Incompressible Navier-Stokes equations

Transcription

1 Incompressible Navier-Stokes equations Jean-Yves Chemin Laboratoire J.-L. Lions, Case 87 Université Pierre et Marie Curie, 75 3 Paris Cedex 5, France chemin@ann.jussieu.fr February 3,

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3 Introduction This text is notes of a two lectures given at Saint Étienne de Tinée near Nice during the winter. The purpose is to give a self contained introduction to recent results about global smooth solution for the tridimensionnal incompressible Navier-Stokes equations in the whole space R 3. A different version with some additionnal chapter will be published as Lectures Notes of the Beijing Academy of Sciences.The notes are organized as follows: In the first part, we first present the now classical theory of globall wellposedness for small initial data in the framework of Kato s method. The understanding of what small means in the core of the subject. Then, we present a recent result where global wellposedness is obtained without the hypothesis of smallness on the initial data, but for the relative smallness of the first iterate The partcicular structure of the Navier-Stokes equation is pointed out. In this part, we insists also on the key role of oscillations in the stabilization of the system. In the second part, we investigate the case of initial data which vary slowly in one direction. This is in some sense the dual case of fast oscillating data. The first case is the so called well prepared case. Here, the initial data converges to a bidimensionnal divergence free vector field. The variation is respect to the vertical variable is slow enough, then the solution exists globally. The second case is the so called ill prepared case. Here, the horizontal part is a divergence of size. Because of the divergence free condition, the size of the vertical component of the vector died is the inverse of the speed of variation in the vertical direction, this very large. A rescaling in the vertical variable leads to a problem which looks like an ill posed. Using a global Cauchy- Kovalevska method, which is explain is a model case, we can proved global existence a smooth solution of this initial data which are very large. This method consists in the control of the decay of the radius of analyticity of the solution. Acknowlegments I want to thank very much Gérard Iooss and Laurent Stolovitch for inviting me to give this lectures at Saint Etienne de Tinée and for the organization of this very nice and very interesting meeting. 3

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5 Contents Global wellposedness and Kato theory. 7. Weak solutions and Kato s approach The Kato theory is the L p framework Besov spaces of negative index The endpoint space for Picard s scheme An abstract non linear smallness condition Main steps of the proof A particular case of large oscillating data The role of the special structure: a Navier-Stokes type equation which blows up 3.8 References and remarks Slowly varying vector fields 35. Ill prepared data: the vertical rescaling Study of a model problem The global Cauchy-Kovalevska method Main steps of the proof Proof of the theorem assuming the two propositions

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7 Chapter Global wellposedness and Kato theory.. Weak solutions and Kato s approach Let us define the concept of weak solutions of the incompressible Navier-Stokes system. Let us first recall what the incompressible Navier Stokes system is. We consider as unknown the speed u = u, u, u 3 a time dependant divergence free vector field on R 3 and the pressure p. We consider the system NS t u + u u u = p + f in R + R 3 div u = u t= = u. The notion of C solution i.e. classical solution is not efficient because singularity can appear here and also we can be interested in rough initial data. This has been pointed out by C. Ossen see [5] and [5] that another concept of solution must be used. This has been formalized by J. Leray in 934 in his seminal work [45]. Let us define the notion of weak solution that we shall denote simply solution in all that follows. Definition.. A time-dependent vector field u with components in L loc [, T ] Rd is a weak solution simply a solution in this paper of NS if for any smooth compactly supported divergence free vector field Ψ, ut,, Ψt, = u, Ψ, + 3 j= u j t, x t Ψ j t, x + Ψt, x dt dx R 3 + u j u k t, x j k Ψt, x dt dx. j,k R 3 This definition is too weak in the sense there is not enough constraints on the solution. In particular it ignores the fundamental concept of energy. J. Leray introduced in his seminal paper [45] the concept of turbulent solution we shall not use in the notes Definition.. A turbulent solution of NS is a divergence free vector field u which is a weak solution, has component is L T L L T H and satisfies in addition the energy 7

8 inequality ut L + ut L dt u L + R 3 ft, x ut, xdt dx.. Remark For a turbulent solution, Definition.. of a weak solution becomes ut, x Ψt, x dx = u x Ψ, x dx + ft, Ψt dt R d R d u : Ψ u u : Ψ u t Ψ t, x dx dt. In [45], J. Leray proved the following theorem. R d Theorem.. Let u be a divergence free vector field in L R d. Then a turbulent solution u exists on R + R 3. The proof of this theorem relies on compactness methods, We shall not develop this notion here. We are going to focuse on results of existence and thus uniqueness that can to prove by a fixed point theorem. In order to do so, let us follow Kato s approach. Let us define the biinear operator Bu, v as the solution of the evolution Stokes problem t B j B j = 3 k u j v k + v j u k j p. k= div B = and B t= =. In weak formulation, this writes ut,, Ψt, = 3 j= u j t, x t Ψ j t, x + Ψt, x dt dx R 3 + j,k R 3 k u j v k + v j u k t, x j Ψ k t, x dt dx..3 It is obvious that u a solution of NS if and only if u satisfies u = e t u + Bu, u. Solving globally NS is equivalent to find a fixed point for the map u e t u + Bu, u. Now let us assume that we have a Banach space X of functions locally in L on R + R 3 such that B is a bilinar map from X X into X. Then Picard s fixed point theorem implies the existence of a unique solution. Such a space X will be called adapted. Let us remark there is a strong constrain on X due to the scaling property. If u is a solution of NS on [, T ] R 3, then for any positive λ, the vector field u λ t, x def = λuλ t, λx is also a solution of NS on [, λ T ] R 3. Thus, if X adapted, it must be scaling invariant and also translation invariant in the sense that λ >, a R 3, u X uλ a X and u X uλ a X. Let us give a first example of an adapted spaces: the space L 4 R + ; Ḣ. Let us define an operator which will be of some use later one. 8

9 Definition..3 We denote by L the operator defined by the fact that L f is the solution of { The key lemma is the following. t L f L f = f p div L f = and L f t= =. Lemma.. The operator L maps continously L R 3 ; Ḣ into L 4 R + ; Ḣ. Proof. In Fourier space, we can write that FL ft, ξ = e t t ξ Pξ ft, ξdt where Pξ is the orthogonal projection in R 3 orthogonal of ξ. Thus, we get ξ FL ft, ξ e t t ξ ξ 3 θt, ξ ft, Ḣ dt with θt, L = for any t of R +. Taking the L in ξ norm in the above inequality gives L ft, Ḣ ft, t t 3 Ḣ dt. 4 Then Hardy-Littlewood-Sobolev inequality allows to conclude the proof. As a corollary, we get Corollary.. The operator B maps L 4 R +, Ḣ L 4 R +, Ḣ into L 4 R +, Ḣ. Proof. Let us obseve that, thank to diverngence free condition, we have 3 k= k u j v k + v j u k = u v + v u. Thank you the Sobolev embeddings and its dual, we have u v + v ut, Ḣ u v + v ut, 3 L ut L t L. Thus u v + v u L R + ;Ḣ u L 4 R + ;Ḣ v L 4 R + ;Ḣ. As we have e t u L 4 R + ;Ḣ u Ḣ we get that if u Ḣ is small enough, then a unique global solution exists in L 4 R + ; Ḣ. This is Fujita-Kato theorem. 9

