Thomas Alazard

Nicolas Burq

Claude Zuily

Strichartz Estimates

N Ambrose

Masmoudi

B Schweiser

T Iguchi

D Coutand

S Shkoller

J Shatah

C Zeng

M Ming

Z Zhang

F Rousset

N Tzvetkov

STRICHARTZ ESTIMATES FOR WATER WAVES

ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction

In a time-dependent domain Ω t ⊂ R d+1 which is located between a free hypersurface Σ t and a fixed known bottom Γ, consider a potential flow v = ∇ x,y φ, with ∆ x,y φ = 0 in Ω t , ∂ n φ = 0 on Γ.

The surface-tension water-waves problem is given by two equations: a kinematic condition (which states that the free surface moves with the fluid), and a dynamic condition (that expresses a balance of forces across the free surface). The system reads (1.1)

   ∂ t η = ∂ y φ -∇η • ∇φ on Σ t = {y = η(t, x)}, ∂ t φ + 1 2 |∇ x,y φ| 2 + gη = H(η) on Σ t ,
where ∇ = ∇ x , g > 0 is the acceleration of gravity and

H(η) = div ∇η 1 + (∂ x η) 2 .
is the mean curvature of the free surface.

1.1. Assumptions. -We work in a fluid domain such that there is uniformly a minimum depth of water, more precisely we assume that for each time t one has

Ω t = Ω 1,t ∩ Ω 2
where Ω 1,t is the half space located below the free surface Σ t ,

Ω 1,t = { (x, y) ∈ R d × R : y < η(t, x) } (d ≥ 1)
for some unknown function η and Ω 2 contains a fixed strip around Σ t , that means that there exists h > 0 such that,

(1.2) {(x, y) ∈ R d × R : η(t, x) -h ≤ y ≤ η(t, x)} ⊂ Ω 2 ,
for all t ∈ [0, T ]. We shall also assume that the domain Ω 2 (and hence the domain

Ω t = Ω 1,t ∩ Ω 2 ) is connected.
We emphasize that no regularity assumption is made on the bottom Γ = ∂Ω t \ Σ t . We consider both cases of infinite depth and bounded depth bottoms (and all cases inbetween). Finally, we could consider the cases where the free surface is a graph over a given smooth hypersurface and the bottom is time dependent.

1.2. Main results. -Following Zakharov we reduce the system to a system on the free surface. If ψ = ψ(t, x) ∈ R is defined by ψ(t, x) = φ(t, x, η(t, x)), then φ(t, x, y) is the unique variational solution of (1.3) ∆φ = 0 in Ω t , φ(t, x, η(t, x)) = ψ(t, x).

The Dirichlet-Neumann operator is then defined by (G(η)ψ)(t, x) = 1 + |∇η| 2 ∂ n φ| y=η(t,x) = ∂ y φ -∇η • ∇φ

   y=η(t,x)
.

(we refer to Section 2 in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] for a precise construction). Then (η, φ) is solution of the water-waves system (1.1) if and only if (η, ψ) solves the system (1.4)

     ∂ t η -G(η)ψ = 0, ∂ t ψ + gη -H(η) + 1 2 |∂ x ψ| 2 - 1 2
∂ x η • ∂ x ψ + G(η)ψ
The purpose of this work is precisely, in the case d = 1, to investigate the dispersive properties of these solutions. Our results are twofold: first we prove Strichartz type estimates with loss of derivatives at the very same level of regularity we were able to construct the solutions in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF]. On the other hand, for smoother initial data, we prove that the solutions enjoy the optimal Strichartz estimates (i.e, without loss of regularity compared to the system linearized at (η = 0, ψ = 0)). More precisely, our main results are the following.

Theorem 1.1. -Let s > 5/2 and T > 0. Consider a solution (η, ψ) of (1.4) on the time interval I = [0, T ] such that Ω t satisfies (1.2) for t ∈ I. If

(η, ψ) ∈ C 0 I, H s+ 1 2 (R) × H s (R) , then (η, ψ) ∈ L 4 I, W s+ 1 4 ,∞ (R) × W s-1 4 ,∞ (R) .
Theorem 1.2. -Let s > 11/2, T > 0 and p, q, σ be such that

2 p + 1 q = 1 2 , 2 ≤ q < +∞.
Consider a solution (η, ψ) of (1.4) on the time interval I = [0, T ] such that Ω t satisfies (1.2) for t ∈ I. If

(η, ψ) ∈ C 0 I, H s+ 1 2 (R) × H s (R) ,
then (η, ψ) ∈ L p I, W

s+ 3 8 + 1 4q ,q (R) × W s-1 8 + 1
4q ,q (R) .

Remark 1.3. -(i) Theorem 1.1 was obtained recently under the assumption s ≥ 15 by Christianson-Hur-Staffilani [START_REF] Christianson | Strichartz estimates for the water-wave problem with surface tension Preprint[END_REF] .

(ii) Let s > 5/2 and (η 0 , ψ 0 ) ∈ H s+ 1 2 (R) × H s (R) satisfying dist(Σ 0 , Γ) ≥ c > 0, we proved in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] that there exist T > 0 and a solution (η, ψ) ∈ C 0 [0, T ]; H s+ 1 2 (R) × H s (R) satisfying dist(Σ t , Γ) ≥ c > 0.

(iii) Letting q tend to infinity we see that the result in Theorem 1.2 exhibits a gain of 1/8 derivatives with respect to Theorem 1.1 (iv) For the end point (p, q) = (4, +∞) we prove in fact, under the assumptions in Theorem 1.2, that (η, ψ) ∈ L 4 I, B

s+ 3 8 ∞,2 (R) × B s-1 8 ∞,2 (R) .
where B σ ∞,2 is the standard Besov space (see Section 6) (v) The gain of regularity exhibited in Theorem 1.2 is optimal as can be seen at the level of the linearized system around the trivial solution (η, ψ) = (0, 0) which reads (when g = 0),

∂ t η -|D x | ψ = 0, ∂ t ψ -∆η = 0. Indeed u = |D x | 1 2
η + iψ is a solution of the equation i∂ t u -|D x | 3 2 u = 0, for which one can prove the optimal estimate exp(-it

|D x | 3 2 )u 0 L 4 I,W s-1 8 ,∞ (R) ≤ C u 0 H s ,
which gives the desired regularity on (η, ψ).

(vi) It is most likely that Theorem 1.1 remains valid when R is replaced by the one dimensional torus T. Indeed, our proof relies on a semi-classical parametrix (on time intervals taylored to the frequency) which exhibits finite speed of propagation and which can consequently be easily localized in space.

(vii) Notice that the dispersive estimates proved in this paper can be combined with our previous work to improve the regularity threshold obtained in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] and give local well posednesss for initial data below the s = 2 + 1 2 threshold. This will be the matter of a forthcoming paper (including the 3-d water-waves system) [START_REF] Alazard | Dispersive estimates for the 3-d water-waves system and applications[END_REF].

(viii) Notice finally that dispersive properties of the operator linearized at (η = 0, ψ = 0) were used recently by Wu [START_REF] Wu | Almost global wellposedness of the 2-D full water wave problem[END_REF][START_REF] Wu | Global well-posedness of the 3-D full water wave problem preprint[END_REF] and Germain-Masmoudi-Shatah [START_REF] Germain | Global Solutions for the Gravity Water Waves Equation in Dimension 3[END_REF] to prove global existence results.

1.3. Strategy of the proofs. -Following the approach in Alazard-Métivier [START_REF] Alazard | Paralinearization of the Dirichlet to Neumann operator, and regularity of three dimensional water waves[END_REF], after suitable paralinearizations, we have shown in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] that the water waves system can be arranged into an explicit paradifferential symmetric equation of Schrödinger type, and we deduced the smoothing effect for the 2-d surface tension water waves. Here, we will also take benefit of this paralinearization reduction, and this reduced system will be our starting point. The guiding line for the rest of our proof is very classical: construction of a parametrix to prove dispersion (L 1 -L ∞ estimates), and then T T * argument.

There are two main difficulties in the analysis of this equation. First the coefficients of the operator are time dependent and consequently we cannot get rid of the lower order terms by simple conjugation arguments (see Burq-Planchon [START_REF] Burq | On well-posedness for the Benjamin-Ono equation[END_REF]). Second the coefficients enjoy poor regularity, and finally, whereas the principal part in the operator is of order 3/2, the subprincipal part in the operator is of order 1 which gives only a 1/2 difference compared to the usual 1 difference encountered for magnetic Schrödinger operators. As will be shown in our analysis, the presence of such subprincipal parts will produce non trivial oscillations which here have to be taken into account in the analysis.

The first common step for both theorems is to perform several reductions for the paradifferential equation. The first one is to use Alinhac's para-composition theory [START_REF] Alinhac | Paracomposition et opérateurs paradifférentiels[END_REF] (see also where a similar idea was used) to reduce the matters to the study of a Schrödinger type operator with constant coefficients principal part. This is particular to space dimension 1 and reflects the fact that there is only one metric on R. The second reduction, inspired by works by Smith [START_REF] Smith | A parametrix construction for wave equations with C 1,1 coefficients[END_REF] and Bahouri-Chemin [START_REF] Bahouri | Equations d'ondes quasi-linaires et estimations de Strichartz[END_REF], consists in smoothing out the coefficients of the operator.

Once this reduction has been achieved, we can construct the parametrix, for which the natural time is the semi-classical one: s = t|ξ| -1/2 . Here the differences between our two theorems appear. Indeed, in the proof of Theorem 1.1, following the strategy in Burq-Gérard-Tzvetkov [START_REF] Burq | Strichartz inequalities and the nonlinear Schrdinger equation on compact manifolds[END_REF], we construct the parametrix on small times |s| ≤ c) and the main difficulty is to handle sharp regularity threshold (for smooth enough initial data the proof would be much simpler). In the proof of Theorem 1.2 the difficulties are different: first we have to handle the oscillations generated by the subprincipal part and furthermore we have to prove very large time asymptotics (|s| ≤ c|ξ| 1/2 ) in the high frequency regime |ξ| → +∞. Notice that, even for initial data with arbitrarily large smoothness, the analysis would be non trivial. Finally, once the parametrix is constructed, the dispersion estimate is obtained by using non classical stationary phase lemmas involving precise controls on the remainder terms.

Preliminaries

In this section we recall some notations and results from [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] which will be used in the sequel.

2.1. Paradifferential calculus. -In this paragraph we review classical facts about Bony's paradifferential calculus.

For ρ ∈ N, according to the usual definition, we denote by W ρ,∞ (R) the Sobolev spaces of L ∞ functions whose derivatives of order ≤ ρ are in L ∞ . For ρ ∈]0, +∞[\N, we denote by W ρ,∞ (R) the space of bounded functions whose derivatives of order [ρ] are uniformly Hölder continuous with exponent ρ -

[ρ]. Definition 2.1. -Given ρ ≥ 0 and m ∈ R, Γ m ρ (R)
denotes the space of functions a(x, ξ) on R × (R \ 0), which are C ∞ with respect to ξ and such that, for all α ∈ N and all ξ = 0, the function x → ∂ α ξ a(x, ξ) belongs to W ρ,∞ (R) and there exists a constant C α such that,

(2.1) ∀ |ξ| ≥ 1 2 , ∂ α ξ a(•, ξ) W ρ,∞ (R) ≤ C α (1 + |ξ|) m-|α| . Definition 2.2. -Σ m ρ (R) denotes the space of symbols a(x, ξ) such that a = 0≤j<ρ a (m-j) (j ∈ N),
where a (m-j) ∈ Γ m-j ρ-j (R) is homogeneous of degree mj with respect to ξ.

Given a symbol a, we define the paradifferential operator T a by

(2.2) T a u(ξ) = (2π) -d χ(ξ -η, η) a(ξ -η, η)ψ(η) u(η) dη,
where a(θ, ξ) = e -ix•θ a(x, ξ) dx is the Fourier transform of a with respect to the first variable, χ, ψ are two fixed C ∞ functions such that

ψ(η) = 0 for |η| ≤ 1, ψ(η) = 1 for |η| ≥ 2,
χ(θ, η) is homogeneous of degree 0 and satisfies, for 0 < ε 1 < ε 2 small enough,

χ(θ, η) = 1 if |θ| ≤ ε 1 |η| , χ(θ, η) = 0 if |θ| ≥ ε 2 |η| .
We shall use quantitative results from Métivier [START_REF] Métivier | Para-differential calculus and applications to the Cauchy problem for nonlinear systems[END_REF] about operator norms estimates in symbolic calculus. To do so we introduce the following semi-norms.

Definition 2.3. -For m ∈ R, ρ ≥ 0 and a ∈ Γ m ρ (R), we set (2.3) M m ρ (a) = sup |α|≤ d 2 +1+ρ sup |ξ|≥1/2 (1 + |ξ|) |α|-m ∂ α ξ a(•, ξ) W ρ,∞ (R)
.

The main features of symbolic calculus for paradifferential operators are given by the following theorems.

Definition 2.4. -Let m ∈ R. An operator T is said of order ≤ m if, for all µ ∈ R, it is bounded from H µ (R) to H µ-m (R). Theorem 2.5. -Let m ∈ R. If a ∈ Γ m 0 (R), then T a is of order ≤ m.
Moreover, for all µ ∈ R there exists a constant K such that

(2.4) T a H µ →H µ-m ≤ KM m 0 (a). Theorem 2.6 (Composition). -Let m ∈ R and ρ > 0. If a ∈ Γ m ρ (R) and b ∈ Γ m ′ ρ (R) then T a T b -T a#b is of order ≤ m + m ′ -ρ, where a#b = |α|<ρ 1 i |α| α! ∂ α ξ a∂ α x b.
Moreover, for all µ ∈ R there exists a constant K such that

(2.5) T a T b -T a#b H µ →H µ-m-m ′ +ρ ≤ KM m ρ (a)M m ′ ρ (b).
If a = a(x) is a function of x only, the paradifferential operator T a is a called a paraproduct. Paraproducts can also be defined using the Littlewood-Paley decomposition of the frequency space. Indeed, let φ : R → R be a smooth even function with φ(t) = 1 for |t| ≤ 1 and φ(t) = 0 for |t| ≥ 2. For k ∈ N, we introduce the symbol φ k (ξ) = φ ξ 2 k , and then the operators S k and ∆ k defined by

S k f (ξ) := φ k (ξ) f (ξ), ∆ k f (ξ) := (φ k (ξ) -φ k-1 (ξ)) f (ξ) For all f ∈ S ′ (R), the spectrum of ∆ k f satisfies spec ∆ k f ⊂ {ξ : 2 k-1 ≤ |ξ| ≤ 2 k+1 }. Hence ∆ j ∆ k = 0 if |j -k| ≥ 2.
Moreover we have the Littlewood-Paley decomposition:

f = S 0 f + k∈N * ∆ k f.
With this decompositon, paraproducts can be defined by

T a f = k≥4 S k-3 (a)∆ k f.
Notice that the difference between paraproducts defined in these two ways is a smoothing operator. Namely, if a ∈ W ρ,∞ (R) for some ρ > 0 then the difference is of order -ρ.

Theorem 2.7. -Let α, β ∈ R be such that α + β > 0. If a ∈ H α (R) and b ∈ H β (R) then ab -T a b -T b a ∈ H α+β-1 2 (R) and ab -T a b -T b a H α+β-1 2 (R) ≤ K a H α (R) b H β (R)
for some positive constant K independent of a, b.

We use the following result which is a consequence of (2.5) with m = m ′ = 0, ρ = 1.

Lemma 2.8.

