the three-dimensional euler equations: recent advances ... · in the two-dimensional case the euler...

The Three-dimensional Euler Equations:Recent Advances Through Examples

Edriss S. Titi

Weizmann Institute of Science andThe University of California - Irvine

Geometry and Fluids - Clay Mathematics Institute, OxfordApril 7–11, 2014

Joint work with Claude Bardos

Claude Bardos & Edriss S. Titi Euler Equations: Recent Advances

Overview

1 BackgroundEuler equations

Classical results of local well-posednessNon-uniqueness of weak solutions in 3d Euler equations

2 Shear FlowA shear-flow example of DiPerna-MajdaIll-posedness of 3d Euler equations in C0,α

Vanishing viscosity limit as a ruling out principle - Theshear-flow and symmetric flows as examples

3 Vortex sheets induced by three-dimensional shear flowsExamples for non-smooth vortex sheets in R3

The 2d verses the 3d Kelvin-Helmholtz (Birkhoff-Rott)problems

4 Numerical investigation of formation of singularityThe effect of hyper-viscosityThe effect of boundary conditionsDeos the advection term deplete singularity?

Overview

1 BackgroundEuler equationsClassical results of local well-posedness

Non-uniqueness of weak solutions in 3d Euler equations

Overview

1 BackgroundEuler equationsClassical results of local well-posednessNon-uniqueness of weak solutions in 3d Euler equations

Overview

2 Shear FlowA shear-flow example of DiPerna-Majda

Ill-posedness of 3d Euler equations in C0,α

Overview

3 Vortex sheets induced by three-dimensional shear flows

Examples for non-smooth vortex sheets in R3

Overview

4 Numerical investigation of formation of singularity

The effect of hyper-viscosityThe effect of boundary conditionsDeos the advection term deplete singularity?

Overview

4 Numerical investigation of formation of singularityThe effect of hyper-viscosity

The effect of boundary conditionsDeos the advection term deplete singularity?

Overview

4 Numerical investigation of formation of singularityThe effect of hyper-viscosityThe effect of boundary conditions

Deos the advection term deplete singularity?

Overview

Euler equations

We consider the Euler Equations of inviscid incompressiblefluid in Ω = R3, the whole space, or Ω = (R/Z)3, the threedimensional torus.

ut + (u · ∇)u +∇p = 0, x ∈ Ω, t > 0

∇ · u = 0,

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

where the velocity field u = (u1, u2, u3) and pressure p areunknowns.

Vorticity formulation

Denoting by ω = ∇× u, the voriticity.

The Euler equations areequivalent to evolution equation of the vorticity:

∂tω + u · ∇ω = ω · ∇u,

the velocity can be recovered from the vorticity via theBiot-Savart law in R3

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

Denoting by ω = ∇× u, the voriticity. The Euler equations areequivalent to evolution equation of the vorticity:

∂tω + u · ∇ω = ω · ∇u,

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

Denoting by ω = ∇× u, the voriticity. The Euler equations areequivalent to evolution equation of the vorticity:

∂tω + u · ∇ω = ω · ∇u,

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

Classical well-posedness results

In the two-dimensional case the Euler equations haveglobal existence and uniqueness for initial data ω0 ∈ L∞.

This result is due to Yudovich (1963).

For initial data in the space C1,α the Euler equations arewell-posed for short time in C1,α. Moreover, the solutionconserves energy.

This result was originally proved by Lichtenstein (1925).

The same result holds the context of the Sobolev spacesHs, s > 5

2 (Ebin-Marsden, Kato-Lai, Temam).

Question: Does there exist a regular solution (say in C1,α)of the 3d Euler equations that becomes singular in a finitetime (blows up problem)?

A trivial blowup criterion

For any transport equation∂θ

∂t+ u · ∇θ = 0

The solution remains regular over [0,T], provided∫ T

0‖∇u(·, t)‖L∞dt <∞.

Recall the vorticity dynamics for the three-dimensionalEuler equations:

∂tω + u · ∇ω = ω · ∇u,

Trivial TheoremThe solution for the three-dimensional Euler equations remainsregular over [0,T], iff

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

∂t+ u · ∇θ = 0

The solution remains regular over [0,T], provided

0‖∇u(·, t)‖L∞dt <∞.