10 . The Kato theory is the L p framework Let us first define operatros that will be of some use later on. Definition.. For j {,, 3}, we denote by L j the operator defined by the fact that L j f is the solution of { t L j f L j f = j f p div L j f = and L j f t= =. Let us introduce spaces adpated to theses operators. Definition.. For p in [, ] and for positive σ, we denote by Kp σ the space of functions on R + R 3 such that def u K σ p = sup t σ ut L p <. t> Let us remark that when σ = 3 p, the space Kσ p is the following Proposition.. For p > 3 the operator L j maps continuously K is scaling invariant. THe key proposition 3 p p into K 3 p p. Proof. It relies on the explicite computation of the operator L j. We have the following lemma. Lemma.. Let σ be an homogeneous function of degree α > d. Then a constant C exists such that, for any positive t, F σξe t ξ x C t + x α+d Proof. By changing variable in the integral Ix = e ix ξ σξe t ξ dξ R d it is enough to prove the estimate in the case whent =. As α > d, it is obvious that the function is bounded. Let us write that Ix = e ix ξ χλ ξσξe t ξ dξ + e ix ξ χλ ξ σξe t ξ dξ. R d R d where χ is a smooth cut-off function. We have e ix ξ χλ ξσξe t ξ dξ λ α+d. R d Using that x N e ix ξ = ξ N e ix ξ, we infer that, for small λ, x N e ix ξ χλ ξ σξe t ξ dξ λ α+d N. R d Choosing λ x = gives the result.

11 As a consequence, we get L j f k t, x = l Γ k j,l t t, f l t, dt.4 where the functions Γ k j,l satisfy Γk j,l τ, z τ + z 4 Proof of Proposition.. Young inequality ans definition of the spaces K σ p gives L j ft, L p Γt t, L p ft, L p dt f K p 3 p t 3 p f K p 3. p t t + 3 p t 3 p dt This proves the proposition. As K 3 p p K 3 p p K 3 p p, we proves that B maps continuously K 3 p p K 3 p p into K 3 p p. Thus, we have the following theorem Theorem.. If e t u K 3 p p of NS in K 3 p p. is small enough, then there is a unique global solution Now we have to understand what is the space of distribution u such that e t u belongs to K 3 p p. This is the family of Besov spaces..3 Besov spaces of negative index In this section, we interpret Theorem.. in term of Besov spaces. Let us introduce the following spaces. Definition.3. Let s be be in ], ] and p, r in [, ]. We define the space Bp,r s as the space of tempered distribution u such that def u B s = p,r t s e t u L p L r R +, dt t <. Let us point out that as, for q p, e t u = e t e t u and Thus we get that e τ LL p ;L q Cτ d u Ḃ s+d p q q p q..5 C u Ḃ s p..6 Homogeneous functions of negative degree will also belong to some Besov spaces. We have the following proposition.

12 Proposition.3. Let p be in ], ] and α in ]d/p, d[. Then a homogeneous function f of degree α of R d \{} which is bounded on the sphere of R d belongs to Ḃ α+ d p p. Proof. We have, by changing of variable and because homogeneïty of f that e t fx = t α x t e f As f is bounded on the sphere of R d, we have fx C x α As α belongs to ]d/p, d[, the function f belongs to L + L p. Thus e f belongs to L p. After a change of variable, we get the result. From this, we can infer the following theorem about so called self similar solutions of NS. Corollary.3. Let u be a smooth divergence free vector field on R 3 \{} homogeneous of 3 degree. Then, if u is small enough in Ḃ + p p,, then there exists a unique solution of NS which is self-similar in the sense that it satisfies ut, x = x t U with Ux = u, x. t Proof. Using the scaling invariance of the Navier-Stokes equation, we have that, for any positive λ, λuλ t, λx is the global solution with initial data λu λx which is equal u x because of the homogeneity. Thus, for any positive λ, we have ut, x = λuλ t, λx Choosing λ = t gives the result. Proposition.3. Let φ be a function of SR d, p in ], [ and s in in ], ] [, [, let us define ], d p [. For ε, Λ φ ε,λ x def = e i x ε φx, Λx, x 3. Then, we have φ ε,λ Ḃ s C φ ε s Λ p and φ ε,λ p, Ḃ d p p, C φ ε s Λ p. Proof. Using.5, we get that Thus we infer ε t s e t φ ε,λ L p C φ t s Λ p. t s e t dt φ ε,λ L p t C φε s Λ p..7 Now, let us assume that t ε. We can assume without loss of generality Using that ω =,,,. We have iε k e i x ω ε = k e i x ω ε.

13 We get after k integrations by parts and Leibnitz formula, t s e t φ ε,λ = iε k t s k Ck l l= t d + l f l t e ix ω ε φ k l x, Λ x, x 3 where f l and φ k l are functions of SR d. As Λ is assume to be greater than, convolution inequalities give t s e t φ ε L p C φ,k Λ p k l= { t l s, t l s+ d } p min. Let us assume that k > s. If k l > s, we use ε k If l s, then we use that ε k If k is large enough, we get the result. t + ε t + s l dp ε Now, let us prove a bound from below. s l dt ε s. dt ε k+s l d p. Proposition.3.3 Let φ be a function of SR 3 and s in ], 3[. For ε, Λ in ], ] [, [, let us define φ ε,λ x def = e i x ε φx, Λx, x 3. Then, if Λε is small enough, we have φ ε,λ Ḃ s, C φε s. Proof. Let us first observe that, as the space of smooth compactly supported functions is dense in S and the Fourier transform is continuous on S. Thus, for any positive η, a function ϕ exists, the Fourier transform of which is smooth and compactly supported such that, denoting as before θ ε,λ x = e i x 3 ε θx, Λx, x 3, φ ε θ ε Ḃ σ, ηεσ and φ θ L η..8 As the support of the Fourier transform of θ is included in the ball B, R for some positive R, that of θx, Λx, x 3 is included in the ball B, RΛ. Then the support of Fθ ε,λ is included in the ball Bε,,, ΛR which can be written as B,,, ΛεR ε If λε is small enough, we can assume that this set is included in ε C where C denotes a fixed ring. By definition of B s,, we have θ ε,λ Ḃ s = sup t s, e t θ ε,λ L t> Cε s e ε θ ε,λ L. 3

