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Large data wave operator for the generalized Korteweg-de Vries equations

Rapha¨ el Cˆ ote

Abstract

We consider the generalized Korteweg-de Vries equations : ut+ (uxx+up)x= 0, t, x∈R,

forp∈(3,∞). LetU(t) be the associated linear group. GivenV in the weighted Sobolev spaceH2,2={f ∈L2| k(1 +|x|)2(1−∂x2)fkL2<∞}, possibly large, we construct a solutionu(t) of the generalized Korteweg- de Vries equation such that :

t→∞lim ku(t)−U(t)VkH1 = 0.

We also prove uniqueness of such a solution in an adequate space.

In theL2-critical case (p= 5), this result can be improved to any possibly large functionV in L2 (with convergence inL2).

1 Introduction.

1.1 Recall of known results.

We consider the generalized Korteweg-de Vries equation :

u

t

+ (u

xx

+ u

p

)

x

= 0, t, x, ∈

R

, (1) where p ≥ 2. The case p = 2 corresponds to the original equation introduced by Korteweg and de Vries [6] in the context of shallow water waves. For both p = 2 and p = 3, this equation has many applications to Physics : see for example Miura [14], Lamb [7].

There are two formally conserved quantities for solutions to (1) :

Z

u

2

(t) =

Z

u

2

(0) (L

2

mass), (2)

E(u(t)) = 1 2

Z

u

2x

(t) − 1 p + 1

Z

u

p+1

(t) = E(u(0)) (energy). (3)

Mathematical Subject Classification : Primary 35Q53, Secondary 35B40.

Keywords : Generalized Korteweg-de Vries equations, wave operator, large data.

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The local Cauchy problem for (1) has been intensively studied by many au- thors. Kenig, Ponce and Vega [5] proved the following existence and unique- ness result in H

1

(

R

) : for u

0

∈ H

1

(

R

), there exist T = T(ku

0

k

H1

) > 0 and a solution u ∈ C([0, T ], H

1

(

R

)) to (1) satisfying u(0) = u

0

, which is unique in some class Y

T

⊂ C([0, T ], H

1

(R)). For such solution, one has conservation of mass and energy. Moreover, if T

1

denotes the maximal time of existence for u, then either T

1

= +∞ (global solution) or T

1

< ∞ and ku(t)k

H1

→ ∞ as t ↑ T

1

(blow-up solution).

For p = 2 and p = 3, equation (1) is completely integrable, and thus has very special features. The inverse scattering transform method allows to solve the Cauchy problem in an appropriate space (for example if k(1 +

|x|

2

)

5

u

0

k

C4

< ∞) and to find the asymptotic behavior of solutions as t → ±∞ : see for example Schuur [16], Eckhaus and Schuur [2], Miura [14]. However, if p 6= 2 or 3, the inverse scattering transform method does not longer apply, and the description of solutions as t → +∞ in the non- integrable case is an open problem. Let us recall some results which are not based on the inverse scattering transform method.

In the case 2 ≤ p < 5, all solutions in H

1

are global and uniformly bounded due to the conservations laws and the Gagliardo-Nirenberg inequal- ity :

∀v ∈ H

1

(

R

),

Z

|v|

p+1

≤ C(p)

Z

v

2

p+34 Z

v

x2

p−14

.

The case p = 5 is L

2

-critical, in the sense that the mass remains unaffected by scaling. If :

u

t

+ (u

xx

+ u

5

)

x

= 0, t, x, ∈

R

. (4) Then u

λ

(t, x) = λ

1/6

u(λt, λ

1/3

x) is also a solution to (4), and ku

λ

k

L2

= kuk

L2

. In this case, the local existence result of [5] is improved to initial data in L

2

(instead of H

1

). However, existence of finite time blow-up solutions was proved by Merle [13] and Martel and Merle [10]. Therefore p = 5 also appears as a critical exponent for the long time behavior of solutions to (1).

Another problem which was studied by many authors is scattering for small initial data in an appropriate functional space, see for example [17], [15], [1], [4]. Let us recall the result of Hayashi and Naumkin [4]. Introduce the following weighted Sobolev spaces :

H

s,m

= {φ ∈ S

0

| kφk

Hs,m

= k(1 + |x|

2

)

m/2

(1 − ∂

x2

)

s/2

φk

L2

< ∞}. (5) Let p > 3. Given u

0

small enough in H

1,1

, the out-coming solution u(t) is global in time, and there is scattering, in the sense that there exists a function V ∈ L

2

so that :

ku(t) − U (t)V k

L2

→ 0.

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where U (t) denotes the linear operator for the KdV equation, i.e. v(t) = U (t)V satisfies v

t

+ v

xxx

= 0, v(0) = V . This is the description of solutions around with initial data around 0 (in H

1,1

).

Now, there exist also special explicit traveling wave solutions called soli- tons. If Q denotes the unique solution (up to translation) of :

Q > 0, Q ∈ H

1

(

R

), Q

xx

+ Q

p

= Q, i.e. Q(x) = p + 1 2 cosh

2

(

p−12

x)

!p−11

, then for c > 0, the soliton

R

c,x0

= c

p−11

Q( √

c(x − x

0

− ct)) is a solution to (1).

