Unconditional uniqueness for the modified Korteweg de Vries equation on the line

Transcription

1 ev Mat Iberoam (online first) doi /rmi/1036 c European Mathematical Society Unconditional uniqueness for the modified Korteweg de Vries equation on the line Luc Molinet, Didier Pilod and Stéphane Vento Abstract We prove that the modified Korteweg de Vries (mkdv) equation is unconditionally well-posed in H s () fors>1/3 Our method of proof combines the improvement of the energy method introduced recently by the first and third authors with the construction of a modified energy Our approach also yields aprioriestimates for the solutions of mkdv in H s (), for s>0, and enables us to construct weak solutions at this level of regularity 1 Introduction We consider the initial value problem associated to the modified Korteweg de Vries (mkdv) equation (11) { t u + u 3 + κ (u 3 )=0, u(, 0) = u 0, where u = u(, t) is a real function, κ =1or 1,, t In the seminal paper [16], Kenig, Ponce and Vega proved the well-posedness of (11) inh s () fors 1/4 his result is sharp on the H s -scale in the sense that the flow map associated to mkdv fails to be uniformly continuous in H s () if s<1/4 inboththefocusingcaseκ = 1 (cf Kenig, Ponce and Vega [17]) and the defocusing case κ = 1 (cf Christ, Colliander and ao [3]) Global well-posedness (GWP) for mkdv was proved in H s () fors>1/4 by Colliander, Keel, Staffilani, akaoka and ao [5] byusingthei-method (see also [9], [18] for the GWP at the end point s =1/4) We also mention that another proof of the local wellposedness result for s 1/4 was given by ao by using the Fourier restriction norm method [28] On the other hand, if one eits the H s -scale, Grünrock [7] and Mathematics Subject Classification (2010): Primary 35A02, 35E15, 35Q53; Secondary 35B45, 35D30 Keywords: Modified Korteweg de Vries equation, unconditional uniqueness, well-posedness, modified energy

2 2 L Molinet, D Pilod and S Vento then Grünrock Vega [8] proved that the Cauchy problem is well-posed in Ĥr s for 1 <r<2ands 1/2 1/(2r), where u 0 Ĥr := ξ s û 0 s L r with 1/r +1/r =1 ξ ote that Ĥ1 0 is critical for scaling considerations and thus the result in [8] isnearly optimal on this scale whereas the inde 1/4 intheh s -scale is far above the critical inde which is 1/2 he proof of the well-posedness result in [16] relies on the dispersive estimates associated with the linear group of (11), namely the Strichartz estimates, the local smoothing effect and the maimal function estimate A normed function space is constructed based on those estimates and allows to solve (11) viaafiedpoint theorem on the associated integral equation Of course the solutions obtained in this way are unique in this resolution space he same occurs for the solutions constructed by ao which are unique in the space X s,1/2+ he question to know whether uniqueness holds for solutions which do not belong to these resolution spaces turns out to be far from trivial at this level of regularity his kind of question was first raised by Kato [11] in the Schrödinger equation contet We refer to such uniqueness in C([0,]:H s ()), or more generally in L (]0,[: H s ()), without intersecting with any auiliary function space as unconditional uniqueness his ensures the uniqueness of the weak solutions to the equation at the H s -regularity his is useful, for instance, to pass to the limit on perturbations of the equation as the perturbative coefficient tends to zero (see for instance [23] for such an application) Unconditional uniqueness was proved for the KdV equation to hold in L 2 (), see [29], and in L 2 (), see [1]; and for the mkdv in H 1/2 (), see [20] he aim of this paper is to propose a strategy to show the unconditional uniqueness for some dispersive PDEs and, in particular, to prove the unconditional uniqueness of the mkdv equation in H s () fors>1/3 ote that, doing so, we also provide a different proof of the eistence result Before stating our main result, we give a precise definition of our notion of solution Definition 11 Let > 0 We will say that u L 3 (]0,[ ) isasolution to (11) associated with the initial datum u 0 H s () ifu satisfies (11) inthe distributional sense, ie for any test function φ Cc (],[ ), there holds [ (12) (φt + φ)u 3 + φ u 3] d dt + φ(0, )u 0 d =0 0 emark 12 ote that L (]0,[: H s ()) L 3 (]0,[ ) assoonass 1/6 Moreover, for u L (]0,[:H s ()), with s 1/6, u 3 is well-defined and belongs to L (]0,[: L 1 ()) herefore (12) forces u t L (]0,[:H 3 ()) and ensures that (11) is satisfied in L (]0,[:H 3 ()) In particular, u C([0,]:H 3 ()) and (12) forces the initial condition u(0) = u 0 ote that, since u L (]0,[: H s ()), this actually ensures that u C w ([0,]:H s ()) and that u C([0,]: H s ()) for any s <s Finally, we notice that this also ensures that u satisfies the Duhamel formula associated with (11) inc([0,]:h 3 ())

3 Unconditional uniqueness for mkdv 3 heorem 13 Let s>1/3 be given (Eistence) For any u 0 H s (), there eists = ( u 0 H s) > 0 and a solution u of the initial value problem (11) such that (13) u C([0,]:H s ()) L 4 L Xs 1,1 (Uniqueness) he solution is unique in the class (14) u L (]0,[:H s ()) X s 7/8,15/16 Moreover, the flow map data-solution: u 0 u is Lipschitz from H s () into C([0,]:H s ()) emark 14 We refer to Section 22 for the definition of the norms u X s,b Our technique of proof also yields aprioriestimates for the solutions of mkdv in H s () fors>0 It is worth noting that aprioriestimates in H s () were already proved by Christ, Holmer and ataru for 1/8 <s<1/4 in[4] heir proof relies on short time Fourier restriction norm method in the contet of the atomic spaces U, V and the I-method Although our result is not as strong as Christ, Holmer and ataru s one, we hope that it still may be of interest due to the simplicity of our proof heorem 15 Let s>0 and u 0 H () hen there eists = ( u 0 H s) > 0 such that the solution u to (11) emanating from u 0 satisfies 1 (15) + u u L H s X s 1,1 + u L 4 L u 0 H s Moreover, for any u 0 H s (), there eists a solution u L Hs L 4 L to (11) emanating from u 0 that satisfies (15) emark 16 ote that for u 0 L 2 (), the eistence of weak solutions of (11), in the sense of Definition 11, is well-known by making use of the so-called Kato smoothing effect Such solution belongs to L t L 2 L 2 t,loc H1 loc Our result indicates that if u 0 belongs to H s (), s>0, instead of L 2 (), then we can ask the weak solution to satisfy also (15) and, in particular, to propagate the H s -regularity on some time interval o prove heorems 13 and 15, we derive energy estimates on the dyadic blocks P u 2 H by taking advantage of the resonant relation and the fact that s any solution enjoys some conormal regularity his approach has been introduced by the first and the third authors in [25] ote however that, here, to bound some Bourgain s norm of a solution, we need first to bound its L 4 L -norm his norm is in turn controlled by using a refined Strichartz estimate derived by chopping the time interval in small pieces whose length depends on the spatial frequency ote that it was first established by Koch and zvetkov [19] (see also Kenig and Koenig [13] for an improved version) in the Benjamin Ono contet 1 See Section 22 for the definition of the L Hs -norm

4 4 L Molinet, D Pilod and S Vento he main difficulty to estimate d dt P u 2 H is to handle the resonant term s, typical of the cubic nonlinearity (u 3 ) When u is the solution of mkdv, writes = ( P+ up + up u ) P ud Actually, it turns out that we can always put the derivative appearing in on a low frequency product by integrating by parts 2, as it was done in [10] for quadratic nonlinearities his allows us to derive the aprioriestimate of heorem 15 in H s () fors > 0 Unfortunately, this is not the case anymore for the difference of two solutions of mkdv due to the lack of symmetry of the corresponding equation o overcome this difficulty we modify the H s -norm by higher order terms up to order 6 hese higher order terms are constructed so that the contribution of their time derivatives coming from the linear part of the equation will cancel out the resonant term he use of a modified energy is well-known to be a quite powerful tool in PDE s (see for instance [22] and[14]) ote however that, in our case, we need to define the modified energy in Fourier variables due to the resonance relation associated to the cubic nonlinearity his way to construct the modified energy has much in common with the way to construct the modified energy in the I-method (cf [5]) Finally let us mention that the tools developed in this paper together with some ideas of [27] and[26] enabled us in [24] to get the unconditional well-posedness of the periodic mkdv equation in H s () fors 1/3 We also hope that the techniques introduced here could be useful in the study of the Cauchy problem at low regularity of other cubic nonlinear dispersive equations such as the modified Benjamin Ono equation and the derivative nonlinear Schrödinger equation he rest of the paper is organized as follows In Section 2, we introduce the notations, define the function spaces and state some preliminary estimates he multilinear estimates at the L 2 -level are proved in Section 3 hose estimates are used to derive the energy estimates in Section 4 Finally, we give the proofs of heorems 13 and 15 respectively in Sections 5 and 6 2 otation, function spaces and preliminary estimates 21 otation For any positive numbers a and b, the notation a b means that there eists a positive constant c such that a cb We also denote a b when a b and b a Moreover, if α, α +, respectively α, will denote a number slightly greater, respectively lesser, than α Let us denote by D = { >0: =2 n for some n Z} the dyadic numbers Usually, we use n i, j i, m i to denote integers and i =2 ni, L i =2 ji and M i =2 mi to denote dyadic numbers For 1, 2 D, we use the notation 1 2 =ma{ 1, 2 } and 1 2 = min{ 1, 2 } Moreover, if 1, 2, 3 D, wealsodenoteby ma med min the maimum, sub-maimum and minimum of { 1, 2, 3 } 2 For technical reason we perform this integration by parts in Fourier variables

