On well-posedness and wave operator for the gKdV equation

Bull. Sci. math. 137 (2013) 229–241 www.elsevier.com/locate/bulsci

On well-posedness and wave operator for the gKdV equation Luiz G. Farah a,1 , Ademir Pastor b,∗,2 a ICEx, Universidade Federal de Minas Gerais, Av. Antônio Carlos, 6627, Caixa Postal 702, 30123-970,

Belo Horizonte-MG, Brazil b IMECC-UNICAMP, Rua Sérgio Buarque de Holanda, 651, 13083-859, Campinas-SP, Brazil

Received 30 March 2012 Available online 25 April 2012

Abstract We consider the generalized Korteweg–de Vries (gKdV) equation ∂t u + ∂x3 u + μ∂x (uk+1 ) = 0, where k > 4 is an integer number and μ = ±1. We give an alternative proof of the Kenig, Ponce and Vega result in Kenig, Ponce and Vega (1993) [9], which asserts local and global well-posedness in H˙ sk (R), with sk = (k − 4)/2k. A blow-up alternative in suitable Strichartz-type spaces is also established. The main tool is a new linear estimate. As a consequence, we also construct a wave operator in the critical space H˙ sk (R), extending the results of Côte (2006) [2]. © 2012 Elsevier Masson SAS. All rights reserved.

1. Introduction In this paper, we consider the generalized Korteweg–de Vries (gKdV) equation    ∂t u + ∂x3 u + μ∂x uk+1 = 0, x ∈ R, t > 0, u(x, 0) = u0 (x), where μ = ±1 and k > 4 is an integer number. * Corresponding author.

E-mail addresses: [email protected] (L.G. Farah), [email protected] (A. Pastor). 1 Partially supported by FAPEMIG, CNPq and CAPES/Brazil. 2 Partially supported by FAPESP and CNPq/Brazil.

0007-4497/$ – see front matter © 2012 Elsevier Masson SAS. All rights reserved. http://dx.doi.org/10.1016/j.bulsci.2012.04.002

(1.1)

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In the particular case k = 1, this equation was derived by Korteweg and de Vries [11] in their study of waves on shallow water. Here, we are mainly interested in the case k > 4 (the L2 supercritical case), which is a generalization of the model proposed in [11]. Well-posedness for the Cauchy problem (1.1) is now well understood. We first recall the scaling argument: if u is a solution of (1.1), then, for any λ > 0, uλ (x, t) = λ2/k u(λx, λ3 t) is also a solution with initial data uλ (x, 0) = λ2/k u0 (λx). Moreover,   uλ (·, 0) ˙ s = λs+2/k−1/2 u0  ˙ s . H H Thus, for each k fixed, the scale-invariant Sobolev space is H˙ sk (R), sk = 1/2 − 2/k. As a consequence, the natural Sobolev spaces to study Eq. (1.1) are H s (R) with s  sk . The well-posedness theory to the Cauchy problem (1.1) was developed by Kenig, Ponce and Vega [9] (see also Kato [7] for a previous result on this direction). Concerning the small data global theory in the critical Sobolev space H˙ sk (R), it was proved in [9, Theorem 2.15] the following result. Theorem 1.1. Let k > 4 and sk = (k − 4)/2k. Then there exists δk > 0 such that for any u0 ∈ H˙ sk (R) with  s  D k u0  < δk x 2 there exists a unique solution u(·) of the IVP (1.1) satisfying   u ∈ C R : H˙ sk (R) ,  s  D k u 5 10 < ∞, x L L  s x t D k ux  ∞ 2 < ∞, x L L x

and

t

 α βk  D k Dt u pk qk < ∞, x L L x

t

(1.2) (1.3) (1.4)

(1.5)

where 1 2 − , 10 5k 2 1 1 = + , pk 5k 10

αk =

3 6 − , 10 5k 3 1 4 = − . qk 10 5k

βk =

(1.6) (1.7)

