Talk:Navier-Stokes equations

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The r-component of the Equations of Motion in cylindrical coordinates have errors, uz should be replaced with ur.

1 The continuity equation is flat wrong
2 Removed equation again
3 Why are the equations missing?
4 This article needs a complete rewrite
5 This article nedds quite some editing
6 Historical background and actual relevance
7 Too much sorry
8
9 Just an opinion about Math tech.
10 Math in Physics About The Substantive Derivative
11 Stars in a galaxy
12 Wilkinson equations
13 $1,000,000 prize problem
14 This page should be split
15 New Paper
16 Most important case missing
17 How we can classify Nav.. eqn as parabolic or hyperbolic or elliptic?

[edit] 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:

$\int_V\frac{d\rho}{dt}dV =\int_A \rho\mathbf{v\cdot n}dA$

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; : $ρ$ 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)

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)

[edit] Removed equation again

$-\nabla p + \mu \left( \nabla^2 \mathbf{u} + {1 \over 3} \nabla (\nabla \cdot \mathbf{u} ) \right) + \rho \mathbf{u} = \rho \left( { \partial\mathbf{u} \over \partial t } + \mathbf{u} \cdot \nabla \mathbf{u} \right)$

where:

p is pressure

$\mathbf{u}$ is fluid velocity

μ

is viscosity

ρ

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)

[edit] Why are the equations missing?

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 (ugh), 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 [1] 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)

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)

I think that the (simplified) Navier Stokes equations should be put somewhere in the beginning of the page. We can just add a small note that this is a simplified version and there are some other verisons that account for stuff like thermal gradients and magnetic fluids. I think that most of the readers to know what the formula looks like. We don't want to force these people to read the entire derivation. Therefore the equations should be placed first and the derivation (or whatever) should be placed last. (anonymous)

[edit] This article needs a complete rewrite

Most of the equations here are just plain wrong! Phys 19:26, 30 August 2005 (UTC)

Ok. What equations are wrong? I mean, the expanded forms are a bit painful to read, but they are correct...CyrilleDunant 21:38, 30 August 2005 (UTC)

I corrected most of them Phys 17:06, 31 August 2005 (UTC)

I am sorry to contradict you. You did not correct them, you just swapped a set of conventions for another (I know, I checked in a textbook which used the "other" convention.). Wich is sort of OK, just a bit pointless, but it would have been more useful to eradicate.

I disagree with the statement saying "The nature of the trace of $\mathbb{P}$ is known, it is thrice the pressure, thus:"

$\frac{\sigma_{xx}+\sigma_{yy}+\sigma_{zz}}{3} = - p$

as opposed to the diagonal part of $\nabla\mathbb{T}$ is the gradient of pressure. Mostly because the gradient of pressure has some physical sense to it unlike the trace of the tensor...

Also, but again, this is purely conventional, the stress components are divided in shear and non-shear, symbolised by

τ

and

σ

. I am at loss to understand what the hell

τ i i

might be ( $\frac{\partial d_i}{\partial x_i}=\tau_{ii}$ derivative of the displacement i along axis i normal to i ??). Of course they are nil...

I will not revert brutally (because I suppose you checked what you wrote, and I you think the way it is now is better, well, why not :)) But it would be nice to discuss before making radical changes...CyrilleDunant 12:39, 1 September 2005 (UTC)

$\nabla\mathbb{T}$ is a tensor of rank 3 whereas $\nabla\cdot\mathbb{T}$ is a vector. By the diagonal, I assume you mean the trace. Neither of them will give a matrix, since neither of them are of rank 2. And it simply isn't covariant to split a tensor of rank 2 like this:

$\begin{pmatrix}\sigma_{xx}&\tau_{xy}&\tau_{xz}\\ \tau_{yx}&\sigma_{yy}&\tau_{yz}\\ \tau_{zx}&\tau_{zy}&\sigma_{zz}\end{pmatrix}$

Phys 19:15, 1 September 2005 (UTC)

The decomposition of a tensor of rank 2 into a traceless part and its trace goes as