We should notice that for p > 3, solitons do not appear in the small data analysis of [4], as :

kR

c,x0

k

H1,1

≥ c

0

, for some uniform constant c

0

> 0.

Indeed :

Z

(R

c,x0

)

2x

= c

p+3 2(p−1)

Z

Q

2x

, and

Z

(xR

c,x0

)

2

= c

7−3p 2(p−1)

Z

(xQ)

2

. For p > 3,

2(p−1)p+3

> 0 but

2(p−1)7−3p

< 0, so that if kR

c,x0

k

H1

→ 0, then c → 0 and thus kR

c,x0

k

H0,1

→ ∞. (Notice that H

1,1

is not sharp from this point of view).

Description of solutions around a sum of decoupled solitons is available : Martel and Merle [9], Martel, Merle and Tsai [11] proved stability in H

1

and asymptotic stability (in L

2

(x ≥ ct) for c > 0) of a sum of decoupled solitons, in the sub-critical case 2 ≤ p < 5 (in the critical case p = 5, one has blow-up around a soliton [10]).

Our goal is to construct solutions to (1) with a given linear asymptotic behavior : that is the construction of a wave operator with respect to the free KdV operator U (t). This problem is reciprocal to (linear) scattering for small initial data.

Let p > 3, and V ∈ H

2,2

(without smallness assumption). We construct a solution u(t) to (1), defined for large enough times, and such that u(t) − U (t)V → 0 in H

1

as t → ∞. Furthermore, u(t) is unique in an adequate space.

In the L

2

-critical case (p = 5), we obtain an optimal result, in the sense that for V ∈ L

2

, we construct u(t) solution to (4) such that u(t) −U (t)V → 0 in L

2

. Again V need not be small in L

2

.

The philosophy underlying these results is the following : the tools needed

to prove global existence for small data can be applied successfully to con-

struct solutions with a given linear profile, small or large.

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In the case of non-linear Schr¨ odinger equations, there are many results concerning the construction of wave operators. For a review, see e.g. Ginibre and Velo [3].

1.2 Statement of the results.

There are two main results : one in the general case p > 3 which uses the framework of Hayashi and Naumkin [4], and one in the critical case p = 5, using the framework of Kenig, Ponce and Vega [5].

Let p > 3. Fix once for all the three constants : γ ∈ (0, min{1/2, (p− 3)/3}), α = 1

2 −γ ∈ (0, 1/2) and δ = p − 3 − 2γ 3 > 0.

(6) Following the framework of Hayashi and Naumkin [4], we will use the nota- tion D = ∂

x

=

∂x

for the partial differentiation with respect to the space variable x, and :

D

α

φ = F

−1

ξ

α

e

−(iπ/2)(1+α)

φ. ˆ As in [4], we will use the following two operators :

J

t

φ = U (t)xU(−t)φ = (x − 3t∂

x2

)φ, and I

t

φ = xφ + 3t

Z x

−∞

t

φ(t, y)dy.

We write J

t

and I

t

to emphasize that we will always consider norms at a fixed time t although J

t

and I

t

are space-time operators.

Our working spaces will be defined through the time dependent M

0t

norm :

H

t

= {φ ∈ L

2

(

R

)| M

0t

(φ) = kφk

H1

+ kDJ

t

φk

L2

+ kD

α

J

t

φk

L2

< ∞}.

J

t

only appears in the norm, since it is convenient to do linear estimates (see [4], Lemma 2.3). But we introduced I

t

because it is easier to handle when doing energy methods estimates. Notice that M

00

is very similar to k · k

H1,1

.

Different positive constants might be denoted by the same letter C.

We now state our results.

Theorem 1 (Large data wave operator).

Let p > 3, and V ∈ H

2,2

. Then :

1. There exist T

0

= T

0

(kV k

H2,2

) ≥ 1 and a unique u ∈ C([T

0

, ∞), H

t

) solution to (1) so that :

M

0t

(u(t) − U (t)V ) → 0 as t → ∞.

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2. Moreover, there exists a constant C independent of time and V , so that :

∀t ≥ T

0

, M

0t

(u(t) − U (t)V ) ≤ C(1 + kV k

p

H2,2

)t

−δ

. In particular, ku(t) − U (t)V k

H1

≤ Ct

−δ

.

Remark.

The scattering result of Hayashi and Naumkin [4] is the following : for small initial data in H

1,1

, the associated solution u(t) is global in time, satisfies ku(t)k

L

≤ Ct

−1/3

(linear decay rate), and there exists a scattering function V such that u(t) − U (t)V → 0 in L

2

as t → ∞.

Our point here is the reciprocal problem : we construct a solution u with a given scattering state U (t)V . We do not need any smallness assumption ; however some integration by parts do not work as well as in [4], so we need to assume V ∈ H

2,2

, with basically a convergence result in H

1,1

.

