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NON-UNIQUENESS OF WEAK SOLUTIONS TO HYPERVISCOUS NAVIER-STOKES EQUATIONS - ON

SHARPNESS OF J.-L. LIONS EXPONENT

TIANWEN LUOAND EDRISS S. TITI

Abstract. Using the convex integration technique for the three-dimensional Navier-Stokes equations introduced by T. Buckmaster and V. Vicol, it is shown the existence of non-unique weak solutions for the 3D Navier-Stokes equations with fractional hyperviscosity (−∆)θ, whenever the exponentθ is less than J.-L. Lions’ exponent 5/4, i.e., whenθ <5/4.

1. Introduction

In this paper we consider the question of non-uniquness of weak solutions to the 3D Navier-Stokes equations with fractional viscosity (FVNSE) onT3

(∂tv+∇ ·(v⊗v) +∇p+ν(−∆)θv= 0,

∇ ·v= 0, (1)

where θ ∈ R is a fixed constant, and for u∈ C(T3) the fractional Laplacian is defined via the Fourier transform as

F((−∆)θu)(ξ) =|ξ|F(u)(ξ), ξ∈Z3.

Definition(weak solutions). A vector field v∈Cweak0 (R;L2(T3))is called a weak solution to the FVNSE if it solves (1)in the sense of distribution.

Whenθ = 1, FVNSE (1) is the standard Navier-Stokes equations. J.-L. Lions first considered FVNSE (1) in [17], and showed the existence and uniqueness of weak solutions to the initial value problem, which also satisfied the energy equality, for θ ∈ [5/4,∞) in [18]. Moreover, an analogue of the Caffarelli-Kohn-Nirenberg [6] result was established in [16] for the FVNSE system (1), showing that the Hausdorff dimension of the singular set, in space and time, is bounded by 5−4θfor θ∈(1,5/4). The existence, uniqueness, regularity and stability of solutions to the FVNSE have been studied in [15, 20, 22, 23] and references therein. Very recently, using the method of convex integration introduced in [11], Colombo, De Lellis and

2010Mathematics Subject Classification. 35Q30.

Key words and phrases. Non-uniqueness, Weak solutions, Wild solutions, Navier-Stokes equa- tions, Hyperviscosity, Convex integration.

Yau Mathematical Sciences Center, Tsinghua University, China. twluo@mail.tsinghua.edu.cn.

Department of Mathematics, Texas A&M University, 3368 TAMU, College Station, TX 77843-3368, USA. Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Wilberforce Road, Cambridge CB3 0WA, UK. Department of Computer Sci- ence and Applied Mathematics, The Weizmann Institute of Science, Rehovot 76100, Israel.

titi@math.tamu.edu; Edriss.Titi@damtp.cam.ac.uk; edriss.titi@weizmann.ac.il.

1

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De Rosa in [7] showed the non-uniquenss of Leray weak solutions to FVNSE (1) for θ∈(0,1/5).

In the recent breakthrough work [5], Buckmaster and Vicol obtained non-uniqueness of weak solutions to the three-dimensional Navier-Stokes equations. They develope- d a new convex integration scheme in Sobolev spaces using intermittent Beltrami flows which combined concentrations and oscillations. Later, the idea of using in- termittent flows was used to study non-uniqueness for transport equations in [19], employing scaled Mikado waves.

The schemes in [5, 19] are based on the convex integration framework in H¨older spaces for the Euler equations, introduced by De Lellis and Sz´ekelyhidi in [11], subsequently refined in [2, 3, 9, 13], and culminated in the proof of the second half of the Onsager conjecture by Isett in [14]; also see [4] for a shorter proof. For the first half of the Onsager conjecture, see, e.g., [1, 8], and the references therein.

The main contribution of this note is to show that the results in Buckmaster- Vicol’s paper hold for FVNSE (1) forθ <5/4:

Theorem 1. Assume that θ ∈ [1,5/4). Suppose u is a smooth divergence-free vector field, define on R+×T3, with compact support in time and satisfies the condition

ˆ

T3

u(t, x)dx≡0.

Then for any givenε0>0, there exists a weak solutionv to the FVNSE (1), with compact support in time, satisfying

kv−ukL

t Wx2θ−1,1 < ε0.

As a consequence there are infinitely many weak solutions of the FVNSE (1)which are compactly supported in time; in particular, there are infinitely many weak solu- tions with initial values zero.

