divergence of $[\hat\epsilon.\vec{E}]$?

Question

What is the identity of divergence of $[\hat\epsilon.\vec{E}]$, where $\hat\epsilon$ is tensor 3x3, and E is vector 1x3

$\nabla.[\hat\epsilon.\vec{E}]=?$

$\nabla.[\phi\vec{E}]=\phi\nabla.\vec{E}+\vec{E}.\nabla\phi $ for scalar $\phi$

Answer 1

The product of tensors can be expanded as follow， $$\hat\epsilon\cdot\vec{E}=\begin{pmatrix} \epsilon_{11}\hat x \hat x&\epsilon_{12}\hat x \hat y &\epsilon_{13}\hat x \hat z\\ \epsilon_{21}\hat y \hat x&\epsilon_{22}\hat y \hat y &\epsilon_{23}\hat y \hat z\\ \epsilon_{31}\hat z \hat x&\epsilon_{32}\hat z \hat y &\epsilon_{33}\hat z \hat z \end{pmatrix} \cdot \begin{pmatrix} E_1 \hat x\\E_2 \hat y\\E_3 \hat z\end{pmatrix}=\begin{pmatrix} (E_1 \epsilon_{11}+E_2 \epsilon_{12}+E_3 \epsilon_{13} )\hat x\\ (E_1 \epsilon_{21}+E_2 \epsilon_{22}+E_3 \epsilon_{23} ) \hat y\\ (E_1 \epsilon_{31}+E_2 \epsilon_{32}+E_3 \epsilon_{33} ) \hat z\end{pmatrix} $$ then we can get $$\nabla \cdot [\hat\epsilon \cdot \vec{E}]=\sum_{j=1,2,3} ( \frac{\partial E_j}{\partial x}\epsilon_{1j}+\frac{\partial \epsilon_{1j}}{\partial x}E_j)+\sum_{j=1,2,3} ( \frac{\partial E_j}{\partial y}\epsilon_{2j}+\frac{\partial \epsilon_{2j}}{\partial y}E_j)+\sum_{j=1,2,3} ( \frac{\partial E_j}{\partial z}\epsilon_{3j}+\frac{\partial \epsilon_{3j}}{\partial z}E_j)$$

If $\hat\epsilon$ is not a function of position$(x,y,z)$, then $$\frac{\partial \epsilon_{1j}}{\partial x}=\frac{\partial \epsilon_{2j}}{\partial y}=\frac{\partial \epsilon_{3j}}{\partial z}=0 $$ and the expressions can be simplified as $$\nabla \cdot [\hat\epsilon \cdot \vec{E}]=\sum_{j=1,2,3} \frac{\partial E_j}{\partial x}\epsilon_{1j}+\sum_{j=1,2,3} \frac{\partial E_j}{\partial y}\epsilon_{2j}+\sum_{j=1,2,3} \frac{\partial E_j}{\partial z}\epsilon_{3j}$$