Talk:Navier–Stokes equations/Archive 1

message: If you look carefully you will see that that the left hand side of the equation contains a time derivative of a scalar field, which is a scalar field (i.e. has a dimension of 1)
:\frac{D\rho}{Dt} 
plus a space derivative of a vector field, which is a tensor field (i.e. has the dimension of a 3x3 matrix)
:\rho\nabla\mathbf{v} 
Therefore, dimensions in the terms of the equation do not agree, i.e.,  it is not possible to add a scalar to a matrix. The problem seems to be that one of the  authors  is under the impression that the continuity equation is derived from the substantial derivative expression. The fact is that the continuity equation is derived from considering a volume fixed in space (does not move with the fluid). Therefore changes in mass inside  such a volume are stated in the following way:
:\int_V\frac{d\rho}{dt}dV =\int_A \rho\mathbf{v\cdot n}dA
the sign of the right hand side depends on the direction of the surface normal n. normally
it's because n pointing away from the surface
:\int_V\frac{d\rho}{dt}dV = - \int_A \rho\mathbf{v\cdot n}dA
Where V is the control volume and A its area; :\rho is the density of the fluid and \mathbf{v} its velocity. The left hand side represents changes in density inside the volume whereas the right han side represents advective transport of mass (mass carried out of the boundary of the volume by the moving fluid). This equation states simply that the density of the fluid inside the volume must change (expand or compress) in response to either loss or addition of mass through the boundaries of the volume. Now applying the divergence theorem we have
:\int_V(\frac{d\rho}{dt} - \nabla\cdot(\rho\mathbf{v}) ) = 0
Where the differential operator :\nabla\cdot() is the divergence. Eliminating the integral, one finally finds 
:\frac{d\rho}{dt} - \nabla\cdot(\rho\mathbf{v})= 0
This equation is the "continuity equation" for a moving fluid (see eq. 1.2 in p. 2 of  Landau L.D., and Lifshitz E.M., 1987, Fluid Mechanics, 2nd Edition, Pergamon Press, pp. 539).  Now,  the divergence of a vector field :\nabla\cdot(\rho\mathbf{v}) is a scalar field. Thus dimensions agree for all terms in the equation. 
If density is constant, the  continuity equation is reduced  to
:\nabla\cdot(\mathbf{v})= 0 
-----------------
The equation pretending to be the Navier-Stokes equation contains two terms of different physical dimension, rho u and rho d/dt(rho). This cannot be right.
-----------------
Formula removed from article, thus:
: -\nabla p + 
\mu \left( \nabla^2 \mathbf{u} + {1 \over 3} \nabla (\nabla \cdot \mathbf{u} ) \right) +
\rho \mathbf{F}
= \rho \left( { \partial\mathbf{u} \over \partial t }  +
\mathbf{u} \cdot \nabla \mathbf{u}
\right) 
Equations without defined variables and context are useless.
Note that the article in its current form gives the impression that there is "a Navier-Stokes equation", rather than a set of them: we should either give them all, or none of them.
-- The Anome 11:05, 23 Dec 2003 (UTC)
replies: text: Look, I don't want to sound nasty or anything, but
* the variables in this article are self-consistant
* they are defined
* the article does state that NS are a 'system' of equations and that many specific form are derived from the general form, of which there is a single form.CyrilleDunant 16:38, 25 August 2005 (UTC)
thread_title:  The continuity equation is curved correct 

message: : -\nabla p + 
\mu \left( \nabla^2 \mathbf{u} + {1 \over 3} \nabla (\nabla \cdot \mathbf{u} ) \right) +
\rho \mathbf{u}
= \rho \left( { \partial\mathbf{u} \over \partial t }  +
\mathbf{u} \cdot \nabla \mathbf{u}
\right) 
where:
:p is pressure
:\mathbf{u} is fluid velocity
:\mu is viscosity
:\rho is density
Problems include: no sign of which one this is, no derivation or motivation, and what's with the factor of one-third? -- The Anome 19:28, 19 Jul 2004 (UTC)
thread_title:  Removed equation again 

