The boundary condition of Hall-MHD model

Hello, everyone.

Now I’m using freefem++ to solve the incompressible hall-MHD equations, but I’ve encountered a problem regarding the boundary conditions of the magnetic field.

\mu\nabla\times (\nabla\times\mathbf{B})+(\mathbf{u}^n\cdot\nabla)\mathbf{B}^n-(\mathbf{B}^n\cdot\nabla)\mathbf{u}^n+\eta\nabla\times\left((\nabla\times\mathbf{B}^n)\times\mathbf{B}^n\right)=\mathbf{g}
\nabla\cdot\mathbf{B}=0
Before solving, assume that the velocity field \mathbf{u}^n,\mathbf{B}^n is already known (I decoupled the equation).

I use the boundary condition of magnetic
\mathbf{B}\times \mathbf{n} = \mathbf{0} or \mathbf{B}\times \mathbf{n} = \mathbf{B}_e , where \mathbf{B}_e is the analytical solution that I set, and \mathbf{n} is the unit external normal vector of the boundary. The domain is [0,1]^3.

solve Magnetic(Bx1, By1, Bz1, wx, wy, wz, solver = GMRES) 
        = int3d(Th)( mu*curl(Bx1, By1, Bz1)'*curl(wx, wy, wz) + mu*div(Bx1, By1, Bz1)*div(wx, wy, wz) )
        + int3d(Th)( UgradV(ux0, uy0, uz0, Bx0, By0, Bz0)'*[wx, wy, wz] - UgradV(Bx0, By0, Bz0, ux0, uy0, uz0)'*[wx, wy, wz] 
            + eta*crossProduct(dy(Bz0)-dz(By0), dz(Bx0)-dx(Bz0), dx(By0)-dy(Bx0), Bx0, By0, Bz0)'*curl(wx, wy, wz) 
            - (g1t*wx + g2t*wy + g3t*wz))
    ;

with

    macro Grad(ux, uy, uz) [(dx(ux)), (dy(ux)), (dz(ux)), (dx(uy)), (dy(uy)), (dz(uy)), (dx(uz)), (dy(uz)), (dz(uz))] //
    macro div(ux, uy, uz) ( dx(ux) + dy(uy) + dz(uz) ) // 
    macro UgradV(ux, uy, uz, vx, vy, vz) [(ux)*(dx(vx)) + (uy)*(dy(vx)) + (uz)*(dz(vx)), (ux)*(dx(vy)) + (uy)*(dy(vy)) + (uz)*(dz(vy)), (ux)*(dx(vz)) + (uy)*(dy(vz)) + (uz)*(dz(vz))] //
    macro curl(Bx, By, Bz) [dy(Bz) - dz(By), dz(Bx) - dx(Bz), dx(By) - dy(Bx)] //
    macro crossProduct(ux, uy, uz, vx, vy, vz) [(uy)*(vz) - (uz)*(vy), (uz)*(vx) - (ux)*(vz), (ux)*(vy) - (uy)*(vx)] // 

But, the result is wrong. Maybe I should add a stabilization term, but I do not how to do

To be consistent with the boundary condition B\times n=0 you should add a penalty boundary term
+int2d(Th)(1.e15*crossProduct(Bx1, By1, Bz1, N.x, N.y, N.z)'*crossProduct(wx, wy, wz, N.x, N.y, N.z))
For a non homogeneous condition B\times n = B_e you should add also
-int2d(Th)(1.e15*[Bex,Bey,Bez]'*crossProduct(wx, wy, wz, N.x, N.y, N.z))

I understand. But is there any requirement for the penalty parameter? I see you gave 1.e15

The value is not important, it needs just to be much larger than the other terms, from 1.e8 to 1.e30 should do the job. The standards of FFEM uses tgv=1.e30.

OK, I understan.

Finally, I need to solve a Poisson equation for pressure

solve poisson(p, pp, solver = CG) = int3d(Th)( dx(p)*dx(pp) + dy(p)*dy(pp) + dz(p)*dz(pp) )
        - int3d(Th)( (dx(p0)*dx(pp) + dy(p0)*dy(pp) + dz(p0)*dz(pp)) - div(ux, uy, uz)*pp/dt )
        + int2d(Th)( 1.e-8*p*pp );

Can I set the parameter to 1.e-8?

Yes it should work with 1.e8 (not 1.e-8). Alternatively you can use the built-in command
+on(1,2,3,4,5,6,p=0)
1,2,3,4,5,6 refers to the boundary labels. This built-in command is fully equivalent to
+int2d(Th,1,2,3,4,5,6)(tgv*p*pp)

It should be 1e-8 (I try to set 1.e8, but it is wrong)


This is my equation, which has the artificial Newman boundary condition.

Now there is another problem.The convergence order of the L2 norm of the magnetic field I calculated is not correct (as the space is divided finer, the order decreases), but the convergence order of the H1 semi norm is no problem. I don’t know if this is due to the code.

h = [1./2, 1./4, 1./8, 1./12, 1./14, 1./16, 1./18, 1./20, 1./22, 1./24];

Convergence B_L2: 0     2.937716417     3.003305649     2.935914756     2.751860597
        2.933047063     2.438786889     1.90737733      4.622726824     0.3392563845
Convergence B_H1_semi: 0     1.885937569     1.974759866     1.985809507     1.985307756
        1.983228133     1.998874224     1.943471302     2.039907908     2.000346527

The 1.e8 is for homogeneous Dirichlet boundary condition. If you want Neumann, you just don’t put the penalty term!
About the order of convergence I don’t know, but your results look pretty good.