I understand the computational implementation of the PC and PI bcs in cartesian case. In terms of the magnetic field components for PI, we have $B_x = B_y = B_z = 0$, and for PC, we have $B_x = \partial_x B_y = \partial_x B_z = 0$, given that $x$ is the normal direction to the boundary. Let's say you have a domain bounded by two cylindrical surfaces at radii $R_1$ and $R_2$, in cylindrical coordinates $(r,\theta,z)$ what would the corresponding bcs on $B_r$, $B_{\theta}$ and $B_z$ be? The cylindrical setup for mhd problems is an idealized system to study angular momentum transport (with magnetorotational instability) in accretion discs. Hence relevant to this group.
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$\begingroup$ Is it not a simple coordinate transformation here? $\endgroup$– Kyle KanosCommented Sep 20, 2023 at 13:40
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If you are familiar with fluid mechanics, there is a natural analogy between the magnetic field $B$ and the velocity field $u$. The PI boundary boundary condition correspond to no slip boundary condition i.e. the vector field cancels at the boundary: $$ B\cdot n = 0 \\ B\times n = 0 $$ with $n$ the normal vector. Physically, the first equation is due to zero flux of $B$ from Maxwell's equation. The second equation is due to the zero surface currents since the boundary is insulating.
The PC boundary condition corresponds to the free slip with no penetration boundary condition i.e. the tangential component of the rotational of the field cancels as well as the normal component: $$ B\cdot n = 0 \\ (\nabla \times B)\times n = 0 $$ with $n$ the normal vector. The first equation is still justified the same way. The second equation comes from the tangential boundary condition for the electric field that you can express in terms of magnetic field using Ohm's law.
This general expression of the conditions thus generalise to any geometry of the boundary. For more information on the boundary conditions (mathematically well-posed etc.), and the equations of MHD in general, check out Mathematical Methods for the Magnetohydrodynamics of Liquid Metals by Gerbeau, Le Bris, and Lelièvre.
In the case of a cylindrical boundary, the PI boundary condition becomes: $$ B_r = B_\phi = B_z = 0 $$ The PC boundary condition becomes: $$ B_r = \partial_r B_z = \frac{1}{r}\partial_r(rB_\phi) = 0 $$
Hope this helps.
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$\begingroup$ Hi LPZ! Thank you for your response. Analogy to fluid mechanics NS and FS conditions makes it easy for me. The "magneto-" part in MHD is new to me. So to summarize, does the PC bc mean the normal component of the magnetic field together with the tangential components of the current vanish at the boundary? i.e., $B_r = 0$ and $J_t = (\nabla \times B) \cdot t = 0$. For PI, I know the currents are not allowed to enter or leave the domain. so $J_r = (\nabla \times B) \cdot r = 0$, r being the normal to the surface. $\endgroup$– myreshCommented Sep 20, 2023 at 18:31
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$\begingroup$ I don't think that your physical interpretation with currents is accurate. I've added how to justify the boundary conditions are usually justified. $\endgroup$– LPZCommented Sep 21, 2023 at 9:02
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$\begingroup$ Thank you very much, LPZ! Your help is greatly appreciated! I got a good grasp of the problem now! Also, thank you for the book reference! $\endgroup$– myreshCommented Sep 21, 2023 at 16:06