Is Stokes equation a reduction of Navier-Stokes equations? The following Stokes problem:
$$\begin{cases}-\nu\Delta u+\nabla p=f&,\textrm{in }\Omega\\ \nabla\cdot u=0&, \textrm{in } \Omega\end{cases}$$
is a reduction of the Navier--Stokes equations?
$$\begin{cases}-\nu\Delta u+(\nabla u)\,u+\nabla p=f&,\textrm{in }\Omega\\ \nabla\cdot u=0&, \textrm{in } \Omega\end{cases}$$
If the answer is "yes", What is the difference from the physical point of view? That is, what represents the term $(\nabla u)\,u$?
 A: Yes it is a reduction in the sense that:
$$|\nu\Delta u|\gg|(\nabla u)\,u|$$
and therefore essentially the inertial term $(\nabla u)\,u$ is negligible.
This physically states that the viscous term $\nu\Delta u$ dominates over the inertial term. This is often quantified as the analysis by @Thomas shows as $Re\ll1$.
A: This is an expansion for small velocities, or to be more precise, an expansion in Reynolds number 
$$
\frac{u(\nabla u)}{\nu \nabla^2 u} \sim \frac{u}{\nu\nabla} 
\sim \frac{uL}{\nu} \sim {\it Re} .
$$
A: We write the Navier-Stokes equation in Cartesian coordinates $$\frac{\partial V_l}{\partial t}+\sum_{k=1}^3V_k\frac{\partial V_l}{\partial x_k}=-\frac{\partial p}{\rho \partial x_l}+\nu \sum_{k=1}^3\frac{\partial^2 V_l}{\partial x_k^2}$$ Multiply it by $a^3/Re_{cr}\nu^2$. Get the equation in a dimensionless form. $$\frac{\partial Re_l}{\partial \tau}+\sum_{k=1}^3Re_k\frac{\partial Re_l}{\partial y_k}=-\frac{\partial T}{\partial y_l}+Re_{cr}\sum_{k=1}^3\frac{\partial^2 Re_l}{\partial y_k^2}$$,where $x_l=y_l a/Re_{cr}$,$t=\tau a^2/Re_{cr}\nu$,$p=T\rho \nu^2/Re_{cr}a^2$,$Re_l=\frac{V_l a}{\nu}$
Derivatives in the dimensionless form have the same value, and the nonlinear term $\sum_{k=1}^3Re_k\frac{\partial Re_l}{\partial y_k}$ can be neglected in comparison with the Laplace operator $Re_{cr}\sum_{k=1}^3\frac{\partial^2 Re_l}{\partial y_k^2}$ in the case $Re_k<Re_{cr}$ of the laminar mode, when the Reynolds number of the stream $Re_k$ is less than the critical one $Re_{cr}$. In the case of a circular pipe, the critical Reynolds number is $Re_{cr}=2300$. In the case of the turbulent regime, the Reynolds number of the flow is greater than the critical one $Re_k>Re_{cr}$ and the non-linear term cannot be neglected.
