Lecture Notes on Fluid Dynamics

<strong>Lecture</strong> <strong>Notes</strong> on Fluid DynamicsAlberto Bressan and Carlotta Donadello26th October 20051 Review of differential operatorsGiven a scalar function φ : R N ↦→ R, its gradient ∇φ is the vector of partial derivatives∇φ = . ( ∂ ∂∂)(φ x1 , φ x2 , . . . , φ xN ) = φ , φ , . . . , φ .∂x 1 ∂x 2 ∂x NFor a vector field u : R N ↦→ R N , its divergence isThe Laplace operator is defined asdiv u . =N∑u i x i.i=1∆φ . =N∑i=1∂ 2 φ∂x 2 i= div ∇φ .Ňow consider a velocity field u : R × R N ↦→ R N , so that u(t, x) = (u 1 , . . . , u N )(t, x) denotesthe velocity of the particle located at the point x = (x 1 , . . . , x N ) at time t. We then define theMaterial Derivative of a function φ = φ(t, x) asD t φ . = ∂ ∂t φ + N ∑i=1u i∂φ .∂x iThis provides the time-derivative of φ along particle trajectories. ˇWe recall that the wedgeproduct of two vectors v, w ∈ R 3 can be obtained as follows. Let e 1 , e 2 , e 3 be the vectors in thestandard orthonormal basis of R 3 . Then⎛⎞e 1 e 2 e 3v ∧ w = det ⎝ v 1 v 2 v 3⎠ .w 1 w 2 w 3Given a velocity field u : R 3 ↦→ R 3 , its vorticity ω = curl u is defined asNote that, formally, one can writeω = curl u . = ( ∂ 2 u 3 − ∂ 3 u 2 , ∂ 3 u 1 − ∂ 1 u 3 , ∂ 1 u 2 − ∂ 2 u 1) .⎛curl u = ∇ ∧ u = det ⎝⎞e 1 e 2 e 3∂ 1 ∂ 2 ∂ 3⎠ .u 1 u 2 u 31

2 Euler and Navier-Stokes equationsfor a homogeneous incompressible fluidThe flow of a homogeneous incompressible, non-viscous fluid in R N is modelled by the systemof Euler equations⎧⎨ D t u(t, x) = −∇p(t, x), balance of momentumdiv u(t, x) ≡ 0,incompressibility condition(1)⎩u(0, x) = u 0 (x),initial dataWherelatex fluid.text is time,x ∈ R N is the Eulerian space variable,u = (u 1 , . . . , u N ) is the fluid velocity,p is the scalar pressure,D t = ∂ t + u · ∇ x is the material derivative.When viscosity is present, an additional diffusion term is present in the momentum equation.This leads to the Navier-Stokes equations:⎧⎨ D t u(t, x) = −∇p(t, x) + ν∆u, balance of momentumdiv u(t, x) ≡ 0,incompressibility condition(2)⎩u(0, x) = u 0 (x),initial dataThe positive constant ν is the coefficient of kinematic viscosity. Its reciprocal is called theReynolds number: Re = 1/ν.Ǐf the fluid motion takes place in a bounded set Ω ⊂ R N , one must add suitable boundaryconditions. In the case of the inviscid Euler equations, one requires that the velocity u of thefluid is tangential to the boundary ∂Ω. Calling n the unit outer normal, we thus haven · u = 0 x ∈ ∂Ω . (3)In the presence of viscosity, instead of (3), the Navier-Stokes equations are supplemented bythe non-slip boundary conditionsu = 0 x ∈ ∂Ω . (4)2.1 Derivation of Euler equation from the balance law of momentumConsider any region W 0 ⊂ R n . Denote by W t the region occupied at time t > 0 by those fluidparticles which are initially in W 0 . Assume that the only force acting on the fluid through theboundary ∂W t is the pressure. Then the balance law of momentum takes the formddt∫W tu dV = [total forces across the boundary]∫= − p n dA.∂W tIf e is any fixed vector in R N by the divergence theorem we get∫∫p e · n dA = div(p e) dV∂W t W∫ t= ∇p · e dV.W t2(5)(6)