5 Unconditional uniqueness for mkdv 5 For u = u(, t) S ( 2 ), Fu will denote its space-time Fourier transform, whereas F u = û, respectively F t u, will denote its Fourier transform in space, respectively in time For s, we define the Bessel and iesz potentials of order s, J s and D,by s J s u = F 1 ( (1 + ξ 2 ) s/2 F u ) and Du s = F 1 ( ξ s F u ) We also denote by U(t) =e t 3 the unitary group associated to the linear part of (11), ie, U(t)u 0 = e t 3 u0 = F 1 ( e itξ 3 F (u 0 )(ξ) ) hroughout the paper, we fi a smooth cutoff function χ such that χ C 0 (), 0 χ 1, χ [ 1,1] = 1 and supp(χ) [ 2, 2] We set φ(ξ) :=χ(ξ) χ(2ξ) For l Z, we define φ 2 l(ξ) :=φ(2 l ξ), and, for l Z [1, + ), ψ 2 l(ξ,τ) =φ 2 l(τ ξ 3 ) By convention, we also denote φ 0 (ξ) =χ(2ξ) and ψ 0 (ξ,τ) :=χ(2(τ ξ 3 )) Any summations over capitalized variables such as, L, K or M are presumed to be dyadic Unless stated otherwise, we will work with non-homogeneous dyadic decompositions in, L and K, ie, these variables range over numbers of the form D nh = {2 k : k } {0}, whereas we will work with homogeneous dyadic decomposition in M, ie, these variables range over D WecallthenumbersinD nh nonhomogeneous dyadic numbers hen, we have that φ (ξ) =1, supp (φ ) I := {/2 ξ 2}, 1, and supp (φ 0 ) I 0 := { ξ 1} Finally, let us define the Littlewood Paley multipliers P, K and Q L by P u = F 1 ( φ F u ), K u = F 1 ( φk F t u ) and Q L u = F 1( ψ L Fu ), t P := K P K, P := K P K, Q L := K L Q K, Q L := K L Q K Sometimes, for the sake of simplicity and when there is no risk of confusion, we also denote u = P u

6 6 L Molinet, D Pilod and S Vento 22 Function spaces For 1 p, L p () is the usual Lebesgue space with the norm L p For s, the Sobolev space H s () denotes the space of all distributions of S () whose usual norm u H s = J s u L 2 is finite If B is one of the spaces defined above, 1 p and >0, we define the space-time spaces L p t B, L p B, L p t B and L p B equipped with the norms ( ) 1/p, u L p = f(,t) p t B B dt with obvious modifications for p =, and u Lp t B = ( ) 1/2, P u 2 L p t B ( ) 1/p u L p = f(,t) p B B dt 0 u Lp B = ( ) 1/2 P u 2 L p B For s, b, we introduce the Bourgain spaces X s,b related to the linear part of (11) as the completion of the Schwartz space S( 2 ) under the norm ( ) 1/2, (21) u X s,b := τ ξ 3 2b ξ 2s F(u)(ξ,τ) 2 dξ dτ 2 where := 1 + By using the definition of U, it is easy to see that (22) u X s,b U( t)u H s,b,t where u H s,b = JJ s t b u,t L 2,t We define our resolution space Y s = X s 1,1 X s 7/8,15/16 L t associated norm H s,withthe (23) u Y s = u X s 1,1 + u X s 7/8,15/16 + u L t H s It is clear from the definition that L Hs L Hs, ie, (24) u L H s u L H s, u L Hs ote that this estimate still holds true if we replace by t However, the reverse inequality is only true if we allow a little loss in space regularity Let s,s be such that s <s hen, (25) u L H s u L H s, u L H s Finally, we will also use a restriction in time versions of these spaces Let >0beapositive time and F be a normed space of space-time functions he restriction space F will be the space of functions u: [0,] satisfying u F =inf { ũ F : ũ: and ũ [0, ] = u } <

7 Unconditional uniqueness for mkdv 7 23 Etension operator We introduce an etension operator ρ which is a bounded operator from L Hs X s 1,1 X s 7/8,15/16 L 4 L into L t H s X s 1,1 X s 7/8,15/16 L 4 t L Definition 21 Let 0 < 1andu: [0,] be given We define the etension operator ρ by (26) ρ (u)(t) :=U(t)χ(t) U( μ (t)) u(μ (t)), where χ is the smooth cut-off function defined in Section 21 and μ is the continuous piecewise affine function defined by 0 for t<0, μ (t) = t for t [0,], for t> It is clear from the definition that ρ (u)(, t) =u(, t) for(, t) [0,] Lemma 22 Let 0 < 1, s, α, θ, b such that α s+1/4 and 1/2 <b 1 hen, ρ : L Hs Xθ,b L4 W α, L t Hs Xθ,b L 4 t W α, u ρ (u) is a bounded linear operator, ie, (27) ρ (u) L t H s + ρ (u) X θ,b + ρ (u) L 4 t W α, u L H s + u X θ,b + u L 4 W α,, for all u L Hs Xθ,b L4 W α, Moreover, the implicit constant in (27) can be chosen independent of 0 < 1, s, α, θ and 1/2 <b 1 Proof First, the unitarity of the free group U( ) inh s () easily leads to ρ (u) L t H s u(μ ( )) L t H s u L H s + u(0) H s + u( ) H s ow, since b>1/2, it is well known (see for instance [6]), that X θ,b C([0,]: H θ ()) herefore, u C([0,]:H θ ()) L Hs C([0,]:H θ ()) L Hs and we claim that (28) u(0) H s u L H s and u( ) H s u L H s Indeed, if it is not the case, assuming for instance that u(0) H s > u L H s, there would eist ɛ>0 and a decreasing sequence {t n } (0,) tending to 0 such that for any n, u(t n ) H s u(0) H s ɛ he continuity of u with values in H θ () then ensures that u(t n ) u(0) in H s (), which forces u(0) H s

8 8 L Molinet, D Pilod and S Vento lim inf u(t n ) H s and yields the contradiction herefore, we conclude by using (24) that (29) ρ (u) L t H s u L H s Second, according to classical results on etension operators (see eg [21]), for any 1/2 <b 1, f χf(μ ( )) is linear continuous from H b ([0,]) into H b () with a bound that does not depend on >0 hen, the definition of the X θ,b -norm leads, for 1/2 <b 1andθ, to (210) ρ (u) X θ,b = χu( μ ( ))u(μ ( )) H θ,b,t Finally, for α, U( )u H b ([0, [;H θ ) u X θ,b J α ρ (u) L 4 t L χu( )J α u(0) L 4 (],0[;L ) + J α u L 4 L + χu( )J α U( )u( ) L 4 (],+ [;L ) ow by using the Strichartz estimate related to the unitary group U (see estimate (214) in the net subsection), we deduce that χu( )J α u(0) L 4 (],0[;L ) U( )J α u(0) L 4 (] 2,0[;L ) u(0) H s, since α s 1/4, and in the same way χu( )U( )J α u( ) L 4 (],+ [;L ) U( )u( ) H s = u( ) H s his ensures by using (28) that (211) J α ρ (u) L 4 t L J α u L 4 + L u L H s herefore, we conclude the proof of (27) gathering (29) (211) emark 23 In the following, we will work with the resolution space Y s While it follows clearly from the definition of Y s that (212) u L H s + u X s 1,1 + u X s 7/8,15/16 u Y s, u Y s, the reverse inequality is not straightforward However, it can be proved by using the etension operator ρ Indeed, it follows from Lemma 22 that (213) u Y s ρ (u) Y s u L H s + u X s 1,1 In particular, this proves that Y s = L Hs X s 1,1 + u X s 7/8,15/16 X s 7/8,15/16

