# A KAM Theorem for Lower Dimensional Elliptic Invariant Tori of Nearly Integrable Symplectic Mappings

Authors: Shunjun Jiang
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

This paper develops a new KAM theorem for a class of lower dimensional elliptic invariant tori of nearly integrable symplectic mappings with generating functions but without assuming any nondegenerate condition.


---

## Body

## 1. Introduction and Main Results

The research on nondegeneracy condition of Hamiltonian systems is the fundamental problem of KAM theory because of small divisor problem. There are a significant number of results on nondegeneracy condition. We refer to Kolmogorov [1], Bruno [2], Rüssmann [3, 4], and Xu et al. [5]. The proofs of classic KAM theorems [1, 6–8] are based on the KAM iteration procedure, which include some nondegeneracy conditions. Can the nondegeneracy condition be separated from KAM iteration procedure? In 2016, Xu and Lu [9] provided a new KAM technique to prove some general KAM theorems for nearly integrable Hamiltonian systems without any nondegeneracy condition. Xu and Lu also generalize some classic KAM theorems [1, 6–8] by the general KAM theorems in [9] combining some nondegeneracy conditions. We note that although the proofs of the KAM theorems in [9] did not need nondegeneracy condition, the assumption of nondegeneracy condition is still necessary for the application of those theorems; otherwise they may have no dynamical consequences. We refer to [9] for more details.As one part of the classical KAM theory, the persistence of invariant tori of nearly integrable twist mappings was investigated by lots of mathematicians. The first work was due to Moser [7, 8]. Since then, a large body of KAM results for mappings have been proved. We refer to [10–17]. As discrete Hamiltonian systems, symplectic mappings are special among all the mappings because they have symplectic structures, and hence they attract many mathematicians’ attention. In 2015, Lu et al. [18] proved a KAM theorem on lower dimensional elliptic invariant tori for nearly integrable symplectic mappings. We refer readers to [19–21] for more results on symplectic mappings.Motivated by [9, 18, 22], we will extend the result [9] to symplectic mappings and prove a KAM theorem of symplectic mappings with generating functions but without assuming any nondegenerate condition.We consider the following parameterized symplectic mappingΦx,u,y,v;ξ such that (1)Φ:x,u,y,v∈Tn×W×O×W⟶x^,u^,y^,v^∈Tn×Rm×Rn×Rm,and ξ∈Π⊂O is parameter where O⊂Rn is a bounded closed connected domain and Π a subset of O such that dist∂O,Π=infs∈Π,t∈∂Os-t>0. Moreover, Φx,u,y,v;ξ is implicitly defined by a generating function(2)Hx,u,y^,v^;ξ=N+P,where(3)Nx,u,y^,v^;ξ=x+ωξ,y^+Au,v^+12Bu,u+12Cv^,v^withA,B and C being constant matrices. Moreover, B and C are symmetric.There are a great number of results in symplectic mappings and Hamiltonian systems, which are parallel and almost identical, but the proofs are different. The reason lies in special properties of symplectic mappings. For instance, the generating functions, which decide the symplectic mappings and the relation of variables, take on sophisticated implicit forms. So the relation of variables in symplectic mappings is not easily understood, which makes KAM estimates more complicated.Without assuming any nondegenerate condition, we will give a formal KAM theorem for symplectic mappings. The idea of the proof is to separate the nondegenerate conditions from the KAM iteration, which was introduced in [9]. It is noted that the proofs of the classic KAM theorems [1, 6–8] usually need nondegenerate conditions, which can assure that the small divisor conditions hold at each KAM-step; thus the Diophantine constants αn in the n-KAM steps are chosen decreasing and the existence of invariant tori can be guaranteed. However, in this paper, since we do not have any nondegenerate condition, the existence of invariant tori depends on whether the final frequencies ω∗ meet the small divisor conditions, where ω∗=limn→∞ωn. The key lies in an explicit extension of small divisors to the parameter definition domain and the choice of the Diophantine parameter αn. In our KAM iteration we require αn to increase as n→∞. Moreover, in our problem there are not only tangential frequency ω∗ but also normal frequency θ∗; this makes our conditions more completed than in [9]. Although no nondegenerate condition is assumed in our theorem, it is necessary in its application. In fact, we usually use some nondegenerate condition to guarantee the final frequencies ω∗ and θ∗ to satisfy the nonresonance and so the existence of invariant tori. In particular, the previous KAM results on symplectic mappings can be consequences of our theorem under various kinds of nondegenerate conditions. Without the assumption of nondegenerate conditions, our theorem is only formal and may have no dynamical consequences.Before giving the main result, we introduce some assumptions.Assumption 1 (elliptic condition).
Set(4)A=sec⁡θ,B=C=tan⁡θ,where θ=θ1,…,θm, sec⁡θ=diagsec⁡θ1,…,sec⁡θm.tan⁡θ,sin⁡θ, and cos⁡θ have similar notations.Remark 2.
Consider the symplectic mappingΦx,u,y^,v^;ξ generated by Hx,u,y^,v^;ξ in (2). If A is nonsingular, Φ can be expressed explicitly as(5)x^=x+ωy,y^=y,u^=A-CAT-1Bu+CAT-1v,v^=-AT-1Bu+AT-1v.We define(6)Ω=A-CAT-1BCAT-1-AT-1BAT-12m×2m.By Assumption 1, we have(7)Ω=cos⁡θsin⁡θ-sin⁡θcos⁡θ.It is easy to verify Ω has the eigenvalues e±iθ1,…,e±iθm. Since e±iθi=1 and e±iθ1≠1,i=1,2,…,m, we call the lower dimensional invariant torus determined by the above symplectic mapping Φ elliptic.Assumption 3 (nonresonance conditions).
Supposeω satisfies the following: for k∈Zn and 1≤i,j≤n,(8)k,ω-2πl≥α1+kτ,k,ω-θj-2πl≥α1+kτ,k,ω-θi-θj-2πl≥α1+kτ,k,ω+θi-θj-2πl≥α1+kτ,k+i-j≠0.
Set(9)Ts=x∈Cn2πZn:Im⁡x∞≤s,Br=y∈Cn:y1≤r2,Wr=w∈Cm:w2≤r.Denote Ds,r=Ts×Wr×Br×Wr. Here, x∞=max1≤j≤nxj,y1=∑1≤j≤n‍yj and w2=∑1≤j≤m‍wj21/2. In this paper, we will use the norms defined in [18].Theorem 4.
Consider the parameterized symplectic mappingΦ·;ξ generated by H·;ξ defined in (2). Let H be real analytic in Ds,r and Cm-smooth in ξ on Π. Suppose that Assumptions 1 and 3 hold. Let(10)Oα,τ=ω,θ∈Rn×Rm:k,ω-s1θi-s2θj-2πl≥α1+kτ,k+i-j≠0,∀k∈Zn∖0,θ=θ1,…,θm,s1+s2⩽2,si⩽1,si∈Z,i=1,2.Then, for any 0<α⩽1, τ>n-1 and m⩾0, there exists sufficiently small ϵ>0 such that if XPr;Ds,r×Π⩽ϵ the following results hold:
(i) There exist a family of symplectic mappings{Ψ∗·;ξ:ξ∈Π} with (11)Ψ∗-idDr;s/2,r/2×Π≤cϵ,such that Φ∗·;ξ=Ψ∗-1∘Φ∘Ψ∗ is generated by functions H∗=N∗+P∗ as in (2), where (12)N∗x,u,y^,v^;ξ=x+ω∗,y^+A∗u,v^+12B∗u,u+12C∗v^,v^,P∗x,u,y^,v^;ξ=∑i+j+2l≥3Plijx;ξy^luiv^j.Let A∗=sec⁡θ∗ and B∗=C∗=tan⁡θ∗. One has (13)ω∗-ωCmΠ⩽2ϵ,θ∗-θCmΠ⩽cϵ.
(ii) Ifω∗ξ,θ∗ξ∈Oα,τ, the symplectic mapping Φ·;ξ has an invariant torus Ψ∗Tn,0;ξ with frequencies ω∗(ξ).Remark 5.
By Theorem4, for all ξ∈ξ∈Π:ω∗ξ,θ∗ξ∈Oα,τ, the original symplectic mapping Φ·;ξ is transformed into the conjugate mappings Φ∗·;ξ under the compatible transformation Ψ∗·;ξ. Hence the mapping Φ·;ξ has an invariant torus with the frequency ω∗ξ. If ξ∈Π:ω∗ξ,θ∗ξ∈Oα,τ=∅, the conjugate mapping Φ∗·;ξ has nothing to do with the symplectic mapping Φ·;ξ. Then our result is with no sense. To avoid this situation, some nondegeneracy conditions are necessary for Φ·;ξ to satisfy with ξ∈Π:ω∗ξ,θ∗ξ∈Oα,τ≠∅. But Theorem 4 itself does not need any nondegeneracy condition.