Furthermore, the map u0 → u(t) form {u0 ∈ H˙ sk (R): Dxsk u0 2 < δk } into the class defined by (1.2)–(1.5) is Lipschitz. The method to prove Theorem 1.1 combines smoothing effects and Strichartz-type estimates together with the Banach contraction principle. This result shows to be sharp in view of the work due to Birnir, Kenig, Ponce, Svanstedt and Vega [1]. One of the main goals of this paper is to reprove the above theorem without using any norm that involves derivative in the time variable. To this end, we introduce a new linear estimate (see Lemma 2.5 below) which allows us to obtain the following result. Theorem 1.2. Let k  4 and sk = (k − 4)/2k. Given u0 ∈ H˙ sk (R), assume  s  D k u0  2  K < ∞. x L x

(1.8)

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

231

There exists δ = δ(K) > 0 such that if   U (t)u0  5k/4 5k/2 < δ, Lx

(1.9)

Lt

then there exists a unique solution u of the integral equation t u(t) = U (t)u0 − μ

     U t − t  ∂x uk+1 t  dt 

(1.10)

0

such that uL5k/4 L5k/2  2δ x

and

t

 s  D k u x

L5x L10 t

< 2cK.

(1.11)

The solution also satisfies  s  D k u ∞ 2 < 2cK. x

(1.12)

Lt Lx

Moreover, there exist f± ∈ H˙ sk (R) such that    lim D sk u(t) − U (t)f±  2 = 0. t→±∞

x

(1.13)

Lx

It is worth to mention that the question of how small the initial data should be to imply global well-posedness in the energy space H 1 (R) has been recently addressed by Farah, Linares and Pastor [4] where sufficient conditions have been obtained. Without imposing any smallness restriction on the initial data, a local version of Theorem 1.1 is also available in [9, Theorem 2.17]. Here, following the same strategy of Theorem 1.2, we are able to prove the following local well-posedness result. Theorem 1.3. Let k  4 and sk = (k − 4)/2k. Given u0 ∈ H˙ sk (R) there exist T = T (u0 ) and a unique solution u of the integral equation (1.10) satisfying  s    D k u ∞ 2 + u 5k/4 5k/2 + D sk u 5 10 < ∞. (1.14) x

L[0,T ] Lx

Lx

L[0,T ]

x

Lx L[0,T ]

Furthermore, given T  ∈ (0, T ) there exists a neighborhood V of u0 in H˙ sk (R) such that the map u0 → u(t) ˜ from V into the class defined by (1.14) in the time interval [0, T  ] is Lipschitz. Note that the previous result asserts that the existence time depends on the initial data itself and not only on its norm. Our next theorem concerns the behavior of the local solution near the possible blow-up time. This is inspired by the results in [10, Theorem 1.2]. Let T ∗ = T ∗ (u0 ) be the maximum time of existence for the unique solution u of the integral equation (1.10) with initial data u0 ∈ H˙ sk (R). When T ∗ < ∞, we prove that always uL5k/4 L5k/2 = ∞. On the other x

[0,T ∗ ]

hand, a direct application of Theorem 1.3 implies that either the H˙ sk (R)-norm of u(t) blows-up in time or the H˙ sk (R)-limt↑T ∗ u(t) does not exist. However, even in the case when the H˙ sk (R)norm of the solution u(t) does not blow-up (this is the case when k = 4 by the mass conservation law) we established blow-up for the Strichartz norm that appears in (2.28). Our result reads as follows. Theorem 1.4. Assume k  4 and sk = (k − 4)/2k. Suppose u0 ∈ H˙ sk (R) and T ∗ = T ∗ (u0 ) be the maximum time of existence given in Theorem 1.3. If T ∗ < ∞ then uL5k/4 L5k/2 = ∞. x

[0,T ∗ ]

(1.15)

