Lecture Notes on the Theory of

Incompressible Fluid Motion

Alberto Bressan and Carlotta Donadello

15th November 2007

Contents

1 The equations of fluid motion 11.1 Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Euler and Navier-Stokes equations . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Symmetry groups for the Euler and Navier-Stokes equations . . . . . . . . . . . . 31.4 Particle trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Vorticity transport formula for the Euler equation . . . . . . . . . . . . . . . . . 71.7 Vortex lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8 Conserved quantities for the Euler equation . . . . . . . . . . . . . . . . . . . . . 101.9 Leray formulation of the Navier-Stokes equations . . . . . . . . . . . . . . . . . . 121.10 Vorticity-stream formulation of the Navier-Stokes Equation . . . . . . . . . . . . 141.11 Lagrangian formulation of the Euler equation . . . . . . . . . . . . . . . . . . . . 17

2 Construction of solutions 192.1 Uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 The Sobolev spaces Hm(RN ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3 Local-in-time existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 Global weak solutions of the Navier-Stokes equations . . . . . . . . . . . . . . . . 272.5 Construction of finite dimensional approximate solutions . . . . . . . . . . . . . . 29

3 Vortex sheets 323.1 Velocity behavior near a vortex sheet . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 The Birkhoff-Rott equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Kelvin-Helmholtz instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 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

∇u.= (ux1 , ux2 , . . . , uxN ) =

( ∂

∂x1u ,

∂

∂x2u , . . . ,

∂

∂xNu

).

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

divu.=

N∑

i=1

uixi

.

The Laplace operator is defined as

∆u.=

N∑

i=1

∂2u

∂x2i

= div∇u .

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

Dtφ.=

∂

∂tφ +

N∑

i=1

ui ∂

∂xiφ .

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. Lete1, e2, e3 be the vectors in the standard orthonormal basis of R3. Then

v ∧ w = det

e1 e2 e3

v1 v2 v3

w1 w2 w3

.

1

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

ω = curlu .=(∂2u

3 − ∂3u2 , ∂3u

1 − ∂1u3 , ∂1u

2 − ∂2u1).

Note that, formally, one can write

curlu = ∇∧ u = det

e1 e2 e3

∂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 ahomogeneous, incompressible, non-viscous fluid in RN is modelled by the Euler equations

ut + u · ∇u = −∇p balance of momentum,divu ≡ 0 incompressibility condition.

(1.1)

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

t is time,x = (x1, . . . , xN ) ∈ RN 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

ut + u · ∇u = −∇p + ν∆u balance of momentum,divu ≡ 0, incompressibility condition.

(1.2)

The positive constant ν is the coefficient of kinematic viscosity. Many authors refer to it as theinverse 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 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 have

n · u = 0 x ∈ ∂Ω . (1.4)

In the presence of viscosity, instead of (1.4), the Navier-Stokes equations are supplementedby 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

Consider any region W0 ⊂ Rn. Denote by Wt the region occupied at time t > 0 by thosefluid particles which are initially in W0. Assume that the only force acting on the fluid throughthe boundary ∂Wt is the pressure. Then the balance law of momentum takes the form

d

dt

∫

Wt

u dV = [total forces across the boundary]

= −∫

∂Wt

pn dA .

(1.6)

If e is any fixed vector in RN by the divergence theorem we get∫

∂Wt

p e · n dA =∫

Wt

div(p e) dV

=∫

Wt

∇p · e dV.

(1.7)

Then:d

dt

∫

Wt

u dV =∫

Wt

Dtu dV = −∫

Wt

∇p dV, (1.8)

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-Stokesequations. Further solutions can then be obtained by various variable transformations.

• Translation Invariance: For any constant vector c in RN

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

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 > 0the functions

uλ, τ (t, x) .= λτ u( t

τ , xλ),

pλ, τ (t, x) .= λ2

τ2 p( tτ , x

λ),(1.9)

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( tτ , x

τ1/2 ),

pτ (t, x) .= τ−1 p( tτ , x

τ1/2 ).(1.10)

provide a 1-parameter family of solutions to the Navier-Stokes equation for τ ∈ R. Observethat (1.9) coincides with (1.10) when λ = τ1/2

1.4 Particle trajectories

Given a fluid flow with velocity field u (not necessarily incompressible), the particle trajectorymapping

X : [0,∞)× RN 7→ RN

t , α 7→ X(t, α)(1.11)

describes the trajectory of the particle which is initially located at point α = (α1, . . . , αN ) 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.

An initial domain W0 ⊂ RN 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, sothat (

∇αX(t, α))

ij=

∂Xi

∂αj(t, α) .

The evolution of ∇αX along a particle trajectory is described by the linear evolution equation

∂

∂t

(∇αX(t, α)

)=

(∇xu(t, X(t, α)

)· ∇αX(t, 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 isJ(0, α) ≡ 1. If u is a smooth velocity field then the time evolution of J is given by

∂

∂tJ(t, α) =

[divu(t,X(t, α))

]J(t, α) .

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 particletrajectory mapping of a given smooth velocity field u.For any smooth function f = f(t, x) we have

d

dt

∫

Wt

f dx =∫

Wt

[ft + div (fu)] dx.

Proofd

dt

∫

Wt

f dx =d

dt

∫

Wf(t,X(t, α))J(t,X(t, α)) dα

=∫

W

(Dtf J + f Jt

)dα

=∫

W

(ft + u · ∇f + f divu

)J dα

=∫

Wt

[ft + div(fu)] dx.

¤

Definition 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 obtains

Proposition 1 The following facts are equivalent:

• The fluid flow is incompressible,

• divu = 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 fieldω as the curl of the velocity:

ω = curlu .=(∂2u

3 − ∂3u2 , ∂3u

1 − ∂1u3 , ∂1u

2 − ∂2u1).

Proposition 2 (Local behavior of an incompressible flow). In a neighborhood of anypoint 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 anddeformation velocities.

5

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

u(x0 + h) = u(x0) +∇u(x0) · h +O(h2).

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

∇u = D︸︷︷︸Symm.

+ Ω︸︷︷︸Antisymm.

whereD = 1

2

(∇u + (∇u)t

), hence trD = 0

Ω = 12

(∇u − (∇u)t

)hence Ωh = 1

2ω ∧ h .

Observe that, since every symmetric matrix can be diagonalized and the trace is invariant underorthogonal trasformations, D can be written as

D =

γ1 0 00 γ2 00 0 −(γ1 + γ2)

.

The Taylor expansion thus takes the form

u(x0 + h) = u(x0)︸ ︷︷ ︸translation

+12ω ∧ h

︸ ︷︷ ︸rotation

+ Dh︸ ︷︷ ︸deformation

+O(h2).

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=curlu evolves accordingto the equation

Dtω = (ω · ∇)u + ν∆ω. (1.12)

Proof We use the vector identity:

12∇|u|2 = u ∧ curl u + (u · ∇)u, (1.13)

in order to replace the term (u · ∇)u in the Navier-Stokes equation:

∂t u +12∇|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. (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 − udivω. (1.15)

6

Since divu ≡ 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 thepresent section we will show that the inviscid vorticity equation, i.e. the vorticity evolutionequation associated to Euler equation,

Dtω = (ω · ∇)u, (1.17)

can be integrated by means of the particle trajectory equation

d

dtX(t, α) = u(t, X(t, α)).