-Let s > 2 + 1 2 and a ∈ W 1,∞ (R).
Then for all σ ∈ R there exists a constant C > 0 such that for all j ∈ N,

[∆ j , T a ]u H σ+1 (R) ≤ C a W 1,∞ (R) u H σ (R) . 2.2. The Dirichlet-Neumann operator. - Lemma 2.9. -Let s > 2 + 1
2 and 1 ≤ σ ≤ s. Then there exists an increasing function

C : R + → R + such that for all (η, ψ) ∈ H s+ 1 2 (R) × H s (R) G(η)ψ H σ-1 (R) ≤ C( η H s+ 1 2 (R) ) ψ H σ (R) . Furthermore, if (η, ψ) ∈ L ∞ (I; H s+ 1 2 (R) × H s (R)) is a solution of (1.4). Then (2.6) ∂ t (G(η)ψ) = G(η)(∂ t ψ -B∂ t η) -div(V ∂ t η)
where

(2.7) B(t, x) := ∂ x ψ∂ x η + G(η)ψ 1 + |∂ x η| 2 , V (t, x) := ∂ x ψ -B∂ x η 2.3. Symmetrization.
-We consider a solution (η, ψ) of (1.4) on the time interval I = [0, T ] with 0 < T < +∞, satisfying the assumption (1.2) for all t ∈ I and such that

(η, ψ) ∈ C 0 I, H s+ 1 2 (R) × H s (R) ,
for some s > 5 2 . Then we set (2.8)

U = ψ -T B η.
where B has been defined in (2.7). It follows from the analysis in [START_REF] Alazard | On the Cauchy problem for the water waves with surface tension[END_REF] that we have the following symmetrization of the equations.

Lemma 2.10 ([2, Corollary 4.9]). -Let c, c 1 be defined by

c = 1 + (∂ x η) 2 -3 4 , c 1 = 1 + (∂ x η) 2 -1 2 .
There exists an elliptic symbol p ∈ Σ 1/2

s-1 such that the complex-valued unknown (2.9)

Φ = T p η + iT c 1 U
satisfies a scalar equation of the form

(2.10) ∂ t Φ + T V ∂ x Φ + i |D x | 3 4 T c |D x | 3 4 Φ = F,
where V has been defined in (2.7) and F ∈ L ∞ (I, H s (R)).

Reductions

3.1. Change of variables. -Our aim in this section is to simplify the equation (2.10) by a change of variable. To compute the effect of a change of variable we shall use Alinhac's paracomposition operators and we refer to [START_REF] Alinhac | Paracomposition et opérateurs paradifférentiels[END_REF] for the general theory .

Let κ be a C 1 diffeomorphism from R to R. We define the operator κ * by,

(3.1) κ * u = u • κ -T (∂xu)•κ κ.
One of the main properties of κ * is that there is a symbolic calculus theorem which allows to compute the equation satisfied by κ * u in terms of the equation satisfied by u (in analogy with the paradifferential calculus).

Theorem 3.1. -Let d ≥ 1, m ∈ R, r > 1, ρ > 0 and set σ := inf{ρ, r -1}. Consider a C r (R d )-diffeomorphism χ and set κ = χ -1 . Let a be symbol in Σ m ρ (R d ). Then there exists a * ∈ Σ m σ (R d ) such that κ * T a -T a * κ * is order ≤ m -σ.
Moreover one can give an explicit formula for a * . If a = a m-k , then

(3.2) a * (χ(x), η) = α 1 i |α| α! ∂ α ξ a m-k (x, t χ ′ (x)η)∂ α y (e iΨx(y)•η )| y=x ,
where the sum is taken over all α ∈ N d such that the summand is well defined, χ ′ (x) is the differential of χ, t denotes transpose and

(3.3) Ψ x (y) = χ(y) -χ(x) -χ ′ (x)(y -x).
We note that it is easy to obtain regularity results on u given results on κ * u. Namely, we have the following lemma.

Lemma 3.2. -Let ρ ≥ 0, κ ∈ W ρ,∞ (R) and u ∈ W 1,∞ (R). If κ * u ∈ W ρ,∞ (R) then u ∈ W ρ,∞ (R).
Proof. -This follows from the fact that u = (κ * u)

• χ + (T (∂xu)•κ) κ) • χ where χ = κ -1 ∈ W ρ,∞ (R).
We are now ready to simplify (2.10). Define χ by

(3.4) χ(t, x) = x 0 c(t, y) -2 3 dy = x 0 1 + (∂ y η(t, y)) 2 dy, so that ∂ x χ(t, x) = 1 + (∂ x η(t, x)) 2 = c(t, x) -2 3 .
Then for each t ∈ [0, T ], x → χ(t, x) is a diffeomorphism from R to R. Introduce its inverse

(3.5) κ = χ -1 .
3.1.1. Notations:-We shall set I = [0, T ] and we shall denote

(3.6) A = C (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R)
where C : R + → R + is an increasing function which may change from line to line. Moreover we shall denote by f • κ the function

(3.7) (f • κ)(t, x) = f (t, κ(t, x)).
3.1.2. Estimates of χ and κ.-From (3.4), the equation ∂ t η = G(η)ψ, the Lemma 2.9, the Hölder inequality and the fact that s > 2 + 1 2 we deduce,

(3.8) ∂ t χ L ∞ (I×R) ≤ A. Now since ∂ x χ(t, x) = 1 + f (∂ x η), f ∈ C ∞ (R), f (0) = 0,
we deduce from the assumption s > 2 + 1 2 and the Sobolev embedding that, (3.9)

∂ x χ(t, x) -1 L ∞ (I,H s-1 2 (R)) + ∂ x χ L ∞ (I×R) ≤ A.
Let us consider the function κ. Since ∂ t κ = -∂tχ ∂xχ • κ we have, using (3.8), (3.10)

∂ t κ L ∞ (I×R) ≤ A.
On the other hand we have

∂ x κ = 1 + f (∂ x η) where f ∈ C ∞ (R), f (0) = 0. It follows that, (3.11) ∂ x κ -1 L ∞ (I,H s-1 2 (R)) ≤ A.
It is clear from the definition that we have, (3.12)

|∂ x κ(t, x)| ≤ 1, ∀(t, x) ∈ I × R.
It follows then by induction that for every p ∈ N we have,

(3.13) κ L ∞ (I,W p,∞ (R)) ≤ C η L ∞ (I,W p,∞ (R)) .
To go further we shall need the following elementary lemma.

Lemma 3.3. -Let p ∈ N * and κ : R → R be a diffeomorphism such that ∂ x κ ∈ W p-1,∞ (R). Set χ = κ -1 . Then for all F ∈ H µ (R) with 0 ≤ µ ≤ p we have F •κ ∈ H µ (R) and
(3.14) F • κ H µ (R) ≤ χ ′ L ∞ (R) C ∂ x κ W p-1,∞ (R) F H µ (R)
where C is an increasing function from R + to R + . Now in our case for (almost all) fixed t and all ε > 0 we have,

(3.15) ∂ x κ(t, •) W p-1,∞ (R) ≤ C( η(t, •) W p,∞ (R) ) ≤ C( η(t, •) H p+ 1 2 +ε (R)
).

We deduce then from Lemma 3.3 that for 0 ≤ µ ≤ s -1 and F ∈ L ∞ (I, H µ (R)) we have,

(3.16) F • κ L ∞ (I,H µ (R)) ≤ A F L ∞ (I,H µ (R)) .
Coming back to the regularity of χ we deduce from (3.4) that,

∂ 2 x χ = (∂ x η)(∂ 2 x η) (1 + (∂ x η) 2 ) 1 2
.

It follows from (3.16) that,

(3.17) (∂ 2 x χ) • κ L ∞ (I,H s-3 2 (R))
≤ A.

On the other hand we have,

∂ x ∂ t χ = (∂ x η)∂ x (G(η)ψ) (1 + (∂ x η) 2 ) 1 2
.

So using Lemma 2.9 and (3.16) we obtain,

(3.18) (∂ x ∂ t χ) • κ L ∞ (I,H s-2 (R)) ≤ A.
Now we would like to estimate ∂ 2 t χ. Since ∂ t η = G(η)ψ we have,

∂ 2 t χ(t, x) = - x 0 [∂ x η∂ x (G(η)ψ)] 2 (1 + (∂ x η) 2 ) 3 2 dy + x 0 [∂ x (G(η)ψ)] 2 (1 + (∂ x η) 2 ) 1 2 dy (3.19) + x 0 ∂ x η∂ x ∂ t (G(η)ψ) (1 + (∂ x η) 2 ) 1 2 dy. (3.20) Since s > 2 + 1
2 , the Hölder inequality and Lemma 2.9 show that the first two terms are pointwise bounded by A. By the Holder inequality the last term can be pointwise bounded by

∂ x η L ∞ (I,L 2 (R)) ∂ x ∂ t G(η)ψ L ∞ (I,L 2 (R)) .
Using (2.6) and the equation satisfied by (η, ψ) we find, if

s > 3 + 1 2 , ∂ x ∂ t G(η)ψ L ∞ (I,L 2 (R)) ≤ A. Therefore if s > 3 + 1 2 we obtain, (3.21) ∂ 2 t χ L ∞ (I×R) ≤ A.
Finally let us estimate the term ∂ x ∂

W = V • κ(∂ x χ • κ) + ∂ t χ • κ, (3.24) Φ * = κ * Φ = Φ • κ -T (∂xΦ)•κ κ.
Then we have the following result.

(3.25) (∂ t + T W ∂ x + i |D x | 3 2 )u = F, with F ∈ L ∞ (I, H s (R d ))
and W is defined by (3.23).

Proof. -We apply the operator κ * to the equation (2.10). We first show that

(3.26) κ * (∂ t + T V ∂ x )Φ = (∂ t + T W ∂ x )Φ * + R(Φ) R(Φ) L ∞ (I,H s (R)) ≤ C( (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R) ) Φ L ∞ (I,H s (R)) .
We begin by showing that

(3.27) κ * (∂ t Φ) = (∂ t -T (∂tχ)•κ )Φ * + R 1 (Φ)
where R 1 satisfies the estimate in (3.26).

We have

κ * (∂ t Φ) = (∂ t Φ) • κ -T (∂x∂tΦ)•κ κ = ∂ t (Φ • κ) -(∂ t κ)(∂ x Φ • κ) -T (∂x∂tΦ)•κ κ, therefore, (3.28) κ * (∂ t Φ) = ∂ t (κ * Φ) + B 1 + B 2 , B 1 = T (∂ 2 x Φ•κ)∂tκ κ B 2 = T (∂xΦ)•κ ∂ t κ -(∂ t κ)(∂ x Φ • κ)
Let us consider the term B 1 in (3.28) and let us set a = ∂ t κ(∂ 2

x Φ • κ). We have,

T a κ = j≥4 S j-3 (a)∆ j (κ) = j≥4 2 -j S j-3 (a) φ(2 -j D)(∂ x κ) = j≥4 g j where φ ∈ C ∞ (R), supp φ ⊂ { 1 2 ≤ |ξ| ≤ 2}. Since ∂ x κ = 1 + f (∂ x η) with f (0) = 0, we have ∆ j (∂ x κ) = ∆ j (f (∂ x η)) so, g j L 2 (R) ≤ 2 -j a L ∞ (R) 2 -j(s-1 2 ) c j C( η H s+ 1 2 (R)
), (c j ) ∈ l 2 .

On the other hand using (3.8), (3.9) we can write,

a L ∞ (I×R) ≤ Φ L ∞ (I,H s (R)) ∂ t χ(∂ x χ) -1 L ∞ (I×R) ≤ Φ L ∞ (I,H s (R)) C( (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R)
).

It follows that,

(3.29) B 1 ≤ C( (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R) ) Φ L ∞ (I,H s (R)) .
Let us consider the term B 2 .

We have,

∂ t κ = ab where a = ∂ t χ ∈ Γ 0 1 , b = ∂ x κ ∈ Γ 0 1 . It follows from Theorem 2.6 that a#b = ab and T ab -T a T b is of order -1. Let us set (3.30) B 21 = (T ab -T a T b )(∂ x Φ • κ) L ∞ (I,H s (R)) .
Using (2.5) we obtain,

B 21 ≤ ∂ t χ L ∞ (I,W 1,∞ (R)) ∂ x κ L ∞ (I,W 1,∞ (R)) (∂ x Φ • κ) L ∞ (I,H s-1 (R)) .
Since s -3 2 > 1, using (3.16) with µ = s -1 we obtain, (3.31)

B 21 ≤ C η L ∞ (I,H s+ 1 2 (R)) Φ L ∞ (I,H s (R)) .
Therefore using (3.28), (3.29), (3.31) we obtain,

(3.32) κ * (∂ t Φ) = ∂ t κ * Φ + T ∂tχ T ∂xκ ∂ x Φ • κ + R 2 (Φ),
where R 2 satisfies (3.26). Now let us set

a = ∂ x κ ∈ L ∞ (I, H s-1 2 (R)), b = ∂ x Φ • κ ∈ L ∞ (I, H s-1 (R)). It follows from (2.7) that (3.33) ab -T a b -T b a L ∞ (I,H 2s-2 (R)) ≤ a L ∞ (I,H s-1 2 (R)) b L ∞ (I,H s-1 (R)) .
Therefore we obtain

(3.34) κ * (∂ t Φ) = ∂ t (κ * Φ) -T ∂tχ ∂ x (Φ • κ) + T ∂tχ T ∂xΦ•κ ∂ x κ + R 3 ,
where R 3 satisfies (3.26). Using (3.1) we obtain

κ * (∂ t Φ) = (∂ t -T ∂tχ ∂ x )((κ * Φ) -T ∂tχ ∂ x (T ∂xΦ•κ ) + T ∂tχ T ∂xΦ•κ ∂ x κ + R 3 ,
where R 3 satisfies (3.26). It follows that

κ * (∂ t Φ) = (∂ t -T ∂tχ ∂ x )((κ * Φ) -T (∂ 2 x Φ•κ)∂xκ κ + R 3 . Now the term T (∂ 2
x Φ•κ)∂xκ κ can be estimated exactly by the same method as the term B 1 , therefore we obtain

κ * (∂ t Φ) = (∂ t -T ∂tχ ∂ x )(κ * Φ) + R 4
, where R 4 satisfies (3.26). This is precisely (3.27). Now we claim that

(3.35) κ * (T V ∂ x Φ) = T (V ∂xχ)•κ ∂ x κ * Φ + R 5 (Φ),
where R 5 satisfies (3.26). But this is precisely a consequence of Theorem 3.1. Indeed we have for (almost all) fixed t, a(x, ξ) = iV (t, x)ξ ∈ Σ 1 s-1 , and the diffeomorphism κ is in Let us consider now the principal part. Applying again Theorem 3.1 we find that,

W s-3 2 (R), so σ = s-
κ * (|D x | 3 4 T c |D x | 3 4 Φ) = |D x | 3 2 κ * Φ + T a κ * Φ,
where a is of order 1 2 . Finally, it remains to reduce to the case where a = 0. Indeed, let g be a real-valued symbol such that ∂ x g ∈ Γ 0 s-3/2 (R) and {|ξ| 3/2 , g} = -a,

then if we set (3.36) u = T e ig Φ * ,
we obtain by symbolic calculus that u satisfies

(∂ t + T W ∂ x + i |D x | 3 2 + iT a + T b )u = F, with F ∈ L ∞ (I, H s (R d )
) and b = i{|ξ| 3/2 , g}. This completes the proof of Proposition 3.4.

Regularity of W.

-The following result gives some informations on the function W defined in (3.23).

Lemma 3.5.

-Let I = [0, T ], E = L ∞ (I × R), F = L ∞ (I, H s-2 (R)). 1. If s > 2 + 1 2 , we have W ∈ E, ∂ x W ∈ F , and W E + ∂ x W F ≤ C( (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R))
).

