Implementation of Neumann Boundary for HTS-Superconductor H-Formulation

Hello,

I am trying to implement the common H-Formulation, used to model magnetic fields in superconductors, in MATLAB Partial Differential Equation Toolbox. I am starting as a 2-D Problem to understand to functionality.

The H-Formulation generally looks like

where H is the magnetic field, ρ is the material resistivity and μ the material permiability. To transform theis into the form

used in MATLAB Partial Differential Equation Toolbox for the 2-D case (only and ) I get

m = 0;

d = mu;

a = 0;

f = 0;

c = [0, 0, 0, 0;...

0, rho, -rho, 0;...

0, -rho, rho, 0;...

0, 0, 0, 0];

To model the H-Formulation I have to model the superconductor domain and the air around it. So I have two domains that are seperated by a shared interface. As far as my understanding on this shared interface a Neumann boundary of the form

is used. My problem is that with the equation for a Neumann boundary in Matlab Documentation given as

where c is the same as in the PDE formulation, I do not see how this represents the typical Neumann boundary condition.

Can someone explain to me how I can implement the correct Neumann boundary condition in such a PDE problem?

Answer 1

[Torsten]

Set g = q = 0 to get the condition you want - whether you set n*( c*grad H ) = 0 or n * grad H = 0 makes no difference.

But if it's a condition at the interface between two solution domains, you usually have to set two transmission conditions, not only one.

And I'm not sure whether the PDE toolbox supports transmission conditions at all.

[Martin Rupp]

Can you explain to me why n*( c*grad H ) = 0 is equal to n * grad H = 0? I fail to understand this. If I expand my equation I get

while

.

Maybe I am missing something?

[Torsten]

I assumed c was a diagonal matrix with constant equal entries. In this case it does not matter whether you impose

n1*c11*dH/dx + n2*c22*dH/dy = 0 

or

n1*dH/dx + n2*dH/dy = 0

because c11 = c22 and you can divide the first equation by c11 to get the second.

But now looking at the c you derived, there are off-diagonal elements (although I don't understand why c is 4x4 and not 2x2).

So you are correct: The two conditions are not equivalent.

[Martin Rupp]

Hi,

c is a 4x4 as we have a 2-D Problem wit two eqautions. I am sorry I was not clear in my original post H is a vector and not a scalar. If you expand the PDGL from my original post for 2-D you get

from MATLAB Dcumentation https://de.mathworks.com/help/pde/ug/c-coefficient-for-systems-for-specifycoefficients.html

For 2-D Systems c is a tensor with 4 * N^2 Elements. N is the number of equations.

Maybe you have an idea how I can solve this problem?