message: The equaitons are correct (yes, the units are correct) except the div(u) term is zero, because constant density is assumed for the "official" Navier-Stokes equations. Also, to address the other concern, this is not one equation, it is a set of equations. u is a vector and has three components. This set of equations should be added to the text with the variables defined.
:The problem is that there is no single "official" version of the Navier-Stokes equation or equations. Different ways of posing the problem (streamline/turbulent, incompressible/compressible...) yield different but related sets of equations. Some people prefer to write everything out in components, others like vectors. Some like the Reynolds-approximation version, some want to do the whole thing right down to thermal terms and shearing. There are lots of different choices for variable names. Just pulling an  equation out of a hat and saying "this is the Navier-Stokes equation" is not useful, because it does not provide insight for the reader into what the equation(s) mean(s). What would be useful would be a derivation from scratch, and then some special case versions for simple cases. See some of the references, particularly  for a good example of this.
:Just for comparison, the reference above gives the (simplified) Navier-Stokes equations as:
::{{\partial \rho} \over {\partial t}} + \nabla \cdot ( \rho \mathbf{u}) = 0
::{ \partial  \mathbf{u} \over \partial t} + (  {\mathbf{u}}\cdot \nabla )  \mathbf{u} = -{1 \over \rho} \mathbf{ \nabla}p -  \mathbf{ \nabla}\phi + {\mu \over \rho } \nabla ^2  {\mathbf{u}},
::\rho \left( { \partial \varepsilon \over \partial t} +  {\mathbf{u}}\cdot \nabla \varepsilon \right) - \nabla \cdot ( K_H \nabla T) + p \nabla \cdot  {\mathbf{u}}= 0.
:-- The Anome 07:45, 21 Jul 2004 (UTC)
replies: text: Is a full derivation appropriate here?  That seems to be more of a wikibooks thing.  However, if we really want one, I'm sure I can cobble one up.  -- Kaszeta 21:16, 28 Oct 2004 (UTC)
thread_title:  Why are the equations missing? 

message: Perhaps the expanded forma ought to simply die. They are illegible, and apparently are there only to make the following point : "look! those equations are horrible !"
: I would second that. All it does is drawn the physical meaning of the equations into an indigest ocean of trivial calculations. Rama 12:58, 1 September 2005 (UTC)
thread_title:  This article nedds quite some editing 

message: Outside of these mathematical concerns, I think it would be very nice to have some historical background on the equations. When were they v��
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tUser:CyrilleDunant|CyrilleDunant]] 21:45, 20 November 2005 (UTC)
thread_title:  Historical background and actual relevance 

message: Never too much, needs much expanding
Sorry, I'm a fresh man here,and not able to use this systems here very much. Well,I decide to apology to all men. I will try my best to learn.
For the followings,I would tell something to discuss. '''''
《Quote:》
Original subject:
              The continuity equation is flat wrong 
If you look carefully you will see that that the left hand side of the equation contains a time derivative of a scalar field, which is a scalar field (i.e. has a dimension of 1)
:\frac{D\rho}{Dt} 
plus a space derivative of a vector field, which is a tensor field (i.e. has the dimension of a 3x3 matrix)
:\rho\nabla\mathbf{v} 
Therefore, dimensions in the terms of the equation do not agree, i.e., it is not possible to add a scalar to a matrix. The problem seems to be that one of the authors is under the impression that the continuity equation is derived from the substantial derivative expression. The fact is that the continuity equation is derived from considering a volume fixed in space (does not move with the fluid). Therefore changes in mass inside such a volume are stated in the following way:
:\int_V\frac{d\rho}{dt}dV =\int_A \rho\mathbf{v\cdot n}dA
 
Where V is the control volume and A its area; :\rho is the density of the fluid and \mathbf{v} its velocity.:
《Reply:》
Sorry but I'd deleted the other words of your talking about. I 
recently just use 
:\frac{D}{Dt}=\frac{\partial}{\partial t}+{\mathbf u}
\cdot\nabla
to correct to 
        \frac
{D\rho}{Dt}  which is a scalar field (it's a vector one)
Actually, you would see that it(the math formula) does not 
belong scarlar field,but vector one does. Because of
:\frac{D}{Dt}u=\frac{\partial{u}}{\partial t}+......
(we just take the front terms to analyse) which is operated to acceralations. Hence it is not a scalar field.
Totally I want to say is :
           \frac{D}{Dt}{\rho}*u=\frac{\partial{{\rho}*u}}{\partial t}+...... refers to vectors * vectors = tensors
I'm from Taiwain. If I have any mistake opinions, tell me. Thank you.
:Well... The Continuity equation is 
:\frac{D\rho}{Dt}  = \frac{\partial \rho}{\partial t} + \mathbf{v}\cdot\nabla\rho  = 0  
in indicial notation:  
:\frac{\partial \rho}{\partial t} + v_i\frac{\partial \rho}{\partial x_i} \| x \in \{1,2,3\} 
which is all scalar. The original discussion point was wrong :)CyrilleDunant 20:23, 19 December 2005 (UTC)
thread_title:  Too much sorry 