9 Unconditional uniqueness for mkdv 9 24 efined Strichartz estimates First, we recall the Strichartz estimates associated to the unitary Airy group derived in [15] For all u 0 L 2 () (214) e t 3 D 1/4 u 0 L 4 t L u 0 L 2, and for all g L 4/3 t L 1, (215) t 0 e (t t ) 3 D 1/2 g(t ) dt L 4 t L g 4/3 L t L 1 ote that these two estimates are equivalent thanks to the -argument Following the arguments in [13] and[19], we derive a refined Strichartz estimate for the solutions of the linear problem (216) t u + 3 u = F Proposition 24 Assume that > 0 and δ 0 Let u beasmoothsolution to (216) defined on the time interval [0,] hen, (217) u L 4 L (δ 1)/4+θ (δ+1)/2+θ J u L L 2 + J F L 4 L, 1 for any θ>0 Proof Let u be solution to (216) definedonatimeinterval[0,] We use a nonhomogeneous Littlewood Paley decomposition, u = u where u = P u, is a nonhomogeneous dyadic number and also denote F = P F hen, we get from the Minkowski inequality that u L 4 L u L 4 L sup θ u L 4 L, for any θ>0 ecall that P 0 corresponds to the projection in low frequencies, so that we set 0 θ = 1 by convention Since the Hölder and Bernstein inequalities easily yield P 0 u L 4 L 1/4 P 0 u L L 2, it is enough to prove that (218) u L 4 L D(δ 1)/4 u L L 2 + D (δ+1)/2 F L 4 L, 1 for any δ 0 and any dyadic number {2 k : k } Let δ be a nonnegative number We chop out the interval in small intervals of δ In other words, we have that [0,]= j J I j,wherei j =[a j,b j ], I j δ and #J δ Sinceu is a solution to the integral equation t u (t) =e (t aj) 3 u (a j )+ e (t t ) 3 F (t ) dt a j

10 10 L Molinet, D Pilod and S Vento for t I j, we deduce from (214) (215) that u L 4 L ( j δ/4 D 1/4 D (δ 1)/4 ) 1/4 ( D 1/4 u (a j ) 4 L 2 + ( u L L 2 + ( ( u L L 2 + I j 2 j j j D 1/2 I j D 1/2 D (δ 1)/4 u L L 2 + D (δ+1)/2 F L 4 L, 1 F 4 L 4/3 I j L 1 ) 1/4 F (t ) 4/3 dt ) 3 ) 1/4 L 1 D 1/2 F (t ) 4 L dt ) 1/4 1 I j which concludes the proof of (218) 3 L 2 multilinear estimates In this section we follow some notations of [28] For k Z + and ξ, letγ k (ξ) denote the k-dimensional affine hyperplane of k+1 defined by Γ k (ξ) = { (ξ 1,,ξ k+1 ) k+1 : ξ ξ k+1 = ξ }, and endowed with the obvious measure F = F (ξ 1,,ξ k+1 ):= F ( ξ 1,,ξ k,ξ (ξ 1 + +ξ k ) ) dξ 1 dξ k, Γ k (ξ) Γ k (ξ) k for any function F :Γ k (ξ) C When ξ =0,wesimplydenoteΓ k =Γ k (0) with the obvious modifications Moreover, given > 0, we also define = [0,]andΓ k =Γk [0,] with the obvious measures u := u(, t) d dt [0, ] and F := F ( ξ 1,,ξ k,ξ (ξ ξ k ),t ) dξ 1 dξ k dt Γ k k [0, ] 31 L 2 trilinear estimates Lemma 31 Let f j L 2 (), j =1,,4 and M D hen it holds that (31) φ M (ξ 1 + ξ 2 ) Γ 3 f j (ξ j ) M f j L 2

11 Unconditional uniqueness for mkdv 11 Proof Let us denote by I 3 M (f 1,f 2,f 3,f 4 ) the integral on the left-hand side of (31) We can assume without loss of generality that f i 0fori =1,,4 hen, we have that (32) I 3 M (f 1,f 2,f 3,f 4 ) J M (f 1,f 2 ) sup f 3 (ξ 3 )f 4 ( (ξ 1 + ξ 2 + ξ 3 )) dξ 3, ξ 1,ξ 2 where (33) J M (f 1,f 2 )= φ M (ξ 1 + ξ 2 ) f 1 (ξ 1 ) f 2 (ξ 2 ) dξ 1 dξ 2 2 Hölder s inequality yields (34) J M (f 1,f 2 )= φ M (ξ 1 )(f 1 f 2 )(ξ 1 )dξ 1 M f 1 f 2 L M f 1 L 2 f 2 L 2 Moreover, the Cauchy Schwarz inequality yields (35) f 3 (ξ 3 ) f 4 ( (ξ 1 + ξ 2 + ξ 3 )) dξ 3 f 3 L 2 f 4 L 2 herefore, estimate (31) follows from (32) (35) For a fied 1 dyadic, we introduce the following disjoint subsets of D 3 : M low 3 = { (M 1,M 2,M 3 ) D 3 : M min 1/2 and M med 2 9 }, M med 3 = { (M 1,M 2,M 3 ) D 3 : 1/2 <M min M med 2 9 }, M high 3 = { (M 1,M 2,M 3 ) D 3 :2 9 } <M med, where M min M med M ma denote respectively the minimum, sub-maimum and maimum of {M 1,M 2,M 3 } We will denote by φ M1,M 2,M 3 the function φ M1,M 2,M 3 (ξ 1,ξ 2,ξ 3 )=φ M1 (ξ 2 + ξ 3 ) φ M2 (ξ 1 + ξ 3 ) φ M3 (ξ 1 + ξ 2 ) et,westateausefultechnicallemma Lemma 32 Let (ξ 1,ξ 2,ξ 3 ) 3 satisfy ξ j j for j =1, 2, 3 and ξ 1 + ξ 2 + ξ 3 Let(M 1,M 2,M 3 ) M low 3 M med 3 hen it holds that M ma if (ξ 1,ξ 2,ξ 3 ) supp φ M1,M 2,M 3, Proof Without loss of generality, we can assume that M 1 M 2 M 3 Let (ξ 1,ξ 2,ξ 3 ) supp φ M1,M 2,M 3 hen, we have ξ 2 + ξ 3 and ξ 1 + ξ 3, so that 1 2 since ξ 1 + ξ 2 + ξ 3 On one hand 3 would imply that M 1 M 2 which is a contradiction On the other hand, 3 would imply that ξ 1 + ξ 2 + ξ 3 which is also a contradiction herefore, we must have 3 Finally, M 1 implies that ξ 2 ξ 3 < 0andM 2 implies ξ 1 ξ 3 < 0 hus, ξ 1 ξ 2 > 0, so that M 3

12 12 L Molinet, D Pilod and S Vento For η L, let us define the trilinear pseudo-product operator Π 3 η,m 1,M 2,M 3 in Fourier variables by ( (36) F Π 3 η,m1,m 2,M 3 (u 1,u 2,u 3 ) ) (ξ) = (ηφ M1,M 2,M 3 )(ξ 1,ξ 2,ξ 3 ) û j (ξ j ) Γ 2 (ξ) It is worth noticing that when the functions u j are real-valued, the Plancherel identity yields (37) Π 3 ( ) η,m 1,M 2,M 3 (u 1,u 2,u 3 ) u 4 d = ηφm1,m 2,M 3 (ξ1,ξ 2,ξ 3 ) Γ 3 Finally, we define the resonance function of order 3 by û j (ξ j ) (38) Ω 3 (ξ 1,ξ 2,ξ 3 )=ξ ξ ξ 3 3 (ξ 1 + ξ 2 + ξ 3 ) 3 = 3(ξ 1 + ξ 2 )(ξ 1 + ξ 3 )(ξ 2 + ξ 3 ) In the following proposition, we give suitable estimates for the pseudo-product Π 3 M 1,M 2,M 3 when (M 1,M 2,M 3 ) M high 3 Proposition 33 Let i, i =1,,4, and denote nonhomogeneous dyadic numbers Assume that 0 < 1, η is a bounded function and u i are realvalued functions in Y 0 = X 1,1 X 7/8,15/16 L t L 2 with spatial Fourier support in I i for i =1,,4 Assume also that 1, (M 1,M 2,M 3 ) M high 3 and M min 1 hen (39) [0, ] Π 3 η,m 1,M 2,M 3 (u 1,u 2,u 3 ) u 4 d dt ma(m 1 min 1) 1/16 u i Y 0, where ma =ma{ 1, 2, 3 } and η = ηφ (ξ 1 + ξ 2 + ξ 3 ) Moreover, the implicit constant in estimate (39) only depends on the L -norm of the function η Before giving the proof of Proposition 33, we state some important technical lemmas whose proofs can be found in [25] Lemma 34 Let L be a nonhomogeneous dyadic number hen the operator Q L is bounded in L t L 2 uniformly in L Inotherwords, (310) Q L u L t L 2 u L t L2, for all u L t on L L 2 and the implicit constant appearing in (310) does not depend Proof See Lemma 23 in [25] For any 0 < 1, let us denote by 1 the characteristic function of the interval [0,] One of the main difficulty in the proof of Proposition 33 is that the operator of multiplication by 1 does not commute with Q L o handle this