This result can be extended to a more general non-linearity. For example :

Corollary 1.

Let us consider the equation :

u

t

+ (u

xx

+ f (u))

x

= 0. (7) where f :

R

R

is C

1

, with f (0) = 0 and f

0

(x) = O(x

p−1

) as x → 0 (and p > 3). Given V ∈ H

2,2

, there exist T

0

= T

0

(kV k

H2,2

) ≥ 1 and a unique u ∈ C([T

0

, ∞), H

t

) solution to (7) so that :

M

0t

(u(t) − U (t)V ) → 0 as t → ∞.

Moreover, there exists a constant C independent of time and V , so that :

∀t ≥ T

0

, M

0t

(u(t) − U (t)V ) ≤ C(1 + kV k

pH2,2

)t

−δ

.

The proof of the corollary follows from that of Theorem 1. In particular, although our proof is done for the focusing power case f (x) = x

p

, it is also true for the defocusing case f (x) = −|x|

p−1

x.

In the critical case p = 5, we prove an analogous result for V ∈ L

2

:

Theorem 2 (Wave operator in the critical case).

Let p = 5. For any V ∈ L

2

, there exist T

0

= T

0

(V ) ∈

R

and u ∈ C

0

([T

0

, ∞), L

2

) solution to the critical KdV equation (4), so that :

ku(t) − U (t)V k

L2

→ 0.

u is unique in the class {u|u ∈ L

t

L

2x

∩ L

5x

L

10t

and ∂

x

u ∈ L

x

L

2t

}.

Remark.

The theorem is true if one weakens the hypothesis to V be such that kU (t)V k

L5

xL10t

+ k∂

x

U (t)V k

L

x L2t

< ∞.

As previously, our proof extends to the defocusing case u

t

+(u

xx

−u

5

)

x

=

0.

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The proofs of the two results strongly rely on the scattering analysis of [4] and [5]. However, the arguments developed in each case are of completely different nature, this is why the proofs will be presented in a separate way : Section 2 gives the strategy of the proofs, Section 3 is devoted to the proof of Theorem 1, and Section 4, to that of Theorem 2.

Acknowledgment

I would like to express my gratitude to my advisor Frank Merle for his availability and his constant encouragements, and to Luis Vega and Yvan Martel for enlightening discussions. I would also thank the referee for his careful reading of the manuscript.

2 Strategy of the proofs.

2.1 The general case p > 3.

We shall always consider w(t) = u(t) − U (t)V . Thus we are looking for w satisfying the equivalent conditions :

w

t

+ w

xxx

+ (w + U (t)V )

px

= 0,

M

0t

(w(t)) → 0 as t → ∞. (8)

To construct w, we will use the following approximation scheme. Let (S

n

)

n∈N

be an increasing sequence in

R

such that S

n

→ ∞ as n → ∞.

Define w

n

(t) as a solution to :

w

nt

+ w

nxxx

+ (w

n

+ U (t)V )

px

= 0,

w

n

(S

n

) = 0, (9)

defined on a maximal interval of the form (I

n

, S

n

] : it corresponds to u

n

solution to (1) with initial condition u

n

(S

n

) = U (S

n

)V . Since V ∈ H

2,2

, U (S

n

)V ∈ H

2

, so u

n

exists and is the unique H

2

solution : the same is true for w

n

.

Our method is then to prove that in fact :

1. I

n

≤ T

0

independent of n : we can define w

n

on an interval [T

0

, S

n

] whose lower bound T

0

is fixed.

2. We have uniform (in n) decay estimates for w

n

on the interval [T

0

, S

n

].

(Of course, if 3 < p < 5, there is global existence in H

1

, and thus

I

n

= −∞ is automatic). To prove this, we will make an intensive use

of the tools developed by Hayashi and Naumkin : this is the heart of

the proof, and it is done in Proposition 1.

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3. We then prove that the sequence w

n

(t) converges to a certain w(t) (in C

t0

L

2x

).

4. By weak limit, we improve the regularity of w(t), to conclude that w is a strong solution to (8).

This does the existence part of the theorem. For the uniqueness part, we study again the L

2

difference of a solution with the one we constructed, and show, with a Gronwall type argument, that these two solutions coincide where they are defined.

This scheme of proof is very similar to that of [8] and [12]. In [8], Martel also study the problem of constructing solutions with a given asymptotic behavior : there it is proved that given a sum of solitons R(t), there exist a unique u(t) solution to (1) with 2 ≤ p ≤ 5 such that ku(t) − R(t)k

H1

→ 0 as t → ∞, and furthermore, the convergence takes place in all the Sobolev spaces H

s

with an exponential decay. Remark that Martel deeply used ideas of the stability for a sum of decoupled solitons [11].

2.2 The L

2

-critical case (p = 5).

In the critical case, we do not need H

1,1

regularity. Since Kenig, Ponce and Vega [5] obtained global existence for small data in L

2

, we use their setting.

