# Traveling Waves in the Underdamped Frenkel–Kontorova Model

Authors: Hengyan Li; Shaowei Liu
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

This paper studies a damped Frenkel–Kontorova model with periodic boundary condition. By using Nash–Moser iteration scheme, we prove that such model has a family of smooth traveling wave solutions.


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## Body

## 1. Introduction

The present work concerns the existence of traveling wave solutions for the following underdamped Frenkel–Kontorova model:(1)ψj′′+Γψj′+sin⁡ψj=F+Kψj+1-2ψj+ψj-1,∀j∈Z,with periodic boundary condition(2)ψj+nt=ψjt+2mπ,∀j∈Z,where the parameters Γ>0, K>0, F>0, and m≥1.In the last decades, there has been large growth in the study of the existence and stability of traveling wave solutions for lattice systems including Frenkel–Kontorova model (discrete sine-Gordon equations), which arises from many physical systems, such as circular arrays of Josephson junctions, glassy materials, sliding friction, adsorbate layer on the surface of a crystal, ionic conductors, and mechanical interpretation as a model for a ring of pendula coupled by torsional springs (see [1–4]). When Γ=0, system (4) is conservative. Baesens and MacKay [5] proved the existence and also global stability of traveling waves. When Γ>0, system (4) is dissipative. Under the condition Γ>22K+1 and ϵ=1, Baesens and MacKay [6] showed that the traveling wave solution is globally stable if and only if (4) and (2) do not have stationary solutions. Levi in [7] pointed out that the local stability of traveling waves can be obtained by the monotonicity method in [8]. Under the condition Γ>22K and F≥F0>1, Qin et al. [9] investigated the stability of single-wave-form for the underdamped Frenkel–Kontorova model (4) by the monotonicity method.Recently, by using Schauder fixed point theorem, Mirollo and Rosen [10] and Katriel [11] have obtained a series of results about the existence of traveling waves for (4) with periodic boundary condition (2). Katriel [11] proved the following:( 1 ) Fixing any Γ>0 and K>0 and given any velocity v>0, there exists a traveling wave solution of (4) and (2) with velocity v for an appropriate F>0.( 2 ) For any F>1 there exists a traveling wave solution of (4) and (2).( 3 ) Assume that n does not divide m. Fixing any F~>0 and Γ~>0, for all K sufficiently large there exists a traveling wave solution of (4) and (2) for any F≥F~>0 and Γ≥Γ~>0.( 4 ) Fixing any F~>0 and Γ~>0, for all Γ>0 sufficiently small there exists a traveling wave solution of (4) and (2) for any F≥F~>0 and 0<K≤K~.In the final of Katriel’s paper, he gave several open problems. One of them is the following: Is it true that, fixingΓ>0 and K>0, for sufficiently small F>0 and small applied force, a traveling wave does exist? If n divides m, what is the situation of the existence of traveling waves for (4) with periodic boundary condition (2)? In fact, if n divides m, there appears the “small divisor.” Then, the problem is difficult. Levi et al. [12] showed that, for fixing Γ>0, (4) possesses a traveling wave only when F exceeds a positive critical value.In this paper, we will construct a new Nash–Moser iteration to answer the open problem mentioned above. This method has been used in solving the existence of periodic solutions for nonlinear elliptic equations [13], nonlinear wave equations [14–18], and standing waves [19]. Here, we try to use this method to study the existence of traveling wave solutions for dissipative and conservative lattice systems.Instead of looking for solutions of (1) in a shrinking neighborhood of zero, it is a convenient device to perform the rescaling (3)ψ⟶ϵψ,ϵ>0having(4)ψj′′+Γψj′+ϵ-1sinϵψj=ϵ-1F+Kψj+1-2ψj+ψj-1,∀j∈Z.To overcome the “small divisor” problem, we need the following nonresonance conditions: (5)NR1Oγ,τ≔ω∈Ω:kvω-1≥γkτorkvω+1≥γkτ∀k∈Z,k≠0≠∅,NR2O¯γ,τ≔ω∈Ω:kvω±1≥γkτorkvω±i1-4K≥γkτ∀k∈Z,k≠0≠∅,where Ω⊂R is a bounded region.It is shown in [20] that if, for some l>0, (6)vlω≥d>0onω∈Ω,then the Lebesgue measure (7)Ω∖Oγ,τ=Oγ1/l⟶0,Ω∖O¯γ,τ=Oγ1/l⟶0asγ⟶0.Now, we state our main result.Theorem 1.
Under the assumption (NR1), fixing anyK>0 and sufficient small F>0, there exist Γ0>0, ϵ0>0, and 0<γ<1≤τ, such that, for any ζ≔vΓ∈[0,Γ0], ϵ∈[0,ϵ0], and ω∈Oγ,τ, (4) with periodic boundary condition (2) possesses a unique traveling wave solution u(θ;ζ)+θ, where θ∈T≔R/2π.WhenΓ=0, (4) is(8)ψj′′+ϵ-1sin⁡ϵψj=ϵ-1F+Kψj+1-2ψj+ψj-1,∀j∈Z.The corresponding Hamiltonian of (8) is (9)H=∑j12dψjdt2+ϵ-11-cos⁡ϵψj+ϵ-1Fψj+Wψj,where the nearest-neighbor coupling potential is (10)Wψj=K2ψj+1-ψj2.We have the following result about the existence of traveling waves for (8).Theorem 2.
Under the assumption (NR2), fixing anyK>0, there exist F∗>0, ϵ0>0, and 0<γ<1≤τ, such that, for any ϵ∈[0,ϵ0] and ω∈O¯γ,τ, (4) with periodic boundary condition (2) possesses a unique traveling wave solution u(θ;ζ)+θ, where θ∈T≔R/2π.This paper is organized as follows. In Section2, we first establish a Nash–Moser theorem for the case of Γ>0. Then, we apply this result to prove our main results. The case of Γ=0 is also considered.


