Divergence of a vector tensor product/outer product: $ u \bullet \nabla u = \nabla \bullet (u \otimes u) $ I'm currently studying the derivation of the RANS (Reynolds Averaged Navier Stokes) equations, used in the study of turbulence, and I've stumbled upon a step wich I don't understand very well.
The problem is on the following identity (one of the steps of the formal derivation of the RANS equations, present in many online texts):
$$ u \bullet \nabla u = \nabla \bullet (u \otimes u) $$
, where $ u = \{ u \; v \; w \} $ is a 3D vector and $ u \otimes u $ is the tensor dot product / outer product of vector $u$ by itself:
$$ u \otimes u = 
        \begin{pmatrix}
        u^2 & uv & uw \\
        vu & v^2 & vw \\
        wu & wv & w^2 \\
        \end{pmatrix}
$$
My question is: where can I find a proof of the previous equality? 
When I try to derive it by hand I don't end up with the result above (I'm probably commiting an error somewhere).
Can someone derive it (in a simple way) here by hand?
I've searched online, but I haven't found anything this specific.
 A: 
You are missing a term, we have$\def\tensor{\otimes}$
  $$ \nabla \cdot (u \tensor u) = (\nabla \cdot u)u + (u\cdot \nabla) u$$

By definition of $\nabla \cdot (u \tensor u)$, the $i$-th component of this vector is the divergence of the $i$-th row of $u \tensor u$. So 
\begin{align*}
  [\nabla \cdot (u \tensor u)]_i &= \nabla \cdot (uu_i)\\
          &= (\nabla \cdot u)u_i + (u \cdot \nabla)u_i\\
          &= [(\nabla \cdot u)u + (u \cdot \nabla)u]_i
\end{align*}
A: You can use Einstein's summation
$$ \nabla \cdot (u \otimes u ) = \dfrac{\partial}{\partial x_j} (u_i u_j) = u_i \dfrac{\partial}{\partial x_j} (u_j) + u_j \dfrac{\partial}{\partial x_j} (u_i) = u(\nabla \cdot u) + u \cdot \nabla(u) $$
Using these rules:
Gradient of vector $\vec{V}$: $\boxed{\nabla V = \dfrac{\partial}{\partial x_j} (V_i)}$
Divergence of vector $\vec{V}$: $\boxed{\nabla \cdot V = \dfrac{\partial}{\partial x_j} (V_j)}$
Divergence of tensor(Rank 2) $V_{ij}$: $\boxed{\nabla \cdot V_{ij} = \dfrac{\partial}{\partial x_j} (V_{ij})}$
And in case of incompressible flow, $\nabla \cdot u = 0$
$\implies \nabla \cdot (u \otimes u ) = u \cdot \nabla(u)$
PS : In the derivation of RANS equations, we know the all-important term is the Reynolds Stress Term ($-\overline{u' \otimes u'}$). To derive this term, it's easier to use the conservation form $(\nabla \cdot (u \otimes u ))$ rather than the convective form $(u \cdot \nabla(u))$ although both are identical for incompressible flow
