I have the following expression:
$\nabla \cdot (\vec{\mathbf{v}} \mathbf{A})$.
Where: $\vec{\mathbf{v}}$ is vector [1x3] and $\mathbf{A}$ a tensor [3x3]
I would like to know if the following identity holds:
$\nabla \cdot (\vec{\mathbf{v}} \mathbf{A}) = \vec{\mathbf{v}} \cdot \nabla \mathbf{A} + (\nabla \cdot \vec{\mathbf{v}} )\mathbf{A}$
If not, is there any identity for the divergence of a vector multiplied by a tensor?
Best regards
You can be sure that your answer's not quite correct because the second term is not a scalar. Let's compute it out in cartesian coordinates: \begin{align} \nabla\cdot(\mathbf{v}\mathbf{A}) &= \nabla\cdot(v_i A_{ij}\mathbf{e_j})\\ &= (v_i A_{ij})_{,j}\\ &= v_{i,j} A_{ij} + v_i A_{ij,j}\\ &= (v_{i,j}\ \mathbf{e_i} \otimes \mathbf{e_j}):\mathbf{A} + \mathbf{v} \cdot (A_{ij,j} \mathbf{e_i})\\ &= \nabla\mathbf{v} : \mathbf{A} + \mathbf{v} \cdot (\nabla\cdot\mathbf{A}) \end{align} Note that $\mathbf{e_i}$ are the $3 \times 1$ basis vectors for $\mathbb{R}^3$. In light of this, $\mathbf{v}\mathbf{e_i}$ is already an inner product, but I wrote $\mathbf{v} \cdot \mathbf{e_i}$ just to be extra clear there.
You also need to be a little bit careful about how you define the divergence and gradient of your vectors/tensors (some books may show this result off by a transpose), but hopefully the definitions I've used are clear from the line immediately before the final result.
