A KAM Theorem for Hamiltonian Partial Differential Equations with Unbounded Perturbations

Transcription

1 A KAM Theorem for Hamiltonian Partial Differential Equations with Unbounded Perturbations Jianjun Liu, Xiaoping Yuan School of Mathematical Sciences and Key Lab of Math. for Nonlinear Science, Fudan University, Shanghai 2433, P R China September 3, 2 Abstract: We establish an abstract infinite dimensional KAM theorem dealing with unbounded perturbation vector-field, which could be applied to a large class of Hamiltonian PDEs containing the derivative x in the perturbation. Especially, in this range of application lie a class of derivative nonlinear Schrödinger equations with Dirichlet boundary conditions and perturbed Benjamin-Ono equation with periodic boundary conditions, so KAM tori and thus quasi-periodic solutions are obtained for them. Introduction and Main Results Consider a Hamiltonian partial differential equation (HPDE) ẇ = Aw + F(w), where Aw is linear Hamiltonian vector-field with d := ord A > and F(w) is nonlinear Hamiltonian vector-field with d := ord F and is analytic in the neighborhood of the origin w =. When d = ord F, the vector-field F is called bounded perturbation. For example, in this case lie a class of nonlinear Schrödinger equations (with d = ) iu t + u xx +V (x)u + u 2 u = and a class of nonlinear wave equations (with d = ) u tt u xx +V (x)u + u 3 =. For the existence of KAM tori of the PDEs with bounded perturbations has been deeply and widely investigated by many authors. In this field of study there are too many references to list here. We give just two survey papers by Kuksin [8] and Bourgain [4]. When d = ord F >, the vector-field F is called unbounded perturbation. According to a wellknown example, due to Lax [] and Klainerman [5] (See also [7]), it is reasonable to assume d d in order to guarantee the existence of KAM tori for the PDE. The quantity d d measures the strength of nonlinearity of the PDE. The smaller the d d is, the stronger the nonlinearity is. When d d =, the nonlinearity of the PDE is the strongest. Supported by NNSFC and 973 Program (No. 2CB3279) and the Research Foundation for Doctor Programme.

2 For the PDE with unbounded Hamiltonian perturbation, the only previous KAM theorem is due to Kuksin [7] where it is assumed that d d >. Kuksin s theorem is in [7] used to prove the persistence of the finite-gap solutions of KdV equation, as well as its hierarchy, subject to periodic boundary conditions. See also Kappeler-Pöschel []. Another KAM theorem with unbounded linear Hamiltonian perturbation is due to Bambusi-Graffi [2] where the spectrum property is investigated for the time dependent linear Schrödinger equation iẏ(t) = (A + εf(t))y(t) with d d > also. The assumption d d > excludes a large class of interesting partial differential equations such as a class of derivative nonlinear Schrödinger (DNLS) equations iu t + u xx M σ u + i f (u,ū)u x = subject to Dirichlet boundary conditions, and perturbed Benjamin-Ono (BO) equation u t + H u xx uu x + perturbation = subject to periodic boundary conditions, where H is the Hilbert transform. In both cases, d d =. In the present paper, we will construct a KAM theorem including d d > and limiting case d d = and show the existence of KAM tori for DNLS and perturbed BO equations. If the Hamiltonian operator A has continuous spectra, there is usually no KAM tori for the HPDE. Therefore, we should assume that the Hamiltonian operator A is of pure point spectra before stating our theorems. Taking the eigenfunctions of A as a basis and changing partial coordinates into actionangle variables, from the HPDE one can usually obtain a small perturbation H = N +P of an infinite dimensional Hamiltonian in the parameter dependent normal form on a phase space N = j (ξ )y j + j nω 2 Ω j (ξ )(u 2 j + v 2 j) (.) j P a,p = T n R n l a,p l a,p (x,y,u,v) with symplectic structure j n dy j dx j + j du j dv j, where l a,p is the Hilbert space of all real sequences w = (w,w 2, ) with w 2 a,p = e 2a j j 2p w j 2 <, j where a and p. The tangential frequencies ω = (ω,,ω n ) and normal frequencies Ω = (Ω,Ω 2, ) are real vectors depending on parameters ξ Π R n, Π a closed bounded set of positive Lebesgue measure, and roughly Ω j (ξ ) = j d +. The perturbation term P is real analytic in the space coordinates and Lipschitz in the parameters, and for each ξ Π its Hamiltonian vector field X P = (P y, P x, P v,p u ) T defines near T := T n {y = } {u = } {v = } a real analytic map X P : P a,p P a,q, where p q = d. 2

3 In the whole of this paper the parameter a is fixed. Moreover, throughout this paper, for convenience, we will adopt lots of notations and definitions from []. We denote by P a,p C the complexification of Pa,p. For s,r >, We introduce the complex T - neighborhoods in P a,p C D(s,r) : Imx < s, y < r 2, u a,p + v a,p < r, (.2) and weighted norm for W = (X,Y,U,V ) P a,q C W r,a,q = X + Y r 2 + U a,q r + V a,q, r where denotes the sup-norm for complex vectors. Furthermore, for a map W : D(s,r) Π P a,q C, for example, the Hamiltonian vector field X P, we define the norms W r,a,q,d(s,r) Π = W lip r,a,q,d(s,r) Π = sup sup D(s,r) Π sup ξ,ζ Π,ξ ζ D(s,r) W r,a,q, ξ ζ W r,a,q, ξ ζ where ξ ζ W = W( ;ξ ) W( ;ζ ). In a completely analogous manner, the Lipschitz semi-norm of the frequencies ω and Ω are defined as ω lip Π = sup ξ,ζ Π,ξ ζ ξ ζ ω ξ ζ, Ω lip δ,π = sup ξ,ζ Π,ξ ζ j j δ ξ ζ Ω j sup ξ ζ (.3) for any real number δ. Theorem.. Suppose the normal form N described above satisfies the following assumptions: (A) The map ξ ω(ξ ) between Π and its image is a homeomorphism which is Lipschitz continuous in both directions, i.e. there exist positive constants M and L such that ω lip Π M and ω lip ω(π) L; (B) There exists d > such that Ω i Ω j m i d j d (.4) for all i j uniformly on Π with some constant m >. Here Ω = ; (C) There exists δ d such that the functions ξ Ω j(ξ ) are uniformly Lipschitz on Π for j δ j, i.e. there exist a positive constant M 2 such that Ω lip δ,π M 2; (D) We additionally assume where E = ω Π := sup ξ Π ω(ξ ). 4ELM 2 m, (.5) Set M = M + M 2. Then for every β >, there exists a positive constant γ, depending only on n,d,δ,m, the frequencies ω and Ω, s > and β, such that for every perturbation term P described above with d = p q d (.6) and for some r > and < α <, there exist ε := X P r,a,q,d(s,r) Π + α M X P lip r,a,q,d(s,r) Π (αγ)+β (.7) 3

4 () a Cantor set Π α Π with Π \ Π α c α, (.8) where denotes Lebesgue measure and c > is a constant depends on n, ω and Ω; (2) a Lipschitz family of smooth torus embeddings Φ : T n Π α P a,p satisfying: for every non-negative integer multi-index k = (k,...,k n ), where k x := k x (Φ Φ ) r,a,p,t n Π α + α M k x (Φ Φ ) lip r,a,p,t n Π α c 2 ε +β /α, (.9) k x k xn kn with k := k + + k n, Φ : T n Π T, (x,ξ ) (x,,,) is the trivial embedding for each ξ, and c 2 is a positive constant which depends on k and the same parameters as γ; (3) a Lipschitz map φ : Π α R n with φ ω Πα + α φ ω lip Π M α c 3 ε, (.) where c 3 is a positive constant which depends on the same parameters as γ, such that for each ξ Π α the map Φ restricted to T n {ξ } is a smooth embedding of a rotational torus with frequencies φ(ξ ) for the perturbed Hamiltonian H at ξ. In other words, t Φ(θ +tφ(ξ ),ξ ), t R is a smooth quasi-periodic solution for the Hamiltonian H evaluated at ξ for every θ T n and ξ Π α. This theorem can be applied to the class of derivative nonlinear Schrödinger equations mentioned above. However, the assumption (.5) excludes the perturbed Benjamin-Ono equation mentioned above. Thus, in the following, we give a modified version of the above theorem: Theorem.2. The above theorem also holds true with, respectively, replacing the assumption (D) and conclusion () by the assumption (D*) and conclusion (*) below: (D*) For every k Z n and l Z with l 2 (here l = j l j ), the resonance set {ξ Π : k,ω(ξ ) + l,ω(ξ ) = } has Lebesgue measure zero; Moreover, if δ = d, we additively assume that there exist δ < d, a partition Ω = Ω + Ω 2, positive constant M 3 and M 4 with such that Ω lip δ,π M 3, Ω 2 lip δ,π M 4; 8ELM 4 m, (.) (*) a Cantor set Π α Π with Π \ Π α as α, where denotes Lebesgue measure. This paper is organized as follows: In section 2 we give an outline of the proof of the above theorems, and some new difficult and ideas compared with [7] [] are exhibited. In sections 3-6 the above theorems are proved in detail. The proof of Theorem.2 is the same as that of Theorem. except the measure estimate. Thus, sections 3-5 and subsection 6. are devoted to the proof of Theorem., while the measure estimate for Theorem.2 is given in subsection 6.2. In sections 7-8 Theorem. and Theorem.2 are applied to derivative nonlinear Schrödinger equations and perturbed Benjamin-Ono equation, respectively. Finally, a technical lemma is listed in section 9. 4