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Moreover, if supt∈[0,T ∗ ) Dxsk u(t)L2x = K < ∞, we also have  2/3k  Dx u 3k/2 3k/2 = ∞. L L x

[0,T ∗ ]

(1.16)

Remark 1.5. In the limit case k = 4 we recover the result in [10, Theorem 1.2]. Next we turn to the construction of the wave operator associated to Eq. (1.1). This is the reciprocal problem of the scattering theory, which consists in constructing a solution with a prescribed scattering state. Roughly speaking, for a given profile V (regardless of its size), one looks for a solution of the nonlinear problem u(t), defined for large enough t , such that   lim u(t) − uV (t)Y = 0, t→∞

where uV (t) is the solution of the linear problem with initial data V , and Y stands for a suitable Banach space. Solving the latter problem is also known as the construction of a wave operator. This question was studied in Besov spaces for the generalized Boussinesq equation in [3] and in Sobolev or weighted Sobolev spaces for Schrödinger equations by Ginibre, Ozawa and Velo [5,6] and for generalized Korteweg–de Vries equations by Côte [2]. In this last paper, the author introduced two different approaches to deal with this problem. The case k > 4 was treated in the weighted Sobolev space setting. On the other hand, the case k = 4 (L2 -critical KdV equation) is simpler since the fixed point problem is in fact very similar to that of the Cauchy problem treated by Kenig, Ponce and Vega [9] in their small data global existence theorem (see Theorem 1.1). In this paper, using the new linear estimate (Lemma 2.5 below), we show that we can also apply the same approach to the case k > 4, extending Côte’s result to the classical Sobolev spaces. Our theorem is the following: Theorem 1.6. Let k  4 and sk = (k − 4)/2k. For any v ∈ H˙ sk (R), let uv (t) be the solution of the linear problem associated with (1.1) with initial data v. Then, there exist T0 = T0 (v) and u ∈ C([T0 , ∞) : H˙ sk (R)) solution of the IVP (1.1) satisfying   lim u(t) − uv (t)H˙ sk (R) = 0.

t→∞

(1.17)

Moreover, u(·) is unique in the class  5k/4 5k/2  sk 5 10 ˙ sk u ∈ L∞ . LT t H (R)/Dx u ∈ Lx LT , u ∈ Lx The plan of this paper is as follows. In the next section we introduce some notation and prove the linear estimates related to our problem. In Section 3 we prove Theorems 1.2 and 1.4. Next, in Section 4, we show Theorem 1.6. 2. Notation and preliminaries Let us start this section by introducing the notation used throughout the paper. We use c to denote various constants that may vary line by line. Given any positive numbers a and b, the notation a  b means that there exists a positive constant c such that a  cb.

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

233

We use  · Lp to denote the Lp (R)-norm. If necessary, we use subscript to inform which q q q variable we are concerned with. The mixed norms Lt Lrx , L[a,b] Lrx and LT Lrx of f = f (x, t) are defined, respectively, as +∞

1/q b

1/q  q  q     f (·, t) r dt f (·, t) r dt , f  q f  q r = r = Lt Lx

L[a,b] Lx

Lx

−∞

and f Lq Lr T

x

Lx

a

+∞

1/q  q f (·, t) r dt = L x

T

with the usual modifications when q = ∞ or r = ∞. Similarly, we also define the norms in the q q q spaces Lrx Lt , Lrx L[a,b] and Lrx LT . The spatial Fourier transform of f (x) is given by  ˆ f (ξ ) = e−ixξ f (x) dx. R

