STRICHARTZ ESTIMATES FOR SCHRODINGER EQUATIONS WITH¨ VARIABLE COEFFICIENTS · 2018-12-20 ·...

MEMOIRES DE LA SMF 101/102

STRICHARTZ ESTIMATES FORSCHRODINGER EQUATIONS WITH

VARIABLE COEFFICIENTS

Luc Robbiano

Claude Zuily

Societe Mathematique de France 2005Publie avec le concours du Centre National de la Recherche Scientifique

L. RobbianoUniversité de Versailles, UMR 8100, Bât. Fermat, 45, Avenue des États-Unis,78035 Versailles.

C. ZuilyUniversité Paris Sud, UMR 8628, Département de Mathématiques, Bât. 425,91406 Orsay Cedex.

2000Mathematics Subject Classification. — 35A17, 35A22, 35Q40, 35Q55.Key words and phrases. — Strichartz estimates, Schrödinger equations, dispersive es-timates, FBI transform, Sjöstrand’s theory.

STRICHARTZ ESTIMATES FOR SCHRÖDINGEREQUATIONS WITH VARIABLE COEFFICIENTS

Luc Robbiano, Claude Zuily

Abstract. — We prove the (local in time) Strichartz estimates (for the full range ofparameters given by the scaling unless the end point) for asymptotically flat and nontrapping perturbations of the flat Laplacian in Rn, n ! 2. The main point of theproof, namely the dispersion estimate, is obtained in constructing a parametrix. Themain tool for this construction is the use of the FBI transform.

Résumé (Inégalités de Strichartz pour l’équation de Schrödinger à coefficients va-riables)

On démontre les inégalités de Strichartz (locales en temps) pour l’ensemble desindices donnés par l’invariance d’échelle (sauf le point final) pour des perturbationsasymptotiquement plates et non captantes du laplacien usuel de Rn, n ! 2. Le pointprincipal de la preuve, à savoir l’estimation de dispersion, est obtenu en construisantune paramétrixe. L’outil principal de cette construction est la théorie de la transfor-mation de FBI construite par Sjöstrand.

c! Mémoires de la Société Mathématique de France 101/102, SMF 2005

CONTENTS

1. Introduction and statement of the result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2. Preliminaries and reduction to the case of a small perturbation ofthe Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2. Reduction to a small perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3. Study of the flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2. The flow for short time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3. The forward flow from points in S+ and backward from S! . . . . . . . . . . . . . 193.4. Precisions on the flow in the general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5. The flow from points in (S+ ! S!)c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4. The phase equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.1. Statement of the result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2. The preparation theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.3. The case of outgoing points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.4. The case of incoming points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.5. The phase for small ! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5. The transport equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.1. Statement of the result and preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.2. The case of outgoing points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1085.3. The case of incoming points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1175.4. The amplitude for short time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6. Microlocal localizations and the use of the FBI transform . . . . . . . . . . 1496.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1496.2. The microlocalization procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1516.3. The one sided parametrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1566.4. Conclusion of Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

vi CONTENTS

7. The dispersion estimate and the end of the proof of Theorem 1.0.1 1717.1. The dispersion estimate for the operators K±(t) . . . . . . . . . . . . . . . . . . . . . . . . 1717.2. End of the proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.1. The Faa di Bruno Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.2. Proof of Proposition 3.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195A.3. Proof of Proposition 3.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196A.4. Proof of Lemma 5.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207


CHAPTER 1

INTRODUCTION AND STATEMENT OF THE RESULT

The purpose of this work is to provide a proof of the full (local in time) Strichartzestimates for the Schrodinger operator related to a non trapping asymptotically flatperturbation of the usual Laplacian in Rn.

Let "0 be in ]0, 1[. We introduce a space of symbols which decay like "x#!1!!0

where "x# = (1 + |x|2)1/2. More precisely we set(1.0.1)

B!0 =!a $ C"(Rn) : %# $ Nn, &C" > 0 : |$"a(x)| " C"

"x#1+|"|+!0, %x $ Rn

"

Let P be a second order di!erential operator,

(1.0.2) P =n#

j,k=1

Dj

$gjk(x)Dk

%+

n#

j=1

(Dj bj(x) + bj(x)Dj) + V (x), Dj =1i

$

$xj,

with principal symbol p(x, %) =&n

j,k=1 gjk(x) %j %k. (Here gjk = gkj).We shall make the following assumptions.

'(((()

((((*

(i) The coe"cients gjk, bj , V are real valued, 1 " j " k " n.

(ii) There exists "0 > 0 such that gjk ' &jk $ B!0 , bj $ B!0 .Here &jk is the Kronecker symbol.

(iii) V $ L"(Rn).

(1.0.3)

There exists ' > 0 such that for every (x, %) in Rn ( Rn, p(x, %) ! ' |%|2.(1.0.4)

Then P has a self-adjoint extension with domain H2(Rn).Now we associate to the symbol p the bicharacteristic flow given by the following

equations for j = 1, . . . , n,

(1.0.5)

'(()

((*

xj(t) =$p

$%j(x(t), %(t)), xj(0) = xj ,

%j(t) = ' $p

$xj(x(t), %(t)), %j(0) = %j .

We shall denote by (x(t, x, %), %(t, x, %)) the solution, whenever it exists, of the system(1.0.5). In fact it is an easy consequence of (1.0.3) and (1.0.4) that this flow exists for

2 CHAPTER 1. INTRODUCTION AND STATEMENT OF THE RESULT

all t in R. Indeed by (1.0.4) we have

' |%(t)|2 " p(x(t), %(t)) = p(x, %),

and it follows from (1.0.4) that

|xj(t)| " 2n#

k=1

|gjk(x) %k(t)| " C |%(t)| " C '!1/2 p(x, %)1/2.

Our last assumption will be the following.

(1.0.6) For all (x, %) in T #Rn ! {0} we have limt$±"

|x(t, x, %)| = +).

This means that the flow is not trapped backward nor forward. Now let us denote bye!itP the solution of the following initial value problem

(1.0.7)

')

*i$u

$t' Pu = 0

u(0, ·) = u0.

Then the main result of this work is the following.

Theorem 1.0.1. — Assume that the operator P satisfies the conditions (1.0.3),(1.0.4), (1.0.6). Let T > 0 and (q, r) be a couple of real numbers such that q > 2 and2q = n

2 ' nr . Then there exists a positive constant C such that

(1.0.8) *e!itP u0*Lq([!T,T ],Lr(Rn)) " C *u0*L2(Rn),

for all u0 in L2(Rn).

Such estimates are known in the literature under the name of Strichartz estimates.They have been proved for the flat Laplacian by Strichartz [Str] when p = q = 2n+4

n

and extended to the full range of (p, q) given by the scaling by Ginibre-Velo [GV]and Yajima [Y]. The limit case q = 2 (the end point) when n ! 3 is due to Keel-Tao[KT]. These estimates have been a key tool in the study of non linear equations. Veryrecently several works appeared showing a new interest for such estimates in the caseof variable coe"cients. Sta"lani-Tataru [ST] proved Theorem 1.0.1 under conditions(1.0.4) and (1.0.6) for compactly supported perturbations of the flat Laplacian. In [B]Burq gave an alternative proof of this result using the work of Burq-Gerard-Tzvetkov[BGT]. In the same work Burq announced without proof that if you accept to replacein the right hand side of (1.0.8) the L2 norm by an H# norm, for any small ( > 0,then you can weaken the decay hypotheses on the coe"cients of P in the sense thatyou may replace in the definition (1.0.1) of B!0 the power |#| + 1 + "0 by |#| + "0.We have also to mention a recent work of Hassell-Tao-Wunsch [HTW1] who provedin dimension n = 3 a weaker form of our result corresponding to the case whereq = 4, r = 3, under conditions similar to ours. Still more recently these three authorsannounced the same result as ours under hypotheses on the coe"cients similar to ours(see [HTW2]).


CHAPTER 1. INTRODUCTION AND STATEMENT OF THE RESULT 3

It is also worthwhile to mention the work of Burq-Gerard-Tzvetkov who investigatethe Strichartz estimates on compact Riemannian manifolds. In that case they showthat such estimates hold with the L2 norm replaced by the H1/q norm. In the samepaper these authors show that the same result holds on Rn when the coe"cients oftheir Laplacian (and its derivatives) are merely bounded. Let us note also that theseestimates concern also the wave equation and many works have been devoted to thiscase. However we would like to emphasize that, due to the finite speed of propagation,the extension to the variable coe"cients case appear to be much less technical (see[SS]).

Let us now give some ideas on the proof. It is by now well known that a proofof the Strichartz estimates can be done using a dispersion result, duality argumentsand the Hardy-Littlewood-Sobolev lemma. This has been formulated as an abstractresult in the paper [KT] as follows. Assume that for every t $ R we have an operatorU(t) which maps L2(Rn) to L2(Rn) and satisfies,

')

*(i) *U(t) f*L2(Rn) " C *f*L2(Rn), % t $ R, C independent of t,

(ii) *U(s)(U(t))# g*L!(Rn) " C |t ' s|!n/2 *g*L1(Rn), t += s,

then the Strichartz estimates (1.0.5) hold for U(t). It is not di"cult to see that theserious estimate to be proved is (ii). In the case when U(t) = eit!0 (the flat Laplacian)this estimate is obtained by the explicit formula giving the solution in term of thedata u0. In the variable coe"cients case such a formula is of course out of hope andthe better we can have is a parametrix. However due to strong technical di"culties(which we try to explain below) which seem to be serious we are not able to writea parametrix for e!itP so we have to explain what we do instead. First of all let)0 $ C"

0 (Rn) be such that )0(x) = 1 if |x| " 32 and supp)0 , ['1, 1]. With a large

R > 0 we write

e!itP u0(x) = )0

+ x

R

,e!itP u0(x) +

+1 ' )0

+ x

R

,,e!itP u0(x) = v + w.

It is not di"cult to see that the Strichartz estimates for v will be ensured by the resultof Sta"lani-Tataru [ST] while the same estimate for w leads to consider an operatorwhich is a small perturbation of the Laplacian (see Chapter 2).

Now it is not a surprise that microlocal analysis is strongly needed in our proof.So let %0 $ Rn, |%0| = 1 be a fixed direction. Let *0 $ C"(R), *0(s) = 1 if s " 3

4 ,*0(s) = 0 if s ! 1, 0 " *0 " 1 and let us set *+(x) = *0

$'x · %0/&1

%, *!(x) =

*0

$x · %0/&1

%, &1 > 0. We set U+(t) = *+ e!itP , U!(t) = *! e!itP . Now since

*+(x) + *!(x) ! 1 for all x in Rn then Strichartz estimates separately for U+(t) andU!(t) will give the result. It is therefore su"cient to prove the estimate (ii) abovefor U+(s) (U+(t))# = *+ ei(s!t)P *+ (and for U!(s)(U!(t))#). In our proof we shallconstruct a parametrix for these operators.

SOCIETE MATHEMATIQUE DE FRANCE 2005


Our construction relies heavily on the theory of FBI transform (see Sjostrand [Sj]and Melin-Sjostrand [MS]) viewed as a Fourier integral operator with complex phase.One of the reason of our choice is that in our former works on the analytic smoothinge!ect [RZ2] we have already done similar constructions (but only near the outgoingpoints: see below). Let us explain very roughly the main ideas. The standard FBItransform is given by

(1.0.9) Tv(#, +) = cn +3n/4

-

Rn

ei$(y!"x)·"!!"2 |y!"x|2+ "

2 |"!|2 v(y) dy

where # = (#x, #%) $ Rn ( Rn and cn is a positive constant.Let us note that the phase can be written i+)0 where )0(y, #) = i

2 (y'(#x+i#%))2.Then T maps L2(Rn) into the space L2(R2n, e!$|"!|2 d#). The adjoint T # of T is givenby a similar formula (see (6.1.2)) and we have,

(1.0.10) T #T is the identity operator on L2(Rn).

We embed the transform T into a continuous family of FBI transform

(1.0.11)

')

*T&v(#, +) = +3n/4

-

Rn

ei$'(&,y,")a(!, y, #) v(y) dy with

)(0, y, #) = 12 (y ' (#x + i#%))2, a(0, y, #) = cn.

Let us set U(!, t, #, +) = T&[K±(t)u0](#, +), where K±(t) = *± e!itP *±. Then it isshown that if ) satisfies the eikonal equation,

(1.0.12).$)

$!+ p+x,

$)

$x

,/(!, x, #) = 0,

and if the symbol a satisfies appropriate transport equations then U is a solution ofthe following equation +$U

$t+ +

$U

$!

,(!, t, #, +) - 0.

It follows that essentially we have, U(!, t, #, +) = V (! ' +t, #, +). In particular thisshows that U(0, t, #, +) = U('+t, 0, #, +). Written in terms of the transformations T&

this readsT [K±(t)u0](#, +) = T!$t[*2

± u0](#, +).Applying T # to both members and using (1.0.10) we obtain

K±(t)u0(x) = T #{T!$t[*2± u0](·, +)}(t, x).

Thus we have expressed the solution in terms of the data through a Fourier integraloperator with complex phase.

This short discussion shows that as usual the main point of the proof is to solvethe eikonal and transport equations. Let us point out the main di"culties whichoccur in solving these equations. They are of three types: the bad behavior of theflow from incoming points and for large time, the global (in !, x) character of allour constructions and the mixing of C" coe"cients and complex variables (comingfrom the non real character of our phase). Let us discuss each of them. First of all


CHAPTER 1. INTRODUCTION AND STATEMENT OF THE RESULT 5

whatever the method you use to solve an eikonal equation (symplectic geometry oranother one) a precise description of the flow of the symbol p is needed. Let us recall(see (1.0.5)) that our flow (x(t, x, %), %(t, x, %)), issued from the point (x, %) $ T #Rn!0,is defined for all t $ R. In the case of the flat Laplacian we have %(t, x, %) = % andx(t, x, %) = x + 2t%. Let now (x, %) $ T #Rn ! {0} and assume that x · % ! 0. Thenit is easy to see that |x(t, x, %)|2 ! |x|2 + 4t2 |%|2 for t ! 0 so that |x(t, x, %)| becomeslarger and larger while x(t, x, %) may vanish for a large t < 0. Such a point is called“outgoing for t ! 0” and “incoming for t < 0”. In the case of a perturbed Laplacianthis distinction between the directions is very important. Indeed although the flowfrom outgoing points for t ! 0 is very well described for t ! 0 and has very similarproperties to the flat case, it has a bad behavior for t < 0 in what concerns itsderivatives with respect to (x, %). For instance (xj

(%k(t, x, %) does not behave at all as

2t &jk. This is of great importance and causes some trouble in the proof. Howeverstill when t < 0, the flow behaves correctly as long as the point (x(t, x, %), %(t, x, %))is outgoing for t ! 0. Roughly speaking that is the reason why we are not able toconstruct a parametrix for e!itP while it is possible for the operator *± e!itP *±. TheChapter 3 is entirely devoted to a careful study of the flow. Let us now describe ourmethod of resolution of the eikonal equation. The classical method uses the ideasof symplectic geometry. Roughly speaking the manifold constructed from the flowis a Lagrangian manifold on which the symbol , + p(x, %) is constant. If it projects(globally) and clearly on the basis then it is a graph of some function ) which is thedesired phase. However this general method leads immediately to a di"culty in ourcase. Indeed since we want that for ! = 0 the phase ) coincides with the phase )0 ofthe FBI transform (see (1.0.9)) which is non real, we should take, in solving the flow,data which are non real, so the flow itself would be non real; but our symbol has merelyC" coe"cients. To circumvent this di"culty a method has been proposed by Melin-Sjostrand [MS] which uses the almost analytic machinery. Another method, di!erentin spirit, that the one described above and known under the name of “Lagrangianideals”, has been introduced by Hormander [H]. Here the initial data in the flow arekept real. Let us set uj(x, %) = %j ' ('0

(xj(x, %) = %j '#j

% ' i(xj '#jx). Then obviously

we have {uj, uk} = 0 if j += k (where {, } denotes the Poisson bracket). Now let usset vj(!, x, %) = uj(x('!, x, %), %('!, x, %)), j = 1, . . . , n. Then for every ! in R thePoisson bracket of vj and vk still vanishes if j += k. Thus the ideal generated by thevj ’s is closed under the Poisson bracket. The main step in Hormander’s method is toshow that this ideal is generated by functions of the form %j ' #j(!, x, #). This willimply that one can find a function ) = )(!, x, #) such that ('

(xj(!, x, #) = #j(!, x, #)

and it turns out that ) is the desired phase. To achieve its main step, Hormanderuses a precise version of the Malgrange preparation theorem which is discussed in [H],tome 1. This is the way we chose to use in our case. It occupies all Chapter 4 of thepaper. The proof is made separately for outgoing and incoming points. Since the vj ’sare defined by mean of the backward flow, in both cases we encounter the di"culty



caused by the bad behavior of the flow from incoming points. As it can be seen manytechnical di"culties arise in the procedure.

The next step in the proof is the resolution of the transport equations. Here alsothe cases of outgoing and incoming points have to be separated. We have also tobe careful since these are first order equations with non real C" coe"cients. Thefirst case is easier. Indeed due to the good behavior of the flow and the decay of theperturbation one can cut the Taylor expansion of the coe"cients of the vector field tosome order and thus reduce ourselves to the case of polynomial coe"cients. Then byclassical holomorphic methods one can solve the equations modulo flat terms whichwill be enough for our purpose. In the second case there is no more such an asymptoticand the situation is much more intricate. So we use the classical idea which consists instraightening the vector field. This forces us to enter in the almost analytic machineryof Melin-Sjostrand [MS] (see Chapter 5). Of course all the constructions made aboveare done microlocally and in a neighborhood of the bicharacteristic. Therefore todefine the general FBI transform T& (see (1.0.11)) as well as to pass from the standardT to T!$t we have to insert many microlocal cut-o!. Of course we have to check ateach microlocalization that the remainder leads to an acceptable error. This is thegoal of Chapter 6. At this stage of the proof the operator K±(t) = *± e!itP *± iswritten as

K±(t)u0(x) =-

k±(t, x, y)u0(y) dy

wherek±(t, x, y) =

-ei$F (!$t,x,y,") a(+t, x, y, #) d#.

Thus the dispersion estimate would follow from the bound

|k±(t, x, y)| " C

|t|n/2

for 0 < |t| " T .Here we have two regimes according to the fact that |+t| ! 1 or |+t| " 1. In the

first case on the support of a(+t, x, y, #) we could be very far from the critical pointof F . Fortunately the phase F has enough convexity to produce the desired boundof k±. In the second regime we are close to the critical point of F so we expect astationary phase method to work. However since the phase F is non real and since thedeterminant of its Hessian in # degenerates in some direction when |+t| . 0 we cannotapply the standard results as they appear in [H]. Instead, after a careful study ofthe phase F we use merely an integration by part method with an appropriate vectorfield to conclude. This is done in Chapter 7. The rest of this part is devoted, usingthe Littlewood-Paley theory, to the end of the proof of our main Theorem.

Finally an Appendix gathers the proofs of some technical results used in the paper.

Acknowledgments. — We would like to thank Nicolas Burq for useful discussions atan earlier stage of the work.


CHAPTER 2

PRELIMINARIES AND REDUCTION TO THE CASE OFA SMALL PERTURBATION OF THE LAPLACIAN

2.1. Preliminaries

We begin by recalling several earlier results which will be used in the sequel.The first result concerns the case of compactly supported perturbations of the

Laplacian.

Theorem 2.1.1 (Sta"lani-Tataru [ST]). — Let P be defined by (1.0.2). Assumethat P satisfies (1.0.4), (1.0.6) and

(2.1.1) for j, k = 1, . . . , n, gjk ' &jk, bj , V are compactly supported.

Then the Strichartz estimates (1.0.8) hold.

The second result which we recall is the extension to the variable coe"cients caseby Doı [D] of the Kato smoothing e!ect. Let us introduce the following space. Weset for s, µ in R

Hsµ(Rn) = {u $ S% : "x#µ (I ' $)s/2 u $ L2(Rn)}

with its standard norm.

Theorem 2.1.2 (Doı [D]). — Let P be defined by (1.0.2) and assume it satisfies theconditions (1.0.3), (1.0.4), (1.0.6). Then for all T > 0 and all " > 1

2 one can find aconstant C ! 0 such that,

(2.1.2) *e!itP u0*L2([!T,T ],H1/2"# (Rn))

" C *u0*L2(Rn),


We shall also use the following result.

Lemma 2.1.3 (Keel-Tao [KT]). — Let (X, dx) be a measure space, H a Hilbert spaceand T > 0. Suppose that for each time t $ ['T, T ] we have an operator U(t) : H .L2(X) which satisfies the following estimates.

8 CHAPTER 2. PRELIMINARIES

(i) There exists C1 ! 0 such that for all t $ ['T, T ] and all f $ H,

*U(t) f*L2(X) " C1 *f*H .

(ii) There exists C2 ! 0 such that for all t, s $ ['T, T ], t += s and all g $ L1(X),

*U(t)(U(s))# g*L!(X) " C2 |t ' s|!n/2 *g*L1(X).

Let (q, r) be a couple of real numbers such that q ! 2, r < +) and 2q = n

2 ' nr . Then

there exists C ! 0 such that for all f in H

*U(t) f*Lq([!T,T ],Lr(X)) " C *f*H .

This result will be used in the sequel with H = L2(Rn), X = Rn.Finally let’s recall the following technical lemma.

Lemma 2.1.4 (Christ-Kiselev [CK]). — Let X, Y be two Banach spaces and K(t, s)be a continuous function taking its values in B(X, Y ), the space of bounded linearmappings from X to Y . Let ') " a < b " +) and set

Sf(t) =- b

aK(t, s) f(s) ds

Wf(t) =- t

aK(t, s) f(s) ds.

Let 1 " p < q " +). Then if we can find a constant C > 0 such that

*Sf*Lq((a,b),Y ) " C *f*Lp((a,b),X)

it follows that

*Wf*Lq((a,b),Y ) " 2!2( 1p! 1

q ) · 2C

1 ' 2!( 1p! 1

q )*f*Lp((a,b),X).

Using these results we shall see that Theorem 1.0.1 will be a consequence of thefollowing Theorem.

Theorem 2.1.5. — Let us set $g =&n

j,k=1(

(xj

$gjk (

(xk

%and assume that the con-

ditions (1.0.3), (1.0.4), (1.0.6) are satisfied by $g. Let T > 0 and (q, r) be a coupleof real numbers such that q > 2 and 2

q = n2 ' n

r . Then there exists a positive constantC such that 00eit!g u0

00Lq([!T,T ],Lr(Rn))

" C *u0*L2(Rn)


Let us show how Theorem 2.1.5 implies Theorem 1.0.1.Let us set I = [0, T ]. (The case I = ['T, 0] is symmetric). Using (1.0.2) we can

write

(2.1.3) i $t u + $gu = '+ n#

j=1

(Dj bj) + V,

u ' 2n#

j=1

bj Dju =: F = F1 + F2.


2.1. PRELIMINARIES 9

It follows from Duhamel formula that

(2.1.4) e!itP u0 = eit!g u0 + i

- t

0ei(t!s)!g [F (s, ·)] ds.

Using Theorem 2.1.5 we obtain

(2.1.5)00eit!g u0

00Lq(I,Lr(Rn))

" C *u0*L2(Rn).

Let us set now

(2.1.6) Sf(t) =- T

0ei(t!s)!g [f(s, ·)] ds.

Since Sf(t) = eit!g1 T0 e!is!g [f(s, ·)] ds we can use Theorem 2.1.5 to write

*Sf(t)*Lq(I,Lr(Rn)) " C000- T

0e!is!g [f(s, ·)] ds

000L2(Rn)

" C

- T

0

00e!is!g [f(s, ·)]00

L2(Rn)ds

" C

- T

0*f(s, ·)*L2(Rn) ds = C *f*L1(I,L2(Rn)).

Using Lemma 2.1.4 with p = 1, q > 2, Y = Lr(Rn), X = L2(Rn) we deduce that000- t

0ei(t!s) !g [F1(s, ·)] ds

000Lq(I,Lr(Rn))

" C *F1*L1(I,L2(Rn))

where F1 = '$&n

j=1(Dj bj)+V )u. Since&n

j=1 |Dj bj|+|V | is bounded (by condition(1.0.3)) we have

*F1*L1(I,L2(Rn)) " C

- T

0*u(s, ·)*L2(Rn) ds " C% T *u0*L2 .

Therefore we have

(2.1.7)000- t

0ei(t!s)!g [F1(s, ·)] ds

000 " C(T ) *u0*L2(Rn).

Let us look to the term corresponding to F2 in (2.1.3), (2.1.4). Let us fix " =12 + 1

2 "0. Then by Theorem 2.1.2 the operator eit !g is continuous from L2(Rn) toL2(I, H1/2

!! (Rn)). Its adjoint is defined by

((eit !g u0, f)) = (u0, U#f)L2(Rn)

where (( , )) denotes the duality between L2(I, H1/2!! ) and L2(I, H!1/2

! ). It satisfiesthe estimate

*U#f*L2(Rn) " C *f*L2(I,H

"1/2# (Rn))

.

A straightforward computation shows that

U#f(x) =- T

0e!is !g [f(s, ·)] ds.



Using Theorem 2.1.5 for $g we see that the operator S introduced in (2.1.6) satisfiesthe estimate

*Sf(t)*Lq(I,Lr(Rn)) " C *f*L2(I,H"1/2

# (Rn)).

Using Lemma 2.1.4 with p = 2, q > 2, Y = Lr(Rn), X = H!1/2! (Rn) we see that

(2.1.8)000- t

0ei(t!s) !g [F2(s, ·)] ds

000Lq(I,Lr(Rn))

" C *F2*L2(I,H"1/2# (Rn))

where F2 = '2&n

j=1 bj Dj u. If we set, with $ =&n

j=1(2

(x2j,

(2.1.9) A = "x#! (I ' $)!1/4n#

j=1

bj Dj(I ' $)!1/4 "x#!

then we can write

(2.1.10) *F2*2L2(I,H"1/2

# (Rn))= 4

- T

0

00A"x#!! (I ' $)1/4 u(s, ·)002

L2(Rn)ds.

Let us consider the metric on the cotangent space

G =dx2

"x#2 +d%2

"%#2 .

It is a Hormander’s metric and we have "x#! $ OpS("x#! , G), (I ' $)!1/4 $OpS

$"%#!1/2, G

%, bj $ Op S("x#!2! , G), Dj $ Op S("%#, G). It follows that the oper-

ator A introduced in (2.1.9) belongs to OpS(1, G) and therefore is L2 continuous. Itfollows then from (2.1.8), (2.1.10) that

000- t

0ei(t!s) !g [F2(s, ·)] ds

000Lq(I,Lr(Rn))

" C. - T

0*u(s, ·)*2

H1/2"# (Rn)

ds/1/2

.

Using Theorem 2.1.2 for P we deduce that

(2.1.11)000- t

0ei(t!s)!g [F2(s, ·)] ds

000Lq(I,Lr(Rn))

" C% *u0*L2(Rn).

Gathering the informations given by (2.1.4), (2.1.5), (2.1.7) and (2.1.11) we obtainthe conclusion of Theorem 1.0.1. So we are left with the proof of Theorem 2.1.5.

2.2. Reduction to a small perturbation

The purpose of this Section is to show that, using the result of 2.1 one can reducethe proof of Theorem 2.1.5 to the case of a small perturbation of the flat Laplacian.

Let ) be in C"0 (Rn). We write eit !gu0 = u and

(2.2.1) u = )u + (1 ' ))u = v + w.


2.2. REDUCTION TO A SMALL PERTURBATION 11

(i) Estimate of v. — Since v = )u it follows from (1.0.7) that (i$t+$g) v = [$g, )] u.Let )1 $ C"

0 (Rn) be such )1 = 1 on the support of ) then setting $ =&n

j=1(2

(x2j

one can write

(2.2.2) (i $t + $g) v = (i $t + $ + )1($g ' $))1) v = [$g, )] u

and )1($g ' $) is a compactly supported perturbation of the flat Laplacian. Let usset 2P = '$ ' )1($g ' $))1. We have, from (2.2.2)

(2.2.3) v = e!it eP )u0 +- t

0e!i(t!s) eP [f(s, ·)] ds

where f = [$g, )] u.It follows from Theorem 2.1.1 that

(2.2.4)00e!it eP )u0

00Lq([!T,T ],Lr(Rn))

" C *u0*L2(Rn).

To estimate the second term in the right-hand side of (2.2.3) we shall use Lemma 2.1.4with a = 'T , b = T , Y = Lr(Rn), p = 2, X = H!1/2(Rn). For this one first remarkthat if U = e!it eP then Theorem 2.1.2 shows that U is continuous from L2(Rn) toL2(['T, T ], H1/2

loc (Rn)). Then it is easy to see that U# : L2(['T, T ], H!1/2c (Rn)) .

L2(Rn) is continuous and is given by U#f(x) =1 T0 e!is eP [f(s, ·)] ds. It follows that

000- T


000Lq([!T,T ],Lr(Rn))

= *U U#f*Lq([!T,T ],Lr(Rn)).

Then, using again Theorem 2.1.1 and the above continuity of U# we get

*U U#f*Lq([!T,T ],Lr(Rn)) " C *U#f*L2(Rn) " C% *f*L2([!T,T ],H"1/2(Rn))

since f = [$g, )] u has compact support in x.Now we use Lemma 2.1.4 to deduce that

000- t


000Lq([!T,T ],Lr(Rn))

" C%% *f*L2([!T,T ],H"1/2(Rn))

since f(s, ·) has compact support in x and q > 2.Moreover since [$g, )] is first order we have, using again Theorem 2.1.2,

*f*L2([!T,T ],H"1/2(Rn)) " C *-u*L2([!T,T ],H1/2(Rn)) " C% *u0*L2(Rn)

where - $ C"0 (Rn), - = 1 on the support of ). This gives the estimate of the

second term in the right hand side of (2.2.3) which, together with (2.2.4) shows thatv satisfies the Strichartz estimate.

(ii) Estimate of w. — We shall take the function ), introduced above, of the followingform. Let R > 0 (which will be chosen large enough) and )0 $ C"

0 (Rn) such that)0(x) = 1 if |x| " 3

2 , supp )0 , ['2, 2]. We shall take )(x) = )R(x) = )0

$x/R

%.

Let 2)0 $ C"0 (Rn) be such that 2)0(x) = 1 if |x| " 1

2 , supp 2)0 , ['1, 1] and let usset 2)R(x) = 2)0

$xR

%.



Let w = (1 ' )R)u be the second term in the right hand side of (2.2.1). Since1 ' 2)R = 1 on the support of 1 ' )R we have according to (1.0.2)

(2.2.5) (i $t + $g)w =+i$t + $ +

n#

j,k=1

$

$xj

.(1 ' 2)R) bjk

$

$xk

/,w = '[$g, )R] u

where bjk = gjk ' &jk.Now if we denote by f one of the coe"cients bjk we claim that we have

(2.2.6) |$"x [(1 ' 2)R) f ](x)| " 1

R!0/2

C"

"x#|"|+1+#02

, %x $ Rn.

Using (1.0.1) and denoting by A the left hand side of (2.2.6) we see that

A "+1 ' 2)

+ x

R

,,|$"

x f(x)| +#

0<)!"

3#.

41

R|)|

555($)x 2))

+ x

R

,555|$"!)x f(x)|

A "+1 ' 2)

+ x

R

,, C"

"x#|"|+1+!0+#

0<)!"

C%")

R|)|

555($)x 2))

+ x

R

,5551

"x#|"|!|)|+1+!0.

Now, on the support of 1 ' 2)(x/R) we have "x# > |x| ! 32 R so the first term is

bounded byC%

"

R!0/2

1"x#|"|+1+

#02

.

On the support of $) 2)( xR ), with . += 0, we have 1

2 R " |x| " R so "x# "/

2R ifR > 1. Therefore the second term is bounded by

1R!0/2

#

0<)!"

C%%")

1"x#|)|!

#02 +|"|!|)|+1+!0

" 1R!0/2

C%%"

"x#|"|+1+#02

.

It follows from (2.2.6) that we can work in the rest of the paper with a non negativeself adjoint operator P such that

(2.2.7)

'()

(*

P = '$ + ( Q, where Q =&

|)|!2 a#) D) ,

( is a small constant and |D"x a#

)(x)| " C"/"x#|"|+1+!0/2, %# $ Nn,

uniformly for x $ Rn with C" independent of (.

Since the estimates on the coe"cients are uniform in ( we shall write a) instead ofa#

) . The principal symbol p of P will be written as

p(x, %) = |%|2 + ( q(x, %), q(x, %) =n#

j,k=1

bjk(x) %j %k

and we shall take ( so small that910

|%|2 " p(x, %) " 1110

|%|2.

Finally without loss of generality we shall take "0 instead of !02 in (2.2.7).

We assume that P satisfies the condition (1.0.3) and (1.0.6). Let T > 0.


2.2. REDUCTION TO A SMALL PERTURBATION 13

Theorem 2.2.1. — Let (q, r) be such q ! 2 and 2q = n

2 ' nr . If ( is small enough

then there exists C > 0 such that

*e!itP v0*Lq([!T,T ],Lr(Rn)) " C *v0*L2(Rn)

for all v0 $ L2(Rn).

Let us assume that we have proved this result. Then we can applied it to theoperator occurring in (2.2.5) with R large enough. We have,

w = e!itP (1 ' )R)u0 +- t

0e!i(t!s)P [fR(s, ·)] ds.

It follows from Theorem 2.2.1 that

(2.2.8) *e!itP (1 ' )R)u0*Lq([!T,T ],Lr(Rn)) " C *u0*L2(Rn)

and the same argument as used in the estimate of v, namely the use of Theorem2.2.1,and Lemma 2.1.4 shows that

(2.2.9)000- t

0e!i(t!s)P [fR(s, ·)] ds

000Lq([!T,T ],Lr(Rn))

" C(R)*u0*L2(Rn).

Then using (2.2.8) we see that the second term w in the right hand side of (2.2.1)satisfies the Strichartz estimate which completes the proof of Theorem 2.1.5.

Our goal now is to prove Theorem 2.2.1. The first step is to make a careful studyof the flow.


CHAPTER 3

STUDY OF THE FLOW

3.1. Preliminaries

Let p(x, %) = |%|2 + ( q(x, %), q(x, %) =&n

j,k=1 bjk(x) %j %k where,

(3.1.1)

6there exists "0 > 0 such that for every / $ N one can find A* > 0

such that&

|"|=*

&nj,k=1 |$"

x bjk(x)| " A*/"x#1+*+!0 for all x in Rn.

We introduce the equations of the bicharacteristic flow issued from a point (x, %)in T #Rn ! {0}. They are given for j = 1, . . . , n, by

(3.1.2)

'(()

((*

xj(t) =$p

$%j(x(t), %(t)), xj(0) = xj ,

%j(t) = ' $p

$xj(x(t), %(t)), %j(0) = %j ,

and we denote by (x(t, x, %), %(t, x, %)) the solution of (3.1.2) whenever it exists (or(x(t), %(t)) for short if no confusion is possible).

Let us remark that when p(x, %) = |%|2 then

(3.1.3)7

x(t, x, %) = x + 2t%%(t, x, %) = %

In general case assuming ( so small that ( A0 " 110 we see that 9

10 |%|2 " p(x, %) "

1110 |%|

2. It follows that910

|%(t, x, %)|2 " p(x(t), %(t)) = p(x, %) " 1110

|%|2,

so that

(3.1.4) |%(t, x, %)| " 2 |%|.

Using the first equation of (3.1.2) we see then, that the solution of (3.1.2) exists forall t in R and is a C" function with respect to (x, %). Moreover we have the followinglemma.

16 CHAPTER 3. STUDY OF THE FLOW

Lemma 3.1.1. — For all t in R we have

x(t, x, %) · %(t, x, %) = x · % + 2t p(x, %) + f(t, x, %)

where

|f(t, x, %)| " 4( A1 |%|2555- t

0

ds

"x(s)#1+!0

555 " 4( A1 |t| |%|2.

Proof. — We have by (3.1.2)

d

dt[x(t) · %(t)] = %(t) · $p

$%(x(t), %(t)) ' ( x(t) · $q

$x(x(t), %(t)).

Using Euler’s identity we obtain

%(t) · $p

$%(x(t), %(t)) = 2p(x(t), %(t)) = 2p(x, %).

We set f(t, x, %) = '(1 t0 x(s) · (q

(x (x(s), %(s)) ds. Now since555$q

$x(x(s), %(s))

555 " A1

"x(s)#2+!0|%(s)|2,

it follows from (3.1.4) that

|f(t, x, %)| " 4( A1 |%|2555- t

0

ds

"x(s)#1+!0

555 " 4( A1 |t| |%|2.

We shall use later on the result given by the following lemma.For t $ R and (x, %) $ T #Rn ! {0} let us set,

(3.1.5) 0(t, x, %) = (x(t, x, %), %(t, x, %)).

Lemma 3.1.2. — We have the following identities for j, k = 1, . . . n,.

$xj

$xk(t, x, %) =

$%k

$%j('t, 0(t, x, %))

$xj

$%k(t, x, %) = '$xk

$%j('t, 0(t, x, %))

$%j

$xk(t, x, %) = ' $%k

$xj('t, 0(t, x, %))

$%j

$%k(t, x, %) =

$xk

$xj('t, 0(t, x, %)).

Proof. — For j = 1, . . . , n we have7

xj('t; 0(t; x, %)) = xj

%j('t; 0(t, x, %)) = %j


3.1. PRELIMINARIES 17

Di!erentiating both sides with respect to xk and %k we obtainn#

*=1

$xj

$x*('t; 0(t; x, %))

$x*

$xk(t; x, %) +

n#

*=1

$xj

$%*('t; 0(t; x, %))

$%*

$xk(t; x, %) = &jk

n#

*=1

$xj

$x*('t; 0(t; x, %))

$x*

$%k(t; x, %) +

n#

*=1

$xj

$%*('t; 0(t; x, %))

$%*

$%k(t; x, %) = 0

n#

*=1

$%j

$x*('t; 0(t; x, %))

$x*

$xk(t; x, %) +

n#

*=1

$%j

$%*('t; 0(t; x, %))

$%*

$xk(t; x, %) = 0

n#

*=1

$%j

$x*('t; 0(t; x, %))

$x*

$%k(t; x, %) +

n#

*=1

$%j

$%*('t; 0(t; x, %))

$%*

$%k(t; x, %) = &jk

where &jk is the Kronecker symbol.If we set

M(t; 0) =

8$$xj/$xk

%(t; 0)

$$xj/$%k

%(t; 0)

$$%j/$xk

%(t; 0)

$$%j/$%k

%(t; 0)

9

then the above relations can be written

(3.1.6) M('t; 0(t; x, %)) · M(t; x, %) =3

In 00 In

4

where In denotes the n ( n identity matrix.Let us introduce for s $ R the following matrix.

(3.1.7) A(s; 0) =

8t$$%j/$%k

%(s; 0) 't

$$xj/$%k

%(s; 0)

't$$%j/$xk

%(s; 0) t

$$xj/$xk

%(s; 0)

9.

We claim that for s $ R and 0 $ T #Rn we have

(3.1.8) A(s; 0)M(s; 0) = I2n

where I2n is the 2n ( 2n identity matrix.Indeed let us set A(s; 0) · M(s; 0) = (C"))1!",)!2n. We have for j, k = 1 . . . n,

(3.1.9)

'((((((((((((()

(((((((((((((*

Cj,k =n#

*=1

$$%*

$%j

$x*

$xk' $x*

$%j

$%*

$xk

%(s; 0)

Cj,k+n =n#

*=1

$$%*

$%j

$x*

$%k' $x*

$%j

$%*

$%k

%(s; 0)

Cj+n,k =n#

*=1

$$x*

$xj

$%*

$xk' $x*

$xk

$%*

$xj

%(s; 0)

Cj+n,k+n =n#

*=1

$$x*

$xj

$%*

$%k' $%*

$xj

$x*

$%k

%(s; 0).

Let us remark that Cj+n,k+n = Ck,j .



Now we recall that for every s $ R the map (x, %) 0. 0(s; x, %) is symplectic whichmeans that

(3.1.10)n#

*=1

d(%*(s; x, %)) 1 d(x*(s; x, %)) =n#

j=1

d%j 1 dxj .

Writing u(s) = u(s; x, %) for short we have

(1) =n#

*=1

d(%*(s)) 1 d(x*(s)) =n#

*=1

3 n#

j=1

+ $%*

$xj(s) dxj +

$%*

$%j(s) d%j

,

1n#

k=1

+ $x*

$xk(s) dxk +

$x*

$%k(s) d%k

,4

It follows that

(1) =#

j!k

3 n#

*=1

+ $%*

$xj(s)

$x*

$xk(s) ' $%*

$xk(s)

$x*

$xj(s),4

dxj 1 dxk+

n#

j,k=1

3 n#

*=1

+$%*

$%j(s)

$x*

$xk(s) ' $%*

$xk(s)

$x*

$%j(s),4

d%j 1 dxk+

#

j!k

3 n#

*=1

+$%*

$%j(s)

$x*

$%k(s) ' $%*

$%k(s)

$x*

$%j(s),4

d%j 1 d%k.

Using (3.1.9) and (3.1.10) we see easily that

Cj,k = &jk, Cj,k+n = Cj+n,k = 0, Cj+n,k+n = Ck,j = &jk.

This proves (3.1.8).It follows from (3.1.6) and (3.1.8) that

(3.1.11) M(t; x, %) = A('t; 0(t; x, %))

which by (3.1.7) proves the Lemma 3.1.2.

3.2. The flow for short time

Here is a description of the flow for short time.

Proposition 3.2.1. — Let us set6r(t, x, %) = x(t, x, %) ' (x + 2t%)

1(t, x, %) = %(t, x, %) ' %.

Let T > 0. Then for all A, B in Nn one can find CA,B > 0 such that6

(i) |$Ax $B

% Z(t, x, %)| " CA,B ( |t|(ii) |$t $A

x $B% Z(t, x, %)| " CA,B (

if Z = r or 1, for all |t| " T and all (x, %) $ T #Rn with |%| " 3.


3.3. THE FORWARD FLOW FROM POINTS IN S+ AND BACKWARD FROM S" 19

Proof. — See Appendix, Paragraph A.2.

We introduce now the following definition which distinguish microlocally the pointsin the cotangent bundle.

Definition 3.2.2. — LetS+ =

:(x, %) $ T #Rn ! {0} : x · % ! ' 1

4 "x# |%|;

S! =:(x, %) $ T #Rn ! {0} : x · % " 1

4 "x# |%|;.

Then S+ (resp. S!) is called the set of outgoing points for t ! 0 (resp. t " 0).

Of course the constant 14 in the above definition is unimportant and could be

replaced by any fixed small constant. The reason for this definition is the following.If (x, %) $ S+ then, for t ! 0

1 + |x + 2t%|2 ! 12

("x#2 + t2 |%|2).

Since x + 2t% will be an approximation of x(t; x, %), then S+, will be the set of points(x, %) for which the projection of the bicharacteristic goes to +) when t . +) instaying away from the origin.

3.3. The forward flow from points in S+ and backward from S!

Our goal is to obtain for these points a nice global representation of the flowtogether with precise estimates of its derivatives with respect to x and %.

Proposition 3.3.1. — There exists (0 > 0 depending on the constants A0, A1 in(3.1.1) such that for ( in ]0, (0[ the solution of (3.1.2) with (x, %) in S+ (resp. S!)and 1

2 " |%| " 2 can be written for all t ! 0 (resp. t " 0)

(3.3.1)

6x(t; x, %) = x + 2t %(t; x, %) + z(t; x, %)

%(t; x, %) = % + 1(t; x, %)

with

(3.3.2) |zj(t; x, %)| " 2 · 102

"0( max(A0, A1), |1j(t; x, %)| " 2 · 102

"0( max(A0, A1),

where A0A1 are the constants arising in (3.1.1) and j = 1, . . . , n.Moreover for all t ! 0 (resp. t " 0) we have

(3.3.3)13

" 1 + |x(t; x, %)|2

1 + |x|2 + t2" 40.

Proof. — Let

I =!T > 0 : |zj(t)| " 2 · 102

"0( max(A0, A1), |1j(t)| " 2 · 102

"0( max(A0, A1)

for j = 1, . . . , n and all t $ [0, T ]"

.



Then I is an interval which is non empty by the local Cauchy-Lipschitz Theorem. LetT # = sup I. If T # = +) we are done. Otherwise let T < T #. Since 1

2 " |%| " 2 wehave for t $ [0, T ], 1

3 " |%(t)| " 3 if ( max(A0, A1) is small enough. Indeed we have

(1 ' ( A0)|%|2 " p(x, %) = p(x(t), %(t)) " (1 + ( A0)|%(t)|2

(1 ' ( A0)|%(t)|2 " p(x(t), %(t)) = p(x, %) " (1 + ( A0)|%|2.

Now, for t in [0, T ] we have

1 + |x(t)|2 = "x#2 + 4t2 |%|2 + 4t2 |1(t)|2 + |z(t)|2 + 4t x · %< => ?(1)

+ 4t x · 1(t)< => ?(2)

+ 2x · z(t)< => ?(3)

+ 8t2 % · 1(t)< => ?(4)

+ 4t % · z(t)< => ?(5)

+ 4t 1(t) · z(t)< => ?(6)

.

Since (x, %) $ S+ we have for t ! 0, (1) ! ' 12 ("x#2 + t2 |%|2). Now, by the definition

of I we have on [0, T ] if ( max(A0, A1) is small enough.

|(2)| " C1(n) t |x| ( max(A0, A1) " 10!2 (|x|2 + t2)

|(3)| " C2(n) |x| ( max(A0, A1) " 10!2 "x#2

|(4)| " C3(n) t2 ( max(A0, A1) " 10!2 t2

|(5)| " C4(n) t ( max(A0, A1) " 10!2 (1 + t2)

|(6)| " C5(n) t(( max(A0, A1))2 " 10!2 (1 + t2).

It follows that

"x(t)#2 ! 12("x#2 + t2) ' 4 · 10!2("x#2 + t2) ! 1

3("x#2 + t2).

The same computation shows that "x(t)#2 " 40("x#2 + t2). It follows that on [0, T ]we have

(3.3.4)1/6

(1 + t) " 1/3

("x#2 + t2)1/2 " "x(t)# " 7("x#2 + t2)1/2.

Now it follows from (3.1.2) that (z(t), 1(t)) satisfy the equations

(3.3.5)

'(()

((*

zj(t) = '($q

$%j(x(t), %(t)) + 2t (

$q

$xj(x(t), %(t))

1j(t) = '($q

$xj(x(t), %(t))

with zj(0) = 1j(0) = 0.We deduce from (3.1.1), (3.3.4) and the bounds 1

3 " |%(t)| " 3 that555$q

$%j(x(t), %(t))

555 " 3 A0

"x(t)#1+!0" 3(

/6)1+!0A0

(1 + t)1+!0" 12 A0

(1 + t)1+!0,

555$q

$xj(x(t), %(t))

555 " 9 A1

"x(t)#2+!0" 9(

/6)1+!0

/3A1

(1 + t)1+!0 "t# " 60 A1

(1 + t)1+!0 "t# ,


3.4. PRECISIONS ON THE FLOW IN THE GENERAL CASE 21

(since we may assume that 1 + "0 < 32 and (

/6)1+!0 < 4). It follows from (3.3.5)

that

|zj(t)| " 132 (

(1 + t)1+!0max(A0, A1), |1j(t)| " 60 (

(1 + t)1+!0max(A0, A1).

Therefore we have on [0, T ]

|zj(t)| " 132"0

( max(A0, A1), |1j(t)| " 60"0

( max(A0, A1).

Since z(t) and 1(t) exist for all t ! 0 and are smooth we still have the above estimateson [0, T #]. By continuity it will exist 2 > 0 such that |zj(t)| " 2·102

!0( max(A0, A1)

and |1j(t)| " 2·102

!0( max(A0, A1) on [0, T # + 2]. This contradicts the maximality of

T # and proves that T # = +).

Now we estimate the derivatives of the flow with respect to (x, %).

Proposition 3.3.2. — With the notations of Proposition 3.3.1, for every integer kone can find a positive constant Mk such that for all (A, B) $ Nn ( Nn such that|A| + |B| " k, all t ! 0 (resp. t " 0) and (x, %) in S+ (resp. S!) we have,

'()

(*

55$Ax $B

% z(t, x, %)55 " ( Mk

"x#|A|+!0,

55$Ax $B

% 1(t, x, %)55 " ( Mk

"x#1+|A|+!0.

Proof. — See Appendix A.3.

Corollary 3.3.3. — Keeping the notations of Proposition 3.3.1 we have, for allt ! 0 (resp. t " 0) and all (x, %) $ S+ (resp. S!)

$xj

$%k(t, x, %) = 2t &jk + O(("t#), $xj

$xk(t, x, %) = &jk + O(("t#)

$%j

$%k(t, x, %) = &jk + O((),

$%j

$xk(t, x, %) = O((), j, k = 1, . . . , n,

where &jk is the Kronecker symbol and O(() means “bounded by C ( where C is inde-pendent of (x, %)”. In particular we have

(3.3.6)$%j

$%k(t, x, %) ' i

$xj

$%k(t, x, %) = (1 ' 2it) &jk + O(("t#), j, k = 1, . . . , n .

3.4. Precisions on the flow in the general case

The results obtained above allow us to give a rough form of the flow through anypoint in T #Rn ! {0} for t $ R.

Proposition 3.4.1. — Let (x, %) $ T #Rn ! {0} with |%| " 2. Then,(i) the function s 0. "x(s, x, %)#!(1+!0) belongs to L1(R),



(ii) for t $ R we have,6

x(t, x, %) = x + 2t % + r(t, x, %),

%(t, x, %) = % + 1(t, x, %),

where |r(t, x, %)| " C ( "t#, |1(t, x, %)| " C ( with C independent of (x, %).

Before going into the proof let us note that in general we do not have good estimateson the derivatives of r with respect to (x, %) (in the spirit of those given in Proposition3.3.1 for instance). In particular we do not have a good control of (xj

(%k(t, x, %). This

occurs for instance for points (x, %) such that |x| is very large and the bicharacteristiccrosses back a neighborhood of the origin. That’s why we used the term rough forthis description.

Proof of Proposition 3.4.1. — If |x · %| " ' 14 "x# |%| then Proposition 3.3.1 gives the

claimed description of the flow for t ! 0 and t " 0. If x · % " ' 14 "x# |%| the same

Proposition applies for t " 0 so we are left with the case t ! 0. (The case x · % !' 1

4 "x# |%| is symmetric). It follows from Lemma 3.1.1 that, if ( A1 is small enough,we have limt$+" x(t) · %(t) = +). Since x · % " 0 one can find t# > 0 such thatx(t#, x, %) · %(t#, x, %) = 0. If we set x# = x(t#, x, %) %# = %(t#, x, %) then, according toDefinition 3.2.2, we have (x#, %#) $ S+ !S! so we can use Proposition 3.3.1 for t $ R.Now we have by the flow property for t ! 0,

x(t, x, %) = x(t ' t#, x#, %#).

Using Proposition 3.3.1 we deduce the following lower bound

(3.4.1) "x(t, x, %)# = "x(t ' t#, x#, %#)# ! 1/3"t ' t##.

This proves the part (i) in Proposition 3.4.1. To prove part (ii) we use the formulas(3.1.2) for the flow. Then we see that for t ! 0,

%*(t, x, %) = %* + 1*(t, x, %), 1*(t, x, %) = '(

- t

0

n#

jk=1

$bjk

$x*(x(s)) %j(s) %k(s) ds.

Then using (3.1.1), (3.4.1), (3.1.4) and the fact that |%| " 2 we see that |1*(t, x, %)| "C (, where C depends only on A1.

On the other hand we have

xj(t, x, %) = 2%j + 21j(t, x, %) + 2(n#

k=1

bjk(x(t, x, %)) %k(t, x, %).

Integrating between 0 and t and using the above estimates we obtain the claimeddescription of x(t, x, %).


3.5. THE FLOW FROM POINTS IN (S+ & S")c 23

3.5. The flow from points in (S+ ! S!)c

We study now, more carefully the flow from points (x, %) $ T #Rn ! {0}, such that,

(3.5.1) |x · %| > c0 "x# |%| and12

" |%| " 2.

Even if, as we said before, we do not have a nice representation of the flow for all t inR we shall see that such a representation is available for limited values of t.

Since the description is symmetric, we shall assume that

(3.5.2) x · % " 'c0 "x# |%|.

Then (x, %) $ S! and Proposition 3.3.1 give a good description of the flow for t " 0.

Definition 3.5.1. — Let (x, %) satisfying 3.5.2. We set

I+ =!

t ! 0 : x(t, x, %) · %(t, x, %) " 14"x(t, x, %)# |%(t, x, %)|

".

In other words I+ is the set of t ! 0 such that (x(t, x, %), %(t, x, %)) belongs to S!.

The main result of this Section is the following description of the flow on I+.

Proposition 3.5.2. — Let (x, %) satisfying 3.5.2. Then for t in I+ we have

x(t, x, %) = x + 2t% ' z('t, x(t, x, %), %(t, x, %)),

%(t, x, %) = % ' 1('t, x(t, x, %), %(t, x, %)),

where z and 1 have been defined in Proposition 3.3.1. Moreover for j, k = 1, . . . , n wehave,

$xj

$xk(t, x, %) = &jk + O((),

$xj

$%k(t, x, %) = 2t &jk + O(("t#)

$%j

$%k(t, x, %) = &jk + O(("t#), $%j

$xk(t, x, %) = O(()

where &jk is the Kronecker symbol and O(A) means “bounded by CA” with C inde-pendent of (x, %). In particular we have

$%j

$%k(t, x, %) ' i

$xj

$%k(t, x, %) = (1 ' 2it) &jk + O(("t#).

Proof. — As said before, for t $ I+ the point 0(t, x, %) = (x(t, x, %), %(t, x, %)) belongsto S!. Therefore we can apply Proposition 3.3.1 for ! " 0. We get

x(!, 0(t, x, %)) = x(t, x, %) + 2! %(!, 0(t, x, %)) + z(!, 0(t, x, %))

%(!, 0(t, x, %)) = %(t, x, %) + 1(!, 0(t, x, %)).

Taking ! = 't with t ! 0 we obtain

x = x(t, x, %) ' 2t% + z('t, 0(t, x, %)),

% = %(t, x, %) + 1('t, 0(t, x, %)) .

This proves the first part of Proposition 3.5.2. To prove the claim on the derivativeswe use Lemma 3.1.2 and Corollary 3.3.3.



Remark 3.5.3. — Since the points (x, %) satisfying 3.5.2 belong to S!, Propositions3.3.1 and 3.5.2 provide a description of the flow on ('), 0) 2 I+.


CHAPTER 4

THE PHASE EQUATION

The goal of this section is to solve approximatively the phase equation$)

$!+ p(x,

$)

$x) = 0, )(0, x, #) = )0(x, #),

(see Theorem 4.1.2). In the case of the flat Laplacian this problem can be solvedexactly and we have

)(!, x, #) =(x ' #x)#% + i

2 |x ' #x|2 + 12i |#%|2

1 + 2i!

In the general case the classical method using the symplectic geometry leads to amajor di"culty. Indeed the symbol p has C" coe"cients but since )0 has to be nonreal we must deal with a non real flow. Instead we use here a method introduced byHormander [H] called method of ”Lagrangian ideals” which keeps real the data of theflow. It is briefly described in the Introduction (section 1).

The main result of this section is Theorem 4.1.2 whose proof is fairly long andcould be skipped in a first lecture. One of the reasons for the length of the proof isthat we have to consider separately the cases of outgoing and incoming points andthen to match them. Moreover in the case of incoming points the flow behaves badlyfor large time which leads to serious di"culties.

4.1. Statement of the result

Let p(x, %) = |%|2 + ( q(x, %), q(x, %) =&n

j,k=1 bjk(x) %j %k where the coe"cients bjk

satisfy the condition (3.1.1).In this Section # = (#x, #%) will be a fixed point in T #Rn such that 1

2 " |#%| " 2.Let us recall that (x(t, #), %(t, #)) denotes the flow of p starting for t = 0 at thepoint #.

We introduce now several sets.

Definition 4.1.1. — Let & > 0, c0 > 0, c1 > 0 be small constants (chosen later on).

26 CHAPTER 4. THE PHASE EQUATION

(i) If |#x · #%| " c0 "#x# |#%| we set,

(4.1.1) %+ =:(!, x) $ R ( Rn : |x ' x(!, #)| " & "!#

;.

(ii) If #x · #% > c0 "#x# |#%| we set,

(4.1.2) %+ =:(!, x) $ R ( Rn : |x ' x(!, #)| " & "!#, x · #% ! 'c1 "x# |#%|

;.

(iii) If #x · #% < 'c0 "#x# |#%| we set,

(4.1.3) %+ =:(!, x) $ R ( Rn : |x ' x(!, #)| " & "!#, x · #% " c1 "x# |#%|

;.

Let us give some explanations on this Definition.Taking c0 and c1 small with respect to 1

4 we see from Definition 3.2.2 that the case(i) corresponds to points (#x, #%) which are outgoing for ! ! 0 and ! " 0. Then %+

is simply a conic neighborhood of the projection of the bicharacteristic. In the case(ii) the point (#x #%) is outgoing for ! ! 0 and %+ can be written as follows.

%+ =:(!, x) $ (0, +)) ( Rn : |x ' x(!, #)| " & "!#

;2:(!, x) $ ('), 0) ( Rn :

|x ' x(!, #)| " & "!# and x · #% ! 'c1 "x# |#%|;.

(4.1.4)

Indeed if |x ' x(!, #)| " & "!# and ! ! 0 we have by Proposition 3.4.1, x · #% =(x' x(!, #)) ·#% + #x ·#% + 2! |#%|2 +O(( "!#). Since |#%| ! 1

2 and we are in case (ii)we deduce that x · #% ! c0 "#x# |#%| + 1

2 ! ' C(( + &)"!# ! 0 if ( + & is small enough.Therefore when ! ! 0 the condition x · #% ! 'c1 "x# |#%| is automatically satisfied.

In the case (iii) we have the same discussion changing ! ! 0 to ! " 0.The purpose of this Section is to prove the following result.

Theorem 4.1.2. — There exist & > 0, c0 > 0, c1 > 0 such that for any # $ T #Rn

with 12 " |#%| " 2 one can find a function ) = )(!, x, #) on %+ which is C" and

satisfies the following.

(i) )(0, x, #) = (x ' #x) · #% +i

2|x ' #x|2 +

12i

|#%|2 + g(x, #)

where |g(x, #)| " CN |x ' #x|N for all N $ N.(ii) For any N $ N there exists CN ! 0 such that

555$)

$!(!, x, #) + p

+x,

$)

$x(!, x, #)

,555 " CN

+ |x ' x(!, #)|"!#

,N

for all (!, x) in %+.Moreover for (!, x) in %+ we have

(iii)555$)

$x(!, x, #) ' #%

555 " C(( +/

&).