Then:∫ ∫∫du dV = D t u dV = − ∇p dV, (7)dt W t W t W tfor any region W t in the fluid at time t. Hence the equality holds also in the differential form:D t u = −∇pEuler equation.2.2 Symmetry groups for the Euler and Navier-Stokes equationsAssume that u = u(t, x) and p = p(t, x) provide a solution to Euler or Navier-Stokes equation.Further solutions can then be obtained by various variable transformations.• Translation Invariance: For any constant vector c in R N{ u c (t, x) := u(t, x − ct) + c,p c (t, x) := p(t, x − ct),is another solution.• Rotation Invariance: For any orthogonal matrix Q{ u Q (t, x) := Q T u(t, Qx),p Q (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 > 0the functions⎧⎨ u λ, τ (t, x) := λ τ u( t τ , x λ ),(A)⎩p λ, τ (t, x) := λ2τp( t 2 τ , x λ ),provide a 2-parameters family of solutions to the Euler equation.If u, p are a solution to the Navier-Stokes equations, then for all τ > 0 the functions⎧⎨ u τ (t, x) := τ −1/2 u( t τ , x),τ 1/2(B)⎩p τ (t, x) := τ −1 p( t τ ,xτ 1/2 ).provide a 1-parameter family of solutions to the Navier-Stokes equation for τ ∈ R. Observethat (A) coincides with (B) when λ = τ 1/23 Particle trajectoriesGiven a fluid flow with velocity field u (not necessarily incompressible), the particle trajectorymappingX : [0, ∞) × R N −→ R N(8)t , α ↦→ X(t, α)describes the trajectory of the particle which is initially located at point α = (α 1 , . . . , α N ) attime t = 0. The Lagrangian variable α, can be regarded as a particle marker. The functionX(t, α) is determined by solving the Cauchy problem⎧⎨⎩∂∂tX(t, α) = u(t, X(t, α)),X(0, α) = α.3

An initial domain W 0 ⊂ R N in the fluid evolves in time toX(t, W 0 ) = W t = {X(t, α)| α ∈ W 0 }.We shall denote by ∇ α X the Jacobian matrix of first order partial derivatives of X w.r.t. α, sothat(∇ α X(t, α))ij = ∂Xi (t, α) .∂α jThe evolution of ∇ α X along a particle trajectory is described by the linear evolution equation) ()D t(∇ α X(t, α) = ∇ x u(t, X(t, α) · ∇ α X(t, α) .The Jacobian determinant of X(t, ·) is denoted asJ(t, α) = det ( ∇ α X(t, α) ) .Clearly J(0, α) ≡ 1. If u is a smooth velocity field then the time evolution of J is given by∂[]∂t J(t, α) = div u(t, X(t, α)) J(t, α) .In particular, J(t, X(t, α)) ≡ 1 if the flow is incompressible.Theorem 1 The Trasport FormulaLet W be an open, bounded domain in R N with smooth boundary, and let X be the particletrajectory mapping of a given smooth velocity field u.For any smooth function f = f(t, x) we have∫ ∫df dx = [f t + div (fu)] dx.dt W t W tProof∫df dx = d ∫f(t, X(t, α))J(t, X(t, α)) dαdt W tdt W∫( )= Dt f J + f J t dα∫W(= ft + u · ∇f + f div u ) J dα∫W= [f t + div(fu)] dx.W tDefinition 1 A flow X is incompressible if for any subregion W with smooth boundary, andfor any time t > 0, X is volume preserving:Vol ( X(t, W ) ) = Vol ( W ) .Applying the transport formula with f ≡ 1 one obtainsProposition 1 The following are equivalent:• The fluid flow is incompressible,• div u = 0,• J(t, α) ≡ 1□4