14 For any function h such that the support of ĥ is included in ε C, we have F e ε ξ h L C h L. Applied with h = e ε θ ε,λ, this inequality gives θ ε,λ L C e ε θ ε,λ L and thus θ ε,λ Ḃ s, C ε s θ ε,λ L = C ε s g L. Now let us write that φ ε Ḃ s, θ ε,λ Ḃ s, ηεs Together with.8, this gives the proposition. C ε s φ L η. Let us give another example of estimates of Besov norm of negative index in the case of slowly varying functions. Proposition.3.4 Let f, g be in SR and SR respectively. Let us define We have, if ε is small enough, h ε x h, x 3 def = fx h gεx 3. h ε Ḃ, R 3 4 f Ḃ, R g L R. Proof. By the definition of Ḃ, R 3, we have to bound from below et h ε L R 3. Let us write that e t h ε t, x = e t h ft, x h e t 3 gε t, εx 3. Let us consider a positive time t such that Then we have t e t h f L R f Ḃ, R. t e t h ε L R 3 = t e t h f L R et 3 gε t, ε L R f Ḃ, R eε t 3 g L R. As lim e ε t 3 g = g in L R, the proposition is proved. ε As a conclusion of this section about Besov space, let us prove the following property. Proposition.3.5 If σ is a homogeneous Fourier multiplier of order m and s a positive number such that s + m >. Then σd maps continuously Bp,q s into Bp,q s m. Proof. For the sake of simplicity, let us write it only for m =. as for any function v, we have v = τ s s + e τ dτ 4

15 Using the semi group law on the heat flow and splitting s into two parts, we infer that t s e t σdu = C s t s σd s 4 e t Using Lemma.., we have t τ t s s 4 + e τ e τ udτ..9 σd s 4 e t a L p t s 4 a L p and s 4 + e τ a L p τ s 4 a L p. Plugging this into.9 gives t t s e t σdu L p t s 4 t τ s τ 4 e dτ u L p τ Using Hölder inequality with respet to the measure τ dτ, we get τ s 4 + s τ 8 e u L pτ s 8 dτ q t qs8 τ t τ 3qs 8 e τ u q L p dτ τ Thus we infer that t qs e t u q dt R + L p t R + τ t qs 8 R + τ qs e τ u q L p dτ τ dt τ 3qs τ 8 e u q dτ L t p τ This concludes the proof of the proposition..4 The endpoint space for Picard s scheme The following proposition guarantees that it is hopeless to go beyond the space B,. Proposition.4. Let E be a Banach space continuously embedded in the set S R 3. Assume that, for any λ, a in R + R 3, Then E is continuously embedded in Ḃ,. fλ a E = λ f E. Proof. As B is continuously included in S, we have that f, e C f B. Then by dilation and translation, we deduce that This proves the proposition. f Ḃ = sup t, e t f L C f B. t> It turns out however that Ḃ, is too large a space. The mayine reason why is that if we want to solve the problem using an iterative scheme then we need that e t u belongs to L loc R+ R 3 so that Be t u, e t u makes sense. Taking into consideration the scaling and the translation invariance thus leads to the following definition. 5

16 Definition.4. We denote by X the space of tempered distributions u such that u X def = u Ḃ, + sup x R 3 R> R 3 P x,r e t uy dy dt < where P x, R = [, R ] Bx, R and Bx, R denotes the ball of R 3 of center x and radius R. We denote by X be the space of functions f on R + R 3 such that def f X = sup t ft L + sup t> x R 3 R> R 3 P x,r ft, y dy dt We denote by Y the space of functions on R + R 3 such that def f Y = sup t ft L + sup R 3 ft, y dy dt <. t> x R 3 P x,r R> < Proposition.4. For any p in ]3, [, the space Ḃ + 3 p p, is continuously embedded in X. Proof. Let us notice that for any x R 3 and R >, we have R Bx,R e t u y dy dt µbx, R p R Bx,R e t u y p p dy dt. 3 By definition of the space Ḃ + p p, we have R Bx,R e t u y dy dt u Ḃ + 3 p p µbx, R p R t + 3 p dt which obviously entails the announced embedding. Proposition.4.3 The space Ḃ, is included in X. Proof. As Ḃ, is included in B,, we have sup t e t u L C u B.. t>, Moreover, we have R R 3 e t fy dydt Bx,R e t f L dt C u Ḃ., Together with., this proves the proposition. 6

17 As X X Y, the fact that we have a global unique solution for small initial data in X will follow from the following proposition. Theorem.4. Let us define the operators L j jn by { t L j f L j f = j f p div v = and L j t= =. The operators L j maps continuously Y into X. Proof. Using Lemma.4 page, we get that L j f k t, x = with for all positive real number R, with Γ def R τ, ζ = ζ R ζ 4 3 Γ k j,l t t, x yf l t, y dt dy l= Γ k j,l τ, ζ C τ + ζ 4 C Γ R τ, ζ + Γ R τ, ζ and Γ R τ, ζ def = ζ R τ + ζ 4 The operators of convolution with functions Γ R the following proposition. and Γ R Lemma.4. There exists a constant C such that, for any R >, Proof. Let us decompose Γ R Γ R ft, x may be bounded according to Γ R f L [,R ] R 3 C R f Y,. Γ R f L [R, [ R 3 C R f Y.. ft, x as a sum of integrals on annulus: p= R B, p+ R\B, p R p+3 p+ R 3 p= As p is nonnegative, we have for t R, Γ R ft, x C R C R p p+ R 3 p= p= By definition of Y, Inequality. is proved. y 4 ft, x y dy dt ft, x y dy dt. B, p+ R P x, p+ R ft, z dt dz p sup R > R 3 ft, z dt dz. P x,r 7

18 In order to prove the second inequality, let us observe that for all x R 3 and t R, we have Γ R ft, x Γ R t, x + Γ R t, x with minr Γ R t, x def, t = Γ R t, x def = minr, t B,R B,R For bounding Γ R t, x, we use that t t t. We get Γ R t, x C R3 so that, for any t R and x in R 3, t B,R t t + y 4 ft, x y dy dt, t t + y 4 ft, x y dy dt. R R 3 ft, x y dt dy B,R Γ R t, x C f t Y..3 In order to estimate Γ R, let us use that t t and that, for any a >, dy a + y 4 dz a + z 4 This enables us to write that Γ R t, x minr, t C f Y C f Y t t/ B,R R 3 t t + y 4 ft, L dy dt, dt t t + R tr 3 t As R t, this concludes the proof of the lemma. Proof of Lemma.4. continued. yields t +. µb, R dt R t t Note that applying the above proposition with R = t L j ft, L C f t Y..4 Hence, it suffices to estimate L j f L P x,r for an arbitrary x R 3. Using translations and dilations, we can assume that x = and R =. Let us write Observing that, for any y B,, we have and using Inequality., we get L j f = L j c B,f + L j B, f. L j c B,ft, y CΓ c B, f t, y, L j c B,f L P, C f Y. 8,

