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Modified heat equation steady state

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    determine the steady state equation for the given heat equation and boundary conditions


    2. Relevant equations
    Ut=Uxx-4(U-T)
    U(0,T)=T U(4,T)=0 U(x,0)=f(x)


    3. The attempt at a solution
    I put Ut=0
    so 0=UInf''-4(Uinf-T)
    then once I tried to integrate I ended up with a Uinf thats impossible to isolate is there a specific method used for this kind of heat equation?
     
  2. jcsd
  3. Jul 17, 2011 #2

    hunt_mat

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    The steady state is usually when the system is time independent, so:
    [tex]
    \frac{\partial U}{\partial t}=0
    [/tex]
    So this will turn your equation into and ODE, which I see you have already done and your ODE is:
    [tex]
    \frac{\partial^{2}U}{\partial x^{2}}-4U+4T=0
    [/tex]
    I am not too sure what T is, is it a constant? A function?
     
    Last edited: Jul 17, 2011
  4. Jul 17, 2011 #3

    LCKurtz

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    What do you mean by "impossible to isolate". Calling what you call Uinf u(x) for simplicity the equation you have written is just

    u''(x) - 4u(x) = -4T
    u(0) = u(T) = 0

    A second order constant coefficient ODE.
     
  5. Jul 18, 2011 #4
    I believe T is constant. so I tried using variation of parameters to solve (Uinf=X) X''-4X=-4T
    and came up with a particular solution of
    c1e2x+c2e-2x-T then solved for the constants using boundary conditions. is this method right as all the example I have seen involve just integrating it twice.
     
  6. Jul 18, 2011 #5

    LCKurtz

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    That should work, but I'm guessing it gets a bit messy solving for the constants. Here's a little trick you might like to learn. You probably know that instead of using the pair {e2x,e-2x} for the general solution of the homogeneous equation, you could have used {sinh(2x),cosh(2x)}, which simplifies things a bit when you calculate X(0) = 0. But an even better choice is the pair {sinh(x),sinh(T-x}. Try evaluating the constants when you write your solution

    X(x) = Asinh(x)+Bsinh(T-x) + T

    and you will see what I mean.
     
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