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Homework Help: Solutions to 2D wave equation using 1D equation solution.

  1. Oct 20, 2012 #1
    http://imageshack.us/a/img824/1121/asdasdaw.png [Broken]

    I am having trouble completely understanding what the question wants. I know it is quite clear but the part I am having trouble is the following.

    It says 'pretend' w(x,t) is a solution to the 2D equation, just independent of y, then to substitute it into the integrate,

    That means that u_t(x,y,0) = w_t(x,0) = ψ(x)

    But I am trouble seeing how you can perform this integration without knowing what ψ(x) actually is as it would be,

    ∫∫ψ(x)/(stuff) dx dy over D.

    But ψ(x) is unknown, so how can you integrate?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 20, 2012 #2


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    Hi Greger! :smile:
    No, ψ(x) is given, as an initial boundary condition! :wink:
  4. Oct 20, 2012 #3
    Thanks for replying,

    Yea u_t(x,y,0) = w_t(x,0) = ψ(x),

    But what I mean is there is no explicate form of ψ(x), for example ψ(x) = x, ψ(x) = cos(x), its juts left arbitrarily as ψ(x).

    What would ∫ψ(x)dxdy be if there is no explicate form for ψ(x)?
  5. Oct 21, 2012 #4


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    it's just a formula for you to use in the future, when you do know ψ(x), isn't it? :confused:

    not an actual problem?
  6. Oct 21, 2012 #5
    Yea it's an actual problem, haha I see what you mean, it looks like instructions.

    The question asks to actually preform the integration given,

    And in the integral you have u_t(x,y,0) and the question says substitute:

    u_t(x,y,0) = w_t(x,0) = ψ(x)

    so now the integral becomes ∫∫ ψ(x)/stuff dx dy.

    Then asks to actually compute the integral,

    The integration is over x and y and ψ(x) is a function of x, so you will have to know how to integrate ψ(x) right? Which means you need to have an explicit form of ψ(x) right?

    I understand what you are saying, that if its left arbitrarily as ψ(x) you perform the integration when you know ψ(x), but the question is actually asking to integrate as if you already know what ψ(x) is.

    There must be some kind of trick that I'm just not seeing to be able to perform the integration, or a way to find the expression for ψ(x), or maybe there is just information missing from the question?
  7. Oct 21, 2012 #6


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    Have you tried doing the integration over y first? Since the initial data doesn't depend on y you may be able to do that, and then perhaps it will be clearer what to do with the x integral.
  8. Oct 21, 2012 #7


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    if there is, i'm not seeing it either :redface:
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