## 2. Proof of the Main Results

### 2.1. The Case ofΓ>0

In numerical simulations or experimental works on (4) with periodic boundary condition (2), it is observed that solutions often converge to a traveling wave (11)ψjt=φt+jmnT,where the waveform φ:R→R is a function satisfying(12)φt+T=φt+2π,∀t∈R.φ is a waveform if and only if it satisfies (12) and(13)φ′′t+Γφ′t+ϵ-1sin⁡ϵφt=ϵ-1F+Kφt+mnT-2φt+φt-mnT.Hence, as in [11], we investigate the traveling wave of the type(14)φt=uvt+vt,where the wave velocity v=2π/T=2πω and u satisfies (15)uθ+2π=uθ,∀θ∈R.Inserting (14) into (13), we get(16)v2u′′θ+ζu′θ+ϵ-1sin⁡ϵθ+ϵuθ=Kuθ+2πmn-2uθ+uθ-2πmn+ϵ-1F-ζ.Write (17)sin⁡ϵθ+ϵuθ=sin⁡ϵθ+2ϵcos⁡ϵθu+gu.We consider the following space:(18)Xσ≔uθ≔∑k∈Z2ukeik·θ∣uσ2≔∑k∈Zuk2e2σk<∞,where uk denotes the kthe Fourier coefficient.Obviously, for a nested family of Banach spaces{Xσ:σ≥0}, there holds (19)Xσ2⊂Xσ1,uσ1≤uσ2∀0≤σ1≤σ2.Forσ≥0, the space Xσ is Banach algebra with respect to multiplication of functions; that is, if u1,u2∈Xσ, then u1u2∈Xσ and there exists a positive constant C, such that (20)u1u2σ≤Cu1σu2σ.It is obviously that each function in Xσ has a bounded analytic extension in the complex multistrip Imθ<σ, where θ∈C. By the definition of the space Xσ, the following inequality holds: (21)∂θhuσ≤∂θluσ,∀h,l∈N:h≤l.For uniqueness, we assume that u satisfies(22)u≔∫Tusds=0,T=R2π.Now we define a function space with zero average by (23)Xσ0≔u∈Xσs.t.u0=u=0,as the closed subspace of Xσ.Letσ~>σ>0. Then, we defineY σ denotes the set of functions u∈C∞(T×[0,Γ0]) such that, for all ζ∈[0,Γ0], θ→u(θ;ζ)∈Xσ0.W σ denotes the set Yσ×C∞([0,Γ0]).Denote operatorA≔′. Then (16) can be written as(24)v2A2uθ+ζAuθ+sin⁡θ+uθ=F-ζ+Kuθ+2πmn-2uθ+uθ-2πmn.We define an operatorL:Xσ→Xσ by (25)Lu≔v2A2uθ+ζAuθ-Kuθ+2πmn-2uθ+uθ-2πmn+1-iΓ.Then, (24) can be written as(26)Fu,ζ=Lu+ϵsin⁡θ+uθ+iΓ-1uθ+ζ-F=0.We have the following properties about operator L.Lemma 3.
Fix the followingΓ>0 and K>0. The “diagonal” operator L (on Fourier spaces) satisfies the following:
(1)∀u∈Xσ0, (27)Lu=L∑k∈Zaleikθ=∑k∈Zλaleikθ,where (28)λ≔kv-1iΓ-1-kv-2Kcos⁡2mkπn-1.
(2) Let0≤σ~<σ and v∈Oγ,τ. The operator L is bounded and invertible, and L-1 maps Xσ0 onto Xσ~0,(29)L-1u=L-1∑k∈Zakeikθ=∑k∈Zλ-1akeikθ∈Xσ~0.
Ifn divides m, that is, m/n∈Z, then,(30)L-1uσ~≤μσ-σ~uσ,where (31)μσ≔μσ;v,Γ≔supk∈Z∖0vk-1-1iΓ-1-vk-1e-σk.
Ifn does not divide m, then(32)L-1uσ~≤μ¯σ-σ~uσ,where (33)μ¯σ≔μ¯σ;v,Γ≔supk∈Z∖0Γ-1vk-1-1e-σk.
Furthermore,(34)μσ≤1σ2τγ22τe2τ,μ¯σ≤1Γγσττeτ.Proof.
By the definition of operatorL, we can easily get (1) and (29). Now we prove (30), (32), and (34).
Ifn divides m, that is, m/n∈Z and 2K(cos(2mkπ/n)-1)=0, then we have (35)L-1uσ~=∑k∈Z∖0vk-1-1iΓ-1-vk-1e-σ-σ~kakekσ≤μσ-σ~uσ.Since(36)iΓ-1-vk≥1+vk≥γkτ,vk-1≥γkτ,and supx>0(xae-x)=(a/e)a, ∀a≥0, we obtain(37)μσ≤1σ2τγ22τe2τ.
Ifn does not divide m, that is, 2K(cos(2mkπ/n)-1)≠0, then operator L is invertible and no “small divisor” appears. We have (38)L-1uσ~=∑k∈Z∖0iΓvk-1+1-v2k2-2Kcos⁡2mkπn-1-1e-σ-σ~kakekσ≤∑k∈Z∖0Γvk-1-1e-σ-σ~kakekσ≤μ¯σ-σ~uσ.
Bysupx>0(xae-x)=(a/e)a, ∀a≥0, and (36), we obtain(39)μ¯σ≤1Γγσττeτ.This completes the proof.Remark 4.
