Navier-Stokes Equations I

Navier-Stokes equations I

Sauli Lindberg

University of Helsinki, Fall 2019

Sauli Lindberg Navier-Stokes equations I Fall 2019 1 / 168 Learning objectives of the course

• A knowledge of the basics of ﬂuid dynamics, especially the homogeneous incompressible Navier-Stokes equations ∂u + u · ∇u = −µ∆u − ∇p + F , div u = 0 ∂t

• An understanding of what the Navier-Stokes equations say and what is the role of each variable and each term

• A knowledge of the basic mathematical tools used in ﬂuid dynamics

• A solidiﬁed understanding of vector calculus–we will use cross products, divergence, rotation/curl, gradient, Stokes Theorem etc. in a natural context

• Geometric and physical intuition on the concepts studied

Sauli Lindberg Navier-Stokes equations I Fall 2019 2 / 168 Fluid dynamics

What is fluid dynamics? Study of the motion of a fluid (gas, liquid or plasma) under the action of forces applied on it: https://www.grc.nasa.gov/www/k-12/airplane/state.html The motion of fluids is called flow In principle, one could describe fluid flow by describing the motion of each molecule. The molecules are in constant, random motion, colliding with each other (and possibly with the walls of a container). However, a molecular description is very complicated Fluid dynamics does not neglect the molecular structure of the fluid but takes molecular properties into account by averaging over a high number of molecules For instance, when asking questions such as "How much water flows out of the pipe per second?" we are only interested in the average behaviour of large numbers of atoms, not in their individual behaviour Emphasis is on the features of the fluid that are relevant to the question at hand. For instance, electromagnetic forces and relativistic effects are only considered when they are not negligibly small

Sauli Lindberg Navier-Stokes equations I Fall 2019 3 / 168 The continuum hypothesis

Fluids dynamics assumes the continuum hypothesis.

The ﬂuid is modeled as a continuous substance

Properties such as velocity and pressure are assumed to have a deﬁnite value in every point of a domain in space-time. We assume that these values and their derivatives of all orders vary continuously

Greatly simplifies the analysis of fluid flow

On the scale of atoms, highly inaccurate

However, often extremely accurate at a macroscopic (e.g. visible) scale

The "piece of ﬂuid occupying a point x in space" is called a ﬂuid particle

(Not to be confused with the continuum hypothesis of set theory.)

Sauli Lindberg Navier-Stokes equations I Fall 2019 4 / 168 Some areas, applications and problems of ﬂuid dynamics

Fluid dynamics is divided into Aerodynamics: the study of ﬂow of air and other gases, used e.g. in aeroplane design Hydrodynamics: the study of ﬂow of liquids, used e.g. in naval engineering

Fluid dynamics is used in numerous areas of science and engineering, e.g. meteorology and climate science oceanography plasma physics (solar physics, nuclear fusion etc.)

Fluid dynamics is studied via numerical simulation as well as experimental and theoretical work. Some of the major open research problems: The existence and uniqueness of smooth solutions of the Navier-Stokes equations (Clay Institute Millennium Prize Problem) Turbulence–the most important unsolved problem of classical physics

Sauli Lindberg Navier-Stokes equations I Fall 2019 5 / 168 The mathematical/physical basis

We will follow an axiomatic approach where we use the following axioms.

The continuum assumption.

Mass is neither created nor destroyed.

Newton’s second law: the rate of change of the momentum of a body equals the net force F on the body, F = d(mv)/dt.

Energy is neither created nor destroyed.

Extra assumptions are introduced and modiﬁed later as needed.

Sauli Lindberg Navier-Stokes equations I Fall 2019 6 / 168 The main quantities of interest

Let D ⊂ R3 be a domain (connected open set) ﬁlled with a ﬂuid, and let −∞ < T1 < T2 < ∞.

We assume that the following quantities are smooth functions of (x, t) ∈ D × [T1, T2]: velocity u(x, t) ∈ R3, pressure p(x, t) ∈ R, ﬂuid density %(x, t) ∈ (0, ∞). They are expressed in the following units: u in m/s, p in N/m2 = Pa, % in kg/m3.

Questions: What kinds of equations do u, p and % satisfy? How are those equations derived from the axioms of the previous slide? How do the solutions of the equations behave?

Sauli Lindberg Navier-Stokes equations I Fall 2019 7 / 168 Review of basic notions of vector analysis I

We denote the standard basis vectors of R3 by

e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1).

3 3 Let a = (a1, a2, a3) ∈ R and b = (b1, b2, b3) ∈ R . The inner product of a and b is

3 X a · b = ai bi ∈ R, i=1 and the cross product of a and b is

a × b = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1)

= (a2b3 − a3b2)e1

+ (a3b1 − a1b3)e2 3 + (a1b2 − a2b1)e3 ∈ R .

Sauli Lindberg Navier-Stokes equations I Fall 2019 8 / 168 Review of basic notions of vector analysis II

∞ Let f ∈ C (D × [T1, T2])–here and elsewhere, we denote by f a generic smooth (= inﬁnitely diﬀerentiable) function.

The ﬁrst order partial derivatives of f at (x, t) ∈ D × (T1, T2) are

∂f f (x1 + h, x2, x3, t) − f (x1, x2, x3, t) (x, t) = ∂1f (x) = lim , ∂x1 h→0 h ∂f f (x1, x2 + h, x3, t) − f (x1, x2, x3, t) (x, t) = ∂2f (x) = lim , ∂x2 h→0 h ∂f f (x1, x2, x3 + h, t) − f (x1, x2, x3, t) (x, t) = ∂3f (x) = lim , ∂x3 h→0 h ∂f f (x1, x2, x3, t + h) − f (x1, x2, x3, t) (x, t) = ∂t f (x) = lim ; ∂t h→0 h the gradient of f at (x, t) is ∂f ∂f ∂f 3 ∇f (x, t) = , , (x, t) ∈ R . ∂x1 ∂x2 ∂x3

Sauli Lindberg Navier-Stokes equations I Fall 2019 9 / 168 Review of basic notions of vector analysis III

∞ 3 Suppose then v = (v1, v2, v3) ∈ C (D × [T1, T2], R ). (We denote by v a generic infinitely differentiable vector field.)

The divergence of v at (x, t) is deﬁned by

3 3 X ∂vi X div v(x, t) = ∇ · v(x, t) = (x, t) = ∂i vi (x, t) ∈ R. ∂xi i=1 i=1

The curl of v at (x, t) is ∂v2 ∂v3 ∂v3 ∂v1 ∂v1 ∂v2 3 ∇ × v(x, t) = − , − , − (x, t) ∈ R . ∂x3 ∂x2 ∂x1 ∂x3 ∂x2 ∂x1

Sauli Lindberg Navier-Stokes equations I Fall 2019 10 / 168 Admissible volumes of ﬂuid

We wish to consider volumes W ⊂ D ⊂ R3 of ﬂuid where standard tools of vector analysis are available. Deﬁnition 1 We call W ⊂ D an admissible volume if: W is open and simply connected, and W¯ ⊂ D is compact, R R the Riemannian integral W f dV = W f (x, t)dx1dx2dx3 exists for every ∞ f ∈ C (D × [T1, T2]) and every t ∈ [T1, T2], R R the surface integral ∂W v · n dA = ∂W v(x, t) · n(x)dA exists for every ∞ 3 v ∈ C (D × [T1, T2], R ) and t ∈ [T1, T2]. (Here n(x) is the outward unit normal to ∂W .) R R The Divergence theorem (Gauss’s theorem) ∂W v · n dA = W div v dV holds ∞ for every v ∈ C (D × [T1, T2]) and t ∈ [T1, T2].

Examples: W = (a, b) × (c, d) × (e, f ). W = B(0, r) = {x ∈ R3 : |x| = (x 2 + y 2 + z2)1/2 < r}. 3 2 2 W = {x ∈ R : x1 + x2 < 1, 0 < x3 < 1}. Sauli Lindberg Navier-Stokes equations I Fall 2019 11 / 168 Conservation of mass I

Here and elsewhere we denote by W ⊂ D an admissible volume. Furthermore, let T1 < t < T2.

Deﬁnition 2 R The mass of the ﬂuid in W ⊂ D at time t is m(W , t) = W %dV .

The Principle of conservation of mass says that the rate of increase of mass in W equals the rate of net mass crossing ∂W in the inward direction.

(Note that there might be mass ﬂowing in from some portion of ∂W and mass ﬂowing out from another.)

The rate of change of mass in W is

d d Z (∗) Z ∂% m(W , t) = % dV = dV . dt dt W W ∂t Our next task is to justify equality (∗).

Sauli Lindberg Navier-Stokes equations I Fall 2019 12 / 168 Diﬀerentiating under the integral sign I

Theorem 3

∞ Let f ∈ C (D × [0, T ]) and T1 < t < T2. Then d Z Z ∂f f dV = dV . dt W W ∂t Proof. Let > 0. Our aim is to show the existence of δ > 0 such that R R Z W f (x, t + h) dx − W f (x, t) dx ∂f − (x, t) dV < h W ∂t whenever 0 < |h| < δ. Let 0 < |h| < δ, where δ > 0 is chosen later in such a way that (t − δ, t + δ) ⊂ (T1, T2). Assume h > 0, the case h < 0 being similar. Let x ∈ W . Use the Mean Value Theorem on the function s 7→ g(s) := f (x, s) to write ∂f f (x, t + h) − f (x, t) = g(t + h) − g(t) = hg 0(ξ) = h (x, ξ) ∂t for some ξ ∈ (t, t + h).

Sauli Lindberg Navier-Stokes equations I Fall 2019 13 / 168 Diﬀerentiating under the integral sign II

Now ∂f /∂t is continuous in the compact set W¯ × [t, t + h] and therefore it is uniformly continuous there. Thus there exists δ > 0 such that

∂f ∂f (x, ξ) − (x, t) < R ∂t ∂t W dV whenever |ξ − t| < δ. By possibly making δ smaller, we may assume that (t − δ, t + δ) ⊂ (T1, T2). For such δ, whenever 0 < |h| < δ, we have Z Z f (x, t + h) − f (x, t) ∂f ∂f ∂f − (x, t) dV = (x, ξ) − (x, t) dV W h ∂t W ∂t ∂t Z ∂f ∂f ≤ (x, ξ) − (x, t) dV W ∂t ∂t Z < R dV = . W dV W

Sauli Lindberg Navier-Stokes equations I Fall 2019 14 / 168 Conservation of mass II

Axiom 4 (Principle of conservation of mass) The rate of change of mass in W equals the net rate of mass crossing ∂W in the inward direction, i.e., d Z Z % dV = − % u · n dA (T1 < t < T2). dt W ∂W

The surface integral above is a somewhat diﬃcult object. The Divergence theorem helps us to turn it into an integral over W : Z Z − % u · n dA = − div (% u)dV . ∂W W We thus arrive at the integral form of the continuity equation:

Z ∂% + div (% u) dV = 0 W ∂t for all admissible volumes W ⊂ D.

Sauli Lindberg Navier-Stokes equations I Fall 2019 15 / 168 Conservation of mass III

The integral form of the continuity equation implies the diﬀerential form of the continuity equation: ∂% + div (% u) = 0. (1) ∂t (Exercise 2.3.)

Thus, the rate of change of the density % is minus the divergence of % u. In the Lectures of 11/9/2019 we study what divergence means geometrically and physically.

Sauli Lindberg Navier-Stokes equations I Fall 2019 16 / 168 3 × 3 determinants

Let a, b, c ∈ R3. We write  T    a a1 a2 a3 T det b  = det b1 b2 b3 T c c1 c2 c3

= a1(b2c3 − b3c2) + a2(b3c1 − b1c3) + a3(b1c2 − b2c1) 3 X = ijk ai bj ck , i,j,k=1 where  1, (ijk) ∈ {(123), (231), (312)},  ijk = −1, (ijk) ∈ {(132), (213), (321)}, 0, otherwise  1, (ijk) is an even permutation of (1, 2, 3),  = −1, (ijk) is an odd permutation of (1, 2, 3), 0, (ijk) has a repeated index.

Here, ijk are called Levi-Civita symbols. Sauli Lindberg Navier-Stokes equations I Fall 2019 17 / 168 Basic properties of Levi-Civita symbols

If we do a transposition (change two indices), the sign changes: e.g. 123 = 1 and 213 = −1. In general, jik = −ijk , kji = −ijk and ikj = −ijk .

If an index is repeated, then the symbol is zero: 113 = 0, 222 = 0 etc.

If we do two transpositions, the sign changes back: 312 = −213 = 123. In general, ijk = jki = kij .

Sauli Lindberg Navier-Stokes equations I Fall 2019 18 / 168 Cross product in terms of Levi-Civita symbols

Recall that when a, b ∈ R3, we have

3 a × b = (a2b3 − a3b2)e1 + (a3b1 − a1b3)e2 + (a1b2 − a2b1)e3 ∈ R . Using Levi-Civita symbols, we write this as

3 X (a × b)i = ijk aj bk , i = 1, 2, 3, j,k=1

so that 3 X a × b = ijk ei aj bk . i,j,k=1 Note that

3 3 3 X X X a × b = ijk ei aj bk = ijk ei bk aj = − ikj ei bk aj = −b × a. i,j,k=1 i,j,k=1 i,j,k=1

More properties of the cross product are found in Exercise set 1.

Sauli Lindberg Navier-Stokes equations I Fall 2019 19 / 168 Surface integrals

3 Recall from yesterday: D ⊂ R is open and connected, and −∞ < T1 < T2 < ∞. The set W is open and W¯ ⊂ D is compact, and the Divergence Theorem holds in W .

∞ 3 ∞ We studied a velocity u ∈ C (D × (T1, T2), R ), density % ∈ C (D × (T1, T2)) ∞ and pressure p ∈ C (D × (T1, T2)). Recall:

Axiom 5 (Principle of conservation of mass) The rate of change of mass in W equals the net rate of mass crossing ∂W in the inward direction, i.e., d Z Z % dV = − % u · n dA (T1 < t < T2). dt W ∂W R R We used the Divergence Theorem to write − ∂W % u · n dA = − W div (% u). Our next aim is to study outer unit normals, surface integrals and the divergence operator.

Sauli Lindberg Navier-Stokes equations I Fall 2019 20 / 168 Outer unit normal to the upper hemisphere I

We wish compute an outer unit normal to the unit ball 3 P3 2 B(0, 1) = {x = (x1, x2, x3) ∈ R : i=1 |xi | < 1}. P3 2 The upper hemisphere S1 = {x: i=1 |xi | = 1, x3 > 0} is parametrised by q 2 2 x = r(x1, x2) = x1, x2, 1 − x1 − x2 .

2 2 2 Fix x ∈ S1, so that x1 + x2 = 1 − x3 < 1.

Consider the path γ = (γ1, γ2, γ3):(x1 − δ, x1 + δ) → S1 deﬁned by 2 2 γ(t) = r(t, x2). (Choose δ > 0 so small that t + x2 < 1 for t ∈ (x1 − δ, x1 + δ).) The derivative 0 0 0 0 ∂r1 ∂r2 ∂r3 γ (x1) = (γ1(x1), γ2(x1), γ3(x1)) = (x1, x2), (x1, x2), (x1, x2) ∂x1 ∂x1 ∂x1 ! x = 1, 0, − 1 p 2 2 1 − x1 − x2

0 is perpendicular to x: x · γ (x1) = x1 − x1 = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 21 / 168 Outer unit normal to the upper hemisphere II

We denote ∂r ∂r1 ∂r2 ∂r3 (x1, x2) = (x1, x2), (x1, x2), (x1, x2) . ∂x1 ∂x1 ∂x1 ∂x1 Similarly, ∂r ∂r1 ∂r2 ∂r3 (x1, x2) = (x1, x2), (x1, x2), (x1, x2) ∂x2 ∂x2 ∂x1 ∂x1 ! x = 0, 1, − 2 p 2 2 1 − x1 − x2

satisﬁes x · ∂r/∂x2(x1, x2) = x2 − x2 = 0. Now ∂r ∂r ∂r ∂r ∂r ∂r (x1, x2)· (x1, x2)× (x1, x2) = (x1, x2)· (x1, x2)× (x1, x2) = 0 ∂x1 ∂x1 ∂x2 ∂x2 ∂x1 ∂x2 (Exercise 1.2). Note the order of the operations: a · b × c = a · (b × c).

Sauli Lindberg Navier-Stokes equations I Fall 2019 22 / 168 Outer unit normal to the upper hemisphere III

Now ! ∂r ∂r x x x (x , x )× (x , x ) = 1 , 2 , 1 = ∂x 1 2 ∂x 1 2 p 2 2 p 2 2 p 2 2 1 2 1 − x1 − x2 1 − x1 − x2 1 − x1 − x2

is a normal vector to S1 at x. We deﬁne the outer unit normal vector by setting

∂r ∂r (x1, x2) × (x1, x2) n(x) = ∂x1 ∂x2 = x. ∂r ∂r (x1, x2) × (x1, x2) ∂x1 ∂x2

P3 2 On the lower hemisphere S2 = {x: i=1 |xi | = 1, x3 < 0} the outer unit normal vector is also of the form n(x) = x. However, by repeating the construction above p 2 2 with the parametrisation r(x1, x2) = (x1, x2, − 1 − x1 − x2 ), 2 2 S2 = {r(x1, x2): x1 + x2 < 1} one gets the inner unit normal n(x) = −x.A diﬀerent parametrisation gives the outer unit normal. (See Exercise 2.4.)

