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Laplace's equation on an annulus

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I am looking at Laplace's equation on an annulus (just a circle where a < r < b, where "a" and "b" are constants). The boundaries of the annulus at r=a and r=b are kept at a certain temperature, which is theta-independent!

    Using my physical intuition, of course the solution to Laplace's equation n polar coordinates must be theta-independent as well. But is there a mathematical way of proving it?
     
  2. jcsd
  3. Oct 3, 2008 #2

    gabbagabbahey

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    What is the general solution [itex]T(r,\theta)[/itex] to Laplace's equation in plane polar coordinates? You are essentially given the boundary conditions that [itex]T(a,\theta)=T_a=constant[/itex] and [itex]T(b,\theta)=T_b=constant[/itex] use these conditions to find the solution.
     
  4. Oct 3, 2008 #3

    Dick

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    The boundary conditions are theta independent, as you observed. IF the solution is unique, given the boundary conditions, doesn't that mean the solution is theta independent as well? Is the solution of the Laplace equation unique?
     
  5. Oct 4, 2008 #4
    It is easy to find the solution to Laplace's equation, when the solution is theta-independent. My problem is that I need to justify that it is.

    I can't quite follow your line of thought: "IF the solution is unique, given the boundary conditions, doesn't that mean the solution is theta independent as well?". Is this an intuitive argument?
     
  6. Oct 4, 2008 #5

    gabbagabbahey

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    If you find a solution that satisfies both Laplace's solution and the given boundary conditions, then there is a uniqueness theorem that guarantees it is the only solution. So, if you can find a solution which is theta independent, you are guaranteed that it is the only solution and hence the solution is theta independent.
     
  7. Oct 4, 2008 #6
    Ahh, I see. I have encountered this uniqueness-theorem before in an electromagnetics-class. I remember it now!

    So you guys are telling me that there is no other way to see that the solution must be theta-independent other than using physical reasoning?
     
  8. Oct 4, 2008 #7

    gabbagabbahey

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    Well, you could do it explicitly by using separation of variables to find the general solution [itex]T(r,\theta)=R(r) \Theta (\theta)[/itex]. When you substitute your boundary conditions, You will find that [itex]\Theta (\theta)[/itex] has to be a constant, and so there is no theta dependence.
     
  9. Oct 4, 2008 #8
  10. Oct 4, 2008 #9

    gabbagabbahey

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    In that derivation, the boundary conditions are [itex]u(a,\theta)=f(\theta)[/itex] and [itex]u(b,\theta)=g(\theta)[/itex] but your boundary conditions are that [itex]f(\theta)[/itex] and [itex]g(\theta)[/itex] are constants. If you use that, theta independence will follow.
     
  11. Oct 4, 2008 #10
    Ok, I arrive at the exact same expression as in 282 (in the link). I still can't see how I can set a_n = b_n = c_n = d_n = 0 - for r = const., I just get the Fourier-series for a constants function, which is legal?
     
  12. Oct 4, 2008 #11

    gabbagabbahey

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    The document is loading properly for me right now. Can you post equation 282 for me?
     
  13. Oct 4, 2008 #12
    [tex]
    \frac{1}{2}(a_0 + b_0 \ln b) + \sum\limits_{n = 1}^\infty {((a_n b^n + b_n b^{ - n} )\cos n\theta + (c_n b^n + d_n b^{ - n} )\sin n\theta )} = g(\theta )
    [/tex]

    where g(theta) is one of the boundary conditions (which is constant in our case), and b is the upper radius of the annulus.
     
  14. Oct 4, 2008 #13

    gabbagabbahey

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    Okay, it's working again. Take a look at equations 283,284,285 if [itex]g(\theta)=\beta[/itex] is a constant and similarily, for 287,288,289 for [itex]f(\theta)=\alpha[/itex]. They will give you a system of equations to solve for where you should get [itex]a_n=b_n=c_n=d_n=0[/itex]
     
  15. Oct 4, 2008 #14
    You are right. It really does work!

    Thanks for being patient with me.
     
  16. Oct 4, 2008 #15

    gabbagabbahey

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