The proof goes in this way : denoting as usual w(t) = u(t) − U (t)V , we write formally that an eventual w should solve a fixed point problem with the condition w(t) → 0 as t → ∞.

The fixed point problem is in fact very similar to that of the Cauchy problem, so that we can reuse the linear estimates proved by Kenig, Ponce and Vega [5] for their global existence theorem for small data.

The fact that w(t) → 0 is our smallness condition, which allows us to have a contracting map, and thus a fixed point.

3 Proof of Theorem 1 : p > 3.

We start by reminding the linear estimates of [4] which will be used through- out the proof.

3.1 Linear estimates.

Let p > 3. Remind our notations : three fixed constants γ ∈]0, min{(p − 3)/3, 1/2}[, α = 1/2 − γ, δ = (p − 3 − 2γ)/3 > 0, the operator J

t

φ = xφ − 3t∂

x2

φ = U (t)xU(−t)φ, and our working norm:

M

0t

(φ) = kφk

H1

+ kD

α

J

t

φk

L2

+ kDJ

t

φk

L2

.

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First a few remarks on M

0t

. Of course M

00

(φ) ≤ Ckφk

H1,1

. Second, note that J

t

U (t)V = U (t)xV (and U(t) is a H

s

isometry), so that if V ∈ H

1,1

, we have the uniform control in t :

M

0t

(U (t)V ) ≤ CkV k

H1,1

. (10) We now remind the linear results obtained in [4] (Lemma 2.2), in a slightly improved form.

Lemma 1.

Let t > 0 and φ be a function so that M

0t

(φ) is bounded. Then for r > 4 :

kφk

Lr

≤ C

(1 + t)

1/3−1/(3r)

M

0t

(φ).

And one also has the pointwise inequalities :

|φ(x)| ≤ CM

0t

(φ) (1 + t)

1/3

1 +

x t

1/3

14

, |φ

x

(x)| ≤ CM

0t

(φ) t

2/3

1 +

x t

1/3

14

. As a simple consequence, for V ∈ H

1,1

, we have similar decay estimates on U (t)V .

Proof. See [4], Lemma 2.2 and its proof (especially inequalities 2.16, 2.17.

and 2.18). For completeness, the proof is given in Appendix A.

We will also need the polarized version of Lemma 2.3 of [4] :

Lemma 2.

Let h, k :

R

R

. Then the following inequalities are valid if their right-hand side is bounded :

kD

α

k

p

k

L2

≤ Ckk

p−1

k

L2

(kkk

x

k

1/2L

+ kkk

L

kkk

x

k

(1−3γ)/2L

), kD

α

|k|

p−1

h

x

k

L2

≤ C(kD

α

hk

L2

+ kh

x

k

L2

)(kkk

p−3L

kkk

x

k

L

+kkk

p−3−2γL

kkk

L2

kkk

x

k

L

+ kkk

p−3+2γL

kkk

x

k

1−γL

).

Proof. See [4], Lemma 2.3 and its proof (case σ = 0).

3.2 Uniform estimates on w

n

.

Recall that w

n

is the solution to the problem :

w

nt

+ w

nxxx

+ (w

n

+ U (t)V )

px

= 0,

w

n

(S

n

) = 0, (9)

where S

n

→ ∞ is an increasing sequence of times. Using estimates developed

in [4], we have the following proposition for w

n

:

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Proposition 1 (Uniform estimates on

w

n).

Let p > 3. There exists T

0

≥ 1 independent of n, so that for all n ∈

N

such that S

n

≥ T

0

, w

n

∈ C([T

0

, S

n

], H

1

(

R

)), and :

∀t ∈ [T

0

, S

n

], M

0t

(w

n

(t)) ≤ C(1 + kV k

pH2,2

)t

−δ

(11) where δ = (p − 3 + 2γ)/3 > 0 and C are independent of n and V .

Of course Proposition 1 is the heart of the proof of Theorem 1.

Proof. Define I

n

≥ 1 minimal so that :

∀t ∈ [I

n

, S

n

], M

0t

(w

n

(t)) ≤ 1. (12)

Lemma 3.

Suppose we can prove :

∀t ∈ [I

n

, S

n

], M

0t

(w

n

(t)) ≤ C(1 + kV k

pH2,2

)t

−δ

. Then I

n

≤ T

0

independent of n and Proposition 1 holds true :

∀t ∈ [T

0

, S

n

], M

0t

(w

n

(t)) ≤ C(1 + kV k

pH2,2

)t

−δ

.

Proof of Lemma 3. This follows from a continuity argument. First, due to Theorem 3, proved in Appendix B, t 7→ M

0t

(u

n

(t)) is upper semi-continuous : since M

0t

(w

n

(S

n

)) = 0, I

n

< S

n

.

Let T

0

≥ 1 be such that C(1 + kV k

p

H2,2

)T

0−δ

≤ 1. If I

n

> T

0

, C(1 + kV k

pH2,2

)I

n−δ

< 1, our hypothesis gives that :

∀t ∈ [I

n

, S

n

], M

0t

(w

n

(t)) ≤ C(1 + kV k

pH2,2

)t

−δ

≤ C(1 + kV k

pH2,2

)I

n−δ

< 1.