If s > 4, we have

∂ t W, ∂ 2 x W, ∂ t ∂ x W ∈ E and ∂ t W E + ∂ 2 x W E + ∂ t ∂ x W E ≤ C( (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R))
).

Proof. -Let us recall that we have set,

(3.37) A = C (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R)
where C : R + → R + is an increasing function which may change from place to place. Since s > 2 + 1 2 using (2.7) we obtain, (3.38)

V E ≤ ∂ x ψ L ∞ (I,H s-2 (R)) + B L ∞ (I,H s-2 (R)) ∂ x η L ∞ (I,H s-2 (R)) ≤ A.
Then the estimate W E ≤ A follows fom (3.8) and (3.9) . Now we have

(3.39) ∂ x W = ∂ x V • κ + V • κ(∂ 2 x χ • κ)∂ x κ + (∂ t ∂ x χ • κ)∂ x κ. Using (3.16) we see that, (3.40) V • κ F + ∂ x V • κ F ≤ A V L ∞ (I,H s-1 (R)) ≤ A.
Now using (3.11), (3.17) and the fact that H s-2 (R) is an algebra we deduce,

(3.41) V • κ(∂ 2 x χ • κ)∂ x κ F ≤ A. Then the estimate ∂ x W F ≤ A follows from (3.

18) and (3.11).

Let us now prove 2. We have

(3.42) ∂ t W = ∂ t V • κ(∂ x χ • κ) + ∂ x V • κ(∂ x χ • κ)∂ t κ + V • κ(∂ t ∂ x χ • κ) + V • κ(∂ 2 x χ • κ)∂ t κ -∂ 2 t χ • κ -∂ t ∂ x χ • κ(∂ x κ) =: 6 i=1 B i .
It follows from (3.16),(3.9),(3.10),(3.17), (3.18), and the Sobolev embedding that

(3.43) |B 2 | + |B 3 | + |B 4 | + |B 6 | ≤ A. Now we have ∂ t V = ∂ x ∂ t ψ -(∂ t B)∂ x η -B∂ x ∂ t η.
So using the equations satisfied by (η, ψ), the Sobolev embedding and Lemma 2.9 we obtain

(3.44) ∂ t V L ∞ (I,H s-5 2 (R)) ≤ A.
It follows that

(3.45) |B 1 | ≤ A.
The term B 5 is estimated by A using (3.21). Therefore using (3.43) and (3.45) we deduce that

∂ t W E ≤ A.
The claim on ∂ 2

x W follows from the first part of the Lemma and the Sobolev embedding since s > 3 + 1 2 . It remains to consider the quantity

∂ t ∂ x W. We go back to (3.42). The term ∂ t V • κ(∂ x χ • κ) is bounded by A in L ∞ (I, H s-5 2 (R)). The third term V • κ(∂ t ∂ x χ • κ) is bounded by A in L ∞ (I, H s-2 (R)). The term ∂ t ∂ x χ • κ(∂ x κ) is bounded by A in L ∞ (I, H s-2 (R)). Therefore the ∂ x derivative of these three terms are bounded by A in L ∞ (I, H s-7 2 (R))
. By (3.8) we have,

∂ x V • κ(∂ x χ • κ)∂ t κ L ∞ (I×R) ≤ A ∂ x V • κ(∂ x χ • κ) L ∞ (I×R) ≤ A ∂ x V • κ(∂ x χ • κ) L ∞ (I,H s-2 ) ≤ A.
We can apply the same argument for the term

V •κ(∂ 2 x χ•κ)∂ t κ. Finally we bound the term V • κ(∂ x ∂ 2 tt χ • κ) in the space L ∞ (I, H s-7 2 (R)
) by using (3.16) and (3.22).This completes the proof of our Lemma.

Symbol

Smoothing. -In this section we follow an idea of Smith [START_REF] Smith | A parametrix construction for wave equations with C 1,1 coefficients[END_REF] (see also ), and we are going to smooth out the coefficients of the function W with respect to x. As already mentioned, here is the main place where the idea of allowing loss in remainder terms enters. We define for 0 < δ ≤ 1,

T δ W = j≥4 S [δ(j-3)] (W )∆ j ,
The key difference between T W and T δ W is made clear below.

Lemma 3.6. -The operator T W -T δ W is of order -δ(s -3 2 ).
Proof. -Since for almost all fixed t we have,

∂ x W (t, •) ∈ H s-2 (R) we have, S j (W ) -S [δj] (W ) L ∞ (R) ≤ j n=[δj] ∆ n (W ) L ∞ (R) ≤ K j n=[δj] 2 -n(s-3 2 ) ≤ K2 -δj(s-3 2 ) .
In the sequel we shall set

(3.46)        h = 2 -j , j ∈ N, W δ h = S [δ(j-3)] (W ), a(ξ) = χ 0 (ξ)|ξ| 3 2 , where χ 0 ∈ C ∞ 0 (R), supp χ 0 ⊂ { 1 4 ≤ |ξ| ≤ 4}, χ 0 = 1 in { 1 2 ≤ |ξ| ≤ 2}. Lemma 3.7. -Let s > 2 + 1 2 . There exist δ < 1 2 , ǫ > 0 and f h ∈ L ∞ (I, H s+ǫ-1 2 (R)) such that (3.47) f h L ∞ (I,H s+ǫ-1 2 (R)) ≤ C (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R)) , supp( f h ) ⊂ { 1 2 h -1 ≤ |ξ| ≤ 2h -1 }
and the functions u h = ∆ j u satisfy

(3.48) (∂ t + 1 2 (W δ h ∂ x + ∂ x W δ h ) + ia(D x ))u h = f h Proof.
-First of all we remark that we have

|D x | 3 2 u h = a(D x )u h . Now, applying the operator ∆ j to (3.25), we obtain (3.49) (∂ t + T W ∂ x + i |D x | 3 2 )u h = ∆ j f -[∆ j , T W ]∂ x u := g 1 h
Since by Lemma 3.5 we have

∂ x W ∈ L ∞ (I, H s-2 (R)), it follows from Lemma 2.8 that g 1 h satisfies (3.47) (for any 0 < ǫ ≤ 1 2
). Then we can replace T W by T δ W , which gives

(3.50) (∂ t + |k-j|≤1 S [δ(k-3)] (W )∆ k ∂ x + i |D x | 3 2 )u h = g 1 h + (T δ W -T W )∂ x u h := g 1 h + g 2 h
where, according to Lemma 3.6,

g 2 h satisfies (3.47) with ǫ = δ(s -3 2 ) -1 2 > 0 if δ < 1 2 is chosen close enough to 1 2 . Now, we have S [δ(j-3)] (W )∂ x u h = |k-j|≤1 S [δ(j-3)] (W )∆ k ∂ x u h .
Consequently, we obtain

(3.51) (∂ t + S [δ(j-3)] (W )∂ x + i |D x | 3 2 )u h = g 1 h + g 2 h + |k-j|≤1 (S [δ(j-3)] (W ) -S [δ(k-3)] (W ))∆ k ∂ x u h = g 1 h + g 2 h + g 3 h ,
and using that for |k -j| ≤ 1,

S [δ(k-3)] (W ) -S [δ(j-3)] (W ) L ∞ ≤ C2 -jδ(s-3 2 ) ,
we obtain that g 3 h satisfies (3.47). Finally, we obtain

(∂ t + 1 2 (W δ h ∂ x + ∂ x W δ h ) + i |D x | 3 2 )u h = g 1 h + g 2 h + g 3 h + g 4 h ,
where

g 4 h = 1 2 S [δ(j-3)] (∂ x W )u h satisfies (3.47) (for any 0 < ǫ ≤ 1 2 ). Lemma 3.8. -Let s > 11 2 and set δ = 1 s -3 2 ∈]0, 1 4 [.

Then there exists

f h ∈ L ∞ (I, H s (R)) such that (3.52) f h L ∞ (I,H s (R)) ≤ C (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R)) , supp( f h ) ⊂ { 1 2 h -1 ≤ |ξ| ≤ 2h -1 }
and the functions u h = ∆ j u satisfy

(3.53) (∂ t + 1 2 (W δ h ∂ x + ∂ x W δ h ) + ia(D x )u h = f h Proof.
-The proof is identical to that of Lemma 3.7, the only difference beeing that now we take δ such that δ(s -3 2 ) = 1.

Semi-classical parametrix

Following [START_REF] Burq | Strichartz inequalities and the nonlinear Schrdinger equation on compact manifolds[END_REF] we shall reduce the analysis to establishing semi-classical estimates.

Recall that 2 -j = h and

W δ h = S [δ(j-3)] (W ) = γ(h δ D x )W , γ ∈ C ∞ 0 (R). Theorem 4.1. -Let χ ∈ C ∞ 0 (R ) with supp χ ⊂ {ξ : 1 2 ≤ |ξ| ≤ 2} and t 0 ∈ R. For any initial data u 0,h = χ(hD x )u 0 , where u 0 ∈ L 1 (R), let U h := S(t, t 0 , h)u 0,h be the solution of (4.1) ∂ t U h + 1 2 (W δ h ∂ x + ∂ x W δ h )U h + ia(D x )U h = 0, U h | t=t 0 = u 0,h .
Then there exists ε > 0 such that for any 0 < h ≤ 1 and any |t -

t 0 | ≤ h 1 2 -ε , (4.2) S(t, t 0 , h)u 0,h L ∞ (R) ≤ C h 1/4 |t -t 0 | 1/2 u 0,h L 1 (R) .
To prove this result, we shall follow a very classical trend and construct a parametrix. Notice that our assumptions being time-translation invariant we can assume t 0 = 0. The parametrix will take the following form,

(4.3) U h (t, x) = 1 2πh e i h (Φ(t,x,ξ,h)-zξ) B(t, x, z, ξ, h)u 0,h (z)dzdξ,
where Φ will satisfy the eikonal equation and

(4.4) B(t, x, z, ξ, h) = B(t, x, ξ, h)ζ(x -z -th -1 2 a ′ (ξ)),
where B will satisfy the transport equations and

ζ ∈ C ∞ 0 (R), ζ(s) = 1 if |s| ≤ 1, ζ(s) = 0 if |s| ≥ 2.
In addition to χ we shall use two more cut-off functions

χ j ∈ C ∞ 0 (R), j = 1, 2, such that (4.5)      supp χ 1 ⊂ {ξ : 1 3 ≤ |ξ| ≤ 3}, χ 1 = 1 on supp χ, supp χ 0 ⊂ {ξ : 1 4 ≤ |ξ| ≤ 4}, χ 0 = 1 on supp χ 1 .
4.1. The eikonal and transport equations. -We introduce some space of symbols in which we shall solve our equations.

Definition 4.2. -For small ε, h 0 to be fixed, we introduce the sets

Ω ε = (t, x, ξ, h) ∈ R 4 : h ∈ (0, h 0 ), |t| < h 1 2 -ε , 1 < |ξ| < 3 , O ε = (σ, x, ξ, h) ∈ R 4 : h ∈ (0, h 0 ), |σ| < h -ε , 1 < |ξ| < 3 . If m ∈ R and ̺ ∈ R + , we denote by S m ̺,ε (Ω ε ) (resp.S m ̺,ε (O ε ))
the set of all functions f on Ω ε which are C ∞ with respect to (t, x, ξ) (resp.(σ, x, ξ)) and satisfy the estimate

(4.6) |∂ α x f (t, x, ξ, h)| (resp. |∂ α x f (σ, x, ξ, h)|) ≤ C α h m-̺α , for all (t, x, ξ, h) ∈ Ω ε (resp.(σ, x, ξ, h) ∈ O ε ). Remark 4.3. -(i) If f ∈ S m ̺,ε , g ∈ S m ′ ̺,ε then f g ∈ S m+m ′ ̺,ε ; if f ∈ S m ̺,ε , (m ≥ 0) and F ∈ C ∞ (C) then F (f ) ∈ S m ̺,ε ; if f ∈ S m ̺,ε , (m ≤ 0) and F ∈ C ∞ b (C) then F (f ) ∈ S 0 ̺-m,ε . Let f ∈ S µ ̺,ε , then ∂ x f ∈ S µ-̺ ̺,ε . Moreover S m ̺,ε ⊂ S m ̺ ′ ,ε if ̺ ≥ ̺ ′ (ii) Let W be such that ∂ x W ∈ H s-2 (R) with s > 2 + 1 2 and set W ̺ h = γ(h ̺ D x )W where γ ∈ S(R). Then ∂ x W ̺ h ∈ S 0 ̺,ε .
Let δ ∈ (0, 1 2 ). We fix (4.7)

µ 0 = 1 2 1 2 -δ , ε ∈ (0, µ 0 5 ).
Until the end of this section, for the simplicity of notations we shall drop the index ε and denote by S m ̺ (Ω) the space S m ̺,ε (Ω ε ). Finally we set, (4.8)

L 0 = ∂ t + 1 2 (W δ h ∂ x + ∂ x W δ h ) + iχ 0 (hD x ) |D x | 3 2 , a(ξ) = χ 0 (ξ) |ξ| 3 2 .
The main result of this section is the following.

Proposition 4.4. -There exist a phase Φ of the form

Φ(t, x, ξ, h) = xξ -h -1 2 ta(ξ) + h 1 2 Ψ(t, x, ξ, h) with ∂ x Ψ ∈ S -ε δ (Ω)
and an amplitude B ∈ S 0 δ+3ε (Ω) such that, with B defined in (4.4), (4.9)

L 0 e i h Φ B = e i h Φ R h .
and for all N ∈ N we have,

(4.10) e i h (Φ(t,x,ξ,h)-zξ) R h (t, x, z, ξ, h)u 0,h (z) dz dξ H 1 (Rx) ≤ C N h N u 0,h L 1 (R) , for all t in [0, h 1 2 -ε ].
Proof. -We set, (4.11)

t = h 1 2 σ, ϕ(σ, x, ξ, h) = Φ(σh 1 2 , x, ξ, h), b(σ, x, ξ, h) = B(σh 1 2 , x, ξ, h), V h (σ, x, h) = W δ h (σh 1 2 , x, h), L = h∂ σ + 1 2 h 1 2 V h (h∂ x ) + h∂ x V h + ia(hD x ).
Multiplying (4.9) by h

e i h (ϕ(σ,x,ξ,h)-zξ) r(σ, x, z, ξ, h)u 0,h (z) dz dξ H 1 (Rx) ≤ C N h N u 0,h L 1 (R) .
In the proof of (4.12), z, ξ, h will be considered as parameters.

