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Homework Help: Heat Equation + 2 Robin Boundary Conditions

  1. Sep 22, 2009 #1


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    1. The problem statement, all variables and given/known data

    Find the temperature distribution in the long thin bar −a ≤ x ≤ a with a
    given initial temperature u(x,0) = f(x).
    The side walls of the bar are insulated, while heat radiates from the ends into
    the surrounding medium whose temperature is u = 0.
    The radiation is taken to obey Newton’s Law.

    2. Relevant equations

    [tex]u_{t} = \alpha^{2}u_{x}_{x}[/tex]
    [tex]u_{x}(-a,t) = (h/k)u(-a,t)[/tex]
    [tex]u_{x}(a,t) = -(h/k)u(a,t)[/tex]
    [tex]u(x,0) = f(x)[/tex]

    h and k are constants.

    3. The attempt at a solution

    My main concern is: I do not know if according to the problem description, the BCs should be as written above OR
    u_{x}(-a,t) = -(h/k)u(-a,t)[/tex]
    [tex]u_{x}(a,t) = (h/k)u(a,t)[/tex]

    I had to construct the BCs myself, they were not given explicitly in equation form.

    notice that the only difference is that the negative sign has gone from one equation to the other. I would appreciate it if someone could tell me where the negative sign belongs, and perhaps also explain why, I did not fully grasp the explanation I was given in class. Thank You.
  2. jcsd
  3. Sep 22, 2009 #2


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    Hi PAR! :smile:

    (have an alpha: α and try using the X2 and X2 tags just above the Reply box :wink:)
    Physics tells us that if u > 0, then the bar is hottest at x = 0 …

    so at the +a end, the temperature is decreasing in the + direction (and at the -a end, the temperature is increasing in the + direction).

    In other words, at the +a end, ux is negative. :wink:

    Does that help? :smile:
  4. Sep 22, 2009 #3


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    If u[tex]_{x}[/tex] is negative in the +x direction at x=+a that means that there should a negative sign in front of the h/k, otherwise that side of the equation wouldn't be negative, right?
  5. Sep 22, 2009 #4


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    (wrong tag! :wink:)
    s'right! :biggrin:
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