By our upper semi-continuity argument (again Theorem 3), there would be a t

0

< I

n

so that for t ∈ [t

0

, S

n

], M

0t

(w

n

(t)) ≤ 1. This contradicts the minimality of I

n

, and so I

n

≤ T

0

.

The decay estimate follows immediately.

So it is enough to prove :

∀t ∈ [I

n

, S

n

], M

0t

(w

n

(t)) ≤ C(1 + kV k

pH2,2

)t

−δ

, assuming (12). Recall :

M

0t

(φ) = kφk

L2

+ kDφk

L2

+ kD

α

J

t

φk

L2

+ kDJ

t

φk

L2

.

We will now estimate each one of the 4 norms involved. Let us denote K = 1 + kV k

H2,2

. We will successively prove that for t ∈ [I

n

, S

n

], we have :

(i) kw

n

k

L2

≤ CK

p

t

−(p−3)/3

.

(ii) kw

n

k

H1

≤ CK

p

t

−(p−3)/3

.

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(iii) kD

α

I

t

w

n

k

L2

≤ CE

p

t

−δ

. (iv) kI

t

w

nx

k

L2

≤ CK

p

t

−(p−3)/3

.

(v) M

0t

(w(t)) ≤ CK

p

t

−δ

.

Remark that we first do estimates on I

t

(step (iii) and (iv)), since it easier to handle when doing energy methods estimates. In step (v) we shall go back to estimates on J

t

.

Before we do this, we have to notice that Lemma 1 applies for t ∈ [I

n

, S

n

] to give :

|w

n

(t, x)| ≤ Ct

−1/3

1 + |x|

t

1/3 14

, |w

nx

(t, x)| ≤ Ct

−2/3

1 + |x|

t

1/3 14

. Then we can add up U (t)V in these estimates to get :

|(U (t)V + w

n

(t))(x)| ≤ CKt

−1/3

1 + |x|

t

1/3 14

, (13)

|(U (t)V + w

n

(t))

x

(x)| ≤ CKt

−2/3

1 + |x|

t

1/3 1

4

, (14)

kU (t)V + w

n

(t)k

Lr

≤ CKt

−1/3−1/(3r)

, r > 4. (15) Proof of (i).

Let us multiply (9) by w

n

and integrate in x : 1

2 d dt

Z

w

2n

= −

Z

(U(t)V +w

n

)

px

w

n

= −p

Z

(U (t)V +w

n

)

x

(U (t)V +w

n

)

p−1

w.

So (after simplification by kw

n

k

L2

) : d

dt kw

n

k

L2

≤ kU (t)V + w

n

k

L2

k(U (t)V + w

n

)(U (t)V + w

n

)

x

k

L

kU (t)V + w

n

k

p−3L

. For t ∈ [I

n

, S

n

], we get :

d

dt kw

n

k

L2

≤ C(K + kw

n

(t)k

L2

) · K

2

t · K

p−3

t

(p−3)/3

≤ CK

p

t

p3

. Thus, by integration in time on [t, S

n

], using w

n

(S

n

) = 0, this gives :

kw

n

(t)k

L2

≤ C(p)K

p

t

−(p−3)/3

− S

n−(p−3)/3

≤ CK

p

t

−(p−3)/3

. (16) Proof of (ii).

Let us differentiate (9) with respect to x :

w

nxt

+ w

nxxxx

+ (U (t)V + w

n

)

pxx

= 0.

(11)

Now, multiply by w

nx

and integrate in x. After an integration by part, we get :

1 2

d dt

Z

w

n2

x

=

Z

(U (t)V + w

n

)

px

w

nxx

= p

Z

w

nxx

(U (t)V + w

n

)

x

(U (t)V + w

n

)

p−1

. The point is to rule out w

nxx

, term that we cannot control : for this we will split the second term and do integrations by parts. First :

2

Z

w

nxx

w

nx

(U (t)V + w

n

)

p−1

= −p

Z

w

n2

x

(U (t)V + w

n

)

x

(U (t)V + w

n

)

p−2

, which we can easily control. Indeed :

p

Z

w

nxx

w

nx

(U (t)V + w

n

)

p−1

≤ pkw

nx

k

2L2

k(U (t)V + w

n

)

x

(U (t)V + w

n

)k

L

kU (t)V + w

n

k

p−3L

≤ CK

p−1

kw

nx

(t)k

2L2

t

p3

≤ CK

p

kw

nx

k

L2

t

p3

. (as kw

n

k

H1

≤ M

0t

(w(t)) ≤ 1). For the second term :

Z

w

nxx

(U (t)V )

x

(U (t)V + w

n

)

p−1

= −

Z

w

nx

(U (t)V )

xx

(U(t)V + w

n

)

p−1

Z

w

nx

(U (t)V )

x

(U (t)V + w

n

)

p−2

(U (t)V + w

n

)

x

. The first integral would be troublesome (with the double derivative on U (t)V ), this is why we made the hypothesis V ∈ H

