Lecture Notes on the Theory of Incompressible Fluid Motion

Alberto Bressan and Carlotta Donadello

15th November 2007 Contents

1 The equations of ﬂuid motion 1 1.1 Diﬀerential operators ...... 1 1.2 The Euler and Navier-Stokes equations ...... 2 1.3 Symmetry groups for the Euler and Navier-Stokes equations ...... 3 1.4 Particle trajectories ...... 4 1.5 Vorticity ...... 5 1.6 Vorticity transport formula for the Euler equation ...... 7 1.7 Vortex lines ...... 9 1.8 Conserved quantities for the Euler equation ...... 10 1.9 Leray formulation of the Navier-Stokes equations ...... 12 1.10 Vorticity-stream formulation of the Navier-Stokes Equation ...... 14 1.11 Lagrangian formulation of the Euler equation ...... 17

2 Construction of solutions 19 2.1 Uniqueness of solutions ...... 19 2.2 The Sobolev spaces Hm(RN ) ...... 20 2.3 Local-in-time existence of solutions ...... 22 2.4 Global weak solutions of the Navier-Stokes equations ...... 27 2.5 Construction of ﬁnite dimensional approximate solutions ...... 29

3 Vortex sheets 32 3.1 Velocity behavior near a vortex sheet ...... 32 3.2 The Birkhoﬀ-Rott equation ...... 34 3.3 Kelvin-Helmholtz instability ...... 35 3.4 Spiralling solutions ...... 36

i Chapter 1

The equations of ﬂuid motion

1.1 Diﬀerential operators

Given a scalar function u : RN 7→ R, its gradient ∇u is the vector of partial derivatives ³ ´ . ∂ ∂ ∂ ∇u = (ux1 , ux2 , . . . , uxN ) = u , u , . . . , u . ∂x1 ∂x2 ∂xN

For a vector ﬁeld u : RN 7→ RN , its divergence is

XN . i div u = uxi . i=1

The Laplace operator is deﬁned as

N . X ∂2u ∆u = = div ∇u . ∂x2 i=1 i

Consider a velocity ﬁeld u : R × RN 7→ RN , so that u(t, x) = (u1, . . . , uN )(t, x) denotes the velocity of the particle located at the point x = (x1, . . . , xN ) at time t. We then deﬁne the material derivative of a function φ = φ(t, x) as

N . ∂ X ∂ D φ = φ + ui φ . t ∂t ∂x i=1 i

This provides the time-derivative of φ along particle trajectories.

We recall that the wedge product of two vectors v, w ∈ R3 can be obtained as follows. Let 3 e1, e2, e3 be the vectors in the standard orthonormal basis of R . Then   e1 e2 e3 v ∧ w = det  v1 v2 v3  . w1 w2 w3

1 Given a velocity ﬁeld u : R3 7→ R3, its vorticity ω is deﬁned as

. ¡ 3 2 1 3 2 1¢ ω = curl u = ∂2u − ∂3u , ∂3u − ∂1u , ∂1u − ∂2u .

Note that, formally, one can write   e1 e2 e3 curl u = ∇ ∧ u = det  ∂1 ∂2 ∂3  . u1 u2 u3

1.2 The Euler and Navier-Stokes equations

Let u = u(t, x) denote the velocity of a fluid, at time t and at a point x in space. The flow of a homogeneous, incompressible, non-viscous fluid in RN is modelled by the Euler equations ½ u + u · ∇u = −∇p balance of momentum, t (1.1) div u ≡ 0 incompressibility condition.

Notice that ut + u · ∇u = Dtu is the material derivative of the vector ﬁeld u. Here

t is time, N x = (x1, . . . , xN ) ∈ R is the Eulerian space variable, u = (u1, . . . , uN ) is the ﬂuid velocity, p is the scalar pressure.

When viscosity is present, an additional diﬀusion term is present in the momentum equation. This leads to the Navier-Stokes equations ½ u + u · ∇u = −∇p + ν∆u balance of momentum, t (1.2) div u ≡ 0, incompressibility condition.

The positive constant ν is the coeﬃcient of kinematic viscosity. Many authors refer to it as the inverse of the Reynolds number: ν = 1/Re. The equations (1.1) or (1.2) are naturally supplemented by initial data

u(0, x) = u0(x) . (1.3)

If the ﬂuid motion takes place in a bounded set Ω ⊂ RN , one must add suitable boundary conditions. In the case of the inviscid Euler equations, one requires that the velocity u of the ﬂuid is tangential to the boundary ∂Ω. Calling n the unit outer normal, we thus have

n · u = 0 x ∈ ∂Ω . (1.4)

In the presence of viscosity, instead of (1.4), the Navier-Stokes equations are supplemented by the non-slip boundary conditions

u = 0 x ∈ ∂Ω . (1.5)

We now show how the Euler equation can be derived from the balance law of momentum.

2 n Consider any region W0 ⊂ R . Denote by Wt the region occupied at time t > 0 by those ﬂuid particles which are initially in W0. Assume that the only force acting on the ﬂuid through the boundary ∂Wt is the pressure. Then the balance law of momentum takes the form Z d u dV = [total forces across the boundary] dt Wt Z (1.6) = − p n dA . ∂Wt

If e is any ﬁxed vector in RN by the divergence theorem we get Z Z p e · n dA = div(p e) dV ∂Wt ZWt (1.7) = ∇p · e dV. Wt

Then: Z Z Z d u dV = Dtu dV = − ∇p dV, (1.8) dt Wt Wt Wt for any region Wt in the ﬂuid at time t. Hence the equality holds also in the diﬀerential form:

Dt u = −∇p Euler equation.

1.3 Symmetry groups for the Euler and Navier-Stokes equations

Assume that u = u(t, x) and p = p(t, x) provide a solution to the Euler or the Navier-Stokes equations. Further solutions can then be obtained by various variable transformations.

• Translation Invariance: For any constant vector c in RN ½ . uc(t, x) = u(t, x − ct) + c, . pc(t, x) = p(t, x − ct),

is another solution.

• Rotation Invariance: For any orthogonal matrix Q ½ . uQ(t, x) = QT u(t, Qx), . pQ(t, x) = p(t, Qx),

is another solution.

• Scale Invariance: If u, p provide a solution to the Euler equations, then for any λ, τ > 0 the functions   λ, τ . λ t x u (t, x) = τ u( τ , λ ), (1.9)  λ, τ . λ2 t x p (t, x) = τ 2 p( τ , λ ), provide a 2-parameters family of solutions to the Euler equation.

3 If u, p are a solution to the Navier-Stokes equations, then for all τ > 0 the functions  .  uτ (t, x) = τ −1/2 u( t , x ), τ τ 1/2 . (1.10)  pτ (t, x) = τ −1 p( t , x ). τ τ 1/2

provide a 1-parameter family of solutions to the Navier-Stokes equation for τ ∈ R. Observe that (1.9) coincides with (1.10) when λ = τ 1/2

1.4 Particle trajectories

Given a fluid flow with velocity field u (not necessarily incompressible), the particle trajectory mapping X : [0, ∞) × RN 7→ RN (1.11) t , α 7→ X(t, α) describes the trajectory of the particle which is initially located at point α = (α1, . . . , αN ) at time t = 0. The Lagrangian variable α, can be regarded as a particle marker. The function X(t, α) is determined by solving the Cauchy problem

 ∂  ∂t X(t, α) = u(t, X(t, α)),  X(0, α) = α.

N An initial domain W0 ⊂ R in the ﬂuid evolves in time to . Wt = X(t, W0) = {X(t, α)| α ∈ W0}.

We shall denote by ∇αX the Jacobian matrix of ﬁrst order partial derivatives of X w.r.t. α, so that ³ ´ ∂Xi ∇αX(t, α) = (t, α) . ij ∂αj

The evolution of ∇αX along a particle trajectory is described by the linear evolution equation ∂ ³ ´ ³ ´ ∇ X(t, α) = ∇ u(t, X(t, α) · ∇ X(t, α) . ∂t α x α The Jacobian determinant of X(t, ·) is denoted as ¡ ¢ J(t, α) = det ∇αX(t, α) .

At the initial time t = 0, ∇αX(0, α) = I is the identity matrix, hence its determinant is J(0, α) ≡ 1. If u is a smooth velocity ﬁeld then the time evolution of J is given by

∂ h i J(t, α) = div u(t, X(t, α)) J(t, α) . ∂t In particular, J(t, X(t, α)) ≡ 1 if the ﬂow is incompressible.

4 Theorem 1 The Trasport Formula Let W be an open, bounded domain in RN with smooth boundary, and let X be the particle trajectory mapping of a given smooth velocity ﬁeld u. For any smooth function f = f(t, x) we have Z Z d f dx = [ft + div (fu)] dx. dt Wt Wt

Proof Z Z d d f dx = f(t, X(t, α))J(t, X(t, α)) dα dt dt Wt Z W ¡ ¢ = Dtf J + f Jt dα ZW ¡ ¢ = ft + u · ∇f + f div u J dα ZW = [ft + div(fu)] dx. Wt ¤

Deﬁnition 1 A ﬂow X is incompressible if for any subregion W with smooth boundary, and for any time t > 0, X is volume preserving: ¡ ¢ ¡ ¢ Vol X(t, W ) = Vol W .

