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Math Help - Quick PDEs Question! BVPs: time-dependent heat gen in a bar

  1. #1
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    Exclamation Quick PDEs Question! BVPs: time-dependent heat gen in a bar

    Quick question, what would be the general method for solving this type of problem;

    (d/dt)u - (d2/dx2)u = h(x, t)

    with homogeneous BCs:
    u (0, t) = 0
    u (L, t) = 0,

    and IC u (x, 0) = f(x)

    Just not sure how to deal with the RHS h(x,t) being dependent on t as well as x, (corresponding to time-dependent heat generation I think??)
    What method should be used?

    Thanks,

    Squiggles
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  2. #2
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    Quote Originally Posted by Squiggles View Post
    Quick question, what would be the general method for solving this type of problem;

    (d/dt)u - (d2/dx2)u = h(x, t)

    with homogeneous BCs:
    u (0, t) = 0
    u (L, t) = 0,

    and IC u (x, 0) = f(x)

    Just not sure how to deal with the RHS h(x,t) being dependent on t as well as x, (corresponding to time-dependent heat generation I think??)
    What method should be used?

    Thanks,

    Squiggles
    If h(t,x)=0, the we would use the usual separation of variables. Because of the fixed endpoints we would obtain

    u = \sum_{n=1}^\infty a_n e^{- \frac{n^2\pi^2}{L^2}t} \sin \frac{n \pi}{L} x where a_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi}{L} x\,dx

    If h(t,x) \ne 0, since we still have fixed endpoints we will assume a solutuion of the form

    u = \sum_{n=1}^\infty T_n(t) \sin \frac{n \pi}{L} x

    and assume that h(x,t) has a fourier since series

    h(x,t) = \sum_{n=1}^\infty h_n(t) \sin \frac{n \pi}{L} x

    Substitute into thet PDE and equate like sin terms. This will give a series of ODEs for T_n.
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