We shall take ϕ of the form

(4.14) ϕ(σ, x, ξ, h) = xξ -σa(ξ) + h 1 2 ψ(σ, x, ξ, h),
where ψ is the solution of the problem

(4.15) ∂ σ ψ + a ′ (ξ)∂ x ψ = -ξV h , ψ| σ=0 = 0.
Differentiating (4.15) with respect to x and ξ, using an induction on k and the fact that

∂ x V h ∈ S 0 δ (O), we see easily that, (4.16) |∂ k ξ ∂ α x ψ(σ, x, ξ, h)| ≤ C kα |σ|h -kε h -δ(α+k-1) + , for every (σ, x, ξ, h) ∈ O, where a + = sup(a, 0). It follows in particular that ∂ x ψ ∈ S -ε δ (O), ∂ σ ψ ∈ S -ε δ (O) . Now, since b = b ζ we have, (4.17) e -i h ϕ h∂ σ + h 3 2 2 (V h ∂ x + ∂ x V h ) (e i h ϕ b) = i[h 1 2 ξV h -a(ξ) + h 1 2 ∂ σ ψ + hV h ∂ x ψ] b + h[∂ σ b + h 1 2 V h ∂ x b + 1 2 h 1 2 (∂ x V h )b]ζ + h[-a ′ (ξ) + h 1 2 V h ]b ζ ′ .
On the other hand recall that we have for all M ∈ N * (see the appendix),

(4.18) e -i h ϕ a(hD x ) e i h ϕ b = A + r 1 + r 2 ,
where

(4.19) A = M -1 k=0 h k i k k! ∂ k y (∂ k ξ a) (ρ(x, y)) b(y)    y=x . with (4.20) ρ(x, y) = 1 0 ∂ϕ ∂x (σ, λx + (1 -λ)y, ξ, h) dλ,
and the remainder r 1 , r 2 are given by, (4.21)

r 1 = c h M -1 1 0 e i h (x-y)η κ 0 (η)(1 -λ) M -1 ∂ M y [a (M ) (λη + ρ((x, y)) b(y)]dλdydη and (4.22) r 2 = M -1 k=0 c k,M h M +k 1 0 z M κ0 (z)(1 -λ) M -1 ∂ M +k y [a (k) (θ) b]| y=x-λhz dλdz, where c M , c k,M ∈ C, κ 0 ∈ C ∞ 0 (R), κ 0 = 1 in a neighborhood of the origin. Now since b(σ, x, z, ξ, h) = b(σ, x, ξ, h)ζ(x -z -σa ′ (ξ)), writing for simplicity b(y) = b(σ, y, ξ, h) and ζ = ζ(x -z -σa ′ (ξ)) we have, (4.23)                        A = ( M -1 k=0 A k )ζ + r 3 , A k = h k i k k! ∂ k y (∂ k ξ a) (ρ(x, y)) b(y) | y=x , r 3 = M -1 k=1 k j=1 c jk h k ∂ k-j y (∂ k ξ a)(ρ(x, y))b(y) | y=x ζ (j) .
The term A 0 in (4.23) is equal to a(ξ + h

1 2 ∂ x ψ)b. Then (4.24) A 0 =   2 j=0 1 j! a (j) (ξ)(h 1 2 ∂ x ψ) j + (h 1 2 ∂ x ψ) 3 2 1 0 (1 -λ) 2 ∂ 3 ξ a(ξ + λh 1 2 ∂ x ψ) dλ   b.
The term A 1 in (4.23) can be written as

A 1 = h i a ′ (ξ + h 1 2 ∂ x ψ)∂ x b + 1 2 h 1 2 (∂ 2 x ψ)a ′′ (ξ + h 1 2 ∂ x ψ)b .
Therefore (4.25)

A 1 = h i a ′ (ξ) + h 1 2 ∂ x ψ 1 0 a ′′ (ξ + λh 1 2 ∂ x ψ) dλ ∂ x b + 1 2 h 1 2 (∂ 2 x ψ)a ′′ (ξ + h 1 2 ∂ x ψ)b . Since h ε ∂ x ψ ∈ S 0 δ , h ε+δ ∂ 2
x ψ ∈ S 0 δ and 5ε ≤ µ 0 , we deduce from (4.24) and (4.25) that (4.26)

A 0 + A 1 = a(ξ) + h 1 2 a ′ (ξ)∂ x ψ + h 2 a ′′ (ξ)(∂ x ψ) 2 b + h i a ′ (ξ)∂ x b + hh µ 0 h ε (c 1 b + c 2 h 3ε+δ ∂ x b)
for some c j ∈ S 0 δ . Now, consider the term A k with k ≥ 2. We have

A k = h k i k k! k k 1 =0 k k 1 ∂ k 1 y (∂ k ξ a)(ρ(x, y))   y=x ∂ k-k 1 x b.
Since h

1 2 ∂ x ψ ∈ S 0 δ , we obtain, c k,k 1 := h k 1 δ ∂ k 1 y (∂ k ξ a)(ρ(x, y))   y=x ∈ S 0 δ .
It follows that the generic term in A k can be written as

hh k-1 h -k 1 δ c k,k 1 h -ε h -(δ+3ε)(k-k 1 ) h ε (h δ+3ε ∂ x ) k-k 1 b.
We have, since 3ε < µ 0 and k ≥ 2,

(4.27) k -1 -k 1 δ -ε -(δ + 3ε)(k -k 1 ) ≥ k(1 -δ -3ε) -1 -ε ≥ 2(1 -δ -3ε) -1 -ε ≥ 2( 1 2 -δ -3ε) -ε ≥ 1 2 -δ -ε ≥ µ 0 so that A k = hh µ 0 h ε k ℓ=0 c ℓ (h δ+3ε ∂ x ) ℓ b, c ℓ ∈ S 0 δ .
We deduce from (4.26) that (4.28)

M -1 k=0 A k =   2 j=0 1 j! a (j) (ξ)(h 1 2 ∂ x ψ) j   b + h i a ′ (ξ)∂ x b + hh µ 0 h ε M -1 ℓ=0 d ℓ (h δ+3ε ∂ x ) ℓ b
with d ℓ ∈ S 0 δ . Then it follows from (4.12),(4.17),(4.18) and (4.28) that

r = i -a(ξ) + h 1 2 ξV h + h 1 2 ∂ σ ψ + hV h ∂ x ψ b ζ + h[∂ σ b + h 1 2 V h ∂ x b + 1 2 h 1 2 (∂ x V h )b]ζ + i a(ξ) + h 1 2 a ′ (ξ)∂ x ψ + h 2 a ′′ (ξ)(∂ x ψ) 2 b ζ + ha ′ (ξ)∂ x b ζ + hh µ 0 h ε M -1 ℓ=0 d ℓ (h δ+3ε ∂ x ) ℓ b ζ + 3 j=1 r j .
Gathering the terms in powers of h, noting that the coefficient of h 0 vanishes and using the eikonal equation to see that the coefficient in h 1 2 vanishes, we are left with

(4.29) r = h ∂ σ b + a ′ (ξ)∂ x b + if b + h µ 0 h ε M -1 ℓ=0 e ℓ (h δ+3ε ∂ x ) ℓ b ζ + i 4 j=1 r j .
where

f = V h ∂ x ψ + a ′′ (ξ)(∂ x ψ) 2 is real-valued, e ℓ ∈ S 0 δ and (4.30) r 4 = 1 i h[-a ′ (ξ) + h 1 2 V h ]b ζ ′ .
It follows from (4.16) that

(4.31) |∂ α x ∂ k ξ f (σ, x, ξ, h)| ≤ C kα σh -ε h -kε h -δ(α+k) ,
for every (σ, x, ξ, h) ∈ O. In particular f ∈ S -2ε (O). Now we shall seek b under the form

(4.32) b = J-1 j=0 h jµ 0 b j .
where the b ′ j s are the solutions of the following problems (4.33)

   ∂b 0 ∂σ + a ′ (ξ) ∂b 0 ∂x + if b 0 = 0, b 0 | σ=0 = χ 1 (ξ),
where χ 1 ∈ C ∞ 0 (R) has been introduced in (4.5) and (4.34)

       ∂b j ∂σ + a ′ (ξ) ∂b j ∂x + if b j = -h ε M -1 ℓ=0 e ℓ (h δ+3ε ∂ x ) ℓ b j-1 , b j | σ=0 = 0.
It is easy to see that for all j we have,

(4.35) b j (σ, x, ξ, h) = χ 1 (ξ)c j (σ, x, ξ, h).
For the estimates we shall use the following elementary lemma.

Lemma 4.5.

-If u is a solution of the problem

∂ σ u + a ′ (ξ)∂ x u + if u = g, u| σ=0 = z ∈ C,
where f be real-valued, then it satisfies the estimate

|u(σ, x, ξ, h)| ≤ |z| + σ 0 |g(σ ′ , x + (σ ′ -σ)a ′ (ξ), ξ, h)|dσ ′ for every (σ, x, ξ, h) ∈ O.
Proof. -Indeed, the solution is given by

u(σ, x, ξ, h) = e -i σ 0 f (σ ′ ,x+(σ ′ -σ)a ′ (ξ),ξ,h) dσ ′ × × z + h ε σ 0 e i σ ′ 0 f (t,x+(t-σ)a ′ (ξ),ξ,h) dt g(σ ′ , x + (σ ′ -σ)a ′ (ξ), ξ, h) dσ ′ .
Using this lemma we deduce the following.

Lemma 4.6. -The problems (4.33), (4.34) have unique solutions b j = χ 1 (ξ)c j where the c j satisfy the estimates

(4.36) |∂ α x ∂ k ξ c j (σ, x, ξ, h)| ≤ C αkj h -kε h -(α+k)(δ+3ε
) for all (σ, x, ξ, h) ∈ O, all α, k ∈ N, and all j = 0, ..., M.

In particular c = M j=0 c j belongs to S 0 δ+3ε (O).

∂U ∂σ + a ′ (ξ) ∂U ∂x + if U = -i α l=1 C α l (∂ l x f )∂ α-l x c 0 .
Using (4.31), Lemma 4.5 and the induction we deduce that

|U | ≤ Ch -ε α l=1 h -2ε h -l(δ+3ε) h -(α-l)(δ+3ε) ≤ Ch -α(δ+3ε) .
This proves (4.36) for k = 0 and all α. Then using an induction on k we differentiate the equation (4.37) k times wit respect to ξ we use again (4.31), Lemma 4.

(4.39) x -z -σa ′ (ξ) |r j (σ, x, z, ξ, h)| ≤ C N h N |χ 1 (ξ)|.
To prove (4.39) we write in the integral giving r 1 (resp.r 2 ) , xzσa ′ (ξ) = (xy) + (yzσa ′ (ξ)), (resp. = (xzλhzσa ′ (ξ)) + λhz) we integrate by parts using the fact that h i ∂ η e with c ∈ S 0 δ+3ε and δ + 3ε < 1. Then (4.13) follows from (4.39) and Young's inequality. The terms corresponding to r 3 and r 4 defined in (4.23) and (4.30) will be treated in the same manner and will use the fact that on the support of a derivative of the function ζ one has |xzσa ′ (ξ)| ≥ 1. Since by (4.16) we have h

1 2 |∂ ξ ψ| ≤ Ch 1 2 |σ| ≤ Ch 1 2 -ε we deduce from (4.14) that |∂ ξ (ϕ(σ, x, ξ, h) -zξ)| ≥ 1
2 if h is small enough. Therefore we can integrate N times by parts using the vector field

L = h i(∂ ξ (ϕ(σ, x, ξ, h) -zξ)) ∂ ξ .
Finally the estimate for the term r 5 follows easily from the fact that the convolution of

L 1 (R) with L 2 (R) is contained in L 2 (R).
The proof of Proposition 4.4 is complete. In the variable σ = th -1 2 we have

K(t, x, z, h) = K(σ, x, z, h)
where

K(σ, x, z, h) = 1 2πh e i h (ϕ(σ,x,ξ,h)-zξ) b(σ, x, z, ξ, h) dξ,
where ϕ and b have been determined in (4.14), (4.33) and (4.34).

Proposition 4.7. -There exists C > 0 such that

(4.43) |K(σ, x, z, h)| ≤ C h h σ 1 2
,

for all (σ, x, z, h) in ]0, h -ε [×R × R×]0, h 0 [. Proof.
-Since b ∈ S 0 δ is bounded with compact support in ξ, the estimate (4.43) is trivial for |σ| ≤ Ch. Let us assume that |σ| ≥ Ch. We have by (4.11),

L = h∂ σ + h 1 2 W δ h (h 1 2 σ, x)(h∂ x ) 1 2 h 3 2 (∂ x W δ h ) + i |hD x | 3 2
By a scaling argument we can assume without loss of generality that σ ≥ 1. Indeed, otherwise, setting

τ = σ σ 0 , x = x σ 0 , h = h σ 0 ,
we see that in the new variables, the operator reads

L = h∂ τ + h 1/2 W δ h h∂ x + h 3 2 (∂ x W δ h ) + i| hD x | 3/2 where W δ h (τ, x) = σ 1/2 0 W δ h (σ 0 τ, σ 0 x
) and consequently we have

W δ h ∈ L ∞ (H s-1 ), ∂ x W δ h ∈ S 0 δ
with bounds uniform with respect to σ 0 . Assume now that the dispersion estimate has been proved for the kernel of the operator L and σ = 1. Since we have

S h (σ)u 0 (x) = ( S h ( σ σ 0 ) u 0 )( x σ 0 ), u 0 ( x σ 0 ) = u 0 (x),
we can write

(4.44) S h (σ 0 )u 0 L ∞ (R) = S h ( σ σ 0 ) u 0 L ∞ (R) ≤ C | h| 1/2 u 0 L 1 (R) ≤ C|σ 0 | 1/2 |h| 1/2 |σ 0 | u 0 L 1 (R) ≤ C |hσ 0 | 1/2
which is the dispersion estimate for the kernel of the operator L and σ = σ 0 .

Let us set

(4.45) θ(x, y, ξ, h) = ϕ(σ, x, ξ, h) -zξ = (x -z)ξ + a(ξ)σ + h 1 2 ψ(σ, x, ξ, h). Then ∂ 2 ξ θ(x, y, ξ, h) = ∂ 2 ξ a(ξ)σ + h 1 2 ∂ 2 ξ ψ(σ, x, ξ, h
). Now by (4.5) and (4.8), on the support of χ 1 we have a(ξ

) = |ξ| 3 2 . Therefore ∂ 2 ξ a(ξ) = 3 4 |ξ| -1 2 ≥ c 0 > 0.
On the other hand from (4.16) we have |∂ 2 ξ ψ| ≤ Cσh -δ-2ε which implies h

1 2 |∂ 2 ξ ψ| ≤ Cσh 1 2 -δ-2ε ≤ Cσh µ 0 . It follows that on the support of χ 1 one can find a constant c 1 > 0 such that (4.46) 0 < c 1 σ ≤ ∂ 2 ξ θ(x, y, ξ, h) ≤ 1 c 1 σ, if h 0 is small enough.
In the sequel we shall omit to note (x, z, h) which are fixed taking care of the fact that all the constants are independent of (x, z, h) ∈ R × R×]0, h 0 [. Let us denote by [α, β] ⊂ [ 1 3 , 3] the support of χ 1 .We deduce from (4.46) that the function ξ → ∂ ξ θ(ξ) is increasing on [α, β]. Therefore one can find ρ ∈ [α, β] such that

∂ ξ θ(ξ) ≤ 0 for ξ ∈ [α, ρ], ∂ ξ θ(ξ) ≥ 0 for ξ ∈ [ρ, β].
Noting b(σ, x, ξ, h) = b(ξ) and assuming that ]ρ, β[ is non empty, we shall estimate

K + (σ, x, ξ, h) = 1 2πh β ρ e i h θ(ξ) b(ξ)ζ(x -z -a ′ (ξ))dξ,
the estimate corresponding to the intervall [α, ρ] being similar. We write for small h, (4.47)

                   K + = 1 2πh (I 1 + I 2 ), I 1 = ρ+( h σ ) 1 2 ρ e i h θ(ξ) b(ξ)ζ(x -z -a ′ (ξ))dξ I 2 = β ρ+( h σ ) 1 2 e i h θ(ξ) b(ξ)ζ(x -z -a ′ (ξ))dξ
We have obviously, (4.48)

|I 1 | ≤ C( h σ ) 1 2
In the integral I 2 using (4.46) and the Taylor formula we see that, (4.49)

∂ ξ θ(ξ) ≥ c 1 σ h σ 1 2 = C 1 (hσ) 1 2 , ∀ξ ∈ [ρ + ( h σ ) 1 2 , β].
Let us estimate the integral I 2 . We introduce the following notation. We shall write

A ⋊ ⋉ B if |A-B| ≤ C h σ 1 2
where C is a constant depending only q. Then we can state the following lemma which is a refined version of the well known Van der Corput Lemma.