Applying the transport formula with f ≡ 1 one obtains

Proposition 1 The following facts are equivalent:

• The ﬂuid ﬂow is incompressible,

• div u = 0,

• J(t, α) ≡ 1

1.5 Vorticity

In the following, we consider a smooth velocity field u : R3 7→ R3. We define the vorticity field ω as the curl of the velocity:

. ¡ 3 2 1 3 2 1¢ ω = curl u = ∂2u − ∂3u , ∂3u − ∂1u , ∂1u − ∂2u .

Proposition 2 (Local behavior of an incompressible flow). In a neighborhood of any point x0, up to higher order terms w.r.t. (|x − x0|), a smooth incompressible velocity field u = u(x) can be written in a unique way as the sum of an infinitesimal translation, rotation and deformation velocities.

5 N Proof Let x0 be a given point in R . A ﬁrst order Taylor expansion of u yields

2 u(x0 + h) = u(x0) + ∇u(x0) · h + O(h ).

The Jacobian matrix ∇u, can be written as the sum of its symmetric and antisymmetric parts:

∇u = |{z}D +|{z} Ω Symm. Antisymm. where ³ ´ 1 t D = 2 ∇u + (∇u) , hence tr D = 0 ³ ´ 1 t 1 Ω = 2 ∇u − (∇u) hence Ωh = 2 ω ∧ h . Observe that, since every symmetric matrix can be diagonalized and the trace is invariant under orthogonal trasformations, D can be written as   γ1 0 0 D =  0 γ2 0  . 0 0 −(γ1 + γ2)

The Taylor expansion thus takes the form

1 2 u(x0 + h) = u(x0) + ω ∧ h + Dh +O(h ). | {z } | 2 {z } | {z } deformation translation rotation

We now study how the vorticity of the ﬂuid evolves in time.

Proposition 3 (Evolution equation for the vorticity) Let the velocity ﬁeld u = u(t, x) provide a solution to the Navier-Stokes equations. Then the vorticity ω=curl u evolves according to the equation Dtω = (ω · ∇)u + ν∆ω. (1.12)

Proof We use the vector identity: 1 ∇|u|2 = u ∧ curl u + (u · ∇)u, (1.13) 2 in order to replace the term (u · ∇)u in the Navier-Stokes equation: 1 ∂ u + ∇|u|2 − u ∧ ω = −∇p + ν∆u. t 2 We now recall that curl(∇ϕ) = 0 for every ϕ. Taking the curl of both sides of the above identity one obtains ∂t ω − curl (u ∧ ω) = ν∆ω. (1.14) In order to write esplicitly the term curl (u ∧ ω) we recall the following general vector identity:

curl (u ∧ ω) = (ω · ∇)u − ω div u − (u · ∇)ω − u div ω. (1.15)

6 Since div u ≡ 0, this reduces to

curl (u ∧ ω) = (ω · ∇)u − (u · ∇)ω .

Therefore, from (1.14) we obtain

∂t ω + (u · ∇)ω = (ω · ∇)u + ν∆ω. (1.16)

1.6 Vorticity transport formula for the Euler equation

The map t 7→ X(t, α) describes the trajectory of the particle that was in α at t = 0. In the present section we will show that the inviscid vorticity equation, i.e. the vorticity evolution equation associated to Euler equation,

Dtω = (ω · ∇)u, (1.17) can be integrated by means of the particle trajectory equation

d X(t, α) = u(t, X(t, α)). dt

Throughout the following, we call ω0(α) = ω(0, α) the initial vorticity at time t = 0.

Proposition 4 (Vorticity transport formula). Let X(t, α) be the smooth particle trajectory mapping corresponding to the smooth divergence-free vector ﬁeld u(t, x). Then ¡ ¢ ω t, X(t, α) = ∇αX(t, α)ω0(α) solves (1.17).

Proof We observe that the vorticity equation has the same form of the evolution equation satisfied by a first order perturbation. Let h(t, x) be a smooth vector field in RN , and consider the perturbed flow

X(t, α + h) = X(t, α) + ∇ X(t, α) h +O(h2). | α {z } (?)

Here (?) is the term which tell us how the perturbation evolves in time. Then the evolution equation of a ﬁrst order perturbation takes the form

Dt h = h · ∇u.

Notice that the right hand side represents the directional derivative of the velocity u in the direction of the vorticity ω. The following lemma applies also to the more general case in which the velocity ﬁeld u is not divergence free.

7 Lemma 1 Let u = u(t, x) be a smooth vector field with associated particle trajectory mapping X(t, α). The smooth vector field h = h(t, x) satisfies

Dt h = h · ∇u (1.18) if and only if h(t, X(t, α)) = ∇αX(t, α)h(0, α). (1.19)

Proof The particle trajectory mapping is the solution of

 d  dt X(t, α) = u(t, X(t, α)),  X(0, α) = α.

By diﬀerentiation with respect to α we obtain

d ∇ X(t, α) = (∇ u) ∇ X(t, α), dt α x |x=X(t,α) α and then d ∇ X(t, α)h(0, α) = (∇ u) ∇ X(t, α)h(0, α). dt α x |x=X(t,α) α If (1.19) holds then d h(t, X(t, α)) = (∇ u) h(t, X(t, α)). dt x |x=X(t,α) Notice that the right hand side represents the directional derivative of the velocity u in the direction of the vector h(t, X(t, α)). This implies (1.18). Viceversa, if (1.18) holds, then the map t 7→ h(t, X(t, α)) satisﬁes

d h(t, X(t, α)) = (∇u) h(t, X(t, α)), dt |x=X(t,α) and, since h(t, X(t, α)) and ∇αX(t, α)h(0, α) both satisfy the same linear ODE with initial data h(0, α), they coincide. ¤ Proposition 4 is now proved by applying the lemma to the case h = ω. ¤

Corollary 1 For a planar ﬂow, the vorticity is constant along particle trajectories:

ω(t, X(t, α)) = ω0(α). (1.20)

Indeed, if the ﬂow lies on the plane R2, then the velocity ﬁeld is u = (u1, u2, 0), the vorticity is 2 1 ω = (0, 0, ω3) = (0, 0, ux1 − ux2 ), and therefore   dX1 dX1 dα dα 0      1 2  0 0  dX dX  ∇αX(t, α) ω =  2 2 0   0  =  0  .  dα1 dα2  ω3 ω3 0 0 1

8 1.7 Vortex lines

At a ﬁxed time t, the integral curves of the vorticity ﬁeld ω = curl u are called vortex lines:

Definition 2 A smooth curve s 7→ γ(s) ∈ RN is a vortex line at time t if d γ(s) = λ(s)ω(t, γ(s)), for some λ(s) 6= 0. ds Corollary 2 In a non-viscous fluid flow, the vortex lines move with the fluid.

Proof Let γ0 be a vortex line at time t = 0. At time t > 0, the ﬂuid particles which are initially along γ0 will have moved to the curve . γt = {X(t, γ0(s))}.

We claim that γt is also a vortex line, at time t. Indeed, d d γ (s) = ∇ X(t, γ (s)) γ (s) ds t α 0 ds 0 = ∇αX(t, γ0(s))λ(s)ω(0, γ0(s)) (1.21) = λ(s)ω(t, X(t, γ0(s)))

= λ(s)ω(t, γt(s)) . ¤

Definition 3 Let u be a smooth divergence-free vector field and let C be a smooth oriented closed curve in the fluid. We call circulation of u around C the line integral I

ΓC = u · dl. C . Lemma 2 Transport formula for curves. Let Ct = X(t, C) be a curve transported by the ﬂow. The following identity holds: I I d u · dl = Dtu · dl. (1.22) dt Ct Ct

Proof Let s 7→ C(s) be a parametrization of C, and let s 7→ X(t, C(s)) = Ct(s) be the parametrization of Ct. I Z d d 1 ∂ u · dl = u(t, X(t, C(s))) X(t, C(s)) ds dt C dt 0 ∂s t Z 1 ∂ = D u(t, X(t, C(s))) X(t, C(s)) ds t ∂s 0 Z (1.23) 1 ∂ ∂ + u(t, X(t, C(s))) X(t, C(s)) ds ∂t ∂s I 0 Z 1 ∂ |u(t, X(t, C(s)))|2 = Dtu · dl + ds. Ct 0 ∂s 2

Since Ct is a loop the last term of the right hand side is null and we obtain the desired identity (1.22). ¤

9 Theorem 2 Kelvin’s circulation theorem. For an inviscid, incompressible ﬂuid the circulation of the velocity u around any closed curve Ct moving with the ﬂuid is constant in time.

Proof Using the transport formula (1.22) one obtains I I d u · dl = D u · dl dt t Ct CIt (1.24) = − ∇p · dl = 0 Ct because the line integral of a gradient along a loop is zero. ¤

Corollary 3 The vorticity ﬂux through a 2-dimensional surface moving with the ﬂuid is constant in time.

Proof Indeed, consider any 2-dimensional surface Σt whose boundary is a closed curve Ct moving with the ﬂuid. At any point of the surface, call n the perpendicular unit vector. Using Stokes’ formula one obtains Z I ω · ndA = u · dl . Σt Ct The result thus follows from Kelvin’s circulation theorem. ¤ An example of this phenomena is provided by tornadoes. If the surface of a horizontal section decreases proportionally to the height h from the ground, then as h → 0 the angular velocity around the vertical axis must increase.

