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More PDE fun

  1. Dec 4, 2005 #1
    [tex] \frac{\partial u}{\partial t} - k \frac{\partial^2 u}{\partial x^2} = 0 [/tex]
    for 0 <x < pi, t> 0

    [itex] u(0,t) = u(\pi,t) = 0 [/itex]
    [itex] u(x,0) = x (\pi - x) [/itex]

    OK i know the boring part of getting u(x,t) = X(x) T(t)
    the infinite series part is hard part
    the coefficient [tex] c_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin(n(\pi -x)) \sin(nx) = \frac{-2}{\pi} \left(\frac{-1 + (-1)^n}{n}\right) \sin(nx) [/tex]
    the -1^n is from the Cos n pi term taht would coem from the integration

    thus n must be odd
    [tex] c_{n} = \frac{4}{\pi} \frac{1}{2n-1} \sin(nx) [/tex]
    is this good so far?
     
  2. jcsd
  3. Dec 4, 2005 #2
    I am confused with your IC's. Is it u(x,0) = x(pi - x) or sin(n(pi - x))?
     
  4. Dec 4, 2005 #3
    x(pi -x ) is the initial condition
    and i think i have mistaken this for X(pi -x)
    wich is not the same thing
     
  5. Dec 4, 2005 #4
    Ok, so if x(pi -x) is your IC, then you will have to integrals to do. xpi* sin(nx)
    and x^2 sin (nx). Its a bit more involved than the last one.

    - harsh
     
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