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Question about heat equation

  1. Apr 7, 2012 #1

    So if I start with the boundary conditions

    [tex]U(0,t) = T1 [/tex] and [tex] U(L,t) = T2 [/tex]

    and T1 does not equal T2, it seems that you are supposed to look at the 'steady state solution' (solution as t goes to infinity)?

    which satisfies

    [tex] T''(x) = 0 [/tex]

    so the solutions are

    [tex]T(x) = Ax + B[/tex]

    and then you fit the BCs to this 'steady-state solution'

    Why is this the 'steady state solution'? And why does the second derivative have to be 0? After a long period of time, wouldn't the entire bar (minus the ends) be the same temperature, so then just T'(x) = 0?

  2. jcsd
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