4 VorticityIn the following, we consider a smooth velocity field u : R 3 ↦→ R 3 . We define the vorticity fieldω = as the curl of the velocity:ω = curl u . = ( ∂ 2 u 3 − ∂ 3 u 2 , ∂ 3 u 1 − ∂ 1 u 3 , ∂ 1 u 2 − ∂ 2 u 1) .4.1 Local behavior of an incompressible flowIn a neighborhood of any point x 0 , we now show that, up to higher order terms w.r.t. |x − x 0 |,a smooth incompressible velocity field u = u(x) can be written in a unique way as the sum ofinfinitesimal translation, rotation and deformation field.Indeed, let x 0 be a given point in R 3 . A first order Taylor expansion of u yieldsu(x 0 + h) = u(x 0 ) + ∇u(x 0 ) · h + O(h 2 ).The Jacobian matrix ∇u, can be written as the sum of its symmetric and antisymmetric parts:∇u =}{{} D + }{{} ΩSymm. Antisymm.WhereD = 1 2Ω = 1 2(∇u + (∇u) t ), hence tr D = 0(∇u − (∇u) t ), hence Ωh = 1 2 ω ∧ h.Observe that, since every symmetric matrix can be diagonalized and the trace is invariant underorthogonal trasformations, D can be written as⎛D = ⎝The Taylor expansion thus takes the formγ 1 0 00 γ 2 00 0 −(γ 1 + γ 2 )⎞⎠u(x 0 + h) = u(x 0 ) + 1 } {{ } 2 ω ∧ h + Dh +O(h 2 ).} {{ }} {{ }translationdeformationrotation byvector ω4.2 Evolution equation for the vorticityProposition 2 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 equationProof We use the vector identity:D t ω = (ω · ∇)u + ν∆ω. (9)12 ∇|u|2 = u ∧ curl u + (u · ∇)u5

in order to replace the term (u · ∇)u in the Navier-Stokes equation:∂ t u + 1 2 ∇|u|2 − u ∧ ω = −∇p + ν∆u.We now recall that curl(∇ϕ) = 0 for every ϕ. Taking the curl of both sides of the above identityone obtains∂ t ω − curl (u ∧ ω) = ν∆ω. (10)In order to write esplicitly the term curl (u ∧ ω) we recall the following general vector identity:Since div u ≡ 0, this reduces toTherefore, from (10) we obtaincurl (u ∧ ω) = (ω · ∇)u − ω div u − (u · ∇)ω + u div ω .curl (u ∧ ω) = (ω · ∇)u − (u · ∇)ω .∂ t ω + (u · ∇)ω = (ω · ∇)u + ν∆ω. (11)□5 Vorticity transport formula for the Euler equationThe map t ↦→ X(t, α) describes the trajectory of the particle that was in α at t = 0. Considerany vector h 0 ∈ R 3 . At a given time t, the position of the particle which is initially at α + εhcan be described byX(t, α + εh) = X(t, α) + ε · ∇ α X(t, α) · h + O(ε 2 ) .In the present section we will show that the inviscid vorticity equation, i.e. the vorticityevolution equation associated to Euler equation,can be integrated by means of the particle trajectory equationD t ω = (ω · ∇)u, (12)dX(t, α) = u(t, X(t, α)).dtLet h(t, x) be a smooth vector field in R N , and consider the perturbed flowX(t, α + h) = X(t, α) + ∇ α X(t, α) h +O(h 2 ).} {{ }(⋆)(⋆) is the term which tell us how the perturbation evolves in time. Then the evolution equationof a first order perturbation takes the formD t h = h · ∇u.The following lemma applies also to the more general case in which the velocity field u is notdivergence free.6