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 trajectorymapping 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 equationsatisfied 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) +∇αX(t, α) h︸ ︷︷ ︸(?)

+O(h2).

Here (?) is the term which tell us how the perturbation evolves in time. Then the evolutionequation 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 thedirection of the vorticity ω. The following lemma applies also to the more general case in whichthe velocity field u is not divergence free.

7

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

Dt h = h · ∇u (1.18)

if and only ifh(t,X(t, α)) = ∇αX(t, α)h(0, α). (1.19)

Proof The particle trajectory mapping is the solution of

ddtX(t, α) = u(t, X(t, α)),

X(0, α) = α.

By differentiation with respect to α we obtain

d

dt∇αX(t, α) = (∇xu)|x=X(t,α)∇αX(t, α),

and thend

dt∇αX(t, α)h(0, α) = (∇xu)|x=X(t,α)∇αX(t, α)h(0, α).

If (1.19) holds thend

dth(t, X(t, α)) = (∇xu)|x=X(t,α)h(t,X(t, α)).

Notice that the right hand side represents the directional derivative of the velocity u in thedirection 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

dth(t,X(t, α)) = (∇u)|x=X(t,α)h(t,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 datah(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 isω = (0, 0, ω3) = (0, 0, u2

x1− u1

x2), and therefore

∇αX(t, α) ω =

dX1dα1

dX1dα2

0

dX2dα1

dX2dα2

0

0 0 1

00ω3

=

00ω3

.

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

dsγ(s) = λ(s)ω(t, γ(s)), for some λ(s) 6= 0.

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 areinitially 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

dsγt(s) = ∇αX(t, γ0(s))

d

dsγ0(s)

= ∇αX(t, γ0(s))λ(s)ω(0, γ0(s))= λ(s)ω(t,X(t, γ0(s)))= λ(s)ω(t, γt(s)) .

(1.21)

¤

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

ΓC =∮

Cu · dl.

Lemma 2 Transport formula for curves. Let Ct.= X(t, C) be a curve transported by the

flow. The following identity holds:

d

dt

∮

Ct

u · dl =∮

Ct

Dtu · dl. (1.22)

Proof Let s 7→ C(s) be a parametrization of C, and let s 7→ X(t, C(s)) = Ct(s) be theparametrization of Ct.

d

dt

∮

Ct

u · dl =d

dt

∫ 1

0u(t,X(t, C(s))) ∂

∂sX(t, C(s)) ds

=∫ 1

0Dtu(t,X(t, C(s))) ∂

∂sX(t, C(s)) ds

+∫ 1

0u(t,X(t, C(s))) ∂

∂t

∂

∂sX(t, C(s)) ds

=∮

Ct

Dtu · dl +∫ 1

0

∂

∂s

|u(t,X(t, C(s)))|22

ds.

(1.23)

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 circu-lation 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

d

dt

∮

Ct

u · dl =∮

Ct

Dtu · dl

= −∮

Ct

∇p · dl = 0(1.24)

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 constantin 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. UsingStokes’ formula one obtains ∫

Σt

ω · ndA =∮

Ct

u · dl .

The result thus follows from Kelvin’s circulation theorem. ¤

An example of this phenomena is provided by tornadoes. If the surface of a horizontal sectiondecreases proportionally to the height h from the ground, then as h → 0 the angular velocityaround 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)

where t ≥ 0 is the time variable, x = (x1, . . . , xN ) ∈ RN is the space variable, u = (u1, . . . , uN )is a vector in RN and F is a smooth map defined on a convex neighborhood of the origin inRN with values in RN . The components of u are called the conserved quantities while thecomponents of F , are the fluxes.

Let W ⊂ RN be a closed region with smooth boundary. Integrating (1.25) on W we obtain

d

dt

∫

Wu(t, x) dx =

∫

Wut(t, x) dx

= −∫

Wdiv F

(u(t, x)

)dx

= −∫

∂WF

(u(t, x)

) · n dA

= [flow across the boundary].

10

Remark. Notice that

d

dt

∫

RN

u(t, x) dx = 0 if |F | = o(|x|1−N ).

Indeed

d

dt

∫

RN

u(t, x) dx = limR−→∞

d

dt

∫

|x|≤Ru(t, x) dx = − lim

R−→∞

∫

|x|=RF (u(t, x)) · n dA, = 0 ,

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 bal-ance law. Then, if u is a solution to the Euler equation its components satisfy

uit +

N∑

j=1

ujuixj

= −pxi i = 1, . . . , N.

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

uit + div (ui 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

ωit + (u∇)ωi = (ω∇)ui.

These equations also yield a system of conservation laws

ωit + div (ωi u − ui ω) = 0 i = 1, . . . , N .

• Energy. Taking the scalar product of both sides of the Euler equation by u, we obtainthe energy equation

( |u|22

)t

+N∑

i,j=1

uiujuixj

+N∑

i=1

uipxi = 0.

It can be written as a conservation law

( |u|22

)t

+ div( |u|2

2u + p v

)= 0.

• Helicity. The helicity measures the component of velocity in the direction of thevorticity:

H.= u · ω.

11

Differentiating w.r.t. time we obtain

∂

∂tH = ut · ω + u · ωt

= [−(u · ∇)u − ∇p ] · ω + u · [−(u · ∇)ω + (ω · ∇)u ]

= −N∑

i,j=1

ujuixj

ωi −N∑

i=1

pxiωi −

N∑

i,j=1

uiujωixj

+N∑

i,j=1

uiωjuixj

.

This equation can be written in conservation form, namely

Ht + div[ N∑

i=1

ωiuiu −N∑

i=1

|ui|22

ω + pω]

= 0.

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, andobtain an evolution equation for the velocity u alone. Assume that, at time t = 0, divu = 0.Moreover, let u satisfy the equation

du

dt= −(u · ∇)u + ν∆u−∇p (1.26)

How can we determine the pressure function p = p(t, x), so that the divergence of u remainszero st all positive times?

To set the ideas, suppose u ∈ L2, so that

‖u‖L2 =

(∫|u|2 dx

)1/2

< +∞.

We introduce the subspace E ⊂ L2 containing all vector fields whose divergence (in distri-bution sense) vanishes. Notice that, if u is a C1 vector field with zero divergence, then for everysmooth scalar function ϕ with compact support we have

∫u · ∇ϕ dx =

∫div (ϕu) dx = 0 .

We can thus define the subspace E by setting

E.=

u ∈ L2 |

∫u · ∇ϕdx = 0 for every ϕ ∈ C∞c

. (1.27)

If u(t) ∈ E for all t ≥ 0, the time derivative ∂tu(t) must remain in E for all times. Lookingat the equation (1.26) we observe that

divu = 0 ⇒ div (∆u) = 0,divu = 0 ; div (u · ∇u) = 0.

