Boundary conditions for inhomogeneous non-sepearable 3D PDE

  • Thread starter vibe3
  • Start date
  • #1
46
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!
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,879
527
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
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.
 

Related Threads on Boundary conditions for inhomogeneous non-sepearable 3D PDE

  • Last Post
Replies
2
Views
736
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
783
Replies
2
Views
2K
Replies
6
Views
2K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
5
Views
4K
Replies
8
Views
1K
Top