Lemma 1 Let u = u(t, x) be a smooth vector field with associed particle trajectory mappingX(t, α). The smooth vector field h = h(t, x) satisfiesD t h = h · ∇u (13)if and only ifh(t, X(t, α)) = ∇ α X(t, α)h(0, α). (14)Proof The particle trajectory mapping is the solution of⎧d⎨ dtX(t, α) = u(t, X(t, α)),⎩X(0, α) = α.By differentiation with respect to α we obtainddt ∇ αX(t, α) = ∇ x u(t, X(t, α)) ∇ α X(t, α),henceddt ∇ αX(t, α) h(0, α) = ∇ x u(t, X(t, α)) ∇ α X(t, α) h(0, α) ,If (14) holds thenddt h(t, X(t, α)) = ∇ xu(t, X(t, α)) h(t, X(t, α))This implies (13). Viceversa, if (13) holds, then for a fixed α the vector h(t, X(t, α)) satisfiesddt h(t, X(t, α)) = ∇ xu(X(t, α)) h(t, X(t, α)).Since t ↦→ h(t, X(t, α)) and t ↦→ ∇ α X(t, α)h(0, α) satisfy the same linear ODE with initial datah(0, α), they coincide for all times t.□Proposition 3 Vorticity Transport FormulaLet X(t, α) be the smooth particle trajectory mapping corresponding to the smooth divergencefreevector field u(t, x). Thensolves (12).ω ( t, X(t, α) ) = ∇ α X(t, α)ω 0 (α)Proof We observe that the vorticity equation has the same form of the evolution equationsatisfied by a first order perturbation.The proof of Proposition 3 follows from the application of the lemma in the case h = ω. □Definition 2 The smooth curveis a vortex line at time t ifγ = {γ(s) ∈ R N | 0 < s < 1}dγ(s) = λ(s)ω(t, γ(s)), for some λ(s) ≠ 0.ds7

Corollary 1 In a non-viscous fluid flow, the vortex lines move with the fluid.Proof Let γ 0 be a vortex line at time t = 0. As a set of particles γ 0 evolves in time to γ tγ t := {X(t, γ 0 (s)) | 0 < s < 1},and there holdsdds γ t(s) = ∇ α X(t, γ 0 (s)) d ds γ 0(s)= ∇ α X(t, γ 0 (s))λ(s)ω(0, γ 0 (s))= λ(s)ω(t, X(t, γ 0 (s)))= λ(s)ω(t, γ t (s))(15)□Definition 3 Let u be a smooth divergence-free vector field and let γ be a smooth oriented closedcurve in the fluid. The circulation of u around γ the line integral∮u · dl.γLemma 2 Transport theorem for curvesLet γ t := X(t, γ 0 ) be a closed curve transported by the flow. The following identity holds:∮ ∮du · dl = D t u · dl.dt γ t γ tProof Let s ↦→ γ 0 (s) be a parametrization of γ 0 , and let s ↦→ X(t, γ 0 (s)) = γ t (s) be theparametrization of γ t .∮du · dl = d dt γ tdt=∫ 10∫ 10u(t, X(t, γ 0 (s))) ∂ ∂s X(t, γ 0(s)) dsD t u(t, X(t, γ 0 (s))) ∂ ∂s X(t, γ 0(s)) ds∫ 1+ u(t, X(t, γ 0 (s))) ∂ ∂0∂t ∂s X(t, γ 0(s)) ds∮∫ 1∂ |u(t, X(t, γ 0 (s)))| 2= D t u · dl +ds.γ t∂s 2Since γ t is a loop the last term of the right hand side is null and we obtain the equality∮ ∮du · dl = D t u · dl.dt γ t γ t0(16)□Theorem 2 Kelvin’s circulation theoremFor an inviscid, incompressible fluid the circulation of the velocity u around any closed curveγ t moving with the fluid is constant in time.8