I 2 ⋊ ⋉ (-1) k h i k β ρ+( h σ ) 1 2 e h i θ(ξ) 1 (∂ ξ θ(ξ)) k ∂ k ξ q(ξ)dξ where q(ξ) = b(ξ)ζ(x -z -a ′ (ξ).
Proof. -Let us denote by J k the term in the right hand side of (4.50). The Lemma is true for k = 0. Assume it is true up to the order k. Using the fact that

h i∂ ξ θ(ξ) ∂ ξ e i h θ(ξ) = e i h θ(ξ)
and integrating by parts in J k we obtain,

J k = (-1) k+1 h i k+1 β ρ+( h σ ) 1 2 e h i θ(ξ) ∂ ξ 1 (∂ ξ θ(ξ)) k+1 ∂ k ξ q(ξ)dξ + (-1) k+1 h i k+1 β ρ+( h σ ) 1 2 e h i θ(ξ) 1 (∂ ξ θ(ξ)) k+1 ∂ k+1 ξ q(ξ)dξ + (-1) k+1 h i k+1 [e h i θ(ξ) 1 (∂ ξ θ(ξ)) k+1 ∂ k ξ q(ξ)] β ρ+( h σ ) 1 2 = J 1 k + J 2 k + J 3 k .
First of all we have J 2 k = J k+1 . Now using (4.36) and (4.49) we can write,

|J 3 k | ≤ Ch k+1 h -k(δ+4ε) (hσ) k+1 2 ≤ C h σ 1 2 h k( 1 2 -δ-4ε) ≤ C h σ 1 2 ,
since σ ≥ 1. Now using again (4.36) we obtain,

|J 1 k | ≤ Ch k+1-k(δ+4ε) β ρ+( h σ ) 1 2 ∂ ξ 1 (∂ ξ θ(ξ)) k+1 dξ.
Since by (4.46) the function ∂ ξ θ is increasing we have

∂ ξ 1 (∂ ξ θ(ξ)) k+1 = -∂ ξ 1 (∂ ξ θ(ξ)) k+1 .
Therefore we can write,

|J 1 k | ≤ Ch k+1-k(δ+4ε) [- 1 (∂ ξ θ(ξ)) k+1 ] β ρ+( h σ ) 1 2
.

We deduce exactly as for J 3 k that,

|J 1 k | ≤ C h σ 1 2 .
It follows that J k ⋊ ⋉ J 2 k which proves our induction. Now using Lemma 4.8, (4.36) and (4.49) we can write,

|J k | ≤ Ch k 1 (hσ) k 2 h -k(δ+4ε) ≤ C h σ 1 2 h k( 1 2 -δ-4ε)-1 2 ≤ C h σ 1 2 h kµ 0 -1 2 h k(µ 0 -4ε) ,
so taking k such that kµ 0 ≥

∂ t U h + 1 2 (W δ h ∂ x + ∂ x W δ h ) U h + ia(D x ) U h = F h , U h | t=0 = U h (0, x), with (4.52) sup s∈Jε F h (s, •) H 1 (R) ≤ C N h N u 0,h L 1 (R) .
We claim that,

(4.53) U h (0, •) = u 0,h + v 0,h , v 0,h H 1 (R) ≤ C N h N u 0,h L 1 (R) .
Indeed using (4.3),(4.11),(4.14),(4.33),(4.34) and (4.35) we see that,

(4.54) v 0,h (x) = (2πh) -1 e i h (x-z)ξ (1 -ζ(x -z))χ 1 (ξ)u 0,h (z)dzdξ.
Since on the support of 1ζ(xz) we have |x -z| ≥ 1 we can integrate by parts as much as we want to obtain that for all N ≥ 1,

v 0,h (x) = c N h N -1 e i h (x-z)ξ [ 1 -ζ(x -z) (x -z) N ](∂ N ξ χ 1 )(ξ)u 0,h (z)dzdξ.
Using the Hölder inequality we deduce that,

|v 0,h (x)| 2 ≤ C N h N -1 1 -ζ(x -z) (x -z) N 2 |u 0,h (z)|dz u 0,h L 1 (R)
from which we deduce that,

v 0,h L 2 (R) ≤ C N h N -1 u 0,h L 1 (R) .
Differentiating (4.54) with respect to x and using the same trick we obtain the estimate in (4.53). Now by (4.51), the Duhamel formula and the definition in Proposition 4.1 we can write, (4.55)

S(t, 0, h)u 0,h = D 1 + D 2 + D 3 where D 1 = U h (t, x), D 2 = -S(t, 0, h)v 0,h (x), D 3 = - t 0 S(t, s, h)[F h (s, x)]ds.
First of all the estimate (4.56)

D 1 (t) L ∞ (R) ≤ C h 1/4 |t| 1/2 u 0,h L 1 (R)
follows from Proposition 4.7 and (4.41).

Let us estimate D 2 . We have by Sobolev inequality,

D 2 (t) L ∞ (R) ≤ C 1 D 2 (t) H 1 (R) ≤ C 2 v 0,h H 1 (R) ,
therefore by (4.53), (4.57)

D 2 (t) L ∞ (R) ≤ C N h N u 0,h L 1 (R) .
Let us look now to the term D 3 . We have,

D 3 (t) L ∞ (R) ≤ C Jε S(t, s, h)F h (s, •) H 1 (R) ds ≤ C ′ Jε F h (s, •) H 1 (R) ds,
from which we deduce, (4.58)

D 3 (t) L ∞ (R) ≤ C N h N u 0,h L 1 (R) .
Then Theorem 4.1 follows from (4.55), (4.56), (4.57), and (4.58). Proposition 4.9. -There exist ε > 0, C > 0 such that for any 0 < h < 1 and any initial data u 0,h = χ(hD x )u 0 , we have (4.59) S(t, 0, h)u 0,h L 4 ((0,h

1 2 -ε ),L ∞ (R)) ≤ C u 0,h H 1 8 (R)
.

Proof. -Indeed, applying the usual T T * argument, it suffices to prove that the operator

h 1 2 -ε 0 S(t, 0, h)S(s, 0, h) * f (s)ds maps continuously L 4 3 ((0, h 1 2 -ε ), L 1 (R)) to L 4 ((0, h 1 2 -ε ), L ∞ (R)). But a direct calculation shows that since 1 2 (W δ h ∂ x + ∂ x W δ h ) is self adjoint, one has S(s, 0, h) * = S(0, s, h),
and consequently, Proposition 4.9 follows from the classical Hardy-Littlewood-Sobolev inequality and the dispersion estimate (4.2).

Corollary 4.10. -Let u be a solution of the problem

∂ t u + 1 2 (W δ h ∂ x + ∂ x W δ h )u + ia(D x )u = f, u| t=0 = 0 with supp f ⊂ { 1 2 h -1 ≤ |ξ| ≤ 2h -1 }. Then we have, u L 4 ((0,h 1 2 -ε ),L ∞ (R)) ≤ Kh -1/8 f L 1 ((0,h 1 2 -ε ),L 2 (R))
.

Proof. -Indeed we have,

u(t, •) = t 0 S(t, 0, h)S * (s, 0, h)f (s, •) ds. Let us set J ε = [0, h 1 2 -ε ].
It follows from Proposition 4.9 that,

u L 4 (Jε,L ∞ (R)) ≤ C h 1 2 -ε 0 S(s, 0, h) * f (s, •) H 1 8 (R) ds ≤ C ′ h 1 2 -ε 0 f (s, •) H 1 8 (R) ds ≤ C ′′ h -1/8 f L 1 (Jε,L 2 (R)) , since f is supported in { 1 2 h -1 ≤ |ξ| ≤ 2h -1 }.
4.5. Gluying the estimates. -It remains to glue the estimates which up to now have been proved on small time intervals of size h 1 2 -ε . Recall that from Lemma 3.7 we have

∂ t u h + 1 2 (W δ h ∂ x + ∂ x W δ h )u h + i |D x | 3 2 u h = f h ∈ L ∞ ((0, τ ); H s+ε-1 2 ). Let ϕ ∈ C ∞ 0 (0, 2), equal to 1 on ( 1 2 , 3 2 ). For -1 ≤ k ≤ h ε-1 2 , define u h,k = ϕ t -kh 1 2 -ε h 1 2 -ε u h , which satisfies (4.60) ∂ t u h,k + 1 2 (W δ h ∂ x u + ∂ x W δ h )u h,k + i |D x | 3 2 u h,k = ϕ t -kh 1 2 -ε h 1 2 -ε f h + h ε-1 2 ϕ ′ t -kh 1 2 -ε h 1 2 -ε u h , u h,k | t=kh 1
2 -ε = 0, As a consequence, using Corollary 4.10 we obtain, (4.61) u h,k L 4 ((kh

1 2 -ε ,(k+2)h 1 2 -ε ), L ∞ (R)) ≤ h -1 8 ϕ t -kh 1 2 -ε h 1 2 -ε f h + h ε-1 2 ϕ ′ t -kh 1 2 -ε h 1 2 -ε u h L 1 ((kh 1 2 -ε ,(k+2)h 1 2 -ε ), L 2 (R)) ≤ Ch 1 2 -ε-1 8 f h L ∞ (0,τ ), L 2 (R)) + h ε-1 2 u h L ∞ ((0,τ ), L 2 (R)) ≤ Ch s-1 8
where in the last inequality we used the assumption u ∈ L ∞ ((0, τ ), H s (R)) and (3.47). Eventually, noticing that,

u h 4 L 4 ((0,τ ), L ∞ (R)) ≤ h ε-1 2 k=-1 u h,k 4 L 4 ((kh 1 2 -ε ,(k+2)h 1 2 -ε ), L ∞ (R))
,

we obtain u h 4 L 4 ((0,τ ), L ∞ (R)) ≤ Ch ε-1 2 h 4(s-1 which implies u h L 4 ((0,τ ), W s-1 4 ,∞ (R)) ≤ Ch ε/4
and consequently, since h = 2 -j , (4.62) u

L 4 ((0,τ ), W s+ ε 8 -1 4 ,∞ (R)) ≤ ∞ j=0 u 2 -j L 4 ((0,τ ), W s+ ε 8 -1 4 ,∞ (R)) ≤ C ∞ j=0
2 -jε/4 < +∞.

Classical time parametrix

In this section we take s > 11 2 and we prove the usual Strichartz estimates. The main step is, as before, the dispersion estimate. To do so, we seek a parametrix. The main difference with respect to the previous section is that (in the semi-classical framework), we are looking for a large (O(h -1/2 )) time parametrix. As a consequence, the lower order term T W ∂ x induces oscillations. This is reflected in the fact that the new eikonal equation will be quasi-linear.

We begin by an analogue of Theorem 4.1.

Theorem 5.1. -Let χ ∈ C ∞ 0 (R ) with supp χ ⊂ {ξ : 1 2 ≤ |ξ| ≤ 2} and t 0 ∈ R.
For any initial data u 0,h = χ(hD x )u 0 where u 0 ∈ L 1 (R) let us denote by S(t, t 0 , h)u 0,h := U h the solution of (5.1)

∂ t U h + 1 2 (W δ h ∂ x + ∂ x W δ h )U h + ia(D x )U h = 0, U h | t=t 0 = u 0,h .
Then there exists τ 0 > 0 such that for any 0 < h ≤ 1 and any |t -

t 0 | ≤ τ 0 , (5.2) S(t, t 0 , h)u 0,h L ∞ (R) ≤ C h 1/4 |t -t 0 | 1/2 u 0,h L 1 (R) .
According to Lemma 3.8, Theorem 5.1, the Duhamel formula and the same T T * argument as in Section 4, we deduce Corollary 5.2. -With the notations of Lemma 3.8, we have

(5.3) u h L 4 ((0,T ),L ∞ (R)) = ∆ j u L 4 ((0,T ),L ∞ (R)) ≤ C2 j( 1 8 -s) c j , c j ∈ ℓ 2 .
In the remaining of this section, we shall prove Theorem 5.1. We need first to refine the constructions in Section 4 to handle large times. An important point in the construction of the phase function is that handling large times leads us to non linear geometric optics.

Our parametrixe will be of the form (4.3),(4.4) that is,

(5.4) U h (t, x) = 1 2πh e i h (Φ(t,x,ξ,h)-zξ) B(t, x, z, ξ, h)u 0,h (z)dzdξ,
where Φ will satisfy the eikonal equation and

(5.5)

B(t, x, z, ξ, h) = B(t, x, ξ, h)ζ(x -z -th -1 2 a ′ (ξ)),
where B will satisfy the transport equations and As before we shall set 2 -j = h, where j ∈ N and we shall work with the semiclassical time σ = th -1 2 . In addition to the function χ introduced in Theorem 5.1, we shall use two more cut-off functions χ j ∈ C ∞ 0 (R), j = 1, 2, such that, (5.6)

ζ ∈ C ∞ 0 (R), ζ(s) = 1 if |s| ≤ 1, ζ(s) = 0 if |s| ≥ 2.
     supp χ 1 ⊂ {ξ : 1 3 ≤ |ξ| ≤ 3}, χ 1 = 1 on the support of χ, supp χ 0 ⊂ {ξ : 1 4 ≤ |ξ| ≤ 4}, χ 0 = 1 on the support of χ 1 .
Recall that we have

W ∈ L ∞ ([0, T ], W 2,∞ (R)), ∂ t W ∈ L ∞ ([0, T ], W 1,∞ (R)) (Lemma 3.5), (i) W δ h = S [δ(j-3)] (W ) satisfies ∂ α x W δ h L ∞ t,x ≤ C α ∂ α x W L ∞ t,x , (ii) a(ξ) = χ 0 (ξ) |ξ| 3 2 , (iii) Definition 5.3.
-For small h 0 , τ 0 to be fixed, we introduce the sets

Ω = (t, x, ξ, h) ∈ R 4 : h ∈ (0, h 0 ), |t| < τ 0 , 1 < |ξ| < 3 O = (σ, x, ξ, h) ∈ R 4 : h ∈ (0, h 0 ), |σ| < τ 0 h -1 2 , 1 < |ξ| < 3 .
If m ∈ R and ̺ ∈ R + , we denote by S m ̺ (Ω) (resp.S m ̺ (O)) the set of all functions f on Ω (resp.O) which are C ∞ with respect to (t, x, ξ) (resp.(σ, x, ξ) and satisfy the estimate

(5.7) |∂ α x f (t, x, ξ, h)| (resp. |∂ α x f (σ, x, ξ, h)|) ≤ C α h m-̺α , for all (t, x, ξ, h) ∈ Ω (resp.(σ, x, ξ, h) ∈ O).
5.2. The eikonal and transport equations. -In all this section we keep the notations of (4.11),(4.12), and (4.17) to (4.23).

The main result is the following.

Proposition 5.4. -There exist a phase Φ of the form

Φ(t, x, ξ, h) = xξ -h -1 2 ta(ξ) + h 1 2 Ψ(t, x, ξ, h) with ∂ x Ψ ∈ S 0 δ (Ω)
and an amplitude B ∈ S 0 δ Ω) such that, with B defined in (5.5),

(5.8)

L 0 e i h Φ B = e i h Φ R h .
and for all N ∈ N we have, (5.9) e i h (Φ(t,x,ξ,h)-zξ) R h (t, x, z, ξ, h)u 0,h (z) dz dξ

H 1 (Rx) ≤ C N h N u 0,h L 1 (R) , for all t in [0, τ 0 ].
Proof. -According to (4.23) we have,

     A 0 = a(x, ξ + h 1 2 ∂ x ψ)b, A 1 = h i a(x, ξ + h 1 2 ∂ x ψ)∂ x b + h 3 2 i (∂ 2 x ψ)a ′ (x, ξ + h 1 2 ∂ x ψ)b.
We deduce that L 0 (e

r = i h 1 2 ∂ σ ψ -a(ξ) + a(x, ξ + h 1 2 ∂ x ψ) + hW δ h ∂ x ψ + h 1 2 ξW δ h b ζ + h ∂ σ b + a(x, ξ + h 1 2 ∂ x ψ)∂ x b + h 1 2 W δ h ∂ x b + 1 2 h 1 2 (∂ x W δ h )b + h 1 2 (∂ 2 x ψ)a ′ (x, ξ + h 1 2 ∂ x ψ)b + i h M -1 k=2 A k ζ + i 4 j=1 r j ,
where r 1 , r 2 are defined in (4.21), (4.22), r 3 in (4.23) and

(5.11)

r 4 = h i -a ′ (ξ) + h 1 2 W δ h b ζ ′ .