## 2. Proof of Main Results

### 2.1. KAM-Step

Lemma 6 (KAM iteration lemma).
Consider the symplectic mappingΦ·;ξ defined in Theorem 4. Let α⩽α∗⩽2α,τ>n-1,m⩾0. Assume ω∈CmΠ and P satisfies XPr;Ds,r×Π≤ϵ=η2α2ν¯ρ2νE, where 0<α<1,0<ρ<s/5,0<η<1/8,η2e-Kρ=E with ν¯=4m+1,ν=4τm+1+m+n. Then the following conclusions hold:
(i)∀ξ∈Π there exists a symplectic diffeomorphism Ψ·;ξ with (14)Ψ-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨ-Idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1,such that conjugate mapping Φ+·;ξ=Ψ-1∘Φ∘Ψ is generated by H+=N++P+, where (15)N+=x+ω+ξ,y^+A+u,v^+12B+u,u+12C+v^,v^with A+=sec⁡θ+, B+=C+=tan⁡θ+. Moreover, one has(16)ω+-ωCmΠ≤ϵ,θ+-θCmΠ≤cϵ.P+ is a smaller term with the estimate (17)XPr+;Ds+,r+×Π≤cϵ+=cη+2α+2ν¯ρ+2νE+.Here (18)s+=s-5ρ,ρ+=σρ,η+=E+,r+=ηr,E+=E4/3,α⩽α+⩽2α.
(ii) Letα+=α+3Kτ+1ϵ and (19)Oα,K=ωξ,θξ∈Rn×Rm:ω,k-s1θi-s2θj-2πl⩾α1+kτ,∀k∈Zn,0<k≤K,s1+s2⩽2,si⩽1,si∈Z,i=1,2.Define(20)U=ξ∈Π:ωξ,θξ∈Oα,K,U+=ξ∈Π:ω+ξ,θ+ξ∈Oα+,K+,where K+>K satisfies e-K+ρ+/η+2=E+. Then one has U+⊂U.


#### 2.1.1. Generating Functions of Conjugate Mappings

For the convenience, letp=x,u and q=y,v. p^ and q^ have similar meaning. The symplectic structure becomes dp∧dq on Rn+m×Rn+m. Consider a symplectic mapping Φ:p,q→p^,q^ generated by(21)p^=∂q^Hp,q^,q=∂pHp,q^.The generating function is Hp,q^=Np,q^+Pp,q^, where N is main term and P is a small perturbation.We need a symplectic transformationψ1:p+,q+→p,q, generated by(22)q=q++F1p,q+,p+=p+F2p,q+.The generating function is p,q++Fp,q+, with F being a small function. So we have ψ1→I. At the same time, ψ1 is also satisfied with p^+,q^+→p^,q^ by(23)q^=q^++F1p^,q^+,p^+=p^+F2p^,q^+.By (21) to (23), we have a conjugate mapping Φ¯=ψ1-1∘Φ∘ψ1:p+,q+→p^+,q^+ implicitly by(24)p^+=H2p,q^+F2p^,q^+,q+=H1p,q^-F1p,q+.We will prove that there is a function H¯p+,q^+ generating Φ¯.Lemma 7 (Lemma2.1 in [18]).
The conjugate symplectic mappingΦ¯ can be implicitly determined by a generating function H¯p+,q^+ through(25)p^+=∂q^+H¯p+,q^+,q+=∂p+H¯p+,q^+,where(26)H¯p+,q^+=Hp,q^+H1p,q^F2p,q+-H2p,q^F1p^,q^++Fp^,q^+-Fp,q+-F1p,q+F2p,q+,where p,p^,q^,q+ depend on (p+,q^+) as explained above.
Moreover, setz=(p+,q^+), Δz=-F2p,q+,F1p^,q^+, and then we have(27)H¯z=Hz+FN2z,q^+-Fp+,N1z+Qz.The small term Qz has the estimate(28)XQr;Ds-5ρ,r/16×Π≤cϵ2α2ν¯ρ2νwith ν¯=4m+1, ν=m+n+4τm+1.


#### 2.1.2. Homological Equations

Now we will solve homological equations forF. Let(29)P=R+P~,where(30)Rp,q^=P000x+P100x,y^+P010x,u+P001x,v^+P011xu,v^+12P020xu,u+12P002xv^,v^,with (31)Plij=∂l+i+jP∂y^l∂ui∂v^ju=0,y^=0,v^=0,2l+i+j≤2.Let F(p,q^) possess the same form as (30).We will use the idea in [22] to solve homological equations: (32)Np+,q^++Rp+,q^+-Fp+,Npp+,q^++FNqp+,q^+,q^+=N¯p+,q^+.For simplicity, here and below we drop the subscripts “+” in p+ and q^+.Letx+ω=x~. Since (33)p^=Nq^p,p^=x~,Au+Cv,q=Np,q^p,p^=y^,Av^+Bu,we have (34)FNqp+,q^+,q^+-Fp+,Npp+,q^+=L0+L1+L2,where L0,L1,L2 indicate the 0, 1, 2 order terms of u,v^, respectively, with (35)L0=F000x~-F000x+F100x~-F100x,y^,L1=ATF010x~-F010x-BF001x,u+CF010x~+F001x~-AF001x,v,L2=F011x~A-AF011x+CF020x~A-AF002xBu,v^+12ATF020x~A-F020x-BF002xB-BF011x-F011TxBu,u+12CF020x~C+F002x~-AF002xAT+F011x~C+CF011Tx~v^,v^.Let A=diaga1,…,an and B=C=diag(b1,…,bn) with aj=sec⁡θj and bj=tan⁡θj. Now we solve the following equations:(36)L0=R000x-R000+R100x-R100L1=R010x+R001xL2=R011x-A^+R020-B^+R002-C^,where A^,B^, and C^ will be determined later.Firstly, forFj00,j=0,1, we consider the equation (37)Fj00x+ω-Fj00x=Rj00x-Rj00.We now expand the following functions as Fourier series: Fj00x=∑k∈Zn‍Fkj00ei〈k,x〉 and Rj00x=∑k∈Zn‍Rkj00ei〈k,x〉. Then we get(38)Fkj00=1ek-1Rkj00,with ek=eik,ω,k≠0.Next we try to getF010 and F001. Let F010=(F0101,…,F010m) and F001=(F0011,…,F001m). We expand the following functions as Fourier series: F0i′j′lx=∑k∈Zn‍Fk0i′j′leik,x and R0i′j′lx=∑k∈Zn‍Rk0i′j′leik,x with l=1,2,…,m and i′,j′=0,1,1,0.To get the relations betweenF0i′j′lx and R0i′j′lx, we need the following equation set:(39)aek-1-bbekek-a·Fk010lxF001x·eik,x=Rk010lRk001l,with a=sec⁡θl, b=c=tan⁡θl. Then we have(40)Fk0i′j′l=R~i′j′l2sec⁡θjekcos⁡k,ω-cos⁡θj,where R~i′j′l are linear combination of R010lx and R001lx.Thirdly, we solve the third equation of (36). For this purpose, we solve(41)L2=R011x+R020+R002.LetF0i′j′=F0i′j′ij1≤i,j≤m with i′,j′=1,1,2,0,0,2. We expand the following functions: F0i′j′ij=∑k∈Zn‍Fk0i′j′ei〈k,x〉 and R0i′j′ij=∑k∈Zn‍Rk0i′j′ei〈k,x〉.To get the relations betweenFk0i′j′ij and R010ijx, we need the following equation set:(42)0ekaj-aiekbiaj-aibjekai-aj0ekaibj-biaj-bj-biekaiaj-1-bibjekbiekbjekbibjek-aiaj·Fk011jiFk011ijFk020ijFk002ij·eik,x=Rk011jiRk011ijRk020ijRk002ij.Then we have(43)Fk0i′j′ij=R~ij4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,with R~ij being linear combination of Rk0i′j′ij.


