Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about heat equation

  1. Apr 7, 2012 #1
    Hi,

    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?

    Thanks
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Question about heat equation
  1. Heat Equation Question (Replies: 10)

Loading...