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Given the wave equation

$$u_{xx}(x,t)=\frac {1}{c^2}u_{tt}(x,t) $$

with initial conditions:

IC1: $$u(x,0)= f(x)$$

IC2: $$u_{t}(x,0)= g(x)$$

Why isn't $g(x)$ always equal to $f_t(x)$?
For example, if $t=0$ is the time that a snapshot is taken of a freely traveling wave it seems to me that it must be true that $g(x)=f_t(x)$ Then IC1 would be the only initial condition needed since IC2 could be derived from IC1.

My question: Then why isn't only one initial condition needed?

Maybe if the wave was not freely traveling $g(x)$ could be forced to be something else--but that's not obvious to me. Physical examples would be great. ( I know that mathematically since the equation is second order it needs two initial conditions but I don't understand it intuitively or physically.)

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Note that f(x) and g(x) are functions of x alone [not "x and t"].

f(x) is what the string looks like on a photo [taken at t=0]...
and you don't have access to other snapshots.
That is, "f" doesn't have information on how the string is moving.

g(x) is what the velocity profile would look like at t=0.

Analogously, for a particle, you need to know "where it is" at t=0 and "what its velocity is" at t=0 to predict its future.

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  • $\begingroup$ Remember g is the transverse velocity, not in the direction of the waves motion. $\endgroup$ – user45664 Mar 9 '18 at 20:47
  • $\begingroup$ Refer to D'Alemberts Formula to see that that is true. $\endgroup$ – user45664 Mar 9 '18 at 20:57
  • $\begingroup$ My interpretation of f(x) and g(x) is about the piece of the string at point x (where it is and how it is moving [transverse to the propagation]) ... not about how the disturbance moves along the string. $\endgroup$ – robphy Mar 10 '18 at 20:29
  • $\begingroup$ Why can't $g$ always be found by taking the time derivative of $f$ ?? Then only a initial condition for $f$ would be needed. This is the point of my question. $\endgroup$ – user45664 Mar 10 '18 at 20:46
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    $\begingroup$ As I said in my answer, f(x) is a photo of the string at t=0. There is no further information available to form that time-derivative... no information about about the time-step before t=0 or the time-step after t=0. $\endgroup$ – robphy Mar 11 '18 at 14:47
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consider a wave on a rope, if you choose to define the height of a point on this wave at x, there will be two possible conditions for this wave. The two conditions are that the point may move either up or down, so you need to define the velocity in order to know the motion of the wave.

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  • $\begingroup$ Why not 3 or 4 conditions? $\endgroup$ – user45664 Mar 9 '18 at 22:48
  • $\begingroup$ can you justify why 3 or 4? $\endgroup$ – Omar Ali Mar 9 '18 at 23:24
  • $\begingroup$ Do not see why it would be limited to two. E.G. a taylor series requires a unlimited number of derivatives to represent the waves motion. $\endgroup$ – user45664 Mar 10 '18 at 0:14
  • $\begingroup$ ocw.mit.edu/courses/mathematics/… this might help $\endgroup$ – Omar Ali Mar 10 '18 at 2:36
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    $\begingroup$ The height alone is not enough. Let the IC for the height to be for example the peak of a wave. That point may move up or down based on the direction of movement of the wave. So we need the velocity. $\endgroup$ – Omar Ali Mar 10 '18 at 21:02

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