Proof∮du · dl =dt γ tsince the line integral of a gradient on a closed loop is zero.∮∮D t u · dl = −γ t∇p dl .γ t(17)□Corollary 2 The vorticity flux through a 2-dimensional surface moving with the fluid is constantin time.Consider the 2-dimensional surface Σ t whose boundary is a closed curve γ t moving with thefluid. By Stokes formula we obtain∮ ∫u · dl = ω · dA,γ t Σ twhere n is a vector, ‖u‖ = 1, normal to Σ tAn example of this behavior is given by tornadoes: the surface of a horizontal section decreasesas the distance from the soil, then the angular velocity around the vertical axis has toincrease.6 Conserved quantities for the Euler equationA conservation law in N space dimensions takes the form:∂ t u(t, x) + div x F (u(t, x)) = 0 , (18)where t 0 is the time variable, x = (x 1 , . . . , x N ) ∈ R N is the space variable, u = (u 1 , . . . , u n )is a vector in R N and F is a smooth map defined on a convex neighborhood of the origin in R Nwith values in R N . The components of u are called the conseved quantities while the componentsof F , are the fluxes.Let W ⊂ R N be a closed region with smooth boundary. Integrating (18) on W we obtain:ddt∫W∫u(t, x) dx = u t (t, x) dxW∫= − div F ( u(t, x) ) dxW∫= − F ( u(t, x) ) · n dA∂W= [Flow through the boudary].Observation 1 Notice that∫du(t, x) dx = 0dt R N if |F | = ≀(|x| 1−N ).In fact∫∫ddu(t, x) dx = lim u(t, x) dx = limdt R N R→∞ dt− F (u(t, x)) · n dA,|x|RR→∞∫|x|=Ris zero if |F | = o(|x| 1−N ).9

The following are conserved quantities for the Euler equations.• Components of momentum The Euler equation is equivalent to the momentum balancelaw. Then, if u is a solution to the Euler equation its components satisfyNu i t + ∑u j u i x j= −p xi ,which can be written (div u=0) as a system of conservation law:j=1u i t + div (ui u + P i ) = 0,i = 1, . . . , Nwhere P i ≡ (0, . . . , 0, p i , 0, . . . , 0).• Vorticity The i-th component of the vorticity equation isωt i + (u∇)ω i = (ω∇)u i .It can be written as a system of conservation laws::ωt i + div (ωi u + u i ω) = 0.• The energy By multipling Euler equation by u we obtain the energy equation( |u|22)t N + ∑u i u j u i x j+i,j=1N∑u i p xi = 0.i=1It can be written as a conservation law( |u|22)t + div ( |u|22 u + p v )= 0.• Helicity The helicity measures the component of velocity in the direction of the vorticity:H . = u · ω.∂∂t H = u t · ω + u · ω t= [−(u · ∇)u − ∇p ] · ω + u · [−(u · ∇)ω + (ω · ∇)u ]N∑N∑N∑N∑= − u j u i x jω i − p xi ω i − u i u j ωx i j+ u i ω j u i x ji,j=1i=1i,j=1i,j=1This can be written asH t + div[(ω · u)u − |u|22 ω + p ω ]= 0.10

7 Leray formulation of Navier-StokesequationOur aim in this section will be to eliminate the pressure from the explicit formulation of Navier-Stokes equation in order to obtain a system of closed evolution equations for u. Let’s assumethat u satisfies the equationdu= −(u · ∇)u + ν∆u − ∇p (19)dtThen u solves (2) if and only if the function p has been chosen in such a way that div u(t, x) = 0for all t 0.We define a subspace E ⊂ L 2 (R 3 ; R 3 ) containing all vector fields whose divergence (indistributional sense) vanishes identically.E . = { u ∈ L 2 ; div u = 0 } .Assume that t ↦→ u(t) ∈ E is a smooth solution of (19) Observe thatdiv u = 0 ⇒ div (∆u) = 0,div u = 0 div (u · ∇u) = 0.(20)More precisely, if the i-th component of (u · ∇)u is[(u · ∇)u] i = ∑ ju j u i x j,we can writediv (u · ∇)u = ∑ i,j(ujxiu i x j+ u j u i )x i x j .} {{ }A i,jSince div u = ∑ i ui x i= 0, we get ∑ i,j A i,j = 0. Therefore, in general,div (u · ∇)u = ∑ i,ju j x iu i x j= tr(∇u) 2 ≠ 0.We now need to choose p so that the sumlies in E.Sincedudt= −(u · ∇)u + ν∆u} {{ } }{{} −∇p/∈E∈Ediv [ (u · ∇)u + ∇p] = ∑ i,ju j x iu i x j+ ∆pthe pressure p satisfies the relation= tr(∇u) 2 + ∆p,if and only if it solves the elliptic equation:div [ (u · ∇)u + ∇p] = 0on the whole space R N .We now recall a solution formula for linear elliptic equations∆p = −tr(∇u) 2 (21)11

