We will invoke the KAM iteration technique to prove the following normal form theorem.
Proof.
The proof is the standard KAM step and we divide it into several parts.
(A) Truncation. Let R=R0+R1x with R0=P|x=0 and R1=Px|x=0. It follows easily that
(54)|||R0|||m,Δr,s×M≤ϵ, |||R1|||m,Δr,s×M≤ϵr.
Hence |||R|||m,Δr,s×M≤2ϵ. Let
(55)R=∑Λ∈τ,suppk⊂Λ,k∈Z∞Rk(x;p)ei〈k,ωt〉,RK=∑Λ∈τ,suppk⊂Λ,|k|≤KRk(x;p)ei〈k,ωt〉.
By definition, we have
(56)|||R-RK|||m,Δr,s-ρ×M≤2ϵe-Kρ.
(B) Construnction of the Transformation. Define the transformation ϕ1:x=u(t)+y, where u satisfies
(57)∂ωu=AΩu+R0K-[R0].
From Lemma 12, we have
(58)|||u(t)|||m-(m/2),s-2ρ ≤Γ(m/2)Γ(2ρ)α|||R0(t)|||m,s ≤Γ(m/2)Γ(2ρ)αϵ<crE.
By the transformation ϕ1, the equation becomes
(59)u˙+y˙=N(p)+AΩ(p)(u+y)+ϵa(t)Ω(p)(u+y)+P(u+y,t,p),(60)y˙=N(p)+[R0]+(AΩ(p)+R1K)y+ϵa(t)Ω(p)y+R(u+y,t)-RK(u+y,t)+R1Ku+ϵa(t)Ω(p)u+P(u+y,t)-R(u+y,t);
then
(61)y˙=N+(p)+(AΩ(p)+R1K)y+ϵa(t)Ω(p)y+R(u+y,t)-RK(u+y,t)+R1Ku+ϵa(t)Ω(p)u+P(u+y,t)-R(u+y,t).
Define the transformation ϕ2:y=(1+ϵQ)x+; Q satisfies
(62)Q˙=a(t)Ω(p).
We assume
(63)Q(t)=∑Λ∈τQΛ(t), QΛ(t)=∑suppk⊂Λ,QΛkei〈k,ωt〉,a(t)=∑Λ∈τaΛ(t), aΛ(t)=∑suppk⊂Λ,aΛkei〈k,ωt〉.
Then
(64)QΛk={0,|k|=0, suppk⊂Λ,aΛkΩi〈k,ω〉,|k|>0, suppk⊂Λ.
Similar to Lemma 12, we have
(65)|||Q(t)|||m-(m/2),s-2ρ ≤Γ(m/2)Γ(2ρ)α|||a(t)|||m,s|Ω| ≤cΓ(m/2)Γ(2ρ)α,|||ϵQ(t)|||m-(m/2),s-2ρ≤crE.
By the transformation ϕ2, the equation becomes
(66)(1+ϵQ)x˙++ϵQ˙x+ =N+(p)+(AΩ(p)+R1K)(1+ϵQ)x+ +ϵa(t)Ω(p)(1+ϵQ)x+ +R(u+(1+ϵQ)x+,t)-RK(u+(1+ϵQ)x+,t) +R1Ku+ϵa(t)Ω(p)u +P(u+(1+ϵQ)x+,t)-R(u+(1+ϵQ)x+,t);
then
(67)x˙+=N+(p)+(AΩ(p)+R1K)x++ϵ2a(t)Ω(p)Q1+ϵQx++(11+ϵQ-1)N++11+ϵQR(u+(1+ϵQ)x+,t)-11+ϵQRK(u+(1+ϵQ)x+,t)+11+ϵQR1Ku+ϵa(t)Ω(p)u1+ϵQ+11+ϵQP(u+(1+ϵQ)x+,t)-11+ϵQR(u+(1+ϵQ)x+,t).