$\begin{pmatrix}\tau_{xx}&\tau_{xy}&\tau_{xz}\\ \tau_{yx}&\tau_{yy}&\tau_{yz}\\ \tau_{zx}&\tau_{zy}&\tau_{zz}\end{pmatrix}+\begin{pmatrix}\sigma&0&0\\0&\sigma&0\\0&0&\sigma\end{pmatrix}$

where

τ x x + τ y y + τ z z = 0

. Phys 19:19, 1 September 2005 (UTC)

Yes, but this is not pure math: there is also some physics behind... The goal is not to get a traceless part, but to make physically meaningful terms appear. Such, for example as pressure. Or the constraints on the fluid particle. What you do is correct, but not very didactic or indeed physicall relevant at this point in the article. Again Tau_xx has _no_sense_ CyrilleDunant 20:19, 1 September 2005 (UTC)

Let's define $\mathbb{T}$ as

$\mathbb{T}=\begin{pmatrix}\tau_{xx}&\tau_{xy}&\tau_{xz}\\\tau_{yx}&\tau_{yy}&\tau_{yz}\\\tau_{zx}&\tau_{zy}&\tau_{zz}\\\end{pmatrix}$

and $\mathbb{T}'$ as

$\mathbb{T}=\begin{pmatrix}0&\tau_{xy}&\tau_{xz}\\\tau_{yx}&0&\tau_{yz}\\\tau_{zx}&\tau_{zy}&0\\\end{pmatrix}$

$\nabla\cdot\mathbb{T}$ and $\nabla\cdot\mathbb{T}'$ are NOT equivalent. One gives

$\begin{pmatrix}\frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}\\\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yy}}{\partial y}+\frac{\partial \tau_{yz}}{\partial z}\\\frac{\partial \tau_{zx}}{\partial x}+\frac{\partial \tau_{zy}}{\partial y}+\frac{\partial \tau_{zz}}{\partial z}\\\end{pmatrix}$

and the other gives

$\begin{pmatrix}\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}\\\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yz}}{\partial z}\\\frac{\partial \tau_{zx}}{\partial x}+\frac{\partial \tau_{zy}}{\partial y}\\\end{pmatrix}$

$\nabla\cdot\mathbb{P}$ is equal to

$-\begin{pmatrix}\frac{\partial p}{\partial x}\\\frac{\partial p}{\partial y}\\\frac{\partial p}{\partial z}\\\end{pmatrix}+\begin{pmatrix}\frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}\\\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yy}}{\partial y}+\frac{\partial \tau_{yz}}{\partial z}\\\frac{\partial \tau_{zx}}{\partial x}+\frac{\partial \tau_{zy}}{\partial y}+\frac{\partial \tau_{zz}}{\partial z}\\\end{pmatrix}$

but not

$-\begin{pmatrix}\frac{\partial p}{\partial x}\\\frac{\partial p}{\partial y}\\\frac{\partial p}{\partial z}\\\end{pmatrix}+\begin{pmatrix}\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}\\\frac{\partial \tau_{yx}}{\partial x}+\frac{\partial \tau_{yz}}{\partial z}\\\frac{\partial \tau_{zx}}{\partial x}+\frac{\partial \tau_{zy}}{\partial y}\\\end{pmatrix}$

And yes, a symmetric tensor (which has 6 independent components) decomposes into a shear part (which is 5 components, not three!!!) and a trace (which has 1 component) if we insist upon a covariant decomposition. It does not decompose into an off-diagonal part (with three components) and a diagonal part (with three components) covariantly. And stress isn't a constraint. Phys 19:41, 30 September 2005 (UTC)

[edit] This article nedds quite some editing

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)

[edit] Historical background and actual relevance

Outside of these mathematical concerns, I think it would be very nice to have some historical background on the equations. When were they written ?

Good question. Around 1830ish, I would guess.CyrilleDunant 21:45, 20 November 2005 (UTC)

Also, what is their significance today ? Do we have solutions to them or just approximations, and for which cases (turbulent, laminar, etc) ?