(iv)555 Im )(!, x, #) ' 1

2|x ' x(!, #)|2

1 + 4!2+

12|#%|2

555 " C(( +/

&)|x ' x(!, #)|2

"!#2 .

(v) |$Ax )(!, x, #)| " CA, for every A in Nn ! {0}

where C, CN and CA are independent of (!, x, #).


4.2. THE PREPARATION THEOREM 27

The proof of this result is based on the theory of Lagrangian ideals of L. Hormander([H], vol 4, chap. XXV). It will require several steps. The first one is a slight extensionof Theorem 7.5.4 in [H], vol. 1 to the case of higher dimensions.

4.2. The preparation theorem

The aim of this Section is to prove the following result.

Lemma 4.2.1. — Let g $ S(Rn% ) and z $ Cn. Then there exist functions qj(%, z, g),

j = 1, . . . , n, r(z, g) which are C" with respect to % and z, which depend linearly ong such that

(4.2.1) g(%) =n#

j=1

qj(%, z, g)(%j + zj) + r(z, g)

(4.2.2)

'((()

(((*

|$"% $)

z qj(%, z, g)| " C")

#

|,|!|"|+|)|+4n

-|$,

- g(2)| d2

|$)z r(z, g)| " C)

#

|,|!|)|+3n

-|$, g(2)| d2.

Proof. — We proceed by induction on the dimension n. If n = 1 this follows fromTheorem 7.5.4 of [H]. Let n ! 2 and let us set %% = (%1, . . . , %n!1). For fixed %% $ Rn!1

we apply Theorem 7.5.4 of [H] to the function %n 0. g(%%, %n). We get

(4.2.3) g(%%, %n) = q(%n, zn, g(%%, ·))(%n + zn) + r(zn, g(%%, ·)).

Let us set Qn(%, zn, g) = q(%n, zn, g(%%, ·)) and 2r(zn, %%, g) = r(zn, g(%%, ·)). Since ris linear in g we have $"

%# 2r(zn, %%, g) = r(zn, $"%# g(%%, ·)) and the estimates (4.2.2) for

n = 1 show that %% 0. 2r(zn, %%, g) is in S(Rn!1). Therefore we can apply, by theinduction, the Lemma to the function %% 0. 2r(zn, %%, g) and to z% = (z1, . . . , zn!1). Weobtain the existence of qj , j = 1, . . . , n'1 and R satisfying the estimates (4.2.2) suchthat

2r(zn, %%, g) =n!1#

j=1

qj(%%, z%, 2r(zn, ·, g)(%j + zj) + R(z%, 2r(zn, ·, g)).

Using (4.2.3) we obtain therefore

g(%) = Qn(%, zn, g)(%n + zn) +n#

j=1

qj(%%, z%, 2r(zn, ·, g))(%j + zj) + R(z%, 2r(zn, ·, g)).

If we set

(4.2.4)

6Qj(%, z, g) = qj(%%, z%, 2r(zn, ·, g)), j = 1, . . . , n ' 1

r(z, g) = R(z%, 2r(zn, ·, g))



we obtain (4.2.1) at the level n. Moreover Qj and r are linear in g since qj , R arelinear in 2r(zn, ·, g) and 2r is linear in g. Let us look to the estimate (4.2.2) for r. Wehave

$)#

z# $)nzn

r(z, g) = $)#

z# R$z%, $)n

zn2r(zn, ·, g)

%

so,

|$)z r(z, g)| " C

#

|,#|!|)#|+3(n!1)

- 55$,#

%# $)nzn2r(zn, %%, g)

55 d%%.

Now |$,#

%# $)nzn2r(zn, %%, g) = $)n

znr(zn, $,#

%# g(%%, ·)) and from the case n = 1 we have55$)n

znr(zn, $,#

%# g(%%, ·))55 " C

#

|,n|!)n+3

- 55$,#

%n$,#

%# g(zn, %%, %n)55 d%n.

It follows that

(4.2.5) |$)z r(z, g)| "

#

|,#|!|)#|+3(n!1)

#

|,n|!)n+3

- 55$,#

%# $,n

%ng(%%, %n)

55 d%% d%n.

The proof of the estimates for the q%js is the same.

Remark 4.2.2. — Let us set z = a + ib and let us write r(z, g) = r(a, b, g) andqj(%, z, g) = qj(%, a, b, g). If we take in (4.2.1) b = 0, % = 'a we obtain

(4.2.6) r(a, 0, g) = g('a).

If we di!erentiate (4.2.1) with respect to bk and then take z = a $ Rn, % = 'a, weget

(4.2.7)$r

$bk(a, 0, g) = 'i qk('a, a, 0, g), k = 1, . . . , n.

Finally if we di!erentiate (4.2.1) with respect to %* and then take z = a $ Rn, % = 'awe obtain

(4.2.8)$g

$%*('a) = q*('a, a, 0, ), / = 1, . . . , n.

We introduce now the following notations which will be used in the next sections.

Notation 4.2.3. — Let # = (#x, #%) $ T # Rn ! 0. We introduce

(4.2.9)

'()

(*

)0(x, #) = (x ' #x)#% +i

2(x ' #x)2 +

12i

#2%,

uj(x, %, #) = %j '$)0

$xj(x, #) = %j ' #j

% ' i(xj ' #jx).

Let p(x, %) = |%|2 + ( q(x, %) ; we denote by Hp its hamiltonian and we introduce thepull-back by the backward flow of the function uj. We set

(4.2.10)

6vj(!; x, %, #) = uj(exp('! Hp)(x, %))

= %j('!; x, %) ' #j% ' i(xj('!; x, %) ' #j

x),

where (x, %) is close to (x(!; #), %(!, #)).


4.3. THE CASE OF OUTGOING POINTS 29

In the flat case we find

vj = (1 + 2i!).%j '

#j% ' i(xj ' #j

x)1 + 2i!

/

We split the proof of Theorem 4.1.2 according to the di!erent values of # describedin Definition 4.1.1.

4.3. The case of outgoing points

Let us set

(4.3.1) S =:# $ R2n :

12

" |#%| " 2, |#x · #%| " c0 "#x# |#%|;.

We shall use the following notations

(4.3.2)

62%+ = {(!, y) $ R ( Rn : |y| " & "!#}sgn ! = 1 (resp. ' 1) if ! > 0 (resp. ! < 0).

Let now *0 $ C"0 (Rn), *1 $ C"

0 (R) be such that,

*0(t) = 1 if |t| " 1, *0(t) = 0 if |t| ! 2 and 0 " *0 " 1,

*1(!) = 1 if |!| " 1, *1(!) = 0 if |!| ! 2 and 0 " *1 " 1.

Then we can state the following result.

Theorem 4.3.1. — There exist small positive constants µ0, & such that if we set for# $ S, ! $ R, y $ Rn, 2 $ Rn, j = 1, . . . , n,

gj(2) = *0

+ 1µ0

2,

vj

+!, y+x(!, #), 2 *1(!)+(1'*1(!))

. 2

"!# +12

sgn !

"!# y/+%(!, #), #

,

there exist smooth functions aj = aj(!, y, #), bj = bj(!, y, #) defined on 2%+ such that,with a = (aj)j=1,...,n, b = (bj)j=1,...,n we have for 2 in Rn and (!, y) $ 2%+,

(i) gj(2) =n#

k=1

qk(2, a, b, gj)(2k + ak(!, y, #) + i bk(!, y, #))

where the q%ks have been introduced in Lemma 4.2.1.Moreover in the set 2%+ we have

(ii) |a(!, y, #)| " 10|y|"!# ,

55b(!, y, #) +K(!)

1 + 4!2y55 "

/&|y|"!# ,

where K(!) = '&('&(.1(&)+1!.1(&) ,

1)5"!# " K(!) " "!#.

On the other hand we have, uniformly with respect to (!, y) $ 2%+ and # $ S,

(iii) |$Ay a(!, y, #)| + |$"

y b(!, y, #)| " CA

"!#|A| , A $ Nn.

Moreover for j = 1, . . . , n, k = 1, . . . , n,

(iv)55qk(2, a, b, gj) ' (1 + 2i!)

k(!)"!# &jk

55 " C (( + &), if |2| " &

where k(!) = "!#*1(!) + 1 ' *1(!).(v) |$A

(a,b) $B- qk(2, a, b, gj)| " C(µ0), if |A| + |B| ! 1, |2| " µ0, 1 " j, k " n.



Let us make some comments about the above Theorem. First of all we have skippedfor convenience the dependence of gj with respect to (!, y, #) which are consideredhere as parameters. The main object to be considered is vj(!, x, %). Since x has to beclose to x(!, #) we have set x = y + x(!, #). For small |!| (*1 3 1) % has to be closeto %(!, #) that’s why we have set % = 2 + %(!, #). For large |!| we have *1 3 0. Inthis region for technical reasons we need to renormalize the variable 2 and to isolate

the first term of a. This is the reason for the expression2

"!# +12

sgn !

"!# y. Finally

the two regimes |!| small and |!| large have to be matched. This is the reason forintroducing *1.

Proof. — According to Definition 3.2.2 we have S , S+ ! S!. Moreover

(4.3.3) B(#, c0) := {2# $ T # Rn : |# ' 2#| " c0} , S+ ! S!.

We start with the following Lemma.

Lemma 4.3.2. — There exists a small positive constant µ0 such that for all (!, y)with ! $ R, |y| " µ0 "!# and all 2 in Rn such that |2| " 2 µ0 there exists a unique. = .(!, y, #, 2) in B(#, c0) such that

(4.3.4)

')

*

x(!, .) = y + x(!, #)

%(!, .) = *1(!) 2 + (1 ' *1(!))@ 2

"!# +12

sgn !

"!# yA+ %(!, #).

Moreover we have

(4.3.5)

'(((((()

((((((*

.x = #x +@1 ' |!|

"!# (1 ' *1(!))Ay ' 2!

"!# 2(*1(!)"!# + 1 ' *1(!))

+z(!, #) ' z(!, .)

.% = #% +2

"!# [*1(!)"!# + (1 ' *1(!))] +12

1 ' *1(!)"!# sgn ! y

+1(!, #) ' 1(!, .)

where z and 1 have been introduced in Proposition 3.3.1 and

(4.3.6)

'((((((((((()

(((((((((((*

(i) |. ' #| " 10$|2| + |y|

"!#%

(ii)55$.

$2(!, y, #, 2)

55 " C

(iii)555+$.j

%

$2k' i

$.jx

$2k

,(!, y, #, 2) ' 1 + 2i!

"!# ("!#*1(!) + (1 ' *1(!))) &jk

555

" C (

(iv) |$"- $B

y .(!, y, #, 2)| " CAB (, if |A| + |B| ! 2.



Proof. — The system (4.3.4) with . $ B(#, c0) , S+!S! is equivalent by Proposition3.3.1 to the following

(4.3.7)

')

*

.x + 2!%(!, .) + z(!, .) = y + #x + 2!%(!, #) + z(!, #)

.% + 1(!, .) = *1(!) 2 +1 ' *1(!)

"!#@2 +

12

sgn ! yA+ #% + 1(!, #).

Using again (4.3.4) the left hand side of the first line of (4.3.7) can be written

.x + 2! *1(!) 2 +2!

"!# (1 ' *1(!)) 2 +|!|"!# (1 ' *1(!)) y + 2!%(!, #) + z(!, .).

Finally (4.3.7) is equivalent to

(4.3.8)

'(((((()

((((((*

.x = #x +@1 ' |!|

"!# (1 ' *1(!))Ay ' 2!

"!# 2(*1(!)"!# + 1 ' *1(!))

+z(!, #) ' z(!, .)

.% = #% +2

"!# [*1(!)"!# + (1 ' *1(!))] +12

1 ' *1(!)"!# sgn ! y

+1(!, #) ' 1(!, .).

Writing this system .x = #x(.) .% = #%(.) and setting #(.) = (#x(.), #%(.)) weare going to solve it using the fixed point theorem in B(#, c0).

(i) # maps B(#, c0) in itself.We have 0 < 1 ' |&|

'&( " 1'&(2 , |!|*1(!) " 2, *1(!)"!# + 1 ' *1(!) "

/5, |y| " µ0 "!#,

|2| " 2 µ0, |z|+|1| " C ( by Proposition 3.3.2. It follows that |#(.)'#| " 20 µ0+C ( "c0 if µ0 and ( are small enough.

(ii) Let ., .% be in B(#, c0). Then t . + (1 ' t).% $ B(#, c0) , S+ ! S! for all t in(0, 1). It follows that

|#(.) ' #(.%)| " |z(!, .) ' z(!, .%)| + |1(!, .) ' 1(!, .%)| " C ( |. ' .%|

by Proposition 3.3.2. Here C depends only on the constants A0, A1 in (3.1.1). Taking( so small that C ( < 1 we see that we can apply the fixed point theorem in B(#, c0).This proves the existence of . satisfying (4.3.4) and (4.3.5) by (4.3.8). Now (i) in(4.3.6) follows from (4.3.5) taking ( small enough since |Z(!, #)'Z(!, .)| " C ( |#'.|if Z = z or 1. The claim (ii) is obtained by di!erentiating the equations (4.3.5) withrespect to 2k and using Proposition 3.3.2. Then (iii) follows easily from (4.3.5) and(ii). Finally (iv) is obtained by an induction on |A| + |B|.

From now on we fix the constant µ0 occurring in Lemma 4.3.2.Now for j = 1, . . . , n let gj be the function introduced in the statement of Theorem

4.3.1. Then, according to Lemma 4.3.2 and (4.2.10) we have for |y| " µ0 "!# and2 $ R,

(4.3.9) gj(2) = *0

+ 2

µ0

,@.j

%(!, y, #, 2) ' #j% ' i(.j

x(!, y, #, 2) ' #jx)A,



since f('!, x(!, .), %(!, .)) = .f for f = x and %. It follows from Lemma 4.2.1 thatthe existence of aj , bj in Theorem 4.3.1 will be proved if we can solve the equations

(4.3.10) r(a, b, gj(·)) = 0, j = 1, . . . , n.

Let us now take (!, y) $ 2%+, that is ! $ R, |y| " & "!# where 0 < & < 12 µ0 is to be

chosen. We look for a solution (a, b) of the system (4.3.10) in the set

(4.3.11)

'(()

((*

E =!(a, b) $ Rn ( Rn : |a| " 10 |y|

"!# ,555b +

K(!)1 + 4 !2

y555 "

/&|y|"!#

"

where K(!) ="!#

"!#*1(!) + (1 ' *1(!)),

1/5"!# " K(!) " "!#.

We shall first give equivalent equations to (4.3.10) in the set E. We write,

r(a, b, gj) = r(a, 0, gj) +n#

k=1

$r

$bk(a, 0, gj) bk +

n#

p,q=1

Hjp,q(!, y, #, a, b) bp bq

where

(4.3.12) Hjp,q(!, y, #, a, b) =

- 1

0(1 ' t)

$2r

$bp $bq(a, t b, gj(·)) dt.

It follows from (4.2.6), (4.2.7) and (4.2.8) that

(4.3.13) r(a, b, gj(·)) = gj('a) ' in#

k=1

$gj

$2k('a) bk +

n#

p,q=1

Hjp,q(!, y, #, a, b) bp bq.

Now if (a, b) $ E we have |a| " 12 |y|'&( " 12 & " µ0. Therefore *0

$' a/µ0

%= 1,

*%0

$' a/µ0

%= 0. Then by (4.3.9) and (4.3.5) we obtain,

(4.3.14) gj('a) =12

1 ' *1(!)"!# sgn ! yj '

aj

"!# [*1(!)"!# + 1 ' *1(!)]

+ 1j(!, #) ' 1j(!, .) ' i+1 ' |!|

"!# (1 ' *1(!)),

yj

' 2i!

"!# aj [*1(!)"!# + 1 ' *1(!)] ' i(zj(!, #) ' zj(!, .)

(4.3.15)$gj

$2k('a) =

1 + 2i!

"!# [*1(!)"!# + (1 ' *1(!))]

' $1j(!, .(!, y, #,'a))$.

$2k(!, y, #,'a)

+ i $zj(!, .(!, y, #,'a)) · $.

$2k(!, y, #,'a)

where $ = ($x, $%) and . = (.x, .%).



On the other hand we deduce from (4.3.12) and (4.2.2) that

(4.3.16)55$A

(a,b) $By Hj

pq(!, y, #, a, b)55 " CAB

#

|,|!|A|+3n+2

-|$,

- $By gj(2)| d2.

Using (4.3.6) and (4.3.9) we obtain

(4.3.17)55$A

(a,b) $By Hj

pq(!, y, #, a, b)55 " C%

AB(µ0).

It follows from (4.3.13), (4.3.14), (4.3.15) that (4.3.10) is equivalent to

' aj

"!# [*1(!)"!# + 1 ' *1(!)] '2i!

"!# aj [*1(!)"!# + 1 ' *1(!)] +12

1 ' *1(!)"!# sgn ! yj

' i+1 ' |!|

"!# (1 ' *1(!)),

yj + 1j(!, #) ' 1j(!, .) ' i(zj(!, #) ' zj(!, .))

' i1 + 2i!

"!# [*1(!)"!# + 1 ' *1(!)] bj + F j1 (!, y, #, a) · b + i F j

2 (!, y, #, a) · b

+ Hj1(!, y, #, a, b) b · b + i Hj

2(!, y, #, a, b) b · b = 0,

where

(4.3.18)

'(((((()

((((((*

. = .(!, y, #,'a),

F j1 (!, y, #, a) = $ zj(!, .) · $.

$2(!, y, #,'a),

F j2 (!, y, #, a) = $ 1j(!, .) · $.

$2(!, y, #,'a),

Hj = (Hjpq) = Hj

1 + i Hj2 .

Taking the real and the imaginary parts we are led to the system

aj ' 2! bj = K(!).12

1 ' *1(!)"!# sgn ! yj + 1j(!, #) ' 1j(!, .) + F j

1 b + Hj1 b · b

/

2! aj + bj = K(!).'+1 ' |!|

"!# (1 ' *1(!)),

yj ' (zj(!, #) ' zj(!, .)) + F j2 b + Hj

2 b · b/

where K(!) = '&(.1(&)'&(+1!.1(&) .

Inverting this system we are led to solve

(4.3.19)

'(((()

((((*

aj =K(!)

1 + 4 !2

+12

1 ' *1(!)"!# sgn ! ' 2!

+1 ' |!|

"!# (1 ' *1(!)),,

yj

+ Zj1(!, #) ' Zj

1(!, .) + F j3 b + Hj

3 b · b =: #j1(a, b)

bj = ' K(!)1 + 4!2

yj + Zj2(!, #) ' Zj

2(!, .) + F j4 b + Hj

4 b · b =: #j2(a, b)



where

(4.3.20)

'((((((((()

(((((((((*

Zj1(!, ·) =

K(!)1 + 4!2

(1j(!, ·) ' 2! zj(!, ·))

Zj2(!, ·) = ' K(!)

1 + 4!2(2! 1j(!, ·) + zj(!, ·))

F j3 =

K(!)1 + 4!2

(F j1 + 2! F j

2 ), F j4 =

K(!)1 + 4!2

('2! F j1 + F j

2 )

Hj3 =

K(!)1 + 4!2

(Hj1 + 2! Hj

2), Hj4 =

K(!)1 + 4!2

('2! Hj1 + Hj

2).

Let us set #j = (#j1, #

j2) (see (4.3.19)) and # = (#j)j=1,...,n. We have shown that

our initial system (4.3.10) is equivalent in E to the equation #(a, b) = (a, b). We aregoing to show that this equation has a unique solution in E by using the fixed pointtheorem.

(i) #(E) , E.We have 2 |!|

$1' |&|

'&(%

" 1/"!#, 2 |&|2'&( *1(!) " 8/"!# and by (4.3.20), (4.3.18), (4.3.6)

we see that |F j3 | + |F j

4 | " C (. Moreover we deduce from (4.3.17) and (4.3.20) that|Hj

3 | + |Hj4 | " C(µ0). Finally in E we have |b| " 2 |y|

'&( " 2&. It follows then from(4.3.19) that

|#1(a, b)| " 192

|y|1 + 4!2

+ C ( |# ' .| + C% (|y|"!# + C(µ0) &

|y|"!# .

Now using (4.3.6) (i) we see that when (a, b) belongs to E we have

|.(!, y, #,'a) ' #| " 10+|a| + |y|

"!#

," 110

|y|"!# .

Therefore

|#1(a, b)| "+19

2+ 110 C ( + C% ( + C(µ0) &

, |y|"!# " 10 |y|

"!# ,

if ( and & are small enough.By the same estimates we obtain,

555#2(a, b) +K(!)

1 + 4!2y555 " C ( |. ' #| + C (

|y|"!# + C(µ0) &

|y|"!# ;

so if ( and & are small enough we can bound the right hand side by/

& |y|/"!#. Thisshows that # maps E to E.

(ii) # is a contraction.



Let now (a1, b1), (a2, b2) be two points in E. Then

|#(a1, b1) ' #(a2, b2)| "n#

j=1

2#

*=1

55Zj* (!, .(!, y, #,'a1)) ' Zj

* (!, .(!, y, #,'a2))< => ?(1)

55

+n#

j=1

4#

*=3

!|F j

* (!, y, #, a1) · b1 ' F j* (!, y, #, a2) · b2|< => ?

(2)

+|Hj* (!, y, #, a1, b1) b1 · b1 ' Hj

* (!, y, #, a2, b2) b2 · b2|< => ?(3)

".

Using (4.3.20) and Proposition 3.3.2 we can write

(1) " C ( |.(!, y, #,'a1) ' .(!, y, #,'a2)|.

Then (4.3.6) gives (1) " C% ( |a1 ' a2|.To handle the therm (2) we use (4.3.18), (4.3.20), (4.3.6) and Proposition 3.3.2.

We obtain(2) " C ((|a1 ' a2| + |b1 ' b2|).

Finally using (4.3.17), (4.3.18), (4.3.20) and the fact that in E we have |b| " 2 |y|'&( " 2&

we see easily that(3) " C (( + &)(|a1 ' a2| + |b1 ' b2|).

It follows then that

|#(a1, b1) ' #(a2, b2)| " C (( + &)(|a1 ' b1| + |a2 ' b2|)

where C is an absolute constant depending only on the dimension and a finite numberof A* appearing in (3.1.1). Thus we can take ( and & so small that C (( + &) < 1.

Therefore we can apply the fixed point theorem to solve (4.3.19) which is equivalentto (4.3.10). This proves the claims (i) and (ii) in Theorem 4.3.1.

Let us now prove the point (iii). We state a Lemma.

Lemma 4.3.3. — There exists C0 > 0 such that for every A $ Nn there exist CA ! 0,C%

A ! 0 such that with . defined in Lemma 4.3.2,

a)55$A

y [.(!, y, #,'a(!, y, #))]55 " C0 |$A

y a(!, y, #)| + CA

"!#|A| ,

b) |$Ay a(!, y, #)| + |$A

y b(!, y, #)| " C%A/"!#|A|, for all (!, y) in 2%+.

Proof. — We shall use an induction on |A|, starting with the formulas (4.3.19). Butbefore we need some intermediate results. We introduce the following space of func-tions.

Let f = f(!, y, #) be a function from 2%+ ( R2n to C. We shall say that f $ G± ifwe can write

(4.3.21) f(!, y, #) = G(!, .(!, y, #,'a(!, y, #)))



where G : R±& ( R2n

X . C is smooth in X and satisfies

(4.3.22) supR±*R2n

|$,X G(!, X)| " C, (, % 3 $ Nn.

For example Proposition 3.3.2 shows that the functions

(!, y, #) 0'. zj(!, .(!, y, #,'a(!, y, #))

(and 1j) belong to G± if # $ S±. (Here we have the sign + if # $ S+ and ' if # $ S!).Then we have the following claim.

Claim 1. — For all ' $ Nn, |'| ! 1, j = 1, . . . , n we have

(4.3.23) $/- .j

x(!, y, #, 2) = Gj/(!, .(!, y, #, 2))

where Gj/ has all derivatives uniformly bounded on R± ( R2n. The same is true for

$/- .j

% .

Proof of the claim. — We proceed by induction on |'| beginning with |'| = 1. Letus set k(!) = "!#*1(!) + 1 ' *1(!). Then 1 " k(!) "

/5 since |!| " 2 on supp*1. It

follows from (4.3.5) that for fixed k in {1, 2, . . . , n} we have

$.jx

$2k(!, y, #, 2) = ' 2!

"!# k(!) &jk 'n#

*=1

+$zj

$x*(!, .(!, y, #, 2))

$.*x

$2k(!, y, #, 2)

+$zj

$%*(!, .(!, y, #, 2))

$.j%

$2k(!, y, #, 2)

,

$.j%

$2k(!, y, #, 2) =

k(!)"!# &jk '

n#

*=1

+ $1j

$x*(!, .(!, y, #, 2))

$.*x

$2k(!, y, #, 2)

+$1j

$%*(!, .(!, y, #, 2))

$.*%

$2k(!, y, #, 2)

,.

Let us set

Xj =$.j

x

$2k, &j =

$.j%

$2k, U =

B

CCCCCCD

X1

&1

...Xn

&n

E

FFFFFFG, Mj =

B

CCD

$zj

$x1

$zj

$%1. . .

$zj

$xn

$zj

$%n

$1j

$x1

$1j

$%1

$1j

$xn

$1j

$%n

E

FFG ,

M =

B

CCCD

M1

M2

...Mn

E

FFFGand F (!) =

k(!)"!#

B

CCCCCCD

'2! &1k

&1k

...'2! &nk

&nk

E

FFFFFFG.

Then the above equation imply that

U(!, y, #, 2) = F (!) ' M(!, .(!, y, #, 2))U(!, y, #, 2).



Now by Proposition 3.3.2 the entries of the matrix M are O((). It follows that thematrix I + M is invertible. Therefore we obtain

U(!, y, #, 2) = (I + M(!, .(!, y, #, 2)))!1 F (!).

This proves the case |'| = 1.Assume now that our claim is proved for |'| " N ' 1. Then

$/- .j

x(!, y, #, 2) = Gj/(!, .(!, y, #, 2)).

Then for k = 1, . . . , n

$

$2k$/

- .jx(!, y, #, 2) =

n#

*=1

$Gj/

$X*(!, .(!, y, #, 2))

$.*x

$2k(!, y, #, 2)

+2n#

*=n+1

$Gj/

$X*(!, .(!, y, #, 2))

$.*%

$2k(!, y, #, 2).

Using (4.3.23) with |'| = 1 we obtain the claim up to the order |'| = N .

Consequence 4.3.4. — With the notations of (4.3.18), (4.3.20) and (4.3.21) Thefunctions (!, y, #) 0. Zj

* (!, .(!, y, #,'a(!, y, #)), / = 1, 2, F jp , p = 3, 4 belong to G+.

Let us now go back to the proof of Lemma 4.3.3. We begin by the case |A| = 1.For convenience we shall set

(4.3.24) f(y) = .(!, y, #,'a(!, y, #)).

It follows from (4.3.5) that

$f jx

$yk(y) =

+1 ' |!|

"!# (1 ' *1(!)),

&jk +2!

"!# k(!)$aj

$yk'

n#

*=1

+ $zj

$x*(!, f(y))

$f jx

$yk(y)

' $zj

$%*(!, f(y))

$f j%

$yk(y),

$f j%

$yk(y) =

12

(1 ' *1(!))sgn !

"!# &jk ' 1"!# k(!)

$aj

$yk'

n#

*=1

+ $1j

$x*(!, f(y))

$f jx

$yk(y)

' $1j

$%*(!, f(y))

$f j%

$yk(y),.

It follows from Proposition 3.3.2 that,555$f

$yk(y)555 " C

+ 1"!# +

555$a

$yk(!, y, #)

555+ (555$f

$yk(y)555,

where C depends only on the constants A0, A1 appearing in (3.1.1). Therefore taking( so small that C ( " 1

2 we obtain the point a) in Lemma 4.3.3 for |A| = 1. Let us



prove the point b). First of all we deduce from a) and the Consequence 4.3.4 that

(4.3.25)

'()

(*

$

$yk[Zj

* (!, f(y))], / = 1, 2,$

$yk[F j

p ], p = 3, 4

are bounded by C ($ 1"!# +

55 $a

$yk(!, y, #)

55%.

Now we claim that for p = 3, 4,

(4.3.26)555

$

$yk[Hj

p(!, y, #, a, b)]555 " C

+555$a

$yk(!, y, #)

555+555

$b

$yk(!, y, #)

555+1"!#

,.

Indeed the left hand side of (4.3.26) can be written

$Hjp

$a· $a

$yk+

$Hjp

$b· $b

$yk+

$Hjp

$yk= (1) + (2) + (3).

Now using (4.3.20), (4.3.18) and (4.3.17) we see that (1) and (2) can be bounded bythe right hand side of (4.3.26). To handle the term (3) we use (4.3.16) with A = 0,|B| = 1. We obtain

|(3)| " C#

|,|!3n+2

- 555$

$yk$,

- gj(2)555 d2.

Now we use (4.3.9) and (4.3.23). We obtain

$,- gj(2) = $,

-

.*0

+ 1µ0

2)/.

(.j% ' i .j

x)(!, y, #, 3) ' (#j% + i #j

x)/

+#

|,#|<|,|

$,#

-

.*0

+ 1µ0

2,/

Gj,#(!, .(!, y, #, 2))

where Gj,# satisfy (4.3.22). Since by (4.3.5) we have

555$.

$yk(!, y, #, 2)

555 " C

"!# .

We obtain|(3)| " C

"!#which proves (4.3.26).

Now we use (4.3.19), (4.3.25), (4.3.26) and the fact that |b| " 2 |y|'&( " 2&. We obtain

with aj = aj(!, y, #), bj = bj(!, y, #),555$aj

$yk

555+555$bj

$yk

555 " 12"!# + C (( + &)

+ 1"!# +

555$a

$yk

555+555

$b

$yk

555,.

Taking ( and & small enough we obtain the point b) in Lemma 4.3.3 when |A| = 1.Assume now that a) and b) in Lemma 4.3.3 are true when |A| " N and let |A| =

N + 1. It follows from the induction that

(4.3.27)55$B

y [.(!, y, #,'a(!, y, #))]55 " CB

"!#|B| , if |B| " N.



Claim 1. — If |A| = N + 1 ! 2,

(4.3.28)55$A

y [.(!, y, #,'a(!, y, #))]55 " C0

55$Ay a(!, y, #)

55+ CA

"!#|A| .

Indeed, according to (4.3.5) we have, setting for short f(y) = .(!, y, #,'a(!, y, #))and k(!) = "!#*1(!) + 1 ' *1(!),'(()

((*

fx(y) = #x +$1 ' |!|

"!# (1 ' *1(!))%y +

2!

"!# k(!) a(!, y, #) + z(!, #) ' z(!, f(y))

f%(y) = #% +12

(1 ' *1(!))sgn !

"!# y ' k(!)"!# a(!, y, #) + 1(!, #) ' 1(!, f(y)).

Di!erentiating both side A times with respect to y we obtain since |A| ! 2,

|$Ay f(y)| " 5 |$A

y a(!, y, #)| + |$Ay [z(!, f(y))]| + |$A

y [1(!, f(y))]|.

We use now the Faa di Bruno formula (see Appendix A.1). Let Z be z or 1 then

$Ay [Z(!, f(y))] =

n#

j=1

! $Z

$xj(!, f(y)) $A

y f jx(y) +

$Z

$%j(!, f(y)) $A

y f j% (y)

< => ?(1)

"+ (2)

where (2) is a finite sum of terms of the form

($/X Z)(!, f(y))

sH

j=1

$$*j

y f(y)%kj

where X = (x, %), 2 " |'| " |A|, |/j| ! 1, |kj | ! 1 ands#

j=1

kj = ',s#

j=1

|kj | /j = A.

The term (1) can be bounded by C0 ( |$Ay f(y)| (where C0 depends on A0, A1 in

(3.1.1)). Since |'| ! 2 it is easy to see that |/j | " |A| ' 1. We can therefore use(4.3.27) and Proposition 3.3.2 to write

|(2)| " CA (sH

j=1

1"!#|*j | |kj |

" CA

"!#|A| .

Thus (4.3.28) is proved which implies the part a) of Lemma 4.3.3 when |A| = N + 1.

Claim 2. — If F $ G± (see (4.3.21)) and |A| = N + 1 we have

(4.3.29) |$Ay F (!, y, #)| " (

+C0 |$A

y a(!, y, #)| + CA

"!#|A|

,.

Let us set for convenience as in (4.3.24),

f(y) = .(!, y, #,'a(!, y, #)).



We know by assumption that F (!, y, #) = G(!, f(y)) where G satisfies (4.3.22). Bythe Faa di Bruno formula we have

(4.3.30) $Ay F (!, y, #) =

2n#

i=1

$G

$Xi(!, f(y)) $A

y f(y)< => ?

(1)

+(2)

where (2) is a finite sum of terms of the form

($/X G)(!, f(y))

sH

j=1

$$*j

y f(y)%kj

where 2 " |'| " |A|, |/j | ! 1, |kj | ! 1, 1 " s " |A| and

(4.3.31)s#

j=1

kj = ',s#

j=1

|kj | /j = A.

Now by the Claim 1 we have

(4.3.32) |(1)| " (+C0 |$A

y a(!, y, #)| + CA

"!#|A|

,.

On the other hand in the term (2) it is easy to see that |/j | " |A| ' 1. Indeed if wehad a j0 such that |/j0 | = |A| it would follow from (4.3.31) that j0 = 1, s = 1 and|k1| = 1 ; but then |k1| = 1 = |'| which is in contradiction with |'| ! 2. Therefore wecan use (4.3.27), (4.3.31) to write,

|(2)| " CA (sH

j=1

+ 1"!#|*j |

,|kj |= CA (

1"!#|A| .

Then (4.3.29) follows from (4.3.30) and (4.3.32).

Claim 3. — If |A| = N + 1, j = 1, . . . , n, / = 3, 4 we have,

(4.3.33)55$A

y (Hj* (!, y, #, a, b))

55 " C0

$|$A

y a(!, y, #)| + |$Ay b(!, y, #)|

%+

CA

"!#|A| ,

where Hj* is defined in (4.3.20), (4.3.18), (4.3.13).

The proof is exactly the same as in the Claim 2. We use the Faa di Bruno formula,the estimates on a, b given by the induction, the estimate (4.3.16) to obtain (4.3.33).We are ready now to prove the part b) of Lemma 4.3.3 when |A| = N + 1.

We use the equations (4.3.19), (4.3.20) which we di!erentiate |A| times with respectto y. Since Zj

* (!, .) and F jk belong to G± we use the Claim 2 to estimate them ; the

term Hj* b ·b is handled by (4.3.33), the Leibniz formula and the induction hypothesis.

Finally we obtain since |b| " 2 |y|'&( " 2&,

|$Ay aj | " (( + &)C0

$|$A

y a| + |$Ay b|%

+CA

"!#|A| .

Taking ( and & small enough we obtain the part b) of Lemma 4.3.3.



So far we have proved the points (i), (ii), (iii) in Theorem 4.3.1.Let us prove (iv). It follows from (4.2.1), (4.3.6) and (4.3.9) that

555$qk

$2*(2, a, b, gj

555+555$qk

$b*(2, a, b, gj)

555 " C

where C is an absolute constant.Since |2| " &, |a| " 10 |y|

'&( " 10 &, |b| " 2 |y|'&( " 2& we can write

(4.3.34)55qk(2, a, b, gj) ' qk('a, a, 0, gj)

55 " C% &.

Now (4.2.8) gives qk('a, a, 0, gj) = (gj

(-k('a). It follows then from (4.3.15), (4.3.6)

and Proposition 3.3.2 that555qk('a, a, 0, gj) ' (1 + 2i!)

k(!)"!# &jk

555 " C (

which combined with (4.3.34) gives the point (iv).Finally (v) follows easily from (4.2.2), (4.3.9) and (4.3.6). This ends the proof of

Theorem 4.3.1.

Corollary 4.3.5. — Let us set k(!) = "!#*1(!) + (1' *1(!)) and for j = 1, . . . , n,

2gj(2) = *0

+2 ' 12 (1 ' *1(!)) (sgn !) y

µ0 k(!)

,vj

+!, y + x(!, #),

2

"!# + %(!, #), #,.

Then we can write

(4.3.35) 2gj(2) =n#

*=1

2q*(2,2a,2b, 2gj)+2* '

12

(1 ' *1(!)) (sgn !)y + (2a* + i2b*)(!, y, #),

where

2q*(2,2a,2b, 2gj) =1

k(!)q*

+2 ' 12 (1 ' *1(!)) (sgn !)y

k(!),2a

k(!),2b

k(!), 2gj

,

and 2q*, 2a*, 2b* satisfy

(i) |2a(!, y, #)| " 10/

5|y|"!# ,

552b(!, y, #) +"!#

1 + 4!2y55 "

/5&

|y|"!# .

(ii) |$Ay 2a(!, y, #)| + |$A

y2b(!, y, #)| " CA

"!#|A| , A $ Nn.

(iii)552q*(2,2a,2b, 2gj) '

(1 + 2i!)"!# &jk

55 " C (( + &) if |2| " &.

(iv) |$A(a,b) $B

- 2q*(2,2a,2b, 2gj)| " C(µ0) if |A| + |B| ! 1, |2| " µ0, 1 " j, / " n

uniformly with respect to (!, y) $ 2%+ and # $ S.

Proof. — We have

*1(!) 2 + (1 ' *1(!))@ 2

"!# +12

1 ' *1(!)"!# (sgn !) y

A=

k(!)"!# 2 +

12

1 ' *1(!)"!# (sgn !) y.



So let us set in the statement of Theorem 4.3.1 22 = k(!) 2+ 12 (1'*1(!)) (sgn !) y ; then

we obtain the decomposition of 2gj in Corollary 4.3.5 with 2a* = k(!) a*, 2b* = k(!) b*

and the estimates on 2q*, 2a*, 2b* follow easily from the correspondent one for q*, a*, b*

stated in Theorem 4.3.1.

Lagrangian ideals and the phase equation. — We pursue here the proof of Theorem4.1.2. Let us set

(4.3.36) O =!(!, y, 2) $ R(Rn(Rn : |y| < & "!#,

|2 ' 12 (1 ' *1(!)) (sgn !) y|

k(!)< &".

We introduce now a space of families (f(·, #))"+S of functions on O.

Definition 4.3.6. — We say that (f(·, #))"+S belongs to H if(i) For all # in S, (!, y, 2) 0. f(!, y, 2, #) belongs to C"(O).(ii) For every A, B in Nn there exists CAB > 0 independent of # such that

sup(&,y,-)+O

|$Ay $B

- f(!, y, 2, #)| " CAB, for all # $ S.

Remark 4.3.71) H is closed under multiplication and derivation with respect to (y, 2).2) If we set, with notation (4.2.10)

f(!, y, 2, #) = vj(!, y + x(!, #),2

"!# + %(!, #), #)

then (f(·, #))"+S belongs to H. This follows from (4.3.6).

Definition 4.3.8. — Let F = (F (·, #))"+S . We say that F $ J if we can write

F (!, y, 2, #) =n#

j=1

fj(!, y, 2, #) vj

+!, y + x(!, #),

2

"!# + %(!, #), #,

for all (!, y, 2, #) in O ( S with (f(·, #))"+S $ H.

Example 4.3.9. — Let us set F = (F (·, #))"+S where

F (!, y, 2, #) = 2k ' 12

(1 ' *1(!) (sgn !) yk + (2ak + i2bk)(!, y, #), k $ {1, 2, . . . , n}.

Then F $ J .

This follows from Corollary 4.3.5. Indeed the matrix (2q*(2,2a,2b, 2gj))j,* is invertibleif ( + & is small enough and according to the estimate (v) if we set (d*j) = (2q(· · · ))!1

then (djk(·, #))"+S belongs to H so our claim follows from Remark 4.3.7.Now if F = (F (·, #))"+S , G = (G(·, #))"+S we set

(4.3.37) {F, G} =+ n#

j=1

+ $F

$2j

$G

$yj' $F

$yj

$G

$2j

,(·, #)

,

"+S.

Then we have the following result.



Lemma 4.3.10. — J is closed under the Poisson bracket (4.3.37).

Proof. — Recall (see (4.2.10)) that vj(!, x, %, #) = uj 4 *!&(x, %) where uj(x, %, #) =%j ' #j

% ' i(xj ' #jx) and *!&(x, %) = (x('!; x, %), %('!; x, %)) is the symplectic map

defined by the flow. Since {uj, uk} = 0 we have, denoting by { , } the Poisson bracketin (x, %), {vj , vk} = {uj 4 *!&, uk 4 *!&} = {uj, uk} 4 *!& = 0. It follows that, in thecoordinates (y, 2),

!vj(!; y + x(!; #),

2

"!# + %(!; #)), vk(!; y + x(!; #),2

"!# + %(!; #))"

= 0.

Let F =$&

j fj vj(·, #)%

"+S , G =$&

k gk vk(·, #)%

"+S two elements of J . Then if{ , } is the Poisson bracket in (y, 2), we have,!#

j

fj vj ,#

k

gk vk

"

=#

k

+#

j

fj{vj , gk},

vk +#

k

+#

j

{fj, gk} vj

,vk +

#

j

+#

k

{fj, vk} gk

,vj .

Since fj , gk, vj belong to H it follows from Remark 4.3.7 (i) and Definition 4.3.8 that{F, G} $ J .

Here is an important lemma which is a generalization to our context of Lemma7.5.10 of [H].

According to Corollary 4.3.5, we shall set(4.3.38)

-k(!, y, #) =12

(1 ' *1(!)) (sgn !) yk ' (2ak(!; y, #) + i2bk(!; y, #)), k = 1, . . . , n.

Lemma 4.3.11. — Let R = (R(·, #))"+S $ J where R(!, y, #) is independent of 2.Then for every N $ N one can find CN > 0 such that for every (!, y) in 2%+ (see(4.3.2)) and # in S we have

|R(!, y, #)| " CN | Im -(!, y, #)|N .

Proof. — We are going to show by induction on N $ N# that we can write for (!, y, #)in 2%+ ( S,

(4.3.39) R(!, y, #) =#

0<|,|<N

h,(!, y, #)(2 ' -), +#

|,|=N

w,(!, y, 2, #)(2 ' -), ,

where (h,(·, #))"+S and (w,(·, #))"+S belong to H.For N = 1 the first sum in (4.3.39) is empty and by assumption we have

R(!, y, #) =n#

j=1

fj(!, y, 2, #) vj

+!; y + x(!; #),

2

"!# + %(!; #),,



where (fj(·, #))"+H. Now we use Corollary 4.3.5. Since *0

$-! 12 (1!.1(&)) (sgn &) y

k(&) µ0

%= 1

if (!, y, 2) $ O we can write

R(!, y, #) =n#

k=1

rk(!, y, 2, #)(2k ' -k(!, y, #))

where

rk(!, y, 2, #) =n#

j=1

fj(!, y, 2, #) 2qk(2,2a,2b, 2gj).

Now it follows from Corollary 4.3.5 and Remark 4.3.7 that (rk(·, #))"+S belongs toH. Therefore (4.3.39) holds when N = 1. Assume now that (4.3.39) is true at thelevel N . Then apply Lemma 4.2.1 to the function

g,(!, y, 2, #) = *0

+2 ' 12 (1 ' *1(!)) (sgn !) y

µ0 k(!)

,w,(!, y, 2, #), |3| = N,

with z = '-(!, y, #). It follows then that

(4.3.40) g,(!, y, 2, #) =n#

k=1

qk(2, a, b, g,)(2k ' -k(!, y, #)) + r(a, b, g,).

For the q%ks and r we have the estimates (4.2.2). If we set

(4.3.41)

6h,(!, y, #) = r(a(!, y #), b(!, y, #), g,(!, y, ·, #))

w,j(!, y, 2, #) = qj(2, a(!, y, #), b(!, y, #), g,(!, y, ·, #)).

We deduce from (4.2.2) and Corollary 4.3.5 that (h,(·, #))"+S and (w,j(·, #))"+Sbelong to H. Then using (4.3.39) at the level N and (4.3.40), (4.3.41) we obtain(4.3.39) at the level N+1. Let us take now in (4.3.39) 2 = Re -(!, y, #)+s Im -(!, y, #)where s $ [0, 1]. We obtain(4.3.42)

|R(!, y, #) '#

0<|,|<N

h,(!, y, #)(Im -(!, y, #)), (s ' i), | " CN | Im -(!, y, #)|N

where CN is independent of (!, y, #) $ 2%+ ( S.Using an interpolation formula we deduce that the coe"cients of the polynomial

in (s' i) in the left hand side of (4.3.42) satisfy the same estimate which proves thatR has the claimed bound.

Corollary 4.3.12. — For every N $ N there exists a constant CN > 0 such that555+$-j

$yk' $-k

$yj

,(!, y, #)

555 " CN | Im -(!, y, #)|N

for every (!, y) in 2%+ and # in S.



Proof. — According to (4.3.38) and Example 4.3.9 we have 2k ' -k(!, y, #) $ J . Itfollows from Lemma 4.3.10 that

Rjk(!, y, #) :=+$-j

$yk' $-k

$yj

,(!, y, #) = {2j ' -j , 2k ' -k}(!, y, #)

defines an element of J . Since Rjk does not depend on 2 we can apply Lemma 4.3.11and the conclusion follows.

So far we have worked in the coordinates (y, 2). Let us go back to the originalcoordinates (x, %) and let us summarize the results already obtained.

We set x = y + x(!, #), % = -'&( + %(!, #). Then (!, x) $ %+ (see (4.3.1)). Let us

recall that,

vj(!, x, %, #) = %j('!, x, %) ' #j% ' i(xj('!, x, %) ' #j

x),

(see (4.2.10)).Now let us introduce

(4.3.43) #k(!, x, #) = %k(!, #)

+12 (sgn !)(1 ' *1(!))(x ' x(!, #)) ' (2ak + i2bk)(!, x ' x(!, #), #)

"!#

where 2ak, 2bk have been introduced in Corollary 4.3.5. Then we have,

Theorem 4.3.13. — We can write

(i) %k ' #k(!, x, #) =n#

j=1

djk(!, x, %, #) vj(!, x, %, #)

where djk are smooth functions defined for (!, x) $ %+ and

555% ' %(!, #) ' 12

(1 ' *1(!)) (sgn !)x ' x(!, #)

"!#

555 " &

"!# .

Moreover we have in this set,

(ii) |$Ax dj

k(!, x, %, #)| " CA

"!# , A $ Nn,

(iii) |#k(!, x, #) ' #%| " C(( + &),

(iv)555 Im #k(!, x, #) ' xk ' xk(!, #)

1 + 4!2

555 "/

5&|x ' x(!, #)|

"!#2 .

(v) |$Ax #k(!, x, #)| "

6CA "!#!|A| if |A| " 1CA "!#!|A|!1 if |A| ! 2

.

(vi) #k(!, x(!, #), #) = %(!, #).

(vii)555+$#j

$xk' $#k

$xj

,(!, x, #)

555 " CN

"!#

+ |x ' x(!, #)|"!#

,N, %N $ N

where the constants CA, C0, CN are independent of (!, x, %, #).



Proof. — It follows from Corollary 4.3.5 that when55% ' %(!, #) ' 1

2(sgn !) (1 ' *1(!))

x ' x(!, #)"!#

55 " &

"!#then

vj(!, x, %, #) =n#

k=1

cjk(!, x, %, #)(%k ' #k(!, x, #))

with

cjk(!, x, %, #) = "!# qk

+"!#(% ' %(!, #)) ' 1

2(sgn !) (1 ' *1(!))(x ' x(!, #)),2a,2b, 2gj

,

where 2qk is defined in Corollary 4.3.5 (i). By (v) of the same result we have,

(4.3.44)55cjk(!, x, %) ' (1 + 2i !) &jk

55 " C (( + &) "!#.

It follows then that the matrix (cjk) is uniformly invertible. Let us set (djk(!, x, %)) =

(cjk(!, x, %))!1. Then we obtain (i) in Theorem 4.3.13. The estimate (ii) follows thenfrom Corollary 4.3.5 (v) and (4.3.44).

Let us now prove the claimed properties of #k. First of all since %(!, #) = #%+O((),|x!x(&,")|

'&( " &, |2ak|+|2bk| " C &. We deduce easily from (4.3.43) that |#k(!, x, %)'#% | "C (( + &). On the other hand it follows from (4.3.43) and Corollary 4.3.5 (i)

Im #k(!, x, %) = ' 1"!#2bk(!, x ' x(!, #), #) =

xk ' xk(!, #)1 + 4 !2

+ R

where |R| " |x ' x(!, #)|/"!#2.The point (v) in Theorem 4.3.13 follows easily from (4.3.43) and Corollary 4.3.5

(ii) ; the point (vi) is obvious since 2ak(!, 0, #) = 2bk(!, 0, #) = 0. Finally for the point(vii) we remark that according to (4.3.38) and (4.3.43) we have

(1) =+$#j

$xk' $#k

$xj

,(!, x, #) =

1"!#

+$-j

$yk' $-k

$yj

,(!, x ' x(!, #), #).

Using Corollary 4.3.12 and the point (iv) we obtain

|(1)| " CN

"!# | Im -(!, x ' x(!, #), #)|N =CN

"!#$"!# | Im #(!, x, #)|

%N

|(1)| " CN

"!#

+ |x ' x(!, #)|"!#

,N, for all N $ N.

We need now to introduce the definition of Lagrangian ideals in the coordinates(x, %).

Let (F (·, #))"+S a family of functions F (!, x, %, #) defined for (!, x) in %+ and55% ' %(!, #) ' 12 (sgn !) (x!x(&,"))

'&(55 < &/"!#.

Definition 4.3.14. — We shall say that the family (F (·, #))"+S belongs to J(x,%) if

(4.3.45)+"!#F (!, y + x(!, #),

2

"!# + %(!, #), #),

"+S$ J

where J has been introduced in Definition 4.3.8.



For example, (%j '#j(!, x, #))"+S $ J(x,%). As a consequence of Lemma 4.3.11 wehave the following result.

Lemma 4.3.15. — Let (R(·, #)"+S be in J(x,%) and assume that R is independent of %then for every N $ N there exists CN > 0 such that,

(4.3.46) |R(!, x, #)| " CN

"!#

+ |x ' x(!, #)|"!#

,N

We can now pursue the proof of the existence of a phase as described in Theorem4.1.2.

Lemma 4.3.16. — With # defined in (4.3.43) we have

+' $p

$xk(x, %) ' $#k

$!(!, x, #) '

n#

j=1

$#k

$xj(!, x, #)

$p

$%j(x, %)

,

"+S$ J(x,%).

Proof. — We know from (4.3.43), (4.3.38) and Example 4.3.9, with % = -'&( + %(!, #)

that,

%k ' #k (!, x, #) =1"!#$2k ' -k(!, x ' x(!, #), #)

%

=1"!#

n#

j=1

fjk(!, x ' x(!, #), "!#(% ' %(!, #)), #) vj (!, x, %, #)

where (fjk(·, #))"+S = ((qk(2 ' 12 (sgn !)(1 ' *1(!)) y,2a,2b, gj))!1)"+S $ H (see Defi-

nition 4.3.6).Recall now that vj(!, x, %, #) = uj 4*!&(x, %) where uj(X, &) = &j'#j

%'i(Xj'#jx)

(see (IV.2.9)). Let us set

*!&(x, %) = (X, &) 56 x(!, X, &) = x, %(!, X, &) = %.

It follows that

(4.3.47) %k(!, X, &) ' #k(!, x(!, X, &), #)

=1"!#

n#

j=1

fjk

$!, x(!, X, &) ' x(!, #), "!# (%(!, X, &) ' %(!, #)), #

%uj(X, &).

Let us set'()

(*

M(!, X, &, #) = (!, x(!, X, &) ' x(!, #), "!# (%(!, X, &) ' %(!, #))),

0(!, X, &) = (x(!, X, &), %(!, X, &)),

0(!, #) = (x(!, #), %(!, #)).



Now we di!erentiate (4.3.47) with respect to ! using the equation of the flow givenin (3.1.2). We obtain

' $p

$xk(0(!, X,&)) ' $#k

$!(!, x(!, X, &), #) '

n#

j=1

$#k

$xj(!, x(!, X, &), #) · $p

$%j(0(!, X, &))

=n#

j=1

I' !

"!#3 fjk(M(!, X, &, #)) +1"!#

$fjk

$!(M(!, X, &, #))

+1"!#

n#

*=1

.+ $p

$%*(0(!, X, &)) ' $p

$%*(0(!, #))

, $fjk

$y*(M(!, X, &, #))

+! !

"!# (%*(!, X, &) ' %*(!, #)) ' "!#+ $p

$x*(0(!, X, &) ' $p

$x*(0, #)

,"

$fjk

$2*(M(!, X, &, #))

/Juj(X, &)

We can now go back to the coordinates (x, %) = 0(!, X, &) and then to (y, 2) wherey = x ' x(!, #), % ' %(!, #) = -

'&( . We obtain

'. $p

$xk' $#k

$!'

n#

j=1

$#k

$xj

$p

$%j

/+!, y + x(!, #),

2

"!# + %(!, #),

=n#

j=1

I'!

"!#3 fjk(!, y, 2, #) +1"!#

$fjk

$!(!, y, 2, #)

+1"!#

n#

*=1

+ $p

$%*(y + x(!, #),

2

"!# + %(!, #),' $p

$%*(0(!, #))

$fjk

$y*(!, y, 2, #)

+n#

*=1

+ !

"!#3 2* '7

$p

$x*(y + x(!, #)

2

"!# + %(!, #)) ' $p

$x*(0(!, #))

K, $fjk

$2*(!, y, 2, #)

J

vj

+!, y + x(!, #),

2

"!# + %(!, #),

=:n#

j=1

Gj(!, y, 2, #) vj

+!, y + x(!, #)

2

"!# + %(!, #),.

According to Definition 4.3.14 the Lemma will be proved if we show that("!#Gj(!, y, 2, #))"+S belongs to H that is all the derivatives with respect to(y, 2) are uniformly bounded when |y| " & "!# and

552 ' 12 sgn !(1 ' *1(!)) y| " &.

Recall that (fjk) = ((qk(2 ' 12 sgn !(1 ' *1(!)) y,2a,2b, gj))!1). Using (4.3.5) we see

that ()(& $ H. Then di!erentiating (4.3.19) with respect to ! we see that ("

(& , (b(& $ H.

It follows from (4.3.9) that (gj

(& $ H. Then we deduce from the estimates (4.2.2) that((&

@qk(2 ' 1

2 sgn !(1 ' *1(!)) y,2a,2b, gj)A

belongs to H and we deduce from Corollary4.3.5 (iii) that (

(&

@(qk(2 ' 1

2 sgn !(1 ' *1(!)) y,2a,2b, gj))!1A$ H. Thus ((fjk

(& ) $ H.Since fjk, (fjk

(& , (fjk

(y$, (fjk

(-$belong to H and since H is closed under multiplication it

remains to prove that the functions h(!, y, 2, #) = (p(%$

(y + x(!, #), -'&( + %(!, #)) or



"!# (p(x$

(y + x(!, #), -'&( + %(!, #)) belong to H. Since -$

'&( + %*(!, #) has all its deriva-tives in 2 uniformly bounded, we are led to prove that if gjk(x) are the coe"cients ofp(x, %) then $A

x gjk(y + x(!, #)) for A $ Nn and "!# $Bx gjk(y + x(!, #)) for B $ Nn,

|B| ! 1 are uniformly bounded when |y| " & "!#. This is obvious if A = 0 and if|B| ! 1 condition (3.1.1) shows that

|$Bx gjk(y + x(!, #))| " CB

"y + x(!, #)#|B|+1+!0=

CB

"x(!, .)#|B|+1+!0

" CB

"!#|B|+1+!0

since . $ S , S+ ! S! (see (3.3.3)).

Corollary 4.3.17. — With # defined in (4.3.43) we have for k = 1, . . . , n,+' $p

$xk(x, #(!, x, #))'$#k

$!(!, x, #)'

n#

j=1

$#k

$xj(!, x, #)

$p

$%j(x, #(!, x, #))

,

"+S$ J(x,%).

Proof. — First of all we show that

(1) =+ $p

$xk(x, #(!, x, #)) ' $p

$xk(x, %)

,

"+S$ J(x,%).

Denoting by gij the coe"cients of p we can write

$gij

$xk(x)#i #j '

$gij

$xk(x) %i %j =

$gij

$xk(x) (#i ' %i) %j

< => ?(a)

' $gij

$xk(x)#i(%j ' #j)

< => ?(b)

.

To see that this belongs to J(x,%) we use (4.3.45). In the coordinates (y, 2) we have

"!# (a) = '$gij

$xk(y + x(!, #))

+ 2j

"!# + %j(!, #),(2i ' -i(!, y, #)).

Since 2i ' -i(!, y, #) $ J (see Example 4.3.9 and (4.3.38)) and$ -j

'&( + %j(!, #)%·

(gij

(xk(y + x(!, #)) belongs to H we have (a) $ J(x,%).

A similar argument shows that (b) belongs to J(x,%).We show now that

(2) =+$#k

$xj(!, x, #)

. $p

$%j(x, #(!, x, #)) ' $p

$%j(x, %)

/,

"+S$ J(x,%).

The coe"cients of p can be written gij(x) = &ij + cij(x). It follows that

(2) =+2

$#k

$xj(!, x, #)

+#j(!, x, #) ' %j +

n#

*=1

cj*(x)(#*(!, x, #) ' %*),,

.

Now ("k(xj

= 1'&(

(0k

(y$$ H and cj*(y +x(!, #)) $ H. It follows that (2) $ J(x,%) and the

Corollary follows from Lemma 4.3.16.



Corollary 4.3.18. — For every N $ N there exists CN > 0 such that555'

$p

$xk(x, #(!, x, #)) ' $#k

$!(!, x, #) ' $#k

$x(!, x, #) · $p

$%(x, #(!, x, #))

555

" CN

"!#

+ |x ' x(!, #)|"!#

,N.

Proof. — This follows from Lemma 4.3.15 and Corollary 4.3.17 since the left handside does not depend of %.

Proposition 4.3.19. — Let # $ S (see (4.3.1)). Let us set for (!, x) $ %+,

(4.3.48) )(!, x, #) =- 1

0(x'x(!, #))·#(!, s x+(1's)x(!, #), #) ds+! p(#)+

12i

|#%|2.