5 2 Outline of The Proof and more Remarks The above theorems generalize Kukisn s theorem from d < d to d d such that the range of application is extended to a class of derivative nonlinear Schrödinger equations and perturbed Benjamin-Ono equation. Here we would like to compare the proof of our theorems with that of Kuksin s theorem. By and large, as any KAM theorem, in both cases Newton iteration is used to overcome the notorious small divisor difficult. Therefore, our proof is mainly based on Kuksin s approach in [7]. (Also see []). There is, however, some essential differences between the proof of our theorems and that of Kuksin s theorem. In order to see clearly the differences, let us give the basic procedure of the proof of KAM theorem from [7] and [], which consists of the following steps. 2.. Derivation of the homological equations. For convenience, introduce complex variables z = (u iv)/ 2 and z = (u+iv)/ 2. Assume we are now in the ν-th KAM iterative step. Write the integrable part N ν of the Hamiltonian H ν N ν = ω,y + j= Ω j z j z j and develop the perturbation P ν into Taylor series in (y,z, z): P ν = ε ν R ν + O( y 2 + y z a,p + z 3 a,p), where ε ν goes to zero very fast, for example, taking ε ν ε (5/4)ν, and R ν = R x (x) + R y,y + R z (x),z + R z (x), z + R zz (x)z,z + R z z (x) z, z + R z z (x)z, z = O(). A key point is, very roughly speaking, to search for a Hamiltonian function of the same form as R ν : F ν = F x (x) + F y,y + F z (x),z + F z (x), z + F zz (x)z,z + F z z (x) z, z + F z z (x)z, z which satisfies {N ν,f ν } + R ν =, (2.) where {, } is the Poisson bracket with respect to the symplectic structure j n dy j dx j i dz j d z j. j By Φ ν := X ε ν F ν denote the time- map of the Hamiltonian vector field X εν F ν. It is a symplectic transformation. A simple calculation shows that Φ ν changes H ν = N ν + P ν into H ν+ = H ν Φ ν = N ν + R ν+ + O( y 2 + y z a,p + z 3 a,p) (2.2) with R ν+ = 2 ε2 ν{{n ν,f ν },F ν } + ε 2 ν{r ν,f ν } +. Our task is now to search for F ν satisfying {N ν,f ν } + R ν = which is a set of the first order partial differential equations: ω x F x (x) = R x (x), ω x F y (x) = R y (x), ω x F z z (x) + iλf z z (x) if z z (x)λ = R z z (x), 5

6 where Λ = diag (Ω j : j =,2,...). Let us consider the last equation which is the most difficulty one. By F i j (x) and R i j (x), denote the matrix elements of the operators F z z and R z z, respectively. Then the last equation becomes or iω x F i j (x) + (Ω i Ω j )F i j (x) = ir i j (x), i, j =,2,... (2.3) ( k,ω + Ω i Ω j ) F i j (k) = i R i j (k), where F i j (k) and R i j (k) are the k-th Fourier coefficients of F i j (x) and R i j (x), respectively. One can assume R j j () =, otherwise R j j () is put into Ω j as a modification of the normal form N ν. Thus, (2.3) can be solved by R i j (k) F i j (k) = k,ω + Ω i Ω j under the non-resonant conditions k,ω + Ω i Ω j unless k =,i = j. This is actually the KAM iterative procedure for bounded perturbation P. However, the thing is not so simple when the perturbation P is unbounded, i.e., d = p q >. When X P : P a,p P a,q, d = p q >, one has, very roughly, R i j (k) i d + j d, as i + j. This leads usually to F i j (k). In other words, the solution F ν or the transformation Φ ν = X ε ν F ν would be unbounded. One should note that the coordinate transformations Φ ν = X ε ν F ν must be bounded even if the the perturbation P is unbounded in order that the domains of the KAM iteration are always in the same phase space P a,p and that the KAM iterative procedure can work. In order to guarantee the bounded-ness of F ν, it is required in [7] and [] that or rather roughly, Together with d d, one has that for i j, Ω i Ω j i d + j d, i j k,ω + Ω i Ω j i d + j d. F i j (k) i d + j d i d = O(). + j d It is clear that this estimate fails when i = j. To avoid this plight, Kuksin[7] smartly put the whole R j j (x) rather than R j j () into Ω j as a modification of the normal form N ν so that it is not necessary to solve the equation for F j j (x). In doing so, the term N ν becomes into the generalized normal form (i.e. normal frequencies depend on the angle variable x) Ñ ν := N ν + j= R j j (x)z j z j = ω,y + the homological equation (2.) is modified into j= (Ω j + R j j (x))z j z j, {Ñ ν,f ν } + R ν =, (2.4) 6

7 and the remaining term R ν+ is changed into R ν+ = 2 ε2 ν{{ñ ν,f ν },F ν } + ε 2 ν{r ν,f ν } +. (2.5) Accordingly, the homological equation (2.3) becomes into iω x F i j (x) + (Ω i Ω j + R ii (x) R j j (x))f i j (x) = ir i j (x), i j. (2.6) Let u = F i j (x), λ = Ω i Ω j, µ(x) = R ii (x) R j j (x) and r(x) = ir i j (x). Then this equation can be abbreviated as an abstract equation iω x u + (λ + µ(x))u = r(x). (2.7) Since R ii and R j j are large, µ(x) is usually large. And the coefficient µ(x) involves the angle variable. The equation of this type is called small-denominators equations with large variable coefficients by Kuksin [6]. Remark that, for simplicity, the modification of ω is omitted here Solving the homological equations. In order to make KAM iterative procedure work, the existence domain of the solution u should be the strip-type neighborhood of T n with some width s > : D(s) := {x C n /2πZ n : Im x s}. Assume µ(x) C γ, where C is some small constant and where γ should be usually a large magnitude. Since (2.7) is scalar, it can be solved directly and estimated by sup u(x) e 2C γ r s, σ < s, (2.8) x D(s σ) where r s := sup x D(s) r(x). This estimate is, however, not good enough to support the KAM iteration procedure, since the solution u becomes too large as µ is large. In fact, in the ν-th KAM iteration step, γ 2 ν. Thus, u exp(2 ν ) goes to infinity very rapidly, which makes the coordinate transformations essentially unbounded. When d < d, Kuksin s lemma [6] can solve this problem. Following Kuksin [6], we assume there are constant C > and < θ < such that λ θ C µ s, (2.9) Kuksin s Lemma states that under suitable non-resonant conditions on ω, the solution u satisfies the estimate: θ u s σ C exp(c 2 C 3 ) r s, (2.) where C and C 2 are positive constants depending on only n and σ, and C 3 > is a constant depending on the non-resonant conditions. When this estimate is applied to (2.6), one needs to take λ = Ω i Ω j i d j d i d + j d, γ i d + j d, i, j =,2,3,... From this, we see that there is indeed a constant < θ < such that (2.9) holds true if d < d. Therefore, the solution u of the homological equation (2.7) has a uniform bound independent of the size of µ. This makes the coordinate transformation bounded. We see also that d = d leads to θ = in (2.9). In this case, the estimate (2.) is invalid. (The right hand side of (2.) is equal to.) Now it is clear that we need some new estimate for the solution u covering not only the case d < d but also the limit case d = d. The new estimate has been obtained in our recent paper [2]: u s σ C 4 e C γs r s, (2.) 7