#### 2.1.3. Extension of Small Divisors

This part is critical for this paper. Letϕt be a C∞R-smooth function with (44)ft=0,t⩽12,1,t⩾1.For h⩾0, let fht=ft/h. Then fht∈C∞R with (45)dλdtλfht⩽cλhλ,∀t∈R,∀λ⩾0,where cλ is a constant depending on λ.Firstly, we extendFkj00 from U to the whole set Π. Let (46)h=α1+kτ,t1kξ=ωξ,k-2πl,g1kξ=fht1kξeiωξ,k-1,∀ξ∈Π.By the definition of U, we have (47)g1kξ=1eiωξ,k-1,∀ξ∈U.Note that even if U=∅, the extension of g1kξ is still well defined on Π. Furthermore, g1kξ∈CmΠ with estimate (48)∂βg1k∂ξβ⩽ckβλβ+1,ξ∈Π,∀β⩽m.Noticing (38), we now extend Fkj00 from U to whole set Π by setting (49)F~kj00ξ=g1kξRkj00ξ=fqtkξeiωξ,k-1Rkj00ξ,j=0,1,0<k⩽K.Let(50)F~j00ξ;x,y=∑0<k⩽KF~kj00ξ;yeiωξ,k.Then we have the following estimates: (51)F~j00s-ρ×Π≤cRj00sαm+1ρm+n+τm+1.Secondly, we extendFk0i′j′l from U to the whole set Π, where (i′,j′)=(0,1),(1,0). Let (52)t2k=ωξ,k-θ-2πl,(53)g2kξ=fht2kξ2sec⁡θjekcos⁡k,ω-cos⁡θj.Then we have (54)g2kξ=12sec⁡θjekcos⁡k,ω-cos⁡θj,∀ξ∈U.Noticing (40), we extend Fk0i′j′l by setting (55)F~k0i′j′lξ=g2kξR~i′j′l=fhtkξ2sec⁡θjekcos⁡k,ω-cos⁡θjR~i′j′l,with l=1,2,…,m and (i′,j′)=0,1,1,0,0<|k|⩽K. Let (56)F~0i′j′ξ;x,y=∑0<k⩽KF~k0i′j′ξ;yeiωξ,k.Then we have (57)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα2m+2ρ2τm+1+m+nwith i′,j′=0,1,1,0.At last, we expendFk0i′j′ij, where 1≤i,j≤m and i′,j′=1,1,2,0,0,2. Let (58)t3k=ωξ,k-θi-θj-2πlωξ,k-θi+θj-2πl,g3kξ=fh2t3kξ4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj.So we have (59)g3kξ=14sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,∀ξ∈U. Noting (43), we now extend Fk0i′j′ij to the whole set Π by setting (60)F~k0i′j′ij=g3kξR~ij,0<k<K.Let (61)F~0i′j′ij=∑0<k⩽KF~k0i′j′ijeiωξ,k.Recalling the third equation of (36) and the conclusions of (41), we let(62)ω^=diagω^1,…,ω^n,A^=diagA^1,…,A^m,B^=diagB^1,…,B^m,C^=diagC^1,…,C^m,with (63)ω^j=R100jj,A^j=R011jj,B^j=R020jj,C^j=R002jj.We sum up the above discussion and arrive at (64)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα4m+4ρ4τm+1+m+nwith i′,j′=1,1,2,0,0,2.By the above discussion, we get a functionF~ defined on Π which is the extension of F. Moreover F~ not only has the same form as R and F, but also inherits all properties of F when ξ∈U. So F~ satisfies L0+L1+L2F~=R-N^, where(65)N^p,q^=ω^,y^+A^u,v^+12B^u,u+12C^v^,v^.So we have (66)XF~r;Ds-ρ,r×Π≤cϵαν¯ρν.Let χ:p,q→-F~y+,F~x. Since ψ1=id+χ, we combine Cauchy estimate to obtain (67)ψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,Dψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.


#### 2.1.4. Normal Form

From the above discussion, we get a conjugate mappingΦ¯·;ξ=ψ1-1∘Φ∘ψ1 generated by H¯=N¯1+N¯2+P¯ where N¯1=x+ω+ω^,y^, N¯2=A¯u,v^+1/2B¯u,u+1/2C¯v^,v^, and P¯=P~+Q with (68)ω^=ω+-ω,A¯=A0+A^,B¯=B0+B^,C¯=C0+C^.We note that the normal form depends only onu,v, so we consider N¯2. Since B=C and A2=B2+C2 may not be satisfied, N¯2 may not be a normal form and hence we need normalize N¯2 into a new normal form.Lemma 8.
There exists a symplectic mappingψ2 generated by 〈u,v^〉+Gu,v+, such that the corresponding conjugate mapping Φ+·;ξ=ψ¯2-1∘Φ¯∘ψ¯2:u+,v+→u^+,v^+ is generated by (69)N~2+=A+u,v^+12B+u,u+12C+v^,v^,where A+2=B+2+1, B+=C+. Let A+=sec⁡θ+. One has(70)θ+-θCmΠ<cϵ,and hence(71)ψ¯2-id=Oϵ.Remark 9.
We note thatψ¯2 can normalize N¯2 into a new normal form. Using the same idea and method, we can find a new symplectic mapping ψ2 normalizing N¯1x,y^+N¯2u,v^ into (72)N+=x+ω+,y^+A+u,v^+12B+u,u+12C+v^,v^.In fact, ψ2=p,q^+Gp,q+ with ψ2-id=O(ϵ).LetΨ=ψ1∘ψ2. Similar to the estimates of ψ1 and ψ2, we have (73)Ψ-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨ-Idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.Moreover, H+=N++P+, where N+ is normal form. H+ can be shown just like H¯, and we omit the details.


#### 2.1.5. Estimates for New Perturbation and Error Terms

We aim at the estimate ofP+ and some error terms. By (29), we have(74)XP~ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2.By (28) and (74), we have(75)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν.Moreover, for the error terms ω^=ω+-ω and θ^=θ+-θ, we have ω^Cm(Π)≤ϵ and θ^Cm(Π)≤cϵ.