2,2

(in [4], this phe- nomenon is avoided because the integration by parts works better). The second integral is fine. So we can do the estimate (note that we use the L

decay estimate (16) on (U (t)V )

xx

= (U (t)V

x

)

x

, as V

x

∈ H

1,1

), for all t ∈ [I

n

, S

n

] :

p

Z

w

nxx

(U (t)V )

x

(U (t)V + w

n

)

p−1

≤ Ckw

nx

k

L2

k(U (t)V + w

n

)(U (t)V

x

)

x

k

L

kU (t)V + w

n

k

L2

kU (t)V + w

n

k

p−3L

+Ckw

nx

k

L2

k(U (t)V )

x

k

L2

k(U (t)V + w

n

)(U (t)V +w

n

)

x

k

L

kU (t)V +w

n

k

p−3L

≤ CK

p

kw

nx

k

L2

t

−p/3

. (we used estimate (16), and kw

n

k

H1

≤ 1). So finally, after simplifying by kw

nx

k

L2

:

d

dt kw

nx

(t)k

L2

≤ CK

p

t

−p/3

.

(12)

After integration between t and S

n

(w

n

(S

n

) = 0) : kw

nx

(t)k

L2

≤ CK

p

t

−(p−3)/3

− S

n−(p−3)/3

.

And so, we neglect the S

n

term, and when adding up with (16), we obtain : kw

n

(t)k

H1

≤ C(p)K

p

t

−(p−3)/3

. (17) Proof of (iii).

We now turn to the estimates on I

t

w

n

. Let us denote L = ∂

t

+ ∂

xxx

the linear KdV operator, and remind some commutation relations of the different operators involved . Remind the definition of the dilation operator I

t

φ = xφ + 3t

Rx

−∞

φ

t

dx and of J

t

φ = xφ − 3t∂

xx

φ. Then : I

t

φ − J

t

φ = 3t

Z x

−∞

Lφdx.

We have the following commutation relations : [L, J

t

] = 0, [L, I

t

]φ = 3

Z x

−∞

Lφdx, [J

t

, ∂

x

] = [I

t

, ∂

x

] = −Id.

Again notice that I

t

U (t)V − J

t

U(t)V = 3t

Rx

−∞

LU (t)V dx = 0 so that kD

α

I

t

U (t)V k

L2

+ kDI

t

U (t)V k

L2

≤ CkV k

H1,1

.

Let f :

R

R

be a C

1

function and φ be such that f

0

(φ)I

t

φ

x

has a sense. Then :

I

t

(f (φ)

x

) = xf (φ)

x

+ 3tf(φ)

t

= xf

0

(φ)φ

x

+ 3tf

0

(φ)φ

t

= f

0

(φ)I

t

φ

x

. We will use this formula for f (x) = x

p

and φ = U (t)V + w

n

(t).

We compute : LI

t

w

n

= I

t

Lw

n

+ 3

Z x

−∞

Lw

n

= −I

t

(U (t)V + w

n

)

px

− 3(U (t)V + w

n

)

p

= −p(U (t)V + w

n

)

p−1

I

t

(U (t)V + w

n

)

x

− 3(U (t)V + w

n

)

p

= −p(U (t)V + w

n

)

p−1

(I

t

(U (t)V + w

n

))

x

− (3 − p)(U (t)V + w

n

)

p

. Apply the operator D

α

, multiply by D

α

I

t

w

n

and integrate in space (remind that [D

α

, L] = 0 and (Lφ, φ) =

12dtd R

φ

2

) : d

dt kD

α

I

t

w

n

k

2L2

≤ kD

α

I

t

w

n

k

L2

kD

α

(U (t)V + w

n

)

p−1

(I

t

(U (t)V + w

n

))

x

k

L2

+ CkD

α

I

t

w

n

k

L2

kD

α

(U (t)V + w

n

)

p

k

L2

. (18)

(13)

We now apply Lemma 2 with k = U (t)V + w

n

and h = I

t

(U (t)V + w

n

). So that (along with the linear estimates (13), (14), (15)) :

kkk

x

k

L

≤ CK

2

t

−1

, kkk

L2

≤ CK, kkk

L

≤ CKt

13

, kk

p−1

k

L2

= kkk

p−1

L2(p−1)

≤ CK

p−1

t

(p−1)(−13+6(p−1)1 )

≤ CK

p−1

t

−(2p−3)/6

, and :

kD

α

hk

L2

≤ CK, kh

x

k

L2

≤ CK.

We can compute :

kD

α

(U (t)V + w

n

)

p−1

(I

t

(U (t)V + w

n

))

x

k

L2

≤ CK

p

(t

−(p−3)/3+1

+ t

−(p−3−2γ)/3+1

+ t

−(p−3+2γ)/3−1+γ

)

≤ CK

p

t

−(p−2γ)/3

.