19 As the volume of P, is finite, we infer that L j c B,f L P, C f Y..5 Now the proof of Lemma.4. is reduced to the proof the following Lemma. Lemma.4. For any function f : [, ] R 3 R such that ft, be supported in B, for all t [, ], we have L j ft, L [,] R 3 C f Y Proof. Let us point out that, for any t, L j ft = P L j ft where L j is the solution of As P is an orthogonal projection in L, we get t Lj f L j f = j f and L j f t= =. L j ft, L L j ft, L..6 Let us decompose f into low and high frequencies in the sense of the heat flow: f = f + f with f t, def = F θt ξ ft, ξ where θ denotes a function such that θ be compactly supported and with value near the origin. Let us write that f L [,];Ḣ = π 3 θt ξ [,] R 3 t ξ t ft, ξ dt dξ C t ft, [,] R 3 L dt C f L [,] R 3 sup t ft, L. t> So using the energy estimate on the heat equation and.6, we end up with L j f L [,] R 3 C f Y..7 Now let us estimate L j f L [,] R 3. Let us first observe that, by definition of L j and f, we have F L j f t, ξ = iξ j e t t ξ f t, ξ dt = iξ j e t ξ F f t, ξ dt with F f t, ξ def = e t ξ θt ξ ft, ξ. Let us notice that, by definition of θ, we have that f t, = t 3 θ ft, t with θ SR 3..8 Thus, using.6, we get 3 L j f L [,] R 3 j= N f with N f def = e t f t dt L dt. 9

20 By symmetry, we can write N f = A A e t f t e t f t L dt dt dt with def = {t, t, t [, ] 3 / t t t}. By integration by parts and because e t is self-adjoint on L, we get e t f t e t f t L = e t f t, f t. For any positive t and t such that t t, the function f t, and f t, have Fourier transform with compact support in a ball of radius C/ t. In this space denoted in FL t, we have, in the sense of LFL t, We infer that t, t ], ], By integration, we deduce that e t = d dt et. e t f t e t f t L = t e t f t e t f t L dt = Thanks to Fubini theorem, we deduce that N f = e t e t d e t f t dt f t. L d e t f t f t dt dt L = e t e f t f t L By definition of the L inner product, we infer that N f f L [,] R 3 sup t [,] f t dt f t L dt e t e f t dt L. Using.8, we infer that N f f L [,] R 3 sup t [,] e t e f t dt L..9 First of all, let us notice that, using.8 and the fact that operator e maps L R 3 in L R 3, we have e f t dt L C f L [,] R 3. Thanks to.8, we have, as f is supported in the ball B,, Thus, we get t [, ],, f t, L R 3 C f Y.. e Let us admit for a while the following two lemmas. f t dt L C f Y..

21 Lemma.4.3 Let θ in SR 3. Let us define f θ t, def = t 3 θt ft, We have f θ Y C f Y. Lemma.4.4 Let us define Eft = e t ft dt. Then a constant C exists such that for any function f in Y and supported in [, ] B,, we have Ef L [,] R 3 C f Y Conclusion of the proof of Theorem.4. These two lemmas imply that sup e t t [,] f t dt L C f Y. Inequalities.4,.7 and. allows to conclude the proof of the theorem. Proof of Lemma.4.3 Let us first obverse that, for any t, we have Thus we have ft, L θ L ft, L. t ft, L θ L t ft, L.. Now, let us write that, for any x in the ball of center and radius R, we have f θ t, x t 3 x y B,R θ y ft, y dy R 3 t + Ct 3 t 4 ft, y dy R 3 + x y R t Thus, we infer t 3 R 3 f θ L P,R C R 3 This proves Lemma.4.3. θt B,R ft, P,R x + C R sup t> t ft, L. ft, y dt dy + C sup t ft, L. t> Proof of Lemma.4.4 Because of translation invariance, it is enough to bound I f t def = e t ft, dt. We decompose the space R 3 as a disjoint union of cubes of center n t and radius t where n = n, n, n 3 is the generic point of Z 3. This gives I f t = e y n Z 3 t 8πt 3 8t ft, ydt dy. [,T ] Bn t,

22 As y belongs to Bn t, t, we have y t n Thus we get I f t e n 8 n Z 3 f Y f Y. t 3 e n 8 n Z 3 P n t, t ft, y dt dy This concludes the proof of Lemma.4.4. Of course, this proves the following theorem Theorem.4. If e t u is small enough then there is a unique global solution to NS associated with u..5 An abstract non linear smallness condition Let us define a space which is a good space for being an external force that Theorem.4.. Definition.5. We shall denote by E the space of functions f in L R + ; Ḃ, such that sup t ft B is finite equipped with the norm, t> f E def = f L R + ;Ḃ, + sup t> t ft B,. Theorem.5. There is a constant C such that the following result holds. Let u be a divergence free vector field in B,. Let us assume that e t u e t u E C exp C U 4 with U def = u + u Ḃ 4 4, Ḃ,4.3 Let us define Ut = e t u L + t et u 4 L 4 and v λ t, x def = exp λ Ut dt vt, x. Then there is a unique global solution to NS such that ut e t u λ X C exp CU 4. The proof of this theorem is the purpose of all this section.

23 .5. Main steps of the proof Let us start by remarking that in the case when u is small then there is nothing to be proved, so in the following we shall suppose that u Ḃ is not small, say u Ḃ.,, We search for the solution u under the form Then R is the solution of u L + R where u L t def = e t u. NS R = Bu L, u L + Bu L, R + BR, R. To prove the global existence of u, we are reduced to prove that MNS has a global solution. We use the following easy lemma, the proof of which is omitted. Lemma.5. Let X be a Banach space, let L be a continuous linear map from X to X, and let B be a bilinear map from X X to X. Let us define def L LX = sup Lx and def B BX = sup Bx, y. x = x = y = If L LX <, then for any x in X such that x X < L LX 4 B BX, the equation x = x + Lx + Bx, x has a unique solution in the ball of center and radius L LX B BX Let us introduce the functional space for which we shall apply the above lemma. We define the quantity Ut def = u L t L + t u Lt 4 L, which satisfies, tby definition of Besov spaces, Utdt C u Ḃ + C u 4 Ḃ,,4 C u 4 Ḃ,.4 recalling that we have supposed that u Ḃ to simplify the proof., For all λ, let us denote by X λ the set of functions on R + R 3 such that def v λ = sup t vλ t L + sup t> x R 3 R> R 3 P x,r v λ t, y dy <,.5 where v λ t, x def = vt, x exp λ Ut dt 3

24 while P x, R = [, R ] Bx, R and Bx, R denotes the ball of R 3 of center x and radius R. Let us point out that, in the case when λ =, this is exactly the space of Definition.4. page 6. For any non negative λ and that for any λ we have due to.4, v λ v X C v λ exp Cλ u 4 Ḃ,..6 As B maps continously X X into X, we infer that Bv, w λ C v λ w λ exp Cλ u 4 Ḃ..7, Theorem.5. follows from the following two lemmas we admit for a while. Lemma.5. There is a constant C > such that the following holds. For any non negative λ, for any t and any f E, we have e t t ft dt λ C f E. Lemma.5.3 Let u Ḃ, be given, and define u Lt = e t u. There is a constant C > such that the following holds. For any λ, for any t and any v X λ, we have Bu L, vt λ C v λ λ. 4 Conclusion of the proof of Theorem.5. Let us apply Lemma.5. to Equation NS satisfied by R, in a space X λ. We choose λ so that according to Lemma.5.3, Bu L, t LXλ 4 Then according to Lemma.5., there is a unique solution R to NS in X λ as Bu L, u L satisfies Bu L, u L Xλ 6 B BXλ as soon But.7 guarantees that So, if B BXλ exp λ u 4 Ḃ,, Bu L, u L Xλ C exp λ u 4 Ḃ,, by Lemma.5., there is a unique solution of MNS. The above condition is exactly re is precisely condition.3 of Theorem.5., so under assumption.3, there is a unique small in the sense of λ solution R to NS. Proof of Lemma.5. Thanks to.6, it is enough to prove Lemma.5. for λ =. Let us start by proving that L f def = observe that L ft, L e t t ft dt belongs to L R + ; L. Let us e t t ft L dt. 4