(1.28)

More precisely, if the i-th component of (u · ∇)u is

[(u · ∇)u]i =∑

j

ujuixj

,

12

we can writediv (u · ∇)u =

∑

i,j

(uj

xiui

xj+ ujui

xi xj︸ ︷︷ ︸Ai,j

).

Since divu =∑

i uixi

= 0, we get∑

i,j Ai,j = 0, so, in general,

div (u · ∇)u =∑

i,j

ujxi

uixj

= tr(∇u)2 6= 0.

We now seek a scalar function p such that the sum

du

dt= −(u · ∇)u︸ ︷︷ ︸

/∈E

+ ν∆u︸︷︷︸∈E

−∇p

lies in the subspace E. Observing that

div[(u · ∇)u +∇p] =

∑

i,j

ujxi

uixj

+ ∆p = tr(∇u)2 + ∆p,

we conclude that the identitydiv

[(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 thefollowing lemma.

Lemma 3 Solution of the Poisson equation. Let f be a smooth function in RN , vanishingsufficiently rapidly as |x| −→ ∞. Then the solution to the Poisson equation

∆z = f,∇z −→ 0 as |x| −→ ∞,

is given by

z(x) = NN ∗ f(x) =∫

RN

NN (x− y)f(y) dy ,

where N is the N -dimensional Newton potential

NN (x) = 1

2π ln |x|, if N = 2,1

(2−N) σN|x|2−N , if N ≥ 3 (1.30)

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

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

p(t, x) =∫

RN

NN (x− y) tr(∇u(t, y))2 dy.

The gradient of p takes the form

∇p(t, x) = ∇NN ∗ tr(∇u)2 (x) = CN

∫

RN

x− y

|x− y|N tr(∇u(t, y))2 dy,

for a suitable constant CN . Using this formula, we can eliminate the pressure from the standardformulation of the Navier-Stokes equation and obtain a system of closed evolution equations foru. This is the Leray formulation of the Navier-Stokes equations:

Dtu(t, x) = ν∆u − CN

∫

RN

x− y

|x− y|N tr(∇u(t, y))2 dy. (1.31)

Remark. According to (1.27), the subspace E of vector fields with zero divergence can beregarded 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 thefluid velocity u. In order to obtain an equation involving only the vorticity field ω, we need toexpress the vector field u in terms of ω = curlu.

The 2-dimensional case.If the flow lies in a plane, the velocity field is u = (u1, u2, 0) and the vorticity field ω =

(0, 0, u2x1− u1

x2). We can denote them respectively as

u =(

u1

u2

), and ω = u2

x1− u1

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 systemof PDEs

ψx1 = a(x1, x2),ψx2 = b(x1, x2),

has a solution if and only if ax2 = bx1 . Since divu = u1x1

+ u2x2

= 0, there exists a so-calledstream function ψ, such that

ψx1 = u2(x1, x2),ψx2 = −u1(x1, x2).

(1.33)

By (1.33) it follows

u = ∇⊥ψ.=

( −ψx2

ψx1

), (1.34)

ω = curlu = curl (∇⊥ψ) = ∆ψ . (1.35)

Let ω : R2 7→ R be a given vorticity function. We can find the corresponding stream function ψby solving the Poisson equation (1.35) as in Lemma 3, namely

ψ(x) = N2 ∗ ω(x) =12π

∫

R2

ln |x− y| · ω(t, y) dy .

In turn, the velocity field u : R2 7→ R2 is now computed by (1.34). In the end, we see that u canbe recovered from ω by the Biot-Savart formula

u(x) = K2 ∗ ω(x) =∫

R2

K2(x− y)ω(y) dy. (1.36)

Here the kernel K2 is the fundamental solution of the equation

curl z = δ0 ,

with δ0 a Dirac mass at the origin. More precisely,

K2(x) =1

2π |x|2( −x2

x1

). (1.37)

Notice that K2 has a singularity at the origin, namely

|K2(x)| = 12π |x| .

Observe that, in this second formulation of the Navier-Stokes equations, the pressure can beobtained 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

curlu = ω,divu = 0,

(1.38)

admits a solution if and only if divω = 0.In the positive case, the solution is constructed by means of the equation

u = −curlψ. (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

ψ(x) = N3 ∗ ω(x) =12π

∫

R3

1|x− y| ω(t, y) dy .

The explicit solution to (1.39) is then provided by the Biot-Savart formula

u(x) = K3 ∗ ω =∫

R3

K3(x− y) ω(y) dy,

where the kernel K3 is defined by the relation

K3(x) · h =14π

x ∧ h

|x|3 h ∈ R3. (1.41)

ProofNecessity : If ω = curlu then divω = div (curlu) = 0.Sufficiency : Start with the vector identity

∆ψ = ∇(divψ) − curl (curlψ) (1.42)

and consider the vector fieldsω

.= ∆ψ,

k.= ∇(divψ),

h.= − curl (curlψ).

Assuming that divω = 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-freevector 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 scalarfunction such that

|w(x)| |q(x)| = o(|x|1−N ) as |x| −→ ∞.

Then w and ∇q are orthogonal: ∫

RN

w · ∇q dx = 0.

16

Indeed,∫

RN

w · ∇q dx = limR−→∞

∫

|x|≤Rdiv (w q) dx = lim

R−→∞

∫

|x|=R(w q) · n dA = 0 .

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

〈ω, k〉 = 〈k, k〉 + 〈h, k〉 = 〈k, k〉.

Therefore,

〈k, k〉 =∫

RN

|∇(divψ)|2 dx = 0 .

This shows that ψ is divergence-free and

ω = ∆ψ = − curl (curlψ).

Definingu

.= − curlψ

we obtain ω = curlu, and of course divu = div (− curlψ) = 0. ¤

1.11 Lagrangian formulation of the Euler equation

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

ddtX(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

u(t, x) =∫

RN

KN (x− y) ω(t, y) dy.

Putting together these equations we obtain

u(t, X(t, α)) =d

dtX(t, α) =

∫

RN

KN (X(t, α)− y) ω(t, y) dy .

Consider the change of integration variable given by

y = X(t, α′),

17

for a suitable value of the Lagrangian variable α′. Then

u(t, X(t, α)) =∫

RN

KN (X(t, α)−X(t, α′))ω(t,X(t, α′)) dα′

=∫

RN

KN (X(t, α)−X(t, α′))∇αX(t, α′)ω0(α′) dα′.(1.43)

In the 3-dimensional case the kernel K3 is defined by (1.41). We remark that, because of thegrowth 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 almosteverywhere.

In the 2-dimensional case the situation is much simpler. Indeed, according to (1.20), thevorticity is constant along particle trajectories. The integral formula (1.43) thus reduces to

u(t,X(t, α)) =∫

R2

K2(X(t, α)−X(t, α′))ω0(α′) dα′,

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 theNavier-Stokes equation

ut + (u · ∇)u = −∇p + ν∆u, (2.1)

divu ≡ 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 sufficientlyfast as |x| −→ ∞. In particular, we assume v1, v2 ∈ L2(RN ). If p1 and p2 are the correspondingpressures, we define

v.= v1 − v2,

p.= p1 − p2.