The correct choice of the outer unit normal is studied in diﬀerential geometry. In this course, the choice is intuitively obvious from context.

Sauli Lindberg Navier-Stokes equations I Fall 2019 23 / 168 Surface integrals on parametrised surfaces I

Above we studied the upper hemisphere as the graph of the function p 2 2 f (x1, x2) = 1 − x1 − x2 . We next consider more general parametrised surfaces in R3.

Suppose that U ⊂ R2 is open and that Φ ∈ C 0(U¯, R3) and Φ ∈ C ∞(U, R3). Denote S = {(Φ1(p, q), Φ2(p, q), Φ3(p, q)): (p, q) ∈ U¯}. For every (p, q) ∈ U set

∂Φ ∂Φ ∂Φ ∂Φ (p, q) = 1 (p, q), 2 (p, q), 3 (p, q) , ∂p ∂p ∂p ∂p ∂Φ ∂Φ ∂Φ ∂Φ (p, q) = 1 (p, q), 2 (p, q), 3 (p, q) . ∂q ∂q ∂q ∂q

Suppose that ∂Φ ∂Φ (p, q) × (p, q) 6= 0 at every (p, q) ∈ U. ∂p ∂q

Sauli Lindberg Navier-Stokes equations I Fall 2019 24 / 168 Surface integrals on parametrised surfaces II

Whenever (p, q) ∈ U, at x = Φ(p, q) ∈ S we deﬁne

∂Φ ∂Φ ∂p (p, q) × ∂q (p, q) n(x) = n(Φ(p, q)) = . ∂Φ (p, q) × ∂Φ (p, q) ∂p ∂q

Now, if v ∈ C 0(S, R3), then the surface integral of v over S is deﬁned by Z Z ∂Φ ∂Φ v(x) · n(x) dA = v(Φ(p, q)) · (p, q) × (p, q) dp dq. S U ∂p ∂q

(We should, in fact, choose Φ in a suitable way to make n(x) the outer unit normal in our applications. See Exercise set 2.)

Sauli Lindberg Navier-Stokes equations I Fall 2019 25 / 168 An example of a surface integral

Consider W = (a, b) × (c, d) ⊂ (e, f ) ⊂ R3 and v ∈ C 0(W¯ , R3). One of the six faces of the rectangle W is S = {(x1, x2, f ): a ≤ x1 ≤ b, c ≤ x2 ≤ d}.

We parametrise S as

S = {Φ(x1, x2):(x1, x2) ∈ [a, b] × [c, d]}

= {(x1, x2, f ):(x1, x2) ∈ [a, b] × [c, d]}.

Now ∂Φ ∂Φ (x1, x2) = (1, 0, 0) and (x1, x2) = (0, 1, 0) ∂x1 ∂x2

at each (x1, x2) ∈ (a, b) × (c, d), so that ∂Φ ∂Φ n(x) = (x1, x2) × (x1, x2) = (0, 0, 1), ∂x1 ∂x2 Z Z b Z d v(x) · n(x) dA = v3(x1, x2, f ) dx1 dx2. S a c

Sauli Lindberg Navier-Stokes equations I Fall 2019 26 / 168 Elementary regions I

Note that we can write x ∈ B¯(0, 1) as

− 1 ≤ x1 ≤ 1, q q 2 2 − 1 − x1 ≤ x2 ≤ 1 − x1 , q q 2 2 2 2 − 1 − x1 − x2 ≤ x3 ≤ 1 − x1 − x2 .

We call an open set W ⊂ R3 an elementary region of type 1 if W¯ consists of points satisfying

a ≤ x1 ≤ b,

f1(x1) ≤ x2 ≤ f2(x1),

g1(x1, x2) ≤ x3 ≤ g2(x1, x2), where −∞ < a < b < ∞,

f1 and f2 are continuous in [a, b] and infinitely differentiable in (a, b), g1 and g2 are continuous in {(x1, x2): a ≤ x1 ≤ b, f1(x1) ≤ x2 ≤ f2(x2)} and infinitely differentiable in {(x1, x2): a < x1 < b, f1(x1) < x2 < f2(x2)}.

Sauli Lindberg Navier-Stokes equations I Fall 2019 27 / 168 Elementary regions II

Example: Let a = −1 and b = 1, and deﬁne q q 2 2 f1(x1) = − 1 − x1 , f2(x1) = 1 − x1 , q q 2 2 2 2 g1(x1, x2) = − 1 − x1 − x2 , g2(x1, x2) = 1 − x1 − x2 .

Then W¯ = B¯(0, 1).

Example: Let a < b, and deﬁne

f1(x1) = c, f2(x1) = d, g1(x1, x2) = e, g2(x1, x2) = f .

Then W¯ = [a, b] × [c, d] × [e, f ].

Example: Let a = −1 and b = 1. Deﬁne q q 2 2 f1(x1) = − 1 − x1 , f2(x1) = 1 − x1 , g1(x1, x2) = −1, g2(x1, x2) = 1.

¯ 2 2 Then W is the cylinder {(x1, x2): x1 + x2 ≤ 1} × [−1, 1].

Sauli Lindberg Navier-Stokes equations I Fall 2019 28 / 168 Elementary regions III

Recall that an open set W ⊂ R3 is an elementary region of type 1 if W¯ consists of points satisfying

a ≤ x1 ≤ b, (2)

f1(x1) ≤ x2 ≤ f2(x1), (3)

g1(x1, x2) ≤ x3 ≤ g2(x1, x2). (4)

Now the boundary ∂W¯ consists of points where one of the six inequalities in (2)–(4) is an equality.

Thus ∂W is the union of the following six sets:

S1 = {(a, x2, x3): f1(a) ≤ x2 ≤ f2(a), g1(a, x2) ≤ x3 ≤ g2(a, x2)},

S2 = {(b, x2, x3): f1(b) ≤ x2 ≤ f2(b), g1(b, x2) ≤ x3 ≤ g2(b, x2)},

S3 = {(x1, f1(x1), x3): a ≤ x1 ≤ b, g1(x1, f1(x1)) ≤ x3 ≤ g2(x1, f1(x1))},

S4 = {(x1, f2(x1), x3): a ≤ x1 ≤ b, g1(x1, f2(x1)) ≤ x3 ≤ g2(x1, f2(x1))},

S5 = {(x1, x2, g1(x1, x2)): a ≤ x1 ≤ b, f1(x1) ≤ x2 ≤ f2(x1)},

S6 = {(x1, x2, g2(x1, x2)): a ≤ x1 ≤ b, f1(x1) ≤ x2 ≤ f2(x1)}.

Sauli Lindberg Navier-Stokes equations I Fall 2019 29 / 168 Divergence Theorem in symmetric elementary regions

Similarly, W is said to be an elementary region of type 2 if W¯ consists of points satisfying

a ≤ x2 ≤ b,

f˜1(x2) ≤ x3 ≤ f˜2(x2),

g˜1(x2, x3) ≤ x1 ≤ g˜2(x2, x3).

And so on.

We call W a symmetric elementary region if it is an elementary region of all types (1-6).

Theorem 6 Suppose W is a symmetric elementary region and W¯ ⊂ D. Let ∞ 3 v ∈ C (D × (T1, T2), R ). Then

6 Z X Z Z v(x, t)·n(x) dA = v(x, t)·n(x)dA = div v(x, t) dx (T1 < t < T2). ∂W i=1 Si W

Sauli Lindberg Navier-Stokes equations I Fall 2019 30 / 168 Divergence Theorem on a rectangle

Sketch of the proof of the case W = (a, b) × (c, d) × (e, f ).

Recall that div v = ∂v1/∂x1 + ∂v2/∂x2 + ∂v3/∂x3. Consider the third term: by the Fundamental theorem of calculus,

Z b Z d Z f ∂v3 (x1, x2, x3, t) dx1 dx2 dx3 a c e ∂x3 Z b Z d = (v3(x1, x2, f , t) − v3(x1, x2, e, t)) dx1 dx2 a c Z Z = v(x, t) · n(x) dA + v(x, t) · n(x) dA. S5 S6

By doing a similar computation on ∂v1/∂x1 and ∂v2/∂x2 we get

6 Z X Z div v(x, t) dV = v(x, t) · n(x)dA. W i=1 Si

Sauli Lindberg Navier-Stokes equations I Fall 2019 31 / 168 Eulerian and Lagrangian frames of reference

Eulerian frame of reference: Consider the particle moving through x at time t ∈ (T1, T2). Denote its velocity by u(x, t). This is the frame of reference of an observer ﬁxed in space.

When t1 < t2, u(x, t1) and u(x, t2) are typically velocities of two diﬀerent particles. The ﬁrst particle occupies the point x ∈ D at time t1 and the second one occupies x at time t2.

Lagrangian frame of reference: Suppose a particle is at x ∈ D at time t = T1. 3 Denote its position at time t ∈ [T1, T2] by ϕ(x, t) ∈ R . Thus ϕ(x, T1) = x. The velocity of the ﬂuid particle at time t is ∂ϕ/∂t(x, t).

On the other hand, the particle occupies the point ϕ(x, t) at time t. We therefore assume that ϕ and u satisfy ∂ [ϕ(x, t)] = u(ϕ(x, t), t). (5) ∂t We call ϕ the ﬂow map. We assume furthermore: ∞ ϕ = (ϕ1, ϕ2, ϕ3) ∈ C (D × [T1, T2], D).

x 7→ ϕ(x, t): D → D is a bijection at every t ∈ [T1, T2].

Sauli Lindberg Navier-Stokes equations I Fall 2019 32 / 168 Interpretation of the conditions on the ﬂow map

The condition ϕ(x, T1) = x is made in order that ϕ follow the trajectory of the ﬂuid particle that was at x at time T1.

The condition that x 7→ ϕ(x, t) maps D into D says that the ﬂuid is contained in a container with a solid wall ∂D (impenetrable to the ﬂuid).

The injectivity of x 7→ ϕ(x, t) says that two ﬂuid particles never occupy the same point in space.

The surjectivity of x 7→ ϕ(x, t) says that no vacuum is created by the ﬂow.

The equation ∂ [ϕ(x, t)] = u(ϕ(x, t), t) ∂t says that u(ϕ(x, t), t) gives the velocity of the particle at time t.

Sauli Lindberg Navier-Stokes equations I Fall 2019 33 / 168 An example: constant-velocity ﬂow in an inﬁnite pipe

We illustrate equation (5) in a simple case. Set

2 3 D = {(x1, x2, x3): x1 ∈ R, x2 + x2 < 1}. Thus, D is a horisontal pipe of inﬁnite length. Suppose the ﬂuid is moving in the x1-direction at constant velocity,

u(x, t) = (1, 0, 0) (6)

at every (x, t) ∈ D × [T1, T2]. Substituting this velocity ﬁeld u into (5), note that the ﬂow map ϕ must satisfy ∂ϕ/∂t(x, t) = (1, 0, 0) at every (x, t) ∈ D × (T1, T2) as well as ϕ(x, T1) = x for all x ∈ D. Now ϕ(x, t) = x + (t − T1, 0, 0) (7) has all the desired properties.

We interpret the results geometrically. Formula (6) says that the ﬂuid is moving at the same velocity in the whole space-time domain, while (7) says that each ﬂuid particle moves in the x1-direction by one unit of length per a unit of time.

Sauli Lindberg Navier-Stokes equations I Fall 2019 34 / 168 Acceleration of a ﬂuid particle

Our next aim is to express Newton’s second law in the context of ﬂuid dynamics. We start by studying the acceleration of the ﬂuid particles.

Consider a ﬂuid particle that was at x ∈ D at time T1. Its position at time t ∈ (T1, T2) is ϕ(x, t). Recall that the velocity of the ﬂuid particle is ∂ϕ (x, t) = u(ϕ(x, t), t). (8) ∂t The acceleration of the particle is ∂2ϕ ∂ (x, t) = [u(ϕ(x, t), t)]. ∂t2 ∂t For each j ∈ {1, 2, 3} we use the Chain rule and (8): 3 ∂ ∂uj X ∂uj ∂ϕk [uj (ϕ(x, t), t)] = (ϕ(x, t), t) + (ϕ(x, t), t) (x, t) ∂t ∂t ∂xk ∂t k=1 3 ∂uj X ∂uj = (ϕ(x, t), t) + (ϕ(x, t), t)uk (ϕ(x, t), t) ∂t ∂xk k=1 ∂u = j + u · ∇u (ϕ(x, t), t). ∂t j Sauli Lindberg Navier-Stokes equations I Fall 2019 35 / 168 The material derivative I

3 We deﬁne u · ∇u: D × [T1, T2) → R by

(u · ∇u)j = u · ∇uj .

Then the acceleration of the ﬂuid particle at time t is

∂2ϕ ∂ ∂u (x, t) = [u(ϕ(x, t), t)] = + u · ∇u (ϕ(x, t), t). ∂t2 ∂t ∂t

We denote the so-called material derivative of u by Du ∂u = + u · ∇u. Dt ∂t The acceleration of the ﬂuid particle at time t is thus ∂ Du [u(ϕ(x, t), t)] = (ϕ(x, t), t). ∂t Dt

Sauli Lindberg Navier-Stokes equations I Fall 2019 36 / 168 The material derivative II

More generally:

Deﬁnition 7 ∞ Let f ∈ C (D × [T1, T2]). The material derivative of f at (x, t) is Df ∂f (x, t) = (x, t) + u(x, t) · ∇f (x, t). Dt ∂t

∞ 3 Similarly, if v ∈ C (D × [T1, T2], R ), the material derivative of v at (x, t) is Dv ∂v (x, t) = (x, t) + u(x, t) · ∇v(x, t). Dt ∂t Sometimes Df /Dt is called the convective derivative or the substantial derivative.

Sauli Lindberg Navier-Stokes equations I Fall 2019 37 / 168 A geometric interpretation of the material derivative

∞ Question: What does the material derivative of a function f ∈ C (D × [T1, T2]) ∞ 3 or vector ﬁeld v ∈ C (D × [T1, T2], R ) measure?

When x ∈ D and T1 ≤ t1 < t2 ≤ T2,

Z t2 ∂ Z t2 Df f (ϕ(x, t2), t2) − f (ϕ(x, t1), t1) = [f (ϕ(x, t), t)] dt = (ϕ(x, t), t) dt. t1 ∂t t1 Dt Thus Df /Dt measures the rate of change of f in the frame of reference of the moving ﬂuid particle.

We have Df /Dt = 0 if and only if the quantity f remains constant in time for each ﬂuid particle.

E.g. Du/Dt = 0 if and only if the ﬂuid particles do not perceive acceleration.

In turn, ∂f /∂t = 0 means that at each point x ∈ D, the value f (x, t) is constant in t.

Sauli Lindberg Navier-Stokes equations I Fall 2019 38 / 168 An example: rotation around the x3-axis

We illustrate the material derivative in the case of the velocity ﬁeld

u(x, t) = (−x2, x1, 0),

2 2 D = {x: x1 + x2 < 1, 0 < x3 < 1}, studied in Exercise 2.6. Suppose %(x, t) = 1 for all (x, t) ∈ D × [T1, T2].

Clearly ∂u/∂t = 0. However, Du/Dt does not vanish:

u · ∇u  (−x , x , 0) · (0, −1, 0) Du 1 2 1 = u · ∇u = u · ∇u = (−x , x , 0) · (1, 0, 0) = (−x , −x , 0). Dt  2  2 1  1 2 u · ∇u3 (−x2, x1, 0) · (0, 0, 0)

Thus, the fluid particles perceive acceleration towards the axis of rotation. (However, to an observer fixed in space, the flow looks identical at every time t ∈ [T1, T2].)

Sauli Lindberg Navier-Stokes equations I Fall 2019 39 / 168 Volumes moving with the ﬂuid

Given t ∈ [T1, T2], we denote

ϕt (x) = ϕ(x, t),

Wt = ϕt (W ) = {ϕt (x): x ∈ W }.

We wish to formulate Newton’s second law (rate of change of momentum equals total force) on the moving volumes Wt .

