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Math Help - PDE by Fourier soltion giving zero coeffiecant

  1. #1
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    PDE by Fourier solution giving zero coeffiecant

    Hi all,

    I am working through a PDE and get through to the general solution, but get a seemingly zero Fourier coefficient.

    \frac {\partial u}{\partial t} = \frac {\partial^2 u}{\partial x^2}

    Using seperable variables method and setting the general solution u(x,t) = X(x)T(t) = XT and substituting into original equation:

    \frac{X''}{X} = \frac {T'}{T} = -p^2 where -p^2 is an arbitary constant. Solving the ODE's given yields:
    X = ACos(px) + Bsin(px) and T = Ce^{-p^2t}

    u(x,t) = XT = (YCos(px) + ZSin(px))e^{-p^2t}

    ->Condition u(0,t) =0 applied => 0 = (Y +0) e^0=>  Y = 0

    ->now apply second condition u(\pi,t) = 0

    0 = sin(\pi p) => n\pi = p\pi => n = p since sin(n/pi} always satisfies the zero condition

    n= 1,2,3, 4 etc etc rewrite infinite solution as a fourier series.

    u(x,t) = \sum_{n =1,2,3, etc}^{\infty} QrSin(nx)e^{-p^2t}

    u(x,0) = sin^2(x) ;  0<x<\pi evaluate Qr

    Qr = \frac{\pi}{2} \int_{o}^{\pi} sin^2(x)sin(2nx)

    I evaluated the sin terms to simplify first:

    sin^2(x) = 1 - cos^2(x) subn

    = sin(2nx) - Cos^2(x)sin(2nx)

    but cos^2{x} = 1/2 + 1/2cos(2x) subn

    = sin(2nx) -1/2sin(2nx)- 1/2cos(2x)sin(2nx)

    but 1/2cos(2x)sin(2nx) = 1/4(sin(2x-2nx) -sin(2x+2nx))  subn

    = 1/2 sin(2nx) - 1/4(sin(2x-2nx) -sin(2x+2nx))  now back into integral:

    Qr = \frac{\pi}{2} \int_{o}^{\pi} 1/2 sin(2nx) - 1/4(sin(2x-2nx) -sin(2x+2nx))

    = \frac{\pi}{2} [-ncos(2nx) +1/4(2n-2)cos((2n-2)x)) - 1/4(2n+2)cos((2+2n)x)]_{0}^{\pi}

    It is this last part I always seem to have the expression collapsing to zero; since cos(0) and cos(n\pi) are always the same with the limits of 0;\pi
    I know I must have made a mistake somewhere, I suspect in changing the period length and associated limits ,but can't see it!
    Many thanks for reading.
    Last edited by padawan; August 16th 2010 at 07:00 AM.
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  2. #2
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    Your Qr should be

    \dfrac{2}{\pi}\displaystyle{\int}_0^{\pi} \sin^2 x \sin nx\,dx = - \dfrac{8}{\pi n(n^2-4)} for n odd and zero otherwise.
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  3. #3
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    Worked through it and finally got the solution.. as always, thank you very much for the pointer Danny.
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