Remark 1. In the above theorem we assume thatθ∈[1,5/4). However, using the constructions in [5] with a slightly different choice of parameters, one can actually show that Theorem 1.2 and Theorem 1.3 in [5] hold for the 3D FVNSE, i.e., there exist non-unique weak solutions v∈Ct0Wxβ,2, with a differentβ >0, depending on θ. However, in this paper we choose to prove a weaker result, Theorem 1, in order to simplify the presentation while retaining the main idea.

Remark 2. For the case θ ∈ (−∞,1), the same construction also yields weak solutions v∈Ct0L2x∩Ct0Wx1,1 with a suitable choice of parameters.

We now make some comments on the analysis in this paper. Using the technique in [5], we adapt a convex integration scheme with intermittent Beltrami flows as the building blocks. The main difficulty in a convex integration scheme for (FVNSE), is the error induced by the frictional viscosityν(−∆)θv, which is greater for a larger exponent θ. This error is controlled by making full use of the concentration effect of intermittent flows introduced in [5]. As it is shown in the crucial estimate (36), the error is controllable only for θ <5/4. Compared with [5], since our goal is to construct weak solutionsv∈Ct0L2x,weak∩Lt Wx2θ−1,1, we adapt a slightly simpler cut-off function and prove only estimates that are sufficient for this purpose.

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2. Outline

2.1. Iteration lemma. Following [5], we consider the approximate system (∂tv+∇ ·(v⊗v) +∇p+ν(−∆)θv=∇ ·R,

∇ ·v= 0, (2)

whereR is a symmetric 3×3 matrix.

Lemma 1 (Iteration Lemma forL2weak solutions). Let θ∈(−∞,5/4). Assume (vq, Rq)is a smooth solution to (2)with

kRqkL

t L1x ≤δq+1, (3)

for some δq+1 >0. Then for any given δq+2 > 0, there exists a smooth solution (vq+1, Rq+1)of (2) with

kRq+1kL

t L1x ≤δq+2, (4)

and supptvq+1∪supptRq+1⊂Nδq+1(supptvq∪supptRq). (5) Here for a given set A⊂R, the δ-neighborhood ofA is denoted by

Nδ(A) ={y∈R:∃y0 ∈A,|y−y0|< δ}.

Furthermore, the increment wq+1=vq+1−vq satisfies the estimates

kwq+1kLt L2x ≤Cδ1/2q+1, (6) kwq+1kL

t Wx2θ−1,1 ≤δq+2, (7)

where the positive constantC depends only onθ.

Proof of Theorem 1. Assume Lemma 1 is valid. Letv0=u. Then ˆ

T3

tv0(t, x)dx= d dt

ˆ

T3

v0(t, x)dx≡0.

Let

R0=R(∂tv0+ν(−∆)θv0) +v0⊗v0+p0I, p0=−1 3|v0|2,

whereRis the symmetric anti-divergence operator established in Lemma 5, below.

Clearly (v0, R0) solves (2). Set

δ1=kR0kL

t L1x,

δq+1= 2−qε0, forq≥1.

Apply Lemma 1 iteratively to obtain smooth solution (vq, Rq) to (2). It follows from (6) that

Xkvq+1−vqkL

t L2x =X

kwq+1kL

t L2x ≤CX

δ1/2q+1<∞.

Thusvq converge strongly to somev ∈Ct0L2x. Since kRq+1kLt L1x →0, as q→ ∞, v is a weak solution to the FVNSE (1). Estimate (7) leads to

kv−v0k

L t W2θ−1,1

x

X

q=1

kwqk

L t W2θ−1,1

x

X

q=1

δq+1≤ε0.

Furthermore, it follows from (5) that supptv⊂ ∪q≥0supptvq⊂NP

q≥0δq+1(supptu)⊂Nδ10(supptu).

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Now we show the existence of infinitely many weak solutions with initial values zero. Let u(t, x) = ϕ(t)P

|k|≤Nakeik·x with ak 6= 0, ak ·k = 0, a−k = ak for all

|k| ≤N, andϕ∈Cc(R+). Thus ∇ ·u= 0 satisfies the conditions of the theorem.

Hence there exists a weak solutionv to (1) close enough touso thatv6≡0.

3. Iteration scheme

3.1. Notations and Parameters. For a complex numberζ∈C, we denote byζ its complex conjugate. Let us normalize the volume

|T3|= 1.

For smooth functionsu∈C(T3) ands∈R, we define F(|∇|su)(ξ) =|ξ|sF(u)(ξ), ξ∈Z3. ForM, N ∈[−∞,+∞], denote the Fourier projection ofuby

F(P[M,N)u) =

(u(ξ), M ≤ξ < N, ξ∈Z3, 0, otherwise.

