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Submitted on 6 Oct 2006

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Global well-posedness for the KP-I equation on the background of a non localized solution

Luc Molinet, Jean-Claude Saut, Nikolay Tzvetkov

To cite this version:

Luc Molinet, Jean-Claude Saut, Nikolay Tzvetkov. Global well-posedness for the KP-I equation on the

background of a non localized solution. Communications in Mathematical Physics, Springer Verlag,

2007, 272 (3), pp.775-810. �hal-00104398�

(2)

ccsd-00104398, version 1 - 6 Oct 2006

THE BACKGROUND OF A NON LOCALIZED SOLUTION

L. MOLINET, J. C. SAUT, AND N. TZVETKOV

Abstract. We prove that the Cauchy problem for the KP-I equation is globally well-posed for initial data which are localized perturbations (of arbitrary size) of a non-localized (i.e. not decaying in all directions) traveling wave solution (e.g.

the KdV line solitary wave or the Zaitsev solitary waves which are localized in x and y periodic or conversely).

1. Introduction

We study here the initial value problem for the Kadomtsev-Petviashvili (KP-I) equation

(1) (u t + u xxx + uu x ) x − u yy = 0, where u = u(t, x, y), (x, y) ∈ R 2 , t ∈ R , with initial data (2) u(0, x, y) = φ(x, y) + ψ c (x, y),

where ψ c is the profile 1 of a non-localized (i.e. not decaying in all spatial directions) traveling wave of the KP-I equation moving with speed c 6 = 0. This ψ c could be for instance the line soliton of the Korteweg- de Vries (KdV) equation

(3) ψ c (x, y) = 3c cosh −2

c x 2

In (3) the KdV soliton is of course considered as a two dimensional (constant in y) object. Another possibility to see (3) as a solution of KP-I is to consider (1) posed on R × T . Global solutions of (1) for data on R × T , including data close to (3) were recently constructed in a work by Ionescu-Kenig [12]. In (2), the function ψ c may also be the profile of the Zaitsev [36] traveling waves (see also [30]) which is localized in x and periodic in y :

(4) ψ c (x, y) = 12α 2 1 − β cosh(αx) cos(δy) (cosh(αx) − β cos(δy)) 2 ,

1 This means that ψ(x − ct, y) solves (1).

1

(3)

where

(α, β) ∈ ]0, ∞ [ × ] − 1, 1[, and the propagation speed is given by

c = α 2 4 − β 2 1 − β 2 .

Let us observe that the transform α → iα, δ → iδ, c → ic produces solutions of (1) which are periodic in x and localized in y. The profiles of these solutions are also admissible in (2), under the assumption | β | > 1. Notice that for β = 0, (4) coincides with (3).

The global well-posedness of (1)-(2) with data given by (3) which will be proved in this paper can be viewed as a preliminary step towards the rigorous mathemat- ical justification of the (conjectured) nonlinear instability of the KdV soliton with respect to transversal perturbations governed by the KP-I flow. This question is, as far as we know, still an open problem (see however [1] for a linear analysis of the instability and [10] for a linear instability analysis in the framework of the full Euler system). The instability scenario of the line soliton seems to be a symmetry breaking phenomenon : the line soliton should evolve towards the Zaitsev solitary wave (4). Note that Haragus and Pego [11] have shown that this solution is the only one close to the line soliton which is periodic in y and decays to zero as x → ∞ .

The question of solving (1) together with the initial data (2) when ψ is the profile of the KdV line soliton, has been recently addressed by Fokas and Pogrobkov [8], by the inverse scattering transform (IST) techniques. However, the Cauchy problem is not rigorously solved in [8] and it is unlikely that it could be solved for an arbitrary large data φ using IST since the Cauchy problem with purely localized data has been solved by IST techniques only for small initial data (see [35, 38]).

On the other hand, PDE techniques have been recently fruitfully used to obtain

the global solvability of the KP equation with arbitrary large initial data, starting

with the pioneering paper of Bourgain [4] on the KP-II equation (that is (1) with

+u yy instead of − u yy ). In [4], the global well-posedness of the KP-II equation for

data in H s ( R 2 ), s ≥ 0 is established. The result in [4] is obtained by performing the

Picard iteration scheme to an equivalent integral equation in the Fourier transform

restriction spaces of Bourgain. The situation for the KP-I equation turned out to be

more delicate. We showed in [27] that the Picard iteration scheme can not be applied

in the context of the KP-I equation as far as one considers initial data in Sobolev

spaces. Sobolev spaces are natural, since the conservation laws for the KP-I equation

(4)

control Sobolev type norms. In [22], a quite flexible method is introduced that allows to incorporate the dispersive effects in a context of a compactness method for proving the well-posedness. The work [22] was in turn inspired by the considerations in [5]

in the context of the Nonlinear Schr¨ odinger equation on a compact manifold which is solved in [5] as a semi-linear problem (i.e. by the Picard iteration scheme). The main point in [22] is to realize that the idea of [5] can also be used in the context of a quasi-linear problem. The method of [22] turned out to be useful in the context of the KP-I equation (and some other models such as the Schr¨ odinger maps [15, 19]) and the first global well-posedness result for the KP-I equation has been obtained by the authors of the present paper in [26]. This result has been improved (i.e. the space of the allowed initial data is larger) by Kenig [17]. Together with the idea of [22], a new commutator estimate for the KP-I equation is used in [17]. The main point in Kenig’s result is the proof that KP-I is locally well-posed for data in the space

{ u ∈ L 2 ( R 2 ) : ∂ x −1 u y ∈ L 2 ( R 2 ), | D x | s u ∈ L 2 ( R 2 ), s > 3/2 }

All papers [4], [26] and [17] consider the KP equations in spaces of “localized” (zero at infinity) functions.

The main goal of this paper is to prove that, for a large class of ψ c , the Cauchy problem (1), (2) is globally well-posed for data φ ∈ Z, where

Z := { u ∈ L 2 ( R 2 ) : ∂ x −2 u yy ∈ L 2 ( R 2 ), u xx ∈ L 2 ( R 2 ) } .

(notice that u ∈ Z implies u y ∈ L 2 ( R 2 ) and ∂ x −1 u y ∈ L 2 ( R 2 )). More precisely, we have the following result.

