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Homework Help: Wave equation problem

  1. May 1, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Show that the function [tex]u(x,y,z,t)=f(\alpha x + \beta y + \gamma z \mp vt)[/tex] where [tex]\alpha ^2 + \beta ^2 + \gamma ^2 =1[/tex] satisfies the tridimensional wave equation if one assume that f is differentiable twice.


    2. Relevant equations
    [tex]\frac{\partial ^2 u}{\partial t ^2}-c^2 \triangle u=0[/tex].


    3. The attempt at a solution
    I'm not sure how to use the chain rule.
    [tex]\frac{\partial u}{\partial t}=\frac{\partial f}{\partial t}=\mp v \frac{\partial f}{\partial \sigma}[/tex] where [tex]\sigma =vt[/tex]. Thus [tex]\frac{\partial ^2 u}{\partial t ^2}=v^2 \left ( \frac{\partial f}{\partial \sigma} \right ) ^2[/tex].
    I'm 98% sure it's not right.
    Am I approaching well the problem?
     
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  3. May 1, 2010 #2

    gabbagabbahey

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    You're not too far off. The chain rule tells you that [itex]\frac{\partial}{\partial t}f(q)=f'(q)\frac{\partial q}{\partial t}[/tex], so....
     
  4. May 2, 2010 #3

    fluidistic

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    Thank you for your help.
    My confusion resides in "[tex]f'(q)[/tex]". Is it equal to [tex]\frac{\partial f}{\partial \sigma}[/tex] where [tex]\sigma = \mp vt[/tex] in the exercise?
     
  5. May 2, 2010 #4

    fluidistic

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    Second attempt:
    [tex]\frac{\partial u}{\partial t}=\mp v u '[/tex]; [tex]\frac{\partial ^2 u}{\partial t^2}=v^2 u'[/tex].
    [tex]\nabla u = \frac{\partial u}{\partial x} \hat i + \frac{\partial u}{\partial y} \hat j + \frac{\partial u}{\partial z} \hat k = u'(\alpha \hat i + \beta \hat j + \gamma \hat k)[/tex] thus [tex]\triangle u = u''(\alpha ^2 + \beta ^2 + \gamma ^2)=u''[/tex].
    Using the wave equation, it follows that the wave equation holds if and only if [tex]v^2=c^2[/tex] so [tex]v=\pm c[/tex].
    Does this mean that the speed of the wave given is the one of the light? What kind of wave is that? Ellipsoidal or so?
    Also, I didn't understand what I've done to reach this result. Each time I wrote u'. To me the u' appearing in [tex]\frac{\partial u}{\partial t}[/tex] is different from the one appearing in [tex]\nabla u[/tex].
    There is more than 1 variable, so writing u' is confusing to me.
     
  6. May 2, 2010 #5

    gabbagabbahey

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    No, the argument of [itex]u[/itex] is [itex]\alpha x+\beta y+\gamma z \mp vt[/itex], so that's what you would set [itex]q[/itex] to (or [itex]\sigma[/itex] if you prefer).

    The more general wave equation has [itex]c[/itex] as the speed of the wave, not necessarily the speed of light. So, I would say that [itex]u[/itex] satisfies the wave equation, with speed [itex]v[/itex].

    The shape of the wave at any given time will depend on the exact functional form of [itex]u[/itex], as well as the constants [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex].

    [itex]u[/itex] has the same argument in both derivatives. Just name the variable [itex]q=\alpha x+\beta y+\gamma z \mp vt[/itex]
     
  7. May 2, 2010 #6

    fluidistic

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    Thank you once again, it's clearer to me now. At least the argument q is much clearer and so is u'. So [tex]u'=\frac{\partial u}{\partial q}[/tex] where [tex]q=\alpha x+ \beta y + \gamma z \mp vt[/tex]. Is this right?
    Ok but I found out that [tex]v=\pm c[/tex]. So if v is the speed of the wave, then it's equal to c, which is the speed of light or I'm missing your point?

    Ah I see. Say if u is the identity and [tex]\alpha = \beta = \gamma[/tex] then it's a spherical wave?
    And in the hypothetical case of u is worth the identity and [tex]\alpha ^2 + \beta ^2 + \gamma ^2 = 1[/tex] such that [tex]\alpha \neq \beta \neq \gamma[/tex], would the wave be ellipsoidal?
     
  8. May 2, 2010 #7

    gabbagabbahey

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    Yup:smile:

    You're missing the point; the [itex]c[/itex] in your wave equation isn't necessarily the speed of light, just the speed of the wave. For electromagnetic waves, in vacuum, [itex]c[/itex] will be the speed of light.

    Not really. For a spherical wave, you would expect a solution of the form [itex]u=\frac{1}{r}g(r-vt)[/itex].
     
  9. May 2, 2010 #8

    fluidistic

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    Ah I get it now! Ok.


    Ah you're right, it must decay as 1/r to be a spherical wave. Not sure what happens if it decays as 1/r^2 but I get the point. Thanks for all.
    Problem solved.
     
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