message: I find it hard to believe that the motion of stars in a galaxy would be governed by Navier-Stokes.
Keithdunwoody 01:51, 18 February 2006 (UTC)
replies: text: in the same way that a fluid is composed of particles that don't actually touch themselves, but interact through electro-magnetic forces, the stars in a galaxy interact through gravity. This creates a form of viscosity. Note that for accurate modeling, you would have to add relativistic behaviour. CyrilleDunant 09:43, 18 February 2006 (UTC)
replies: text: I'd be interested to read more about this.  Do you have some references?  All I could find on google was either 1) Navier-Stokes for protoplanitary systems, 2) University speaker series where both Navier-Stokes and galaxies were mentioned, but not in the same talk, or 3) Non-linear PDE course webpages.  Thanks, 154.20.174.63 17:04, 18 February 2006 (UTC)
replies: text: Well, aside from the size differences, a proto-planetary system and a galaxy would be pretty similar, as far as dynamics are concerned :) But no, I don't have references.CyrilleDunant 18:45, 18 February 2006 (UTC)
thread_title:  Stars in a galaxy 

message: From the article:
The equations can be converted to Wilkinson equations for the secondary variables vorticity and stream function. Solution depends on the fluid properties (such as viscosity, specific heats, and thermal conductivity), and on the boundary conditions of the domain of study.
I tried to make a Wikilink to Wilkinson equations, but there is no Wiki-page. I tried to search wikipedia,
but there is no reference to Wilkinson equations. I tried to let Google search for Wilkinson equations,
but still no useful results...
Is this Wilkinson important? If so, one should make a Wikipage, if not, it shouldn't be mentioned... Pie.er 09:59, 28 April 2006 (UTC)
thread_title:  Wilkinson equations 

message: It should perhaps be mentioned that $1,000,000 prize problem concerns the incompressible Navier-Stokes equations.
:Just noticed that the article has two references to the Clay Institute Millenium prize problem, one in the intro section and another in the turbulence section.  One should probably be removed. --Chetvorno 01:22, 3 October 2007 (UTC)
thread_title:  $1,000,000 prize problem 

message: I just stumbled over a brand new paper:
* http://comet.lehman.cuny.edu/sormani/others/SmithNavierStokes.html
* http://www.math.columbia.edu/~woit/wordpress/?p=470
I dont even understand the problem, but im sure someone here will be able to check if its relevant. MillKa 10:27, 6 October 2006 (UTC)
replies: text: I think we should stick to peer-reviewed papers. User A1 (talk) 14:06, 26 April 2010 (UTC)
thread_title:  New Paper 

message: Quoting the Inertia article:
 Physics and mathematics appear to be less inclined to use the original concept of inertia as
 "a tendency to maintain momentum" and instead favor the mathematically useful definition of
 inertia as the measure of a body's resistance to changes in momentum or simply a body's inertial mass.
 
 [...]
 
 This meaning of a body's inertia therefore is altered from the original meaning as
 "a tendency to maintain momentum" to a description of the measure of how difficult it
 is to change the momentum of a body.
A "resistance to changes in momentum" is exactly what the left side of the NS eqns contains. I think that it's ok that inertia isn't technically a quantity, the left side of the eqn represents inertia just fine. I can't think of a better phrase and I don't think anyone else can, hence absurdities like "force density" (which is something else) or "quantity of motion" (which I've tried looking up and am now very confident that it was made up on the spot).
In science we use non-quantities all the time when it's logical to do so. For example,
:\lim_{t \to \infty} t = \infty
is technically very incorrect since infinity is not a number (though it can be considered an "element". See extended real numbers). Everyone uses it anyway since it'd be impractical and confusing to do something else.
I've asked three fluid mechanics junkies in the old ME department about this. They say it's called inertia.
-Ben pcc 16:29, 12 April 2007 (UTC)
replies: text: I resent that, it was not made on the spot, it came from a bad translation :) . It was, however, wrong. The quantity on the left has indeed the dimension of inertia (a tensor of Newtons per metre cube) -- inertia is actually a quantity, a very useful one at that . I apologise for my unfortunate edit. CyrilleDunant 17:06, 12 April 2007 (UTC)
thread_title:  Inertia 