By the estimate (39), we have that Γ≠0 in our Nash–Moser algorithm. However, when ω=0, we have v=0 and ζ≔vΓ=0.Our method of finding traveling waves comes from the idea of Newton scheme, which is an approximation method. If we choose first step(u0,ζ0) suitable, by finding a “quadratically better approximation,” we can move forward a single step to our target. Hence, the critical point is to construct “second step,” that is, to get (u1,ζ1); then, the method of making “next step” is the same. Finally, our solution of (26) can be written as(40)u,ζ=u0+∑s=1∞us,ζ0+∑s=1∞ζs.For convenience, we define(41)Tu0,ζ0≔u0+u1,ζ0+ζ1,foru0,ζ0∈Wσ,(42)E≔Fu0,ζ0,E1≔Fu0+u1,ζ0+ζ1=FTu0,ζ0.Now, we construct the “first step approximation” to find(u1,ζ1).Lemma 5.
Fix anyK>0, F>0, and Γ0>0. Assume that ω∈Oγ,τ. Then, for any ζ∈[0,Γ0], one obtains the “first step approximation”:(43)u1≔-L-1E+L-1E,ζ1≔iΓ-1L-1E.Proof.
We define(44)R≔ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+iΓ-1u1.Then we have (45)Fu0+u1,ζ0+ζ1=Lu0+u1+ϵsin⁡θ+u0+u1+iΓ-1u0+u1+ζ0+ζ1=Lu0+ϵsin⁡θ+u0+iΓ-1u0+ζ0+Lu1+ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+iΓ-1u1+ζ1=E+Lu1+ζ1+R.Based on our approximation method, we need to solve the following equation:(46)E+Lu1+ζ1=0.
Ifn divides m, that is, m/n∈Z, operator L is not invertible, the “small divisor” appears. Therefore, the removing of a “small set” (in Lebesgue measure sense) is needed; that is, we require ω∈Oγ,τ. Then, we construct (47)u1≔-L-1E+L-1E,ζ1≔iΓ-1L-1E.
Ifn dose not divide m, operator L is invertible. Then we can also construct (u1,ζ1) as the same form.
It is easy to verify that(u1,ζ1) is the solution of (46) and satisfies condition (22). This completes the proof.Remark 6.
In fact, to obtainsth step approximation (us,ζs)(s≥1), we need to solve(48)Es+Lus+ζs=0,where (49)Es=FTsu0,ζ0.
By the method in Lemma3, we can construct sth step solution for (48) as (50)us≔-L-1Es+L-1Es,ζs≔iΓ-1L-1Es.Now, in order to prove the convergence of our algorithm, we need the following KAM estimates.Lemma 7 (KAM estimates).
Assume that(u0,ζ0)∈Wσ. Then there exist β≔β(τ)>1 and C0≔C0(Γ0,τ)>1 such that, for any 0<α<σ and any Γ∈[0,Γ0], the following estimates hold:(51)u1σ-2/3α,ζ1,E1σ-α≤C0α-βEσ.Proof.
We first estimate the case thatn divides m. It follows from (30) that(52)supTL-1E≤L-1E0≤μσEσ.
By the definition of(u1,ζ1) and (52), we have(53)u1σ-2/3α≤L-1eσ-2/3α+μα3Eσ≤2μα3Eσ≤C1α-6τEσ,(54)ζ1≤C2σ-τEσ≤C2α-τEσ.By (53) and the definition E1 in (42), we get(55)E1σ-α=Rσ-α≤ϵsin⁡θ+u0+u1-sin⁡θ+u0σ-α+Γ2+1u1σ-α≤C3u1σ-α≤C3σ-6τEσ.
By (52), (53), and (55), there exist β≔β(τ)>1 and C0≔C0(Γ0,τ)>1 such that (56)u1σ-2/3α,ζ1,E1σ-α≤C0α-βEσ.
For the case ofn not dividing m, we can also get the estimate (51). The method is the same. So we omit it. This completes the proof.Now, we will give a sufficient condition on the convergence of our algorithm. Fors≥0 and 0<σ¯<σ<σ~, we set (57)σs≔σ¯+σ-σ¯2s,αs+1≔σs-σs+1=σ-σ¯2s+1.Then, we have the following result about the convergence of Nash–Moser algorithm.Lemma 8.
Assume thatC4≥C04β(σ-σ¯)-β and C4Eσ≤ι<1. Then, (u,ζ)∈Wσ¯ defined in (40) is a solution of (26); that is, F(u;ζ)=0, ∀ζ∈[0,Γ0].Proof.
We claim that, fors≥1,(58)us,ζs≔Tsu0,ζ0∈Wσs,Esθ≔FTsu0,ζ0∈Xσs,max⁡usσs,ζs,Esσs≤C4Eσ2s-12β.In fact, if (58) holds, then by the decay of (C4Eσ)2s, we obtain that (59)Tsu0,ζ0⟶u,ζ∈Wσsuniformly,ass⟶∞,Fu,ζ=lims→∞FTsu0,ζ0=lims→∞Es=0.
In the following, we will prove (58) by induction. Firstly, we check (58) for the case of s=1. Let α1≔α and σ1≔σ-α. By (51), we have (60)u1σ1,ζ1,E1σ1≤C0e-βEσ=C02βEσ,which implies that (u1,ζ1)∈Wσ1, so, (58) holds for s=1.
Lets′≥1. Assume that (58) holds true for 1≤s≤s′. Now we will prove that it also holds for s=s′+1. Let α≔αs+1 and σ≔σs. Note that Cαs+1-β=(C/2β)2sβ. By (51), we get (61)us+1σs+1,ζs+1,Es+1σs+1≤C4αs+1-βEsσs=C42β2sβEsσs≤C4Esσs2s2β,which shows that (us+1,ζs+1)∈Wσs+1. Hence, our claim holds. This completes the proof.Remark 9.