Thus, by the transformation Φ=ϕ1∘ϕ2:x=u+(1+ϵQ)x+, the equation is transformed to
(68)x˙+=N++Ω+x++ϵ2S+x++P+(x+,t;p),
where
(69)P+=(11+ϵQ-1)N++11+ϵQR(u+(1+ϵQ)x+,t)-11+ϵQRK(u+(1+ϵQ)x+,t)+11+ϵQR1Ku+ϵa(t)Ω(p)u1+ϵQ+11+ϵQP(u+(1+ϵQ)x+,t)-11+ϵQR(u+(1+ϵQ)x+,t),S+=a(t)Ω(p)Q1+ϵQ.
With the estimates of u and Q, we have
(70)|||Φ-id|||m,Δr,s-2ρ×M+ ≤r|||ϵQ|||m,Δr,s-2ρ×M++|||u|||m,Δr,s-2ρ×M+ ≤c(r+r2)E,|||DxΦ-1|||m,Δr,s-2ρ×M+≤|||ϵQ|||m,Δr,s-2ρ×M+≤crE.
Let η=E1/2≤1/4, s+=s-2ρ, and r+=ηr. Then the transformation Φ:x+∈D(0,r+)→D(0,2r+) is analytic almost-periodic on 𝕋s+ with respect to t and affine in x+.
(C) Estimates of Error Terms. Because |||ϵQ|||m+,Δr+,s+×M+<1, then 1/(1+ϵQ)=1-ϵQ+(ϵQ)2-(ϵQ)3+⋯+(-ϵQ)n+⋯. Thus,
(71)|||11+ϵQ|||m+,Δr+,s+×M+ ≤1+|||ϵQ|||m+,Δr+,s+×M++|||ϵQ|||m+,Δr+,s+×M+2 +⋯+|||ϵQ|||m+,Δr+,s+×M+n+⋯ ≤2.
Let η=E1/2. Then, it follows that
(72)|||(11+ϵQ-1)N+|||m+,Δr+,s+×M+ ≤∥|ϵQN+|∥m+,Δr+,s+×M+≤cϵE,|||11+ϵQ(R∘Φ-RK∘Φ)|||m+,Δr+,s+×M+≤cϵe-Kρ,|||11+ϵQR1Ku|||m+,Δr+,s+×M+≤2ϵrcrE≤cϵE,|||ϵa(t)Ωu1+ϵQ|||m+,Δr+,s+×M+≤cϵE,|||11+ϵQ(P∘Φ-R∘Φ)|||m+,Δr+,s+×M+ ≤cϵr2r2η2=cη2ϵ.
So
(73)|||P+|||m+,Δr+,s+×M+ ≤cϵE+cϵe-Kρ+cη2ϵ=cϵE =αr+E+Γ(m+/2)Γ(ρ+)=ϵ+,
where ρ+=(1/2)ρ, r+=ηr, E+=cE3/2, and m+=(1/2)m. Thus we have proved Lemma 13.
Iteration. Now we choose some suitable parameters so that the above KAM step can be iterated infinitely. At the initial step, let
(74)s0=s, r0=r, E0>0, η0=E01/2,ρ0=s08, ϵ0=αr0E0Γ(m0/2)Γ(ρ0), m0=s08.
Let K0 satisfy E0=e-K0ρ0 and σ0=δ. Inductively, we define
(75)sj=sj-1-2ρj-1, ηj=Ej1/2, ρj=12ρj-1, rj=ηj-1rj-1, Ej=cEj-13/2, mj=12mj-1,ϵj=αrjEjΓ(mj/2)Γ(ρj), σj=σj-14.
And Kj satisfies Ej=e-Kjρj.
