1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differential Equations
  1. Differential Equation (Replies: 7)

  2. Differential Equations (Replies: 3)

  3. Differential equations (Replies: 5)

  4. Differential Equations (Replies: 4)