Divergence of Material Derivative Let $u : \Bbb{R}^n\times \Bbb{R}  \to \Bbb{R}^n\times \Bbb{R} $ be a divergence free vector field. Then the material derivative $D $  is given by: $$ \frac { \partial u_j}{\partial t}+\sum_{i=1}^{n} u_i \frac { \partial u_j}{\partial x_i} $$ Taking the divergence of $Du_j $ one obtains : $$\sum_{j=1}^{n}\sum_{i=1}^{n} \frac {\partial}{\partial x_j}u_i \frac { \partial u_j}{\partial x_i}=\sum_{j=1}^{n}\sum_{i=1}^{n}\left ( \frac {\partial u_i}{\partial x_j} \frac { \partial u_j}{\partial x_i}+u_i \frac { \partial^2 u_j}{\partial x_i \partial x_j}\right)$$ and rearranging terms, $$=\left(\sum_{j=1}^{n} \frac {\partial u_j}{\partial x_j}\right)^2 -\sum_{i\neq j} \left (\frac {\partial u_i}{\partial x_i}\frac {\partial u_j}{\partial x_j}-\frac {\partial u_i}{\partial x_j}\frac {\partial u_j}{\partial x_i}\right)+ \sum_{j=1}^{n}u_j \frac {\partial}{\partial x_j}\sum_{i=1}^{n}\frac {\partial u_i}{\partial x_i}$$ from which it follows that for divergence free vector fields $$div (Du_j)=-\sum_{i\neq j} \left (\frac {\partial u_i}{\partial x_i}\frac {\partial u_j}{\partial x_j}-\frac {\partial u_i}{\partial x_j}\frac {\partial u_j}{\partial x_i}\right)$$ which corresponds to a sum of Jacobian determinants. Is this correct?
 A: I have a few comments first, then a proposed solution. Firstly, you can't take the divergence of a scalar, so you really want to take the divergence of the material derivative of the vector $u$.
It is easier (for me) to avoid the summation signs, and use Einsteins summation notation instead. The material derivative is
$$ Du_j=\dot u_j+u_iu_{j,i}$$
where I'm using a dot to denote differentiation with respect to time, and $,i$ to represent partial differentiation with respect to $x_i$. Now, the divergence is
\begin{align} \left(Du_j\right)_{,j}=&\left(\dot u_j+u_iu_{j,i}\right)_{,j} \\
=&\dot u_{j,j}+u_{i,j}u_{j,i}+u_iu_{j,ij}\end{align}
and we can interchange the derivatives on the first term to write it as $$\frac{\partial u_{j,j}}{\partial t}$$
or, the time derivative of the divergence of $u$, which is zero if $u$ is divergence-free. Likewise, the last terms can be written as $u_i\left(u_{j,j}\right)_{,i}$ which is again zero because $u$ is solenoidal.
So we are left with 
$$\nabla\cdot Du=u_{i,j}u_{j,i}=\sum_{i,j}\frac{\partial u_i}{\partial x_j}\frac{\partial u_j}{\partial x_i}.$$