Then we have

(i) )(0, x, #) = (x ' #x) · #% +i

2(x ' #x)2 +

12i

|#%|2 + O(|x ' #x|N ).For every N $ N there exists CN > 0 such that

(ii)555$)

$x(!, x, #) ' #(!, x, #)

555 " CN

+ |x ' x(!, #)|"!#

,N.

(iii)555$)

$!(!, x, #) + p

+x,

$)

$x(!, x, #)

,555 " CN

+ |x ' x(!, #)|"!#

,N

uniformly with respect to (!, x) $ %+ and # $ S.Moreover, uniformly with respect to (!, x, #) $ %+ ( S, we have

(iv)555$)

$x(!, x, #) ' #%

555 " C (( +/

&).

(v) |$Ax )(!, x, #)| " CA, %A $ Nn.

(vi)555 Im )(!, x, #) ' 1

2|x ' x(!, #)|2

1 + 4 !2+

12|#%|2

555 " C (( +/

&)|x ' x(!, #)|2

"!#2 .

Proof. — If we can prove that for j = 1, . . . , n,

#j(0, s x + (1 ' s)#x, #) = #j% + i s(xj ' #j

x) + O(sN |x ' #x|N )

then (i) will follow according to (4.3.48). By (4.3.43), Corollary 4.3.5 and Theorem4.3.1 we have

#j(0, x, #) = #j% ' (2aj + i 2bj)(0, x ' #x, #) = #j

% ' (ak + i bk)(0, x ' #x, #).

Now Example 4.3.9 (for ! = 0) shows that 2j + aj(0, y, #) + i bj(0, y, #) belongs tothe ideal J introduced in Definition 4.3.8. On the other hand, since by (4.2.10) for! = 0 gj(2) = *0( 1

µ02)(2j ' i yj) it follows that 2j ' i yj belongs also to J . Thus the

di!erence aj(0, y, #) + i bj(0, y, #) + i yj belongs also to J and does not depend on 2.It follows from Lemma 4.3.11 that for all N $ N,

aj(0, y, #) = O(| Im -(0, y, #)|N )

bj(0, y, #) + yj = O(| Im -(0, y, #)|N ).



Now (4.3.38) and Theorem 4.3.1 (ii) show that | Im -j(0, y, #)| = |bj(0, y, #)| " C |y|.Thus for all N $ N,

aj(0, y, #) = O(|y|N ), bj(0, y, #) = 'yj + O(|y|N ).

It follows that for all N $ N,

#j(0, y, #) = #j% + i(x ' #j

x) + O(|x ' #x|N )

which proves our claim.Let us prove (ii). We have, by (4.3.48),

$)

$xj(!, x, #) =

- 1

0#j(!, s x + (1 ' s)x(!, #), #) ds +

n#

k=1

- 1

0s(xk ' xk(!, #))·

·$#k

$xj(s x + (1 ' s)x(!, #), #) ds

Now we use Theorem 4.3.13 (vii) and the fact that |x ' x(!, #)| " & "!#. We deducethat

555$)

$xj(!, x, #) '

- 1

0

.#j + s

n#

k=1

(xk ' xk(!, #))$#j

$xk

/(!, s x + (1 ' s)x(!, #), #) ds

555

" CN

+ |x ' x(!, #)|"!#

,N

where CN is independent of (!, x, #).It follows that

555$)

$xj(!, x, #) '

- 1

0#j(!, s x + (1 ' s)x(!, #), #) ds

'- 1

0s

d

ds

@#j(!, s x + (1 ' s)x(!, #), #)

Ads555 " CN

+ |x ' x(!, #)|"!#

,N.

Integrating by parts in the second integral above, we obtain (ii). As a consequence of(ii) we have the estimate

(4.3.49)555p(x, #(!, x, #)) ' p

+x,

$)

$x(!, x, #)

,555 " CN

+ |x ' x(!, #)|"!#

,N.

Let us prove (iii). We deduce from (4.3.48) that

(4.3.50)$)

$!(!, x, #) = (1) + (2) + (3) + (4)



where

(1) = 'n#

k=1

- 1

0xk(!, #)#k(!, s x + (1 ' s)x(!, #), #) ds

(2) =n#

k=1

- 1

0(xk ' xk(!, #))

$#k

$!(!, s x + (1 ' s)x(!, #), #) ds

(3) =n#

k=1

- 1

0(xk ' xk(!, #))

n#

j=1

$#k

$xj(!, s x + (1 ' s)x(!, #), #)(1 ' s) xj(!, #) ds

(4) = p(#).

Let us consider the term (2). We use Corollary 4.3.18 to get

(2) =n#

k=1

- 1

0(xk'xk(!, #))

.' $p

$xk(s x+(1's)x(!, #), #(!, s x+(1's)x(!, #), #))

'n#

j=1

$#k

$xj(!, s x + (1 ' s)x(!, #), #)

$p

$%j(s x + (1 ' s)x(!, #),

#(!, s x + (1 ' s)x(!, #), #))/ds

+ O++ |x ' x(!, #)|

"!#

,N ,.

Now, by Theorem 4.3.13, (vii),

(4.3.51)555+$#k

$xj' $#j

$xk

,(!, s x + (1 ' s)x(!, #), #)

555 " CNsN

"!#|x ' x(!, #)|N

"!#N

and sN " 1. Therefore,

(2) = '- 1

0

d

ds

@p(s x + (1 ' s)x(!, #), #(!, s x + (1 ' s)x(!, #), #))

Ads

+ O+ |x ' x(!, #)|N

"!#N,.

Therefore we obtain

(4.3.52)555(2) + p(x, #(!, x, #)) ' p(x(!, #), #(!, x(!, #), #))

555 " CN

+ |x ' x(!, #)|"!#

,N.

Let us consider the term (3). Using again (4.3.51) we get

(3) =n#

j=1

- 1

0(1 ' s) xj(!, #)

d

ds(#j(!, s x + (1 ' s)x(!, #), #)) + O

+ |x ' x(!, #)|N

"!#N,.

Integrating by part we obtain

(3) = 'x(!, #) · #(!, x(!, #), #) +n#

j=1

- 1

0xj(!, #)#j(!, s x + (1 ' s)x(!, #), #) ds.


4.4. THE CASE OF INCOMING POINTS 53

Comparing with the term (1) we obtain,55(1) + (3) + x(!, #)#(!, x(!, #), #)

55 " CN

+ |x ' x(!, #)|"!#

,N.

Now by Theorem 4.3.13 (vi) and the Euler relation we have,

x(!, #) · #(!, x(!, #), #) = %(!, #)$p

$%(x(!, #), %(!, #))

= 2p(x(!, #), %(!, #)) = 2p(#).

It follows that

(4.3.53)55(1) + (3) + 2p(#)

55 " CN

+ |x ' x(!, #)|"!#

,N.

Since in (4.3.52) we have p(x(!, #), #(!, x(!, #), #)) = p(x(!, #), %(!, #)) = p(#), wededuce from (4.3.50), (4.3.52) and (4.3.53) that

(4.3.54)555$)

$!(!, x, #) + p(x, #(!, x, #))

555 " CN

+ |x ' x(!, #)|"!#

,N.

Therefore (iii) follows from (4.3.54) and (4.3.49).Finally (iv), (v), (vi) follow easily from Theorem 4.3.13.

Remark 4.3.20. — Assume that # is such that

(4.3.55)12

" |#%| " 2 and #x · #% " c0 "#x# |#%|

(so # $ S! instead of # $ S). Then Theorem 4.3.1, Corollary 4.3.5 and Proposition4.3.19 are true for ! " 0.

By the same way if 12 " |#%| " 2 and #x · #% ! 'c0 "#x# |#%| (which imply that

# $ S+) the above results hold for ! ! 0.

4.4. The case of incoming points

We are going to prove Theorem 4.1.2 when

|#x · #%| > c0 "#x# |#%| and12

" |#%| " 2.

Since the problem is entirely symmetric we can without loss of generality assume that

(4.4.1)12

" |#%| " 2, #x · #% < 'c0 "#x#|#%|.

It follows from Definition 4.1.1 and the discussion after, that

2%+ =:(!, y) $ R ( Rn : ! " 0, |y| " & "!#

;2:(!, y) $ R ( Rn : ! ! 0, |y| " & "!#

and (y + x(!, #)) · #% " c1 "y + x(!, #)# |#% |;.



Let now *0 $ C"0 (Rn), *1 $ C"(Rn) be such that,

*0(t) = 1 if |t| " 1, *0(t) = 0 if |t| ! 2 and 0 " *0 " 1,

*1(!) = 1 if ! ! '1, *1(!) = 0 if ! " '2 and 0 " *1 " 1.

For j = 1, . . . , n we introduce

(4.4.2) gj(2) = *0

+ 1µ0

2,

vj(!, y + x(!, #), 2 *1(!) + (1 ' *1(!)). 2

"!# +12

sgn !

"!# y/

+ %(!, #), #)),

where µ0 is a small constant to be chosen, (!, y) $ 2%+, # satisfies (4.4.1) and vj hasbeen introduced in (4.2.9).

Remark 4.4.1(i) According to Remark 4.3.20 since (4.4.1) implies (4.3.55) the phase has been

already constructed when (!, x) belongs to the first part of 2%+ where ! " 0. Thereforewe are left with the case ! ! 0.

(ii) If # satisfies (4.4.1) and (!, y) $ 2%+, ! ! 0, then the point (y+x(!, #), 2+%(!, #))belongs to S!. Indeed recall that 1

2 " |#%| " 2 implies 14 " 1

2 |#%| " |2 + %(!, #)| "2 |#%| if |2| " 2 µ0 and µ0, ( are small enough. Therefore setting x = x(!, #) we canwrite

x · (2 + %(!, #)) = x · (#% + 2 + 1(!, #)) " c1 "x# |#%| + |x| (µ0 + ()

" 2 c1 "x# |2 + %(!, #)| + 4(µ0 + () "x# |2 + %(!, #)|

" 14"x# |2 + %(!, #)|

if c1, µ0, ( are small enough.

Our first step will be the proof of the following result.

Theorem 4.4.2. — There exist small positive constants µ0, & and C" functions a =a(!, y, #), bk = bk(!, y, #), k = 1, . . . , n, defined on 2%+ with ! ! 0 such that, witha = (ak), b = (bk) we have for 2 $ Rn,

(i) gj(2) =n#

k=1

qk(2, a, b, gj)(2k + ak(!, y, #) + i bk(!, y, #))

where the q%k s have been introduced in Lemma 4.2.1.Moreover we have for (!, y) $ 2%+, ! ! 0 and k = 1, 2, . . . , n,

(ii)555ak(!, y, #) +

2! yk

1 + 4 !2

555 "/

& inf(1, |y|),555b(!, y, #) +

yk

1 + 4 !2

555 "/

&|y|"!#2 .

(iii) If we set

2a(!, y, #) = a(!, y, #) +2! y

1 + 4 !2and 2b(!, y, #) = "!#

$b(!, y, #) +

y

1 + 4 !2

%



then for every A $ Nn, |A| ! 1 one can find CA ! 0 such that with x = y + x(!, #),! ! 0,

|$Ay 2a(!, y, #)| + |$A

y2b(!, y, #)| " CA

. (

"x#!0

+ 1"x# +

1"!#

,|A|+1+

&

"!#|A|+1

/

(iv) |qk(2, a, b, gj) ' (1 + 2i!) &jk| " C(( +/

&) "!# if |2| "/

&.(v) |$B

(a,b) $,- qk(2, a, b, gj)| " CB,, "!#, if B $ Nn, 3 $ Nn, 1 " k " n.

Proof. — We use the same method as in Theorem 4.3.1. According to Lemma 4.2.1,the claim (i) is equivalent to solve the system of equations

(4.4.3) r(a, , gj(!, y, #; ·) = 0, j = 1, . . . , n.

We shall solve this system in the set(4.4.4)

E =!(a, b) $ Rn ( Rn :

555a +2!y

1 + 4!2

555 "/

& inf(1, |y|),555b +

y

1 + 4!2

555 "/

&|y|"!#2

"

where 0 < & 7 1.First of all we give equivalent equations to (4.4.3) in the set E. We write as in

(4.3.13)

(4.4.5) r(a, b, gj) = gj('a) ' in#

k=1

$gj

$%k('a) bk +

n#

p,q=1

Hjpq(!, y, #, a, b) bp bq

where

(4.4.6) Hjpq(!, y, #, a, b) =

- 1

0

$2r

$bp $bq(a, t b, gj(!, y, #; ·))(1 ' t) dt.

By (4.2.2) we have

(4.4.7) |$"(a,b) $B

y r(a, t b, gj(· · · ))| " CAB

#

|,|!|A|+3n

-|$,

% $By gj(!, y, #, %)| d%.

Since for ! ! 0 we have,

gj(2) = *+ 2

µ0

,@(%j ' i xj)('!; y + x(!; #), 2 + %(!; #)) + i(#x + i #%)

A

we deduce from Propositions 3.3.1 and 3.3.2 that |$,- $B

y gj(!, y, #, 2)| is bounded onthe support of *, by

'(((()

((((*

C "!# if B = 0,

C+1 +

("!#"x#2+!0

,if |B| = 1,

CB(

"x#|B|+!0

+1 +

"!#"x#

,if |B| ! 2.



It follows that

(4.4.8) |$A(a,b) $B

y Hjp,q(!, y, #, a, b)| "

'(((()

((((*

C "!# if |B| = 0

C$1 +

( "!#"x#2+!0

%if |B| = 1

CB(

"x#|B|+!0

$1 +

"!#"x#%

if |B| ! 2

Since *('a) = 1 and *%('a) = 0 we see that (4.4.3) is equivalent in E to the vectorialequation,

(4.4.9).% ' i x ' i

n#

k=1

+ $%

$%k' i

$x

$%k

,bk

/('!; y + x(!; #), %(!; #) ' a)

+ i(#x + i #%) +n#

p,q=1

Hpq(!, y, #, a, b) bp bq = 0.

To shorten the notations we shall set

(4.4.10)

'()

(*

0(!; #) = (x(!; #), %(!; #))

0y(!; #) = (y + x(!; #), %(!; #))

0y,a(!; #) = (y + x(!; #), %(!; #) ' a).

Since, by assumption, the point 0y,a(!; #) belongs to S! (the outgoing set for ! " 0)we can use the Proposition 3.3.1 to write

x('!; 0y,a(!; #)) = y + x(!; #) ' 2! %('!; 0y,a(!; #)) + z('!; 0y,a(!; #))

%('!; 0y,a(!; #)) = 'a + %(!; #) + 1('!; 0y,a(!; #)).

It follows that (4.4.9) is equivalent to

(1 + 2i !) %('!; 0y,a(!; #)) ' i y ' i x(!; #) ' i z('!; 0y,a(!; #))

' i.(1 + 2i !)

n#

k=1

$%

$%k('!; 0y,a(!; #)) bk ' i

n#

k=1

$z

$%k('!; 0y,a(!; #)) bk

/

+ i(#x + i #%) +n#

p,q=1

Hpq(· · · ) bp bq = 0.

Taking the real and the imaginary parts, we are led to the 2n real equations

%j('!; 0y,a(!; #)) + 2! bj + 2!$1j

$%('!; 0y,a(!; #)) · b

' $zj

$%('!; 0y,a(!; #)) · b ' #j

% + Hj1 b · b = 0

' 2! %j('!; 0y,a(!; #)) + bj + yj + xj(!; #) ' #jx + zj('!; 0y,a(!; #))

+$1j

$%('!; 0y,a(!; #)) · b + Hj

2 b · b = 0



where$1j

$%· b =

n#

k=1

$1j

$%k· bk, Hj b · b =

n#

p,q=1

Hjp,q bp bq, Hj = Hj

1 + i Hj2 .

Setting X =$ %j(!&;1y,a(&;"))

bj

%, A =

$1 2&

!2& 1

%, our system can be written AX = F .

Since A!1 = 11+4&2

$1 !2&2& 1

%, it is equivalent to the following system.

%('!; 0y,a(!; #)) =2!y

1 + 4!2+

11 + 4!2

.2!(x(!; #) ' #x) + 2! z('!; 0y,a(!; #))

+ #% +$z

$%('!; 0y,a(!; #)) · b ' (H1 ' 2! H2) b · b

/.

b ='y

1 + 4!2' 1

1 + 4!2

@x(!; #) ' #x ' 2! #% + z('!; 0y,a(!; #)) + (2! H1 + H2) b · b

A

' $1

$%('!; 0y,a(!; #)) · b +

2!

1 + 4!2

$z

$%('!; 0y,a(!; #)) · b.

Finally, since %('!; 0y,a(!; #)) = 'a + %(!; #) + 1('!; 0y,a(!; #)) the system (4.4.3) isequivalent to

a ='2! y

1 + 4!2+ %(!; #) ' 1

1 + 4!2

.#% + 2!(x(!; #) ' #x) + 2! z('!; 0y,a(!; #))

+$z

$%('!; 0y,a(!; #)) · b

/+ 1('!; 0y,a(!; #)) + H3 b · b

b ='y

1 + 4!2' 1

1 + 4!2

@x(!; #) ' #x ' 2! #% + z('!; 0y,a(!; #))

A

' $1

$%('!; 0y,a(!; #)) · b +

2!

1 + 4!2

$z

$%('!; 0y,a(!; #)) · b + H4 b · b

where according to (4.4.8) H*, / = 3, 4, are two matrices which entries satisfy thefollowing estimates

(4.4.11) |$A(a,b) $B

y Hj*,p,q(!, y, #, a, b)| "

'(((()

((((*

C, if |B| = 0C

"!#$1 +

( "!#"x#2+!0

%, if |B| = 1

CB(

"!# "x#|"|+1+!0

$1 +

"!#"x#%, if |B| ! 2.

Let us set(4.4.12)'(((((((()

((((((((*

#1(a, b) ='2! y

1 + 4!2+ %(!; #) ' 1

1 + 4!2

.#% + 2!(x(!; #) ' #x) + 2! z('!; 0y,a(!; #))

+$z

$%('!; 0y,a(!; #)) · b

/+ 1('!; 0y,a(!; #)) + H3 b · b

#2(a, b) ='y

1 + 4!2' 1

1 + 4!2

@x(!; #) ' #x ' 2! #% + z('!; 0y,a(!; #))

A

' $1

$%('!; 0y,a(!; #)) · b +

2!

1 + 4!2

$z

$%('!; 0y,a(!; #)) · b + H4 b · b.



Then the system (4.4.3) to be solved is equivalent in E to the equation

(#1(a, b), #2(a, b)) = (a, b).

Let us set #(a, b) = (#1(a, b), #2(a, b)). We shall use the fixed point theorem in E(see (4.4.4)).

(i) #(E) , E.Let us recall that (y + x(!; #)) · #% " c0 "y + x(!; #)# |#% |.

Case 1. — Assume that

(4.4.13) x(!; #) · #% ! 2 c0 "x(!; #)# |#% |.

It follows that

(4.4.14) |y| ! c0

2"x(!; #)# ! c0

2.

Indeed one can write

2 c0 "x(!; #)# |#% | " (x(!; #) + y) · #% ' y · #%

" c0 "y + x(!; #)#|#% | + |y| · |#%|" c0 "x(!; #)# |#% | + c0 |y| · |#%| + |y| · |#%|.

Therefore c0 "x(!; #)# " 2 |y|. Here we have used the inequality "a + b# " "a# + |b|.Let !# ! 0 be such that x(!#; #) · #% = 0 (this is possible since #x · #% " 0 and

x(!; #) · #% ! #x · #% + ! |#%|2 . +) if ! . +)). Then the point (x(!#; #), #%) isoutgoing for ! ! 0 and ! " 0.

We can write by Proposition 3.3.1,

x(!; #) = x(! ' !#; x(!#; #), %(!#; #))

= x(!#; #) + 2(! ' !#) %(! ' !#; x(!#; #), %(!#; #)) + z(! ' !#, · · · )= x(!#; #) + 2(! ' !#) %(!; #) + z(! ' !#; x(!#; #), %(!#; #)).

It follows that

x(!; #) · #% = 2(! ' !#) |#%|2 + O(() |! ' !#| + z(! ' !#; x(!#, #), %(!#; #)) · #%.

Since |z(! ' !#, · · · )| " C ( |! ' !#| we deduce the estimate,

2 |! ' !#| |#%|2 " |x(!, #)| · |#%| + C% ( |! ' !#|.

Therefore if ( is small enough we obtain

(4.4.15) |! ' !#| " 5 |x(!, #)|.

Now let us introduce

(4.4.16) u(!) = x(!; #) ' #x ' 2! #%.

We claim that

(4.4.17) |u(!)| " C ( "! ' !##.



Indeed we have u(0) = 0 and for all ! in R,

u(!) = x(!; #) ' 2 #% = 2(%(!; #) ' #%) + 2( b(x(!; #)) · %(!; #).

It follows from Proposition 3.4.1 that

|u(!)| " C (.

Now since x(!#; #) · #% = 0 it follows from Proposition 3.5.2 that

x(!#; #) = #x + 2!# #% ' z('!#; x(!#; #), %(!#; #)).

This implies that |u(!#)| " C1 (. Now we write

|u(!) ' u(!#)| "555- &

&$|u(s)| ds

555 " C2 ( |! ' !#|

and|u(!)| " C1 ( + C2 ( |! ' !#| " C3 ( "! ' !##.

It follows then from (4.4.14) to (4.4.17) that,

|x(!; #) ' #x ' 2! #%| " C3 ( "! ' !## " 5 C3 ( "x(!; #)#

" 10 C3

c0( |y|.

Using (4.4.12) we see that555#2(a, b) +

y

1 + 4!2

555 " |x(!; #) ' #x ' 2! #%|1 + 4!2

< => ?(1)

+|z('!; 0y,a)|

1 + 4!2< => ?

(2)

+555$1

$%('!, 0y,a)

555< => ?

(3)

·|b| + 1"2!#

555$z

$%('!; 0y,a)

555< => ?

(4)

·|b| + *H4* · |b|2< => ?(5)

.

We have (1) " C4 ( |y|/"!#2, (2) " C (/"!#2 " C% ( |y|/"!#2 since by (4.4.14) |y| ! c0/2.Moreover

(3) " C ( |b| " C% ( |y|"!#2 , (4) " C (

"2!# |b| " C% ( |y|"!#2

and, by (4.4.11), (5) " C |y|'&(2

|y|'&(2 . Since |y| " & "!# it follows that (5) " C & |y|/"!#2.

Summing up we obtain

(4.4.18)555#2(a, b) +

y

1 + 4!2

555 " C (( + &)|y|

1 + 4!2

so we take (, & so small that C (( + &) "/

&.Let us look to the term

(II) =555#1(a, b) +

2! y

1 + 4!2

555.



We have

#1(a, b) +2! y

1 + 4!2= %(!; #) ' #%< => ?

(1)

' 2!

1 + 4!2[x(!; #) ' #x ' 2! #%]

< => ?(2)

' 2!

1 + 4!2z('!; 0y,a)

< => ?(3)

' 11 + 4!2

$z

$%('!; 0y,a) · b

< => ?(4)

+ 1('!; 0y,a)< => ?(5)

+ H3 b · b< => ?(6)

.

We have |(1)| " C (, |(2)| " C # &'&!&$(1+4&2 " C% ( (see (4.4.17)), |(3)| " C #

'&( (by Proposi-tion 3.3.2), |(4)| " C #

'&(2 |b| " C%% (, |(5)| " C (, |(6)| " C &2. It follows that if ( and &are small enough we have,

(4.4.19) (II) " C (( + &2) " C (( + &2)2c0

inf(1, |y|) "/

& inf(1, |y|)

since |y| ! c0/2. It follows from (4.4.18) and (4.4.19) that # maps E into E.We show now that one can find a constant k $ ]0, 1[ such that

(4.4.20) |#(a, b) ' #(a%, b%)| " k |(a, b) ' (a%, b%)|, % (a, b), (a%, b%) $ E.

We have

|#1(a, b) ' #1(a%, b%)| " 11 + 4!2

|z('!; 0y,a(!; #)) ' z('!; 0y,a#(!, #))|< => ?

(1)

+1

1 + 4!2

555$z

$%('!; 0y,a(!; #)) · b ' $z

$%('!; 0y,a#(!; #)) · b%

< => ?(2)

555

+ |1('!; 0y,a(!; #)) ' 1('!; 0y,a#(!; #))< => ?

(3)

| + |H3 b · b ' H %3 b% · b%< => ?

(4)

|.

Since the point (y +x(!; #), #%) is outgoing we can use Proposition 3.3.2, (4.4.11) andthe fact that |b| " 2 |y|

'&(2 " 2&, to write

(1) " C !

1 + 4!2( |a ' a%|, (2) " C ! (

1 + 4!2(|b ' b%| + |a ' a%|)

(3) " C ( |a ' a%|, (4) " C & (|a ' a%| + |b ' b%|).

It follows that

(4.4.21) |#1(a, b) ' #1(a%, b%)| " C (( + &)(|a ' a%| + |b ' b%|).



Now we have

|#2(a, b) ' #2(a%, b%)| "555$1

$%('!; 0y,a(!; #)) · b ' $1

$%('!; 0y,a#(!; #)) · b%

555

+2!

1 + 4!2

555$z

$%('!; 0y,a(!; #)) · b ' $z

$%('!; 0y,a#(!; #)) · b%

555

+ |H4 b · b ' H %4 b% · b%| + 1

1 + 4!2|z(('!; 0y,a(!; #)) ' z('!; 0y,a#(!; #))|.

The same estimates as those used in the first case show that

(4.4.22) |#2(a, b) ' #2(a%, b%)| " C (( + &)(|a ' a%| + |b ' b%|).

Thus (4.4.20) is proved if ((+&) is small enough. Then the fixed point theorem showsthat the system (4.4.3) has a unique solution in E.

Case 2. — Assume that

x(!; #) · #% " 2 c0 "x(!; #)# |#% |.

It follows that we can apply Proposition 3.5.2 with (y, 2) = # which allows us to write6

x(!; #) = #x + 2! #% ' z('!; 0(!; #))

%(!; #) = #% ' 1('!; 0(!; #))

where 0(!; #) = (x(!; #), %(!; #)).Using (4.4.12) we obtain the following expressions of #1, #2.

#1(a, b) ='2! y

1 + 4!2' 2!

1 + 4!2

@z(('!; 0y,a(!; #)) ' z('!; 0(!; #))

< => ?(1)

]

' 11 + 4!2

$z

$%('!; 0y,a(!; #)) · b

< => ?(2)

+ 1('!; 0y,a(!; #)) ' 1('!; 0(!; #))< => ?

(3)

+ H3 b · b< => ?(4)

.

#2(a, b) = ' y

1 + 4!2' 1

1 + 4!2

@z('!; 0y,a(!; #)) ' z('!; 0(!; #))

< => ?(5)

A

' $1

$%('!; 0y,a(!; #)) · b

< => ?(6)

+2!

1 + 4!2

$z

$%('!; 0y,a(!; #)) · b

< => ?(7)

+ H4 b · b< => ?(8)

.

Let us show that

(4.4.23)

'()

(*

55#1(a, b) +2! y

1 + 4!2

55 "/

& inf(1, |y|)55#2(a, b) +

y

1 + 4!2

55 "/

& |y|'&(2 .



If |y| ! c0 then inf(1, |y|) ! c0. It follows that

|(1)| " 2!

1 + 4!2C ( " C (

c0inf(1, |y|),

|(2)| " C ( " C (

c0inf(1, |y|),

|(3)| " C ( |b| " 2C ( |y|"!#2 " 2C ( & " 2C ( &

c0inf(1, |y|),

|(4)| " C &

c0inf(1, |y|),

|(5)| " C (

"!#2 " C (

c0

|y|"!#2 ,

|(6)| " C ( |b| " C% (|y|"!#2 ,

|(7)| " C (|y|"!#2 ,

|(8)| " C &|y|"!#2 .

These estimates imply (4.4.23).Assume now that |y| " c0. It follows that for every t in [0, 1] the point (t y +

x(!; #), %(!; #) ' t a) is outgoing for ! " 0 (i.e. belongs to S!). Indeed we have

%(!; #) ' t a = #% + O(( + &)|x(!; #)| " |y| + |y + x(!; #)| " 1 + |y + x(!; #)|

so

(t y + x(!; #)) · #% " |y| · |#%| + 2 c0 "x(!; #)# |#% |" 3 c0 "x(!; #)# |#% |" 6 c0 "y + x(!; #)# |#% |.

Then we have the following estimates.

|(1)| "- 1

0

.|y|555$z

$y

555+ |a|555$z

$%

555/('!, t y + x(!; #), %(!; #) ' ta) dt

|(1)| " C ( |y| since |a| " C% | y|.

By the same way we have |(3)| " C ( |y|. Moreover |(2)| " C ( |y|, |(4)| " C & |y| since|b| " 2 |y|

'&(2 " 2&.On the other hand we have

|(5)| " C ( |y|"!#2 , |(6)| " C ( |b| " C%( |y|

"!#2

|(7|) " C ( |b| " C% (|y|"!#2 , |(8)| " C &

|y|"!#2 .



These estimates imply (4.4.23) since in this case |y| = inf(1, |y|). Summing up wehave proved that # maps E into itself.

We show now the estimate (4.4.22). But it is easy to see that the proof givenin Case 1 works also in Case 2. Using again the fixed point theorem we see thatthe system (4.4.3) has a unique solution in E. This proves the points (i) and (ii) ofTheorem 4.4.2.

To prove (iii) we use an induction on |A| starting with |A| = 1. Let us set for fixed(!, #)

(4.4.24) 3y,ea =+y + x(!, #), %(!, #) ' 2a(!, y, #) +

2! y

1 + 4!2

,.

Using (4.4.12) we see that (2a,2b) satisfy the system

(4.4.25) 2a = %(!, #) ' #% + 2!(x(!, #) ' #x)1 + 4!2

' 2!

1 + 4!2z('!, 3y,ea)

< => ?(1)

' 11 + 4!2

$z

$%('!, 3y,ea)

1"!#2b

< => ?(2)

+1

1 + 4!2

$z

$%('!, 3y,ea)

y

1 + 4!2< => ?

(3)

+ 1('!, 3y,ea)< => ?(4)

+1

"!#22H32b ·2b

< => ?(5)

+2"!#

2H3,2by

1 + 4!2

< => ?(6)

+ 2H3y

1 + 4!2

y

1 + 4!2< => ?

(7)

.

(4.4.26) 2b = '"!#(x(!, #) ' #x ' 2! #%)1 + 4!2

' "!#1 + 4!2

z('!, 3y,ea)< => ?

(8)

' $1

$%('!, 3y,ea)2b

< => ?(9)

+$1

$%('!, 3y,ea)

y"!#1 + 4!2

< => ?(10)

+1"!#

2H42b ·2b

< => ?(11)

+ 2 2H42b

y

1 + 4!2< => ?

(12)

+ "!# 2H4y

1 + 4!2· y

1 + 4!2< => ?

(13)

+2!

1 + 4!2

$z

$%('!, 3y,ea)2b

< => ?(14)

' 2!

1 + 4!2

$z

$%('!, 3y,ea)

y"!#1 + 4!2

< => ?(15)

where for j = 3, 4 2Hj = Hj

$!, y, #,2a' 2& y

1+4&2 , "!#2b' y1+4&2

%and Hj satisfies (4.4.11).

We claim that, for j = 3, 4,

(4.4.27) |$y[ 2Hj ]| " C+|$y 2a| + "!# |$y

2b| + 1"!# +

(

"x#2+!0

,.



Indeed, skipping the index j for convenience, we have

$

$yk[ 2H ] =

$H

$yk+

$H

$a

$2a$yk

' 2!

1 + 4!2

$H

$ak+ "!# $H

$2b$2b$yk

' 11 + 4!2

$H

$bk.

Now we use (4.4.11). The first term in the right hand side is bounded by C'&( + C #

'x(#0+2 ,

the second by C |82a|, the third by C'&( , the fourth by C "!#|8y

2b| and the last one by"!#!2.

For / $ N let us introduce the following space(4.4.28)

F* =!F $ C"(Rn ( Rn) : |$A

x $B% F (x, %)| " CAB (

"x#|A|+*+!0, % (x, %) $ Rn ( Rn

".

For example 1, (2(% $ F1, z, (z

(% $ F0 according to Proposition 3.3.2.Let us set now,

(4.4.29)

6g(y) = y + x(!; #)

h(y) = %(!; #) ' 2a(!, y, #) + 2& y1+4&2 .

Then for F $ F* and k = 1, . . . , n we have,

(4.4.30)555

$

$yk[F (g(y), h(y))]

555 " C (+ 1"x#*+1+!0

+1

"x#*+!0 "!# + |8y 2a|,

where x = y + x(!; #).Let us prove (iii) for |A| = 1. We di!erentiate the equations (4.4.25), (4.4.26) with

respect to yk and we use (4.4.27), (4.4.30) and the fact that |2b| " 3 |y|'&( " 3& " 1. We

have, with the notations in (4.4.25), (4.4.26),

|$yk (1)| + |$yk (8)| " C ( |8y 2a| + C ( 1'x(#0

1'&(

+1'x( + 1

'&(

,

|$yk (2)| " C ((|8y 2a| + |8y2b|) + C #

'x(#0'&(3

+1'x( + 1

'&(

,

|$yk (3)| " C (|8y 2a| + C #'x(#0'&(3

+1'x( + 1

'&(

,

|$yk (4)| " C ( |8y 2a| + C #'x(#0+1

+1'x( + 1

'&(

,

|$yk (5)| + |$yk (6)| + |$yk (7)| " C &(|8 2a| + |82b|) + C #'x(2+#0'&(2 + C +

'&(3

|$yk (9)| " C ((|82a| + |82b|) + C #'x(#0+1

+1

'x( + 1'&(

,

|$yk (10)| " C (|82a| + C #'x(#0

+1'x( + 1

'&(

,2

|$yk (11)| + |$yk (12)| + |$yk (13)| " C & (|82a| + |82b|) + C #'x(2+#0'&( + C +

'&(2

|$yk (14)| + |$yk (15)| " C (( + &)(|8y 2a| + |82b|) + C #'x(#0

+1

'x( + 1'&(

,2.

It follows from (4.4.25), (4.4.26), that

|8y 2a| + |8y2b| " C (( + &)(|82a| + |82b|) +

C (

"x#!0

+ 1"x# +

1"!#

,2+

C &

"!#2

which is (iii) for |A| = 1.



Let us assume that (iii) is true for 1 " |A| " k and let |A| = k + 1 ! 2.We claim that for F $ F* and |B| " k we have

(4.4.31)55$B

y [F (g(y), h(y))]55 " CB(

"x#!0

+ 1"x# +

1"!#

,|B|+*.

Indeed the term we want to estimate is a finite sum of terms of the following form(see Section A.1).

(4.4.32) R), s = ($)x $,

% F )(g(y), h(y))sH

j=1

$$*j

y g%k#

j$$*j

y h%kj

where 1 " s " |B|, 1 " |.| + |3| " |B|,&s

j=1 k%j = .,

&sj=1 kj = 3,

&sj=1(|kj | +

|k%j |) /j = B, /j += 0, (k%

j , kj) += 0, j = 1, . . . , s.Let us write {1, . . . , s} = I1 2 I2 where

I1 = {j : |/j | = 1}, |I2| = {j : |/j | ! 2}.

For j $ I1 we have $*jy gk = O(1), $

*jy hk = '$

*jy 2a + O

$1'&(%. For j $ I2 we have

$*jy gk 3 0. Therefore the only terms which are present are those for which k%

j = 0. Itfollows that

&sj=1 k%

j =&

j+I1k%

j = .. Moreover $*jy h = '$

*jy 2a. It follows from these

facts and the definition of F* that

|R), s| " C),(

"x#|)|+*+!0

L(C ()

Psj=1 |kj |

"x#!0Ps

j=1 |kj |

+ 1"x# +

1"!#

,Psj=1 |kj |(|*j|+1)

+(C &)

Psj=1 |kj |

"!#Ps

j=1 |kj |(|*j|+1)

M.

Now |.| =&s

j=1 |k%j | =

&sj=1 |k%

j | |/j| since |/j| = 1 in I1 and k%j = 0 in I2. It follows

that

|R), s| " C (

"x#!0

+ 1"x# +

1"!#

,Psj=1 |k#

j | |*#j |+*+ 1

"x# +1"!#

,Psj=1 |kj | |*j |

.

The result follows then, since&s

j=1(|kj | + |k%j |)|/j | = |B|.

On the other hand, for F $ F* and |A| = k + 1 ! 2, we have

(4.4.33)55$A

y (F (g(y), h(y))55 " C0( |$A

y 2a| +CA(

"x#!0

+ 1"x# +

1"!#

,|A|+*.

Indeed

$Ay (F (g(y), h(y)) '

n#

k=1

+ $F

$xk(g(y), h(y)) $A

y gk +$F

$%k(g(y), h(y)) $A

y hk)

is a finite sum of terms R), s given by (4.4.32) where 1 " s " |A|, 2 " |.|+ |3| " |A|,&sj=1 k%

j = .,&s

j=1 kj = 3,&s

j=1(|kj | + |k%j |) /j = A, |/j | ! 1, |kj | + |k%

j | ! 1,j = 1, . . . , s. Since |.| + |3| ! 2 we have |/j| " |A|' 1 so the term R), s is bounded,using the induction, by

CA(

"x#!0

+ 1"x# +

1"!#

,|A|+*.

On the other hand, since |A| ! 2 we have $Ay gk = 0 and $A

y hk = '$Ay 2ak. By (4.4.28)

we have55 (F(%k

(g(y), h(y))55 " C ( so (4.4.33) is proved. Note that C is independent

of A.



Let us now prove the last step of the induction. We apply $Ay to both members of

(4.4.25), (4.4.26). Then we estimate each term of the right hand side using (4.4.31),(4.4.33). We obtain

|$Ay (1)| + |$A

, (8)| " C0( |$Ay 2a| +

CA(

"x#!0

1"!#

+ 1"x# +

1"!#

,|A|.

|$Ay (4)| " C0( |$A

y 2a| +CA(

"x#!0

+ 1"x# +

1"!#

,|A|+1.

|$"y (2)| + |$A

y (3)| " C0($|$A

y 2a| + |$Ay2b|%

+CA(

"x#!0

1"!#3

+ 1"x# +

1"!#

,|A|.

|$Ay (9)| + |$A

y (10)| " C0(($|$A

y 2a| + |$Ay2b|%

+CA(

"x#!0

+ 1"x# +

1"!#

,|A|+1.

Finally

|$Ay (5)| + |$A

y (6)| + |$Ay (7)| + |$A

y (11)| + |$Ay (12)| + |$A

y (13)|

" C0&$|$A

y ,2a| + |$Ay2b|%

+CA(

"x#!0

+ 1"x# +

1"!#

,|A|+1+

CA&

"!#|A|+1.

If ( + & is small enough (compared with a finite number of fixed constants) we canabsorb the term C0(( + &)(|$A

y 2a| + |$Ay2b|) by the left hand side and we obtain the

estimate given in (iii).Let us now prove (iv). First of all since the point (y + x(!, #), 2 + %(!, #)) belongs

to S! and '! " 0 we deduce from Proposition 3.3.2 that

|$,% %('!, y + x(!, #), 2 + %(!, #))| + |$,

% x('!, y + x(!, #), 2 + %(!, #)| " C, "!#.

It follows then, from (4.4.2) and Lemma 4.2.1 that555$qk

$%j(2, a, b, gj)

555+555$qk

$b*(2, a, b, gj)

555 " C "!#.

Therefore we will have,

|qk(2, a, b, gj) ' qk('a, a, 0, gj)| " C "!# (|2 + a| + |b|).

Now if |2| "/

& we will have |2 + a| + |b| " 5/

&. On the other hand (4.2.7) showsthat, when |2| "

/& < 1

2 c0,

qk('a, a, 0, gj) =$gj

$%k('a) =

+ $%j

$%k' i

$xj

$xk

,('!, y + x(!, #),'a + %(!, #)).

By Corollary 3.3.3 we have

qk('a, a, 0, gj) = (1 + 2i!) &jk + O(( "!#).

Finally we obtain,

|qk(2, a, b, gj) ' (1 + 2i!) &jk| " C (( +/

&) "!#,

which is precisely the claim of point (iv).The last point (v) can be easily deduced from Proposition 3.3.2 and Lemma 4.2.1.

This ends the proof of Theorem 4.4.2.



We continue the proof of Theorem 4.1.2 in the case (4.4.1). We follow basically thesame method as in Section 4.3 with small changes. For the convenience of the readerwe give some details.

Let us set,

(4.4.34)

'()

(*

O =:(!, (y, 2)) $ R+ ( R2n :

|y| < &"!#, (y + x(!, #)) · #% " c1 "y + x(! #)#|#% |; |2| "/

&;

2' =:# $ T #Rn : 1

2 " |#%| " 2, #x · #% " 'c0 "#x#|#%|} .

We consider families (f(·, #))"+e# of function on O.

Definition 4.4.3. — We say that (f(·, #))"+e# belongs to H if

(i) for all # $ 2', (!, y, 2) 0. f(!, y, 2, #) is C" on O.(ii) For every A, B in Nn there exists CAB > 0 such that

sup(&,y,")+O*e#

|$Ay $B

- f(!, y, 2, #)| " CAB.

Remark 4.4.41) H is closed under multiplication and derivation with respect to (y, 2).2) If we set, with the notation (4.2.10),

f(!, y, 2, #) =1"!# vj(!, y + x(!, #), 2 + %(!, #))

=1"!#@%j('!, y + x(!, #), 2 + %(!, #)) ' #j

%

' i(x('!, y + x(!, #), 2 + %(!, #)) ' #jx)A

then (f(·, #))"+e# $ H. This is a consequence of Proposition 3.3.2.

Definition 4.4.5 (Lagrangian ideals). — The Lagrangian ideal J is defined as theset of families F = (F (·, #))"+e# which can be written as

F (!, y, 2, #) =n#

j=1

fj(!, y, 2, #)1"!# vj(!, y + x(!, #), 2 + %(!, #))

for all (!, y, 2) in O and # in 2', where (f(·, #))"+e# $ H.

Example 4.4.6. — Let us set

F (!, y, 2, #) = 2k ' -k(!, y, #)

with

(4.4.35) -k(!, y, #) = '(ak + i bk)(!, y, #),

where ak, bk are those given in Theorem 4.4.2. Then (F ((·, #))"+# $ J .



Indeed if |2| "/

& < 12 c0 then *0(2) = 1 so it follows from (4.4.2) and Theorem

4.4.2 that we have

vj(!, y + x(!, #), %(!, #) + 2) =n#

k=1

qk(2, a, b, gj)(2k ' -k(!, y, #)).

Since qk(2, a, b, gj) = (1+2i!) &jk +O(((+/

&)"!#) (by (v)) it follows that the matrix(qk(2, a, b, gj))!1 = (djk(!, y, 2, #))1!j,k!n exists. Moreover ("!# djk(·, #))"+e# $ H.Now we have

2k ' -k(!, y, #) =n#

j=1

"!# djk(!, y, #) · 1"!# vj(!, y + x(!, #), 2 + %(!, #)).

This proves our claim.

Lemma 4.4.7. — For F and G in J let us define

{F, G}(!, y, 2, #) =n#

j=1

+ $F

$2j

$G

$yj' $F

$yj

$G

$2j

,(!, y, 2, #).

Then {F, G} $ J .

Proof. — Since vj(!, y + x(!, #), 2 + %(!, #) = uj 4 *!&(y, 2) where uj(x, %, #) = %j '#j

%'i(xj '#jx) and *!&(y, 2) = (x('!, y, 2), %('!, y, !)) is the symplectic map defined

by the flow we have

{vj, vk}(!, y, 2, #) = {uj, uk}(*!&(y, 2)) = 0

because {uj, uk} 3 0.Let F =

&fj

1'&( vj , G =

&gk

1'&( vk be two elements of J with fj $ H, gk $ H.

Then a straightforward computation and the Remark 4.4.4 give the conclusion (seethe proof of Lemma 4.3.10).

Lemma 4.4.8. — Let R = (R(·, #))"+e# $ J and assume that R(·, #) does not dependon 2. Then for every N $ N one can find CN > 0 such that for every (!, y) in 2%+

and # in 2' we have|R(!, y, #)| " CN | Im -(!, y, #)|N .

Proof. — We are going to show by induction on N ! 1 that we can write

(4.4.36) R(!, y, #) =#

0<|,|<N

h,(!, y, #)(2 ' -), +#

|,|=N

g,(!, y, #, 2)(2 ' -),

where (h,(·, #))"+e# and (g,(·, #))"+e#) belong to H.For N = 1 the first sum in the right hand side of (4.4.36) is empty and by assump-

tion we have

R(!, y, #) =n#

j=1

fj(!, y, 2, #)1"!# vj(!, y + x(!, #), 2 + %(!, #)).



Using Theorem 4.4.2 we obtain since *0(2) = 1 when |2| "/

&,

R(!, y, #) =n#

k=1

+ n#

j=1

1"!# fj(!, y, 2, #) qk(2, a, b, gj)

,(2k ' -k(!, y, #)).

Since fj and 1'&( qk belong to H this shows that (4.4.36) is true when N = 1. Assume

now it is true up to the order N . We can apply Lemma 4.2.1 to the function

2g,(!, y, 2, #) = *0(2) g,(!, y, 2, #), |3| = N

with zj = '-j(!, y, #). It follows that

(4.4.37) 2g,(!, y, 2, #) =n#

k=1

qk(2, a, b, 2g,)(2k ' -k(!, y, #)) + r(a, b, 2g,).

For the q%ks and r we have the estimates (4.2.2). Let us set

(4.4.38)

6h,(!, y, #) = r(a(!, y, #), b(!, y, #), 2g,(h, y, ·, #))

g,(!, y, 2, #) = qk(2, a(!, y, #), b(!, y, #), 2g,(!, y, ·, #)).

It follows from (4.2.2) and Theorem 4.4.2 that (h,(·, #))" and (g, k(·, #))" belong toH. Using (4.4.36) at the level N and (4.4.37), (4.4.38) we deduce that (4.4.36) holdsat the level N + 1.

Now let us take in (4.4.36) 2 = (Re - + s Im -)(!, y, #) when s $ [0, 1]. Then thesame argument as in the end of the proof of Lemma 4.3.11 gives the result.

Corollary 4.4.9. — For every N $ N there exists a constant CN > 0 such that555+$-j

$yk' $-k

$yj

,(!, y, #)

555 " CN | Im -(!, y, #)|N

for every (!, y) in 2%+ and # in 2'.

Proof. — Identical to the proof of Corollary 4.3.12.

Now we go back to the original coordinates

x = y + x(!, #), % = 2 + %(!, #)

and we set for k = 1, . . . , n,

(4.4.39) #k(!, x, #) = -k(!, x ' x(!, #), #) = %k(!, #) ' (ak + i bk)(!, x ' x(!, #), #)

where ak, bk have been described in Theorem 4.4.2.Then we can state the following result.

Theorem 4.4.10. — We can write for (!, x) $ %+ and |% ' %(!, #)| "/

&,

(i) %k ' #k(!, x, #) =n#

j=1

ejk(!, x, %, #) vj(!, x, %)

where ejk are smooth functions which satisfy



(ii) |$*& $A

x ejk(!, x, %, #)| " CA

"!# for all # $ N and / = 0, 1.

Moreover we have for (!, x) in %+ and |% ' %(!, #)| </

&,(iii) |#k(!, x, #) ' #%| " C0 (( +

/&).

(iv)555 Im #k(!, x, #) ' xk ' xk(!, #)

1 + 4!2

555 "/

&|x ' x(!, #)|

"!#2 .

(v) |#(!, x, #)| " C0,

|$Ax #(!, x, #)| " CA

+ 1"!#|A|+1

+1

"x#|A|+1+!0

,if A $ Nn, |A| ! 1.

(vi) #k(!, x(!, #), #) = %k(!, #).

(vii)555+$#k

$xj' $#j

$xk

,(!, x, #)

555 " CN

+ 1"x#3/2

+1

"!#3/2

, |x ' x(!, #)|N

"!#2N, N $ N

where the constants CA, C0, CN are independent of (!, x, %, #).

Proof. — (i) follows immediately from the computations made in Example 4.4.6 aswell as (ii). The point (iii) is obvious since %(!, #) = #% +O(() and |ak|+ |bk| = O(

/&)

by (4.4.4). Then (iv) follows from Theorem 4.4.2 (ii) as well as (v). The point (vi) isobvious since ak = bk = 0 when y = 0. To prove (vii) we use Corollary 4.4.9, (4.4.35),(4.4.39) and (iv) of Theorem 4.4.10. We obtain

555+$#k

$xj' $#j

$xk

,(!, x, #)

555 " CN|x ' x(!, #)|N

"!#2N.

Now by (v) of the present theorem we have,

|$j #(!, x, #)| " C+ 1"!#2 +

1"x#2

,.

Writing |a| = |a|3/4 |a|1/4 and using the above estimates we obtain (vii).

Definition 4.4.11. — Let (F (·, #))"+e# be a family of C" functions for (!, x) in %+

and |% ' %(!, #)| </

&. We shall say that F $ J(x,%) if we can write

F (!, x, %, #) =n#

j=1

fj(!, x, %, #)1"!# vj(!, x, %, #)

where|$A

x $B% fj(!, x, %, #)| " CAB

for all (!, x) in %+, |% ' %(!, #)| </

&, # $ 2' where CAB is independent of (!, x, %, #).

Then exactly as in Lemma 4.4.7 J(x,%) is closed under the Poisson bracket in (x, %)and we have the analogue of Lemma 4.4.8. In fact J(x,%) is just the image of J underthe di!eomorphism x = y + x(!, #), % = 2 + %(!, #). Then we have

Theorem 4.4.12. — With # defined in (4.4.39) we have for k = 1, . . . , n,+' $p

$xk(x, #(!, x, #)) ' $#k

$!(!, x, #) ' $#k

$x(!, x, #)

$p

$%(x, #(!, x, #))

,

"$ J(x,%).



Proof. — We follow word by word the proof of Lemma 4.3.16 and Corollary 4.3.17.Let us just sketch the proof.

We begin by the proof of the theorem when #(!, x, #) is replaced by %. By Theorem4.4.10 (i) we have

%k ' #k(!, x, #) =#

ejk (!, x, %, #) vj(!, x, %).

Then we set x('!, x, %) = X , %('!, x, %) = & that is x(!, X, &) = x, %(!, X, &) = %.The above identity reads

%k(!, X, &) ' #k(!, x(!, X, &), #) =n#

k=1

ejk(!, x(!, X, &), %(!, X, &), #)uj(X, &).

We di!erentiate this equality with respect to ! using the equations of the flow given by(3.1.2). Then we use Theorem 4.4.10 (ii) and we come back to the original coordinates(x, %). Finally we write % = % ' #(!, x, #) + #(!, x, #) as in the proof of Corollary4.3.17. Details are left to the reader.

Corollary 4.4.13. — For every N $ N one can find CN > 0 such that555'

$p

$xk(x, #(!, x, #)) ' $#k

$!(!, x, #) ' $#k

$x(!, x, #)

$p

$%(x, #(!, x, #))

555

" CN

+ |x ' x(!, #)|"!#2

,N.

Proof. — Use Theorem 4.4.12 and Lemma 4.4.8 in the coordinates (x, %).

We are ready now to define the phase ), as we did in Proposition 4.3.19 for theoutgoing points, but we find here a slight problem. Indeed if we look to formula(4.3.48) we see that ) is defined by mean of #(!, s x + (1 ' s)x(!, #)), s $ [0, 1].In the present case when ! ! 0, #(!, z, #) is defined for z · #% " c0 "z# |#%| and|z ' x(!, #)| " & "!# and it is easily seen that the point z = s x + (1 ' s)x(!, #)does not satisfy these conditions. Therefore we have to modify the expression of ) in(4.3.48) to take care of this problem. We split the discussion into several cases givingin each of them a di!erent expression of ). Our purpose is to prove the followingresult.

Let us set,

(4.4.40) O+ =!(!, x) $ R+ ( Rn : x · #% " c0

10"x# |#% |, |x ' x(!, #)| " &

10"!#"

.

Proposition 4.4.14. — There exists a smooth function ) = )(!, x, #) defined onO+ such that,

(i) )(0, x, #) = (x ' #x)#% +i

2(x ' #x)2 +

12i

#2% + O(|x ' #x|N ), %N $ N.

For every N $ N there exists CN > 0 such that

(ii)555$)

$x(!, x, #) ' #(!, x, #)

555 " CN

+ |x ' x(!, #)|"!#

,N



(iii)555$)

$!(!, x, #) + p

+x,

$)

$x(!, x, #)

,555"CN

+ |x ' x(!, #)|"!#

,Nuniformly in (!, x, #).

Moreover(iv)

555$)

$x(!, x, #) ' #%

555 " C (( +/

&).

(v) |$Ax )(!, x, #)| " CA, %A $ Nn, |A| ! 1.

(vi)555 Im )(!, x, #) ' 1

2|x ' x(!, #)|2

1 + 4!2+

12

#2%

555 " C (( +/

&)|x ' x(!, #)|2

"!#2 .

We split the proof into several cases which are summarized in the following figures.

Case 1 (see Figure 1). — Let (!, x) $ O+ be such that

x(!, #) · #% " c0

3"x(!, #)# |#% | and |x ' x(!, #)| " "x(!, #)#.

Lemma 4.4.15. — In the case 1 we have s x+(1' s)x(!, #) $ %+, for s $ [0, 1], thatis

(s x + (1 ' s)x(!, #)) · #% " c0 "s x + (1 ' s)x(!, #)# |#% |

and

|s x + (1 ' s)x(!, #) ' x(!, #)| " & "!#.

Proof. — We use the following elementary lemma.

Lemma 4.4.16. — Let a, b $ Rn be such that |a ' b| " "a#. Then for all s in [0, 1]

(1 ' s)|a| + s |b| "/

2 "(1 ' s) a + s b#.

Proof. — Since |a ' b|2 " |a|2 + 1 we have 2a · b ! '1, it follows that

|(1 ' s) a + s b|2 = (1 ' s)2 |a|2 + 2s(1 ' s) a · b + s2 |b|2 ! (1 ' s)2 |a|2

+ s2 |b|2 ' s(1 ' s) ! 12

((1 ' s) |a| + s |b|))2 ' 12.

Therefore "(1 ' s) a + s b#2 ! 12 ((1 ' s)|a| + s|b|)2.

Let us now apply Lemma 4.4.16 to a = x(!, #), b = x. Using our hypotheses weobtain

(s x + (1 ' s)x(!, #)) · #% " c0

3(s "x# + (1 ' s)"x(!, #)#) |#% |

" c0

3(s + s |x| + (1 ' s) + (1 ' s)|x(!, #)|) |#% |

" c0

3(1 +

/2 "s x + (1 ' s)x(!, #)#) · |#%|

" c0 "s x + (1 ' s)x(!, #)# |#% |,

On the other hand |s x + (1 ' s)x(!, #) ' x(!, #)| = s |x ' x(!, #)| " & "!#.



C#x•

x(!, #)•

• x (case 1)

#%• 0

(case 2) • x

H

Figure 1. C = {y, |y ' x(!, #)| = &"!#} et H = {y, y.#% = c0"y#|#%|}

C #x•x(!#, #)

•

• x(!, #)x • (case 3.1)

#%

• 0

• x (case 3.2)H

Figure 2. C = {y, |y ' x(!, #)| = &"!#} et H = {y, y.#% = c0"y#|#%|}

In the set defined in case 1 we can therefore define ) by the same formula as inProposition 4.3.19. We set

(4.4.41) )(!, x, #) =- 1

0(x'x(!, #)) ·#(!, s x+(1's)x(!, #), #) ds+! p(#)+

12i

#2%.

The proof of the points (i) to (vi) is exactly the same as the corresponding points inProposition 4.3.19 using Theorem 4.4.10.

Case 2. — Let (!, x) $ O+ be such that

x(!, #) · #% " c0

3"x(!, #)# |#% | and |x ' x(!, #)| ! 1

2|x(!, #)|.

(See Figure 1).



In this set we have

(4.4.42)

6|x(!, #)| " 2 |x ' x(!, #)| " & "!#,|x| " 3 |x ' x(!, #)| " & "!#.

Moreover for y $ [0, x] 2 [0, x(!, #)] the point (!, y) belongs to the set %+ on which# is defined. Indeed if s $ [0, 1] we have s x · #% " s c0

2 "x# |#% | and |s x ' x(!, #)| "s |x ' x(!, #)| + (1 ' s)|x(!, #)| " 2 |x ' x(!, #)| " & "!# by (4.4.42). On the otherhand, s x(!, #) · #% " s c0

3 "x(!, #)# |#% | " c0 "x(!, #)# |#% | and |s x(!, #) ' x(!, #)| =(1's) |x(!, #)| " & "!#. Therefore we can define the phase ) by the following formula.

(4.4.43) )(!, x, #)

=- 1

0x · #(!, s x, #) ds '

- 1

0x(!, #) · #(!, s x(!, #), #) ds + ! p(#) +

12i

#2% .

Let us show that ) satisfies the conditions of Proposition 4.4.14. It follows fromTheorem 4.4.2 and (4.4.39) that

#(0, z, #) = #% + i (z ' #x) + O(|z ' #x|N ).

Therefore

)(0, x, #) =- 1

0

$x · #% + i x(s x ' #x) ' #x #% ' i #x(s #x ' #x)

55x| O(|s x ' #x|N )

+ |#x| O(|s x ' #x|N )) ds +12i

#2%

= (x ' #x) · #% +i

2(x ' #x)2 +

12i

#2% + O(|x ' #x|N )

because |x| " |x ' #x| + |#x|, |s x ' #x| " |x ' #x| and |#x| " |x ' #x|. Thus (i) isproved. Now

$)

$xj(!, x, #) =

- 1

0#j(!, s x, #) ds +

n#

k=1

- 1

0s xk

$#k

$xj(!, s x, #) ds.

Using Theorem 4.4.10, (vii) we obtain

555$)

$xj(!, x, #) '

- 1

0#j(!, s x, #) ds '

- 1

0s

d

ds(#j(!, s x, #)) ds

555

" CN

- 1

0s |x|

+ 1"s x#3/2

+1

"!#3/2

, |s x ' x(!, #)|N

"!#2Nds.

Now s |x| " "s x#3/2 and by (4.4.42), s |x| " & "!#, |s x ' x(!, #)| " s |x ' x(!, #)| +(1' s)|x(!, #)| " 2 |x'x(!, #)|. Therefore the right hand side of the above inequalityis bounded by CN |x ' x(!, #)|N/"!#2N . Integrating by parts in the second integralof the left hand side we obtain (ii).



Now we have

$)

$!=- 1

0x

$#$!

(!, s x, #) ds

< => ?(1)

'- 1

0x(!, #) · #(!, s x(!, #), #) ds

< => ?(2)

'- 1

0x(!, #) · $#

$!(!, s x(!, #), #) ds

< => ?(3)

'n#

k,*=1

- 1

0xk(!, #)

$#k

$x*(!, s x(!, #), #) s x*(!, #) ds

< => ?(4)

+p(#).