These two functions v, p satisfy the evolution equation

vt + (v1 · ∇) v + (v · ∇) v2 = −∇p + ν∆v. (2.3)

Observe that this equation is linear w.r.t. v and p. Taking the inner product of both sides of(2.3) with v we obtain

∫

RN

v · vt dx+∫

RN

v · (v1 · ∇)v dx

︸ ︷︷ ︸(?)

+∫

RN

v · (v · ∇)v2 dx =

−∫

RN

v · ∇p dx

︸ ︷︷ ︸(? ?)

− ν

∫

RN

v ·∆v dx

(2.4)

We observe that the two terms (?) and (? ?) both vanish. Indeed, (? ?) is the scalar product inL2 of a divergence free vector field with the gradient of a scalar function. Moreover, integrating(?) by parts one obtains

(?) =N∑

i=1

∫

RN

v1 · ∇( |vi|2

2

)dx =

N∑

i=1

∫

RN

div(v1 · |v

i|22

)dx = 0

19

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

d

dt

∫

RN

|v|22

dx = −∫

RN

v · (v · ∇)v2 dx − ν

∫

RN

|∇v|2 dx

︸ ︷︷ ︸≤0

.

Therefored

dt‖v‖2

L2 ≤ ‖v‖2L2‖∇v2‖L∞ . (2.5)

Thanks to (2.5), we can now prove that v ≡ 0 by a direct comparison with the correspondingODE. Indeed, the function

z(t) = exp( ∫ t

02‖∇v2(τ)‖L2 dτ

)z0

solves the Cauchy problem

z(t) = 2‖∇v2(t)‖L∞ z(t),

z(0) = z0 = ‖v(0)‖2L2 .

By comparison, for any time t ≥ 0 we have

‖v(t)‖2L2 ≤ z(t).

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

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

∫ T

0‖∇v2(t)‖L∞ dt < ∞ ,

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 inL2(RN ) for every multindex α, 0 ≤ |α| ≤ m, where Dα denotes the distributional derivative∂α1

1 · · · ∂αNN . The space Hm(RN ) is equipped with the norm

‖v‖Hm.=

( ∑

0≤|α|≤m

‖Dαv‖2L2

)1/2.

Theorem 4 (Sobolev embedding theorem). If m > k + N2 then Hm(RN ) is continuously

embedded in Ck, i.e. there exist a constant c > 0 such that

‖v‖Ck ≤ c‖v‖Hm .

20

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

‖Dαv‖L2m/k ≤ C‖v‖1−k/mL∞ ‖v‖k/m

Hm

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

‖u v‖Hm ≤ Cm

‖u‖L∞ ‖v‖Hm + ‖v‖L∞ ‖u‖Hm

. (2.6)

Proof For any multi-index α with |α| = m, using the Leibnitz differentiation formula and thenHolder’s inequality, we obtain

‖Dα(u v)‖L2 ≤ cα

∑

β≤α

‖DβuDα−βv‖L2 ≤ cα

∑

β≤α

‖Dβu‖L2m/|β| ‖Dα−βv‖L2m/|α−β| .

Using the Gagliardo-Nirenberg inequality with k = |β| and k = |α− β|, we obtain

‖Dα(u v)‖L2 ≤ C ·∑

β≤α

‖u‖1−|β|/mL∞ ‖u‖|β|/m

Hm ‖v‖1−|α−β|/mL∞ ‖v‖|α−β|/m

Hm

≤ C ·∑

β≤α

(‖u‖L∞ ‖v‖Hm

)|α−β|/m (‖v‖L∞ ‖u‖Hm

)|β|/m.

For convenience, here and in the sequel the numerical value of the constant C is allowed tochange 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‖Dα(u v)‖L2 ≤ C ·

‖u‖L∞ ‖v‖Hm + ‖v‖L∞ ‖u‖Hm

.

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, thereexists a constant c such that

‖uv‖Hm ≤ ‖u‖Hm ‖v‖Hm . (2.7)

Corollary 5 (A further inequality). For |α| ≤ m, one has

‖Dα(u v) − uDαv‖L2 ≤ c′α‖∇u‖L∞ ‖v‖Hm−1 + ‖v‖L∞ ‖∇u‖Hm−1

. (2.8)

Indeed,

‖Dα(u v)−uDαv‖L2 ≤ cα

∑

β≤α ,|β|>0

‖DβuDα−βv‖L2 ≤ cα

∑

|β′|+|α−β′|≤m−1

‖Dβ′(∇u)Dα−β′v‖L2 .

(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-Stokesequation

ut + (u · ∇)u = −∇p + ν∆u,

div u ≡ 0,

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

(2.10)

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 suchthat, as long as the norms ‖u(t)‖Hm and ‖∇u(t)‖L∞ remain bounded, one has

d

dt‖u(t)‖2

Hm ≤ cm‖∇u(t)‖L∞‖u(t)‖2Hm . (2.11)

Proof Applying the differential operator Dα to both sides of the Navier-Stokes equation(2.1) and then multiplying by Dαu, we obtain

DαuDαut + DαuDα((u · ∇)u

)= −DαuDα∇p + νDαu∆Dαu.

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| −→ ∞.

∫

RN

Dαu (u · ∇)Dαu dx = limR−→∞

∫

|x|=Rdiv

( |Dαu|22

u)

dx = 0.

The above yields

d

dt

∫

RN

|Dαu|22

dx +∫

RN

Dαu[Dα

((u · ∇)u

) − (u · ∇)Dαu]dx =

= −∫

RN

Dαu∇Dαp dx

︸ ︷︷ ︸=0

+ ν

∫

RN

Dαu∆Dαu dx

d

dt

∫

RN

|Dαu|22

dx + ν

∫

RN

|∇Dαu|2 = −∫

RN

Dαu[Dα

((u · ∇)u

)− (u · ∇)Dαu]dx (2.12)

Using Cauchy’s inequality, and then (2.8), we finally obtain

∫RN

∣∣∣Dαu[Dα

((u · ∇)u

) − (u · ∇)Dαu]∣∣∣ dx ≤ ‖Dαu‖L2‖Dα(u · ∇u) − uDα∇u‖L2

≤ cm ‖Dαu‖L2 ‖∇u‖L∞ ‖u‖Hm .(2.13)

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

Corollary 6 Consider a solution of (2.10), with initial data u0 ∈ Hm. If the map t 7→‖∇u(t)‖L∞ is integrable on [0, T ], then the norm ‖u(t)‖Hm of the solution remains uniformlybounded for t ∈ [0, T ].

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

z(t) = cm‖∇u(t)‖L∞ z(t), z(0) = ‖u(0)‖Hm .

By (2.11), a comparison argument yields

‖u(t)‖2Hm ≤ z(t) = exp

( ∫ t

0cm‖∇u(t)‖L∞ dt

)‖u(0)‖2

Hm .

¤

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

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

‖∇u‖L∞ ≤ C ‖u‖Hm .