Theorem 1. Let ψ c (x − ct, y) be a solution of the KP-I equation such that ψ c : R 2 −→ R

is bounded with all its derivatives 2 . Then for every φ ∈ Z there exists a unique global solution u of (1) with initial data (2) satisfying for all T > 0,

[u(t, x, y) − ψ c (x − ct, y)] ∈ C([0, T ]; Z ), ∂ x [u(t, x, y) − ψ c (x − ct, y)] ∈ L 1 T L xy . Furthermore, for all T > 0, the map φ 7→ u is continuous from Z to C([0, T ]; Z)).

Since the KP-I equation is time reversible, a similar statement to Theorem 1 holds for negative times as well. Let us notice that the assumptions on ψ c in Theorem 1 are clearly satisfied by the line or the Zaitsev solitary wave. In the proof of Theorem 1, we write the solution u of (1), (2) as

u(t, x, y) = ψ c (x − ct, y) + v(t, x, y)

2 The bounds can of course depend on the propagation speed c.

(5)

where v is localized. This v satisfies the equation

(5) (v t + v xxx + vv x + ∂ xc v)) x − v yy = 0, v(0, x, y) = φ(x, y).

Our strategy is then to adapt the proof of [26, 17]. Starting from the local well- posedness result, we implement a compactness method based on “almost conserva- tion laws”. New terms occur with respect to [26] but they are controlled since ψ and its derivatives are bounded. It is of importance for our analysis that equation (5) does not contain a source term.

We refer to the work by Gallo [9] and the references therein, where non vanishing at infinity solutions to one dimensional dispersive models are constructed.

Let us notice that the framework considered in Theorem 1 is also a convenient one for a rigorous study of the interaction of a line and lump solitary waves (see e.g. [6]).

The rest of this paper is organized as follows. In the next section, using a compact- ness method, we prove a basic well-posedness result for (5). In Section 3, inspired by the formal KP-I conservation laws, we provide bounds for some Sobolev type norms of the local solutions. These bounds are however not sufficient to get global solu- tions. For that reason, in Section 4 we prove a Strichartz type bound. This bound is then used in Section 5 to get a first global well-posedness result. In Sections 6 and 7 we extend the well-posedness to the class Z introduced above. The last section is devoted to the “usual” KP-I equation. We show how the estimates of Section 4 can be used to give a slight improvement of the Kenig well-posedness result [17].

2. Local well-posedness for data in H −1 s ( R 2 ), s > 2

In this section, we prove a basic local well-posedness result for (5). The proof fol- lows a standard compactness method. We however need to work in Sobolev spaces of integer indexes, in order to make work the commutator estimates related to the term ∂ xc v).

By H s ( R 2 ), s ∈ R , we denote the classical Sobolev spaces. The local existence result for (6) will be obtained in the spaces H −1 s ( R 2 ) equipped with the norm

k u k H

s1

(R

2

) = k (1 + | ξ | −1 ) h| ξ | + | η |i s u(ξ, η) b k L

2ξη

,

where h·i = (1+ |·| 2 )

12

and u b denotes the Fourier transform of u. The spaces H −1 s ( R 2 )

are adapted to the specific structure of the KP type equations. In the estimates it

(6)

will appear W r,∞ -norms of ψ c defined for any integer r ≥ 0 by k ψ c k W

r,∞

= X

0≤|α|=|(α

1

2

)|≤r

k ∂ x α

1

y α

2

ψ c k L

xy

.

Consider the “integrated” equation (5)

(6) u t + u xxx − ∂ x −1 u yy + uu x + ∂ x (ψu) = 0, with initial data

(7) u(0, x, y) = φ(x, y).

For the solutions we study in this section, (5) may be substituted by (6). For conciseness we skip the c of ψ c in (6) and we suppose that c = 1 in the sequel. Of course, the case c 6 = 1 can be treated in exactly the same manner. We have the following local well-posedness result for (6).

Proposition 1. Let s > 2 be an integer. Then for every φ ∈ H −1 s ( R 2 ) there exists T & (1+ k φ k H

s

) −1 and a unique solution u to (6) on the time interval [0, T ] satisfying

u ∈ C([0, T ]; H s ( R 2 )) , u ∈ L ([0, T ]; H −1 s ( R 2 )) . In addition, for t ∈ [0, T ],

(8) k u(t, · ) k H

s

(R

2

) ≤ C k φ k H

s

(R

2

) exp

c k∇ x,y u k L

1T

L

xy

+ cT k ψ k W

s,∞

.

Moreover if φ ∈ H −1 σ ( R 2 ) where σ > s is an integer then u ∈ C([0, T ]; H −1 σ ( R 2 )).

Finally, the map φ 7→ u φ is continuous from H s ( R 2 ) to C([0, T ]; H s ( R 2 )).

Proof of Proposition 1. The process is very classical (see [13] for a closely related result). For ε > 0 we look at the regularized equation

(9) u ε t + ε∆ 2 u ε t = − u ε xxx + ∂ x −1 u ε yy − u ε u ε x − ∂ x (ψu ε )

where ∆ = ∂ x 2 + ∂ y 2 is the Laplace operator. Equation (9) with initial condition φ ε = (1 − √

ε∆) −1 φ can be rewritten under the form

(10) u ε (t) = L ε (t)φ ε − Z t

0

L ε (t − τ )(1 + ε∆ 2 ) −1 u ε (τ )u ε x (τ ) + ∂ x (ψ(τ )u ε (τ ) dτ , where

L ε (t) := exp

− t(1 + ε∆ 2 ) −1 (∂ x 3 − ∂ x −1y 2 )

.

(7)

Notice that, thanks to the regularization effect of (1 + ε∆ 2 ) −1 , for ε 6 = 0, the map u −→ (1 + ε∆ 2 ) −1 u u x + ∂ x (ψu

is locally Lipschitz from H s ( R 2 ) to H s ( R 2 ), provided s > 2. The operator L ε (t) is clearly bounded on H s ( R 2 ) and therefore by the Cauchy-Lipschitz-Picard theorem there is a unique local solution

(11) u ε ∈ C([0, T ]; H s ( R 2 )) of (10) with

T & (1 + k φ ε k H

s

) −1 ≥ (1 + k φ k H

s

) −1 .