We use Corollary 4.4.13 to write

(1) =n#

k=1

- 1

0xk

$p

$xk(s x, #(!, s x, #)) ds

'n#

k,*=1

- 1

0xk

$#k

$x*(!, s x, #) · $p

$%*(s x, #(!, s x, #)) ds + R0

with

|R0| "- 1

0|x| |s x ' x(!, #)|N

"!#2N.

By (4.4.42) we have |x| " & "!#; since "!#2N!1 ! "!#N if N ! 1, and

|s x ' x(!, #)| " s |x ' x(!, #)| + (1 ' s)|x(!, #)| " 2 |x ' x(!, #)|,

we obtain |R0| " CN |x ' x(!, #)|N /"!#N if N ! 1. Now |x ' x(!, #)| " & "!# so thesame estimate is valid for N = 0. Finally

(4.4.44) |R0| " CN|x ' x(!, #)|N

"!#N , %N ! 0.

Using Theorem 4.4.10 (vii) we obtain

(1) = '- 1

0

d

ds(p(s x, #(!, s x, #))) ds + R1

where R satisfies (4.4.44). Therefore

(4.4.45)55(1) + p(#(!, x, #) ' p(0, #(!, 0, #))

55 " CN|x ' x(!, #)|N

"!#N .

Let us look the term (4); we use Theorem 4.4.10 (vii) again and we obtain

(4) =n#

*=1

- 1

0s x*(!, #)

d

ds(#*(!, s x(!, #), #)) ds + R2,



with

|R2| "- 1

0s |x(!, #)|

+ 1"s x(!, #)#3/2

+1

"!#3/2

,+ |x(!, #)|N

"!#2N

,(s ' 1)N ds.

It follows from (4.4.42) that R2 satisfies (4.4.44). Therefore integrating by parts inthe above integral we obtain

(4) = '- 1

0x(!, #)·#(!, s x(!, #), #)) ds+x(!, #)#(!, x(!, #), #)+O

+ |x ' x(!, #)|N

"!#N,.

Using Theorem 4.4.10, (vi) and the Euler identity we can write

x(!, #)#(!, x(!, #), #) = %(!, #) · $p

$%(x(!, #), %(!, #)) = 2p(#).

It follows that

(4.4.46) |(2) + (4) ' 2p(#)| " CN|x ' x(!, #)|N

"!#N .

Now, by Corollary 4.4.13 we have

(3) = '- 1

0x(!, #) · $p

$x(s x(!, #), #(!, s x(!, #), #)) ds

'n#

*,k=1

- 1

0xk(!, #) · $#k

$x*(!, s x(!, #), #)

$p

$%*(s x(!, #), #(!, s x(!, #), #) ds + R3

where

|R3| " CN |x(!, #)|- 1

0

|x(!, #)|N |s ' 1|N

"!#2Nds.

If N ! 1 we have |x(&,")|'&(2N " 1

'&(N so R3 satisfies (4.4.44) using (4.4.42).Using again Theorem 4.4.10 (vii) we obtain

(3) = '- 1

0

d

ds

@p(s x(!, #), #(!, s x(!, #), #)

Ads + O

+ |x ' x(!, #)|N

"!#N,

so55(3) + p(x(!, #), #(!, x(!, #), #) ' p(0, #(!, 0, #)

55 " CN|x ' x(!, #)|N

"!#N .

Finally we obtain

(4.4.47)55(3) + p(#) ' p(0, #(!, 0, #))

55 " CN|x ' x(!, #)|N

"!#N .

Since ('(& (!, x, #) = (1) ' (2) ' (3) ' (4) + p(#) we deduce from (4.4.45) to (4.4.47)

that 555$)

$!(!, x, #) + p(x, #(!, x, #))

555 " CN|x ' x(!, #)|N

"!#N .

Using the point (ii) already proved in Proposition 4.4.14 we obtain the point (iii).The point (iv) follows easily from (ii) since by (4.4.39) and Theorem 4.4.10 we have

#(!, x, #) = #% + O(( +/

&).



Let us prove (v). To bound $Ax ) when |A| ! 1 we have to bound, accord-

ing to (4.4.43) the quantities (1) =1 10 s|A

#|($A#

x #)(!, s x, #) ds, |A%| " |A| ' 1 and(2) =

1 10 |x| |$A

x #(!, s x, #)| ds. Using Theorem 4.4.10, (v), we see easily that (1) isuniformly bounded and

|(2)| " CA |x| 1"!#|A| + CA |x|

- 1

0

ds

"s x#|A|+1+!0.

By (4.4.42) we have |x| " 2& "!# " 2& "!#|A| since |A| ! 1 and setting t = |x| s inthe integral above we see that (2) is uniformly bounded in (!, x, #). This shows (v).Finally by (4.4.43),

Im )(!, x, #) =- 1

0x · Im #(!, s x, #) ds '

- 1

0x(!, #) · Im #(!, s x(!, #), #) ds ' 1

2#2

%.

Using Theorem 4.4.10, (iv) and (4.4.42) we obtain (vi). This completes the proof ofProposition 4.4.14 in the case 2.

Case 3. — We consider here the case where

(4.4.48) (!, x) $ O+ and x(!, #) · #% >c0

3"x(!, #)# |#% |.

(See Figure 2).Let us recall that we are dealing in this Section 4.4 with the case where

#x · #% " 'c0 "#x# |#%|, (see (4.4.1)).(1) The continuous function t 0. x(t, #) · #% is then strictly negative for t = 0 and

strictly positive for t = !. It follows that

(4.4.49) there exists !# $ ]0, ![ depending only on # such that x(!#, #) · #% = 0.

Then we have the following Lemma.

Lemma 4.4.17

(i)32|! ' !#| |#%| " |x(!, #) ' x(!#, #)| " 3 |! ' !#| |#%|,

(ii) |! ' !#| ! c0

50,

(iii) |x ' x(!, #)| " |x ' x(!#, #)| + |x(!#, #) ' x(!, #)| " 5 |x ' x(!, #)|,(iv) K1 "!# " "!## " K2 "!#.

Proof. — It follows from (4.4.48) and Definition 3.2.2 that the point

0# = (x(!#, #), %(!#, #))

belongs to S+ ! S!. By the group property and Proposition 3.3.1 we have

x(!, #) = x(! ' !#, 0#) = x(!#, #) + 2(! ' !#)#% + O(( |! ' !#|) + O(().

It follows that

(4.4.50) x(!, #) ' x(!#, #) = 2(! ' !#)#% + O(( |! ' !#|) + O(().



Now we deduce from (4.4.48) and (4.4.49) that

(x(!, #) ' x(!#, #)) · #% >c0

3"x(!, #)# |#% | ! c0

3|#%| ! c0

6since 1

2 " |#%| " 2. So by (4.4.50),

2(! ' !#) |#%|2 ! c0

6' C1 ( |! ' !#|' C2 (.

Taking ( small compared to c0 and C1 we obtain (ii). Then (i) follows easily from(4.4.50) if ( 7 c0. Now the first inequality in (iii) being obvious, let us prove thesecond one. We write

(4.4.51)

'()

(*

|x ' x(!, #)|2 = (1) + (2)

(1) = |x ' x(!#, #)|2 + |x(!#, #) ' x(!, #)|2

(2) = 2(x ' x(!#, #))(x(!#, #) ' x(!, #)).

If we use (4.4.50), (i) and (ii) we obtain,

(2) = '4(! ' !#)(x ' x(!#, #)) · #% + O(((1)),

so by (4.4.49),(2) = '4(! ' !#)x · #% + O(( (1)).

Now since (!, x) belongs to O+ (see (4.4.40)) we have

x · #% " c0

10"x# |#%| " c0

10"x(!, #)# |#% | +

c0

10|x ' x(!, #)| |#% |.


x · #% " 310

x(!, #) · #% +c0

10|x ' x(!, #)| |#% |,

and we deduce from (4.4.49) that

x · #% " 310

(x(!, #) ' x(!#, #)) · #% +c0

10|x ' x(!, #)| |#% |,

x · #% " 310

|x(!, #) ' x(!#, #)| |#%| +c0

10|x ' x(!, #)| |#% |,

x · #% "+ 3

10+

c0

10

,|x(!, #) ' x(!#, #)| |#%| +

c0

10|x ' x(!#, #)| |#%|,

(2) ! '4+ 3

10+

c0

10

,|x(!, #) ' x(!#, #)| |! ' !#| |#%|

' 2c0

5|x ' x(!#, #)| |! ' !#| |#%| + O(( (1)).

Using the first inequality in (i) we obtain

(2) ! '+4

5+

4c0

15

,|x(!, #)'x(!# , #)|2' 4c0

15|x'x(!#, #)| |x(!, #)'x(!# , #)|+O(( (1)).

Finally

(2) ! '+4

5+

12 c0

15+ C (

,(1).



If c0 and ( are small enough we find (1) + (2) ! 16 (1) so (iii) is proved using (4.4.51).

Finally (iv) follows from (i) and (iii) taking & small enough.

Then we split the case 3 in two subcases.

Case 3.1. — (!, x) $ O+, x(!, #)·#% > c03 "x(!, #)# |#% | and |x'x(!#, #)| " "x(!#, #)#.

It follows then that

(4.4.52) s x + (1 ' s)x(!#, #) $ %+ for all s $ [0, 1].

Indeed, using Lemma 4.4.16 with a = x(!#, #), b = x we obtain

s |x| "/

2 "s x + (1 ' s)x(!#, #)#

so if (!, x) $ O+ we get

(s x + (1 ' s)x(!#, #)) · #% = s x · #% " c0

10s "x# |#%| " c0 "s x + (1 ' s)x(!#, #)# |#%|.

Moreover by Lemma 4.4.17,

|s x + (1 ' s)x(!#, #) ' x(!, #)| " s |x ' x(!, #)| + (1 ' s) |x(!#, #) ' x(!, #)|" s |x ' x(!, #)| + (1 ' s) 5 |x ' x(!, #)|" 5 |x ' x((!, #)| " & "!#

since in O+, |x ' x(!, #)| " +10 "!#.

Therefore we can define ) on this part of O+ by the following formula.

(4.4.53) )(!, x, #) =- 1

0(x ' x(!#, #)) · #(!, s x + (1 ' s)x(!#, #), #) ds

'- &

&$p(x(!#, #), #(s, x(!#, #), #) ds + !# p(#) +

12i

|#%|2.

Our goal now is to show that ) satisfies the claims (i) to (vi) of Proposition 4.4.14.The point ! = 0 does not belong, by (4.4.48), to this part of O+. Thus the claim

(i) is empty. Let us check (ii). We have

$)

$xk(!, x, #) =

- 1

0#k(!, s x + (1 ' s)x(!#, #), #) ds

+n#

*=1

- 1

0s(x* ' x*(!#, #))

$#*

$xk(!, s x + (1 ' s)x(!#, #), #) ds.

Using Theorem 4.4.10, (vii) we see easily that

(4.4.54)$)

$xk(!, x, #) =

- 1

0#k(!, s x + (1 ' s)x(!#, #), #) ds

+- 1

0s

d

ds[#k(!, s x + (1 ' s)x(!#, #), #)] ds + R



with

|R| " CN |x ' x(!#, #)|- 1

0

|s x + (1 ' s)x(!#, #) ' x(!, #)|N

"!#2Nds.

It follows from Lemma 4.4.17, (iii) that

(4.4.55) |R| " CN|x ' x(!, #)|N

"!#N , N ! 0.

Integrating by parts in the second integral of the right hand side of (4.4.54) we obtainthe claim (ii). Let us prove (iii). We have

$)

$!(!, x, #) =

- 1

0(x ' x(!#, #)) · $#

$!(!, Xs, #) ds ' p(x(!#, #), #(!, x(!#, #), #)

where Xs = s x + (1 ' s)x(!#, #).Using Corollary 4.4.13 we obtain

$)

$!(!, x, #) = '

- 1

0(x ' x(!#, #)) · $p

$x(Xs, #(!, Xs, #)) ds

'n#

k,*=1

- 1

0(xk ' xk(!#, #))

$#k

$x*(!, Xs, #)

$p

$%*(Xs, #(!, Xs, #)) ds

' p(x(!#, #), #(!, x(!#, #), #)) + R

where R satisfies (4.4.55). We use again Theorem 4.4.10, (vii), and we obtain

$)

$!(!, x, #) = '

- 1

0

d

ds[p(Xs, #(!, Xs, #))] ds ' p(x(!#, #), #(!, x(!#, #), #) + R%

where R% satisfies also (4.4.55). This implies immediately (iii).The points (iv), (v) follow easily from Theorem 4.4.10. Let us check (vi). According

to (4.4.53) we can write

(4.4.56)

'()

(*

)(!, x, #) = A + B with,

A =- 1

0(x ' x(!#, #)) · #(!, s x + (1 ' s)x(!#, #), #) ds.

Using Theorem 4.4.10, (iv) and Lemma 4.4.17 (iii) we see that(4.4.57)'()

(*

Im A =12

11 + 4!2

[(x ' x(!#, #))2 + 2(x ' x(!#, #))(x(!# , #) ' x(!, #))] + R

|R| " C/

&|x ' x(!, #)|2

"!#2 .

To check the term B let us set x(t) = x(t, #) and

F (t) = '- &

tp(x(t), #(s, x(t), #)) ds + t p(#).



Then

F %(t) = p(x(t), #(t, x(t), #))< => ?(1)

'- &

t

n#

k=1

$p

$xj(x(t), #(s, x(t), #)) xk(t) dt

< => ?(2)

'- &

t

n#

k,*=1

$p

$%*(x(t), #(s, x(t), #))

$#*

$xk(s, x(t), #) xk(t) ds

< => ?(3)

+p(#).

By Theorem 4.4.10 (vi) we have,

(1) = p(x(t), %(t)) = p(#).

By the point (vii) of the same theorem we have,

(3) =n#

k,*=1

- &

t

$p

$%*(x(t), #(s, x(t), #))

$#k

$x*(s, x(t), #) xk(t) ds + R0

with

|R0| " CN

- &

t

|x(t) ' x(s)|N

"s#2N.

Since |x(t) ' x(s)| "1 s

t |x(")| d" " C(s ' t) we obtain

(4.4.58) |R0| " C%N

- &

t

(s ' t)N

"s#2Nds, !# " t " !.

Using Corollary 4.4.13 we obtain

(3) = 'n#

k,*=1

- &

t

$p

$xk(x(t), #(s, x(t), #), #) x(t) ds

'n#

k=1

- &

t

$#k

$s(s, x(t), #) xk(t) ds + R1

where R1 satisfies (4.4.58).It follows that

(3) = '(2) 'n#

k=1

xk(t)(#k(!, x(t), #) ' #k(t, x(t), #)) + R1.

Nown#

k=1

xk(t)#k(t, x(t), #) =n#

k=1

%k(t)$p

$%k(x(t), %(t)) = 2p(#).

Therefore we obtain,

F %(t) = (1) ' (2) ' (3) + p(#) =n#

k=1

xk(t)#k(!, x(t), #) + R1.



Now by Theorem 4.4.10 (iv),

Im #k(!, x(t), #) =xk(t) ' xk(!)

1 + 4!2+ O (

/&)

|xk(t) ' xk(!)|"!#2 .

Since x(t) is uniformly bounded we deduce that

Im F %(t) =12

d

dt

|x(t) ' x(!)|2

1 + 4!2+ G(t) with

|G(t)| " C/

&! ' t

"!#2 + CN

- &

t

(s ' t)N

"s#2Nds.

Integrating between !# and ! we obtain

(4.4.59)555 Im F (!#) ' 1

2|x(!#) ' x(!)|2

1 + 4!2

555

" C%/

&(! ' !#)2

"!#2 + CN

- &

&$

- &

t

(s ' t)N

"s#2Nds dt.

Let us call I (resp. II) the first (resp. the second) term in the right hand side of(4.4.59). By Lemma 4.4.17 we have

(4.4.60) |I| " C/

&|x ' x(!, #)|2

"!#2 .

Now

|II| " C%N

- &

&$

+- s

&$(s ' t)N dt

, ds

(1 + s)2N" C%%

N

- &

&$

(s ' !#)N+1

(1 + s)2Nds.

Now it follows from Lemma 4.4.17 and (4.4.40) that !'!# " 2& "!# " 2& (1+!) whichmeans that (1 ' 2&) ! " !# + 2&. Since 2& " 1/2 we have ! " 2!# + 1. It is then easyto see that the function s 0. (s ' !#)N+1/(1 + s)2N is increasing on (!#, !). Therefore

|II| " CN(! ' !#)N+2

"!#2N" CN (! ' !#)2

(2&)N

"!#N .

Taking N = 2 and using Lemma 4.4.17 (i) and (iii) we obtain

(4.4.61) |II| " C &2 |x ' x(!, #)|2

"!#2 .

It follows from (4.4.59) to (4.4.61) and from (4.4.53), (4.4.56), (4.4.57) that

Im )(!, x, #) =12|x ' x(!, #)|2

1 + 4!2' 1

2|#%|2 + O

+/&|x ' x(!, #)|2

"!#2,

which is precisely the claim of point (vii) of Proposition 4.4.14.



Case 3.2. — (!, x) $ O+, x(!, #)·#% > c03 "x(!, #)#|#% | and |x'x(!#, #)| ! 1

2 |x(!#, #)|.According to Lemma 4.4.17 (iii) we have

(4.4.62)

6|x(!#, #)| " 10 |x ' x(!, #)| " 2& "!#|x| " |x ' x(!#, #)| + |x(!#, #)| " 15 |x' x(!, #)| " 3

2 & "!#.

On the other hand if y belongs to the union of the two segments [0, x] and [0, x(!#, #)]then (y, !) belongs to %+, the set (defined in (4.1.4)) on which # is defined. Indeed,by (4.4.40), if s $ (0, 1) then s x · #% " s · c0

10 "x#|#%| " c0 "s x#|#%|. Moreover

|s x ' x(!, #)| " |s x ' x(!#, #)| + |x(!#, #) ' x(!, #)|" s|x ' x(!#, #)| + (1 ' s)|x(!#, #)| + |x(!#, #) ' x(!, #)|.

Since we are in case 3.2 we have by Lemma 4.4.17, |s x' x(!, #)| " 10 |x' x(!, #)| "& "!#. On the other hand, if s $ (0, 1) we have, by (4.4.49), s x(!#, #) · #% = 0.Moreover

|s x(!#, #) ' x(!, #)| " |x(!#, #) ' x(!, #)| + (1 ' s)|x(!#, #)|" |x(!#, #) ' x(!, #)| + 2(1 ' s)|x ' x(!#, #)|" 10 |x ' x(!, #)| " & "!#,

by Lemma 4.4.17, (iii).Therefore in the present case we can set

(4.4.63) )(!, x, #) =- 1

0x · #(!, s x, #) ds '

- 1

0x(!#, #) · #(!, s x(!#, #), #) ds

'- &

&$p(x(!#, #), #(s, x(!#, #), #)) ds + !# p(#) +

12i

#2% .

Our goal is to show that ) satisfies all the requirements of Proposition 4.4.14.The point (i) is empty since ! = 0 does not belong to this part of O+. Let us check

(ii). We have

$)

$xj(!, x, #) =

- 1

0#j(!, x, #) ds +

- 1

0

n#

k=1

s xk$#k

$xj(!, s x, #) ds.

By Theorem 4.4.10 and (4.4.62) we have555$#k

$xj(!, s x, #) ' $#j

$xk(!, s x, #)

555 " CN|s x ' x(!, #)|N

"!#2N

" C%N

|x ' x(!, #)|N

"!#2N.

(4.4.64)

It follows that

$)

$xj(!, x, #) =

- 1

0#j(!, s x, #) ds+

- 1

0s

d

ds(#j(!, s x, #)) ds+O

+ |x ' x(!, #)|N+1

"!#2N

,.



Integrating by parts and using the bound |x ' x(!, #)| " & "!# we obtain555$)

$xj(!, x, #) ' #j(!, x, #)

555 " CN|x ' x(!, #)|N

"!#N , %N $ N.

Thus (ii) is proved. Let us prove (iii). We have

$)

$!(!, x, #) =

- 1

0x

$#$!

(!, s x, #) ds

< => ?(1)

'- 1

0x(!#, #)

$#$!

(!, s x(!#, #), #) ds

< => ?(2)

' p(x(!#, #), # (!, x(!#, #), #)< => ?(3)

.

(4.4.65)

Using Corollary 4.4.13 we can write

(1) = 'n#

k=1

- 1

0xk

$p

$xk(s x, #(!, s x, #)) ds

+n#

k,*=1

- 1

0xk (!, s x, #) · $p

$x*(s x, #(!, s x, #))

$#k

$xk(!, s x, #) ds.

Using again (4.4.64) we obtain,

(1) = '- 1

0

d

ds

$p(s x, #(!, s x, #))

%ds + O

+ |x ' x(!, #)|N+1

"!#2N

,.

Finally, since |x ' x(!, #)| " & "!#, we have

(4.4.66) (1) = p(0, #(!, 0, #)) ' p(x, #(!, x, #)) + O+ |x ' x(!, #)|N

"!#N,.

By exactly the same computation (using (4.4.62)) we obtain

(4.4.67) (2) = p(0, #(!, 0, #)) ' p(x(!#, #), #(!, x(!#, #), #) + O+ |x ' x(!, #)|N

"!#N,.

So using (4.4.65) to (4.4.67) we derive the point (iii). The last non trivial point to beproved is the point (vi).

Using the expression of Im # given by Theorem 4.4.10, (4.4.62) and (4.4.59) to(4.4.61) which are valid also in the Case 3.2, we can write

Im )(!, x, #) =12

11 + 4!2

$|x ' x(!, #)|2 ' |x(!, #)|2 ' |x(!#, #) ' x(!, #)|2

+ |x(!, #)|2 + |x(!#, #) ' x(!, #)|2%

+ O+/

&|x ' x(!, #)|2

"!#2,

which is exactly what is needed.To finish the proof of Proposition 4.4.14 we must show that the phases ) which

have been constructed by the formulas (4.4.41), (4.4.43), (4.4.53), (4.4.63) in di!erentregions can be matched in only one phase. We begin by a Lemma.



Lemma 4.4.18. — Let I =@

c010 , c0

2

Aand let us consider the function on [0, +)[(I,

g(s, c) = x(s, #) · #% ' c "x(s, #)# |#% |.

(i) For all c in I the function [0, +)[(R, s 0. g(s, c) is strictly increasing.(ii) For all c in I there exists a unique !(c) > 0 such that

g(!(c), c) = 0.

(iii) The function I . [0, +)[, c 0. !(c) is strictly increasing.Moreover we have the following estimates

(iv)32|!(c) ' !#| |#%| " |x(!(c), #) ' x(!#, #)| " 3 |!(c) ' !#| |#%|.

(v) |!(c) ' !#| ! c0/120.(vi) For all x in O+ and c in

@c03

c02

A,

|x ' x(!(c), #)| " |x ' x(!#, #)| + |x(!#, #) ' x(!(c), #)| " 4 |x ' x(!(c), #)|.

(vii) If c $@

c03

c02

Awe havec

10"x(!#, #)# " |!(c) ' !#| " 4c "x(!#, #)#.

Proof(i) We have

$g

$s(s, c) = x(s, #) · #% ' c

x(s, #) · x(s, #)"x(s, #)# |#%| = 2 |#%|2 + O(( + c0).

Thus (g(s (s, c) ! 1

10 if ( and c0 are small enough.(ii) It follows from above that g(s, c) ! 1

10 s + g(0, c) so g(s, c) . +) if s . +).Moreover g(0, c) = #x ·#% ' c "#x# |#%| " #x ·#% " 'c0 "#x# |#%| < 0. Therefore thereexists a unique !(c) such that g(!(c), c) = 0 and c 0. !(c) is C". Di!erentiating thisequality with respect to c we obtain

!%(c)$g

$s(!(c), c) +

$g

$c(!(c), c) = 0.

By the above computation of (g(s we can write

!%(c) ="x(!(c), #)# |#% |

2 |#%|2 + O(( + c0)

which proves (iii). Now we have

x(!(c), #) ' x(!#, #) =- &(c)

&$x(s, #) ds = 2#%(!(c) ' !#) + O((|!(c) ' !#|)

from which (iv) follows easily. Let us prove (v). By definition of !(c) and !# we canwrite

(4.4.68) (x(!(c), #) ' x(!#, #)) · #% = c "x(!(c), #)# |#% | ! c0

20



so by (iv),c0

20" 3 |!(c) ' !#| · |#%| " 6 |!(c) ' !#|.

The first inequality in (vi) being trivial let us prove the second one. We write

|x ' x(!, #)|2 = |x ' x(!#, #)|2 + |x(!#, #) ' x(!(c), #)|2< => ?(1)

+ 2(x ' x(!#, #)) · (x(!#, #) ' x(!(c), #))< => ?(2)

.

We have(2) = 2(x ' x(!#, #))[2(!# ' !(c))#% + O(( |!# ' !(c)|].

It follows from (iv) that

(2) = '4(!(c) ' !#)x · #% + O(((1)), where !(c) ' !# ! 0.

Now in O+ we have x · #% " c010 "x# |#%|. It follows that

x · #% " c0

10"x(!(c), #)# |#% | +

c0

10|x ' x(!(c), #)| |#% |.

Using (4.4.68) we obtain

x · #% " c0

10c(x(!(c), #) ' x(!#, #)) · #% +

c0

10|x ' x(!(c), #)| |#% |,

so we will have

(2) ! '2c0

5c|x(!(c), #) ' x(!#, #)| |#%| |!(c) ' !#|

' 2c0

5|x ' x(!(c), #)| |#% | |!(c) ' !#|'O(((1))

(2) ! '4c0

15c|x(!(c), #) ' x(!#, #)|2

' 8c0

30|x ' x(!(c), #)| |x(!(c), #) ' x(!#, #)|'O(((1))

(2) !+' 4c0

15c' 4c0

30

,|x(!(c), #) ' x(!#, #)|2

' 215

c0 |x ' x(!(c), #)|2 'O(((1)).

If c ! c0/3 then 4c0/15c " 4/5 so we obtain

(1) + (2) ! 110$|x(!(c), #) ' x(!#, #)|2 + |x ' x(!, #)|2

%

if c0 is small enough. This implies (vi). Let us prove (vii). We have

x(!(c), #) = x(!#, #) + 2(!(c) ' !#) · #% + O(().

It follows that

c "x(!(c), #)# |#% | = x(!(c), #) · #% = 2(!(c) ' !#)|#%|2 + O(()



so12|!(c) ' !#| " c "x(!(c), #)# " 5 |!(c) ' !#|.

Moreover we have

"x(!(c), #)# " "x(!#, #)# + |x(!#, #) ' x(!(c), #)|" "x(!#, #)# + 3 |!(c) ' !#| |#%|" "x(!#, #)# + 6c "x(!(c), #)# |#% |" "x(!#, #)# + 12c "x(!(c), #)#

which implies that "x(!(c), #)# " 2 "x(!#, #)# if c0 is small enough. By the same way"x(!#, #)# " 2 "x(!(c), #)#. Thus we obtain (vii).

Now let us set

(4.4.69)

62O+(!) =

:x $ Rn : x · #% " c0

10 "x# |#% |, |x ' x(!, #)| " +40 "!#

;

! = !(c0/2), !% = !(c0/10).

In the beginning of this Section we have constructed the di!erent ) assuming x ·#% "c010 "x# |#%|, |x ' x(!, #)| " +

5 "!#.In the proof of Proposition 4.4.14 we have constructed

)1 when ! $ [0, !] and |x ' x(!, #)| " "x(!, #)#, (case 1)

)2 when ! $ [0, !] and |x ' x(!, #)| ! 12|x(!, #)|, (case 2)

)4 when ! $ [!% + )[ and |x ' x(!#, #)| " "x(!#, #)#, (case 3.1)

)5 when ! $ [!% + )[ and |x ' x(!#, #)| ! 12|x(!#, #)|, (case 3.2).

We are going first to match )1 and )2, )4 and )5. The matched phase will be definedon a smallest set than O+ defined in (4.4.40) namely for (!, x) where x $ 2O+(!) (see(4.4.69)). We show first that the point (!, 0) belongs to the sets where )1 and )2

are defined. According to (4.4.40) and what we recalled above it will be the caseif |x(!, #)| " +

5 "!#. We may assume that the domain where )2 is defined containspoints (!, x) where x $ 2O+(!) otherwise we don’t match )1 and )2 and we take only)1. So let (!, x) be such |x ' x(!, #)| " +

40 "!# and |x(!, #)| " 2 |x ' x(!, #)|. Then|x(!, #)| " +

40 "!# which implies our claim.Now it follows from (4.4.41) and (4.4.43) that

(4.4.70) )1(!, 0, #) = )2(!, 0, #).

By the same way we may assume that the domain where )5 is defined containspoints (!, x) where x $ 2O+(!). So let x be such that |x ' x(!, #)| " +

40 "!# and



|x ' x(!#, #)| ! 12 |x(!#, #)|. Then we write, using Lemma 4.4.18,

|x(!, #)| " |x(!#, #)| + |x(!#, #) ' x(!, #)|,

|x(!, #)| " 2(|x ' x(!#, #)| + |x(!#, #) ' x(!, #)|) " 8 |x ' x(!, #)| " &

5"!#.

So the point (!, 0) belongs also to the sets where )4 and )5 are defined and by (4.4.53),(4.4.63) we have

(4.4.71) )4(!, 0, #) = )5(!, 0, #).

Let us match )1 and )2. Let x $ 2O+, ! $ [0, !] be such that

(4.4.72)12|x(!, #)| " |x ' x(!, #)| " "x(!, #)#.

We are going to show then that

(4.4.73) %N $ N &CN > 0 : |)1(!, x, #) ' )2(!, x, #)| " CN|x ' x(!, #)|N+1

"!#N .

Indeed let 3(", x) be a regular path such that

(4.4.74) 3(0, x) = 0, 3(1, x) = x

and there exists K ! 0 such that for all " in [0, 1],555$3

$"(", x)

555 " K |x ' x(!, #)|(4.4.75)

3(", x) · #% " c0

10"3(", x)# |#% |(4.4.76)

|3(", x) ' x(!, #)| " &

5"!#(4.4.77)

6if |x ' x(!, #)| ! |x(!, #)| then,

|x(!, #)| " |3(", x) ' x(!, #)| " |x ' x(!, #)|(4.4.78)

6if |x ' x(!, #)| " |x(!, #)| then,

|x ' x(!, #)| " |3(", x) ' x(!, #)| " |x(!, #)|.(4.4.79)

The construction of this path will be made at the end of this Section.It follows from (4.4.78) or (4.4.79) that

12|x(!, #)| " |3(", x) ' x(!, #)| " "x(!, #)#.

We write for j = 1, 2,

)j(!, x, #) = )j(!, 0, #) +- 1

0

$3

$"(", x)

$)j

$x(!, 3(", x), #) d".



Using Proposition 4.4.14 (ii) and (4.4.75) we obtain'(()

((*

)j(!, x, #) = )j(!, 0, #) +- 1

0

$3

$"(", x)#(!, 3(", x), #) d" + Rj ,

|Rj | " CN

- 1

0|x ' x(!, #)| |3(", x) ' x(!, #)|N

"!#N d".

Then by (4.4.72), (4.4.78) or (4.4.79) and (4.4.70) we obtain (4.4.73).Now let *0 $ C"

0 (R) be such that 0 " *0 " 1 and *0(") = 1 if |"| " 12 , *0(") = 0

if |"| ! 1. Let us set

*1(!, x) = *0

+x ' x(!, #)"x(!, #)#

,.

Now for x in 2O+(!) we set

(4.4.80) )3(!, x, #) = *1(!, x))1(!, x, #) + (1 ' *1(!, x)))2(!, x, #).

On the support of *1 we have |x'x(!, #)| " "x(!, #)# thus )1 is well defined. On thesupport of 1 ' *1(!, x) we have |x ' x(!, #)| ! 1

2 "x(!, #)# ! 12 |x(!, #)| so )2 is well

defined. Therefore )3 is well defined when x $ 2O+(!). We show now that )3 satisfiesall the conditions in Proposition 4.4.14. We have

(4.4.81)$)3

$x(!, x, #) =

.*1

$)1

$x+ (1 ' *1)

$)2

$x+

$*1

$x()1 ' )2)

/(!, x, #).

On the support of (.1(x we have |()1 ' )2)(!, x, #)| " CN

|x!x(&,")|N+1

'&(N by (4.4.73).Moreover we have by (4.4.72)

555$*1

$x(!, x)

555 " C

"x(!, #)# " C

|x ' x(!, #)|and555.*1

$)1

$x+ (1 ' *1)

$)2

$x' #/(!, x, #)

555 " |*1|555$)1

$x(!, x, #) ' #(!, x, #)

555

+ (1 ' *1)555$)2

$x(!, x, #) ' #(!, x, #)

555.

It follows that the claim (ii) in Proposition 4.4.14 holds for )3. The point (iv) followsfrom (4.4.81) and (4.4.73) for N = 1 which gives |()1')2)(!, x, #)| " C &. The points(v) and (vi) are straightforward. Let us show (iii). We have

$*1

$!(!, x) =

.' x(!, #)

"x(!, #)# 'x(!, #) x(!, #)"x(!, #)#3 (x ' x(!, #))

/ $*0

$"(· · · ).

Since x(!, #) is bounded we deduce from (4.4.73) that555$*1

$!(!, x)

555 " C

|x ' x(!, #)| .

It follows then that5555

3$)3

$!'+*1

$)1

$!+ (1 ' *1)

$)2

$!

,4(!, x, #)

5555 " CN|x ' x(!, #)|N

"!#N .



Then (iii) follows easily.Let us now take ! $ [!%, +)[ and x $ 2O+(!). Assume that

(4.4.82)12|x(!#, #)| " |x ' x(!#, #)| " "x(!#, #)#.

We shall show that

(4.4.83) |()4 ' )5)(!, x, #)| " CN|x ' x(!, #)|N

"!#N |x ' x(!#, #)|.

Let us take a path 3 satisfying (4.4.74), (4.4.76), (4.4.77) and6

if |x ' x(!#, #)| ! |x(!#, #)| then

|x(!#, #)| " |3(", x) ' x(!#, #)| " |x ' x(!#, #)|,(4.4.84)

6if |x ' x(!#, #)| " |x(!#, #)| then

|x ' x(!#, #)| " |3(", x) ' x(!#, #)| " |x(!#, #)|,(4.4.85)

555$3

$"(", x)

555 " K |x ' x(!#, #)|.(4.4.86)

Let us remark that in the two cases (4.4.84) or (4.4.85) we have

|3(", x) ' x(!#, #)| " 2 |x ' x(!#, #)| " 8 |x ' x(!, #)|

by Lemma 4.4.18.Then using the same method as before we obtain easily (4.4.83). To match )4 and

)5 we set

*1(x) = *0

+x ' x(!#, #)"x(!#, #)#

,

and we deduce from (4.4.83) that55(.1

(x (x)55 " C

|x!x(&$,")| . Then we set

(4.4.87) )6(!, x, #) = [*1 )4 + (1 ' *1))5](!, x, #).

It is then easy to see that )6 satisfies all the requirements of Proposition 4.4.14.Our last step is to match )3 and )6. With the notation !(c) introduced in Lemma

4.4.18 let us set

!1 = !+11 c0

30

,, !2 = !

+12 c0

30

,, !3 = !0

+14 c0

30

,.

We have therefore according to (4.4.69)

!% < !1 < !2 < !3 < !.

Using (4.4.87), the fact that *1(x(!#, #)) = 1 and (4.4.53) we get(4.4.88)

)6(!2, x(!#, #), #) = '- &2

&$p(x(!#, #), #(s, x(!#, #), #) ds + !# p(#) +

12i

|#%|2.


(4.4.89) p(x(!#, #), #(s, x(!#, #), #)) = '$)3

$s(s, x(!#, #), #) + A + B



where

A = p$x(!#, #), #(s, x(!#, #), #)

%' p+x(!#, #),

$)3

$x(s, x(!#, #), #)

,

B = p+x(!#, #),

$)3

$x(s, x(!#, #), #)

,+

$)3

$s(s, x(!#, #), #).

By the estimates proved in Proposition 4.4.14 we have

(4.4.90) |A| + |B| " CN|x(!#, #) ' x(s, #)|N

"s#N " CN|x(!#, #) ' x(!2, #)|N

"!##N .

Here we used the fact that for s $ [!%, !] we have "s# - "!## (see Lemma 4.4.17 (ii))and |x(!#, #) ' x(s, #)| - |x(!#, #) ' x(!2, #)| by Lemma 4.4.18.

It follows from (4.4.88) and (4.4.89) that

)6(!2, x(!#, #), #) =- &2

&$

$)3

$s(s, x(!#, #), #) ds+!# p(#)+

12i

|#%|2+- &2

&$(|A|+|B|) ds.

Therefore we obtain

(4.4.91) )6(!2, x(!#, #), #) = )3(!2, x(!#, #), #)

' )3(!#, x(!#, #), #) + !# p(#) +12i

|#%|2 + R

where by (4.4.90)

(4.4.92) |R| " CN|x(!#, #) ' x(!2, #)|N+1

"!##N .

Now using (4.4.80) and (4.4.53) we have

)3(!#, x(!#, #), #) = )1(!#, x(!#, #), #) = !# p(#) +12i

|#%|2,

so we obtain

(4.4.93) )6(!2, x(!#, #), #) = )3(!2, x(!#, #), #) + R

where R satisfies (4.4.92).Now let ! $ [!1, !3]. We set

O%+(!) =

!x $ Rn : x · #% " c0

20"x# |#%|, |x ' x(!, #)| " &

40"!#"

.

Let x $ O%+(!). We can find a path 3 joining x to x(!#, #)) such that 3 , O%

+(!) andthere exists K ! 0 such that

(4.4.94)555$3

$"(", x)

555 " K |x ' x(!#, #)|, |3(", s) ' x(!2, #)| " K|x ' x(!2, #)|

Indeed if |x ' x(!#, #)| " |x(!#, #)| we set

3(", x) = " x + (1 ' ")x(!#, #)



and Lemma 4.4.16 show that

3(", x)) · #% = " x · #% " c0

20"(1 + |x|) |#%|

" c0

20(" + " |x| + (1 ' ")|x(!#, #)|)) |#% |

" c0

20/

2 (1 + |3(", x)|)|#% | " c0

10"3(", x)# |#% |.

If |x ' x(!#, #)| > |x(!#, #)| we take 3 to be the union of the two segments joining xand x(!#, #) to 0 and we obtain with y = x or y = x(!#, #) for t $ [0, 1],

t y · #% " tc0

20"y# |#%| " c0

20"t y# |#%|.

Since 0 belongs to O%+(!) these two segments are contained in O%

+(!).Let us prove the estimate on 3 given in (4.4.94).If |x ' x(!#, #)| " |x(!#, #)| we have

|" x + (1 ' ")x(!#, #) ' x(!2, #)| " "|x ' x(!#, #)| + (1 ' ")|x(!#, #) ' x(!2, #)|" K |x ' x(!2, #)|.

If |x ' x(!#, #)| > |x(!#, #)| we have

|t x ' x(!2, #)| " t |x ' x(!#, #)| + (1 ' t) |x(!#, #)| + |x(!#, #) ' x(!2, #)|" |x ' x(!#, #)| + |x(!#, #) ' x(!2, #)|" K |x ' x(!2, #)|

again by Lemma 4.4.18. Moreover

|t x(!#, #) ' x(!2, #)| " |x(!2, #) ' x(!#, #)| + (1 ' t) |x(!#, #)|" K |x ' x(!2, #)|.

Concerning the estimate on (,(! , if |x ' x(!#, #)| " |x(!#, #)| it is straightforward by

Lemma 4.4.18. If |x(!#, #)| " |x ' x(!#, #)| the same Lemma shows that |x(!#, #)| "K |!# ' !2| " K |x(!#, #) ' x(!2, #)| and |x| " |x ' x(!2, #)| + |x(!2, #) ' x(!#, #)| +|x(!#, #)| " K % |x ' x(!2, #)|. Thus (4.4.94) is entirely proved.

Now for j = 3 or 6 we can write

)j(!2, x, #) = )j(!2, x(!#, #), #) +- 1

0

$3

$"(", x)

$)j

$x(!2, 3(", x), #) d"

= )j(!2, x(!#, #), #) +- 1

0

$3

$"(", x)#(!2, 3(", x), #) d" + Rj

where

|Rj | " CN|x ' x(!2, #)|N+1

"!2#Nby Proposition 4.4.14 and (4.4.94). By (4.4.91) and (4.4.92) we have

(4.4.95) |()3 ' )6)(!2, x, #)| " CN|x ' x(!2, #)|N+1

"!2#N.



Now for ! $ [!1, !3] and j = 3 or 6 we can write

)j(!, x, #) = )j(!2, x, #) +- 1

0(! ' !2)

$)j

$!("! + (1 ' ") !2, x, #) d"

so by Proposition 4.4.14 and Lemma 4.4.18 we have

(4.4.96) )j(!, x, #) = )j(!2, x, #) '- 1

0(! ' !2) p(x, #("! + (1' ") !2, x, #)) d" + A,

where

|A| " CN|x ' x("! + (1 ' ") !2, #)|N

""! + (1 ' ") !2#N .

Proposition 4.4.19

(4.4.97)

')

*

For all ! in [!1, !3] and all " in [0, 1] we have(i) ""! + (1 ' ") !2# ! K1 "!#(ii) |x ' x("! + (1 ' ") !2, #)| " K2 |x ' x(!, #)|.

Proof

Case 1: ! $ [!2, !3]. — We write with !! = "! + (1 ' ") !2,

(4.4.98)

')

*

|x ' x(!, #)|2 = I + II whereI = |x ' x(!! , #)|2 + |x(!! , #) ' x(!, #)|2

II = 2(x ' x(!!, #)) · (x(!! , #) ' x(!, #)).

Since x(!, x) ' x(!!, #) =1 &

&# x(s, #) ds = 2(! ' !!)#% + O(( |! ' !!|) we have II ='4(! ' !!)(x ' x(!! , #)) · #% + O(( I). Now in O%

+ we have x · #% " c020 "x# |#% | ;

moreover by Lemma 4.4.18 (i) we have x(!!, #) · #% ! 12 c030 "x(!! , #)# |#%| since !! !

!2 = !$

12 c030

%. It follows that II ! ' c0

5 (! ' !!)"x# |#% | + 8 c05 "x(!! , #)# |#% |'O(( I).

Therefore we obtain

II ! 'c0

5(! ' !!)"x(!! , #)# |#% |

' c0

5(! ' !!) |x ' x(!! , #)| |#%| +

8 c0

5"x(!! , #)# |#% |'O(( I).

The second term in the right hand side can be bounded by c010 I. Using (4.4.98) we

obtain

|x ' x(!, #)|2 = I + II !+1 ' c0

10' ( K) I +

7 c0

5"x(!! , #)# |#%|.

Taking c0 and ( small enough we obtain I " 2 |x ' x(!, #)|2 which implies since|! ' !!| " 2 |x(!, #) ' x(!! , #)| " 2

/I, that |! ' !!| " 2 |x ' x(!, #)| " 2& "!# so

"!# " "!!# + 2& "!# and therefore "!!# ! 12 "!# since & is small. This proves the claim

(i) of (4.4.97).To prove (ii) we just use the fact that

|x(!, #) ' x(!! , #)| " 3 |! ' !!| " 6 |x ' x(!, #)|.



Case 2: ! $ [!1, !2]. — The point (i) in (4.4.97) is obvious in this case since "! +(1 ' ") !2 ! !.

By the same computation as above, since ! ! !1, we will have

(4.4.99)

6|x ' x(!, #)| ! 1

2 (|x ' x(!1, #)| + |x(!1, #) ' x(!, #)|),12 |! ' !1| " |x(!, #) ' x(!1, #)| " 6 |! ' !1|.

On the other hand we claim that we have

(4.4.100)9 c0

30"x(!, #)# |#% | " 3 |x ' x(!, #)|.

Indeed we have

(x ' x(!, #)) · #% = x · #% ' x(!, #) · #% " c0

20"x# |#% |'

11 c0

30"x(!, #)# |#% |

by Lemma 4.4.18 (i) since ! ! !1 = !$

11 c030

%. Thus

(x ' x(!, #)) · #% " c0

20"x(!, #)# |#% | +

c0

20|x ' x(!, #)| |#% |'

11 c0

30"x(!, #)# |#% |.

It follows that9 c0

30"x(!, #)# |#% | " c0

20|x ' x(!, #)| |#% |' (x ' x(!, #)) · #%

from which (4.4.100) follows easily since |#%| " 2 and c020 |#%| " 1. Now !! = "! +

(1 ' ") !2 belongs to [!1, !2] for " $ [0, 1]. Since by Lemma 4.4.18 (iii) the function!(c) is strictly increasing there exists a unique c! $

@11 c030 , 12 c0

30

Asuch that !! = !(c!).

Now we have |x ' x(!!, #)| " |x ' x(!1, #)| + |x(!1, #) ' x(!! , #)| which implies

(4.4.101) |x ' x(!!, #)| " |x ' x(!1, #)| + 6555!+11 c0

30

,' !(c!)

555.

We claim that

(4.4.102)555!+11 c0

30

,' !(c!)

555 " c0

15sup

s+[&1,&2]"x(s, #)#.

To see this we compute !%(c). Recall (see Lemma 4.4.18) that g(!(c), c) = 0 forc $@

c010 , c0

2

A.It follows that (g

(s (!(c), c) !%(c)+ (g(c (!(c), c) = 0. Now we have (g

(c (s, c) ='"x(s, #)# |#% | and (g

(s (s, c) = x(s, #) · #% ' c x(s,")·x(s,")'x(s,")( |#%|, which shows that

(g(s (s, c) = 2(|#%|2 +O(( + c). Therefore we have |!%(c)| " 2 "x(!(c), ## and we obtain(4.4.102). The last step consists in showing that

(4.4.103) sup&+[&1,&2]

"x(s, #)# " 2 "x(!, #)#.

To see this let us set h(s) = "x(s, #)#. Then h%(s) = x(s,")·x(s,")'x(s,")( . Thus h%(s) =

2x(s,")·"!

'x(s,")( + O((). Now since s $ [!1, !2] we have 11 c030 "x(s, #)# |#% | " x(s, #) ·

#% " 12 c030 "x(s, #)# |#% | so 0 < h%(s) " 2 c0 + O(() and therefore if s1, s2 $ [!1, !2],



|h(s1)'h(s2)| " (2 c0+O(()) |s1's2|. Let us take s1 = !(c), s2 = !(c%) with 11 c030 " c,

c% " 12 c030 . Then

(4.4.104) |"x(!(c), #)# ' "x(!(c%), #)#| " (2 c0 + O(())|!(c) ' !(c%)|.

On the other hand

x(!(c), #) ' x(!(c%), #) = 2(!(c) ' !(c%)) · #% + O(( |!(c) ' !(c%)|)

which implies that

x(!(c), #) · #% ' x(!(c%), #) · #% = 2(!(c) ' !(c%)) |#%|2 + O(( |!(c) ' !(c%)|

and, by definition of !(c) (see Lemma 4.4.18)@c"x(!(c), ## ' c% "x(!(c%), #)#

A|#%| = 2(!(c) ' !(c%)) |#%|2 + O(( |!(c) ' !(c%)|).

Combining with (4.4.104) we obtain

|"x(!(c), #) ' "x(!(c%), #)#|" 2(2 c0 + O(())

@c |"x(!(c), #)# ' "x(!(c%), ##| + |c ' c%|"x(!(c%), ##

A.

Since c0 and ( are small enough we obtain

|"x(!(c), #)# ' "x(!(c%), #)#| " 2 |c ' c%|"x(!(c%), ## " c0

15"x(!(c%), #)#

which shows that all the "x(!(c), #)# are equivalent in [!1, !2], more precisely takings = !(c%), ! = !(c) we obtain

sups+[&1,&2]

"x(s, #)# " 2 "x(!, #)#,

which is (4.4.103).Finally using (4.4.101), (4.4.99), (4.4.102), (4.4.103) and (4.4.100) we obtain

|x ' x(!! , #)| " K |x ' x(!, #)|

which is Proposition 4.4.19 (iii) in the case 2.

Now using Proposition 4.4.19, (4.4.95), (4.4.96) we obtain

|()3 ' )6)(!2, x, #)| " CN|x ' x(!, #)|N+1

"!#N(4.4.105)'(()

((*

)j(!, x, #) = )j(!2, x, #) '- 1

0p(x, #(!! , x, #) d" + A, j = 3, 6,

|A| " CN|x ' x(!, #)|N+1

"!#N .(4.4.106)

Let now *2 $ C"(R) be such that *2(s) = 1 if s ! 1, *2(s) = 0 if s " 0 and set*3(!) = *2

$&!&1&3!&1

%. Then let us set

(4.4.107) )(!, x, #) = *3(!))3(!, x, #) + (1 ' *3(!)))6(!, x, #).



We have$)

$!(!, x, #) = *3(!)

$)3

$!(!, x, #)+(1'*3(!))

$)6

$!(!, x, #)+

$*3

$!(!)()3')6)(!, x, #).

Now we deduce from (4.4.105) and (4.4.106) that on the support of (.3(& we have

|()3 ' )6)(!, x, #)| " CN|x ' x(!, #)|N+1

"!#N .

By Proposition 4.4.14 for )3 and )6 we have for j = 3 or 6,555$)j

$x(!, x, #) ' #(!, x, #)

555 " CN|x ' x(!, #)|N

"!#N

therefore this is also true for ) and since for j = 3 or 6,555$)j

$!(!, x, #) ' p(x, #(!, x, #))

555 " CN|x ' x(!, #)|N

"!#N ,

the function ) defined in (4.4.107) satisfies all the requirements of Proposition 4.4.14.The proof of Proposition 4.4.14 will be therefore complete when we will construct

the path 3(", x) satisfying (4.4.74) to (4.4.79).

Construction of 3(", x). — Let us set a = x(!, #). We first show that we can joinany point x to a point a ' |x ' a|#% by path remaining in the set

!y $ Rn : y · #% " c0

10"y# |#%|, |y ' a| = |x ' a|

".

Making rotations we may without loss of generality assume that "!

|"!| = ('1, 0, . . . , 0),a = (a1, a2, 0, . . . , 0), x = (x1, x2, x3, 0, . . . , 0). Therefore it will be su"cient to restrictourselves to the dimension three. We will construct our path on planes so we beginby the dimension two. Let us set with D $ ]0, 1[, k > 0,

C =:y $ R2 : |y ' a|2 = |x ' a|2

;,

H =:y $ R2 : 'y1 = D

Nk2 + y2

2

;,

D =:y $ R2 : 'y1 " D

Nk2 + y2

2

;.

Lemma 4.4.20. — Dc = R2 ! D is strictly convex.

Proof. — This follows easily from the strict convexity of the function g(t) =/

k2 + t2.

Lemma 4.4.21

(i) Let b $ D and u = (1, y) with |y| " 1. Then for all t > 0 we have b+t u $ D!H.(ii) Let b $ Dc and v = ('1, y) with |y| " 1. Then for all t > 0 we have b+t v $ Dc.



Proof(i) Let b = (b1, b2) and h(t) = b1 + t + D

Ok2 + (b2 + t y). Then

h(0) = b1 + DN

k2 + b22 ! 0 and h%(t) = 1 +

D(b2 + t y) yOk2 + (b2 + t y)2

.

Since D |y| < 1 we have D |b2 + t y| |y|PO

k2 + (b2 + t y)2 < 1. It follows that h%(t) > 0so h(t) > h(0) ! 0.

The proof of (ii) is the same.

Assume that C !H contains at least two di!erent points (otherwise C ! (C ! H)would be connected). Let us set

(4.4.108)

'(()

((*

M& = a + |x ' a|3

cos !sin !

4, ! $ [0, 24[,

!1 = inf{! $ [0, 24[: M& $ C !H},!2 = sup{! $ [0, 24[: M& $ C !H}.

Remark 4.4.22(i) If ! $

@0, 3

4

A2@24 ' 3

4 , 24A

we have'''.a M& = |x ' a|

$cos &sin &

%with cos ! > 0 and55 sin &

cos &

55 " 1. Since a $ D Lemma 4.4.21 (i) implies that M& $ D ! H. It follows thatwe have

4

4< !1 < !2 < 24 ' 4

4.

(ii) We cannot have !1 $A

34 , 3

2

Aand !2 $

@332 , 24 ' 3

4

@. Indeed if this was true

then by Lemma 4.4.20 the segment ]M&1 , M&2[ would be in Dc. But sin !1 > 0 andsin !2 < 0 so there exists t $ ]0, 1[ such that t sin !1 + (1 ' t) sin !2 = 0 ; then

Nt = t M&1 + (1 ' t)M&2 = a + |x ' a|3

t cos !1 + (1 ' t) cos !2

0

4= a + #

310

4

with # > 0 since cos !1 > 0 and cos !2 > 0. By Lemma 4.4.21 (i) Nt $ D since a $ Dwhich is in contradiction with Nt $ ]M&1 , M&2 [, Dc.

(iii) If !1 $A

34 , 3

2

Athen for all ! in ]!1, 4] we have M& $ Dc which implies !2 $

A4, 33

2

Aby (ii). Indeed we have

''''.M&1M& = |x ' a|

$cos &!cos &1sin &!sin &1

%; since for ! $ [0, 4],

cos ! is decreasing we have cos ! ' cos !1 < 0 and555sin ! ' sin !1

cos ! ' cos !1

555 =555 cotg

+! + !1

2

,555 " 1

since 34 " &+&1

2 " 334 . Then Lemma 4.4.21 (ii) implies that M& $ Dc.

It follows from Remark 4.4.22, (i), (ii), (iii) that we have else !1 $A

32 , 4@

or!2 $

A4, 33

2

@. By symmetry it is enough to consider one case. Therefore we shall

assume in the sequel that 32 < !1 < !2 < 24 ' 3

4 , !1 $A

32 , 4@. We claim that

(4.4.109) M& $ Dc for all ! in ]!1, !2[.

We split the proof in two cases.



Case 1: !2 " 332 . — Since sin ! is decreasing on

@32 , 33

2

Awe have

– if 32 < !1 < !2 " 33

2 , sin !1 > sin ! > sin !2,– if 3

2 < !1 < ! < 34 ' !2 " 332 , sin !1 > sin ! > sin(34 ' !2) = sin !2.

Let us set

Nt = t M&1 + (1 ' t)M&2 = a + |x ' a|3

t cos !1 + (1 ' t) cos !2

t sin !1 + (1 ' t) sin !2

4.

Now there exists t $ ]0, 1[ such that

t sin !1 + (1 ' t) sin !2 = sin !

and since ! $@

32 , 33

2

Awe have ! = Arc sin(t sin !1 + (1 ' t) sin !2) + 4. Then

cos ! = ' cos(Arc sin(t sin !1 + (1 ' t) sin !2)) = 'O

1 ' (t sin !1 + (1 ' t) sin !2)2,

cos ! < 'tN

1 ' sin2 !1 ' (1 ' tN

1 ' sin2 !2 " t cos !1 + (1 ' t) cos !2.

Here we have used the strict convexity of the function/

1 ' x2. Since Nt $ Dc andM& = Nt +#

$!10

%where # > 0 (see (4.4.108)) we deduce from Lemma 4.4.21 (ii) that

M& $ Dc which proves (4.4.109) in case 1.

Case 2: !2 > 332 and ! < 34'!2. — Since !2 < 24 we have ! > 4. Now by (4.4.108),

''''.M&2M& = |x ' a|

3cos ! ' cos !2

sin ! ' sin !2

4.

Since cos ! is increasing for ! $ [4, 24] we have cos ! ' cos !2 < 0. Moreover,55 sin &!sin &2cos &!cos &2

55 =55 cotg &+&2

2

55 " 1 since &+&22 ! 33

2 , ! " 24 ' 34 , !2 " 24 ' 3

4 so&+&2

2 " 24 ' 34 . It follows from Lemma 4.4.21 (ii) that M& $ Dc which proves

(4.4.109) in case 2.We conclude that if x $ C 2 (D ! H) then x = a + |x ' a|

$cos &sin &

%with ! /$]!1, !2[

and there exists a path joining the point x to the point a + |x ' a|$

10

%with length

less than 24 |x ' a|.

Construction of the path in dimension 3. — We have #% = ('1, 0, 0), a = (a1, a2, 0),x = (x1, x2, x3) and 'a " D0

O1 + |a|2, 'x1 " D0

O1 + |x|2 with D0 = c0

10 . We firstconstruct a path in the plane y3 = x3. We set

D =!(y1, y2, x3) : 'y1 " D0

N1 + |x3|2 + y2

1 + y22

"

=!(y1, y2, x3) : 'y1 " D0O

1 ' D20

N1 + |x3|2 + y2

2

".

Since c0 is small enough we have D0/1!D2

0< 1. By the same way we see that the

point a is such that 'a1 " D0/1!D2

0

O1 + a2

2. Therefore 2a = (a1, a2, x3) $ D. Since

x $ D, by the construction made in two dimensions there exists a path lying in the


4.5. THE PHASE FOR SMALL & 99

set:y : y3 = x3, |y ' 2a| = |x ' 2a| =

O(x1 ' a1)2 + (x2 ' a2)2

;joining x to z =$

a1 +O

(x1 ' a1)2 + (x2 ' a2)2, a2, x3

%of length smaller than 24 |x'2a| " 24 |x' a|.

Let us construct now a path in the plane y2 = a2. Let us set

D =!y = (y1, a2, y3) : 'y1 " D0

N1 + y2

1 + a22 + y2

3

"

=!y = (y1, a2, y3) : 'y1 " D0O

1 ' D20

N1 + a2

2 + y23

".

We have z $ D, a $ D. There exists a path joining z to (a1+ |x'a|, a2, 0) lying in theset {y = (y1, y2, y3) : y2 = a2, |y ' a| = |x ' a|} with length smaller than 24 |x ' a|.

Now to join 0 to x we join 0 to z1 = (a1+|a|, a2, 0) then x to z2 = (a1+|x'a|, a2, 0)and since the segment [z1, z2] is included in D by Lemma 4.4.21 (i), the path joins0 to x and its length is smaller than C(|x ' a| + |a|). Now by (4.4.78) we have|a| " |x ' a| or |x ' a| " |a| " 2 |x ' a|, so the length of the path is smaller thanC |x ' a| = c |x ' x(!, #)|. Moreover |3(", x) ' a| " 2 |x ' a| " +

20 "!# so (4.4.77) issatisfied. Finally (4.4.78) and (4.4.79) are obviously satisfied.

This ends the proof of Proposition 4.4.14.