#### 2.1.6. The Choice of Parameters in KAM Iteration Lemma

We choose a weighted error0<E<1 and set (76)η=E,ϵ=α2ν¯ρ2νEη2,E=e-Kρη2,h=α2Kτ+1T.Fix σ∈0,1, for the next step, we define (77)ρ+=σρ,s+=s-5ρ,r+=ηr,α+=α+31+Kτ+1ϵ,ϵ+=cηϵ,E+=cE4/3.By (75) and α<2α+, we have(78)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν≤cηϵ=cα2ν¯ρ2νE4≤cα+2ν¯ρ+2νE+3.Setting ϵ+=cα+2ν¯ρ+2νE+3, we arrive at (79)XP+r+;Ds+,r+×Π≤ϵ+.At last, let α+=α+=α+31+Kτ+1ϵ, and then we have U+⊂U.


### 2.2. Iteration

In this section, we will summarize the above results on parameters so that KAM-step can iterate infinitely. At the initial step set(80)s0=s,ρ0=1-σs10,r0=r,α0=α2,E0=E,η0=E0,ϵ0=α02ν¯ρ02νE0η02.Let (81)e-K0ρ0η02=E0,ω0ξ=ωξ.Assume that ρj,sj,rj,Ej,αj are well defined for the jth step. Then ηj,Kj,ϵj are defined as follows: (82)ηj=Ej,ϵj=αj2ν¯ρj2νEjηj2,e-Kjρjηj2=Ej.Define inductive sequences: (83)ρj+1=σρj,sj+1=sj-5ρ,rj+1=ηjrj,αj+1=αj+31+Kjτ+1ϵj,Ej+1=cEj4/3.Similarly, we can define ηj+1,Kj+1,ϵj+1.Similar to the proof of KAM iteration Lemma, we haveUj+1⊂Uj. Moreover, there exists a sequence of {Ψj}={ψ1j∘ψ2j}, where ψ1j is generated by p,q++Fjp,q+ and ψ2j is generated by p,q++Gjp,q+. Moreover, we have (84)Ψj-idrj;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨj-Idrj;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.There exists a sequence of symplectic mappings Φj which are well defined on Dsj,rj and satisfy(85)Φj+1=Ψj-1∘Φj∘Ψj,where Ψj=Ψ1∘Ψ2∘⋯∘Ψj. The symplectic mappings Φj are generated by Hj=Nj+Pj, where (86)Nj=x+ωjξ+Aju,v^+12Bju,u+12Cjv^,v^with Aj=sec⁡θj, Bj=Cj=tan⁡θj. Moreover we have(87)ωj+1-ωjCmΠ≤cϵj,θj+1-θjCmΠ≤cϵj∀j≥1.


### 2.3. Convergence of the Iteration

Now we prove the convergence of KAM iteration. In the same way as [22, 23], we have (88)Ψj-Ψj-1rj;Dsj,rj×Π≤cαj-12ν¯ρj-12νEj-13,DΨj-Ψj-1rj;Dsj,rj×Π≤αj-12ν¯ρj-12ν-1Ej-13.Note that sj→s/2 and rj→0 as j→∞. Let Ψ∗=limj→∞Ψj. Since Ψj is affine in y, we have Ψj converging to Ψ∗ on Dr/2,s/2 with (89)Ψ∗-idr/2;Ds/2,r/2×Π≤cα,ρ,σE03,∀β∈Zn+.LetP∗=limj→∞Pj and ω∗=limj→∞ωj. Then P∗ is real analytic in x,y on D(r/2,s/2) and Cm in ξ on Π. Note that ωj=ω+∑i=0j-1‍ω^j. Then we have (90)ω∗-ωjCmΠ⩽∑i=j∞ϵj⩽2ϵj.Particularly, (91)ω∗-ωCmΠ⩽2ϵ.Correspondingly, letHj=Nj+Pj→H∗=N∗+P∗ satisfy (92)N∗=x+ω∗ξ,y^+A∗ξu,v^+12B∗ξu,u+12C∗ξv^,v^,where A∗=sec⁡θ∗ and B∗=C∗=tan⁡θ∗ with estimates (93)θ∗-θCmΠ≤cϵ.Moreover, P∗p,q^=∑2|l|+|i|+|j|≥3‍Plij(x)y^luiv^j.Let(94)Π∗=ξ∈Π:ω∗ξ,θ∗ξ∈Oα,τ.In the sequel we will prove Π∗⊂Uj for all j⩾0, which is equal to proving α⩾αj+4ϵjKjτ+1. By the choice of parameters, we have Kj+1τ+1ϵj+1/Kjτ+1ϵj⩽cEj and K0=-3ln⁡E0/ρ0, so (95)Kjτ+1ϵj⩽E0c0K0τ+1ϵ0⩽cE0c0+3ln⁡E0ρ0τ+1αρ0τ′.Let E0 be small enough such that 8Kjτ+1ϵj⩽α/2, and then we have (96)αj+4ϵjKjτ+1=α2+4∑l=1jKlτ+1ϵl⩽α2+8Kjτ+1ϵj⩽α.So Π∗⊂Uj and hence Π∗⊂⋂j⩾0Uj, for all j⩾0. This completes the proof of Theorem 4.


## 2.1. KAM-Step

Lemma 6 (KAM iteration lemma).
Consider the symplectic mappingΦ·;ξ defined in Theorem 4. Let α⩽α∗⩽2α,τ>n-1,m⩾0. Assume ω∈CmΠ and P satisfies XPr;Ds,r×Π≤ϵ=η2α2ν¯ρ2νE, where 0<α<1,0<ρ<s/5,0<η<1/8,η2e-Kρ=E with ν¯=4m+1,ν=4τm+1+m+n. Then the following conclusions hold:
(i)∀ξ∈Π there exists a symplectic diffeomorphism Ψ·;ξ with (14)Ψ-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨ-Idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1,such that conjugate mapping Φ+·;ξ=Ψ-1∘Φ∘Ψ is generated by H+=N++P+, where (15)N+=x+ω+ξ,y^+A+u,v^+12B+u,u+12C+v^,v^with A+=sec⁡θ+, B+=C+=tan⁡θ+. Moreover, one has(16)ω+-ωCmΠ≤ϵ,θ+-θCmΠ≤cϵ.P+ is a smaller term with the estimate (17)XPr+;Ds+,r+×Π≤cϵ+=cη+2α+2ν¯ρ+2νE+.Here (18)s+=s-5ρ,ρ+=σρ,η+=E+,r+=ηr,E+=E4/3,α⩽α+⩽2α.
(ii) Letα+=α+3Kτ+1ϵ and (19)Oα,K=ωξ,θξ∈Rn×Rm:ω,k-s1θi-s2θj-2πl⩾α1+kτ,∀k∈Zn,0<k≤K,s1+s2⩽2,si⩽1,si∈Z,i=1,2.Define(20)U=ξ∈Π:ωξ,θξ∈Oα,K,U+=ξ∈Π:ω+ξ,θ+ξ∈Oα+,K+,where K+>K satisfies e-K+ρ+/η+2=E+. Then one has U+⊂U.