Thanks to the perfect derivative structure of the integral term in (10), we also obtain that

u ε ∈ C([0, T ]; H −1 s ( R 2 )).

Notice also that, thanks to the assumption s > 2, (1 − √

ε∆)u ε belongs to H s ( R 2 ).

We next study the convergence of u ε as ε → 0. For that purpose, we establish a priori bounds, independent 3 of ε on k u ε (t, · ) k H

s

on time intervals of size of order (1 + k φ k H

s

) −1 .

Multiplying (9) with u ε , after an integration by parts, we obtain that

(12) d

dt h

k u ε (t, · ) k 2 L

2

+ ε k ∆u ε (t, · ) k 2 L

2

i

. k ψ x k L

k u ε (t, · ) k 2 L

2

. Let us recall a classical commutator estimate (see e.g. [16]).

Lemma 1. Let ∆ be the Laplace operator on R n , n ≥ 1. Denote by J s the operator (1 − ∆) s/2 . Then for every s > 0,

k [J s , f ]g k L

2

(R

n

) . k∇ f k L

(R

n

) k J s−1 g k L

2

(R

n

) + k J s f k L

2

(R

n

) k g k L

(R

n

) . Using Lemma 1, we obtain for fixed y,

k [∂ x s , u(x, · )]u x (x, · ) k L

2x

. k u x (x, · ) k L

x

k J x s u(x, · ) k L

2x

where J x s = (1 − ∂ x 2 ) s/2 . Squaring, integrating over y and integrating by parts, it yields

(13)

Z

R

2

x s (uu x )∂ x s u . k u x k L

xy

k J x s u k 2 L

2xy

.

3 Notice that the bounds on ku

ε

(t, ·)k

Hs

resulting from the Cauchy-Lipschitz theorem applied to

(9) are unfortunately very poor (depending on ε).

(8)

On the other hand, by Leibniz rule and integration by parts (recall that s is an integer), we get

(14)

Z

R

2

x s (ψu) xx s u . k ψ x k W

s−1,∞

k J x s u k 2 L

2xy

. Therefore applying ∂ s x to (9), multiplying it with ∂ x s u ε gives (15) d

dt h

k ∂ s x u ε (t, · ) k 2 L

2

+ ε k ∂ x s ∆u ε (t, · ) k 2 L

2

i . .

k u ε x (t, · ) k L

+ k ψ x k W

s1,

k J x s u ε (t, · ) k 2 L

2

(R

2

) Next we estimate the y derivatives. In the same way, using Lemma 1, we obtain that for a fixed x

(16) k [∂ y s , u(x, · )]u x (x, · ) k L

2y

. .

k u x (x, · ) k L

y

+ k u y (x, · ) k L

y

k J y s−1 u x (x, · ) k L

2y

+ k J y s u(x, · ) k L

2y

, where J y s = (1 − ∂ y 2 ) s/2 . Squaring (16), integration over x and an integration by parts yield

Z

R

2

y s (uu x )∂ y s u . k∇ x,y u k L

xy

k u k 2 H

s

(R

2

) . On the other hand, by Leibniz rule and integration by parts,

Z

R

2

y s (ψu) xy s u . k ψ k W

s,

k u k 2 H

s

(R

2

) . Therefore applying ∂ s y to (9) and multiplying it by ∂ y s u ε gives (17) d

dt h

k ∂ s y u ε (t, · ) k 2 L

2

+ ε k ∂ y s ∆u ε (t, · ) k 2 L

2

i . .

k∇ x,y u ε (t, · ) k L

+ k ψ k W

s,

k u ε (t, · ) k 2 H

s

(R

2

) . Since for s integer

k u k H

s

≈ k u k L

2

+ k ∂ x s u k L

2

+ k ∂ y s u k L

2

, combining (12), (15) and (17) gives

d dt h

k u ε (t, · ) k 2 H

s

(R

2

) + ε k ∆u ε (t, · ) k 2 H

s

(R

2

)

i . .

k∇ x,y u ε (t, · ) k L

+ k ψ k W

s,∞

k u ε (t, · ) k 2 H

s

(R

2

) and therefore using that

k φ ε k 2 H

s

(R

2

) + ε k ∆φ ε k 2 H

s

(R

2

) ≤ k φ k 2 H

s

(R

2

)

(9)

by the Gronwall lemma for every T > 0 on the time of existence of u ε , (18) k u ε k L

T

H

s

(R

2

) ≤ k φ k H

s

(R

2

) exp

c k∇ x,y u ε k L

1T

L

xy

+ cT k ψ k W

s,

. which is the key inequality. Since s > 2, using the Sobolev embedding, we get (19)

Z T

0 k∇ x,y u ε (τ, · ) k L

dτ ≤ CT k u ε (t, · ) k L

T

H

s

(R

2

) .

Using (18), (19) and the continuity of u ε (t) with respect to time (see (11)), we obtain that there exists C > 0 such that if

T . (1 + k φ k H

s

(R

2

) ) −1 then

(20)

Z T

0 k∇ x,y u ε (τ, · ) k L

(R

2

) dτ ≤ C and

(21) k u ε k L

T

H

s

(R

2

) ≤ C k φ k H

s

(R

2

) .

We next estimate the anti-derivatives of u ε . Let v ε := ∂ x −1 u ε . Then, using (10), we obtain that v ε solves the equation

v ε (t) = L ε (t)∂ x −1 φ ε − Z t

0

L ε (t − τ )(1 + ε∆ 2 ) −1 1

2 (u ε (τ )) 2 + ψ(τ )u ε (τ ) dτ . Therefore, since s > 2, using the Leibniz rule and the Sobolev inequality, we get the bound

(22) k v ε k L

T

H

s

(R

2

) ≤ k ∂ x −1 φ k H

s

(R

2

) + CT k u ε k L

T

H

s

(R

2

)

k u ε k L

T

H

s

(R

2

) + k ψ k W

s,

. Coming back to the equation (9), we infer from (22) that the sequence (∂ t (u ε )) is bounded in a weaker norm, say in L T H s−2005 ( R 2 ). Therefore from the Aubin- Lions compactness theorem (see e.g. [23]), we obtain that u ε converges, up to a subsequence, to some limit u in the space L 2 loc ((0, T ) × R 2 ) which satisfy (8) and (23)

Z T

0 k∇ x,y u(τ, · ) k L

(R

2

) dτ ≤ C .