1.8 Conserved quantities for the Euler equation

A conservation law in N space dimensions takes the form:

∂tu(t, x) + divxF (u(t, x)) = 0 , (1.25)

N 1 N where t ≥ 0 is the time variable, x = (x1, . . . , xN ) ∈ R is the space variable, u = (u , . . . , u ) is a vector in RN and F is a smooth map defined on a convex neighborhood of the origin in RN with values in RN . The components of u are called the conserved quantities while the components of F , are the fluxes. Let W ⊂ RN be a closed region with smooth boundary. Integrating (1.25) on W we obtain Z Z d u(t, x) dx = u (t, x) dx dt t W WZ ¡ ¢ = − div F u(t, x) dx ZW ¡ ¢ = − F u(t, x) · n dA ∂W = [flow across the boundary].

10 Remark. Notice that Z d u(t, x) dx = 0 if |F | = o(|x|1−N ). dt RN

Indeed Z Z Z d d u(t, x) dx = lim u(t, x) dx = − lim F (u(t, x)) · n dA, = 0 , dt RN R−→∞ dt |x|≤R R−→∞ |x|=R provided that |F |/|x|N−1 → 0 as |x| → ∞.

The following are conserved quantities for the Euler equations.

• Components of momentum. The Euler equation is equivalent to the momentum balance law. Then, if u is a solution to the Euler equation its components satisfy

XN i j i ut + u uxj = −pxi i = 1,...,N. j=1

The above equations can be written as a system of conservation laws, namely

i i ut + div (u u + Pi) = 0, i = 1,...,N,

where Pi ≡ (0,..., 0, p, 0,..., 0).

• Vorticity. The equation for the i-th component of the vorticity takes the form

i i i ωt + (u∇)ω = (ω∇)u .

These equations also yield a system of conservation laws

i i i ωt + div (ω u − u ω) = 0 i = 1,...,N.

• Energy. Taking the scalar product of both sides of the Euler equation by u, we obtain the energy equation ³ 2 ´ XN XN |u| i j i i + u u ux + u pxi = 0. 2 t j i,j=1 i=1 It can be written as a conservation law

³|u|2 ´ ³|u|2 ´ + div u + p v = 0. 2 t 2

• Helicity. The helicity measures the component of velocity in the direction of the vorticity: . H = u · ω.

11 Diﬀerentiating w.r.t. time we obtain ∂ H = u · ω + u · ω ∂t t t = [−(u · ∇)u − ∇p ] · ω + u · [−(u · ∇)ω + (ω · ∇)u ] XN XN XN XN j i i i i j i i j i = − u uxj ω − pxi ω − u u ωxj + u ω uxj . i,j=1 i=1 i,j=1 i,j=1 This equation can be written in conservation form, namely

h XN XN |ui|2 i H + div ωiuiu − ω + p ω = 0. t 2 i=1 i=1

1.9 Leray formulation of the Navier-Stokes equations

In this section, our goal is to eliminate the pressure term from the Navier-Stokes equations, and obtain an evolution equation for the velocity u alone. Assume that, at time t = 0, div u = 0. Moreover, let u satisfy the equation du = −(u · ∇)u + ν∆u − ∇p (1.26) dt How can we determine the pressure function p = p(t, x), so that the divergence of u remains zero st all positive times? To set the ideas, suppose u ∈ L2, so that Ã Z !1/2 2 kukL2 = |u| dx < +∞.

We introduce the subspace E ⊂ L2 containing all vector ﬁelds whose divergence (in distribution sense) vanishes. Notice that, if u is a C1 vector ﬁeld with zero divergence, then for every smooth scalar function ϕ with compact support we have Z Z u · ∇ϕ dx = div (ϕ u) dx = 0 .

We can thus deﬁne the subspace E by setting ½ Z ¾ . 2 ∞ E = u ∈ L | u · ∇ϕ dx = 0 for every ϕ ∈ Cc . (1.27)

If u(t) ∈ E for all t ≥ 0, the time derivative ∂tu(t) must remain in E for all times. Looking at the equation (1.26) we observe that div u = 0 ⇒ div (∆u) = 0, (1.28) div u = 0 ; div (u · ∇u) = 0. More precisely, if the i-th component of (u · ∇)u is X i j i [(u · ∇)u] = u uxj , j

12 we can write X ¡ j i j i ¢ div (u · ∇)u = uxi uxj + u uxi xj . i,j | {z } Ai,j P i P Since div u = i uxi = 0, we get i,j Ai,j = 0, so, in general, X j i 2 div (u · ∇)u = uxi uxj = tr(∇u) 6= 0. i,j

We now seek a scalar function p such that the sum du = −(u · ∇)u + ν∆u −∇p dt | {z } |{z} ∈/E ∈E lies in the subspace E. Observing that X £ j i 2 div (u · ∇)u + ∇p] = uxi uxj + ∆p = tr(∇u) + ∆p, i,j we conclude that the identity £ div (u · ∇)u + ∇p] = 0 holds if and only if the function p solves the elliptic equation

∆p = −tr(∇u)2 (1.29) on the whole space RN . A solution formula for the Poisson equation on RN is given in the following lemma.

Lemma 3 Solution of the Poisson equation. Let f be a smooth function in RN , vanishing suﬃciently rapidly as |x| −→ ∞. Then the solution to the Poisson equation ½ ∆z = f, ∇z −→ 0 as |x| −→ ∞, is given by Z

z(x) = NN ∗ f(x) = NN (x − y)f(y) dy , RN where N is the N-dimensional Newton potential ½ 1 ln |x|, if N = 2, N (x) = 2π (1.30) N 1 |x|2−N , if N ≥ 3 (2−N) σN

N By σN we denote here the surface area of a unit sphere in R .

We observe that the kernel NN is the distributional solution of the equation

∆N = δ0

13 where δ0 is the Dirac mass concentrate upon the origin.

By applying Lemma 3, we can solve the equation (1.29) and compute the pressure Z 2 p(t, x) = NN (x − y) tr(∇u(t, y)) dy. RN The gradient of p takes the form Z 2 x − y 2 ∇p(t, x) = ∇NN ∗ tr(∇u) (x) = CN N tr(∇u(t, y)) dy, RN |x − y| for a suitable constant CN . Using this formula, we can eliminate the pressure from the standard formulation of the Navier-Stokes equation and obtain a system of closed evolution equations for u. This is the Leray formulation of the Navier-Stokes equations: Z x − y 2 Dtu(t, x) = ν∆u − CN N tr(∇u(t, y)) dy. (1.31) RN |x − y|

Remark. According to (1.27), the subspace E of vector ﬁelds with zero divergence can be regarded as the space perpendicular to all gradient vector ﬁelds (with suitable decay as |x| → ∞). The Navier-Stokes equation can be equivalently written as

ut = ΠE(u · ∇)u + ν∆u, where ΠE is the perpendicular projection operator on the space E.

1.10 Vorticity-stream formulation of the Navier-Stokes Equa- tion

Taking the curl of both sides of the Navies-Stokes equation we obtain the vorticity equation

ωt + u · ∇ω = ω · ∇u + ν∆ω , (1.32) in which the pressure does not appear. However, both sides of the equation (1.32) contain the fluid velocity u. In order to obtain an equation involving only the vorticity field ω, we need to express the vector field u in terms of ω = curl u.

The 2-dimensional case. If the flow lies in a plane, the velocity field is u = (u1, u2, 0) and the vorticity field ω = 2 1 (0, 0, ux1 − ux2 ). We can denote them respectively as µ ¶ u1 u = , and ω = u2 − u1 . u2 x1 x2

In this case the vorticity equation becomes a scalar equation

∂tω + (u · ∇)ω = ν∆ω.

14 We now show how to recover the vector ﬁeld u, knowing its vorticity ω. Recall that the system of PDEs ½ ψx1 = a(x1, x2),

ψx2 = b(x1, x2), 1 2 has a solution if and only if ax2 = bx1 . Since div u = ux1 + ux2 = 0, there exists a so-called stream function ψ, such that ½ 2 ψx1 = u (x1, x2), 1 (1.33) ψx2 = −u (x1, x2).

By (1.33) it follows µ ¶ . −ψ u = ∇⊥ψ = x2 , (1.34) ψx1

ω = curl u = curl (∇⊥ψ) = ∆ψ . (1.35) Let ω : R2 7→ R be a given vorticity function. We can ﬁnd the corresponding stream function ψ by solving the Poisson equation (1.35) as in Lemma 3, namely Z 1 ψ(x) = N2 ∗ ω(x) = ln |x − y| · ω(t, y) dy . 2π R2

In turn, the velocity ﬁeld u : R2 7→ R2 is now computed by (1.34). In the end, we see that u can be recovered from ω by the Biot-Savart formula Z u(x) = K2 ∗ ω(x) = K2(x − y)ω(y) dy. (1.36) R2

Here the kernel K2 is the fundamental solution of the equation

curl z = δ0 , with δ0 a Dirac mass at the origin. More precisely, µ ¶ 1 −x2 K2(x) = 2 . (1.37) 2π |x| x1

Notice that K2 has a singularity at the origin, namely 1 |K (x)| = . 2 2π |x|

Observe that, in this second formulation of the Navier-Stokes equations, the pressure can be obtained from the the Poisson equation

∆p = −tr(∇u)2.

The 3-dimensional case. In the 3-dimensional case, the term ω·∇u in the vorticity equation does not vanish. Therefore we have to recover both u and ∇u from the vorticity ω.