Lemma 3 Solution of the Poisson equationLet f be a smooth function in R N , vanishing sufficiently rapidly as|x| → ∞. Then the solution to the Poisson equation{ ∆z = f,∇z → 0, as |x| → 0,is given by∫z(x) = N N ∗ f(x) = N N (x − y)f(y) dy,R Nwhere the fundamental solution N is the Newton potential{ 1N N (x) =2πln |x|, if N = 2,1(2−N) σ N|x| 2−N , if N 3(22)and σ N denotes the surface area of a unit sphere in R N .Observation 2 N N is the solution in R N of∆N = δ 0where δ 0 is the Dirac mass concentrated at the origin.By applying this Lemma we can solve the equation (21)∫p(t, x) = N N (x − y) tr(∇u(t, y)) 2 dy.R NThe gradient of p takes the form:∇p(t, x) = ∇N N ∗ tr(∇u) 2 (x) = C N∫R Nx − y|x − y| N tr(∇u(t, y))2 dy,for a suitable constant C N . Using this result we can eliminate the pressure from the explicitformulation of Navier-Stokes equation and we obtain a system of closed evolution equations foru. This is the Leray’s formulation of Navier-Stokes equation∫x − yD t u(t, x) = ν∆u − C N|x − y| N tr(∇u(t, y))2 dy. (23)Observation 3 The Navier-Stokes equation can be written:R Nu t = Π E (u · ∇)u + ν∆u,where Π E is the perpendicular projection operator on the space E.To prove this claim, we have to show that, for any scalar function p vanishing sufficientlyfast as |x| → ∞, the gradient ∇p is orthogonal to all functions v ∈ E.Lemma 4 Let w be a smooth, divergence-free vector field in R N and let q be a smooth scalarfunction such that|w(x)||q(x)| = o(|x| 1−N ) as |x| → ∞.Then w and ∇q are orthogonal in L 2 :∫R N w · ∇q dx = 0.12

Proof A direct computation yields∫div (w q) dx =|x|≤R∫lim (w q) · n dA = 0R→∞ |x|=R□8 Vorticity-stream formulation of the Navier-Stokes equationsTaking the curl of both sides of Navies-Stokes equation we obtain the vorticity evolution equation,in which the pressure does not appear. This provides an alternative way to close oursystem. In this case, the key step is to recover u from ω = . curl u).8.1 The 2-dimensional caseIf the flow lies on a plane, the velocity field has components u = (u 1 , u 2 , 0) and the vorticityfield is ω = (0, 0, u 2 x 1− u 1 x 2). We can denote them respectively asu =( u1u 2 )and ω = u 2 x 1− u 1 x 2.In this case the evolution of the vorticity is reduced to a scalar equation∂ t ω + (u · ∇)ω = ν∆ω.To recover the velocity u from the scalar function ω, we first recall that the system of PDE’s{ψx1 = a(x 1 , x 2 ),ψ x2 = b(x 1 , x 2 ),has a solution if and only if a x2 = b x1 . Since div u = u 1 x 1+ u 2 x 2= 0, there exist a function ψ,called stream function , which solves{ψx1 = u 2 (x 1 , x 2 ),ψ x2 = −u 1 (x 1 , x 2 ).We then haveu =( −ψx2ψ x1).=∇ ⊥ ψ, (24)ω = curl u = curl (∇ ⊥ ψ) = ∆ψ. (25)Let ω : R 2 ↦→ R be a given vorticity function. We can determine ψ by solving the Poissonequation (25) as in lemma 3ψ(t, x) = N 2 ∗ ω(x) = 1 ∫ln(|x − y|)ω(t, y) dy .2π R 2We then obtain u from (24),u(t, x) = K 2 ∗ ω(x) =∫13R 2 K 2 (x − y)ω(y) dy. (26)

