Boundary conditions for inhomogeneous non-sepearable 3D PDE

vibe3
Messages
39
Reaction score
1
Hello, I am looking to solve the 3D equation in spherical coordinates
<br /> \nabla \cdot \vec{J} = 0<br />
using the Ohm's law
<br /> \vec{J} = \sigma \cdot (\vec{E} + \vec{U} \times \vec{B})<br />
where \sigma is a given 3x3 nonsymmetric conductivity matrix and U,B are given vector fields. I desire the electric potential \Phi where \vec{E} = -\nabla \Phi. This leads to the inhomogeneous elliptic PDE:
<br /> \nabla \cdot (\sigma \cdot \nabla \Phi) = f<br />
where the right hand side f is known and is f = \nabla \cdot (\sigma \cdot \vec{U} \times \vec{B}).

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:
<br /> a \Phi + b \hat{n} \cdot \nabla \Phi = g<br />

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

Now I know that at the lower boundary,
<br /> \vec{J}(r_1,\theta,\phi) = 0<br />
which means
<br /> \sigma \cdot \nabla \Phi(r_1,\theta,\phi) = (\sigma \cdot (\vec{U} \times \vec{B}))(r_1,\theta,\phi) = g(r_1,\theta,\phi)<br />
where g 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!
 
Physics news on Phys.org
vibe3 said:
Hello, I am looking to solve the 3D equation in spherical coordinates
<br /> \nabla \cdot \vec{J} = 0<br />
using the Ohm's law
<br /> \vec{J} = \sigma \cdot (\vec{E} + \vec{U} \times \vec{B})<br />
where \sigma is a given 3x3 nonsymmetric conductivity matrix and U,B are given vector fields. I desire the electric potential \Phi where \vec{E} = -\nabla \Phi. This leads to the inhomogeneous elliptic PDE:
<br /> \nabla \cdot (\sigma \cdot \nabla \Phi) = f<br />
where the right hand side f is known and is f = \nabla \cdot (\sigma \cdot \vec{U} \times \vec{B}).

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:
<br /> a \Phi + b \hat{n} \cdot \nabla \Phi = g<br />

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

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

Your g is a vector.

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

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

Otherwise you may need to find a different solution method.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top