Show that $ \nabla . ({\bf T}^T {\bf v}) = {\bf T} : \nabla {\bf v} + {\bf v} \cdot (\nabla . {\bf T})$

Question

Here is my feeble attempt:

$$ \begin{equation}\begin{aligned} \nabla .({\bf T}^T {\bf v}) & = \frac{\partial {\bf T}_{ji}}{\partial x_j} {\bf v}_i + {\bf T}_{ji} \frac{\partial {\bf v}_{i}}{\partial x_j} \\ & = (\nabla . {\bf T}^T) \cdot {\bf v} + {\bf T}^T \, \nabla {\bf v} \end{aligned}\end{equation} \\ $$

Source: page 2 of these lecture notes.

Answer 1

It's a matter of unwrapping the definitions very carefully. Even if you are already familiar with the objects, it is always convenient to go back to the definitions because different authors have conflicting conventions and this often results in different formulas.

From module 1 lecture 4, we have $R:S=R_{ij}S_{ij}$.

Also $Te_j=T_{ij}e_i$, so $$Tv={\textstyle\sum}Te_jv_j={\textstyle\sum}T_{ij}e_iv_j,$$ This means the components of $Tv$ are given by $(Tv)_i=T_{ij}v_j$. Thus we have $(T^tv)_i=T_{ji}v_j$.

From module 2 lecture 2, we have

• $\nabla G=\partial_kG_{ij}e_i\otimes e_j\otimes e_k$, so the compontents of $\nabla G$ are given by $$(\nabla G)_{ijk}=\partial_kG_{ij}$$ (the new index is put at the last position).
• $\nabla v = \partial_jv_ie_i\otimes e_j$, so the components of $\nabla v$ are given by $$(\nabla v)_{ij}=\partial_jv_i$$ (again, the new index is put at the last position).
• $\nabla \cdot G = \partial_jG_{ij}e_i$, so the components of $\nabla\cdot G$ are given by $$(\nabla\cdot G)_i=(\nabla G)_{ijj}=\partial_jG_{ij}$$ (the derivative is summed over with the second index of $G$).

Puting all of this together, we have $$\begin{aligned} \nabla \cdot (T^tv) &= \partial_i(T^tv)_i \\ &=\partial_i(T_{ji}v_j) \\ &=\partial_iT_{ji}v_j+T_{ji}\partial_iv_j \\ &=(\nabla\cdot T)_jv_j+T_{ji}(\nabla v)_{ji} \\ &=(\nabla\cdot T)\cdot v+T :\nabla v \end{aligned}$$

as you wanted. Hope this helps.