15 Lemma 4 Let ω : R3 −→ R3 be a smooth vector ﬁeld with a suﬃciently fast decay as |x| −→ ∞. The over-determined elliptic system ½ curl u = ω, (1.38) div u = 0, admits a solution if and only if div ω = 0. In the positive case, the solution is constructed by means of the equation

u = −curl ψ. (1.39) where the stream function ψ solves the Poisson equation

∆ψ = ω, (1.40)

Remark. The equation (1.40) is solved using Lemma 3 Z 1 1 ψ(x) = N3 ∗ ω(x) = ω(t, y) dy . 2π R3 |x − y| The explicit solution to (1.39) is then provided by the Biot-Savart formula Z u(x) = K3 ∗ ω = K3(x − y) ω(y) dy, R3 where the kernel K3 is deﬁned by the relation 1 x ∧ h K (x) · h = h ∈ R3. (1.41) 3 4π |x|3 Proof Necessity: If ω = curl u then div ω = div (curl u) = 0. Suﬃciency: Start with the vector identity

∆ψ = ∇(div ψ) − curl (curl ψ) (1.42) and consider the vector fields . ω = ∆ψ, . k = ∇(div ψ), . h = − curl (curl ψ). Assuming that div ω = 0, we will show that there exists a vector field u that solves the system (1.38). As a preliminary, we prove that gradient vector fields are perpendicular to divergence-free vector fields, in the L2(R3) inner product.

Lemma 5 Let w be a smooth, divergence-free vector ﬁeld in RN and let q be a smooth scalar function such that |w(x)| |q(x)| = o(|x|1−N ) as |x| −→ ∞.

Then w and ∇q are orthogonal: Z w · ∇q dx = 0. RN

16 Indeed, Z Z Z w · ∇q dx = lim div (w q) dx = lim (w q) · n dA = 0 . RN R−→∞ |x|≤R R−→∞ |x|=R

We now return to the proof of the theorem. Taking the inner product of the vector identity (1.42) with k, by Lemma 5 it follows

hω, ki = hk, ki + hh, ki = hk, ki.

Therefore, Z hk, ki = |∇(div ψ)|2 dx = 0 . RN This shows that ψ is divergence-free and

ω = ∆ψ = − curl (curl ψ).

Deﬁning . u = − curl ψ we obtain ω = curl u, and of course div u = div (− curl ψ) = 0. ¤

1.11 Lagrangian formulation of the Euler equation

In this section we show that the Euler equation can be reformulated as an integro-differential equation for the particle trajectories. Given a smooth velocity field u, the particle trajectory mapping X satisfies the following ODE, for each value of the Lagrangian variable α ∈ RN :

( d dt X(t, α) = u(t, X(t, α)), X(0, α) = α.

Our goal here is to recover u from X and ∇X. We recall the vorticity transport formula

ω(t, X(t, α)) = ∇αX(t, α) ω0(α), and the Biot-Savart formula Z

u(t, x) = KN (x − y) ω(t, y) dy. RN Putting together these equations we obtain Z d u(t, X(t, α)) = X(t, α) = KN (X(t, α) − y) ω(t, y) dy . dt RN Consider the change of integration variable given by

y = X(t, α0),

17 for a suitable value of the Lagrangian variable α0. Then Z 0 0 0 u(t, X(t, α)) = KN (X(t, α) − X(t, α )) ω(t, X(t, α )) dα N ZR (1.43) 0 0 0 0 = KN (X(t, α) − X(t, α )) ∇αX(t, α )ω0(α ) dα . RN

In the 3-dimensional case the kernel K3 is deﬁned by (1.41). We remark that, because of the growth of K3 as x → 0, the integrand is not absolutely integrable. The right hand side of (1.43) must therefore be interpreted as a principal value of the integral. This is well deﬁned almost everywhere.

In the 2-dimensional case the situation is much simpler. Indeed, according to (1.20), the vorticity is constant along particle trajectories. The integral formula (1.43) thus reduces to Z 0 0 0 u(t, X(t, α)) = K2(X(t, α) − X(t, α )) ω0(α ) dα , R2 with K2 deﬁned at (1.37).

18 Chapter 2

Construction of solutions

In this chapter we study the existence, uniqueness, and regularity properties of solutions to the Navier-Stokes equation ut + (u · ∇)u = −∇p + ν∆u, (2.1) div u ≡ 0 , (2.2) assuming that the initial data is smooth enough.

2.1 Uniqueness of solutions

Let v1 and v2 be two smooth solutions of the same Navier-Stokes equation, vanishing suﬃciently 2 N fast as |x| −→ ∞. In particular, we assume v1, v2 ∈ L (R ). If p1 and p2 are the corresponding pressures, we deﬁne . v˜ = v − v , . 1 2 p˜ = p1 − p2. These two functionsv, ˜ p˜ satisfy the evolution equation

v˜t + (v1 · ∇)v ˜ + (˜v · ∇) v2 = −∇p˜ + ν∆˜v. (2.3) Observe that this equation is linear w.r.t.v ˜ andp ˜. Taking the inner product of both sides of (2.3) withv ˜ we obtain Z Z Z v˜ · v˜t dx + v˜ · (v1 · ∇)˜v dx + v˜ · (˜v · ∇)v2 dx = RN | RN {z } RN (?) Z Z (2.4) − v˜ · ∇p˜ dx − ν v˜ · ∆˜v dx | RN {z } RN (??) We observe that the two terms (?) and (??) both vanish. Indeed, (??) is the scalar product in L2 of a divergence free vector ﬁeld with the gradient of a scalar function. Moreover, integrating (?) by parts one obtains Z Z XN ³|v˜i|2 ´ XN ³ |v˜i|2 ´ (?) = v1 · ∇ dx = div v1 · dx = 0 N 2 N 2 i=1 R i=1 R

19 because of the fast decay of v1 as |x| −→ ∞. Integrating by parts the other terms the equation (2.4), we now obtain

Z 2 Z Z d |v˜| 2 dx = − v˜ · (˜v · ∇)v2 dx − ν |∇v˜| dx . dt N 2 N N R R | R {z } ≤0 Therefore d 2 2 kv˜k ≤ kv˜k k∇v k ∞ . (2.5) dt L2 L2 2 L Thanks to (2.5), we can now prove thatv ˜ ≡ 0 by a direct comparison with the corresponding ODE. Indeed, the function Z ¡ t ¢ z(t) = exp 2k∇v2(τ)kL2 dτ z0 0 solves the Cauchy problem   z˙(t) = 2k∇v2(t)kL∞ z(t),

 2 z(0) = z0 = kv˜(0)kL2 . By comparison, for any time t ≥ 0 we have

2 kv˜(t)kL2 ≤ z(t).

Therefore, if the solutions v1 and v2 coincide at t = 0, then they coincide for all t ≥ 0. We have thus proved

Theorem 3 (Uniqueness of solutions). Let v1, v2 be solutions of the Navier-Stokes equations, having bounded energy. If v1(0) = v2(0) and if the gradient of v2 satisﬁes the bound

Z T k∇v2(t)kL∞ dt < ∞ , 0 then v1(t) = v2(t) for all t ∈ [0,T ].

2.2 The Sobolev spaces Hm(RN )

Toward a proof of existence of solutions, we recall here some basic facts about Sobolev spaces. The Hilbert-Sobolev space Hm(RN ) consists of all functions v in L2(RN ) such that Dαv is in L2(RN ) for every multindex α, 0 ≤ |α| ≤ m, where Dα denotes the distributional derivative α1 αN m N ∂1 ··· ∂N . The space H (R ) is equipped with the norm

³ X ´1/2 . α 2 kvkHm = kD vkL2 . 0≤|α|≤m

N m N Theorem 4 (Sobolev embedding theorem). If m > k + 2 then H (R ) is continuously embedded in Ck, i.e. there exist a constant c > 0 such that

kvkCk ≤ ckvkHm .

20 Lemma 6 (Gagliardo-Nirenberg). Let α be a multi-index, with |α| = k ≤ m. Then

α 1−k/m k/m kD vkL2m/k ≤ CkvkL∞ kvkHm

Lemma 7 (An inequality in Sobolev spaces). For every m ≥ 1 there exists a constant Cm such that, for every u, v ∈ L∞ ∩ Hm(RN ), there holds n o ku vkHm ≤ Cm kukL∞ kvkHm + kvkL∞ kukHm . (2.6)

Proof For any multi-index α with |α| = m, using the Leibnitz diﬀerentiation formula and then H¨older’sinequality, we obtain X X α β α−β β α−β kD (u v)kL2 ≤ cα kD u D vkL2 ≤ cα kD ukL2m/|β| kD vkL2m/|α−β| . β≤α β≤α Using the Gagliardo-Nirenberg inequality with k = |β| and k = |α − β|, we obtain X α 1−|β|/m |β|/m 1−|α−β|/m |α−β|/m kD (u v)kL2 ≤ C · kukL∞ kukHm kvkL∞ kvkHm β≤α X ³ ´|α−β|/m ³ ´|β|/m ≤ C · kukL∞ kvkHm kvkL∞ kukHm . β≤α For convenience, here and in the sequel the numerical value of the constant C is allowed to change from one line to the next. We now apply the following Geometric-Arithmetic Inequality. Let α, β be positive numbers, and let θ ∈ [0, 1]. Then αθ β1−θ ≤ θα + (1 − θ)β ≤ α + β.