In order to avoid repetition, we study the following general problem: Question 8 ∞ R If f ∈ C (D × [Tt , T2]), what is the rate of change (d/dt) f dV ? Wt

∞ 3 R If v ∈ C (D × [Tt , T2], ), what is the rate of change (d/dt) v dV ? R Wt

Sauli Lindberg Navier-Stokes equations I Fall 2019 40 / 168 Change of variables in an integral

Denote ∂ϕ1 ∂ϕ1 ∂ϕ1 

 ∂x1 ∂x2 ∂x3  3 ∂ϕ2 ∂ϕ2 ∂ϕ2  X ∂ϕ1 ∂ϕ2 ∂ϕ3 Jϕ = det Dϕ = det   = ijk .  ∂x ∂x ∂x  ∂xi ∂xj ∂xk  1 2 3  i,j,k=1 ∂ϕ3 ∂ϕ3 ∂ϕ3  ∂x1 ∂x2 ∂x3

Theorem 9 (Change of variables in an integral)

R R ∞ f (x, t)dV = f (ϕ(x, t), t)Jϕ(x, t)dV for every f ∈ C (D × [T1, T2]). Wt W Using Theorem 9 and the Product rule ∂(gh)/∂t = g∂h/∂t + h∂g/∂t, we can write d Z Z ∂ Z ∂Jϕ f dV = [f (ϕ(x, t), t)]Jϕ(x, t) dV + f (ϕ(x, t)) (x, t) dV dt Wt W ∂t W ∂t Z Df Z ∂Jϕ = (ϕ(x, t), t)Jϕ(x, t) dV + f (ϕ(x, t)) (x, t) dV W Dt W ∂t = I1 + I2. (completed in the slide "Transport theorem")

Sauli Lindberg Navier-Stokes equations I Fall 2019 41 / 168 Euler’s identity I

Theorem 10 (Euler’s identity)

For all (x, t) ∈ D × (T1, T2), ∂Jϕ t (x) = [div u](ϕ (x), t) Jϕ (x). ∂t t t Proof: Using Levi-Civita symbols and the Product rule, we compute

3 3 2 ∂Jϕt ∂ X ∂ϕ1 ∂ϕ2 ∂ϕ3 X ∂ ϕ1 ∂ϕ2 ∂ϕ3 = ijk = ijk ∂t ∂t ∂xi ∂xj ∂xk ∂t∂xi ∂xj ∂xk i,j,k=1 i,j,k=1

3 2 X ∂ϕ1 ∂ ϕ2 ∂ϕ3 + ijk ∂xi ∂t∂xj ∂xk i,j,k=1

3 2 X ∂ϕ1 ∂ϕ2 ∂ ϕ3 + ijk . ∂xi ∂xj ∂t∂xk i,j,k=1

Sauli Lindberg Navier-Stokes equations I Fall 2019 42 / 168 Euler’s identity II

Now, using the equation ∂ϕ/∂t = u(ϕ, t) and the Chain rule,

3 2 3 X ∂ ϕ1 ∂ϕ2 ∂ϕ3 X ∂ ∂ϕ1 ∂ϕ2 ∂ϕ3 ijk = ijk ∂t∂xi ∂xj ∂xk ∂xi ∂t ∂xj ∂xk i,j,k=1 i,j,k=1 3 X ∂ ∂ϕ2 ∂ϕ3 = ijk [u1(ϕ, t)] ∂xi ∂xj ∂xk i,j,k=1 3 3 X X ∂u1 ∂ϕ` ∂ϕ2 ∂ϕ3 = ijk (ϕ, t) ∂x` ∂xi ∂xj ∂xk i,j,k=1 `=1 3 3 X ∂u1 X ∂ϕ` ∂ϕ2 ∂ϕ3 = (ϕ, t) ijk ∂x` ∂xi ∂xj ∂xk `=1 i,j,k=1 3 X ∂u1 = (ϕ, t)J(ϕ`, ϕ2, ϕ3) ∂x` `=1 ∂u1 = (ϕ, t)J(ϕ1, ϕ2, ϕ3). ∂x1 The claim follows by repeating the calculation above for the other two sums. Sauli Lindberg Navier-Stokes equations I Fall 2019 43 / 168 Forces acting on a ﬂuid I

The forces acting on a volume of ﬂuid are divided into two types:

• Stresses, whereby the rest of the continuum acts on a volume of ﬂuid W by forces across its surface ∂W . (E.g. pressure.) • Body forces (external forces) which generate a force per unit volume on the ﬂuid. (E.g. gravity.)

A simple (but useful) model: Deﬁnition 11

An ideal fluid is a fluid with the following property: there exists a function ∞ p ∈ C (D × [T1, T2]), called the pressure, with the following property: if W is an admissible volume, then the total force exerted on the fluid inside W by means of stress on ∂W at time t is Z 3 S∂W (t) = − p(x, t)n(x)dA ∈ R . ∂W

On ∂W , the force −p(x, t)n(x) only acts in the direction of n; no tangential forces act on ∂W . (The ﬂuid has no internal friction.)

Sauli Lindberg Navier-Stokes equations I Fall 2019 44 / 168 Forces acting on a ﬂuid II

R We next wish to write S∂W (t) = − ∂W p n dA as a volume integral over W .

If e ∈ R3, then the Divergence theorem yields Z Z e · S∂W (t) = − p(x, t)e · n(x, t) dA = − div (p(x, t)e) dx ∂W W Z = − ∇p(x, t) · e dx. W R Thus S∂W (t) = − W ∇p(x, t) dx by Exercise 1.1.

Deﬁnition 12 If b(x, t) denotes the net body force per unit mass at (x, t), the total body force R on W is B(t) = W %(x, t) b(x, t) dx.

Sauli Lindberg Navier-Stokes equations I Fall 2019 45 / 168 Newton’s second law I

We express Newton’s second law in this context: Axiom 13 (Law of balance of momentum) For an ideal ﬂuid, d Z Z Z % u dV = − ∇p dV + % b dV (9) dt Wt Wt Wt for all admissible volumes W . Now (9) involves integrals over domains that move in time. We wish to put it into a more manageable form. Using a change of variables and the shorthand notation ϕ = ϕ(x, t), d Z d Z % u dV = %(ϕ, t)u(ϕ, t)Jϕ dV dt Wt dt W Z ∂ = [%(ϕ, t)u(ϕ, t)Jϕ] dV W ∂t Z ∂ Z ∂Jϕ = [%(ϕ, t)u(ϕ, t)]JϕdV + %(ϕ, t)u(ϕ, t) dV W ∂t W ∂t = I1 + I2. Sauli Lindberg Navier-Stokes equations I Fall 2019 46 / 168 Newton’s second law II

We write Z ∂ ∂ I1 = %(ϕ, t) [u(ϕ, t)] + [%(%, t)]u(%, t) Jϕ dV W ∂t ∂t Z Du D% = %(ϕ, t) (ϕ, t) + (ϕ, t)u(%, t) Jϕ dV . W Dt Dt We next note that the continuity equation ∂%/∂t + div (% u) = 0 can be written as

D% ∂% ∂% + % div u = + u · ∇% + % div u = + div (% u) = 0. Dt ∂t ∂t

Thus, using Euler’s identity on I2, Z Du I1 = %(ϕ, t) (ϕ, t) − %(ϕ, t)[div u(ϕ, t)] u(ϕ, t) Jϕ dV , W Dt Z I2 = %(ϕ, t)[div u(ϕ, t)] u(ϕ, t)Jϕ dV . W

Sauli Lindberg Navier-Stokes equations I Fall 2019 47 / 168 Newton’s second law III

Our aim was to ﬁnd a convenient form for d Z Z Z % u dV = − ∇p dV + % b dV . (10) dt Wt Wt Wt We have computed d Z Z Du % u dV = I1 + I2 = %(ϕ, t) (ϕ, t)Jϕ dV , dt Wt W Dt and by a change of variables, Z Z Z − ∇p dV + % b dV = [−∇p(ϕ, t) + %(ϕ, t)b(ϕ, t)]Jϕ dV , Wt Wt W so that (10) can be written as

Z Du Z %(ϕ, t) (ϕ, t)Jϕ dV = [−∇p(ϕ, t) + %(ϕ, t)b(ϕ, t)]Jϕ dV . (11) W Dt W

Sauli Lindberg Navier-Stokes equations I Fall 2019 48 / 168 Cauchy momentum equation

The following theorem follows from (11) (see Exercise 3.6):

Theorem 14

Suppose a ﬂuid is ideal. Then, in D × (T1, T2), Du % = −∇p + %b. (12) Dt

Equation (12) is (a special case of) the Cauchy momentum equation or the law of balance of momentum.

If there is no net force on an ideal ﬂuid, then Du/Dt = 0, i.e., the ﬂuid particles do not perceive acceleration.

Sauli Lindberg Navier-Stokes equations I Fall 2019 49 / 168 A recap of our progress so far

First, by using the axiom of conservation of mass, we obtained the continuity equation ∂% + div (%u) = 0. ∂t If the fluid is ideal, then Newton’s second law implies that Du % = −∇p + %b. Dt Different extra assumptions then lead to different models of fluid dynamics.

We will work towards a speciﬁc model, the incompressible Euler equations. (They will eventually be replaced by the incompressible Navier-Stokes equations.)

Sauli Lindberg Navier-Stokes equations I Fall 2019 50 / 168 Transport theorem

We ﬁnish the computation of the slide "Change of variables in an integral": d Z f dV dt Wt Z Df Z ∂Jϕ = (ϕ(x, t), t)Jϕ(x, t) dV + f (ϕ(x, t)) (x, t) dV W Dt W ∂t Z ∂f = (ϕ(x, t), t) + [u · ∇f ](ϕ(x, t), t) + [f div u](ϕ(x, t), t) Jϕ(x, t) dV W ∂t Z ∂f = + u · ∇f + f div u dV (change of variables) Wt ∂t Z ∂f = + div (f u) dV . (product rule) Wt ∂t

Theorem 15 (Transport theorem) ∞ For every f ∈ C (D × [T1, T2]),

d Z Z ∂f f dV = + div (f u) dV . dt Wt Wt ∂t

Sauli Lindberg Navier-Stokes equations I Fall 2019 51 / 168 Volumes moving with the ﬂuid

Given t ∈ [T1, T2], we denote

ϕt (x) = ϕ(x, t),

Wt = ϕt (W ) = {ϕt (x): x ∈ W }.

Yesterday we studied the following general problem: Question 16 ∞ R If f ∈ C (D × [Tt , T2]), what is the rate of change (d/dt) f dV ? Wt

∞ 3 R If v ∈ C (D × [Tt , T2], ), what is the rate of change (d/dt) v dV ? R Wt

Denoting (ϕ, t) = (ϕ(x, t), t), we found the expressions

d Z Z Df f dV = (ϕ, t) + [f div u](ϕ, t) Jϕ dV (13) dt Wt W Dt Z ∂f = + div (f u) dV . (Transport theorem) (14) Wt ∂t

Sauli Lindberg Navier-Stokes equations I Fall 2019 52 / 168 Transport theorem with density

Corollary 17 (Transport theorem with density) ∞ Let f ∈ C (D × [T1, T2]). Then d Z Z Df % f dV = % dV . dt Wt Wt Dt

Proof. Inserting % f into (14) and using the Continuity equation (1),

d Z Z ∂(% f ) % f dV = + div (f % u) dV dt Wt Wt ∂t Z ∂f ∂% = % + f + f div (% u) + % u · ∇f dV Wt ∂t ∂t Z Df = % dV . Wt Dt

Sauli Lindberg Navier-Stokes equations I Fall 2019 53 / 168 The rate of change of speciﬁc quantities

We computed the rate of change of momentum of Wt as d Z Z Du % u dV = %(ϕ, t) (ϕ, t)Jϕ dV dt Wt W Dt Z = [−∇p(ϕ, t) + %(ϕ, t) b(ϕ, t)]Jϕ dV . (ideal ﬂuid) W

We next study the following questions:

R How does the mass % dV of Wt vary in time? Wt R How does the volume dV of Wt vary in time? Wt R R How does the average density % dV / dV in Wt vary in time? Wt Wt

Sauli Lindberg Navier-Stokes equations I Fall 2019 54 / 168 Conservation of mass for moving volumes of ﬂuid

Recall that the continuity equation ∂%/∂t + div (%u) = 0 can be written as

D% ∂% ∂% + % div u = + u · ∇% + % div u = + div (% u) = 0. (15) Dt ∂t ∂t

By (13) and (15), the rate of change of mass of Wt is

d Z Z D% % dV = (ϕ, t) + [% div u](ϕ, t) Jϕ dV = 0 (16) dt Wt W Dt (as expected).

Contrast (16) with the formula

d Z Z % dV = − % u · ndA dt W ∂W where W ⊂ D does not vary in time.

Sauli Lindberg Navier-Stokes equations I Fall 2019 55 / 168 The rate of change of volume for moving volumes of ﬂuid

The rate of change of volume of Wt is d Z d Z Z ∂Jϕ dV = Jϕ(x, t) dV = (x, t) dV (17) dt Wt dt W W ∂t Z Z = div u(ϕ, t)Jϕ(x, t) dV = div u(x, t) dV . (18) W Wt (Here we used a change of variables and Euler’s formula

∂Jϕ = div u(ϕ, t) Jϕ.) ∂t Thus, div u controls the change of volume of the ﬂuid.

Sauli Lindberg Navier-Stokes equations I Fall 2019 56 / 168 Examples of divergence

3 In the following examples, the ﬂow domain is R × [T1, T2]. P3 Let v(x, t) = (x1, 0, 0). Then div v = j=1 ∂uj /∂xj = ∂x1/∂x1 = 1.

The ﬂow is expanding the ﬂuid in the x1 direction.

P3 Let v(x, t) = (x1, x2, x3). Then div v = j=1 ∂xj /∂xj = 3. The ﬂow is expanding the ﬂuid as it moves away from the origin.

Let v(x, t) = (−x1, −x2, −x3). Then div v = −3. The ﬂow is compressing the ﬂuid towards the origin.

Let v(x, t) = (−x2, x1, 0). Then div v = 0. The ﬂow is neither expanding nor compressing the ﬂuid.

Sauli Lindberg Navier-Stokes equations I Fall 2019 57 / 168 Geometric intuition on divergence I

∞ 3 Let v ∈ C (D × [T1, T2], R ) and (y, t) ∈ D × [T1, T2]. Deﬁne a cube δ δ δ δ δ δ Q(y, δ) = y − , y + × y − , y + × y − , y + ⊂ D. 1 2 1 2 2 2 2 2 3 2 3 2

The center of Q(y, δ) is y and the volume is δ3.

The continuity of div v implies that 1 Z div v(y, t) = lim 3 div v(x, t) dV . δ→0 δ Q(y,δ)

Next, by the Divergence theorem, 1 Z 1 Z lim 3 div v(x, t)dV = lim 3 v(x, t) · n(x)dA. δ→0 δ Q(y,δ) δ→0 δ ∂Q(y,δ) R Here ∂Q(y,δ) v(x, t) · n(x)dA is the flux of fluid flowing out of Q(y, δ).

Sauli Lindberg Navier-Stokes equations I Fall 2019 58 / 168 Geometric intuition on divergence II

Suppose div v(y, t) > 0. Then, for small δ > 0, there is more ﬂuid ﬂowing out of Q(y, δ) than into Q(y, δ).

In other words, the ﬂow is expanding the ﬂuid near y at time t.

The point y is called a source (of x 7→ v(x, t)).

If div v(y, t) < 0, then for small δ > 0, there is more ﬂuid ﬂowing out of Q(y, δ) than into Q(y, δ).

In other words, the ﬂow is compressing the ﬂuid near y at time t.

The point y is called a sink (of x 7→ v(x, t)).

Sauli Lindberg Navier-Stokes equations I Fall 2019 59 / 168 Fluid density and divergence

The rate of change of the average density in Wt is, by (16)–(18),

=0 z Z }| { R d R d R R ! dV % dV − % dV dV d % dV Wt dt Wt dt Wt Wt = Wt dt R dV (R dV )2 Wt Wt >0 zZ }| { R − %(x, T1) dV div u(x, t)dV Wt = W . (R dV )2 Wt

If div u > 0, the average density in Wt decreases in time. (If div u > 0, the flow is expanding the volume of the fluid while the mass of the fluid remains constant.)

If div u < 0, the the average density in Wt increases in time (because the ﬂow is compressing the ﬂuid).

If div u = 0, the the average density in Wt is constant in time.

Sauli Lindberg Navier-Stokes equations I Fall 2019 60 / 168 Incompressibility I

Deﬁnition 18 We call a ﬂow incompressible if for every admissible W ⊂ D, Z volume of Wt = dV = constant in t. Wt

Theorem 19 The following conditions are equivalent: (i)The ﬂow is incompressible, (ii) div u = 0, (iii) Jϕ = 1. (iv) D%/Dt = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 61 / 168 Incompressibility II

Proof of the direction (i) ⇒ (ii): By assumption,

d Z Z dV = div u(ϕ, t)Jϕ dV = 0 dt Wt W for every admissible volume W ⊂ D. Thus, by Exercise 2.3, div u(ϕ, t)Jϕ = 0.

By Exercise 3.6, we have Jϕ(x, t) > 0 at every (x, t) ∈ D × [T1, T2]. Thus div u(ϕ(x, t), t) = 0 at every (x, t) ∈ D × [T1, T2].