In this lemma, we do not care whethern divides m or not. Because the convergence of Nash–Moser algorithm is the same.Lemma 10 (uniqueness).
Assume that(u¯,ζ¯) and (u~,ζ~) are solutions of (26) in the domain C54βτσ¯-βτuσ¯<1. Then, (u¯,ζ¯)=(u~,ζ~); that is, the solution of (26) is unique.Proof.
Letu≔u¯-u~ and ζ≔ζ¯-u~. Then(62)Fu+u~,ζ+ζ~=Lu+u~+ϵsin⁡θ+u+u~+iΓ-1u+u~+ζ+ζ~=Lu~+ϵsin⁡θ+u~+iΓ-1u~+ζ~+Lu+ϵsin⁡θ+u+u~-ϵsin⁡θ+u~+iΓ-1u+ζ=E+Lu+ζ+R¯,where(63)R¯=ϵsin⁡θ+u+u~-ϵsin⁡θ+u~+iΓ-1u.
Note thatF(u¯,ζ¯)=0. Therefore, by (62), we have (64)Lu+ζ+R¯=0.It follows from condition (22) that(65)ζ=iΓ-1L-1R¯,(66)u≔-L-1R¯+L-1R¯.
Note thatR¯σ≤C6uσ, so we have that (67)ζ≤C7α-βuσ.
Next we will estimate (66). By (63) and the similar estimates in (53), for 0<α<σ≤σ¯, we have(68)uσ-α=-L-1R¯+L-1R¯σ-α≤C8α-βτR¯σ≤C8α-βτuσ.
If we takeσs≔σ¯/2s and αs≔σ¯/2s+1(s≥0). Then, it follows from (68) that(69)uσs+1≤C5αs-βτuσs=C5σ¯-βτ2βτs+1uσs≤absuσs,where a≔C52βτσ¯-βτ and b≔2βτ.
By (69), we have (70)abs+1uσs+1b≤abs+12uσs=abs+1absuσsb≤abs+1absabs-1uσs-1b≤⋯≤abs+1abs⋯abuσ¯=as+1bs+12uσ¯,which shows that(71)uσs+1≤asbs2+s-1uσ¯.
There existsχ≔χ(s)∈C1(1,∞) such that (72)χ′s>0,χ′′s=0,fors≥1,lims→∞χs=∞.
Then, it follows from (71) that(73)u0≤uσs≤abuσ¯1/χss2+s-1.
Note our assumptionC54βτσ¯-βτuσ¯<1 and u0=∑k∈Zuk. Therefore, by (73), we obtain (74)u0=0,which implies that (75)
u
=
0
,
t
h
a
t
i
s
.
,
u
¯
=
u
~
.This together with (65) means the uniqueness of solutions for (26). This completes the proof.The following result can be seen as a Nash–Moser theorem for dissipative lattice systems.Theorem 11.
Let0<γ<1≤τ, 0<σ¯<σ≤1, ω∈Oγ,τ, and ζ∈[0,Γ0] for some Γ0>0. Assume that “initial approximate solution” (u0,ζ0)∈C∞(T×[0,Γ0])×C∞[0,Γ0] and θ→u0(θ,Γ)∈Xσ0, ϱEσ≤1, ∀ζ∈[0,Γ0], and ϱ>1. Then, (26) possesses solutions (u,ζ)∈C∞(T×[0,Γ0])×C∞([0,Γ0]) and u∈Xσ¯0. Moreover, if (u¯,ζ¯)∈C∞(T×[0,Γ0])×C∞([0,Γ0]) is also the solution of (26) and satisfies ϱu-u¯σ¯<1, then, (u,ζ)=(u¯,ζ¯); that is, the solution of (26) is unique.Proof.
This result is the conclusion of Lemmas5–14. Let β(τ) and C0 be defined in Lemma 7. We choose ϱ such that C04β(σ-σ¯)-β≤ϱ≤min{C4,C54βτσ¯-βτ}. Then, by our assumption, Lemmas 8 and 10, we can get the existence and uniqueness of solutions of (26). This completes the proof.Remark 12.
In fact, in this abstract result, we do not need any assumption onϵ>0 in the case of ϵ=1. Then, the problem of finding traveling wave solutions for (4) with periodic boundary condition (2) is another open problem in [11]. By Theorem 11, we can see that, for fixing K>0, Γ>0 and sufficient small F>0, there is a unique traveling wave solution for (4). However, it is difficult to find the initial approximation solution (u0,ζ0) which must make the error function E satisfying ϱEσ≤1(ϱ>1).Now, we will use Theorem11 to prove our main result.Proof of Theorem1.
Let0<σ¯<σ, ∀σ¯,σ∈R+. We choose the initial approximation solution (76)u0,ζ0=0,0.
Letϵ0≤1/ϱ-F and ϵ∈[0,ϵ0] (ϱ>1). Then, the error function E defined in (42) is given by (77)E≔F0,0=ϵsin⁡θ-F,Eσ≤ϵ+F≤1ϱ.Here, we require that F>0 be sufficiently small so that 1/ϱ-F>0.
It follows from Theorem11 that our result holds. This completes the proof.Remark 13.
By the proof of Theorem1, we can see that our result also holds for the case of F=0. It suffices to take ϵ0≤1/ϱ and ϵ∈[0,ϵ0].


