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I was reading up on the Clay Institute's Millenium prizes in mathematics.

And I noticed the Navier-Stokes equations were described as minimally understood.

As far as I was taught in physics a few weeks ago(SCQF Level 6), they are used but solutions to them are hard to find in three dimensions because they require large amounts of computational power due to the complexity of the equations and so approximations are used.

How were the equations discovered in the first place if we can't solve them?

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    $\begingroup$ Navier-Stokes equations are just a transcription of Newton's 2nd law to continuum mechanics. $\endgroup$ – Qmechanic Oct 14 '17 at 12:27
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    $\begingroup$ They are very far from being minimally understood! The (both mathematical and physical) literature is very rich and treats many subjects of NS. Numerics can then solve DNS (without approximations) in some very intersting turbulent cases with billions of grid points. $\endgroup$ – Vladimir F Oct 14 '17 at 15:01
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    $\begingroup$ @VladimirF without approximations? Numerics are approximations themselves, by definition — regardless of how fine the grid is. $\endgroup$ – Ruslan Oct 14 '17 at 15:11
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    $\begingroup$ @Ruslan Sure, I thought that it is so obvious in the term numerical solution that I thought that can be kept implicit, but... I missed the word additional when typing. The point is there are no additional approximations put into the equations about some unresolved scales in DNS. $\endgroup$ – Vladimir F Oct 14 '17 at 16:25
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    $\begingroup$ Being able to express something as a general equation (or system of equations) doesn't say anything about the ease of solving that equation. For instance, Newton's law of gravity is pretty simple, but there's still no general solution to the 3 body problem. $\endgroup$ – jamesqf Oct 15 '17 at 5:15
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I just wanted to give a more concrete idea of how we know these equations even though we have trouble proving analytical theorems about them.

Stuff moving in space

Consider any stuff (as in, any conserved quantity) distributed over space. We know that we can describe this with a time-dependent density field $\rho(x,y,z,t)$ such that any little volume $dV$ has some amount of stuff $\rho~dV$ at that point. We also know that this stuff might be flowing around over time and we formally treat this by saying that we want to know the flow through a little flat surface of area $dA,$ which is oriented in the $\hat n$ direction: that is, the surface is normal to $\hat n$ and "positive" flow will be in the $+\hat n$ direction. Combined together this is a vector $d\mathbf A = \hat n~dA$ and there is some vector field $\mathbf J(x,y,z,t)$ such that the amount of stuff which flows through this area over a time $\delta t$ is $\delta t~d\mathbf A\cdot\mathbf J(x,y,z,t).$ With $\rho$ and $\mathbf J$ we know almost everything. Since the stuff is conserved, we can say that in this box of volume $dV,$ if the amount of stuff in the box changes, it is either because there was a net flow into or out of the sides of the box, so we are doing some $\iint d\mathbf A\cdot \mathbf J$ which turns out by Gauss's theorem to be just $dV~\nabla\cdot\mathbf J,$ or else it came from outside the system we're studying, so there is some term $dV~\Phi$. Equating that to the change in the box $dV~(\partial\rho/\partial t)$ gives the simple starting equation $${\partial \rho\over\partial t} = -\nabla\cdot \mathbf J + \Phi.$$Now when we've got a flow field $\mathbf v(x,y,z,t)$ dictating how a fluid flows, the most dominant transport term is that the box flows downstream, $\mathbf J = \rho~\mathbf v + \mathbf j$ for some deviation $\mathbf j.$ Usually the principal deviation then comes from Fick's law, that there is a flow proportional to the difference in density between adjacent points, $\mathbf j = -D~\nabla \rho,$ but there may be more complex terms there; in particular we shall see pressure here.

Conservation of momentum

The key point here is that $p_x$, the momentum in the $x$-direction, is a stuff. It is a known conserved quantity. It is conserved as a direct result of Newton's third law which turns out, under Emmy Noether's celebrated theorem, to be the same as the statement that the laws of physics are the same at position $x$ as they are at position $x+\delta x$, for a suitable definition of "laws of physics." We are pretty sure about this, and we are pretty sure that the momentum of the fluid itself in the $x$-direction must therefore also be conserved, and this is $\rho~v_x$ where I am shifting definitions a bit on you: $\rho$ now refers to the mass density field and $v_x$ still refers to the fluid velocity in the $x$-direction.