Since x 7→ ϕ(x, t): D → D is a bijection for every t, we get div u = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 62 / 168 Incompressibility III

Proof of the direction (ii) ⇒ (iii): Let (x, t) ∈ D × [T1, T2]. Since div u = 0, Euler’s formula gives ∂Jϕ (x, s) = div u(ϕ(x, s), s)Jϕ(x, s) = 0 ∂t

for all s ∈ (T1, t). We use the Fundamental theorem of calculus on s 7→ Jϕ(x, s):

∂x1 ∂x1 ∂x1  ∂x1 ∂x2 ∂x3  Z t ∂Jϕ ∂x ∂x ∂x  Jϕ(x, t) = Jϕ(x, T ) + (x, s) ds = det  2 2 2  + 0 = 1. 1 ∂t ∂x ∂x ∂x  T1  1 2 3  ∂x3 ∂x3 ∂x3  ∂x1 ∂x2 ∂x3

Sauli Lindberg Navier-Stokes equations I Fall 2019 63 / 168 Incompressibility IV

Proof of the direction (iii) ⇒ (i): If Jϕ = 1, we compute

d Z d Z d Z dV = Jϕ dV = dV = 0. dt Wt dt W dt W

Proof of the direction (ii) ⇒ (iv): Recall that

D% = −% div u. Dt

Since we have assumed that ϕ(x, t) > 0 for every (x, t) ∈ D × [T1, T2], we clearly have D% = 0 ⇐⇒ div u = 0. Dt

Sauli Lindberg Navier-Stokes equations I Fall 2019 64 / 168 Homogeneous ﬂuids I

Recall ∂% + % div u + u · ∇% = 0. (continuity equation) (19) ∂t For ideal ﬂuids, we have proved ∂u % + % u · ∇u + ∇p = % b (balance of momentum). ∂t

Deﬁnition 20 The ﬂuid is said to be homogeneous if

∂% ∂% ∂% ∇% = , , = 0, ∂x1 ∂x2 ∂x3 i.e., density is independent of the position x ∈ D.

If the ﬂuid is homogeneous and incompressible (div u = 0), then (19) gives ∂%/∂t = 0, so that density is constant in D × [T1, T2].

Typically one sets the constant % = 1 in D × [T1, T2]. Sauli Lindberg Navier-Stokes equations I Fall 2019 65 / 168 Homogeneous, incompressible Euler equations

Deﬁnition 21 The homogeneous, incompressible Euler equations are given by ∂u + u · ∇u + ∇p = b, (20) ∂t div u = 0. (21)

Typically, one sets

b(x, t) = g(x, t) = (0, 0, −g) = ∇(−g x3) with g > 0 so that b is gravity.

In the case above, p˜(x, t) = p(x, t) + gx3 satisﬁes ∂u + u · ∇u + ∇p˜ = 0, (22) ∂t div u = 0. (23) One often writes homogeneous, incompressible Euler equations without the b term as in (22)–(23).

Sauli Lindberg Navier-Stokes equations I Fall 2019 66 / 168 The Cauchy problem for Euler equations

We will discuss the Cauchy problem where the initial data u(·, T1) = u0 is divergence-free.

One typically sets one of the following boundary conditions: Impenetrability (of the container wall): u(x, t) · n(x) = 0 on ∂D if ∂D 6= ∅. Periodic conditions: u(x + k, t) = u(x, t) for all x ∈ R3 and k ∈ Z3. Decay conditions: u(x, t) → 0 in some sense when |x| → ∞ in the case D = R3.

The fundamental mathematical problem (when D is bounded and b = 0) is:

Question 22 Suppose ∞ 3 u0 ∈ C (D, R ), div u0 = 0, u0 · n|∂D = 0. ∞ 3 ∞ Do there exist u ∈ C (D × [T1, T2], R ) and p ∈ C (D × [T1, T2]) such that

(20)–(21) hold, u · n|∂D = 0, u(x, T1) = u0(x) for all x ∈ D?

The answer is still unknown 264 years after the formulation of the Euler equations.

Sauli Lindberg Navier-Stokes equations I Fall 2019 67 / 168 Static ﬂuids

We will for now consider just the Euler equations (20)–(21) without initial and boundary values. We will work from simpler to more complicated cases.

Suppose ﬁrst the ﬂuid is at rest, i.e., u = 0.

The Euler equations now reduce to −∇p + b = 0, that is, the pressure gradient ∇p and the body force b cancel each other.

Note that the force −∇p points towards the direction where p is reducing the fastest.

Consider the case where b = ∇(−gx3) is gravity. Then −∇p + b = −∇(p + g x3) = 0 says that p + g x3 is constant in space. Thus, pressure is lower "up" than "down". The force −∇p is called buoyancy.

Sauli Lindberg Navier-Stokes equations I Fall 2019 68 / 168 Stationary ﬂows

Next suppose that ∂u/∂t = 0; we then say that u is a stationary ﬂow. The Euler equations reduce to u · ∇u = −∇p + b.

Assume again that b = −∇(gx3) is gravity, so that the Euler equations reduce to

u · ∇u = −∇(p + gx3), (24) div u = 0. (25)

Equation (24) is hard to analyse because the term u · ∇u is nonlinear in u. In particular, if (u1, p1) and (u2, p2) are two solutions of (24)–(25), then (u1 + u2, p1 + p2) is, in general, not a solution.

However, it turns out that

|u|2 u · ∇u = ∇ − u × (∇ × u). (26) 2

2 Thus, if div u = 0 and ∇ × u = 0, then setting p = −g x3 − |u| /2 gives a solution (u, p) of (24)–(25).

Sauli Lindberg Navier-Stokes equations I Fall 2019 69 / 168 Products of Levi-Civita symbols I

Our next aim is to prove (26) via Levi-Civita symbols.

∞ 3 For a general v ∈ C (D × [T1, T2], R ) we write

3 3 3 X X X ∂vm v × (∇ × v) = ijk ei vj (∇ × v)k = ijk ei vj k`m ∂x` i,j,k=1 i,j,k=1 `,m=1 3 3 ! X X ∂vm = ijk `mk ei vj . ∂x` i,j,`,m=1 k=1

P3 We thus wish to compute k=1 ijk `mk .

Sauli Lindberg Navier-Stokes equations I Fall 2019 70 / 168 Products of Levi-Civita symbols II

When i, j ∈ {1, 2, 3}, deﬁne the Kronecker delta δij by ( 1, i = j, δij = 0, i 6= j.

P3 3 Recall that det [a b c] = i,j,k=1 ijk ai bj ck for all a, b, c ∈ R . Thus det ei ej ek = ijk .

Lemma 23 Let i, j, `, m ∈ {1, 2, 3}. Then

3 X ijk `mk = δi`δjm − δimδj`. (27) k=1

Sauli Lindberg Navier-Stokes equations I Fall 2019 71 / 168 Products of Levi-Civita symbols III

Proof of Lemma 23. If i = j or ` = m, both sides of (27) vanish.

Assume then i 6= j. Now iji = ijj = 0. Choose k such that {i, j, k} = {1, 2, 3} so no index is repeated. By using properties of determinants,

 T   T  ei ei T T ijk `mk = det ej  det e` em ek = det ej  e` em ek T T ek ek   ei · e` ei · em ei · ek = det ej · e` ej · em ej · ek  ek · e` ek · em ek · ek   ei · e` ei · em 0 = det ej · e` ej · em 0 = (ei · e`)(ej · em) − (ei · em)(ej · e`) ek · e` ek · ek 1

= δi`δjm − δimδj`.

Sauli Lindberg Navier-Stokes equations I Fall 2019 72 / 168 A formula for the convective term I

Proposition 24

∞ 3 Let v ∈ C (D × [T1, T2], R ). Then

|v|2 v · ∇v = ∇ − v × (∇ × v). (28) 2 For the proof, recall that

3 3 ! X X ∂vm v × (∇ × v) = ijk `mk ei vj ∂x` i,j,`,m=1 k=1 3 X ∂vm = (δi`δjm − δimδj`)ei vj . ∂x` i,j,`,m=1

Sauli Lindberg Navier-Stokes equations I Fall 2019 73 / 168 A formula for the convective term II

Proof. Let i ∈ {1, 2, 3}. We compute the ith component

3 3 X ∂vm X ∂vm (v × (∇ × v))i = δi`δjmvj − δimδj`vj ∂x` ∂x` j,`,m=1 j,`,m=1 3 3 X ∂vj X ∂vi = vj − vj ∂xi ∂xj j=1 j=1 3 2 3 X 1 ∂(vj ) X ∂vi = − vj 2 ∂xi ∂xj j=1 j=1 1 ∂ |v|2 = − v · ∇vi . 2 ∂xi

Thus v × (∇ × v) = ∇ |v|2 /2 − v · ∇v.

Sauli Lindberg Navier-Stokes equations I Fall 2019 74 / 168 Irrotational, solenoidal vector ﬁelds as solutions

Deﬁnition 25 If v ∈ C ∞(D) satisﬁes div v = 0, then v is said to be solenoidal.

If v ∈ C ∞(D) satisﬁes ∇ × v = 0, then v is said to be irrotational.

Corollary 26 Suppose ∂u/∂t = 0, div u = 0 and ∇ × u = 0. Then u and 2 p(x) = −g x3 − |u(x)| solve the Euler equations with b(x) = ∇(−g x3).

Question 27 What kinds of stationary vector ﬁelds are both solenoidal and irrotational?

We give some examples in the following four slides; further examples are presented in coming lectures.

Sauli Lindberg Navier-Stokes equations I Fall 2019 75 / 168 Potential ﬂows

Suppose u = ∇ψ, where ∆ψ = 0 (i.e. ψ is harmonic). Then

div u = div (∇ψ) = ∆ψ = 0, ∇ × u = ∇ × (∇ψ) = 0. (Exercise 3.2)

Then u is called a potential ﬂow.

Sauli Lindberg Navier-Stokes equations I Fall 2019 76 / 168 Solenoidal, irrotational 2D ﬂows

Deﬁnition 28 We call v ∈ C ∞(R3, R3) a stationary 2D ﬂow if v is of the form

v(x1, x2, x3) = (v1(x1, x2), v2(x1, x2), 0). (29)

(Sometimes the third component v3 = 0 is omitted in the notation.) If v ∈ C ∞(R3, R3) is a stationary 2D ﬂow, then the conditions div v = 0 and ∇ × v = 0 read as ∂v ∂v div v = 1 + 2 = 0, (30) ∂x1 ∂x2 ∂v ∂v ∇ × v = 0, 0, 2 − 1 = 0. (31) ∂x1 ∂x2 Note that (30)–(31) are the Cauchy-Riemann equations on the complex function v1 + iv2 : C → C.

∞ If v1 + iv2 ∈ C (C, C) is an analytic function, then (29) defines a solenoidal, irrotational flow v ∈ C ∞(R3, R3). Thus v is a stationary solution of the Euler equations. Sauli Lindberg Navier-Stokes equations I Fall 2019 77 / 168 Beltrami flows

Deﬁnition 29 We call v ∈ C ∞(R3, R3) a Beltrami ﬂow if there exists λ ∈ C ∞(R3) such that

3 ∇ × v(x) = λ(x)v(x) for all x ∈ R .

Proposition 30 Suppose u ∈ C ∞(R3, R3) is a Beltrami ﬂow with λ(x) 6= 0 constant. Then u satisﬁes the stationary Euler equations u · ∇u = −∇p and div u = 0 with p(x) = − |u|2 /2.

Proof. By (28),

|v|2 |v|2 |v|2 v · ∇v = ∇ − v × (∇ × v) = ∇ − λv × v = ∇ . 2 2 2 Furthermore, div v = λ−1 div (∇ × v) = 0 (Exercise 3.2).

Sauli Lindberg Navier-Stokes equations I Fall 2019 78 / 168 ABC ﬂows

Definition 31 The ABC flows (Arnold-Beltrami-Childress flows) v ∈ C ∞(R3, R3) are defined by   A sin x3 + C cos x2 v(x) = B sin x1 + A cos x3  , A, B, C ∈ R. C sin x2 + B cos x1

ABC ﬂows satisfy ∇ × v = v.

Thus, ABD ﬂows satisfy the stationary Euler equations in R3.

ABC ﬂows are 2π-periodic: v(x + 2πk) = v(x) for all x ∈ R3 and k ∈ Z3.

Sauli Lindberg Navier-Stokes equations I Fall 2019 79 / 168 Irrotational ﬂows and gradients I

We next consider irrotational vector ﬁelds.

If v ∈ C ∞(D, R3) is of the form v = ∇ψ for some ψ ∈ C ∞(D), then necessarily ∇ × v = ∇ × (∇ψ) = 0 (Exercise 3.2).

Question 32 If ∇ × v = 0, does there exist ψ ∈ C ∞(D) such that v = ∇ψ?

The answer depends on the domain D ⊂ R3.

Sauli Lindberg Navier-Stokes equations I Fall 2019 80 / 168 Irrotational ﬂows and gradients II

Theorem 33

Suppose D ⊂ R3 is open and simply connected. If v ∈ C ∞(D, R3) with ∇ × v = 0, there exists ψ ∈ C ∞(D) such that ∇ψ = v.

We prove the case where D = R3; the same construction works e.g. for a cube but needs to be modiﬁed in general. We build ψ step-by-step. First note that, by the Fundamental theorem of calculus,

Z x1 ψ˜(x) = v1(s, x2, x3) ds 0 has the correct ﬁrst partial derivative: using the assumption ∇ × v = 0,

Z x1 Z x1 ∂v1 ∂v1 ∇ψ˜(x) = v1(x), (s, x2, x3) ds, (s, x2, x3) ds 0 ∂x2 0 ∂x3 Z x1 Z x1 ∂v2 ∂v3 = v1(x), (s, x2, x3) ds, (s, x2, x3) ds 0 ∂x1 0 ∂x1

= (v1(x), v2(x) − v2(0, x2, x3), v3(x) − v3(0, x2, x3)).

Sauli Lindberg Navier-Stokes equations I Fall 2019 81 / 168 Irrotational ﬂows and gradients III

We modify ψ˜ to get the correct second partial derivative as well: setting

x x ˜ Z 1 Z 2 ψ˜(x) = v1(s, x2, x3) ds + v2(0, s, x3) ds 0 0 we get, similarly,

Z x2 ˜ ∂v2 ∇ψ˜(x) = v1(x), v2(x), v3(x) − v3(0, x2, x3) + (0, s, x3) ds 0 ∂x3 Z x2 ∂v3 = v1(x), v2(x), v3(x) − v3(0, x2, x3) + (0, s, x3) ds 0 ∂x2

= (v1(x), v2(x), v3(x) − v3(0, 0, x3)).

Finally, setting

Z x1 Z x2 Z x3 ψ(x) = v1(s, x2, x3) ds + v2(0, s, x3) ds + v3(0, 0, s) ds 0 0 0 we obtain ∇ψ = v.

Sauli Lindberg Navier-Stokes equations I Fall 2019 82 / 168 Conservative vector ﬁelds

Definition 34 Suppose v ∈ C ∞(D) and there exists ψ ∈ C ∞(D) such that ∇ψ = v. Then v is said to be a conservative vector field (or an exact vector field).

The function ψ is called a scalar potential of v.

Sauli Lindberg Navier-Stokes equations I Fall 2019 83 / 168 A decomposition of ﬂuid ﬂow I

We next use curl in the characterisation of ﬂuid ﬂows.

Whenever v ∈ C ∞(D, R3), we write   ∂1v1 ∂2v1 ∂3v1 ∇v = ∂1v2 ∂2v2 ∂3v2 . ∂1v3 ∂2v3 ∂3v3

Theorem 35 ∞ 3 Suppose v ∈ C (D, R ) and x0 ∈ D. Then 1 v(x) = v(x ) + D(x )(x − x ) + [∇ × v(x )] × (x − x ) + O(|x − x |2) 0 0 0 2 0 0 0

when x → x0; here 1 D(x ) = (∇v(x ) + [∇v(x )]T ). 0 2 0 0

Sauli Lindberg Navier-Stokes equations I Fall 2019 84 / 168 A decomposition of ﬂuid ﬂow II

Proof: By Taylor’s theorem, for i ∈ {1, 2, 3},

2 vi (x) = vi (x0) + ∇vi (x0) · (x − x0) + O(|x − x0| ).

This can be written as

2 v(x) = v(x0) + ∇v(x0)(x − x0) + O(|x − x0| ).