### 2.2. The Case ofΓ=0

We now focus on the proof of Theorem2 by the same method.By strong monotonicity arguments, Baesens and MacKay have obtained the existence and stability of traveling waves for (8) with periodic boundary condition (2). Here, we will use Nash–Moser iteration to study the existence and uniqueness of traveling wave solutions for (8) with periodic boundary condition (2).Note that a waveformφ satisfies the following equation:(78)φ′′t-Kφt+mnT-2φt+φt-mnT+ϵsin⁡φt=F.Hence, as in [11], we investigate the traveling wave of the type(79)φt=uvt+vt,where the wave velocity v=2π/T=2πω and u satisfies (80)uθ+2π=uθ,∀θ∈R.Inserting (79) into (78), we get(81)v2u′′θ-Kuθ+2πmn-2uθ+uθ-2πmn+ϵsin⁡θ+uθ=F.Define the operatorM:Xσ→Xσ as (82)M≔v2A2uθ-Kuθ+2πmn-2uθ+uθ-2πmn+1.Then, (81) can be written as(83)Gu,F=Mu+ϵsin⁡θ+uθ-uθ-F=0.We will also use the idea of Newton scheme to obtain the solution of (83). Firstly, we need to give some notations:(84)T¯u0,F0≔u0+u1,F0+F1,foru0,F0∈Wσ,E¯≔Gu0,F0,E¯s≔G∑i=0sui,∑i=0sFi=GT¯su0,F0,where (us,Fs) denotes the sth step approximation solution.Next, the spectrum analysis of operatorM is essential.Lemma 14.
FixK>0. The “diagonal” operator M (on Fourier spaces) satisfies the following:
(1)∀u∈Xσ0, (85)Mu=M∑k∈Zaleikθ=∑k∈Zλ¯aleikθ,where (86)λ¯≔-k2v2+1-2Kcos⁡2mkπn-1.
(2) Let0≤σ~<σ and v∈O¯γ,τ. The operator M is bounded and invertible, and M-1 maps Xσ0 onto Xσ~0, (87)M-1u=M-1∑k∈Zakeikθ=∑k∈Zλ¯-1akeikθ∈Xσ~0.
Ifn divides m, that is, m/n∈Z, then(88)M-1uσ~≤ϑσ-σ~uσ,where (89)ϑσ≔supk∈Z∖0vk+1-1vk-1-1e-σk.
Ifn does not divide m, then(90)M-1uσ~≤ϑ¯σ-σ~uσ,where (91)ϑ¯σ≔supk∈Z∖0vk-i1-4K-1e-σk.
Furthermore,(92)ϑσ,ϑ¯σ≤1σ2τγ22τe2τ.Proof.
The idea of this proof is similar to the proof of Lemma3. Here, we only need to verify (88), (90), and (92).
Note thatsupx>0(xae-x)=(a/e)a, ∀a≥0, and (93)kvω±1,kvω±i1-4K≥γkτ,∀k∈Z.Hence, in the case that n divides m, (94)M-1uσ~=∑k∈Z∖0vk+1-1vk-1-1e-σ-σ~kakekσ≤ϑσ-σ~uσ;in the case that n does not divide m, (95)M-1uσ~=∑k∈Z∖0-v2k2+1-2Kcos⁡2mkπn-1-1e-σ-σ~kakekσ≤∑k∈Z∖0vk-i1-4K-1vk+i1-4K-1e-σ-σ~kakekσ≤ϑ¯σ-σ~uσ,where (96)ϑσ,ϑ¯σ≤1σ2τγ22τe2τ.This completes the proof.Lemma 15.
Fix anyK>0 and F0>0. Assume that ω∈O¯γ,τ. Then, for any F∈[0,F0], one obtains the “first step approximation”: (97)u¯1≔-M-1E¯+M-1E¯,F1≔M-1E¯.Proof.
Define(98)R~≔ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+u1θ.Then we have (99)Gu0+u1,ζ0+ζ1=Mu0+u1+ϵsin⁡θ+u0+u1+u0+u1+F0+F1=Mu0+ϵsin⁡θ+u0+u0+F0+Mu1+ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+u1+F1=E¯+Mu1+F1+R~.For getting (u1,F1), we need to solve the following equation: (100)E¯+Mu1+F1=0.By condition (22), we can construct “the first approximation solution”: (101)u¯1≔-M-1E¯+M-1E¯,F1≔M-1E¯.This completes the proof.Remark 16.
In fact, we can construct thesth step approximation solution as (102)u¯s≔-M-1E¯s+M-1E¯s,Fs≔M-1E¯s,by solving the following equation: (103)E¯s+Mus+Fs=0.Lemma 17 (KAM estimates).
Assume that(u0,F0)∈Wσ. Then, there exist β¯≔β¯>1 and C9≔C9(F∗,τ,γ)>1 such that, for any 0<α<σ and any F∈[0,F∗], the following estimates hold: (104)u1σ-2/3α,F1,E¯1σ-α≤C9α-βE¯σ.Proof.
The proof is the same as Lemma7, so we omitted it.Lemma 18.
Assume thatC10≥C94β(σ-σ¯)-β and C10Eσ≤ι<1. Then, (u,F)=(∑s=0∞us,∑s=0∞Fs)∈Wσ¯ is a solution of (83); that is, F(u;F)=0, ∀F∈[0,F∗]. Furthermore, in the domain (105)u,F∈C∞T×0,F∗×C∞0,F∗∣C114βτσ¯-βτuσ¯<1,(83) admits a unique solution (u,F).Proof.
This proof is also similar to Lemmas8 and 10, so we omitted it.Based on Lemma18, we show the following Nash–Moser theorem for the conservative lattice systems.Theorem 19.
Let0<γ<1≤τ, 0<σ¯<σ≤1, ω∈O¯γ,τ, and F∈[0,F∗] for some F∗>0. Assume that “initial approximate solution” (u0,F0)∈C∞(T×[0,Γ0])×C∞[0,F∗] and θ→u0(θ,F)∈Xσ0, ϱ¯E¯σ≤1, ∀F∈[0,F∗], and ϱ¯>1. Then, (83) possesses solutions (u,F)∈C∞(T×[0,F∗])×C∞([0,F∗]) and u∈Xσ¯0. Moreover, the solution of (83) is unique in the domain {(u,F)∈C∞(T×[0,F∗])×C∞([0,F∗])∣ϱ¯uσ<1,∀ϱ¯>1}.Proof of Theorem2.
Let0<σ¯<σ, ∀σ¯,σ∈R+. We choose the initial approximation solution (106)u0,F0=0,0.
Letϵ0≤1/ϱ¯ and ϵ∈[0,ϵ0] (ϱ¯>1). Then, the error function E¯ defined in (84) is given by (107)E¯≔G0,0=ϵsin⁡θ,E¯σ≤ϵ≤1ϱ.
It follows from Theorem19 that our result holds. This completes the proof.


