A Summary of the Course Navier-Stokes Equations I

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A summary of the course Navier-Stokes equations I

Sauli Lindberg

University of Helsinki, Fall 2019

Sauli Lindberg Navier-Stokes equations I Fall 2019 1 / 34 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 2 / 34 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, ∞).

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 3 / 34 Conservation of mass I

We consider volumes W ⊂ D ⊂ R3 of ﬂuid where standard tools of vector analysis are available (and call them admissible volumes).

Axiom 1 (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 Z % dV = − % u · n dA = − div (% u)dV (T1 < t < T2). dt W ∂W W

As a consequence, we get the continuity equation ∂% + div (% u) = 0. (1) ∂t

Sauli Lindberg Navier-Stokes equations I Fall 2019 4 / 34 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.

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 (x, t) = u(ϕ(x, t), t). (2) ∂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 5 / 34 Acceleration of ﬂuid particles

The acceleration of a ﬂuid particle 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 More generally: Deﬁnition 2 ∞ Let f ∈ C (D × [T1, T2]). The material derivative of f is Df ∂f = + u · ∇f . Dt ∂t

∞ 3 Similarly, if v ∈ C (D × [T1, T2], R ), the material derivative of v is Dv ∂v = + u · ∇v. Dt ∂t

Sauli Lindberg Navier-Stokes equations I Fall 2019 6 / 34 A geometric interpretation of the material derivative

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.

Sauli Lindberg Navier-Stokes equations I Fall 2019 7 / 34 Volumes moving with the ﬂuid I

Given t ∈ [T1, T2], we denote

ϕt (x) = ϕ(x, t),

Wt = ϕt (W ) = {ϕt (x): x ∈ W }.

We will 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

How does the momentum R % u dV vary in time? Wt

How does the kinetic energy R % |u|2 /2 dV vary in time? Wt

Sauli Lindberg Navier-Stokes equations I Fall 2019 8 / 34 Volumes moving with the ﬂuid II

We consider the following general question: Question 3 ∞ R If f ∈ C (D × [Tt , T2]), what is the rate of change (d/dt) f dV ? Wt How about (d/dt) R % f dV ? Wt

Theorem 4 (Transport theorem) ∞ For every f ∈ C (D × [T1, T2]),

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

Corollary 5 (Transport theorem with density) ∞ Let f ∈ C (D × [T1, T2]). Then d Z Z Df % f dV = % dV . dt Wt Wt Dt

The proofs proceed via a change of variables from Wt into W . Sauli Lindberg Navier-Stokes equations I Fall 2019 9 / 34 Change of variables in an integral

The proofs are based on a change of variables. Denote

∂ϕ1 ∂ϕ1 ∂ϕ1   ∂x1 ∂x2 ∂x3  ∂ϕ ∂ϕ ∂ϕ   2 2 2  Jϕ = det Dϕ = det   .  ∂x1 ∂x2 ∂x3  ∂ϕ3 ∂ϕ3 ∂ϕ3  ∂x1 ∂x2 ∂x3

Theorem 6 (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

Theorem 7 (Euler’s identity)

For all (x, t) ∈ D × (T1, T2), ∂Jϕ t (x) = [div u](ϕ (x), t) Jϕ (x). ∂t t t

Sauli Lindberg Navier-Stokes equations I Fall 2019 10 / 34 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.)

Deﬁnition 8

An ideal fluid is a fluid with the following property: the total force exerted on the fluid inside W by means of stress on ∂W at time t is Z Z − p(x, t)n(x)dA = − ∇p(x, t) dV . ∂W W

Deﬁnition 9 If b(x, t) denotes the net body force per unit mass at (x, t), the total body force R on W is W %(x, t) b(x, t) dx.