We also denoteP≤k =P[−∞,k)andP≥k =P[k,+∞).

Following the notation in [5], we introduce here several parametersσ, r, λ, with 0< σ <1< r < λ < µ < λ2, σr <1, (8) where λ = λq+1 ∈ 5N is the ‘frequency’ parameter; σ with 1/σ ∈ N is a small parameter such thatλσ∈Nparameterizes the spacing between frequencies;r∈N denotes the number of frequencies along edges of a cube; µmeasures the amount of temporal oscillation.

Laterσ, r, µwill be chosen to be suitable powers ofλq+1. We also fix a constant p >1 which will be chosen later to be close to 1. The constants implicitly in the notation ‘.’ may depend onpbut are independent of the parametersσ, r, λ.

3.2. Intermittent Beltrami flows. We use intermittent Beltrami flows intro- duced in [5] as the building blocks. Recall some basic facts of Beltrami waves.

Proposition 1. ( [5, Proposition 3.1]) Givenξ∈S2∩Q3, letAξ ∈S2∩Q3 be such that

Aξ·ξ= 0, |Aξ|= 1, A−ξ =Aξ.

Let Λ be a given finite subset of S2 such that −Λ = Λ, and λ ∈ Z be such that λΛ⊂Z3. Then for any choice of coefficientsaξ ∈Cwith a

ξ =a−ξ the vector field W(x) =X

ξ∈Λ

aξBξeiλξ·x, withBξ= 1

√2(Aξ+iξ×Aξ), is real-valued, divergence-free and satisfies

∇ ×W =λW, ∇ ·(W ⊗W) =∇|W|2 2 . Furthermore,

hW⊗Wi:=

T3

W ⊗W dx=X

ξ∈Λ

1

2|a(ξ)|2(Id−ξ⊗ξ).

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Let Λ,Λ+⊂S2∩Q3 be defined by Λ+={1

5(3e1±4e2),1

5(3e2±4e3),1

5(3e3±4e1)}, Λ=−Λ+, Λ = Λ+∪Λ. Clearly we have

5Λ∈Z3, and min

ξ0,ξ∈Λ,ξ0+ξ6=0

0+ξ| ≥1

5. (9)

Also it is direct to check that 1 8

X

ξ∈Λ

(Id−ξ⊗ξ) = Id.

In fact, representations of this form exist for symmetric matrices close to the iden- tity. We have the following simple variant of [5, Proposition 3.2].

Proposition 2. LetBε(Id) denote the ball of symmetric matrices, centered at the identity, of radius ε. Then there exist a constant εγ > 0 and smooth positive functionsγ(ξ)∈C(Bεγ(Id)), such that

(1) γ(ξ)(−ξ);

(2) for eachR∈Bεγ(Id)we have the identity R=1

2 X

ξ∈Λ

γ(ξ)(R)2

(Id−ξ⊗ξ).

Define the Dirichlet kernel Dr(x) = 1

(2r+ 1)3/2 X

ξ∈Ωr

eiξ·x, Ωr={(j, k, l) :j, k, l∈ {−r,· · ·, r}}.

It has the property that, for 1< p≤ ∞,

kDrkLp.r3/2−3/p, kDrkL2 = (2π)3.

Following [5], forξ∈Λ+, define a directed and rescaled Dirichlet kernel by η(ξ)(t, x) =ηξ,λ,σ,r,µ(t, x) =Dr(λσ(ξ·x+µt, Aξ·x,(ξ×Aξ)·x)), (10) and forξ∈Λ, define

η(ξ)(t, x) =η−(ξ)(t, x).

Note the important identity 1

µ∂tη(ξ)(t, x) =±(ξ· ∇)η(ξ)(t, x), ξ∈Λ±. (11) Since the mapx7→λσ(ξ·x+µt, Aξ·x,(ξ×Aξ)·x) is the composition of a rotation by a rational orthogonal matrix mapping{e1, e2, e3}to{ξ, Aξ, ξ×Aξ}, a translation, and a rescaling by integers, for 1< p≤ ∞, we have

T3

η(ξ)(t, x)2(t, x)dx= 1, kη(ξ)kL

t Lpx(T3).r3/2−3/p. LetW(ξ) be the Beltrami plane wave at frequencyλ,

W(ξ)=Wξ,λ(x) =Bξeiλξ·x.