This yields n o α kD (u v)kL2 ≤ C · kukL∞ kvkHm + kvkL∞ kukHm . Of course, an estimate of the same type holds for lower order derivatives Dα(uv), for 0 ≤ α < m. This proves (2.6). ¤ From (2.6) and the Sobolev embedding Hm(RN ) ⊂ L∞(RN ) for m > N/2, we deduce

Corollary 4 If m > N/2, then the space Hm(RN ) is closed under multiplication. Namely, there exists a constant c such that kuvkHm ≤ kukHm kvkHm . (2.7) Corollary 5 (A further inequality). For |α| ≤ m, one has n o α α 0 kD (u v) − uD vkL2 ≤ cα k∇ukL∞ kvkHm−1 + kvkL∞ k∇ukHm−1 . (2.8)

Indeed, X X α α β α−β β0 α−β0 kD (u v)−uD vkL2 ≤ cα kD uD vkL2 ≤ cα kD (∇u)D vkL2 . β≤α ,|β|>0 |β0|+|α−β0|≤m−1 (2.9) Applying (2.6) with m replaced by m − 1 to each term on the right hand side of (2.9), we obtain (2.8). ¤

21 2.3 Local-in-time existence of solutions

In this section, our goal is to construct a solution to the Cauchy problem for the Navier-Stokes equation   ut + (u · ∇)u = −∇p + ν∆u,  div u ≡ 0, (2.10)   u(0, x) = u0(x), assuming that the initial data is smooth enough. The key step in the proof is to derive an a-priori estimate showing that the Hm norm of the solution remains bounded, at least during a short interval of time [0,T ].

Lemma 8 Let u be a solution to the Navier-Stokes equations. There exists a constant cm such that, as long as the norms ku(t)kHm and k∇u(t)kL∞ remain bounded, one has

d 2 2 ku(t)k m ≤ c k∇u(t)k ∞ ku(t)k m . (2.11) dt H m L H Proof Applying the diﬀerential operator Dα to both sides of the Navier-Stokes equation (2.1) and then multiplying by Dαu, we obtain

α α α α¡ ¢ α α α α D u D ut + D u D (u · ∇)u = −D u D ∇p + νD u ∆D u.

After we integrate this equation, it will be convenient to subtract the following integral term, which vanishes because of the decay of u as |x| −→ ∞. Z Z ³|Dαu|2 ´ Dαu (u · ∇)Dαu dx = lim div u dx = 0. RN R−→∞ |x|=R 2 The above yields Z Z d |Dαu|2 £ ¡ ¢ ¤ dx + Dαu Dα (u · ∇)u − (u · ∇)Dαu dx = dt N 2 N R Z R Z = − Dαu ∇Dαp dx + ν Dαu ∆Dαu dx | RN {z } RN =0

Z Z Z d |Dαu|2 £ ¡ ¢ ¤ dx + ν |∇Dαu|2 = − Dαu Dα (u · ∇)u − (u · ∇)Dαu dx (2.12) dt RN 2 RN RN Using Cauchy’s inequality, and then (2.8), we ﬁnally obtain R ¯ £ ¡ ¢ ¤¯ ¯ α α α ¯ α α α ¯D u D (u · ∇)u − (u · ∇)D u ¯ dx ≤ kD uk 2 kD (u · ∇u) − uD ∇uk 2 RN L L (2.13) α ≤ cm kD ukL2 k∇ukL∞ kukHm . Since the above holds for all multi-indices α with |α| ≤ m, from (2.12) and (2.13) we obtain (2.11). ¤ The inequality (2.11) has two important consequences.

22 m Corollary 6 Consider a solution of (2.10), with initial data u0 ∈ H . If the map t 7→ k∇u(t)kL∞ is integrable on [0,T ], then the norm ku(t)kHm of the solution remains uniformly bounded for t ∈ [0,T ].

Indeed, let t 7→ z(t) be the solution to the linear O.D.E.

z˙(t) = cmk∇u(t)kL∞ z(t), z(0) = ku(0)kHm .

By (2.11), a comparison argument yields

³ Z t ´ 2 2 ku(t)kHm ≤ z(t) = exp cmk∇u(t)kL∞ dt ku(0)kHm . 0 ¤

m N Corollary 7 Let t 7→ u(t) be a solution of (2.10), with initial data u0 ∈ H (R ), with m > 1 + N/2. Then the norm ku(t)kHm remains bounded, on some initial interval [0,T ].

Indeed, by the Sobolev embedding theorem, there exists a constant C such that

k∇ukL∞ ≤ C kukHm .

Using this in (2.11) we obtain

d 2 3 ku(t)k m ≤ C ku(t)k m , (2.14) dt H m H for a suitable constant Cm . Call t 7→ z(t) the solution to the O.D.E.

3/2 2 z˙ = Cm z , z(0) = ku0kHm . (2.15) Observe that z is well deﬁned and remains bounded on a suitably small interval [0,T ]. By a comparison argument, from (2.14) we now obtain

2 ku(t)kHm ≤ z(t) t ∈ [0,T ] . ¤ Before we state a local existence result, we wish to clarify what type of solutions will be constructed. As in (1.27), let E ⊂ L2(RN ) be the subspace of vector ﬁelds u having zero 2 divergence, in distribution sense, and call PE : L 7→ E the perpendicular projection. By a solution of the initial value problem (2.10), we mean a Lipschitz continuous map u : [0,T ] 7→ E such that (u · ∇)u ∈ L2 and ∆u ∈ L2 for a.e. time τ ∈ [0,T ], and moreover

Z t n o u(t) = u0 + PE [(u(τ) · ∇)u(τ)] + ν∆u(τ) dτ (2.16) 0 for every t ∈ [0,T ].

Theorem 5 (Existence of solutions to the Navier-Stokes equations, locally in time). m Let the initial data satisfy u(0, ·) = u0 ∈ H , with m > 2 + N/2. Then the equations (2.10) admit a solution deﬁned for t ∈ [0,T ], with T > 0 suﬃciently small.

23 We sketch the proof, consisting of various steps:

ε 1. First, one constructs a family of approximate solutions (u )ε>0, by a molliﬁcation procedure. ∞ N Consider a smooth, radially symmetric function ρ = ρ(|x|) in C0 (R ) satisfying Z ρ ≥ 0 , and ρ dx = 1 . RN

Define the family of mollifiers ³ ´ . x ρ (x) = ε−N ρ ε ε and the associate family of mollification operators Jε Z ³ ´ . −N x − y Jε u(x) = ρε ∗ u(x) = ε ρ u(y) dy. RN ε It is useful to recall here some basic properties of mollifiers:

(i) Uniform convergence. If v ∈ C0, then

kJεvkL∞ ≤ kvkL∞ and lim Jεv(x) = v(x) (2.17) ε→0 uniformly on bounded sets.

(ii) Commuting with derivatives. If v ∈ Hm, then for every multi-index α with |α| ≤ m there holds α α D (Jεv) = Jε(D v) (2.18)

(iii) Commuting with integrals. For every u ∈ Lp, v ∈ Lq with p−1 + q−1 = 1, there holds Z Z (Jεu) v dx = u (Jεv) dx . (2.19) RN RN

(iv) Rate of convergence. If v ∈ Hm, then for some constant C one has

lim kJεv − vkHm = 0 , kJεv − vkHm−1 ≤ Cε kvkHm . (2.20) ε→0

(v) Regularization. For every v ∈ Hm and k ≥ 0,

−k k −(N/2+k) kJεvkHm+k ≤ Cmkε kvkHm , kJεD vkL∞ ≤ Ckε kvkL2 . (2.21)

2. Consider the equation obtained by applying the ε-mollifier operator to Navier-Stokes equation. ε £ ε ε ¤ ε ε ut + Jε (Jε u ) · ∇(Jε u ) = −∇p + νJε(Jε ∆u ). This mollification procedure transforms the unbounded differential operators into bounded op- m erators on the Sobolev space H . Observing that the perpendicular projection operator PE

24 commutes with the derivation and with the molliﬁcation operators, we can write the previous equation as an O.D.E. in the Banach space E ⊂ L2, namely d £ ¤ uε = − P J (J uε) · ∇(J uε) + νJ (J ∆uε). (2.22) dt E ε ε ε ε ε The right hand side of this equation is Lipschitz continuous, because of the presence of the ε molliﬁers. Hence, given the initial data u (0) = u0, for each ε > 0 this O.D.E. has a unique solution, which we denote as t 7→ uε(t).

3. Following the proof of Lemma 8, one can show that *the inequality

d ε 2 ε 3/2 ku (t)k m ≤ C · ku (t)k dt H Hm remains valid for approximate solutions of (2.22), with some constant C independent of ε. Hence, ε on a suitably small interval [0,T ], the norms ku (t)kHm remain uniformly bounded. In turn, since we are assuming m > 2 + N/2, this yields an uniform a priori bounds on the norms ε ku (t)kC2 , independent of ε > 0.

4. Next, we claim that the family of mappings uε : [0,T ] 7→ E ⊂ L2 is a Cauchy, as ε → 0. Indeed, let ε > ε0 > 0. We need to estimate

d 1 0 kuε − uε k2 = A + A , (2.23) dt 2 L2 1 2 where D E . ε0 ε0 ε ε ε ε0 A1 = PEJε0 [(Jε0 u ) · ∇(Jε0 u )] − PEJε[(Jεu ) · ∇(Jεu )] , u − u , D E . 2 ε 2 ε0 ε ε0 A2 = ν Jε ∆u − Jε0 ∆u , u − u .

2 N ε ε0 Here h·, ·i denotes the inner product in L (R ). Observing that u , u ∈ E and PE is a perpendicular projection, we can estimate D E ε ε ε ε0 A = (J 0 − J )[(J u ) · ∇(J u )] , u − u 1 D ε ε ε ε E ε ε ε ε0 + J 0 [(J 0 − J )u · ∇(J u )] , u − u D ε ε ε ε E ε0 ε ε ε ε0 + J 0 [(J 0 (u − u ) · ∇(J u )] , u − u D ε ε ε E ε0 ε ε ε0 + J 0 [(J 0 u ) · ∇((J 0 − J )u )] , u − u D ε ε ε ε E ε0 ε0 ε ε ε0 + Jε0 [(Jε0 u ) · ∇(Jε0 (u − u ))] , u − u

= A11 + A12 + A13 + A14 + A15 .