4.5. The phase for small !

We shall need the following precision on the phase when |!| " 1.

Theorem 4.5.1. — Let ) be the phase given by Theorem 4.1.2. Then one can findpositive constants such that for |!| " 1, |x ' x(!, #)| " & "!# and |#%| " 2 one canwrite

)(!, x, #) =(x ' #x) · #% ' ! |#%|2 + i

2 |x ' #x|2

1 + 2i!+

12i

|#%|2 + R(!, x, #);

where555$R

$#x

555 " C (( + &)(|x ' #x|2 + |!|),555$R

$#%

555 " C (( + &) |!|,555$2R

$#2x

555 " C (( + &)(|x ' #x|2 + |!|),555

$2R

$#x $#%

555 " C (( + &) |!|,555$2R

$#2%

555 " C (( + &) |!|,

and

|$A1"x

$A2"!

R(!, x, #)| "6

CA1 if A2 = 0CA1,A2 |!| if |A2| ! 1.

Proof. — Let us introduce the following space of functions.(4.5.1)6

E =:Z $ C"(R ( Rn ( Rn) : |$*

t $A1x $A2

% Z(t, x, %)| " C*,A1,A2 ( |t|1!*, for all

Aj $ Nn, / = 0, 1, |t| " 1, x $ Rn, % $ Rn with |%| " 2 and Z(0, x, %) = 0;



Let us also recall that Proposition 3.2.1 gives the following description of the flow for|t| " 1, x $ Rn, % $ Rn with |%| " 3.

(4.5.2)

')

*

x(t, x, %) = x + 2t % + r(t, x, %),%(t, x, %) = % + 1(t, x, %),z, 1 $ E .

It follows that, with f = x or %, we have

(4.5.3)55$*

t $A1x $A2

% f(t, x, %)55 " C*,A1,A2 if / + |A1| + |A2| ! 1.

Let us set now

gj(2) = *0

$ 2

µ0

%@(%j ' i xj)('!, y + x(!, #), 2 + %(!, #)) ' (#j

% ' i #jx)A

where *0 $ C"0 (Rn), *0(2) = 1 if |2| " 1

2 , *0(2) = 0 if |2| ! 1 and |y| " &. Settingx = y + x(!, #), % = 2 + %(!, #) and using (4.5.2) we obtain

(4.5.4) gj(2) = *0(2)@(1 + 2i !) 2j ' i yj + (1 + 2i !) 1j(!, #) ' i rj(!, #)

+ (1 + 2i!) 1j('!, x, %) ' i zj('!, x, %)A.

We claim that we have the following estimates for / = 0, 1,

(4.5.5) |$*& $,

y $µ- $A

" gj | "6

C*,,,µ if A = 0,C*,,,µ,A ( |!|1!* if |A| ! 1.

These estimates are obvious for the four first terms of gj . So we are left with theestimate of

(1) = $*& $,

y $µ- $A

"

@Z('!, y + x(!, #), 2 + %(!, #))

A, Z $ E .

To handle this term we shall make use of the Faa di Bruno formula given in Ap-pendix A.1, with F = Z, Y = (!, y, 2, #), U1(Y ) = '!, U1+j(Y ) = yj + xj(!, #),U1+n+j(Y ) = 2j + %j(!, #), j = 1, . . . , n. Since Z $ E we find easily, using (4.5.3) that(1) " C ( |!|1!* which proves our claim.

Another property of gj which will be used in the sequel is the following.

(4.5.6) For ! = 0, gj(2) = *0(2)(2j ' i yj) is independent of #.

Now according to our procedure we have solved the equations (see (4.3.14)),

(4.5.7) 0 = r(a, b, gj) = gj('a) ' in#

k=1

$gj

$2k('a) bk +

n#

p,q=1

Hjpq(!, y, #, a, b) bp bq

in the set

E =!

(a, b) $ Rn ( Rn :555a +

2! y

1 + 4!2

555 "/

&|y|"!# ,

555b +y

1 + 4 !2

555 "/

&|y|"!#2

".



Let us recall that we have the following bounds on Hjpq (see (4.3.13) and (4.3.16))

55$*& $,

y $A" $M

(a,b) Hjp,q

55 "#

|µ|!|M|+3n+2

- 55$*& $,

y $A" $µ

- gj(2)55 d2, / = 0, 1.

Here we have used the fact that r(a, b, gj) is linear with respect to gj. It follows from(4.5.5), since gj has compact support in 2, that

(4.5.8) |$*& $,

y $A" $M

(a,b) Hjpq| "

6C*,,,M if A = 0,

C*,,,A,M ( |!|1!* if |A| ! 1.

Using (4.5.4) we see easily that the equations (4.5.7) are equivalent to the followingsystem

(4.5.9)

'(((((()

((((((*

aj ='2! yj

1 + 4!2+ Za

j,1(!, #) + Zaj,2('!, y + x(!, #),'a + %(!, #))

+ Zaj,3('!, y + x(!, #),'a + %(!, #)) b + Ha

j (!, y, #, a, b) b · b

bj ='yj

1 + 4!2+ Zb

j,1(!, #) + Zbj,2('!, y + x(!, #),'a + %(!, #))

+ Zbj,3('!, y + x(!, #),'a + %(!, #)) b + Hb

j (!, y, #, a, b) b · b

where the Z %s belong to the space E defined in (4.5.1) and the H %js satisfy the estimates

(4.5.8).According to (4.5.6), (4.4.2), Theorem 4.4.2 and Theorem 4.3.1 for ! = 0 aj and

bj do not depend on # and moreover we have,

(4.5.10)

6aj(0, y, #) = 2aj(y) = O(|y|N ),

bj(0, y, #) = 2bj(y) = 'yj + O(|y|N ),

for every N $ N and |y| " &.Let us set

(4.5.11) G#j (!, y, #) = H#

j (!, y, #, a(!, y, #), b(!, y, #)) ·b(!, y, #) ·b(!, y, #), 9 = a or b.

Then, since the Z %js vanish for ! = 0, (4.5.10) implies that

(4.5.12) G#j (0, y, #) = 2G#

j (y) = O(|y|N ), %N $ N.

Therefore we can write

(4.5.13) G#j (!, y, #) = G#

j (y) +- &

0

d

d!G#

j (", y, #) d".

We claim that we have the following estimates on aj, bj . Let us set for conveniencefj = aj or bj .

(4.5.14)

'(()

((*

555$fj

$!(!, y, #)

555 " C (( + &)

|$*& $,

y $A" fj(!, y, #)| "

6C*,, if A = 0,C*,,,A (( + &) |!|1!* if |A| ! 1, / = 1.



To prove the first estimate we di!erentiate both sides of (4.5.9) with respect to !.Since the terms in Z belong to E and using (4.5.8), the fact that x(!, #), %(!, #) arebounded, we obtain

555$aj

$!

555+555$bj

$!

555 " C1 |yj | + C2 ( + C3(( + &)+555

$a

$!

555+555$b

$!

555,.

Taking ( + & small enough and since |y| " & we obtain our first claim. To provethe second estimate we use the Faa di Bruno formula (see Appendix A.1) and aninduction procedure.

Let us set Y = (!, y, #), ' = (/, 3, A) and let us apply the operator $#Y to both

sides of (4.5.9). We have

$#Y f0 =

60(1) if A = 0,0 if |A| ! 1.

, f0 ='2! yj

1 + 4!2or

'yj

1 + 4!2

|$#Y Z(!, #) " C ( |!|1!*.

Assume now that our estimate is true for |'| " k and let |'| = k + 1. Then

$#Y

@Z('!, y + x(!, #),'a + %(!, #)) b(!, y, #)

A= (1) + (2) + (3)

where

(1) = Z('!, · · · ) $#Y b(Y )

(2) = $#Y

@Z('!, y + x(!, #),'a(Y ) + %(!, #))

Ab(Y )

(3) =#

#1+#2=##j ,=0

3''1

4$#1

Y [Z('!, · · · )] $#2Y b.

Using the Faa di Bruno formula in the terms (2) and (3) we see that

|$#Y [Z('!, · · · )]| " C ( |!|(|$#

Y a| + |$#Y b|) +

60(1) if A = 0,0((( + &)|!|1!*) if |A| += 0.

By the same way, using (4.5.8) we obtain

|$#Y [H(Y, a(Y ), b(Y )) · b(Y ) · b(Y )|] " C (( + &)|!|(|$#

Y a(Y )|

+ |$#Y b(Y )|) +

60(1) if A = 0,

0((( + &)|!|1!*) if |A| ! 1.

Taking ( + & small enough we obtain the second estimate of (4.5.14).



Now it follows from (4.5.9), (4.5.11), (4.5.12) and (4.5.13) that we can write, withY = (!, y, #),(4.5.15)'((((((((()

(((((((((*

(aj + i bj)(Y ) = ' yj

2! ' i+ Uj(Y ) where

Uj(Y ) = Zj,1(!, #) + Zj,2('!, y + x(!, #),'a(Y ) + %(!, #))

+Zj,3('!, y + x(!, #),'a(Y ) + %(!, #)) · b(Y )

+ 2Gj(y) +- &

0

!.$H

$!+

$H

$a· $a

$!+

$H

$b

$b

$!

/b · b + 2H b · $b

$!

"(", y, #) d"

with |$,y2Gj(y)| " CN |y|N for all N $ N and 3 $ Nn.

Using (4.5.1), (4.5.3), (4.5.8), (4.5.14) we deduce the following estimates

(4.5.16)

'()

(*

|U(Y )| +55(U

(y (Y )55 " CN |y|N + C (( + &) |!| " C% (( + &)

|$,y U(Y )| " CN,, |y|N + C, |!| if |3| ! 2

|$,y $A

" U(Y )| " C,,A (( + &) |!| if |3| ! 0 and |A| ! 1.

Now recall that,

(4.5.17)

'()

(*

#j(!, x, #) = %j(!, #) ' (aj + i bj)(!, x ' x(!, #), #)

)(!, x, #) =1 10 (x ' x(!, #)) · #(!, s x + (1 ' s)x(!, #), #) ds

+! p(#) + 12i |#%|2.

It follows that,

)(!, x, #) = (x ' x(!, #))< => ?(1)

·%(!, #) +12|x ' x(!, #)|2

2! ' i< => ?(2)

'- 1

0(x ' x(!, #)) · U(!, s(x ' x(!, #)), #) ds + !p(#) +

12i

|#%|2.

We have,

(4.5.18)

'()

(*

(1) = (x ' #x)#% ' 2! |#%|2 ' r(!, #) · #% + (x ' x(!, #)) · 1(!, #)

(2) = 12

12&!i

@(x ' #x)2 ' 4!(x ' #x) · #% + 4!2 |#%|2

+ 2(x ' #x ' 2! #%) · r(!, #) + |r(!, #)|2A.

Let us consider the term in ) which does not contain any error term. It can be writtenas

(x ' #x) · #% ' 2! |#%|2 +(x ' #x)2 ' 4!(x ' #x) · #% + 4!2 |#%|2

2(2! ' i)+ ! |#%|2 +

12i

|#%|2

which is equal to

(x ' #x) · #% + i2 (x ' #x)2 ' ! |#%|2

1 + 2i !+

12i

|#%|2.



It follows that

)(!, x, #) =(x ' #x) · #% + i

2 (x ' #x)2 ' ! |#%|2

1 + 2i!+

12i

|#%|2 + R(!, x, #)

where

R(!, x, #) = 'r(!, #) · #%< => ?(1)

+ (x ' x(!, #)) · 1(!, #)< => ?(2)

(x ' #x ' 2! #%) · r(!, #) + 12 |r(!, #)|2

2! ' i< => ?(3)

+ !(p(#) ' |#%|2)< => ?(4)

'- 1

0(x ' x(!, #)) · U(!, s(x ' x(!, #)), #) ds< => ?

(5)

.

We are ready now to show that the remainder R satisfies the estimates given inTheorem 4.5.1.

First of all, since r and 1 belong to E (see (4.5.1)), since the functions x ' x(!, #)and x ' #x ' 2! #% are bounded together with all their derivatives with respect to #and since p(#) = |#%|2 + (

&nj,k=1 bjk(#x)#j

% #k% we have

(4.5.19) |$A" (i)| " CA ( |!| for i = 1, 2, 3, 4.

So we are left with the term (5). Let us note that if we set f0(!, x, #) = x ' x(!, #)then,

(4.5.20)

')

*|f0| " 2&, |f0| " |x ' #x| + C |!|,

55 $f0

$#x

55 " C,55 $f0

$#%

55 " C |!|

|$A" f0| " C ( |!| if |A| ! 2, uniformly in (x, !, #).

With this notation one has

(5) = f0(!, x, #)- 1

0U(!, s f0(!, x, #), #) ds.

Then, with i = x or %,

$

$#i(5) =

$f0

$#i

- 1

0U(!, s f0, #) ds + f0

- 1

0

+s

$f0

$#i

$U

$y+

$U

$#i

,(!, s f0, #) ds.

Now it follows from (4.5.18) and (4.5.20) that

(4.5.21)

'()

(*

55 $

$#x(5)55 " C (( + &)(|x ' #x|2 + |!|),

55 $

$#%(5)55 " C (( + &) |!|),



since | (f0("!

(5)| " C |!| and |x ' #x| " &. Now, with i, j = x or %, we have

$2

$#i $#j(5) =

$2 f0

$#i $#j

- 1

0U(!, s f0, #) ds +

$f0

$#i

- 1

0

+s

$f0

$#j

$U

$y+

$U

$#j

,ds

+$f0

$#j

- 1

0

+s

$f0

$#i

$U

$y+

$U

$#i

,ds + f0

- 1

0

.s

$2f0

$#i $#j

$U

$y+ s

$f0

$#i

+ $f0

$#j

$2U

$y2+

$U

$#j

,

+$2U

$#i $y

$f0

$#j+

$2U

$#i $#j

/ds.

Using (4.5.16) and (4.5.20) we check easily that

(4.5.22)

'(((((()

((((((*

555$2

$#2x

(5)555 " C (( + &)(|x ' #x|2 + |!|)

555$2

$#x $#%(5)555 " C (( + &)|!|

555$2

$#2%

(5)555 " C (( + &)|!|.

Combining (4.5.19), (4.5.21) and (4.5.22) we obtain the claimed estimates on the twofirst derivatives of R.

Finally using again (4.5.18) and (4.5.20) we deduce the following estimates of thehigher derivatives

|$A1"x

$A2"!

(5)| "6

CA1 if A2 = 0,CA1 A2 |!| if |A2| ! 1.

The gain of |!| when |A2| ! 1 coming from the fact that a derivative of x ' x(!, #)and U with respect to #% makes appear a !. Thus the proof of Theorem 4.5.1 iscomplete.


CHAPTER 5

THE TRANSPORT EQUATIONS

5.1. Statement of the result and preliminaries

Let P =&n

j,k=1 gjk(x)Dj Dk be a second order di!erential operator of the form(2.2.7) We shall denote by tP the transposed operator.

In Chapter 4 we have constructed a phase function for P . The purpose now is toconstruct an amplitude.

Recall that the set %+ have been introduced in Definition 4.1.1.The main result of this Section is the following.

Theorem 5.1.1. — For every # $ T #Rn with 12 " |#%| " 2, every N $ N and every

+ ! 1 one can find an amplitude eN (!, y, #, +) which is C" on 2%+ such that(i) eN(0, y, #, +) = 1.

(ii)+i+

$

$!' i+

n

2!

1 + !2' tP

,$ei$'(&,x,") eN (!, x ' x(!, #), #, +)

%

= RN (!, x, #, +) ei$'(&,x,"), where

|RN (!, x, #, +)| " CN

++!N + +2 |x ' x(!, #)|N

"!#N,

for every (!, x, #) in %+, + ! 1 and CN is independent of (!, x, #, +).(iii) |$A

x eN (!, x ' x(!, #), #, +)| " CN,A uniformly with respect to (!, x, #, +).

Corollary 5.1.2. — For every # $ T #Rn with 12 " |#%| " 2, every N $ N and

every + ! 1 one can find an amplitude aN (!, x, #, +) which is C" on %+ such that(i) aN(0, x, #, +) = 1,(ii)

$i+ (

(& ' tP%$

ei$'(&,x,") aN (!, x, #, +)%

= R%N (!, x, #, +) ei$'(&,x,") where

|R%N (!, x, #, +)| " C%

N

++!N + +2 |x ' x(!, #)|N

"!#N,

uniformly with respect to (!, x, #, +).(iii) |$A

x aN (!, x, #, +)| " CN,A "!#!n/2, uniformly with respect to (!, x, #, +).

108 CHAPTER 5. THE TRANSPORT EQUATIONS

In flat case RN and R%N are bounded by CN+!N , but no explicit formula for eN is

available.

Proof. — We have just to set aN(!, x, #, +) = "!#!n/2 eN (!, x ' x(!, #), #, +) whereeN has been defined in Theorem 5.1.1.

Proof of Theorem 5.1.1. — We have tP =&n

j,k=1 gjk(x)Dj Dk +&n

j=1 gj(x)Dj +g0(x), where gjk = &jk + ( bjk and bjk $ B1

!0, gj $ B2

!0, 1 " j, k " n, g0 $ B3

!0where,

(5.1.1) B*!0

=!g $ C"(Rn) : |$A

x g(x)| " CA

"x#|A|+*+!0, %x $ Rn, %A $ Nn

".

A straightforward computation shows that

(5.1.2)+i+

$

$!' i+

n

2!

1 + !2' tP

,(ei$'f)

= ei$'

I' +2

+$)

$!+ p+x,

$)

$x

,,f + i+

+$f

$!+ 2

n#

j,k=1

gjk $)

$xj

$f

$xk

+n#

j,k=1

gjk $2)

$xj $xkf ' n

2!

1 + !2f + i

n#

j=1

gj$)

$xjf,' tPf

J.

According to Theorem 4.1.2 the coe"cient of +2 in the right hand side of (5.1.2) isbounded by CN

$ |x!x(&,")|'&(

%N , for any N . Therefore if we set

(5.1.3)

'(()

((*

I = e!i$'+i+

$

$!' i+

n

2!

1 + !2' tP

,$ei$'f

%

X =$

$!+ 2

n#

j,k=1

gjk $)

$xj· $

$xk

we obtain

(5.1.4)5555I ' i+Xf ' i+

+ n#

j,k=1

gjk $2)

$xj $xk' n

2!

1 + !2+ i

n#

j=1

gj$)

$xj

,f ' tPf

5555

" CN +2+ |x ' x(!, #)|

"!#

,N.

To pursue the proof we consider separately the two cases.

5.2. The case of outgoing points

For convenience we shall set

2S+ =:# $ T #Rn :

12

" |#%| " 2, #x · #% ! 'c0 "#x# |#%|;

2S! =:# $ T #Rn :

12

" |#%| " 2, #x · #% " c0 "#x# |#%|;.



Here we assume # $ 2S±. Let us set Rj = ('(xj

' #j . It follows from Theo-rem 4.3.13 (v) and (4.3.48) that |$A

x Rj(!, x, #)| " CA/"!#|A| for A $ Nn and fromProposition 4.3.19 (ii) that |Rj(!, x, #)| " CN

$ |x!x(&,")|'&(

%N for all N $ N.First of all, according to (4.3.43) and Taylor’s formula, we have

(5.2.1)

'(((((((()

((((((((*

$)

$xj(!, x, #) = %j(!, #)

+12 sgn !(1 ' *1(!))(xj ' xj(!, #)) ' (aj + i bj)(!, x ' x(!, #), #)

"!#+Rj(!, x, #),

|$Ax Rj(!, x, #)| " CA,N

+ |x ' x(!, #)|"!#

,N.

Using Theorem 4.3.1 (iii) we deduce,

(5.2.2)

'(((((((()

((((((((*

$2)

$xj $xk(!, x, #) =

12

sgn !

"!# (1 ' *1(!)) &jk + djk(!, x ' x(!, #), #)

+Rjk(!, x, #),

|$Ax djk(!, x ' x(!, #), #)| " CA

"!#|A|+2,

|$Ax Rjk(!, x, #)| " CAN

+ |x ' x(!, #)|"!#

,N

uniformly with respect to (!, x, #).Now if g $ B!0 (see (5.1.1)) and (!, x) $ %+ we can write

|$Ax g(x)| = |($A

x g)(y + x(!, #))| " CA

"y + x(!, #)#|A|+1+!0.

Since |y| " &"!# we can use Proposition 3.3.1 to write

"y + x(!, #)# ! "x(!, #)# ' |y| ! 1/3"!# ' &"!# ! 1

2"!#.

It follows that

(5.2.3) |$Ax g(x)| " CA

"!#|A|+1+!0.

This can be applied to the functions gjk ' &jk, gj , g0, 1 " j, k " n. It follows from(5.2.2) that

n#

j,k=1

gjk(x)$2)

$xj $xk(!, x, #) =

n

2sgn !

"!# (1 ' *1(!)) + d(!, x ' x(!, #), #) + R(!, x, #)

where d and R satisfy the same estimates as in (5.2.2). Now we have

(5.2.4)

12

sgn !

"!# (1 ' *1(!)) '12

!

1 + !2=

12

sgn !

"!# (1 ' *1(!))+1 ' |!|

"!#

,' ! *1(!)

2"!#2

=12

sgn !

"!#21 ' *1(!)"!# + |!| = O

+ 1"!#3

,.



Summing up, we have proved

(5.2.5)

'(((((((((((((()

((((((((((((((*

n#

j,k=1

gjk(x)$2)

$xj $xk(!, x, #) ' n

2!

1 + !2= D1(!, x ' x(!, #), #)

+R1(!, x, #),n#

j=1

gj(x)$)

$xj(!, x, #) = D2(!, x ' x(!, #), #) + R2(!, x, #),

|$Ax Dj(!, x ' x(!, #), #)| " CA

"!#|"|+1+!0, j = 1, 2,

|Rj(!, x, #)| " CN

+ |x ' x(!, #)|"!#

,N, j = 1, 2, %N $ N.

We are going now to simplify the vector fields X introduced in (5.1.3). Let us set

(5.2.6)7

s = !y = x ' x(!, #).

Since xk(!, #) = (p(%k

(x(!, #), %(!, #)) = 2&n

j=1 gjk(x(!, #))%k(!, #), we obtain

X =$

$s' 2

n#

j,k=1

!gjk(x(s, #)) %j(s, #)' gjk(y + x(s, #))

$)

$xj(s, y + x(s, #), #)

" $

$yk.

Now using (5.2.1) and gjk = &jk + ( bjk, bjk $ B!0 we can write

X =$

$s+

sgn s

"s# (1 ' *1(s))n#

j=1

yj$

$yj

' 2(n#

j,k=1

:bjk(x(s, #)) ' bjk(y + x(s, #))

;%j(s, #)

$

$yk

' 2n#

j=1

(aj + i bj)(s, y, #)"s# · $

$yj+ 2(

n#

j,k=1

bjk(y + x(s, #))1"s#

+12

sgn s yj

' (aj + i bj)(s, y, #), $

$yk+ 2

n#

j,k=1

gjk(y + x(s, #))Rj(s, y + x(s, #), #)$

$yk.

Definition 5.2.1. — We shall say that a function f = f(s, y, #) on 2%+ ( 2S± belongsto E if

(5.2.7)

')

*

f(s, 0, #) = 0

|$Ay f(s, y, #)| " CA

"s#|A|+1, A $ Nn

uniformly when (s, y) $ 2%+ and # $ 2S±.



According to (5.2.4), (5.2.3) and Theorem 4.3.1, (ii), (iii) we have,

(5.2.8)

'((((((()

(((((((*

(1 ' *1(s))sgn s

"s# yj 's

1 + s2yj $ E

((bjk(x(s, #) ' bjk(y + x(s, #)))) %j(s, #) $ E

( bjk(y + x(s, #))1"s#

+12· sgn s · yj ' (aj + i bj)(s, y, #)

,$ E

1"s# (aj + i bj)(s, y, #) $ E .

Then we have,

(5.2.9)

'(()

((*

X =$

$s+

s

1 + s2

n#

j=1

yj$

$yj+

n#

j=1

Ej(s, y, #)$

$yj+

n#

j=1

R%j(s, y, #)

$

$yj

where Ej $ E and |R%j(s, y, #)| " CN

+|y|'s(

,N.

Now we perform another change of variables. We set

(5.2.10)

')

*! = s

z =y

"s# .

Then we have ((& = (

(s + s1+s2

&nj=1 yj

((yj

. It follows that

(5.2.11)

'((((((((((()

(((((((((((*

(i) X =$

$!+

n#

j=1

hj(!, z, #)$

$zj+

n#

j=1

2Rj(!, z, #)$

$zj,

(ii) hj(!, 0, #) = 0,

(iii) |$Az hj(!, z, #)| " CA

"!#2 , A $ Nn,

(iv) | 2Rj(!, z, #)| " CN

"!# |z|N , for all N in N, uniformly when

! ! 0 (resp. ! " 0), |z| " &, # $ '±, j = 1, . . . , n.

Moreover, since ((yj

= 1'&(

((zj

we have by (5.2.3),

(5.2.12)

'(()

((*

tP =#

|/|!2

k/(!, z, #) $/z ,

|$,z k/(!, z, #)| " C,

"!#1+!0, 3 $ Nn.

Let us set

(5.2.13) X0 =$

$!+

n#

j=1

hj(!, z, #)$

$zj,

where hj satisfies (5.2.11).



It follows from (5.1.4), (5.2.5), (5.2.6), (5.2.10) and (5.2.11) (i) that

(5.2.14)

'()

(*

555I ' i++X0 f + d(!, z, #) f ' i

+tPf,555 " CN +2 |z|N (|f | + |8zf |)

|$,z d(!, z, #)| " C,

"!#1+!0, 3 $ Nn .

Now let us fix an integer N0 large enough depending only on the dimension n (andchosen later on). For the coe"cients hj, k/ , d in (5.2.13), (5.2.12) and (5.2.14) wewrite

(5.2.15)

'((()

(((*

F (!, z, #) = FN0(!, z, #) + rN0(!, z, #) where,

FN0(!, z, #) =#

|,|!N0!1

$,z F (!, 0, #)

z,

3!,

|rN0(!, z, #)| " CN0 |z|N0 .

Let us set

(5.2.16)

'((()

(((*

L =$

$!+

n#

j=1

hN0j (!, z, #)

$

$zj+ dN0(!, z, #),

Q =#

|/|!2

kN0/ (!, z, #) $/

z .

Using (5.2.14) to (5.2.16) we see that

(5.2.17)555I ' i+

+L f ' i

+Q f,555 " CN0 +2 |z|N0

#

|,|!2

|$,z f(!, z, #)|.

Now we have the following result.

Lemma 5.2.2. — There exist functions A* = A*(!, z, #), / = 0, . . . , N0 + 1 which areC" in (!, z) in the set O = {(!, z) : ! $ R±, |z| " &} such that

(i) A0(0, z, #) = 1, A*(0, z, #) = 0, / = 1, . . . , N0 + 1,(ii) |$,

z A*(!, z, #)| " C*,, , uniformly in O ( 2S±, (/ $ N, 3 $ Nn),(iii) L A0 = 0, L A* = i Q A*!1, / = 1, . . . , N0 + 1.

Let us assume for a moment this lemma proved. Let us set

(5.2.18) f = fN0 = A0 +1+

A1 + · · · + 1+N0+1

AN0+1.

Then Lemma 5.2.2 shows that

(5.2.19)

'((()

(((*

fN0(0, z, #, +) = 1

|$,z fN0(!, z, #, +)| " C,,N0 if (!, z) $ O, # $ 2S±, + ! 1555L fN0 '

i

+Q fN0

555 " +!N0!2 |Q AN+1| " CN0 +!N0!2.


(5.2.20) |I| " C%N0

+2 |z|N0 + C%%N0

+!N0!1.



Coming back to the variables (!, x) we set

eN0(!, x ' x(!, #), #, +) = fN0

+!,

x ' x(!, #)"!# , #, +

,.

Then it follows from (5.2.20), (5.2.19) that eN0 satisfies the conditions (i), (ii), (iii) inTheorem 5.1.1.

So we are left with the proof of Lemma 5.2.2.

Proof of Lemma 5.2.2. — We are going to straighten the principal part of the oper-ator L given by (5.2.16). Recall that we have L = L0 + dN0(!, z, #) with

L0 =$

$!+

n#

j=1

hN0j (!, z, #)

$

$zj.

Moreover, according to (5.2.11) and (5.2.15) we have

(5.2.21)

'(((((((()

((((((((*

(i) hN0j (!, z, #) =

n#

k=1

$hj

$zk(!, 0, #) zk + gj(!, z, #),

(ii) gj(!, z, #) =#

2!|,|!N0!1

13!

$,z hj(!, 0, #) z, ,

(iii)n#

j=1

|$,z hj(!, 0, #)| " C,

"!#2 , % 3 $ Nn.

In that follows all the objects will depend on # $ 2S± but all the estimates will beuniform with respect to #.

Let us set

(5.2.22) H(!) =+$hj

$zk(!, 0, #)

,

1!j,k!n.

If !0 $ R± we shall denote by Y (!, !0) the unique n(n matrix solution of the problem

(5.2.23)

6Y (!, !0) = H(!)Y (!, !0), ! $ R±,

Y (!0, !0) = Id .

Since by (5.2.21) (iii) the entries of the matrix H(!) are bounded by C/"!#2, theGronwall inequality shows that there exists M0 ! 1 such that

(5.2.24) *Y (!, !0)* " M0, for all !, !0 $ R± and # $ 2S±.

Moreover since Y (!, !0)!1 = Y (!0, !) we have also,

(5.2.25) *Y (!, !0)!1* " M0, for all !, !0 $ R± and # $ 2S±.

Now using (5.2.21) we see that the problem

(5.2.26)

6zj(!) = hN0

j (!, z(!), #), ! $ R±, 1 " j " n,

zj(0) = yj ,



is equivalent, setting z = (zj)1!j!n, g = (gj)1!j!n, to

(5.2.27) z(!) = Y (!, 0) y +- &

0Y (!, t) g(t, z(t), #) dt.

Then we have the following Lemma.

Lemma 5.2.3. — One can find 2 > 0 such that for all y $ Cn such that |y| " 2,the problem (5.2.27) has a unique global solution z such that |z(!)| " 2M02 for all! $ R±. This solution will be denoted by z(!, y). Moreover one can find a constantC(N0, M0) ! 0 such that

(i)00$ (zj

(yk

%(!, y) ' Y (!, 0)

00 " C(M0, N0) 2, and for every 3 $ Nn, one can find aconstant C, ! 0 such that

(ii) |$,y z(!, y)| " C, , for all ! ! 0 and |y| " 2.

Proof. — Let 2 > 0 (to be chosen). Assume |y| " 2 and set A = {T > 0 such that(5.2.27) has a solution for ! $ [0, T ] satisfying |z(!)| " 2M0 2}. Since (5.2.27) (which isequivalent to (5.2.26)) has a continuous solution for small ! and since |z(0)|| = |y| " 2there exists (0 > 0 such that (0 $ A. Thus A is non empty and it is obviously aninterval. Let T # = sup A. If T # = +) we are done so assume T # < +). Let us takeT $ ]0, T #[. Then on [0, T ] we have

|z(!)| " |Y (!, 0) y| +- &

0*Y (!, t)* |g(t, z(t), #)| dt.

Now by (5.2.21) (ii) and (iii) we have

|g(t, z(t), #)| " KN0

"t#2 2M0 2 |z(t)|

if 2M0 2 " 1, where KN0 =&

2!|,|!N0!1C%

, !.

It follows from (5.2.24) that,

|z(!)| " M0 2 +- &

0

2M20 KN0 2

"t#2 |z(t)| dt.

Then the Gronwall inequality implies that

|z(!)| " M0 2 exp.2M2

0 KN0 2

- +"

0

dt

"t#2/.

Therefore taking 2 small (compared to M0 and KN0) we can achieve that|z(!)| " 3

2 M0 2 for all ! $ [0, T ]. A classical argument shows that z(T #) can bedefined and |z(T #)| " 3

2 M0 2. Then solving again (5.2.26) with data z(T #) we seeeasily that this contradicts the definition of T # as the supremum of A. ThereforeT # = +).

Now di!erentiating (5.2.27) with respect to yk yields

(5.2.28)$z

$yk(!, y) ' Y (!, 0) ek +

- &

0Y (!, t)

n#

*=1

$g

$z*(t, z(t), #)

$z*

$yk(t, y) dt.



First of all (5.2.21) (ii) show that

(5.2.29)n#

*=1

555$g

$z*(t, z(t), #)

555 "#

2!|,|!N0!1

C%,

"t#2 |z(t, y)||,|!1 " CN0 M0 2

"t#2

if 2M0 2 " 1. It follows that555

$z

$yk(!, y)

555 " M0 +- &

0

C%N0

M20 2

"t#2555

$z

$yk(t, y)

555 dt.

The Gronwall inequality shows that one can find K = K(N0, M0) such that55 (z(yk

(!, y)| " K, for all ! $ R± and |y| " 2.Using again (5.2.28) and (5.2.29) we see that

555$z

$yk(!, y) ' Y (!, 0) ek

555 "- &

0

CN0 M20 · K2

"t#2 dt " C(M0, N0) 2.

Finally the estimate on $,y z, which is true for |3| = 0, 1 by the above results, can be

easily obtained by induction on |3| using (5.2.27), (5.2.21) and the Gronwall Lemma.

In the sequel we shall take 2 so small that C(M0, N0) 2 " 12 .

Let us now consider the map

(5.2.30)7

# : R± ( {y $ Cn : |y| " 2} '. R± ( Cn,(!, y) 0'. (!, z(!, y)).

We claim that # is injective. Indeed for a fixed ! $ R± if we have yj , j = 1, 2 suchthat |yj| " 2 and z(!, y1) = z(!, y2) then

0 =n#

k=1

- 1

0

$z

$yk(!, t y1 + (1 ' t) y2)(yk

1 ' yk2 ) dt.

Since |t y1+(1't) y2| " 2 when t $ [0, 1] we can use the estimate given in Lemma 5.2.3to ensure that

|Y (!, 0)(y1 ' y2)| " C%(N0, M0) 2 |y1 ' y2|.According to (5.2.25) this implies that y1 = y2 if 2 is small enough.

It follows that # is bijective on its range. We show now that

(5.2.31)7

If & is small enough we haveR± ( {z $ Cn : |z| " &} , #(R± ( {y $ Cn : |y| " 2}).

This equivalent to show that for fixed ! $ R±,

(5.2.32)7

for all z $ Cn such that |z| " & there exists y $ Cn

such that |y| " 2 and z(!, y) = z.

According to (5.2.27) the equation to solve is equivalent to the equation y = F (y)where

(5.2.33) F (y) = Y (!, 0)!1 z ' Y (!, 0)!1

- &

0Y (!, t) g(t, z(t, y), #) dt.



Let B = {y $ Cn : |y| " 2}. We shall show that if & and 2 are small enoughcompared to N0, M0 then F maps B into B and there exists (0 < 1 such that|F (y1) ' F (y2)| " (0 |y1 ' y2| for all y1, y2 in B. Then (5.2.32) will follow from thefixed point Theorem. Since Y (!, 0)!1 = Y (0, !) and Y (0, !)Y (!, t) = Y (0, t) it followsfrom (5.2.24) and (5.2.21) that if |z| " & we have

|F (y)| " M0 & +555- &

0

C(M0) 22

"t#2 dt555

since |z(t, y)| " 2M0 2 by Lemma 5.2.3. Then |F (y)| " 2 if 2 is small enough in termsof M0 and M0 & " 1

2 2.Moreover if y1, y2 belong to B we have

|F (y1) ' F (y2)| "555- &

0

C(M0, N0)"t#2 2 |z(t, y1) ' z(t, y2)| dt

555.

Since by Lemma 5.2.3 we have |z(t, y1)' z(t, y2)| " C%(M0)|y1 ' y2| we obtain finally

|F (y1) ' F (y2)| " C%(M0, N0) 2 |y1 ' y2|.

Taking 2 small enough we obtain (5.2.32).We can now straighten the vector field L0 which is the principal part of L given

in (5.2.16). Let us make the change of variables, (!%, y) 0. (!, z(!, y)). Then we have,according to (5.2.26)

$

$!%=

$

$!+

n#

j=1

zj(!, y)$

$zj=

$

$!+

n#

j=1

hN0j (!, z(!, y))

$

$zj= L0.

In the new coordinates (!%, y) the operator L has therefore the form

L =$

$!%+ dN0(!%, z(!%, y), #) =

$

$!%+ 2d(!%, y, #).

Now we note that$

$!%

+e

R &#0

ed(t,y,")dt u(t, y, #),

= eR &#0

ed(t,y,")dt Lu(t, y, #).

It follows that the problem

L 2A0 = 0, 2A0(0, y, #) = 1

has the (unique) solution 2A0(!%, y, #) = e!R &#0

ed(t,y,") dt. By the same way the problems

L 2A* = i 2Q 2A*!1, 2A*(0, y, #) = 0, / = 1, . . . , N0 + 1,

are solved by

2A*(!%, y, #) = e!R &#0

ed(t,y,") dt

- &#

0i( 2Q 2A*!1)(t, y, #) e

R t0

ed(s,y,") ds dt.

To end the proof of Lemma 5.2.2 we are left with the uniform estimates (ii).



First of all, using the estimate in (5.2.14), (5.2.15), Lemma 5.2.3 (ii) and the Faadi Bruno formula we see that,

(5.2.34) |$,y (dN0(!%, z(!%, y), #)| " C,

"!%#2 .

Denoting by 5(!, z) the inverse map of y 0. z(!, y), that is 5(!, z(!, y)) = y and usingLemma 5.2.3 we see that,

(5.2.35) |$,z 5(!, z)| " C, for ! ! 0 and |z| " &.

Then let us set for / = 0, . . . , N0 + 1

A*(!, z, #) = 2A*(!, 5(!, z), #).

Using (5.2.34), (5.2.35), (5.2.16), (5.2.15) and the estimate in (5.2.12) we see that(A*)*=0,...,N0+1 satisfy all the requirements of Lemma 5.2.2. This ends the proof ofTheorem 5.1.1 in the case of outgoing points.

We consider now the case of incoming points.

5.3. The case of incoming points

We assume here that # $ T #Rn is such that 12 " |##| " 2 and

(5.3.1) #x · #% " 'c0 "#x# |#%|.

Since such points belong to S! (see Definition 3.2.2) the case where ! " 0 is coveredby the Section 5.2. We focus now on the case ! ! 0. Here the method used inSection 5.2 does not work for many technical reasons. For instance, when |#x| is verylarge, #% = '#x/|#x| and ! = 1

2 |#x| we can see that "x(!, #)# is of magnitude one.Therefore we are far from the estimate "x(!, #)# ! 1)

2"!# used for instance to get

(5.2.3). Here also we shall use the method which consists to straighten the vectorfield X , defined in (5.1.3). This is done by a change of variables in (!, x) deducedfrom the flow of X . The problem here is that X has non real coe"cients (because of('(xj

) which are merely C". Therefore we are led to push the problem in the complexdomain by extending all the functions almost analytically as in [MS] for instance.So we begin our Section by a Lemma on almost analytic extensions adapted to oursituation. In that follows we shall consider together two cases. Case 1: % = Rn

x , case2: % = %+ (see Definition 4.1.1). We shall denote by X the variable in % that isX = x in the first case, X = (!, x) in the second one.



Lemma 5.3.1. — Let f be a function defined on % which is C" in X and satisfiesfor all X in %, in case 1 (resp. case 2),

(5.3.2)

'(()

((*

|f(X)| " M0

"x#!1

+resp. M0

+ 1"x#!1

+1

"!#!2

,,

#

|,|=k

|$,x f(X)| " Mk

"x#k+!3

+resp. Mk

+ 1"x#k+!3

+1

"!#k+!3

,, k ! 1

where (Mk)k"0 is an increasing sequence in ]0, +)[ and 0 " "1 " "3, 0 " "2 " "3.Then there exists F = F (X, y) defined on %(Rn

y which is C" in (X, y) and satisfiesfor all (X, y) in % ( Rn

y ,

(i) F (X, 0) = f(X).

(ii) |F (X, y)| " C0

"x#!1(resp. C0

$ 1"x#!1

+ 1'&(#2

%).

(iii) For every A, B in Nn with |A| + |B| ! 1 there exists CAB > 0 such that

|$Ax $B

y F (X, y)| " CAB

"x#|A|+|B|+!3(resp. CAB

$ 1"x#|A|+|B|+!3

+ 1'&(|A|+|B|+#3

%).

(iv) For every N $ N there exists CN > 0 such that for j = 1, . . . , n,

|$j F (X, y)| " CN

$ |y|"x#%N · 1

"x#1+!3(resp. CN |y|N

@$ 1"x# + 1

'&(%N$ 1

"x#1+!3+

1"!#1+!3

%A)

where $j = 12

$(

(xj+ i (

(yj

%.

Proof. — See Section A.4 in the Appendix.

Now recall that for # $ T #Rn such that #x · #% " 'c0 "#x#|##| and 12 " |##| " 2

we have constructed in Theorem 4.4.10 a function # = #(!, x, #) uniformly boundedon the set %+. By Lemma 5.3.1 we can extend # almost analytically as a function,which we denote by #(!, z, #), on the set(5.3.3)

%C+ =

:(!, z) $ R ( Cn : |z ' x(!, #)| " & "!#, Re z · #% " c0 "Re z#|#%|, | Im z| " &

;

and #(!, z, #) is still uniformly bounded on this set.Again by Lemma 5.3.1 one can extend almost analytically the coe"cients of our

symbol p, keeping the bounds of its coe"cients. In that follows for z $ Cn we shalldenote by X(t, !, z) the solution, whenever it exists, of the following problem.

(5.3.4)

')

*X(t, !, z) =

$p

$%(X(t, !, z), #(t, X(t, !, z), #)),

X(!, !, z) = z.

Our aim is to prove the following result.



Theorem 5.3.2. — One can find positive constants c1, &1, K, 2K, with c1 7 c0,&1 7 &, such that for all x $ Rn such that

|x ' x(!, #)| " &1 "!#, x · #% " c1 "x# |#%|,

the solution of (5.3.4) exists on [0, !] and satisfies the estimates,

(5.3.5)

'(((((((()

((((((((*

(i) |X(t, !, x) ' x(t, #)| " K |x ' x(!, #)| "t#"!# ,

(ii) | ImX(t, !, x)| " K|x ' x(!, #)|

"!#(iii) "x# + "! ' t# " K "Re X(t, !, x)#,

(iv) ReX(t, !, x) · #% " 12K"Re X(t, !, x)#|#%|,

uniformly for t $ [0, !].

Let us remark that the estimates (5.3.5) ensure in particular that if &1 is smallenough we have (t, X(t, !, #)) $ %C

+ . With 0 < c1 7 c2 7 c0 to be chosen, we dividethe proof in three cases.

– Case 1: x ·#% " c2 "x#|#%|, x(!, #) ·#% " c2 "x(!, #)#|#% |, |x'x(!, #)| " |x(!, #)|.– Case 2: x ·#% " c2 "x#|#%|, x(!, #) ·#% " c2 "x(!, #)#|#% |, |x'x(!, #)| > |x(!, #)|.– Case 3: x · #% " c1 "x#|#% |, x(!, #) · #% > c2 "x(!, #)#|#% |.Here is the geometrical interpretation of case 1 and 2. We denote by [a, b] the

segment joining two points a, b $ Rn.

Lemma 5.3.3. — Let c2 > 0 and assume that x $ Rn is such that x ·#% " c2 "x#|#%|,|x ' x(!, #)| " & "!# and that x(!, #) · #% " c2 "x(!, #)#|#% |. Then we have:

(i) either |x ' x(!, #)| " |x(!, #)| and then,

% y $ [x, x(!, #)], y · ## " 2 c2 "y#|#%|,

(ii) or |x ' x(!, #)| > |x(!, #)| and then,

[0, x(!, #)] 2 [0, x] , {y $ Rn : |y ' x(!, #)| " &0 "!# and y · #% " c2 "y#|#%|}.

Moreover|x| + |x(!, #)| " 3 |x ' x(!, #)| " 3 (|x| + |x(!, #)|).

Proof. — In the first case applying Lemma 4.4.16 we obtain for t $ [0, 1]

(tx + (1 ' t)x(!, #)) · #% " c2(t"x# + (1 ' t)"x(!, #)#)|#% |" c2(1 + t|x| + (1 ' t)|x(!, #)|)|#% |

" c2(1 +/

2|tx + (1 ' t)x(!, #)|)|#% |" 2 c2"tx + (1 ' t)x(!, #)#|#% |.



Assume now that |x ' x(!, #)| > |x(!, #)|. Then |x(!, #)| " &0"!#. Therefore [0, x] 2[0, x(!, #)] , B(x(!, #), &0"!#). Now if t $ [0, 1] and Z = x or x(!, #) we havet Z · #% " t c2"Z#|#%| " c2 "t Z#|#%|. Moreover

3 |x ' x(!, #)| = |x ' x(!, #)| + 2 |x ' x(!, #)| ! |x|' |x(!, #)| + 2 |x(!, #)|= |x| + |x(!, #)|.

1) Proof of Theorem 5.3.2 in case 1 and 2. — Let us take c2, &2 such that0 < c2 7 c0, 0 < &2 7 &. Let A be the set of T $ [0, !] such that for everyz $ Cn such that |z ' x(!, #)| " &2 "!#, Re z · #% " c2 "Re z#|#%|, | Im z| " &2,x(!, #) · #% " c2"x(!, #)#|#% | the problem (5.3.5) has a unique solution on [T, !] whichsatisfies for t $ [T, !], in case 1:

|X(t, !, z) ' x(t, #)| " M1 |z ' x(!, #)| "t#"!#(5.3.6)

"! ' t# + "Re(sz + (1 ' s)x(!, #))#(5.3.7)

" M1"s Re X(t, !, z) + (1 ' s)x(t, #)#, s $ [0, 1]

(s Re X(t, !, z) + (1 ' s)x(!, #)) · #%(5.3.8)

" M2 "s Re X(t, !, z) + (1 ' s)x(!, #)# |#% |

| Im X(t, !, z)| " M3

+ |z ' x(!, #)|"!# + | Im z|

,(5.3.9)

in case 2:

|X(t, !, z)' X(t, !, 0)| " M1 |z|"t#"!#(5.3.10)

"! ' t# + "Re(sz)# " M1"s Re X(t, !, z) + (1 ' s) Re X(t, !, 0)#, s $ [0, 1](5.3.11)$s Re X(t, !, z) + (1 ' s) Re X(t, !, 0)

%· #%(5.3.12)

" M2 "s Re X(t, !, z) + (1 ' s) Re X(t, !, 0)#

and (5.3.9).Our aim is to show that if M1, M2, M3 are correctly chosen then A = [0, !].Let us show that the set A is not empty. Indeed if t = ! the estimates (5.3.6) to

(5.3.12) are satisfied with strict inequalities if M1 > 1, M2 > 2 C2, M3 > 1 (usingLemma 5.3.3). It follows that they still hold for T = ! ' (, if ( is small enough.

On the other hand A is an interval. Let T# = inf A. If T# = 0 then the theorem5.3.2 is proved. Assume then that T# > 0 and let T ! T#. Then on [T, !], (5.3.6) to(5.3.12) hold.

Remark 5.3.4. — If the case 2 is not empty then the point z0 = 0 satisfies all therequirements of case 1. Indeed if there exists z1 such that |z1 ' x(!, #)| > |x(!, #)|then |x(!, #)| < &2 "!# so |0 ' x(!, #)| < &2 "!# and the other requirements are trivial.Therefore if the case 2 is not empty then X(t, !, 0) is well defined on [T, !] and satisfies(5.3.6) to (5.3.9).



Let us show that we have (t, X(t, !, z)) $ %C+ (see (5.3.3)). This is the case if

M1 &2 " &, M2 " c0, 2 M3 &2 " &. Indeed the only non trivial point is to prove that|X(t, !, z) ' x(t, #)| " & "t# in case 2. We have

|X(t, !, z) ' x(t, #)| " |X(t, !, z) ' X(t, !, 0)| + |X(t, !, 0) ' x(t, #)| = (1) + (2).

It follows from (5.3.10) that (1) " M1 |z| 't('&( and from (5.3.6) with z = 0, that

(2) " M1 |x(!, #)| 't('&( . Now by Lemma 5.3.3 (ii) we have |Re z| + |x(!, #)| < 3 |Re z

'x(!, #)|; since | Im z| " &2 we will have (1) + (2) " M1(&2 + 3 |Re z ' x(!, #)|) 't('&( .

Since |Re z ' x(!, #)| " &2 "!# we obtain finally (1) + (2) " 4 M1 &2 "t# " & "t#.In the sequel we shall denote by C or O(1) the constants which may depend on

bounds of p, # but are independent of M1, M2, M3. Moreover for the sake of simplicitywe shall write

(5.3.13)

6X(t) = X(t, !, z)2X(t) = x(t, #) in case 1, X(t, !, 0) in case 2.

In particular 2X(!) = x(!, #) in case 1 and 2X(!) = 0 in case 2. Our goal is to show thatthe estimates (5.3.6) to (5.3.12) hold on [T, !] with better constants than M1, M2, M3.

a) Improvement of (5.3.7) and (5.3.11). — By Theorem 4.4.10 (iii) we have#(!, x, #) ' #% = O(( + &) if (!, x) $ %+ and by Lemma 5.3.1 this estimate stillhold on %C

+ ; it follows that #(t, X(t), #) ' #% = O(( + &). On the other hand(p(% (x, %) ' 2% = O(()|%| which also extends for z $ Cn, | Im z| " &2. It follows thenfrom (5.3.4) that X(t) = 2## + O(( + &). Therefore

(5.3.14)

6X(t) = z ' 2(! ' t)#% + O(( +

/&)(! ' t)

2X(t) = 2X(!) ' 2(! ' t)#% + O(( +/

&)(! ' t).

Now for s $ [0, 1]

(1) = |Re(sX(t) + (1 ' s) 2X(t))|2 = |s Re z + (1 ' s) Re 2X(!)|2 + 4(! ' t)2 |#%|2

' 4(! ' t)(s Re z + (1 ' s) Re 2X(!)) · #%

+ O((( + &)@(! ' t)2 + |s Re z + (1 ' s) Re 2X(!)|2

A.

It follows from the conditions on z and the definition of 2X(!) that

(s Re z + (1 ' s) Re 2X(!)) · ## " 2c2"s Re z + (1 ' s) Re 2X(!)# |#% |

so

(1) ! 12|s Re z+(1's) Re 2X(!)|2+3(!'t)2 |#%|2'8c2"s Re z+(1's) Re 2X(!)#(!'t)|#% |

if ( + & is small enough. It follows that

(1) !+1

2' 16 c2

,|s Re z + (1 ' s) Re 2X(!)|2 + (3 ' 16 c2)(! ' t)2 |#%|2 ' 16 c2.



If c2 has been chosen small enough we obtain in particular(5.3.15)'()

(*

(i) |Re(s X(t) + (1 ' s) 2X(t))|2 ! 14 |Re(s z + (1 ' s) 2X(!)|2

+2(! ' t)2 |#%|2 ' 12

(ii) "Re(s X(t) + (1 ' s) 2X(t))#2 ! 110 ["! ' t#2 + "Re(s z + (1 ' s) 2X(!))#2].

This improves (5.3.7) and (5.3.11) if M1 > 4.

b) Improvement of (5.3.8), (5.3.12). — It follows from (5.3.14) that

(2) = Re(s X(t) + (1 ' s) 2X(t)) · #%

= Re(s z + (1 ' s) 2X(!)) · #% ' 2(! ' t)|#%|2 + O(( + &)(! ' t).

Applying Lemma 5.3.3 we obtain if ( + & is small,

(2) " 2 c2 "Re(s z + (1 ' s) 2X(!))# ' (! ' t)|#%|2.

Using (5.3.15) (i) we obtain, (2) " 4 c2"Re(s X(t) + (1' s) 2X(t)#. Taking 16 c2 " M2

we deduce finally that

(5.3.16) Re(s X(t) + (1 ' s) 2X(t)) · #% " 12

M2"Re(s X(t) + (1 ' s) 2X(t))# |#%|.

This improves (5.3.8) and (5.3.12).

c) Improvement of (5.3.6) and (5.3.10). — We have'()

(*

X(t) =$p

$%(X(t), #(t, X(t), #)),

2X(t) =$p

$%( 2X(t), #(t, 2X(t), #)),

the second equation being true in the case 1 according to the fact that the identity#(t, x(t, #), #) = %(t, #). Let us set

(5.3.17) Z(t) = X(t) ' 2X(t).

Then

Z(t) = 2@#(t, X(t), #) ' #(t, 2X(t), #)

A+

$q

$%(X(t), #(t, X(t), #))

' $q

$%( 2X(t), #(t, 2X(t), #)),

since p = |%|2 + q.Now we use (4.4.39) and Theorem 4.4.2 (i). It follows after extending almost

analytically 2a,2b and the coe"cients of q by Lemma 5.3.1,

(5.3.18) #(t, z, #) = %(t, #) +z ' x(t, #)

2t ' i'+2a +

i

"t#2b,(t, z, #) .



It follows then that(5.3.19)'(((((()

((((((*

Z(t) =2Z(t)2t ' i

'$2a(t, X(t), #) ' 2a(t, 2X(t), #)

%' i

"t#$2b(t, X(t), #) '2b(t, 2X(t), #)

%

+$q

$%

$X(t), #(t, X(t), #)

%' $q

$%( 2X(t), #(t, X(t), #)) +

$q

$%( 2X(t), #(t, X(t), #))

' $q

$%

$ 2X(t), #(t, 2X(t), #)%.

We have the following lemma.

Lemma 5.3.5. — One can find a positive constant C such that555Z(t) ' 2Z(t)

2t ' i

555 " C(( + &) |Z(t)|+ 1"! ' t#2 +

1"t#2,.

Proof(i) Estimation of (1) = 2a(t, X(t), #) ' 2a(t, 2X(t), #). We have

(1) =- 1

0

$2a$z

(t, s X(t) + (1 ' s) 2X(t), #)(X(t) ' 2X(t))ds

+- 1

0

$2a$z

(t, s X(t) + (1 ' s) 2X(t), #) (X(t) ' 2X(t)) ds.

Using the estimates on 2a given in Theorem 4.4.2 and Lemma 5.3.1 with "3 = 1 wefind

+555$2a$z

555+555$2a$z

555,(t, · · · ) " C(( + &)

+ 1"Re(s X(t) + (1 ' s) 2X(t))#2

+1

"t#2,.

Using (5.3.7) and (5.3.12) we deduce that555$2a$z

555+555$2a$z

555(t, · · · ) " C(M1)(( + &)+ 1"! ' t#2 +

1"t#2,.

It follows that,

(5.3.20) |(1)| " C(M1)(( + &) |Z(t)|+ 1"! ' t#2 +

1"t#2,.

Here C(M1) is a constant depending only on M1.(ii) Setting (2) = 1

't( (2b(t, X(t), #) '2b(t, 2X(t), #)) we have exactly by the same way

(5.3.21) |(2)| " C(M1)(( + &)|Z(t)|"t#

+ 1"! ' t#2 +

1"t#2,.

(iii) Estimation of (3) = (q(% (X(t), #(t, X(t), #)) ' (q

(% ( 2X(t), #(t, X(t), #)). We have

(3) =- 1

0

$2q

$z $%(s X(t) + (1 ' s) 2X(t), #(t, X(t), #))(X(t) ' 2X(t)) ds

+ analogue term with$2q

$z $%.



Now the coe"cients of q say bjk, extended by Lemma 5.3.1 satisfy555$bjk

$z(z)555+555$bjk

$z(z)555 " C (

"z#2+!0.

Using again (5.3.7) and (5.3.13) we obtain

(5.3.22) |(3)| " C (|Z(t)|"! ' t#2+!0

.

(iv) Estimation of (4) = (q(% ( 2X(t), #(t, X(t), #)) ' (q

(% ( 2X(t), #(t, 2X(t), #)). We have,by (5.3.18),

|#(t, X(t), #) ' #(t, 2X(t), #)| " C|Z(t)|"t# + |(1)| + |(2)|.

On the other hand (5.3.7), (5.3.11) with s = 0 imply that M1 "Re 2X(t)# ! "! ' t#.Therefore using the decay of the coe"cients bjk of q and the estimates (5.3.20), (5.3.21)we obtain

|(4)| " C (

"! ' t#1+!0|Z(t)|

+ 1"t# +

1"! ' t#2

,.

It follows then that,

(5.3.23) |(4)| " C ( |Z(t)|+ 1"t#2 +

1"! ' t#2

,.

Gathering the estimates (5.3.20) to (5.3.23) we obtain the claim of the Lemma.

Next we state the following Lemma.

Lemma 5.3.6. — Let 0 " T < !. Let Y (t) = (Y1(t), . . . , Yn(t)) $ Cn be such thatY $ C1([T, !]) and satisfies on [T, !] the inequality

555Y (t) ' 22t ' i

Y (t)555 " |h(t)| |Y (t)| + |g(t)|,

for some continuous functions h, g. Then for all t in [T, !] we have

|Y (t)| "+ "2t#"2!# |Y (!)| + "2t#

- &

T

|g(s)|"2s# ds

,exp+- &

T|h(s)| ds

,.

Proof. — Let us set W (t) = Y (t)2t!i . Then |W (t)| = |Y (t)|

'2t( ,

W (t) =Y (t)2t ' i

' 2Y (t)(2t ' i)2

=1

2t ' i

+Y (t) ' 2Y (t)

2t' i

,.

It follows that |W (t)| " 1'2t( (|h(t)| |Y (t)| + |g(t)|) " |h(t)| |W (t)| + |g(t)|

'2t( . Then, fort ! T and " $ [t, !],

|W (")| " |W (!)| +- &

!|h(s)| |W (s)| ds +

- &

t

|g(s)|"2s# ds.



By the Gronwall Lemma we obtain

|W (t)| "+|W (!)| +

- &

T

|g(s)|"2s# ds

,exp+- &

T|h(s)| ds

,.

Coming back to Y (t) we obtain the claim of the Lemma.

Corollary 5.3.7. — Let Z(t) be defined by (5.3.17). Then if (+& is small comparedto M1 we have

|Z(t)| " 2"2t#"2!# |z ' 2X(!)|.

Proof. — We apply Lemma 5.3.6 and Lemma 5.3.5 with

')

*g(t) = 0

h(t) = C(( + &)$ 1"! ' t#2 +

1"t#2%.

Then- &

T|h(s)| ds " C(( + &)

-

R

d"

""#2 .

It follows that

|Z(t)| " eC#(#++) · "2t#"2!# |Z(!)|.

Since Z(!) = X(!) ' 2X(!) = z ' 2X(!) our lemma follows.