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Define the intermittent Beltrami waveW(ξ)as

W(ξ)(t, x) :=Wξ,λ,σ,r,µ(t, x) =η(ξ)(t, x)W(ξ)(x). (12) It follows from the definitions and (9) that

P[λ2,2λ)W(ξ)=W(ξ), (13) P[λ5,4λ)(W(ξ)⊗W0)) =W(ξ)⊗W0), ξ06=−ξ. (14) The following properties are immediate from the definitions.

Proposition 3. ( [5, Proposition 3.4]) Letaξ∈Cbe constants witha

ξ =a−ξ. Let W(x) =X

ξ∈Λ

aξW(ξ)(x).

ThenW(x)is real valued. Moreover, for eachR∈Bεγ(Id)we have X

ξ∈Λ

γ(ξ)(R)2

T3

W(ξ)⊗W(−ξ)=X

ξ∈Λ

γ(ξ)(R)2

Bξ⊗B−ξ =R.

Proposition 4. ( [5, Proposition 3.5]) For any1< p≤ ∞, N≥0, K≥0:

k∇NKt W(ξ)kL

t LpxN(λσrµ)Kr3/2−3/p, (15) k∇NtKη(ξ)kL

t Lpx.(λσr)N(λσrµ)Kr3/2−3/p. (16) 3.3. Perturbations. Letψ(t) be a smooth cut-off function such that

ψ(t) = 1 on supptRq, suppψ(t)⊂Nδq+1(supptRq), |ψ0(t)| ≤2δ−1q+1. (17) Take a smooth increasing functionχsuch that

χ(s) =

(1, 0≤s <1 s, s≥2 , and set

ρ(t, x) =ε−1γ δq+1χ(δq+1−1 |Rq(t, x)|)ψ2(t).

whereεγ is the constant in Proposition 2. Then clearly

supptρ⊂Nδq+1(supptRq). (18) It follows from the above definition that

|Rq|/ρ=εγ

|Rq|

δq+1χ(δ−1q+1|Rq(t, x)|)ψ2 ≤εγ =⇒ Id−Rq/ρ∈Bεγ(Id) on suppRq. Therefore, the amplitude functions

a(ξ)(t, x) :=ρ1/2(t, x)γ(ξ)(Id−ρ(t, x)−1Rq(t, x))

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are well-defined and smooth. Define the velocity perturbation to bew=wq+1: w=w(p)+w(c)+w(t),

w(p)=X

ξ∈Λ

a(ξ)W(ξ)=X

ξ∈Λ

a(ξ)(t, x)η(ξ)(t, x)Bξeiλξ·x,

w(c)= 1 λq+1

X

ξ∈Λ

a(ξ)η(ξ)

×W(ξ),

w(t)= 1 µ

X

ξ∈Λ+

PLHP6=0

a2(ξ)η2(ξ)ξ ,

wherePLH= Id− ∇∆−1div is the Leray-Helmholtz projection into divergence-free vector field, and P6=0f = f −ffl

T3f dx. It is well-known that PLH is bounded on Lp,1< p <∞(see , e.g., [12]). It follows from Proposition 3 that

X

ξ∈Λ

a2(ξ)

T3

W(ξ)⊗W(−ξ)dx=ρId−Rq. (19) 3.4. Estimates for perturbations.

Lemma 2. The following bounds hold:

kρkLt L1x≤Cδq+1, (20) kρ−1kC0(suppRq)−1q+1, (21)

kρkCN

t,x≤C(δq+1,kRqkCN), (22)

ka(ξ)kLt L2x.kρk1/2L

t L1xq+11/2, (23)

ka(ξ)kCN

t,x≤C(δq+1,kRqkCN). (24)

Proof. It follows from (3) that kρ(t,·)kL1

x= ˆ

|Rq|≤δq+1

ρ+ ˆ

|Rq|>δq+1

ρ.δq+1+ ˆ

|Rq|>δq+1

|Rq|

≤Cδq+1.

It is direct to verify (21) and (23), while (22) and (24) follow from (17) and (21).

Now we can estimate the time support ofwq+1:

supptwq+1⊂supptρ⊂suppψ⊂Nδq+1(supptRq). (25) We need the following Lemma, which is a variant of [5, Lemma 3.6].

Lemma 3. ( [19, Lemma 2.1]) Let f, g ∈ C(T3), and g is (T/N)3 periodic, N ∈N. Then for1≤p≤ ∞,

kf gkLp≤ kfkLpkgkLp+CpN−1/pkfkC1kgkLp. Let us denote

CN =C(sup

ξ∈Λ

ka(ξ)kCN

t,x) (26)

to be some polynomials depending on supξ∈Λka(ξ)kCN t,x.

