III.23 The Euler and Navier-Stokes Equations

Charles Feffermcm

The Euler and Navier-Stokes equations describe the motion of an idealized fluid. They are important in science and engineering, yet they are very poorly understood. They present a major challenge to mathematics.

To state the equations we work in Euclidean space ^d, with d equal to 2 or 3. Suppose that, at position x = (x1, . . . , x_d) ∈^d and at time t ∈ , the fluid is moving with a velocity vector u(x, t) =(u₁ (x, t) . . . , u_d(x, t)) ∈ ^d, and the pressure in the fluid is p(x, t) ∈ . The Euler equation is

for all (x, t); and the Navier-Stokes equation is

for all (x, t). Here, v > 0 is a coefficient of friction called the “viscosity” of the fluid.

In this article we restrict our attention to incompressible fluids, which means that, in addition to requiring that they satisfy (1) or (2), we also demand that

for all (x, t). The Euler and Navier-Stokes equations are nothing but Newton’s law F = ma applied to an infinitesimal portion of the fluid. In fact, the vector

is easily seen to be the acceleration experienced by a molecule of fluid that finds itself at position x at time t.

The forces F leading to the Euler equation arise entirely from pressure gradients (e.g., if the pressure increases with height, then there is a net force pushing the fluid down). The additional term

in (2) arises from frictional forces.

The Navier-Stokes equations agree very well with experiments on real fluids under many and varied circumstances. Since fluids are important, so are the Navier-Stokes equations.

The Euler equation is simply the limiting case v 0 of Navier-Stokes. However, as we shall see, solutions of the Euler equation behave very differently from solutions of the Navier-Stokes equation, even when v is small.

We want to understand the solutions of the Euler equations (1) and (3), or the Navier-Stokes equations (2) and (3), together with an initial condition

where u⁰(x) is a given initial velocity, i.e., a vector-valued function on ^d. For consistency with (3), we assume that

Also, to avoid physically unreasonable conditions, such as infinite energy, we demand that u⁰ (x), as well as u(x, t) for each fixed t, should tend to zero “fast enough” as |x| → ∞. We will not specify here exactly what is meant by “fast enough,” but we assume from now on that we are dealing only with such rapidly decreasing velocities.

A physicist or engineer would want to know how to calculate efficiently and accurately the solution to the Navier-Stokes equations (2)-(4), and to understand how that solution behaves. A mathematician asks first whether a solution exists, and, if so, whether there is only one solution. Although the Euler equation is 250 years old and the Navier-Stokes equation well over 100 years old, there is no consensus among experts as to whether Navier-Stokes or Euler solutions exist for all time, or whether instead they “break down” at a finite time. Definitive answers supported by rigorous proofs seem a long way off.

Let us state more precisely the problem of “breakdown” for the Euler and Navier-Stokes equations. Equations (1)-(3) refer to the first and second derivatives of u(x, t). It is natural to suppose that the initial velocity u⁰ (x) in (4) has derivatives

of all orders, and that these derivatives tend to zero “fast enough” as |x| → ∞. We then ask whether the Navier-Stokes equations (2)-(4), or the Euler equations (1), (3), and (4), have solutions u(x, t), p(x, t), defined for all x ∈ ^d and t > 0, such that the derivatives

and p(x,t) of all orders exist for all x ∈^d, t ∈ [0, ∞) (and tend to zero “fast enough” as |x| ∈, ∞). A pair u and p with these properties is called a “smooth” solution for the Euler or Navier-Stokes equations. No one knows whether such solutions exist (in the three-dimensional case). It is known that, for some positive time T = T(u⁰) > 0 depending on the initial velocity u⁰ in (4), there exist smooth solutions u(x, t), p(x, t) to the Euler or Navier-Stokes equations, defined for x ∈ and t ∈ [0, T).

In two space dimensions (one speaks of “2D Euler” or “2D Navier-Stokes”), we can take T = +∞; in other words, there is no “breakdown” for 2D Euler or 2D Navier-Stokes. In three space dimensions, no one can rule out the possibility that, for some finite T = T(u⁰) as above, there is an Euler or Navier-Stokes solution u(x, t), p(x, t), which is defined and smooth on

such that some derivative | u(x,t)| or |p(x,t)| is unbounded on Ω. This would imply that there is no smooth solution past time T. (We say that the 3D Navier-Stokes or Euler solution “breaks down” at time T.) Perhaps this can actually happen for 3D Euler and/or Navier-Stokes. No one knows what to believe.

Many computer simulations of the 3D Navier-Stokes and Euler equations have been carried out. Navier-Stokes simulations exhibit no evidence of breakdown, but this may mean only that initial velocities u⁰ that lead to breakdown are exceedingly rare. Solutions of 3D Euler behave very wildly, so that it is hard to decide whether a given numerical study indicates a breakdown. Indeed, it is notoriously hard to perform a reliable numerical simulation of the 3D Euler equations.

It is useful to study how a Navier-Stokes or Euler solution behaves if one assumes that there is a breakdown. For instance, if there is a breakdown at time T < ∞ for the 3D Euler equation, then a theorem of Beale, Kato, and Majda asserts that the “vorticity”

grows so large as t → T that the integral

diverges. This has been used to invalidate some plausible computer simulations that allegedly indicated a breakdown for 3D Euler. It is also known that the direction of the vorticity vector ω(x, t) must vary wildly with x, as t approaches a finite breakdown time T.