message: Alot of people have been suggesting that this page be split. So it's done. I've made a subarticle concerning the derivation and other "non encyclopedic" details which I think are important but indeed out of place in the main article. Compare this to mathematic proofs, which use this cool template: Template:Proof.
:I had structured it like that a long time ago, and it seemed a good idea. The logic behind it was that it is not so much a derivation (well, ok the part about the derivatives and control volume is) as a description of the different component of the equations (all the conservation laws). Otherwise, you are left with a system of equation and a bunch of simplified forms. 
:On the other hand, if you feel it is better split, well, I am not going to fight you :)CyrilleDunant 07:06, 14 April 2007 (UTC)
replies: text: I think it is a terrible idea: the substantive derivative is really the time derivative in lagrangian coordinates -- in any case it is central to the derivation and understanding of NS.CyrilleDunant 07:06, 14 April 2007 (UTC)

text: OK, have fun CyrilleDunant 07:06, 14 April 2007 (UTC)

text: I think that the "See Also" section should be properly expanded, there are many other related articles on this matter. Anyway, I find introduction very nice.Xerenio (talk) 18:58, 23 May 2008 (UTC)
thread_title:  Article splitted 

message: Solid mechanics does not dictate position, please stop saying that it does. It dictates displacement; the difference is important. Wikipedia doesn't have a lot of info on the equations of solid mechanics, but have look in any introductory (structural) finite element analysis text; you'll find equations describing displacement. Solid mechanics, as with fluid mechanics and rheology, can't appropriately describe position (well, it can, but it sucks, and that's no different for fluid mechanics); even the wave equation honestly describes displacement.
On the other hand, look up classical mechanics, for example this, this, this, or this; these include Newtonian, Hamiltonian, and Lagrangian formulations of classical mechanics and they almost always lead to finding the position. It's true that NS is classic, but it's not true that classical mechanics MUST deal with position, or that "solid mechanics" can substitute "classic".  —Preceding unsigned comment added by Ben pcc (talk • contribs) 19:27, 19 September 2007 (UTC)
thread_title:  Classical mechanics vs Solid mechanics 

message: What is this convective flow that is mentioned in this article (e.g. in the badly written section on non-linearity)? ...and is it differnet from any fluid flow?  Tt261 (talk) 12:54, 5 April 2008 (UTC)
replies: text: Convective flow means that momentum convection has a significant influence on the dynamics of the flow. In, for instance, acoustics the convective terms may in most cases be negligible, compared to temporal acceleration and pressure gradients. Crowsnest (talk) 23:19, 5 April 2008 (UTC)

text: Stokes flow is another limit of the Navier-Stokes equations where convection may be neglected, so that is another example of a fluid flow which is not a convective flow. Etc. Crowsnest (talk) 23:22, 5 April 2008 (UTC)
replies: text: Ah, thanks. Perhaps "very low Re flow" wld make it clearer.  Is it not also better to call it "advective flow" to make it clear it does not refer specifacally to flows with thermal convection currents?  86.159.211.92 (talk) 12:29, 8 April 2008 (UTC)
replies: text: Actually, in wave motion it is not the Reynolds number which is determining the relative importance of convection terms as compared to temporal acceleration, but the Keulegan-Carpenter number (no WP article here, yet): KC = UT/L, with U a characteristic velocity amplitude, T the oscillation period and L a characteristic length scale. This dimensionless number is in the context of oscillatory flow around bluff bodies: I am not aware of other names of such a dimensionless number for other types of oscillatory flow. Even at large Reynolds numbers, e.g. O(106), but low-enough KC number, convective terms may be neglected. Crowsnest (talk) 12:41, 8 April 2008 (UTC)

text: "Very low Re"? I think you mean very high. High Re flows are called inertial flows. The phrase "convective flow" refers to a flow where the inertial change in velocity is important, even if it isn't very high Re. The radial flow at the bottom of the article is an example of this. Even for Re barely above zero, the convective acceleration becomes important. — Ben pcc (talk) 00:35, 10 April 2008 (UTC)
replies: text: Is "barely" not just another phrase for "very low"? Crowsnest (talk) 07:52, 10 April 2008 (UTC)
replies: text: Yep. — Ben pcc (talk) 17:56, 10 April 2008 (UTC)
thread_title:  Convective Flow 