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Lemma 4. Suppose the parameters satisfy (8)and

r3/2≤µ. (27)

Then the following estimates for the perturbations hold:

kw(p)q+1kL

t L2xq+11/2 + (λq+1σ)−1/2C1, (28) kwq+1kL

t Lpx .r3/2−3/pC1, (29) kwq+1(c) kL

t Lpx+kw(t)q+1kL

t Lpx .(σr+µ−1r3/2)r3/2−3/pC1, (30) k∂twq+1(p)kL

t Lpx+k∂tw(c)q+1kL

t Lpxq+1σµr5/2−3/pC2, (31) k|∇|Nwq+1kL

t Lpx .r3/2−3/pλNq+1CN+1, (32) for1< p <∞, N≥1.

Proof. SinceW(ξ)is (T/λσ)3periodic, it follows from (15), (23), and Lemma 3 that kwq+1(p)kL

t L2x.X

ξ∈Λ

(ka(ξ)kL

t L2x+ (λq+1σ)−1/2ka(ξ)kC1)kW(ξ)kL

t L2x

1/2q+1+ (λq+1σ)−1/2C1. In view of (8), (15) and (16) yield that

kwq+1(p)kL

t Lpx.X

ξ∈Λ

ka(ξ)kC0kW(ξ)kL

t Lpx .r3/2−3/pC0, kwq+1(c) kL

t Lpx−1q+1X

ξ∈Λ

(ξ)kL

t Lpx+k∇η(ξ)kL

t Lpx

ka(ξ)kC1kW(ξ)kL

t Lpx

.(σr)r3/2−3/pC1, kwq+1(t) kL

t Lpx−1 X

ξ∈Λ+

ka2(ξ)η2

(ξ)ξkL

t Lpx−1 X

ξ∈Λ+

ka2(ξ)kC0(ξ)k2L t L2px

−1r3−3/pC0,

where the boundedness of PLH and P6=0 onLp, for 1< p <∞, is used in the first inequality of the estimate forkw(t)q+1kL

t Lpx. In the same way, we can estimate k∂twq+1(p)kL

t Lpx.X

ξ∈Λ

k∂ta(ξ)kC0kW(ξ)kL

t Lpx+ka(ξ)kC0k∂tW(ξ)kL

t Lpx

q+1σµr5/2−3/pC1, k∂twq+1(c) kL

t Lpx−1q+1X

ξ∈Λ

ka(ξ)kC2

t,x

(ξ)kL

t Lpx+k∇η(ξ)kL

t Lpx+k∂tη(ξ)kL

t Lpx

+k∂t∇η(ξ)kL

t Lpx

.σrλq+1σµr5/2−3/pC2q+1σµr5/2−3/pC2.

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ForN ≥1, Using (15) and (16), we obtain that k∇Nw(p)q+1kL

t Lpx .X

ξ∈Λ N

X

k=0

k∇ka(ξ)kC0k∇N−kW(ξ)kL

t Lpx

Nq+1r3/2−3/pCN, k∇Nw(c)q+1kL

t Lpx−1q+1X

ξ∈Λ N

X

m=0 m

X

k=0

λNq+1−mk∇k+1a(ξ)kC0k∇m−kη(ξ)kL

t Lpx

−1q+1X

ξ∈Λ N

X

m=0 m

X

k=0

λN−mq+1 k∇ka(ξ)kC0k∇m−k+1η(ξ)kL

t Lpx

Nq+1r3/2−3/pCN+1, k∇Nw(t)q+1kL

t Lpx−1X

ξ∈Λ N

X

m=0

k∇N−m(a2(ξ))kC0

m

X

k=0

k∇kη(ξ)kL

t L2px k∇m−kη(ξ)kL t L2px

Nq+1r3/2−3/p(σr)Nr3/2

µ CNNq+1r3/2−3/pCN,

where we use (8) and (27).

3.5. Estimates for the stress. Let us recall the following operator in [11].

Lemma 5(symmetric anti-divergence). There exists a linear operatorR, of order

−1, mapping vector fields to symmetric matrices such that

∇ · R(u) =u−

T3

u, (33)

with standard Calderon-Zygmund estimates, for1< p <∞,

kRkLp→W1,p .1, kRkC0→C0 .1, kRP6=0ukLp .k|∇|−1P6=0ukLp. (34) Proof. Supposeu∈C(T3,R3) is a smooth vector field. Define

R(u) = 1

4 ∇PLHv+ (∇PLHv)T +3

4 ∇v+ (∇v)T

−1

2(∇ ·v)Id wherev∈C(T3,R3) is the unique solution to ∆v=u−ffl

T3uwithffl

T3v= 0.

