Confused how the expression like $ \nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v} $ is expanded

Question

I'm quite confused to figure out how this expression is expanded:

$ \nabla \cdot (\rho \textbf{v} \otimes \textbf{v}) \cdot \textbf{v} \stackrel{?}{=} \nabla \cdot [\rho (\textbf{v} \cdot \textbf{v}) \textbf{v}] - \rho \textbf{v} \cdot [(\textbf{v} \cdot \nabla) \textbf{v}] $

$\rho$ is a scalar (density), $\textbf{v}$ is a vector (velocity).

from: The Finite Volume Method in Computational Fluid Dynamics, page 60

Edit: Edited the title expression.

Edit2: Edited the right hand side

Answer 1

This is relatively straightforward to expand out in Cartesian index notation, using the product rule. We have \begin{align*} \nabla \cdot [\rho (\textbf{v} \cdot \textbf{v}) \textbf{v}] &= \partial_i \left( \rho v_j v_j v_i \right) \\ &= \partial_i \left[ \left(\rho v_j v_i\right) v_j \right] \\ &= \rho v_j v_i \partial_i v_j + \left[\partial_i \left( \rho v_j v_i\right)\right] v_j \\ &= \rho v_j (v_i \partial_i v_j) + \left[\partial_i \left( \rho v_i v_j\right)\right] v_j \\ &= \rho \mathbf{v} \cdot \left[ (\mathbf{v} \cdot \nabla) \mathbf{v}\right] + \left[\nabla \cdot (\rho \mathbf{v} \otimes \mathbf{v}) \right]\cdot \mathbf{v}. \end{align*} The requested identity then follows immediately by rearranging terms to the other side of the equation.

Answer 2

This follows from the identity

$$\nabla\cdot(A \otimes B) = (\nabla\cdot A)B + (A\cdot \nabla)B$$

see also: Divergence of a tensor product