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Boundary conditions for inhomogeneous non-sepearable 3D PDE

  1. Jan 31, 2014 #1
    Hello, I am looking to solve the 3D equation in spherical coordinates
    [tex]
    \nabla \cdot \vec{J} = 0
    [/tex]
    using the Ohm's law
    [tex]
    \vec{J} = \sigma \cdot (\vec{E} + \vec{U} \times \vec{B})
    [/tex]
    where [itex]\sigma[/itex] is a given 3x3 nonsymmetric conductivity matrix and [itex]U,B[/itex] are given vector fields. I desire the electric potential [itex]\Phi[/itex] where [itex]\vec{E} = -\nabla \Phi[/itex]. This leads to the inhomogeneous elliptic PDE:
    [tex]
    \nabla \cdot (\sigma \cdot \nabla \Phi) = f
    [/tex]
    where the right hand side [itex]f[/itex] is known and is [itex]f = \nabla \cdot (\sigma \cdot \vec{U} \times \vec{B})[/itex].

    Now my question relates to how to express the boundary conditions. Many existing PDE software require inputs of Robin-type boundary conditions, which would be of the form:
    [tex]
    a \Phi + b \hat{n} \cdot \nabla \Phi = g
    [/tex]

    For my particular problem, I am using a spherical region
    [tex]
    \Omega = [r_1,r_2] \times [\theta_1,\theta_2] \times [0, 2 \pi]
    [/tex]
    which is like a spherical shell with the top and bottom cut off at some [itex]\theta_1,\theta_2[/itex]

    Now I know that at the lower boundary,
    [tex]
    \vec{J}(r_1,\theta,\phi) = 0
    [/tex]
    which means
    [tex]
    \sigma \cdot \nabla \Phi(r_1,\theta,\phi) = (\sigma \cdot (\vec{U} \times \vec{B}))(r_1,\theta,\phi) = g(r_1,\theta,\phi)
    [/tex]
    where [itex]g[/itex] is known.

    What I can't see easily is now to convert this into the Robin-type equation above so it can be input into a PDE software. Does anyone have any ideas?

    Many thanks in advance!
     
  2. jcsd
  3. Feb 1, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Your [itex]g[/itex] is a vector.

    You need somehow to solve
    [tex]
    \sigma \cdot \nabla \Phi = \vec g
    [/tex]
    for the radial component of [itex]\nabla \Phi[/itex], which is [itex]\vec n \cdot \nabla\Phi[/itex] for this boundary. That in general is possible only if [itex]\sigma[/itex] is invertible on the boundary ([itex]\det \sigma \neq 0[/itex]), so that
    [tex]
    \frac{\partial \Phi}{\partial r} = \hat r \cdot (\sigma^{-1} \cdot \vec g).
    [/tex]

    (Actually for [itex]r = r_1[/itex] we have [itex]\vec n \cdot \nabla\Phi = - \dfrac{\partial \Phi}{\partial r}[/itex], so on that boundary
    [tex]
    -\frac{\partial \Phi}{\partial r} = \hat r \cdot (\sigma^{-1} \cdot \vec g)
    [/tex]
    and on [itex]r = r_2[/itex]
    [tex]
    \frac{\partial \Phi}{\partial r} = \hat r \cdot (\sigma^{-1} \cdot \vec g).
    [/tex]

    Otherwise you may need to find a different solution method.
     
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