Divergence of the Gradient of a Vector Field
Formula
Let the vector function $\mathbf{u} : \mathbb{R}^{n} \to \mathbb{R}^{n}$ be smooth so that $\mathbf{u} \in C^{2} \left( \mathbb{R}^{n} \right)$. The divergence of the transpose $\left( \nabla \mathbf{u} \right)^{T}$ of $\mathbf{u}$ is given by $$ \nabla \cdot \left( \nabla \mathbf{u} \right)^{T} = \nabla \left( \nabla \cdot \mathbf{u} \right) $$
Explanation
This formula is used in the derivation of the Navier–Stokes equations to compute the divergence of the symmetric gradient.
The point of this post is to show concretely what $\nabla \left( \nabla \cdot \mathbf{u} \right)$ actually looks like, and to make clear in the derivation how the transpose $^{T}$ disappears.
Note that, unlike $\nabla \cdot \nabla \mathbf{u} = \nabla^{2} \mathbf{u}$, an assumption of being twice continuously differentiable is required. Physically this is typically satisfied, but it does not always hold, so bear that in mind.
Derivation
$$ \nabla \mathbf{u} = \nabla \left( u_{1} , u_{2} , u_{3} \right) = \begin{bmatrix} {\frac{ \partial u_{1} }{ \partial x_{1} }} & {\frac{ \partial u_{1} }{ \partial x_{2} }} & {\frac{ \partial u_{1} }{ \partial x_{3} }} \\ {\frac{ \partial u_{2} }{ \partial x_{1} }} & {\frac{ \partial u_{2} }{ \partial x_{2} }} & {\frac{ \partial u_{2} }{ \partial x_{3} }} \\ {\frac{ \partial u_{3} }{ \partial x_{1} }} & {\frac{ \partial u_{3} }{ \partial x_{2} }} & {\frac{ \partial u_{3} }{ \partial x_{3} }} \end{bmatrix} = \begin{bmatrix} \partial_{1} u_{1} & \partial_{2} u_{1} & \partial_{3} u_{1} \\ \partial_{1} u_{2} & \partial_{2} u_{2} & \partial_{3} u_{2} \\ \partial_{1} u_{3} & \partial_{2} u_{3} & \partial_{3} u_{3} \end{bmatrix} $$
Without loss of generality (../2720) we will perform the calculation by brute force only for the case $n = 3$. $\partial_{i}$ denotes the partial derivative operator with respect to the $i$-th variable. In fact, this sort of straightforward calculation is often more useful in fluid mechanics, etc.
$$ \begin{align*} \nabla \cdot \left( \nabla \mathbf{u} \right)^{T} =& \nabla \cdot \begin{bmatrix} \partial_{1} u_{1} & \partial_{1} u_{2} & \partial_{1} u_{3} \\ \partial_{2} u_{1} & \partial_{2} u_{2} & \partial_{2} u_{3} \\ \partial_{3} u_{1} & \partial_{3} u_{2} & \partial_{3} u_{3} \end{bmatrix} \\ = & \begin{bmatrix} \partial_{1} \partial_{1} u_{1} + \partial_{2} \partial_{1} u_{2} + \partial_{3} \partial_{1} u_{3} \\ \partial_{1} \partial_{2} u_{1} + \partial_{2} \partial_{2} u_{2} + \partial_{3} \partial_{2} u_{3} \\ \partial_{1} \partial_{3} u_{1} + \partial_{2} \partial_{3} u_{2} + \partial_{3} \partial_{3} u_{3} \end{bmatrix} \\ = & \begin{bmatrix} \partial_{1} \partial_{1} u_{1} + \partial_{1} \partial_{2} u_{2} + \partial_{1} \partial_{3} u_{3} \\ \partial_{2} \partial_{1} u_{1} + \partial_{2} \partial_{2} u_{2} + \partial_{2} \partial_{3} u_{3} \\ \partial_{3} \partial_{1} u_{1} + \partial_{3} \partial_{2} u_{2} + \partial_{3} \partial_{3} u_{3} \end{bmatrix} & \because \mathbf{u} \in C^{2} \left( \mathbb{R}^{n} \right) \\ = & \begin{bmatrix} \partial_{1} \left( \partial_{1} u_{1} + \partial_{2} u_{2} + \partial_{3} u_{3} \right) \\ \partial_{2} \left( \partial_{1} u_{1} + \partial_{2} u_{2} + \partial_{3} u_{3} \right) \\ \partial_{3} \left( \partial_{1} u_{1} + \partial_{2} u_{2} + \partial_{3} u_{3} \right) \end{bmatrix} \\ = & \nabla \left( \partial_{1} u_{1} + \partial_{2} u_{2} + \partial_{3} u_{3} \right) \\ = & \nabla \left( \nabla \cdot \mathbf{u} \right) \end{align*} $$