We observe that A11,A12 and A14 contain the term Jε0 − Jε . Using the second inequality in (2.20), we obtain the bound ³ ´ 0 ε 2 ε0 2 ε ε0 |A11| + |A12| + |A14| ≤ C(ε + ε ) ku kHm + ku kHm ku − u kL2 .

Moreover, we have ε ε ε0 2 |A13| ≤ Cku kHm ku − u kL2 .

25 Finally, an integration by parts yields D E Z ³ ´ ε0 ε0 ε ε ε0 1 ε0 ε ε0 2 A15 = (Jε0 u ) · ∇(Jε0 (u − u )) ,Jε0 (u − u ) = − Jε0 u · ∇ |Jε0 (u − u )| dx = 0 . 2 RN The last term in (2.23) is estimated by D E D E 2 2 ε ε ε0 2 ε ε0 ε ε0 A2 = (Jε − Jε0 )∆u , u − u + Jε0 ∆(u − u ) , u − u 0 ε ε ε0 ε ε0 ≤ C(ε + ε )ku kH3 ku − u kL2 − kJε0 ∇(u − u )kL2 . Combining all the previous estimates, from (2.23) we obtain ³ ´ h i d ε ε0 ε ε0 0 ε ε0 ku − u k 2 ≤ C 1 + ku k m + ku k m · (ε + ε ) + ku − u k 2 . dt L H H L By construction, at time t = 0 we have uε(0) = uε0 (0). Moreover, by step 3 the Hm norms of the approximate solutions remain uniformly bounded, for t ∈ [0,T ]. This implies an estimate of the form ε ε0 0 Kt ku (t) − u (t)kL2 ≤ (ε + ε ) e t ∈ [0,T ] . showing that as ε → 0, the family of approximations uε is Cauchy.

5. By the previous step, there exists a continuous map u : [0,T ] 7→ L2(RN ) such that

ε lim ku (t) − u(t)kL2 = 0 ε→0 uniformly on the time interval [0,T ]. We claim that all the derivatives Dαuε, with |α| < m, converge to the corresponding derivatives of u. Indeed, for any 0 < k < m the following interpolation inequality holds:

1−k/m k/m kvkHk ≤ CkkvkL2 kvkHm . (2.24) An application of this inequality yields

ε ε 1−k/m ε k/m lim ku − ukHk ≤ lim Ckku − uk 2 ku − uk m = 0. ε→0 ε→0 L H In particular, if m > 2 + N/2, we have the convergence

ε ε ε k(u · ∇)u − (u · ∇)ukL2 → 0 , k∆u − ∆ukL2 → 0 , uniformly for t ∈ [0,T ]. From the relations

Z t Z t ε ε ε 2 ε u (t) = u0 − PEJε [(Jεu (τ)) · ∇(Jεu (τ))] dτ + ν Jε ∆u (τ) dτ , 0 0 ε letting ε → 0 and using the convergence ku − ukHm → 0, one checks that the limit function u satisﬁes (2.16). ¤

Remarks. By choosing even larger exponent m, one can prove the local existence of classical solutions with Ck regularity, for any given k ≥ 1. One should also note that the above proof did not use in any way the regularization eﬀect produced by the diﬀusion term ν∆u. Indeed, the previous estimates remain valid if ν = 0.

26 2.4 Global weak solutions of the Navier-Stokes equations

The regular solutions obtained in the previous section are only locally defined. Indeed, their Hm norm may in principle blow up in finite time. In the present section we study the existence of globally defined distributional solutions, on a bounded domain Ω ⊂ R3 with smooth boundary. Consider the equations  u + (u · ∇)u = −∇p + ν∆u,  t   div u ≡ 0, (2.25)  u(t, x) = 0, for all x ∈ ∂Ω ,    u(0, x) = u0(x).

Notice that the boundary condition is appropriate because ν > 0.

Deﬁnition 4 A distributional solution u of the Navier-Stokes equation (2.25) is characterized by the following properties: • The energy of u has to be ﬁnite Z 2 |u(x, t)| dx ≤ E0, for a.e. t ∈ [0,T ] . (2.26) R3

• The loss of energy due to the viscosity has to be also ﬁnite

Z T Z 2 |∇u(x, t)| dx dt ≤ E0 . (2.27) 0 R3

• The function u = u(t, x) satisﬁes the evolution equation and the incompressibility¡ condition¢ ∞ 3 ∞ 3 in distributional sense. Namely, for any ψ ∈ C0 (R ) and for any φ ∈ C0 (−∞,T ]×Ω; R we have Z u∇ψ dx = 0 for a.e. t ∈ [0,T ], R3 Z T Z n o Z (2.28) u[φt + ν∆φ] − (u · ∇u)φ + p divφ dx dt + u0(x)φ(0, x) dx = 0 0 R3 R3

The proof of existence of a solution will be given in three steps. First, we derive some uniform a priori estimates on exact solutions to the Navier-Stokes equations. In a second step, by a Galerkin method, we construct a sequence of approximate solutions satisfying the same a priori estimates. In turn, these estimates allow us to use a compactness argument, and obtain a subsequence converging to a weak solution, in suitable function spaces.

Proposition 5 Assume u to be a solution of (2.25), we claim that the following estimates hold: ¡ ¢ ∞ 2 • u ∈ L [0,T ], L (Ω) , i.e. there exists a constant E0 such that Z 2 |u(t, x)| dx ≤ 2E0 for a.e. t ∈ [0,T ] . (2.29) Ω

27 ¡ ¢ 2 1 • u ∈ L [0,T ], H0(Ω) , i.e.

Z T Z 2 2 |∇u(t, x)| dx dt ≤ E0 (2.30) 0 Ω ν ¡ ¢ 4/3 −1 • du/dt ∈ L [0,T ], H , i.e. there exists a constant C2 such that

Z T 4/3 kut(t)kH−1 dt ≤ C2 . (2.31) 0

Proof If u is a solution of the Navier-Stokes equation, then

hut, ui + hu, u · ∇ui − νhu, ∆ui + hu, ∇pi = 0 .

By integrating over Ω we obtain Z Z Z Z d |u|2 X3 dx + uiujui dx + u · ∇p dx + ν |∇u|2 dx = 0 dt 2 xj Ω Ω i,j=1 Ω Ω Z Z Z Z d |u|2 ³|u|2 ´ dx + div · u dx + div(u · p) dx + ν |∇u|2 dx = 0 (2.32) dt 2 2 ZΩ ΩZ Ω Ω d |u|2 dx + ν |∇u|2dx = 0 dt Ω 2 Ω

The last identity shows that the L2 norm of the solution u decreases in time. Indeed, Z Z 2 2 |u(t, x)| dx ≤ |u0(x)| dx, for all t ≥ 0. Ω Ω

Moreover Z Z Z Z T |u (x)|2 |u(T, x)|2 ν |∇u(t, x)|2dx dt ≤ 0 dx − dx. 0 Ω Ω 2 Ω 2

This proves the ﬁrst two estimates, taking E0 to be the total energy at the initial time t = 0. In order to prove (2.31) we recall that

ut = PE (ν∆u − u · ∇u) (2.33)

2 3 where PE denotes the perpendicular projection on on the subspace E ⊂ L (Ω; R ) consisting of divergence free vector ﬁelds. The H−1 norm of (2.33) can be estimated as follows: Z

kν∆uk −1 = sup −ν∇u · ∇w dx ≤ νkuk 1 , H H0 kwk 1 ≤1, div w≡0 Ω H0 Z . ku · ∇ukH−1 = sup (u · ∇u) w dx . kwk 1 ≤1, div w≡0 Ω H0

28 1 1 1 Observing that 3 + 2 + 6 = 1, using a generalized H¨olderinequality and a Sobolev embedding theorem we obtain: Z Z ¯ ¯ ¯ X3 ¯ ¯ ¯ ¯ i j j¯ ¯ (u · ∇u) w dx¯ ≤ u uxi w dx Ω Ω i,j=1 (2.34)

≤ K1kukL3 k∇ukL2 kwkL6 ≤ K2kukL3 k∇ukL2 . By Sobolev embedding and interpolation there holds

1/2 1/2 kukL3 ≤ CkukH1/2 ≤ CkukH0 kukH1 . Hence 1/2 3/2 ku · ∇ukH−1 ≤ CkukL2 k∇ukL2 Z T ° ° Z T ³ ´ °du°4/3 4/3 ¡ 1/2 3/2¢4/3 ° ° dt ≤ C kuk 1 + kukL2 + k∇ukL2 dt dt H−1 H0 0 0 (2.35) ³ Z T ´ 2 ≤ C 1 + k∇ukL2 dt ≤ C1 0 ¤

2.5 Construction of ﬁnite dimensional approximate solutions

In this section we establish the following global existence theorem

3 2 Theorem 6 Let Ω ⊂ R be a bounded domain with smooth boundary. Let u0 ∈ L (Ω) be a vector ﬁeld with zero divergence. Then the equations (2.25) admit a weak solution, deﬁned for all times t ≥ 0.

