by vibe3
Tags: solution
 P: 19 I am trying to solve the following equation in spherical coordinates: $$\left( \nabla f \right) \cdot \vec{B} = g$$ where $g$ is a known scalar function, $\vec{B}$ is a known vector field, and $f$ is the unknown function. I think the best way to approach this is to expand everything into a spherical harmonic basis: $$f(r,\theta,\phi) = \sum_{lm} f_{lm}(r) Y_{lm}(\theta,\phi)$$ $$g(r,\theta,\phi) = \sum_{lm} g_{lm}(r) Y_{lm}(\theta,\phi)$$ $$\vec{B}(r,\theta,\phi) = \sum_{lm} \left[ B_{lm}^r(r) \vec{Y}_{lm} + B_{lm}^{(1)} \vec{\Psi}_{lm} + B_{lm}^{(2)} \vec{\Phi}_{lm} \right]$$ where $\vec{Y}_{lm}, \vec{\Psi}_{lm}, \vec{\Phi}_{lm}$ are the vector spherical harmonics (VSH) defined here: http://en.wikipedia.org/wiki/Vector_spherical_harmonics Then, to evaluate the dot product between $\nabla f$ and $\vec{B}$, it is necessary to integrate over the unit sphere since the VSH orthogonality relations are defined in terms of integrals over $d\Omega$. So, integrating the original equation over $d\Omega$ will yield the following ODE equation for the unknown $f_{lm}(r)$: $$B_{lm}^r(r) \frac{d}{dr} f_{lm}(r) + \frac{l(l+1)}{r} B_{lm}^{(1)}(r) f_{lm}(r) = c_{lm} g_{lm}(r)$$ with $$c_{lm} = \int d\Omega Y_{lm} e^{-im\phi}$$ This ODE should be straightforward to solve numerically. However, my question is the ODE equation will determine $f_{lm}$ values which satisfy the equation: $$\int d\Omega \left( \nabla f \right) \cdot \vec{B} = \int d\Omega g$$ Is it true that these $f_{lm}$ will also satisfy the original equation?
 HW Helper P: 1,584 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.
 P: 19 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.
P: 280