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Solving Laplace equation in a hollow sphere

  1. Jan 8, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Determine the stationary temperature distribution in the hollow sphere a<r<b where r=a is kept at T1 and r=b is kept at T2.


    2. Relevant equations
    [itex]\triangle u =0[/itex].
    But with the Laplacian in spherical coordinates.

    3. The attempt at a solution
    I think I must solve the given equation (Laplace equation) and use the spherical expression of the Laplacian.
    My boundary conditions are [itex]u(r=a, \theta, \phi )=T1[/itex] and [itex]u(r=b, \theta, \phi )=T2[/itex]. However I'm not sure this information is enough. I don't know how to mathematically describe the fact that the "sphere" is finite and therefore ends for r outside of [a,b].
    I've found a document on the Internet treating a similar problem (with a normal sphere), but when I try to follow every steps, I'm stuck at one.
    I attach the document to this post.
    The right hand side of equation 6 is [itex]\frac{\lambda}{\sin \theta }[/itex]. But to me it looks like equation 6 is equation 4 divided by [itex]\sin ^2 \theta[/itex], so that the right hand side of equation 6 should be [itex]\frac{\lambda}{\sin ^2 \theta }[/itex]. The author then continues the document with [itex]\frac{\lambda}{\sin \theta }[/itex].
    How did he reach this?!
     

    Attached Files:

  2. jcsd
  3. Jan 8, 2012 #2

    Dick

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    I think you are making this way too complicated. Your boundary conditions aren't a function of the two angles. They are only a function of r. I think you just have to solve the radial equation in the separation of variables problem.
     
  4. Jan 8, 2012 #3

    vela

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    He made a mistake, but he undoes it when he gets to equation (10).
     
  5. Jan 9, 2012 #4

    fluidistic

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    Ok guys thank you very much.
    Shouldn't I also solve the azimuthal equation to get that [itex]\mu =n(n+1)[/itex] where n is an integer?
    Because without doing it, solving only the radial equation I get [itex]f(r)=c_1 r^{-\frac{1}{2}} \ln (r)+c_2 r^{-\frac{1}{2}}[/itex] for [itex]\mu =-1/4[/itex].
    [itex]f(r)=c_1 r^{-\frac{1}{2}+\frac{\sqrt{1-4\mu }}{2}}+c_2 r^{-\frac{1}{2}-\frac{\sqrt{1-4\mu }}{2}}[/itex] if [itex]\mu >-1/4[/itex].
    And [itex]f(r)=c_1 r^{-\frac{1}{2}} \cos \left ( \frac{\sqrt{1-4\mu }}{2} \ln (r) \right )+c_2 r^{-\frac{1}{2}} \sin \left ( \frac{\sqrt{1-4\mu }}{2} \ln (r) \right )[/itex] if [itex]\mu <-1/4[/itex].
    Now if I cheat and look into the document, [itex]f(r)=c_1r^n[/itex].
     
  6. Jan 9, 2012 #5

    vela

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    As Dick mentioned, the problem has spherical symmetry, so the solution can depend only on r. In particular, that means the derivatives with respect to θ or φ vanish.

    How did you get this? What was the radial equation you solved?
     
  7. Jan 9, 2012 #6

    Dick

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    Take the angular parts to be constant. That means taking n (and hence mu) to be zero.
     
  8. Jan 9, 2012 #7

    fluidistic

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    Hmm ok guys, not so sure what I must solve.
    What I solved (I think I made a sign mistake for the solution when [itex]\mu <-1/4[/itex] for the cos and sin arguments) was [itex]\frac{1}{f} \frac{d}{dr} (r^2 f')=cste=\mu[/itex]. I reached the Euler-Cauchy equation [itex]r^2f''+2rf'-\mu f=0[/itex]. I solved this equation via wikipedia's method (assuming a solution of the form [itex]f(r)=r^\alpha[/itex]).
    Now if I understand well, you are telling me to solve [itex]r^2f''+2rf'=0[/itex]?
     
  9. Jan 9, 2012 #8

    Dick

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    Yes. Set mu=0.
     
  10. Jan 9, 2012 #9

    fluidistic

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    In this case I used reduction of order (substitution v=f') and reached [itex]f(r)=\frac{c}{r}[/itex]. I checked out if this satisfies mathematically the DE and it does. However I see that when r tends to 0 there's a big problem.
    P.S.: Anyway r cannot tend to 0 but to a as a lower limit, according to the problem. Unless a tends to 0 also.
     
