Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|improve this question
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

3 Answers 3

$(u \cdot \nabla)u$ 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.

share|improve this answer

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.

share|improve this answer

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).

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.