We can now show the improvement of (5.3.6) and (5.3.10). In the case 1 wehave 2X(!) = x(!, #) and in the case 2, 2X(!) = 0. Therefore in case 1 we find byCorollary 5.3.7,

|X(t, !, z)' x(t, #)| " 4"t#"!# |z ' x(!, #)|,

and in case 2,

|X(t, !, z) ' X(t, !, 0)| " 4"t#"!# |z|.

Taking M1 > 4 this shows that (5.3.6) and (5.3.10) have been improved.



d) Improvement of (5.3.9). — Let us set(5.3.24)'((((((((((((((((((((((((((((((((((()

(((((((((((((((((((((((((((((((((((*

U(t) = Im(X(t) ' 2X(t)),

(1) =4t

1 + 4t2U(t),

(2) =2 Re(X(t) ' 2X(t))

1 + 4t2,

(3) = ' Im[2a(t, X(t), #) ' 2a(t, 2X(t), #)],

(4) = ' 1"t# Re[2b(t, X(t), #) '2b(t, 2X(t), #)],

(5) = Im.$q

$%(Re X(t), #(t, X(t), #)) ' $q

$%(Re X(t), #(t, 2X(t), #))

/,

(6) = Im.$q

$%(Re X(t), #(t, 2X(t), #)) ' $q

$%(Re 2X(t), #(t, 2X(t), #))

/,

(7) = Im- 1

0

$2q

$% $z(Re X(t) + is Im X(t), #(t, X(t), #)) ds (i Im X(t)),

(8) = Im- 1

0

$2q

$% $z(Re X(t) + is Im X(t), #(t, X(t), #)) ds ('i Im X(t)),

(9) = ' Im- 1

0

$2q

$% $z(Re 2X(t) + is Im 2X(t), #(t, 2X(t), #)) ds (i Im 2X(t)),

(10) = ' Im- 1

0

$2q

$% $z(Re 2X(t) + is Im 2X(t), #(t, 2X(t), #)) ds ('i Im 2X(t)).

Then it follows from (5.3.17) and (5.3.19) that,

(5.3.25) U(t) =4t

1 + 4t2U(t) +

10#

i=2

(i).

Lemma 5.3.8. — With the above notations, if ( + & is small enough we have555U(t) ' 4t

1 + 4t2U(t)

555 " 3 M1 |z ' x(!, #)|"!#"t# +

|z ' x(!, #)|"!#

+ 1"! ' t#1+!0

+1

"t#2,

++ 1"! ' t#2 +

1"t#2,|U(t)|.

Proof. — We use (5.3.25) and (5.3.24). We estimate the terms (i) for i = 2, . . . , 10.(i) Estimation of (2). It follows from (5.3.6) and (5.3.10), since 2X(!) = x(!, #) in

case 1 and 2X(!) = 0 in case 2, that

|X(t) ' 2X(t)|1 + 4t2

" M1"t#

"!#(1 + 4t2)|z ' 2X(!)| " M1

"!#"t#

6|z ' x(!, #)|, case 1|z|, case 2

.

But in case 2 according to Lemma 5.3.3 (ii) we have |z| " 3 |z ' x(!, #)|. It followsthat in both cases we have

(5.3.26) |(2)| " 3 M1 |z ' x(!, #)|"!#"t# .



(ii) Estimation of (3) and (4). We note that 2a(t, Re X(t), #) and 2a(t, Re 2X(t), #)are real. It follows that

(3) = ' Im- 1

0

$2a$z

(t, Re X(t) + s i Im X(t), #) ds(i Im X(t))

' Im- 1

0

$2a$z

(idem)ds('i Im X(t))

+ Im- 1

0

$2a$z

(t, Re 2X(t) + s i Im 2X(t), #) ds (i Im 2X(t))

+ Im- 1

0

$2a$z

(idem) ds ('i Im 2X(t)) .

Now, according to Theorem 4.4.2 and Lemma 5.3.1 we have555$2a$z

(t, w, #)555+555$2a$z

(t, w, #)555 " C (( + &)

+ 1"Re w#2 +

1"t#2,.

We use this estimate with w = ReX(t) + i s Im X(t) and w = Re 2X(t) + i s Im 2X(t).By (5.3.7) (with s = 1) and (5.3.11) (with s = 0) we have "Re w# ! 1

M1"! ' t#.

Moreover in case 1, Im 2X(t) = Im x(t, #) = 0 and in case 2,

(5.3.27) | Im 2X(t)| " M3|z|"!# " 3M5 |z ' x(!, #)|

"!# .

Summing up we obtain

(5.3.28) |(3)| " C (( + &)+ 1"t#2 +

M2+!01

"! ' t#2,+

|U(t)| + M5|z ' x(!, #)|

"!#

,,

since | Im X(t)| " |U(t)| + | Im 2X(t)|.For the term (4), due to the factor 1

't( we have a better estimate. Indeed by (5.3.21),(5.3.17), (5.3.6) and (5.3.10) we have

(5.3.29) |(4)| " C (M1) (( + &)|z ' x(!, #)|

"!#

+ 1"! ' t#2 +

1"t#2,.

(iii) Estimation of (5). We note here that (q(% (x, %) is linear in % and real if

(x, %) $ Rn ( Rn. It follows that

(5) =$q

$%

$Re X(t), Im(#(t, X(t), #) ' #(t, 2X(t), #))

%.


Im(#(t, X(t), #) ' #(t, 2X(t), #)) =Re(X(t) ' 2X(t))

1 + 4t2+

2t

1 + 4t2Im(X(t) ' 2X(t))

' Im@2a(t, X(t), #) ' 2a(t, 2X(t), #)

A' 1

"t# Re@2b(t, X(t), #) '2b(t, 2X(t), #)

A.



By (5.3.6) and (5.3.10) we have |X(t) ' 2X(t)| " 3M1 |z ' x(!, #)| 't('&( . Moreover

we can use (5.3.28) and (5.3.29). Finally we use the fact that the coe"cients of(q(% (Re X(t), · · · ) are bounded by C #

'Re X(t)(1+#0 which by (5.3.7) (with s = 1) and(5.3.11) (with s = 1) can be estimated by C #

'&!t(1+#0 . Gathering these informations wesee that

(5.3.30) |(5)| " C (

"! ' t#1+!0

+ 1"t#2 +

1"! ' t#2

,|U(t)| + C(M1 (|z ' x(!, #)|

"! ' t#1+!0"!#

+ C ( C(M1, M3)(( + &)+ 1"t#2 +

1"! ' t#2

, |z ' x(!, #)|"!# .

(iv) Estimation of (6). Since again (q(% (x, %) is linear in % and real when (x, %) $ Rn(Rn

we can write

|(6)| "555$q

$%(Re X(t), Im #(t, 2X(t), #))

555 +555$q

$%(Re 2X(t), Im #(t, 2X(t), #))

555.

Since the coe"cients of (q(% are bounded by C #

'&!t(1+#0 we obtain

|(6)| " C (

"! ' t#1+!0| Im #(t, 2X(t), #)|.

In the case 1, 2X(t) = x(t, #) which implies that Im#(t, 2X(t), #) = 0. In the case 2,| Im #(t, 2X(t), #)| " M3

|x(&,")|'&( " 3 M3

|z!x(&,")|'&( . Therefore

(5.3.31) |(6)| " C M3 ( |z ' x(!, #)|"! ' t#1+!0 "!# .

(v) Estimation of (7), (8), (9), (10). Using (5.3.7) and (5.3.11) and the estimates onthe coe"cients of q we find that

|(7) + (8) + (8) + (10)| " C (

"! ' t#2+!0(|U(t)| + | Im 2X(t)|).

Using (5.3.28) we obtain finally

(5.3.32) |(7) + (8) + (9) + (10)| " C (

"! ' t#2+!0(|U(t)| + 3M3 |z ' x(!, #)|

"!#

,.

Gathering the estimates given by (5.3.26) to (5.3.32) and taking (+ & small comparedto M1, M3 we obtain the conclusion of Lemma 5.3.8.

Lemma 5.3.9. — Let Y (t) = (Y, (t), . . . , Yn(t)) be a C1 function from [T, !] to Rn

which satisfies555Y (t) ' 4t

1 + 4t2Y (t)

555 " |h(t)| |Y (t)| + |g(t)| + K

"2t#for some continuous functions h, g and K ! 0. Then

|Y (t)| "+ "2t#"2!# |Y (!)| +

- &

T|g(s)| ds + K

,exp+- &

T|h(s)| ds

,.



Proof. — Let us set Z(t) = Y (t)'2t( . Then Z(t) = Y (t)

'2t( ' 4t'2t(3 Y (t). It follows that

|Z(t)| " |h(t)| |Z(t)| + |g(t)|"2t# +

K

1 + 4t2.

Therefore for " $ [t, !], t ! T we have

|Z(")| " |Z(!)| +- &

!|h(s)| |Z(s)| ds +

- &

t

|g(s)|"2s# ds + K

- &

t

ds

1 + 4s2.

Now we have,- +"

t

ds

1 + 4s2" 1

"2t# ,- &

t

|g(s)|"2s# ds " 1

"2t#

- &

t|g(s)| ds.

Using Gronwall’s Lemma we obtain

|Z(")| "+|Z(!)| + 1

"2t#

- &

t|g(s)| ds +

K

"2t#

,exp+ - &

t|h(s)| ds

,.

Taking t = T and " = t we obtain, since Y (t) = "2t#|Z(t)|, the claim of the Lemma.

Corollary 5.3.10. — With U(t) = Im(X(t)' 2X(t)) introduced in (5.3.24) we have

|U(t)| " C+|U(!)| + (6 M1 + C)

|z ' x(!, #)|"!#

,.

Proof. — This follows from Lemmas 5.3.8 and 5.3.9.

We can now finish the proof of the improvement of (5.3.9).Indeed we have U(!) = Im(X(!, !, z) ' 2X(!, !, 0)) = Im z. Therefore Corol-

lary 5.3.10 and Remark 5.3.4 show that if C < M1 and (6 M1 + C) · C < M3 then(5.3.9) is improved.

End of the proof of Theorem 5.3.2 in the cases 1 and 2. — The estimates (5.3.5) to(5.3.12) improved are true for t $ [T, !] for all T > T#. By continuity they continueto hold on [T#, !]. Now we consider problem (5.3.5) with data at t = T# equal toX(T#, !, #). For this problem the estimates (5.3.6) to (5.3.12) hold on [T# ' (0, T#]which contradicts the fact that T# = inf A. Therefore A = [0, !] which implies Theo-rem 5.3.2 in this case.

2) Proof of Theorem 5.3.2 in case 3. — Here we shall take x $ Rn such thatx · #% " c1 "x#|#% | and |x ' x(!, #)| " &1 "!#, with 0 < c1 7 c2, 0 < &1 7 &2.

Let us recall (see (4.4.49)) that there exists a unique !# $ [0, !] such that x(!#, #) ·#% = 0. We shall make use of Lemma 4.4.17. To prove the claim of Theorem 5.3.2we shall use the same method as in the cases 1 and 2.



We introduce first the set A of T ! !# such that the problem (5.3.4) has a solutionon [T, !] which satisfies

|X(t, !, x) ' x| " M4 |t ' !|(5.3.33)

ReX(t, !, x) · #% " M5 "Re X(t, !, x)# |##|(5.3.34)

"x# + "t ' !# " M4 "Re X(t, !, x)#(5.3.35)

| Im X(t, !, x)| " M6|x ' x(!, #)|

"!# .(5.3.36)

If M4 is large enough, M5 > c1, M6 > 0 one can find (0 > 0 such that !' (0 $ A. LetT# = inf A. We want to prove that T# = !# if M4, M5, M6 are correctly chosen; letus assume T# > ! and let t ! T#. Then on [T, !] we have a solution X(t, !, z) whichsatisfies (5.3.33) to (5.3.36). Let us show that this implies that (t, X(t, !, x)) $ %C

+ fort $ [T, !] (see (5.3.3)) if &1 is small enough.

From (5.3.36) we have | Im X(t, !, z)| " M6 &1 " & if &1 is small enough. Moreoverby (5.3.34) we have

Re X(t, !, x) · #% " M5 "Re X(t, !, x)# |##| " c0"Re X(t, !, x)# |##|

if M5 " c0. Finally,

(1) = |X(t, !, x) ' x(t, #)| " |X(t, !, x) ' x| + |x ' x(!#, #)| + |x(!#, #) ' x(t, #)|.

From (5.3.33) we have |X(t, !, x) ' x| " M4 |t ' !| " M4(! ' !#) since t ! !#. Nowwe use Lemma 4.4.17 to write

|X(t, !, x) ' x| " 10 M4 |x ' x(!, #)| " 10 M4 &1"!# " 10 M4

K1&1"!## " 10 M4

K1&1"t#.

Again by Lemma 4.4.17,

|x ' x(!#, #)| " 6|x ' x(!, #)| " 6&1"!# " 6&1

K1"!## " 6&1

K1"t#.

Finally |x(t, #)'x(!#, #)| "1 t

&$ |x(s, #)| ds " 5(t' !#) if ( is small enough. It followsfrom Lemma 4.4.17 that

|x(t, #) ' x(!#, #)| " 5(! ' !#) " 50 |x ' x(!, #)| " 50 &1"!# " 50 &1

K1"t#.

Summing up we find that if &1 is small enough,

(5.3.37) (1) " max+10M4

K1,

56K1

,&1"t# " &"t#.

As in the cases 1 and 2 our goal is to prove that one can improve the estimates (5.3.33)to (5.3.36).



(i) Improvement of (5.3.33). — We have by (5.3.5), X(t, !, x) = 2#% + O(( + &).Therefore

(5.3.38) X(t, !, x) = x ' 2(! ' t)#% + O((( + &)(! ' t)).

It follows that |X(t, !, x) ' x| " 5(! ' t) if ( + & is small enough. We shall take M4

so that 5 " 12 M4 and then, (5.3.33) will be improved.

(ii) Improvement of (5.3.35). — We deduce from (5.3.38) that

1+ |ReX(t, !, z)|2 = 1+ |x|2+4(!'t)2 |#%|2+O(((+&)(|x|2+(!'t)2))'2(!'t)x·#%.

Since x · ## " c1"x#|##|, taking ( + & small enough we obtain, 1 + |Re X(t, !, z)|2 !1 + 1

2 |x|2 + 1

2 (! ' t)2 ' 4c1(! ' t)"x#, so

(5.3.39) 1 + |Re X(t, !, z)|2 ! 14("x#2 + (! ' t)2),

if c1 " 110 . In particular 3"Re X(t, !, x)# ! "! ' t# + "x#, so "! ' t# + "x# "

12 M4 "Re X(t, !, x)# if M4 ! 6.

(iii) Improvement of (5.3.34). — From (5.3.38) we have

Re X(t, !, x) · #% = x · #% ' 2(! ' t) |#%|2 + O((( + &)(! ' t)).

It follows that

ReX(t, !, x) · #% " c1"x# |#%|'(! ' t)

4" 10 c1 "Re X(t, !, x)# |#%|,

by (5.3.39). We shall take 10 c1 " 12 M5 and (5.3.34) will be improved.

(iv) Improvement of (5.3.36). — Let us set X(t, !, x) = X(t) = Y1(t) + i Y2(t) whereY1, Y2 are real.

Lemma 5.3.11. — There exists positive constants C, K independent of (, & and Tsuch that for all t $ [T, !] we have

|Y2(t)| " C M6

+ &1

"!# + (( + &)|x ' x(!, #)|

"!# g(t),

+&1

"!#

where g is a continuous positive function satisfying1 &

T g(s) ds " K.

Proof. — From (5.3.36) and (5.3.37) we get

(5.3.40) |X(t) ' x(t, #)| " C (M4) &1 "t#, |Y2(t)| " M6 &1.

Now (5.3.18) shows that

Im #(t, X(t), #) = Im.X(t) ' x(t)

2t ' i' 2a(t, X(t, X(t), #) ' i

"t#2b(t, X(t), #)

/

where x(t) = x(t, #). First of all we have,

ImX(t) ' x(t)

2t ' i=

2t Y2(t)1 + 4t2

+Y1(t) ' x(t)

1 + 4t2.



Using (5.3.40) we deduce, since "!# - "t#,

(5.3.41)555 Im

X(t) ' x(t)2t' i

555 " C M6&1

"!# + C&1

"!# .

On the other hand we can write with f = 2a or 2b,

f(t, X(t), #) = f(t, Y1(t), #) +- 1

0

$f

$z(t, Y1(t) + is Y2(t), #) ds(i Y2(t))

+- 1

0

$f

$z(t, Y1(t) + is Y2(t), #) ds ('i Y2(t)).

Since 2a(t, Y1(t), #) is real, using the estimates on the derivatives of 2a and 2b given byTheorem 4.4.2 and their extensions to the complex domain proved in Lemma 5.3.1we obtain

(5.3.42) | Im2a(t, X(t), #)| " C M6 (( + &)+ 1"! ' t#2 +

1"t#2, |x ' x(!, #)|

"!# .

Here we have used the estimate in (ii) and (5.3.36).Moreover we have by Theorem 4.4.2 (ii) and (5.3.37),

|2b(t, Y1(t), #)| "/

&|Y1(t) ' x(t)|

"t# "/

& C(M4) &1 " &1

if & is small enough. Therefore

(5.3.43)555 Im

i2b(t, X(t), #)"t#

555 " C&1

"!# + C M6

+ 1"! ' t#2 +

1"t#2, |x ' x(!, #)|

"!# .

We deduce from (5.3.41) to (5.3.43) that

(5.3.44) | Im #(t, X(t), #)| " C &1

"!# + C M6&1

"!# + M6|x ' x(!, #)|

"!# g(t)

where g(t) = C$

1'&!t(2 + 1

't(2%.


|Y2(t)| " 2 | Im#(t, X(t), #)| +555$q

$%(Y1(t), Im #(t, X(t), #))

555

+555- 1

0

$2q

$% $z(Y1(t) + is Y2(t), #(t, X(t), #)) ds

555 |Y2(t)|

+555- 1

0

$2q

$%$z(idem) ds

555 |Y2(t)|

" C | Im #(t, X(t), #| + C(

"! ' t#2+!0|Y2(t)|.

This estimate together with (5.3.36), (5.3.44) prove the Lemma.



We can now improve (5.3.36). Indeed, by Lemma 5.3.11, we have, sinceX(!, !, x) = x is real

|Y2(t)| "- &

t|Y2(s)| ds " C M6

.&1(! ' t)"!# + (( + &)K

|x ' x(!, #)|"!#

/+ &1

! ' t

"!# .

Moreover by Lemma 4.4.17 we have

! ' t " ! ' !# " C |x ' x(!, #)|.

Taking &1, &, ( small enough we obtain |Y2(t)| " 12 M6

|x!x(&,")|'&( which improves

(5.3.36).The improvements (i) to (iv) show that the set A where (5.3.33) to (5.3.36) are

true is equal to [!#, !].We can now give the proof of Theorem 5.3.2 in the case 3. Indeed (5.3.34) to

(5.3.36) imply the estimates (ii) to (iv) in this Theorem. To prove (i) we just remarkthat

|X(t, !, x) ' x(t, #)| " |X(t, !, x) ' x| + |x ' x(!, #)|

" (10M4 + 1)|x ' x(!, #)| " C |x ' x(!, #)| "t#"!#

since "t# - "!# when t $ [!#, !]. Therefore we are done for t $ [!#, !]. For t $ [0, !#]we first remark that

X(t, !, x) = X(t, !#, X(!#, !, x)).

We would like to apply the cases 1 and 2 already done, with ! = !# and z = X(!#, !, x).So we have to prove that

(i) x(!#, #) · #% " c2 "x(!#, #)# |#%|,(ii) |z ' x(!#, #)| " &2 "!##,(iii) Re z · #% " c2 "Re z# |#%|,(iv) | Im z| " &2.

First of all (i) is trivial since x(!#, #) · ## = 0. Now we have,

|X(!#, !, x) ' x(!#, #)| " |X(!#, !, x) ' x| + |x ' x(!#, #)| = (1) + (2).

By (5.3.33) and Lemma 4.4.17 we have if &1 7 &2

(1) " M4(! ' !#) " 10 M4 |x ' x(!, #)| " 10 M4 &1"!# " C &1"!## " &2

2"!##

(2) " 5 |x ' x(!, #)| " C% &1"!## " &2

2"!##

since "!# - "!##. Thus (ii) is satisfied. Now (iii) is also satisfied if M5 " c2. Thisis possible since the only constraint on M5 (see (iii) improvement of (5.3.34)) wasM5 ! 20 c1. Finally by (5.3.36), | Im X(!#, !, x)| " &1 M6 " &2 if &1 7 &2. Therefore



X(t, !#, X(!#, !, x)) satisfies the estimates (5.3.5) to (5.3.9) in case 1 and (5.3.9) to(5.3.12) in case 2. Therefore we have the following estimate,

(1) = |X(t, !#, X(!#, !, x)) ' x(t, #)| " 3 M1 |X(!#, !, x) ' x(!#, #)| "t#"!##

(1) " 3 M1"t#"!## (|X(!#, !, x) ' x| + |x ' x(!#, #)|)

(1) " 3 M1"t#"!## (M4(! ' !#) + |x ' x(!#, #)|)

(1) " C M1(1 + M4)|x ' x(!, #)| "t#"!# .

Here we have used (5.3.33), Lemma 4.4.17 and "!## - "!#. Therefore we obtain (i) of(5.3.5) if K ! C M1(1 + M4). Now (5.3.9) implies that

(2) = | Im X(t, !#, X(!#, !, x))| " M3

+ |X(!#, !, x) ' x(!#, #)|"!## + | Im X(!#, !, x)|

,.

Using (5.3.36) and the same argument as in the term (1) we obtain

(2) " C(M1, M4, M6)|x ' x(!, #)|

"!# .

Thus (ii) in (5.3.5) is satisfied if K ! C(M1, M4, M6).From (5.3.15) (ii) with s = 1 we have

(3) = "Re X(t, !#, X(!#, !, x))# ! 15

["!# ' t# + "Re X(!#, !, x)#].

So using (5.3.35) we obtain

(3) ! 15

."!# ' t# +

1M4

"!# ' !#/

! C(M4)"! ' t#,

and (iii) satisfied K · C(M4) ! 1.Finally let us set (4) = Re X(t, !#, X(!#, !, x)) ·#% . Using (5.3.8) and (5.3.12) with

s = 1 we can write,(4) " M2 "Re X(t, !#, X(!#, !, x))#.

This shows that (5.3.5) (iv) holds if 2K ! 1M2

and completes the proof of Theorem 5.3.2.

Having proved in Theorem 5.3.2 the existence of the solution X(t, !, x) of (5.3.4)we want to give estimates on its derivatives with respect to (!, x).

Proposition 5.3.12. — The solution given by Theorem 5.3.2 is C" with respect toy = (!, x) and satisfies the following estimates,

(5.3.45) |$Ay X(t, !, x)| "

'(()

((*

C"t#"!# if |A| = 1,

CA"t#"!#

+ 1"x#|A|+!0

+1

"!#|A|!1

,if |A| ! 2.

.

uniformly in (t, !, x) $ [0, !] ( %+.



To prove this result we need a Lemma.

Lemma 5.3.13. — For j = 1, . . . , n let us set Lj(t, z) = (p(%j

(z, #(t, z, #)). Then forany integer N > 0 one can find CN > 0 such that for j = 1, . . . , n and all (t, !, x) in[0, !] ( %+ we have

555$Lj

$zk(t, X(t, !, x)) ' 2&jk

2t ' i

555 " C1

+ 1("x# + "! ' t#)2+!0

+1

"t#2,

(5.3.46)

555$Lj

$zk(t, X(t, !, x))

555 " CN

+ 1("x# + "! ' t#)2+!0

+1

"t#2,+ |x ' x(!, #)|

"!#

,N.(5.3.47)

For any µ, ' $ Nn, such that |µ| + |'| = N ! 2, j = 1, . . . , n,

(5.3.48)555$µ+/ Lj

$zµ $z/ (t, X(t, !, x))555 " Cµ/

+ 1("x# + "! ' t#)|µ|+|/|+1+!0

+1

"t#|µ|+|/|+1

,.

Proof of Lemma 5.3.13. — We have

Lj(t, z) = 2 #j(t, z, #) + 2(n#

*=1

bj*(z)#*(t, z, #).


$#j

$zk(t, z, #) =

&jk

2t ' i'+$2aj

$zk+

i

"t#$2bj

$zk

,(t, z, #).

Then (5.3.46) follows easily from the estimates on 2a,2b given in Theorem 4.4.2, theestimates on the coe"cients bj* and from the inequality (iii) in Theorem 5.3.2.

The estimate (5.3.47) follows from the same arguments and Lemma 5.3.1 (iv),Theorem 5.3.2 (ii), (iii). The same method can also be used to prove (5.3.48).

Proof of Proposition 5.3.12. — Let us set for k = 1, . . . , n, q ! 1,

Y qk (t) = $A

x Xk(t, !, x)

where |A| = q.We begin by the case q = 1. Di!erentiating one time (5.3.4) with respect to y we

obtain

(5.3.49) Y 1k (t) =

n#

j=1

.$Lk

$zj(t, X(t, !, x))Y 1

j (t) +$Lk

$zj(t, X(t, !, x))Y 1

j (t)/.

Using (5.3.46) and (5.3.47) we see that Y 1(t) = (Y 11 (t), . . . , Y 1

n (t)) satisfies thehypotheses of Lemma 5.3.6 with g 3 0 and h(t) = 1

('x(+'&!t()2+#0 + 1't(2 . Since

(Xj

(xk(!, !, x) = &jk and (Xj

(& (!, !, x) is bounded we obtain (5.3.45) when |A| = 1. Letus consider the case |A| = 2. Di!erentiating (5.3.49) with respect to y we see thatY 2

k (t) satisfies the equation

Y 2k (t) =

n#

j=1

.$Lk

$zj(t, X(t, !, x))Y 2

j (t) +$Lk

$zj(t, X(t, !, x))Y 2

j (t)/

+ Zk(t, !, x)



where, by (5.3.48) and (5.3.45) for |A| = 1, Zk(t) is estimated as follows

|Zk(t, !, x)| " C"t#2

"!#2+ 1

("x# + "! ' t#)3+!0+

1"t#3,.

We want to use Lemma 5.3.6 (with T = 0) so we are led to estimate the quantity(1) =

1 &0

|Zk(s,&,x)|'2s( ds. Using the above estimation we see that

(1) " C% 1"!#2

- &

0

+ ""#("x# + "! ' "#)3+!0

+1

""#2,

d".

By a straightforward computation we see that we have

(5.3.50)- &

0

""#*

("x# + "! ' t#)k+!0d" " C

"!#*

"x#k!1+!0, k, / ! 1.

It follows that

(1) " C

"!#

+ 1"x#2+!0

+1"!#

,.

Using Lemma 5.3.6 and the fact that $Ax X(!, !, x) = 0, since |A| ! 2, we obtain

(5.3.45) when |A| = 2.Now we proceed by induction on q ! 2. Let |A| = q + 1 and let us di!erentiate the

equation Xk(t, !, x) = Lk(t, X(t, !, x)) |A| times with respect to x. Using the Faa diBruno formula and the notation Y q+1

k = $Ax Xk we obtain the equation

Y q+1k (t) =

n#

j=1

.$Lk

$zj(t, X(t, !, x))Y q+1

j (t) +$Lk

$zj(t, X(t, !, x))

/Y q+1

j (t) + Zk(t)

where Zk(t) is a finite linear combination of terms of the form

(2) =$$)(z,z) Lj

%(t, X(t, !, x))

sH

*=1

$$L$

x X(t, !, x)%K$

where 2" |.|"q + 1, 1"s"q + 1, |K*|!1, |L*|!1,&s

*=1 K* =.,&s

*=1 |K*|L* = A.It follows that |L*| " |A|' 1 = q.

Since by (5.3.45) we have di!erent estimates for |L*| = 1 and |L*| ! 2 we mustseparate these two cases. So let us write {1, . . . , s} = I1 2 I2, I1 = {/ : |L*| = 1},|I2| = {/ : |L*| ! 2}.

Now let us use (5.3.48) and the induction. We obtain

|(2)| " C+ 1

("x# + "! ' t#)|)|+1+!0+

1"t#|)|+1

, H

*+I1

+ "t#"!#

,|K$|

·H

*+I2

I"t#"!#

+ 1"x#|L$|+!0

+1

"!#|L$|!1

,J|K$|.



Since&s

i=1 |Ki| = |.| we have

H

*+I1

$ "t#"!#%|K$| H

*+I2

$ "t#"!#%|K$| =

"t#|)|

"!#|)|.

It follows from (5.3.50) that- &

0

|(2)|""# d" " C

"!#

+ 1"x#|)|+!0

+1

"!#|)|!1

, H

*+I2

+ 1"x#|L$|+!0

+1

"!#|L$|!1

,|K$|.

Now we haveH

*+I2

+ 1"x#|L$|+!0

+1

"!#|L$|!1

,|K$|"H

*+I2

+ 1"x#|L$|!1

+1

"!#|L$|!1

,|K$|

" CH

*+I2

+ 1"x#|K$|(|L$|!1)

+1

"!#|K$|(|L$|!1)

,

" C%I+ 1

"x#

,P$%I2

|K$|(|L$|!1)++ 1"!#

,P$%I2

|K$|(|L$|!1)J

" C%I+ 1

"x#

,|A|!|)|++ 1"!#

,|A|!|)|J

.

Indeed

|A|' |.| =s#

*=1

|K*| |L*|'s#

*=1

|K*| =#

*+I1

|K*| +#

*+I2

|K*| |L*|'#

*+I1

|K*|'#

*+I2

|K*|

=#

*+I2

|K*|(|L*|' 1).

It follows that- &

0

|(2)|""# d" " C

"!#

+ 1"x#|)|+!0

+1

"!#|)|!1

,+ 1"x#|A|!|)| +

1"!#|A|!|)|

,

" C%

"!#

+ 1"x#|A|+!0

+1

"!#|A|!1

,.

Then using Lemma 5.3.6 and the fact that $Ax X(!, !, x) = 0 since |A| ! 2 we obtain

(5.3.45) for |A| = q + 1.

We need another lemma. Let us recall that we have set Lj(!, x) = (p(%j

(x, #(!, x, #)).

Lemma 5.3.14. — Let uj(t) = (Xj

(& (t, !, x) +&n

k=1(p(%k

(x, #(!, x, #)) (Xj

(xk(t, !, x).

Then for every integer N > 0 one can find a constant CN > 0 such that for allt $ [0, !] and all (!, x) in %+ we have

|uj(t)| " CN

+ |x ' x(!, #)|"!#

,N.



Proof. — First of allwe claim that uj(!) = 0. Indeed since Xj(!, !, x)+ (Xj

(& (!, !, x)=0we have (Xj

(& (!, !, x) = 'Lj(!, x). Then our claim follows from the fact that(Xj

(xk(!, !, x) = &jk. Now

uj(t) =$Xj

$!(t, !, x) +

n#

k=1

Lk(!, x)$Xj

$xk(t, !, x).


uj(t) =n#

µ=1

.$Lj

$zµ(t, X(t, !, x))

$Xµ

$!(t, !, x) +

$Lj

$zµ(t, X(t, !, x))

$Xµ

$!(t, !, x)

/

+n#

k=1

Lk(!, x)n#

µ=1

.$Lj

$zµ(t, X(t, !, x))

$Xµ

$xk(t, !, x)

+$Lj

$zµ(t, X(t, !, x))

$Xµ

$xk(t, !, x)

/

uj(t) =n#

µ=1

$Lj

$zµ(t, X(t, !, x))uµ(t)

+n#

µ=1

$Lj

$zµ(t, X(t, !, x))

.$Xµ

$!(t, !, x) +

n#

k=1

Lk(!, x)$Xµ

$xk(t, !, x)

/.

It follows then from (5.3.45) that with u(t) = (u1(t), . . . , un(t)),555uj(t) '

2 uj(t)2t ' i

555 " |h(t)| |u(t)| + |g(t)|,

where '(((()

((((*

h(t) =n#

µ,j=1

555$Lj

$zµ(t, X(t, !, x)) ' 2 &jk

2t ' i

555,

g(t) = Cn#

µ,j=1

555$Lj

$zµ(t, X(t, !, x))

555.

Now using (5.3.46) and (5.3.47) we have1 &0 h(t) dt " C and

- &

0g(t) dt " C CN

|x ' x(!, #)|N

"!#N ,

so Lemma 5.3.14 follows from Lemma 5.3.6 since uj(!) = 0.

To solve the transport equations we need to introduce some notations. First of allwe shall set,(5.3.51)

D =!(!, z) $ R(Cn : |z'x(!, #)| " &1

K"!#, | Im z| " &1

K, Re z ·## " c1 "Re z# |##|

"

where &1, c1, K have been introduce in the statement of Theorem 5.3.2.



Let a $ C"(D). We introduce some possible estimates.7%µ, ' $ Nn, &Cµ,/,* ! 0 such thatfor all (!, z) $ D, |$µ

z $/z a(!, z)| " Cµ,/

(5.3.52)')

*

%N $ N, &CN ! 0 : % j = 1, . . . , n, % (!, z) $ D555$a

$zj(!, z)

555 " CN | Im z|N(5.3.53)

')

*

&"0 > 0 : %µ, ' $ Nn, &Cµ,/ ! 0 : % (!, z) $ D55$µ

z $/z a(!, z)

55 " Cµ,/

"Re z#1+!0.

(5.3.54)

We first state the following result.

Proposition 5.3.15. — Let u0 = u0(z) be a C" function in a neighborhood ofD0 = {z $ Cn : |z ' #x| " &1} such that for any N $ N one can find CN ! 0 suchthat for every j = 1, . . . , n and z $ D0,

555$u0

$zj(z)555 " CN | Im z|N .

For (!, x) $ R ( Rn, (!, x) $ D we set u(!, x) = u0(X(0, !, x)). Then for any N ! 0we can find C%

N ! 0 such that

(5.3.55)555$u

$!(!, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$u

$xk(!, x)

555 " C%N

+ |x ' x(!, #)|"!#

,N

(5.3.56) u(0, x) = u0(x)

(5.3.57)7

For any 3 $ Nn one can find C, ! 0 such that|$,

x u(!, x)| " C, for every (!, x) in D ! R ( Rn.

Proof. — First of all by (5.3.5) (i) we have for (!, x) $ D ! R ( Rn,

|X(0, !, x) ' #x| " K|x ' x(!, #)|

"!# " K&1

K= &1.

Therefore u(!, x) = u0(X(0, !, x)) is well defined and satisfies (5.3.57) by Proposi-tion 5.3.12, the fact that u0 is C" in a neighborhood of D0 and the Faa di Brunoformula (Chapter 7.2). Now since X(0, 0, x) = x, (5.3.56) is obvious. Let us check(5.3.55). We set

(1) =$u

$!(!, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$u

$xk(!, x).



Then

(1) =n#

j=1

$u0

$zj(X(0, !, x))

.$Xj

$!(0, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(0, !, x)

/

+n#

j=1

$u0

$zj(X(0, !, x))

.$Xj

$!(0, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(0, !, x)

/

and we write (1) = (A) + (B).By Lemma 5.3.14 with t = 0, the term (A) satisfies (5.3.55). By the hypothesis

made on u0 and (5.3.5) (ii) we have555$u0

$zj(X(0, !, x))

555 " CN | Im X(0, !, x)|N " C%N

+ |x ' x(!, #)|"!#

,N.

Using (5.3.45) and the fact that # and the coe"cients of p are bounded we deducethat the term (B) satisfies also (5.3.55).

Proposition 5.3.16. — Let a $ C" on D which satisfies (5.3.53), (5.3.54). Let usset

A(s, !, x) =- s

&a(", X(", !, x)) d"

for s $ [0, !] and (!, x) $ D ! R ( Rn. Then'(((((()

((((((*

for any N $ N one can find CN ! 0 such that555$A

$!(s, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$A

$xk(s, !, x) + a(!, x)

555

" CN

+|x!x(&,")|

'&(

,N

for all s $ [0, !] and (!, x) $ D ! R ( Rn.

(5.3.58)

A(!, !, x) = 0.(5.3.59)7

For every 3 $ Nn, there exists C, ! 0 such that|$,

x A(s, !, x)| " C, on [0, !] ( D ! R ( Rn.(5.3.60)

Proof. — The claim (5.3.59) is trivial, (5.3.60) follows from Proposition 5.3.12,(5.3.54) and (5.3.5) (iii). Let us show (5.3.58). We set

(1) =$A

$!(s, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$A

$xk(s, !, x) + a(!, x).

Then

(1) =- s

&

n#

j=1

$a

$zj(", X(", !, x))

L$Xj

$!(", !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(", !, x)

Md"

+- s

&

n#

j=1

$a

$zj(", X(", !, x))

L$Xj

$!(", !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(", !, x)

Md"

= (A) + (B).



By (5.3.54) and (5.3.5)(iii) we have55 (a(zj

(", X(", !, x))55 " C/"! ' "#1+!0 . Therefore

using Lemma 5.3.14 we obtain (5.3.58) for the term (A).Now it follows from (5.3.53) and (5.3.54) by interpolation that

555$a

$zj(!, z)

555 " C%%N,!0

| Im z|N

"Re z#1+!0/2.

Thus using (5.3.45), (5.3.5) (ii), (iii), we obtain (5.3.58) for the term (B) since # andthe coe"cients of p are uniformly bounded.

This is the last result before the final one solving the transport equations.

Proposition 5.3.17. — Let b be C" on D satisfying (5.3.52) and (5.3.53). Let usset B(s, !, x) = b(s, X(s, !, x)), s $ [0, !], (!, x) $ D ! R ( Rn. Then,

'((((()

(((((*

for every N ! 0 there exists CN ! 0 such that555$B

$!(s, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$B

$xk(s, !, x)

555

" CN

+ |x ' x(!, #)|"!#

,N,

(5.3.61)

B(!, !, x) = b(!, x),(5.3.62)7

for every 3 $ Nn, /, m $ N there exists C, ! 0 such that|$,

x B(s, !, x)| " C, , % (s, !, x) $ [0, !] ( D ! R ( Rn.(5.3.63)

Proof. — The claim (5.3.62) is obvious and (5.3.63) follows from Proposition 5.3.12and (5.3.52). Let us show (5.3.61). The left hand side of (5.3.61) can be written,

(1) =n#

j=1

$b

$zj(s, X(s, !, x))

L$Xj

$!(s, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(s, !, x)

M

+n#

j=1

$b

$zj(s, X(s, !, x))

L$Xj

$!(s, !, x) +

n#

k=1

$p

$%k(x, #(!, x, #))

$Xj

$xk(s, !, x)

M

= (A) + (B).

The estimation of (A) follows from Lemma 5.3.14 and (5.3.52). Now from (5.3.53),(5.3.5) (ii) and Proposition 5.3.12 we deduce the estimation of (B) since # and thecoe"cients of p are uniformly bounded.

Theorem 5.3.18. — Let a = a(!, z) be a C" function on D satisfying (5.3.53),(5.3.54). Let b = b(!, z) be a C" function on D satisfying (5.3.52), (5.3.53). Letu0 = u0(z) be a C" function on D0 satisfying the hypothesis of Proposition 5.3.15.With the notations of Propositions 5.3.15, 5.3.16 and 5.3.17 we set

v(!, x) =- &

0eA(s,&,x) B(s, !, x) ds + eA(0,&,x) u(!, x).



Then'((((((()

(((((((*

for every N ! 1 there exists CN ! 0 such that555$v

$!(!, x) +

n#

j=1

$p

$%j(x, #(!, x, #))

$v

$xk(!, x) + a(!, x) v(!, x) ' b(!, x)

555

" CN|x ' x(!, #)|N

"!#N!1

for all (!, x) $ D ! R ( Rn,

(5.3.64)

v(0, x) = u0(x),(5.3.65)7

for all 3 $ Nn, there exists C, ! 0 such that|$/

x v(!, x)| " C,"!# for all (!, x) $ D ! R ( Rn.(5.3.66)

Proof. — (5.3.65) is obvious, (5.3.66) follows from (5.3.57), (5.3.60) and (5.3.63). Letus show (5.3.64). We set

L =$

$!+

n#

k=1

$p

$%k(x, #(!, x, #))

$

$xk.

Then

Lv(!, x) + a(!, x) v(!, x) ' b(!, x) = b(!, x)

+- &

0eA(s,&,x) [LB(s, !, x) + LA(s, !, x)B(s, !, x)] ds

+ eA(0,&,x) (u(!, x)LA(0, !, x) + Lu(!, x)) + a(!, x) eA(0,&,x) u(!, x)

+ a(!, x)- &

0eA(s,&,x) B(s, !, x) ds ' b(!, x).

So

Lv(!, x) + a(!, x) v(!, x) ' b(!, x)

=- &

0eA(s,&,x)

@LB(s, !, x) + (LA(s, !, x) + a(!, x))B(s, !, x)

Ads

+ eA(0,&,x)@LA(0, !, x) + a(!, x)

Au(!, x) + eA(0,&,x) Lu(!, x).

The Propositions 5.3.15, 5.3.16 and 5.3.17 show that

|LB(s, !, x)| + |LA(s, !, x) + a(!, x)| + |Lu(!, x)| " CN|x ' x(!, #)|N

"!#N

and |A(s, !, x)| " C. Then (5.3.64) follows.



Proof of Theorem 5.1.1 (continued). Case of incoming points. — Let us set

(5.3.67)

'(((()

((((*

L =$

$!+

n#

j=1

$p

$%j(x, #(!, x, #))

$

$xj

a(!, x) =n#

j,k=1

gjk(x)$#j

$xk(!, x, #) ' n

2!

1 + !2+ i

n#

j=1

gj(x)#j(!, x, #).

By Proposition 4.4.14 (ii) we have555$)

$x(!, x, #) ' #(!, x, #)

555 " CN

3|x ' x(!, #)|

"!#

4N

.

Therefore using (5.1.3), (5.1.4) we see that to prove Theorem 5.1.1 it will be su"cientto construct a smooth symbol f = f(!, x, +) (with all derivatives in x bounded) suchthat 555Lf + af +

1+

tPf555 " CN +!N .

We shall take f on the form

f(!, x, +, #) =N#

k=0

+!k fk(!, x, #),

where the f %ks are the solutions of the problems,

(5.3.68)

6Lf0 + af0 = 0, f0|&=0 = 1,

Lfk + afk = 'tP fk!1, fk|&=0 = 0, k ! 1.

Since # is fixed, we shall skip it in writing the fj’s. By Theorem 5.3.18 we have,

(5.3.69)

'((((((()

(((((((*

f0(!, x) = eA(0,&,x),

fk(!, x) =- &

0eA(s,&,x) Bk(s, !, x) ds where,

A(s, !, x) = '- &

sa(", X(", !, x)) d",

Bk(s, !, x) = 'tP fk!1(s, X(s, !, x)), k ! 1.

Our aim is to prove, by induction on k ! 0 that,

(5.3.70) |$,x fk(!, x)| " Ck,,

31"x# +

1"!#

4|,|.

We claim that for all / $ Nn,

(5.3.71) |$*x A(s, !, x)| " C*

31"x# +

1"!#

4|*|,

uniformly with respect to s $ [0, !] and (!, x) $ %+.



Indeed using (5.3.18) and the estimates on 2a,2b given in Theorem 4.4.2 we see easilythat

(5.3.72) |$)(z,z) a(", z)| " C)

31

"Re z#|)|+2+!0+

1""#|)|+2

4.

Moreover by Theorem 5.3.2 (iii) we have

(5.3.73) "Re X(", !, x)# ! 1K

("x# + "! ' "#).

Using the Faa di Bruno formula we see that $*x A(s, !, x) is bounded by a finite sum

of terms of the following form.

(1) =- &

0

555$)(z,z) a(", X(", !, x))

sH

j=1

($*jx X(", !, x))kj

555 d"

where 1 " |.| " |/|, 1 " s " |.|,&s

j=1 kj = .,&s

j=1 |kj | /j = /.Setting I1 = {j $ {1, 2, . . . , s} : |/j | = 1}, I2 = {j $ {1, 2, . . . , s} : |/j | ! 2} and

using (5.3.72), (5.3.73) and (5.3.45) we can write

(1)"C

- &

0

I1

("x# + "! ' "#)|)|+2+!0+

1""#|)|+2

J""#|)|

"!#|)|d"H

j+I2

+ 1"x#|*j |!1

+1

"!#|*j|!1

,|kj |

since&s

j=1 kj = ..Using (5.3.50) and the fact that

&j+I2

|kj | |/j | = |/|' |.| we obtain

(1) " C

31"x# +

1"!#

4|)| 3 1"x# +

1"!#

4|*|!|)|

which proves (5.3.71).As a consequence of (5.3.71) we claim that

(5.3.74)55$,

x(eA(s,&,x))55 " C,

31"x# +

1"!#

4|,|.

Indeed by the Faa di Bruno formula the left hand side of (5.3.74) can be bounded bya finite sum of terms of the form

(2) =555eA(s,&,x)

sH

j=1

$$*j

x A(s, !, x)%kj555

where 1 " s " |3|, 1 "& |kj | " /,&s

j=1 |kj | /j = |3|.Then (5.3.74) follows easily from (5.3.71).Now (5.3.70) for k = 0 follows from (5.3.74) (take s = 0). On the other hand by

the Faa di Bruno formula $,x fk(!, x, #) can be bounded by a finite sum of terms of

the form

(3) =- &

0

55$,1x (eA(s,&,x))

55 $,2x

@tP fk!1(s, X(s, !, x))

Ads.


5.4. THE AMPLITUDE FOR SHORT TIME 145

Setting for convenience bk = tP fk!1 it follows from the induction and Lemma 5.3.1that

(5.3.75)55$)

(z,z) bk(s, z)55 " C)

31

"Re z# +1"s#

4|)|+2

.

Then the Faa di Bruno formula shows that the term $,2x [tP fk!1(s, X(s, !, x))] can be

estimated by a finite sum of terms of the form

(4) =555$$)(z,z) bk

%(s, X(s, !, x))

sH

*=1

($*jx X)kj

555

where 1 " |.| " |32|, 1 " s " |32|,&s

j=1 kj = .,&s

j=1 |kj | /j = 32. Then using(5.3.75), (5.3.73), (5.3.45) we see that

(4) " C

31

("x# + "! ' s#)|)|+2+

1"s#|)|+2

4"s#|)|

"!#|)|H

j+I2

31"x# +

1"!#

4|kj |(|*j |!1)

.

Now by (5.3.71) we can write

(3) " C

- &

0

I1

("x# + "! ' s#)|)|+2+

1"s#|)|+2

J"s#|)|

"!#|)|ds

31"x# +

1"!#

4|,1|+|,2|!|)|

since&

j+I2|kj |(|/j |' 1) = |32|' |.|.


(3) " C

31

"x#|)|+

1"!#|)|

431"x# +

1"!#

4|,1|+|,2|!|)|" C%

31"x# +

1"!#

4|,|

since |31| + |32| = |3|. This proves (5.3.70) for all k.

5.4. The amplitude for short time

We shall need the following precision on the amplitude when |!| " 1.

Proposition 5.4.1. — Let aN be the amplitude defined in Corollary 5.1.2. Then forevery 3 $ N2n one can find a constant C, ! 0 such that

|$,"[aN (!, x, #, +)]| " C,

for all |!| " 1, |x ' x(!, #)| " &"!#, + ! 1 and # $ T #Rn such that 12 " |##| " 2.



Proof. — When ! " 1 we can use the method of Section 5.2 no matter is #, providedthat 1

2 " |#%| " 2. Let us recall how the amplitude aN is produced. We have

(5.4.1)

'(((((((((((((()

((((((((((((((*

aN (!, x, #, +) = "!#!n/2 eN(!, x ' x(!, #), #, +),

eN (!, x ' x(!, #), #, +) = fN

+!,

x ' x(!, #)"!# , #, +

,,

fN (!, z, #) =N+1#

*=0

+!* A*(!, z, #),

L A0 = 0, L A* = i Q A*!1, / = 1, . . . , N + 1,

A0(0, z, #) = 1, A*(0, z, #) = 0,

|$)z A*(!, z, #)| " C) , %. $ Nn, |!| " 1, |z| " &,

12

" |#%| " 2.

Now, according to Proposition 3.2.1 for every 3 $ N2n such that |3| ! 1 one can findC%

, ! 0 such that

(5.4.2) |$," x(!, #)| + |$,

" %(!, #)| " C%,

if |!| " 1, # $ T #Rn, 12 " |#%| " 2.

Assume that we show that for all . $ Nn, 3 $ N2n one can find C), ! 0 such that

(5.4.3) |$)z $,

" A*(!, z, #)| " C),

if |!| " 1 and 12 " |#%| " 2. It will follow from (5.4.1) to (5.4.3) and the Faa di Bruno

formula that

(5.4.4) |$," aN (!, x, #, +)| " C,

if |!| " 1, |x ' x(!, #)| " & "!#, 12 " |#%| " 2, which is the claim of Proposition 5.4.1.

So we are left with the proof of (5.4.3). By (5.2.8), (5.2.9) and (5.4.2) for all µ $ Nn,3 $ N2n there exists Cµ, ! 0 such that

|$µy $,

" Ej(s, y, #)| " Cµ, ,

for all |s| " 1, |y| " & "s# and 12 " |#%| " 2. It follows from (5.2.10), (5.2.11) that for

all . $ Nn, 3 $ N2n there exists C), ! 0 such that

(5.4.5) |$)z $,

" hj(!, z, #)| " C), ,

for all |!| " 1, |z| " &, 12 " |#%| " 2. And we see easily from (5.2.5), (5.4.2), (5.2.15),

(5.2.16) that hN0j , dN0 , kN0

/ satisfy also the bound (5.4.5). By induction on the sizeof derivation, using the Faa di Bruno formula and the Gronwall Lemma we see easilythat the solution z = z(!, y, #) of (5.2.26) satisfies the bound

(5.4.6) |$µy $,

" z(!, y, #)| " Cµ,

uniformly for |!| " 1, |y| " 2, 12 " |#%| " 2. Moreover by Lemma 5.2.3, if we denote

by 5(!, z) the inverse map of z(!, y) we have also by (5.4.6),

(5.4.7) |$)z $,

" 5(!, z, #)| " C),


5.4. THE AMPLITUDE FOR SHORT TIME 147

uniformly for |!| " 1, |z| " &, 12 " |#%| " 2. Finally we have set for / = 0, 1, . . . , N0+1,

(5.4.8) A*(!, z, #) = 2A*(!, 5(!, z, #), #)

where

2A0(!, y, #) = exp.'- &

0dN0(t, z(t, y, #), #) dt

/

2A*(!, y, #) = exp.'- &

0dN0(t, z(t, y, #), #) dt

/ - &

0i( 2QA*!1)(t, y, #) dt

so using (5.4.6), (5.4.7), (5.4.8) and the estimates (5.4.5) for dN0 , kN0/ we see easily

that (5.4.3) holds, which completes the proof of Proposition 5.4.1.


CHAPTER 6

MICROLOCAL LOCALIZATIONS AND THE USE OFTHE FBI TRANSFORM

In this Chapter using the phase and the amplitude constructed in Chapters 4and 5 we shall define general FBI transforms which will lead to a parametrix for theSchrodinger equation. These constructions will be microlocal so we will need severalmicrolocal localizations.

6.1. Preliminaries

6.1.1. The semi-classical calculus. — We shall work with semi-classical pseudo-di!erential operators (p.d.o) and we shall use the Weyl calculus described by Hor-mander. We refer to [H] for notations and details.

Let p $ Sm1,0(Rn) (the usual class of symbol of order m) and let us set

a(x, %) = p$x, %

$

%, + ! 1. It is easy to see that a $ S(M, g) where,

g = dx2 +d%2

+2 + |%|2 , M = +!m(+2 + |%|2)m/2.

The p.d.o associated to the symbol a is denoted by p$x, D

$

%. Then we have the

following symbolic calculus.i) Let p $ Sm

1,0, q $ Sm#

1,0. Then one can find /$ $ Sm+m#

1,0 such that

p+x,

D

+

,4 q+x,

D

+

,= /$

+x,

D

+

,.

The semi norms of /$ are uniformly bounded when + ! 1 and for any N $ N# wehave

/$(x, %) =#

|"|!N!1

1#!

1i|"|

1+|"| $"

% p(x, %) $"x q(x, %) +

1+N

rN (+, x, %)

where rN $ Sm+m#!N1,0 uniformly for + ! 1.

150 CHAPTER 6. MICROLOCAL LOCALIZATIONS

ii) Let p $ S01,0. Then there exists C > 0 such that

000p+x,

D

+

,u000

L2" C *u*L2

for every u $ L2(Rn) and + ! 1. As a consequence, for all s $ R and all p $ Sm1,0 one

can find a constant C > 0 such that for every u $ S(Rn),000+I ' $

+2

,sp+x,

D

+

,u000

L2" C

000+I ' $

+2

,s+mu000

L2

for all + ! 1, where $ =&n

j=1 $2j .

6.1.2. The FBI transform. — We recall here the definition of the classical FBItransform as described in Sjostrand [Sj]. We set for # = (#x, #%) $ T #Rn, + ! 1,u $ L2(Rn) and cn = 2!n/24!3n/4,

(6.1.1) Tu(#, +) = cn +3n/4

-

Rn

ei$(y!"x)·"'!"2 |y!"x|2+ "

2 |"!|2 u(y) dy.

Then T maps continuously the space L2(Rn) to the space of functions v = v(#) suchthat e!

"2 |"!|2 v $ L2(R2n).

The adjoint of T is then given by the formula

(6.1.2) T #v(x, +) = cn +3n/4

-e!i$(x!"x)·"!!"

2 |x!"x|2!"2 |"!|2 v(x) d#.

Then we have

(6.1.3) T #T is the identity operator of L2(Rn).

We shall need also the expressions of T and T # by means of the Fourier transform.We have,

(6.1.4)

'()

(*

Tu(#, +) =++

4

,n/4-

ei!·"x!"2 |"!+ #

" |2+ "2 |"!|2 Qu(") d"

RT #v(%, +) = c%n +n/4

-e!i%·"x!"

2 |"!+ !" |2!"

2 |"!|2 v(#) d#.

Let us consider now a self adjoint operator,

(6.1.5) P =n#

j,k=1

Dj(gjk Dk) +n#

j=1

(Dj bj + bj Dj) + b0

with gjk = &jk + ( bjk, where ( is a small positive constant, &jk is the Kroneckersymbol and,

#

|"|=k

|$"x bjk(x)| " Ak

"x#k+1+!0, k = 0, 1, . . . , x $ Rn, "0 > 0.

Then by interpolation and duality we can prove the following estimates. For all s $ Rthere exists C ! 1 such that for all u $ C"

0 (Rn)

(6.1.6)1C*u*Hs " *(I + P )s/2 u*L2 " C *u*Hs .


6.2. THE MICROLOCALIZATION PROCEDURE 151

For all s ! 0 there exists C ! 1 such that for all + ! 1 and u $ C"0 (Rn),

1C

+!s *u*Hs "000+I + P

+x,

D

+

,,s/2u000

L2" C *u*Hs(6.1.7)

1C

*u*H"s "000+I + P

+x,

D

+

,,!s/2u000

L2" C +s*u*H"s .(6.1.8)

6.2. The microlocalization procedure

We begin by introducing several cut-o! functions. Generally speaking we shalldenote by * (resp. -) cut-o! functions in the space (resp. frequency) variables.

Let %0 $ Rn, |%0| = 1 be fixed.Let *0 $ C"(R) be such that,

(6.2.1) *0(s) = 1 if s " 34, *0(s) = 0 if s ! 1, 0 " *0 " 1.

With &1 > 0 to be chosen later on we set

(6.2.2)

'()

(*

*+1 (x) = *0

+' x · %0

&1

,, *+

2 (x) = *0

+' x · %0

2&1

,, *+

3 (x) = *0

+' x · %0

3&1

,

*!1 (x) = *0

+x · %0

&1

,, *+

2 (x) = *0

+x · %0

2&1

,, *+

3 (x) = *0

+x · %0

3&1

,.

These cut-o! functions will correspond to outgoing and incoming points. Now forconvenience we shall set,

(6.2.3) a =610

, b =1910

,

and with &2 " 1/100 (chosen later on) we introduce the following cut-o! functions.Let -0 $ C"(Rn) be such that 0 " -0 " 1 and

(6.2.4)

'()

(*

-0(%) = 1 if555

%

|%| ' %0

555 " &2 and |%| ! 2&2

supp -0 ,!

% :555

%

|%| ' %0

555 " 2&2 and |%| ! &2

"

Let -1 $ C"0 (Rn) be such that, 0 " -1 " 1 and

(6.2.5)

6-1(%) = 1 if a ' &2 " |%| " b + &2,

supp -1 ,:% : a ' 2&2 " |%| " b + 2&2

;.

We shall set

(6.2.6) -2(%) = -0(%)-1(%).

Now we introduce for t $ R the operators,

(6.2.7)

'()

(*

U+(t) = *+1 (x)-2

+D

+

,e!itP ,

U!(t) = *!1 (x)-2

+D

+

,e!itP .



It follows that if we denote by U# the adjoint of U then we have

(6.2.8)

'(((()

((((*

U±(t1)U±(t2)# = K±(t1 ' t2), where

K+(t) = *+1 (x)-2

+D

+

,e!itP -2

+D

+

,*+

1 (x),

K!(t) = *!1 (x)-2

+D

+

,e!itP -2

+D

+

,*!

1 (x).

Let now -3 $ C"0 (Rn) be such that 0 " -3 " 1 and

(6.2.9)

'()

(*

-3(%) = 1 if555

%

|%| + %0

555 " 3 &2 and a ' 3&2 " |%| " b + 3&2

supp -3 ,!% :555

%

|%| + %0

555 " 4&2 and a ' 4&2 " |%| " b + 4&2

".

The first localization result requires to introduce the following Definition.

Definition 6.2.1. — We shall call R the set of families of operators R = (R$(t))depending on + $ [1, +)[ and t $ ['T, T ] such that for every N in N one can find aconstant CN ! 0 such that for every u $ S(Rn),

(6.2.10) *R$(t)u*H2N (Rn) " CN *u*H"2N (Rn)

uniformly with respect to (+, t) $ [1, +)[(['T, T ].

Then we can state the following result.

Theorem 6.2.2. — Let T > 0. For every t $ ['T, T ] and + ! 1 one can write

(6.2.11) K+(t) = *+1 -2

+D

+

,*+

2 T #"$x *+

3 (#x)-3(#%)Ty$"

.e!itP *+

2 -2

+D

+

,*+

1

/

+ R+$ (t)

where (R+$ (t)) $ R. The same formula is true with the sign ' instead of +.

The proof of this Theorem requires several steps.

Lemma 6.2.3. — There exist a constant C > 0 such that000-2

+D

+

,T #@(1 ' -3(#%)) v

A000L2(Rn)

" C e!"8 +2

200e!