### 2.1.1. Generating Functions of Conjugate Mappings

For the convenience, letp=x,u and q=y,v. p^ and q^ have similar meaning. The symplectic structure becomes dp∧dq on Rn+m×Rn+m. Consider a symplectic mapping Φ:p,q→p^,q^ generated by(21)p^=∂q^Hp,q^,q=∂pHp,q^.The generating function is Hp,q^=Np,q^+Pp,q^, where N is main term and P is a small perturbation.We need a symplectic transformationψ1:p+,q+→p,q, generated by(22)q=q++F1p,q+,p+=p+F2p,q+.The generating function is p,q++Fp,q+, with F being a small function. So we have ψ1→I. At the same time, ψ1 is also satisfied with p^+,q^+→p^,q^ by(23)q^=q^++F1p^,q^+,p^+=p^+F2p^,q^+.By (21) to (23), we have a conjugate mapping Φ¯=ψ1-1∘Φ∘ψ1:p+,q+→p^+,q^+ implicitly by(24)p^+=H2p,q^+F2p^,q^+,q+=H1p,q^-F1p,q+.We will prove that there is a function H¯p+,q^+ generating Φ¯.Lemma 7 (Lemma2.1 in [18]).
The conjugate symplectic mappingΦ¯ can be implicitly determined by a generating function H¯p+,q^+ through(25)p^+=∂q^+H¯p+,q^+,q+=∂p+H¯p+,q^+,where(26)H¯p+,q^+=Hp,q^+H1p,q^F2p,q+-H2p,q^F1p^,q^++Fp^,q^+-Fp,q+-F1p,q+F2p,q+,where p,p^,q^,q+ depend on (p+,q^+) as explained above.
Moreover, setz=(p+,q^+), Δz=-F2p,q+,F1p^,q^+, and then we have(27)H¯z=Hz+FN2z,q^+-Fp+,N1z+Qz.The small term Qz has the estimate(28)XQr;Ds-5ρ,r/16×Π≤cϵ2α2ν¯ρ2νwith ν¯=4m+1, ν=m+n+4τm+1.


### 2.1.2. Homological Equations

Now we will solve homological equations forF. Let(29)P=R+P~,where(30)Rp,q^=P000x+P100x,y^+P010x,u+P001x,v^+P011xu,v^+12P020xu,u+12P002xv^,v^,with (31)Plij=∂l+i+jP∂y^l∂ui∂v^ju=0,y^=0,v^=0,2l+i+j≤2.Let F(p,q^) possess the same form as (30).We will use the idea in [22] to solve homological equations: (32)Np+,q^++Rp+,q^+-Fp+,Npp+,q^++FNqp+,q^+,q^+=N¯p+,q^+.For simplicity, here and below we drop the subscripts “+” in p+ and q^+.Letx+ω=x~. Since (33)p^=Nq^p,p^=x~,Au+Cv,q=Np,q^p,p^=y^,Av^+Bu,we have (34)FNqp+,q^+,q^+-Fp+,Npp+,q^+=L0+L1+L2,where L0,L1,L2 indicate the 0, 1, 2 order terms of u,v^, respectively, with (35)L0=F000x~-F000x+F100x~-F100x,y^,L1=ATF010x~-F010x-BF001x,u+CF010x~+F001x~-AF001x,v,L2=F011x~A-AF011x+CF020x~A-AF002xBu,v^+12ATF020x~A-F020x-BF002xB-BF011x-F011TxBu,u+12CF020x~C+F002x~-AF002xAT+F011x~C+CF011Tx~v^,v^.Let A=diaga1,…,an and B=C=diag(b1,…,bn) with aj=sec⁡θj and bj=tan⁡θj. Now we solve the following equations:(36)L0=R000x-R000+R100x-R100L1=R010x+R001xL2=R011x-A^+R020-B^+R002-C^,where A^,B^, and C^ will be determined later.Firstly, forFj00,j=0,1, we consider the equation (37)Fj00x+ω-Fj00x=Rj00x-Rj00.We now expand the following functions as Fourier series: Fj00x=∑k∈Zn‍Fkj00ei〈k,x〉 and Rj00x=∑k∈Zn‍Rkj00ei〈k,x〉. Then we get(38)Fkj00=1ek-1Rkj00,with ek=eik,ω,k≠0.Next we try to getF010 and F001. Let F010=(F0101,…,F010m) and F001=(F0011,…,F001m). We expand the following functions as Fourier series: F0i′j′lx=∑k∈Zn‍Fk0i′j′leik,x and R0i′j′lx=∑k∈Zn‍Rk0i′j′leik,x with l=1,2,…,m and i′,j′=0,1,1,0.To get the relations betweenF0i′j′lx and R0i′j′lx, we need the following equation set:(39)aek-1-bbekek-a·Fk010lxF001x·eik,x=Rk010lRk001l,with a=sec⁡θl, b=c=tan⁡θl. Then we have(40)Fk0i′j′l=R~i′j′l2sec⁡θjekcos⁡k,ω-cos⁡θj,where R~i′j′l are linear combination of R010lx and R001lx.Thirdly, we solve the third equation of (36). For this purpose, we solve(41)L2=R011x+R020+R002.LetF0i′j′=F0i′j′ij1≤i,j≤m with i′,j′=1,1,2,0,0,2. We expand the following functions: F0i′j′ij=∑k∈Zn‍Fk0i′j′ei〈k,x〉 and R0i′j′ij=∑k∈Zn‍Rk0i′j′ei〈k,x〉.To get the relations betweenFk0i′j′ij and R010ijx, we need the following equation set:(42)0ekaj-aiekbiaj-aibjekai-aj0ekaibj-biaj-bj-biekaiaj-1-bibjekbiekbjekbibjek-aiaj·Fk011jiFk011ijFk020ijFk002ij·eik,x=Rk011jiRk011ijRk020ijRk002ij.Then we have(43)Fk0i′j′ij=R~ij4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,with R~ij being linear combination of Rk0i′j′ij.