25 Let us observe that, after a change of variable L f L R + ;L sup e τ ft L ϕt + τdτ dt. ϕ L R + R + By Cauchy-Schwarz inequality and by definition of the norm on B,, we get L f L R + ;L e τ ft L dt f L R + ;B,. In order to estimate t ft L, let us write that t L ft, L t t e t t ft L dt + t ft B, f B, + sup ft B., t> t t t t t t e t t ft L dt dt + sup ft B, t> t t t t dt This proves Lemma.5.. Proof of Lemma.5.3 From Proposition.4 page, we have Bv, wt, x = R 3 kt t, yvt, x ywt, x ydydt = k vwt, x with kτ, ζ The proof relies now mainly on the following proposition. C τ + ζ 4 Proposition.5. Let u Ḃ, be given, and define u Lt = e t u. There is a constant C such that the following holds. Consider, for any positive R and for τ, ζ R + R 3, the following functions: K def R τ, ζ = ζ R ζ 4 and K R τ, ζ def = ζ R τ + ζ 4 Then for any λ and any R >, e λ Ut dt K R u Lv L [,R ] R 3 C λ R v λ..8 Moreover, for any λ and any R >, e λ Ut dt K R u Lv L [R,R ] R 3 C λ 4 R v λ..9 5

26 Proof. Let us write that V λ t, x def = e λ Ut dt K R c B,R u Lvt, x y 4 e λ t Ut dt u L t, L v λ t, x y dt dy. By the Cauchy-Schwarz inequality and by definition of U, we infer that V λ t, x c B,R C t λr y 4 e λ t Ut dt u L t, L dt dy c B,R c B,R Now let us decompose the integral on the right on rings; this gives c B,R y 4 v λt, x y dt dy = As t R and p is non negative, we have c B,R y 4 v λt, x y dt dy y 4 v λt, x y dt dy..3 p= R B, p+ R\B, p R p+3 p+ R 3 p= y 4 v λt, x y dt dy C R By definition of λ, we infer that c B,R C R B, p+ R p p+ R 3 p= p= Then, using.3, we conclude the proof of.8. y 4 v λt, x y dt dy v λ t, x y dtdy. P x, p+ R v λ t, z dtdz p sup R > R 3 v λ t, z dtdz. P x,r y 4 v λt, x y dt dy C R v λ. In order to prove the second inequality, let us observe that K R t, x def = K R t, x def = e λ Ut dt K t B,R B,R R u Lvt, x K t, x + K t, x with R t t + y 4 e λ t Ut dt u L t, L v λ t, x y dt dy, t t + y 4 e λ t Ut dt u L t, L v λ t, x y dt dy. R 6

27 Using the Cauchy-Schwarz inequality, as t [R, R ] and t t t, we infer that K R t, x so that C λ C tλ B,R B,R R 3 t t + y 8 e λ t Ut dt u L t, L dt dy v λ t, x y dt dy B,R dy R + y 8 v λ t, x y dt dy B,R R B,R In order to estimate K R, let us write that K R t, x R 3 t t + y 4 e λ t C v λ t v λ t, x y dt dy, K C R t, x v tλ λ..3 t Ut dt u L t, L v λ t, L dt dy t t e λ t Ut dt u Lt, L By definition of U and using the fact that t t, Hölder s inequality implies that K R t, x C t C λ 4 t v λ v λ. t dt. e 4λ t Ut dt t u L t, 4 L dt 4 Together with.3, this concludes the proof of the proposition. From this proposition, we infer immediately the following corollary. This corollary proves directly one half of Lemma.5.3, as it gives a control of Bu L, v in the first norm out of the two entering in the definition of X λ. Corollary.5. Under the assumptions of Proposition.5., we have Proof. Let us write that t e λ Ut dt Bu L, vt, L C v λ λ. 4 k u L v t, x = k u L c Bx, t v t, x + k u L Bx, t v t, x. From Proposition.5., we infer that e λ Ut dt k u L c Bx, t v t, x e λ Ut dt K u t L Bx, t v t, x C v tλ λ. 7

28 Moreover, thanks to Proposition.5., we have also e λ Ut dt k u L Bx, t v t, x e λ Ut dt K u t L Bx, t v t, x C v λ 4 t λ. This proves the corollary. Proof of Lemma.5.3 continued Let us estimate k u L v L P x,r, for an arbitrary x in R 3. Let us write that k u L v = k u L c Bx,Rv + k u L Bx,R v. Observing that, for any y Bx, R, we have k u L c Bx,Rvt, y CK R and using Inequality. of Proposition.5., we get u L c Bx,R v t, y, e λ Ut dt k u L c Bx,Rv L P x,r C λ R v λ. As the volume of P x, R is proportional to R 5, we infer that k u L c Bx,Rv λ L P x,r C λ R 3 v λ..3 Now let us prove the equivalent of Lemma.4.. Lemma.5.4 For any T, we have Bu L, v λ L [,T R 3 C v λ L λ [,T R 3 Proof. Let us write that FBu L, v λ t, ξ exp t t ξ λ Ut 4 dt Fu L t v λ t ξ dt. t We can write that Fu L t v λ t ξ at, ξ u L t L v λ t L with a t, ξdξ =. R 3 Thus using Cauchy-Schwarz inequality, we infer that T T FBu L, v λ t, ξ dt exp t t ξ λ Ut 4 dt t T λ 4λ at, ξ u L t L v λ t L dt dt exp λ Ut 4 dt u L t L dt t t ξ exp t t ξ a t, ξ v λ t L dt dt T T T An integration with respect to ξ gives the result. t ξ exp t t ξ dt a t, ξ v λ t L dt a t, ξ v λ t L dt 8

29 Conclusion the proof of Lemma.5.3 Together with Corollary.5. and.3, this concludes the proof of Lemma A particular case of large oscillating data It is not obvious that Theorem.5. is not empty. Of course, the non linear smallness condition is satisfied in the case when u Ḃ is small. Let us first state the theorem that presents a, class of large oscillating initial data satisfying hypotheses of Theorem.5.. Proposition.6. Let ϕ be a function in SR 3. Let us consider P = ε, Λ in ], ] [, [ such that ελ is small enough. Let us define ϕ ε x = cos x3 ε ϕx, Λx, x 3. If A Λ 3 C exp C A 4, the divergence free vector field u x = A ελ ϕ P x, ϕ P x, satisfies and the hypotheses of Theorem.5.. u Ḃ c A.33, Proof. The fact that.33 is satisfied is an obvious consequence of Proposition.3.3. As P is a Fourier multiplier operator of order, we have P e t u e t u E C e t u e t u E. Let us observe that e t u e t u + e t u e t u = e t u e t u + e t u e t u = A e t f P e t g P and.34 ε A ε Λ et fp e t g P.35 where f, g, f and g are functions in SR 3. Now, the main lemma is the following. Lemma.6. There is a constant C such that the following result holds. Let in Ḃ, Ḣ. Then we have e t fe t g E C f Ḃ, g Ḃ, 3 f Ḃ g Ḃ,, 3 f and g be Proof. Let us first prove the following interpolation result. Lemma.6. A constant C exists such that, for a and b in L L, we have ab B a, L b 3 L a L b L 3. 9