13 Unconditional uniqueness for mkdv 13 situation, we follow the arguments introduced in [25] and use the decomposition (311) 1 =1 low, +1 high,, with F ( ( ) t 1 low,) (τ) =χ(τ/)ft 1 (τ), for some >0tobefiedlater Lemma 35 For any >0 and >0 it holds (312) 1 high, L 1 1, and (313) 1 low, L 1 Proof See Lemma 24 in [25] Lemma 36 Assume that >0, >0 and L hen, it holds (314) Q L (1 low, u) L 2,t Q Lu L 2,t, for all u L 2 ( 2 ) Proof See Lemma 25 in [25] ProofofProposition33 Given u i,1 i 4, satisfying the hypotheses of Proposition 33, let G 3 M 1,M 2,M 3 = G 3 M 1,M 2,M 3 (u 1,u 2,u 3,u 4 ) denote the left-hand side of (39) We use the decomposition in (311) and obtain that (315) G 3 M 1,M 2,M 3 = G 3,low M 1,M 2,M 3, + G3,high M 1,M 2,M 3,, where G 3,low M = 1,M 2,M 3, 1 low, Π3 η,m 1,M 2,M 3 (u 1,u 2,u 3 )u 4 d dt 2 and G 3,high M = 1,M 2,M 3, 1 high, Π3 η,m 1,M 2,M 3 (u 1,u 2,u 3 )u 4 d dt 2 We deduce from Hölder s inequality in time, (31), (37) and(312) that G 3,high M 1,M 2,M 3, 1 high, L 1 Π 3 η,m 1,M 2,M 3 (u 1,u 2,u 3 )u 4 d which implies that 1 M min u i L t L 2, L t (316) G 3,high M 1,M 2,M 3, 1 ma(m min 1) 1/16 u i L t L 2 if we choose = M min (M min 1) 1/16 ma

14 14 L Molinet, D Pilod and S Vento o deal with the term G 3,low M 1,M 2,M 3,, we decompose with respect to the modulation variables hus, G 3,low M, = Π 3 η,m 1,M 2,M 3 (Q L1 (1 low, u 1),Q L2 u 2,Q L3 u 3 )Q L4 u 4 d dt 2 L 1,L 2,L 3,L 4 Moreover, we observe from the resonance relation in (38) and the hypothesis (M 1,M 2,M 3 ) M high 3 that (317) L ma M min 2 ma, where L ma =ma{l 1,L 2,L 3,L 4 } In the case where ma, (317) isclear from the definition of M high In the case where ma med, we claim that M ma M med ma Indeed, denote {ξ 1,ξ 2,ξ 3 } = {ξ ma,ξ med,ξ min },where ξ min ξ med ξ ma hen we compute, using also the hypothesis ξ 4, ξ ma + ξ min = ξ 4 ξ med med and ξ med + ξ min = ξ 4 ξ ma ma,which proves the claim In particular, (317) implies that L ma = M min (M min 1) 1/16 ma, since ma 1andM min 1 1 ma In the case where L ma = L 1, we deduce from (31), (37), (310), (314) and (317) that G 3,low M 1,M 2,M 3, which implies that (318) L 1 M min 2 ma ma 1 1/16 (1 M G 3,low M 1,M 2,M 3, M min Q L1 (1 low,u 1 ) L 2,t min Q Li u i L t i=2 )( u 1 X 1,1 + u 1 X 7/8,15/16) 1 1/16 (1 M ) u i Y 0 ma min L 2 u i L t L 2, We can prove arguing similarly that (318) still holds true in all the other cases, ie, L ma = L 2, L 3 or L ote that for those cases we do not have to use (312) but we only need (313) herefore, we conclude the proof of estimate (39) gathering (315), (316) and(318) 32 L 2 5-linear estimates Lemma 37 Let f j L 2 (), j =1,,6 and M 1,M 4 D hen it holds that (319) Γ 5 φ M1 (ξ 2 + ξ 3 )φ M4 (ξ 5 + ξ 6 ) i=2 6 6 f j (ξ j ) M 1 M 4 f j L 2

15 Unconditional uniqueness for mkdv 15 If moreover f j are localized in an annulus { ξ j } for j =5, 6, then (320) Γ 5 φ M1 (ξ 2 + ξ 3 ) φ M4 (ξ 5 + ξ 6 ) 6 f i (ξ i ) M 1 M 1/2 4 1/4 5 1/4 6 6 f i L 2 Proof Let us denote by I 5 = I 5 (f 1,,f 6 ) the integral on the right-hand side of (319) We can assume without loss of generality that f j 0, j =1,,6 We have by using the notation in (34) that ( (321) I 5 J M1 (f 2,f 3 ) J M4 (f 5,f 6 ) sup f 1 (ξ 1 )f 4 ξ 2,ξ 3,ξ 5,ξ 6 6 ) ξ j dξ 1 hus, estimate (319) follows applying (34) and the Cauchy Schwarz inequality to (321) Assuming furthermore that f j are localized in an annulus { ξ j } for j = 5, 6, then we get arguing as above that (322) I 5 M 1 J M4 (f 5,f 6 ) f j L 2 From the Cauchy Schwarz inequality, ( ) ( ) J M4 (f 5,f 6 ) f 5 (ξ 5 ) dξ 5 f 6 (ξ 6 ) dξ 6 1/2 5 1/2 6 f 5 L 2 f 6 L 2, which together with (322) yields (323) I 5 M 1 1/2 5 1/2 6 6 f i L 2 j 4 herefore, we conclude the proof of (320) interpolating (319) and(323) For a fied 1 dyadic, we introduce the following subsets of D 6 : M low 5 = { (M 1,,M 6 ) D 6 :(M 1,M 2,M 3 ) M med 3, M min(5) 2 9 M med(3) and M med(5) 2 9 }, M med 5 = { (M 1,,M 6 ) D 6 :(M 1,M 2,M 3 ) M med 3 and 2 9 M med(3) <M min(5) M med(5) 2 9 }, M high 5 = { (M 1,,M 6 ) D 6 :(M 1,M 2,M 3 ) M med 3 and 2 9 } <M med(5), where M ma(3) M med(3) M min(3), respectively M ma(5) M med(5) M min(5), denote the maimum, sub-maimum and minimum of {M 1,M 2,M 3 }, respectively {M 4,M 5,M 6 } We will also denote by φ M1,,M 6 the function defined on 6 by φ M1,,M 6 (ξ 1,,ξ 6 )=φ M1,M 2,M 3 (ξ 1,ξ 2,ξ 3 ) φ M4,M 5,M 6 (ξ 4,ξ 5,ξ 6 )

16 16 L Molinet, D Pilod and S Vento For η L, let us define the operator Π 5 η,m 1,,M 6 in Fourier variables by ( F Π 5 η,m1,,m 6 (u 1,,u 5 ) ) (ξ) = (ηφ M1,,M 6 )(ξ 1,,ξ 5, 5 ξ j) 5 (324) ûj(ξ j ) Γ 4 (ξ) Observe that, if the functions u j are real valued, the Plancherel identity yields 6 (325) Π 5 η,m 1,,M 6 (u 1,,u 5 ) u 6 d = (ηφ M1,,M 6 ) û j (ξ j ) Γ 5 Finally, we define the resonance function of order 5 for ξ (5) =(ξ 1,,ξ 6 ) Γ 5 by (326) Ω 5 ( ξ (5) )=ξ1 3 + ξ2 3 + ξ3 3 + ξ4 3 + ξ5 3 + ξ6 3 It is worth noticing that a direct calculus leads to (327) Ω 5 ( ξ (5) )=Ω 3 (ξ 1,ξ 2,ξ 3 )+Ω 3 (ξ 4,ξ 5,ξ 6 ) In the following proposition, we give suitable estimates for the pseudo-product Π 5 M 1,,M 6 when (M 1,,M 6 ) M high 5 in the non resonant case M 1 M 2 M 3 M 4 M 5 M 6 Proposition 38 Let i, i =1,,6 and denote nonhomogeneous dyadic numbers Assume that 0 < 1, η is a bounded function and u i are functions in X 1,1 L t L 2 with spatial Fourier support in I i for i =1,,6 If 1 and (M 1,,M 6 ) M high 5 satisfies the non resonance assumption M 1 M 2 M 3 M 4 M 5 M 6,then Π 5 η,m 1,,M 6 (u 1,,u 5 ) u 6 d dt (328) [0, ] M min(3) 1 ma(5) 6 ( u i X 1,1 + u i L t L 2 ), where ma(5) =ma{ 4, 5, 6 } and η = ηφ (ξ 1 + ξ 2 + ξ 3 ) Moreover, the implicit constant in estimate (328) only depends on the L - norm of the function η Proof he proof is similar to the proof of Proposition 33 We may always assume M 1 M 2 M 3 and M 4 M 5 M 6 Since ξ 1 + ξ 2 + ξ 3 = ξ 4 + ξ 5 + ξ 6 and (M 1,,M 6 ) M high 5,weget from Lemma 32 that 1 2 3, sothat ma(5) ma{ 1,, 6 } Moreover, it follows arguing as in the proof of (317) thatm 4 M 5 M 6 M 4 ma(5) 2 Hence, we deduce from identities (327) and(38) and the non resonance assumption that (329) L ma ma(m 1 M 2 M 3,M 4 M 5 M 6 ) M 4 M 5 M 6 M 4 ma(5) 2 Estimate (328) follows then from estimates (329) and(319) arguing as in the proof of Proposition 33