∂tω + u · ∇ω = ω · ∇u,

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

∂t+ u · ∇θ = 0

0‖∇u(·, t)‖L∞dt <∞.

∂tω + u · ∇ω = ω · ∇u,

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

∂t+ u · ∇θ = 0

0‖∇u(·, t)‖L∞dt <∞.

∂tω + u · ∇ω = ω · ∇u,

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

∂t+ u · ∇θ = 0

0‖∇u(·, t)‖L∞dt <∞.

∂tω + u · ∇ω = ω · ∇u,

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

∂t+ u · ∇θ = 0

0‖∇u(·, t)‖L∞dt <∞.

∂tω + u · ∇ω = ω · ∇u,

∫ T0 ‖∇u(·, t)‖L∞dt <∞.

The Beal-Kato-Majda blowup criterion

Using the Calderon-Zygmund singular integral theory forBiot-Savart law in R3

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

A more sophisticated criterion was established:

Theorem Beal-Kato-MajdaThe solution for the three-dimensional Euler equations remainsregular over [0,T], iff∫ T

0‖ω(·, t)‖L∞dt <∞.

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

0‖ω(·, t)‖L∞dt <∞.

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

0‖ω(·, t)‖L∞dt <∞.

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

Theorem Beal-Kato-MajdaThe solution for the three-dimensional Euler equations remainsregular over [0,T], iff

0‖ω(·, t)‖L∞dt <∞.

u(x, t) =1

(x− y) ∧ ω(y)

|x− y|3dy .

0‖ω(·, t)‖L∞dt <∞.

Constantin-Fefferman-Majda Criterion

Instead of looking at the maximum of the amplitude of thevorticity Constantin-Fefferman-Majda established anothercriterion in terms of the vorticity direction

ξ =ω

Constantin-Fefferman-Majda Criterion

Instead of looking at the maximum of the amplitude of thevorticity Constantin-Fefferman-Majda established anothercriterion in terms of the vorticity direction

ξ =ω

New Blow up Criterion in Bounded Domain

Consider the Euler equations in a smooth bounded domain Ωwith boundary condition u ·~n|∂Ω = 0.

Let θ, q and B be defined as follows:

∂t+ u · ∇θ = 0, q = ω · ∇θ, and B = ∇q×∇θ.

Theorem (Gibbon-Titi [J. Nonlinear Science 2013])

Let B(x, 0) ∈ L∞(Ω) satisfying |B(x, 0)| > 0, and suppose thatu(x, t) is a smooth solution of the 3d Euler equations on theinterval [0,T]. Then |B(x, t)| > 0, for all t ∈ [0,T].

Consider the Euler equations in a smooth bounded domain Ωwith boundary condition u ·~n|∂Ω = 0.Let θ, q and B be defined as follows:

∂t+ u · ∇θ = 0, q = ω · ∇θ, and B = ∇q×∇θ.

Let B(x, 0) ∈ L∞(Ω) satisfying |B(x, 0)| > 0,

and suppose thatu(x, t) is a smooth solution of the 3d Euler equations on theinterval [0,T]. Then |B(x, t)| > 0, for all t ∈ [0,T].

∂t+ u · ∇θ = 0, q = ω · ∇θ, and B = ∇q×∇θ.

Let B(x, 0) ∈ L∞(Ω) satisfying |B(x, 0)| > 0, and suppose thatu(x, t) is a smooth solution of the 3d Euler equations on theinterval [0,T].

Then |B(x, t)| > 0, for all t ∈ [0,T].

∂t+ u · ∇θ = 0, q = ω · ∇θ, and B = ∇q×∇θ.

Lack of uniqueness of weak solutions

Theorem DeLellis - Szekelyhidi

There exist a set of initial data u0 ∈ L2(Ω) (not explicitlyconstructed) for which the Cauchy problem has, for the sameinitial data, an infinite family of weak solutions of thethree-dimensional Euler equations: a residual set in the spaceC(Rt; L2

weak(Ω)) .

Remark: Earlier results were established by Shnirelmanand by V. Sheffer.

Lack of uniqueness of weak solutions

Theorem DeLellis - Szekelyhidi

There exist a set of initial data u0 ∈ L2(Ω) (not explicitlyconstructed) for which the Cauchy problem has, for the sameinitial data, an infinite family of weak solutions of thethree-dimensional Euler equations: a residual set in the spaceC(Rt; L2

weak(Ω)) .