Sauli Lindberg Navier-Stokes equations I Fall 2019 11 / 34 Newton’s second law I

We express Newton’s second law for ideal ﬂuids:

Axiom 10 (Law of balance of momentum) For an ideal ﬂuid, the rate of change of momentum d Z Z Z % u dV = − ∇p dV + % b dV . (3) dt Wt Wt Wt

Via the Transport theorem with density we obtain the following easier form:

Theorem 11

Suppose a ﬂuid is ideal. Then, in D × (T1, T2), Du % = −∇p + % b. (4) Dt

Equation (4) is (a special case of) the Cauchy momentum equation or the law of balance of momentum.

Sauli Lindberg Navier-Stokes equations I Fall 2019 12 / 34 Rates of change

Under the assumptions of the course, we have

(d/dt) R % dV = 0, Wt

(d/dt) R dV = R div u dV , Wt Wt

(d/dt)(R % dV / R dV ) = − R % dV R div u dV /(R dV )2, Wt Wt Wt Wt Wt

(d/dt) R %u dV = − R ∇p dV + R % b dV (for ideal ﬂuids), Wt Wt Wt

(d/dt) R % |u|2 /2 dV = R (−∇p + ρb) · u dV (for ideal ﬂuids). Wt Wt

In particular, divergence controls the rate of change of volume. This phenomenon is captured in the notion of incompressibility.

Sauli Lindberg Navier-Stokes equations I Fall 2019 13 / 34 Incompressibility

Deﬁnition 12 We call a ﬂow incompressible if for every admissible W ⊂ D, Z volume of Wt = dV = constant in t. Wt

Theorem 13 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 14 / 34 Homogeneous, incompressible Euler equations

A ﬂuid is said to be homogeneous if the density % is independent of the position x ∈ D. Typically one sets % = 1.

Deﬁnition 14 The homogeneous, incompressible Euler equations are given by ∂u + u · ∇u + ∇p = b, (5) ∂t div u = 0. (6)

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.

Sauli Lindberg Navier-Stokes equations I Fall 2019 15 / 34 Stationary solutions of Euler equations

Suppose that ∂u/∂t = 0 and b = −∇(gx3) is gravity, so that the Euler equations reduce to

u · ∇u = −∇(p + gx3), (7) div u = 0. (8)

Equation (7) is hard to analyse because the term u · ∇u is nonlinear in u. However,

|u|2 u · ∇u = ∇ − u × (∇ × u). (9) 2

2 Thus, if div u = 0 and ∇ × u = 0, then setting p = −g x3 − |u| /2 gives a solution (u, p) of (7)–(8).

The remark above (among other things) motivates us to study irrotational ﬁelds v (i.e. ones satisfying ∇ × v = 0).

Sauli Lindberg Navier-Stokes equations I Fall 2019 16 / 34 Solenoidal, irrotational vector ﬁelds

Deﬁnition 15 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.

Suppose v = ∇ψ, where ∆ψ = 0. Then

div v = div (∇ψ) = ∆ψ = 0, ∇ × v = ∇ × (∇ψ) = 0

and v is called a potential ﬂow.

Sauli Lindberg Navier-Stokes equations I Fall 2019 17 / 34 Irrotational ﬂows and gradients

More generally, If v ∈ C ∞(D, R3) is of the form v = ∇ψ for some ψ ∈ C ∞(D), then ∇ × v = ∇ × (∇ψ) = 0. Question 16 If ∇ × v = 0, does there exist ψ ∈ C ∞(D) such that v = ∇ψ?

The answer depends on the domain D ⊂ R3. Theorem 17

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.

Proposition 18 3 Let D = R \{(0, 0, x3): x3 ∈ R} and deﬁne −x2 x1 v(x) = 2 2 , 2 2 , 0 for all x ∈ D. x1 + x2 x1 + x2 Then there exists no ψ ∈ C ∞(D) such that ∇ψ = v.

Sauli Lindberg Navier-Stokes equations I Fall 2019 18 / 34 Line integrals

Deﬁnition 19 0 Let a = c1 < c2 < ··· < cn = b, and suppose γ ∈ C ([a, b], D) 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] → D is closed if γ(a) = γ(b).