## 2.1. The Case ofΓ>0

In numerical simulations or experimental works on (4) with periodic boundary condition (2), it is observed that solutions often converge to a traveling wave (11)ψjt=φt+jmnT,where the waveform φ:R→R is a function satisfying(12)φt+T=φt+2π,∀t∈R.φ is a waveform if and only if it satisfies (12) and(13)φ′′t+Γφ′t+ϵ-1sin⁡ϵφt=ϵ-1F+Kφt+mnT-2φt+φt-mnT.Hence, as in [11], we investigate the traveling wave of the type(14)φt=uvt+vt,where the wave velocity v=2π/T=2πω and u satisfies (15)uθ+2π=uθ,∀θ∈R.Inserting (14) into (13), we get(16)v2u′′θ+ζu′θ+ϵ-1sin⁡ϵθ+ϵuθ=Kuθ+2πmn-2uθ+uθ-2πmn+ϵ-1F-ζ.Write (17)sin⁡ϵθ+ϵuθ=sin⁡ϵθ+2ϵcos⁡ϵθu+gu.We consider the following space:(18)Xσ≔uθ≔∑k∈Z2ukeik·θ∣uσ2≔∑k∈Zuk2e2σk<∞,where uk denotes the kthe Fourier coefficient.Obviously, for a nested family of Banach spaces{Xσ:σ≥0}, there holds (19)Xσ2⊂Xσ1,uσ1≤uσ2∀0≤σ1≤σ2.Forσ≥0, the space Xσ is Banach algebra with respect to multiplication of functions; that is, if u1,u2∈Xσ, then u1u2∈Xσ and there exists a positive constant C, such that (20)u1u2σ≤Cu1σu2σ.It is obviously that each function in Xσ has a bounded analytic extension in the complex multistrip Imθ<σ, where θ∈C. By the definition of the space Xσ, the following inequality holds: (21)∂θhuσ≤∂θluσ,∀h,l∈N:h≤l.For uniqueness, we assume that u satisfies(22)u≔∫Tusds=0,T=R2π.Now we define a function space with zero average by (23)Xσ0≔u∈Xσs.t.u0=u=0,as the closed subspace of Xσ.Letσ~>σ>0. Then, we defineY σ denotes the set of functions u∈C∞(T×[0,Γ0]) such that, for all ζ∈[0,Γ0], θ→u(θ;ζ)∈Xσ0.W σ denotes the set Yσ×C∞([0,Γ0]).Denote operatorA≔′. Then (16) can be written as(24)v2A2uθ+ζAuθ+sin⁡θ+uθ=F-ζ+Kuθ+2πmn-2uθ+uθ-2πmn.We define an operatorL:Xσ→Xσ by (25)Lu≔v2A2uθ+ζAuθ-Kuθ+2πmn-2uθ+uθ-2πmn+1-iΓ.Then, (24) can be written as(26)Fu,ζ=Lu+ϵsin⁡θ+uθ+iΓ-1uθ+ζ-F=0.We have the following properties about operator L.Lemma 3.
Fix the followingΓ>0 and K>0. The “diagonal” operator L (on Fourier spaces) satisfies the following:
(1)∀u∈Xσ0, (27)Lu=L∑k∈Zaleikθ=∑k∈Zλaleikθ,where (28)λ≔kv-1iΓ-1-kv-2Kcos⁡2mkπn-1.
(2) Let0≤σ~<σ and v∈Oγ,τ. The operator L is bounded and invertible, and L-1 maps Xσ0 onto Xσ~0,(29)L-1u=L-1∑k∈Zakeikθ=∑k∈Zλ-1akeikθ∈Xσ~0.
Ifn divides m, that is, m/n∈Z, then,(30)L-1uσ~≤μσ-σ~uσ,where (31)μσ≔μσ;v,Γ≔supk∈Z∖0vk-1-1iΓ-1-vk-1e-σk.
Ifn does not divide m, then(32)L-1uσ~≤μ¯σ-σ~uσ,where (33)μ¯σ≔μ¯σ;v,Γ≔supk∈Z∖0Γ-1vk-1-1e-σk.
Furthermore,(34)μσ≤1σ2τγ22τe2τ,μ¯σ≤1Γγσττeτ.Proof.
By the definition of operatorL, we can easily get (1) and (29). Now we prove (30), (32), and (34).
Ifn divides m, that is, m/n∈Z and 2K(cos(2mkπ/n)-1)=0, then we have (35)L-1uσ~=∑k∈Z∖0vk-1-1iΓ-1-vk-1e-σ-σ~kakekσ≤μσ-σ~uσ.Since(36)iΓ-1-vk≥1+vk≥γkτ,vk-1≥γkτ,and supx>0(xae-x)=(a/e)a, ∀a≥0, we obtain(37)μσ≤1σ2τγ22τe2τ.
Ifn does not divide m, that is, 2K(cos(2mkπ/n)-1)≠0, then operator L is invertible and no “small divisor” appears. We have (38)L-1uσ~=∑k∈Z∖0iΓvk-1+1-v2k2-2Kcos⁡2mkπn-1-1e-σ-σ~kakekσ≤∑k∈Z∖0Γvk-1-1e-σ-σ~kakekσ≤μ¯σ-σ~uσ.
Bysupx>0(xae-x)=(a/e)a, ∀a≥0, and (36), we obtain(39)μ¯σ≤1Γγσττeτ.This completes the proof.Remark 4.
By the estimate (39), we have that Γ≠0 in our Nash–Moser algorithm. However, when ω=0, we have v=0 and ζ≔vΓ=0.Our method of finding traveling waves comes from the idea of Newton scheme, which is an approximation method. If we choose first step(u0,ζ0) suitable, by finding a “quadratically better approximation,” we can move forward a single step to our target. Hence, the critical point is to construct “second step,” that is, to get (u1,ζ1); then, the method of making “next step” is the same. Finally, our solution of (26) can be written as(40)u,ζ=u0+∑s=1∞us,ζ0+∑s=1∞ζs.For convenience, we define(41)Tu0,ζ0≔u0+u1,ζ0+ζ1,foru0,ζ0∈Wσ,(42)E≔Fu0,ζ0,E1≔Fu0+u1,ζ0+ζ1=FTu0,ζ0.Now, we construct the “first step approximation” to find(u1,ζ1).Lemma 5.
Fix anyK>0, F>0, and Γ0>0. Assume that ω∈Oγ,τ. Then, for any ζ∈[0,Γ0], one obtains the “first step approximation”:(43)u1≔-L-1E+L-1E,ζ1≔iΓ-1L-1E.Proof.
We define(44)R≔ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+iΓ-1u1.Then we have (45)Fu0+u1,ζ0+ζ1=Lu0+u1+ϵsin⁡θ+u0+u1+iΓ-1u0+u1+ζ0+ζ1=Lu0+ϵsin⁡θ+u0+iΓ-1u0+ζ0+Lu1+ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+iΓ-1u1+ζ1=E+Lu1+ζ1+R.Based on our approximation method, we need to solve the following equation:(46)E+Lu1+ζ1=0.
Ifn divides m, that is, m/n∈Z, operator L is not invertible, the “small divisor” appears. Therefore, the removing of a “small set” (in Lebesgue measure sense) is needed; that is, we require ω∈Oγ,τ. Then, we construct (47)u1≔-L-1E+L-1E,ζ1≔iΓ-1L-1E.
Ifn dose not divide m, operator L is invertible. Then we can also construct (u1,ζ1) as the same form.
It is easy to verify that(u1,ζ1) is the solution of (46) and satisfies condition (22). This completes the proof.Remark 6.
In fact, to obtainsth step approximation (us,ζs)(s≥1), we need to solve(48)Es+Lus+ζs=0,where (49)Es=FTsu0,ζ0.
By the method in Lemma3, we can construct sth step solution for (48) as (50)us≔-L-1Es+L-1Es,ζs≔iΓ-1L-1Es.Now, in order to prove the convergence of our algorithm, we need the following KAM estimates.Lemma 7 (KAM estimates).