8 where C 4 is a constant depending the non-resonant conditions on ω and where C is a positive constant small enough. Since the parameter γ (which measures the magnitude of the perturbation µ(x)) goes into the exponential in the right hand side of (2.), in this sense, the upper bound of this new estimate looks weaker than that of the original Kuksin s lemma. However, compared with (2.8), there is some essential improvement in (2.): there is an s in the exponential in (2.). This number s will be used crucially in the following manner: In the ν-th KAM iteration, we let s = 2 ν. One will find that there is no the small divisor problem in the homological equation (2.7) when γ > some large constant K 2 ν (5/4) ν lnε. In this case, the homological equation (2.7) can be solved by the implicit function theorem. So the non-trivial case is when γ K. At this time, we find that e C γs ε C ν with ε ν := ε (5/4)ν and the constant C. Thus, by the new estimate (2.) and noting r s ε ν, we have u s σ ε C ν r s ε C ν ε ν ε ν. (2.2) In the usual KAM iteration, one would have obtained u s σ 2 ν ε ν ε ν. Although here ε C ν ε ν 2 ν ε ν, inequality (2.2) can guarantee the KAM procedure to be iterated. Therefore, although the new estimate (2.) is weaker than that of the original Kuksin s Lemma, it can covers both d < d and the limit case d = d, and it is sufficient for the proof of the KAM theorems of the present paper Estimate of the remaining terms R ν+, etc. Recall R ν = O(). By (2.), F ν = O(ε C ν ) with < C. Thus, {F ν,r ν } = O(ε C ν ). By (2.4), Thus, Consequently, by (2.5), very roughly, {Ñ ν,f ν } = R ν = O(). {{Ñ ν,f ν },F ν } = {O(),O(ε C ν )} = O(ε C ν ). R ν+ = 2 ε2 ν{{ñ,f ν },F ν } + ε 2 ν{r ν,f ν } + = O(ε 2 C ν ) = O(ε ν+ ) := ε ν+ R ν+, where R ν+ = O(). Therefore, we can rewrite H ν+ as H ν+ = Ñ ν+ + ε ν+ R ν+ + O( y 2 + y z a,p + z 3 a,p). (2.3) 2.4. Convergence of the iterative procedure. Repeating the above procedure and letting ν and noting ε ν very fast, one can finally get H := lim ν H Φ Φ ν = Ñ + O( y 2 + y z a,p + z 3 a,p), where Ñ = lim ν Ñ ν = ω,y + j Ω, j (x)z j z j. The system corresponding to H is It is clearly seen that ẋ = H y = ω + O( y + z a,p ) ẏ = H x = O( y 2 + y z a,p + z 2 a,p) ż j = i H z j = iω, j z j + O( y + z 2 a,p), j. T := {(x,y,z, z) P a,p : x = ω t,y =,z = z =,t R} 8

9 forms an invariant torus of the hamiltonian vector field X H. Going back to the original vector field X H, then (lim ν Φ Φ ν )T is an invariant torus of the Hamiltonian system defined by H Estimate of the measure of the parameters. When d = d, another new difficult also arises in search for F. It is under suitable non-resonant conditions that either (2.) or (2.) can hold true. In other words, one has to remove some resonant sets consisting of bad parameters ξ, equivalently, to remove the bad parameters ω when ω = ω(ξ ) depends on ξ in some non-degenerate way. For example, we need to eliminate the resonant set {ξ Π : k,ω(ξ ) + Ω i (ξ ) Ω j (ξ ) is small} where i j N and k Z n. Clearly we hope that the Lebesgue measure of the set is small. To that end we need to verify that k,ω(ξ ) + Ω i (ξ ) Ω j (ξ ) is twisted with respect to ξ Π, equivalently, twisted with respect to ω ω(π), that is, we need to show ( ) := k,ω + Ω i (ξ (ω)) Ω j (ξ (ω)) lip ω(π) >, where ω(ξ (ω)) = ω. Recall Ω i Ω j m i d j d. Thus, k,ω + Ω i Ω j is not small if k C i d j d with some constant C depending on m and ω Π. Now we assume k C i d j d. At the ν-th KAM step, because of the modification of frequencies from unbounded perturbation, we have Ω j lip ω(π) = O( jδ ) + O( j d ). By a small trick (See 3.2 below.), we can let δ = d. Therefore, there exists a constant C such that ( ) k C (i + j ) (2.4) 2 C (i d + j d ) C (i + j ) (2.5) ( 4 C (i d d + j d d ) C )(i d + j d ), (2.6) which is similar to (.5) in [7] (See also page 74 in []). When d < d, it follows from (2.6) that ( ) > if ( 4 C ) d max(i, j) > d := K. (2.7) C Therefore, if d < d it remains to verify the twist condition ( ) > just for only finite number of cases (k,i, j) : C i d j d k C (i d + j d ) and i, j K. (2.8) Note that the constant K is independent of i, j and k, so is the number of the cases. For these cases, the measure estimate of the resonant sets can be dealt with by some initial assumptions (For example, see Proposition 22.2 in []). When d = d, K = +. However, it follows directly from (2.5) that ( ) > ( 2 C C )(i d + j d ) > if C > 2 C. This completes the measure estimate of the resonant sets for Theorem.. One can verify that the condition C > 2 C is indeed satisfied by DNLS equations. However, the inequality C > 2 C is not satisfied by BO equation. The procedure of the measure estimate of the resonant set is modified as follows: Assume that Ω can be split into two part: Ω j = Ω j + Ω2 j such that Ω j lip ω(π) C 2 j δ with δ < d, and Ω 2 j lip ω(π) C 3 j d with C 3 suitably small. Thus, ( ) ( ) k C 2 (i δ + j δ ) C 3 (i d + j d ) ( 2 C C 3 )(i d + j d ) C 2 (i δ + j δ ). 9

10 It follows from δ < d that there is a constant K = K (d,δ, C, C 2, C 3 ) > such that ( ) > if max{i, j} > K and 2 C > C 3. The latter inequality is satisfied by BO equation with d = 2 and δ =. We also mention that the partition of Ω into Ω + Ω 2 is rather natural. In fact, Ω is usually regarded as the initial frequency vector, while Ω 2 corresponds to the modification in KAM iteration steps. Remarks.. In [2] we mentioned Theorem. above and its application to DNLS equation without proof. In this paper we give the proof and add Theorem.2 and a new application to Benjamin-Ono equation. 2. The proofs of both (2.) and (2.) depends heavily on the fact that the homological equation (2.7) is -dimensional so that the solution can be expressed explicitly. This dimensional restriction requires that the normal frequency Ω j s must be simple, i.e., Ω j =. Therefore, the range of application of Theorems. and.2 and Kuksin s KAM theorem ([7]) lies in those PDEs with simple frequencies such as DNLS equation subject to Dirichlet boundary conditions, KdV and BO equations with periodic boundary conditions. For a class of DNLS equation iu t + u xx M σ u + i( u 2 u) x = subject to periodic boundary conditions, the multiplicity Ω j = 2. And for Kadomtsev-Petviashvili (KP) equation (u t + u xxx + uu x ) x ± u yy =, u = u(t,x,y), (x,y) T 2, its frequency multiplicity Ω j as j. There is nothing to know about the existence of KAM tori for these two classes of PDEs with perturbations. In particular, the existence of KAM tori for KP equation is a well-known open problem by Kuksin. See [8],[4]. 3 The Homological Equations 3. Derivation of Homological Equations The proof of Theorem. employs the rapidly converging iteration scheme of Newton type to deal with small divisor problems introduced by Kolmogorov, involving infinite sequence of coordinate transformations. At the ν-th step of the scheme, a Hamiltonian H ν = N ν + P ν is considered, as a small perturbation of some normal form N ν. A transformation Φ ν is set up so that H ν Φ ν = N ν+ + P ν+ with another normal form N ν+ and a much smaller perturbation P ν+. We drop the index ν of H ν, N ν, P ν, Φ ν and shorten the index ν + as +. Using convenient complex notation z = (u iv)/ 2 and z = (u+iv)/ 2, the generalized normal form reads N = ω(ξ ),y + Ω j (x;ξ )z j z j. (3.) j Let R be 2-order Taylor polynomial truncation of P, that is, R = R x + R y,y + R z,z + R z, z + R zz z,z + R z z z, z + R z z z, z, (3.2) where, is formal product for two column vectors and R x, R y, R z, R z, R zz, R z z, R z z depend on x and ξ. For a function u on T n, let [u] = (2π) n u(x)dx. T n