Deﬁnition 20 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

Proposition 21 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 19 / 34 A geometric interpretation of curl

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

Let v ∈ C ∞(D, R3). By using the continuity of ∇ × v and Stokes’ theorem, we showed in the lectures thet R v(x) · ds ∂Sr ∇ × v(x0) · n = lim . r→0 πr 2

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

Sauli Lindberg Navier-Stokes equations I Fall 2019 20 / 34 Vorticity

Definition 22 ∞ 3 When u ∈ C (D × [T1, T2], R ) satisfies the Euler equations with some ∞ p ∈ C (D × [T1, T2]), the vector field

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

Theorem 23 ∞ 3 Suppose u ∈ C (D × [T1, T2], R ) solves the Euler equations with b = −∇ψ. Then the vorticity ω satisﬁes ∂ω Dω − ω · ∇u + u · ∇ω = − ω · ∇u = 0 ∂t Dt and

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

Sauli Lindberg Navier-Stokes equations I Fall 2019 21 / 34 Kinetic energy conservation in Euler equations

Theorem 24

Assume: D is bounded, u ∈ C ∞(D, R3) ∩ C 0(D¯, R3) and p ∈ C ∞(D) ∩ C 0(D¯), u and p satisfy the Euler equations with a conservative body force b = −∇ψ,

u · n|∂D = 0. Then 1 Z the kinetic energy |u|2 dV is constant in time. 2 D

Sauli Lindberg Navier-Stokes equations I Fall 2019 22 / 34 d’Alembert’s paradox

We will consider stationary potential flow around a fixed obstacle Λ in R3. We assume that v · n|∂Λ = 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, in some cases, the solutions are incompatible with experiments. A famous instance of this is called d’Alembert’s paradox.

Theorem 25

Let Λ ⊂ R3 be a bounded domain such that the Divergence theorem holds for D = R3 \ 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 23 / 34 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.

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 24 / 34 The Cauchy stress tensor

Axiom 26 If W ⊂ D is an admissible volume, then

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

where σ(x, t) ∈ R3×3 is called the Cauchy stress tensor.

For ideal ﬂuids, σ(x, t, n) = −p(x, t)n(x).

σ(x, t)n(x) need not be parallel to n(x). In particular, momentum transfer across a surface is allowed.

Sauli Lindberg Navier-Stokes equations I Fall 2019 25 / 34 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 27 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 26 / 34 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. (10) 0 νu1(x3) 0 0

Sauli Lindberg Navier-Stokes equations I Fall 2019 27 / 34 An example: the deviatoric stress tensor of a shear ﬂow II

We interpret the result in (10):

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 28 / 34 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 = σn 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.

Computing ∇ · σ, we get the following equations:

Deﬁnition 28 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 29 / 34 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.

Sauli Lindberg Navier-Stokes equations I Fall 2019 30 / 34 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 2D 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. (11) 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 (11).)

Sauli Lindberg Navier-Stokes equations I Fall 2019 31 / 34 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.

At every (x1, x2) ∈ D we have limt→∞ u(x, t) = 0.

Sauli Lindberg Navier-Stokes equations I Fall 2019 32 / 34 Kinetic energy in Navier-Stokes equations

Theorem 29 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 Z |u|2 dV = −ν k∇uk2 dV = −ν |ω|2 dV . (12) dt 2 D D D

Corollary 30 (energy equality for Navier-Stokes equations)

Under the assumptions of Theorem 29, for all t ∈ [T1, T2], 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

Sauli Lindberg Navier-Stokes equations I Fall 2019 33 / 34 Uniqueness of smooth solutions I

Theorem 31

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 32

Let u1 and u2 satisfy the assumptions of Theorem 31 with b1 = b2 and u1(·, 0) = u2(·, 0). Then u1 = u2.

Sauli Lindberg Navier-Stokes equations I Fall 2019 34 / 34