By Ej=cEj-13/2, we have Ej=c-2(c2E0)(3/2)j. Thus, if E0 is sufficiently small, we have cEj≤1 and ηj=Ej1/2≤1/4. Moreover, by definition it follows that
(76)(ϵj+1/Ej+1)(ϵj/Ej)=Γ(mj/2)Γ(ρj)Γ(mj+1/2)Γ(ρj+1)Ej1/2≤Ej1/2≤14≤1.
Thus ϵj+1/Ej+1≤ϵj/Ej, j≤0.
Now we prove that, for E0 sufficiently small, 4ϵj/Ej≤σj hold for all j≤0.
Let Gj=4ϵj/Ejσj; from (21) we have G0≤1. Moreover, we have
(77)Gj+1Gj=σjϵj+1Ejσj+1ϵjEj+1=4αrj+1Ej+1Γ(mj+1/2)Γ(ρj+1)·Γ(mj/2)Γ(ρj)αrjEj·EjEj+1≤4ηjEj+1Ej·EjEj+1=4Ej1/2≤1,
for all j≤0. Thus, Gj≤G0≤1. So the inequalities in the assumption 2 of Lemma 13 hold for all j≤0.
Let M0=M, N0=(A+ϵ[a])ξ2n+1-λ, h0=0, Ω0=AΩ, S0=aΩ, and P0=P. By Lemmas 11, 12, and 13, we have a sequence of closed domains Mj with Mj+1⊂Mj and a sequence of affine transformations
(78)Φj:D(0,rj+1)⟶D(0,2rj+1)⊂D(0,rj),Φj:xj=(1+ϵ2jQj)xj+1+uj.
We also have
(79)|||Φj-id|||mj+1,Δrj+1,sj+1×Mj+1≤c(rj+rj2)Ej,|||DxΦj-1|||mj+1,Δrj+1,sj+1×Mj+1≤crjEj.
Let Φj=Φ0∘Φ1∘⋯Φj-1 with Φ0=id. Then, after the transformation Φj, (19) is changed to
(80)x˙=Nj(p)+Ωj(p)x+ϵ2jSjx+Pj(x,t;p).
By the inductive assumptions of KAM iteration, we have |||Pj|||mj,Δrj,sj×Mj≤ϵj.
The correction terms h^j and Ω^j satisfy
(81)∥h^j∥Mj≤ϵj, ∥Ω^j∥Mj≤ϵjrj.
By Lemma 11, we have dist(Mj+1,∂Mj)≥ϵj-1/4Ej-1. For Cauchy’s estimate we have
(82)∥Dph^j∥Mj+1≤4ϵjEj-1ϵj-1.
Noting that M0=B(I×J,σ0) and T0:λ=(A+ϵ[a])ξ2n+1, |ξ|≤δ≤1. Since σ0≥4ϵ0/E0 and M1=B(T0,3ϵ0/E0)∩(Iσ0/2+ϵ0/E0×ℂ), it follows that M1⊂M0 and dist(M1,∂M0)≥σ0/4. For Cauchy’s estimate we have
(83)∥Dph^0∥M1≤4ϵ0σ0.
Let Fj=ϵjEj-1/ϵj-1rj; then
(84)Fj=Ej-1rj·αrjEjΓ(mj-1/2)Γ(ρj-1)αrj-1Ej-1Γ(mj/2)Γ(ρj)≤Ej-1rj·αrjEjαrj-1Ej-1=cEj-12rj=Rj.
So Rj+1/Rj=cEj1/6. Obviously, we can choose E0 sufficiently small so that Rj+1/Rj≤1/2, Rj+1≤(1/2)Rj. Noting that h0=0 and ϵ0/r0σ0≤1/16, we have
(85)∥Dphj∥Mj+1≤∥Dph^0∥Mj+1+∑l=1j-1∥Dph^l∥Mj+1≤4ϵ0σ0+∑l=1j-14rjRj≤14+cE02≤12.
So condition (29) holds for all j≤0.