The eikonal equation.

-As already mentioned, an important point in the construction of the phase function is that handling large times leads us to non linear geometric optics. Namely, we determine ψ by solving the following nonlinear problem, (5.12)

     ∂ σ ψ + a(ξ + h 1 2 ∂ x ψ) -a(ξ) h 1 2 + h 1 2 W δ h (h 1 2 σ, x)∂ x ψ = -W δ h (h 1 2 σ, x)ξ, ψ(0, x, ξ, h) = 0.
In this system, ξ and h are seen as parameters. We begin by establishing that the solutions exist for a time interval of size h -1 2 and satisfy some uniform estimates.

Proposition 5.5. -There exists τ 0 > 0 such that the problem (5.12) has a unique

C ∞ solution ψ in the set O such that ∂ x ψ and ∂ 2 x ψ are uniformly bounded on O by C (η, ψ) L ∞ (I,H s+ 1 2 (R)×H s (R))
where C is an increasing function from R + to itself.

Proof. -Let us differentiate the equation (5.12) with respect to x and let us set

ψ 1 = ∂ x ψ.
Then ψ 1 is solution of the quasi-linear equation (5.13)

∂ σ ψ 1 + A(σ, x, h, ξ, ψ 1 )∂ x ψ 1 = B(σ, x, h, ξ, ψ 1 ), ψ 1 (0, x, h, ξ) = 0.
where

(5.14) A(σ, x, h, ξ, z) = a ′ (ξ + h 1 2 z) + h 1 2 W δ h (h 1 2 σ, x), B(σ, x, h, ξ, z) = -h 1 2 (∂ x W δ h )(h 1 2 σ, x)z -ξ(∂ x W δ h )(h 1 2 σ, x).
We shall solve (5.13) by the method of characteristics. The characteristics are given by the system (5.15)

     σ(s) = 1, σ(0) = 0, Ẋ(s) = A(s, X(s), h, ξ, Z(s)), X(0) = x, Ż(s) = B(s, X(s), h, ξ, Z(s)), Z(0) = 0.
Since A is uniformly bounded and |B| ≤ C 1 + C 2 |z|, the above system has a unique global solution (i.e defined for s ∈ [0, +∞[).

Properties of the flow.-(i)

We have, (

)

∃τ 0 > 0, c 1 > 0, c 2 > 0 : c 1 ≤ | Ẋ(s)| ≤ c 2 , 0 ≤ s ≤ τ 0 h -1 2 .
We show first that (5.17)

h 1 2 |Z(s)| ≤ Cτ 0 ∂ x W L ∞ exp τ 0 ∂ x W L ∞ , 0 ≤ s ≤ τ 0 h -1 2 .
To see this we integrate the equation satisfied by Z and use (5.14). We obtain

(5.18) |Z(s)| ≤ C ∂ x W L ∞ t,x |s| + h 1 2 ∂ x W L ∞ t,x s 0 |Z(σ)|dσ, 0 ≤ s ≤ τ 0 h -1 2 .
Then (5.17) follows from the Gronwall inequality.

On the other hand, setting m(s) = (s, X(s), h, ξ, Z(s)), we have

A(m(s)) = a ′ (ξ) + h 1 2 Z(s) 1 0 a ′′ (ξ + λh 1 2 Z(s))dλ + h 1 2 W δ h (h 1 2 σ, X(s))dσ := a ′ (ξ) + R where |R| ≤ τ 0 C(τ 0 , ∂ x W L ∞ t,x ) a ′′ L ∞ + h 1 2 W L ∞ t,x ) Since for 1/2 ≤ |ξ| ≤ 3 we have |a ′ (ξ)| ≥ 2c 1 > 0 we obtain |A(s, X(s), h, ξ, Z(s))| ≥ c 1 when 0 ≤ s ≤ τ 0 h -1
2 , (τ 0 and h small enough). This proves (5.16). (ii) We have,

(5.19) | Ẍ(s)| ≤ h 1 2 C ∂ x W L ∞ t,x + ∂ t W L ∞ t,x , 0 ≤ s ≤ τ 0 h -1 2 .
Indeed let us set m(s) = (s, X(s), h, ξ, Z(s)).Then we have,

Ẍ(s) = (∂ s A)(m(s)) + (∂ x A)(m(s)) Ẋ(s) + (∂ z A)(m(s)) Ż(s).
Moreover we have,

(∂ s A)(m(s)) = h(∂ s W δ h )(σh 1 2 , x), (∂ x A)(m(s)) = h 1 2 (∂ x W δ h )(σh 1 2 , x) (∂ z A)(m(s)) = a ′′ (ξ + h 1 2 Z(s)).
Then (5.19) follows from the expressions of Ẋ(s), Ż(s) and (5.17).

(iii) We improve now (5.17). We have, (5.20)

|Z(s) ≤ C |α|≤1 ∂ α W L ∞ t,x , ∂ = (∂ t , ∂ x ).
Indeed we can write (5.21)

Z(s) = -ξ s 0 (∂ x W δ h )(σh 1 2 , X(σ))dσ -h 1 2 s 0 (∂ x W δ h )(σh 1 2 , X(σ))Z(σ)dσ
Now, using (5.16) we have,

(5.22) (∂ x W δ h )(σh 1 2 , X(σ)) = ∂ σ [W δ h (σh 1 2 , X(σ))] Ẋ(σ) - h 1 2 (∂ s W δ h )(σh 1 2 , X(σ)) Ẋ(σ)
.

After an integration by parts we obtain,

I =: s 0 (∂ x W δ h )(σh 1 2 , X(σ))dσ = W δ h (sh 1 2 , X(s)) Ẋ(s) - W δ h (0, X(0)) Ẋ(0) + s 0 Ẍ(σ) ( Ẋ(σ)) 2 W δ h (σh 1 2 , X(σ))dσ -h 1 2 s 0 1 Ẋ(σ) (∂ s W δ h )(σh 1 2 , X(σ))dσ. Using (5.16),(5.19)
we deduce that for 0 ≤ s ≤ τ 0 h -1 2 we have,

|I| ≤ C |α|≤1 ∂ α W L ∞ t,x ∂ = (∂ s , ∂ x ).
It follows from (5.21) that,

|Z(s)| ≤ C |α|≤1 ∂ α W L ∞ t,x + ∂ x W δ h L ∞ t,x h 1 2 s 0 |Z(σ)dσ,
which using Gronwall inequality proves (5.20). We are going now to give some estimates on the x-derivative of the flow. We claim that,

(5.23) | ∂X ∂x (s)| + | ∂Z ∂x (s)| ≤ C ∂ t W W 1,∞ + W W 2,∞ 0 ≤ s ≤ τ 0 h -1 2 (5.24) | ∂X ∂x (s) -1| ≤ 1 2 , 0 ≤ s ≤ τ 0 h -1 2 , if τ 0 C ∂ t W W 1,∞ + W W 2,∞
) is small enough.

Indeed using (5.15) we can write,

∂ X ∂x (s) = h 1 2 (∂ x W δ h )(sh 1 2 , X(s)) ∂X ∂x (s) + h 1 2 a ′′ (ξ + h 1 2 Z(s)) ∂Z ∂x (s), ∂ Z ∂x (s) = -ξ(∂ 2 x W δ h )(sh 1 2 , X(s)) ∂X ∂x (s) -h 1 2 (∂ 2 x W δ h )(sh 1 2 , X(s)) ∂X ∂x (s)Z(s) -h 1 2 (∂ x W δ h )(sh 1 2 , X(s)) ∂Z ∂x .
From the first equation we deduce

(5.25) | ∂X ∂x (s)| ≤ 1 + h 1 2 (∂ x W δ h ) L ∞ t,x s 0 | ∂X ∂x (σ)|dσ + h 1 2 a ′′ L ∞ s 0 | ∂Z ∂x (σ)|dσ.
From the second equation we deduce, (

                         ∂Z ∂x (s) = I 1 = I 2 + I 3 5.26)
where,

I 1 = -ξ s 0 (∂ 2 x W δ h )(σh 1 2 , X(σ)) ∂X ∂x (σ)dσ, I 2 = -h 1 2 s 0 (∂ 2 x W δ h )(σh 1 2 , X(σ)) ∂X ∂x Z(σ)dσ,
I 3 = -h 1 2 s 0 (∂ x W δ h )(σh 1 2 , X(σ)) ∂Z ∂x dσ.
We have easily,

(5.27)

|I 3 | ≤ h 1 2 ∂ x W L ∞ t,x s 0 | ∂Z ∂x (σ)|dσ
Moreover using (5.20) we get,

(5.28)

|I 2 | ≤ h 1 2 C |α|≤1 ∂ α W L ∞ t,x ∂ 2 x W L ∞ t,x s 0 | ∂X ∂x (σ)|dσ.
We are left with the estimate of I 1 . We use (5.22) applied to ∂ x W δ h .We obtain

I 1 = -ξ s 0 ∂ σ [∂ x W δ h (σh 1 2 , X(σ))] Ẋ(σ) ∂X ∂x (σ)dσ (5.29) -h 1 2 ξ s 0 (∂ x ∂ s W δ h )(σh 1 2 , X(σ)) Ẋ(σ) ∂X ∂x (σ)dσ =: A + B. (5.30)
We see easily that

(5.31) |B| ≤ Ch 1 2 ∂ s ∂ x W L ∞ t,x s 0 | ∂X ∂x (σ)|dσ.
Let us consider the term A. After an integration by parts one can write,

A = -ξ (∂ x W δ h )(sh 1 2 , X(s)) Ẋ(s) ∂X ∂x (s) - (∂ x W δ h )(0, x) Ẋ(0) + ξ s 0 (∂ x W δ h )(σh 1 2 , X(σ)) ∂ X ∂x (s) dσ Ẋ(σ) -ξ s 0 (∂ x W δ h )(σh 1 2 , X(σ)) Ẍ(σ) ( Ẋ (σ)) 2 ∂X ∂x (σ)dσ.
Using (5.24), (5.16), (5.19) and the equation satisfied by ∂X ∂x we obtain

(5.32) |A| ≤ C ∂ x W L ∞ t,x 1 + h 1 2 s 0 (| ∂X ∂x (σ)| + | ∂Z ∂x (σ)|)dσ .
Using (5.26) to (5.32) we obtain

| ∂Z ∂x (s)| ≤ C ∂ x W L ∞ t,x + C ∂ t W L ∞ t W 1,∞ x + W L ∞ t W 2,∞ x s 0 ∂X ∂x (σ) + ∂Z ∂x (σ)
dσ so using (5.25) and the Gronwall inequality we obtain (5.23).

Then coming back to the equation satisfied by ∂X ∂x we deduce

∂X ∂x (s) -1 ≤ C ∂ t W L ∞ t W 1,∞ x + W L ∞ t W 2,∞ x h 1 2 |s| for 0 ≤ s ≤ τ 0 h -1 2 .
Therefore taking τ 0 small enough we obtain (5.24).

Resolution of the eikonal equation.-

We claim now that for all s in [0, τ 0 h -1 2 ] the map x → X(s, x) is a global diffeomorphism from R to R. Indeed we have for such fixed s,

X(s, x) = x + s 0 A(σ, X(σ), h, ξ, Z(s))dσ. Since A is bounded by a ′ L ∞ + h 1 2 W L ∞ t,
x we have lim |x|→+∞ |X(s, x)| = +∞. Moreover by (5.24) we have, ∂X ∂x (s, x) = 0 for all 0 ≤ s ≤ τ 0 h -1 2 and all x ∈ R. Therefore our claim follows from a well known result by Hadamard. Then (5.33)

X(s, x) = y ⇔ x = Y (s, x), x, y ∈ R,
and the function (s, y) → Y (s, y) is C ∞ by the implicit function theorem. Let us consider then the set S = {(s, X(s, x), Z(s, x), 0 ≤ s ≤ τ 0 h -1 2 , x ∈ R}. It follows from (5.33) that S is a submanifold of R 3 of dimension two to which the vector field L = ∂ ∂σ + A(σ, x, h, ξ, z) ∂ ∂x + B(σ, x, h, ξ, z) ∂ ∂z is tangent. It follows then from (5.33) that the function ψ 1 (s, y, h, ξ) = Z(s, Y (s, y)) is the solution of our eikonal equation (5.13).

Then ψ 1 ∈ L ∞ . Moreover we have

∂ψ 1 ∂y = ∂Z ∂x (s, Y (s, y)) ∂Y ∂y (s, y) = ∂Z ∂x (s, Y (s, y))( ∂X ∂x (s, Y (s, y))) -1 ,
so, since ∂Z ∂x is bounded and from (5.24) we deduce,

| ∂ψ 1 ∂y | ≤ C ∂ t W W 1,∞ + W W 2,∞ .
It follows that the solution ψ of our eikonal equation (5.12) is such that

∂ψ ∂x ∈ L ∞ t,x , ∂ 2 ψ ∂x 2 ∈ L ∞ t,x ,
uniformly in h, ξ.

Properties of the solution.

-We investigate in this section futher regularity of the solution ψ.