The class of Schwartz functions is represented by S(R). We shall also define Dxs and Jxs to be, respectively, the Fourier multiplier with symbol |ξ |s and ξ s = (1 + |ξ |)s . In this case, the norms in the Sobolev spaces H s (R) and H˙ s (R) are given, respectively, by         f H˙ s ≡ Dxs f L2 = |ξ |s fˆ L2 . f H s ≡ Jxs f L2 =  ξ s fˆ L2 , x

x

ξ

ξ

Let us present now some useful lemmas and inequalities. In what follows, U (t) denotes the linear propagator associated with the gKdV equation, that is, for any function u0 , U (t)u0 is the solution of the linear problem  ∂t u + ∂x3 u = 0, x ∈ R, t ∈ R, (2.18) u(x, 0) = u0 (x). We begin by recalling the results necessary to prove Theorem 1.2. Lemma 2.1. Let p, q, and α be such that −α + Then

3 1 1 + = , p q 2

1 1 − α . 2 q

(2.19)

  α D U (t)u0  q p  u0  2 . Lx x L L t

(2.20)

x

Also, if 1 1 1 + = , p 2q 4 then,

α=

2 1 − , q p

1 1  p, q  ∞, −  α  1, 4

  α D U (t)u0  p q  u0  2 . Lx x L L x

(2.22)

t

Proof. See [8, Theorem 2.1] and [10, Proposition 2.1].

(2.21)

2

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Remark 2.2. Note that when (p, q) = (∞, 2) in (2.22) we have the sharp version of the Kato smoothing effect   ∂x U (t)u0  ∞ 2  u0  2 . (2.23) Lx L L t

x

Moreover, the dual version of (2.23) reads as follows:   t         ∂x U t − t  g ·, t  dt    gL1x L2t  2 

(2.24)

Lx

0

(see [9, Theorem 3.5]). Lemma 2.3. Assume 1 1 α< , 4 2 Then,

1 1 = − α. p 2

(2.25)

 −α  D U (t)u0  p ∞  u0  2 . Lx x L L x

(2.26)

t

Proof. See [9, Lemma 3.29].

2

We can also obtain the following particular cases of the Strichartz estimates in the critical Sobolev space H˙ sk (R). Corollary 2.4. Let k  4 and sk = (k − 4)/2k. Then  −1/k    Dx U (t)u0  k ∞  D sk u0  2 L L L x

and

t

(2.27)

x

  2/3k Dx U (t)u0 

3k/2 3k/2 Lx Lt

   D sk u0 L2 .

(2.28)

x

Proof. The estimate (2.27) is a particular case of (2.26). On the other hand, by Sobolev embedding and inequality (2.20) with q = 3k/2 and α = 2/3k, we obtain  2/3k      Dx U (t)u0  3k/2 3k/2  Dx2/3k U (t)D sk u0  3k/2 p  D sk u0  2 , x L L L L L x

where

1 p

=

1 2

+

2 3k

−

2 k,

t

t

which implies (2.28).

x

x

2

Our next lemma is the fundamental tool to prove Theorem 1.2. Lemma 2.5. Let k  4 and sk = (k − 4)/2k. Then     U (t)u0  5k/4 5k/2  D sk u0  2 . x L L L x

t

(2.29)

x

Proof. Interpolate the inequalities (2.28) and (2.27). Next, we recall the following integral estimates.

2

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

235

Lemma 2.6. If pi , qi , αi , i = 1, 2, satisfy (2.21), then   t          α1   U t − t g ·, t dt   Dx−α2 g  p2 q2 , Dx   p1 q1 Lx Lt Lx Lt

0

where p2 and q2 are the Hölder conjugate of p2 and q2 , respectively. Moreover, assume (p1 , q1 , α1 ) satisfies (2.21) and (p2 , α2 ) satisfies (2.25). Then,   t          −α2   U t − t g ·, t dt   D −α1 g  p1 q1 . (2.30) D   p2 ∞ Lx Lt Lx Lt

0

Proof. Use the duality and T T ∗ arguments combined with Lemmas 2.1 and 2.3 (see also [10, Proposition 2.2]). 2 Applying the same ideas as the previous lemma together with Lemma 2.5 and (2.23) we also have: Corollary 2.7. Let k  4 and sk = (k − 4)/2k, then the following estimate holds:  t              Dxsk g L1 L2 . ∂x U t − t g ·, t dt  x t   5k/4 5k/2 Lx