Remark: Earlier results were established by Shnirelmanand by V. Sheffer.

Existence of Weak solutions to the Cauchy Problem

Theorem Wiedemann (2011).There exists a family (non-uniqueness) of weak solutions to theCauchy problem of the 3D Euler.

Ruling out Principle of wild weak solutions

Ruling out PrincipleAny wild weak solution of Euler equations that cannot beachieved as a vanishing viscosity limit of the Navier-Stokesequations should be ruled out.

Does 2D Flow Remain 2D?

Theorem Bardos, Lopes-Filho,Nussenzveig-Lopes, Niu, Titi–[SIAM, Jour. Math. Analysis (2013)].

Let u0 be a function of (x, y), then the Leray-Hopf weak solutionof the 3D Navier-Stokes remains a function of only (x, y).

Similar result for axi-symmetric initial data, or helical initial data.

Let u0 be a function of (x, y) only, then the weak solution of the3D Euler might become a function of (x, y, z).Also, if the initialdata is axi-symmetric or helical symmetric, the weak solutionsof Euler might break the symmetry.

Remark: Ruling out principle: all the wild weak solutions ofEuler equations that do not obey the two-dimensionalsymmetry of the initial data should be ruled out. Becausethey cannot be obtained as vanishing viscosity limit ofNavier-Stokes solutions.

Let u0 be a function of (x, y), then the Leray-Hopf weak solutionof the 3D Navier-Stokes remains a function of only (x, y).Similar result for axi-symmetric initial data, or helical initial data.

Let u0 be a function of (x, y) only,

then the weak solution of the3D Euler might become a function of (x, y, z).Also, if the initialdata is axi-symmetric or helical symmetric, the weak solutionsof Euler might break the symmetry.

Let u0 be a function of (x, y) only, then the weak solution of the3D Euler

might become a function of (x, y, z).Also, if the initialdata is axi-symmetric or helical symmetric, the weak solutionsof Euler might break the symmetry.

Let u0 be a function of (x, y) only, then the weak solution of the3D Euler might become a function of (x, y, z).

Also, if the initialdata is axi-symmetric or helical symmetric, the weak solutionsof Euler might break the symmetry.

Let u0 be a function of (x, y) only, then the weak solution of the3D Euler might become a function of (x, y, z).Also, if the initialdata is axi-symmetric or helical symmetric,

the weak solutionsof Euler might break the symmetry.

Shear Flow

We consider in the whole space R3 or on the periodic box(R/Z)3 :

The shear flow

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2)).

For u1, u3 ∈ C1, the above shear flow is a classical solutionof the Euler equations with pressure p = 0.Moreover, in the case of the periodic box (R/Z)3 the aboveshear flow conserves the energy.

Shear Flow

The shear flow

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2)).

For u1, u3 ∈ C1, the above shear flow is a classical solutionof the Euler equations with pressure p = 0.

Moreover, in the case of the periodic box (R/Z)3 the aboveshear flow conserves the energy.

Shear Flow

The shear flow

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2)).

For u1, u3 ∈ C1, the above shear flow is a classical solutionof the Euler equations with pressure p = 0.Moreover, in the case of the periodic box (R/Z)3 the aboveshear flow conserves the energy.

Some history

This example of shear flow was used by DiPerna andMajda (1987) to show that a weak limit of oscillatingsmooth solutions of the three-dimensional Euler equationsmay not be a solution of Euler equations.

Theorem DiPerna-Lions For every p ≥ 1, T > 0 and M > 0there exists a smooth shear flow solution for which‖u(·, 0)‖W1,p = 1 and ‖u(·,T)‖W1,p > M.

Idea of the proof

∂x2u3(x1 − tu1(x2)) = −t∂x2u1(x2)∂x1u3(x1 − tu1(x2))

If ∂x2u1, ∂x1u3 ∈ Lp this does not imply that ∂x2u3 ∈ Lp.

Some history

This example of shear flow was used by DiPerna andMajda (1987) to show that a weak limit of oscillatingsmooth solutions of the three-dimensional Euler equationsmay not be a solution of Euler equations.Theorem DiPerna-Lions For every p ≥ 1, T > 0 and M > 0there exists a smooth shear flow solution for which‖u(·, 0)‖W1,p = 1 and ‖u(·,T)‖W1,p > M.