### 2.1.3. Extension of Small Divisors

This part is critical for this paper. Letϕt be a C∞R-smooth function with (44)ft=0,t⩽12,1,t⩾1.For h⩾0, let fht=ft/h. Then fht∈C∞R with (45)dλdtλfht⩽cλhλ,∀t∈R,∀λ⩾0,where cλ is a constant depending on λ.Firstly, we extendFkj00 from U to the whole set Π. Let (46)h=α1+kτ,t1kξ=ωξ,k-2πl,g1kξ=fht1kξeiωξ,k-1,∀ξ∈Π.By the definition of U, we have (47)g1kξ=1eiωξ,k-1,∀ξ∈U.Note that even if U=∅, the extension of g1kξ is still well defined on Π. Furthermore, g1kξ∈CmΠ with estimate (48)∂βg1k∂ξβ⩽ckβλβ+1,ξ∈Π,∀β⩽m.Noticing (38), we now extend Fkj00 from U to whole set Π by setting (49)F~kj00ξ=g1kξRkj00ξ=fqtkξeiωξ,k-1Rkj00ξ,j=0,1,0<k⩽K.Let(50)F~j00ξ;x,y=∑0<k⩽KF~kj00ξ;yeiωξ,k.Then we have the following estimates: (51)F~j00s-ρ×Π≤cRj00sαm+1ρm+n+τm+1.Secondly, we extendFk0i′j′l from U to the whole set Π, where (i′,j′)=(0,1),(1,0). Let (52)t2k=ωξ,k-θ-2πl,(53)g2kξ=fht2kξ2sec⁡θjekcos⁡k,ω-cos⁡θj.Then we have (54)g2kξ=12sec⁡θjekcos⁡k,ω-cos⁡θj,∀ξ∈U.Noticing (40), we extend Fk0i′j′l by setting (55)F~k0i′j′lξ=g2kξR~i′j′l=fhtkξ2sec⁡θjekcos⁡k,ω-cos⁡θjR~i′j′l,with l=1,2,…,m and (i′,j′)=0,1,1,0,0<|k|⩽K. Let (56)F~0i′j′ξ;x,y=∑0<k⩽KF~k0i′j′ξ;yeiωξ,k.Then we have (57)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα2m+2ρ2τm+1+m+nwith i′,j′=0,1,1,0.At last, we expendFk0i′j′ij, where 1≤i,j≤m and i′,j′=1,1,2,0,0,2. Let (58)t3k=ωξ,k-θi-θj-2πlωξ,k-θi+θj-2πl,g3kξ=fh2t3kξ4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj.So we have (59)g3kξ=14sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,∀ξ∈U. Noting (43), we now extend Fk0i′j′ij to the whole set Π by setting (60)F~k0i′j′ij=g3kξR~ij,0<k<K.Let (61)F~0i′j′ij=∑0<k⩽KF~k0i′j′ijeiωξ,k.Recalling the third equation of (36) and the conclusions of (41), we let(62)ω^=diagω^1,…,ω^n,A^=diagA^1,…,A^m,B^=diagB^1,…,B^m,C^=diagC^1,…,C^m,with (63)ω^j=R100jj,A^j=R011jj,B^j=R020jj,C^j=R002jj.We sum up the above discussion and arrive at (64)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα4m+4ρ4τm+1+m+nwith i′,j′=1,1,2,0,0,2.By the above discussion, we get a functionF~ defined on Π which is the extension of F. Moreover F~ not only has the same form as R and F, but also inherits all properties of F when ξ∈U. So F~ satisfies L0+L1+L2F~=R-N^, where(65)N^p,q^=ω^,y^+A^u,v^+12B^u,u+12C^v^,v^.So we have (66)XF~r;Ds-ρ,r×Π≤cϵαν¯ρν.Let χ:p,q→-F~y+,F~x. Since ψ1=id+χ, we combine Cauchy estimate to obtain (67)ψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,Dψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.


### 2.1.4. Normal Form

From the above discussion, we get a conjugate mappingΦ¯·;ξ=ψ1-1∘Φ∘ψ1 generated by H¯=N¯1+N¯2+P¯ where N¯1=x+ω+ω^,y^, N¯2=A¯u,v^+1/2B¯u,u+1/2C¯v^,v^, and P¯=P~+Q with (68)ω^=ω+-ω,A¯=A0+A^,B¯=B0+B^,C¯=C0+C^.We note that the normal form depends only onu,v, so we consider N¯2. Since B=C and A2=B2+C2 may not be satisfied, N¯2 may not be a normal form and hence we need normalize N¯2 into a new normal form.Lemma 8.
There exists a symplectic mappingψ2 generated by 〈u,v^〉+Gu,v+, such that the corresponding conjugate mapping Φ+·;ξ=ψ¯2-1∘Φ¯∘ψ¯2:u+,v+→u^+,v^+ is generated by (69)N~2+=A+u,v^+12B+u,u+12C+v^,v^,where A+2=B+2+1, B+=C+. Let A+=sec⁡θ+. One has(70)θ+-θCmΠ<cϵ,and hence(71)ψ¯2-id=Oϵ.Remark 9.
We note thatψ¯2 can normalize N¯2 into a new normal form. Using the same idea and method, we can find a new symplectic mapping ψ2 normalizing N¯1x,y^+N¯2u,v^ into (72)N+=x+ω+,y^+A+u,v^+12B+u,u+12C+v^,v^.In fact, ψ2=p,q^+Gp,q+ with ψ2-id=O(ϵ).LetΨ=ψ1∘ψ2. Similar to the estimates of ψ1 and ψ2, we have (73)Ψ-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨ-Idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.Moreover, H+=N++P+, where N+ is normal form. H+ can be shown just like H¯, and we omit the details.


### 2.1.5. Estimates for New Perturbation and Error Terms

We aim at the estimate ofP+ and some error terms. By (29), we have(74)XP~ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2.By (28) and (74), we have(75)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν.Moreover, for the error terms ω^=ω+-ω and θ^=θ+-θ, we have ω^Cm(Π)≤ϵ and θ^Cm(Π)≤cϵ.


### 2.1.6. The Choice of Parameters in KAM Iteration Lemma

We choose a weighted error0<E<1 and set (76)η=E,ϵ=α2ν¯ρ2νEη2,E=e-Kρη2,h=α2Kτ+1T.Fix σ∈0,1, for the next step, we define (77)ρ+=σρ,s+=s-5ρ,r+=ηr,α+=α+31+Kτ+1ϵ,ϵ+=cηϵ,E+=cE4/3.By (75) and α<2α+, we have(78)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν≤cηϵ=cα2ν¯ρ2νE4≤cα+2ν¯ρ+2νE+3.Setting ϵ+=cα+2ν¯ρ+2νE+3, we arrive at (79)XP+r+;Ds+,r+×Π≤ϵ+.At last, let α+=α+=α+31+Kτ+1ϵ, and then we have U+⊂U.


## 2.1.1. Generating Functions of Conjugate Mappings

For the convenience, letp=x,u and q=y,v. p^ and q^ have similar meaning. The symplectic structure becomes dp∧dq on Rn+m×Rn+m. Consider a symplectic mapping Φ:p,q→p^,q^ generated by(21)p^=∂q^Hp,q^,q=∂pHp,q^.The generating function is Hp,q^=Np,q^+Pp,q^, where N is main term and P is a small perturbation.We need a symplectic transformationψ1:p+,q+→p,q, generated by(22)q=q++F1p,q+,p+=p+F2p,q+.The generating function is p,q++Fp,q+, with F being a small function. So we have ψ1→I. At the same time, ψ1 is also satisfied with p^+,q^+→p^,q^ by(23)q^=q^++F1p^,q^+,p^+=p^+F2p^,q^+.By (21) to (23), we have a conjugate mapping Φ¯=ψ1-1∘Φ∘ψ1:p+,q+→p^+,q^+ implicitly by(24)p^+=H2p,q^+F2p^,q^+,q+=H1p,q^-F1p,q+.We will prove that there is a function H¯p+,q^+ generating Φ¯.Lemma 7 (Lemma2.1 in [18]).
The conjugate symplectic mappingΦ¯ can be implicitly determined by a generating function H¯p+,q^+ through(25)p^+=∂q^+H¯p+,q^+,q+=∂p+H¯p+,q^+,where(26)H¯p+,q^+=Hp,q^+H1p,q^F2p,q+-H2p,q^F1p^,q^++Fp^,q^+-Fp,q+-F1p,q+F2p,q+,where p,p^,q^,q+ depend on (p+,q^+) as explained above.
Moreover, setz=(p+,q^+), Δz=-F2p,q+,F1p^,q^+, and then we have(27)H¯z=Hz+FN2z,q^+-Fp+,N1z+Qz.The small term Qz has the estimate(28)XQr;Ds-5ρ,r/16×Π≤cϵ2α2ν¯ρ2νwith ν¯=4m+1, ν=m+n+4τm+1.


