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Physical derivation of Navier-Stokes Equations, Solution of ODE Global attractors, Banach and Hilbert spaces, Linear functions and dual spaces, sobolev spaces, stokes operator, navier-strokes equations, global attractors for NSE, finite dimension attractors
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In this course we will study the Navier-Stokes equations that govern fluid flow. Having derived the equations we will introduce various tools from func- tional analysis that we need in order to treat them mathematically, and then look in some detail at the mathematical questions of existence and unique- ness of solutions. That done we will concentrate of the two-dimensional equations, viewing them as a dynamical system, and draw some physical conclusions about ‘two-dimensional turbulence’.
The main reference throughout is ‘[R]’, which also contains many addi- tional references:
J.C. Robinson (2001) Infinite-Dimensional Dynamical Systems (Cambridge University Press)
Other useful books are
P. Constantin & C. Foias (1988) Navier-Stokes Equations (University of Chicago Press)
C.R. Doering & J.D. Gibbon (1995) Applied Analysis of the Navier-Stokes Equation (Cambridge University Press)
M. Renardy & R.C. Rogers (1992) An Introduction to Partial Differential Equations, Texts in Applied Mathematics Volume 13 (Springer Verlag, New York).
ular scale (∼ 10 −^8 m), but significantly smaller than the lengths associated with the ‘large-scale motion’ of the fluid (e.g. 1 cm).
For example, Figure 1.1 shows schematically the result of measuring the average density over volumes whose sides have differing lengths. For a range of L away from the smallest and largest scales, these quantities appear to be essentially constant and well-defined.
Figure 1.1: ‘Local averages’ of the density appear well-defined over a range of length scales. (After a figure in G.K. Batchelor’s Fluid Dynamics)
1.2 Langrangian & Eulerian pictures
If we are prepared to accept this continuum hypothesis there are now two ways to think of the fluid.
In the Lagrangian picture we think of the fluid as a collection of ‘fluid particles’. Suppose that we label each particle with its initial position a ∈ D. As the fluid moves, each fluid particle will move, with some velocity v(a, t);
we can find the position x(a, t) of ‘particle a’ at time t by solving the ordinary differential equation dx dt
= v(a, t).
If we assume that the particles all retain their own distinct identity then we can also invert a 7 → x(a, t) if we want to find the initial position of the fluid particle that is now at x, we denote this by a(x, t).
An alternative point of view, which is more useful for us, is to consider quantities referred to coordinates that are fixed in the domain D. This approach, ‘the Eulerian picture’, was pioneered by Euler and led to the first derivation of the equations of motion for fluids. We will denote by u(x, t) the fluid velocity at the point x, i.e.
u(x, t) = v(a(x, t), t).
Suppose that f (x, t) is any function of time and the Eulerian space coor- dinates. If we look how f changes as we move with a fluid particle then we need to consider
∂ ∂t|a
f (x(a, t), t) =
∂f ∂t |x
∑^ d
j=
∂f ∂xj
∂xj ∂t
∂f ∂t
∑^ d
j=
uj
∂f ∂xj
∂f ∂t
The final expression here is the ‘convective derivative’ of f ,
Df Dt
∂f ∂t
reflecting both how f changes in time, and how it changes because the fluid is in motion.
1.3 Conservation of mass
The first equation we need embodies the conservation of mass – as the fluid flows around, mass in neither created nor destroyed. Suppose that V is a
The forces on Vt arise due to the influence of the surrounding fluid, and will act on the boundary of the volume Vt. We now show that these forces can be derived from a symmetric stress tensor σij (x, t), so that the force on an area A is given by
f =
A
σ · n dA,
i.e.
fi =
A
σij nj dA.
To see this, consider a small tetrahedral volume in the fluid, with three of its faces aligned with the coordinate axes, as in Figure 1.2, with sides that have length O(l).
Figure 1.2: A small tetrahedral element in the fluid, and the forces on it
The mass of this little volume will be O(l^3 ), while the area of the faces will be O(l^2 ). Since the magnitude of the forces is proportional to the area
of the faces, if the acceleration is not to become infinite as l → 0, the forces on this volume must cancel. So we should have
f (n)δA + f (−e 1 )δA 1 + f (−e 2 )δA 2 + f (−e 3 )δA 3 = 0
However, since δAj = nj δA, and using Newton’s third law (f (n) = −f (−n)) it follows that f (n) = f (e 1 )n 1 + f (e 2 )n 2 + f (e 3 )n 3 ,
or fi(n) = σij nj
for some appropriate stress tensor σij.
