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PDE for temperature distribution in rectangle

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A rectangular chip of dimensions a by b is insulated on all sides and at t=o temperature u=0. The chip produces heat at a constant rate h. Find an expression for u(x,y,t)

    2. Relevant equations
    δu/δt = h + D(δ2u/δx2 + δ2u/δy2) x∈(0,a), y∈(0,b)

    3. The attempt at a solution
    I'm just wondering what effect the insulation on all sides will have on the equation. The temperature goes up by h per unit time throughout the chip but will the other term be affected?
     
  2. jcsd
  3. Oct 27, 2014 #2

    pasmith

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    "Insulated" means that there is no heat flux into or out of the rectangle. Here this reduces to the requirement that the component of [itex]\nabla u[/itex] normal to the boundary must vanish. This gives you the boundary condition you need in order to solve the PDE.
     
  4. Oct 27, 2014 #3
    Why just the normal component? If its insulated won't the tangential component also vanish? So δu/δx=δu/δy=0 at x=0,a and y=0,b.

    The question actually says that there is no need to solve the pde to get to u(x,y,t) - that there is a way to guess the answer without solving it. I can't think of any such way. Any ideas?
     
  5. Oct 27, 2014 #4

    pasmith

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    The physical constraint is "no heat flux across the boundary". The heat flux is [itex]D\nabla u[/itex], and the heat flux across the boundary is the component of [itex]D\nabla u[/itex] normal to the boundary. Flux parallel to the boundary is not a problem.

    Is there any reason why the temperature should vary with position here, given that the heat source doesn't depend on position and the boundary conditions are consistent with a uniform temperature?
     
  6. Oct 27, 2014 #5
    Alright...got it.

    There's another related PDE: d2u/dx2 + d2u/dy2 = a*u with u = 0 on all edges for some eigenvalues a. I'm not sure how to go about this. I can start by solving the steady state equation but what do I do about the condition?
     
  7. Oct 27, 2014 #6

    pasmith

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    Seek a separable solution [itex]u(x,y) = X(x)Y(y)[/itex]. You will find that there are particular values of [itex]a[/itex] (eigenvalues) for which [itex]u(x,y) \equiv 0[/itex] is not the only such solution.
     
  8. Oct 27, 2014 #7
    I'm not clear what you mean by eigenvalues for which u=0 is not the only solution. Could you please elaborate?
     
  9. Oct 28, 2014 #8

    pasmith

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    Simple example: Let [itex]X'' = kX[/itex] subject to [itex]X(0) = X(\pi) = 0[/itex]. The general solution of that ODE with satisfies [itex]X(0) = 0[/itex] is [tex]
    X(x) = \begin{cases} A\sin(\sqrt{|k|}x), & k < 0 \\
    Ax, & k = 0 \\
    A\sinh(\sqrt{k}x), & k > 0\end{cases}
    [/tex] In the second and third alternatives, the only way to satisfy [itex]X(\pi) = 0[/itex] is to take [itex]A = 0[/itex], which means that [itex]X(x) = 0[/itex]. However in the first case we can set [itex]k = -n^2 > 0[/itex] for integer [itex]n[/itex] and then [itex]X(x) = A \sin(nx)[/itex] satisfies [itex]X(\pi) = 0[/itex] for any value of [itex]A[/itex]. Thus [itex]\{-n^2 : n \in \mathbb{N}\}[/itex] are the eigenvalues, and [itex]\sin(nx)[/itex] is the corresponding eigenfunction. We don't need [itex]n < 0[/itex] because [itex]\sin(-|n|x) = -\sin(|n|x)[/itex].
     
  10. Oct 30, 2014 #9
    Maybe I'm being dense but I'm not able to separate the variables, I get X''/X + Y''/Y = a. So how do I get separate equations for x and y?
     
  11. Oct 30, 2014 #10

    pasmith

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    X''/X is a function of x only. Y''/Y is a function of y only. Their sum is a constant. Therefore each of them is constant.
     
  12. Oct 30, 2014 #11
    So X''/X=m (a cpnstant) and Y''/Y = n (another constant) and m + n = a?
     
  13. Oct 30, 2014 #12

    pasmith

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    Yes.
     
  14. Oct 30, 2014 #13
    Ok...so do I break the equation down into 2 parts? A steady state one where d2u/dx2 + d2u/dy2 = 0 and another where it's = a*v and then add the 2 solutions? Or do I just do the latter equation?
     
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