We use the shorthand notation ∇v = ∇v(x0) etc. We decompose ∇v into symmetric and antisymmetric parts:

∇v = D + S,

where ∇v + (∇v)T ∇v − (∇v)T D = , S = . 2 2

It only remains to show that Sh = (∇ × v) × (x − x0)/2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 85 / 168 A decomposition of ﬂuid ﬂow III

Note that     ∂1v1 ∂2v1 ∂3v1 ∂1v1 ∂1v2 ∂1v3 2S = ∂1v2 ∂2v2 ∂3v2 − ∂2v1 ∂2v2 ∂2v3 ∂1v3 ∂2v3 ∂3v3 ∂3v1 ∂3v2 ∂3v3   0 −(∇ × v)3 (∇ × v)2 =  (∇ × v)3 0 −(∇ × v)1 . −(∇ × v)2 (∇ × v)1 0

Thus (∇ × v) (x − x ) − (∇ × v) (x − x )  1 2 0 3 3 0 2 S(x − x ) = (∇ × v) (x − x ) − (∇ × v) (x − x ) 0 2  3 0 1 1 0 3 (∇ × v)1(x − x0)2 − (∇ × v)2(x − x0)1 1 = [∇ × v(x )] × (x − x ), 2 0 0 ﬁnishing the proof.

Sauli Lindberg Navier-Stokes equations I Fall 2019 86 / 168 Vorticity

Definition 36 ∞ 3 When u ∈ C (D × [T1, T2], R ) satisfies the Euler equations with some ∞ p ∈ C (D × [T − 1, T2]), the vector field

∞ 3 ω = ∇ × u ∈ C (D × [T1, T2], R ) is called the vorticity of u.

Question 37 How does vorticity evolve in time?

Taking the curl of both sides of the Cauchy momentum equation, ∂ω + ∇ × (u · ∇u) = 0. ∂t Using the formula |u|2 u · ∇u = ∇ − u × (∇ × u), 2 we are lead to compute ∇ × [u × (∇ × u)].

Sauli Lindberg Navier-Stokes equations I Fall 2019 87 / 168 The curl of a cross product I

Proposition 38 ∞ 3 Let v, w ∈ C (D × [T1, T2], R ). Then ∇ × (v × w) = (div w)v − (div v)w + w · ∇v − v · ∇w.

Proof: Let i ∈ {1, 2, 3}. Using Levi-Civita symbols,

3 3 3 X X X [∇ × (v × w)]i = ijk ∂j (v × w)k = ijk ∂j k`mv`wm j,k=1 j,k=1 `,m=1 3 3 ! X X = ijk `mk ∂j (v`wm) j,`,m=1 k=1 3 3 X X = δi`δjm(v`∂j wm + wm∂j v`) − δimδj`(v`∂j wm + wm∂j v`) j,`,m=1 j,`,m=1 3 3 X X = (vi ∂j wj + wj ∂j vi ) − (vj ∂j wi + wi ∂j vj ). j=1 j=1

Sauli Lindberg Navier-Stokes equations I Fall 2019 88 / 168 The curl of a cross product II

We continue,

3 3 X X (vi ∂j wj +wj ∂j vi )− (vj ∂j wi +wi ∂j vj ) = (div w)vi +w·∇vi −v·∇wi −(div v)wi , j=1 j=1

ﬁnishing the proof.

Corollary 39 ∞ 3 Suppose u ∈ C (D × [T1, T2], R ) solves the Euler equations with b = 0 or b = ∇(−g x3). Then the vorticity ω satisﬁes ∂ω Dω − ω · ∇u + u · ∇ω = − ω · ∇u = 0. ∂t Dt

Sauli Lindberg Navier-Stokes equations I Fall 2019 89 / 168 Evolution of vorticity of 2D ﬂows

Corollary 40 ∞ 3 Suppose u ∈ C (D × [T1, T2], R ) solves the Euler equations with b = 0. Assume that u and p are two-dimensional,

u(x, t) = (u1(x1, x2, t), u2(x1, x2, t), 0), p(x, t) = p(x1, x2, t).

Then the vorticity ω satisﬁes Dω = 0. Dt

Sauli Lindberg Navier-Stokes equations I Fall 2019 90 / 168 Irrotational ﬂows and gradients II

Recall some terminology: v ∈ C ∞(D; R3) is irrotational if ∇ × v = 0, v ∈ C ∞(D, R3) is conservative if v = ∇ψ for some ψ ∈ C ∞(D).

By Exercise 3.2, conservative vector ﬁelds are irrotational.

By Theorem 33, in simply connected domains, irrotational vector ﬁelds are conservative.

Our next goal is twofold: show that irrotational vector ﬁelds are not in general conservative when the domain is not simply connected, give a geometric interpretation of curl via line integrals.

Sauli Lindberg Navier-Stokes equations I Fall 2019 91 / 168 Line integrals

Deﬁnition 41 0 3 Let a = c1 < c2 < ··· < cn = b, and suppose γ ∈ C ([a, b], R ) and ∞ γ ∈ C (ci , ci+1) for 1 ≤ i ≤ n − 1. Then γ is called a piecewise smooth path.

A piecewise smooth path γ :[a, b] → R3 is closed if γ(a) = γ(b).

Suppose now γ([a, b]) ⊂ D.

Deﬁnition 42 If v ∈ C ∞(D, R3), then the line integral of v is deﬁned as

Z Z b 3 Z b X 0 0 v(x) · ds(x) = vj (γ(s))γj (s) ds = v(γ(s)) · γ (s) ds. γ a j=1 a

Sauli Lindberg Navier-Stokes equations I Fall 2019 92 / 168 Example: line integrals of conservative vector ﬁelds

Theorem 43 Suppose f ∈ C ∞(D) and γ :[a, b] → D is a piecewise smooth path. Then Z ∇f (x) · ds(x) = f (γ(b)) − f (γ(a)). γ

Proof. By the Chain rule and the Fundamental theorem of calculus,

Z Z b Z b d ∇f (x)·ds(x) = ∇f (γ(s))·γ0(s) ds = [f (γ(s))] ds = f (γ(b))−f (γ(a)). γ a a ds

Corollary 44 If f ∈ C ∞(D) and γ :[a, b] → D is a piecewise smooth closed path, then Z ∇f (x) · ds(x) = 0. γ Sauli Lindberg Navier-Stokes equations I Fall 2019 93 / 168 An irrotational, non-conservative vector ﬁeld

Proposition 45 3 Let U = R \{(0, 0, x3): x3 ∈ R}, and deﬁne −x2 x1 v(x) = 2 2 , 2 2 , 0 for all x ∈ U. x1 + x2 x1 + x2 Then there exists no ψ ∈ C ∞(U) such that ∇ψ = v.

Proof. Deﬁne γ :[0, 2π] → U by γ(s) = (cos s, sin s, 0). Then

Z Z 2π v(s) · ds(x) = v(γ(s)) · γ0(s) ds γ 0 Z 2π −(sin s)(− sin s) + (cos s)(cos s) = ds 2 2 0 cos s + sin s Z 2π = ds = 2π 6= 0. 0

Sauli Lindberg Navier-Stokes equations I Fall 2019 94 / 168 Stokes’ theorem

Suppose U ⊂ R2 is open and Green’s theorem holds for U. Let Φ ∈ C 0(U¯, R3) and Φ ∈ C ∞(U, R3). Denote

S = {(Φ1(p, q), Φ2(p, q), Φ3(p, q)): (p, q) ∈ U¯}.

(Suppose ∂S = γ([a, b]) for some piecewise-smooth closed path.)

Theorem 46 (Stokes’ theorem) Let v ∈ C ∞(S, R3). Then Z Z ∇ × v(x) · n(x) dA = v(x) · ds S γ

(when the direction in which γ travels along ∂S is chosen suitably).

Sauli Lindberg Navier-Stokes equations I Fall 2019 95 / 168 A geometric interpretation of curl I

3 Let now x0 ∈ D. Denote by Sr ⊂ R a disc of radius r centred at x0 and with unit 3 normal n(x) = n ∈ R at every x ∈ Sr .

For example, n = e3 and

3 2 2 Sr = {x ∈ R :(x1 − (x0)1) + (x2 − (x0)2) ≤ r, x3 = (x0)3}.

Let v ∈ C ∞(D, R3). By Stokes’ theorem, Z Z ∇ × v(x) · n(x) dA = v(x) · ds. Sr ∂Sr

By the Mean value theorem, there exists xr ∈ Sr such that

R ∇ × v(x) · n(x) dA R v(x) · ds ∇ × v(x ) · n = Sr = ∂Sr . r R dA πr 2 Sr

Sauli Lindberg Navier-Stokes equations I Fall 2019 96 / 168 A geometric interpretation of curl II

Now, by continuity,

R v(x) · ds ∂Sr ∇ × v(x0) · n = lim ∇ × v(xr ) · n = lim . r→0 r→0 πr 2

R R 2π 0 We call v · ds = v(γ(s)) · γ (s) ds the circulation of v around ∂Sr . It is ∂Sr 0 the net velocity around ∂Sr .

Thus ∇ × v(x0) · n measures the circulation of v around the axis n.

(Note that ∇ × v(x0) · n is highest when n and ∇ × v(x0) point in the same direction.)

We illustrate circulation by two examples.

Sauli Lindberg Navier-Stokes equations I Fall 2019 97 / 168 An example

3 3 Consider v(x) = (−x2, x1, 0) for all x ∈ R . Fix x0 ∈ R and consider

3 2 2 Sr = {x ∈ R :(x1 − (x0)1) + (x2 − (x0)2) ≤ r, x3 = (x0)3}, n = e3.

3 Now ∂Sr is parametrised by γ :[0, 2π] → R ,

γ(s) = x0 + (r cos s, r sin s, 0) = ((x0)1 + r cos s, (x0)2 + r sin s, (x0)3).

We compute

Z Z 2π v · ds = v(γ(s)) · γ0(s) ds γ 0 Z 2π = [−((x0)2 + r sin s)(−r sin s) + ((x0)1 + r cos s)(r cos s)] ds 0 Z 2π 2 = (r + (x0)2r sin s + (x0)1r cos s) ds 0 2 2 = 2πr = πr ∇ × v(x0) · n,

as expected.

Sauli Lindberg Navier-Stokes equations I Fall 2019 98 / 168 Another example

Next consider a shear ﬂow of the form v(x) = (x2, 0, 0). Now ∇ × v(x) = (0, 0, −1) for all x ∈ R3. We compute the circulation around the same disk as above: Z Z 2π v · ds = v(γ(s)) · γ0(s) ds γ 0 Z 2π = ((x0)2 + r sin s) · (−r sin s) ds 0 Z 2π 2 2 = [−r(x0)2 sin s − r sin s] ds. 0

R 2π s=2π Now 0 [−r(x0) − 2 sin s] ds = −r(x0)2(− cos s)|s=0 = 0 and Z 2π Z 2π 2 2 2 2 2 [−r sin s] ds = −r sin s ds = −πr = ∇ × v(x0) · n 0 0

R 2π 2 R 2π 2 (because 0 sin s ds = 0 cos s ds by integration by parts and furthermore R 2π 2 2 R 2π 0 (sin s + cos s) ds = 0 dV = 2π).

Sauli Lindberg Navier-Stokes equations I Fall 2019 99 / 168 Kelvin’s circulation theorem I

Deﬁnition 47 Suppose γ :[0, 1] → D be a piecewise-smooth closed path and denote C = γ([a, b]) ⊂ D. We call C a loop.

Denote by Ct = ϕt (C) = {ϕ(x, t): x ∈ C}, t ∈ [T1, T2] the loop carried by the ﬂuid ﬂow.

Note that Ct is parametrised by s 7→ ϕ(γ(s), t):[0, 1] → D.

Theorem 48 (Kelvin’s circulation theorem) Suppose u solves the Euler equations with or without gravity. Then the circulation R ΓC = u · ds is constant in time. t Ct

For the proof we ﬁnd a formula for the rate of change of Γt .

Sauli Lindberg Navier-Stokes equations I Fall 2019 100 / 168 Kelvin’s circulation theorem II

Lemma 49

For all t ∈ (T1, T2), d Z Z Du u · ds = · ds dt Ct Ct Dt

Proof: By the deﬁnition of the line integral,

d Z d Z 1 ∂ u · ds = u(ϕ(γ(s), t), t) · [ϕ(γ(s), t)] ds dt Ct dt 0 ∂s Z 1 ∂ ∂ = [u(ϕ(γ(s), t), t)] · [ϕ(γ(s), t)] ds 0 ∂t ∂s Z 1 ∂ ∂ + u(ϕ(γ(s), t), t) · [ϕ(γ(s), t)] ds 0 ∂t ∂s = I1 + I2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 101 / 168 Kelvin’s circulation theorem III

By the deﬁnition of material derivative,

Z 1 Du ∂ Z Du I1 = (ϕ(γ(s), t), t) · [ϕ(γ(s), t)] ds = · ds. 0 Dt ∂s Ct Dt

For I2, note that ∂ ∂ ∂ ∂ ∂ [ϕ(γ(s), t)] = [ϕ(γ(s), t)] = [u(ϕ(γ(s), t)), t)], ∂t ∂s ∂s ∂t ∂s so that Z 1 ∂ I2 = u(ϕ(γ(s), t), t) · [u(ϕ(γ(s), t), t)] ds 0 ∂s Z 1 1 ∂ = [u · u](ϕ(γ(s), t), t) ds = 0 0 2 ∂s since ϕ(γ(1), t) = ϕ(γ(0), t).

Sauli Lindberg Navier-Stokes equations I Fall 2019 102 / 168 Kelvin’s circulation theorem IV

Recall: Theorem 50 (Kelvin’s circulation theorem) Suppose u solves the Euler equations with or without gravity. Then the circulation R ΓC = u · ds is constant in time. t Ct

Proof of Kelvin’s circulation theorem. We compute d Z Z Du Z u · ds = · ds = − ∇(p + g x3) · ds = 0. dt Ct Ct Dt Ct

Sauli Lindberg Navier-Stokes equations I Fall 2019 103 / 168 Helmholtz decomposition of vector ﬁelds I

Theorem 51 Let v ∈ C ∞(D, R3) ∩ C 0(D¯, R3). Then there exist unique w ∈ C ∞(D, R3) ∩ C 0(D¯, R3) and ∇q ∈ C ∞(D) such that

v = w + ∇q, div w = 0, w · n|∂D = 0.

Sketch of proof: Solve the Neumann problem

∆q = div v in D, ∇q · n = v · n on ∂D.

Now set w = v − ∇q. Then

div w = div (v − ∇q) = div v − ∆q = 0, w · n = v · n − ∇q · n = 0.

We next show that w and ∇q are unique.

Sauli Lindberg Navier-Stokes equations I Fall 2019 104 / 168 Helmholtz decomposition of vector ﬁelds II

Seeking a contradiction, suppose

v = w1 + ∇q1 = w2 + ∇q2,

where wi and qi have the sought properties. Then

0 = w1 − w2 + ∇q1 − ∇q2. (32)

Taking the inner product with w1 − w2 and integrating over D we get Z 0 = [w1 − w2] · [w1 − w2 + ∇(q1 − q2)] dV D Z Z 2 = |w1 − w2| dV + [w1 − w2] · ∇(q1 − q2) dV D D Z 2 = |w1 − w2| dV (Exercise 5.2). D

Therefore, w1 = w2 (as in Exercise 2.3). Then (32) gives ∇q1 = ∇q2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 105 / 168 Evolution of vorticity I

We next continue the study of the evolution of vorticity. Recall that   ∂1ϕ1 ∂2ϕ1 ∂3ϕ1 ∇ϕ = ∂1ϕ2 ∂2ϕ2 ∂3ϕ2 . ∂1ϕ3 ∂2ϕ3 ∂3ϕ3

Theorem 52

Suppose u and p satisfy the Euler equations in D × [T1, T2] with b = 0 or b(x, t) = ∇(−g x3). Then the vorticity satisﬁes

ω(ϕ(x, t), t) = ∇ϕ(x, t) ω(x, T1) for all (x, t) ∈ D × [T1, T2].

Corollary 53 In Theorem 52, if u and p are of the 2D form u(x, t) = (u1(x1, x2, t), u2(x1, x2, t), 0) and p(x, t) = p(x1, x2, t), then

ω(ϕ(x, t), t) = ω(x, T1)

for all (x, t) ∈ D × [T1, T2].

Sauli Lindberg Navier-Stokes equations I Fall 2019 106 / 168 Evolution of vorticity II

Proof: Denote

G(x, t) = ω(ϕ(x, t), t), H(x, t) = ∇ϕ(x, t) ω(x, T1);

we will prove Theorem 52 by showing that G = H.

We intend to show that

∂t G(x, t) = G(x, t) · ∇u(ϕ(x, t), t), ∂t H(x, t) = H(x, t) · ∇u(ϕ(x, t), t)

for all (x, t) ∈ D × [T1, T2]. Then, since G(x, T1) = H(x, T1) for all x ∈ D, we will get G = H in D × [T1, T2].

We compute Dω ∂ G(x, t) = ∂ [ω(ϕ(x, t), t)] = (ϕ(x, t), t) = G(x, t) · ∇u((ϕ(x, t), t), t t Dt ∂t H(x, t) = ∇∂t ϕ(x, t) ω(x, T1) = ∇[u(ϕ(x, t), t)] ω(x, T1)

= ∇u(ϕ(x, t), t)[∇ϕ(x, t)ω(x, T1)] = H(x, t) · ∇u(ϕ(x, t), t).