17 Unconditional uniqueness for mkdv L 2 7-linear estimates Lemma 39 Let f i L 2 (), i =1,,8 and M 1,M 4,M 6 and M 7 D henit holds that (330) Γ 7 φ M1 (ξ 2 +ξ 3 )φ M6 (ξ 4 +ξ 5 )φ M7 (ξ 7 +ξ 8 ) 8 8 f i (ξ i ) M 1 M 6 M 7 f i L 2 and (331) ( φ M1 (ξ 2 +ξ 3 )φ M4 ξ j )φ M7 (ξ 7 +ξ 8 ) f i (ξ i ) M 1 M 4 M 7 f i L 2 Γ 7 If moreover f j is localized in an annulus { ξ j } for j =7, 8, then (332) and (333) Γ 7 φ M1 (ξ 2 + ξ 3 ) φ M6 (ξ 4 + ξ 5 ) φ M7 (ξ 7 + ξ 8 ) 8 f i (ξ i ) M 1 M 6 M 1/2 7 1/4 7 1/4 8 ( 4 ) 8 φ M1 (ξ 2 + ξ 3 ) φ M4 ξ j φ M7 (ξ 7 + ξ 8 ) f i (ξ i ) Γ 7 M 1 M 4 M 1/2 7 1/4 7 1/4 8 8 f i L 2 8 f i L 2 Proof Let us denote by I 7 = I 7 (f 1,,f 8 ) the integral on the right-hand side of (330) We can assume without loss of generality that f j 0, j =1,,8 We have by using the notation in (34) that (334) I 7 J M1 (f 2,f 3 ) J M6 (f 4,f 5 ) J M7 (f 7,f 8 ) ( sup f 1 (ξ 1 )f 6 ξ 2,ξ 3,ξ 4,ξ 5,ξ 7,ξ 8 8 ) ξ j dξ 1 j 6 hus, estimate (330) follows applying (34) and the Cauchy Schwarz inequality to (334) Assuming furthermore that f j are localized in an annulus { ξ j } for j = 7, 8, then we get arguing as above that (335) I 7 M 1 M 6 J M7 (f 7,f 8 ) 6 f j L 2

18 18 L Molinet, D Pilod and S Vento From the Cauchy Schwarz inequality, ( ) ( (336) J M7 (f 7,f 8 ) f 7 (ξ 7 ) dξ 7 which together with (335) yields (337) I 7 M 1 M 6 1/2 7 1/2 8 f 8 (ξ 8 ) dξ 8 ) 8 f j L 2 1/2 7 1/2 8 f 7 L 2 f 8 L 2, herefore, we conclude the proof of (332) interpolating (330) and(337) ow, we prove estimate (331) Let us denote by Ĩ7 = Ĩ7 (f 1,,f 8 ) the integral on the right-hand side of (331) We can assume without loss of generality that f j 0, j =1,,8 Let us define J M (f 1,f 2 )(ξ) = φ M (ξ 1 + ξ 2 + ξ) f 1 (ξ 1 ) f 2 (ξ 2 ) dξ 1 dξ 2 2 Hence, we have, by using the notation in (33), J M (f 1,f 2 )(0) = J M (f 1,f 2 ) Moreover, it follows from Young s inequality on convolution that sup J M (f 1,f 2 )(ξ) (338) ξ =sup φ M (ξ 1 )f 1 f 2 (ξ 1 ξ) dξ 1 M f 1 f 2 L M f 1 L 2 f 2 L 2 ξ (339) By using this notation and the fact that 4 ξ j = 8 j=5 ξ j,wehave Ĩ 7 J M1 (f 2,f 3 ) sup ξ 7,ξ 8 J M4 (f 5,f 6 )(ξ 7 + ξ 8 ) J M7 (f 7,f 8 ) sup ξ 2,ξ 3,ξ 5,ξ 6,ξ 7,ξ 8 f 1 (ξ 1 )f 4 ( 8 ) ξ j dξ 1 Hence, we conclude the proof of (331) by applying (34), (338) andthecauchy Schwarz inequality to (339) he proof of (333) follows arguing as above and using (336) to estimate J M7 (f 7,f 8 ) For a fied 1 dyadic, we introduce the following subsets of D 9 : M low 7 = { (M 1,,M 9 ) D 9 :(M 1,,M 6 ) M med 5, M min(7) 2 9 M med(5) and M med(7) 2 9 }, M med 7 = { (M 1,,M 9 ) D 9 :(M 1,,M 6 ) M med 5, 2 9 M med(5) <M min(7) M med(7) 2 9 }, M high 7 = { (M 1,,M 9 ) D 9 :(M 1,,M 6 ) M med 5, 2 9 } <M med(7), where M ma(7) M med(7) M min(7) denote respectively the maimum, submaimum and minimum of {M 7,M 8,M 9 } j 4

19 Unconditional uniqueness for mkdv 19 We will denote by φ M1,,M 9 the function defined on Γ 7 by (340) φ M1,,M 9 (ξ 1,,ξ 7,ξ 8 )=φ M1,,M 6 (ξ 1,,ξ 5, 5 ξ j) φ M7,M 8,M 9 (ξ 6,ξ 7,ξ 8 ) For η L, let us define the operator Π 7 η,m 1,,M 9 in Fourier variables by ( (341) F Π 7 η,m1,,m 9 (u 1,,u 7 ) ) 7 (ξ)= (ηφ M1,,M 9 )(ξ 1,,ξ 7, ξ) û j (ξ j ) Γ 6 (ξ) Observe that, if the functions u j are real valued, the Plancherel identity yields (342) Π 7 η,m 1,,M 9 (u 1,,u 7 ) u 8 d = (ηφ M1,,M 9 ) Γ 7 8 û j (ξ j ) We define the resonance function of order 7 for ξ (7) =(ξ 1,,ξ 8 ) Γ 7 by (343) Ω 7 ( ξ (7) )= 8 ξj 3 Again it is direct to check that (344) Ω 7 ( ξ (7) )=Ω 5 (ξ 1,,ξ 5, 5 ξ i)+ω 3 (ξ 6,ξ 7,ξ 8 ) In the following proposition, we give suitable estimates for the pseudo-product Π 7 M 1,,M 9 when (M 1,,M 9 ) M high 7 in the nonresonant case M 4 M 5 M 6 M 7 M 8 M 9 Proposition 310 Let i, i =1,,8 and denote nonhomogeneous dyadic numbers Assume that 0 < 1, η is a bounded function and u j are functions in X 1,1 L t L 2 with spatial Fourier support in I j for j =1,,8 (a) Assume that 1 and (M 1,,M 9 ) M high 7 satisfies the non resonance assumption M 4 M 5 M 6 M 7 M 8 M 9 hen Π 7 η,m 1,,M 9 (u 1,,u 7 ) u 8 d dt (345) [0, ] M min(3) M min(5) 1 ma(7) where ma(7) =ma{ 6, 7, 8 } and η = ηφ (ξ 1 + ξ 2 + ξ 3 ) (b) Assume that 1 and (M 1,,M 9 ) M med 7 hen Π 7 η,m 1,,M 9 (u 1,,u 7 ) u 8 d dt (346) [0, ] where η = ηφ (ξ 1 + ξ 2 + ξ 3 ) M min(3)m min(5) M med(7) 8 ( u j X 1,1 + u j L t L 2 ), 8 ( u j X 1,1 + u j L t L 2 ),