From Lemma 11, Nj(p)=(A+ϵ[a])ξ2n+1-λ+hj(p)=0 defines implicitly a real analytic curve Tj⊂Mj:λ=λj(ξ), ξ∈Iσj/2, satisfying
(86)|λj+1(ξ)-λj(ξ)|≤2ϵj, ∀ξ∈Iσj/2.
Furthermore, ∥Nj+1∥Mj+1≤8ϵj/Ej.
Convergence of KAM Iteration. Now we prove the convergence of KAM iteration. By the definition of Ej, if E0 is sufficiently small, it follows that
(87)|||DxΦj|||mj,Δrj,sj×Mj≤∏i=0j-1(1+criEi)≤2.
Therefore, we have
(88)|||Φj-Φj-1|||mj,Δrj,sj×Mj≤c(rj-1+rj-12)Ej-1,|||Dx(Φj-Φj-1)|||mj,Δrj,sj×Mj≤crj-1Ej-1.
Let
(89)Δ*=⋂j=0∞Δrj,sj=Δ0,s0/2, M*=⋂j≥0Mj,Φ=limj→∞Φj.
We have
(90)|||Φj-Φ0|||mj,Δrj,sj×Mj ≤|||Φj-Φj-1|||mj,Δrj,sj×Mj +|||Φj-1-Φj-2|||mj,Δrj,sj×Mj +⋯+|||Φ1-Φ0|||mj,Δrj,sj×Mj ≤c(rj-1+rj-12)Ej-1+c(rj-2+rj-22)Ej-2 +⋯+c(r0+r02)E0 ≤c(r0+r02)E0.
Thus
(91)|||Φ-id|||m*,Δr*,s*×M*≤c(r0+r02)E0,|||DxΦ-id|||m*,Δr*,s*×M*≤cr0E0.
So we have the convergence of Φj to Φ on Δr0/2,s0/2.
From (86) it is easy to show that λj is convergent on I. In fact, ϵj+1/ϵj≤cEj≤1/2. For i>j, it follows that
(92)|λi(ξ)-λj(ξ)|≤∑l=ji-12ϵl≤4ϵj, ∀ξ∈I(1/2)σj.
Let λi(ξ)→λ(ξ)=(A+ϵ[a])ξ2n+1+N^(ξ), ξ∈I. For λ0(ξ)=(A+ϵ[a])ξ2n+1, we have
(93)|λi(ξ)-(A+ϵ[a])ξ2n+1|≤4ϵ0, ∀ξ∈I(1/2)σj,|N^(ξ)|=|λ(ξ)-(A+ϵ[a])ξ2n+1|≤4ϵ0, ∀ξ∈I.
Moreover, by Cauchy’s estimate we have
(94)|λj+1′(ξ)-λj′(ξ)|≤4ϵjσj, ∀ξ∈I.
Let Lj=4ϵj/σj; then Lj+1/Lj≤cEj. Thus, it is easy to prove that {λj′(ξ)} is convergent uniformly on I, and so λ(ξ) is differentiable on I. In fact, in the same way as in [7], we can prove that λ(ξ) is C∞-smooth on I.
Since Ti⊂Mi⊂Mj, for all i≥j, letting i→∞ we have T={(ξ,λ)∣λ=λ(ξ), ξ∈I⊂Mj} and T=M*=∩j≤0Mj. Obviously, Nj(p)→0, for p∈T. Let Ωj→Ω* and let Pj→P*. By Cauchy’s estimate we have
(95)|DxPj|x=0|≤ϵjrj=αEjΓ(mj/2)Γ(ρj)⟶0.
Thus P*|x=0=0 and DxP*|x=0=0. Hence P*(x,t;p)=O(x2) for p∈T.
Noting that ∥Ω^j∥Mj≤ϵj/rj and Ωj=AΩ+∑i=1j-1Ω^j, we have
(96)|Ω*(p)-AΩ(p)|≤2ϵ0r0, ∀p∈Γ.
The proof of Theorem 10 is complete.