Sauli Lindberg Navier-Stokes equations I Fall 2019 107 / 168 Evolution of vorticity III

By the Fundamental theorem of calculus,

=0 z }| { G(x, t) − H(x, t) = G(x, T1) − H(x, T1) (33) Z t + (G(x, s) − H(x, s)) · ∇u(ϕ(x, s), s) ds. (34) T1 Denote α(t) = |G(x, t) − H(x, t)| , β(t) = |∇u(ϕ(x, t), t)| . 3×3 (Recall that |A| = max|y|=1 |Ay| for all A ∈ R .) Now (33)–(34) gives

Z t α(t) ≤ α(s)β(s) ds for all t ∈ [T1, T2]. T1 The claimed equality G = H, i.e. α = 0, follows from the following classical result.

Sauli Lindberg Navier-Stokes equations I Fall 2019 108 / 168 Grönwall’s inequality I

Lemma 54 (Grönwall’s inequality) 0 Suppose α, β ∈ C ([T1, T2], [0, ∞)) and A ≥ 0 satisfy Z t α(t) ≤ A + α(s)β(s) ds (35) T1

for all t ∈ [T1, T2]. Then R t β(s) ds α(t) ≤ Ae T1

for all t ∈ [T1, T2].

Proof: Suppose ﬁrst A > 0. Let t ∈ (T1, T2). Then (35) gives

d Z t A + α(s)β(s) ds = α(t)β(t) dt T1 Z t ≤ β(t) A + α(s)β(s) ds . T1

Sauli Lindberg Navier-Stokes equations I Fall 2019 109 / 168 Grönwall’s inequality II

Thus

d R t Z t A + α(s)β(s) ds d dt T1 log A + α(s)β(s) ds = R t ≤ β(t). dt T1 A + α(s)β(s) ds T1 Therefore, by the Fundamental theorem of calculus, Z t Z t log A + α(s)β(s) ds − log A ≤ β(s) ds T1 T1

for all t ∈ [T1, T2]. Using (35) and exponentiating, Z t log A+R t β(s) ds R t β(s) ds α(t) ≤ A + α(s)β(s) ds ≤ e T1 = A e T1 , T1 proving the case A > 0.

If A = 0, choose a sequence of constants Aj > 0 with Aj → 0. Then R t α(t) ≤ Aj + α(s)β(s) ds for all j ∈ . Now, by the case A > 0, T1 N R t β(s) ds α(t) ≤ Aj e T1 for every j ∈ N. Consequently, α(t) = 0. Sauli Lindberg Navier-Stokes equations I Fall 2019 110 / 168 Conservation of enstrophy

Theorem 55 Suppose u and p form a 2D solution of the Euler equations. Assume that D ⊂ R2 0 ¯ 3 is bounded and that u, ω ∈ C (D × [T1, T2], R ) with u · n|∂D = 0. Then the R 2 enstrophy D |ω(x, t)| dV is conserved in time.

Proof.

Recall that Dω/Dt = ∂t ω + u · ∇ω = 0 and ω = (0, 0, ω3). We may therefore compute Z Z Z d 2 |ω| dV = 2 ω · ∂t ω dV = −2 ω · (u · ∇)ω dV dt D D D   Z 2 2 Z X X 2 = −2 ω3 ·  uj ∂j ω3 dV = − uj ∂j |ω| dV D j=1 j=1 D Z = − u · ∇ |ω|2 = 0 D by Exercise 5.2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 111 / 168 Kinetic energy conservation in Euler equations I

Theorem 56 Assume that D is bounded u ∈ C ∞(D, R3) ∩ C 0(D¯, R3) and p ∈ C ∞(D) ∩ C 0(D¯) satisfy the Euler equations with or without gravity. Suppose u · n|∂D = 0. Then R 2 the kinetic energy D |u| /2 dV is constant in time.

Proof: Using the assumption that ∂u + u · ∇u = −∇p + ∇(−g x ) ∂t 3 we compute

d Z |u|2 Z ∂ |u|2 Z ∂u dV = dV = u · dV dt D 2 D ∂t 2 D ∂t Z = u · [−u · ∇u − ∇(p + g x3)] dV D Z = − u · [u · ∇u] dV (Exercise 5.2). D

Sauli Lindberg Navier-Stokes equations I Fall 2019 112 / 168 Kinetic energy conservation in Euler equations II

Now

3 3 3 X X 1 X 1 u2 u · [u · ∇u] = u (u · ∇u ) = u u ∂ u = u ∂ [u2] = u · ∇ . i i i j j i 2 j j i 2 2 i=1 i,j=1 i,j=1

Thus, by Exercise 5.2,

Z 1 Z |u|2 − u · [u · ∇u] dV = − u · ∇ dV = 0. D 2 D 2

Sauli Lindberg Navier-Stokes equations I Fall 2019 113 / 168 Other conservation laws in 3D

Theorem 57 ∞ 3 3 ∞ 3 Suppose u ∈ C (R × [T1, T2], R ) and p ∈ C (R × [T1, T2]) solve the Euler equations. Assume that |u|, |∇u| and |p| tend to zero rapidly enough when −N |x| → ∞ (e.g. |u(x, t)| + |∇u(x, t)| + |p(x, t)| = O(|x| ) for a suitable N ∈ N). The following quantities are conserved in time: R 3 u dV (total flux of velocity), R R 3 ω dV (total flux of vorticity), R −1 R 2 2 3 |u| dV (kinetic energy), R R 3 u · ω (helicity), R −1 R 2 3 x × ω dV (fluid impulse), R −1 R 3 3 x × (x × ω) dV (moment of fluid impulse). R

Sauli Lindberg Navier-Stokes equations I Fall 2019 114 / 168 Other conservation laws in 2D

Theorem 58 ∞ 3 3 ∞ 3 Suppose u ∈ C (R × [T1, T2], R ) and p ∈ C (R × [T1, T2]) solve the Euler equations and are of the 2D form

u(x, t) = (u1(x1, x2, t), u2(x1, x2, t), 0), p(x, t) = p(x1, x2, t).

−N Assume |u(x, t)| + |∇u(x, t)| + |p(x, t)| = O(|x| ) for a suitable N ∈ N. The following quantities are conserved in time: R 2 u dV (total flux of velocity), R R 2 ω dV (total flux of vorticity), R −1 R 2 2 2 |u| dV (kinetic energy), R R p 1/p p ( 2 |ω| ) , 1 ≤ p < ∞ (L norm of vorticity), R −1 R 2 −3 2 |x| ω dV (moment of fluid impulse). R

Sauli Lindberg Navier-Stokes equations I Fall 2019 115 / 168 Flow around an obstacle

We will consider stationary ﬂow around a ﬁxed obstacle such as a ball in R3.

Now v · n|∂B = 0. In order to have a unique potential ﬂow with such boundary values, we need e.g. to specify the velocity at inﬁnity.

The potential ﬂows solve the Euler equations and thus are mathematically correct. However, it turns out that in some cases, the solutions are incompatible with experiments. A famous instance of this is called d’Alembert’s paradox.

First, however, we consider potential ﬂow in bounded, simply connected domains.

Sauli Lindberg Navier-Stokes equations I Fall 2019 116 / 168 Potential ﬂow in a bounded, simply connected container

Proposition 59 Suppose D is bounded and simply connected. Suppose v = ∇ϕ for some ψ ∈ C ∞(D) with div v = ∆ϕ = 0 and ∂ϕ/∂n = ∇ϕ · n = v · n = 0. Then v = 0.

Proof. The Neumann problem

∆ϕ = 0, ∂ϕ | = 0 ∂n ∂D has the unique solution ϕ = 0.

Thus, it is impossible to extend to the interior of a bounded, simply connected domain a smooth, divergence-free vector ﬁeld with vanishing normal component on the boundary without creating vorticity.

The case of unbounded domains is very diﬀerent.

Sauli Lindberg Navier-Stokes equations I Fall 2019 117 / 168 An example: potential ﬂow around the unit ball

In D = R3 \ B(0, 1), consider ϕ ∈ C ∞(D) deﬁned by ! 1 ϕ(x) = 1 + x · u∞, 2 |x|3

3 where u∞ ∈ R . Now ! 3x 1 u(x) = ∇ϕ(x) = − x · u∞ + 1 + u∞. 2 |x|5 2 |x|3 When |x| → ∞, we have ! 1 u(x) = u∞ + O . |x|3

Now, if x ∈ ∂D = ∂B(0, 1), recall that the unit normal vector n(x) = x/ |x|. Then, since |x| = 1, we get ! 3 1 u(x) · n(x) = − (x · x)(x · u∞) + 1 + u∞ · x = 0. 2 |x|5 2 |x|3

Sauli Lindberg Navier-Stokes equations I Fall 2019 118 / 168 d’Alembert’s paradox I

We then consider a potential flow around a more general obstacle than the unit ball. We assume that as above, at infinity, the fluid velocity has a limit 3 lim|x|→∞ u(x) = u∞ ∈ R .

Theorem 60

Let Λ ⊂ R3 be a bounded domain such that the Divergence theorem holds for

3 D = R \ Λ.

Let u = ∇ϕ be a potential ﬂow in D such that u · n|∂D = 0 and 3 lim|x|→∞ u(x) = u∞ ∈ R . Then the ﬂuid produces zero net force on Λ, that is, Z F = − p n dA = 0. ∂Λ

Sauli Lindberg Navier-Stokes equations I Fall 2019 119 / 168 d’Alembert’s paradox II

Sketch of proof: We deﬁne

η(x) = ϕ(x) − u∞ · x

so that ϕ(x) = u∞ · x + η(x), u(x) = u∞ + ∇η(x). The function η has the following properties: ∆η = 0 in D,

∇η(x) = ∇ϕ(x) − u∞ = u(x) − u∞ → 0 as |x| → ∞,

∂η/∂n|∂D = −u∞ · n|∂D . Now η is given by the Newtonian potential

1 Z f (y) η(x) = dV (y), (36) 4π Λ |x − y|

where f is (somehow) determined by ∂η/∂n|∂D = −u∞ · n|∂D .

Sauli Lindberg Navier-Stokes equations I Fall 2019 120 / 168 d’Alembert’s paradox III

We show that, indeed, ∆η = 0 in D. Fix x ∈ D and let i ∈ {1, 2, 3}. Then

∂2 1 Z ∂2 1 2 η(x) = f (y) 2 dV (y), ∂xi 4π Λ ∂xi |x − y| where

− 1 − 3 3 ! 2 3 ! 2 ∂ 1 ∂ X 2 X 2 = (xk − yk ) = −(xi − yi ) (xk − yk ) ∂xi |x − y| ∂xi k=1 k=1 x − y = − i i , |x − y|3 − 3 − 5 3 ! 2 3 ! 2 ∂2 1 X X = − (x − y )2 + 3(x − y )2 (x − y )2 ∂x 2 |x − y| k k i i k k i k=1 k=1 1 (x − y )2 = − + 3 i i . |x − y|3 |x − y|5

3 P3 2 Thus ∇(1/ |x − y|) = −(x − y)/ |x − y| and ∆η(x) = i=1 ∂i η(x) = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 121 / 168 d’Alembert’s paradox IV

Our next aim is to show that u(x) − u∞ = ∇η(x) and p(x) tend to zero rapidly as |x| → ∞.

First, using properties of Newtonian potentials (see e.g. Evans: Partial Diﬀerential Equations) and the Divergence theorem, Z Z Z f (y) dV (y) = − ∆η(y) dV (y) = − ∇η(y) · n(y)dA(y) (37) Λ Λ ∂Λ Z Z = u∞ · n(y) dA(y) = div u∞ dV (y) = 0. (38) ∂Λ ∂Λ We denote the positive and negative parts of f by f + and f −, so that

f = f + − f −.

By (37)–(38), Z Z f +(y) dV (y) = f −(y) dV (y). Λ Λ

Sauli Lindberg Navier-Stokes equations I Fall 2019 122 / 168 d’Alembert’s paradox V

Now, since ∇(1/ |x − y|) = −(x − y)/ |x − y|3, we have

1 Z f +(y) 1 Z f +(y)(x − y) ∇ dV (y) = − 3 dV (y) 4π Λ |x − y| 4π Λ |x − y| Z ! x + 1 = − 3 f (y) dV (y) + O 3 , 4π |x| Λ |x| 1 Z f −(y) 1 Z f −(y)(x − y) ∇ dV (y) = − 3 dV (y) 4π Λ |x − y| 4π Λ |x − y| Z ! x − 1 = − 3 f (y) dV (y) + O 3 4π |x| Λ |x|

as |x| → ∞. Combining the equalities above with (37)–(38) we get ! ! x Z 1 1 ∇η(x) = − 3 f (y) dV (y) + O 3 = O 3 4π |x| Λ |x| |x|

as |x| → ∞.

Sauli Lindberg Navier-Stokes equations I Fall 2019 123 / 168 d’Alembert’s paradox VI

We conclude that ! 1 u(x) − u∞ = ∇η(x) = O |x|3 as |x| → ∞.

Recall that u and p(x) = − |x|2 /2 solve the stationary Euler equations in D. The pressure p satisﬁes

|u(x)|2 1 p(x) = − = − (|u |2 + (u(x) − u ) · (u(x) + u )) 2 2 ∞ ∞ ∞ ! |u |2 1 = − ∞ + O 2 |x|3

as |x| → ∞.

Sauli Lindberg Navier-Stokes equations I Fall 2019 124 / 168 d’Alembert’s paradox VII

The force that the ﬂuid exerts on the obstacle Λ is Z F = − p(y) n(y) dA(y). ∂Λ We now wish to evaluate F . Consider a ball B(0, R) ⊃ Λ and denote D0 = B(0, R) \ Λ. By the Euler equations, Z I = [u · ∇u + ∇p] dV = 0. D0

On the other hand, by the Divergence theorem and the assumption u · n|∂Λ = 0, Z Z 0 = I = [(u · n)u + pn] dA − p n dA. ∂B(0,R) ∂Λ

We conlude that Z Z 1 Z 1 F = − p n dA = − [(u·n)u+pn] dA ≤ O 3 dA = O . ∂Λ ∂B(0,R) R ∂B(0,R) R

Letting R → ∞ we get F = 0, proving d’Alembert’s paradox. Sauli Lindberg Navier-Stokes equations I Fall 2019 125 / 168 A 2D analogue

Theorem 61 (Kutta-Joukowski) In 2D, under assumptions analogous to those of Theorem 60, we have Z F = − p n dA = −Γ |u∞| e, ∂Λ

where Γ is the circulation of u around Λ and e is a unit vector orthogonal to u∞.

Sauli Lindberg Navier-Stokes equations I Fall 2019 126 / 168 An example: ideal channel ﬂow with stationary pressure I

Consider a 2D channel ﬂow in the direction x1 in

2 D = R × (0, 1) ⊂ R . This time it is more convenient to write u as having two components etc. Suppose b = 0 and the pressure p(x1, x2, t) = p(x1) is larger at x1 = 0 than x1 = L > 0.

We ﬁrst look for a stationary solution of the Euler equations,

u(x1, x2, t) = (u1(x1, x2), 0), p(x, t) = p(x1).

By incompressibility, we must have

div u = ∂1u1 = 0.

The Cauchy momentum momentum equation ∂t u + u · ∇u = −∇p becomes

∂t u1 = −∂1p 6= 0,

so that u cannot be stationary.

Sauli Lindberg Navier-Stokes equations I Fall 2019 127 / 168 An example: ideal channel ﬂow with stationary pressure II

With the same pressure p, we then look for a solution u of the form

u(x1, x2, t) = (u1(x1, x2, t), 0).

Again, incompressibility gives ∂1u1 = 0 and the Cauchy momentum equation ∂t u + u · ∇u = −∇p is of the form

∂t u1 = −∂1p. (39) 2 Therefore, ∂1 p = −∂t ∂1u1 = 0. Thus p(x1) = ax1 + b for some a, b ∈ R. We solve p(L) − p(0) p(x ) = p(0) + x . 1 L 1 Substituting into (39) and recalling that ∂1u1 = 0 we get p(0) − p(L) u (x , x , t) = t + c(x ). (40) 1 1 2 L 2 Thus the velocity increases indeﬁnitely.

Physically, the solution (40) is not entirely satisfactory. We will find a more realistic velocity field by taking into account friction. This will lead to the Navier-Stokes equations. Sauli Lindberg Navier-Stokes equations I Fall 2019 128 / 168 Momentum transport across a fluid surface

Consider a velocity ﬁeld u and rectangles B = [a, b] × [c, d] × [e, f ] and B0 = [a, b] × [c, d] × [f , g]. Now the contact surface S = [a, b] × [c, d] × {f } has an outer unit normal (for B) of the form n(x) = e3.

Assume that the horizontal component of u, i.e. (u1(x, t), u2(x, t), 0), is much 0 larger in B than in B. (E.g. consider u(x, t) = (g(x3), 0, 0) with g rapidly increasing.)