"2 |"!|2 v

00L2(R2n)

for every v such that e!"2 |"!|2v $ L2(R2n).

Proof. — We claim that on the support of -2(%)(1 ' -3(#%)) we have

(6.2.12) |% + #%| ! 12

&2 |#%|.

Indeed, according to (6.2.4) to (6.2.6) and (6.2.9) we have on this support55 %|%| ' %0| " 2&2, a ' 2&2 " |%| " b + 2&2 and either |#%| " a ' 3&2 or |#%| ! b + 3&2 or



55 "!

|"!| + %0

55 ! 3&2. In the first case we have |% + #%| ! |%| ' |#%| ! &2. In the secondcase we have |% + #%| ! |#%|' |%| ! &2 and in the third one we have

555#%

|#%|+

%

|%|

555 !555

#%

|#%|+ %0

555'555

%

|%| ' %0

555 ! &2.

Therefore55#% + |"!|

|%| %55 ! &2 |#%|. It follows that

|#% + %| !555#% +

|#%||%| %

555'555% '

|#%||%| %

555 ! &2 |#%|'55|%|' |#%|

55 ! &2 |#%|' |#% + %|

so |#% + %| ! 12 &2 |#%| and our claim is proved.

Now using (6.1.4) we can write

(1) =: F+-2

+D

+

,T #@(1 ' -3(#%))

Av,(%)

= cn +n4 -2

+ %

+

,-

R2n

e!i%·"x!"2 |"!+ !

" |2!"2 |"!|2(1 ' -3(#%)) v(#) d#.

Let us set

w(%, #%) =-

Rn

e!i%·"x!"2 |"!|2 v(#) d#x = F"x

$e!

"2 |"!|2 v

%(%, #%).

Using Cauchy-Schwartz inequality we obtain

|(1)|2 " C +n/2

3-

Rn

e!$|"!+ !" |2 -2

2

+ %

+

,$1 ' -3(#%)

%2d#%

43-|w(%, #%)|2 d#%

4

so by (6.2.12)

|(1)|2 " C +$/n e!"8 +2

2

3-e!

"2 |"!+ !

" |2d#%

43-|w(%, #%)|2 d#%

4.

It follows that|(1)|2 " C% e!

"8 +2

2

-|w(%, #%)|2 d#%.

Integrating with respect to % and using Parseval identity we obtain000-2

+D

+

,T #@(1 ' -3(#%)) v

A000L2

" C e!"8 +2

2

-

R2n

e!"2 |"!|2 |v(#)|2 d#.

Corollary 6.2.4. — We have for t $ ['T, T ] and + ! 1,

K+(t) = *+1 -2

+D

+

,T # -3(#%)T e!itP -2

+D

+

,*+

1 + R+$ (t)

where (R+$ (t)) $ R. The same is true for the sign '.

Proof. — Using (6.1.3) we write Id = T #T = T # -3(#%)T + T #(1 ' -3(#%))T . So wehave to prove that the family of operators

R+$ (t) = *+

1 -2

+D

+

,T #(1 ' -3(#%))T e!itP -2

+D

+

,*+

1

belong to R.



Let 2- $ C"0 (Rn ! 0) be such that 2-(%)-2(%) = -2(%). Then writing

-2

$D$

%= 2-$

D$

%-2

$D$

%, using Lemma 6.2.3 and the fact that

00 2-$

D$

%v00

H2N"C +2N*v*L2

we obtain

*R+$ (t) v*H2N " C +2N e!

18 $ +2

2

000e!"2 |"!|2 T e!itP -2

+D

+

,*+

1 u000

L2.

Since T is continuous from L2 to the space of v such that e!"2 |"!|2 v $ L2 and using

the conservation of L2 norm we obtain

*R+$ (t) v*H2N " C +2N e!

18 $ +2

2

000-2

+D

+

,*+

1 u000

L2" C% +4n e!

18 $ +2

2 *u*H"2N

so our claim is proved.

Lemma 6.2.5. — Let *+2 be defined in (6.2.2). Then we have for t $ ['T, T ] and

+ ! 1,

K+(t) = *+1 -2

+D

+

,*+

2 T # -3(#%)T e!itP *+2 -2

+D

+

,*+

1 + R+$ (t)

where (R+$ (t)) $ R.

Proof. — Since by (6.2.2) the support of *+1 and 1 ' *+

2 are disjoint, the symboliccalculus shows that the operators *+

1 -2

$D$

%(1 ' *+

2 ) and (1 ' *+2 )-2

$D$

%*+

1 belongto 1

$M S!M1,0 for any M $ N. It follows from (VI.1.7) that if M ! 2N ,

000*+1 -2

+D

+

,(1 ' *+

2 )T # -3(#%)T e!itP -2

+D

+

,*+

1 u000

H2N

" C +2N000+I ' $

+2

,N*+

1 -2

+D

+

,(1 ' *+

2 )T # · · ·000

L2

" C +2N!M000T # -3(#%)T e!itP -2

+D

+

,*+

1 u000

L2

" C +2N!M000e!

"2 |"!|2 -3(#%)T e!itP -2

+D

+

,*+

1 u000

L2(R2n)

" C +2N!M000e!itP -2

+D

+

,*+

1 u000

L2" C +2N!M

000-2

+D

+

,*+

1 u000

L2

" C +4N!M *u*H"2N .

Taking M ! 4N we conclude that the remainder under consideration belong to R.By the same way

000*+1 -2

+D

+

,*+

2 T # -3(#%)T e!itP (1 ' *+2 )-2

+D

+

,*+

1 u000

H2N

" C +2N000+I ' $

+2

,N-2

+D

+

,*+

2 T # · · ·u000

L2

" C +2N000(1 ' *+

2 )-2

+D

+

,*+

1

+I ' $

+2

,N +I ' $

+2

,!Nu000

L2

" C% +2N!M000+I ' $

+2

,!Nu000

L2" C%% +4N!M *u*H"2N .

The proof is complete.



Lemma 6.2.6. — Let -a $ C"0 (R) and *a(x), *b(#x) be C" functions such that one

can find µ > 0 such that |x ' #x| ! µ if (x, #x) belongs to supp[*a(x)(1 ' *b(#x))].Then one can find ( > 0, C > 0 such that

00*a T #[-a(#%)(1 ' *b(#x)) v]00

L2(Rn)" C e!#$

00e!"2 |"!|2 v

00L2(R2n

( )

for all v such that the right hand side norm is finite.

Proof. — It follows from (6.1.2) that

*a(x)T #(-a(#%)(1 ' *b(#x)) v)(x) =-

K(x, #) e!"2 |"!|2 v(#) d#

where

K(x, #) = cn +3n/4 e!i$(x!"x)·"!!"2 |x!"x|2 -a(#%)*a(x)(1 ' *b(#x)).

Using our assumption we can write

|K(x, #)| " C +3n/4 e!"4 µ2

e!"4 |x!"x|2 -a(#%).

Therefore one can find ( > 0 such that,

supx

-|K(x, #)| d# " C e!#$, sup

"

-|K(x, #)| dx " C e!#$,

so the Lemma follows from the well known Schur Lemma.

Corollary 6.2.7. — We have

K+(t) = *+1 -+D

+

,*+

2 T # -3(#%)*+3 (#x)T e!itP *+

2 -2

+D

+

,*+

1 + R+$ (t)

where R $ R and the same is true for the sign '.

Proof. — We have to show that the operator

R+$ (t) = *+

1 -2

+D

+

,*+

2 T # -3(#%) (1 ' *+3 (#x))T e!itP *+

2 -2

+D

+

,*+

1

belongs to R.We apply Lemma 6.2.6 with *a = *+

2 , -a = -3, *b = *+3 . Then according to

(6.2.2) we have on the support of *+2 (x)(1 ' *+

3 (#x)), 'x · %0 " 2&1, '#x · %0 ! 94 &1.

Therefore we have, 94 &1 " '#x · %0 " (x ' #x) · %0 ' x · %0 " |x ' #x| · |%0| + 2&1, so

|x ' #x| ! µ > 0. Then using Lemma 6.2.6 we can write

*R$(t)u*H2N " C +2N e!#$000e!

"2 |"!|2 T e!itP *+

2 -2

+D

+

,*+

1 u000

L2

" C% +2N e!#$000-2

+D

+

,*+

1 u000

L2" C%% +4N e!#$ *u*H"2N .

Proof of Theorem 6.2.2. — It follows immediately from Corollary 6.2.7.



6.3. The one sided parametrix

The purpose of this Section is to use the results of Chapters 4, 5 and 6.2 to showthat the operators K+(t) and K!(t) introduced in (6.2.5) can be written as Fourierintegral operators with complex phase functions. We shall take the expression ofK±(t) given by Theorem 6.2.2 and we begin by considering the expression

T.e!itP *+

2 -2

+D

+

,*+

1 u/(#, +) (see (6.1.1)).

Let us first introduce some other cut-o! functions.

(6.3.1)

'((((((((()

(((((((((*

If |#x · #%| " c0

2"#x# |#%| we set *±

4 (y) 3 1.

If |#x · #%| >c0

2"#x# |#%| we set,

(i) in the + case, *+4 (y) = *0

+' y · %0

5&1

,,

(ii) in the ' case, *!4 (y) = *0

+y · %0

5&1

,,

where *0 has been defined (in (6.2.1)).

In all cases let *5 $ C"0 (Rn) be such 0 " *5 " 1 and

(6.3.2) *5(y) = 1 if |y| " &

2, supp *5 , {y : |y| " &}

where & is the small constant introduced in Theorem 4.1.2. For the convenience ofthe reader let us recall the main properties of the phase and the symbol constructedin Corollary 5.1.2, and Theorem 4.1.2. First of all the phase ) is defined on the set%+ where,

(i) if |#x · #%| " c0 "#x#|#%| then

%+ =:(!, y) $ R ( Rn : |y ' x(!, #)| < &"!#

;.

(ii) if #x · #% > c0"#x#|#%| then,

%+ = {(!, y) $ (0, +)) ( Rn : |y ' x(!, #)| < &"!#} 2 {(!, y) $ ('), 0) ( Rn :

|y ' x(!, #)| < &"!# and y · #% ! 'c1 "y#|#%|}.

(iii) if #x · #% < 'c0"#x#|#%| then,

%+ = {(!, y) $ ('), 0) ( Rn : |y ' x(!, #)| < &"!#} 2 {(!, y) $ (0, +)) ( Rn :

|y ' x(!, #)| < &"!# and y · #% " c1 "y#|#%|}.

Moreover on this domain

(6.3.3) Im)(!, y, #) ! 14|y ' x(!, #)|2

1 + 4!2' 1

2|#%|2.

Now if we set, with the notations of Theorem 5.1.1, for N $ N,

(6.3.4) a(!, y, #, +) = (1 + !2)!n/4 eN(!, y ' x(!, #), #, +),


6.3. THE ONE SIDED PARAMETRIX 157

then a is defined on %+ for + ! 1 and satisfies,

(6.3.5)

'(((((()

((((((*

a(0, y, #, +) = 1,

|a(!, y, #, +)| " c (1 + !2)!n/4,+i+

$

$!+ tP

,$ei$'(&,y,") a(!, y, #, +)

%= bN(!, y, #, +) ei$'(&,y,"),

with |bN(!, y, #)| " cN (1 + !2)!n/4++!N + +2

+ |y ' x(!, #)|"!#

,N,.

Let us introduce now the following set.

(6.3.6) W± =!# $ T #Rn :

12

" |#%| " 2, |#x · #%| " c0"#x#|#%|"

2:# $ T #Rn : |#x · #%| > c0 "#x#|#%|, (#x, #%) $ supp(*±

3 (#x) · -3(#%));

where *±3 and -3 have been defined in (6.2.2) and (6.2.9).

Then we can state the main result of this Section.

Theorem 6.3.1. — We have for t $ ['T, T ], and # $ W+,

T@e!itP *+

2 vA(#, +)

= +3n/4

-ei$'(!$t,y,") a('+t, y, #, +)*+

4 (y)*5

+y ' x('+t, #)"+t#

,[*+

2 v](y) dy

+ J+$ (t) v(#)

where the operator J+$ is such that, for any M $ N one can find a constant CM > 0

such that for all + ! 1 and t $ ['T, T ]00e!

"2 |"!|2 J+

$ (t) v00

L2(W+)" CM

+M*v*L2(Rn),

and the same is true with the minus sign.

Proof. — Let us introduce the following family of operators. We set for # $ W+

(6.3.7) Sv(!, t, #, +)

= +3n/4

-

Rn

ei$'(&,y,") a(!, y, #, +)*+4 (y)*5

+y ' x(!, #)"!#

,@e!itP *+

2 vA(y) dy.

We must verify that the right hand side is indeed well defined.On the support of *5 we have |y'x(!, #)| < &"!# (which is one of the conditions for

(!, y) to be in %+). If # $ W+ then either |#x ·##| " c0"#x#|##| or |#x ·##| > c0"#x#|##|and (#x, ##) $ supp(*+

2 ·-3). In the first case by (6.3.1) *+4 (y) = 1 but (!, y) $ %+ for

! $ R. In the second case since #x $ supp *+3 and #% $ supp -3 we have, by (6.2.1),

(6.2.2) and (6.2.9), #x · %0 ! '3&1 and55 "!

|"!| + %0

55 " 4&2. It follows that

(6.3.8) #x · #%

|#%|= '#x · %0 + #x ·

+ #%

|#%|+ %0

," 7 &1 "#x#.



On the other hand we have |#x · #%| > c0 "#x#|#%|. According to (6.3.8) we cannothave #x · #% > c0 "#x#|#%| if &1 7 c0 ; thus we must have #x · #% < 'c0"#x#|#%|. Sowe are in the case (iii) for the definition of %+. Since in the integral, on the supportof *+

4 (y) we have by (6.3.1), y · %0 ! '5 &1 ! '5 &1 "y#, we deduce that

y · #%

|#%|= 'y · %0 + y ·

+ #%

|#%|+ %0

," 9 &1"y# < c0"y#.

Therefore (!, y) $ %+ and the right hand side of (6.3.7) is well defined. Of course thesame argument is valid in the case of the minus sign.

Now let us set for s $ [0, t] if t > 0 (resp. s $ [t, 0] if t < 0)

(6.3.9) g(s) = Sv(+(s ' t), s, #, +).

Since g%(s) = (+$& + $t)(Sv)(+(s ' t), s, #, +) we obtain

(6.3.10)

'((((((()

(((((((*

Sv(0, t, #, +) = Sv('+t, 0, #, +)

+- t

0(+$& + $t)(Sv)(+(s ' t), s, #, +) ds if t > 0

Sv(0, t, #, +) = Sv('+t, 0, #, +)

'- t

0(+$& + $t)(Sv)(+(s ' t), s, #, +) ds if t < 0.

Let us set

(6.3.11) U(t, y, +) =@e!itP *+

2 vA(y).

Then according to (6.3.7) we can write

(6.3.12) (+$& + $t)(Sv)(!, t, #, +) = +3n/4 (A1 + A2 + A3)

where

(6.3.13)

'(((((((((()

((((((((((*

A1 =-

+$&

$ei$'(&,y,") a(!, y, #, +)

%*+

4 (y)*5

+y ' x(!, #)"!#

,U(t, y, +) dy

A2 =-

ei$'(&,y,") a(!, y, #, +)*+4 (y)*5

+y ' x(!, #)"!#

,('iP )U(t, y, +) dy

A3 =n#

j=1

-ei$'(&,y,") a(!, y, #, +)*+

4 (y) ($j *5)+y ' x(!, #)

"!#

,

· $&

+yj!xj(&,")

'&(

,U(t, y, +) dy.

Integrating by parts in the integral giving A2 we obtain

A2 =-

('i tP )@ei$'(&,y,") a(!, y, #, +)

A*+

4 (y)*5

+y ' x(!, #)"!#

,U(t, y, +) dy

+#

1!|)|!2)=)1+)2

-ei$'(&,y,")b)(!, y, #, +)($)1

y *+4 )(y)"!#!|)2|

· ($)2y *5)

+y ' x(!, #)"!#

,U(t, y, +) dy

where b) is a symbol satisfying the same estimates as a(!, y, #, +) (see (6.3.5)).



Combining the term A1 with the first part of A2 and using (6.3.5) we can write

(6.3.14) (+$& + $t)(Sv)(!, t, #, +) = +3n/4(B1 + B2 + B3)

where

(6.3.15)

'((((((((((((((()

(((((((((((((((*

B1(!, t, #, +) =-

ei$'(&,y,") bN (!, y, #, +)*+4 (y)

*5

+y ' x(!, #)"!#

,U(t, y, +) dy

B2(!, t, #, +) =#

|)1+)2|!2)2 ,=0

-ei$'(&,y,") C)1)2(!, y, #, +)($)1

y *+4 )(y)"!#!|)2|

($)2y *5)

+y ' x(!, #)"!#

,U(t, y, +) dy

B3(!, t, #, +) =&

1!|,|!2

1ei$'(&,y,") d,(!, y, #, +)($,

y *+4 )(y)

*5

+y ' x(!, #)"!#

,U(t, y, +) dy

where C)1,)2 and d, are bounded symbols. Here we used the estimate55$&

$yj!xj(&,")'&(

%55 "C/"!#. Now we state a Lemma.

Lemma 6.3.2. — One can find constants C > 0, 2 > 0 and for every M ! 0 aconstant CM > 0 such that for all !, t, + such that + ! 1, |!| " |+t|, |t| " T we have

00e!"2 |"!|2 B1(!, t, ·, +)

00L2(W+)

" CM +!M *v*L2(Rn)

00e!"2 |"!|2 B2(!, t, ·, +)

00L2(W+)

" C e!-$ *v*L2(Rn)

for all v in L2(Rn).

Proof. — Let us consider the term B1. Using (6.3.5) and (6.3.3) we obtain555ei$'(&,y,")!"

2 |"!|2 bN(!, y, #, +)*5

+y ' x(!, #)"!#

,*+

4 (y)1W+(#)555

" CN e! "

16|y"x(&,()|2

&&'2 "!#!n/2++!N + +2

+ |y ' x(!, #)|"!#

,N,1W+(")

=: K(!, y, #, +).


e!"2 |"!|2 B1 =

-K(!, y, #, +)U(t, y, +) dy.

We want to apply Schur Lemma to this integral operator. First of all making thechange of variables y ' x(!, #) = "!# z/

/+ we can write

-K(!, y, #, +) dy " CN "!#n/2

-e!

|z|216 +!n/2(+!N + +2 +!N/2 |z|N) dz.

Since "!# " 1 + +T we obtain finally for all M ! 0 and + ! 1,1

K(!, y, #, +) dy "CM +!M .



In order to estimate (1) =1

K(!, y, #, +) d# we make the change of variable . =(x(!, #), %(!, #)) Since this transformation is symplectic we have d# = d.. Thereforewe obtain,

(1) " CN

-e! "

16|y")x|2

&&'2 "!#!n/2 1 12 !|%(!&,))|!2 (.)

3+!N + +2

+ |y ' .x|"!#

,N4

d..

Since %('!, .) = .% + O(() we have 13 " |.%| " 3. Setting as before y ' .x = '&()

$we

deduce easily that (1) " CM +!M for every M ! 0. It follows from these estimates,the Schur Lemma and (6.3.11) that,00e!

"2 |"!|2 B1

00L2(W+)

" CM +!M00e!itP -2

+ P

+2

,*+

2 v00

L2(Rn)" C%

M +!M *v*L2(Rn).

To deal with the term B2 we use exactly the same computations and the fact that onthe support of ($)2

y *5)$ y!x(&,")

'&(%, for |.2| ! 1 we have +

2 " |y!x(&,")|'&( " &.

Proof of Theorem 6.3.1 in the following cases

(6.3.16)

6(i) |#x · #%| " c0 "#x# |#%| and t $ ['T, T ],

(ii) |#x · #%| > c0 "#x# |#%| and t $ [0, T ].

In both these cases we are going to show, with the notations of (6.3.15), that B3 3 0.In the case (i) this is obvious since by (6.3.1) we took *+

4 (y) 3 1. Now ift > 0, by (6.3.9) we have ! = +(s ' t) " 0. In the case (ii) we have seen thatin the case + we must have #x · ## " 'c0 "#x# |##|. Now on the support of*5

$y!y(&,")'&(

%we have |y ' x(!, #)| " &"!# and it follows from Proposition 3.4.1 that

x(!, #) = #x + 2! #% + O(("!#). Then we write

y · %0 = (y ' x(!, #)) · %0 + (#x + 2! #%) · %0 + O(("!#).

On the support of -3(#%) we have55 "!

|"!| + %0

55 " 4 &1 so

y · %0 = (#x + 2! #%) ·+%0 +

#%

|#%|

,' (#x + 2! #%) ·

#%

|#%|+ O((( + &)"!#)

y · %0 ! c0 "#x# + 2 |!| |#%|' 4 &1 "#x# ' C(& + ( + &1)"!#.

Since &, (, &1 are small compared to c0 we deduce that y · %0 ! c02 "!# in the integral

defining B3 in (6.3.15). Since the support of $,y *+

4 (y), for 3 += 0, is contained in34 " !y·%0

5 +1" 1 we deduce that B3 3 0 in this case, (see (6.3.1) and (6.2.2)). It follows

from (6.3.10) and (6.3.14) that

(6.3.17) Sv(0, t, #, +) = Sv('+t, 0, #, +) +- t

0(B1 + B2)(s) ds.

It follows from Lemma 6.3.2 that,

(6.3.18)000e!

"2 |"!|2

- t

0(B1 + B2)(s) ds

000L2(W+)

" C

+M*v*L2(Rn).



Now we have'()

(*

)(0, y, #) = )0(y, #) + g(y ' #x), where

)0(y, #) = (y ' #x) · #% + i2 |y ' #x|2 + 1

2i |#%|2,|g(x)| " CN |x|N for every N $ N.

Let * $ C"0 (Rn) be such that *(x) = 1 if |x| " 1, supp* , {x : |x| " 2} and let us

fix N ! 3. We can write

(6.3.19) Sv(0, t, #, +) = A1 ' A2 ' A3 + A4

where

(6.3.20)

'((((((((((((((()

(((((((((((((((*

A1 = cn +3n/4

-ei$'0(y,") *+

4 (y)*5(y ' #x)U(t, y, +) dy

A2 = cn +3n/4

-ei$'0(y,") (1 ' *(CN+|y ' #x|N ))

· *+4 (y)*5(y ' #x)U(t, y, +) dy

A3 = cn +3n/4

-ei$'0(y,") (1 ' ei$g(y!"x))*(CN +|y ' #x|N )*+

4 (y)

· *5(y ' #x)U(t, y, +) dy

A4 = cn +3n/4

-ei$('0(y,")+g(y!"x)) (1 ' *(CN +|y ' #x|N ))*+

4 (y)

· *5(y ' #x)U(t, y, +) dy.

We claim that we have for j = 2, 3, 4,

(6.3.21)00e!"

2 |"!|2 Aj

00L2(W+)

" CMN

+MN*v*L2(Rn), MN '. +) if N '. +).

(i) Term A2

On the support 1 ' *(CN+|y ' #x|N ) we have |y ' #x| ! C%N +!1/N . So,

555e!"2 |"!|2 ei'0(y,") (1 ' *(CN+|y ' #x|N ))

555 " C e!"4 |y!"x|2 e!C##

N $1" 2N .

Using the Schur Lemma and the inequality *U(t, ·, #)*L2 " C *v*L2 (see (6.3.11)) weobtain (6.3.21) for A2.

(ii) Term A3

On the support of *(CN+|y ' #x|N ) we have +|g(y ' #x)| " CN+|y ' #x|N " 2.Therefore we have |1' ei$g(y!"x)| " C + |g(y '#x)| " C%

N +|y '#x|N . It follows that555e!

"2 |"!|2 ei$'0(y,") (1 ' ei$g(y!"x)*(CN +|y ' #x|N )

555 " C%N e!

"2 |y!"x|2 · +|y ' #x|N

" C%N

+N2 !1

(+1/2|y ' #x|)N e!"4 |y!"x|2 e!

"4 |y!"x|2

" C%N

+N2 !1

e!"4 |y!"x|2 .

The Schur Lemma shows again that A3 satisfies (6.3.21).(iii) Term A4



On the support of *5(y'#x) we have, according to (6.3.2), |y'#x| " &. It followsthat |g(#x ' y)| " & C3|y ' #x|2 so if & is small enough,

'12|#%|2 ' Im )0(y, #) ' Im g(y ' #x) " '1

4|y ' #x|2

and, as for A2, the Schur Lemma implies that A4 satisfies (6.3.21).Using (6.3.19) and (6.3.21) we see that

Sv(0, t, #, +) = cn +3n/4

-ei$'0(y,") *+

4 (y)*5(y ' #x)U(t, y, +) dy(6.3.22)

+J+$ (t) v(#).

00e!"2 |"!|2 J+

$ (t) v00 " CM +!M *v*L2(Rn).(6.3.23)

Now, on the support of *5(y ' #x)' 1 we have |y ' #x| ! +2 so modulo a term which

satisfies (6.3.23) we can remove *5(y ' #x) in the right hand side of (6.3.22). Let usremove *+

4 . When # $ W+ we have |#x ·##| " c0"#x#|#%| and *+4 (y) 3 1 (see (6.3.1))

(so there is nothing to remove) or #x $ supp *+3 (see (6.3.6)) that is '#x · %0 " 3&1.

In the later case on the support of 1 ' *+4 (y) we have !y·%0

5+1! 3

4 (see (6.2.1), (6.3.1))so |y ' #x| ! #x · %0 ' y · %0 ! 3

4 &1. The corresponding term, again by the SchurLemma, satisfies (6.3.23).

Using (6.1.1) we see therefore that

(6.3.24) Sv(0, t, #, +) = T [e!itP *+2 v](#, +) + J+

$ (t) v(#),

where J+$ (t) satisfies (6.3.23).

Gathering the informations given by (6.3.17), (6.3.18), (6.3.24) and (6.3.23) weobtain the claim of Theorem 6.3.1 in the case (6.3.16).

Proof of Theorem 6.3.1 in the following case

(6.3.25) |#x · #%| > c0"#x#|#%| and t $ ['T, 0].

According to (6.3.14), (6.3.15) and Lemma 6.3.2, we must prove that for all N $ None can find CN > 0 such that for + ! 1,

(6.3.26)00e!"

2 |"!|2 B3(!, t, ·, +)00

L2(W+)" CN +!N *v*L2(Rn).

Here ! = +(s ' t) > 0 since s $ [t, 0].Let us introduce a new cut-o! function. Let -4 $ C"

0 (Rn) be such that 0 " -4 " 1and,

(6.3.27)

')

*-4(%) = 1 if

555 %|%| + %0

555 " 5&2, a ' 5&2 " |%| " b + 5&2

supp -4 ,!

% :555 %|%| + %0

555 " 6&2, a ' 6&2 " |%| " b + 6&2

".

We state a Lemma.



Lemma 6.3.3. — Let us set(6.3.28)SW+ =

7# :

12

" |#%| " 2, |#x · #%| > c0"#x#|#%|, (#x, #%) $ supp(*+3 (#x)-3(#%))

K.

Let k(!, y, #, +) be a symbol and let us set

F (!, #, +) = 1fW+(#) e!"2 |"!|2

-ei$'(&,y,") k(!, y, #, +)*5

+y ' x(!, #)"!#

,

· $,y *+

4 (y).I ' -4

+D

+

,/v(y) dy.

Then for every N $ N one can find CN > 0 such that for + ! 1 and |!| " +T we have

*F (!, ·, +)*L2 " CN +!N *v*L2(Rn).

Proof. — By (6.2.12) we have |% + #%| ! µ > 0 on the support of -3(#%)(1 ' -4(%)).Now recall that Theorem 4.1.2 shows that the phase ) satisfies

(6.3.29)

')

*

555$)

$y(!, y, #) ' #%

555 " C (( +/

&)

|$)y )(!, y, #)| " C) if |.| ! 1

on the support of *5

$y!x(&,")'&(

%$,

y *+4 (y)1fW+(#).

Let g $ C"0 (Rn) be such that g(%) = 1 if |%| " 1. Then

+I ' -4

+D

+

,,v (y) = lim

#$0

+ +

24

,n--

ei$(y!z)·% (1 ' -4(%)) g(( %) v(z) dz d%.

It follows that

(6.3.30)

'((((()

(((((*

F (!, #, +) = lim#$0

-K#(#, z) v(z) dz with

K#(#, z) =+ +

24

,.

--e!

"2 |"!|2 1fW+(#) ei$['(&,y,")+(y!z)·%] k(!, y, #, +)

· *5

+y ' x(!, #)"!#

,$,

y *+4 (y)(1 ' -4(%)) g((%) d% dy.

Let us consider the vector field

X =1

1 + +|y ' z|2+1 +

1i

n#

j=1

(yj ' zj)$

$%j

,.

Then it is easy to see that'(()

((*

X ei$(y!z)·% = ei$(y!z)·%

(tX)N =#

|A|!N

CA(y ' z)A $"%

(1 + +|y ' z|2)N.



Then we can write

(6.3.31)-

ei(y!z)·% (1 ' -4(%)) g((%) d% =-

ei$(y!z)·% (tX)N [(1 ' -4(%)) g(( %)] d%

=#

A=A1+A2|A|!N

-ei$(y!z)·% CA1A2

(y ' z)A

(1 + +|y ' z|2)N$A1

% (1 ' -4(%)) (|A2|

· ($A2% g)(( %) d%.

Now on the support of -3(#%)(1 ' -4(%)) we have |% + #%| ! µ > 0, it follows from(6.3.29) that

555$)

$y(!, y, #) + %

555 ! |#% + %|'555$)

$y(!, y, #) ' #%

555 ! µ ' C(( +/

&) ! µ

2

if ( and & are small enough.If |%| ! 2 sup

55('(y (!, y, #)

55 we have55('

(y (!, y, #) + %55 ! |%|

2 . Therefore in all caseswe have, with 20 > 0,

(6.3.32)555$)

$y(!, y, #) + %

555 ! 20 "%#.

Let us set then

Y =1

i+|% + ('(y (!, y, #)|2

n#

j=1

+ $)

$yj(!, y, #) + %j

, $

$yj.

and T = ('(y (!, y, #) + %.

Then

(6.3.33)

'()

(*

Y ei$('(&,y,")+(y!z)·%) = ei$('(&,y,")+(y!z)·%)

(tY )N =1

(i +)N

7 #

|/|=N

+ T

|T |2,/

$/y +

#

|/|!N!1

P3N!2|/|!1(!, y, #, T, T )|T |4N!2|/| $/

y

K

where Pk(!, y, #, T, T ) is a polynomial in T, T of order " k with C"'bounded coef-ficients.

It follows from (6.3.32) that on the support of -3(#%)(1 ' -4(%)) we have,

(6.3.34)#

|/|=N

5555+ T

|T |2,/5555+

#

|/|!N!1

|P3N!2|/|!1(· · · )||T |4N!2|/| " CN

"%#N .

On the other hand we check by induction that

(6.3.35) $/y

. (y ' z)A

(1 + +|y ' z|2)N

/=

#

j!|/||)|!|A|+j

2j+|A|!|)|+|/|

bN,j,A,)(y ' z)) +j

(1 + +|y ' z|2)N+j.

Now if we insert (6.3.31) into (6.3.30) and if we make integration by parts with respectto Y using (6.3.33) we see using (6.3.34) and (6.3.3) that K#(#, z) is bounded by a



finite sum of integrals of the following type-

+n|y ' z||)| +j

+N "%#N (1 + +|y ' z|2)N+j1fW+(#)|$A1

% (1 ' -4(%))| (|A2| ($A2% g)((%) |$/1

y k|555($/2

y *5)+y ' x(!, #)

"!#

,555"!#!|/2||$,+/3y *+

4 (y)| e!"16

|y"x(&,()|2

&&'2 dy d%

where |.| " |A| + j, '1 + '2 + '3 = ', |'| " N , A = A1 + A2, |A| " N , j " |'|,2j + |A| " |.| + |'|.

Claim. — We have'()

(*

(1) = sup"

-|K#(#, z)| dz " CN

"!#n

+N/2

(2) = supz

-|K#(#, z)| d# " C%

N"!#n

+N/2.

Let us first remark that

(6.3.36)-

+j |x||)|

(1 + +|x|2)N+jdx = C +j! |)|

2 !n2 " C +

N2 !n

2 .

Indeed '|.| " |'|' 2j ' |A| so j ' |)|2 " |/|

2 ' |A|2 " N

2 .To estimate (1) we use the above estimate of K#(#, z) which we integrate with

respect to z. For the integral in z we use (6.3.36). The integral in % is estimatedthanks to the term 1

'%(N where |'| " N , finally the integral in y is bounded by-

e! "

16|y"x(&,()|2

&&'2 dy " C"!#n

+n/2.

Therefore we obtain

(1) " C+n

+N+

N2 !n

2"!#n

+n/2= C

"!#n

+N/2.

To estimate the term (2) we use the change of variables 2# = (x(!, #), %(!, #)) as inthe proof of Lemma 6.3.2 and (6.3.36). This gives, (2) " C '&(n

$N/2 . Since |!| " +T weobtain

(1) + (2) " C +!N2 +n.

We can therefore use the Schur Lemma and (6.3.30) to achieve the proof ofLemma 6.3.3.

Corollary 6.3.4. — With the notations of (6.3.15) and (6.3.11) we have

B3 =#

1!|,|!2

-ei$'(&,y,") d,(!, y, #, +)($,

y *+4 )(y)*5

+y ' x(!, #)"!#

,

· -4

+D

+

,U(t, y, +) dy + J+

$ (t) v(#)

*J+$ (t) v*L2(W+) " CN +!N *v*L2(Rn).(6.3.37)

Now we state the following result.



Lemma 6.3.5. — Let b = b(!, y, #, +) be a bounded symbol. Let us set

G(!, #) =-

ei$'(&,y,") *5

+y ' x(!, #)"!#

,b(!, y, #, +) v(y) dy.

Then one can find C > 0 such that for all |!| " +T and v $ L2(Rn),

00e!"2 |"!|2 G(!, ·)

00L2(W+)

" C"!#n

+n/2*v*L2(Rn).

Proof. — Let us write

G(!, #) =-

K(!, #, y, +) v(y) dy.

Then using the estimate (6.3.3) we see that

e!"2 |"!|2 1W+(#) |K(!, #, y, +)| " C e

! "16

|y"x(&,()|2

&&'2 1 12 !|"!|!2.

From this estimate we can use the Schur Lemma (making the change of variables2# = (x(!, #), %(!, #))) to conclude.

Let now -5 $ C"0 (Rn) be such that 0 " -5 " 1 and

(6.3.38)

')

*-5(%) = 1 if

555 %|%| + %0

555 " 7&2, a ' 7&2 " |%| " b + 7&2

supp -5 ,!555 %

|%| + %0

555 " 8&2

", a ' 8&2 " |%| " b + 8&2

The analogue of Lemma 6.2.3 proves that one can find C > 0, (0 > 0 such that

(6.3.39)000-4

+D

+

,T #

)$y[(1 ' -5(.%))v]000

L2(Rny )

" C e!#0$ *e!"2 |)!|2 v*L2(Rn

) ).


B3 =#

1!|,|!2

-ei$'(&,y,") d,(!, y, #, +)($,

y *+4 )(y)

· *5

+y ' x(!, #)"!#

,T #

)$y[-5(.%)Tz$) U(t, z, +)] dy

+ J+$ (t) v(#)


Proof. — We use Corollary 6.3.4, (6.3.39) and Lemma 6.3.3 to remove -4

$D$

%.

Let now *+6 = *+

6 (.x) $ C"0 (Rn) be such that 0 " *+

6 " 1 and

(6.3.40)

6*+

6 (.x) = 1 if 72 &1 " '.x · %0 " 6 &1

supp *+6 ,

:.x : 17

5 &1 " '.x · %0 " 7 &1

;.



Then we can apply Lemma 6.2.6 with *a = $,y *+

4 , |3| ! 1, *b = *+6 . Indeed on the

support of $,y *+

4 (y)(1 ' *+6 (.x)) we have 15

4 &1 " 'y · %0 " 5 &1 and '.x · %0 " 72 &1

or '.x · %0 ! 6 &1. In the first case we write

|y ' .x| ! .x · %0 ' y · %0 ! 154

&1 '72

&1 =14

&1,

and in the second case we have,

|y ' .x| ! y · %0 ' .x · %0 ! 6 &1 ' 5 &1 = &1.

Therefore we obtain

(6.3.41) *$,y *+

4 (y)T #)$y -5(.%)(1 ' *6(.x))W*L2 " C e!#$ *e!"

2 |)!|2 W*L2 .

Using Corollary 6.3.6 we deduce the following Lemma.

Lemma 6.3.7. — We have

B3 =#

1!|,|!2

-ei$'(&,y,") d,(!, y, #, +)($,

y *+4 )(y)

· *5

+y ' x(!, #)"!#

,T #

)$y[-5(.%)*+6 (.x)Tz$) U(t, ·, +)](y) dy

+ J+$ (t) v(#),


Now on the support of -5(.%)*+6 (.x) we have by (6.3.38), (6.3.40),

|.x · .%| "+|.x · %0| + |.x|

555.%

|.%|+ %0

555,|.%|

" (16 &1 + 8 &2 |.x|) |.%| " c0 ".x# |.%|

if 16 &1 + 8 &2 " c0. Therefore we are in the case (i) of (6.3.17) and since Theorem6.3.1 is already proved in this case for t $ ['T, T ] we can write,

(6.3.42) Tz$) U(t, ·, +) = +3n/4

-ei$'(!$t,z,)) a('+t, z, ., +)*+

4 (z)

· *5

+z ' x('+t, z)"+t#

,(*+

2 v)(z) dz + J+$ (t) v(.),

where J+$ (t) satisfies

(6.3.43)

6for every N $ N one can find CN > 0 such that00e!"

2 |)!|2 -5(.%)*+6 (.x)J+

$ (t)v00

L2 " CN +!N *v*L2(Rn).

From this we can deduce the following result.




Tz$) U(t, ·, +) = +3n/4

-ei'(!$t,z,)) a('+t, z, ., +)

· *5

+z ' x('+t, .)"+t#

,*+

3 (z) (*+2 v)(z) dz + J+

$ (t) v(.)


Proof. — We have just to show that we can replace *+4 by *+

3 in (VI.3.42). But thisis obvious since (see (6.2.1), (6.3.1)) we have *+

4 *+2 = *+

4 (1'*+3 )*+

2 +*+4 *+

3 *+2 and

(1 ' *+3 )*+

2 3 0, *+4 *+

3 = *+3 .

We are ready now to prove (6.3.26).

Lemma 6.3.9. — For all N $ N one can find CN > 0 such that00e!

"2 |"!|2 B3(!, t, ·, +)

00L2(W+)

" CN +!N *v*L2(Rn)

for all + ! 1, ! = +(s ' t) $ [0, +T ] and all v $ L2(Rn).

Proof. — We use first Corollary 6.3.8 and Lemma 6.3.7. On the support of-5(.%)*+

6 (.x)*5

$ z!x(!$t,))'$t(

%we have by (6.3.38), (6.3.40), (6.3.2), since x('+t, .) =

.x ' 2+t .% + O(("t#),

z · %0 " (z ' x('+t, .)) · %0 + (.x ' 2+t .%) · %0 + C ( "+t#

" .x · %0 ' 2+t .% ·+%0 +

.%

|.%|

,+ 2+t |.%| + C (( + &)"+t#

" 72

&1 + 2+t |.%| + C(( + & + &2)"+t#.

Since |.%| ! a ' &2 we obtain

z · %0 " '175

&1 ' 2(a ' &2)+ |t| + C(( + & + &2)"+t#.

Taking (, &, &2 small with respect to &1 and a we obtain z · %0 " ' 103 &1. Now on the

support of *+3 (z) we have by (6.2.1), z · %0 ! '3 &1.

It follows from Corollary 6.3.8 that Tz$) U(t, ·, +) = R+v where R+ satisfies(6.3.37). Now we use Lemma 6.3.5 and we obtain since |!| " +T ,00e!

"2 |"!|2 B3(!, ·, +)

00L2(W+)

" C"!#n/2

+n/2

00T)$y[-5(.%)*+6 (.x)Tz$) U(t, ·, +)]

00L2(Rn)

"00e!"

2 |)!|2 -5(.%)*+6 (.x)Tz$) U(t, ·)

00L2 .

Since Tz$) U(t, ·) = R+v, where R+ satisfies (6.3.37), we obtain the conclusion ofLemma 6.3.9.

To complete the proof of Theorem 6.3.1 in the case (6.3.25) we use (6.3.11), (6.3.14),Lemmas 6.3.2, 6.3.9 and the same argument as in the end of the proof of the case(6.3.17) to remove the cut-o! functions *+

4 (y) and *5(y ' #x).


6.4. CONCLUSION OF CHAPTER 6 169

6.4. Conclusion of Chapter 6

Here we state a result which combines the conclusions of Theorems 6.2.2, 6.3.1 and(6.3.5).

Theorem 6.4.1. — Let K± the operators defined in (6.2.8). Then we can write

K+(t)u(x) = I + II + III

where

I = +3n/2 *+1 (x)-2

+Dx

+

,*+

2 (x)

L --ei$F (!$t,x,y,") a('+t, y, #)*+

2 (y)*+3 (#x)

· -3(#%)*5

+y ' x('+t, #)"+t#

,+-2

+D

+

,*+

1 u,(y) dy d#

M

II = *+1 (x)-2

+Dx

+

,*+

2 (x)T #"$x

I*+

3 (#x)(#x)-3(#%)J+$ (t)

+*+

2 -2

+Dx

+

,*+

1 u,J

III = R+$ (t)u

where

(6.4.1)

'(((()

((((*

F ('+t, x, y, #) = )('+t, y, #) ' (x ' #x) · #% + i2 |x ' #x|2 + i

2 |#%|2,|a('+t, y, #, +)| " C "+t#!n/2,

**+3 (#x)-3(#%) e!

"2 |"!|2 J+

$ (t) v*L2 " C +!N *v*L2, %N $ N,

*R+$ (t)u*H2N " CN *u*H"2N , %N $ N,

and *+i , -j have been defined in (6.2.1) to (6.2.6), (6.2.9), (6.3.1) and (6.3.2). More-

over the same result holds with the minus sign.


CHAPTER 7

THE DISPERSION ESTIMATE AND THE END OFTHE PROOF OF THEOREM 1.0.1

7.1. The dispersion estimate for the operators K±(t)

Let us recall that K±(t) have been introduced in (6.2.8). The purpose of thisparagraph is to prove the following result.

Theorem 7.1.1. — Let T > 0. Then there exists a constant C ! 0 such that

*K±(t)u*L! " C

|t|n/2*u*L1

for all 0 < |t| " T and all u $ L1(Rn).

Proof. — We shall use Theorem 6.4.1 and its notation and we shall consider onlyK+(t). Then we can write

(7.1.1) *K+(t)u*L! " *I*L! + *II*L! + *III*L!.

Let N0 $ N be such that 2N0 > n2 . By the Sobolev embedding and (6.4.1) we have

(7.1.2) *III*L! " C *R+$ (t)u*H2N0 " C%

N0*u*H"2N0 " C%% *u*L1 " C(T )

|t|n/2*u*L1.

Let us consider the term II. We have

*II*L! " C000*+

1 -2

+D

+

,*+

2 T #3

*+3 -3 J+

$

+*+

2 -2

+D

+

,*+

1 u,4000

H2N0

" C% +2N0 *T #(· · · )*L2(Rn)

" C%% +2N0

000e!"2 |"!|2 *+

3 (#x)-3(#%)J+$ (t)

3*+

2 -2

+D

+

,*+

1 u

4000L2(R2n

( )

" CN +2N0!N000*+

2 -2

+D

+

,*+

1 u000

L2(Rn)

" C%N +2N0!N

000-2

+D

+

,(I ' $)!N0 *+

1 u000

L2(Rn)

" C%%N +4N0!N *u*H"2N0 .

172 CHAPTER 7. THE DISPERSION ESTIMATE

Taking N = 4N0 we obtain finally,

(7.1.3) *II*L! " C *u*L1 " C(T )|t|n/2

*u*L1.

So we are left with the estimation of *I*L!. Let us set

(7.1.4) k+(t, x, y, +) = +3n/2

-ei$F (!$t,x,y,") a('+t, y, #, +)*+

2 (y)*+3 (#x)-3(#%)

· *5

+y ' x('+t, #)"+t#

,d#

and

(7.1.5) 2K+(t) v(x) =-

k+(t, x, y, +).-2

+D

+

,*+

1 v/(y) dy.

Then

(7.1.6) I = *+1 -2

+D

+

,*+

22K+(t) v.

Since the operator *+1 -2

$D$

%*+

2 is bounded from L" to L" with bound independentof + we have,

(7.1.7) *I*L! " C * 2K+(t) v*L! .

Assume that the kernel k+ has the following bound,

(7.1.8) |k+(t, x, y, +)| " C

|t|n/2,

with C independent of +. It will follow from (7.1.7), (7.1.5) and (7.1.8) that

*I*L! " C

|t|n/2

000-2

+D

+

,*+

1 v000

L1.

Since the operator -2

$D$

%*+

1 is uniformly bounded on L1 we will have

(7.1.9) *I*L! " C

|t|n/2*v*L1 .

Then Theorem 7.1.1 follows from (7.1.1), (7.1.2), (7.1.3) and (7.1.9).

Proof of (7.1.8). — We divide the proof in three cases: +t ! 1, +t " '1, |+t| " 1.Let us remark first that in the integral in the right hand side of (6.1.4), on the

support of *+3 (#x) · -3(#%) we have '#x · %0 " 3 &1 and

55 "!

|"!| + %0

55 " 4 &2. Therefore

#x · #%

|#%|= #x ·

+ #%

|#%|+ %0

,' #x · %0 " (4 &2 + 3 &1)"#x#

so

(7.1.10) #x · #% " c0 "#x# |#%|,

if &1 and &2 are small compared to c0.


7.1. THE DISPERSION ESTIMATE FOR THE OPERATORS K±(t) 173

Case A: proof of (7.1.8) when + t ! 1. — In this case ! = '+t < 0 and it followsfrom (7.1.10) and Definition 3.2.2 that all the points # in the integral giving k+ areoutgoing for ! < 0. It follows from Corollary 3.3.3 that

(7.1.11)$xj

$#k%

(!, #) = 2! &jk + O(( "!#), 1 " j, k " n.

Now using Theorem 6.4.1 and (7.1.4) we obtain

(7.1.12) |k+(t, x, y, +)| " c +3n/2

-e! "

16|y"x(""t,()|2

&"t'2 !"2 |x!"x|2 -3(#%) "+t#!n/2 d#.

By (7.1.11) we can make the change of variables

2#x = #x, 2#% = x('+t, #)

and55 det (e"

("

55 ! (+t)n if ( is small enough (since "+t# "/

2 |+t|). It follows from(7.1.12) that

|k+(t, x, y, +)| " +3n/2 "+t#!n/2 |+t|!n

--e! "

16|y" e(!|2

&"t'2 e!"2 |x!e"x|2 d2#.

Setting 2#% ' y = 4'$t()$

z1, 2#x ' x =)

2)$

z2, Z = (z1, z2) we obtain

|k+(t, x, y, +)| " C +3n/2 "+t#!n/2 (+t)!n "+t#n +!n

-

R2n

e!|Z|2 dZ

so

|k+(t, x, y, +)| " C

tn/2,

since "+t# "/

2+t. This proves (7.1.8) in this case.

Case B: proof of (7.1.4) when +t " '1. — In the right hand side of (7.1.4) weintegrate on the support of *+

3 (#x) ·-3(#%) on which we have (7.1.10). We divide thissupport in two subsets U1 and U2 where

U1 =!# = (#x, #%) $ supp(*+

3 (#x)-3(#%)) : 'c0 "#x# |#%| " #x · #% " c0"#x# |#%|"

U2 =!# = (#x, #%) $ supp(*+

3 (#x)-3(#%)) : #x · #% " 'c0 "#x# |#%|"

According to Definition 3.2.2 we have U1 , S+ ! S! (which means that the points inU1 are outgoing both for ! ! 0 and ! " 0).

According to Corollary 3.3.3 we have (7.1.11) for ! $ R so in particular for! = '+t ! 1. Therefore the same arguments as those used in case 1 work. It followsthat the part of the integral giving k+ which concerns U1 is bounded by C |t|!n/2.We consider now the integral on U2. Here ! = '+t ! 1 and the points in U2 areincoming for ! ! 0. The needed estimate on k+ will follow from the following result.



Proposition 7.1.2. — One can find a function g = g(!, y, #x) such that for all! ! 1, all # $ U2 and all y $ supp

$*+

2 (y)*5

$y!x(&,")'&(

%%we have,

|#% ' g(!, y, #x)| " C

!|y ' x(!, #)|.

For the proof we need the following Lemma.

Lemma 7.1.3. — Let # $ U2 and ! ! 1. Then for all Y $ Rn such that,')

*(i) Y · #% " 20 c0 "Y # |#%|,

(ii)555Y ' #x

2!' #%

555 " c0,

there exists a unique .%(!, Y, #x) $ Rn such that6|.%(!, Y, #x) ' #%| " 2 c0,

x('!, Y, .%(!, z, #x)) = #x.

Proof. — Let E = {.% $ Rn : |.% ' #%| " 2 c0}. Then we have

Y · .% = Y · #% + Y · (.% ' #%) " 20 c0 "Y # |#%| + 2 c0 "Y # <14"Y # |#%|.

It follows that the point (Y, .%) belongs to S! (see Definition 3.2.2). Since '! < 0 itfollows from Proposition 3.3.1 that the equation x('!, Y, .%) = #x is equivalent to,

Y ' 2!%('!, Y, .%) + z('!, Y, .%) = #x,

that is to the equation, Y '2! .% '2! 1('!, Y, .%)+ z('!, Y, .%) = #x. This equationcan be written,

.% =Y ' #x

2!' 1('!, Y, .%) +

12!

z('!, Y, .%) =: F (.%).

We show now that we can apply the fixed point theorem to F in the set E. First ofall if .% $ E we have,

|F (.%) ' #%| "555Y ' #x

2!' #%

555+ |1('!, Y, .%)| +12!

|z('!, Y, .%)|.

Using our assumption and Proposition 3.3.2 we obtain,

|F (.%) ' #%| " c0 + 2( " 2 c0 if 2( " c0.

Now if .% $ E, .%% $ E we have again by Proposition 3.3.2,

|F (.%) ' F (.%%)| " C ( |.% ' .%

%|.

Taking ( so small that C ( < 1 we obtain the conclusion of the Lemma.

Remark 7.1.4. — Let (!, #, y) as in Proposition 7.1.2. Then they satisfy the con-ditions of Lemma 7.1.3. Indeed we have y $ supp *+

4 that is 'y · %0 " 5 &1 and55 "!

|"!| + %0| " 4 &2. It follows that

y · #%

|#%|" y ·

+ #%

|#%|+ %0

,' y · %0 " 4 &2 |y| + 5 &1 < 10 c0 "y#

since &1 and &2 are small compared to c0.



Moreover we have |y ' x(!, #)| " & "!# "/

2 &!. By Proposition 3.4.1 we havex(!, #) = #x + 2! #% + O(( "!#) so

|y ' #x ' 2! #% + O(("!#)| "/

2 &!.

It follows that 555y ' #x

2!' #%

555 " C (& + () " c0,

if ( + & is small enough.

Now for fixed ! ! 1 and # $ U2 let us set

(7.1.13) A =7

Y $ Rn : Y · #% < 20 c0 "Y # |#%|,555Y ' #x

2!' #%

555 < c0

K,

and for Y $ A let us set

(7.1.14) H(Y ) = %('!, Y, .%(!, Y, #x)).

Lemma 7.1.5. — There exists a constant C > 0 independent of !, # such that000

$H

$Y(Y )000 " C

!

for every Y in A.

Proof. — By Lemma 7.1.3 we have for j = 1, . . . , n,

xj('!, Y, .%(!, Y, #x)) = #jx.

Let us di!erentiate this equality with respect to Yk. We obtain

$xj

$xk('!, Y, .%(!, Y, #x)) +

#

*

$xj

$%*('!, Y, .%(!, Y, #x))

$.*%

$Yk(!, Y, #x) = 0.

Since the point (Y, .%) belongs to S! we have by Corollary 3.3.3

$xj

$xk('!, Y, .%(· · · )) = &jk + O((!),

$xj

$%*('!, Y, .%(· · · )) = '2! &jk + O((!).

It follows that

(7.1.15)000

$.%

$Y(!, Y, #x)

000 " C+1

!+ (,.

Now thanks again to the fact that (Y, .%) $ O! we deduce from Proposition 3.3.1that

#jx = xj('!, Y, .%(!, Y, #x)) = Yj ' 2! Hj(Y ) + zj('!, Y, .%(· · · )).

Di!erentiating with respect to Yk yields

&jk ' 2!$Hj

$Yk(Y ) +

$zj

$xk('!, Y, .%(· · · )) +

n#

*=1

$zj

$%*('!, Y, .%(· · · ))

$.%

$Yk(!, Y, #x) = 0.

Using Proposition 3.3.2 and (7.1.15) we obtain the claim of the Lemma.



Proof of Proposition 7.1.2. — We shall prove that the function g defined by,

(7.1.16) g(!, y, #x) = H(y) = %('!, y, .%(!, y, #x)),

satisfies the claim of the Proposition.To do so we must consider several cases.

Case 1. — Here we assume that

(7.1.17)

6x(!, #) · #% " 10 c0 "x(!, #)# |#% |,|y ' x(!, #)| < |x(!, #)|.

Let us show that this implies that

(7.1.18) t y + (1 ' t)x(!, #) $ A for all t $ [0, 1]

where A has been defined in (7.1.13).First of all using the Remark 7.1.4 and (7.1.17) we can write,

(ty + (1 ' t)x(!, #)) · #% < 10 c0(t "y# + (1 ' t)"x(!, #)#) |#% |" 10 c0[t(1 + |y|) + (1 ' t)(1 + |x(!, #)|)] |#% |.

Now we use Lemma 4.4.16 and we obtain

(ty + (1 ' t)x(!, #)) · #% < 10 c0(1 +/

2 · |ty + (1 ' t)x(!, #)|)|#% |" 20 c0"ty + (1 ' t)x(!, #)# |#% |.


|ty + (1 ' t)x(!, #) ' #x ' 2! #%| " t |y ' x(!, #)| + |x(!, #) ' #x ' 20 #%|" 2t &! + C (! < 2c0 !,

since y $ supp *5

$y!x(&,")'&(

%and (, & are small compared to c0.

In particular (7.1.18) for t = 0 and t = 1 shows that we can apply Lemma 7.1.3 toY = x(!, #), Y = y. Therefore we have

x('!, x(!, #), .%(!, x(!, #), #x)) = #x.

But we have alsox('!, x(!, #), %(!, #)) = #x

and since |%(!, #) ' #%| " C ( " 2 c0 it follows from the uniqueness of .% in the setE (see the proof of Lemma 7.1.3) that .%(!, x(!, #), #x) = %(!, #). Therefore we haveby (7.1.16),

#% = %('!, x(!, #), .%(!, x(!, #), #x)) = H(x(!, #)).Finally we write

|#j% ' gj(!, y, #x)| = |Hj(x(!, #)) ' Hj(y)|

" |y ' x(!, #)|- 1

0

n#

k=1

555$Hj

$Yk(ty + (1 ' t)x(!, #))

555 dt



so using (7.1.18) and Lemma 7.1.5 we obtain

|#% ' g(!, y, #x)| " C

!|y ' x(!, #)|.

Case 2. — Here we assume that

(7.1.19)

6x(!, #) · #% " 10 c0 "x(!, #)# |#% |,|y ' x(!, #)| > |x(!, #)|.

It follows that

(7.1.20) |y| " 2 |y ' x(!, #)|.

We claim that in this case we have

(7.1.21) ty $ A, tx(!, #) $ A for t $ [0, 1].

Indeed we have by Remark 7.1.4

ty · #% " t · 10 c0 "y# |#%| " 10 c0 "ty# |#%||ty ' #x ' 2! #%| " t |y| + |x(!, #)| + |x(!, #) ' #x ' 2! #%|

" 3 |y ' x(!, #)| + C ( "!# " 2(C ( + 3&) ! " 2 c0!.

tx(!, #) · #% " t · 10 c0 "x(!, #)# |#% | " 10 c0 "tx(!, #)# |#% ||tx(!, #) ' #x ' 2! #%| " (1 ' t)|x(!, #)| + |x(!, #) ' #x ' 2! #%| " 2 c0 !.

As before we have

(1) = |#% ' g(!, y, #x)| = |H(x(!, #)) ' H(y)|

and we write

(1) " |H(x(!, #)) ' H(0)| + |H(0) ' H(y)|

" |x(!, #)|- 1

0

000$H

$Y(tx(!, #))

000 dt + |y|- 1

0

000$H

$Y(t, y)

000 dt.

Then we use (7.1.21), Lemma 7.1.5 and (7.1.19), (7.1.20) to conclude that

(1) " C

!|y ' x(!, #)|.

The last case is the following.

Case 3. — We assume that

(7.1.22) x(!, #) · #% > 10 c0 "x(!, #)# |#% |.

Recall that # $ U2 that is in particular #x ·#% " 'c0 "#x# |#%|. By (4.4.49) there exists!# $ ]0, ![ depending only on # such that x(!#, #) · #% = 0. Then by Lemma 4.4.17we have

(7.1.23)

'(()

((*

32 |! ' !#| |#%| " |x(!, #) ' x(!#, #)| " 3 |! ' !#| |#%|

|! ' !#| ! c0

40|y ' x(!, #)| " |y ' x(!#, #)| + |x(!#, #) ' x(!, #)| " 5 |y ' x(!, #)|.



Here again we consider two subcases.

Case 3.1: x(!, #) ·#% > 10 c0 "x(!, #)# |#% |, |y'x(!#, #)| < |x(!#, #)|. — In this case,(7.1.23) ensures that

(7.1.24) ty + (1 ' t)x(!#, #) $ A.

This follows from Lemma 4.4.16, since t |y| "/

2 |ty + (1 ' t)x(!#, #)| and from thefollowing estimates,

|ty + (1 ' t)x(!#, #) ' #x ' 2! #%| " t |y ' x(!#, #)| + |x(!#, #) ' #x ' 2! #%|" 5t |y ' x(!, #)| + 2 |! ' !#| |#%| + O(( !#) " C(& + () ! " 2 c0!.

Then we write

|#% ' g(y, !, #x)| = |H(y) ' #%| " |H(y) ' H(x(!#, #))|< => ?(1)

+ |H(x(!#, #)) ' #%|< => ?(2)

.