If we now consider a small cube in the fluid with sides of O(l) then the couple on the cube must be zero as l → 0 to prevent infinite torque: this implies that σij = σji, i.e. that σ is symmetric. See Figure 1.
Figure 1.3: The couple on a tiny cube must be zero.
We write the stress tensor in the form
σij = −p δij + Tij ,
taking out the isotropic (direction independent) part to define the pressure p, and leaving the ‘deviatoric stress’ Tij.
and when the fluid is incompressible this is just zero. So for an incompressible fluid, the stress tensor is just a term due to the pressure, plus a scalar multiple of the rate of strain tensor:
σ = −pI + μE.
We are now in a position to derive the equation of motion. Returning to (1.4) we have d dt
Vt
ρu dV =
∂Vt
σ · n dS.
Taking the time derivative under the integral on the left-hand side gives us the convective derivative, while using the divergence theorem on the right- hand side will give as a volume integral: ∫
Vt
Dt
(ρu) dV =
Vt
∇ · σ dV.
Now, since σ = μE we have
[∇ · σ]i =
∑^ d
j=
∂σij ∂xj
∂p ∂xi
∑^ d
j=
∂xj
∂uj ∂xi
∂ui ∂xj
= [−∇p]i + μ
∑^ d
j=
∂^2 ui ∂x^2 j
∂xi
∑d
j=
∂uj ∂xj
= [−∇p]i + μ ∆ui + μ
∂xi
[∇ · u]
= [−∇p + μ ∆u]i,
using the incompressibility condition (∇ · u = 0) again, and defining the Laplacian of a vector u by
[∆u]i := ∆ui :=
∑^ d
j=
∂^2 ui ∂x^2 j
We therefore have ∫
Vt
Dt
(ρu) − μ ∆u + ∇p
dV = 0,
and as before, since this should hold for any volume in the fluid we obtain
D Dt
(ρu) − μ ∆u + ∇p = 0. (1.5)
Finally, since we are assuming that the fluid is incompressible we have
D Dt
(ρu) = ρ
∂u ∂t
and so we obtain the momentum equation
ρ
∂u ∂t
− μ ∆u + ∇p = 0. (1.6)
1.5 The Navier-Stokes equations
We have now derived the Navier-Stokes equations for an incompressible, New- tonian fluid:
ρ
∂u ∂t
− μ ∆u + ∇p = 0 and ∇ · u = 0. (1.7)
We will assume in what follows that ρ is constant.
The unknowns are the velocity u(x, t), a d-component vector and the pressure p(x, t). We can specify the density ρ, and the viscosity μ. There are d + 1 equations in (1.7), so we can at least hope that our model is solvable. In what follows we will investigate when we can prove that this model has physically meaningful solutions that are valid for all t ≥ 0.
First, however, we consider the effect of the various terms in the equation. Note that it is very nearly a linear equation. Indeed, without expanding the convective derivative (1.5) appears to linear. All our analytical problems will arise because of the nonlinear term (u · ∇)u.
1.7 Fourier representation
One great advantage of considering flows with periodic boundary conditions is that we can use Fourier series in order to represent the solutions of the equation. We can expand u(x, t) in terms of an infinite sum of complex exponentials, u(x, t) =
k∈ZdL
u ˆ(k, t)eik·x, (1.11)
where ZdL is the set of all k in the form
( 2 πn 1 L
2 πnd L
with nj ∈ Z,
and the coefficients ˆu(k, t) are given by
uˆ(k, t) =
Ld
Q
u(x, t)e−ik·x^ ddx.
Exercise 1.3. If you are unfamiliar with Fourier series, suppose that you can write u(x, t) as in (1.11). Multiply both sides by e−ik·x^ and integrate to find the above expression for uˆ(k, t).
In order for u(x, t) to be real we must have
uˆ(k, t) = ˆu(−k, t).
If we insist that
Q u(x, t) d
dx = 0 then we must have ˆu(0, t) = 0 for all t, so
in fact u(x, t) =
k∈ Z˙dL
u ˆ(k, t)eik·x, (1.12)
where Z˙dL = ZdL \ 0.
The derivatives of such a Fourier series are easy to calculate, assuming that we can differentiate term-by-term: we have
∂ ∂xj
u(x, t) = i
k∈Z ˙dL
kj uˆ(k, t)eik·x.