If the ﬂuid is ideal, then the ﬂuid volume B0 can exert a force on B, but only in the normal direction −n(x) = −e3.

In eﬀect, B and B0 slide on top of each other without friction.

In reality, some of the (faster) molecules in B0 will cross the surface S and increase the momentum of B. Likewise, some of the molecules of B will decrease the momentum of B0.

We therefore discard the assumption that the ﬂuid is ideal.

Sauli Lindberg Navier-Stokes equations I Fall 2019 129 / 168 The Cauchy stress tensor

Axiom 62 If W ⊂ D is an admissible volume, then

force on Wt per unit area at x ∈ ∂Wt is σ(x, t, n),

where σ(x, t, n) ∈ R3 is called the Cauchy stress vector.

For ideal fluids, σ(x, t, n) = −p(x, t)n(x). σ(x, t, n) need not be parallel to n(x). In particular, momentum transfer across a surface is allowed. σ depends on the fluid (and the fluid velocity u). If one just assumes that σ depends continuously on n, then in fact the dependence is linear: σ(x, t, n) = σ(x, t)n(x), where σ(x, t) ∈ R3×3. This is Cauchy’s theorem (see Marsden-Hughes: Mathematical Foundations of Elasticity, p. 134).

We will thus assume that σ(x, t, n) = σ(x, t)n(x); σ(x, t) ∈ R3×3 is called the Cauchy stress tensor.

Sauli Lindberg Navier-Stokes equations I Fall 2019 130 / 168 Newtonian ﬂuids

We decompose σ as σ = −pI + τ:       σ11 σ12 σ13 −p 0 0 τ11 τ12 τ13 σ21 σ22 σ23 =  0 −p 0  + τ21 τ22 τ23 . σ31 σ32 σ33 0 0 −p τ31 τ32 τ33 | {z } | {z } | {z } =σ =−pI =τ

Here τ ∈ R3×3 is called the deviatoric stress tensor. Deﬁnition 63 If the ﬂuid is incompressible and τ is of the form ∂ui ∂uj τij = ν + ∂xj ∂xi

for some ν > 0, then the ﬂuid is said to be Newtonian.

The constant ν is called the viscosity of the ﬂuid.

(Compressible Newtonian fluids have a slightly different definition.)

Sauli Lindberg Navier-Stokes equations I Fall 2019 131 / 168 An example: the deviatoric stress tensor of a shear ﬂow I

We illustrate the deviatoric stress tensor τ in the simple case of a stationary shear flow of the form u(x) = (u1(x3), 0, 0), 0 where u1 > 0 and the derivative u1 > 0. Thus, the higher the fluid layer, the faster the fluid moves (i.e. the larger u1(x3) is).

Consider again the rectangles B = [a, b] × [c, d] × [e, f ] and B0 = [a, b] × [c, d] × [f , g] as well as the contact surface S = [a, b] × [c, d] × {f }.

Now S is a subset of ∂B with outer unit normal n(x) = e3. We compute

 0  0 0 νu1(x3) 0 τ(x) =  0 0 0  , τ(x)n(x) = νu1(x3)e1. (41) 0 νu1(x3) 0 0

Sauli Lindberg Navier-Stokes equations I Fall 2019 132 / 168 An example: the deviatoric stress tensor of a shear ﬂow II

We interpret the result in (41):

The faster fluid in B0 exerts a force on the fluid in B in the positive direction of the x1 axis and so contributes towards acceleration of the slower fluid in B.

This eﬀect is counteracted by the stress on the "ceiling" of B, i.e., S˜ = [a, b] × [c, d] × {e}, where n(x) = −e3.

In order to compute the total eﬀect of the deviatoric stress tensor on B we use the Divergence theorem componentwise, getting Z Z Z 00 τ(x)n(x) dA = ν(u1 (x3), 0, 0) dV = ν∆u dV . ∂B B B

R R 0 R 0 On the other hand, ∂B τ(x)n(x) dA = S νu1(x3)e1 dA − S νu1(x3)e1 dA.

0 Thus the total effect of τ on B depends on whether u1(x3) is larger at S or S˜, that is, whether the velocity difference between adjacent fluid layers is larger at the top or the ceiling of B.

Sauli Lindberg Navier-Stokes equations I Fall 2019 133 / 168 The Cauchy momentum equation for a Newtonian ﬂuid

Suppose the ﬂuid is homogeneous (with % = 1) and incompressible. If W is admissible and t ∈ (T1, T2), Newton’s second law says in this context that d Z Z Z u dV = σ · n dA + b dV . dt Wt ∂Wt Wt Mimicking our reasoning in the case of ideal ﬂuids, this leads to

∂t u + u · ∇u = ∇ · σ + b, where   P3  σ11 σ12 σ13 j=1 ∂j σ1j ∇ · σ = ∇ · σ σ σ = P3 ∂ σ  .  21 22 23  j=1 j 2j  3 σ31 σ32 σ33 P j=1 ∂j σ3j We calculate componentwise

3 3 X X (∇ · σ)i = ∂j σij = (∂j (−∂j p) + ν∂j (∂j ui + ∂i uj )) j=1 j=1

= −∂i p + ν∆ui + ν∂i div u . | {z } =0

Sauli Lindberg Navier-Stokes equations I Fall 2019 134 / 168 Navier-Stokes equations

Deﬁnition 64 The (homogeneous, incompressible) Navier-Stokes equations with body force b are given by

∂t u + u · ∇u = −∇p + ν∆u + b, div u = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 135 / 168 A boundary value condition for Navier-Stokes equations

In Euler equations, one uses the impenetrability condition u · n|∂D = 0. Thus the ﬂuid is allowed to move along the boundary ∂D.

In Navier-Stokes equations, one sets the no-slip boundary condition u|∂D = 0. The same condition is set in the case of 2D ﬂows.

We quote R. Feynman’s book The Feynman Lectures on Physics, Vol. II, p. 41-1: "It turns out–although it is not at all self-evident–that in all circumstances where it has been experimentally checked, the velocity of a ﬂuid is exactly zero at the surface of a solid." (Emphasis his.)

In some experiments, for instance, dye is injected into ﬂow down a pipe and observed near the boundary.

The no-slip condition is also intuitively plausible. Indeed, adjacent fluid layers give each other momentum. Similarly, molecular interaction between the (stationary) container wall and the fluid should give zero velocity to the fluid right adjacent the boundary.

Sauli Lindberg Navier-Stokes equations I Fall 2019 136 / 168 A 2D pipe ﬂow with Navier-Stokes equations

Consider again a 2D channel ﬂow in the direction x1 in D = R × (0, 1). Suppose b = 0 and the pressure p(x1, x2, t) = p(x1) is larger at x1 = 0 than x1 = L > 0. We look for a stationary solution of Navier-Stokes equations,

u(x1, x2, t) = (u1(x1, x2), 0), p(x1, x2, t) = p(x1).

By incompressibility, ∂1u1 = 0. Thus the Cauchy momentum equation ∂t u + u · ∇u = −∇p + ν∆u becomes 2 − ∂1p + ν∂2 u = 0. (42) 2 2 Now (42) gives ∂1 p = ν∂2 ∂1u1 = 0 so that p(L) − p(0) p(x ) = p(0) + x . 1 L 1 Substituting into (42) and imposing the boundary conditions u1(x1, 1) = u1(x1, 0) = 0 we get p(0) − p(L) u (x , x ) = x (1 − x ). 1 1 2 2Lν 2 2 We conclude that viscosity allows the velocity to achieve a stationary state. Sauli Lindberg Navier-Stokes equations I Fall 2019 137 / 168 Stationary irrotational Euler ﬂow in a half-space

2 Consider u(x1, x2, t) = (u∞, 0) in D = {(x1, x2) ∈ R : x2 > 0}. Thus u solves the stationary Euler equations with p = 0 and u · n|∂D = 0. Next consider Navier-Stokes equations in D × [0, T ] with the same initial values

u1(x1, x2) = u∞ whenever x2 > 0,

u2(x1, x2) = 0

and the boundary condition u|∂D = 0. The solution is Z η 2u∞ −y 2 x2 u1(x1, x2, t) = √ e dy, η = η(x2, t) = √ , π 0 2 νt

u2(x1, x2, t) = 0.

Furthermore, 2 u∞ x2 ω(x1, x2, t) = √ exp − 6= 0. (43) πνt 4νt

(If a 2D velocity ﬁeld is written as (u1(x1, x2, t), u2(x1, x2, t), 0), then the vorticity is of the form (0, 0, ω(x1, x2, t)). When writing (u1(x1, x2, t), u2(x1, x2, t)), it is customary to write vorticity as a scalar as in (43).)

Sauli Lindberg Navier-Stokes equations I Fall 2019 138 / 168 Interpretation of the results

Indeed,

2u 2 x 2 ∂ u (x , x , t) = √∞ e−η ∂ η(x , t) = −√ 2 e−η , t 1 1 2 π t 2 νt3/2

2u∞ −η2 u∞ −η2 ∂2u1(x1, x2, t) = √ e ∂2η(x2, t) = √ e , π πνt 2 u∞ −η2 2 u∞x2 −η2 ∂ u1(x1, x2, t) = √ e ∂2(−η ) = −√ e . 2 πνt π(νt)3/2

The Navier-Stokes solution is neither stationary nor irrotational (even though the initial data is irrotational).

R ∞ 2 √ Note that 0 exp(−y ) dy = π/2. Thus, at each t ∈ [0, T ], the Navier-Stokes and Euler√ solutions agree very well outside a boundary layer where, say, 0 < x2 < C νt.

Solutions of Navier-Stokes equations can create vorticity even without the assistance of a non-conservative body force.

Sauli Lindberg Navier-Stokes equations I Fall 2019 139 / 168 Kinetic energy in Navier-Stokes equations I

Theorem 65 Assume: D is bounded, ∞ 3 1 ¯ 3 u ∈ C (D × [T1, T2], R ) ∩ C (D × [T1, T2], R ), ∞ 0 p ∈ C (D × [T1, T2]) ∩ C (D¯ × [T1, T2]),

u|∂D = 0, u and p satisfy the Navier-Stokes equations with or without gravity. Then d 1 Z Z |u|2 dV = −ν k∇uk2 dV . (44) dt 2 D D

2 P3 2 Above, we use the Frobenius norm given by k[aij ]k = i,j=1 |aij | . Note that (44) quantiﬁes the role of viscosity creating friction in the Navier-Stokes equations.

Sauli Lindberg Navier-Stokes equations I Fall 2019 140 / 168 Kinetic energy in Navier-Stokes equations II

Proof.

By the earlier computation on Euler equations and the condition u|∂D = 0 we get

d Z |u|2 Z ∂ |u|2 Z ∂u dV = dV = u · dV dt D 2 D ∂t 2 D ∂t Z = u · [−u · ∇u − ∇(p + g x3) + ν∆u] dV D Z 3 Z X 2 = u · ν∆u dV = ν ui ∂j ui dV D i,j=1 D 3 Z Z X 2 = −ν ∂j ui ∂j ui dV = −ν |∇u| dV . i,j=1 D D

Sauli Lindberg Navier-Stokes equations I Fall 2019 141 / 168 The energy equality I

Corollary 66 (energy equality for Navier-Stokes equations) Under the assumptions of Theorem 65,

Z Z t Z Z 1 2 2 1 2 |u(x, t)| dV + ν k∇u(x, s)k dV ds = |u(x, T1)| dV 2 D T1 D 2 D

for all t ∈ [T1, T2].

Proof. Using the Fundamental theorem of calculus and (44),

Z Z Z t Z 1 2 1 2 d 1 2 |u(x, t)| dV − |u(x, T1)| dV = |u(x, s)| dV ds 2 D 2 D T1 ds 2 D Z t Z = −ν k∇u(x, s)k2 dV ds. T1 D

Sauli Lindberg Navier-Stokes equations I Fall 2019 142 / 168 The energy equality II

In the case of Euler equations, a similar argument, combined with kinetic energy conservation, gives the following result:

Theorem 67 (energy equality for Euler equations) Suppose u satisﬁes assumptions similar to those of Theorem 65 except u solves the Euler equations and u · n|∂D = 0 (instead of u∂D = 0). Then Z Z 1 2 1 2 |u(x, t)| dV = |u(x, T1)| dV 2 D 2 D

for all t ∈ [T1, T2].

Sauli Lindberg Navier-Stokes equations I Fall 2019 143 / 168 Kinetic energy dissipation and enstrophy I

3 3  3   3  2 X 2 X X X |∇ × v| = (∇ × v)i =  ijk ∂j vk   i`m∂`vm i=1 i=1 j,k=1 `,m=1 3 3 ! 3 X X X = ijk i`m ∂j vk ∂`vm = (δj`δkm − δjmδk`)∂j vk ∂`vm j,k,`,m=1 i=1 j,k,`,m=1 3 3 X 2 X = (∂j vk ∂j vk − ∂j vk ∂k vj ) = k∇vk − ∂j vk ∂k vj . j,k=1 j,k=1

Sauli Lindberg Navier-Stokes equations I Fall 2019 144 / 168 Kinetic energy dissipation and enstrophy II

The proof is completed by computing, by the Divergence theorem,

3 3 3 Z X X Z X Z ∂j vk ∂k vj dV = ∂j (vk ∂k vj ) dV − vk ∂k ∂j vj dV D j,k=1 j,k=1 D j,k=1 D

| P3 R {z } = k=1 ∂D vk ∂k v·n dA=0 3 3 X Z X Z = − ∂k (vk (div v)) + ∂k vk div v dV k=1 D k=1 D | R {z } = ∂D (div v)v·n dA=0 Z = |div v|2 dV . D

Note that we showed the pointwise identity |∇ × v|2 = ∇v · ∇v − ∇v · (∇v)T , 3 3 P3 where [aij ]i,j=1 · [bij ]i,j=1 = i,j=1 aij bij .

Sauli Lindberg Navier-Stokes equations I Fall 2019 145 / 168 The Millennium Problem

We present a part of the Millennium Problem on the Navier-Stokes equations.

Problem 69

Assume u0 ∈ C ∞(R3) satisﬁes div u0 = 0 and

0 0 3 u (x + ei ) = u (x) for all x ∈ R and i ∈ {1, 2, 3}.

Do there exist (u, p) ∈ C ∞(R3 × [0, ∞), R3 × R) such that

3 u(x + ei , t) = u(x, t) for all x ∈ R and i ∈ {1, 2, 3} and (u, p) solve the Navier-Stokes equations with initial data u(x, 0) = u0(x)?

Sauli Lindberg Navier-Stokes equations I Fall 2019 146 / 168 Uniqueness of smooth solutions I

Theorem 70

Let D be bounded. Suppose: ∞ 3 1 ¯ 3 u1, u2 ∈ C (D × [0, T ], R ) ∩ C (D × [0, T ], R ). u1 and u2 are solutions of the Navier-Stokes equations in D × [0, T ] with pressures p1 and p2, external forces b1 and b2 and viscosity ν > 0.

u1|∂D = u2|∂D = 0. R 2 1/2 Then, denoting kv(t)kL2 = ( D |v(x, t)| dV ) ,

Z T ! sup ku1(t) − u2(t)kL2 ≤ ku1(0) − u2(0)kL2 + kb1(t) − b2(t)kL2 dt 0≤t≤T 0 Z T · exp k∇u2(t)kC 0 dt. 0

Corollary 71

Let u1 and u2 satisfy the assumptions of Theorem 79 with b1 = b2 and u1(·, 0) = u2(·, 0). Then u1 = u2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 147 / 168 Uniqueness of smooth solutions II

Proof of Theorem 79: We denote ˜ u˜ = u1 − u2, p˜ = p1 − p2, b = b1 − b2.

We take the diﬀerence of the equations

∂t ui + ui · ∇ui = −∇pi + ν∆ui + bi

to write ˜ ∂t u˜ + u1 · ∇u˜ + u˜ · ∇u2 = −∇p˜ + ν∆u˜ + b. (45) 2 With t ∈ (T1, T2) ﬁxed we take the L inner product (i.e. R (v, w) = (v(t), w(t)) = D v(x, t) · w(x, t) dV ) of both sides of (45) with u˜(·, t) to get ˜ (∂t u˜, u˜) + (u1 · ∇u˜, u˜) + (u˜ · ∇u2, u˜) = −(∇p˜, u˜) + ν(∆u˜, u˜) + (b, u˜). (46)

Recall that d 1 (∂ u˜, u˜) = ku˜k2 = ku˜k ∂ ku˜k , (∇p˜, u˜) = 0, ν(∆u˜, u˜) = −ν k∇u˜k2 . t dt 2 L2 L2 t L2 L2

Sauli Lindberg Navier-Stokes equations I Fall 2019 148 / 168 Uniqueness of smooth solutions III

Furthermore, 3 Z 3 Z 2 X X |u˜1| (u · ∇u˜, u˜) = (u · ∇u˜ )˜u dV = u · ∇ = 0. 1 1 1 1 1 2 i=1 D i=1 D We have simpliﬁed (46) into 2 ˜ ku˜kL2 ∂t ku˜kL2 = −(u˜ · ∇u2, u˜) − ν k∇u˜kL2 + (b, u˜). (47) We will ﬁnish the proof via Grönwall’s inequality which we recall: Lemma 72 Suppose α, β ∈ C 0([0, T ], [0, ∞)) and A ≥ 0 satisfy

Z t α(t) ≤ A + α(s)β(s) ds (48) 0 for all t ∈ [0, T ]. Then, or all t ∈ [0, T ],

R t β(s) ds α(t) ≤ Ae 0 . (49)

Sauli Lindberg Navier-Stokes equations I Fall 2019 149 / 168 Uniqueness of smooth solutions IV

Denote Z T ˜ α(t) = ku˜(t)kL2 , β(t) = k∇u2(t)kC 0 , A = ku˜(0)kL2 + kb(t)kL2 dt. (50) 0 Once we show that α, β and A satisfy (48), inequality (49) directly implies Theorem 79. Lemma 73 Deﬁne α, β and A via (50). Then (48) is satisﬁed.