0

Lt

Remark 2.8. In all the above inequalities we can replace the integral estimate will be used in Section 4.

t 0

by

∞ t

. This kind of

Finally, we have the following estimates for fractional derivatives. Lemma 2.9. Let 0 < α < 1 and p, p1 , p2 , q, q1 , q2 ∈ (1, ∞) with Then,

1 p

=

1 p1

(i)  α  D (f g) − f D α g − gD α f  x

x

x

p q

Lx Lt

   Dxα f Lp1 Lq1 gLp2 Lq2 . x

t

The same still holds if p = 1 and q = 2. (ii)  α  D F (f ) x

p q

Lx Lt

     Dxα f Lp1 Lq1 F  (f )Lp2 Lq2 . x

x

t

Proof. See [9, Theorems A.6, A.8, and A.13].

t

2

3. Global well-posedness and the blow-up alternative Our aim in this section is to establish Theorems 1.2 and 1.4.

x

t

+ p12 and

1 q

=

1 q1

+ q12 .

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L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

Proof of Theorem 1.2. As usual, our proof is based on the contraction mapping principle. Hence, we define         k Xa,b = u ∈ C R; H˙ sk (R) : uL5k/4 L5k/2  a and Dxsk uL∞ L2 + Dxsk uL5 L10  b . x

On

k Xa,b

t

t

x

x

t

consider the integral operator t

Φ(u)(t) := U (t)u0 − μ

     U t − t  ∂x uk+1 t  dt  .

(3.31)

0

We need to show that k k Φ : Xa,b → Xa,b

is a contraction, for an appropriated choice of the parameters a > 0 and b > 0. We first estimate the  · L5k/4 L5k/2 -norm. From Corollary 2.7, x t        Φ(u) 5k/4 5k/2  U (t)u0  5k/4 5k/2 + cD sk uk+1  1 2 . x L L L L L L x

x

t

t

x

t

Dxsk (uk+1 )L1 L2 x t

The estimate of the term can be found in [9] (see Eq. (6.1)). For the sake of completeness we will also perform it here. Indeed, applying the fractional derivative rule in Lemma 2.9(i), we obtain  s  k+1         D k u  1 2  uk  5/4 5/2 D sk u 5 10 + uD sk uk  1 2 x x x Lx Lt Lx Lt Lx Lt Lx Lt  s     k k    u 5k/4 5k/2 Dx u L5 L10 + uL5k/4 L5k/2 Dxsk uk Lp0 Lq0 x x t Lx Lt x t t  s  k k    u 5k/4 5k/2 Dx u L5 L10 Lx Lt x t  s    + uL5k/4 L5k/2 Dxk uL5 L10 uk−1 Lp1 Lq1 , (3.32) x

x

t

t

x

t

where 1 1 4 1 4(k − 1) 1 = − =1− − = and p1 p0 5 5k 5 5k 1 1 1 1 2 1 4(k − 1) = − = − − = . q1 q0 10 2 5k 10 10k Therefore     Φ(u) 5k/4 5k/2  U (t)u0  5k/4 5k/2 + ca k b L L L L x

x

t

t

 δ + ca k b.

(3.33)

On the other hand, using the inhomogeneous smoothing effect (2.24), Lemma 2.6 with (p1 , q1 , α1 ) = (5, 10, 0), (p2 , q2 , α2 ) = (∞, 2, 1), and (2.22) with (p, q, α) = (5, 10, 0), estimate (3.32) also implies  s         D k Φ(u) ∞ 2 + D sk Φ(u) 5 10  cK + cD sk U (t)u0  5 10 + cD sk uk+1  1 2 x x x x L L L L L L L L t

x

x

t

x

t

 cK + ca k b. First one chooses b = 2cK and a such that ca k  1/2. Therefore, once for all  s    D k Φ(u) ∞ 2 + D sk Φ(u) 5 10  b. x