11 By [R] denote the part of R in generalized normal form as follows [R] = [R x ] + [R y ],y + diag(r z z )z, z, where diag(r z z ) is the diagonal of R z z. Note that [R x ] and [R y ] are independent of x. In the following, the term [R x ] will be omitted since it does not affect the dynamics. The coordinate transformation Φ is obtained as the time--map X t F t= of a Hamiltonian vector field X F, where F is of the same form as R: F = F x + F y,y + F z,z + F z, z + F zz z,z + F z z z, z + F z z z, z, (3.3) and [F] =. Denote ω = b n ω b x b, Λ = diag(ω j : j ). Then we have H Φ =(N + R) X F + (P R) X F =N + {N,F} + R + {( t){n,f} + R,F} X t Fdt + (P R) X F =N + x Ω j,f y z j z j + [R y ],y + diag(r z z )z, z (3.4) j + ( ω F x + R x ) (3.5) + ω F y + R y [R y ],y (3.6) + ω F z + iλf z + R z,z (3.7) + ω F z iλf z + R z, z (3.8) + ( ω F zz + iλf zz + if zz Λ + R zz )z,z (3.9) + ( ω F z z iλf z z if z z Λ + R z z ) z, z (3.) + ( ω F z z iλf z z + if z z Λ + R z z diag(r z z ))z, z (3.) + {( t){n,f} + R,F} X t Fdt + (P R) X F. (3.2) We wish to find the function F such that (3.5)-(3.) vanish. To this end, F x, F y, F z, F z, F zz, F z z and F z z should satisfy the homological equations: ω F x = R x, (3.3) ω F y = R y [R y ], (3.4) ω Fj z iω jfj z = R z j, j, (3.5) ω Fj z + iω jfj z = R z j, j, (3.6) ω Fi zz j i(ω i + Ω j )Fi zz j = R zz ω F z z i j + i(ω i + Ω j )F z z i j i j = R z z i j, i, j, (3.7), i, j, (3.8) ω Fi z z j + i(ω i Ω j )Fi z z j = R z z i j, i, j, i j. (3.9) 3.2 Solving the Homological Equations Let Ω = (Ω j : j ), Ω = [Ω] and Ω = Ω Ω. Define k = max{, k }, l d = max{, j d l j }. j

12 Moreover, for an analytic function u on D(s), we define u s,τ := k Z n û k k τ e k s, where û k := (2π) n T n u(x)e ik x dx is the k-fourier coefficient of u. Consider the conditions δ d and d = p q d. If δ > d, decreasing q such that δ = d, then for the new q, the inequality (.7) still holds true; if δ < d, increasing δ such that δ = d, then for the new δ, the assumption (C) still holds true, and if d = d, the assumption (D*) is satisfied with Ω 2 =. Thus, without loss of generality we assume δ = d d in the following. Equations (3.3)-(3.9) will be solved under the following conditions: uniformly on Π, k,ω(ξ ) + l, Ω(ξ ) α l d, k, l 2, (3.2) k τ l, Ω(ξ ) m l d, < l 2, (3.2) Ω j s,τ+ αγ j δ, j, (3.22) with constants τ n, d >, < γ /8, m >, and a parameter < α m. We mention that d is the same as in Theorem. and α, m will be the iteration parameters α ν, m ν in the ν-th KAM step. Equations (3.3) (3.4) can be easily solved by a standard approach in classical, finite dimensional KAM theory, so we only give the related results at the end of this subsection. Equations (3.5)-(3.8) are easier than (3.9) and can be solved in the same way as (3.9) done, so we only give the details of solving (3.9) in the following. For any positive number K, we introduce a truncation operator Γ K as follows: (Γ K f )(x) := ˆf k e ik x, f : T n C, k K where ˆf k is the k-fourier coefficient of f. Set C = 2 ω Π /m and K being a positive number which will be the iteration parameter K ν in the ν-th KAM step. () For (i, j) with < i d j d < C K, we solve exactly (3.9): ω F z z i j + i(ω i Ω j )F z z i j = R z z i j ; (3.23) (2) for (i, j) with i d j d C K, we solve the truncated equation of (3.9): ( ) ω Fi z z j + iγ K (Ω i Ω j )F z z i j = Γ K R z z i j, Γ KF z z i j = F z z i j. (3.24) Comparing (3.24) with (3.9), we find that (3.) doesn t vanish. Actually, at this time, (3.) is equivalent to ˆR z z z, z with the matrix elements of ˆR z z being defined by {, i d i j = ( j d < C K, ) ( Γ K ) i(ω i Ω j )Fi z z j + R z z i j, i d j d C K. ˆR z z (3.25) Letting Ω i j = Ω i Ω j = Ω i j + Ω i j and dropping the superscript z z for brevity, (3.23) (3.24) (3.25) become i ω F i j + Ω i j F i j + Ω i j F i j = ir i j, (3.26) 2

13 i ω F i j + Ω i j F i j + Γ K ( Ω i j F i j ) = iγ K R i j, (3.27) {, < i ˆR i j = j d < C K, ( Γ K )( i Ω i j F i j + R i j ), i d j d C K. (3.28) We are now in position to solve the homological equations (3.26) (3.27) by using the following two lemmas, which have been proved in [2] as Theorem.4 and Lemma 2.6 respectively: Lemma 3. ([2]). Consider the first order partial differential equation i ω u + λu + µ(x)u = p(x), x T n, (3.29) for the unknown function u defined on the torus T n, where ω = (ω,,ω n ) R n and λ C. Assume () There are constants α, γ > and τ > n such that k ω α k τ, k Zn \ {}, (3.3) k ω + λ α γ + k τ, k Zn. (3.3) (2) µ : D(s) C is real analytic (here real means µ(t n ) R) and is of zero average: [µ] =. Moreover, assume there is constant C > such that (3) p(x) is analytic in x D(s). µ s,τ+ C γ. (3.32) Then (3.29) has a unique solution u(x) which is defined in a narrower domain D(s σ) with < σ < s, and which satisfies sup x D(s σ) u(x) c(n,τ) αγσ n+τ e2c γs/α sup x D(s) for < σ < min{,s}, where the constant c(n,τ) = (6e + 6) n [ + ( 3τ e )τ ]. p(x) (3.33) Lemma 3.2 ([2]). Consider the first order partial differential equation with the truncation operator Γ K i ω u + λu + Γ K (µu) = Γ K p, x T n, (3.34) for the unknown function u defined on the torus T n, where ω R n, λ C, and < 2K ω λ. Assume that µ is real analytic in x D(s) with ˆµ k e k s λ k Z n 4ι (3.35) for some constant ι, and assume p(x) is analytic in x D(s). Then (3.34) has a unique solution u(x) with u = Γ K u and sup u(x) c(n) x D(s σ) λ σ n sup ( Γ K )(µu)(x) c(n) x D(s σ) for < σ < s, where the constant c(n) = 4(2e + 2) n. sup p(x), (3.36) x D(s) sup ισ n e 9Kσ/ x D(s) p(x) (3.37) 3

