Normally I write the incompressible Newtonian isothermal flow Navier--Stokes equations as follows:

$$\displaystyle \frac{\partial v}{\partial t} -\nu\Delta v +\color{red}{(\nabla v)v} +\frac{1}{\rho}\nabla p = f_b $$

But I'm reading now the followwing

$$\displaystyle \frac{\partial v}{\partial t} -\nu\Delta v +\color{red}{\nabla\cdot( vv)} +\frac{1}{\rho}\nabla p = f_b $$

Are both equations the same? I can't prove that are the same, that is why I'm thinking that are different equations (and is not different notation only).

  • 2
    $\begingroup$ Yes, that's the same thing, because of a vector calculus identity. $\endgroup$ Oct 27, 2020 at 20:22
  • $\begingroup$ thanks, but I don't understand the second operation in red, could you explain me what is $}nabla}cdot (vv)$ $\endgroup$
    – yemino
    Oct 27, 2020 at 21:13
  • 1
    $\begingroup$ After expanding the divergence of the outer product of two vectors and then making use of the incompressibility constraint, the convection term in the second equation simplifies to the one in the first equation. Check the subsection Divergence in the Wiki page for the details. en.wikipedia.org/wiki/Vector_calculus_identities $\endgroup$
    – Chenna K
    Oct 28, 2020 at 0:47

2 Answers 2


The operation in the second equation is the divergence of the dyadic product $vv$.The dyadic product of two vectors is a square matrix. In this case, a $3\times3$ matrix where $(vv)_{ij}=v_iv_j$.

The divergence of this product is given by $$\nabla\cdot(vv)=(\nabla\cdot v)v+v\cdot\nabla v$$ You can find the details of the vector identity here. The first term vanish by virtue of incompressibility constraint, $\nabla\cdot{v}=0$.


Tensorial notations are helpful here as an alternative--and an excellent way to prove-- vector calculus identities mentionned in other answers (if you can't remember them of the top of your head). The Navier-Stokes equations for the i-th component of $v$ are:

$$ \partial_t v_i + \sum_{1\le j \le 3}v_j \partial_j v_i = -\partial_i p + \sum_{1\le j \le 3}\partial_j\partial_j v_i$$

where I use the shorthand notations $\partial_t=\frac{\partial}{\partial t}$ and $\partial_i = \frac{\partial}{\partial x_i}$.

Incompressibility is: $$ \sum_{1\le j \le 3}\partial_j v_j = 0$$

Now let us use this notation to explicit the i-th component of the term in red: $$ \left\{\boldsymbol{\nabla \cdot }( \boldsymbol{vv}) \right\}_i =\boldsymbol{\nabla \cdot }(v_i \boldsymbol{v})= \sum_{1\le j \le 3} \partial_j\left( v_i v_j\right) $$

We can distribute the derivative, and use imcompressibility: $$ \left\{\boldsymbol{\nabla \cdot }( \boldsymbol{vv}) \right\}_i =v_i \underbrace{\sum_{1\le j \le 3} \partial_j v_j}_{=0} + \sum_{1\le j \le 3}v_j \partial_j v_i = \boldsymbol{v \cdot\nabla} v_i$$

This is how, for incompressible flows, advection can be written in these two different (but equivalent) forms. Written as a divergence, energy budgets over control volumes are then easily derived.


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