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Diffusion Equation on a plate - 2 dimensions

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    The edges of a thin plate are held at the temperature described below. Determine the steady-state temperature distribution in the plate. Assume the large flat surfaces to be insulated.

    If the plate is lying along the x-y plane, then one corner would be at the origin. The height of the plate would be 1m along the y-axis and the length would be 2m along the x-axis. The edge along the y-axis is being held at 0 C. The edge along the x-axis is being held at 0 C. The edge parallel to the x-axis is being held at 0 C. The edge parallel to the y-axis is being held at 50sin(pi*y) C.

    2. Relevant equations

    So this question is actually just a diffusion equation. Alpha2uxx=ut
    and
    u(x,t)=X(x)T(t)=(C1coskx+C2sinkx)e-K2alpha2t+C3+C4x

    3. The attempt at a solution

    So the problem I'm having is previously everything was done on a rod. Now I have a plate. I know that for a rod, my boundary conditions would be the temperature at the ends, for example, if this were a rod I would say that u(0,t)=0 and u(2,t)=50sin(piy). BUT!!!! How can I say that if I have corners and a second dimension to deal with?
     
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  3. Jul 31, 2012 #2

    HallsofIvy

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    No, it isn't. There are two space variables, x and y. The equation is [tex]\alpha^2(u_{xx}+ u_{yy})= u_t[/tex].
    (Corrected thanks to LCKurtz.)

     
    Last edited: Aug 1, 2012
  4. Jul 31, 2012 #3

    LCKurtz

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    Halls has a typo there, he means$$
    \alpha^2(u_{xx} + u_{yy})=u_t$$
    in case it isn't obvious.
     
  5. Jul 31, 2012 #4
    Okay, so I'm going to expand the formula from alpha2uxx to alpha2(uxx+uyy)=ut

    Then, u(x,y,t)=(X(x)+Y(y))T(t)

    and

    X'"X+Y"/Y=T'/alpha2T=λ and λ=0

    so, X(x)=ax+b, Y(y)=cy+d and T(t)=e

    Then u(x,y,t)=c(ax+cy+b+d)

    right?
     
  6. Jul 31, 2012 #5

    LCKurtz

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    Assuming that X''' is a typo meaning X'', please show us how you got that equation with that substitution for u(x,y,t). Or are you just guessing?
     
  7. Jul 31, 2012 #6
    I was just guessing. Since u(x,t)=X(x)T(t), I assumed that U(x,y,t)=[X(x)+Y(y)]T(t) because x and y are vectors and T isn't. And yes, I meant X"/X, not X'''/X
     
  8. Jul 31, 2012 #7

    LCKurtz

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    X, Y, and T are not vectors, they are ordinary functions. And it is one thing to guess what the answer might be but surely if you are doing mathematics you would check your guess to see if you got lucky. If you try it, you will not be able to verify your guess.

    I will give you a hint: Try u(x,y,t) = X(x)Y(y)T(t). No guessing from here. Plug it in and see what you get. Work it out.
     
  9. Jul 31, 2012 #8
    I'm confused because this was supposed to be solved with separation of variables. I don't know how to do that with 3 variables.
     
  10. Jul 31, 2012 #9
    alpha2(X"/X+Y"/Y)=T'/T

    X"/X+Y"/Y=T'/(Talpha2)=λ=0

    If I were just doing X"/X=T'/(Talpha2), I would use separation of variables method to solve and get that X(x)=ax+b and T(t)=c

    Since this is not just X"/X, but X"/X+Y"/Y, I don't think I know how to solve the problem.
     
  11. Jul 31, 2012 #10

    LCKurtz

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    There is no reason to assume ##\lambda=0##. Since each fraction uses a different variable you can assume each fraction is constant. So you might start with$$
    \frac{X''} X = \lambda,\ \frac{Y''} Y = \mu,\ \frac{T''}{\alpha^2 T}=\lambda +\mu$$
    Start with the ##Y## equation. Apply u=XYT to the boundary conditions to see what the boundary conditions for Y must be. Then solve the Y eigenvalue problem just like you did for a 1D rod. After that, do the X equation, followed by the T equation.

    I have to leave for the rest of the afternoon but that should get you started.
     
  12. Jul 31, 2012 #11
    Okay, I'm looking more carefully through my book, and I found a problem with a temperature difference on a thin disk. They simplify the problem and explain, "since u does not vary with θ, we dropped the uθθ term and solved the reduced equation. If you have doubts about this step, observe that equation 79 does satisfy the boundary conditions."

    So can I drop the uyy term since the plate have a temperature gradient along x and not y? That would make the whole problem much simpler.
     
  13. Jul 31, 2012 #12
    May I suggest, and I know LCKurtz is doing a fine job helping you, that you find "Basic Partial Differential Equations" by Bleecker and Csordas in the Library and just work through the example? And the reason I suggest this is because you seem to be having trouble with just the concepts, for example separation of variables and it's always nice to first just work through an example in the book, or two or three, then attempt to work on a new one yourself.
     
  14. Aug 1, 2012 #13
    Steady-state means [itex]u_t=0[/itex] so the equation to solve is [itex]u_{xx}+u_{yy}=0[/itex] and [itex]u=u(x,y)[/itex]
     
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