14 Set < σ < min{,s/5}. In what follows the notation a b stands for there exists a positive constant c such that a cb, where c can only depend on n,τ. First, let us consider (3.26) for (i, j) with < i d j d < C K. From (3.2) (3.2) we get From (3.22) we get Applying Lemma 3. to (3.26), we have In view of we get k,ω(ξ ) α k τ, k Zn \ {}, (3.38) k,ω(ξ ) + Ω i j α id j d + k τ, k Z n. (3.39) Ω i j s,τ+ αγ (i δ + j δ ) 2αγ i d j d. (3.4) F i j D(s 2σ) F i j D(s 2σ) α i d j d σ n+τ e4γ i d j d s R i j D(s σ). (3.4) i d j d < C K, α i d j d σ n+τ e4c γ Ks R i j D(s σ). (3.42) Then, let us consider (3.27) for (i, j) with i d j d C K. From (3.22) (3.2) we get Ω i j s, Ω i j s,τ+ αγ (i δ + j δ ) 2αγ Ω i j m i j Now applying Lemma 3.2 to (3.27), we have F i j D(s 2σ) ( Γ K )( Ω i j F i j ) D(s 2σ) Ω i j 4 i j. (3.43) m i d j d σ n R i j D(s σ), (3.44) i j σ n e 9Kσ/ R i j D(s σ). (3.45) For a bounded linear operator from l a,p to l a,q, define its operator norm by a,q,p. As in lemma 9. of [], in view of (3.42) and (3.44), using Lemma 9. below, we get the estimates of F z z : F z z a,p,p,d(s 2σ), F z z a,q,q,d(s 2σ) Multiplying by z, z we then get ασ 2n+τ e4c γ Ks R z z a,q,p,d(s) ασ 2n+τ e4c γ Ks X R r,a,q,d(s,r). (3.46) and finally by Cauchy s estimate we have r 2 Fz z z, z D(s 2σ,r) F z z a,p,p,d(s 2σ), (3.47) X F z z z, z r,a,p,d(s 3σ,r) ασ 2n+τ+ e4c γ Ks X R r,a,q,d(s,r). (3.48) 4

15 To obtain the estimate of the Lipschitz semi-norm, we proceed as follows. Shortening ξ ζ as and applying it to (3.26) and (3.27), one gets that, for (i, j) with < i d j d < C K, i ω ( F i j ) + Ω i j F i j + Ω i j F i j = i ω F i j ( Ω i j )F i j + i R i j := Q i j, (3.49) and that, for (i, j) with i d j d C K, i ω ( F i j ) + Ω i j F i j + Γ K ( Ω i j F i j ) = i ω F i j Γ k (( Ω i j )F i j i R i j ) := Q i j. (3.5) For < i d j d < C K, we have Q i j D(s 3σ) ω σ F i j D(s 2σ) + (i δ + j δ ) Ω δ,d(s) F i j D(s 3σ) + R i j D(s 3σ) e4c γ Ks ασ n+τ+ ( ω + Ω δ,d(s)) R i j D(s σ) + R i j D(s σ) ( e4c γ Ks ω + Ω δ,d(s) σ n+τ+ R i j α D(s σ) + R i j D(s σ) ). (3.5) Again applying Lemma 3. to (3.49), we have e 8C ( γ Ks ω + Ω δ,d(s) F i j D(s 4σ) α i d j d σ 2n+2τ+ R i j α D(s σ) + R i j D(s σ) ). (3.52) For i d j d C K, we have Q i j D(s 3σ) σ 3n ( ω + Ω δ,d(s) R i j m D(s σ) + R i j D(s σ) ). (3.53) Again applying Lemma 3.2 to (3.5), we have, ( ω + Ω δ,d(s) F i j D(s 4σ) m i d j d σ 4n R i j m D(s σ) + R i j D(s σ) ), (3.54) ( ( Γ K )( Ω i j F i j ) D(s 4σ) e 9Kσ/ ω + Ω δ,d(s) i j σ 4n R i j m D(s σ) + R i j D(s σ) ). (3.55) In view of (3.52) and (3.54), applying Lemma 9. below again, we get the estimates of F z z : F z z a,p,p,d(s 4σ), F z z a,q,q,d(s 4σ) ( e8c γ Ks ω + Ω δ,d(s) ασ 3n+2τ+ R z z α a,q,p,d(s) + R z z a,q,p,d(s) ). (3.56) Dividing by ξ ζ = and taking the supremum over Π, we get F z z lip, Fz z lip a,p,p,d(s 4σ) Π a,q,q,d(s 4σ) Π e8c γ Ks ασ 3n+2τ+ ( M α Rz z a,q,p,d(s) Π + R z z lip a,q,p,d(s) Π where M := ω lip Π + Ω lip. Thus, in the same way as (3.48), we get δ,d(s) Π X F z z z, z lip r,a,p,d(s 5σ,r) Π e8c γ Ks ασ 3n+2τ+2 ), (3.57) ( ) M α X R r,a,q,d(s,r) Π + X R lip. (3.58) r,a,q,d(s,r) Π 5

16 For λ, define λ Π = Π + λ lip Π. The symbol λ in Π λ will always be used in this role and never has the meaning of exponentiation. Set λ α/m. From (3.48) (3.58) we get X F z z z, z λ r,a,p,d(s 5σ,r) Π e8c γ Ks ασ 3n+2τ+2 X R λ r,a,q,d(s,r) Π. (3.59) Now considering the homological equations (3.3) (3.4), by a standard approach in finite dimensional KAM theory, we can easily get X F x r,a,p,d(s σ,r), X F y,y r,a,p,d(s σ,r) ασ τ+n X R r,a,q,d(s,r), (3.6) From (3.6) (3.6) we get X F x lip r,a,p,d(s 2σ,r) Π, X F y,y lip r,a,p,d(s 2σ,r) Π ασ 2τ+2n+ (M α X R r,a,q,d(s,r) Π + X R lip r,a,q,d(s,r) Π ). (3.6) X F x λ r,a,p,d(s 2σ,r) Π, X F y,y λ r,a,p,d(s 2σ,r) Π ασ 2τ+2n+ X R λ r,a,q,d(s,r) Π. (3.62) For the other terms of F, i.e. F z,z, F z, z, F zz z,z, F z z z, z, the same results - even better - than (3.59) can be obtained. Thus, we finally get the estimate for F: X F λ r,a,p,d(s 5σ,r) Π 4 The New Hamiltonian From (3.4)-(3.2) we get the new Hamiltonian where N + = (3.4) and e8c γ Ks ασ 3n+2τ+2 X R λ r,a,q,d(s,r) Π. (3.63) H Φ = N + + P +, (4.) P + = ˆR + {( t)( ˆN + ˆR) +tr,f} XFdt t + (P R) XF, (4.2) where ˆR = (3.7) + + (3.) := ˆR z,z + ˆR z, z + ˆR zz z,z + ˆR z z z, z + ˆR z z z, z. The aim of this section is to estimate the new normal form N + and the new perturbation P The New Normal Form In view of (3.4), denote N + = N + ˆN with ˆN = ˆω,y + ˆΩ j z j z j, j where ˆω := [R y ], (4.3) 6

17 ˆΩ j := R j j + x Ω j,f y = R j j + x Ω j,f y. (4.4) From (4.3) we easily get ˆω λ Π X R λ r,a,q,d(s,r) Π. (4.5) In the following, we estimate ˆΩ = ( ˆΩ j : j ). In view of the second estimate of (3.6), Thus, together with we get x Ω j,f y D(s σ) Ω j s,τ+ X F y,y r,a,p,d(s σ,r) γ j δ Applying to ˆΩ j, we have R j j D(s σ) j δ X R r,a,q,d(s,r), σ τ+n X R r,a,q,d(s,r). ˆΩ δ,d(s σ) σ τ+n X R r,a,q,d(s,r). (4.6) ˆΩ j = R j j x Ω j,f y x Ω j, F y. (4.7) Since R j j D(s 2σ) j δ X R r,a,q,d(s,r), we get x Ω j,f y D(s 2σ) σ Ω j D(s σ) X F y,y r,a,p,d(s 2σ,r) Ω δ,d(s) j δ ασ τ+n+ X R r,a,q,d(s,r), x Ω j, F y D(s 2σ) Ω j s,τ+ X F y,y r,a,p,d(s 2σ,r) αγ j δ X F y,y r,a,p,d(s 2σ,r), ˆΩ lip δ,d(s 2σ) Π σ 2n+2τ+ (M α X R r,a,q,d(s,r) Π + X R lip Therefore, from (4.6) (4.8) we get 4.2 The New Perturbation r,a,q,d(s,r) Π ). (4.8) ˆΩ λ δ,d(s 2σ) Π σ 2n+2τ+ X R λ r,a,q,d(s,r) Π. (4.9) We firstly estimate the error term ˆR z z with its matrix elements ˆR i j in (3.28). Split ˆR z z into three parts: ˆR z z = S + S 2 + S 3, such that S,S 2 have their matrix elements as follows: S i j = S 2 i j = {, i f i d j d < C K, ( Γ K )( i Ω i j F i j ), i f i d j d C K, { ( ΓK )(R i j ), i f < i d j d < C K,, i f i = j or i d j d C K, (4.) (4.) and S 3 is the cut-off of the perturbation R z z, that is, In view of (3.45), and using Lemma 9. below, we get S 3 = ( Γ K )(R z z diag(r z z )). (4.2) S a,q,p,d(s 2σ) e 9Kσ/ σ 2n R z z a,q,p,d(s). (4.3) 7