Thanks to (22), we obtain that (up to a subsequence) ∂ x −1 u ε converges in D ((0, T ) × R 2 ) to a limit which can be identified as ∂ x −1 u. By writing the nonlinearity uu x as

1

2 ∂ x (u 2 ), passing into a limit in the equation (9) as ε → 0, we obtain that the function u satisfy the equation (6) in the distributional sense. Moreover thanks to (21) and (22), we obtain that

u ∈ L ([0, T ]; H −1 s ( R 2 )) .

(10)

In addition, thanks to the Lebesgue theorem, φ ε converges in H −1 s ( R 2 ) to φ as ε → 0 and thus u satisfies the initial condition (7). Next, if φ ∈ H −1 σ ( R 2 ) with σ ≥ s, σ ∈ N , then as above, we get the estimate (8) with σ instead of s which, in view of (23), yields the propagation of the H σ ( R 2 ) regularity. We next estimate the anti-derivatives of u in H σ by invoking (22) (with σ instead of s) in the limit ε → 0. The uniqueness is straightforward from the Gronwall lemma and (23). The continuity of the flow map and the fact that the solution is a continuous curve in H s ( R 2 ) can be obtained by the Bona-Smith approximation argument [3]. We do not give the details of this construction in this section since a completely analogous

discussion will be performed later in this paper.

Let us next state a corollary of Proposition 1.

Proposition 2. Let s > 2 be an integer. Then for every φ ∈ H −1 s ( R 2 ) the local solution constructed in Proposition 1 can be extended to a maximal existence interval [0, T [ such that either T = ∞ or

t→T lim

k∇ x,y u k L

1t

L

xy

= ∞ .

Proof. It suffices to iterate the result of Proposition 1 by invoking (8) at each itera-

tion step.

It results from Proposition 2 that the key quantity for the global existence in H −1 s ( R 2 ), s > 2 is k∇ x,y u(t, · ) k L

.

3. A priori estimates using conservation laws

In this section we control the growth of some quantities directly related to the conservation laws of KP I. Recall that the solutions obtained in Proposition 1 satisfy

u t + u xxx + uu x + ∂ x (ψu) − ∂ x −1 u yy = 0.

(24)

In [37], it is shown that the KP-I equation has a Lax pair representation. This in turn provides an algebraic procedure generating an infinite sequence of conservation laws. More precisely, if u is a formal solution of the KP-I equation then

d dt

h Z χ n i

= 0, where χ 1 = u, χ 2 = u + i∂ x −1y u and for n ≥ 3,

χ n = n−2 X

k=1

χ k χ n+1−k

+ ∂ x χ n−1 + i∂ x −1y χ n−1 .

For n = 3, we find the conservation of the L 2 norm, n = 5 corresponds to the energy

functional giving the Hamiltonian structure of the KP-I equation. As we noticed

(11)

in [26], there is a serious analytical obstruction to give sense of χ 9 as far as R 2 is considered as a spatial domain.

Inspired by the above discussion, we define the following functionals M(u) =

Z

R

2

u 2 , E(u) = 1 2

Z

R

2

u 2 x + 1 2

Z

R

2

(∂ x −1 u y ) 2 − 1 6

Z

R

2

u 3 and

F ψ (u) = 3 2

Z

R

2

u 2 xx + 5 Z

R

2

u 2 y + 5 6

Z

R

2

(∂ x −2 u yy ) 2 − 5 6

Z

R

2

u 2x −2 u yy

− 5 6

Z

R

2

u (∂ x −1 u y ) 2 + 5 4

Z

R

2

u 2 u xx + 5 24

Z

R

2

u 4

− 5 3

Z

R

2

ψ u ∂ x −2 u yy − 5 6

Z

R

2

ψ (∂ x −1 u y ) 2 .

Recall that the functionals M and E corresponds to the momentum and energy conservations respectively while the functional F ψ is motivated by the higher order conservation laws for the KP-I equation associated to χ 7 . Let us notice however that the functional F ψ ( · ) contains two supplementary terms involving ψ with respect to the corresponding conservation law of the KP-I equation.

We denote by H −1 ( R 2 ) the intersection of all H −1 s ( R 2 ). The next proposition gives bounds on the quantities M , E and F ψ for data in spaces where the local well-posedness of the previous section holds.

Proposition 3. For every R > 0 there exists C > 0 such that if u ∈ L ([0, T ]; H −1 ( R 2 )) is a solution to (6) corresponding to an initial data φ ∈ Z ∩ H −1 ( R 2 ), k φ k Z ≤ R then E(u(t)) and F ψ (u(t)) are well-defined and ∀ t ∈ [0, T ],

M (u(t)) ≤ C exp(C t)M(φ) (25)

| E(u(t)) | ≤ C exp(C t) | E(φ) | + g 1 (t) (26)

| F ψ (u(t)) | ≤ C exp(C t) | F (φ) | + g 2 (t) (27)

where g 1 , g 2 : R + → R + are continuous bijections depending only on R.

Proof of Proposition 3. Before entering into the proof of Proposition 3, we state an anisotropic Sobolev inequality which will be used in the proof.

Lemma 2. For 2 ≤ p ≤ 6 there exists C > 0 such that for every u ∈ H −1 ( R 2 ), (28) k u k L

p

(R

2

) ≤ C k u k

6−p 2p

L

2

(R

2

) k u x k

p−2 p

L

2

(R

2

) k ∂ x −1 u y k

p−2 2p

L

2

(R

2

) .