message: The Navier–Stokes equations are nonlinear partial differential equations in almost every real situation (exceptions include one dimensional flow and creeping flow). The nonlinearity makes most problems difficult or impossible to solve and is part of the cause of turbulence.
Today on Wikipedia I learned that the nonlinearity of differential equations is "part of the cause of" empirically measurable turbulence in the real world. Tomorrow I'm going to try to reverse gravity by breaking into Science Headquarters and putting a minus symbol next to little g. Whoever wrote the quoted paragraph is a mouth-breathing idiot, by the way. --75.49.223.247 (talk) 13:58, 28 June 2008 (UTC)
thread_title:  Amazing! 

message: Exchanged phi and theta in the spherical and cylindrical equations. It is more common to write phi as the azimuthal angle and theta as the polar angle than vice-versa.  —Preceding unsigned comment added by 130.243.230.54 (talk) 11:21, 2 November 2008 (UTC)
replies: text: It seems to me not all terms have been changed consistently. Are the equations still correct in spherical coordinates? -- Crowsnest (talk) 20:03, 2 November 2008 (UTC)
thread_title:  phi and theta 

message: User:Headbomb suggested this merger.
I Support it; the article is small and would be a better section of the N-S article than a standalone article, I think. Its page could be a redirect to the section of N-S after the merger. Awickert (talk) 18:08, 26 December 2008 (UTC)
replies: text: I am in doubt. Google Scholar finds 11 references to "Wyld diagrams". So it is a very, very specialised topic (but notable). But if included here, they would get an undue weight. Maybe someone will extend this stub in the future. So a weak oppose. -- Crowsnest (talk) 14:56, 8 March 2009 (UTC)
thread_title:  Merge [[Wyld diagrams]] into N-S equations 

message: I don't understand this sentence. Is this supposed to mean that the nonlinearity is the cause of the turbulence? Shouldn't it be the other way round, that turbulence in the real world leads to nonlinearity in the model? Nczempin (talk) 14:22, 28 April 2009 (UTC)
thread_title:  The nonlinearity ... is the main contributor to the turbulence that the equations model. 

message: In Navier–Stokes equations#Derivation and description the article states:
:"...the most general form of the Navier–Stokes equation ends up being:
::\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}\right) = -\nabla p + \nabla \cdot\mathbb{T} + \mathbf{f},\ldots 
Apart from not being in conservation form, so not being "the most general form", it gives the equations of motion before any stress model (relating fluid motion and stresses) is introduced. It is so general that it can be used, if you like to, to describe the motion of any substance either a fluid or solid or whatever other phase of matter in a single-phase continuum description. So these cannot be the Navier–Stokes equations, and no text book does call them so.
According to Batchelor (1967) pp. 147-148; Landau & Lifshitz (1987), 2nd ed., pp. 44-49; Doering & Gibbon (1995) pp. 14-15, the Navier–Stokes equations are the  equations resulting after the inclusion of a specific stress model for fluids: the shear stresses are proportional to the velocity gradient components ∂vi/∂xj and satisfying certain symmetry condititions (Batchelor, pp. 141-147). -- Crowsnest (talk) 09:40, 16 May 2009 (UTC)
replies: text: I think that this form is still useful, though I pretty much agree with Crowsnest. Maybe we should start the article with the equation for a predetermined (maybe Newtonian) rheology, and then have a derivation section where we say "Look! Equation (from above)!" then "Apply rheology - poof!". Awickert (talk) 09:51, 16 May 2009 (UTC)
replies: text: This equation is called by Batchelor simply the 'equation of motion' or 'momentum balance' equation, p. 137.  Doering & Gibbon, p. 14 call it the 'most general form of the equation of motion'. It is just the expression of Newton's 2nd law for any continuum (apart from the cosmetic distinction between pressure and deviatoric stress). -- Crowsnest (talk) 10:11, 16 May 2009 (UTC)