Lt Lx

x

Lx Lt

x

t

(3.34)

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

237

Finally, if one chooses δ = a/2 and a so that ca k−1 b  1/2 then we also have   Φ(u) 5k/4 5k/2  a. Lx

Lt

k → X k is well defined. For the contraction one uses Such calculations establish that Φ : Xa,b a,b similar arguments. The contraction mapping principle then implies the existence of a unique fixed point for Φ, which is a solution of (1.10). The proof is complete with standard arguments. To prove (1.13), one defines

±∞      f ± = u0 + μ U −t  ∂x uk+1 t  dt  . 0

Then, as in (3.32)–(3.34), we have   +∞    s          k+1   D k u(t) − U (t)f+  2  D sk ∂ u t dt U t − t  x x x Lx   2 Lx t  s  k k    u 5k/4 5k/2 Dx u L5 L10 . Lx

Since uL5k/4 L5k/2 < ∞ and x t calculation holds for f− . 2

L[t,+∞)

s Dxk uL5 L10 x t

x

[t,+∞)

< ∞, the above inequality implies (1.13). A similar

Theorem 1.3 follows from similar arguments, so it will be omitted. Next, we prove the blowup result stated in Theorem 1.4. Proof of Theorem 1.4. The proof is based on the arguments in [10, Theorem 1.2]. We first claim that if uL5k/4 L5k/2 < ∞ then Dxsk uL5 L10 ∗ < ∞. Indeed, let ε0 > 0 be an arbitrary small [0,T ∗ ]

x

x

[0,T ]

number. Since the norm uL5k/4 L5k/2 is finite we can split the interval [0, T ∗ ] in 0 = t0 < t1 < [0,T ∗ ]

x

· · · < t = T ∗ such that uL5k/4 L5k/2 < ε0 , where In = [tn , tn+1 ], n = 0, 1, . . . , − 1. Since u is x

In

a fixed point of the integral equation (3.31), from (2.22) with (p, q, α) = (5, 10, 0), we have, for n = 0, 1, . . . , − 1,   t    s   s        D k u 5 10  D k u0  2 + D sk U t − t  ∂x uk+1 t  dt    x x Lx LIn Lx   5 10 Lx LIn

0

   Dxsk u0 L2 x

   D sk u0  x

L2x

  tj +1 n 

  k+1        sk   t dt  + U t − t ∂x u Dx   j =0

L5x L10 In

tj

  t n 

  k+1           sk   t χIj t dt  + U t − t ∂x u Dx   j =0

, (3.35)

L5x L10 t

0

where χIj denotes the characteristic function of the interval Ij . By using Lemma 2.6, with (p1 , q1 , α1 ) = (5, 10, 0) and (p2 , q2 , α2 ) = (∞, 2, 1), we then deduce n

   s  k+1   s   1 2 . D k u D k u 5 10  D sk u0  2 + x x x L L L L L x

In

x

j =0

x

Ij

(3.36)

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L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

Now, inequality (3.32) yields  s  D k u

L5x L10 In

x

n

  s   D k u 5 10 uk 5k/4  Dxsk u0 L2 + x L L x

x

j =0

Lx

Ij

5k/2 j

LI

n

  s   D k u 5 10 .  cDxsk u0 L2 + cε0k x L L x

x

j =0

(3.37)

Ij

Therefore, choosing cε0k < 1/2, we conclude  s  D k u

L5x L10 In

x

n−1

  s   D k u 5 10 .  2cDxsk u0 L2 + 2 x L L x

x

j =0

(3.38)

Ij

s

Inequality (3.38) together with an induction argument implies that Dxk uL5 L10 < ∞ for n = x In 0, 1, . . . , − 1. By summing over the intervals we conclude the claim. Next we prove (1.15). Assume that the solution exists for |t| < T  with uL5k/4 L5k/2 < ∞, by x

the above claim we also have Dxsk uL5 L10 x

[0,T  ]

for all t ∈ [0, T  ]  s    D k u(t) 2  D sk u0  2 + uk 5k/4 x x L L Lx

x

[0,T  ]

< ∞. Moreover, by inequality (3.32), we conclude

5k/2

L[0,T  ]

 s  D k u

L5x L10 [0,T  ]

x

< ∞.