Proposition 5.6. -Let ψ be the solution of (5.12) given by Proposition 5.5 Then we have

ψ ∈ L ∞ (O), ∂ x ψ ∈ L ∞ (O), ∂ 2 x ψ ∈ S 0 δ (O).
Proof. -The first two claims have been proved in Proposition 5.5, let us consider the third one. We shall prove that

∂ x ψ 1 ∈ S 0 δ (O) where ψ 1 = ∂ x ψ. Let us set for 1 2 ≤ |ξ| ≤ 2, (5.34) v(σ, x, ξ, h) = ∂ x ψ 1 (σ, x, ξ, h) - ξ a ′ (ξ) ∂ x W δ h (σh 1 2 , x).
Then according to (5.13) the function v is solution of the equation, (5.35)

∂ σ v + (a ′ (ξ + h 1 2 ψ 1 )+h 1 2 W δ h )∂ x v + h 1 2 a ′′ (ξ + h 1 2 ψ 1 )v 2 +(2h 1 2 a ′′ (ξ + h 1 2 ψ 1 ) ξ a ′ (ξ) ∂ x W δ h + 2h 1 2 ∂ x W δ h )v = f,
where,

(5.36)

f = - ξ a ′ (ξ) h 1 2 ∂ x ∂ t W δ h + ξ a ′ (ξ) ∂ 2 x W δ h a ′ (ξ + h 1 2 ψ 1 ) -a ′ (ξ) -h 1 2 ξ a ′ (ξ) W δ h ∂ 2 x W δ h + h 1 2 ( ξ a ′ (ξ) ) 2 a ′′ (ξ + h 1 2 ψ 1 )(∂ x W δ h ) 2 + 2h 1 2 ξ a ′ (ξ) (∂ x W δ h ) 2 -h 1 2 (∂ 2 x W δ h )ψ 1 .
Let us set Λ = h δ ∂ x .We shall prove by induction on k ∈ N that, (5.37)

Λ j v ∈ L ∞ (O), 0 ≤ j ≤ k.
This will imply our claim since

∂ x W δ h ∈ S 0 δ (O).
Notice that (5.37) is true for k = 0 by Proposition 5.5. Assume it is true up to the order k -1. It follows then, using the Faa-di Bruno formula that, (5.38) Λ l [b(h

1 2 ψ 1 )] ∈ h 1 2 +δ L ∞ (O), 1 ≤ l ≤ k, for any C ∞ -bounded function b from R to R.
Applying the operator Λ k to both sides of (5.35) and using (5.38) and the fact that

∂ α W δ h ∈ S 0 δ (O) for ∂ = (∂ t , ∂ x ), |α| ≤ 2, α = (2, 0) we find that Λ k v is solution of the problem (5.39) (∂ σ + a ′ (ξ + h 1 2 ψ 1 )∂ x + h 1 2 W δ h ∂ x + h 1 2 d(σ, x, ξ, h))Λ k v ∈ h 1 2 L ∞ (O), where d ∈ L ∞ (O ε ).
Let us set ṽk = (Λ k v)(σ, x + a ′ (ξ)σ). Then ṽk is solution of the problem,

∂ σ + h 1 2 1 0 a ′′ (ξ + λh 1 2 ψ1 ) dλ ψ1 ∂ x + h 1 2 W δ h ∂ x + h 1 2 d(σ, x, ξ, h) ṽk ∈ h 1 2 L ∞ (O),
Then the desired conclusion follows from the following Lemma. and c j (t, x) = c j (h -1 2 t, x), j = 1, 2. Then we are led to the problem

P u = h 1 2 F , u| σ=0 = 0 (t ∈ [0, τ 0 ], x ∈ R),
where

P = ∂ t + c 1 (t, x)∂ x + c 2 (t, x). Recall that c 1 ∈ L ∞ , ∂ x c 1 ∈ L ∞ , c 2 ∈ L ∞ .
Then the claim of the lemma follows from the classical method of characteristics. Indeed, the characteristics are given by t(s) = s and Ẋ(s, x) = c 1 (s, X(s, x)), X(0, x) = x.

Then x → X(t, x) is globally invertible for each t ∈ [0, τ 0 ] i.e. X(t, x) = y ⇔ x = Y (t, y) with Y ∈ C 0 ∩ L ∞ . Then d dt [ u(t, X(t, x))] = c 2 (t, X(t, x))u(t, X(t, x)) + F (t, X(t, x)).
Therefore u given by

u(t, y) = exp - t 0 c 2 (t ′ , X(t ′ , Y (t, y)) dt ′ t 0 F (t ′ , X(t ′ , Y (t, y)) dt ′
is our solution.

Corollary 5.8. -Let ψ be defined by Proposition 5.5 and L be the operator

(5.40) L = ∂ σ + a ′ (ξ + h 1 2 ∂ x ψ)∂ x + h 1 2 d 1 ∂ x + h 1 2 d 2 ,
where d 1 , ∂ x d 1 , d 2 are real valued and belong to S 0 δ (O). Then for any F such that Λ j F L ∞ (O) is finite for every j ∈ N, the problem Lu = F, u| σ=0 = 0, has a unique solution which satisfies the estimate,

(5.41) |Λ k u(σ, x, ξ, h)| ≤ C k σ k j=0 Λ j F L ∞ (O) ,
for all (σ, x, ξ, h) ∈ O, where Λ = h δ ∂ x .

Proof. -Since by Proposition 5.6 we have ∂ 2

x ψ ∈ S 0 δ (O) one can write,

L = ∂ σ + a ′ (ξ)∂ x + h 1 2 d 3 ∂ x ψ + h 1 2 d 2 .
where

d 3 , ∂ x d 3 , d 2 belong to S 0 δ (O). Setting U = u(σ, x + σa ′ (ξ), ξ, h), c 1 = d 3 (σ, x + σa ′ (ξ), ξ, h), c 2 = d 2 (σ, x + σa ′ (ξ), ξ, h), we see that c 1 , ∂ x c 1 , c 2 belong to S 0 δ (O)
and U is a solution of the equation (5.42)

L 1 U := (∂ σ + h 1 2 c 1 ∂ x ψ + h 1 2 c 2 )U = F 1 .
We shall prove by induction on k that U satisfies the estimate (5.41). The case k = 0 follows immediately from Lemma 5.7. Assume now that (5.41) is true up to the order k -1, k ≥ 1. By the Leibniz formula we have

LΛ k U + kh 1 2 (∂ x c 1 )Λ k U = -h 1 2 (k-2) + i=0 k i (Λ k-i c 1 )Λ i ∂ x U -h 1 2 k-1 i=0 k i (Λ k-i c 2 )Λ i U + Λ k F 1 := G k .
The sum in the first line can be written, -h

1 2 k-1 i=1 k i -1 (Λ k-i ∂ x c 1 )Λ i U.
According to our assumptions on c 1 , c 2 ,we can apply the Lemma 5.7 to the operator L + kh 1 2 (∂ x c 1 ).We obtain, using the induction and the fact that h

1 2 σ ≤ τ 0 , |Λ k U (σ, x, ξ, h)| ≤ σ Λ k F L ∞ (O) + Cσ k-1 j=0 Λ j F L ∞ (O) ,
which completes the induction. Proposition 5.9. -Let Λ = h δ ∂ x . The solution of (5.12) given by Proposition 5.5 satisfies the estimates, (5.43)

|Λ k ∂ ξ ψ(σ, x, ξ, h)| + |Λ k ∂ x ∂ ξ ψ(σ, x, ξ, h)| ≤ C k σ, |Λ k ∂ 2 ξ ψ(σ, x, ξ, h)| ≤ C k τ 0 h -1 2 σ,
for all (σ, x, ξ, h) ∈ O, where C depends only on (η, ψ)

L ∞ (I,H s+ 1 2 (R)×H s (R))
.

Proof. -Differentiating (5.12) with respect to ξ we find that U = ∂ ξ ψ satisfies the equation (5.44)

∂ σ U + (a ′ (ξ + h 1 2 ∂ x ψ) + h 1 2 W δ h )∂ x U = -(∂ x ψ) 1 0 a ′′ (ξ + h 1 2 λ∂ x ψ)dλ -W δ h .
Then the estimate of the first term in the first line of (5.43) follows from Corollary 5.8.

To estimate the second term we differentiate with respect to ξ the equation (5.13). We find that the fuction

U 1 = ∂ ξ ψ 1 = ∂ x ∂ ξ ψ satisfies the equation ∂ σ U 1 + (a ′ (ξ + h 1 2 ψ 1 ) + h 1 2 W δ h )∂ x U 1 + h 1 2 a ′′ (ξ + h 1 2 ψ 1 )∂ x ψ 1 U 1 = -∂ x W δ h .
The second estimate in the first line of (5.43) follows from Corollary 5.8. Finally to estimate U 2 = ∂ 2 ξ ψ we differentiate (5.44) with respect to ξ and we find that U 2 satisfies the equation

∂ σ U 2 + (a ′ (ξ + h 1 2 ∂ x ψ) + h 1 2 W δ h )∂ x U 2 = F where F = -h 1 2 (∂ x ∂ ξ ψ) 2 a ′′ (ξ + h 1 2 ∂ x ψ) + (∂ x ∂ ξ ψ)a ′′ (ξ + h 1 2 ∂ x ψ) + a ′′ (ξ + h 1 2 ∂ x ψ) -a ′′ (ξ) h 1 2
.

So using the estimate on ∂ x ψ and ∂ x ∂ ξ ψ obtained before, we deduce from Corollary 5.8 the last estimate of (5.43).

The symbol equations.

-According to the formulas (4.18)-(4.23), since the phase ψ now satisfies the eikonal equation (5.12), we are lead to solve the following transport equation (5.45)

∂ σ b + a ′ (ξ + h 1 2 ∂ x ψ)∂ x b + h 1 2 W δ h ∂ x b -h 1 2 (∂ 2 x ψ)a ′′ (ξ + h 1 2 ∂ x ψ)b + 1 2 h 1 2 (∂ x W δ h )b = - i h N k=2 A k , with (5.46) b| σ=0 = χ 1 (ξ),
where χ 1 ∈ C ∞ 0 (R) is equal to one on the support of the function χ given in Theorem 5.1.

Let µ 0 = 1 2 -2δ,
where we recall that δ < 1/4. Let us set (5.47) Λ = h δ ∂ x .

Then according to (4.23) and the Leibniz formula one can write (5.48)

1 h A k = h k(1-δ)-1 k k 1 =0 c k,k 1 (h δ ∂ y ) k-k 1 (∂ k ξ a)(ρ((x, y))   y=x Λ k 1 b, where c k,k 1 ∈ C.
We shall take b of the form

(5.49) b = χ 1 (ξ)e θ , θ = M j=0 θ j , θ j = h jµ 0 θ j .
Then Λ k 1 b is a finite linear combination of terms of the form

Λ α 0 e θ 0 • • • Λ α M e θ M , α 0 + • • • + α M = k 1 . Let ω = α ∈ N M +1 : |α| = k 1 , and, for 0 ≤ p ≤ M , ω p = {α ∈ ω, α = (α 0 , • • • , α p , 0, • • • , 0) with α p = 0} .
Then ω is the disjoint union of the ω p . It follows that where E s,ℓ is a finite linear combintation of terms of the form

s i=1 Λ p i θ ℓ q i where 1 ≤ s i=1 q i ≤ α ℓ , s i=1 p i q i = α ℓ .
Since s i=1 q i ≥ 1 and p(p + 1)/2 ≥ p, it follows from (5.50) and (5.49) that

(5.52) Λ k 1 b = e θ M p=0 h pµ 0 |β|≤k 1 G pβ h, Λ β 0 θ 0 , . . . ,
= 1 2 -2δ > 0 we have (5.53) k(1 -δ) -1 ≥ 2(1 -δ) -1 = 1 2 + µ 0 .
Let us note that this is the only point where we use the fact that δ < 1 4 . On the other hand we have,

(5.54) (h δ ∂ y ) k-k 1 (∂ k ξ a)(ρ((x, y))   y=x ∈ S 0 δ .
It follows then from (5.48), (5.52), (5.53), (5.54) that for k ≥ 2

(5.55)

1 h A k = h 1 2 M p=0 k k 1 =0 |β|≤k 1 h (p+1)µ 0 f k,k 1 H p,β h, Λ β 0 θ 0 , . . . ,
L = ∂ σ + a ′ (ξ + h 1 2 ∂ x ψ)∂ x + h 1 2 W δ h ∂ x . Since b = e θ with θ = M
p=0 h pµ 0 θ p we have Lb = e θ L θ. It follows from (5.55) that the transport equation (5.45) can be written, modulo a remainder,

Lθ 0 -h 1 2 (∂ 2 x ψ) - 1 2 (∂ x W δ h ) + M -1 p=0 h (p+1)µ 0 Lθ p+1 -h 1 2 G p (θ 0 , . . . , θ p ) = 0,
where G j (θ 0 , . . . , θ p ) are polynomials in Λ β θ ℓ for |β| ≤ N . Therefore we shall take for θ p , 0 ≤ p ≤ M -1, the solutions of the problems (5.57)

   Lθ 0 = h 1 2 (∂ 2 x ψ)a ′′ (ξ + h 1 2 ∂ x ψ) - 1 2 (∂ x W δ h ) , θ 0 | σ=0 = 0, Lθ p+1 = h
|Λ k θ p (σ, x, ξ, h)| ≤ C k , |Λ k ∂ ξ θ p (σ, x, ξ, h)| ≤ C k h -δ σ, |Λ k ∂ 2 ξ θ p (σ, x, ξ, h)| ≤ C k h -1 2 -2δ σ.
Proof. -We proceed by induction on p. If p = 0, the estimate of the first term in the first line of (5.58) follows immediately from Proposition 5.6 and Corollary 5.8. Now ∂ ξ θ 0 is solution of the equation

(5.59) L∂ ξ θ 0 = h 1 2 (∂ ξ ∂ 2 x ψ)a ′′ (ξ + h 1 2 ∂ x ψ) + (∂ 2 x ψ)(1 + h 1 2 ∂ x ∂ ξ ψ)a ′′′ (ξ + h 1 2 ∂ x ψ) -(1 + h 1 2 ∂ ξ ∂ x ψ)a ′′ (ξ + h 1 2 ∂ x ψ)∂ x θ 0 := F 0 .
It follows from (5.43) and the first estimate that Λ j F 0 L ∞ (O) ≤ C j h -δ . Using Corollary 5.8 we obtain the estimate of the second term in the first line of (5.58). To estimate ∂ 2 ξ θ 0 we differentiate one more time the equation (5.59) and we find using the same arguments

that L∂ 2 ξ θ 0 = F 1 where F 1 satisfies, Λ j F 1 L ∞ (O) ≤ C j h -1 2 -2δ
. The estimate of the term in the second line of (5.58) follows the from Corollary 5.8. Assuming that (5.58) is true up to the order p the estimate θ p+1 follows from the second equation in (5.57) and the induction.

It follows then from (5.10),(5.12), (5.45) and (5.57) that

r 5 = h 1 2 h (M +1)µ 0 G M (θ 0 , .., θ M )χ 1 (ξ)e θ ζ.
= χ 1 (ξ)c(σ, x, ξ, h)(ζ(x -z -σa ′ (ξ))
with c = e θ = e M p=0 h pµ 0 θp ∈ S 0 δ (Ω) the same arguments as those used in section 4.1.1 give the proof of (5.9).

∈ C 2 ([α, β], C), p ∈ C 1 ([α, β], C) such that ∀ξ ∈ [α, β], |Im φ(ξ)| ≤ h, Im φ ′′ (ξ) ≤ ρ, ρ 2 ≤ Re φ ′′ (ξ) ≤ ρ, we have β α e i h φ(ξ) p(ξ) dξ ≤ 8 p L ∞ + 2 β α p ′ (ξ) dξ h ρ 1 2 .
Proof. -Notice that we can assume that ρ ≥ h (otherwise, the conclusion is straightforward). Notice now that, using the monotonicity assumption of the real part of the phase φ, we can decompose the interval (α, β) into the disjoint union of at most three intervals

(α, β) = I 1 ∪ I 2 ∪ I 3 ,
where I 1 , I 2 or I 3 are possibly empty and satisfy

∀ξ ∈ I 1 , Re(φ ′ (ξ)) ≤ -(ρh) 1/2 , ∀ξ ∈ I 2 , -(ρh) 1/2 ≤ Re(φ ′ (ξ)) ≤ (ρh) 1/2 , ∀ξ ∈ I 3 , Re(φ ′ (ξ)) ≥ (ρh) 1/2 .
Let us first study the contribution of I 3 . Either I 3 = ∅ or I 3 is an interval contained in [δ, β] for some δ ∈ [α, β[. Then, using that

h iφ ′ (ξ) ∂ ξ e i h φ(ξ) = e i h φ(ξ) ,
and integrating by parts, we obtain (5.61)

β δ e i h φ(ξ) p(ξ)dξ = h iφ ′ (ξ) e i h φ(ξ) p(ξ) β δ - β δ e iφ(ξ) h ∂ ξ h iφ ′ (ξ) p(ξ) dξ = h iφ ′ (ξ) e i h φ(ξ) p(ξ) β δ - β δ e iφ(ξ) h h iφ ′ (ξ) ∂ ξ p(ξ)dξ + β δ e iφ(ξ) h hφ ′′ (ξ) φ ′ (ξ) 2 p(ξ)dξ
Clearly, the contributions of the two first terms are easily handled by means of the lower bound on Re(φ ′ (ξ)) on I 3 , and to conclude, it suffices to bound the last term. But according to the assumption on the phase, we now have

Re(φ ′ (δ)) ≥ (ρh) 1/2 ⇒ Re(φ ′ (ξ)) ≥ (ρh) 1/2 + ρ 2 (ξ -δ),
and consequently, the last term is bounded by

p L ∞ β δ 4hρ (2(hρ) 1/2 + ρ(ξ -δ)) 2 dξ ≤ 2h p L ∞ (hρ) 1/2 ,
where the last inequality is obtained by a straightforward computation. Now, of course, the contribution of I 1 is dealt with similarly, and consequently we can focus on the contribution of I 2 . Now remark that according to the assumptions on φ ′′ , the length of I 2 is smaller that 4(h/ρ) 1/2 , which implies

I 2 e i h φ(ξ) p(ξ)dt ≤ 4 p L ∞ h ρ 1/2 .
This completes the proof.

