Let $\mathbf u$ and $\mathbf S$ be smooth fields with $\mathbf u$ vector valued and $\mathbf S$ tensor valued. I would like to prove the following identity:
$$\operatorname{div}\mathbf S\mathbf{u}=\mathbf S^T\cdot\operatorname{grad}\mathbf u+\mathbf u\cdot\operatorname{div}\mathbf S$$
My attempt:
\begin{align}
\operatorname{div}\mathbf S\mathbf{u}=\frac{\partial}{\partial x_i}(S_{ij}u_j)\mathbf{e}_j&=\frac{\partial S_{ij}}{\partial x_i}u_j\mathbf{e}_j+S_{ij}\frac{\partial u_j}{\partial x_i}\mathbf{e}_j\\
&=\frac{\partial S_{ij}}{\partial x_i}u_j\mathbf{e}_j+S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j\\
&=\mathbf u\cdot\operatorname{div}\mathbf S+S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j
\end{align}
Here, I have used the following definition: $\operatorname{div}\mathbf S=\frac{\partial S_{ij}}{\partial x_i}\mathbf{e}_j$.
But I am stuck on the last expression, $S_{ji}^T\frac{\partial u_j}{\partial x_i}\mathbf{e}_j$, since the definition of the gradient of a vector field is $\operatorname{grad}\mathbf u=\frac{\partial u_j}{\partial x_i}\mathbf{e}_i$.
I would appreciate any help or hint. Thank you.
Best Answer
Assuming $\mathbf{S}$ is a matrix and $\mathbf{u}$ is a vector, then $\operatorname{div}\mathbf{Su}$ is a scalar. Hence, we have \begin{align} \operatorname{div}\mathbf{Su} = \frac{\partial}{\partial x_i}(S_{ij}u_j) = \frac{\partial S_{ij}}{\partial x_i}u_j+S_{ij}\frac{\partial u_j}{\partial x_i}. \end{align}
Note that \begin{align} \frac{\partial S_{ij}}{\partial x_i}u_j=\frac{\partial S_{ij}}{\partial x_i}\mathbf{e}_j\cdot u_j\mathbf{e}_j=\mathbf{u}\cdot \operatorname{div}\mathbf{S}. \end{align} Next, observe \begin{align} S_{ij}\frac{\partial u_j}{\partial x_i}= \sum_j \sum_i S_{ij}\frac{\partial u_j}{\partial x_i} = \sum_j \sum_i S_{ji}^Tu_{j, i} = \mathbf{S}^T\cdot \operatorname{grad}\mathbf{u}. \end{align} Here we used the fact that \begin{align} \text{d}\mathbf{u}= (\operatorname{grad}\mathbf{u})^T \end{align} otherwise, we would have the expression \begin{align} \text{d}\mathbf{u}\cdot \mathbf{S}. \end{align}