30 Proof. By definition of the B, norm, we have, for any positive τ, ab B, a L b L Chossing τ = a L b L = τ e τ ab L dτ + a L b L τ τ e τ ab L dτ dτ τ 3 + a L b L τ a L b L + τ a L b L. 3 gives the result. Proof of Lemma.6. continued Applying the above lemma this a = e t f and e t g, this gives by Hölder inequality and by the definition of Besov spaces, e t fε t g dt B, e t f L e t g L 3 τ dτ e t f L e t g L 3 dt e t f L R + ;L et g L R + ;L f B, Along the same lines, let us write that 3 e t f L R + ;L et g L R + ;L 3. e t g 3 B f, B e t g, B, 3 t e t fε t g B, t et fε t g B, This proves Lemma.6. t e t f L t e t g L f B e t g, B, f B, e t g B, 3 t e t f L t e t g L 3 3 f B et g, B, 3 f B e t g 3, B., Conclusion of the proof of Proposition.6. Then, using.34 and.35, we infer that P e t u e t u E A ε f P Ḃ, A ε ε 4 3 ε Λ 3 g P Ḃ, 3 3 f P Ḃ, g P Ḃ, 3 Thus if Λ 3 the hypotheses of Theorem.5. A Λ 3. C A C exp C A 4 Remark Let us point out the importance of the algebraic structure of a non linear term. A term like e t u e t u will produce terms we can bound only in E by A Λ 3. 3

31 .7 The role of the special structure: a Navier-Stokes type equation which blows up The incompressible Navier-Stokes system has three important features: the scaling invariance, the incompressibilty condition and the very special structure of the non linear term. This structure leads to energy estimate, but also appears in relations likes.34 and.35 which has been crucial for proving the global well posedness result of Theorem.6.. The purpose of this section is to study a modifed system which has the first two properties scaling invariance and divergence free condition and with a different structure of the non linear term which will lead to blow up at finite time for some initial data that satisfies the hypotheses of Theorem.6., and the matrix qξ qξ = Eξ ξ E def = { ξ R 3, ξ ξ <, ξ ξ 3 < and ξ < min{ ξ, ξ 3 } }.36 ξ + ξ 3 ξ ξ ξ ξ 3 ξ + ξ 3 ξ ξ ξ ξ 3 ξ + ξ 3 ξ ξ ξ ξ 3 ξ + ξ 3 ξ ξ ξ ξ 3 ξ + ξ 3 ξ ξ ξ ξ 3 ξ + ξ 3 ξ ξ ξ ξ ξ + ξ ξ ξ 3 ξ ξ 3 ξ + ξ ξ ξ 3 ξ ξ 3 ξ + ξ ξ ξ 3 ξ ξ 3 Let us observe that qξ = ξ E ξ Pξ if Pξ denotes the matrix of the Leray projection of divergence free vector field in the Fourier space. Let us consider the following modified incompressible Navier-Stokes system. MNS { t u u = Qu, u with FQ j u, u def = div u =, u t= = u 3 E ξq j,k ξfu j u k. k= The main point of this modified Navier-Stokes system is the following property which plays a key role in the proof of blow up for finite time. Proposition.7. The coefficients of the matrix qξ are non negative. Proof. Let us first notice that on E, ξ ξ 3 is positive.the components of the first line may be written ξ + ξ 3 ξ ξ ξ ξ 3 = ξ ξ ξ + ξ 3 ξ 3 ξ which is also positive since either ξ and ξ 3 are both positive, in which case ξ 3 > ξ, or they are both negative in which case ξ 3 < ξ. Thus the first line of the above matrix is clearly made of positive scalars. The fact that the terms of the second line are non negative is obvious due to the sign condition imposed on the components of ξ. Similarly one has ξ + ξ ξ ξ 3 ξ ξ 3 = ξ + ξ ξ 3 ξ + ξ and either ξ <, ξ >, ξ 3 < and ξ + ξ >, or ξ >, ξ <, ξ 3 > and ξ + ξ <. So the third line is also made of positive real numbers. 3

32 Theorem.7. Let us consider an initial data u such that, for any j {,, 3}, the component û j is a non negative function. Let us assume that for some j, we have û j ξ v ξ with Supp v C r,r..38 where C r,r is some ring of R 3. Let us assume a positive real number m exists such that, for any ξ in the union of the iterated sum of Supp v, If the quantity q j,j ξ m ξ..39 m r R v L is large enough, then the unique solution to MNS associated with the initial date u blows up for finite time. Proof. As the positivity of the components of û is preserved by the flow of MSN, we have that û j ξ = e t ξ û j ξ + e t ξ û j ξ + 3 k= 3 k= e t t ξ q j,kξ û j t û k t, ξdt e t t ξ q j,j ξ û j t û k t, ξdt. As q j,j ξ is non negative, we get that, for any t, u j ξ vξ where v is the solution of t v v = q j,j Dv with v t= = v. We give here a variation of the proof of [5]. Let us define the sequence t k k N by def t = and t k = k R l. We use denote by T its limit which is 4/3R. Let us define the sequence v l l N by Let us notice that v l= def = v and v l def = v v l. Supp v l lc r,r and ξ Supp v l, qξ mrl. Let us make the following induction hypothesis for some sequence A k k N which will be choosen later on: H k t [T, t k ], vt, ξ A k v k ξ. Using the hypothesis on the support of v, we have that, for any t in [, T ], vt, ξ e tξ v ξ e T R v ξ. 3

33 Thus, if we choose A = e T R, H is satisfied. Now let us assume H k. As q j,j and vt, are non negative functions, we have, for any t t k+, vt, ξ = e t ξ v ξ + k t e t t ξ q j,j ξ vt, vt, ξdt e t t ξ dt A k q j,j ξ v k v k ξ. By definition of the sequence v l l N we get using.39, As t t k+, we have t vt, ξ t t k+r e t k k+ R t k. Thus we get dt mr k v k+. vt, ξ e t t k+r t k+ R 4e r R k A k vk+ ξ. dt mr k v k+ ξ Choosing A k+ def = m k A k k with m def = 4e m r R gives H k+. Let us compute A k. By iteration, we find that As k l= A k+ = m k+ k l= lk l A k+. l k l = k, we get that A k+ = 4 m e 8 k+ l 3. As v L = v l L, we infer that, for any t in [T ; T ], we have vt, ξ 4 m e 8 k+ 3 v L Thus, if m e 8 3 v L is large enough, then lim A k = + and thus ût, L k finite time and the theorem is proved. The purpose of this section is the proof of the following theorem. blows up for Theorem.7. Let φ be a function in SR 3 such that its Fourier transform is non negative, even and has its support in the region E. Let P = ε, Λ be in ], ] [, [ such that ελ is small enough, and A a positive real number. Let us consider the initial data If u x def = A ϕ P x, ϕ P x, with ϕ P x def = ελ cos x3 φx, Λx, x 3. ε Λ 4 3 C A exp C A 4 and A ε large enough, then u satisfies the hypothesis of Theorem.5. and the local solution to MNS blows up for finite time. 33