The vector ω in (5) has a natural physical meaning: it indicates how the fluid is rotating about the point x at time t. A small pinwheel placed in the fluid in position x at time t with its axis of rotation oriented parallel to ω(x, t) would be turned by the fluid at an angular velocity |ω(x, t)|.

For the 3D Navier-Stokes equation, a recent result of V. Sverak shows that if there is a breakdown, then the pressure p(x, t) is unbounded, both above and below.

A promising idea, pioneered by J. Leray in the 1930_s, is to study “weak solutions” of the Navier-Stokes equations. The idea is as follows. At first glance, the Navier-Stokes equations (2) and (3) make sense only when u(x, t), p(x, t) are sufficiently smooth: for example, one would like the second derivatives of u with respect to the x_j to exist. However, a formal calculation shows that (2) and (3) are apparently equivalent to conditions that we shall call (2’) and (3’), which make sense even when u(x, t) and p(x, t) are very rough. Let us first see how to derive (2’) and (3’), and then we will discuss their use.

The starting point is the observation that a function F on ⁿ is equal to zero if and only if Fθ dx = 0 for every smooth function θ. Applying this remark to the 3D Navier-Stokes equations (2) and (3) and performing a simple formal computation (an integration by parts), we find that (2) and (3) are equivalent to the following equations:

and

More precisely, given any smooth functions u(x, t) and p(x,t), equations (2) and (3) hold if and only if (2’) and (3’) are satisfied for arbitrary smooth functions θ₁(x, t), θ₂ (x, t), θ³ (x, t), and ϕ (x, t) that vanish outside a compact subset of ³ × (0, ∞).

We call θ₁, θ₂, θ₃, and φ test functions, and we say that u and p form a weak solution of 3D Navier-Stokes. Since all the derivatives in (2’) and (3’) are applied to smooth test functions, equations (2’) and (3’) make sense even for very rough functions u and p. To summarize, we have the following conclusion.

A smooth pair (u, p) solves 3D Navier-Stokes if and only if it is a weak solution. However, the idea of a weak solution makes sense even for rough (u, p).

We hope to use weak solutions, by carrying out the following plan.

Step (i): prove that suitable weak solutions exist for 3D Navier-Stokes on all of ³ × (0, ∞).

Step (ii): prove that any suitable weak solution of 3D Navier-Stokes must be smooth.

Step (iii): conclude that the suitable weak solution constructed in step (i) is in fact a smooth solution of the 3D Navier-Stokes equations on all of ³ × (0, ∞).

Here, “suitable” means “not too big”; we omit the precise definition.

Analogues of the above plan have succeeded for interesting partial differential equations. But for 3D Navier-Stokes, the plan has been only partly carried out. It has been known for a long time how to construct suitable weak solutions of 3D Navier-Stokes, but the uniqueness of these solutions has not been proved. Thanks to the work of Sheffer, of Lin, and of Caffarelli, Kohn, and Nirenberg, it is known that any suitable weak solution to 3D Navier-Stokes must be smooth (i.e., it must possess derivatives of all orders), outside a set E ⊂ ³ × (0, ∞) Of small FRACTAL DIMENSION [III.17]. In particular, E cannot contain a curve. To rule out a breakdown, one would have to show that E is the empty set.

For the Euler equation, weak solutions again make sense, but examples due to Sheffer and Shnirelman show that they can behave very strangely. A two-dimensional fluid that is initially at rest and subject to no outside forces can suddenly start moving in a bounded region of space and then return to rest. Such behavior can occur for a weak solution of 2D Euler.

The Navier-Stokes and Euler equations give rise to a number of fundamental problems in addition to the breakdown problem discussed above. We finish this article with one such problem. Suppose that we fix an initial velocity u⁰(x) for the 3D Navier-Stokes or Euler equation. The energy E₀ at time t = 0 is given by

For v ≥ 0, let u ^(v) (x, t) = denote the Navier-Stokes solution with initial velocity u⁰ and with viscosity v. (If v = 0, then u⁽⁰⁾ is an Euler solution.) We assume that u^(v) exists for all time, at least when v > 0. The energy for u^(v) (x, t) at time t ≥ 0 is given by

An elementary calculation based on (1)-(3) (we multiply (1) or (2) by u_i(x), sum over i, integrate over all x ∈ ³, and integrate by parts) shows that

In particular, for the Euler equation we have v = 0, and (6) shows that the energy is equal to E₀, independently of time, as long as the solution exists.

Now suppose that v is small but nonzero. From (6) it is natural to guess that |(d/dt)E^(v) (t)| is small when v is small, so that the energy remains almost constant for a long time. However, numerical and physical experiments suggest strongly that this is not the case. Instead, it seems that there exists T₀ > 0, depending on u⁰ but independent of v, such that the fluid loses at least half of its initial energy by time T₀, regardless of how small v is (provided that v > 0).

It would be very important if one could prove (or disprove) this assertion. We need to understand why a tiny viscosity dissipates a lot of energy.