic property of linear systems that they conserve not only the value but also the sign of the frequency, i.e. if the input is of the form $f_1(x)e^{i\omega t}$, then the output is of the form $f_2(x)e^{i\omega t}$.

Therefore it's a very bad idea to switch the sign of $\omega$ willy-nilly† between $e^{+i\omega t}$ and $e^{-i\omega t}$. You might think it's fine, because the signals are actually real, so we're going to be taking the real part of everything at the end of the day, so the sign can be whatever. And I admit that you can probably get the right answer that way. But if you do that, you're making your calculations very unnecessarily complicated.

Basically, you want a real function $f(x,t)$ which solves the equation, and you can find it by writing $f(x,t)=\text{Re}[g(x,t)]$, where $g$ is a complex function which solves the same equation. That's what you want, and that's what happens if you keep consistent signs for ω. If you switch the signs of ω willy-nilly, then you are trying to find a complex function $g(x,t)$ that no longer solves the original equation. So that's a step in the wrong direction. There's no easy way to find or interpret such a $g$.

†In certain contexts, like "4-wave-mixing optical phase conjugation", you really are supposed to switch the sign of $\omega$. By "willy-nilly" I mean switching the sign of $\omega$ without a physics-based reason to do so.

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  • $\begingroup$ Thank you. Can you elaborate a bit on what do you mean by a linear system (what is it in the problem at hand that is "a linear system"), what do you mean that such systems conserve the frequency (and its sign) and where do you get that result from? $\endgroup$ – roymend Dec 20 '17 at 20:37
  • $\begingroup$ I mean "linear" as in "linear optics", "linear circuit", etc. - en.wikipedia.org/wiki/Linear_circuit (for completeness, I should have said "linear time-invariant"). Your wave equation as linear because it satisfies the superposition principle - if y1 and y2 are solutions, so is ay1+by2 for any a,b, real or complex. See dsp.stackexchange.com/questions/2915/… $\endgroup$ – Steve Byrnes Dec 20 '17 at 22:07
  • $\begingroup$ Is it correct to say that the system is time-translation symmetric but not symmetric in any spatial reflection because of the boundary conditions? And why exactly does switching the sign imply that we are analysing a non-linear system? Are you referring to down-conversion and such non-linear optical processes? $\endgroup$ – plan Dec 22 '17 at 18:59
  • $\begingroup$ I edited the answer, maybe my comment about "analyzing non-linear" was not a helpful way to think about it. I agree that "the system is time-translation symmetric but not symmetric in any spatial reflections", but I'm not sure why you're bringing that up. There is an analogy: If a system is spatial-translation-invariant, then you shouldn't willy-nilly switch between e^+ikx and e^-ikx. Your system is not spatial-translation invariant, so we expect both e^+ikx and e^-ikx to be present and interrelated. $\endgroup$ – Steve Byrnes Dec 22 '17 at 23:27
  • $\begingroup$ Exactly. The lack of such a spatial symmetry is also important for the reason you just stated. $\endgroup$ – plan Dec 23 '17 at 1:47

Probably just the math. Why you are using just two functions is because $A$ and $A'$ are generally complex numbers, so instead of writing four $cos (\pm\omega t\pm kx)$ you just use two exponentials $e^{i(\omega t\pm kx)}$ with a hope that the real and imaginary parts of your constants will account for any phase shifts. Of course, you need more than one boundary condition to link two real and two imaginary parts together.

Now, I don't know your exact problem, but it seems that the choice of $y(x,t)=Ae^{i(\omega t -kx)}+A^{\prime}e^{i(\omega t+kx)}$ says something about your boundary conditions and the nature of the process you describe.

You may see it when taking your function at the zero moment in time when the original solution gives $Ae^{-ikx} +A'e^{ikx}$ (which, for example, is going to be real for $A=A'$) while your variant leads to a different expression $e^{ikx}(A+A')$.

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  • $\begingroup$ An exact problem in this case would be: given a semi-infinite string extending from $x=-\infty$ to $x=0$ terminated at $x=0$ by a mass $m$ free to move in the $y$-direction, calculate the relative amplitude of the reflected wave if there is a wave propagating in the $+x$ direction. As far as I can tell, the only boundary condition would be the one I mentioned in my first post. What about this boundary condition or the nature of the process would suggest that I have to use $y(x,t)=Ae^{i(\omega t -kx)}+A^{\prime}e^{i(\omega t+kx)}$? $\endgroup$ – Oliver Lunt May 18 '14 at 9:24

We must choose functions containing $e^{ikx}$ and $e^{-ikx}$ because the boundary condition is specified at a certain point in space, at all times. If we had a similar condition at a specific time, the solution would have to include positive and negative angular frequencies.