Proof Fix any time interval [0,T ]. Let E ⊂ L2(Ω ; R3) be the closed subspace of vector ﬁelds whose distributional divergence vanishes. On E, consider the linear diﬀerential operator

Au = −∆u, and the eigenvalue problem ½ Au = λu x ∈ Ω , u = 0 x ∈ ∂Ω . 2 Since the operator A is self-adjoint, there exists an orthonormal set {w1, w2, w3 ...} ⊂ L (Ω) consisting of eigenfunctions of A. We will denote by Em ⊂ E the subspace generated by the ﬁrst m eigenfunctions: . © ª Em = span w1, w2, . . . , wm , and write Pm : E 7→ Em for the orthogonal projection onto Em. Projecting the equation (2.33) into the ﬁnite dimensional subspace Em we obtain the Galerkin system ( d dt um + νAum + Pm(u · ∇)u = 0, (2.36) um(0, x) = Pmu0(x).

29 Notice that (2.36) is a Cauchy problem in the finite dimensional space Em . For each fixed time t, we can write um(t, ·) as linear combination of the functions w1, w2, . . . , wm: Xm um(t, x) = ξi(t)wi(x), i=1 . where ξi(t) = hum(t), wiiL2 . By construction, one has Xm ∆um(t, x) = − λjξj(t)wj(x), i=1 and the problem (2.36) can be written as a system of m ODEs for the coefficients ξ1, . . . , ξm.

d Xm ξ = −νλ ξ + b ξ ξ , (2.37) dt i i i ijk j k j,k=1

The constant coefficients bijk are here computed as Z . 2 bijk = − [(wj · ∇)wk]wi dx = −h(wj · ∇)wk , wiiL . Ω Since the right hand sides of (2.37) are quadratic polynomials, for any given initial data the Cauchy problem has a unique solution defined locally in time. To prove that this solution exists globally, we need to show that its norm remains bounded. Toward this goal, observe that the coefficients bijk satisfy Z Z ³ ¢ bijk + bikj = ∇(wj · wk) · wi dx = div (wj · wk) wi dx = 0. Ω Ω Therefore

d |ξ(t)|2 Xm Xm Xm = b ξ ξ ξ − ν λ ξ2(t) = − ν λ ξ2(t) < 0. (2.38) dt 2 ijk i j k i i i i i,j,k=1 i=1 i=1

Indeed, all the eigenvalues λi of the operator A = −∆ are strictly positive. Hence for all m ≥ 1 we have a well deﬁned approximate solution um : [0,T ] 7→ Em . Repeating the arguments in the proof of Proposition 5, one can show that ¡ ¢ ∞ 2 • um ∈ L [0,T ]; L (Ω) , ¡ ¢ 2 1 • um ∈ L [0,T ]; H0(Ω) , ¡ ¢ 4/3 −1 • (um)t ∈ L [0,T ], H (Ω) .

Moreover, the norms of the approximate solutions um in these various function spaces remain uniformly bounded as m → ∞. Using a compactness lemma, one can extract a subsequence, still called (um)m≥1 , and a limit function u such that ¡ ¢ 2 1 • um * u weakly in L [0,T ]; H0(Ω) ,

30 ¡ ¢ 4/3 −1 • (um)t * ut weakly in L [0,T ], H (Ω) , ¡ ¢ ∗ ∞ 2 • um * u weakly in L [0,T ]; L (Ω) , In turn, by extracting a further subsequence, we can achieve the further convergence ¡ ¢ 2 2 • um → u strongly in L [0,T ]; L (Ω) , ¡ ¢ 0 −1 • um → u in C [0,T ], H (Ω) .

Observe that the uniform bound on the norm of kumkL4/3([0,T ]; H−1) implies that the maps −1 t 7→ um(t) ∈ H are uniformly Holder continuous of exponent 1/4. Indeed, Z t2 1/4 ku(t )−u(t )k −1 ≤ 1·ku k −1 dt ≤ ku k 4/3 −1 ·k1k 1/4 ≤ ku k 4/3 ·|t −t | . 2 1 H t H t L ([t1,t2];H ) L ([t1,t2]) t L 2 1 t1

Since um(0) = Pmu0 → u0 as m → ∞, this uniform continuity implies

lim ku(t) − u0kH−1 = 0 . t→0+

Hence the limit function u takes the correct initial data. 1 Since every function um maps [0,T ] into the closed subspace E ⊂ H0 consisting of divergence- free vector ﬁelds, the same holds for u. It now remains to show that the limit function u satisﬁes the Navier-Stokes equation in a weak sense. Fix an integer k ≥ 1 and consider a smooth test function φ : [0,T ] 7→ Ek with compact support, so that φ(0) = φ(T ) = 0 ∈ Ek . Then for all m ≥ k we have

Z T Z Z (−φt − ν∆φ)um + [(um · ∇)um] · φ dxdt = − (φt + ν∆φ)um + [(um · ∇)φ] · um dxdt = 0 . 0 Ω Ω Taking the limit as m → ∞, this yields Z (φt + ν∆φ)u + [(u · ∇)φ] · u dxdt = 0 . Ω

Observing that, as k → ∞, the set of test functions of the form φ : [0,T ] 7→ Ek becomes dense in the set of all test functions φ : [0,T ] 7→ E, we conclude that u provides a weak solution. ¤

31 Chapter 3

Vortex sheets

3.1 Velocity behavior near a vortex sheet

We have a vortex sheet when vorticity concentrates along a curve γ : R −→ R2 with linear density φ. Let s 7→ γ(s) be a parametrization of the curve γ. Given s1 < s2, the total vorticity on the arc between γ(s1) and γ(s2) is thus given by the integral Z s2 φ(s) ds. s1 If the vorticity is located only along the curve γ, and vanishes outside γ, then the velocity of the ﬂuid at a point x ∈ R2 is computed by Z u(x) = K(x − γ(s)) φ(s) ds.

Here K denotes the two-dimensional Biot-Savart kernel, i.e. µ ¶ 1 −x2 K(x) = 2 . (3.1) 2π |x| x1

To determine how the vorticity changes in time, we should now determine how points on the curve γ move. If x is a point on the curve, i.e. x = γ(s0) for some s0, then the above formula yields Z u(γ(s0)) = K(γ(s0) − γ(s)) φ(s) ds . (3.2)

−1 However, since K(γ(s0)−γ(s)) = O(1)·|s0−s| , the above integral is not absolutely convergent. We now show that, if γ and φ are smooth, then the integral (3.2) can be uniquely deﬁned as 0 a principal value. Indeed, assume that γ (s0) 6= 0. Then γ can be written in the following way

0 γ(s) = γ(s0) +γ ¯ (s)(s − s0), where Z 1 s γ¯0(s) = γ0(σ) dσ. s − s0 s0

32 Similarly, we can write 0 φ(s) = φ(s0) + φ¯ (s)(s − s0). We now compute Z s0+δ K(γ(s0) − γ(s))φ(s) ds s −δ 0 Z s0+δ 0 0 = K(¯γ (s)(s − s0)) · [φ(s0) + φ¯ (s)(s − s0)] ds s −δ Z 0 s0+δ 0 = K(γ (s0)(s − s0))φ(s0) ds (3.3) s −δ 0Z s0+δ h i 0 0 + K(¯γ (s)(s − s0)) − K(γ (s0)(s − s0)) φ(s0) ds s −δ Z 0 s0+δ 0 0 + K(¯γ (s)(s − s0))φ¯ (s)(s − s0) ds. s0−δ Of the three integrals on the right hand side of (3.3), the ﬁrst one vanishes, while the remaining two are absolutely integrable. Indeed,

0 −1 K(¯γ (s)(s − s0)) = O(1) · |s − s0| ,

0 0 2 |γ¯ (s) − γ (s0)|(s − s0) = O(1)(s − s0) , hence 0 0 K(¯γ (s)(s − s0)) − K(γ (s0)(s − s0)) = O(1).

Next, let γ be a vortex sheet. We wish to compute the jump in the velocity of the ﬂuid across γ. To simplify computations, we assume here kγ0(s)| ≡ 1, so that the curve is parametrized by arc length. Call n be the unit normal vector, perpendicular to the curve. The variation of velocity between a point on the curve γ(s0) and a point near the curve γ(s0)+εn is given by the integral Z h i u(γ(s0) + εn) − u(γ(s0)) = K(γ(s0) + εn − γ(s)) − K(γ(s0) − γ(s)) φ(s) ds. (3.4)

To compute the limit as ε → 0+, it suﬃces to restrict the above integral to any interval of the form [s0 − δ, s0 + δ]. Indeed, outside this interval the integrand in (3.4) is a continuous function of ε and goes to 0 as ε −→ 0. Notice that, as s → s0, we have

0 0 γ¯ (s) → γ (s0) , φ(s) → φ(s0) .

