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Question about PDE solution

  1. Feb 15, 2012 #1
    I am trying to solve the following equation in spherical coordinates:
    [tex]
    \left( \nabla f \right) \cdot \vec{B} = g
    [/tex]
    where [itex]g[/itex] is a known scalar function, [itex]\vec{B}[/itex] is a known vector field, and [itex]f[/itex] is the unknown function.

    I think the best way to approach this is to expand everything into a spherical harmonic basis:
    [tex]
    f(r,\theta,\phi) = \sum_{lm} f_{lm}(r) Y_{lm}(\theta,\phi)
    [/tex]
    [tex]
    g(r,\theta,\phi) = \sum_{lm} g_{lm}(r) Y_{lm}(\theta,\phi)
    [/tex]
    [tex]
    \vec{B}(r,\theta,\phi) = \sum_{lm} \left[ B_{lm}^r(r) \vec{Y}_{lm} + B_{lm}^{(1)} \vec{\Psi}_{lm} + B_{lm}^{(2)} \vec{\Phi}_{lm} \right]
    [/tex]
    where [itex]\vec{Y}_{lm}, \vec{\Psi}_{lm}, \vec{\Phi}_{lm}[/itex] are the vector spherical harmonics (VSH) defined here:
    http://en.wikipedia.org/wiki/Vector_spherical_harmonics
    Then, to evaluate the dot product between [itex]\nabla f[/itex] and [itex]\vec{B}[/itex], it is necessary to integrate over the unit sphere since the VSH orthogonality relations are defined in terms of integrals over [itex]d\Omega[/itex].

    So, integrating the original equation over [itex]d\Omega[/itex] will yield the following ODE equation for the unknown [itex]f_{lm}(r)[/itex]:
    [tex]
    B_{lm}^r(r) \frac{d}{dr} f_{lm}(r) + \frac{l(l+1)}{r} B_{lm}^{(1)}(r) f_{lm}(r) = c_{lm} g_{lm}(r)
    [/tex]
    with
    [tex]
    c_{lm} = \int d\Omega Y_{lm} e^{-im\phi}
    [/tex]

    This ODE should be straightforward to solve numerically. However, my question is the ODE equation will determine [itex]f_{lm}[/itex] values which satisfy the equation:
    [tex]
    \int d\Omega \left( \nabla f \right) \cdot \vec{B} = \int d\Omega g
    [/tex]

    Is it true that these [itex]f_{lm}[/itex] will also satisfy the original equation?
     
  2. jcsd
  3. Feb 16, 2012 #2

    hunt_mat

    User Avatar
    Homework Helper

    My first choice of solution method would be method of characteristics, take a 2D case and you can easily see how this works and you can generalise to 3D case.
     
  4. Feb 16, 2012 #3
    Yes its true that the method of characteristics could work, however I am solving this equation numerically on a 3D grid in spherical coordinates. Integrating along characteristic curves would add complexity in that I'd have to pick lots of different starting points for the integration to get a decent grid of solutions, and then interpolate those solutions back to a spherical grid.

    I think the spherical harmonic approach is more elegant and natural for this problem....if only the method I've described above is sound.
     
  5. Feb 16, 2012 #4
    Every time you separate your variables, the resulting functions satisfy the original equations on their own only under special conditions (the obvious one being the other function being equal to 1). As far as I know, it is not something you can generalise.
     
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