20 20 L Molinet, D Pilod and S Vento Moreover, the implicit constants in estimates (345) and (346) only depend on the L -norm of the function η Proof he proof is similar to the proof of Proposition 33 Under the assumptions in (a) ξ 1 + ξ 2 + ξ 3 and (M 1,,M 9 ) M high 7,we get by using twice Lemma 32 that ξ 1 ξ 2 ξ 3 ξ 4 ξ 5 ξ 6 + ξ 7 + ξ 8, sothat ma(7) ma{ 1,, 8 } On the one hand, since (M 1,,M 6 ) M med 5, it is clear that M 4 M 5 M 6 M 1 M 2 M 3 On the other hand, it follows arguing as in the proof of (317) thatm 7 M 8 M 9 M min(7) ma(7) 2 Hence, we deduce from identities (327), (344) and the non resonance assumption that L ma ma(m 4 M 5 M 6,M 7 M 8 M 9 ) M min(7) ma(7) 2 Under the assumptions in (b) ξ 1 + ξ 2 + ξ 3 and (M 1,,M 9 ) M med 7, we get that M 7 M 8 M 9 M 4 M 5 M 6 M 1 M 2 M 3 We also have by applying three times Lemma 32 that 1 8 M ma(7) Hence, we deduce that L ma M min(7) M med(7) Estimates (345) and (346) follow from these claims and estimates (330) and (331), arguing as in the proof of Proposition 33 Indeed, in view of the definition of φ M1,,M 9 in (340), we can always assume by symmetry that M min(3) = M 1 and M min(7) = min(m 7,M 8,M 9 )=M 7 In the case where M min(5) = M 6,weuse(330), whereas in the case where M min(5) = M 4, we use (331) By symmetry, the case where M min(5) = M 5 is equivalent to the case where M min(5) = M 4 4 Energy estimates he aim of this section is to derive energy estimates for the solutions of (11) and the solutions of the equation satisfied by the difference of two solutions of (11) (see equation (53) below) In order to simplify the notations in the proofs below, we will instead derive energy estimates on the solutions u of the more general equation (41) t u + 3 u = c 4 (u 1 u 2 u 3 ), where for any i {1, 2, 3}, u i solves (42) t u i + 3 u i = c i (u i,1 u i,2 u i,3 ) Finally we also assume that each u i,j solves (43) t u i,j + 3 u i,j = c i,j (u i,j,1 u i,j,2 u i,j,3 ), for any (i, j) {1, 2, 3} 2 We will sometimes use u 4,u 4,1,u 4,2,u 4,3 to denote respectively u, u 1,u 2,u 3 Herec j,j {1,,4} and c i,j, (i, j) {1, 2, 3} 2 denote real constants Moreover, we assume that all the functions appearing in (41) (42) (43) are real-valued Also, we will use the notations defined at the beginning of Section 3

21 Unconditional uniqueness for mkdv 21 he main obstruction to estimate d dt P u 2 L at this level of regularity is the 2 resonant term ( ) P+ u 1 P + u 2 P u 3 P ud for which the resonance relation (38) is not strong enough In this section we modify the energy by a fourth order term, whose part of the time derivative coming from the linear contribution of (41) will cancel out this resonant term ote that we also need to add a second modification to the energy to control the part of the time derivative of the first modification coming from the resonant nonlinear contribution of (41) 41 Definition of the modified energy Let 0 =2 9 and be a nonhomogeneous dyadic number For t 0, we define the modified energy at the dyadic frequency by { 1 2 (44) E (t) = P u(,t) 2 L for 2 0, 1 2 P u(,t) 2 L + α E 3 2 (t)+βe5 (t) for > 0, where α and β are real constants to be determined later, E 3 (t) = (M 1,M 2,M 3) M med 3 where ξ (3) =(ξ 1,ξ 2,ξ 3 ), and E 5 (t) = (M 1,,M 6) M med 5 ξ 4 ( ) φ ξ(3) M1,M 2,M 3 φ 2 (ξ 4 ) Γ 3 Ω 3 ( ξ (3) ) û j (t, ξ j ), 4 c j φ M1,,M 6 ( ξ Γ j (5) ) φ 2 ξ 4 ξ j (ξ 4 ) 5 Ω 3 ( ξ j (3) )Ω 5 ( ξ j (5) ) û k (t, ξ k ) û j,l (t, ξ j,l ), with the convention ξ j = 4, ξ k = 3 ξ j,l and the notations with ξ j (5) =( ξ j (3),ξ j,1,ξ j,2,ξ j,3 ) Γ 5 ξ 1(3) =(ξ 2,ξ 3,ξ 4 ), ξ2(3) =(ξ 1,ξ 3,ξ 4 ), ξ3(3) =(ξ 1,ξ 2,ξ 4 ), ξ4(3) =(ξ 1,ξ 2,ξ 3 ) For >0, we define the modified energy by using a nonhomogeneous dyadic decomposition in spatial frequency (45) E s (u) = 2s sup E (t) t [0, ] By convention, we also set E0 s(u) = 2s E (0) he net lemma ensures that, for s 1/4, the energy E s (u) iscoerciveina small ball of H s centered at the origin

22 22 L Molinet, D Pilod and S Vento Lemma 41 (Coercivity of the modified energy) Let s 1/4 and u, u i,u i,j H s hen it holds 4 (46) u 2 L E s H s (u)+ u j + u L H s k u L H s j,l L H s Proof We infer from (45) and the triangle inequality that (47) u 2 L E s H s (u)+ sup E 3 (t) + 0 2s t [0, ] 0 2s sup t [0, ] E 5 (t) We first estimate the contribution of E 3 By symmetry, we can always assume that M 1 M 2 M 3,sothatwehave 1/2 <M 1 M 2 and M 3, since (M 1,M 2,M 3 ) M med 3 hen, we have from Lemma 31, (48) 2s E 3 (t) 1/2 <M 1,M 2 M 3 1/2 2s 2s+1 M 1 M 2 M 3 M 1 P u j (t) H s, P u j (t) L 2 whereweusedthat 1/2 <M 1 1,M 2 M 2 1/2 <M 1 1 M 1 1/2 o estimate the contribution of E 5 (t), we notice from Lemma 32 that for (M 1,,M 6 ) M med 5, the integrand in the definition of E 5 vanishes unless ξ 1 ξ 4 and ξ j,1 ξ j,2 ξ j,3 Moreover, we assume without loss of generality M 1 M 2 M 3 and M 4 M 5 M 6,sothat ξ 4 ξ j 2 Ω 3 ( ξ j (3) )Ω 5 ( ξ j (6) ) M 1 M 2 M 4 M 5 1 M 1 M 2 M 4 M 5 hus we infer from (319) that (49) 4 2s E 5 (t) 1 4s 1/2 M 1 M 2 M 2 M 4 M 5 4 2s M 2 M 5 P u k (t) L 2 P u k (t) H s P u j,l (t) L 2 P u j,l (t) H s Finally, we conclude the proof of (46) by summing (48) and(49) overthe dyadic 0, with s>1/4, and using (47) emark 42 Arguing as in the proof of Lemma 41, wegetthat 4 (410) E0 s (u) u(0) 2 H + u s j (0) H s + u k (0) H s u j,l (0) H s as soon as s 1/4

23 Unconditional uniqueness for mkdv Estimates for the modified energy Proposition 43 Let s>1/3, 0 < 1 and u, u i,u i,j Y s (41), (42) and (43) on ]0,[ hen we have be solutions of (411) E s (u) E s 0(u) m=1 m j 3 u j Y s + u k Y s m=1,m u k Y s 4 u k Y s u j,l Y s l m u j,l Y s u j,l Y s u j,m,n Y s n=1 u m,n Y s n=1 Proof Let 0 <t 1 First, assume that 0 =2 9 By using the definition of E in (44), we have d dt E ( ) (t) =c 4 P u1 u 2 u 3 P ud, which yields after integrating between 0 and t and applying Hölder s inequality that ( ) E (t) E (0) + c 4 P u1 u 2 u 3 P u t E (0) + u i L L 4 E (0) + u i L H 1/4 where the notation t = [0,t] defined at the beginning of Section 3 has been used hus, we deduce after taking the supreme over t [0,] and summing over 0 (recall here that we use a nonhomogeneous dyadic decomposition in ) that (412) 0 2s sup t [0, ] E (t) 0 2s E (0) + u j Y 1/4 et, we turn to the case where 0 As above, we differentiate E with respect to time and then integrate between 0 and t to get 2s E (t) = 2s E (0) + c 4 2s (413) P (u 1 u 2 u 3 )P u + α 2s t t + β 2s d dt E5 (t ) dt 0 =: 2s E (0) + c 4 I + αj + βk t 0 d dt E3 (t ) dt