text: The main question is: which fluids are still covered by the Navier–Stokes equations. Newtonian fluid description of course is, but which non-Newtonian fluids are? Quite often, the equations of motion for fluids with some viscosity description of the deviatoric stresses are labelled as Navier–Stokes equations. Where viscosity may be isotropic or a non-isotropic tensor. -- Crowsnest (talk) 10:24, 16 May 2009 (UTC)
thread_title:  What are the Navier–Stokes equations 

message: Wiki: Navier–Stokes existence and smoothness
Search Wikipedia!
The Navier-Stokes equations are one of the pillars of fluid mechanics. These equations describe the motion of a fluid (that is, a liquid or a gas) in space. Solutions to the Navier-Stokes equations are used in many practical applications. However, theoretical understanding of the solutions to these equations is incomplete. In particular, solutions of the Navier-Stokes equations often include turbulence, which remains one of the greatest unsolved problems in physics despite its immense importance in science and engineering.
Even much more basic properties of the solutions to Navier-Stokes have never been proven. For the three-dimensional system of equations, and given some initial conditions, mathematicians have not yet proved that smooth solutions always exist, or that if they do exist they have bounded kinetic energy. This is called the Navier-Stokes existence and smoothness problem.
Since understanding the Navier-Stokes equations is considered to be the first step for understanding the elusive phenomenon of turbulence, the Clay Mathematics Institute offered in May 2000 a US$1,000,000 prize, not to whoever constructs a theory of turbulence but (more modestly) to the first person providing a hint on the phenomenon of turbulence. In that spirit of ideas, the Clay Institute set a concrete mathematical problem: [1]
Prove or give a counter-example of the following statement: In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier-Stokes equations.
Contents:
1. The Navier-Stokes equations
2. Two settings: unbounded and periodic space
3. Statement of the problem in the whole space
4. Statement of the periodic problem
5. Partial results
6. Notes
7. References
8. External links
Millennium Prize Problems 
P versus NP 
The Hodge conjecture 
The Poincaré conjecture 
The Riemann hypothesis 
Yang-Mills existence and mass gap 
Navier-Stokes existence and smoothness 
The Birch and Swinnerton-Dyer conjecture 
1. The Navier-Stokes equations
Main article: Navier-Stokes equations
In mathematics, it is a system of nonlinear partial differential equations for abstract vector fields of any size. In physics and engineering, it is a system of equations that models the motion of liquids or not-rarefied gases using continuum mechanics. The equations are a statement of the second law of Newton, with the forces modelled according to those in a viscous Newtonian fluid — as the sum of contributions by pressure, viscous stress and an external body force. Since the setting of the problem proposed by the Clay Mathematics Institute is in three dimensions, for an incompressible and homogeneous fluid, we will consider only that case.
Let  be a 3-dimensional vector, the velocity of the fluid, and let  be the pressure of the fluid. [note 1] The Navier-Stokes equations are:
where  is the kinematic viscosity,  the external force,  is the gradient operator and  is the Laplacian operator, which is also denoted by . Note that this is a vector equation, i.e. it has three scalar equations. If we write down the coordinates of the velocity and the external force
then for each  we have the corresponding scalar Navier-Stokes equation:
The unknowns are the velocity  and the pressure . Since in three dimensions we have three equations and four unknowns (three scalar velocities and the pressure), we need a supplementary equation. This extra equation is the continuity equation describing the incompressibility of the fluid:
Due to this last property, the solutions for the Navier-Stokes equations are searched in the set of "divergence-free" functions. For this flow of a homogeneous medium, density and viscosity are constants.
2. Two settings: unbounded and periodic space
There are two different settings for the one-million-dollar-prize Navier-Stokes existence and smoothness problem. The original problem is in the whole space , which needs extra conditions on the growth behavior of the initial condition and the solutions. In order to rule out the problems at infinity, the Navier-Stokes equations can be set in a periodic framework, which implies that we are no longer working on the whole space  but in the 3-dimensional torus  . We will treat each case separately.
3. Statement of the problem in the whole space
3. 1. Hypotheses and growth conditions
The initial condition  is assumed to be a smooth and divergence-free function (see smooth function) such that, for every multi-index  (see multi-index notation) and any , there exists a constant  (i.e. this "constant" depends on  and ) such that
 for all 
The external force  is assumed to be a smooth function as well, and satisfies a very analogous inequality (now the multi-index includes time derivatives as well):
 for all  
For physically reasonable conditions, the type of solutions expected are smooth functions that do not grow large as . More precisely, the following assumptions are made:
1.
2.There exists a constant  such that  for all 
Condition 1 implies that the functions are smooth and globally defined and condition 2 means that the kinetic energy of the solution is globally bounded.
3. 2. The million-dollar-prize conjectures in the whole space
(A) Existence and smoothness of the Navier-Stokes solutions in 
Let . For any initial condition  satisfying the above hypotheses there exist smooth and globally defined solutions to the Navier-Stokes equations, i.e. there is a velocity vector  and a pressure  satisfying conditions 1 and 2 above.
(B) Breakdown of the Navier-Stokes solutions in 
There exists an initial condition  and an external force  such that there exists no solutions  and  satisfying conditions 1 and 2 above.
4. Statement of the periodic problem
4. 1. Hypotheses
The functions we seek now are periodic in the space variables of period 1. More precisely, let  be the unitary vector in the - direction:
Then  is periodic in the space variables if for any  we have that
Notice that we are considering the coordinates modulo 1. This allows us to work not on the whole space  but on the quotient space , which turns out to be the 3-dimensional torus
We can now state the hypotheses properly. The initial condition  is assumed to be a smooth and divergence-free function and the external force  is assumed to be a smooth function as well. The type of solutions that are physically relevant are those who satisfy these conditions:
3.  
4. There exists a constant  such that  for all 
Just as in the previous case, condition 3 implies that the functions are smooth and globally defined and condition 4 means that the kinetic energy of the solution is globally bounded.
4. 2. The periodic million-dollar-prize theorems
(C) Existence and smoothness of the Navier-Stokes solutions in 
Let . For any initial condition  satisfying the above hypotheses there exist smooth and globally defined solutions to the Navier-Stokes equations, i.e. there is a velocity vector  and a pressure  satisfying conditions 3 and 4 above.
(D) Breakdown of the Navier-Stokes solutions in 
There exists an initial condition  and an external force  such that there exists no solutions  and  satisfying conditions 3 and 4 above.
5. Partial results
1.The Navier-Stokes problem in two dimension has already been solved positively since the 60's: there exist smooth and globally defined solutions. [2]
2.If the initial velocity  is sufficiently small then the statement is true: there are smooth and globally defined solutions to the Navier-Stokes equations. [1]
3.Given an initial velocity  there exists a finite time , depending on  such that the Navier-Stokes equations on  have smooth solutions  and . It is not known if the solutions exist beyond that "blowup time" . [1]
4.The mathematician Jean Leray in 1934 proved the existence of so called weak solutions to the Navier-Stokes equations, satisfying the equations in mean value, not pointwise. [3]
6. Notes
1.More precisely,  is the pressure divided by the fluid density, and the density is constant for this incompressible and homogeneous fluid.
7. References
1.^ Official statement of the problem, Clay Mathematics Institute.
2.O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Flows", 2nd edition), Gordon and Breach, 1969.
3.Leray, J. (1934), "Sur le mouvement d'un liquide visqueux emplissant l'espace", Acta Mathematica 63: 193-248, doi:10.1007/BF02547354
8. External links
•The Clay Mathematics Institute's Navier-Stokes equation prize
•Why global regularity for Navier-Stokes is hard — Possible routes to resolution are scrutinized by Terence Tao.
• d •
— Preceding unsigned comment added by 75.54.91.58 (talk • contribs) 02:36, 1 February 2010 (UTC)
thread_title:  Forced Framing 