Well, if we had solutions, they'd be in the article, I guess :). Certain cases, we can solve. As for the significance, let us say that as far as we know, the describe perfectly the behaviour of fluids, if continuity is verifiedCyrilleDunant 21:45, 20 November 2005 (UTC)

All of this I would really like to know ! What do you think ? Olivier

--82.32.65.149 16:19, 20 November 2005 (UTC)

That a section on History would be good. But I have no clue.CyrilleDunant 21:45, 20 November 2005 (UTC)
What I can add is that the article is by Sir George Gabriel Stokes, Fellow of Penbroke College, read april 14, 1845 (this is the correct publication date) and published in the transactions of the Cambridge Philosophical Society volume IIX, 1849 titled:"On the Theories of the internal Friction of Fluids in Motion, and of the Equilibrium and Motion of Elastic Solids" Greetings, Roger Jeurissen

I propose shrinking this article to the following:
1) definition & historical relevance
2) conservation principles and additional assumptions for the general equations
3) the general equations
4) links to specialization

All the details for the substantial derivative need be removed in addition to the special cases. This article is too large and the additional details are completely meaningless to someone who doesn't already know them. Rather then explaining them further, I vote to have them removed. I would like input before I remove large amounts of this article. Josh Quinnell 03:10, 2 December 2005 (UTC)

Actually, it would be a good idea to extend the article by introducing the adimentionnal forms of the equations as well as their classification (parabolic, hyperbolic, elliptic, mixed). Also I believe it is very necessary to have the details on the substantive derivative, as it allows one to understand the equations. But then certainly those sections could be a tad more didactic.CyrilleDunant 16:22, 2 December 2005 (UTC)
This decision to arbitrarily include more information confuses me. There are loads of contributions to NS development that are just as relevant as the substantial derivative. None of these details are pertinent to understanding the NS equations. The details should be linked to their own articles. When inserted into this article they detract from the ability for a lay person to come here and learn about the NS equations. This is not a reference for an expert in NS; we have text books for that. Josh Quinnell 21:03, 2 December 2005 (UTC)
- Should it me merged in a larger section about control volumes ? In fact, I would rather the "full (as in, very large)" version of the equations was eliminated, As it brings nothing else than a statement of the type "look, plenty of confusing signs". Arguably, this is not a vulagarisation article either. The subject is highly technical (in a mathematical sense), and the article reflects that.CyrilleDunant 08:05, 3 December 2005 (UTC)

[edit] Too much sorry

[edit]

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)

$\rho\nabla\mathbf{v}$

$\int_V\frac{d\rho}{dt}dV =\int_A \rho\mathbf{v\cdot n}dA$

Where V is the control volume and A its area; : $ρ$ 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

        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 :

           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)

[edit] Just an opinion about Math tech.

If we consider that dx belongs to a position vector,therefore may exist:

which the operations denote a tensor result.

        【 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:

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,...

[edit] Math in Physics About The Substantive Derivative

         About The Substantive Derivative

We might get some simple derivations by the following descriptions.

A Control Volume is filled with fluids,P represents pressure,with P=P(x,y,z,t) (Note:Pressure can be easily to image.) First we make total differential

$dP=\frac{\partial{P}}{\partial{t}}dt+\frac{\partial{P}}{\partial{x}}dx+\frac{\partial{P}}{\partial{y}}dy+\frac{\partial{P}}{\partial{z}}dz$

The rate of pressure change is

$\frac{dP}{dt}=\frac{\partial{P}}{\partial{t}}+\frac{\partial{P}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{P}}{\partial{y}}\frac{dy}{dt}+\frac{\partial{P}}{\partial{z}}\frac{dz}{dt}$

Hence,

$\frac{dP}{dt}=\frac{\partial{P}}{\partial{t}}+V_{x}\frac{\partial{P}}{\partial{x}}+V_{y}\frac{\partial{P}}{\partial{y}}+V_{z}\frac{\partial{P}}{\partial{z}}$

$\frac{D}{Dt}=\frac{\partial}{\partial t}+{\mathbf V}\cdot\nabla$

therefore,

$\frac{dP}{dt}=\frac{\partial{P}}{\partial{t}}+V_{x}\frac{\partial{P}}{\partial{x}}+V_{y}\frac{\partial{P}}{\partial{y}}+V_{z}\frac{\partial{P}}{\partial{z}}=\frac{DP}{Dt}$

where ${\mathbf V}$ is the fluid velocity, $V (x, y, z)$ is the fluid speed, and $\nabla$ is the differential operator del. ^^