Using this in (2.11) we obtain

d

dt‖u(t)‖2

Hm ≤ Cm‖u(t)‖3Hm , (2.14)

for a suitable constant Cm . Call t 7→ z(t) the solution to the O.D.E.

z = Cm z3/2, z(0) = ‖u0‖2Hm . (2.15)

Observe that z is well defined and remains bounded on a suitably small interval [0, T ]. By acomparison argument, from (2.14) we now obtain

‖u(t)‖2Hm ≤ 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 zerodivergence, in distribution sense, and call PE : L2 7→ E the perpendicular projection. By asolution of the initial value problem (2.10), we mean a Lipschitz continuous map u : [0, T ] 7→ Esuch that (u · ∇)u ∈ L2 and ∆u ∈ L2 for a.e. time τ ∈ [0, T ], and moreover

u(t) = u0 +∫ t

0

PE [(u(τ) · ∇)u(τ)] + ν∆u(τ)

dτ (2.16)

for every t ∈ [0, T ].

Theorem 5 (Existence of solutions to the Navier-Stokes equations, locally in time).Let the initial data satisfy u(0, ·) = u0 ∈ Hm, 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:

1. First, one constructs a family of approximate solutions (uε)ε>0, by a mollification procedure.Consider a smooth, radially symmetric function ρ = ρ(|x|) in C∞

0 (RN ) satisfying

ρ ≥ 0 , and∫

RN

ρ dx = 1 .

Define the family of mollifiersρε(x) .= ε−Nρ

(x

ε

)

and the associate family of mollification operators Jε

Jε u(x) .= ρε ∗ u(x) = ε−N

∫

RN

ρ(x− y

ε

)u(y) dy.

It is useful to recall here some basic properties of mollifiers:

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

‖Jεv‖L∞ ≤ ‖v‖L∞ and limε→0

Jεv(x) = v(x) (2.17)

uniformly on bounded sets.

(ii) Commuting with derivatives. If v ∈ Hm, then for every multi-index α with |α| ≤ mthere holds

Dα(Jεv) = Jε(Dαv) (2.18)

(iii) Commuting with integrals. For every u ∈ Lp, v ∈ Lq with p−1 + q−1 = 1, there holds∫

RN

(Jεu) v dx =∫

RN

u (Jεv) dx . (2.19)

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

limε→0

‖Jεv − v‖Hm = 0 , ‖Jεv − v‖Hm−1 ≤ Cε ‖v‖Hm . (2.20)

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

‖Jεv‖Hm+k ≤ Cmkε−k‖v‖Hm , ‖JεD

kv‖L∞ ≤ Ckε−(N/2+k) ‖v‖L2 . (2.21)

2. Consider the equation obtained by applying the ε-mollifier operator to Navier-Stokes equa-tion.

uεt + Jε

[(Jε uε) · ∇(Jε uε)

]= −∇pε + νJε(Jε ∆uε).

This mollification procedure transforms the unbounded differential operators into bounded op-erators on the Sobolev space Hm. Observing that the perpendicular projection operator PE

24

commutes with the derivation and with the mollification operators, we can write the previousequation as an O.D.E. in the Banach space E ⊂ L2, namely

d

dtuε = −PE Jε

[(Jε uε) · ∇(Jε uε)

]+ νJε(Jε ∆uε). (2.22)

The right hand side of this equation is Lipschitz continuous, because of the presence of themollifiers. Hence, given the initial data uε(0) = u0, for each ε > 0 this O.D.E. has a uniquesolution, which we denote as t 7→ uε(t).

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

d

dt‖uε(t)‖2

Hm ≤ C · ‖uε(t)‖3/2Hm

remains valid for approximate solutions of (2.22), with some constant C independent of ε. Hence,on a suitably small interval [0, T ], the norms ‖uε(t)‖Hm remain uniformly bounded. In turn,since we are assuming m > 2 + N/2, this yields an uniform a priori bounds on the norms‖uε(t)‖C2 , 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 > ε′ > 0. We need to estimate

d

dt

12‖uε − uε′‖2

L2 = A1 + A2 , (2.23)

whereA1

.=⟨PEJε′ [(Jε′u

ε′) · ∇(Jε′uε′)]− PEJε[(Jεu

ε) · ∇(Jεuε)] , uε − uε′

⟩,

A2.= ν

⟨J2

ε ∆uε − J2ε′∆uε′ , uε − uε′

⟩.

Here 〈·, ·〉 denotes the inner product in L2(RN ). Observing that uε, uε′ ∈ E and PE is aperpendicular projection, we can estimate

A1 =⟨(Jε′ − Jε)[(Jεu

ε) · ∇(Jεuε)] , uε − uε′

⟩

+⟨Jε′ [(Jε′ − Jε)uε · ∇(Jεu

ε)] , uε − uε′⟩

+⟨Jε′ [(Jε′(uε′ − uε) · ∇(Jεu

ε)] , uε − uε′⟩

+⟨Jε′ [(Jε′u

ε′) · ∇((Jε′ − Jε)uε)] , uε − uε′⟩

+⟨Jε′ [(Jε′u

ε′) · ∇(Jε′(uε′ − uε))] , uε − uε′⟩

= A11 + A12 + A13 + A14 + A15 .

We observe that A11, A12 and A14 contain the term Jε′ − Jε . Using the second inequality in(2.20), we obtain the bound

|A11|+ |A12|+ |A14| ≤ C(ε + ε′)(‖uε‖2

Hm + ‖uε′‖2Hm

)‖uε − uε′‖L2 .

Moreover, we have|A13| ≤ C‖uε‖Hm‖uε − uε′‖2

L2 .

25

Finally, an integration by parts yields

A15 =⟨(Jε′u

ε′) · ∇(Jε′(uε′ − uε)) , Jε′(uε− uε′)⟩

= −12

∫

RN

Jε′uε′ · ∇

(|Jε′(uε − uε′)|2

)dx = 0 .

The last term in (2.23) is estimated by

A2 =⟨(J2

ε − J2ε′)∆uε , uε − uε′

⟩+

⟨J2

ε′∆(uε − uε′) , uε − uε′⟩

≤ C(ε + ε′)‖uε‖H3‖uε − uε′‖L2 − ‖Jε′∇(uε − uε′)‖L2 .

Combining all the previous estimates, from (2.23) we obtain

d

dt‖uε − uε′‖L2 ≤ C

(1 + ‖uε‖Hm + ‖uε′‖Hm

)·[(ε + ε′) + ‖uε − uε′‖L2

].

By construction, at time t = 0 we have uε(0) = uε′(0). Moreover, by step 3 the Hm norms ofthe approximate solutions remain uniformly bounded, for t ∈ [0, T ]. This implies an estimate ofthe form

‖uε(t)− uε′(t)‖L2 ≤ (ε + ε′) eKt 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

limε→0

‖uε(t)− u(t)‖L2 = 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 followinginterpolation inequality holds:

‖v‖Hk ≤ Ck‖v‖1−k/mL2 ‖v‖k/m

Hm . (2.24)

An application of this inequality yields

limε→0

‖uε − u‖Hk ≤ limε→0

Ck‖uε − u‖1−k/mL2 ‖uε − u‖k/m

Hm = 0.