By (7.1.24), Lemma 7.1.5 and (7.1.23) we have

(7.1.25) (1) " C

!|y ' x(!, #)|.

Now we have

x+' !, x(!#, #),

!#

!%(!#, #)

,= x('!#, x(!#, #), %(!#, #)) = #x.

Therefore

.%(!, x(!#, #), #x) =!#

!%(!#, #).

It follows that

H(x(!#, #)) = %+' !, x(!#, #),

!#

!%(!#, #)

,=

!#

!%('!#, x(!#, #), %(!#, #)) =

!#

!#%.

Therefore

(7.1.26) (2) = |#%||! ' !#|

!" C

!|y ' x(!, #)|

by (7.1.23). The estimates obtained on (1) and (2) show that

|#% ' g(!, y, #x)| " C

!|y ' x(!, #)|

which is the claim of Proposition 7.1.2.The last step concerns the following case.



Case 3.2: x(!, #) · #% > 10 c0 "x(!, #)# |#% |, |x(!#, #)| " |y ' x(!#, #)|. — It followsthen from (7.1.23) that we have

(7.1.27) |y| + |x(!#, #)| " C |y ' x(!, #)|.

Moreover we have

(7.1.28) ty $ A, tx(!#, #) $ A for t $ [0, 1].

Indeed we can write,

ty · #% " t · 10 c0 "y# |#%| " 10 c0 "ty# |#%||ty ' #x ' 2! #%| " t |y| + |x(!#, #)| + |x(!#, #) ' #x ' 2! #%|

" C(|y ' x(!, #)| + |! ' !#| + C ( !#) " 2 c0!

tx(!#, #) · #% = 0

|tx(!#, #) ' #x ' 2! #%| " (1 ' t)|x(!#, #)| + |x(!#, #) ' #x ' 2! #%|" C(1 ' t)|y ' x(!, #)| + C|! ' !#| + C ( !#

" 2 c0 !

by (7.1.23). Then we can write

|#% ' g(!, y, #x)| " |H(y) ' H(x(!#, #)| + |H(x(!#, #) ' #%|" |H(y) ' H(0)| + |H(0) ' H(x(!#, #))| + |H(x(!#, #) ' #%|.

Using (7.1.28), Lemma 7.1.5, (7.1.26) we obtain

|#% ' g(!, y, #x)| " C

!| y ' x(!, #)|.

This completes the proof of Proposition 7.1.2.

End of the proof of (7.1.8) in case B, (+t " '1). — For the part of the integral in(7.1.4) (giving k+) where # $ U2 we use Proposition 7.1.2. Let us call it (1). As in(7.1.12) we have

|(1)| " C +3n/2

-e! 1

16|y"x("" t,()|2

&"t'2 !"2 |x!"x|2 "+t#!n/2 d#

|(1)| " C +3n/2"+t#!n/2

--e!#0

|" t|2"

&"t'2 |"!!g(!$t,y,"x)|2e!

"2 |x!"x|2 d#x d#%

|(1)| " C% +3n/2"+ t#!n/2"+ t#n

+n |+ t|n " C%%

|t|n/2.

Case C: proof of (7.1.8) when |+t| " 1. — In this case the proof made above does notgive the needed result since "+t# : 1. We will use instead a stationary phase method.Let us set ! = '+t and let us recall (see (7.1.4)) that we have to bound the following



kernel

(7.1.29) k+(t, x, y, +) = +3n/2

-ei$F (&,x,y,") a(!, y, #)*+

2 (y)*+3 (#x)-3(#%)

*5

+y ' x(!, #)"!#

,d#.

Let us recall also that, according to the Theorems 6.4.1 and 4.1.2 we have,

(7.1.30) Im F (!, x, y, #) ! 12|x ' #x|2.

Let *6 $ C"0 Rn) be such that,

(7.1.31)

6*6(x) = 1 if |x| " 1

2 ,

*6(x) = 0 if |x| ! 1,

We write in the integral in the right hand side of (7.1.29) 1 = *6(x'#x)+1'*6(x'#x).The part of the integral containing 1'*6(x'#x) can be bounded, using (7.1.30),

by the quantity

C +3n/2 e!"16

--e!

"4 |x!"x|2 |-3(#%)| d#

which is 0(1) uniformly in (t, x, y, +). Setting a1(!, y, #) = a(!, y, #)*+2 (y)*+

3 (#x) wesee therefore that we are left with the bound of the following kernel.

(7.1.32) 2k+(t, x, y, +)

= +3n/2

-ei$F (&,x,y,") a1(!, y, #)*6(x ' #x)-3(#%)*5

+y ' x(!, #)"!#

,d#.

Now, according to Theorem 6.4.1 we have,

F (!, x, y, #) = )(!, y, #) ' (x ' #x) · #% +i

2|x ' #x|2 '

12i

|#%|2.

Using Theorem 4.5.1 we obtain

(7.1.33) F (!, x, y, #)

=(y ' x) · #% ' 2i!(x ' #x) · #% ' ! |#%|2 ' !|x ' #x|2 + i

2 |x ' #x|2 + i2 |y ' #x|2

1 + 2i!+ R(!, y, #)

where

(7.1.34)

'(((((((()

((((((((*

555$R

$#x

555+555$2R

$#2x

555 " C(( + &)(|y ' #x| + |!|),555$R

$#%

555+555$2R

$#2%

555+555

$2R

$#x$#%

555 " C(( + &) |!|,

|$A1"x

$A2"!

R| "6

CA1 if A2 = 0,CA1,A2 |!| if |A2| ! 1.



It follows that we have,

(7.1.35)

'(((((((((((((((((((((((()

((((((((((((((((((((((((*

$F

$#%=

11 + 2i!

[(y ' x) ' 2i!(x ' #x) ' 2! #%] +$R

$#%

$F

$#x=

11 + 2i!

[2i! #% + 2!(x ' #x) ' 2i(x ' #x) ' i(y ' x)] +$R

$#x

$2F

$#j% $#k

%

='2! &jk

1 + 2i!+

$2R

$#j% $#k

%

$2F

$#j% $#k

x

=2i! &jk

1 + 2i!+

$2R

$#j% $#k

x

$2F

$#jx $#k

x

=(2i ' 2!)&jk

1 + 2i!+

$2R

$#jx $#k

x

$|A|F

$#A=

$|A|R

$#Aif |A| ! 3.

Let us recall that (6.2.9) shows that

(7.1.36) supp -3(#%) ,:#% : a ' 4&2 " |#%| " b + 4&2

;.

We shall divide the proof into three cases

(7.1.37)

'((((()

(((((*

case 1:|x ' y|2|!| " a ' 5&2

case 2:|x ' y|2|!| ! b + 5&2

case 3: a ' 5&2 " |x ' y|2|!| " b + 5&2

Case 1. — We have the following result.

Lemma 7.1.6. — When x ' #x $ supp *6, y ' x(!, #) $ supp *5, #% $ supp -3 wehave

Q =:555$F

$#x

5552+

1|!|

555$F

$#%

5552

! C(|x ' #x|2 + |!|),

55$A1"x

$A2"!

F55 "6

CA1 if |A1| ! 1, A2 = 0,

CA1,A2 |!| if |A2| ! 1.

uniformly in (!, x, y, #).

Proof. — Let us set X = x ' #x, Y = #% ' y!x2& . Using (7.1.35) we see that

$F

$#%=

'2!

1 + 2i!(Y + i X) +

$R

$#%.




2|!||1 + 2i!| |Y + i X | "

555$F

$#%

555+ C(( + &)|!|.

Therefore

(7.1.38)4|!|

1 + 4!2(|X |2 + |Y |2) " 2

|!|

555$F

$#%

5552

+ C1(( + &)2 |!|.

By the same way we have$F

$#x=

11 + 2i!

(2i!Y ' 2iX + 2!X) +$R

$#x.

Since |y ' #x| " |y ' x| + |x ' #x| " |x ' #x| + C |!|, we obtain4

1 + 4!2(|!Y ' X |2 + !2 |X |2) " 2

555$F

$#x

5552+ C2(( + &)2(|!| + |X |2).

It follows that

(7.1.39) 2+555

$F

$#x

5552

+1|!|

+555$F

$#%

5552,

+ C3(( + &)2(|!| + |X |2)

! 41 + 4!2

(|!Y ' X |2 + |!| |Y |2).

Since 2 |!||X ||Y | " 23 |X |2 + 3

2 !2 |Y |2 we can write

|!Y ' X |2 + |!||Y |2 ! 13|X |2 ' 1

2!2 |Y |2 + |!| |Y |2 ! 1

3|X |2 +

12|!||Y |2

since |!| " 1. We deduce from (7.1.39) that

Q ! 215

(|X |2 + |!| |Y |2) ' C4(( + &)2(|!| + |X |2).

Now in case 1 we have, according to (7.1.37)

|Y | =555#% '

y ' x

2!

555 ! |#%|'|y ' x|2|!| ! &2.

Taking ( and & small compared to &2 we deduce that Q ! c(|X |2 + |!|) which is thefirst claim of the Lemma. The second claim follows easily from the fact that |Y |, |X |are uniformly bounded and from (7.1.35), (7.1.34).

Case 2. — We have the following result.

Lemma 7.1.7. — When x ' #x $ supp *6, y ' x(!, #) $ supp *5, #% $ supp -3 wehave

Q =:555$F

$#x

5552

+1

|y ' x|

555$F

$#%

5552

! c(|x ' #x|2 + |!|),

|$A"x

F | " CA if |A| ! 1,

|$A1"x

$A2"!

F | " CA1 A2 |y ' x| if |A2| ! 1.



Proof. — Here we set X = x ' #x, Y = y!x|y!x| ' 2! "!

|y!x| . Then we can write,

$F

$#%=

11 + 2i!

(|y ' x|Y ' 2i!X) +$R

$#%,

$F

$#x=

11 + 2i!

(2!X ' i(2X + |y ' x|Y )) +$R

$#x.

It follows that1

1 + 4!2(|y ' x|2 |Y |2 + 4!2 |X |2) " 2

+555$F

$#%

5552

+ C1(( + &)2 |!|2,.

Therefore we have the estimate

|y ' x| |Y |2 " 101

|y ' x|

555$F

$#%

5552+ C2(( + &)2 |!|.

On the other hand since |y ' #x| " |y ' x| + |x ' #x|

|2X + |y ' x|Y |2 " 10555$F

$#x

5552

+ C3(( + &)2(|!|2 + |y ' x|2 + |x ' #x|2).

Summing up we see that

10 Q + C4(( + &)2(|!|2 + |y ' x|2 + |X |2) ! |2X + |y ' x|Y |2 + |y ' x| |Y |2.

Writing y'x = y'x(!, #)+x(!, #)'#x +#x'x, we see that one can find a constantK0 such that |y ' x| " K0. Let 20 > 0 be such that -0

4!-0K0 " 1

2 . Then we have

4 |y ' x| |X | · |Y | " (4 ' 20) |X |2 +4

4 ' 20|y ' x|2 |Y |2.

It follows that

|2X + |y ' x|Y |2 ! 20 |X |2 ' 20

4 ' 20|y ' x|2 |Y |2.

This implies that

10 Q + C4(( + &)2(|!|2 + |y ' x|2 + |X |2) ! 20 |X |2 + |y ' x| |Y |2+1 ' 20

4 ' 20K0

,

! min+20,

12

,(|X |2 + |y ' x| |Y |2).


|Y | ! 1 ' 2 |!| |#%||y ' x| ! 1 ' b + 4&2

b + 5&2=

&2

b + 5&2

because |y ' x| ! 2(b + 5&2)|!|, since we are in case 2. Therefore

10 Q + C5(( + &)(|!| + |y ' x| + |X |2) ! C6(|X |2 + |y ' x|).

Taking ( + & small with respect to C6 we obtain the first part of the Lemma. Thebounds on the derivatives of F can be easily obtained since |!| " (b+5 &2)!1 |y'x|.



Case 3. — Recall that we have a'5 &2 " |x!y|2|&| " b+5 &2. Then we have the following

result.

Lemma 7.1.8. — One can find 00 > 0 such that the equation x(!, x, #%) = y has aunique solution in the set E =

:#% :

55#% ' y!x2&

55 " 00

;.

Proof. — First of all one can find 00 such that if #% $ E then 12 " |#%| " 2. It follows

from Proposition 3.2.1 that x(!, x, #%) = x + 2! #% + r(!, x, #%) where |r| " C ( |!|.The equation to be solved is in E equivalent to the equation #% = y!x

2& + 12& r(!, x, #%).

If we set #(#%) = y!x2& + 1

2& u(!, x, #%) then # maps E in itself if C ( < 00. Moreover,again by Proposition 3.2.1 we have |#(#%) ' #(#%

%)| " C% ( |#% ' #%%|. Taking ( small

enough the Lemma follows from the fixed point theorem.

We shall set

(7.1.40) #c = (x, #c%)

where #c% is the unique solution of x(!, x, #%) = y given by Lemma 7.1.8. Then we

have the following result.

Lemma 7.1.9. — We have$F

$#x(!, x, y, #c) =

$F

$#%(!, x, y, #c) = 0.

Proof. — Let us recall that

(7.1.41)

'(()

((*

$F

$#x= #% ' i(x ' #x) +

$)

$#x$F

$#%= '(x ' #x) + i#% +

$)

$#%.

On the other hand we have (see (4.5.17))

(7.1.42) )(!, x(!, #), #) = ! p(#) +12i

|#%|2.

Therefore

$)

$#jx

(!, x(!, #), #) = !$p

$xj(#) '

n#

k=1

$)

$xk(!, x(!, #), #)

$xk

$xj(!, #).

Moreover we have$)

$xk(!, x(!, #), #) = #(!, x(!, #), #) = %(!, #)

so

(7.1.43)$)

$#jx

(!, x(!, #), #) = !$p

$xj(#) '

n#

k=1

%k(!, #)$xj

$xj(!, #).



Now by the definition of the flow and the Euler relation we have'()

(*

n#

k=1

xk(!, #) %k(!, #) =n#

k=1

%k(!, #)$p

$%k(x(!, #), %(!, #)) = 2p(#)

p(x(!, #), %(!, #)) = p(#).

Di!erentiating these two relations with respect to #jx we obtain

$x

$xj(!, #) · %(!, #) + x(!, #) · $%

$xj(!, #) = 2

$p

$xj(#)

$p

$x(x(!, #), %(!, #)) · $x

$xj(!, #) +

$p

$%(x(!,#), %(!, #))

$%

$xj(!, #)

=$p

$xj(#).

(7.1.44)

Using the equations of the flow the last equality can be written as

'%(!, #) · $x

$xj(!, #) + x(!, #)

$%

$xj(!, #) =

$p

$xj(#).

Combining with (7.1.44) we obtain

$x

$xj(!, #) · %(!, #) + %(!, #) · $x

$xj(!, #) =

$p

$xj(#)

which can be written asd

d!

. $x

$xj(!, #) · %(!, #)

/=

$p

$xj(#).

Integrating both side we obtain$x

$xj(!, #) · %(!, #) = !

$p

$xj(#) + #j

%.

Using (7.1.43) we deduce

$)

$#jx

(!, x(!, #), #) = '#j%.

Since #c = (x, #c%) where x(!, x, #c

%) = y we deduce from (7.1.41) that

$F

$#jx

(!, x, y, #c) = #j% +

$)

$#jx

(!, x(!, #c), #c) = #j% ' #j

% = 0.

Now di!erentiating (7.1.42) with respect to #j% yields

$)

$#j%

(!, x(!, #), #) +$)

$x(!, x(!, #), #) · $x

$#j%

(!, #) = !$p

$%j(#) +

1i

#j%.

As above we see easily that,d

d!

. $x

$#j%

(!, #) · %(!, #)/

=$p

$%j(#),



from which we deduce that (x("j

!

(!, #)·%(!, #) = ! (p(%j

(#). Since ('(x (!, x(!, #), %(!, #)) =

%(!, #), we obtain$)

$#j%

(!, x(!, #), #) =1i

#j%

which implies that (F("j

!

(!, x, y, #c) = 0.

Lemma 7.1.10. — Let us set'((()

(((*

Q =.555

$F

$#x

5552

+1|!|

555$F

$#%

5552/

(!, x, y, #)

X = #x ' x

Y = #% ' #c%

where #c% has been introduced in (7.1.40). Then we have

'()

(*

Q ! C(|X |2 + |!| |Y |2),|$A

"xF | " CA if |A| ! 1,

|$A1"x

$A2"!

F | " CA1A2 |!| if |A2| ! 1,

uniformly in (!, x, y, #) when |!| " 1, x ' #x $ supp *6, y ' x(!, #) $ supp *6,#% $ supp -3.

Proof. — For t $ (0, 1) let us set mt = (!, x, y, t# + (1 ' t)#c). Then usingLemma 7.1.9 we can write for j = 1, . . . , n,

$F

$#jx

(!, x, y, #) =n#

k=1

3- 1

0

I$2F

$#jx$#k

x

(mt)(#kx'xk)+

$2F

$#jx$#k

%

(mt)(#k%'(#c

%)k)k)

Jdt

4

$F

$#j%

(!, x, y, #) =n#

k=1

3- 1

0

I$2F

$#j%$#k

x

(mt)(#kx 'xk)+

$2F

$#j%$#k

%

(mt)(#k% ' (#c

%)k)J

dt

4.


$F

$#jx

=2i ' 2!

1 + 2i!Xj +

2i!

1 + 2i!Yj + O[(( + &)(|X | + |!||Y |]

$F

$#j%

=2i!

1 + 2i!Xj '

2!

1 + 2i!Yj + O[(( + &)|!|(|X | + |Y |)].

Therefore we have

!2 |X |2 + |X ' !Y |2 + |!| |X |2 + |!| |Y |2 " 52

Q + C(( + &)2(|X |2 + |!|2 |Y |2).

Taking (+& small enough we obtain Q ! C(|X |2+|!| |Y |2) which the first claim of theLemma. The other claims follows from (7.1.35) and the fact that |x'y| " 2(b+5 &2)|!|since we are in the case 3.



We shall set in what follows,

(7.1.45)

'((((()

(((((*

L =1i

n#

j=1

+ $F

$#jx

(!, x, y, #)$

$#jx

+1D

$F

$#j%

(!, x, y, #)$

$#j%

,

Q =+555

$F

$#x

5552

+1D

555$F

$#%

5552,

(!, x, y, #)

where D = |!| in the cases 1 and 3, D = |x ' y| in the case 2.

Let us note that, according to Lemmas 7.1.6, 7.1.7 and 7.1.10 L is a vector field whosecoe"cients are uniformly bounded together with their derivatives with respect to #.Moreover we have

(7.1.46) L ei$F = +Q ei$F .

Our first goal is to prove the following result.

Proposition 7.1.11. — For any N in Nn we can write

LN ei$F (&,x,y,") =++N QN +

N!1#

k=1

hk,N (!, x, y, #)+k,

ei$F (&,x,y,")

where hk,N are smooth functions satisfying

|Lj hk,N (!, x, y, #)| " Cj,N Qk, 0 " k " N ' 1, j $ N,

uniformly when 0 < |!| " 1, |x ' #x| " 1, |y ' x(!, #)| " 2&, |#%| " 2.

Proof

Step 1. — Let H be the set of C" functions c = c(!, x, y, #) such that for any3 $ N2n, $,

"c is uniformly bounded when 0 < |!| " 1, |x'#x| " 1, |y ' x(!, #)| " 2&,|#%| " 2. For instance (F

("jx

and 1D

(F("j

!

belong to H.

Let us set T = (F("x

, T = (F("x

, S = 1)D

(F("!

, S = 1)D

(F("!

.Let P be the set of homogeneous polynomials of order 2 in T, T , S, S with coe"-

cients in H. For instance we have Q $ P . We claim that L sends P into P . First ofall if P $ P then (F

("jx

(P("j

x$ P since (F

("jx$ H, (2F

("2 $ H and (F("j

x= T j . On the other

hand 1D

(F("j

!

(P("j

!

$ P since 1D

(F("j

!

$ H, 1)D

(F("j

!

= T j and 1D

(2F("!(" $ H. It follows

that L maps P into P .Now if P $ P then |P | " C Q. It follows that for all j $ N,

(7.1.47) |LjQ| " CjQ

uniformly in (!, x, y, #).

Step 2. — We claim that

(7.1.48)

6for all N $ N, j $ N we have |Lj QN | " CjN QN uniformly when

0 < |!| < 1, |x ' #x| " 1, |y ' x(!, #)| " 2&, |#%| " 2.



Indeed by the Faa di Bruno formula (since L is a homogeneous vector field) Lj QN isa finite linear combination of terms of the form

QN!MsH

i=1

(L*iQ)ki

where 1 " M " N , 1 " s " M ,&s

i=1 ki = M ,&s

i=1 ki /i = j. Each of such termsis bounded, according to (7.1.47) by C QN!M Q

Psi ki = C QN!M QM = C QN which

proves our claim.

Step 3. — Proof of Proposition 7.1.11We use an induction on N . For N = 1 the result follows from (7.1.46). Let us

assume it is true up to the order N . Then

LN+1(ei$F ) =ei$F++N N QN!1 LQ +

N!1#

k=0

(L hk,N )+k + +N+1 QN+1 +N!1#

k=0

Q hk,N +k+1,

= ei$F++N+1 QN+1 +

N#

k=0

hk,N+1 +k,

where '()

(*

hN,N+1 = N QN!1 L Q + Q hN!1,N ,

hk,N+1 = L hk,N + Q hk!1,N , 1 " k " N ' 1,

h0,N+1 = L h0,N .

Now we have

|Lj hN,N+1| " Nj#

i=0

3ji

4|Li QN!1||Lj!iQ| +

j#

i=0

3ji

4|LiQ| |Lj!i hN!1,,N |.

Using (7.1.47), (7.1.48) and the induction hypothesis we deduce that |Lj hN,N+1| "Cj,N QN . The estimates for the other terms are completely analogous.

We need another Lemma.

Lemma 7.1.12

(i) For any N $ N# we can write

LN (ei$F ) = GN (!, x, y, #, +) ei$F .

(ii) There exists a constant KN > 0 such that

|KN + GN (!, x, y, #, +)| ! 12(+N QN + 1).

(iii) For any j $ N, N $ N# we can find a constant Cj,N > 0 such that

|Lj GN (!, x, y, #, +)| " Cj,N |GN (!, x, y, #, +) + KN |.



(iv) For any j $ N, N $ N# we can find a constant C%jN > 0 such that

555(tL)j+ 1

KN + GN

,555 " C%

|KN + GN |where tL denotes the transpose of L.

Proof. — (i) is a consequence of Proposition 7.1.11 with

GN (!, x, y, #, +) = +N QN +N!1#

k=0

hk,N (!, x, y, #)+k

and |hk,N | " CN Qk, 0 " k " N ' 1, CN ! 1.(ii) If +Q " 2N CN then

N!1#

k=0

|hk,N |+k " CN

N!1#

k=0

(+Q)k " CN

N!1#

k=0

(2N CN )k.

If +Q ! 2N CN then

(+Q)N = (+Q)N!k(+Q)k ! 2N CN (+Q)k if 0 " k " N ' 1.

ThenN!1#

k=0

|hk,N |+k " CN

2N CN

+N!1#

k=0

1,(+Q)N =

12(+Q)N .

ThereforeN!1#

k=0

|hk,N |+k " 12(+Q)N + K %

N .

This implies555K %

N +12

+ GN

555 ! K %N +

12

+ +N QN ' 12

+N QN ' K %N ! 1

2(+N QN + 1).

(iii) We have, by (7.1.48) and Proposition 7.1.11

|Lj GN | =555+N Lj QN +

N!1#

k=0

Lj hk,N +k555 " Cj,N +N QN +

N!1#

k=0

C%jN +k Qk

" C%%jN (1 + +N QN) " 2Cj,N |KN + GN |.

(iv) Let us recall that L = (F("x

· (("x

+ 1D

(F("!

· (("!

. According to the Lemmas 7.1.6,7.1.7 and 7.1.10, we see that tL = 'L + c(!, x, y, #) where

|$," c(!, x, y, #)| " C,

uniformly in (!, x, y, #). We claim that for j $ N

(7.1.49)

'()

(*(tL)j =

j#

k=0

cj,k(!, x, y, #)Lk, with

|$," cjk(!, x, y, #)| " cj,k,, .



We saw above that this is true when j = 1. Let us assume that (7.1.49) is true up tothe order j. Then

(tL)j+1 = ('L + c)j#

k=0

cjk(!, x, y, #)Lk =j+1#

k=0

cj+1,k(!, x, y, #)Lk

where '()

(*

cj+1,0 = c cj,0 ' L cj,0

cj+1,k = c cjk ' L cjk ' cj,k!1, 1 " k " j

cj+1,j+1 = 'cj,k.

Then the estimate on |$," cj+1,k| follows from the induction and the fact that the

coe"cients of L have all their derivatives with respect to # bounded uniformly. Itfollows that

(7.1.50)555(tL)j

+ 1KN + GN

,555 " Cj,N

j#

k=0

555Lk+ 1

KN + GN

,555.

Now by the Faa di Bruno formula, Lk$

1KN+GN

%is a finite linear combination of terms

of the form1

(KN + GN )1+)

sH

i=1

(L*i GN )ki

where 1 " . " k, 1 " s " k,&s

i=1 ki = .,&s

i=1 ki /i = k. It follows from (iii) that555Lk+ 1

KN + GN

,555 " Cj,N1

|KN + GN |1+)|KN + GN |

Psi=1 ki " Cj,N

|KN + GN | .

Then (iv) follows from (7.1.50).

We can now state the estimate on the kernel for |!| " 1.

Lemma 7.1.13. — Let 2k+ = 2k+(t, x, y, +) be the kernel defined by (7.1.32). Thenone can find a positive constant C such that

|2k+(t, x, y, +)| " C

|t|n/2

for all (t, x, y, +) such that + ! 1, |+t| " 1, x $ Rn, y $ Rn.

Proof. — The kernel 2k+ can be written as

2k+(t, x, y, +) =-

ei$F (!$t,x,y,") 2a(+t, x, y, #) d#

where

2a(+t, x, y, #) = a1('+t, y, #)*6(x ' #x)-3(#%)*5

+y ' x('+t, #)"+t#

,.

Let us note that all the derivatives of 2a with respect to # are uniformly bounded.



We use Lemma 7.1.12 to write

2k+(t, x, y, +) =-

(KN + LN ) ei$F (!$t,x,y,") 2a(+t, x, y, #)KN + GN ('+t, x, y, #, +)

d#

=-

ei$F (··· )(KN + (tL)N )I

2a(· · · )K+GN (· · · )

Jd#.

Now using (ii), (iv) of Lemma 7.1.12, (7.1.49) and the fact that ImF ! 0 we obtain

(7.1.51) |2k+(t, x, y, +)| " CN +3n/2

- | 2-(#%)|1 + +N QN

d# = CN I

where Q has been introduced in (7.1.45) and 2-(#%) is a smooth function with compactsupport. Now according to the Lemmas 7.1.6, 7.1.7 and 7.1.10 we have

6Q ! C(|x ' #x|2 + |+t|) (case 1 and 2)

Q ! C(|x ' #x|2 + |+t| |#% ' #c%|2) (case 3).

Let us fix the integer N such that N > n. Since 2- has compact support we have incase 1 and 2

I " +3n/2

-

Rn

dX

1 + +N |X |2N + +N |+t|N .

Let us set X = (1+$N |$t|N )1/2N)

$z. Then

I " +3n/2 (1 + +N |+t|N )n/2N

+n/2

11 + +N |+t|N

-

Rn

dz

1 + |z|2N.

If + |+t| " 1, that is +2 " 1|t| , we have,

I " CN +n " CN

|t|n/2.

If + |+t| ! 1, that is 1$2 " |t|, we can write for N > n,

I " CN +n (+ |+t|)n/2

(+ |+t|)N" CN

+2(N!n)|t|n

2 !N " CN |t|N!n+n2 !N " CN

|t|n/2.

It follows from (7.1.51) that |2k+(t, x, y, +)| " C|t|n/2 .

In case 3 we have

I " C +3n/2

-

R2n

d#

1 + (+ |x ' #x|2 + + |+t| |#% ' #c%|2)N

.

Setting X =/

+ (x ' #x), Y =/

+O|+t| (#% ' #c

%) we obtain

I " C +3n/2 +!n/2 +!n/2 |+t|!n/2

-

R2n

dX dY

1 + (|X |2 + |Y |2)N.

It follows that|2k(t, x, y, +)| " C

|t|n/2.

This completes the proof of Lemma 7.1.13 thus the proof of Theorem 7.1.1.



7.2. End of the proof of Theorem 2.2.1

Let us recall that we have set in (6.2.7),

(7.2.1) U±(t) = *+1 -2

+D

+

,e!itP .

It follows from Theorem 7.1.1 that there exists C ! 0 such that

*U±(t1)U±(t2)#f*L!(Rn) " C

|t1 ' t2|n/2*f*L1(Rn)

for all t1, t2 in ['T, T ] with t1 += t2 and all f $ L1(Rn). Moreover there is a conser-vation of the L2 norm for e!itP which implies,

*U±(t)u0*L2(Rn) " C *u0*L2(Rn)

for all |t| " T and u0 $ L2(Rn).We can therefore apply Lemma 2.1.3 with X = Rn, H = L2(Rn) and we obtain

(7.2.2) *U±(·)u0*Lq([!T,T ],Lr(Rn)) " C *u0*L2(Rn)

for all u0 $ L2(Rn), where q ! 2 and 2q = n

2 ' nr . Now it follows from (6.2.2) that we

have*+

1 (x) + *!1 (x) ! 1 for all x $ Rn.

Then (7.2.1) and (7.2.2) show that

(7.2.3)000-2

+D

+

,e!itP u0

000Lq(I,Lr(Rn))

" C *u0*L2(Rn)

for all u0 $ L2(Rn), where I = ['T, T ].Now let us recall (see (6.2.6)) that -2(%) = -0(%)-1(%) where

')

*-0(%) = 1 if

555 %|%| ' %0

555 " &2, |%| ! 2 &2, |%0| = 1,

supp -0 ,!

% : %|%| ' %0

555 " 2 &2

"and |%| ! &2

7-1(%) = 1 if a ' &2 " |%| " b + &2, a = 6

10 , b = 1910 ,

supp -1 , {% : a ' 2 &2 " |%| " b + 2 &2}.By a finite partition of unity we deduce easily from (7.2.3) that

(7.2.4)000-1

+D

+

,e!itP u0

000Lq(I,Lr(Rn))

" C *u0*L2(Rn).

Now let us recall that p(x, %) = |%|2 + (&n

j,k=1 bjk(x) %j %k. Therefore if ( is smallenough we have

(7.2.5)910

|%|2 " p(x, %) " 1110

|%|2.

Let )0(t) $ C"0 (R) be such that

)0(t) = 1 if |t| " 95

)(t) = 0 if |t| ! 3


7.2. END OF THE PROOF OF THEOREM 2.2.1 193

and set

(7.2.6) )(t) = )0(t) ' )0(4t).

It follows that supp ) , {t $ R : 920 " |t| " 3}.

Let 2) $ C"(R) be such that

(7.2.7)

'()

(*

supp 2) ,!t $ R : 4

10 " |t| " 3110

"

2) = 1 on a neighborhood of supp), so)(t) 2)(t) = )(t).

We claim that for every (x, %) $ T #Rn,

(7.2.8) (1 ' -1(%)) 2)(p(x, %)) 3 0.

Indeed on the support of 1 ' -1(%) we have

(i) |%| " 610

' &2 or (ii) |%| ! 1910

+ &2.

In the case (i), (7.2.5) shows that

0 " p(x, %) " 1110

|%|2 " 1110

+ 610

,2<

410

.

In the case (ii) we have

p(x, %) ! 910

+1910

,2>

3110

.

Thus 2)(p(x, %)) = 0 by (7.2.7).Now, with ) introduced in (7.2.6) we claim that

(7.2.9) (1) =000)+ P

+2

,e!itP u0

000Lq(I,Lr(Rn))

" C000)+ P

+2

,u0

000L2(Rn)

for all u0 $ L2(Rn). Indeed we can write

(1) "000-1

+D

+

,)+ P

+2

,

< => ?(2)

e!itP u0

000Lq(I,Lr)

+000+I '-1

+D

+

,,)+ P

+2

,

< => ?(3)

e!itP u0

000Lq(I,Lr)

.

Since )$

P$2

%commutes with e!itP , we deduce from (7.2.4) that

(7.2.10) (2) "000)+ P

+2

,u0

000L2(Rn)

.

Using (7.2.7) we can write

(3) =000+I ' -1

+D

+

,,2)+ P

+2

,e!itP )

+ P

+2

,u0

000Lq(I,Lr)

.

Now, since r ! 2, there exists, by the Sobolev embedding, s ! 0 such thatHs(Rn) 6. Lr(Rn). We fix such an s. Then

(3) "000+I ' -1

+D

+

,,2)+ P

+2

,e!itP )

+ P

+2

,u0

000L!(I,Hs(Rn)

.



Now we use (7.2.8) and Proposition A.1 in [BGT]. It follows that

(3) " C%000e!itP )

+ P

+2

,u0

000L!(I,L2(Rn))

" C%%000)+ P

+2

,u0

000L2(Rn)

,

which, together with (7.2.10) proves our claim (7.2.9). By (7.2.6) we have

)0(t) ++"#

k=1

)(2!2k t) = 1.

Then using Corollary 2.3 in [BGT] we can write

*e!itP u0*Lq(I,Lr(Rn)) " C *u0*L2(Rn) +0000+ +"#

k=1

00e!itP )(2!2kP )u0

002Lr(Rn)

,1/20000

Lq(I)

.

By the Minkowski inequality and (7.2.9) we obtain

*e!itP u0*Lq(I,Lr(Rn)) " C *u0*L2(Rn) ++ +"#

k=1

00)(2!2kP )u0

002L2(Rn)

,1/2.

This implies that 00e!itP u0

00Lq(I,Lr(Rn))

" C% *u0*L2 ,

which is the estimate claimed in Theorem 2.2.1.


APPENDIX

A.1. The Faa di Bruno Formula

We shall make a repeat use of the following formula which can be found in even amore precise form in the paper of Constantin and Savits [CS].

Let m $ N#, F $ Cm(RN , C), Uk $ Cm(Rp, R), k = 1, . . . , N . Then for |'| " mwe have

$#Y [F (U(Y ))] =

N#

k=1

$F

$Uk(U(Y )) $#

Y Uk(Y ) + (1)

where (1) is a finite linear combination of terms of the form

($)UF )(U(Y ))

sH

j=1

$$

Lj

Y U(Y )%Kj

where 2 " |.| " |'|, 1 " s " |'|, |Kj| ! 1, |Lj | ! 1 ands#

j=1

Ki = .,s#

j=1

|Kj |Lj = '.

The precise coe"cients of this sum can be found in [CS].

A.2. Proof of Proposition 3.2.1

The system satisfied by r and 1 is the following.'(((()

((((*

rj(t) = 21j(t) + 2(n#

k=1

bjk(x(t)) %k(t), rj(0) = 0

1j(t) = '(n#

p,q=1

$bpq

$xj(x(t)) %p(t) %q(t), 1j(0) = 0.

We prove our claim by induction on |A| + |B| = k. Moreover it is clear that (ii)implies (i) since r(0) = 1(0) = 0. Setting #(t) = |r(t)| + |1(t)| the above equations

196 APPENDIX

show that #(t) " C0( + C1

1 t0 #(") d". Thus by Gronwall inequality, #(t) " C2(T ) (.

This shows that (i) and (ii) are true for k = 0. Let us assume they are true up tothe order k ' 1. Let us set X = (x, %) and if # = (A, B) $ Nn ( Nn, $"

X = $Ax $B

% . Itfollows from the induction that6

|%(t, x, %)| " C(T )

|$"X x(t, x, %)| + |$"

X %(t, x, %)| " C"(T )

if 1 " |#| " k ' 1 and k ! 2.It follows that if 1 " |#| " k' 1 we have |$"

X [F (x(t))]| " C"(T ) if F = bjk or (bpq

(xj.

Let us now take |#| = k ! 1 and let us set #(t) = |$"X r(t)|+ |$"

X 1(t)|. Then usingthe Leibniz and Faa di Bruno formulas (see Section A.1) and di!erentiating the abovedi!erential system we find

#(t) " C" ( + C%"

- t

0#(") d"

after using the induction. We conclude again by the Gronwall inequality.

A.3. Proof of Proposition 3.3.2

The system satisfied by z and 1 is the following.

(A.3.1)

'(((((()

((((((*

zj(t) = 2(n#

k=1

bjk(x(t)) %k(t) ' 2( tn#

p,q=1

$bpq

$xj(x(t)) %p(t) %q(t)

1j(t) = '(n#

p,q=1

$bpq

$xj(x(t)) %p(t) %q(t)

zj(0) = 1j(0) = 0.

By Proposition 3.3.1 we have "x(t, x, %)# ! C(1 + |x|+ t). This implies that for / ! 1,

(A.3.2)- t

0

ds

"x(s)#*+!0" C*

"x#*!1+!0.

We proceed by induction on k. Let us begin by the case k = 0. We deduce from(A.3.1) that

|zj(t)| " C(A0, A1) (+ 1"x(t)#1+!0

+t

"x(t)#2+!0

," C%(A0, A1) (

"x(t)#1+!0

|1j(t)| " C(A1) (

"x(t)#2+!0

since |%(t, x, %)| " 2 |%| " 4. Then the estimates in Proposition 3.3.2 when k = 0follow from (A.3.2). Assume now that these estimates are true when |A|+ |B| " k'1and let us deduce several facts.

For / $ N let us introduce the space,

B*!0

=!F $ C"(Rn) : |$)F (y)| " C)

"y#*+|)|+!0, for all y $ Rn

".


A.3. PROOF OF PROPOSITION 3.3.2 197

Let us set X = (x, %) and if # = (A, B) $ Nn ( Nn, $"X = $A

x $B% . Then a straightfor-

ward computation shows that for # $ Nn ( Nn,

(A.3.3) |$"X(%j)| " 1

"x#|A| , |$"X(xj)| " 1

"x#|A|!1.

It follows that for |#| " k ' 1,

(A.3.4) |$"X %j(t, x, %)| " 2Mk!1

"x#|A| .

Indeed

|$"X(%j(t, x, %)| " |$"

X(%j)| + |$"X 1j(t, x, %)| " 1

"x#|A| +( Mk!1

"x#|A|+1+!0.

Therefore if |#| " k ' 1

(A.3.5) |$"X(%p(t) %q(t))| " Ck M2

k!1

"x#|A| .

Now if |#| = k we have from (A.3.3)

(A.3.6) $"X %j(t, x, %) = $"

X 1j(t, x, %) + Rj,", |Rj,"| " 1"x#|A|

(A.3.7)

'()

(*

$"X(%p(t) · %q(t)) = %p(t) $"

X %q(t) + %q(t) $"X %p(t) + Rp,q,"

|Rp,q,"| " Ck M2k!1

"x#|A| .

Now we claim that if F $ B*+1!0

and |#| " k ' 1 we have

(A.3.8) |$"X [F (x(t))]| "

Ck M |"|k!1

"x(t)#*+1+!0 "x#|A| .

This estimate is easy if k = 1 and if k ! 2 we use the Faa di Bruno formula (seeSection A.1). It follows that $"

X [F (x(t))] is a finite sum of terms of the following form

($)F )(x(t))sH

j=1

($*j

X x(t))kj , /j $ Nn ( Nn, kj $ Nn

where 1 " |.| " |#|, 1 " s " |#|, |kj | ! 1 ands#

j=1

kj = .,n#

j=1

|kj | /j = #.

If we write /j = (aj , bj), # = (A, B) we have in particular

(A.3.9)s#

j=1

|kj | aj = A.

Now we write {1, 2, . . . , s} = I1 2 I2 2 I3 where

I1 = {j : |/j | ! 2}, I2 = {j : |/j | = 1, /j = (aj , 0)}, I3 = {j : |/j| = 1, /j = (0, bj)}


198 APPENDIX

and we denote by (i the sum&

j+Ii, i = 1, 2, 3. When j $ I1 we have

$*j

X (xj(t)) = 2t $"X(1j(t)) + $"

X(zj(t)).

Since |/j | " |#| " k ' 1 it follows from the induction that

|$"X(x(t))| " 2t ( Mk!1

"x#|aj |+!0+1+

( Mk!1

"x#|aj |+!0=

( Mk!1

"x#|aj |+!0

+1 +

2t

"x#

,.

It follows that

(A.3.10)555H

j+I1

$$

*j

X (x(t))%kj555 " Ck(( Mk!1)$1|kj |

"x#$1|kj ||aj |

+1 +

t$1|kj |

"x#$1|kj |

,.

Now when j $ I2 we have /j = (aj , 0), |aj | = 1. Then55$aj

x (xp(t))55 " 1 +

2( t M1

"x#2+!0+

( M1

"x#1+!0" 2M1

+1 +

t

"x#2+!0

,.

Therefore we have

(A.3.11)555H

j+I2

$$*1

X (x(t))%kj555 " C(2M1)$2|kj |

+1 +

t$2|kj |

"x#$2|kj |(2+!0)

,.

Finally for j $ I3 we have /j = (0, bj), |bj| = 1. Then

|$bj

% (xp(t))| " 2M1 "t#.

It follows that

(A.3.12)555H

j+I3

$$*1

X (x(t))%kj555 " (2M1)$3|kj | "t#$3|kj |, .

Using (A.3.10), (A.3.11) and (A.3.12) we obtain

(1) =555($)F )(x(t))

sH

j=1

$$

*j

X (x(t))%kj555

"C) Mk!1

k!1

"x(t)#|)|+*+1+!0· 1"x#$1|kj ||aj|

+1 +

t$1|kj |

"x#$1|kj |

,+1 +

t$2|kj |

"x#$2|kj |

,"t#$3|kj |.

Now we have

(1 |kj | + (2 |kj | + (3 |kj | = |.|, "x(t)# ! C(t) and "x(t)# ! C "x#.

It follows that

(A.3.13) (1) "Ck Mk!1

k!1

"x(t)#*+1+!0

1"x#$1 |kj ||aj|+$2 |kj |

.

On the other hand,

|A| =s#

j=1

|kj | |aj | = (1 |kj | |aj | + (2 |kj | since |aj | = 1 for t $ I2

and aj = 0 if j $ I3. Therefore (A.3.13) implies (A.3.8) and our claim is proved.


A.3. PROOF OF PROPOSITION 3.3.2 199

Moreover if |#| = k we can write

(A.3.14)

'((()

(((*

$"X [F (x(t))] =

n#

*=1

$F

$y*(x(t)) $"

X x*(t) + R where

|R| "Ck Mk!1

k!1

"x(t)#*+1+!0· 1"x#|A| .

Indeed R is a finite sum of terms of the form ($)F )(x(t))Ts

j=1($*j

X x)kj where2 " |.| " |#|. It follows then that |/j| " |#|' 1 = k ' 1 and the above computationsare valid.

Let us now prove Proposition 3.3.2 for |#| = k. Let us set

Z(t) = $"X z(t), &(t) = "x# $"

X 1(t).

We can write

Zj(t) = 2(n#

j,k=1

7bjk(x(t)) $"

X %k(t)< => ?

(1)

+ $"X [bjk(x(t))] %k(t)< => ?

(2)

+#

"="1+"2"j ,=0

3##1

4$"1

X [bjk(x(t))] $"2X %k(t)

< => ?(3)

K' 2( t

n#

p,q=1

7$"

X

.$bpq

$xj(x(t))

/%p(t) %q(t)

< => ?(4)

+$bpq

$xj(x(t)) $"

X (%p(t) %q(t))< => ?

(5)

+#

"="1+"2"j ,=0

3##1

4$"1

X

.$bpq

$xj(x(t))

/$"2

X (%p(t) %q(t))< => ?

(6)

K

&j(t) = '( "x#n#

p,q=1

7$"

X

.$bpq

$xj(x(t))

/%p(t) %q(t)

< => ?(7)

+$bpq

$xj(x(t)) $"

X(%p(t) %q(t))< => ?

(8)

+#

"="1+"2"j ,=0

3##1

4$"1

X

.$bpq

$xj(x(t))

/$"2

X (%p(t) %q(t))< => ?

(9)

K.

We shall use the fact that bjk $ B1!0

and (bpq

(xj$ B2

!0. We deduce from (A.3.3) that

(A.3.15) |(1)| " C (

"x(t)#1+!0 "x#|A| +C (

"x(t)#1+!0

|&(t)|"x# .

To estimate the term (2) we use (A.3.14) with F = bjk and the equality

$"X xj(t) = $"

X(xj + 2t %j) + 2t&j(t)"x# + Zj(t).

We obtain

(A.3.16) |(2)| " C (

"x(t)#2+!0|Z(t)| + C (

"x(t)#1+!0

|&(t)|"x# +

C (

"x(t)#1+!0 "x#|A| .


200 APPENDIX

To estimate (3) we use (A.3.4) and (A.3.8) ; we obtain,

(A.3.17) |(3)| "Ck ( Mk!1

k!1

"x(t)#1+!0 "x#|A| .

By the same way we have,

(A.3.18) |(6)| "Ck ( Mk!1

k!1

"x(t)#1+!0 "x#|A| .

Then (4) has the same estimate as (2) and (5) as (1) since "x(t)# ! C "t#.To take care of (7) we use (A.3.14) with F = (bpq

(xj$ B2

!0. We obtain

(A.3.19) |(7)| " C (

"x(t)#2+!0|Z(t)| + C (

"x(t)#2+!0|&(t)| + C (

"x(t)#1+!0 "x#|A|

since "x(t)# ! C "x# and "x(t)# ! C "t#. Finally

(A.3.20)

'(()

((*

|(8)| " C (

"x(t)#1+!0|&(t)| + C (

"x(t)#1+!0 "x#|A| ,

|(9)| "C ( Mk!1

k!1

"x(t)#1+!0 "x#|A| .

Gathering the estimates obtained in (A.3.15) to (A.3.20) we obtain

|Z(t)| + |&(t)| " C (

"x(t)#1+!0(|Z(t)| + |&(t)|) +

Ck ( Mk!1k!1

"x(t)#1+!0 "x#|A| .

It follows from Gronwall’s Lemma, (A.3.2) and the estimate "x(t)# ! C "t# that

|Z(t)| + |&(t)| " C(Mk!1) (

"x#|A|+!0

which, according to the definition of Z and & proves Proposition 3.3.2 when |#| = k.

A.4. Proof of Lemma 5.3.1

The proof is the same for the two cases so we shall consider the more general casewhere f = f(x, !).

Let * $ C"0 (Rn), *(%) = 1 if |%| " 1

2 , *(%) = 0 if |%| ! 1. We set

(A.4.1)#

|)|!*

supRn

|$)*| = D*, / $ N.

We want to show that one can find an increasing sequence (Lk)k"1 in ]1, +)[ suchthat if we set for (!, x, y) in % ( Rn

y

(A.4.2)

'(()

((*

F,(!, x, y) = $,x f(!, x)

(iy),

3!*+L|,| y

+ 1"x# +

1"!#

,,, |3| ! 1,

F (!, x, y) = f(!, x) +#

, ,=0

F,(x, !, y)

then F is well defined in % ( Rny and satisfies all the requirements of Lemma 5.3.1.


A.4. PROOF OF LEMMA 5.3.1 201

First of all in the expression of F, , on the support of * we have |y| " 1L|%|

'&('x('&(+'x( .

It follows that

(A.4.3)|y|"x# " 1

L|,|,

|y|"!# " 1

L|,|.

Using (5.3.2) we can write,

|F,(!, x, y)| " M|,|

+ 1"x#|,|+!3

+1

"!#|,|+!3

, |y||,|

3!|*(L|,| y(· · · ))|

so

|F,(!, x, y)| " D0 M|,|

3! L|,||,|

+ 1"x#!3

+1

"!#!3

,.

Taking

(A.4.4) L|,||,| ! D0 M|,|

we deduce that F defined in (A.4.2) is well defined and satisfies

(A.4.5) |F (!, x, y)| " C0

+ 1"x#!1

+1

"!#!2

,

since "3 ! "1 and "3 ! "2.Therefore (ii) in Lemma 5.3.1 is satisfied and (i) follows immediately from (A.4.2).

We shall strengthen the condition (A.4.4) on L|,| to obtain a C" function F . Firstall there exists absolute constants Ci,*, (i $ N, / $ N) such that

(A.4.6)#

|,|=i

5555$,x

.+ 1"x# +

1"!#

,*/5555 " Ci,*

+ 1"x# +

1"!#

,*+i.

Let . $ Nn. For any µ $ Nn one can find an absolute constant K|µ| independent of(Lk)) such that for all (!, x, y) in R ( Rn

x ( Rny we have

(A.4.7)5555$

µx

.($)

% *)+L|,|y

+ 1"x# +

1"!#

,,/5555 " K|µ| D|)|+|µ|

+ 1"x# +

1"!#

,|µ|.

Indeed let us set h(!, x, y) = L|,| y$

1'x( + 1

'&(%. By the Faa di Bruno formula,

$µx [($)*)(h(!, x, y))] is a finite linear combination with absolute coe"cients of terms

of the form

($()+/)% *)(h(!, x, y))

sH

j=1

($*jx h(!, x, y))kj

where 1 " |'| " |µ|, 1 " s " |µ|, |kj | ! 1, |/j| ! 1 ands#

j=1

kj = ',s#

j=1

|kj | /j = µ.

Since |'| + |.| " |µ| + |.| we have

(A.4.8) |$()+/)% *(h(!, x, y))| " D|µ|+|)|.


202 APPENDIX

On the other hand it follows from (A.4.6) that

(1) =:555

sH

j=1

($*jx h(!, x, y))kj

555 "sH

j=1

C|kj ||*j|,1(|y|L|,|)

Ps1 |kj |

+ 1"x# +

1"!#

,Ps1 |kj |(1+|*j|)

.

On the support of * we use the estimates (A.4.3). Moreover we have&s

1 |kj | = |'|,&s1 |kj | |/j| = |µ|. It follows then that,

(A.4.9) (1) " C%|µ|

+ 1"x# +

1"!#

,|µ|.

Then (A.4.7) follows from (A.4.8) and (A.4.9).Now with F, defined in (A.4.2) we can write

$By F,(!, x, y) = i|,|

$,x f(!, x)

3!

#

B1!BB1!,

3BB1

4y,!B1

(3 ' B1)!

.L|,|

+ 1"x# +

1"!#

,/|B|!|B1|

($B!B1% *)

+L|,| y

+ 1"x# +

1"!#

,,.

Then

$Ax $B

y F,(!, x, y)

= i|,|#

B1!BB1!,

#

A1!A

#

A2!A!A1

3BB1

43AA1

43A ' A1

A2

41

(3 ' B1)!y,!B1

L|B|!|B1||,| $,+A1

x f(!, x) $A2x

.+ 1"x# +

1"!#

,|B|!|B1|/

$A!A1!A2x

.($B!B1

% *)+y L|,|

+ 1"x# +

1"!#

,,/.

Now we use (A.4.3), (A.4.6) and (A.4.7). Since |3|+|A1|" |3|+|A|, |A ' A1 ' A2|" |A|,|B ' B1| " |B|, we obtain

|$Ax $B

y F,(!, x, y)|

"#

B1!BB1!,

#

A1!AA2!A!A1

3BB1

43AA1

43A ' A1

A2

42|,|!|B1| L|B1|!|,|

|,|

+ 1"x# +

1"!#

,|B1|!|,|

L|B|!|B1||,| M|,|+|A|

+ 1"x#|,|+|A1|+!3

+1

"!#|,|+|A1|+!3

,C|A2|,|B!B1|

+ 1"x# +

1"!#

,|A2|+|B|!|B1|K|A| D|A|+|B|

+ 1"x# +

1"!#

,|A|!|A1|!|A2|

where |y| satisfies (A.4.3).



It follows that we can find a constant 2C|A|,|B| " 1, depending only on |A|, |B| andthe dimension such that

(A.4.10) |$Ax $B

y F,(!, x, y)|

"2C|A|,|B| D|A|+|B| M|,|+|A|

L|,|!|B||,|

+ 1"x#|A|+|B|+!3

+1

"!#|A|+|B|+!3

,.

Let us set

(A.4.11) 2Ck = max|A|+|B|!k

2C|A|,|B|.

We shall take the increasing sequence (Lk)k"0 such that

(A.4.12) Lk ! max$1, 2k 2Ck Dk M2k

%.

Then we write, according to (A.4.2)

(A.4.13) F (!, x, y) = f(!, x) +#

|,|!|A|+|B|

F,(!, x, y) +#

|,|>|A|+|B|

F,(!, x, y).

The first two terms in the right hand side of (A.4.13) define a C" function. For thethird one we deduce from (A.4.10), (A.4.11) and (A.4.12) that

|$Ax $B

y f,(!, x, y)| "2C|,| D|,| M2|,|

L|,|" 1

2|,|.

This shows that the third term define also a C" function. Thus F is C" in (x, y).Let us prove (iii). According to (A.4.2) if |A| + |B| ! 1 we have

(1) =: $Ax $B

y F (!, x, y) = $Ax $B

y f(!, x) +#

, ,=0

$Ax $B

y F,(!, x, y).

If |B| ! 1 we use (A.4.10), (A.4.13) and (A.4.12). We get

|(1)| " CAB

+ 1"x#|A|+|B|+!3

+1

"!#|A|+|B|+!3

,.

If B = 0 we use furthermore (5.3.2). We obtain

|(1)| " M|A|

+ 1"x#|A|+!3

+1

"!#|A|+!3

,+ C%

AB

+ 1"x#|A|+!3

+1

"!#|A|+!3

,.

This shows that (iii) holds.Finally let us prove (iv). Let us set again

h|,|(!, x, y) = L|,| y+ 1"x# +

1"!#

,.


204 APPENDIX

According to (A.4.2) we have for j $ {1, 2, . . . , n},

$xj F = $xj f(!, x) +#

, ,=0

+$xj $,

x f(!, x)*(h|,|(!, x, y))

'n#

*=1

L|,|xj y*

"x#3 $,x f(!, x) · $*

$%*(h|,|(!, x, y))

, (iy),

3!

$yj F =#

, ,=0

$,x f(!, x)

(iy),

3!L|,|

+ 1"x# +

1"!#

, $*

$%j(h|,|(!, x, y))

+ i#

,j"1

$,x f(!, x)

(iy),!ej

(3 ' ej)!*(h|,|(!, x, y)),

where ej = (0, 0, . . . , 1, . . . , 0). Setting 3% = 3 ' ej the sum above can be written

i $xj f(!, x)*+L1 y

+ 1"x# +

1"!#

,,+ i#

, ,=0

$xj $,x f(!, x)

(iy),

3!*(h|,|+1(!, x, y)).

It follows that

(A.4.14) ($xj F + i $yj F )(!, x, y) = (1) + (2) + (3)

where

(A.4.15)

'((((((((((()

(((((((((((*

(1) =+1 ' *

+L1 y

+ 1"x# +

1"!#

,,$xj f(!, x)

(2) =#

, ,=0

$xj $,x f(!, x)

(iy),

3![*(h|,|(!, x, y)) ' *(h|,|+1(!, x, y))]

(3) =#

, ,=0

$,x f(!, x)

(iy),

3!L|,|

+'

n#

*=1

xj y*

"x#3$*

$%*(h|,|(!, x, y))

+ i+ 1"x# +

1"!#

, $*

$%j(h|,|(!, x, y))

,.

Let us set for convenience

(A.4.16) R = |y|+ 1"x# +

1"!#

,.

On the support of 1 ' *$L1 y

$1'x( + 1

'&(%%

in the term (1) above, we have L1R ! 12 .

Therefore

(A.4.17)(1)RN

" (2L1)N M1

+ 1"x#1+!3

+1

"!#1+!3

,.

On the support of *(h|,|(!, x, y))'*(h|,|+1(!, x, y)) we have 12L|%|+1

" R " 1L|%|

. Nowwe write with N ! 2,

(2) =#

1!|,|!N!1

G,(!, x, y) +#

|,|"N

G,(!, x, y).



When |3| " N ' 1 we have L|,|+1 " LN so R ! 12LN

. It follows that

#

1!|,|!N!1

|G,(!, x, y)| · 1RN

" (2LN)N#

1!|,|!N!1

M|,|+1 ·1

L|,||,|

+ 1"x#1+!3

+1

"!#1+!3

,.

Therefore we obtain

(A.4.18)#

1!|,|!N!1

|G,(!, x, y)| " CN RN+ 1"x#1+!3

+1

"!#1+!3

,.


#

|,|"N

|G,(!, x, y)|

"+ 1"x#1+!3

+1

"!#1+!3

,RN

#

|,|"N

M|,| R|,|!N [*(h|,|) ' *(h|,|+1)]

#

|,|"N

|G,(!, x, y)|

"+ 1"x#1+!3

+1

"!#1+!3

,RN#

)

M|)|+N R|)|[*(h|)|+N) ' *(h|)|+N+1)].

On the support of *(h|)|+N) ' *(h|)|+N+1), we have L|)|+N R " 1. It follows that

(A.4.19)#

|,|"N

|G,(!, x, y)| " CN RN+ 1"x#1+!3

+1

"!#1+!3

,.

Combining (A.4.18) and (A.4.19) we obtain

(A.4.20) |(2)| " C%N RN

+ 1"x#1+!3

+1

"!#1+!3

,.

Finally we consider the term (3) in (A.4.15). We have |xj|'x( " 1 and on the support of

(.(%j

(h|,|(!, x, y)) we have L|,||y|'x( " 1 it follows that (3) is bounded by a finite sum

of terms of the form

(3)% =+ 1"x#1+!3

+1

"!#1+!3

,#

, ,=0

M|,| R|,| L|,|

555$*

$%j(h|,|(!, x, y))

555.

As before we write the above sum as&

1!|,|!N!1 +&

|,|"N . If |3| " N ' 1 thenL|,| " LN!1 so on the support of (.

(%jwe have R ! 1

2L|%|! 1

2LN"1and R " 1

L|%|. It

follows that

1RN

555#

1!|,|!N!1

555 " (2LN!1)N#

1!|,|!N!1

M|,|1

L|,|!1|,|

D1 = CN .


206 APPENDIX

For the second sum we write555#

|,|"N

555 " RN#

|,|"N

M|,| R|,|!N L|,|

555$*

$%j(h|,|)

555

" RN#

)

M|)|+NL|)|+N

L|)||)|+N

D1 = C%N RN .

It follows that

(A.4.21) |(3)| " CN RN+ 1"x#1+!3

+1

"!#1+!3

,.

Using (A.4.14) to (A.4.21) we obtain the part (iv) of Lemma 5.3.1. The proof iscomplete.


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STRICHARTZ ESTIMATES FOR SCHRODINGER EQUATIONS WITH¨ VARIABLE COEFFICIENTS · 2018-12-20 ·...

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