# 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 $$(\nabla u)\,u$$?

• From the physical point of view, $(\nabla u)u$ gives acceleration of fluid. Stokes equation says that inertial force on the fluid particles due to its acceleration is negligible compared to other forces acting on it. – Deep Dec 13 '18 at 6:45

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 $$Re\ll1$$.
This is an expansion for small velocities, or to be more precise, an expansion in Reynolds number $$\frac{u(\nabla u)}{\nu \nabla^2 u} \sim \frac{u}{\nu\nabla} \sim \frac{uL}{\nu} \sim {\it Re} .$$
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 $$Re_k of the laminar mode, when the Reynolds number of the stream $$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.