         (I'm happy on what achived:Mathematics technology from Taiwanese educations)^^

Reference:James R. Welty,Charles E. Wicks,Robert E. Wilson,Gregory Rorrer Foundamentals of Momentum,Heat,and Mass Transfer ISBN 0-471-38149-7

Yes, this is a useful addition which could go under a section named "Control Volume", which certainely lacks in an article about NS :). But please, pay attention to the fact that throughout the article, vector quantities are in bold and scalar not. Also, velocity is $\mathbf{v}$ and pressure

p

. A figure would be nice also :)CyrilleDunant 15:47, 23 December 2005 (UTC)

Thank you first.I like physics very much though I major in C.E. .The culture background of my country Republic of China is conservative but privacy is a little open in some public permitted,so...I talk very much. I will move that article I wrote to Control Volume.

Thanks for the correction by you.[2] Are you the Wikipedia editor here,please?

[edit] Stars in a galaxy

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)

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)

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)

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)

[edit] Wilkinson equations

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)

[edit] $1,000,000 prize problem

It should perhaps be mentioned that $1,000,000 prize problem concerns the incompressible Navier-Stokes equations.

[edit] This page should be split

The main Navier-Stokes page should not be in the "Millenium Prize" cattegory.

The Millenium Prize applies to only one small aspect of the theory around the Navier Stokes Equations, it seems perverse to have links to other Millenium Prize problems rather than other fluid dynamic resources. (Even links to other "famous physical equations" or some such would be better than what is currently there.)

If no one objects I am happy to make this change, but I'm open to criticism.

--cfp 16:00, 17 July 2006 (UTC)

Well. I would tend to agree. But even better would be a fluid dynamics project/category...CyrilleDunant 16:16, 17 July 2006 (UTC)

The new Navier stokes Millennium prize page is at: Navier-Stokes existence and smoothness

The new navigation infobox is at Template:Continuum mechanics, and pages should also have Template:Physics-footer as a footer.

Hope everyone approves.

--cfp 16:26, 18 July 2006 (UTC)

It is an improvement. Good stuff. :) Nephron T|C 21:18, 28 July 2006 (UTC)

Still on the same topic, I find the text in the introduction misleading: "Even though turbulence is an everyday experience it is extremely difficult to find solutions for this class of problems. A $1,000,000 prize (...) in the understanding of this phenomenon." The problem is not just about turbulence...

[edit] New Paper

I just stumbled over a brand new paper:

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)

[edit] Most important case missing

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)

My point is: if 90% of the people coming to this page are needing the Newtonian incompressible form -- and that is missing -- one has a article well suited to 10% of those who come looking. Water, air, salad dressing, usually at low Mach number, these are the fluids of life and most technology. When you suggest "just add it" I'm unsure if you would like me to proceed to edit the article. -- best wishes -- User:Lathrop 21:31, 3 November 2006 (EST)

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)

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)

[edit] How we can classify Nav.. eqn as parabolic or hyperbolic or elliptic?

will somebody explain°140.158.49.62 17:41, 30 January 2007 (UTC) Shoeb

This is still (AFAIK) a not very well defined problem. Basically, depending on various parameters (Reynolds' number, Mach number), various terms of the equations become negligible under certain assumptions. The resulting simplified equations can then be classified as parabolic, hyperbolic or elliptic.

Those considerations are important as they define the type of boundary conditions required for the solving of the problem, and thus the type of numerical code needed, for example.17:05, 2 February 2007 (UTC)

I think the answer to this question is simple: The Navier-Stokes equation is neither a hyperbolic, nor a parabolic, nor an elliptic equation. It is a first-order partial differential equation. Parabolic/hyperbolic/elliptic is only defined for second-order PDE's. Greetings,--Roger Jeurissen 11:54, 26 March 2007 (UTC)

The second order is "hidden" in the expression of stress. Stress is a function of strain, which derived spatially from displacement. The divergence of stress appears, making it a second order term in space. CyrilleDunant 15:45, 26 March 2007 (UTC)

No, the NS equations are second order PDEs. They can't be classified as hyperbolic/parabolic/whatever because they're nonlinear. Like the other guys said, kill the right terms and you can get all kinds of linear equations which you can classify by the conventions. -Ben pcc 16:00, 26 March 2007 (UTC)

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