Assume that(u0,ζ0)∈Wσ. Then there exist β≔β(τ)>1 and C0≔C0(Γ0,τ)>1 such that, for any 0<α<σ and any Γ∈[0,Γ0], the following estimates hold:(51)u1σ-2/3α,ζ1,E1σ-α≤C0α-βEσ.Proof.
We first estimate the case thatn divides m. It follows from (30) that(52)supTL-1E≤L-1E0≤μσEσ.
By the definition of(u1,ζ1) and (52), we have(53)u1σ-2/3α≤L-1eσ-2/3α+μα3Eσ≤2μα3Eσ≤C1α-6τEσ,(54)ζ1≤C2σ-τEσ≤C2α-τEσ.By (53) and the definition E1 in (42), we get(55)E1σ-α=Rσ-α≤ϵsin⁡θ+u0+u1-sin⁡θ+u0σ-α+Γ2+1u1σ-α≤C3u1σ-α≤C3σ-6τEσ.
By (52), (53), and (55), there exist β≔β(τ)>1 and C0≔C0(Γ0,τ)>1 such that (56)u1σ-2/3α,ζ1,E1σ-α≤C0α-βEσ.
For the case ofn not dividing m, we can also get the estimate (51). The method is the same. So we omit it. This completes the proof.Now, we will give a sufficient condition on the convergence of our algorithm. Fors≥0 and 0<σ¯<σ<σ~, we set (57)σs≔σ¯+σ-σ¯2s,αs+1≔σs-σs+1=σ-σ¯2s+1.Then, we have the following result about the convergence of Nash–Moser algorithm.Lemma 8.
Assume thatC4≥C04β(σ-σ¯)-β and C4Eσ≤ι<1. Then, (u,ζ)∈Wσ¯ defined in (40) is a solution of (26); that is, F(u;ζ)=0, ∀ζ∈[0,Γ0].Proof.
We claim that, fors≥1,(58)us,ζs≔Tsu0,ζ0∈Wσs,Esθ≔FTsu0,ζ0∈Xσs,max⁡usσs,ζs,Esσs≤C4Eσ2s-12β.In fact, if (58) holds, then by the decay of (C4Eσ)2s, we obtain that (59)Tsu0,ζ0⟶u,ζ∈Wσsuniformly,ass⟶∞,Fu,ζ=lims→∞FTsu0,ζ0=lims→∞Es=0.
In the following, we will prove (58) by induction. Firstly, we check (58) for the case of s=1. Let α1≔α and σ1≔σ-α. By (51), we have (60)u1σ1,ζ1,E1σ1≤C0e-βEσ=C02βEσ,which implies that (u1,ζ1)∈Wσ1, so, (58) holds for s=1.
Lets′≥1. Assume that (58) holds true for 1≤s≤s′. Now we will prove that it also holds for s=s′+1. Let α≔αs+1 and σ≔σs. Note that Cαs+1-β=(C/2β)2sβ. By (51), we get (61)us+1σs+1,ζs+1,Es+1σs+1≤C4αs+1-βEsσs=C42β2sβEsσs≤C4Esσs2s2β,which shows that (us+1,ζs+1)∈Wσs+1. Hence, our claim holds. This completes the proof.Remark 9.
In this lemma, we do not care whethern divides m or not. Because the convergence of Nash–Moser algorithm is the same.Lemma 10 (uniqueness).
Assume that(u¯,ζ¯) and (u~,ζ~) are solutions of (26) in the domain C54βτσ¯-βτuσ¯<1. Then, (u¯,ζ¯)=(u~,ζ~); that is, the solution of (26) is unique.Proof.
Letu≔u¯-u~ and ζ≔ζ¯-u~. Then(62)Fu+u~,ζ+ζ~=Lu+u~+ϵsin⁡θ+u+u~+iΓ-1u+u~+ζ+ζ~=Lu~+ϵsin⁡θ+u~+iΓ-1u~+ζ~+Lu+ϵsin⁡θ+u+u~-ϵsin⁡θ+u~+iΓ-1u+ζ=E+Lu+ζ+R¯,where(63)R¯=ϵsin⁡θ+u+u~-ϵsin⁡θ+u~+iΓ-1u.
Note thatF(u¯,ζ¯)=0. Therefore, by (62), we have (64)Lu+ζ+R¯=0.It follows from condition (22) that(65)ζ=iΓ-1L-1R¯,(66)u≔-L-1R¯+L-1R¯.
Note thatR¯σ≤C6uσ, so we have that (67)ζ≤C7α-βuσ.
Next we will estimate (66). By (63) and the similar estimates in (53), for 0<α<σ≤σ¯, we have(68)uσ-α=-L-1R¯+L-1R¯σ-α≤C8α-βτR¯σ≤C8α-βτuσ.
If we takeσs≔σ¯/2s and αs≔σ¯/2s+1(s≥0). Then, it follows from (68) that(69)uσs+1≤C5αs-βτuσs=C5σ¯-βτ2βτs+1uσs≤absuσs,where a≔C52βτσ¯-βτ and b≔2βτ.
By (69), we have (70)abs+1uσs+1b≤abs+12uσs=abs+1absuσsb≤abs+1absabs-1uσs-1b≤⋯≤abs+1abs⋯abuσ¯=as+1bs+12uσ¯,which shows that(71)uσs+1≤asbs2+s-1uσ¯.
There existsχ≔χ(s)∈C1(1,∞) such that (72)χ′s>0,χ′′s=0,fors≥1,lims→∞χs=∞.
Then, it follows from (71) that(73)u0≤uσs≤abuσ¯1/χss2+s-1.
Note our assumptionC54βτσ¯-βτuσ¯<1 and u0=∑k∈Zuk. Therefore, by (73), we obtain (74)u0=0,which implies that (75)
u
=
0
,
t
h
a
t
i
s
.
,
u
¯
=
u
~
.This together with (65) means the uniqueness of solutions for (26). This completes the proof.The following result can be seen as a Nash–Moser theorem for dissipative lattice systems.Theorem 11.
Let0<γ<1≤τ, 0<σ¯<σ≤1, ω∈Oγ,τ, and ζ∈[0,Γ0] for some Γ0>0. Assume that “initial approximate solution” (u0,ζ0)∈C∞(T×[0,Γ0])×C∞[0,Γ0] and θ→u0(θ,Γ)∈Xσ0, ϱEσ≤1, ∀ζ∈[0,Γ0], and ϱ>1. Then, (26) possesses solutions (u,ζ)∈C∞(T×[0,Γ0])×C∞([0,Γ0]) and u∈Xσ¯0. Moreover, if (u¯,ζ¯)∈C∞(T×[0,Γ0])×C∞([0,Γ0]) is also the solution of (26) and satisfies ϱu-u¯σ¯<1, then, (u,ζ)=(u¯,ζ¯); that is, the solution of (26) is unique.Proof.
This result is the conclusion of Lemmas5–14. Let β(τ) and C0 be defined in Lemma 7. We choose ϱ such that C04β(σ-σ¯)-β≤ϱ≤min{C4,C54βτσ¯-βτ}. Then, by our assumption, Lemmas 8 and 10, we can get the existence and uniqueness of solutions of (26). This completes the proof.Remark 12.
In fact, in this abstract result, we do not need any assumption onϵ>0 in the case of ϵ=1. Then, the problem of finding traveling wave solutions for (4) with periodic boundary condition (2) is another open problem in [11]. By Theorem 11, we can see that, for fixing K>0, Γ>0 and sufficient small F>0, there is a unique traveling wave solution for (4). However, it is difficult to find the initial approximation solution (u0,ζ0) which must make the error function E satisfying ϱEσ≤1(ϱ>1).Now, we will use Theorem11 to prove our main result.Proof of Theorem1.
Let0<σ¯<σ, ∀σ¯,σ∈R+. We choose the initial approximation solution (76)u0,ζ0=0,0.
Letϵ0≤1/ϱ-F and ϵ∈[0,ϵ0] (ϱ>1). Then, the error function E defined in (42) is given by (77)E≔F0,0=ϵsin⁡θ-F,Eσ≤ϵ+F≤1ϱ.Here, we require that F>0 be sufficiently small so that 1/ϱ-F>0.
It follows from Theorem11 that our result holds. This completes the proof.Remark 13.
By the proof of Theorem1, we can see that our result also holds for the case of F=0. It suffices to take ϵ0≤1/ϱ and ϵ∈[0,ϵ0].