24 24 L Molinet, D Pilod and S Vento We rewrite I in Fourier variable and get I = 2s ( iξ 4 ) φ 2 (ξ 4) û 1 (ξ 1 ) û 2 (ξ 2 ) û 3 (ξ 3 ) û 4 (ξ 4 ) = Γ 3 t (M 1,M 2,M 3) D 3 2s et we decompose I as (414) I = 2s ( M low 3 =: I low + Imed + M med 3 + Ihigh, Γ 3 t + M high by using the notations in Section 3 3 ( iξ 4 ) φ M1,M 2,M 3 ( ξ (3) ) φ 2 (ξ 4) ) Γ 3 t û j (ξ j ) ( iξ 4 )φ M1,M 2,M 3 ( ξ (3) ) φ 2 (ξ 4) û j (ξ j ) Estimate for I low hanks to Lemma 32, the integral in Ilow is non trivial for ξ 1 ξ 2 ξ 3 ξ 4 and M min 1/2 herefore we get from Lemma 31 that I low M min 1/2 M min M med 2s+1 M min since 2s +1/2 < 4s his leads to P u j L L 2 P u j L H s, (415) u j Y s 0 I low Estimate for I high We perform nonhomogeneous dyadic decompositions on u j by writing u j = j P j u j for j =1, 2, 3 We assume without loss of generality that 1 = ma( 1, 2, 3 ) ecall that this ensures that M ma 1 We separate the contributions of two regions that we denote I high,1 and I high,2 M min 1 hen we apply Lemma 31 on the sum over M med and use the discrete Young s inequality to bound I high,1 by M min 1 2s+1 M min (416) 1 1, 2, 3 j=2 P j u j L L 2 P 1 u 1 L 2, P u 4 L 2, ( ) s P1 u 1 L 2 P Hs u 4 L 2 H u 2 s L 1 H 0+ u 3 L H 0+ δ P u 4 L 2 H s u i L H s,

25 Unconditional uniqueness for mkdv 25 with {δ 2 j } l 2 () Summing over this leads to (417) I high,1 u j Y s 0 M min > 1 Forj =1,,4, let ũ j be an etension of u j to 2 such that ũ j Y s 2 u Y s ow, we define u j = P j ũ j and perform nonhomogeneous dyadic decompositions in j,sothati high,2 can be rewritten as I high,2 = 2s+1 Π 3 η,m 1,M 2,M 3 (u 1,u 2,u 3 ) u 4, t j, 4 (M 1,M 2,M 3) M high 3 with η(ξ 1,ξ 2,ξ 3 )=φ 2 (ξ 4)iξ 4 / L (Γ 3 ) hus, it follows from (39) that I high,2 2s ( M 1/16 min + ma j, 4 1 <M min 1 M med Mma ma u 1 Y 0 u 2 Y 0 u 3 Y 0 u 4 Y 0 1<M min med M med Mma ma Proceeding as in (416) (herewesumoverm min 1 by using the factor M 1/16 min and over M min 1byusingthatM min med ), we get (418) I high,2 u j Y s 0 Estimate for c 4 I med + αj + βk Using (41) (42), we can rewrite d dt E3 as the sum of φ M1,M 2,M 3 ( ξ (3) ) φ 2 (ξ 4) iξ 4(ξ1 3 + ξ3 2 + ξ3 3 + ξ3 4 ) Γ 3 Ω 3 ( ξ û j (ξ j ) (3) ) and 4 c j M med 3 M med 3 ξ 4 φ M1,M 2,M 3 ( ξ (3) ) φ 2 (ξ 4) Γ 3 Ω 3 ( ξ (3) ) ( ) û k (ξ k ) F uj,1 u j,2 u j,3 (ξj ) Using (38), we see by choosing α = c 4 that I med is canceled out by the first term of the above epression Hence, 4 (419) c 4 I med + αj = c 4 c j J j, where, for j =1,,4, J j 2s = i φ M1,M 2,M 3 ( ξ (3) ) φ 2 ξ 4 ξ j (ξ 4 ) Ω 3 ( ξ (3) ) M med 3 Γ 5 t û k (ξ k ) û j,l (ξ j,l ), with the convention ξ j = 4 ξ k = 3 ξ j,l and the notation ξ (3) =(ξ 1,ξ 2,ξ 3 ) )

26 26 L Molinet, D Pilod and S Vento ow, we define ξ j (3),forj =1, 2, 3, 4 as follows: ξ 1 (3) =(ξ 2,ξ 3,ξ 4 ), ξ2 (3) =(ξ 1,ξ 3,ξ 4 ), ξ3 (3) =(ξ 1,ξ 2,ξ 4 ), ξ4 (3) =(ξ 1,ξ 2,ξ 3 ) With this notation in hand, we can use the symmetries of M φ med M1,M 3 2,M 3 and Ω 3 to obtain that 2s = i φ M1,M 2,M 3 ( ξ j (3) ) φ 2 (ξ ξ 4 ξ j 4) û k (ξ k ) û j,l (ξ j,l ) J j M med 3 Γ 5 t Ω 3 ( ξ j (3) ) Moreover, observe from the definition of M med 3 in Section 3 that ξ j ξ 4 ξ 1 ξ 2 ξ 3 ξ 4 and Ω 3 ( ξ, (3) ) M min(3) M med(3) on the integration domain of J j Here M ma(3) M med(3) M min(3) denote the maimum, sub-maimum and minimum of {M 1,M 2,M 3 } Since ma( ξ j,1 + ξ j,2, ξ j,1 + ξ j,3, ξ j,2 + ξ j,3 ) on the integration domain of J j, we may decompose j c jj j as 4 c j J j = i 2s( + + ) 4 c j M low 5 M med 5 M high (420) φ M1,,M 6 ( ξ j (5) ) φ 2 (ξ 4 ) Γ 5 t := J low + J med + J high, 5 ξ 4 ξ j Ω 3 ( ξ j (3) ) û k (ξ k ) û j,l (ξ j,l ) where ξ j (5) =( ξ j (3),ξ j,1,ξ j,2,ξ j,3 ) Γ 5 Moreover, we may assume by symmetry that M 1 M 2 M 3 and that M 4 M 5 M 6 Estimate for Jlow In the region Mlow 5,wehavethatM 4 M 2 Moreover, from Lemma 32, the integral in J low is non trivial for ξ 1 ξ 4, ξ j,1 ξ j,2 ξ j,3 and M 3 M 6 herefore by using (319), we can bound J low by 4 2s M 1 M 4 M 1 M 2 so that 1/2 <M 1 M 2 J low 4 M 4 M 2 M 4 M 5 P u k L L 2 P u k L H s P u j,l L L 2, P u j,l L H s

27 Unconditional uniqueness for mkdv 27 since s>1/4 hus, we deduce that (421) J low 0 4 u k Y s u j,l Y s Estimate for J high high Proceeding as for Ihigh, we split J into J high,1 + J high,2 to separate the contributions depending on whether M 4 1 or M 4 > 1 M 4 1 From Lemma 32, the integral in J high,1 is non trivial for ξ 1 ξ 4, M 3, ma(5) =ma{ j,1, j,2, j,3 }, M 4 1 and M 5 M 6 ma(5) herefore by using (319), we can bound J high,1 by 4 2s+1 M 1 M 4 M 1 M 2 j,l so that 1/2 <M 1 M 2 M 4 1 J high,1 since s>1/4 his leads to (422) 4 P u k L L 2 P u k L H s J high,1 0 4 u k Y s P j,l u j,l L L 2, u j,l L H s u j,l Y s M 4 > 1 For 1 k 4, and 1 l 3letũ k and ũ j,l be suitable etensions of u k and u j,l to 2 We define u k = P k ũ k, u j,l = P j,l ũ j,l and perform nonhomogeneous dyadic decompositions in k and j,l We first estimate J high,2 in the resonant case M 1 M 2 M 3 M 4 M 5 M 6 We assume to simplify the notations that M 1 M 2 M 3 and M 4 M 5 M 6 Since we are in M high 5,wehavethatM 5,M 6 and M 1,M 2 which yields M 3 and M 4 M 1M 2 M 5 M 6 his forces j,1 and it follows from (320) that J high,2 4 4 M high 5 2s+1 M 1 M 1/2 4 1/4 j,2 M 1 M 1/4 j,3 2 j,l 1/2 M 1 M 2 s+1/2 (M 1M 2 ) 1/2 M 2 P u k L L 2 P u k L L 2 u j,l L L 2 u j,l L H s

28 28 L Molinet, D Pilod and S Vento Summing over 1/2 M 1, M 2 and 0 and using the assumption s>1/4, we get (423) J high,2 0 4 u k Y s u j,l Y s, in the resonant case By using (328), we easily estimate J high,2 in the non resonant case M 1 M 2 M 3 M 4 M 5 M 6 by 4 J high,2 1/2 M 1 M 2 P ũ k Y 0 1 <M 4 M 5 M 6 ma(5) u j,l Y 0 2s+1 M 1 1 M 1 M ma(5) 2 j,l ecalling that ma(5) =ma{ j,1, j,2, j,3 }, we conclude after summing over that (423) also holds, for s>1/4, in the non resonant case Estimate for αj med + βk Using equations (41) (42) (43) andthereso- dt as nance relation (326), we can rewrite 2s t 0 4 2s c j M med 5 + 2s M med 5,m 4 Γ 5 t c j + 4 2s c j M med 5 û k (ξ k ) d dt E5 φ M1,,M 6 ( ξ j (5) )φ 2 (ξ iξ 4 ξ j 4) Ω 3 ( ξ j (3) ) 4 c m m=1 m j Γ 5 t û k (ξ k ) F (u m,1 u m,2 u m,3 )(ξ m ) 3 c j,m m=1 û k (ξ k ) û j,l (ξ j,l ) φ M1,,M 6 ( ξ j (5) ) φ 2 ξ 4 ξ j (ξ 4 ) Ω 3 ( ξ j (3) )Ω 5 ( ξ j (5) ) Γ 5 t û j,l (ξ j,l ) φ M1,,M 6 ( ξ j (5) ) φ 2 (ξ ξ 4 ξ j 4) Ω 3 ( ξ j (3) )Ω 5 ( ξ j (5) ) û j,l (ξ j,l ) F (u j,m,1 u j,m,2 u j,m,3 )(ξ j,m ) l m := K 1 + K 2 + K 3 By choosing β = α, wehavethat (424) αj med + βk = β(k 2 + K) 3 For the sake of simplicity, we will only consider the contribution of K 3 corresponding to a fied (j, m) {1, 2, 3, 4} {1, 2, 3}, since the other contributions on the right-hand side of (424) can be treated similarly

