While trying to simplify an equation in the integral form (in the context of finite volume method with polyhedral meshes), I came across this term:
$$ \iiint_V{\vec v\nabla\cdot\vec c\,\mathrm dV} \label{1}\tag{1} $$ where $\vec v$ and $\vec c$ are two vector fields. In the other hand, I have the following relation that holds for every control volume $V$ that is fixed (not changing over time): $$ \iiint_V{\nabla\cdot \vec c}\,\mathrm dV \equiv 0 \label{2}\tag{2} $$
My question: is it possible to simplify the expression \eqref{1}? Here is what I am doing:
- From \eqref{2}, we can get: $\nabla\cdot \vec c=0$ because the volume is arbitrary.
- Substituting that in (1), yields:
$$ \iiint_V{\vec v\nabla\cdot\vec c\,\mathrm dV} = \iiint_V{\vec v \cdot 0\,\mathrm dV} = 0 $$
Is that correct? I appreciate your help
If (2) holds for any volume element then $\nabla \cdot \vec{c}$ is identically zero and your claim is true. This is a standard behavior in transport equations (like Navier Stokes). However, you typically also have a term like $(\vec{v} \cdot \nabla) \vec{c}$ (coming from moving along the flow) which is non-trivial in this context.