This might seem like a silly doubt but I am confused about this.

For what kind of waves is the wave equation in 1+1D satisfied?

$$\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 f(x,t)}{\partial t^2},$$

where $f(x,t)$ is the disturbance of the wave at position $x$ and time $t$ and $v$ is the velocity at which the wave is moving.

In particular, is the above equation valid for waves with non-constant profiles?

  • $\begingroup$ What do you mean by "non-constant profiles"? $\endgroup$ – Philip Jun 30 at 15:56
  • $\begingroup$ @Philip "non-constant profiles" means that the shape of the wave changes as it moves with velocity v. I thought the equation was only true for waves of a fixed shape moving at a fixed velocity but apparently this is not true? $\endgroup$ – Harry Potter Jun 30 at 17:04

This solution of one-dimensional wave equation, known as D’Alembert’s solution, can be written in general as

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

where $\psi(x,t)$ is the function that satisfies the wave equation, and $F$ and $G$ are any arbitrary functions. (However, the arguments of these functions must be $x-vt$ and $x+vt$ respectively. )

Physically, these represent two waves travelling in opposite directions with the speeds $v$, whose instantaneous "shapes" is given by the functions $F$ and $G$.

Thus, the functions $F$ and $G$ don't change their "shapes" as time progresses, but $F + G$ certainly would, in general. Now, if your solution was just a function of $x-vt$ (or of $x+vt$), in that case the "profile" wouldn't change with time.

EDIT: The OP also seems to have some questions on the "profile" of the solutions as well as the introduction of damping. I made a couple of simple simulations to explain this.

The form of the solutions: I'm not sure what you mean exactly by "non-constant profiles", but if what you mean that the "shape" of the solution stays exactly the same as what it was at $t=0$, then I would say in general, no. It would depend on the initial conditions as well as the boundary conditions.

For example, consider the following case: I create a "Gaussian" profile on a string like this initially:

enter image description here

It turns out that if you solve the wave equation with these initial conditions, it's equivalent to two little Gaussians moving in either direction. (i.e. both $F$ and $G$ are Gaussians.) I further impose that the string is attached to walls on either side so that the waves reflect with no losses. (This is not necessary, but makes the simulation easier, and it's perfectly physical anyway.)

As you can see below, the initial profile at $t=0$ does not stay the same as time progresses. You get two different Gaussians moving away from each other, reflecting at the walls, and coming back to the initial configuration.

                            Undamped wave on a string

Introducing Damping: Of course, you'll notice that in the above simulation the wave never actually "dies out", as it would if there were some sort of damping in the system. One way to model damping (at least the easiest) is to solve the wave equation with a linear damping term $\propto \frac{\partial \psi}{\partial t}$:

$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} + \gamma \frac{\partial \psi}{\partial t},$$

where the first order derivative introduces dissipation. The "travelling" waves are still Gaussians, but their amplitudes die out with time, as you can see in the following simulation:

                           Damped wave on a string

In general, however, you could introduce more complicated damping terms in which the Gaussians become more "spread" out and so on, but that's much harder to model.

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  • 1
    $\begingroup$ And what about damping? Is the wave equation satisified in the presence of any form of damping? $\endgroup$ – Harry Potter Jun 30 at 16:37
  • $\begingroup$ Your wave equation does not hold if there is damping--there is an extra term. Google it. $\endgroup$ – user45664 Jun 30 at 17:05
  • $\begingroup$ @Philip Ok, now I am confused again. Is there a difference between the words "form" and "shape"? $\endgroup$ – Harry Potter Jun 30 at 17:48
  • $\begingroup$ @HarryPotter I've edited my answer, let me know if this is clearer. $\endgroup$ – Philip Jun 30 at 20:45
  • $\begingroup$ @Philip hmmm.... In that case, it seems to me that almost every wave without damping would satisfy the equation. I am wondering if there are any waves that would not work? $\endgroup$ – Harry Potter Jul 1 at 6:04

We can easily check that the function $$ f_i(x) = A_i \cos(k_i x \pm \omega_i t + \varphi_i) $$ satisfies the wave equation, if $v = \omega_i/k_i$. Therefore, it is straight forward to realise that also a superposition $$ f(x) = \sum_i f_i(x) = \sum_i A_i \cos(k_i x \pm \omega_i t + \varphi_i) $$ satisfies the wave equation. However, since this is a Fourier series and almost every function is representable by a (generalised) Fourier series we conclude that almost every function can be used as initial function at time $t=0s$. For later times, $t >0s$, most functions will change their shape.

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