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Radial solutions to laplace equation

  1. Apr 28, 2012 #1

    When can I assume that the solution the laplace equation (or poisson equation) is radial? That is, when can I look only at (in polar coordinates)

    [tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} = f(r,\theta)[/tex]

    instead of

    [tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} + \frac{1}{r^2}\frac{\partial u^2} {\partial^2 \theta} = f(r,\theta)[/tex]

    With appropriate boundary conditions of course. My guess is any time that the boundary conditions are constant in theta and f depends only on r? But I'm not really sure.

    Thanks for any help!

    Last edited: Apr 28, 2012
  2. jcsd
  3. Apr 28, 2012 #2
    Look at [tex]\frac{\partial^2u} {\partial\theta^2}[/tex] When would this term vanish from the equation...
  4. Apr 28, 2012 #3
    when [tex] u = u(r) [/tex] only?
  5. Apr 28, 2012 #4
    Yes if u(r) only then it would surely drop out. But that is not the only way... For a moment ignore the partial and just think of it like a normal derivative, more specifically notice it is a second derivative. When would the second derivative of some function equal zero?
  6. Apr 28, 2012 #5
    when it's linear in theta? How do you know when the solution is linear in theta?
  7. Apr 28, 2012 #6
    Yes, when its linear in theta. How do you know when its linear in theta...thats a good question. I will have to do a little thinking on that one, this stuff is kinda new to me as well ;). Hopefully that gives you at least some idea and if I come up with something I will keep you updated.
  8. Apr 28, 2012 #7
    I think it has to do with the laplacian being rotationally invariant - hence no theta dependence. See sec 8.3/8.4:


    could anyone else explain this a bit better? Is that true - if a PDE is rotationally invariant, its solutions will have no theta dependence (or some of them do anyway)?
  9. Apr 28, 2012 #8
    After doing some thinking I don't know if 'u' can be linear in theta (at least physically). Because if it was linear in theta then at theta=0 'u' would have some value, while at theta=2pi 'u' would have some different value despite the fact that theta=0 and theta=2pi correspond to the same physical point. Thus if it was linear in 'u' there would be a discontinuity. That being said the only thing that would satisfy it is u=u(r)
  10. Apr 28, 2012 #9


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    wumple, the intuition is really all that you need- if we assume that f has no dependence on theta, we get a differential equation only involving r, so we can find solutions involving only r. Compare that with this proposed differential equation:
    [tex] \theta \frac{\partial f}{\partial r} + f = 0[/tex]

    If you assume that f doesn't depend on theta, then you still have a theta in your differential equation and things get confusing.

    In terms of when your solution is going to be rotationally invariant - if your boundary condition is rotationally invariant and the shape of your domain is rotationally invariant then your solution will be rotationally invariant as well - otherwise, rotating the whole problem yields the same exact problem but a different solution (this is assuming that things are specified to the point where there is a unique solution)
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