Proof: Let t ∈ (0, T ]. By the Fundamental theorem of calculus, R t 0 0 α(t) = α(0) + 0 α (s) ds. Since α :(0, T ) → R is continuous, the set Y = {s ∈ (0, T ): α0(s) 6= 0} ⊂ (0, T )

R t 0 R 0 is open. Note that 0 α (s) ds = Y α (s) ds. We next show that the set Z = {s ∈ Z : α(s) = 0} is discrete. Seeking a contradiction, suppose Z has an accumulation point s. Choose sj ∈ Z \{s} with 0 sj → s and α(sj ) = 0. We therefore get α (s) = 0, which contradicts the assumption that s ∈ Z ⊂ Y . Thus Z must be discrete. Sauli Lindberg Navier-Stokes equations I Fall 2019 150 / 168 Uniqueness of smooth solutions V

We will prove (48) by estimating α0 in Y . Let s ∈ Y . Now (47) and the

Cauchy-Schwarz inequality (v(s), w(s)) ≤ kv(s)kL2 kw(s)kL2 give

0 2 ˜ α(s)α (s) ≤ k∇u2(s)kC 0 α(s) + kb(s)kL2 α(s). Whenever s ∈ Y \ Z, we conclude that

0 ˜ α (s) ≤ k∇u2(s)kC 0 α(s) + kb(s)kL2 , (51) and by continuity and the fact that Z ⊂ Y is discrete, (51) also holds for every s ∈ Z.

We conclude that Z Z 0 ˜ α(t) = α(0) + α (s) ds ≤ α(0) + (k∇u2(s)kC 0 α(s) + kb(s)kL2 ) ds Y Y Z t ≤ A + α(s)β(s) ds. 0 This completes the proof of Lemma 73 and, thereby, Theorem 79.

Sauli Lindberg Navier-Stokes equations I Fall 2019 151 / 168 Uniqueness for Euler equations

Theorem 74 Let D be bounded. Suppose: ∞ 3 0 ¯ 3 u1, u2 ∈ C (D × [0, T ], R ) ∩ C (D × [0, T ], R ). u1 and u2 are solutions of the Euler equations in D × [0, T ] with pressures p1 and p2 and external forces b1 and b2.

u1|∂D = u2|∂D = 0. Then

Z T ! sup ku1(t) − u2(t)kL2 ≤ ku1(0) − u2(0)kL2 + kb1(t) − b2(t)kL2 dt 0≤t≤T 0 Z T · exp k∇u2(t)kC 0 dt. 0

Corollary 75

Let u1 and u2 satisfy the assumptions of Theorem 74 with b1 = b2 and u1(·, 0) = u2(·, 0). Then u1 = u2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 152 / 168 An existence result on Navier-Stokes equations I

Recall from a few weeks ago that we expressed Taylor’s formula in ﬂuid dynamics as follows. Theorem 76 ∞ 3 3 3 Suppose v ∈ C (R , R ) and x0 ∈ R . Then 1 v(x) = v(x ) + D(x )(x − x ) + [∇ × v(x )] × (x − x ) + O(|x − x |2) (52) 0 0 0 2 0 0 0

when x → x0; here 1 D(x ) = (∇v(x ) + [∇v(x )]T ). 0 2 0 0

T Recall that, denoting R(x0) = (∇v(x0) − (∇v(x0))) /2, we have

 0 ∂ v − ∂ v ∂ v − ∂ v  1 2 1 1 2 3 1 1 3 R(x )h = ∂ v − ∂ v 0 ∂ v − ∂ v (x ) h 0 2  1 2 2 1 3 2 2 3 0 ∂1v3 − ∂3v1 ∂2v3 − ∂3v2 0 1 = (∇ × v(x )) × h for all h ∈ 3. 2 0 R

Sauli Lindberg Navier-Stokes equations I Fall 2019 153 / 168 An existence result on Navier-Stokes equations II

Note that the trace (sum of diagonal elements

 2∂ v ∂ v + ∂ v ∂ v + ∂ v  3 1 1 1 2 1 1 2 3 1 1 3 X tr D(x ) = tr ∂ v + ∂ v 2∂ v ∂ v + ∂ v (x ) = ∂ v (x ) 0 2  1 2 2 1 2 2 3 2 2 3 0 i i 0 ∂1v3 + ∂3v3 ∂2v3 + ∂3v2 2∂3v3 i=1

= div v(x0).

Also note that for x0 = 0, (52) reads 1 v(x) = v(0) + D(0)x + [∇ × v(0)] × x + O(|x|2). 2 We next produce solutions of Navier-Stokes and Euler equations with v(0) = 0 and no O(|x|2) term.

Sauli Lindberg Navier-Stokes equations I Fall 2019 154 / 168 An existence result on Navier-Stokes equations III

Theorem 77

Let D ∈ C ∞([0, T ], R3×3) such that D(t) is symmetric and trace-free for each t ∈ [0, T ].

Suppose that the vorticity ω(t) ∈ R3 satisﬁes dω (t) = D(t) ω(t), ω(0) = ω ∈ 3, dt 0 R and deﬁne antisymmetric matrix R(t) via the formula R(t)h = 2−1ω(t) × h.

Then 1 u(x, t) = D(t)x + ω(t) × x = (D(t) + R(t))x, (53) 2 1 p(x, t) = − [∂ D(t) + D2(t) + R2(t)] |x|2 (54) 2 t

are solutions of the 3D Euler and Navier-Stokes equations in R3 × [0, T ].

Sauli Lindberg Navier-Stokes equations I Fall 2019 155 / 168 An existence result on Navier-Stokes equations IV

Proof: We ﬁrst show that u is solenoidal:

3  3 3  X X 1 X div u(x, t) = ∂ d (t)x + ω (t)x i  i` ` 2 i`m ` m i=1 `=1 `,m=1 3 3 X 1 X = dii (t) + i`i ω`(t) = 0 2 |{z} i=1 `=1 =0

P3 since, by assumption, tr D(t) = i=1 dii (t) = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 156 / 168 An existence result on Navier-Stokes equations V

In order to verify the Cauchy momentum equation ∂t u + u · ∇u = −∇p + ν∆u we compute 1 dω 1 ∂ u = ∂ Dx + × x = ∂ Dx + (D ω) × x. t t 2 dt t 2 Next, for every i ∈ {1, 2, 3} we have

=uj =ui z }| { z }| { 3 3 ! 3 ! X X X u · ∇ui = (djk + rjk )xk ∂j (dik + rik )xk j=1 k=1 k=1 3 3 ! X X = (djk + rjk )xk (dij + rij ) j=1 k=1

=[(D+R)(D+R)]ik z }| { 3 3 X X = (dij + rij )(djk + rjk ) xk = [(D + R)(D + R)x]i , k=1 j=1

so that u · ∇u = (D + R)(D + R)x.

Sauli Lindberg Navier-Stokes equations I Fall 2019 157 / 168 An existence result on Navier-Stokes equations VI

−1 2 2 P3 2 Recall that p(x, t) = −2 [∂t D(t) + D (t) + R (t)] i=1 xi . Thus 2 2 −∇p = [∂t D + D + R ]x.

The claim ∂t u + u · ∇u = −∇p therefore reduces to 1 (Dω) × x + (D + R)(D + R)x = (D2 + R2)x, 2 i.e., 1 (Dω) × x + (RD + DR)x = 0. (55) 2 Using the formula Rh = 2−1ω × h we write 1 1 (RD + DR)x = ω × Dx + D (ω × x) . (56) 2 2 In view of (55)–(56), the proof is completed via the following lemma (which can be checked by a tedious but straightforward computation). Lemma 78 Suppose A ∈ R3×3 is symmetric and trace-free, and let b, c ∈ R3. Then A(b × c) + (Ab) × c + b × (Ac) = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 158 / 168 An existence result on Navier-Stokes equations IV

For concreteness, we also compute the curl and divergence of the velocity ﬁeld u:

3 3  3 3  X X X 1 X (∇ × u) (x, t) = ∂ u (x, t) = ∂ d x + ω (t)x i ijk j k ijk j  k` ` 2 k`m ` m j,k=1 j,k=1 `=1 `,m=1 3 3 ! X 1 X = d + ω (t) ijk kj 2 k`j ` j,k=1 `=1 3 3 X 1 X = − d + ω (t)( = − , d = d ) ikj jk 2 ijk `jk ` ijk ikj kj jk j,k=1 j,k,`=1 3 1 X = (2 )ω (t) = ω (t). 2 ijk i i j,k=1

Thus ∇ × u = ω.

We next present some examples that Theorem 77 gives.

Sauli Lindberg Navier-Stokes equations I Fall 2019 159 / 168 Example: jet ﬂows I

Set ω0 = 0 and   −γ1 0 0 D(t) = diag(−γ1, −γ2, γ1 + γ2) =  0 −γ2 0  , γj > 0. 0 0 γ1 + γ2 Thus ω = 0 and   −γ1x1 u(x, t) =  −γ2x2  . (γ1 + γ2)x3 The flow map satisfies       ∂t ϕ1(x, t) u1(ϕ(x, t), t) −γ1ϕ1(x, t) ∂t ϕ2(x, t) = u2(ϕ(x, t), t) =  −γ2ϕ2(x, t)  ∂t ϕ3(x, t) u3(ϕ(x, t), t) (γ1 + γ2)ϕ3(x, t) The flow map is of the form

 −γ1t  e x1 −γ2t ϕ(x, t) =  e x2  (γ1+γ2)t e x3.

Sauli Lindberg Navier-Stokes equations I Fall 2019 160 / 168 Example: jet ﬂows II

We make a few observations.

The ﬂow is irrotational.

The squared distance of ﬂuid particles to the x3-axis

2 2 −2γ1t 2 −2γ2t 2 ϕ1(x, t) + ϕ2(x, t) = e x1 + e x2

decreases exponentially in time.

(γ1+γ2)t The "height" ϕ3((x, t) = e x3 increases exponentially in time (when x3 > 0).

Sauli Lindberg Navier-Stokes equations I Fall 2019 161 / 168 Example: strain ﬂows

Take ω0 = 0 and D(t) = diag(−γ, γ, 0), γ > 0. Then   −γx1 u(x, t) =  γx2  . 0

This time       ∂t ϕ1(x, t) u1(ϕ(x, t), t) −γϕ1(x, t) ∂t ϕ2(x, t) = u2(ϕ(x, t), t) =  γϕ2(x, t)  ∂t ϕ3(x, t) u3(ϕ(x, t), t) 0 with the initial data ϕ(x, 0) = x has the solution

 −γt  e x1 γt ϕ(x, t) =  e x2  . x3

Thus the ﬂow stretches the ﬂuid in the x2 direction and compresses it in the x1 direction. Again, ω = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 162 / 168 Example: a vortex

Choose D = 0 and ω0 = (0, 0, 2). Thus ω(t) = (0, 0, 2) for all t. Thus

u(x, t) = (−x2, x1, 0) .

We have encountered this ﬂow several times in the course.

Sauli Lindberg Navier-Stokes equations I Fall 2019 163 / 168 Example: a rotating jet I

Set

D(t) = diag(−γ1, −γ2, γ1 + γ2), γj > 0,

ω0 = (0, 0, ω0,3).

Now the initial value problem dω/dt(t) = D(t)ω(t), ω(0) = ω0 has the solution

(γ1+γ2)t ω(t) = (0, 0, e ω0,3). The velocity is 1 u(x, t) = D(t)x + ω(t) × x 2 1 1 = −γ x − ω (t)x , −γ x + ω (t)x , (γ + γ )x . 1 1 2 3 2 2 2 2 3 1 1 2 3 Thus the ﬂow map ϕ satisﬁes

1 1 ∂t ϕ1 −γ1 − 2 ω(t) ϕ1 −γ1ϕ1 − 2 ω3(t)ϕ2 = 1 = 1 , ∂t ϕ2 2 ω(t) −γ2 ϕ2 2 ω(t)ϕ1 − γ2ϕ2

(γ1+γ2)t ϕ3(x, t) = e x3.

Sauli Lindberg Navier-Stokes equations I Fall 2019 164 / 168 Example: a rotating jet II

We conclude that the "height" ϕ3(x, t) of the ﬂuid particles increases (or decreases) exponentially in time.

The distance of the ﬂuid particles to the x3-axis satisﬁes 2 2 ∂t ϕ1 ϕ1 2 2 ∂t (ϕ1 + ϕ2) = 2 · = −2γ1ϕ1 − 2γ2ϕ2. ∂t ϕ2 ϕ2

Thus

2 2 2 2 2 2 −2 max{γ1, γ2}(ϕ1 + ϕ2) ≤ ∂t (ϕ1 + ϕ2) ≤ −2 min{γ1, γ2}(ϕ1 + ϕ2).

2 2 Using these bounds we next ﬁnd an estimate for ϕ1(x, t) + ϕ2(x, t) .

Sauli Lindberg Navier-Stokes equations I Fall 2019 165 / 168 Example: a rotating jet III

3 2 2 Fix x ∈ R with x1 + x2 > 0. Denoting

2 2 α(t) = ϕ1(x, t) + ϕ2(x, t), α0(t) f (t) = ∈ [−2 max{γ , γ }, −2 min{γ , γ }] α(t) 1 2 1 2

we use Exercise 3.5 to conclude that α(t) > 0 for every t ∈ [0, T ]. By the Fundamental theorem of calculus,

Z t d Z t α0(s) log α(t) = log α(0) + log α(s) ds = log α(0) + ds 0 ds 0 α(s) Z t = log α(0) + f (s) ds, 0

log α(0)+R t f (s) ds R t f (s) ds so that α(t) = e 0 = α(0)e 0 , giving

−2 max{γ1,γ2}t 2 2 2 2 −2 min{γ1,γ2}t 2 2 e (x1 + x2 ) ≤ ϕ1(x, t) + ϕ2(x, t) ≤ e (x1 + x2 ).

Thus the ﬂuid particles approach the x3-axis at an exponential rate.

Sauli Lindberg Navier-Stokes equations I Fall 2019 166 / 168 The inviscid limit I

Recall from the previous lectures:

Theorem 79 Let D be bounded. Suppose: ∞ 3 1 ¯ 3 u1, u2 ∈ C (D × [0, T ], R ) ∩ C (D × [0, T ], R ), u1 and u2 are solutions of the Navier-Stokes equations

∂t uj + uj · ∇uj = −∇pj + bj ,

div uj = 0,

u1|∂D = u2|∂D = 0. R 2 1/2 Then, denoting kv(t)kL2 = ( D |v(x, t)| dV ) ,

Z T ! sup ku1(t) − u2(t)kL2 ≤ ku1(0) − u2(0)kL2 + kb1(t) − b2(t)kL2 dt 0≤t≤T 0 Z T · exp k∇u2(t)kC 0 dt. 0

Sauli Lindberg Navier-Stokes equations I Fall 2019 167 / 168 The inviscid limit II

Suppose now that for some initial data u0 the Navier-Stokes equations and the Euler equations have a smooth solution,

ν ν ν ν ν ∂t u + u · ∇u = −∇p + ν∆u + b, 0 0 0 0 ∂t u + u · ∇u = −∇u + b, div uν = div u0 = 0, ν 0 u (0) = u (0) = u0.

ν 0 0 Setting b1 = b, u1 = u , b2 = −ν∆u + b and u2 = u in Theorem 79 gives

Z T Z T ν 0 0 0 sup u (t) − u (t) L2 ≤ ν∆u (t) L2 dt exp ∇u (t) C 0 dt. 0≤t≤T 0 0

Hence, ν 0 sup u (t) − u (t) L2 = O(ν) when ν & 0. 0≤t≤T

Similar results hold in the periodic case as well as in R3 (under fast enough decay when |x| → ∞).

Sauli Lindberg Navier-Stokes equations I Fall 2019 168 / 168