From (3.4) we obtain Z s0+δ £ 0 0 ¤ lim [u(γ(s0) + εn) − u(γ(s0))] = lim K(¯γ (s)(s − s0) + εn) − K(¯γ (s)(s − s0)) φ(s) ds ε→0+ ε→0 Zs0−δ s0+δ £ 0 0 ¤ = lim K(γ (s0)(s − s0) + εn) − K(γ (s0)(s − s0)) φ(s0) ds . ε→0 s0−δ

33 Performing the change of variable ξ = (s − s0)/ε, the above yields

Z +∞ £ 0 0 ¤ lim [u(γ(s0) + εn) − u(γ(s0))] = lim K(γ (s0)εξ + εn) − K(γ (s0)εξ) φ(s0)ε dξ. ε→0+ ε−→0 −∞ To compute this integral, we ﬁx the orthonormal basis

0 e1 = γ (s0) , e2 = n and write the Biot-Savart kernel K coordinate-wise w.r.t. this basis. This yields

lim [u(γ(s0) + εn) − u(γ(s0))] ε→0+ Z +∞ µ ¶ µ ¶ φ(s0) −ε εξ εξ = lim 2 2 , 2 2 − 0 , 2 2 ε dξ ε−→0 2π −∞ |ε| (1 + ξ ) |ε| (1 + ξ ) |ε| (1 + ξ ) Z +∞ µ ¶ µ ¶ φ(s0) 1 φ(s0) = − 2 , 0 dξ = − , 0 . 2π −∞ 1 + ξ 2 This shows that the jump of the tangential component of the velocity across a vortex sheet is equal to the vorticity density φ(s0) (up to the sign), while the normal component is continuous.

3.2 The Birkhoﬀ-Rott equation

Assume that the vorticity density is φ(s) is smooth and uniformly positive. Then we can reparametrize γ by means of the vorticity. More precisely, fix a value s0. We then define the new parametrization . σ 7→ z(σ) = γ(s(σ)) ∈ R2 , where the change of variable σ 7→ s(σ) is implicitly defined by the relation

Z s(σ) σ = φ(s) ds. s0 Observe that the new parametrization has the property that the amount of vorticity contained in the arc between z(σ1) and z(σ2) is precisely σ2 − σ1. The Birkhoff-Rott equation is a differential equation defining the motion of the vortex sheet. We now regard the 2-dimensional vector (x1, x2) as a point z = x1 + ix2 of the complex plane. Using (3.2) with φ ≡ 1, we can write

Z +∞ ³ Z +∞ 0 ´ d ¡ 0 ¢ 0 1 dσ z(t, σ) = K z(σ) − z(σ ) dσ = 0 . dt −∞ 2πi −∞ z(σ) − z(σ ) Here the upper bar denotes conjugation. Equivalently, this can be written as Z d 1 +∞ dσ0 z¯(t, σ) = 0 , (3.5) dt 2πi −∞ z(σ) − z(σ ) where the left hand side is the time derivative of the conjugate of z. We remark that all these integrals are not absolutely convergent, and should be interpreted as principal values.

34 3.3 Kelvin-Helmholtz instability

One very simple solution of the Birkoﬀ-Rott equation is the constant function z(t, σ) = σ. In this case the curve γ is precisely the real axis. We wish to understand the stability of this solution w.r.t. small perturbations. Consider the perturbed function

z(t, σ) = σ + εg(t, σ).

This should satisfy Z d d 1 +∞ dσ0 z¯(t, σ) = εg¯(t, σ) = P.V. 0 0 . dt dt 2πi −∞ (σ − σ ) + ε(g(σ) − g(σ )) Performing the expansion µ ¶ µ ¶ 1 1 1 1 b = = 1 − ε + o(ε2) . a + εb a 1 + εb/a a a inside the integral, and letting ε → 0, we obtain

Z +∞ 0 µZ −δ Z +∞¶ 0 d 1 g(σ) − g(σ ) 0 1 g(σ) − g(σ ) 0 g¯(t, σ) = − P.V. 0 2 dσ = lim + 0 2 dσ dt 2πi −∞ (σ − σ ) 2πi δ→0+ −∞ δ (σ − σ ) Integrating by parts and observing that the boundary terms cancel each other as δ → 0+, we obtain Z +∞ 0 0 d 1 g (σ ) 0 1 0 g¯(t, σ) = − P.V. 0 dσ = Hg . (3.6) dt 2πi −∞ σ − σ 2 Here Hg0 denotes the Hilbert transform of g0. We recall that the Hilbert transform of a function f is deﬁned as the principal value Z . 1 +∞ f(y) Hf(x) = P.V. dy. πi −∞ y − x Using the Fourier transform, f can be written as Z f(x) = eiξxfˆ(ξ)dξ .

Setting w = y − x we compute Z Z 1 +∞ eiξ(x−y) 1 ∞ eiξw − e−iξw Heiξx = P.V. dy = eiξx · lim dw πi y πi δ→0+ w Z −∞ Z δ 1 ∞ 2i sin ξw 2 ∞ sin w = eiξx dw = eiξxsign ξ · dw πi 0 w π 0 w = eiξx sign ξ .

Therefore Z Z . Hf(x) = eiξxfˆ(ξ)dσ − eiξxfˆ(ξ)dξ. ξ>0 ξ<0

35 Using a Fourier transform, we can now write

Z +∞ Z +∞ g(t, x) = A(t, ξ)eiξxdξ + B(t, ξ)e−iξxdξ, 0 0 where A(t, ξ) =g ˆ(t, ξ), B(t, ξ) =g ˆ(t, −ξ), ξ > 0. With the above notations, we have

Z +∞ Z +∞ Hg0(t, x) = A(t, ξ)iξeiξxdξ + B(t, ξ)iξe−iξxdξ. 0 0 Recalling (3.6), the time evolution of A(t, ξ) and B(t, ξ) is described by the equation  d iξ ¯  dt A(t, ξ) = − 2 B(t, ξ), (3.7)  d iξ ¯ dt B(t, ξ) = − 2 A(t, ξ). Notice that (3.7) implies

d2 ξ2 d2 ξ2 A(t, ξ) = A(t, ξ), B(t, ξ) = B(t, ξ). dt2 4 dt2 4 The general solution of (3.7) has the form  ξt//2 −ξt/2  A(t, ξ) = A1(ξ)e + A2(ξ)e , (3.8)  ξt/2 −ξt/2 B(t, ξ) = B1(ξ)e + B2(ξ)e .

It is now clear that the coeﬃcient of the Fourier mode eiξσ can grow like e|ξ|t/2. Therefore the solution is very unstable. In order to achieve at least the local existence of solutions, one has to start with initial data whose Fourier transform decreases very fast as |ξ| → ∞, namely analytic data.

3.4 Spiralling solutions

Since a vortex sheet is highly unstable, it very likely that it loses regularity within ﬁnite time. One may conjecture that singularities have the form of an inﬁnite double spiral (vortex roll-up). To get some understanding of these singularities, we can reduce the dimension of the problem using a rescaling symmetry. We recall that, if u solves the equation

ut + PE(u · ∇u) = 0 , (3.9) then λ ³ t x´ uλ,τ (t, x) = u , , τ τ λ 2 is also a solution, for all λ, τ > 0. Here PE the perpendicular projection on the space E ⊂ L of divergence free vector ﬁelds. τ Given a solution u of (3.9), and a constant k, we will call (u )τ>0 the 1-parameter family of solutions deﬁned by setting λ = τ k, namely ³ t x ´ uτ (t, x) = τ k−1u , . τ τ k

36 Suppose that u is invariant with respect to the transformation τ 7→ uτ for all τ > 0. Then ³ x ´ u(t, x) = ut(t, x) = tk−1u 1, . tk In other words, knowing the velocity u at time t = 1, the entire solution is determined for all times t > 0.

We conclude this section by deriving an equation for the self-similar solution of the Birkhoﬀ- Rott equation, in the form of a spiral. Assume that the vorticity is initially concentrated on the real axis, which is parametrized by γ(s) = s. Let the vorticity density be φ(s) = a|s|β , for some constant a. Then, re-parametrizing the curve in term of the vorticity, we obtain Z s a σ(s) = aσβ dσ = sβ+1. 0 β + 1 In terms of the rescaled parameter σ, we have

· ¸ 1 β + 1 β+1 z(σ) = σ . (3.10) a Next, consider the Birkhoﬀ-Rott equation Z d 1 dσ0 z¯(t, σ) = , (3.11) dt 2πi z(σ) − z(σ0) with initial data where the vorticity is concentrated along the real axis, with density as in (3.10), with β = −1/2. Namely 1 z (σ) = · σ2 sign (σ). 0 4ξ2 Here ξ > 0 is an arbitrary constant. Introduce the variable σ τ = . ξ4/3 · t1/3 We seek a special, self-similar solution of (3.11), having the form z(t, σ) = (ξt)2/3ρ(τ), (3.12) where τ 7→ ρ(τ) is a parametrization of the curve, at time t = ξ−1. Notice that, as time increases, the spiral keeps the same shape, while its size is ampliﬁed by t2/3. d d £ ¡ σ ¢¤ z¯(t, x) = (ξt)2/3ρ¯ dt dt ξ4/3t1/3 µ ¶ 2 dρ¯ 1 σ ξ2/3t−1/3ρ¯ + (ξt)2/3 · · − 3 dσ 3 ξ4/3t4/3 2 1 dρ¯ = ξ2/3t−1/3ρ¯ − ξ2/3t−1/3 τ . 3 3 dτ

37 Next, the Birkhoﬀ-Rott equation becomes Z d 1 1 dσ0 z¯(t, x) = · dt 2πi (ξt)2/3 ρ( σ ) − ρ( σ0 ) ξ4/3t1/3 ξ4/3t1/3 Z (3.13) 1 dτ 0 = · ξ2/3t−1/3. 2πi ρ(τ) − ρ(τ 0)

Comparing (3.13) with (3.12) we obtain Z 2 τ d 1 dτ 0 ρ¯ − ρ¯ = , 3 3 dτ 2πi ρ(τ) − ρ(τ 0) hence µ Z ¶ d 1 1 dτ 0 ρ(τ) = 2ρ(τ) + . dτ τ 2πi ρ(τ) − ρ(τ 0) A solution of this equation is the Kaden spiral.