(12)

Proof. We refer to [2] for a proof of (28) and for a systematic study of anisotropic Sobolev embeddings. For a sake of completeness, here we reproduce the proof of (28) given in [33]. Inequality (28) clearly holds for p = 2. By convexity, it suffices thus to prove it for p = 6. Following the Gagliardo-Nirenberg proof of the Sobolev embedding, we write

u 2 (x, y) = 2 Z x

−∞

u x (z, y)u(z, y)dz

and therefore using the Cauchy-Schwarz inequality in the z integration, we obtain that for a fixed y,

h sup

x∈R | u(x, y) | i 4

≤ 4 Z

−∞

u 2 x (z, y)dz Z

−∞

u 2 (z, y)dz . Therefore by writing u 6 = u 4 u 2 , we get

(29) Z

R

2

u 6 (x, y)dxdy ≤ 4 Z

−∞

Z

−∞

u 2 x (z, y)dz Z

−∞

u 2 (z, y)dz 2

dy . Next, using Fubini theorem and an integration by parts, we obtain

Z ∞

−∞

u 2 (z, y)dz = 2 Z ∞

−∞

Z y

−∞

u(z, w)u y (z, w)dwdz

= 2 Z y

−∞

Z ∞

−∞

u(z, w)u y (z, w)dzdw

= − 2 Z y

−∞

Z ∞

−∞

u x (z, w) ∂ −1 x u y (z, w)dzdw . An application of the Cauchy-Schwarz inequality now gives

sup

y∈R

Z

−∞

u 2 (z, y)dz 2

≤ 4 k u x k 2 L

2xy

k ∂ x −1 u y k 2 L

2xy

. Coming back to (29) yields

Z

R

2

u 6 (x, y)dxdy ≤ 16 k u x k 4 L

2xy

k ∂ x −1 u y k 2 L

2xy

which is (28) for p = 6. This completes the proof of Lemma 2.

Let us return to the proof of Proposition 3. Note that since u satisfies (6), one has

u t ∈ C([0, T ]; H s ( R 2 )), ∀ s ∈ N . First, taking the L 2 -scalar product of (6) with u, one obtains

1 2

d dt

Z

R

2

u 2 = Z

R

2

ψuu x = − 1 2

Z

R

2

ψ x u 2

(13)

and thus

(30) k u(t) k L

2

. exp(Ct k ψ x k L

) k φ k L

2

.

We will use, following [24], an exterior regularization of (6) by a sequence of smooth functions ϕ ε that cut the low frequencies. More precisely, let ϕ ε be defined via its Fourier transform as

(31) ϕ ˆ ε (ξ, η) :=

1 if ε < | ξ | < 1 ε and ε < | η | < 1 ε , 0 otherwise.

Note that thanks to the Lebesgue dominated convergence theorem, if u ∈ H −1 s ( R 2 ), s ∈ R then ϕ ε ∗ u converges to u in H −1 s ( R 2 ). Thanks to the Sobolev embedding similar statements hold for u ∈ L p ( R 2 ), 2 ≤ p ≤ ∞ .

Taking the convolution of (6) with ϕ ε , one gets

(32) ϕ ε ∗ u t + ϕ ε ∗ u xxx + ϕ ε ∗ ∂ x (ψu + u 2 /2) − ϕ ε ∗ ∂ x −1 u yy = 0 Setting

u ε = ϕ ε ∗ u, multiplying (32) by

− ϕ ε ∗ u xx + ∂ x −2ε ∗ u yy ) − 1

2 (ϕ ε ∗ u 2 ) and integrating in R 2 , one obtains that

(33) 1 2

d dt

hZ

R

2

(u ε x ) 2 + Z

R

2

(∂ x −1 u ε y ) 2 − Z

R

2

(u ε ) 3 3

i

= 1

2 Z

R

2

h

ϕ ε ∗ u 2 − (u ε ) 2 i u ε t +

Z

R

2

ε x ∗ (ψu))u ε xx

− Z

R

2

ε x ∗ (ψu))∂ x −2 u ε yy + 1 2

Z

R

2

ε x ∗ (ψu))(ϕ ε ∗ u 2 )

:= I + II + III + IV . Our aim is to passe to the limit ε → 0. Let us first estimate I . This argument is very typical for the present analysis and a similar situation will appear frequently in the rest of the proof of Proposition 3. Using equation (32) and the Cauchy-Schwartz inequality, we can write

| I | . k u ε t k L

2

(R

2

) k ϕ ε ∗ u 2 − (u ε ) 2 k L

2

(R

2

)

.

k u k H

31

(R

2

) + k u k 2 H

3

1

(R

2

)

k ϕ ε ∗ u 2 − (u ε ) 2 k L

2

(R

2

) .

(14)

Next, we write using the triangle inequality and the Sobolev inequality, k ϕ ε ∗ u 2 − (u ε ) 2 k L

2

(R

2

) ≤ k ϕ ε ∗ u 2 − u 2 k L

2

(R

2

) + k u 2 − (u ε ) 2 k L

2

(R

2

)

≤ k ϕ ε ∗ u 2 − u 2 k L

2

(R

2

) + k u − u ε k L

2

( k u k L

+ k u ε k L

) . k ϕ ε ∗ u 2 − u 2 k L

2

(R

2

) + k u k H

2

(R

2

) k u − u ε k L

2

.

Since u ∈ H −1 ( R 2 ), we can apply the Lebesgue dominated convergence theorem to conclude that

ε→0 lim k ϕ ε ∗ u 2 − (u ε ) 2 k L

2

(R

2

) = 0 . Therefore the term I tends to zero as ε tends to zero.

Next, we write II =

Z

R

2

(ψu ε ) x u ε xx + Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε xx

= Z

R

2

ψ x u ε u ε xx + Z

R

2

ψu ε x u ε xx + Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε xx

= − 3 2

Z

R

2

ψ x (u ε x ) 2 + 1 2

Z

ψ xxx (u ε ) 2 + Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε xx and

III = − Z

R

2

(ψu ε ) xx −2 u ε yy − Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]∂ −2 x u ε yy

= −

Z

R

2

ψu ε yx −1 u ε y − Z

R

2

ψ y u εx −1 u ε y − Z

R

2

ε ∗ (ψu) y − (ψu ε ) y ]∂ x −1 u ε y

= 1

2 Z

R

2

ψ x (∂ x −1 u ε y ) 2 − Z

R

2

ψ y u εx −1 u ε y − Z

R

2

ε ∗ (ψu) y − (ψu ε ) y ]∂ x −1 u ε y and

IV = 1 2

Z

R

2

(ψu ε ) x (u ε ) 2 + 1 2

Z

R

2

[(ϕ ε x ∗ (ψu))(ϕ ε ∗ u 2 ) − (ψu ε ) x (u ε ) 2 ]

= 1

3 Z

R

2

ψ x (u ε ) 3 + 1 2

Z

R

2

[(ϕ ε x ∗ (ψu))(ϕ ε ∗ u 2 ) − (ψu ε ) x (u ε ) 2 ].

