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Homework Help: Wave equation

  1. Sep 28, 2006 #1
    When doing the initial conditions of the velocity of the wave function, why do they have a position (x) derivative (i.e. cF'(x)-cG'(x)=h(x)).

    It appears in here.

    How someone explain how the c came about and why position derivatives are taken instead of time derivatives?
  2. jcsd
  3. Sep 28, 2006 #2
    Well in the article it just uses the prime notation to represent a derivative, and not [itex]\partial/\partial x[/itex] or [itex]\partial/ \partial t[/itex].

    The first thing to note is that [itex]u(x,t)=F(x-ct)+G(x-ct)[/itex], so if you take a time derivative of one of these functions, you will pull out the constant [itex]c[/itex] because

    \frac{\partial}{\partial t} \equiv c\frac{\partial}{\partial(ct)}.

    The second thing to note is that because the variables [itex]x,t[/itex] in the functions [itex]F\,, G[/itex] appear as a sum or difference, then differentiating with respect to [itex]t[/itex] will give the same functional form as a derivative with respect [itex]x[/itex], up to some constant. For example,

    f(x-ct) = \sin(x-ct)


    \frac{\partial}{\partial x}f(x-ct) = \cos(x-ct) = \frac{\partial}{\partial(-ct)}\sin(x-ct) = \frac{-1}{c}\frac{\partial}{\partial t}\sin(x-ct) = \frac{-1}{c}\frac{\partial}{\partial t}f(x-ct)

    The reason why they write the derivative in this primed notation is because the conditions require that the time derivative is evaluated at [itex]t=0[/itex], which you could write like this:

    \left.\frac{\partial}{\partial t}f(x,t)\right|_{t=0} = h(x).

    The notation is pretty cumbersome so instead they just write:

    f^{\prime}(x,0) = ch(x).

    I think this is right, but maybe someone else could clarify?
    Last edited: Sep 28, 2006
  4. Sep 28, 2006 #3
    Your explanation seems to make sense. It might also got something to do with the (trademark) equality of the second derivative that defines a wave equation. But physically, if you denote the time derivative of the wave with constant speed with c*(spatial derivative), than it means at different locations, it will have different velocities, not just in direction but also in magntitude.

    If you treat the partial time derivative in terms of the x variable than you could integrate it wrt x and get the original wave function back instead of integrating wrt t.
    Last edited: Sep 28, 2006
  5. Sep 29, 2006 #4
    Of course there is an equivalence between the derivatives (up to some constant) because of how the argument of the function is written. This is precisely what I was trying to tell you in my first post.

    I'm not sure what you're saying here about the wave having different velocities?! The velocity is a constant, c, (independent of coordinates).

    By the way in my first post G(x-ct) should be G(x+ct).
    Last edited: Sep 29, 2006
  6. Sep 30, 2006 #5
    What is the physical interpretation of

    \left.\frac{\partial}{\partial t}f(x,t)\right|_{t=0} = h(x).

    if it is not the wave's velocity?
  7. Oct 1, 2006 #6
    The interpration of this is

    "The rate of change of the amplitude of the wave profile for some given [itex]x[/itex] at [itex]t=0[/itex]".

    For example, plot out some wave profile as a function of [itex]x[/itex]. Go to some coordinate [itex]x^{\prime}[/itex] along [itex]x[/itex]. Draw a line from [itex]x^{\prime}[/itex] to [itex]u(x^{\prime})[/itex]. Your initial condition is telling you how much larger or smaller this line will become in some small time interval [itex]\Delta t[/itex]. This is not the velocity of the wave. The velocity of the wave is how far the wave profile propogates along [itex]x[/itex] for some given amount of time [itex]t[/itex].
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