34 Proof. We have u x = A ε cos x3 ϕx, Λx, x 3, ε Λ ϕx, Λx, x 3. First of all, let us check that û j are non negative functions for j {,, 3} and that their support intersects the set ξ j /. Indeed we have û ξ = A ελ ξ ξ ϕ ξ, ξ Λ, ξ 3 ±, ε ± ± ξϕ ξ, ξ Λ, ξ 3 ± ε which gives the non negativity of the Fourier transform of the components of u. Then let us consider a point ω such that ω ω ϕω c and a real number ε such that Let us define v by As we have v ξ def = A ελ w ξ, ξ Λ, ξ 3 ± ε ξ Bω, ε, ξ ξ ϕξ c. with w η def = Bω,ε ηη η ϕη. { A v L c ελ µ ξ R 3 / ξ, ξ } Λ, ξ 3 Bω, ε µbω, ε c A ε, we infer that if A/ε is large enough, the hypotheses of Theorem.7. are satisfied and thus the theorem is proved..8 References and remarks The Koch and Tataru theorem has been originally proved in [4]. The second and third section are adapted from [5]. Let us mention that in the framework of periodic boundary conditions, a type of non linear smallness condition has been introduced in [4]. The methods for proving blow up in section.7 has been introduced in [49] and the modified MNS appeared in [9]. 34

35 Chapter Slowly varying vector fields Introduction In this chapter, we continue to investigate the use of the structure of the incompressible 3D Navier-Stokes equation to construct large global solution. The idea is to work is a situation close to the D situation, i.e. in a situation which vector fields vary slowly in one direction, nameley are of the form u,ε x h, x 3 = v,ε x h, εx 3 where x h belongs to a two dimensionnal domain and x 3 belongs to R and where derivatives of v,ε are bounded with respect to ε in say L space. Of course, the divergence free condition will constrain the properties of the profile v. Indeed, let us write the profile v,ε = v h,ε, v 3,ε = v,ε, v,ε, v 3,ε the divergence free condition on u,ε implies that div h v h,ε + ε 3 v 3,ε =. Thus the form of the initial data is v,εx h h, εx 3, ε v3,εx h, εx 3 where the profile v,ε = v,ε h, v3,ε = v,ε, v,ε, v3,ε is a smooth enough vector field, uniformely with respect to the parameter ε. Because of Proposition.3.4, such initial are very large. For technical reason, it will be necesarry to work in the domain T R. In the firs section, we explain how the problem reduces, after a rescaling with respect to the vertical variable, to a resolution of a system with vanishing viscosity in the vertical variable and a modified gradient of the rescaled pressure. This system looks illposed. In the second section, we present a global Cauchy-Kovalevska mehod in the model case { t u + γu + ADu = u t= = u where γ is a positif real number and AD a Fourier mulptiplier of order. We prove global a priori estimate for small analalytic data. We method used consists in defining an admissible rate of decay of the radius of analyticity. 35

36 In the third section, we introduce, in the case of the real problem the phase function which describe the rate of decay of the radius of analyticity in the vertical variable. The role of the horizontal and the vertical component in this definition is very different. Then, we reduce the proof of the global wellposedness problem to the proof of two propositions which describe how the control the dacay of the radius of analyticity. The rest of the chapter is devoted to the proof of these two propositions.. Ill prepared data: the vertical rescaling The theorem we want to in this case in the following. Theorem.. Let a be a positive number. There are two positive numbers ε and η such that for any divergence free vector field v satisfying e a D 3 v H 4 η, then, for any positive ε smaller than ε, the initial data u,ε x def = v h x h, εx 3, ε v3 x h, εx 3 generates a global smooth solution of NS on T R. Let us notice that for the sake of simplicity, we assume that the profile v does not depend on ε. The theorem is also true in the case when v,ε is a family of profile such that, for all ε, e a D 3 v,ε H 4 η. We look for the solution under the form Let us notice that Moreover, u ε t, x def = vε h t, x h, εx 3, ε v3 εt, x h, εx 3. u ε t, x h, x 3 = ε v ε x h, εx 3 with ε def = + + ε 3 = h + ε 3. u ε fx h, εx 3 = v h h f x h, εx 3 + ε v3 x h, εx 3 3 fx h, εx 3 = v fx h, εx 3. This leads to the following rescaled Navier-Stokes system. RNS ε t v h ε v h + v v h = h q t v 3 ε v 3 + v v 3 = ε 3 q div v = v t= = v Here, we already notice the importance of the structure of the non linear term which prevents from terms of size ε which would appear with terms like u 3 h. 36

37 Let us notice that the system RNS ε is far away from the following system ANS ε For this system, it is proved in [], that if t v h ε v h + v v h = h p t v 3 ε v 3 + v v 3 = 3 q div v = v t= = v v L 3 v L c, with c independent of ε then the system ANS ε is globally wellposed. The fact that the about system satisfied the L energy estimate is crucial. Because of the term ε 3 q, the system RNS ε does not safisty any conservation of energy. Asthere is no boundary, the rescaled pressure q can be computed with the formula ε q = j,k j v k k v j = j,k j k v j v k.. It turns out that when ε goes to, ε looks like h. In the case of R3, for low horizontal frequencies, an expression of the type h ab cannot be estimated in L in general. This is the reason why we work in T R. In this domain, the problem of low horizontal frequencies reduces to the problem of the horizontal average that we denote by Mfx 3 def = fx 3 def = fx h, x 3 dx h. T Let us also define M f def = Id Mf. Notice that, because the vector field v is divergence free, we have v 3. The system RNS ε can be rewritten in the following form. RNS ε t w h ε w h + M v w h + w 3 3 v h = h q t w 3 ε w 3 + M v w 3 = ε 3 M q t v h ε 3v h = 3 Mw 3 w h divv + w = v, w t= = v, w. The problem to solve this sytem is that there is no obvious way to compensate the loss of one vertical derivative which appears in the equation on w h and v and also, but more hidden, in the pressure term. The method we use is inspired by the one introduced in [9] and can be understood as a global Cauchy-Kowalewski result. This is the reason why the hypothesis of analyticity in the vertical variable is required in our theorem. Let us denote by B the unit ball of R 3 and by C the annulus of small radius and large radius. For non negative j, let us denote by L j the space FL Z R j C and by L the space FL Z R B respectively equipped with the semi norms u L j def = π d j C ûξ dξ and u L def = π d B ûξ dξ. Let us now recall the definition of inhomogeneous Besov spaces modeled on L. 37