Similarly to the analysis for I , thanks to the Lebesgue theorem, all commutator type terms involved in II, III , IV tend to 0 as ε → 0. Using Lemma 2, we get the bound

1 6

Z

R

2

| u | 3 ≤ C k u k 3/2 L

2

k u x k L

2

k ∂ x −1 u y k 1/2 L

2

≤ 1

4 k u x k 2 L

2

+ 1

4 k ∂ x −1 u y k 2 L

2

+ C k u k 6 L

2

.

(15)

We therefore obtain that

| E(u) | ≥ 1

4 k u x k 2 L

2

+ 1

4 k ∂ x −1 u y k 2 L

2

− C k u k 6 L

2

. Integrating (33) on (0, t) for t ∈ (0, T ] gives

| E(u ε (t)) − E(u ε (0)) | . Z t

0

( k ψ x k L

+ k ψ y k L

)( | E(u ε (τ )) | + k u ε (τ ) k 6 L

2

)

+( k ψ xxx k L

+ k ψ y k L

) | u ε (τ ) | 2 L

2

dτ +

Z t

0 | A ε (τ ) | dτ, where lim ε→0 A ε (τ ) = 0 for every τ ∈ [0, t] (A ε (τ ) corresponds to I and the commu- tator terms involved in II, III, IV ). Passing to the limit ε → 0, we infer that

| E(u(t)) | − | E(φ) | . Z t

0

( k ψ x k L

+ k ψ y k L

)( | E(u(τ )) | + k u(τ ) k 6 L

2

) +( k ψ xxx k L

+ k ψ y k L

) | u(τ ) | 2 L

2

dτ . By the Gronwall lemma and (30), it follows that

| E(u(t)) | . exp(C t) | E(φ) | + g(t).

where g(t) is an increasing bijection of R + which depends only on k φ k L

2

. We now turn to the bound on F ψ (u(t)). It is worth noticing that

F ψ (u) = F (u) − 5 3

Z

R

2

ψu ∂ −2 x u yy − 5 6

Z

R

2

ψ(∂ x −1 u y ) 2 ,

where F is the corresponding conservation law of the KP-I equation (see [26]). The introduction of two additional terms in F ψ is the main new idea in this paper.

Indeed, if we multiply (32) by the multiplier used in [26], we obtain terms which can not be treated as remainders (see A 1 and A 2 below). The term

− 5 3

Z

R

2

ψu ∂ x −2 u yy

is introduced in the definition of F ψ in order to cancel such “bad” remainders. The second additional term in the definition of F ψ is needed for the proof of the contin- uous dependence of the flow map on the space Z.

As in [26], a difficulty in the sequel comes from the fact that the variational deriv-

ative (F ψ ) (v) contains a term c∂ x −2yy (v 2 + 2ψv). Recall that the formal derivation

of the conservation laws consists in multiplying the KP-I equation with F (u) where

u is a solution. This procedure meets a difficulty since ∂ x −2 acts only on functions

(16)

with zero x mean value which is not a priori the case of v 2 + 2ψv. We overcome this difficulty by introducing the functional F ψ,ε defined by

F ψ,ε (u(t)) := F ψ (u ε (t)) + 5 6

hZ

R

2

(u ε (t)) 2x −2 u ε yy (t) − Z

R

2

ε ∗ u 2 (t))∂ −2 x u ε yy (t) i + 5

3 hZ

R

2

ψu ε (t) ∂ x −2 u ε yy (t) − Z

R

2

ϕ ε ∗ (ψu(t))∂ x −2 u ε yy (t) i . (recall that u ε (t) := ϕ ε ∗ u(t)). We are now in position to state the following lemma.

Lemma 3. Under the assumptions of Proposition 3, d

dt F ψ,ε (u(t)) = − 5 3

Z

R

2

ψ x (∂ x −2 u ε yy )u ε xx + 5 6

Z

R

2

ψ(∂ x −2 u ε yy )((u ε ) 2 + 2ψu ε ) x + 5

3 Z

R

2

ψu ε (∂ x −2 u ε yy ) + 5 3

Z

R

2

ψ y (∂ −1 x u ε y )(∂ x −2 u ε yy ) +G ψ (u ε ) +

Z

R

2

(∂ x −2 u ε yyε (u) + Z

R

2

Λ e ε (u) (34)

where G ψ is a continuous functional on

X = { v ∈ S ( R 2 ) : k v k L

2

+ k ∂ x −1 v y k L

2

+ k v xx k L

2

+ k v y k L

2

< ∞ } Moreover

(35) sup

t∈[0,T ]

Z

R

2

| Λ ε (u(t)) | 2 −→

ε→0 0 and sup

t∈[0,T ]

Z

R

2

| Λ e ε (u(t)) | −→

ε→0 0.

Proof of Lemma 3. After a direct computation, using (32) and the definition of F ψ,ε , we obtain the identity

d

dt F ψ,ε (u) = − 5 3

d dt

Z

R

2

ε ∗ (ψu))(∂ −2 x u ε yy ) − 5 6

d dt

Z

R

2

ψ(∂ x −1 u ε y ) 2 (36)

+ Z

R

2

(A + B )(u ε xxx + ϕ ε ∗ (uu x ) − ∂ −1 x u ε yy ) + 5

3 Z

R

2

u ε (∂ x −1 u ε y )(u ε xxy + ϕ ε ∗ (uu y ) − ∂ x −2 u ε yyy ) + 5

3 Z

R

2

[u ε u ε t − ϕ ε ∗ (uu t )]∂ x −2 u ε yy +

Z

R

2

A(ϕ ε ∗ ∂ x (ψu)) + Z

R

2

B (ϕ ε ∗ ∂ x (ψu)) + 5

3 Z

R

2

u ε (∂ x −1 u ε y )(ϕ ε ∗ (ψu) y )

:= I + II + III + IV + V + V I + V II + V III

(17)

where

A = − 5

3 ∂ x −4 u ε 4y + 5

6 ∂ x −2 ϕ ε,yy ∗ u 2 + 5

3 u εx −2 u ε yy and

B = 5

6 (∂ −1 x u ε y ) 2 − 5

6 (u ε ) 3 − 3u ε 4x + 10u ε yy − 5

2 u ε u ε xx − 5

4 (u ε ) 2 xx . Next one can check that (see [26, Lemma 1] for a similar computation)