Idea of the proof

Some history

Idea of the proof

Some history

Idea of the proof

Some history

Idea of the proof

If ∂x2u1, ∂x1u3 ∈ Lp

this does not imply that ∂x2u3 ∈ Lp.

Some history

Idea of the proof

Weak Limit of Oscillating Initial Data

Original DiPerna–Majda example: Sequence of weak solutionswith energy estimate:

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

RT(uε)(x, t) = limε→0

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

∇ · u = 0 in Ω , u ·~n = 0 on ∂Ω ,

∂tu +∇ · (u⊗ u) +∇ · RT(uε) +∇p = 0 in Ω ,

((uε − u)⊗ (uε − u))

uε(x, t) = (u1(x2

ε), 0, u3(x1 − tu1(

0u1(s)ds = 0

limε→0

weak uε = (0, 0, u3) , u3 =

0u3(x1 − tu1(s))ds

∂tu3 + limε→0∇ · (uεuε3) = 0

limε→0∇ · (uεuε3) = ∂x1

0u1(s)u3(x1 − tu1(s))ds 6= ∂x1(u1u3) = 0

Shear flow is a weak solution that conserves energy

Theorem Bardos-Titi

(i) Let u1, u3 ∈ L2loc(R) then the shear flow

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2))

is a weak solution of the Euler equations, in the sense ofdistribution, in Ω = R3.

(ii) Let u1, u3 ∈ L2(R/Z) then the shear flow defined above is aweak solution of the Euler equations, in the sense ofdistribtions, in Ω = (R/Z)3. Furthermore, in this case theenergy of this solution is constant.

Shear flow is a weak solution that conserves energy

Theorem Bardos-Titi

(i) Let u1, u3 ∈ L2loc(R) then the shear flow

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2))

is a weak solution of the Euler equations, in the sense ofdistribution, in Ω = R3.(ii) Let u1, u3 ∈ L2(R/Z) then the shear flow defined above is aweak solution of the Euler equations, in the sense ofdistribtions, in Ω = (R/Z)3. Furthermore, in this case theenergy of this solution is constant.

Ill-posedness of Euler equations in C0,α

Theorem (i) For u1(x), u3(x) ∈ C1,α, with α ∈ (0, 1], the shearflow solution

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2))

of the three-dimensional Euler equations is in C1,α, for all t ∈ R.

(ii) For u1(x), u3(x) ∈ C0,α, with α ∈ (0, 1), the above shear flowsolution of the three-dimensional Euler equations is always inC0,α2

(iii) There exist shear flow solutions of the above form which, fort = 0, belong to C0,α, for some α ∈ (0, 1), and for t 6= 0, they donot belong to C0,β for any β > α2.

u(x, t) = (u1(x2), 0, u3(x1 − tu1(x2))

Proof of the Theorem

Proof The regularity results concern only the u3 component:

|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|hα2

=|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|

|tu1(x2 + h)− tu1(x2)|α

(|tu1(x2 + h)− tu1(x2)|

)α≤ |t|α||u3||0,α(||u1||0,α)α .

Which proves that the solution belongs to C0,α2.

Next we show that this result is sharp.

|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|hα2

=|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|

|tu1(x2 + h)− tu1(x2)|α

(|tu1(x2 + h)− tu1(x2)|

≤ |t|α||u3||0,α(||u1||0,α)α .

|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|hα2

=|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|

|tu1(x2 + h)− tu1(x2)|α

(|tu1(x2 + h)− tu1(x2)|

)α≤ |t|α||u3||0,α(||u1||0,α)α .

|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|hα2

=|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|

|tu1(x2 + h)− tu1(x2)|α

(|tu1(x2 + h)− tu1(x2)|

)α≤ |t|α||u3||0,α(||u1||0,α)α .

|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|hα2

=|u3(x1 − tu1(x2 + h))− u3(x1 − tu1(x2))|

|tu1(x2 + h)− tu1(x2)|α

(|tu1(x2 + h)− tu1(x2)|

)α≤ |t|α||u3||0,α(||u1||0,α)α .

Proof of the Theorem continue...

Let us introduce two periodic functions u1(s) and u3(s) whichnear the point s = 0 coincide with |s|α then the for t given and x1and x2 small enough u3(x1 − tu3(x2)) coincides with

|x1 − t|x2|α|α .