18 From (3.55) and ( Γ K )(( Ω i j )F i j ) D(s 3σ) e 9Kσ/ σ 2n (i δ + j δ ) Ω δ,d(s 2σ) F i j D(s 2σ) by using Lemma 9. below, we get e 9Kσ/ m i j σ 3n Ω δ,d(s 2σ) R i j D(s σ), (4.4) ( ) S lip a,q,p,d(s 4σ) Π e 9Kσ/ M Rz z σ 5n m a,q,p,d(s) Π + R z z lip. (4.5) a,q,p,d(s) Π Since by Lemma 9. below, we get S 2 i j D(s 2σ) e 9Kσ/ σ n R i j D(s σ), (4.6) S 2 a,q,p,d(s 2σ) e 9Kσ/ σ 2n max{ i j : i 2 j 2 < C K} R z z a,q,p,d(s) C Ke 9Kσ/ R z z a,q,p,d(s) σ 2n Again applying Lemma 9. to S 2, we can get C e 4Kσ/5 σ 2n+ R z z a,q,p,d(s). (4.7) It is obvious that S 2 lip a,q,p,d(s 4σ) Π C e 4Kσ/5 R z z lip a,q,p,d(s) Π. (4.8) σ 2n+ S 3 a,q,p,d(s σ) e 9Kσ/ σ n R z z a,q,p,d(s), (4.9) Thus S 3 lip a,q,p,d(s σ) Π e 9Kσ/ σ n R z z lip a,q,p,d(s) Π. (4.2) X ˆR z z z, z r,a,q,d(s 3σ,r) ( +C )e 4Kσ/5 σ 2n+2 X R r,a,q,d(s,r), (4.2) X ˆR z z z, z lip r,a,q,d(s 5σ,r) Π ( +C )e 4Kσ/5 ( ) M σ 5n+ α X R r,a,q,d(s,r) Π + X R lip. (4.22) r,a,q,d(s,r) Π Therefore, from (4.2) (4.22) we get X ˆR z z z, z λ r,a,q,d(s 5σ,r) Π ( +C )e 4Kσ/5 σ 5n+ X R λ r,a,q,d(s,r) Π. (4.23) For the other terms of ˆR, i.e. ˆR z,z, ˆR z, z, ˆR zz z,z, ˆR z z z, z, the same results - even better - than (4.23) can be obtained. Thus, we finally get the estimate for the error term ˆR: X ˆR λ r,a,q,d(s 5σ,r) Π ( +C )e 4Kσ/5 σ 5n+ X R λ r,a,q,d(s,r) Π. (4.24) 8

19 Now consider the new perturbation (4.2). By setting R(t) = ( t)( ˆN + ˆR) + tr, we have We assume that X P+ = X ˆR + (XF) t [X R(t),X F ]dt + (XF) (X P X R ). (4.25) X P r,a,q,d(s,r) Π λ αη2 e 8C γ Ks B σ (4.26) for λ α/m with some < η < /6 and < σ < min{s,}, where B σ = cσ 9(n+τ+) with c being a sufficiently large constant depending only on n, τ and ω Π. Since R is 2-order Taylor polynomial truncation in y,z, z of P, we can obtain As in lemma 9.3 of [], we obtain X R λ r,a,q,d(s,r) Π X P λ r,a,q,d(s,r) Π, (4.27) X P X R λ ηr,a,q,d(s,4ηr) Π η X P λ r,a,q,d(s,r) Π. (4.28) DX F λ r,a,p,p,d(s 6σ,r) Π, DX F λ r,a,q,q,d(s 6σ,r) Π From (3.63), (4.5), (4.9), (4.24), (4.29) and (4.27) we get X F λ r,a,p,d(s 5σ,r) Π e8c γ Ks ασ 3n+2τ+3 X R λ r,a,q,d(s,r) Π. (4.29) e8c γ Ks ασ 3n+2τ+2 X P λ r,a,q,d(s,r) Π, (4.3) X ˆN λ r,a,q,d(s 2σ,r) Π σ 2n+2τ+ X P λ r,a,q,d(s,r) Π, (4.3) X ˆR λ r,a,q,d(s 5σ,r) Π ( +C )e 4Kσ/5 σ 5n+ X P λ r,a,q,d(s,r) Π, (4.32) DX F r,a,p,p,d(s 6σ,r) Π λ, DX F r,a,q,q,d(s 6σ,r) Π λ e8c γ Ks ασ 3n+2τ+3 X P r,a,q,d(s,r) Π λ. (4.33) Moreover, together with the smallness assumptions (4.26), by properly choosing c, we get X F λ r,a,p,d(s 5σ,r) Π, DX F λ r,a,p,p,d(s 6σ,r) Π, DX F λ r,a,q,q,d(s 6σ,r) Π η2 σ c (4.34) with some suitable constant c. Then the flow X t F of the vector field X F exists on D(s 7σ,r/2) for t and takes this domain into D(s 6σ,r). Similarly, it takes D(s 8σ,r/4) into D(s 7σ,r/2). In the same way as (2.6) in [], we obtain X t F id λ r,a,p,d(s 7σ,r/2) Π X F λ r,a,p,d(s 6σ,r) Π, (4.35) DX t F I λ r,a,p,p,d(s 8σ,r/4) Π DX F λ r,a,p,p,d(s 6σ,r) Π, (4.36) DX t F I λ r,a,q,q,d(s 8σ,r/4) Π DX F λ r,a,q,q,d(s 6σ,r) Π. (4.37) Also in the same way as (2.7) in [], we obtain that for any vector field Y, (DX t F) Y λ ηr,a,q,d(s 9σ,ηr) Π Y λ ηr,a,q,d(s 7σ,4ηr) Π. (4.38) 9

20 From (4.27) (4.3) (4.32) and the assumption ( +C )e 4Kσ/5, we get Moreover, we have X R(t) λ r,a,q,d(s 5σ,r) Π σ 3n+2τ+ X P λ r,a,q,d(s,r) Π. (4.39) [X R(t),X F ] λ r,a,q,d(s 6σ,r/2) Π Hence, also DX R(t) λ r,a,q,p,d(s 6σ,r/2) Π X F λ r,a,p,d(s 6σ,r/2) Π + DX F r,a,q,q,d(s 6σ,r/2) Π λ X R(t) r,a,q,d(s 6σ,r/2) Π λ e 8C γ Ks ( XP ασ 6n+4τ+4 r,a,q,d(s,r) Π λ ) 2. (4.4) [X R(t),X F ] λ ηr,a,q,d(s 6σ,r/2) Π e8c γ Ks ( XP αη 2 σ 6n+4τ+4 r,a,q,d(s,r) Π λ ) 2. (4.4) Together with the estimates of ˆR in (4.32) and X P X R in (4.28), we finally arrive at the estimate X P+ ηr,a,q,d(s 9σ,ηr) Π λ ( Bσ e 8C γ Ks 3 αη 2 X P r,a,q,d(s,r) Π λ + B σ e 4Kσ/5 αη 2 This is the bound for the new perturbation. ) + η X P r,a,q,d(s,r) Π λ. (4.42) 5 Iteration and Convergence Set β = min{ β +β, 4 } and κ = 4 3 β 3. Now we give the precise set-up of iteration parameters. Let ν be the ν-th KAM step. α ν = α (9 + 2 ν ), which is used to dominate the measure of removed parameters, m ν = m (9 + 2 ν ), which is used for describing the growth of external frequencies, E ν = E 9 ( 2 ν ), which is used to dominate the norm of internal frequencies, M,ν = M, 9 ( 2 ν ), M 2,ν = M 2, 9 ( 2 ν ), M ν = M,ν + M 2,ν, which are used to dominate the Lipschitz semi-norm of frequencies, L ν = L 9 ( 2 ν ), which is used to dominate the inverse Lipschitz semi-norm of internal frequencies, J = γ τ+, J ν = J κν, which is used for estimate of measure, s ν = s /2 ν, which dominates the width of the angle variable x, σ ν = s ν /2, which serves as a bridge from s ν to s ν+, B ν = B σν := cσ 9(n+τ+) ν, here c is a large constant only depending on n, τ and E, 2