III + IV + V = Z

R

2

(∂ x −2 u ε yy1 ε (u) + Z

R

2

Λ e 1 ε (u) where

Λ 1 ε (u) = − 5

3 [ϕ ε ∗ (uu t ) − u ε u ε t ] + 5

3 [ϕ ε ∗ (uu x ) − u ε u ε x ]u ε , and

Λ e 1 ε (u) = h

ϕ ε ∗ (uu x ) − u ε u ε x i

×

× 5

6 (∂ −1 x u ε y ) 2 − 5

6 (u ε ) 3 − 3u ε 4x + 25

3 u ε yy − 5

2 u ε u ε xx − 5

4 (u ε ) 2 xx + 5

3

h ϕ ε ∗ (uu y ) − u ε u ε y i

u εx −1 u ε y

Using the Lebesgue dominated convergence theorem, we obtain that Z

R

2

| Λ 1 ε (u) | 2 −→

ε→0 0, Z

R

2

| Λ e 1 ε (u) | −→

ε→0 0, t ∈ [0, T ].

Let us next compute the five other terms in the right hand-side of (36) one by one.

V I = Z

R

2

A(ϕ ε ∗ ∂ x (ψu)) = 5 3

Z

R

2

ε ∗ (ψu))∂ −3 x u ε 4y

− 5 6

Z

R

2

ε ∗ (ψu))(∂ x −1 ϕ ε yy ∗ u 2 ) + 5

3 Z

R

2

ε x ∗ (ψu))u ε (∂ x −2 u ε yy )

:= A 1 + A 2 + A 3 .

(18)

Next I = − 5

3 d dt

Z

R

2

ε ∗ (ψu))(∂ x −2 u ε yy ) = − 5 3

Z

R

2

−2 xy 2 ϕ ε ∗ (ψu) u ε t

− 5 3

Z

R

2

ψ(∂ x −2 u ε yy )u ε t

− 5 3

Z

R

2

ψ t u ε (∂ x −2 u ε yy )

− 5 3

Z

R

2

[(ϕ ε ∗ (ψu) t − (ψu ε ) t ]∂ x −2 u ε yy := C 1 + C 2 + C 3 + C 4

with C 1 = 5

3 Z

R

2

(∂ −2 xy 2 ϕ ε ∗ (ψu))u ε 3x − 5 3

Z

R

2

x −2y 2 ϕ ε ∗ (ψu)

x −1 u ε yy

+ 5 3

Z

R

2

(∂ x −2y 2 ϕ ε ∗ (ψu))(ϕ ε ∗ (uu x )) + 5 3

Z

R

2

(∂ x −2y 2 ϕ ε ∗ (ψu))(ϕ ε ∗ ∂ x (ψu))

= 5

3 Z

R

2

(∂ 2 y ϕ ε ∗ (ψu))u ε x − A 1 − A 2 + 0 .

As mentioned before, here is the crucial cancellation, thanks to the first additional term in F ψ . Next

5 3

Z

R

2

(∂ y 2 ϕ ε ∗ (ψu))u ε x = 5 3

Z

R

2

y 2 (ψu ε )u ε x + 5 3

Z

R

2

h ∂ y 2 ϕ ε ∗ (ψu) − ∂ y 2 (ψu ε ) i u ε x

= 5

6 Z

R

2

ψ x (u ε y ) 2 + 5 3

Z

R

2

(ψ yy u ε + ψ y u ε y )u ε x + 5

3 Z

R

2

h

2 y ϕ ε ∗ (ψu) − ∂ 2 y (ψu ε ) i u ε x

Therefore,

V I + C 1 = 5 6

Z

R

2

ψ x (u ε y ) 2 + 5 3

Z

R

2

ψ yy u ε u ε x + 5 3

Z

R

2

ψ y u ε y u ε x

+ 5 3

Z

R

2

ε x ∗ (ψu))u ε (∂ −2 x u ε yy ) + 5

3 Z

R

2

h ∂ y 2 ϕ ε ∗ (ψu) − ∂ y 2 (ψu ε ) i

u ε x

(19)

On the other hand,

C 2 = 5 3

Z

R

2

ψ(∂ x −2 u ε yy )u ε 3x − 5 3

Z

R

2

ψ(∂ x −2 u ε yy )∂ x −1 u ε yy + 5

6 Z

R

2

ψ(∂ −2 x u ε yy )

ϕ ε x ∗ ((u ε ) 2 + 2ψu ε ) := C 21 + C 22 + C 23 .

First,

C 22 = 5 6

Z

R

2

ψ x (∂ x −2 u ε yy ) 2 .

Next,

C 21 = − 5 3

Z

R

2

ψ(∂ x −1 u ε yy )u ε xx − 5 3

Z

R

2

ψ x (∂ x −2 u ε yy )u ε xx

= 5

3 Z

R

2

ψu ε yy u ε x + 5 3

Z

R

2

ψ x (∂ x −1 u ε yy )u ε x − 5 3

Z

R

2

ψ x (∂ x −2 u ε yy )u ε xx

= 5

6 Z

R

2

ψ x (u ε y ) 2 − 5 3

Z

R

2

ψ x u ε yy u ε

− 5 3

Z

R

2

ψ xx (∂ x −1 u ε yy )u ε − 5 3

Z

R

2

ψ x (∂ −2 x u ε yy )u ε xx − 5 3

Z

R

2

ψ y u ε y u ε x

= 15 6

Z

R

2

ψ x (u ε y ) 2 − 5 6

Z

R

2

ψ xxx (∂ x −1 u ε y ) 2 − 5 3

Z

R

2

ψ x (∂ x −2 u ε yy )u ε xx

− 5 3

Z

R

2

ψ y u ε y u ε x − 5 6

Z

R

2

ψ xyy (u ε ) 2 + 5 3

Z

R

2

ψ xxy (∂ x −1 u ε y ) u ε .

Since ψ t = − ψ x ,

C 3 = 5 3

Z

R

2

ψ x u ε (∂ x −2 u ε yy ) .