In particular, if m > 2 + N/2, we have the convergence

‖(uε · ∇)uε − (u · ∇)u‖L2 → 0 , ‖∆uε −∆u‖L2 → 0 ,

uniformly for t ∈ [0, T ]. From the relations

uε(t) = u0 −∫ t

0PEJε [(Jεu

ε(τ)) · ∇(Jεuε(τ))] dτ + ν

∫ t

0J2

ε ∆uε(τ) dτ ,

letting ε → 0 and using the convergence ‖uε − u‖Hm → 0, one checks that the limit function usatisfies (2.16). ¤

Remarks. By choosing even larger exponent m, one can prove the local existence of classicalsolutions with Ck regularity, for any given k ≥ 1. One should also note that the above proof didnot use in any way the regularization effect produced by the diffusion term ν∆u. Indeed, theprevious 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 ofglobally defined distributional solutions, on a bounded domain Ω ⊂ R3 with smooth boundary.

Consider the equations

ut + (u · ∇)u = −∇p + ν∆u,

divu ≡ 0,

u(t, x) = 0, for all x ∈ ∂Ω ,

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

(2.25)

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

Definition 4 A distributional solution u of the Navier-Stokes equation (2.25) is characterizedby the following properties:

• The energy of u has to be finite∫

R3

|u(x, t)|2 dx ≤ E0, for a.e. t ∈ [0, T ] . (2.26)

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

0

∫

R3

|∇u(x, t)|2 dx dt ≤ E0 . (2.27)

• The function u = u(t, x) satisfies the evolution equation and the incompressibility conditionin distributional sense. Namely, for any ψ ∈ C∞0 (R3) and for any φ ∈ C∞0

((−∞, T ]×Ω; R3

)we have

∫

R3

u∇ψ dx = 0 for a.e. t ∈ [0, T ],∫ T

0

∫

R3

u[φt + ν∆φ]− (u · ∇u)φ + p divφ

dx dt +

∫

R3

u0(x)φ(0, x) dx = 0(2.28)

The proof of existence of a solution will be given in three steps. First, we derive some uniforma priori estimates on exact solutions to the Navier-Stokes equations. In a second step, by aGalerkin method, we construct a sequence of approximate solutions satisfying the same a prioriestimates. In turn, these estimates allow us to use a compactness argument, and obtain asubsequence 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:

• u ∈ L∞([0, T ],L2(Ω)

), i.e. there exists a constant E0 such that∫

Ω|u(t, x)|2 dx ≤ 2E0 for a.e. t ∈ [0, T ] . (2.29)

27

• u ∈ L2([0, T ],H1

0(Ω)), i.e.

∫ T

0

∫

Ω|∇u(t, x)|2 dx dt ≤ 2

νE0 (2.30)

• du/dt ∈ L4/3([0, T ],H−1

), i.e. there exists a constant C2 such that

∫ T

0‖ut(t)‖4/3

H−1 dt ≤ C2 . (2.31)

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

〈ut, u〉+ 〈u, u · ∇u〉 − ν〈u,∆u〉+ 〈u,∇p〉 = 0 .

By integrating over Ω we obtain

d

dt

∫

Ω

|u|22

dx +∫

Ω

3∑

i,j=1

uiujuixj

dx +∫

Ωu · ∇p dx + ν

∫

Ω|∇u|2 dx = 0

d

dt

∫

Ω

|u|22

dx +∫

Ωdiv

( |u|22· u

)dx +

∫

Ωdiv(u · p) dx + ν

∫

Ω|∇u|2 dx = 0

d

dt

∫

Ω

|u|22

dx + ν

∫

Ω|∇u|2dx = 0

(2.32)

The last identity shows that the L2 norm of the solution u decreases in time. Indeed,∫

Ω|u(t, x)|2 dx ≤

∫

Ω|u0(x)|2 dx, for all t ≥ 0.

Moreover

ν

∫ T

0

∫

Ω|∇u(t, x)|2dx dt ≤

∫

Ω

|u0(x)|22

dx −∫

Ω

|u(T, x)|22

dx.

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)

where PE denotes the perpendicular projection on on the subspace E ⊂ L2(Ω; R3) consisting ofdivergence free vector fields.

The H−1 norm of (2.33) can be estimated as follows:

‖ν∆u‖H−1 = sup‖w‖

H10≤1, div w≡0

∫

Ω−ν∇u · ∇w dx ≤ ν‖u‖H1

0,

‖u · ∇u‖H−1.= sup‖w‖

H10≤1, div w≡0

∫

Ω(u · ∇u)w dx .

28

Observing that 13 + 1

2 + 16 = 1, using a generalized Holder inequality and a Sobolev embedding

theorem we obtain:

∣∣∣∫

Ω(u · ∇u) w dx

∣∣∣ ≤∫

Ω

∣∣3∑

i,j=1

uiujxi

wj∣∣ dx

≤ K1‖u‖L3‖∇u‖L2‖w‖L6 ≤ K2‖u‖L3‖∇u‖L2 .

(2.34)

By Sobolev embedding and interpolation there holds

‖u‖L3 ≤ C‖u‖H1/2 ≤ C‖u‖1/2H0 ‖u‖1/2

H1 .

Hence‖u · ∇u‖H−1 ≤ C‖u‖1/2

L2 ‖∇u‖3/2L2

∫ T

0

∥∥∥du

dt

∥∥∥4/3

H−1dt ≤ C

∫ T

0

(‖u‖4/3

H10

+(‖u‖1/2

L2 + ‖∇u‖3/2L2

)4/3)

dt

≤ C(1 +

∫ T

0‖∇u‖2

L2 dt)≤ C1

(2.35)

¤

2.5 Construction of finite dimensional approximate solutions

In this section we establish the following global existence theorem

Theorem 6 Let Ω ⊂ R3 be a bounded domain with smooth boundary. Let u0 ∈ L2(Ω) be avector field with zero divergence. Then the equations (2.25) admit a weak solution, defined forall times t ≥ 0.

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

Au = −∆u,

and the eigenvalue problem Au = λu x ∈ Ω ,u = 0 x ∈ ∂Ω .

Since the operator A is self-adjoint, there exists an orthonormal set w1, w2, w3 . . . ⊂ L2(Ω)consisting of eigenfunctions of A. We will denote by Em ⊂ E the subspace generated by thefirst m eigenfunctions:

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 ddtum + νAum + Pm(u · ∇)u = 0,

um(0, x) = Pmu0(x).(2.36)

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:

um(t, x) =m∑

i=1

ξi(t)wi(x),

where ξi(t).= 〈um(t), wi〉L2 . By construction, one has

∆um(t, x) = −m∑

i=1

λjξj(t)wj(x),

and the problem (2.36) can be written as a system of m ODEs for the coefficients ξ1, . . . , ξm.

d

dtξi = −νλiξi +

m∑

j,k=1

bijkξjξk , (2.37)

The constant coefficients bijk are here computed as

bijk.= −

∫

Ω[(wj · ∇)wk]wi dx = −〈(wj · ∇)wk , wi〉2L .