For (x1, x2, x3) = (0, x2, x3) one has

u3(x1 − tu3(x2)) = |t|α|x2|α2

and the conclusion follows.

Other spaces and optimal spaces

The next part is private communication with Lemarié-Rieusset.

C1,α = B1+α∞,∞ ⊂ B1

∞,1 ⊂ C1 ⊂ F1∞,2 ⊂ B1

∞,∞ ⊂ Bα∞,∞ = C0,α .

Other spaces and optimal spaces

The next part is private communication with Lemarié-Rieusset.

C1,α = B1+α∞,∞ ⊂ B1

∞,1 ⊂ C1 ⊂ F1∞,2 ⊂ B1

∞,∞ ⊂ Bα∞,∞ = C0,α .

Besov and Trieble-Lizorkin spaces

Theorem The 3d Euler equation is well posed in B1∞,1 (Pak and

Park). It is not well posed in B1∞,∞ or in the Triebel-Lizorkin

space ⊂ F1∞,2

B1∞,∞ is the Zygmund class, i.e. bounded functions with

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

They are not Lipschitz but log-Lipschitz

|f (x + h)− f (x)| ≤ C|h| log1|h|.

Now choose a function v(y) smooth outside 0 with

v(y) ∼ y log1|y|

near y = 0

is in the Zygmund class. Then with u1(y) = u3(y) = v(y) andx1 = 0 we have

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

Same proof for ⊂ F1∞,2. More delicate: Construction of a log

Lipschitz function in this space.

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

is in the Zygmund class.

Then with u1(y) = u3(y) = v(y) andx1 = 0 we have

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

supx∈R,h∈R

|f (x + h) + f (x− h)− 2f (x)||h|

|f (x + h)− f (x)| ≤ C|h| log1|h|.

v(y) ∼ y log1|y|

near y = 0

|u3(−tu1(h))− u3(−tu1(0))| ∼ th(log h)2 !!!

Lipschitz function in this space.Claude Bardos & Edriss S. Titi Euler Equations: Recent Advances

Analyticity of Solutions

It is clear that if u1, u3 are analytic functions then the shear-flowsolution is also analytic for a short while.

This is consistent with the classical result of Bardos-Beachour.

Recently Kukavica-Vicol used the shear-flow solution

(sin x2, 0,1

a2 + cos2(x1 − t sin x2))

to provide an explicit example for an analytic solutions whoseradius of analyticity is shrinking with the rate 1

(sin x2, 0,1

Vanishing viscosity a ruling out principle

Theorem[Bardos–Titi–Wiedemann, C.R. Acad. Sci.– Paris,(2012)].Let v0(x) = (v1(x2), 0, v3(x1, x2)), where we assume v1 ∈ L2(T)and v3 ∈ L2(T2).

Then, for every viscosity ν > 0, there exists aunique Leray-Hopf weak solution of the Navier-Stokesequations with viscosity ν and initial data v0, and thesesolutions uν converge weak−∗ in L∞([0,T]; L2(T3)) to the shearflow solution of Euler equations

(v1(x2), 0, v3(x1 − tv1(x2), x2))

corresponding to v0, as ν → 0.

Theorem[Bardos–Titi–Wiedemann, C.R. Acad. Sci.– Paris,(2012)].Let v0(x) = (v1(x2), 0, v3(x1, x2)), where we assume v1 ∈ L2(T)and v3 ∈ L2(T2). Then, for every viscosity ν > 0, there exists aunique Leray-Hopf weak solution of the Navier-Stokesequations with viscosity ν and initial data v0,

and thesesolutions uν converge weak−∗ in L∞([0,T]; L2(T3)) to the shearflow solution of Euler equations

(v1(x2), 0, v3(x1 − tv1(x2), x2))

Theorem[Bardos–Titi–Wiedemann, C.R. Acad. Sci.– Paris,(2012)].Let v0(x) = (v1(x2), 0, v3(x1, x2)), where we assume v1 ∈ L2(T)and v3 ∈ L2(T2). Then, for every viscosity ν > 0, there exists aunique Leray-Hopf weak solution of the Navier-Stokesequations with viscosity ν and initial data v0, and thesesolutions uν converge weak−∗ in L∞([0,T]; L2(T3)) to the shearflow solution of Euler equations