## 2.1.2. Homological Equations

Now we will solve homological equations forF. Let(29)P=R+P~,where(30)Rp,q^=P000x+P100x,y^+P010x,u+P001x,v^+P011xu,v^+12P020xu,u+12P002xv^,v^,with (31)Plij=∂l+i+jP∂y^l∂ui∂v^ju=0,y^=0,v^=0,2l+i+j≤2.Let F(p,q^) possess the same form as (30).We will use the idea in [22] to solve homological equations: (32)Np+,q^++Rp+,q^+-Fp+,Npp+,q^++FNqp+,q^+,q^+=N¯p+,q^+.For simplicity, here and below we drop the subscripts “+” in p+ and q^+.Letx+ω=x~. Since (33)p^=Nq^p,p^=x~,Au+Cv,q=Np,q^p,p^=y^,Av^+Bu,we have (34)FNqp+,q^+,q^+-Fp+,Npp+,q^+=L0+L1+L2,where L0,L1,L2 indicate the 0, 1, 2 order terms of u,v^, respectively, with (35)L0=F000x~-F000x+F100x~-F100x,y^,L1=ATF010x~-F010x-BF001x,u+CF010x~+F001x~-AF001x,v,L2=F011x~A-AF011x+CF020x~A-AF002xBu,v^+12ATF020x~A-F020x-BF002xB-BF011x-F011TxBu,u+12CF020x~C+F002x~-AF002xAT+F011x~C+CF011Tx~v^,v^.Let A=diaga1,…,an and B=C=diag(b1,…,bn) with aj=sec⁡θj and bj=tan⁡θj. Now we solve the following equations:(36)L0=R000x-R000+R100x-R100L1=R010x+R001xL2=R011x-A^+R020-B^+R002-C^,where A^,B^, and C^ will be determined later.Firstly, forFj00,j=0,1, we consider the equation (37)Fj00x+ω-Fj00x=Rj00x-Rj00.We now expand the following functions as Fourier series: Fj00x=∑k∈Zn‍Fkj00ei〈k,x〉 and Rj00x=∑k∈Zn‍Rkj00ei〈k,x〉. Then we get(38)Fkj00=1ek-1Rkj00,with ek=eik,ω,k≠0.Next we try to getF010 and F001. Let F010=(F0101,…,F010m) and F001=(F0011,…,F001m). We expand the following functions as Fourier series: F0i′j′lx=∑k∈Zn‍Fk0i′j′leik,x and R0i′j′lx=∑k∈Zn‍Rk0i′j′leik,x with l=1,2,…,m and i′,j′=0,1,1,0.To get the relations betweenF0i′j′lx and R0i′j′lx, we need the following equation set:(39)aek-1-bbekek-a·Fk010lxF001x·eik,x=Rk010lRk001l,with a=sec⁡θl, b=c=tan⁡θl. Then we have(40)Fk0i′j′l=R~i′j′l2sec⁡θjekcos⁡k,ω-cos⁡θj,where R~i′j′l are linear combination of R010lx and R001lx.Thirdly, we solve the third equation of (36). For this purpose, we solve(41)L2=R011x+R020+R002.LetF0i′j′=F0i′j′ij1≤i,j≤m with i′,j′=1,1,2,0,0,2. We expand the following functions: F0i′j′ij=∑k∈Zn‍Fk0i′j′ei〈k,x〉 and R0i′j′ij=∑k∈Zn‍Rk0i′j′ei〈k,x〉.To get the relations betweenFk0i′j′ij and R010ijx, we need the following equation set:(42)0ekaj-aiekbiaj-aibjekai-aj0ekaibj-biaj-bj-biekaiaj-1-bibjekbiekbjekbibjek-aiaj·Fk011jiFk011ijFk020ijFk002ij·eik,x=Rk011jiRk011ijRk020ijRk002ij.Then we have(43)Fk0i′j′ij=R~ij4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,with R~ij being linear combination of Rk0i′j′ij.


## 2.1.3. Extension of Small Divisors

This part is critical for this paper. Letϕt be a C∞R-smooth function with (44)ft=0,t⩽12,1,t⩾1.For h⩾0, let fht=ft/h. Then fht∈C∞R with (45)dλdtλfht⩽cλhλ,∀t∈R,∀λ⩾0,where cλ is a constant depending on λ.Firstly, we extendFkj00 from U to the whole set Π. Let (46)h=α1+kτ,t1kξ=ωξ,k-2πl,g1kξ=fht1kξeiωξ,k-1,∀ξ∈Π.By the definition of U, we have (47)g1kξ=1eiωξ,k-1,∀ξ∈U.Note that even if U=∅, the extension of g1kξ is still well defined on Π. Furthermore, g1kξ∈CmΠ with estimate (48)∂βg1k∂ξβ⩽ckβλβ+1,ξ∈Π,∀β⩽m.Noticing (38), we now extend Fkj00 from U to whole set Π by setting (49)F~kj00ξ=g1kξRkj00ξ=fqtkξeiωξ,k-1Rkj00ξ,j=0,1,0<k⩽K.Let(50)F~j00ξ;x,y=∑0<k⩽KF~kj00ξ;yeiωξ,k.Then we have the following estimates: (51)F~j00s-ρ×Π≤cRj00sαm+1ρm+n+τm+1.Secondly, we extendFk0i′j′l from U to the whole set Π, where (i′,j′)=(0,1),(1,0). Let (52)t2k=ωξ,k-θ-2πl,(53)g2kξ=fht2kξ2sec⁡θjekcos⁡k,ω-cos⁡θj.Then we have (54)g2kξ=12sec⁡θjekcos⁡k,ω-cos⁡θj,∀ξ∈U.Noticing (40), we extend Fk0i′j′l by setting (55)F~k0i′j′lξ=g2kξR~i′j′l=fhtkξ2sec⁡θjekcos⁡k,ω-cos⁡θjR~i′j′l,with l=1,2,…,m and (i′,j′)=0,1,1,0,0<|k|⩽K. Let (56)F~0i′j′ξ;x,y=∑0<k⩽KF~k0i′j′ξ;yeiωξ,k.Then we have (57)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα2m+2ρ2τm+1+m+nwith i′,j′=0,1,1,0.At last, we expendFk0i′j′ij, where 1≤i,j≤m and i′,j′=1,1,2,0,0,2. Let (58)t3k=ωξ,k-θi-θj-2πlωξ,k-θi+θj-2πl,g3kξ=fh2t3kξ4sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj.So we have (59)g3kξ=14sec⁡θisec⁡θjcos⁡k,ω-cos⁡θi+θjcos⁡k,ω-cos⁡θi-θj,∀ξ∈U. Noting (43), we now extend Fk0i′j′ij to the whole set Π by setting (60)F~k0i′j′ij=g3kξR~ij,0<k<K.Let (61)F~0i′j′ij=∑0<k⩽KF~k0i′j′ijeiωξ,k.Recalling the third equation of (36) and the conclusions of (41), we let(62)ω^=diagω^1,…,ω^n,A^=diagA^1,…,A^m,B^=diagB^1,…,B^m,C^=diagC^1,…,C^m,with (63)ω^j=R100jj,A^j=R011jj,B^j=R020jj,C^j=R002jj.We sum up the above discussion and arrive at (64)F~0i′j′Ds-ρ,r×Π≤cR0i′j′sα4m+4ρ4τm+1+m+nwith i′,j′=1,1,2,0,0,2.By the above discussion, we get a functionF~ defined on Π which is the extension of F. Moreover F~ not only has the same form as R and F, but also inherits all properties of F when ξ∈U. So F~ satisfies L0+L1+L2F~=R-N^, where(65)N^p,q^=ω^,y^+A^u,v^+12B^u,u+12C^v^,v^.So we have (66)XF~r;Ds-ρ,r×Π≤cϵαν¯ρν.Let χ:p,q→-F~y+,F~x. Since ψ1=id+χ, we combine Cauchy estimate to obtain (67)ψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,Dψ1-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.