Now a flow of momentum per unit time, which we said is what $\mathbf J\cdot d\mathbf A$ is, is a force. Therefore $\mathbf J$ naturally takes the form of a force per unit area in this context. Now we know that Newton's expression for viscous forces was in fact to write $F_x = \mu~A~v_x/y$ where I am moving a surface of a fluid at speed $v_x$ at a perpendicular distance $y$ from a place where it is being held still; it will not surprise you at all to see that this is very similar to Fick's law and can be written as just $\mathbf j_\text{viscosity} = -\mu~\nabla v_x.$ To that we also need to add the effects of pressure, as a lowering in pressure also drives a fluid motion; this is a little bit harder to reason out but it takes the form that we can imagine a constant flow in the $x$-direction of $p~\hat x$ and then deviations in this flow would produce the change in momentum per unit time $-\partial p/\partial x$ through this divergence term. (That's a little bit of a sloppy way to show that we are talking about a stress tensor and part of it is $p~\mathbf 1$, the identity matrix multiplied by the pressure.) Combining these two components of $\mathbf j$ we have $${\partial \over\partial t}(\rho~v_x) = -\nabla\cdot (\rho~v_x~\mathbf v - \mu \nabla (v_x)) - \frac{\partial p}{\partial x} + \Phi_x.$$The external contribution $\Phi$ comes from forces influencing the fluid from outside, like gravity.

In the Navier-Stokes equations the Millenium Prize has restricted itself to a considerably simpler case where $\nabla\cdot\mathbf v = 0$ and $\rho$ and $\mu$ are constant, which we call "incompressible flow." This is generally a valid assumption when you're interacting with a fluid at speeds much lower than the speed of sound in that fluid; then the fluid would rather move away from you than be compressed into any one place. In this case we can commute $\rho$ out of all of the spatial derivatives and then divide by it, so that the only impact is to rewrite $\nu=\mu/\rho$ and $\lambda=p/\rho$ and $a_x=\Phi_x/\rho$, eliminating the unit of mass from the equation. For $v_x$ we have specifically, $${\partial v_x\over\partial t} + \mathbf v\cdot\nabla v_x - \nu \nabla^2 v_x = - \frac{\partial \lambda}{\partial x} + a_x,$$ and then we can extend the above analysis to the directions $y,z$ too to find, $$\dot{\mathbf v} + (\mathbf v\cdot\nabla)\mathbf v - \nu \nabla^2 \mathbf v = - \nabla \lambda + \mathbf a.$$This is the version of the Navier-Stokes equations written down in the Millenium Prize; we have a very straightforward explanation of this as "The flow of momentum in a small box flowing downstream in an incompressible homogeneous Newtonian fluid is due entirely to Fick's-law diffusion of the momentum due to the viscosity of the fluid, plus a force due to pressure gradients inside the fluid, plus forces imposed by the external world."

Why this equation?

The understanding of the physics of how we got to this equation is not in question. What's at stake is the mathematics of this equation, in particular this $(\mathbf v \cdot \nabla) \mathbf v$ term which contains $\mathbf v$ twice and thereby makes it a nonlinear partial differential equation: given two flow fields $\mathbf v_{1,2}$ which are valid, in general $\alpha \mathbf v_1 + \beta \mathbf v_2$ will not solve this equation, removing our most powerful tool from our toolbox.

Nonlinearity turns out to be unbelievably hard to solve in general, and essentially the Clay Mathematics institute is giving the million-dollar prize for anyone who cracks nonlinear differential equation theory strongly enough that they can answer one of the more basic mathematical questions about these Navier-Stokes equations, as a "most basic example" for their new theoretical toolkit.

The idea of the Clay prizes is that they are specific problems (which is important for awarding a prize for their solution!) but that they seem to require powerful new general ideas which would allow our mathematics to go into places where it has historically been unable to go. You see this for example in $\text{P} = \text{NP}$, it's a very specific question but to answer it we would seem to need to have a better handle on "here's a classification of set of stuff which computers can do, and here are some things which a computer can't efficiently do" which nobody has yet been able to convincingly present. A new toolbox which could resolve this "stupid little" question would therefore profoundly improve our ability to work on a huge class of related problems in computation.