(as t ≥ I

n

≥ 1; the point being (p − 2γ )/3 > 1). And for the second term : kD

α

(U (t)V + w

n

)

p

k

L2

≤ CK

p

(t

−(2p−3)/6−1/2

+ t

−(2p−3)/6−γ−(1−3γ)/2

)

≤ CK

p

t

−p/3+γ/2

.

Finally, after simplification by kD

α

I

t

w

n

k

L2

, (18) gives : d

dt kD

α

I

t

w

n

k

L2

≤ CK

p

t

−(p−2γ)/3

. And as before, we integrate on [t, S

n

] :

kD

α

I

t

w

n

(t)k

L2

≤ CK

p

t

−(p−3−2γ)/3

. (19)

Proof of (iv).

Again, we compute :

LI

t

(w

nx

) = I

t

Lw

nx

+ 3Lw

n

= −I

t

(U (t)V + w

n

)

pxx

− 3(U (t)V + w

n

)

px

= −(I

t

(U (t)V + w

n

)

px

)

x

− 2(U (t)V + w

n

)

px

= −p(U (t)V + w

n

)

p−1

(I

t

(U (t)V + w

n

)

x

)

x

−p(p − 1)(U(t)V + w

n

)

p−2

(U (t)V + w

n

)

x

I

t

(U (t)V + w

n

)

x

−2p(U (t)V + w

n

)

p−1

(U (t)V + w

n

)

x

. (20) We want to multiply by I

t

w

nx

and integrate in space. There is essentially one troublesome term, the double derivative one. We split it :

(U (t)V + w

n

)

p−1

(I

t

(U (t)V + w

n

)

x

)

x

= (U (t)V + w

n

)

p−1

(I

t

U (t)V

x

)

x

+ (I

t

w

nx

)

x

). (21)

(14)

The second term will have only first order terms after integration by parts, which is fine :

Z

(U (t)V + w

n

)

p−1

(I

t

w

nx

)

x

I

t

w

nx

= (p − 1) 2

Z

(U (t)V + w

n

)

p−2

(U (t)V + w

n

)

x

(I

t

w

nx

)

2

≤ Ck(U (t)V + w

n

)(U (t)V + w

n

)

x

k

L

kU (t)V + w

n

k

p−3L

kI

t

w

nx

k

2L2

≤ CK

p−1

t

−p/3

kI

t

w

nx

k

2L2

.

However the first term on the right hand side of (21) requires extra regularity on V . Indeed

(I

t

U (t)V

x

)

x

= (J

t

U (t)V

x

)

x

= (U (t)xV

x

)

x

,

and (xV

x

∈ H

1,1

), M

0t

(U (t)(xV

x

)) ≤ kV k

H2,2

so that by Lemma 1 :

|(I

t

U (t)V

x

)

x

(x)| ≤ CkV k

H2,2

t

2/3

1 +

x t

1/3

1

4

.

We can now estimate (essentially in the same way that for the H

1

estimate) :

Z

(U (t)V + w

n

)

p−1

(I

t

U (t)V

x

)

x

I

t

w

nx

≤ kU (t)V + w

n

k

p−3L

kU (t)V + w

n

k

L2

k(U (t)V + w

n

)(I

t

U (t)V

x

)

x

k

L

kI

t

w

nx

k

L2

≤ CK

p−1

t

p/3

kI

t

w

nx

k

L2

.

The remaining two terms in (20) are simpler and can be treated directly (after multiplication by I

t

w

nx

and integration in x) :

Z

(U (t)V + w

n

)

p−2

(U (t)V + w)

x

I

t

(U (t)V + w

n

)

x

I

t

w

nx

≤ kU (t)V + w

n

k

p−3L

k(U (t)V + w

n

)(U (t)V + w

n

)

x

k

L

kI

t

(U (t)V + w

n

)

x

k

L2

kIw

nx

k

L2

≤ CK

p

t

p/3

kI

t

w

nx

k

L2

,

Z

(U (t)V + w

n

)

p−2

(U (t)V + w

n

)

x

I

t

w

nx

≤ kU (t)V + w

n

k

p−3L

k(U (t)V + w

n

)(U (t)V + w

n

)

x

k

L

kU (t)V + w

n

k

L2

kI

t

w

nx

k

L2

≤ CK

p

t

p/3

kI

t

w

nx

k

L2

. Hence, (after simplifying by kI

t

w

nx

k

L2

) we get :

d

dt kI

t

w

nx

k

L2

≤ CK

p−1

(K + kI

t

w

nx

k

L2

)t

p/3

≤ CK

p

t

−p/3

.

(15)

And Gronwall’s Lemma (between t and S

n

) gives :

kI

t

w

nx

(t)k

L2

≤ CK

p

t

−(p−3)/3

. (22) Proof of (v).