21 ε ν = ( ε ν µ= ( B µ iteration, α µ ) 3κ µ+ ) κ ν, which dominates the size of the perturbation P ν in ν-the KAM K ν = 5 lnε ν /(4σ ν ), which is the length of the truncation of Fourier series, η 3 ν = ε β ν α ν B ν, r ν+ = η ν r ν, D ν = D(s ν,r ν ), λ ν = α ν M ν. 5. Iterative Lemma Lemma 5.. Suppose that ε ( α γ ) β 8 B 3κ µ µ+, α m µ=. (5.) Suppose H ν = N ν + P ν is regular on D ν Π ν, where N ν is a generalized normal form with coefficients satisfying on Π ν, and P ν satisfies l k,ω ν (ξ ) + l, Ω ν (ξ ) α d ν, k, l 2, (5.2) k τ l, Ω ν (ξ ) m ν l d, < l 2, (5.3) Ω ν, j sν,τ+ (α α ν )γ j δ, j, (5.4) ω ν Πν E ν, ω ν lip Π ν M,ν, ω ν lip ω ν (Π ν ) L ν, (5.5) Ω ν lip δ,d ν Π ν M 2,ν (5.6) X Pν λ ν r ν,a,q,d ν Π ν ε ν. (5.7) Then there exists a Lipschitz family of real analytic symplectic coordinate transformations Φ ν+ : D ν+ Π ν D ν satisfying Φ ν+ id λ ν r ν,a,p,d ν+ Π ν, DΦ ν+ I λ ν r ν,a,p,p,d ν+ Π ν, DΦ ν+ I λ ν r ν,a,q,q,d ν+ Π ν B ν α ν ε β ν, and a closed subset where Π ν+ = Π ν \ k >J ν, l 2 (5.8) R ν+ kl (α ν+ ), (5.9) { R ν+ kl (α) = ξ Π ν : k,ω ν+ (ξ ) + l, Ω ν+ (ξ ) < α l } d k τ, (5.) such that for H ν+ = H ν Φ ν+ = N ν+ + P ν+, the estimate ω ν+ ω ν λ ν Π ν, Ω ν+ Ω ν λ ν δ,d ν+ Π ν B ν ε ν (5.) holds and the same assumptions as above are satisfied with ν + in place of ν. Proof. Setting C,ν = 2E ν /m ν, then it s obvious C,ν 4C,. Thus we have e 8C,ν γ K ν s ν ε β ν (5.2) 2

22 by K ν s ν = 2K ν σ ν = 25 lnε ν and choosing γ small enough such that 8C, γ β. In view of the definition of η ν, namely η 3 ν = ε β ν α ν B ν, the smallness condition (4.26), namely ε ν α ν η 2 ν B ν e 8C,ν γ K ν s ν, is satisfied if ε β ν α ν B ν. (5.3) To verify the last inequality we argue as follows. As B ν and α ν are increasing with ν, ( B ν ) β = ( B ν ) 3(κ ) = ( α ν α ν ( B ν ) ) 3κ µ+ κ ν ( µ=ν α ν ( B µ ) ) 3κ µ+ κ ν. (5.4) µ=ν α µ By the definition of ε ν above, the bound α ν 9α / and the smallness condition on ε in (5.), ε β B ν ν ( ε α ν µ= ( B µ ) ) 3κ µ+ κ ν ( β ) ( γ ) κ ν. (5.5) α µ 72 So the smallness condition (4.26) is satisfied for each ν. In particular, noticing κ 5/4, we have ε β B ν ν γ. (5.6) α ν 2ν+6 Now there exists a coordinate transformation Φ ν+ : D ν+ Π ν D ν taking H ν into H ν+. Moreover, (5.8) is obtained by (4.3) (4.33) (4.35)-(4.37), and (5.) is obtained by (4.3). More explicitly, (5.) is written as ω ν+ ω ν Πν, Ω ν+ Ω ν δ,dν+ Π ν B ν ε ν, (5.7) ω ν+ ω ν lip Π ν, Ω ν+ Ω ν lip δ,d ν+ Π ν M ν α ν B ν ε ν. (5.8) In view of (5.6)-(5.8), by choosing γ properly small, (5.3)-(5.6) are satisfied with ν + in place of ν. As to the diophantine conditions, in view of the definition of Π ν+ in (5.9), only the case k J ν remains to verify. In this case, by the definition of J ν, we have Hence, for k and k J ν, B ν ε ν α νγ κν 2 ν+6 α ν α ν+ 3J τ+ ν. (5.9) k,ω ν+ ω ν + l, Ω ν+ Ω ν k ω ν+ ω ν + 2 l d Ω ν+ Ω ν δ 3 k l d B ν ε ν (α ν α ν+ ) k l d J τ+ ν (α ν α ν+ ) l d k τ (5.2) on D ν+ Π ν. This completes the proof of (5.2) with ν + in place of ν. On the other hand, from (4.42) we get X Pν+ λ ν+ r ν+,a,q,d ν+ Π ν ( Bν e 8C,ν γ K ν s ν 3 α ν η 2 ε ν + B νe 4K ν σ ν /5 ν α ν η 2 + η ν )ε ν ν ( Bν 3 α ν η 2 ε β ν + B ν ν α ν η 2 ε β ν + η ν )ε ν ν = ( B ν ) 3 εν κ α ν = ε ν+. (5.2) 22

23 This completes the proof of iterative lemma. 5.2 Convergence We are now in a position to prove the KAM theorem. To apply iterative lemma with ν =, set N = N, P = P, s = s, r = r and similarly E = E, L = L, M, = M, M 2, = M 2, m = m, α = α and λ = λ = α/m. Define γ in the KAM theorem by setting γ = γ γ s, γ s = 8( µ= B 3κ µ µ+ ) β, (5.22) where γ is the same parameter as before and γ s only depends on n, τ, E, s, β. The smallness condition (5.) of the iterative lemma is then satisfied by the assumption of the KAM theorem: ε := X P λ r,a,q,d Π (αγ) +β (α γ γ s ) β. (5.23) The small divisor conditions (5.2) are satisfied by setting Π = Π \ R kl (α ), (5.24) (k,l) (,), l 2 and the other conditions (5.3)-(5.6) about the unperturbed frequencies are obviously true. Hence, the iterative lemma applies, and we obtain a decreasing sequence of domains D ν Π ν and a sequence of transformations Φ ν = Φ Φ ν : D ν Π ν D, such that H Φ ν = N ν + P ν for ν. Moreover, the estimates (5.8) and (5.) hold. Shorten r,a,p as r and consider the operator norm LW r L r, r = sup. W W r For r r, these norms satisfy AB r, r A r,r B r, r since W r W r. For ν, by the chain rule, using (5.8) (5.6), we get DΦ ν r,r ν,d ν Π ν ν µ= DΦ µ rµ,r µ,d µ Π µ µ= ( + 2 µ+6 ) 2, (5.25) DΦ ν lip r,r ν,d ν Π ν 2 ν DΦ µ lip µ= ν µ= r µ,r µ,d µ Π µ ρ ν,ρ µ DΦ µ I lip r µ,r µ,d µ Π µ 2 DΦ ρ rρ,r ρ,d ρ Π ρ µ= M µ α µ 2 µ+6 M α. (5.26) Thus, with the mean value theorem we obtain Φ ν+ Φ ν r,d ν+ Π ν DΦ ν r,r ν,d ν Π ν Φ ν+ id rν,d ν+ Π ν 2 Φ ν+ id rν,d ν+ Π ν, (5.27) 23

