Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook