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I have to study this system which name is Navier-Stokes. Can you explain please what means that $p$, $u$ and $(u \cdot \nabla)u$. What represents in reality? Tell me please, how should I read the factor: $(u \cdot \nabla)u$? "$u$ multiplied with gradient applied to $u$ " ?

$ (N-S)\begin{cases} -\mu \Delta u +(u \cdot \nabla)u+\nabla{p}=f &\mbox{in } \Omega, \\ \mbox{div }u=0 & \mbox{in } \Omega,\\ u_{\mid{\Gamma}}=0. \end{cases} $

one more question, what happens with with the system if $(u \cdot \nabla )u=0$ ? I found that the system describe the motion of a incompressible viscous fluid and it suppose the the motion is stationary but no slow, what means that stationary and that slow?

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This is all explained in the Wikipedia article about the Navier-Stokes equations. Can you indicate what it is, that is not clear about this? – Bernhard Apr 7 '13 at 21:02
if $(u \cdot \nabla)u=0$ then the system won't be slow ? – Iuli Apr 7 '13 at 21:04
Can you restrict this to a single answerable question? – Sklivvz Apr 7 '13 at 21:13
That not the conventional Navier-Stokes system because it's missing the time derivative. – aberration Apr 25 '13 at 19:22
The time derivative is the (u⋅∇)u term. The partial derivative with respect to time is missing, which means the flow is steady. – jjack Aug 16 '15 at 19:18
up vote 2 down vote accepted

$(u \cdot \nabla)u$ is the so called advective acceleration term which arises when you consider the Navier-Stokes equations in an Eulerian frame of reference. It accounts for the effect that the we are following the particle as it moves around in the fluid, presumably to regions of the flow where the velocity is different. In contrast, if you consider the Navier-Stokes in Lagrangian coordinates, we are by definition tracking individual particles and therefore that term is not present. In large magnitudes, this term is highly-nonlinear and responsible for much of the more interesting behavior we see in fluid motion.

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The term $(u \cdot \nabla)u $ describes non-linear advective acceleration through a fixed point in the stationary frame of reference (Eulerian frame of reference). In your momentum equation, you need to multiply by this term by $\rho$ to have rate of change of momentum per unit volume.

If $(u \cdot \nabla)u = 0$ in your momentum equation, then you have Stokes flow (creeping flow). This applies for very low Reynolds numbers, i.e. $Re << 1$, where inertial forces are small compared with viscous forces.

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The answer to the second part: if $( u. \nabla) u =0$ would mean the flow is uniform in space , i.e. uniform flow. It can still have a time varying part.

x component of : $$( u. \nabla) u = u {du \over dx} + v {du \over dy} + w {du \over dz}$$ Similarly you would have y-component (in terms of v) and z-component (in terms of w).

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Should'n be partial derivatives? – aberration Apr 25 '13 at 19:23
Sorry about that. @aberration is right. I stand corrected. – jadelord May 3 '13 at 5:55

The Navier-Stokes equations represent the flow of a fluid with friction. They consist of the following conservation equations:

  1. Conservation of mass (second line):

    div u=0

  2. Conservation of momentum (first line)


  3. Conservation of energy (probably, in the third line, but I don't know the nomenclature):


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For the first part look up the difference between Lagrangian and Eulerian reference frames and also total derivative. Second bit read about the Reynolds number of a system.

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protected by Qmechanic Aug 16 '15 at 18:53

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