Here the kernel K 2 is the fundamental solution of the equationits explicit form iscurl z = δ 0 ,K 2 (x) = 12π |x| 2 ( −x2x 1),Notice that the singularity at the origin has size|K 2 (x)| = 12π |x| .The formula (26) is called Biot-Savart equation since it takes the same form of the Biot-Savartlaw for the magnetic field. Observe that also in this second reformulation of the Navier-StokesEquation the pressure can be obtained from the the Poisson equation∆p = −tr(∇u) 2 .8.2 The 3-dimensional caseSince in the 3-dimensional case the term ω · ∇u in the vorticity equation is non zero, we haveto recover both u and ∇u from vorticity.Lemma 5 Let ω : R 3 → R 3 be a smooth vector field with a sufficiently fast decay as |x| → ∞.The overdetermined elliptic system { curl u = ω,(27)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 equationwhere the stream function ψ solves the Poisson equationu = −curl ψ. (28)∆ψ = ω . (29)Observation 4 We solve the equation (29) using lemma 3ψ(x) = N 3 ∗ ω(x) = 1 ∫1ω(y) dy,2π R 3 |x − y|and then the explicit solution to (28) is provided by the Biot-Savart formula∫u(x) = K 3 ∗ ω = K 3 (x − y)ω(y) dy,R 3where the kernel K 3 is defined by the relationK 3 (x) · h = 14πx ∧ h|x| 3 h ∈ R 3 .14

ProofNecessity: If ω = curl u then div ω = div (curl u) = 0.Sufficiency: Assume that div ω = 0. Consider the vector identity∆ψ = ∇(div ψ) − curl (curl ψ) . (30)Take the inner product with ∇(div ψ) we obtain∫∫∇(div ψ) · ∆ψ dx = ∇(div ψ) · ∇(div ψ) dx −∫∇(div ψ) · curl (curl ψ) dx .The first and the third integral terms vanish because divergence-free fields are orthogonal togradient fields, as shown in lemma 4. Therefore∫R N |∇(div ψ)| 2 dx = 0 .This proves that ψ is divergence-free andω = ∆ψ = − curl (curl ψ).Settingu . = − curl ψwe obtain ω = curl u and of course div u = div (− curl ψ) = 0.□9 Particle trajectory formulation of the Euler equationIn this section we show that the Euler equations can be reformulated as a family of integrodifferentialequation for the particle trajectories. Given a smooth velocity field u, the particletrajectory mapping X satisfies the following Cauchy problem, for each value of the Lagrangianvariable α ∈ R N{ ddtX(t, α) = u(t, X(t, α)),X(0, α) = α.We would like to recover u from X and ∇X. We begin by recalling the vorticity transportformulaω(t, X(t, α)) = ∇ α X(t, α)ω 0 (α),and the Biot-Savart lawTogether, these equations yieldu(t, x) =∫R N K N (x − y)ω(t, y) dy.u(t, X(t, α)) = d dt X(t, α) = ∫R N K N (X(t, α) − y)ω(t, y) dy .Consider the change of integration variable given byy = X(t, α ′ ), for a suitable Lagrangian variable α ′ ,15