We now have to go back to J

t

w

n

(t) = I

t

w

n

(t) − 3t

Z x

−∞

Lw

n

(t)dx

0

= I

t

w

n

(t) + 3t(U (t)V + w

n

(t))

p

. Using the commutation relations, we get :

kD

α

J

t

w

n

k

L2

+ kDJ

t

w

n

k

L2

≤ kD

α

I

t

w

n

k

L2

+ 3tkD

α

(U (t)V + w

n

(t))

p

k

L2

+ kI

t

w

nx

k

L2

+ kw

n

k

L2

+ 3ptk(U (t)V + w

n

(t))

p−1

(U (t)V + w

n

(t))

x

k

L2

. Now, using again Lemma 2 with k = U (t)V + w

n

(t) :

kD

α

(U (t)V + w

n

(t))

p

k ≤ Ckk

p−1

k

L2

(kkk

x

k

1/2L

+ kkk

L

kkk

x

k

(1−3γ)/2L

)

≤ CK

p

t

−p/3+γ/2

. And as usual :

k(U (t)V + w

n

)

p−1

(U (t)V + w

n

)

x

k

L2

≤ kU (t)V + w

n

k

p−3L

k(U (t)V + w

n

)(U (t)V + w

n

)

x

k

L

kU (t)V + w

n

k

L2

≤ CK

p

t

p/3

. So using these last two inequalities, along with (16), (19) and (22), we get :

kD

α

J

t

w

n

k

L2

+ kDJ

t

w

n

k

L2

≤ CK

p

t

(p−3+2γ)/3

. (23) Recall δ = (p − 3 − 2γ)/3 > 0. So adding up (17) and (23) gives :

∀t ∈ [I

n

, S

n

], M

0t

(w

n

(t)) ≤ CK

p

t

−δ

. 3.3 Construction and uniqueness of u.

Proof of Theorem 1. Existence of u.

Proposition 1 provides us with a sequence w

n

(t) solution to (9) satisfying uniform estimates in n :

∀t ∈ [T

0

, S

n

], M

0t

(w

n

(t)) ≤ CK

p

t

−δ

. In particular, for t ∈ [T

0

, S

n

], estimates (13) and (14) are valid.

Let us prove that for all k ∈

N, (wn

)

n≥k

is a convergent sequence in

C

0

([T

0

, S

k

], L

2

(

R

)). For this we show that (w

n

)

n≥k

is a Cauchy sequence

in C

0

([T

0

, S

k

], L

2

(

R

)) : let us consider v

n,m

= w

n

− w

m

in L

2

. Without

(16)

loss of generality, we can suppose that m > n ≥ k. First : kv

n,m

(S

n

)k

L2

≤ CK

p

S

n−(p−3)/3

(see (16)).

v

n,m

satisfies (we denote v = v

n,m

for simplicity in the computations) : v

t

+ v

xxx

+ (U (t)V + w

n

)

px

− (U (t)V + w

m

)

px

= 0. (24) We multiply by v and integrate in x :

1 2

d dt

Z

v

2

=

Z

v((U (t)V + w

n

)

px

− (U (t)V + w

m

)

px

).

Now, for any functions φ and ψ :

φ

px

− ψ

px

= pφ

p−1

φ

x

− pψ

p−1

ψ

x

= pφ

p−1

(φ − ψ)

x

+ p(φ

p−1

− ψ

p−1

x

, and :

p−1

− ψ

p−1

| ≤ C|ψ − φ|(|φ|

p−2

+ |ψ|

p−2

).

So that with φ = U (t)V + w

n

, ψ = U (t)V + w

m

(after integration by parts on the first term) :

1 2

d dt

Z

v

2

= − p − 1 2

Z

v

2

(U (t)V + w

n

)

p−2

(U (t)V + w

n

)

x

+

Z

v(U(t)V + w

n

)

p−1

− (U (t)V + w

m

)

p−1

)(U(t)V + w

m

)

x

. Treating each term separately (for t ∈ [T

1

, S

n

]) :

Z

v

2

(U (t)V + w

n

)

p−2

(U (t)V + w

n

)

x

≤ CK

p−1

t

−p/3

kvk

2L2

,

Z

v(U (t)V + w

n

)

p−1

− (U (t)V + w

m

)

p−1

)(U (t)V + w

m

)

x

≤ C

Z

v

2

(|U (t)V + w

n

|

p−2

+ |U (t)V + w

m

|

p−2

|(U (t)V + w

m

)

x

|

≤ CK

p−1

t

−p/3

kvk

2L2

. So that :

d

dt kvk

2L2

≤ CK

p−1

t

−p/3

kvk

2L2

. Now, using Gronwall’s Lemma on [t, S

n

] :

kv

n,m

(t)k

L2

≤ Ckv

n,m

(S

n

)k

L2

≤ CS

n−(p−3)/3

.

This proves that (w

n

)

n≥k

is a Cauchy sequence in the space C([T

0

, S

k

], L

2

)

and so converges to a certain w(t, x). Since this can be done for arbitrarily

large n (and S

n

→ ∞), w ∈ C([T

0

, ∞), L

2

) is the only possible weak limit

of (w

n

)

n∈N

.

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