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Math Help - PDE Initial Condition Question

  1. #1
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    PDE Initial Condition Question

    Hello,
    I have the following PDE:

    u_t = k u_xx

    with u(x,0) = f(x), u_x(0,t) = u_x(L,t) = 0

    which I solve using separation of variables by letting u = XT.

    Now, for the conditions we have

    u_x(0,t) = X'(0) T(t) = 0

    so X'(0) = 0 and

    u_x(L,t) = X'(L) T(t) = 0

    X'(L) = 0

    My question is why can't T(t) = 0 in these cases? This assumption is made in every PDE I've solved using sep of vars, but I would like to know why exactly this is.
    Thanks
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  2. #2
    A Plied Mathematician
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    Well, T(t) = 0, when plugged back into your separation guess of u = XT, implies that u = 0, the trivial solution. The trivial solution, I think you'll find, will not work unless f(x) = 0. In addition, most of the time you're interested in a nontrivial solution, which implies that T(t) is not identically zero everywhere. Make sense?
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