5.4. End of the proof of Theorem 5.1. -As in the preceding section we have (5.62)

     S(t, 0, h)u 0,h = D 1 + D 2 + D 3 where D 1 = U h (t, x), D 2 = -S(t, 0, h)v 0,h (x), D 3 = - t 0 S(t, s, h)[F h (s, x)]ds.
The terms D 2 and D 3 are estimated exactly as in section 4.3 while D 1 will be estimated differently using Lemma 5.11 instead of Van der Corput Lemma. Indeed recall that according to (5.5) and (5.49) our amplitude in the parametrix (5.4) is given by b(σ, x, z, ξ, h) = χ 1 (ξ)e θ(σ,x,z,ξ,h) ζ(xzσa ′ (ξ)).

The new fact here is that we shall glue the term e θ with the phase and apply the Lemma 5.11 with the new phase ϕzξ + h i θ. Using (5.4) we can write in the variable σ = th - Let us show that all the hypotheses in this lemma are satisfied. For this we shall use (5.43) and (5.58). First of all, since δ ∈]0, 1 + σ|a ′′ (ξ)||ζ ′ (xzσa ′ (ξ)|dξ .

Since the last integral is bounded by C ′ |ζ ′ (t)|dt, we deduce that the term D 1 satisfies the estimate (5.2) which completes the proof of Theorem 5.1.

Back to estimates for (η, ψ)

Notice that up to now, we only proved estimates for the dyadically localized functions ∆ j u. In this section, we shall show how we can recover estimates for (η, ψ), the solutions of the water-wave system (1.4). Recall that the Besov space B r ∞,2 is defined by

u ∈ B r ∞,2 (R) ⇔ j∈N 2 2jr ∆ j u 2 L ∞ < +∞.
We will use the following elementary lemma Lemma 6.1.

-If the symbol a ∈ Γ m 0 , then the operator T a is bounded from B s ∞,2 (R) to B s-m ∞,2 (R).

We have the slightly stronger result (compared to Theorems 1.1, 1. Notice that Theorem 1.1 follows from the first part in Proposition 6.2, using that,

B σ+ε ∞,2 (R) ⊂ W σ,∞ (R).
On the other hand, using complex interpolation theory (see [START_REF] Bergh | Interpolation spaces. An introduction[END_REF]Theorem 6.4.5 (6)]), we have, with q ≥ 2, (since

H σ 2 = B σ 2 2,2 ), [B σ 1 ∞,2 , H σ 2 ] 2 q = B σ q,2 ⊂ W σ,q , σ = (1 - 2 q )σ 1 + 2 q σ 2 .
Taking σ 1 = s -1 8 and σ 2 = s we obtain σ = s - ≤ C ψ(t, .)

1-2 q B s-1 8 ∞,2 (R) ψ(t, .) 2 q
H s (R) .

It follows that, with (p, q) satisfying 2 p + 1 q = 1 2 , 2 ≤ q < +∞, we have which implies Theorem 1.2 (the estimate for η being similar).

Let us now turn to the proof of Proposition 6.2. For conciseness, we will only prove (6.2), the proof of (6.1) being similar (easier). Recall that the function u is obtained from (η, ψ) through the following steps:

1. u = T e ig Φ * , where the function g is real and satisfies ∂ x g ∈ Γ 0 where R -1 is of order -1 (i.e. bounded from H s (R) to H s+1 (R)) . Since H s+ 1 2 (R) ⊂ B s ∞,2 (R) it follows from Lemma 6.1, the boundedness of Φ * in L ∞ (I, H s (R)) that

Φ * L 4 (I,B s-1 8
∞,2 (R))

< +∞.

Step 2: We have

Φ * = κ * Φ ⇔ Φ • κ = Φ * + T ∂xΦ•κ κ Notice that ∂ x Φ • κ ∈ L ∞ (I × R) and ∂ x κ ∈ L ∞ (I, H s-1 2
). As a consequence, T ∂xΦ•κ κ ∈ L ∞ (I, H s+ 1 2 ) ⊂ L ∞ (I, B s ∞,2 ). We deduce from Step 1

Φ • κ L 4 (I,B s-1 8
∞,2 (R))

< +∞.

We conclude that Φ L 4 (I,B

s-1 8

∞,2 (R))

< +∞, by using the following lemma (with χ = κ -1 , r = s -1 8 and r < σ < s).

Lemma 6.3. -Let σ > 1. Consider χ such that ∂ x χ ∈ W σ-1,∞ (R). Then, for any 0 < r < σ, the map u → u • χ is continuous on B r ∞,2 .

Indeed, a simple calculation shows that for any ρ ≤ σ, the map u → u•χ is continuous on W ρ,∞ , and we conclude by choosing r 1 < r < r 2 ≤ σ with r i / ∈ N (notice that this implies W r i ,∞ = B r i ∞,∞ ) and using the real interpolation result (see [9, Theorem 6.4.5 (1)])

[B r 1 ∞,∞ , B r 2 ∞,∞ ] θ,2 = B r ∞,2 , r = (1θ)r 1 + θr 2 .

Step 3: Separating real and imaginary parts, we obtain

T p η L 4 (I,B s-1 8
∞,2 (R))

+ T c 1 U L 4 (I,B s-1 8
∞,2 (R))

< +∞ and the same proof as in Step 1 (using that p is elliptic, s

-3 2 ≥ 1 ⇒ p -1 ∈ Γ -1 2 1
and for fixed t, c -1 1 (t, .) ∈ W 1,∞ (R) ⊂ Γ 0 1 ) gives

η L 4 (I,B s-1 8 + 1 2 ∞,2 (R)) + U L 4 (I,B s-1 8
∞,2 (R))

< +∞.

Step 4: We have ψ = U + T B η. So using the boundedness of η in L ∞ (I, H s+ 1 2 (R)), of B in L ∞ (I, H s-1 (R)) ⊂ L ∞ (I × R) and Sobolev injections, we obtain,

ψ L 4 (I,B s-1 8 ∞,2 )
< +∞, which completes the proof of Proposition 6.2 and consequently of Theorems 1.1 and 1.2.

Appendix

In this section we give a proof of (

Proposition 3 . 4 . 3 2

 343 -Let s > 2 + 1 2 and I = [0, T ]. There exists a real valued function g such that ∂ x g ∈ Σ 0 s-and the function u = T e ig Φ * satisfies the equation

3 2

 3 and the remainder term is of order less than 1-(s-3 2 ) = 5 2 -s < 0. Then (3.26) follows from (3.27) and (3.35).

3 2

 3 we see that it is equivalent to, (4.12) L e i h ϕ b = e i h ϕ r(σ, x, z, ξ, h), and (4.10) becomes, (4.13)

  Proof. -Let us look to the case j = 0. Then c 0 satisfies the same equation and c 0 | σ=0 = 1. We show first (4.36) for k = 0 and all α. By Lemma 4.5 we have |c 0 | ≤ Ch -ε . So assume that (4.36) is true (for k = 0) up to the order α -1 and let us differentiate the equation (4.33) α time with respect to x. It follows that U = ∂ α x c 0 satisfies the equation (4.37)

5 j=1r 5 =

 55 5 and the induction to prove (4.36) for all k and α. The proof of (4.36) for j ≥ 1 is similar. It follows from (4.29), (4.33), (4.34) that r = h Jµ 0 b J-1 ζ. 4.1.1. End of the proof of Proposition 4.4. -We are left with the proof of (4.13). For r j , j = 1, 2, 5 defined in (4.21), (4.22) and (4.38) we have for all N ∈ N,

ih

  (x-y)η = e i h (x-y)η and we use the fact that (4.40) b(y) = χ 1 (ξ)c(σ, y, ξ, h)ζ(yzσa ′ (ξ))

Lemma 4 . 8 .

 48 -For all k ∈ N we have(4.50)

4. 4 .

 4 The T T * argument. -Having proved the dispersion estimate the Strichartz estimates for the solution of (4.1) follow very classically.

5. 1 .

 1 Notations. -In this section we fix

Lemma 5 . 7 . 1 2 c 1 1 2 c 2 0 F

 5711120 -Let c 1 , c 2 be two real valued functions such that c 1 , ∂ x c 1 , c 2 belong to L ∞ (O) and P = ∂ σ + h (σ, x, ξ, h)∂ x + h (σ, x, ξ, h). Then for any F ∈ L ∞ (O), the problem P u = F, u| σ=0 = 0, has a unique solution u which satisfies the estimate |u(σ, x, ξ, h)| ≤ C σ (s, •, ξ, h) L ∞ (R) ds,for all (σ, x, ξ, h) in O, uniformly in h. Proof of Lemma 5.7. -Let us set t = σh

1

 1

2

 2

  α Λ α 0 e θ 0 • • • Λ αp e θp exp α,p ∈ C. Now by the Faa-di Bruno formula we have for 0 ≤ ℓ ≤ M , (5.51) Λ α ℓ e θ ℓ = e θ ℓ α ℓ s=1 E sℓ

1 2 G 1 2

 21 p (θ 0 , . . . , θ p ), θ p+1 | σ=0 = 0 (0 ≤ p ≤ M -1). We have the following result.Proposition 5.10. -Let Λ = h ∂ x . Then (5.57) has a unique solution (θ 1 , ..., θ M ) such that for 0 ≤ p ≤ M and all integers k ∈ N,(5.58)

where r 1

 1 , r 2 , r 3 , r 4 are defined in (4.21),(4.22),(4.23),(5.11) and

5. 2 . 6 .

 26 End of the proof of Proposition 5.4. -Since we have b

5. 3 .

 3 The stationary phase lemma. -In the sequel we will use the following elementary version of the stationary phase inequality where we allow complex valued phase functions.Lemma 5.11. -For any real numbers α, β, h, ρ such that α < β, 0 < h ≤ 1, ρ > 0 and for any functions φ

1 2 ,D 1 =φ 1 2

 211 K(σ, x, z, ξ, h)u 0h (z) dz, σ,x,ξ,h)-zξ+ h i θ(σ,x,ξ,h)) χ 1 (ξ)ζ(xzσa ′ (ξ)) dξ.Therefore we shall apply the Lemma 5.11 with, = ϕ(σ, x, ξ, h)zξ + h i θ(σ, x, ξ, h), ϕ = xξ + σa(ξ) + h ψ(σ, x, ξ, h), p = χ 1 (ξ)ζ(xzσa ′ (ξ)), ρ = Cσ, C > 0.

2 ) 6 . 2 .

 262 Proposition -Let I = [0, T ]. Under the assumptions of Theorem 1.1, there exists ε > 0 such that (6.1) (η, ψ) ∈ L 4 (I, B

≤ C ψ 4 L 4 ( 4 L

 444 ∞ (I,H s (R)) .

2 s- 1 Step 1 :L 4 (

 2114 * = κ * Φ where (see(3.12)) κ ∈ L ∞ (I, W 2,∞ (R)). 3. Φ = T p η + iT c 1 U , with p ∈ Σ 1/is an elliptic symbol and c 1 = (1 + (∂ x η) 2 ) -1 2 . 4. U = ψ -T B η, where B ∈ L ∞ (I, H s-1 (R)) is defined in (2.7). Starting from Corollary 5.2, we have u symbolic calculus, since e ig ∈ Γ 0 0 , we have u = T e ig Φ * ⇒ Φ * = T e -ig u + R -1 (Φ * )

1 0I 1 =R 1 = 1 0( 1 -

 11111 4.18) to (4.22). Let a ∈ C ∞ 0 (R) with supp a ⊂ {|ξ| ≤ C 0 }, b ∈ C ∞ 0 (R) and ϕ ∈ C ∞ (R) real valued such that sup | ∂ϕ ∂x (x)| ≤ C 0 . Let us set (7.1) I = e -i h ϕ(x) a(hD) b e x-y)ξ+ϕ(y)-ϕ(x)) a(ξ)b(y)dydξ.Moreover we can writeϕ(x)ϕ(y) = (xy)ρ(x, y), ρ(x, y) = ∂ϕ ∂x (λx + (1λ)y)dλ.We have |ρ| ≤ C 0 so, setting η = ξρ(x, y) we obtain,I = (2πh) -1 e i h (x-y)η κ 0 (η)a(η + ρ((x, y))b(y) dydη where κ 0 ∈ C ∞ 0 (R) is such that χ 0 (η) = 1 if |η| ≤ 2C 0 .Using the Taylor expansion of the function a at the point θ(x, y) we obtain I = I 1 +R 1 where, y)η κ 0 (η)a (k) (θ(x, y))b(y)dydη and, c M h -1η M e i(x-y)η h κ 0 (η) λ) M -1 a (M ) (λη + ρ((x, y))b(y)dλdydη.Now we have ηk e i h (x-y)η = (-h i ∂ y ) k e i h (x-y)η, so integrating by parts in the integrals I 1 and R 1 we obtain,I 1 = (2πh) -1 M -1 k=0 h k i k k! e i h (x-y)η κ 0 (η)∂ k y [a (k) (θ(x, y))b(y)]dydη I 1 = (2πh) y [a (k) (θ(x, y))b(y)]dy, R 1 = c ′ h M -1 η)(1λ) M -1 ∂ M y [a(M ) (λη + ρ((x, y))b(y)]dλdydη. Let us set f (x, y) = ∂ k y [a (k) (θ(x, y))b(y)].

  Λ βp θ p , where G p,β (h, ζ 0 , . . . , ζ p ) are bounded in h and polynomial in ζ. Coming back to (5.48) we remark first that since k ≥ 2 and µ 0

  Λ βp θ p e θ , where f k,k 1 ∈ S 0 δ , H p,β (h, ζ 0 , . . . , ζ p ) are bounded in h and polynomial in ζ. Let us denote by L the linear operator appearing in (5.45),

	(5.56)

  1 4 [ we have,| Im φ| = h| Im θ| ≤ Ch, |∂ 2 ξ Im φ| ≤ h|∂ 2 ξ θ| ≤ Chh -1 2 -2δ σ ≤ σ. Re φ| ≤ |a ′′ (ξ)|σ + h 1 2 |∂ 2 ξ ψ| + h|∂ 2 ξ θ| ≤ Cσ. Finally, | Re φ| ≥ |a ′′ (ξ)|σh 1 2 |∂ 2 ξ ψ| -h|∂ 2 ξ θ| ≥ |a ′′ (ξ)|σ -C 1 τ 0 σ -C 2 h

	Moreover,
	|∂ 2 ξ

1 2 -2δ σ ≥ C 3 σ, if τ 0 and h are small enough. It follows the from Lemma (5.11) that, |K| ≤ Ch -1 h σ 1 2

BLAN-0250, is acknowledged.

Now we set in the integral, xy = hz and we write,

Then we use the following equality, which reflects the fact that κ 0 is equal to one near the origin. For j ∈ N we have,

where δ j,0 is the Kronecker symbol. It follows that,

where

Thus we obtain (4.18) to (4.22).