  11. Jan 9, 2012 #10

    Dick

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    There is another solution as well as c/r. You'd expect there to be since it's a second order equation and you'll need a second solution to fit your two boundary conditions. You could have just substituted f(r)=r^n and figured out what values of n work, like you did before. BTW you might notice that c/r is the form for the potential of a charge located at the origin. It's not an accident. The potential due to a distribution of charge also satisfies the Laplace equation outside of the region of charge.
     
    Last edited: Jan 9, 2012
  12. Jan 9, 2012 #11

    fluidistic

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    I reach n or alpha equal to -1 or 0. So [itex]f(r)=c_1+c_2r^{-1}[/itex]. I'm not very confident in that solution but it looks like it is a solution at first glance.
    Hmm about your comment of the analogy with electrostatics... Is it like if the heat source was the center of the hollow sphere? If so, I don't really understand the analogy. In my case I thought it was both outer and inner surfaces of the hollow sphere that were the "heat sources", not a point in the middle. But is it equivalent to the heat source in the center of the hollow sphere? I think yes now... wow.
     
  13. Jan 9, 2012 #12

    Dick

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    Yes, you've got the solution. Now you just need to find c1 and c2. And I wouldn't push the electrostatics analogy too hard. Lots of things have a steady state solution involving Laplace's equation. It's sort of an energy minimizing thing. My point was that you shouldn't be too surprised that c/r is a spherically symmetric solution to Laplace's equation in 3 dimensions.
     
  14. Jan 10, 2012 #13

    fluidistic

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    Ok thanks a lot Dick.
    Is it right to say that [itex]u(r, \theta, \phi)=u(r)=Cf(r)[/itex]?
    If so, I reach [itex]u(r)=T_1+\frac{b(T_1-T_2)}{a-b}+\frac{ab(T_1-T_2)}{r(a-b)}[/itex].
     
  15. Jan 10, 2012 #14

    Dick

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    Sure. u(r) is just a function of r. The general form of your solution looks ok, but check it. Put r=a and b. Do you get u(a)=T1 and u(b)=T2? I think you've left out a sign or something.
     
  16. Jan 10, 2012 #15

    fluidistic

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    I get it now, thanks for all. It was a sign mistake indeed.
    Problem solved.
     
  17. Jan 10, 2012 #16

    fluidistic

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    Hmm I can't grasp something. It seems like there's no difference if the sphere was entirely filled versus the hollow sphere, as long as r=a is kept at T1 and r=b is kept at T2. The sphere could also have an infinite radius or let's say very much bigger than r=b, yet the solution between r=a and r=b would remain unchanged. Is this correct?

    Also if the sphere was totally filled and I wanted to see what is the temperature at r=0... how could I do this?
     
  18. Jan 10, 2012 #17

    Dick

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    I'm not totally sure I get you but there must be something filling the space between r=a and r=b, it doesn't particularly matter what but, sure, the temperature between a and b is determined by the temperature at a and b. It's a Laplace equation with Dirichlet boundary conditions. If you were to remove the inner boundary condition at r=a, then you tell me what the solution should be inside of r=b. It will still have the form u(r)=c1+c2/r, but now you want to say that r=0 must be a regular point of the solution, so what can you say about c2?
     
  19. Jan 10, 2012 #18

    fluidistic

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    c2 must in that case be worth 0 I think.
    So what happens outisde [a,b] has absolutely no effect on what happens in the inside of [a,b]? This is basically my doubt. I know there must be matter inside [a,b] and it looks like outside of it, nothing matters. There can be matter or not, the temperature distribution between [a,b] will remain the same.
     
  20. Jan 10, 2012 #19

    Dick

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    Yes, c2 must be zero. So the temperature distribution inside must be a constant. Isn't that kind of intuitively clear? And I'm not saying effects outside of [a,b] can't affect what happens inside of [a,b], they can. But the only way they can do that are by changing the boundary conditions on [a,b]. Putting a heat source outside of r=b may increase the temperature in [a,b], but it has to do that by raising the temperature at r=b first.
     
  21. Jan 10, 2012 #20

    fluidistic

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    I see now, thank you very much. Everything makes sense.
     
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