Navier-Stokes equation (notation for convection term) Incompressible Navier-Stokes in vector notation is written as
$${\partial U \over \partial t}+(U\cdot\nabla)U =-\frac{1}{\rho} \nabla P + \nu \nabla^2(U),$$
where $U$ is velocity vector field  $U=(u,v)$, $\nu$ is the kinematic viscosity and $(U\cdot\nabla)U$ is the convection term. External forces are neglected.
In another notation, we have:
$${\partial u_i \over \partial t}+  u_j {\partial u_i \over \partial x_j}= -\frac{1}{\rho}{\partial p \over \partial x_i}+\nu \left({\partial^2 u_i \over \partial x_j\partial x_j}\right)$$
Along the $x$ direction the convection term is:
$$u\frac{\partial u}{\partial x}+ v\frac{\partial u}{\partial y}$$
What is written above is what I can understand. What I can't understand is the following notation, which I found in some sources:
$${\partial u_i \over \partial t}+ {\partial (u_i u_j) \over \partial x_j}= -\frac{1}{\rho}{_\partial p \over \partial x_i}+1/Re{\partial^2 u_i \over \partial x_j\partial x_j}$$
My main problem is the 'convection term'. In this notation, it seems that, along the $x$ direction, the convection term now  is:
$(∂ u^2/∂x)+ (∂ uv/∂y)$.
Under which assumptions we can write this equation and how we can reach it?
Also, the velocity used to calculate the Reynolds number $Re$ is $v$, $u$ or something else?
 A: Differentiating by the product rule:  $$\frac{\partial (u_iu_j)}{\partial x_j}=u_j\frac{\partial u_i}{\partial x_j}+u_i\frac{\partial u_j}{\partial x_j}$$But, for an incompressible fluid, the continuity equation reduces to $$\frac{\partial u_j}{\partial x_j}=0$$So, for an incompressible fluid, $$\frac{\partial (u_iu_j)}{\partial x_j}=u_j\frac{\partial u_i}{\partial x_j}$$
The version of the NS equation that includes the Reynolds number has been reduced to dimensionless form.
A: The two forms are strictly equivalent in the incompressible case:
$$
\nabla \cdot (u\otimes u) = (u\cdot \nabla)u+(\nabla\cdot u)u
$$
where $\nabla\cdot u=0$ when the fluid is incompressible.
The reason why this formulation is also natural is due to the convective derivative of momentum density. In general (beyond the incompressible case) the convective derivative for momentum density $\rho u$ is:
$$
\frac{D}{Dt}\rho u = \partial_t(\rho u)+\nabla \cdot (\rho u\otimes u)
$$
and is equivalent to the more usual form:
$$
\frac{D}{Dt}\rho u = \rho\left[\partial_t u+(u\cdot \nabla)u\right]
$$
when using conservation of mass:
$$
\partial_t\rho+\nabla \cdot (\rho u) = 0
$$
Hope this helps.