## 2.2. The Case ofΓ=0

We now focus on the proof of Theorem2 by the same method.By strong monotonicity arguments, Baesens and MacKay have obtained the existence and stability of traveling waves for (8) with periodic boundary condition (2). Here, we will use Nash–Moser iteration to study the existence and uniqueness of traveling wave solutions for (8) with periodic boundary condition (2).Note that a waveformφ satisfies the following equation:(78)φ′′t-Kφt+mnT-2φt+φt-mnT+ϵsin⁡φt=F.Hence, as in [11], we investigate the traveling wave of the type(79)φt=uvt+vt,where the wave velocity v=2π/T=2πω and u satisfies (80)uθ+2π=uθ,∀θ∈R.Inserting (79) into (78), we get(81)v2u′′θ-Kuθ+2πmn-2uθ+uθ-2πmn+ϵsin⁡θ+uθ=F.Define the operatorM:Xσ→Xσ as (82)M≔v2A2uθ-Kuθ+2πmn-2uθ+uθ-2πmn+1.Then, (81) can be written as(83)Gu,F=Mu+ϵsin⁡θ+uθ-uθ-F=0.We will also use the idea of Newton scheme to obtain the solution of (83). Firstly, we need to give some notations:(84)T¯u0,F0≔u0+u1,F0+F1,foru0,F0∈Wσ,E¯≔Gu0,F0,E¯s≔G∑i=0sui,∑i=0sFi=GT¯su0,F0,where (us,Fs) denotes the sth step approximation solution.Next, the spectrum analysis of operatorM is essential.Lemma 14.
FixK>0. The “diagonal” operator M (on Fourier spaces) satisfies the following:
(1)∀u∈Xσ0, (85)Mu=M∑k∈Zaleikθ=∑k∈Zλ¯aleikθ,where (86)λ¯≔-k2v2+1-2Kcos⁡2mkπn-1.
(2) Let0≤σ~<σ and v∈O¯γ,τ. The operator M is bounded and invertible, and M-1 maps Xσ0 onto Xσ~0, (87)M-1u=M-1∑k∈Zakeikθ=∑k∈Zλ¯-1akeikθ∈Xσ~0.
Ifn divides m, that is, m/n∈Z, then(88)M-1uσ~≤ϑσ-σ~uσ,where (89)ϑσ≔supk∈Z∖0vk+1-1vk-1-1e-σk.
Ifn does not divide m, then(90)M-1uσ~≤ϑ¯σ-σ~uσ,where (91)ϑ¯σ≔supk∈Z∖0vk-i1-4K-1e-σk.
Furthermore,(92)ϑσ,ϑ¯σ≤1σ2τγ22τe2τ.Proof.
The idea of this proof is similar to the proof of Lemma3. Here, we only need to verify (88), (90), and (92).
Note thatsupx>0(xae-x)=(a/e)a, ∀a≥0, and (93)kvω±1,kvω±i1-4K≥γkτ,∀k∈Z.Hence, in the case that n divides m, (94)M-1uσ~=∑k∈Z∖0vk+1-1vk-1-1e-σ-σ~kakekσ≤ϑσ-σ~uσ;in the case that n does not divide m, (95)M-1uσ~=∑k∈Z∖0-v2k2+1-2Kcos⁡2mkπn-1-1e-σ-σ~kakekσ≤∑k∈Z∖0vk-i1-4K-1vk+i1-4K-1e-σ-σ~kakekσ≤ϑ¯σ-σ~uσ,where (96)ϑσ,ϑ¯σ≤1σ2τγ22τe2τ.This completes the proof.Lemma 15.
Fix anyK>0 and F0>0. Assume that ω∈O¯γ,τ. Then, for any F∈[0,F0], one obtains the “first step approximation”: (97)u¯1≔-M-1E¯+M-1E¯,F1≔M-1E¯.Proof.
Define(98)R~≔ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+u1θ.Then we have (99)Gu0+u1,ζ0+ζ1=Mu0+u1+ϵsin⁡θ+u0+u1+u0+u1+F0+F1=Mu0+ϵsin⁡θ+u0+u0+F0+Mu1+ϵsin⁡θ+u0+u1-ϵsin⁡θ+u0+u1+F1=E¯+Mu1+F1+R~.For getting (u1,F1), we need to solve the following equation: (100)E¯+Mu1+F1=0.By condition (22), we can construct “the first approximation solution”: (101)u¯1≔-M-1E¯+M-1E¯,F1≔M-1E¯.This completes the proof.Remark 16.
In fact, we can construct thesth step approximation solution as (102)u¯s≔-M-1E¯s+M-1E¯s,Fs≔M-1E¯s,by solving the following equation: (103)E¯s+Mus+Fs=0.Lemma 17 (KAM estimates).
Assume that(u0,F0)∈Wσ. Then, there exist β¯≔β¯>1 and C9≔C9(F∗,τ,γ)>1 such that, for any 0<α<σ and any F∈[0,F∗], the following estimates hold: (104)u1σ-2/3α,F1,E¯1σ-α≤C9α-βE¯σ.Proof.
The proof is the same as Lemma7, so we omitted it.Lemma 18.
Assume thatC10≥C94β(σ-σ¯)-β and C10Eσ≤ι<1. Then, (u,F)=(∑s=0∞us,∑s=0∞Fs)∈Wσ¯ is a solution of (83); that is, F(u;F)=0, ∀F∈[0,F∗]. Furthermore, in the domain (105)u,F∈C∞T×0,F∗×C∞0,F∗∣C114βτσ¯-βτuσ¯<1,(83) admits a unique solution (u,F).Proof.
This proof is also similar to Lemmas8 and 10, so we omitted it.Based on Lemma18, we show the following Nash–Moser theorem for the conservative lattice systems.Theorem 19.
Let0<γ<1≤τ, 0<σ¯<σ≤1, ω∈O¯γ,τ, and F∈[0,F∗] for some F∗>0. Assume that “initial approximate solution” (u0,F0)∈C∞(T×[0,Γ0])×C∞[0,F∗] and θ→u0(θ,F)∈Xσ0, ϱ¯E¯σ≤1, ∀F∈[0,F∗], and ϱ¯>1. Then, (83) possesses solutions (u,F)∈C∞(T×[0,F∗])×C∞([0,F∗]) and u∈Xσ¯0. Moreover, the solution of (83) is unique in the domain {(u,F)∈C∞(T×[0,F∗])×C∞([0,F∗])∣ϱ¯uσ<1,∀ϱ¯>1}.Proof of Theorem2.
Let0<σ¯<σ, ∀σ¯,σ∈R+. We choose the initial approximation solution (106)u0,F0=0,0.
Letϵ0≤1/ϱ¯ and ϵ∈[0,ϵ0] (ϱ¯>1). Then, the error function E¯ defined in (84) is given by (107)E¯≔G0,0=ϵsin⁡θ,E¯σ≤ϵ≤1ϱ.
It follows from Theorem19 that our result holds. This completes the proof.



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*Source: 7081804-2018-04-30.xml*