Since the right hand sides of (2.37) are quadratic polynomials, for any given initial data theCauchy problem has a unique solution defined locally in time. To prove that this solution existsglobally, we need to show that its norm remains bounded.

Toward this goal, observe that the coefficients bijk satisfy

bijk + bikj =∫

Ω∇(wj · wk) · wi dx =

∫

Ωdiv

((wj · wk)wi

)dx = 0.

Therefore

d

dt

|ξ(t)|22

=m∑

i,j,k=1

bijkξiξjξk − νm∑

i=1

λiξ2i (t) = − ν

m∑

i=1

λiξ2i (t) < 0. (2.38)

Indeed, all the eigenvalues λi of the operator A = −∆ are strictly positive. Hence for all m ≥ 1we have a well defined approximate solution um : [0, T ] 7→ Em .

Repeating the arguments in the proof of Proposition 5, one can show that

• um ∈ L∞([0, T ]; L2(Ω)

),

• um ∈ L2([0, T ]; H1

0(Ω)),

• (um)t ∈ L4/3([0, T ],H−1(Ω)

).

Moreover, the norms of the approximate solutions um in these various function spaces remainuniformly bounded as m → ∞. Using a compactness lemma, one can extract a subsequence,still called (um)m≥1 , and a limit function u such that

• um u weakly in L2([0, T ]; H1

0(Ω)),

30

• (um)t ut weakly in L4/3([0, T ],H−1(Ω)

),

• um u weakly∗ in L∞([0, T ]; L2(Ω)

),

In turn, by extracting a further subsequence, we can achieve the further convergence

• um → u strongly in L2([0, T ]; L2(Ω)

),

• um → u in C0([0, T ],H−1(Ω)

).

Observe that the uniform bound on the norm of ‖um‖L4/3([0,T ];H−1) implies that the mapst 7→ um(t) ∈ H−1 are uniformly Holder continuous of exponent 1/4. Indeed,

‖u(t2)−u(t1)‖H−1 ≤∫ t2

t1

1·‖ut‖H−1 dt ≤ ‖ut‖L4/3([t1,t2];H−1) ·‖1‖L1/4([t1,t2]) ≤ ‖ut‖L4/3 ·|t2−t1|1/4.

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

limt→0+

‖u(t)− u0‖H−1 = 0 .

Hence the limit function u takes the correct initial data.Since every function um maps [0, T ] into the closed subspace E ⊂ H1

0 consisting of divergence-free vector fields, the same holds for u. It now remains to show that the limit function u satisfiesthe Navier-Stokes equation in a weak sense.

Fix an integer k ≥ 1 and consider a smooth test function φ : [0, T ] 7→ Ek with compactsupport, so that φ(0) = φ(T ) = 0 ∈ Ek . Then for all m ≥ k we have

∫ T

0

∫

Ω(−φt− ν∆φ)um + [(um · ∇)um] · φdxdt = −

∫

Ω(φt + ν∆φ)um + [(um · ∇)φ] · um dxdt = 0 .

Taking the limit as m →∞, this yields∫

Ω(φt + ν∆φ)u + [(u · ∇)φ] · u dxdt = 0 .

Observing that, as k → ∞, the set of test functions of the form φ : [0, T ] 7→ Ek becomes densein 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 lineardensity φ.

Let s 7→ γ(s) be a parametrization of the curve γ. Given s1 < s2, the total vorticity on thearc between γ(s1) and γ(s2) is thus given by the integral

∫ s2

s1

φ(s) ds.

If the vorticity is located only along the curve γ, and vanishes outside γ, then the velocity ofthe fluid at a point x ∈ R2 is computed by

u(x) =∫

K(x− γ(s))φ(s) ds.

Here K denotes the two-dimensional Biot-Savart kernel, i.e.

K(x) =1

2π |x|2( −x2

x1

). (3.1)

To determine how the vorticity changes in time, we should now determine how points on thecurve γ move. If x is a point on the curve, i.e. x = γ(s0) for some s0, then the above formulayields

u(γ(s0)) =∫

K(γ(s0)− γ(s))φ(s) ds . (3.2)

However, since K(γ(s0)−γ(s)) = O(1)·|s0−s|−1, 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

a principal value. Indeed, assume that γ′(s0) 6= 0. Then γ can be written in the following way

γ(s) = γ(s0) + γ′(s)(s− s0),

whereγ′(s) =

1s− s0

∫ s

s0

γ′(σ) dσ.

32

Similarly, we can writeφ(s) = φ(s0) + φ′(s)(s− s0).

We now compute∫ s0+δ

s0−δK(γ(s0)− γ(s))φ(s) ds

=∫ s0+δ

s0−δK(γ′(s)(s− s0)) · [φ(s0) + φ′(s)(s− s0)] ds

=∫ s0+δ

s0−δK(γ′(s0)(s− s0))φ(s0) ds

+∫ s0+δ

s0−δ

[K(γ′(s)(s− s0))−K(γ′(s0)(s− s0))

]φ(s0) ds

+∫ s0+δ

s0−δK(γ′(s)(s− s0))φ′(s)(s− s0) ds.

(3.3)

Of the three integrals on the right hand side of (3.3), the first one vanishes, while the remainingtwo are absolutely integrable. Indeed,

K(γ′(s)(s− s0)) = O(1) · |s− s0|−1,

|γ′(s)− γ′(s0)|(s− s0) = O(1)(s− s0)2,

henceK(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 acrossγ. To simplify computations, we assume here ‖γ′(s)| ≡ 1, so that the curve is parametrized byarc length.

Call n be the unit normal vector, perpendicular to the curve. The variation of velocitybetween a point on the curve γ(s0) and a point near the curve γ(s0)+εn is given by the integral

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 theform [s0− δ, s0 + δ]. Indeed, outside this interval the integrand in (3.4) is a continuous functionof ε and goes to 0 as ε −→ 0. Notice that, as s → s0, we have

γ′(s) → γ′(s0) , φ(s) → φ(s0) .

From (3.4) we obtain

limε→0+

[u(γ(s0) + εn)− u(γ(s0))] = limε→0

∫ s0+δ

s0−δ

[K(γ′(s)(s− s0) + εn)−K(γ′(s)(s− s0))

]φ(s) ds

= limε→0

∫ s0+δ

s0−δ

[K(γ′(s0)(s− s0) + εn)−K(γ′(s0)(s− s0))

]φ(s0) ds .

33

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

limε→0+

[u(γ(s0) + εn)− u(γ(s0))] = limε−→0

∫ +∞

−∞

[K(γ′(s0)εξ + εn)−K(γ′(s0)εξ)

]φ(s0)ε dξ.

To compute this integral, we fix the orthonormal basis

e1 = γ′(s0) , e2 = n

and write the Biot-Savart kernel K coordinate-wise w.r.t. this basis. This yields

limε→0+

[u(γ(s0) + εn)− u(γ(s0))]

= limε−→0

φ(s0)2π

∫ +∞

−∞

( −ε

|ε|2(1 + ξ2),

εξ

|ε|2(1 + ξ2)

)−

(0 ,

εξ

|ε|2(1 + ξ2)

)ε dξ

=φ(s0)2π

∫ +∞

−∞

(− 1

1 + ξ2, 0

)dξ =

(−φ(s0)

2, 0

).