(v1(x2), 0, v3(x1 − tv1(x2), x2))

Theorem[Bardos–Titi–Wiedemann, C.R. Acad. Sci.– Paris,(2012)].Let v0(x) = (v1(x2), 0, v3(x1, x2)), where we assume v1 ∈ L2(T)and v3 ∈ L2(T2). Then, for every viscosity ν > 0, there exists aunique Leray-Hopf weak solution of the Navier-Stokesequations with viscosity ν and initial data v0, and thesesolutions uν converge weak−∗ in L∞([0,T]; L2(T3)) to the shearflow solution of Euler equations

(v1(x2), 0, v3(x1 − tv1(x2), x2))

Shear flow with vorticity interface

In order for the vorticity of the shear flow solution of thethree-dimensional Euler equations to be concentrated on aninterface one can easily check that the shear flow solution mustbe of the form:

u1(s) =

α1 for s < ξ2β1 for s > ξ2

and u3(s) =

for some fixed real parameters α1, α3, β1, β3, ξ1, ξ2, satisfyingα1 ≥ β1 and α3 6= β3.

Consequently, the corresponding vorticity of the above solutionis concentrated on the singular surface:

Σ(t) = (x1, x2, x3)| x2 = ξ2 ∪(x1, x2, x3)| x1 = ξ1 + tα1, x2 ≤ ξ2∪(x1, x2, x3)| x1 = ξ1 + tβ1, x2 ≥ ξ2 .

u1(s) =

and u3(s) =

u1(s) =

and u3(s) =

u1(s) =

and u3(s) =

2d and 3d Kelvin-Helmholtz problem - a comparison

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.

In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners.

It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d.

There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.

In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity.

For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

In the problem under consideration the vorticity isconcentrated on an orientable curve, in 2d, or a surface in3d.In the first example the density of the vorticity isconcentrated on a surface with corners. It does not seemto be possible to construct the same type of configurationin 2d. There seems to be more room in 3d.In the second example the function x2 7→ u1(x2) does notseem to require more than C1 regularity in order tomaintain this regularity. For the two-dimensionalKelvin-Helmholtz (Birkhoff-Rott) such property is notpossible, while it might be possible in thethree-dimensional case.

Kelvin-Helmholtz (Birkhoff-Rott) Problems

Let Γ(t) be the vortex-sheet, the interface where the vorticity isconcentrated. The velocity outside interface Γ(t) is given by:

In the two-dimensional case:

u(x, t) =1

2πRπ

∫x− r(t, λ′)|x− r(t, λ′)|2

ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′

In the three-dimensional case:

u(x, t) =

− 14π

∫x− r(t, λ′, µ′)|x− r(t, λ′, µ′)|3

∧ ω(t, r(t, λ′, µ′))|∂λ(r(t, λ′)) ∧ ∂µ(r(t, λ′))|dλ′dµ′

u(x, t) =1

2πRπ

∫x− r(t, λ′)|x− r(t, λ′)|2

ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′

u(x, t) =

− 14π

∫x− r(t, λ′, µ′)|x− r(t, λ′, µ′)|3

u(x, t) =1

2πRπ

∫x− r(t, λ′)|x− r(t, λ′)|2

ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′

u(x, t) =

− 14π

∫x− r(t, λ′, µ′)|x− r(t, λ′, µ′)|3

u(x, t) =1

2πRπ

∫x− r(t, λ′)|x− r(t, λ′)|2

ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′

u(x, t) =

− 14π

∫x− r(t, λ′, µ′)|x− r(t, λ′, µ′)|3

u(x, t) =1

2πRπ

∫x− r(t, λ′)|x− r(t, λ′)|2

ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′

u(x, t) =

− 14π

∫x− r(t, λ′, µ′)|x− r(t, λ′, µ′)|3

Vorticity density

When the point x converges to a point r ∈ Γ(t) the velocityu(x, t) converges to two different values u±(r) which satisfy:

u+(r) ·~n = u−(r) ·~n , ω = (u+(r)− u−(r)) ∧~n⊗ δΓ(t)

The vorticity density:

ω = (u+(r)− u−(r)) ∧~n⊗ δΓ(t) = ω(t, r(t, λ′))|∂λ(r(t, λ′))|dλ′