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    $\begingroup$ That was a hell of an answer thanks. I think I'm a little below the level required to understand the maths you've written but I think I have a lot of the general concepts down. I would like to ask what the upside down triangle symbol means though? $\endgroup$ – Douglas Oct 15 '17 at 9:55
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    $\begingroup$ @Ooker Great question. There's nothing in principle wrong with using a Fourier transform but please remember that the Fourier transform of a product is going to be a convolution; in particular $\sum_m~v_m(x_1,x_2,x_3)~\partial_m v_n(x_1,x_2,x_3)$ is going to be a sum of interesting convolutions of $v_m[k_1,k_2,k_3]$ with $k_m~v_n[k_1, k_2, k_3]$. $\endgroup$ – CR Drost Oct 15 '17 at 14:55
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    $\begingroup$ @Ooker in some cases the transform really helps. Spectral methods in turbulence for example. But, a Fourier transformation requires a periodic/infinite domain. So it can't be used for complex geometry. Even simple flow over a wall is limited to using the transformed equations parallel to the wall and has to use the non transformed ones normal to it. These are the pseudo spectral methods. $\endgroup$ – tpg2114 Oct 16 '17 at 0:37
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    $\begingroup$ @tpg2114 Fourier transfor does not always require periodic domain, depends on the equation and boundart condition. You can solve the Poisson condition with zero on the boundary (sines) or zero gradient on the boundary (cosines) with the Fourier transform (even exactly). Spectral (and pseudospactral) method for NS are numerical methods and hence approximate only, due to the nonlinearity and I suppose Ooker meant exact solutions (and see how people complain below the question that i did not stress that obvious fact). $\endgroup$ – Vladimir F Oct 16 '17 at 6:10
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    $\begingroup$ @Ooker I think you've mixed something up: we do Fourier transform to algebraise the differential equations, and this only works out fully if the the equation is linear! — Sometimes it's still useful to use Fourier even on nonlinear equations, but mostly only for small pertubations around a constant density/pressure (in which case the equations for the pertubations are approximately linear, and applying FT quickly gives decent approximations to the exact solution). $\endgroup$ – leftaroundabout Oct 16 '17 at 12:17
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As @QMechanic mentioned in a comment, the Navier-Stokes equations are just $F = ma$, but they look much scarier. Assuming an incompressible fluid, you have:

$$ \rho \frac{D u_i}{D t} = -\frac{\partial \sigma_{ij}}{\partial x_j} + f_b $$

where $\rho$ is the density (mass per unit volume), $D u_i /D t$ is the acceleration (written in the Lagrangian form, rather than Eulerian for clarity), $\sigma$ is the Cauchy stress tensor (generally expanded to take out the pressure from the trace, $\sigma_{ij} = -p \delta_{ij} + \tau_{ij}$ where now $\tau_{ij}$ is the viscous stress tensor), and $f_b$ is the body force (things like gravity).

Everything on the right hand side is the sum of forces, and the left is the mass times the acceleration.

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The Navier Stokes equations are a combination of Newton's 2nd law of motion (differential form) with the 3D version of Newton's law of viscosity (i.e., the mechanical constitutive equation for a Newtonian fluid).

What you were taught in Physics at school was correct. Not many analytic solutions exist to interesting practical problems, and numerical solution (say using computational fluid dynamics (CFD)) is often necessary. But they are certainly well understood.

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To add to tpg124's answer and to answer the implict question in your statetent:

How were the equations discovered in the first place if we can't solve them?

simplicity and clarity of an equation's meaning and difficulty of working out the implications of that equation are two wholly different things. For the Navier-Stokes equation, the meaning is crystal clear (a statement of Newton's laws) and the equations could be written down as soon as the mathematical tools were available to do so. The difficulty with the Navier-Stokes equation can be seen when we expand the Lagrangian form of the derivative on the left hand side of the equation in tpg124's answer; it contains a so-called advection term $(\vec{v}.\nabla)\vec{v}$ that is part of accounting for the the fact that the field co-ordinates of the matter which Newton's laws are being applied to is moving in the field co-ordinates. It's an equation about the flow of a vector field. This advection term is a quadratic nonlinearity, and it is the source of all the strife that is encountered in solving the equation.

