-1
$\begingroup$

Finite volume methods find approximate solutions to equations of the form: $$\frac{\partial \vec{u}}{\partial t}+\nabla\cdot(\vec{f}(\vec{u}))=0.$$ My question is has anyone done any analysis on how well finite volume methods ensure that no flux flows in directions perpendicular to $\vec{f}(\vec{u})$. I ask because I am using a piece of MHD (magnetohydrodynamic) code to study thermal conduction. The conduction is described by: $$\rho\frac{\partial \epsilon}{\partial t}=\nabla\cdot\left(\kappa_0 T^{5/2}\frac{\vec{B}\cdot\nabla T}{B^2}\vec{B}\right)=\nabla\cdot\vec{q},$$ where $\rho$ is the density, $\epsilon$ is the internal energy density, $T$ is the temperature, $\vec{B}$, is the magnetic field, $\kappa_0$ is the thermal conductivity and $\vec{q}$ is the heat flux. It is important that the heat flux, $\vec{q}$, flows only parallel to the magnetic field, which it seems to do pretty well except for when I impose solid, no slippage boundary conditions.

$\endgroup$
  • $\begingroup$ Can you describe your boundary condition and how you implement it in your scheme. $\endgroup$ – cpraveen Jan 12 at 3:47
  • $\begingroup$ I follow the solid boundary conditions described on page 434 of the following book. The code I use is linked here. Basically, the boundary conditions set the velocity equal to zero and the ghost cells are set to mirror the domain. The code uses a staggered grid where the velocity is defined at the cell corner and all other variables are defined at the centre, so only the velocity lies on the boundary. $\endgroup$ – Peanutlex Jan 12 at 10:15
  • $\begingroup$ How do you calculate boundary flux for the internal energy equation ? If $\vec{n}$ is outward normal vector at boundary, then your boundary flux is $\kappa_0 T^{5/2} \frac{\vec{B}\cdot\nabla T}{B^2} \vec{B} \cdot \vec{n}$. You need to estimate $\nabla T$ at the boundary face. Then your flux should have the correct behaviour. $\endgroup$ – cpraveen Jan 13 at 3:59
  • $\begingroup$ I did not make the code, however, looking at conduct.f90 it would seem that they calculate the flux as you describe. $\endgroup$ – Peanutlex Jan 13 at 20:14

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.