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Differential Equations

  1. Sep 19, 2005 #1
    Of the partial kind

    We have just finished solving the one dimensional wave equation for a vibrating string of length L using d'Alembert's solution which is

    [tex] u(x,t) = \frac{1}{2} \displaystyle{[} f(x+ct) + f(x-ct) \displaystyle{]} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds [/tex]

    the question is
    If g(x) satisifes the conditions taht g(x) = -g(-x) and g(x) = -g(2l -x) then
    [tex] \int_{x}^{x+2l} g(s) ds = 0 [/tex]
    Well according to a theorem when a periodic function is integrated over a symmetric interval is zero. Also since the velocity of the wave is ct then ct = 2L

    [tex] \int_{x-ct}^{x+ct} g(s) ds = 0 [/tex]
    [tex] \int_{x-ct}^{x} g(s) ds + \int_{x}^{x+ct} g(s) ds = 0 [/tex]
    [tex] \int_{x-2L}^{x} g(s) ds = -\int_{x}^{x+2L} g(s) ds [/tex]
    but here is where i am stuck. How do i change the limit of integration over the left hand side

    ALso im not even sure is if this is the correct way to go? Is it the right way? Your help and advice would be greatly appreciated!
    Thank you!
  2. jcsd
  3. Sep 19, 2005 #2

    Physics Monkey

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    To convert the integral on the left hand side of your final equaiton into something more useful, try the change of variable [tex] s = s' - 2L [/tex].
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