Rewrite the boundary condition as

$$m^2 v^2 \frac{\partial^2 y(0, t)}{\partial x^2} = T \frac{\partial y(0,t)}{\partial x}$$

where I used the wave equation to substitute for the time derivative. Now, for any $t=t_0=constant$, $y(x,t_0)$ can be expanded in plane waves. We should expect that the boundary condition will involve a relation between waves with both positive and negative wave-vectors. That's why the plane wave expansion $A e^{ikx+i\omega t} + A' e^{-ikx+i\omega t}$ work while $Ae^{ikx+i\omega t} + A' e^{ikx-i\omega t}$ does not!

What you are really doing is expanding in plane waves. The solution is then determined by finding the coefficients of the expansion. We find the coefficients from the constraint by comparing the expansion term-by-term. You are not keeping all the important Fourier terms, so you are effectively comparing the wrong Fourier components.

To understand why, systematically, we solve the wave equation by a Fourier transformation in space:

$$y(x, t) = \frac{1}{2 \pi} \int_{-\infty}^\infty y(k, t) e^{i k x} dk$$

Inserting this into the wave equation, we end up with

$$\frac{\partial^2 y(k,t)}{\partial t^2} = (-v^2 k^2) y(k, t)$$

The general solution to this ordinary differential equation can be written as:

$$y(k, t) = F(k) e^{-ivkt} + G(k) e^{+ivkt}$$

We can then Fourier transform back and end up with the general solution:

$$y(x, t) = F(x-vt) + G(x+vt)$$


$$F(x-vt) = \frac{1}{2 \pi} \int_{-\infty}^\infty F(k) e^{i k (x-vt)} dk$$

$$G(x+vt) = \frac{1}{2 \pi} \int_{-\infty}^\infty F(k) e^{i k (x+vt)} dk$$

These solutions represent travelling waves, and the coefficients $F(k), G(k)$ contains the information about reflection and transmission. There will have to be initial conditions to fully specify the system. Now, using your boundary condition, for $F$ I get for the right side:

$$\frac{\partial}{\partial x} F(-vt) = \frac{1}{2 \pi} \int_{-\infty}^\infty (ik) F(k) e^{-ikvt} dk$$

$$\frac{\partial^2}{\partial t^2} G(+vt) = \frac{1}{2 \pi} \int_{-\infty}^\infty (ik) F(k) e^{ikvt} dk$$

But to compare these, I must first rewrite

$$\frac{\partial}{\partial x} F(-vt) = \frac{1}{2 \pi} \int_{-\infty}^\infty (-ik) F(-k) e^{ikvt} dk$$

If I do this, the boundary condition can be solved term by term, and I will end up with $F(t-x/v)$, $G(t+x/v)$ an equation in $F(-k)$ and $G(k)$ for each value of $k$. This equation does not involve $t$. I may of course do the same thing by a Fourier transform in the time domain - in this case I will end up with a relation between $F(\omega)$ and $G(\omega)$. The reason it works out this way is because the condition is specified for $x=0$ as far as I can tell.

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To summarize your question. You start with a real system (a string). The boundary conditions restrict the motions of the real position of the string. Then you propose that the position of the string $y(t,x)$ is given by the real part of some complex function. Then you state that this complex function should also adhere to this boundary condition. However, it does not necessarily need to. Only the real part of this complex function needs to obey the boundary conditions. Letting the full complex function obey the boundary conditions is a stricter not necessary requirement.

You state that for your 2nd choice of basis functions you find a time dependent equation. However, if you take the real part of this equation the time dependence will drop out and you will obtain the same equation for $\frac{A'}{A}$ as for the regular choice of basis. There is thus no conflict.

If you want to be able to impose boundary conditions on both the real and the imaginary part of the complex functions, you will need a basis of four, rather than 2 functions:

$$f(x,t)=e^{i(\pm \omega t \pm kx)}$$

The reason that in this specific case the first pair of your basis functions gives a nice result while the second does not, is that the boundary condition is specified for a specific position. The changed sign of the $kx$ component in the exponent does not have any influence at the boundary. Were you to choose boundary conditions for a specific time, e.g. $y(x)=0$ for $t=0$, then the second set of basis functions would give a nicer result.

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Your two wave equations can be written as

$y_1(x,t) = e^{ikx}\left ( A' e^{i\omega t} + A e^{-i\omega t}\right)$ and $y_2(x,t) = e^{i\omega t}\left ( A' e^{ikx} + A e^{-ikx}\right)$

You can think of equation $y_1$ as picking a position $x$ and then seeing how the displacement $y_1$ changes with time $t$ at that position $x$ whereas equation $y_2$ is equivalent to picking a time $t$ and seeing how the displacement $y_2$ varies with position $x$ at that time $t$.

Your example has a boundary condition which depends on a position $x(=0)$ and so it is the $y_2$ equation which will give you a connection between $A'$ and $A$ for all values of time $t$ whereas if you had a boundary condition which depended on time $t$ you would use the $y_1$ equation.

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