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!

Some Differential Equation help needed

  1. Sep 20, 2005 #1
    Of the Partial Kind :smile:
    Using d'Alemberts soltuion for the vibrating string in one dimension

    Find u(1/2,3/2), when l-=1, c=1, f(x) = 0, g(x) = x(1-x)
    Now i tried simply substituting this into the solution that is (since f(x)=0)
    [tex] u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} g(x) dx [/tex]
    but it yields the wrong answer.
    Does the length of the string have anything to do with the answer?

    Thank you in advance for your help!
  2. jcsd
  3. Sep 21, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Stunner, to use D'Alembert's forumua, you need to remember to use the "odd extensions" of both f(x) and g(x). Now, I know that's not pretty but that's just how it is. Remember when I said that [itex]Sin[\pi x][/itex] was already an odd-extension and so we didn't have to do anything about it? That's not the case with g(x)=x(1-x) over the interval you're integrating from. Look at the first plot. That's g(x) un-extended. We wish to make an odd periodic function of g(x) over the interval of integration. In your case thats 1/2-3/2 to 1/2+3/2 or the interval [-1,2]. So, first thing is to "odd-extend" what the function looks like in [0,1] to the interval [-1,0]. Well, that's the second plot and the equation for it is:


    The equation for the interval [0,1] is just g(x):


    Now I wish to do that again for the interval [1,2], that is an odd extension of g(x) which would just be flipping it over into the interval [1,2]. The equation for that one would be:


    The third plot is all three. So:

    u(1/2,3/2) &=\frac{1}{2}
    \int_{-1}^{2}\tilde{g_0}(\tau)d\tau \\ &=
    \frac{1}{2}\left(\int_{-1}^0 g_1(\tau)d\tau+\int_0^1 g_2(\tau)d\tau+\int_1^2 g_3(\tau)\tau \right)

    I get -1/12. Is that what you get?

    Edit: Stunner, I initially made a typo on g3 but corrected it above.

    Edit2: Forgot the 1/2 in front of the integral sign. Suppose that's -1/12 now. Sorry.

    Attached Files:

    • g1.JPG
      File size:
      4.8 KB
    • g2.JPG
      File size:
      4.7 KB
    • gtotal.JPG
      File size:
      5.3 KB
    Last edited: Sep 21, 2005
  4. Sep 21, 2005 #3
    it is quite clear that we are not being taught (or the material's presentation) correctly. I did not know how to extend the functions. I understand now... thank you very much!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?