(20)

Now

− 5 6

d dt

Z

R

2

ψ(∂ x −1 u ε y ) 2 = − 5 6

Z

R

2

ψ t (∂ x −1 u ε y ) 2 + 5

3 Z

R

2

ψ(∂ x −1 u ε y )u ε xxy + 5

3 Z

R

2

ψ(∂ x −1 u ε y )(ϕ ε ∗ (uu y )) + 5

3 Z

R

2

ψ(∂ x −1 u ε y )(ϕ ε ∗ (ψu) y )

− 5 3

Z

R

2

ψ(∂ x −1 u ε y )(∂ −2 x u ε 3y ) := D 1 + D 2 + D 3 + D 4 + D 5 .

Since ψ t = − ψ x and X ֒ → L ( R 2 ), by involving two commutators, we infer that D 1 + D 3 + D 4 = G 1 (u ε ) +

Z

R

2

Λ e 2 ε (u) where G 1 is a continuous functional on X and

Z

R

2

| Λ e 2 ε (u) | −→

ε→0 0, t ∈ [0, T ].

On the other hand, D 2 = − 5

3 Z

R

2

ψ x (∂ x −1 u ε y )u ε xy − 5 3

Z

R

2

ψu ε y u ε xy

= 5

3 Z

R

2

ψ xx (∂ x −1 u ε y )u ε y + 5 3

Z

R

2

ψ x (u ε y ) 2 + 5 6

Z

R

2

ψ x (u ε y ) 2

= 15 6

Z

R

2

ψ x (u ε y ) 2 − 5 6

Z

R

2

ψ 3x (∂ x −1 u ε y ) 2 and

D 5 = 5 3

Z

R

2

ψ y (∂ x −1 u ε y )(∂ x −2 u ε yy ) + 5 3

Z

R

2

ψ(∂ x −1 u ε yy )(∂ x −2 u ε yy )

= 5

3 Z

R

2

ψ y (∂ x −1 u ε y )(∂ x −2 u ε yy ) − 5 6

Z

R

2

ψ x (∂ x −2 u ε yy ) 2 .

(21)

Note that the last term above canceled with C 22 . Summarizing, we infer that I + II + V I = − 5

3 Z

R

2

ψ x (∂ x −2 u ε yy )u ε xx + 5 6

Z

R

2

ψ(∂ x −2 u ε yy )((u ε ) 2 + 2ψu ε ) x + 5

3 Z

R

2

ε x ∗ (ψu))u ε (∂ x −2 u ε yy ) + 5 3

Z

R

2

ψ x u ε (∂ x −2 u ε yy ) + 5

3 Z

R

2

ψ y (∂ x −1 u ε y )(∂ x −2 u ε yy ) + G 2 (u ε ) +

Z

R

2

(∂ x −2 u ε yy2 ε (u) + Z

R

2

Λ e 3 ε (u) where G 2 is a continuous functional on X and

Z

R

2

| Λ 2 ε (u) | 2 −→

ε→0 0, Z

R

2

| Λ e 3 ε (u) | −→

ε→0 0, t ∈ [0, T ].

Since clearly

V III = G 3 (u ε ) + Z

R

2

Λ e 4 ε (u), where G 3 is a continuous functional on X and

Z

R

2

| Λ e 4 ε (u) | −→

ε→0 0, t ∈ [0, T ], it remains to estimate V II. We notice that

V II = − 5 6

Z

R

2

(∂ −1 x u ε y ) 2ε x ∗ (ψu)) − 5 6

Z

R

2

(u ε ) 3ε x ∗ (ψu))

− 3 Z

R

2

u ε 4xε x ∗ (ψu)) − 10 Z

R

2

u ε yε ∗ (ψu) y )

− 5 2

Z

R

2

u ε u ε xxε x ∗ (ψu)) − 5 4

Z

R

2

x 2 (u ε ) 2ε x ∗ (ψu)) . We now observe that we can write

V II = G 4 (u ε ) + Z

R

2

Λ e 4 ε (u), where G 4 is continuous on X and and

Z

R

2

| Λ e 4 ε (u) | −→

ε→0 0, t ∈ [0, T ].

(22)

For example,

− 3 Z

R

2

u ε 4xε x ∗ (ψu)) = − 3 Z

R

2

u ε 4xx u ε + ψu ε x )

− 3 Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε 4x

= 3 Z

R

2

u ε 3x (2ψ x u ε x + ψ xx u ε + ψu ε xx )

− 3 Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε 4x

= − 15 2

Z

R

2

ψ x | u ε xx | 2 + 9 2

Z

R

2

ψ 3x | u ε x | 2 − 3 Z

R

2

ψ 3x u ε u ε xx

− 3 Z

R

2

ε ∗ (ψu) x − (ψu ε ) x ]u ε 4x

All other terms in the representation of V II can be treated similarly. This achieves

the proof of Lemma 3.

Now, since

5 6

Z

R

2

ε ∗ (u 2 + 2ψu)]∂ x −2 u ε yy ≤ C( k u k 4 L

4

+ k u k 2 L

2

k ψ k 2 L

) + 5

12 k ∂ x −2 u ε yy k 2 L

2

, there exists a constant C > 0 such that for ε small enough,

(37) F ψ,ε (u(t)) ≥ 5

24 k ∂ x −2 u ε yy (t) k 2 L

2

− C, ∀ t ∈ [0, T ] . We thus deduce from Lemma 3 and (37) that

(38) d

dt F ψ,ε (u(t)) . k ψ k W

3,

| F ψ,ε (u(t)) | + R ψ (u ε ) + Λ ε (t)

where R ψ is continuous on X and | Λ ε | 1 → 0 as ε → 0 uniformly for t ∈ [0, T ]. Here we used that thanks to Lemma 2,

k ∂ x −1 u ε y k L

4

. k ∂ x −1 u ε y k 1/4 L

2

k u ε y k 1/2 L

2

k ∂ x −2 u ε yy k 1/4 L

2

and

k u ε x k L

4

. k u ε x k 1/4 L

2

k u ε xx k 1/2 L

2

k u ε y k 1/4 L

2

and thus, Z

R

2

ψu ε x u ε (∂ −2 x u ε yy ) . k ψ k L

k u ε k L

4

k u ε x k L

4

k ∂ x −2 u ε yy k L

2

. | F ψ,ε (u) | + R ψ (u ε )

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