As a consequence, u(T  ) ∈ H˙ sk (R), which from Theorem 1.3 implies the existence of δ > 0 such that the solution exists for |t|  T  + δ. Thus, if T ∗ < ∞, (1.15) must be true. Next, we turn to proof of (1.16). Let δ > 0 be a small constant to be chosen later. Since uL5k/4 L5k/2 = ∞, we can construct a family of intervals In = [tn , tn+1 ] such that tn < tn+1 , x

[0,T ∗ ]

tn T ∗ , and

uL5k/4 L5k/2 = δ. x

(3.39)

In

From an analytic interpolation, we obtain  2/3k 3/5  −1/k 2/5 u 5k/4 5k/2  Dx u 3k/2 3k/2 Dx u k Lx

LIn

Lx

Lx L∞ In

LIn

(3.40)

.

On the other hand, since u is a fixed point of integral equation (3.31), for t ∈ In we also have t u(t) = U (t − tn )u(tn ) − μ

     U t − t  ∂x uk+1 t  dt  .

(3.41)

tn

Therefore  −1/k    Dx u k ∞  Dx−1/k U (t − tn )u(tn ) k ∞ L L L L x

In

x

In

  t     k+1     −1/k   + Dx U t − t ∂x u dt    tn

.

Lkx L∞ In

To bound the linear part we use Corollary 2.4 and to bound the integral part, we use Lemma 2.6 and (2.30) with (p2 , α2 ) = (k, 1/2 − 1/k) and (p1 , q1 , α1 ) = (∞, 2, 1). Hence,

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

 −1/k  Dx u

Lkx L∞ In

239

      Dxsk U (−tn )u(tn )L2 + Dxsk uk+1 L1 L2 x x In   k  sk   cK + cδ Dx u L5 L10 , x

In

where in the last inequality we have used supt∈[0,T ∗ ) Dxsk u(t)L2x = K and (3.39). s It remains to bound the norm Dxk uL5 L10 . Indeed using again the integral equation (3.41), x In Lemma 2.6 with (p1 , q1 , α1 ) = (5, 10, 0), (p2 , q2 , α2 ) = (∞, 2, 1) and estimate (3.32) we obtain      s  D k u 5 10  D sk U (−tn )u(tn ) 2 + D sk u 5 10 uk 5k/4 5k/2 x x x Lx LIn Lx Lx LIn Lx LIn   k  sk   cK + cδ Dx u L5 L10 . (3.42) x

Hence, choosing −1/k Dx uLkx L∞ I n

cδ k

In

< 1/2 we conclude Dxsk uL5 L10  2cK, which also implies x

In

 2cK. The last two inequalities combining with (3.40) yield

 2/3k  Dx u 3k/2 3k/2  L L x

In

δ 5/3 , (2cK)2/3

and, therefore (1.16) holds.

for all n ∈ N

2

4. The construction of the wave operator In this section, we intend to show Theorem 1.6. Following the ideas introduced by Côte [2], we must look for a fixed point for the operator ∞ Φ : w(t) −→ −μ∂x

     k+1   U t − t w t + U t v dt ,

t

defined in the time interval [T0 , ∞), where T0 > 0 is an arbitrarily large number that will be chosen later. The next proposition says that this fixed point provides a function u satisfying the integral equation (1.1) (we refer to Farah [3] for a proof). Proposition 4.1. Let w be a fixed point of the operator Φ and define u(t) = U (t)v + w(t).