Another physics example that is very similar is the Einstein field equation, in the sense that it has a simple meaning (see my answer here), but the meanings of solutions can be very subtle (requiring a great deal of practice with the notion of general covariance), and is dreadfully nasty to solve, both insofar that there are few exact solutions and numerical solutions can be appallingly badly behaved until a great deal of numerical ingenuity and pure theoretical numerical analysis - quite beyond merely knowledge of relativity physics - are brought to bear on the problems. This simple to understand, difficult to solve situation in this case is also brought about by a nonlinearity: the so called trace-reversing term $-\frac{1}{2}\,R\,g$, where $g$ is the metric tensor and $R$ a scalar that is like a double trace of a commutator (Lie bracket) of derivatives of $g$. So we have a quadratic nonlinearity here too. This nonlinearity is brought about by the simultaneous geometric (Bianchi identity) and energy-momentum conservation meanings encoded into the equation discussed in my other answer. Numerical relativity really only took off in the 1980s - seventy years after the theory was postulated - as the necessary computer power came online for people to make progress on the hard work of learning how to tame the numerical problems.

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  • $\begingroup$ Given a system's equation without computer, can we say anything about its behaviors? Does knowing the equation without solution help us know it any better? $\endgroup$ – Ooker Oct 15 '17 at 7:50
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    $\begingroup$ Yes, sometimes, although it's difficult and often we can say less than we'd like. E.g. asymptotic methods can show the behavior of solutions in the vicinity of special points, such as infinity. Stephen Hawking got his Nobel prize for work of just this sort -- figuring out the general behavior of certain classes of solutions of gravity equations. $\endgroup$ – Anton Tykhyy Oct 15 '17 at 16:40
  • $\begingroup$ @AntonTykhyy ah, you mean something like the characteristic equations? $\endgroup$ – Ooker Oct 15 '17 at 20:26
  • $\begingroup$ @Ooker Of course. For example, in linear systems, we have a huge body of theory - the majority of functional analysis: operator theory, spectral theory, classes of problems known to have fully discrete spectums thus solutions are resolvable into generalized Fourier series, distribution theory ....... the list is endless and the knowledge about the behavior of solutions this brings is immense. Often similar idealized problems have full, analytic solutions. The theory behind the behaviors that arise in nonlinear systems - albeit impressive - is much more piecemeal and the Clay prize .... $\endgroup$ – WetSavannaAnimal Oct 16 '17 at 1:44
  • $\begingroup$ @Ooker ...... is about furthering that knowledge. Note that weak field gravity solutions, post Newtonian theory and the like are all linear approximations to General Relativity that have big uses; Einstein's big 1915 paper made use of a linear approximation to recover Newtonian physics as the limiting form of GTR and to compute the perturbation to Mercury's apsidal precession. I'm much less comfortable with fluid mechanics, but the theory of inompressible, irrotational, linearlized flows are also often useful approximations to the Navier Stokes equations. $\endgroup$ – WetSavannaAnimal Oct 16 '17 at 1:44
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Consider the following example. To describe the motion of a one dimensional mass attached to a spring you first consider all the forces. The force due to the spring is $F_s=-kx$ with $k$ the spring constant and $x$ the extension of the spring and the force due to friction is $F_f=-bv$ with b some constant and $v=\dot x$ the speed of the mass. Writing down the final equation is easy: \begin{align}F_{total}&=ma\\ -bv-kx&=m \ddot x\\ m\ddot x+b\dot x+kx&=0 \end{align} The last equation is a full description of this system in the same way the Navier stokes equations is a full description of fluid motion (for ideal springs and fluids ofcourse). Writing down the differential equations is relatively easy but if you want to find an explicit function $x(t)=\ ...\ $ you would still have to do a lot of work.

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  • $\begingroup$ This answer completely overlooks why Navier-Stokes is fundamentally more difficult than a mass-spring-damper system. Solving for $x(t)$ may be "a lot of work" for a beginner, but you can write a closed-form solution following a well-defined algorithm. This is (for now) not true for Navier-Stokes - we don't even know whether a smooth solution always exists. $\endgroup$ – Sanchises Oct 15 '17 at 10:16
  • $\begingroup$ I forgot to mention that, thanks for pointing that out. I was only trying to give a concrete example to make it easier to distinguish between finding an equation and finding the solution to that equation. $\endgroup$ – AccidentalTaylorExpansion Oct 15 '17 at 19:19

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