## 2.1.4. Normal Form

From the above discussion, we get a conjugate mappingΦ¯·;ξ=ψ1-1∘Φ∘ψ1 generated by H¯=N¯1+N¯2+P¯ where N¯1=x+ω+ω^,y^, N¯2=A¯u,v^+1/2B¯u,u+1/2C¯v^,v^, and P¯=P~+Q with (68)ω^=ω+-ω,A¯=A0+A^,B¯=B0+B^,C¯=C0+C^.We note that the normal form depends only onu,v, so we consider N¯2. Since B=C and A2=B2+C2 may not be satisfied, N¯2 may not be a normal form and hence we need normalize N¯2 into a new normal form.Lemma 8.
There exists a symplectic mappingψ2 generated by 〈u,v^〉+Gu,v+, such that the corresponding conjugate mapping Φ+·;ξ=ψ¯2-1∘Φ¯∘ψ¯2:u+,v+→u^+,v^+ is generated by (69)N~2+=A+u,v^+12B+u,u+12C+v^,v^,where A+2=B+2+1, B+=C+. Let A+=sec⁡θ+. One has(70)θ+-θCmΠ<cϵ,and hence(71)ψ¯2-id=Oϵ.Remark 9.
We note thatψ¯2 can normalize N¯2 into a new normal form. Using the same idea and method, we can find a new symplectic mapping ψ2 normalizing N¯1x,y^+N¯2u,v^ into (72)N+=x+ω+,y^+A+u,v^+12B+u,u+12C+v^,v^.In fact, ψ2=p,q^+Gp,q+ with ψ2-id=O(ϵ).LetΨ=ψ1∘ψ2. Similar to the estimates of ψ1 and ψ2, we have (73)Ψ-idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨ-Idr;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.Moreover, H+=N++P+, where N+ is normal form. H+ can be shown just like H¯, and we omit the details.


## 2.1.5. Estimates for New Perturbation and Error Terms

We aim at the estimate ofP+ and some error terms. By (29), we have(74)XP~ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2.By (28) and (74), we have(75)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν.Moreover, for the error terms ω^=ω+-ω and θ^=θ+-θ, we have ω^Cm(Π)≤ϵ and θ^Cm(Π)≤cϵ.


## 2.1.6. The Choice of Parameters in KAM Iteration Lemma

We choose a weighted error0<E<1 and set (76)η=E,ϵ=α2ν¯ρ2νEη2,E=e-Kρη2,h=α2Kτ+1T.Fix σ∈0,1, for the next step, we define (77)ρ+=σρ,s+=s-5ρ,r+=ηr,α+=α+31+Kτ+1ϵ,ϵ+=cηϵ,E+=cE4/3.By (75) and α<2α+, we have(78)XP+ηr;Ds-5ρ,ηr×Π≤c·ϵη+e-Kρη2+cϵ2η2α2ν¯ρ2ν≤cηϵ=cα2ν¯ρ2νE4≤cα+2ν¯ρ+2νE+3.Setting ϵ+=cα+2ν¯ρ+2νE+3, we arrive at (79)XP+r+;Ds+,r+×Π≤ϵ+.At last, let α+=α+=α+31+Kτ+1ϵ, and then we have U+⊂U.


## 2.2. Iteration

In this section, we will summarize the above results on parameters so that KAM-step can iterate infinitely. At the initial step set(80)s0=s,ρ0=1-σs10,r0=r,α0=α2,E0=E,η0=E0,ϵ0=α02ν¯ρ02νE0η02.Let (81)e-K0ρ0η02=E0,ω0ξ=ωξ.Assume that ρj,sj,rj,Ej,αj are well defined for the jth step. Then ηj,Kj,ϵj are defined as follows: (82)ηj=Ej,ϵj=αj2ν¯ρj2νEjηj2,e-Kjρjηj2=Ej.Define inductive sequences: (83)ρj+1=σρj,sj+1=sj-5ρ,rj+1=ηjrj,αj+1=αj+31+Kjτ+1ϵj,Ej+1=cEj4/3.Similarly, we can define ηj+1,Kj+1,ϵj+1.Similar to the proof of KAM iteration Lemma, we haveUj+1⊂Uj. Moreover, there exists a sequence of {Ψj}={ψ1j∘ψ2j}, where ψ1j is generated by p,q++Fjp,q+ and ψ2j is generated by p,q++Gjp,q+. Moreover, we have (84)Ψj-idrj;Ds-3ρ,r/4×Π≤cϵαν¯ρν,DΨj-Idrj;Ds-3ρ,r/4×Π≤cϵαν¯ρν+1.There exists a sequence of symplectic mappings Φj which are well defined on Dsj,rj and satisfy(85)Φj+1=Ψj-1∘Φj∘Ψj,where Ψj=Ψ1∘Ψ2∘⋯∘Ψj. The symplectic mappings Φj are generated by Hj=Nj+Pj, where (86)Nj=x+ωjξ+Aju,v^+12Bju,u+12Cjv^,v^with Aj=sec⁡θj, Bj=Cj=tan⁡θj. Moreover we have(87)ωj+1-ωjCmΠ≤cϵj,θj+1-θjCmΠ≤cϵj∀j≥1.


## 2.3. Convergence of the Iteration

Now we prove the convergence of KAM iteration. In the same way as [22, 23], we have (88)Ψj-Ψj-1rj;Dsj,rj×Π≤cαj-12ν¯ρj-12νEj-13,DΨj-Ψj-1rj;Dsj,rj×Π≤αj-12ν¯ρj-12ν-1Ej-13.Note that sj→s/2 and rj→0 as j→∞. Let Ψ∗=limj→∞Ψj. Since Ψj is affine in y, we have Ψj converging to Ψ∗ on Dr/2,s/2 with (89)Ψ∗-idr/2;Ds/2,r/2×Π≤cα,ρ,σE03,∀β∈Zn+.LetP∗=limj→∞Pj and ω∗=limj→∞ωj. Then P∗ is real analytic in x,y on D(r/2,s/2) and Cm in ξ on Π. Note that ωj=ω+∑i=0j-1‍ω^j. Then we have (90)ω∗-ωjCmΠ⩽∑i=j∞ϵj⩽2ϵj.Particularly, (91)ω∗-ωCmΠ⩽2ϵ.Correspondingly, letHj=Nj+Pj→H∗=N∗+P∗ satisfy (92)N∗=x+ω∗ξ,y^+A∗ξu,v^+12B∗ξu,u+12C∗ξv^,v^,where A∗=sec⁡θ∗ and B∗=C∗=tan⁡θ∗ with estimates (93)θ∗-θCmΠ≤cϵ.Moreover, P∗p,q^=∑2|l|+|i|+|j|≥3‍Plij(x)y^luiv^j.Let(94)Π∗=ξ∈Π:ω∗ξ,θ∗ξ∈Oα,τ.In the sequel we will prove Π∗⊂Uj for all j⩾0, which is equal to proving α⩾αj+4ϵjKjτ+1. By the choice of parameters, we have Kj+1τ+1ϵj+1/Kjτ+1ϵj⩽cEj and K0=-3ln⁡E0/ρ0, so (95)Kjτ+1ϵj⩽E0c0K0τ+1ϵ0⩽cE0c0+3ln⁡E0ρ0τ+1αρ0τ′.Let E0 be small enough such that 8Kjτ+1ϵj⩽α/2, and then we have (96)αj+4ϵjKjτ+1=α2+4∑l=1jKlτ+1ϵl⩽α2+8Kjτ+1ϵj⩽α.So Π∗⊂Uj and hence Π∗⊂⋂j⩾0Uj, for all j⩾0. This completes the proof of Theorem 4.



---

*Source: 3719395-2017-03-16.xml*