message: If we consider that dx belongs to a position vector,therefore may exist:
          u\frac{\partial{u}}{\partial{t}} = \frac{\partial{x}}{\partial{t}}*( \frac{\partial{u} }{\partial{x} }\frac{\partial{x} }{\partial{t}} )
=\frac{\partial{u}}{\partial{x}}\frac{\partial^2{x}}{\partial{t^2}}
which the operations denote a tensor result.
         【\frac{\partial{u}}{\partial{x}} belongs to a tensor】
I'm the original author(Taiwainese)JhongHuá MínGuó. I now born a new idea for "reclaiming" the continued eqn. .
I'm trying to get "energy conservation" to do. We may use the physical Math something like:'''
           \frac{1}{2} \frac{D}{Dt}({\rho}u^2)=\frac{1}{2}\frac{\partial{({\rho}u^2)}}{\partial t}+......
The official terms' phisical meaning refer to "no fluids missing from some indical systems". And in my created terms of functions, they refer to "energy in some indical systems are saved"(No matter what types of energy exist)
I'm a person of like-creation. Before I come here,I studied some articles from "Riley's Methematical Methods for Physics and Engineering". If I've made any possible mistake originally, please tell me. I want to learn Sciences from America,Britain,Canada,...
— Preceding unsigned comment added by HydrogenSu (talk • contribs) 13:07, 21 December 2005 (UTC)
thread_title:  Just an opinion about Math tech. 