24 Φ ν+ Φ ν lip r,d ν+ Π ν DΦ ν lip r,r ν,d ν Π ν Φ ν+ id rν,d ν+ Π ν + DΦ ν r,r ν,d ν Π ν Φ ν+ id lip r ν,d ν+ Π ν M α Φ ν+ id rν,d ν+ Π ν + 2 Φ ν+ id lip r ν,d ν+ Π ν. (5.28) It follows that From (5.29) (5.8), we get Φ ν+ Φ ν λ r,d ν+ Π ν 3 Φ ν+ id λ ν r ν,d ν+ Π ν. (5.29) Φ ν+ Φ ν λ r,d ν+ Π ν 3 B ν α ν ε β ν. (5.3) For every non-negative integer multi-index k = (k,...,k n ), by Cauchy s estimate we have x k (Φ ν+ Φ ν ) λ r,d ν+2 Π ν 3 B ν ε β k! k n! ν α ν ( s, (5.3) ) k 2 ν+2 the right side of which super-exponentially decay with ν. This shows that Φ ν converge uniformly on D Π α, where D = T n {} {} {} and Π α = ν Π ν, to a Lipschitz continuous family of smooth torus embeddings Φ : T n Π α P a,p, for which the estimate (.9) holds. Similarly, the frequencies ω ν converge uniformly on Π α to a Lipschitz continuous limit ω, and the frequencies Ω ν converge uniformly on D Π α to a regular limit Ω, with estimate (.) holding. Moreover, X H Φ = DΦ X N on D for each ξ Π α, where N is the generalized normal form with frequencies ω and Ω. Thus, the embedded tori are invariant under the perturbed Hamiltonian flow, and the flow on them is linear. Now it only remains to prove the claim about the set Π \ Π α, which is the subject of the next section. 6 Measure Estimate 6. Proof of (.8) We know Π \ Π α = Rkl ν, (6.) ν k >J ν, l 2 where J =, J ν = γ κν /(τ+) and } Rkl {ξ ν = l Π ν : k,ω ν (ξ ) + l, Ω ν (ξ ) < α d ν k τ (6.2) with Π = Π. Here, ω ν and Ω ν are defined and Lipschitz continuous on Π ν, and ω = ω, Ω = Ω are the frequencies of the unperturbed system. Lemma 6.. If γ is sufficiently small and τ n + + 2/(d ), then where c > is a constant depends on n, d, E, L and m. Π \ Π α cα, (6.3) 24

25 Proof. We only need to give the proof of the most difficult case that l has two non-zero components of opposite sign. In this case, rewriting R ν kl as { Rki ν j = i ξ Π ν : k,ω ν (ξ ) + Ω ν,i (ξ ) Ω d j d } ν, j (ξ ) < α ν k τ, i j, (6.4) we only need to estimate the measure of Ξ α := ν k >J ν,i j R ν ki j. (6.5) To estimate the measure of k >J ν,i j Rki ν j, we introduce the perturbed frequencies ζ = ω ν(ξ ) as parameters over the domain Z := ω ν (Π ν ) and consider the resonance zones Ṙki ν j = ω ν(rki ν j ) in Z. Regarding Ω ν as function of ζ, then from the iterative lemma above, we know ζ E ν 9 E, ξ lip Z L ν 9 L, (6.6) Ω ν,i Ω ν, j Z m ν i d j d 9 m i d j d, (6.7) Ω ν, j lip Z L νm 2,ν j δ ( 9 )2 L M 2, j δ, (6.8) where E, L, m, M 2, are just E, L, m, M 2 in theorem.. Now we consider a fixed Ṙ ν ki j. If k < 9m ν E ν i d j d, we get k,ζ < 9m ν id j d. In view of (6.7) and α ν m ν, we know Ṙν ki j is empty. If k 9m ν E ν i d j d, we have k 2 ( 9 )3 m E (iδ + j δ ). Fix w {,} n such that k = k w and write ζ = aw + w 2 with w w 2. As a function of a, for t > s, k,ζ t s = k (t s), (6.9) ( Ω ν,i Ω ν, j ) t s ( 9 )2 LM 2 (i δ + j δ )(t s). (6.) Thus ( ( k,ζ + Ω ν,i Ω ν, j ) t s k (t s) ( 9 ) )2 LM 2 (i δ + j δ )/ k ( k (t s) 2( ELM ) 9 )5 2 m k (t s), (6.) by using the assumption 4ELM 2 m in theorem. and the fact (9/) 6 > /2. Therefore, we get Ṙ ν ki j (diamz)n α ν i d j d k τ+ (2 9 E)n α ( 9 ) 3 E m k τ. (6.2) Going back to the original parameter domain Π ν by the inverse frequency map ω, we get Rki ν j c α k τ, (6.3) 25

26 where c = 5(/9) 2n+2 (2EL) n /m. Consequently, we have that for any k > J ν, Rki ν j i j i d j d (/9) 3 (E/m) k Rki ν j c α 2, (6.4) k τ d 2 where c 2 = c ( 2(/9) 3 (E/m) ) 2/(d ). Moreover, since τ n + + 2/(d ), we have k >J ν,i j Rki ν j c α 2c 3, (6.5) + J ν where c 3 > depends only on n. The sum of the latter inequality over all ν converges, and we finally obtain the estimate of lemma Proof of Theorem.2 Since the proof of the case δ = d < d can be found in [7] or [], we assume δ = d = d in the following. From the assumption (D*) we know Ω = Ω + Ω 2 with Ω lip δ,π M 3 and Ω 2 lip δ,π M 4. Thus in ν-th KAM step, setting Ω ν = Ω + Ων, 2 we know Ω 2 = Ω2 and Ω 2 ν lip δ,π M 4,ν, (6.6) where the iteration parameter M 4,ν = M 4 9 ( 2 ν ). In the following we consider the excluded set of parameters under the assumption (D*) instead of (D). Our aim is to verify the conclusion (*) in Theorem.2. In the same way as [], we write Π \ Π α = Ξ α + Ξ 2 α, where Ξ α = Ξ 2 α = < k J, l 2 ν k >max(j,j ν ), l 2 R kl, (6.7) R ν kl. (6.8) Since k J and for each k there are only finitely many l for which R kl is not empty, the set Ξ α is a finite union of resonance zones. For each of its members we know that R kl as α for l by the first part of assumption (D*), and for l = by elementary volume estimates. Thus Ξ α as α. In the remainder of this section we estimate the measure of Ξ 2 α. Lemma 6.2. If γ is sufficiently small and τ n + + 2/(d ), then where c > is a constant depends on n, d, E, L and m. Ξ 2 α cα, (6.9) Proof. As in Lemma 6., we only need to give the proof of the most difficult case that l has two nonzero components of opposite sign. Seeing (6.4) for the definition of Rki ν j, we only need to estimate the measure of Rki ν j. (6.2) ν k >max(j,j ν ),i j 26

27 For the measure of k >max(j,j ν ),i j Rki ν j, we introduce the perturbed frequencies ζ = ω ν(ξ ) as parameters over the domain Z = ω ν (Π ν ) and consider the resonance zones Ṙki ν j = ω ν(rki ν j ) in Z. The estimate (6.6) (6.7) still hold true. Moreover, for Ω ν, j = Ω j + Ω2 ν, j, we have the estimate Let δ = min(δ δ,δ) and δ = max(,δ /δ ), Choose γ sufficiently small so that Ω j lip Z L νm 3 j δ 9 LM 3 j δ, (6.2) Ω 2 ν, j lip Z L νm 4,ν j δ ( 9 )2 LM 4 j δ. (6.22) L = ( 9 )4 2ELM 3 m, J = 2 9 LM 3L δ. J = γ τ+ J, (6.23) and thus J ν J J for all ν. Now we consider a fixed Ṙ ν ki j with k > J. If k < 9m ν E ν i d j d, then Ṙki ν j is empty. If k 9m ν E ν i d j d, we have k 2 ( 9 )3 m E (iδ + j δ ). (6.24) Fix w {,} n such that k = k w and write ζ = aw +w 2 with w w 2. As a function of a, for t > s, k,ζ t s = k (t s), (6.25) ( Ω i Ω j ) t s 9 LM 3(i δ + j δ )(t s), (6.26) ( Ω 2 ν,i Ω 2 ν, j ) t s ( 9 )2 LM 4 (i δ + j δ )(t s). (6.27) We claim In fact, 9 LM 3(i δ + j δ ) k 2. (6.28) () If i δ + j δ L, in view of (6.24) and then (6.28) follows from 2(i δ + j δ )(i δ + j δ ) i δ + j δ, (6.29) k ( 9 )3 m E (iδ + j δ )(i δ + j δ ) ( 9 )3 m E (iδ + j δ )L = 2 9 LM 3(i δ + j δ ); (6.3) (2) If i δ + j δ < L, then 9 LM 3(i δ + j δ ) 9 LM 3(i δ + j δ ) δ 9 LM 3L δ = J 2 k 2. (6.3) 27