1

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.)

user45664
  • 3,136

2 Answers2

2

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.

robphy
  • 12,829
1

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.

Omar Ali
  • 766