29 Unconditional uniqueness for mkdv 29 hus, for (j, m) fied, we need to bound K := i 2s M med 5 Γ 7 t with the conventions ξ j = where σ( ξ j (5) ) û k (ξ k ) 4 ξ k = û j,l (ξ j,l ) l m 3 ξ j,l and ξ j,m = σ( ξ j (5) )=φ M1,,M 6 ( ξ j (5) ) φ 2 (ξ 4) ow, let us define ξ j,m(7) Γ 7 as follows: We decompose K as and write K = i 2s M7 û j,m,n (ξ j,m,n ), n=1 3 ξ j,m,n and where n=1 ξ 4 ξ j ξ j,m Ω 3 ( ξ j (3) )Ω 5 ( ξ j (5) ) ξ j,1(7) = ( ξj (3),ξ j,2,ξ j,3,ξ j,1,1,ξ j,1,2,ξ j,1,3 ), ξ j,2(7) = ( ξj (3),ξ j,1,ξ j,3,ξ j,2,1,ξ j,2,2,ξ j,2,3 ), ξ j,3(7) = ( ξj (3),ξ j,1,ξ j,2,ξ j,3,1,ξ j,3,2,ξ j,3,3 ) Γ 7 t σ( ξ j,m(7) ) û k (ξ k ) û j,l (ξ j,l ) l m û j,m,n (ξ j,m,n ) n=1 σ( ξ j,m(7) )=φ M7,M 8,M 9 ( ξj,m,1,ξ j,m,2,ξ j,m,3 ) σ( ξj (5) ), (425) K = K low med high + K + K, depending on wether we sum over M low 7, M med 7 or M high 7 Observe from Lemma 32 that the integrand is non trivial for ξ 1 ξ 4 ξ j,1 ξ j,2 ξ j,3 ξ j,m,1 + ξ j,m,2 + ξ j,m,3 Moreover, we have and Hence, M ma(3) M ma(5) 1/2 M min(3) M med(3) M min(5) M med(5) σ( ξj,m(7) ) M min(3) M med(3) M min(5) M med(5) ote that we can always assume by symmetry and without loss of generality that M 1 M 2 M 3 and M 7 M 8 M 9

30 30 L Molinet, D Pilod and S Vento K low low Estimate for In the integration domain of K we have from Lemma 32 that ξ j,m,1 ξ j,m,2 ξ j,m,3 hen it follows applying (330) or(331) (depending on wether M min(5) = M 6 or M min(5) = M 4 or M 5 )onthesumover(m 8,M 9 )that K low 1/2 <M 1 M 2 M 2 M min(5) M med(5) his implies that P u k L L 2 2s+1 M 7 M 2 M med(5) M 7 M med(5) P u j,l L L 2 l m P u j,m,n L L 2 n=1 (426) 0 low K u k L H s u j,l L H s l m u j,m,n L H s, n=1 since 2s +3/2 < 8s s>1/4 K med med Estimate for In the integration domain of K we have from Lemma 32 med that ξ j,m,1 ξ j,m,2 ξ j,m,3 o estimate K, we divide the regions where M 7 1andM 7 1 In the region where M 7 1, we deduce by using (330) or(331) (depending on wether M min(5) = M 6 or M min(5) = M 4 or M 5 )onthesumover(m 8,M 9 )that K med 1/2 <M 1 M 2 M 2 M min(5) M med(5) his implies that P u k L L 2 M 7 1 2s+1 M 7 M 2 M med(5) P u j,l L L 2 l m P u j,m,n L L 2 n=1 (427) 0 med K u k L H s u j,l L H s l m u j,m,n L H s, n=1 since 2s +2< 8s s>1/3 In the region where M 7 1, for 1 k 4, k j, 1 l 3, l m and 1 n 3letũ k,ũ j,l and ũ j,m,n be suitable etensions of u k, u j,l and u j,m,n to 2

31 Unconditional uniqueness for mkdv 31 hen, we deduce from Lemma 32 and (346) that K med his implies that (428) 1/2 <M 1 M 2 M 2 M min(5) M med(5) 0 P ũ k Y 0 med K 2s+1 M 2 M 1 M med(5) M 8 7 M 8 P ũ j,l Y 0 l m u k Y s u j,l Y s l m P ũ j,m,n Y 0 n=1 u j,m,n Y s, since s>1/3 high high Estimate for K We first estimate K in the resonant case M 4 M 5 M 6 M 7 M 8 M 9 Since we are in M high 7,wehavethatM 9 M 8 and M min(5) M med(5) It follows that M ma(5) and M 7 M min(5)m med(5) M 8 M 9 his forces j,m,1 (for eample) and we deduce by using (332) inthecase M min(5) = M 6,and(333) inthecasem min(5) = M 4 or M 5,that K high 1/2 <M 1 M 2 M 9 M 8 j,m,n, j,m,1 M 2 M min(5) M med(5) P u k L L 2 P u j,l L L 2 l m n=1 2s+3/2 M 2 M 1/2 8 M 1/2 9 1/4 j,m,2 1/4 j,m,3 P j,m,n u j,m,n L L 2, which yields summing over 0 and using the assumption s>1/4 that (429) 0 K high u k Y s u j,l Y s l m n=1 u j,m,n Y s ow, in the non resonant case we separate the contributions of the region M 7 1 and M 7 > 1 In the first region, applying (330) or(331) (depending on wether M min(5) = M 6 or M min(5) = M 4 or M 5 )onthesumover(m 8,M 9 ), we get K high 1/2 <M 1 M 2 M 2 M min(5) M med(5) P u k L L 2 M 7 1 n=1 2s+1 M 7 M 2 M med(5) j,m,n P u j,l L L 2 l m P j,m,n u j,m,n L L 2 n=1

32 32 L Molinet, D Pilod and S Vento Observing that ma{ j,m,1, j,m,2, j,m,3 }, we conclude after summing over 0 that high (430) K u k L H s u j,l L H s u j,m,n L H s, 0 since 2s +1< 6s s>1/4 Finally we treat contribution of the region M 7 > 1 For 1 k 4, k j 1 l 3, l m and 1 n 3letũ k,ũ j,l and ũ j,m,n be suitable etensions of u k, u j,l and u j,m,n to 2 We define u k = P k ũ k, u j,l = P j,l ũ j,l, u j,m,n = P j,m,n ũ j,m,n and perform nonhomogeneous dyadic decompositions in k, j,l and j,m,n Byusing(345), we estimate high K 1/2 <M 1 M 2 M 2 M min(5) M med(5) 2s+1 M 2 M med(5) 1 ma(7) K high l m n=1 on this region by 1 M 7 M 8 M 9 ma(7) k j,l j,m,n u k Y 0 u j,l Y 0 l m u j,m,n Y 0, where ma(7) =ma{ j,m,1, j,m,2, j,m,3 } herefore, (429) also holds, for s>1/4, in this region Finally, we conclude the proof of Proposition 43 gathering (412) (429) emark 44 he restriction s>1/3 only appears when estimating the contribution K All the other contributions are estimated with s>1/4 It is likely med that the inde 1/3 may be improved by adding higher order modifications to the energy n=1 43 Estimates for the X s 1,1 and X s 7/8,15/16 norms In this subsection, we eplain how to control the X s 1,1 and X s 7/8,15/16 norms that we used in the energy estimates We start by deriving a suitable Strichartz estimate for the solutions of (41) Proposition 45 Assume that 0 < 1 and let u L (]0,[: H 1/4 ()) be a solution to (41) with u i L (]0,[:H 1/4 ()), i =1, 2, 3 hen, (431) J 1/7 u L 4 L u L H1/4 + u j L H 1/4 Proof Since J 1/7 u is a solution to (41) where we apply J 1/7 on the right-hand side member, we use estimate (217) with F = J 1/7 (u 1 u 2 u 3 )andδ =9/7+ he Hölder and Sobolev inequalities then lead to J 1/7 u L 4 L u L H3/14+ + u 1 u 2 u 3 L 4 L 1 u L H1/4 + u j L H 1/6