38 Definition.. Let s be a nonnegative real number. The space B s is the subspace of L such that u def B s = js u L j j l <. We note that u B s is equivalent to writing u L j Cc j js u B s where c j is a non negative series which belongs to the sphere of l. Let us notice that B 3 is included in FL and thus in the space of continuous bounded functions. Moreover, if we substitute l to l in the above definition, we recover the classical Sobolev space H s. The theorem we actually prove is the following. Theorem.. Let a be a positive number. There are two positive numbers ε and η such that for any divergence free vector field v satisfying e a D 3 v B 7 η, then, for any positive ε smaller than ε, the initial data u,ε x def = v h x h, εx 3, ε v3 x h, εx 3 generates a global smooth solution of NS on T R. Before entering in the technicalities of the proof, let us have a discussion about the hypothesis on analyticity of the intial data. It is known to be somehing unphysical like for instance in the problem of boundary layer of vanishing viscosity see the Prantl problem. Here, for any ε, a positive real number ρ ε exists such that if u u,ε Ḣ < ρ ε then u generates a global smooth solution. Indeed, writing the solution u associated with u as u = u ε + w ε where u ε is given by Theorem.., we have { t w w + w w + w u ε + u ε w = p div w = et w t= = w. Classical Ḣ estimates allows to prove that, as long as wt Ḣ c with c small enough, we have wt + wt dt w H Ḣ exp C Ḣ u ε t L dt w exp C Ḣ ε ea D 3 v 7 B Thus, if w Ḣ c exp C ε ea D 3 v 7 B then u,ε + w generates a global smooth solution. 38

39 . Study of a model problem In order to motivate the functional setting and to give a flavour of the method used to prove the theorem, let us study for a moment the following simplified model problem for RNS ε, in which we shall see in a rather easy way how the same type of method as that of [9] can be used as a global Cauchy-Kovalevska technic: the idea is to control a nonlinear quantity, which depends on the solution itself. So let us consider the equation t u + γu + adu =, where u is a scalar, real-valued function, γ is a positive parameter, and ad is a Fourier multiplier of order one. We shall sketch the proof of the fact that if the initial data satisfies, for some positive δ and some small enough constant c, e δ D u B 3 cγ, then one has a global smooth solution, say in the space B 3 as well as all its derivatives. The idea of the proof is the following: we want to control the same kind of quantity on the solution, but one expects the radius of analyticity of the solution to decay in time. Let us introduce the following notation, which will be used throughout this article. For any locally bounded function Ψ on R + Z R and for any function f, we define f Ψ t def = F e Ψt, ft,. Let us notice that this notation does not make sense for any f; the following can be made rigourous by a cut-off in Fourier space. This will done in the next section, in the proof of Theorem... So let us introduce the function θt which describes the loss of analyticity of the solution. We define θt def = u Φ t B 3 with θ = and Φt, ξ = δ λθt ξ.. The parameter λ will be chosen large enough at the end, and we shall prove that δ λθt remains positive for all times. The computations that follow hold as long as that assumption is true and a bootstrap will prove that in fact it does remain true for all times. Taking the Fourier transform of the equation gives Using the fact that ût, ξ e γt û ξ + C e γt t ξ Fu t, ξ dt. γt + δ λθt ξ γt t λ ξ t θt dt + γt + δ λθt ξ η + δ λθt η, we infer that e γt û Φ t, ξ e δ ξ û ξ + C e λ ξ t θt dt ξ e γt Fu Φ t, ξ dt. 39

40 Thus, for any ξ in j C, we have e γt û Φ t, ξ e δ ξ û ξ + C Taking the L j C, dξ norm gives e γt u Φ t, L j e δ D u L j + C e λj t θt dt j e γt Fu Φ t, ξ dt. e λj t θt dt j e γt u Φt, L j dt..3 Now, we need a lemma of paradifferential calculus type. The statement requires the following spaces, introduced in [7]. Definition.. Let s be a real number. We define the space L T Bs as the subspace of functions f of L T Bs such that the following quantity is finite: f L T B s def = j js f L T L j. Let us notice that L T Bs is obviously included in L T Bs. We shall also use a very basic version of Bony s decomposition: let us define T a b def = F âξ η bηdη and R a b def = F âξ η bηdη. j j C Bξ, j j j C Bξ, j+ We obviously have ab = T a b + R b a. Lemma.. For any positive s, a constant C exists which satisfies the following properties. For any function Ψ satisfying Ψt, ξ Ψt, ξ η + Ψt, η.4 for any function b, a positive sequence c j j Z exists in the sphere of l Z and depending only on T and b such that, for any a and any t [, T ], we have T a b Ψ t L j + R a b Ψ t L j Cc j js a Ψ t B 3 b Ψ L T B s. We prove only the lemma for R, the proof for T being strictly identical. Let us first investigate the case when the function Ψ is identically. We first observe that for any ξ in the annulus j C, we have FR a btξ = ât, ξ η bt, ηdη. j j By definition of, we infer that L T B s Defining c j = j j R a bt L j C at FL j j s c j which satisfies j j C Bξ, j + j j c j j s b L T B s. c j C s, we obtain R a bt L j C c j js at FL b L T B s..5 4

41 As B 3 is included in FL, the lemma is then proved in the case when the function Ψ is identically. In order to treat the general case, let us write that e Ψt,ξ FR a bt, ξ = e Ψt,ξ ât, ξ η bt, η dη j j C Bξ, j e Ψt,ξ η ât, ξ η e Ψt,η bt, η dη. Estimate.5 implies the lemma. Now let us return to.3. We write j j C Bξ, j e γt u Φt = T uφ t e γ t u Φ t + Re γ t u Φ t, u Φ t and Lemma.. gives, for all t T and as long as the function Φ is positive, e γt u Φt, L j Cc j T j 3 uφ t B 3 e γt u Φ t L T B 3. By definition of the function θ, this gives e γt u Φt, L j Cc j T j 3 θt e γt u Φ t L T B 3. Plugging this inequality in.3 after multiplication by j 3 gives, as long as the function Φ is positive, for all t T, j 3 e γt u Φ t, L j j 3 e δ D u L j + Cc j T e γt u Φ t L T B 3 As we get e λj t θt dt j θt dt λ, j 3 e γt u Φ t, L j j 3 e δ D u L j + C λ c jt e γt u Φ t L T B 3. e λj t θt dt j θt dt. Taking the supremum for t T, and summing over j, we get, as long as the function Φ is positive, e γt u Φ t, L T B 3 eδ D u B 3 + C λ eγt u Φ t L T B 3. Thus, choosing λ = C we infer that e γt u Φ t, L T B 3 eδ D u B 3. As a L T B 3 a L T B 3, we get, by definition of θ, as long as the function Φ is positive, which gives γθt e δ D u B 3. If θt e γt e δ D u B 3, e δ D u B 3 δγ 8C, then we get that the function Φ remains positive for all time and the global regularity is proved. 4