How well do finite volume methods ensure that no flux flows perpendicular to the flux vector?

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.

• Can you describe your boundary condition and how you implement it in your scheme. – cpraveen Jan 12 at 3:47
• 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. – Peanutlex Jan 12 at 10:15
• 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. – cpraveen Jan 13 at 3:59
• I did not make the code, however, looking at conduct.f90 it would seem that they calculate the flux as you describe. – Peanutlex Jan 13 at 20:14