Then∫u(t, X(t, α)) = K N (X(t, α) − X(t, α ′ ))ω(t, X(t, α ′ )) dα ′∫R N= K N (X(t, α) − X(t, α ′ ))∇ α X(t, α ′ )ω 0 (α ′ ) dα ′ .R NThe explicit form of this integral depends on N, we will consider the two cases separately.The 3-dimensional caseIn the 3-dimensional case the kernel K 3 is defined by the relationK 3 (x) · h = 14πx ∧ h|x| 3 for x, h ∈ R 3 .Then the integral form of u(t, X(t, α))∫u(t, X(t, α)) = K 3 (X(t, α) − X(t, α ′ ))∇ α X(t, α ′ )ω 0 (α ′ ) dα ′R 3is a singular integral operator in a neighborhood of the origin.The 2-dimensional caseIn this case the situation is simpler because vorticity is constant along particle trajectoriesω(t, X(t, α)) = ω 0 (α).In fact if the flow lies on the plane R 2 , then the velocity field is u = (u 1 , u 2 , 0) t , the vorticity isω = (0, 0, u 2 x 1− u 1 x 2) t , and we get:⎛⎞dX 1 dX 1dα 1 dα 20 ⎛ ⎞ ⎛ ⎞0 0∇ α X(t, α) ω =dX⎜ 2 dX 2⎝dα 1 dα 20 ⎟ ⎝ 0 ⎠ = ⎝ 0 ⎠ .⎠ω 3 ω 30 0 1Then the integral form of u(t, X(t, α)) becomes:∫u(t, X(t, α)) = K 2 (X(t, α) − X(t, α ′ ))ω 0 (α ′ ) dα ′ .R 210 Energy estimatesTaking the scalar product of the Navier-Stokes equation with u and integrating by parts weobtain:d |u|dt∫R 2 ∫ ( |u|2 ) ∫∫N 2 dx + divR N 2 u dx = − div (p u) dx − ν |∇u| 2 dx.R N R NIf we assume, as in the previous sections, that u decays sufficiently fast as |x| → ∞, then thetwo integrals containing the divergence operator are zero because of Stokes theorem. We obtainan equality between the time derivative of the kinetic energy of the fluid and a negative termdepending on the viscosity coefficientd |u|dt∫R 2 ∫N 2 dx = − ν |∇u| 2 dx 0.R NThis equation provides an a priori bound of the L 2 norm of the velocity.16

solves the ODE with initial data:{ ż(t) = 2‖∇v2 ‖ L ∞z(t),11 Uniqueness of solutionsLet v 1 and v 2 be two smooth solutions of the same Navier-Stokes equation, vanishing sufficientlyfast as |x| → ∞, and so, in particular, both in L 2 . If p 1 and p 2 are the associate pressures wedefineṽ . = v 1 − v 2 ,˜p . = p 1 − p 2 .These two functions ṽ, ˜p satisfy the evolution equationṽ t + (v 1 · ∇) ṽ + (ṽ · ∇) v 2 = −∇˜p = ν∆ṽ.Let’s note that the linear terms involve only ṽ and ˜p.Taking the inner product of this equations with ṽ we obtain:∫∫∫ṽ · ṽ t dx + ṽ · (v 1 · ∇)ṽ dx + ṽ · (ṽ · ∇)v 2 dx =R N R} N R{{ }N−∫(⋆)R N ṽ · ∇˜p dx} {{ }(⋆ ⋆)∫− ν ṽ · ∆ṽ dxR NWe observe that (⋆) and (⋆ ⋆) are equal to zero: (⋆ ⋆) is the scalar product in L 2 of a divergencefree vector field and the gradient of a scalar function, while integrating (⋆) by parts we get(⋆) =N∑∫ ( |ṽ i | 2 )v 1 · ∇ dx =R N 2i=1N∑∫div(v 1 · |ṽi | 2R N 2i=1)dx = 0because of the fast decay of v 1 as |x| → ∞.Integrating by parts the other terms the equation (31) becomes:d |ṽ|dt∫R 2 ∫∫N 2 dx = − ṽ · (ṽ · ∇)v 2 dx − ν |∇ṽ| 2 dx .R N R} N {{ }0Then the following estimate holds:Comparison with an ODEThe function defined asddt ‖ṽ‖2 L 2 ‖ṽ‖ 2 L 2 ‖∇v 2 ‖ L ∞.z(T ) = exp ( ∫ T2‖∇v 2 (t)‖ L 2 dt ) z(0)0(31)z(0) = ‖ṽ(0)‖ 2 L, 2and for any instant T 0 there holds:‖ṽ(T )‖ 2 L z(T ).2So if the solutions v 1 and v 2 coicide at t = 0, then they coincide for all t 0.17