This shows that the jump of the tangential component of the velocity across a vortex sheet isequal 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 canreparametrize γ by means of the vorticity. More precisely, fix a value s0. We then define thenew parametrization

σ 7→ z(σ) .= γ(s(σ)) ∈ R2 ,

where the change of variable σ 7→ s(σ) is implicitly defined by the relation

σ =∫ s(σ)

s0

φ(s) ds.

Observe that the new parametrization has the property that the amount of vorticity containedin 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

d

dtz(t, σ) =

∫ +∞

−∞K

(z(σ)− z(σ′)

)dσ′ =

( 12πi

∫ +∞

−∞

dσ′

z(σ)− z(σ′)

).

Here the upper bar denotes conjugation. Equivalently, this can be written as

d

dtz(t, σ) =

12πi

∫ +∞

−∞

dσ′

z(σ)− z(σ′), (3.5)

where the left hand side is the time derivative of the conjugate of z. We remark that all theseintegrals 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. Inthis case the curve γ is precisely the real axis. We wish to understand the stability of thissolution w.r.t. small perturbations. Consider the perturbed function

z(t, σ) = σ + εg(t, σ).

This should satisfy

d

dtz(t, σ) =

d

dtεg(t, σ) =

12πi

P.V.

∫ +∞

−∞

dσ′

(σ − σ′) + ε(g(σ)− g(σ′)).

Performing the expansion

1a + εb

=1a

(1

1 + εb/a

)=

1a

(1− ε

b

a+ o(ε2)

).

inside the integral, and letting ε → 0, we obtain

d

dtg(t, σ) = − 1

2πiP.V.

∫ +∞

−∞

g(σ)− g(σ′)(σ − σ′)2

dσ′ =1

2πilim

δ→0+

(∫ −δ

−∞+

∫ +∞

δ

)g(σ)− g(σ′)

(σ − σ′)2dσ′

Integrating by parts and observing that the boundary terms cancel each other as δ → 0+, weobtain

d

dtg(t, σ) = − 1

2πiP.V.

∫ +∞

−∞

g′(σ′)σ − σ′

dσ′ =12Hg′. (3.6)

Here Hg′ denotes the Hilbert transform of g′. We recall that the Hilbert transform of afunction f is defined as the principal value

Hf(x) .=1πi

P.V.

∫ +∞

−∞

f(y)y − x

dy.

Using the Fourier transform, f can be written as

f(x) =∫

eiξxf(ξ)dξ .

Setting w = y − x we compute

Heiξx =1πi

P.V.

∫ +∞

−∞

eiξ(x−y)

ydy = eiξx · 1

πilim

δ→0+

∫ ∞

δ

eiξw − e−iξw

wdw

= eiξx 1πi

∫ ∞

0

2i sin ξw

wdw = eiξxsign ξ · 2

π

∫ ∞

0

sinw

wdw

= eiξx sign ξ .

ThereforeHf(x) .=

∫

ξ>0eiξxf(ξ)dσ −

∫

ξ<0eiξxf(ξ)dξ.

35

Using a Fourier transform, we can now write

g(t, x) =∫ +∞

0A(t, ξ)eiξxdξ +

∫ +∞

0B(t, ξ)e−iξxdξ,

where A(t, ξ) = g(t, ξ), B(t, ξ) = g(t,−ξ), ξ > 0. With the above notations, we have

Hg′(t, x) =∫ +∞

0A(t, ξ)iξeiξxdξ +

∫ +∞

0B(t, ξ)iξe−iξxdξ.

Recalling (3.6), the time evolution of A(t, ξ) and B(t, ξ) is described by the equation

ddtA(t, ξ) = − iξ

2 B(t, ξ),

ddtB(t, ξ) = − iξ

2 A(t, ξ).(3.7)

Notice that (3.7) implies

d2

dt2A(t, ξ) =

ξ2

4A(t, ξ),

d2

dt2B(t, ξ) =

ξ2

4B(t, ξ).

The general solution of (3.7) has the form

A(t, ξ) = A1(ξ)eξt//2 + A2(ξ)e−ξt/2,

B(t, ξ) = B1(ξ)eξt/2 + B2(ξ)e−ξt/2.(3.8)

It is now clear that the coefficient of the Fourier mode eiξσ can grow like e|ξ|t/2. Thereforethe solution is very unstable. In order to achieve at least the local existence of solutions, onehas to start with initial data whose Fourier transform decreases very fast as |ξ| → ∞, namelyanalytic 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 problemusing a rescaling symmetry. We recall that, if u solves the equation

ut + PE(u · ∇u) = 0 , (3.9)

thenuλ,τ (t, x) =

λ

τu( t

τ,x

λ

),

is also a solution, for all λ, τ > 0. Here PE the perpendicular projection on the space E ⊂ L2

of divergence free vector fields.Given a solution u of (3.9), and a constant k, we will call (uτ )τ>0 the 1-parameter family of

solutions defined by setting λ = τk, namely

uτ (t, x) = τk−1u( t

τ,

x

τk

).

36

Suppose that u is invariant with respect to the transformation τ 7→ uτ for all τ > 0. Then

u(t, x) = ut(t, x) = tk−1u(1,

x

tk

).

In other words, knowing the velocity u at time t = 1, the entire solution is determined for alltimes 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 byγ(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

σ(s) =∫ s

0aσβ dσ =

a

β + 1sβ+1.

In terms of the rescaled parameter σ, we have

z(σ) =[β + 1

aσ

] 1β+1

. (3.10)

Next, consider the Birkhoff-Rott equation

d

dtz(t, σ) =

12πi

∫dσ′

z(σ)− z(σ′), (3.11)

with initial data where the vorticity is concentrated along the real axis, with density as in (3.10),with β = −1/2. Namely

z0(σ) =1

4ξ2· σ2 sign (σ).

Here ξ > 0 is an arbitrary constant. Introduce the variable

τ =σ

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

dtz(t, x) =

d

dt

[(ξt)2/3ρ

( σ

ξ4/3t1/3

)]

23ξ2/3t−1/3ρ + (ξt)2/3 · dρ

dσ·(−1

3

)σ

ξ4/3t4/3

=23ξ2/3t−1/3ρ − 1

3ξ2/3t−1/3 τ

dρ

dτ.

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Next, the Birkhoff-Rott equation becomes

d

dtz(t, x) =

12πi

∫1

(ξt)2/3· dσ′

ρ( σξ4/3t1/3 )− ρ( σ′

ξ4/3t1/3 )

=1

2πi

∫dτ ′

ρ(τ)− ρ(τ ′)· ξ2/3t−1/3.

(3.13)

Comparing (3.13) with (3.12) we obtain

23ρ− τ

3d

dτρ =

12πi

∫dτ ′

ρ(τ)− ρ(τ ′),

henced

dτρ(τ) =

1τ

(2ρ(τ) +

12πi

∫dτ ′

ρ(τ)− ρ(τ ′)

).

A solution of this equation is the Kaden spiral.

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