We can rewrite the Navier-Stokes equations in terms of the Fourier coef- ficients, assuming that
p(x, t) =
k∈ Z˙dL
p ˆ(k, t)eik·x^ and f (x, t) =
k∈ Z˙dL
ˆf (k, t)eik·x.
The momentum balance equation becomes
d dt
uˆ(k, t) + i
k′+k′′=k
[ˆu(k′, t) · k′′]ˆu(k′′, t) + ν|k|^2 ˆk(k, t) + ikpˆ(k, t) = ˆf (k, t)
(1.13) (in the sum we also have k′, k′′^ ∈ Z˙dL) while the incompressibility condition is simply k · uˆ(k, t) = 0.
If we take the dot product of the momentum equation with k then we can use the incompressibility condition to eliminate ˆpk:
ik ·
k′+k′′=k
[ˆu(k′, t) · k′′]ˆu(k′′, t) + i|k|^2 pˆ(k, t) = k · ˆf (k, t),
and so
pˆ(k, t) =
i |k|^2
ik ·
k′+k′′=k
[ˆu(k′, t) · k′′]ˆu(k′′, t) − k · ˆf (k, t)
Note that if f is not divergence free then that part of f is absorbed by the pressure term. In much of what follows we will therefore take f to be divergence-free, i.e. k · ˆf (k, t) = 0. In this case the second term in the expression for ˆp drops out, and the momentum equation is therefore
d dt
uˆ(k, t) + i
kk |k|^2
k′+k′′=k
[ˆu(k′, t) · k′′]ˆu(k′′, t) + ν|k|^2 uˆ(k, t) = ˆf (k, t)
(1.14)
Without the nonlinear term the equation would be very simple. The Laplacian term causes ˆu(k, t) to decay exponentially fast: with no forcing term this would decay to zero. Note that the rate of decay increases as |k| increases – high wavenumbers, corresponding to small lengthscales in the
In this chapter we will consider how to prove the existence and uniqueness of solutions of ordinary differential equations, using the Contraction Mapping Theorem. First we recall the definition of a Banach space which we need to state this theorem precisely [cf. Chapter 1 of [R]].
A norm on a vector space X is a ‘length’ function ‖·‖ : X → [0, ∞) satisfying
(i) ‖x‖ = 0 iff x = 0,
(ii) ‖λx‖ = |λ|‖x‖ for all x ∈ X, λ ∈ R, and
(iii) ‖x + y‖ ≤ ‖x‖ + ‖y‖ for all x, y ∈ X (the “triangle inequality”).
A normed space consists of a vector space X and a norm ‖ · ‖X : strictly a normed space is a pair (X,‖ · ‖X ), but most spaces have a “standard” norm and we tend to drop this slightly pedantic notation.
Example: Rn^ with the standard Euclidean norm | · | is a normed space.
A sequence {xn} ∈ X is Cauchy if for any ≤ > 0 there exists an N such that ‖xn − xm‖ < ≤ for all n, m > N.
A space X is complete if every Cauchy sequence in X converges to another element of X. Rn^ is complete.
A Banach space is simply a complete normed space: Rn^ is the simplest example.
Two norms on X, ‖ · ‖ 1 and ‖ · ‖ 2 are said to be equivalent if there exist constants α 1 and α 2 with 0 < α 1 < α 2 such that
α 1 ‖x‖ 1 ≤ ‖x‖ 2 ≤ α 2 ‖x‖ 1.
If a sequence converges in the ‖ · ‖ 1 norm iff it converges in the ‖ · ‖ 2 norm, hence the idea that they are “equivalent”. (All norms on Rn^ are equivalent, see Theorem 1.1. in [R].)
Suppose that Ω ⊂ Rn, and X is a normed space. Then a function f : Ω → Y is continuous at x if for every ≤ > 0 there exists a δ > 0 such that
|x − x˜| < δ ⇒ ‖f (x) − f (˜x)‖X < ≤,
and f is continuous if it is continuous at every x ∈ Ω.
The collection C^0 (Ω, X) of all continuous functions from Ω to X has a standard norm, ‖f ‖∞ = sup x∈Ω
‖f (x)‖X.
With this norm C^0 (Ω, X) is complete: any Cauchy sequence converges, and its limit (the uniform limit of continuous functions) is also continuous. Thus C^0 (Ω, X) is a Banach space.