It is direct to verify that R(u) is a symmetric matrix field depending linearly on uand satisfies (33). Note that R is a constant coefficient ellitpic operator of order −1. We refer to [12] for the Calderon-Zygmund estimates kRkLp→W1,p .1 andkRP6=0ukLp.k|∇|−1P6=0ukLp. Combining these with Sobolev embeddings, we havekRukCα.kRukW1,4.kukL4 .kukC0, withα= 1/4.

We have the following variant of [5, Lemma B.1] in [5].

Lemma 6. Let a∈C2(T3). For 1< p <∞, and anyf ∈Lp(T3), we have k|∇|−1P6=0(aP≥kf)kLp.k−1(kakL+k∇2akL)kfkLp. (35) Proof of Lemma 6. We follow the proof in [5]. Note that

|∇|−1P6=0(aP≥kf) =|∇|−1P≥k/2(P≤k/2aP≥kf) +|∇|−1P6=0(P≥k/2aP≥kf).

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As direct consequences of the Littlewood-Paley decomposition and Schauder esti- mates we have the bounds (see, for example, [12])

kP≤k/2kL→L .1, k|∇|−1P≥k/2kLp→Lp.k−1, k|∇|−1P6=0kLp→Lp.1.

Combining these bounds with H¨older’s inequality and the embeddingW1,4(T3)⊂ L(T3), we obtain

k|∇|−1P6=0(aP≥kf)kLp.k−1kP≤k/2aP≥kfkLp+kP≥k/2aP≥kfkLp

.k−1(kP≤k/2akL+kkP≥k/2akL)kfkLp.k−1(kakL+kk∇P≥k/2akL4)kfkLp

.k−1(kakL+kk|∇|−1P≥k/2|∇|∇P≥k/2akL4)kfkLp.k−1(kakL+k∇2P≥k/2akL4)kfkLp

.k−1(kakL+k∇2akL)kfkLp.

It follows from the definition ofwq+1 that

ˆ

T3

wq+1dx= ˆ

T3

1 λq+1

X

ξ∈Λ

a(ξ)η(ξ)W(ξ) dx+

ˆ

T3

1 µ

X

ξ∈Λ+

PLHP6=0

a2

(ξ)η2

(ξ)ξ dx= 0.

Hence´

T3ν(−∆)θwq+1dx= 0 and d dt

´

T3wq+1dx= 0. We obtainRq+1by plugging vq+1=vq+wq+1in (2), using (33) and the assumption that (vq, Rq) solves (2):

∇ ·Rq+1=∇ ·h

R(ν(−∆)θwq+1+∂twq+1(p) +∂tw(c)q+1) +vq⊗wq+1+wq+1⊗vq

i

+∇ ·h

(w(c)q+1+w(t)q+1)⊗wq+1+w(p)q+1⊗(wq+1(c) +wq+1(t) )i h∇ ·(wq+1(p) ⊗w(p)q+1−Rq) +∂tw(t)q+1i

+∇(pq+1−pq) :=∇ ·(Relinear+Recorrector+Reoscillation) +∇(pq+1−pq).

It follows from Lemma 4 that kRecorrectorkL

t Lpx.

kw(c)q+1kL

t L2px +kwq+1(t) kL

t L2px kwq+1kL

t L2px +kwq+1(p)kL t L2px

.(σr+µ−1r3/2)r3−3/pC1.

Noting that∇ ×w(p)q+1 λq+1

=w(p)q+1+w(c)q+1, Lemma 4 and (34) yield that kRelinearkL

t Lpx

−1q+1k∂tR∇ ×(w(p)q+1)kL

t Lpx+kR(ν(−∆)θwq+1)kL

t Lpx

+kvq⊗wq+1+wq+1⊗vqkL

t Lpx

−1q+1k∂tw(p)q+1kL

t Lpx+k|∇|2θ−1wq+1kL

t Lpx+kvqkC0kwq+1kL t Lpx

.σµr5/2−3/pC2+r3/2−3/p2θ−1q+1 +kvqkC0)C3. (36) This is the crucial estimate to control the fractional viscosity. If we assume that p∼ 1, r ∼λ−1q+1, we must have θ < 5/4 in order that the second term in (36) is small forλq+1sufficiently large.

