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.


2 Answers 2


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*}

  • $\begingroup$ Is $ \nabla \cdot (u \tensor u) = (\nabla \cdot u)u + (u\cdot \nabla) u$ equal to $2*(\nabla \cdot u)u$ ? $\endgroup$ Apr 7, 2015 at 14:47
  • 3
    $\begingroup$ @AndréAlmeida The term is missing because in incompressible flow $\nabla \cdot \vec{u} = 0$. $\endgroup$ Oct 6, 2017 at 14:01
  • $\begingroup$ @martini: Could you please tell us where can I find the definition "... the ith component of this vector is the divergence of the ith row ..."? Because there is others who say: it is the divergence of the column instead of the row. $\endgroup$
    – user265759
    May 21, 2019 at 15:05
  • $\begingroup$ @Navaro The divergence on $2$-tensors operates rowwise $\endgroup$
    – martini
    Jun 3, 2019 at 6:31
  • $\begingroup$ @martini: Any book any reference? $\endgroup$
    – user265759
    Jun 3, 2019 at 8:08

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


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