message: The Navier-Stokes equation for the case of a Newtonian incompressible fluid are used by more than
90% of researchers in the field.  That this case is not given at all is likely to confuse many students
and lead them to think that this article is irrelevant to their research (or aimed at irrelevant special cases).
I strongly suggest adding these simplfied forms early-on in the article.
:why? the special forms are just that, special forms. And if you want the newtonian incompressible case, just add it: it is trivially the conservation of momentum plus incompressibility, expressed as:
::\nabla\mathbf{v}=0 
:which you would easily get from the article.
:on the other hand, the compressible equations take way too much space :)CyrilleDunant 06:51, 3 November 2006 (UTC)
replies: text: Well, this is WP! Of course you should edit the article if you think information should be added! I will just note that need is perhaps a bit of a strong word for a set of equations...CyrilleDunant 09:58, 4 November 2006 (UTC)
replies: text:  Hi, congratulations to all for the nice work. I just revised the "incompressible" section, perhaps that is what you wanted. But I feel that the rest of the page still needs some rewrite. I kept the full equations with all components etc... I know that students find them very useful.  For Cyrille, "need" is how we teachers see some very useful pieces of math, we sense that students and users of science "need" the equations. You cannot do this type of science or engineering without them. Zaleski 23:54, 6 February 2007 (UTC)
thread_title:  Most important case missing 

message: There's an analogy between the Navier–Stokes equations and Maxwell's equations. It would be good to have a brief section explaining this analogy. JKeck (talk) 05:08, 3 June 2010 (UTC)
thread_title:  Maxwell's Equations Analogy 

message: It's been solved! http://ejde.math.txstate.edu/Volumes/2010/93/jormakka.pdf --188.220.102.130 (talk) 23:08, 21 August 2010 (UTC)
replies: text: The Millennium Prize problem related to these equations might have been solved. Until the paper you linked has been reviewed by others, and third-party sources have reported this, we shouldn't be reporting that the problem has been solved. In any case, we have a separate article, Navier–Stokes existence and smoothness, specifically for the Millennium Prize problem formulation, and any discussion of this should go there. — Gavia immer (talk) 01:50, 22 August 2010 (UTC)
thread_title:  Jormakka 

message: This talks about convective acceleration when it should talk about advective acceleration. 
Convection implies that the flow is buoyancy-driven in some way, while advection refers to tracers (eg temperature/momentum) being carried with the flow.  —Preceding unsigned comment added by 150.203.10.73 (talk) 06:41, 6 April 2011 (UTC)
replies: text: For a previous discussion on this: see Talk:Navier–Stokes equations/Archive 1#Convection Vs Advection. -- Crowsnest (talk) 13:39, 6 April 2011 (UTC)
thread_title:  Convection vs advection 

message: This article has a good description of the uses, derivation, and applicability of the equations, but ironically does not label the equations themselves clearly.  I assume the most general form is this equation in the 'Derivation and description' section:
: \rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v} \right) = -\nabla p + \nabla \cdot\mathbb{T} + \mathbf{f},
except that it is not labelled the "Navier-Stokes equation" but rather the Cauchy momentum equation,  Could someone please clarify this?  Cheers --ChetvornoTALK 22:14, 10 April 2011 (UTC)
replies: text: Because as such, it is not complete: it further requires a constitutive relationship (which will relate T, v and p). NS is the complete set of equations such that the problem can be (possibly) well-posed.CyrilleDunant (talk) 01:40, 11 April 2011 (UTC)

text: The Navier-Stokes equations are not complete in general anyway because they only define the momentum conservation. For incompressible flow the continuity equation is needed; and for compressible flow the continuity equation, energy equation, and an equation of state are all required.  — Preceding unsigned comment added by 164.107.59.74 (talk) 23:24, 26 May 2011 (UTC)
thread_title:  Where are the equations?