(11)

It remains to estimate Reoscillation, which can be handled in the same way as in [5]. It follows from (19) that

∇ ·(w(p)q+1⊗w(p)q+1−Rq) =∇ ·( X

ξ,ξ0∈Λ

a(ξ)a0)Wξ⊗W0)−Rq)

=∇ ·( X

ξ,ξ0∈Λ

a(ξ)a0)P≥λq+1σ/2W(ξ)⊗W0)) +∇ρ

:= X

ξ,ξ0∈Λ

E(ξ,ξ0)+∇ρ.

SinceE(ξ,ξ0)has zero mean, we can split it as E(ξ,ξ0)+E0,ξ)=P6=0

∇(a(ξ)a0))·(P≥λq+1σ/2(W(ξ)⊗W0)+Wξ0⊗W(ξ))) +P6=0

a(ξ)a0)∇ ·(W(ξ)⊗W0)+Wξ0⊗W(ξ)) :=E(ξ,ξ0,1)+E(ξ,ξ0,2).

Using (15), (34) and (35), we obtain kRE(ξ,ξ0,1)kL

t Lpx.k|∇|−1E(ξ,ξ0,1)kL

t Lpx

.(λq+1σ)−1(ka(ξ)a0)kC1+k∇2(a(ξ)a0))kC1)kW(ξ)⊗W0)kL

t Lpx

.(λq+1σ)−1ka(ξ)a0)kC3kW(ξ)kL

t L2px kW0)kL t L2px

.(λq+1σ)−1r3−3/pC3.

Recall the vector identityA·∇B+B·∇A=∇(A·B)−A×(∇×B)−B×(∇×A).

Forξ, ξ0∈Λ, using the anti-symmetry of the cross product, we can write

∇ ·(W(ξ)⊗W0)+W0)⊗W(ξ))

=

W(ξ)⊗W0)+W0)⊗W(ξ)

η(ξ)η0)

(ξ)η0)

W(ξ)· ∇W0)+W0)· ∇W(ξ)

=

W0)· ∇

η(ξ)η0)

W(ξ)+

W(ξ)· ∇

η(ξ)η0)

Wξ0(ξ)η0)

W(ξ)·W0) .

For the termE(ξ,ξ0,2), first consider the caseξ+ξ0 6= 0. It follows from the above identity and (14) that

a(ξ)a0)∇ ·(W(ξ)⊗W0)+W0)⊗W(ξ))

=a(ξ)a0)∇ ·P≥λq+1/10

η(ξ)η0)

W(ξ)⊗W0)+W0)⊗W(ξ)

=a(ξ)a0)P≥λq+1/10

η(ξ)η0)

·

W(ξ)⊗W0)+W0)⊗W(ξ) +a(ξ)a0)P≥λq+1/10

η(ξ)η0)

W(ξ)·W0)

=a(ξ)a0)P≥λq+1/10

η(ξ)η0)

·

W(ξ)⊗W0)+W0)⊗W(ξ) +∇

a(ξ)a0)W(ξ)·W0)

− ∇

a(ξ)a0)

P≥λq+1/10

W(ξ)·W0)

−a(ξ)a0)P≥λq+1/10

W(ξ)·W0)

η(ξ)η0) ,

(12)

where the second term is a pressure, the third can be estimated analogously to E(ξ,ξ0,1). Also note that the first and fourth term can estimated analogously. Using (16), (34) and (35), we obtain

kR

a(ξ)a0)P≥λq+1/10

η(ξ)η0)

·

W(ξ)⊗W0)+W0)⊗W(ξ) kL

t Lpx

−1q+1(ka(ξ)a0)kC1+k∇2(a(ξ)a0))kC1)k∇

η(ξ)η0) kL

t Lpx

.σr4−3/pC3.

Now considerE(ξ,−ξ,2). We can write

∇ ·(W(ξ)⊗W(−ξ)+W(−ξ)⊗W(ξ)) =

W(−ξ)· ∇η2(ξ)

W(ξ)+

W(ξ)· ∇η2(ξ) W(−ξ)

= (Aξ· ∇η(ξ)2 )Aξ+ ((ξ×Aξ)· ∇η(ξ)2 )(ξ×Aξ) =∇ξ2(ξ)−(ξ· ∇η2(ξ))ξ=∇η(ξ)2 − ξ µ∂tη2(ξ),

where we use (11) and the fact that{ξ, Aξ, ξ×Aξ}forms an orthonormal basis of R3. Therefore, we can write

E(ξ,−ξ,2)=P6=0

a2(ξ)∇P≥λq+1σ/2η(ξ)2 −a2(ξ)ξ µ∂tη(ξ)2

=∇ a2

(ξ)P≥λq+1σ/2η2

(ξ)

−P6=0

P≥λq+1σ/22

(ξ))∇a2(ξ)

−µ−1tP6=0

a2(ξ)η2(ξ)ξ

−1P6=0

t a2(ξ)

η2(ξ)ξ .