(4.43)

Then u is a solution of (1.1) in the time interval [T0 , ∞). Proof of Theorem 1.6. Again we use the contraction mapping principle. Given T > 0, define the metric spaces     XT = w ∈ C R; H˙ sk (R) ; |||w|||XT < ∞ and

  XTa = w ∈ XT ; |||w|||XT  a ,

where

  |||w|||XT = wL∞ H˙ sk + Dxsk w L5 L10 + wL5k/4 L5k/2 . T

x

T

x

T

240

L.G. Farah, A. Pastor / Bull. Sci. math. 137 (2013) 229–241

Applying Corollary 2.7 we obtain       Φ(w) 5k/4 5k/2  D sk U (t)v + w(t) k+1  1 2 . x L L L L x

x

T

T

Using the same arguments as the ones used in (3.32), we conclude   s        D k U (t)v + w(t) k+1  1 2  U (t)v + w(t)k 5k/4 5k/2 D sk U (t)v + w(t)  5 10 . x x L L L L L L x

x

T

x

T

T

The other norms can be estimated in the same manner, which implies        Φ(w)  U (t)v + w(t)k 5k/4 5k/2 D sk U (t)v + w(t)  5 10 XT

Lx

LT

x

Lx LT

Lx

LT

x

L5x L10 T

 k  U (t)v + w(t) 5k/4  k  U (t)v  5k/4 Lx

 s   k U (t)v + w(t)  5k/2 D

 s  k 5k/2 D v  x

LT

L2x

  + Dxsk v L2 wkXT + wk+1 XT , x

(4.44)

where in the last inequality we have used Lemma 2.1 and Lemma 2.5. Since U (t)vL5k/4 L5k/2 → 0 as T → ∞ we can find a T0 > 0 large enough and a > 0 small x

T

enough such that Φ : XTa0 → XTa0 is well defined and is a contraction. Therefore, Φ has a unique fixed point, which we denote by w. Moreover, a > 0 can be chosen such that   k  wXT  U (t)v L5k/4 L5k/2 Dxsk v L2 , (4.45) x

x

T

which implies wXT → 0 as T → ∞. Next we show the limit (1.17). By the proposition, u(t) = U (t)v + w(t) satisfies the integral equation in the right hand side of (3.31) in the time interval [T0 , ∞). Therefore     u(t) − U (t)v  ∞ ˙ s = w ∞ ˙ sk = Φ(w) ∞ ˙ s LT H

k

LT H

 k  U (t)v  5k/4 Lx

LT H

 s  k 5k/2 D v 

LT

x

k

L2x

  + Dxsk v L2 wkXT + wk+1 XT . x

The last inequality together with (4.45) implies (1.17), finishing the proof of the theorem.

2

Acknowledgements The authors thank Felipe Linares and an anonymous referee for their helpful comments and suggestions which improved the presentation of the paper. References [1] B. Birnir, C.E. Kenig, G. Ponce, N. Svanstedt, L. Vega, On the ill-posedness of the IVP for the generalized Korteweg–de Vries and nonlinear Schrödinger equations, J. London Math. Soc. 53 (1996) 551–559. [2] R. Côte, Large data wave operator for the generalized Korteweg–de Vries equations, Differential Integral Equations 19 (2006) 163–188. [3] L.G. Farah, Large data asymptotic behaviour for the generalized Boussinesq equation, Nonlinearity 21 (2008) 191– 209. [4] L.G. Farah, F. Linares, A. Pastor, The supercritical generalized KdV equation: Global well-posedness in the energy space and below, Math. Res. Lett. 18 (2011) 357–377. [5] J. Ginibre, T. Ozawa, G. Velo, On the existence of the wave operator for a class of nonlinear Schrödinger equations, Ann. Inst. H. Poincaré Phys. Théor. 60 (1994) 211–239. [6] J. Ginibre, G. Velo, Long range scattering for some Schrödinger related nonlinear systems, preprint, arXiv:math/ 0412430v1, 2004.

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