Using the identity Id−PLH =∇∆−1div , we obtain X

ξ

E(ξ,−ξ,2)+∂tw(t)q+1=∇X

ξ

a2(ξ)P≥λq+1σ/2η(ξ)2

− ∇X

ξ

µ−1−1∇ ·∂t

a2(ξ)η2(ξ)ξ

−X

ξ

P6=0

P≥λq+1σ/2(ξ)2 )∇a2(ξ)

−1X

ξ

P6=0

t

a2(ξ)

η2(ξ)ξ ,

where the first and second terms are pressure terms. Using (16), (34) and (35), we obtain

kRP6=0

P≥λq+1σ/22

(ξ))∇a2(ξ) kL

t Lpx.(λq+1σ)−1(ξ)k2L t L2px C3

.(λq+1σ)−1r3−3/pC3. It follows from (16) and (34) that

µ−1kRP6=0

t

a2(ξ)

η2(ξ)ξ kL

t Lpx−1k∂t

a2(ξ)

η(ξ)2 ξkL

t Lpx

−1r3−3/pC1.

(13)

Let us now give the explicit definition ofReoscillation: Reoscillation= X

ξ,ξ0∈Λ

P6=0

∇(a(ξ)a0))·(P≥λq+1σ/2(W(ξ)⊗W0)+Wξ0⊗W(ξ)))

+ X

ξ,ξ0∈Λ,ξ6=ξ0

a(ξ)a0)P≥λq+1/10

η(ξ)η0)

·

W(ξ)⊗W0)+W0)⊗W(ξ)

− X

ξ,ξ0∈Λ,ξ6=ξ0

a(ξ)a0)

P≥λq+1/10

W(ξ)·W0)

− X

ξ,ξ0∈Λ,ξ6=ξ0

a(ξ)a0)P≥λq+1/10

W(ξ)·W0)

η(ξ)η0)

−X

ξ∈Λ

P6=0

P≥λq+1σ/2(ξ)2 )∇a2(ξ)

−1X

ξ∈Λ

P6=0

t

a2(ξ)

η2(ξ)ξ .

Finally, we estimate the time support ofRq+1. Using (25) we obtain supptRq+1⊂supptwq+1∪supptRq⊂Nδq+1(supptRq).

Now we choose the parametersr, σ, µ. Fixαso that max{0,2

3(2θ−1)}< α <1, which is possible sinceθ∈(−∞,5/4). Fix

r=λαq+1, σ=λ−(α+1)/2q+1 , µ=λ(5α+1)/4q+1 . (37) Clearly (27) is satisfied. Choosep >1 sufficiently close to 1 so that

−α+ 1

2 +5α+ 1

4 +

5 2−3

p

α <0, 3

2 −3 p

α+ max(0,2θ−1)<0,

−5α+ 1

4 +

9 2−3

p

α <0, −1−α

2 +

3−3

p

α <0.

Note thatCN is independent ofλq+1, due to (24). Combining the above estimates with Lemma 4, it is easy to check that, by takingλq+1 sufficiently large, we arrive at (4), (6) and (7). This completes the proof of Lemma 1.

AcknowledgementThe authors would like to thank H. Ibdah and the anony- mous referee for carefully reading the paper and for their constructive suggestions The authors would also like to thank the “The Institute of Mathematical Sciences”, Chinese University of Hong Kong, for the warm and kind hospitality during which part of this work was completed. The work of T.L. is supported in part by NS- FC Grants 11601258. The work of E.S.T. is supported in part by the ONR grant N00014-15-1-2333, the Einstein Stiftung/Foundation - Berlin, through the Einstein Visiting Fellow Program, and by the John Simon Guggenheim Memorial Founda- tion.

References

[1] C. Bardos and E.S. Titi, Onsager’s conjecture for the incompressible Euler equations in bounded domains, Arch. Ration. Mech. Anal.228(2018), 197-207.

[2] T. Buckmaster, C. De Lellis, P. Isett, L. Sz´ekelyhidi, Jr., Anomalous dissipation for1/5- older Euler flows, Ann. of Math.182:1(2015), 127–172.

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