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.

1. Introduction

The present work concerns the existence of traveling wave solutions for the following underdamped Frenkel–Kontorova model:with periodic boundary conditionwhere the parameters , , , and .

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 [14]). When , system (4) is conservative. Baesens and MacKay [5] proved the existence and also global stability of traveling waves. When , system (4) is dissipative. Under the condition and , 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 and , 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:

Fixing any and and given any velocity , there exists a traveling wave solution of (4) and (2) with velocity for an appropriate .

For any there exists a traveling wave solution of (4) and (2).

Assume that does not divide . Fixing any and , for all sufficiently large there exists a traveling wave solution of (4) and (2) for any and .

Fixing any and , for all sufficiently small there exists a traveling wave solution of (4) and (2) for any and .

In the final of Katriel’s paper, he gave several open problems. One of them is the following: Is it true that, fixing and , for sufficiently small and small applied force, a traveling wave does exist? If divides , what is the situation of the existence of traveling waves for (4) with periodic boundary condition (2)? In fact, if divides , there appears the “small divisor.” Then, the problem is difficult. Levi et al. [12] showed that, for fixing , (4) possesses a traveling wave only when 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 [1418], 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 havingTo overcome the “small divisor” problem, we need the following nonresonance conditions: where is a bounded region.

It is shown in [20] that if, for some , then the Lebesgue measure

Now, we state our main result.

Theorem 1. Under the assumption (NR1), fixing any and sufficient small , there exist , , and , such that, for any , , and , (4) with periodic boundary condition (2) possesses a unique traveling wave solution , where

When , (4) is

The corresponding Hamiltonian of (8) is where the nearest-neighbor coupling potential is

We have the following result about the existence of traveling waves for (8).

Theorem 2. Under the assumption (NR2), fixing any , there exist , , and , such that, for any and , (4) with periodic boundary condition (2) possesses a unique traveling wave solution , where

This paper is organized as follows. In Section 2, we first establish a Nash–Moser theorem for the case of . Then, we apply this result to prove our main results. The case of is also considered.

2. Proof of the Main Results

2.1. The Case of

In numerical simulations or experimental works on (4) with periodic boundary condition (2), it is observed that solutions often converge to a traveling wave where the waveform is a function satisfying is a waveform if and only if it satisfies (12) andHence, as in [11], we investigate the traveling wave of the typewhere the wave velocity and satisfies Inserting (14) into (13), we getWrite

We consider the following space: where denotes the the Fourier coefficient.

Obviously, for a nested family of Banach spaces , there holds

For , the space is Banach algebra with respect to multiplication of functions; that is, if , then and there exists a positive constant , such that It is obviously that each function in has a bounded analytic extension in the complex multistrip , where . By the definition of the space , the following inequality holds: For uniqueness, we assume that satisfiesNow we define a function space with zero average by as the closed subspace of .

Let . Then, we define

denotes the set of functions such that, for all , .

denotes the set .

Denote operator . Then (16) can be written as

We define an operator by

Then, (24) can be written asWe have the following properties about operator .

Lemma 3. Fix the following and . The “diagonal” operator (on Fourier spaces) satisfies the following:
(1) , where (2) Let and . The operator is bounded and invertible, and maps onto ,If divides , that is, , then,where If does not divide , thenwhere Furthermore,

Proof. By the definition of operator , we can easily get and (29). Now we prove (30), (32), and (34).
If divides , that is, and , then we have Sinceand , , we obtainIf does not divide , that is, , then operator is invertible and no “small divisor” appears. We have By , , and (36), we obtainThis completes the proof.

Remark 4. By the estimate (39), we have that in our Nash–Moser algorithm. However, when , we have and .

Our method of finding traveling waves comes from the idea of Newton scheme, which is an approximation method. If we choose first step 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 ; then, the method of making “next step” is the same. Finally, our solution of (26) can be written as

For convenience, we define

Now, we construct the “first step approximation” to find .

Lemma 5. Fix any , , and . Assume that . Then, for any , one obtains the “first step approximation”:

Proof. We defineThen we have Based on our approximation method, we need to solve the following equation:If divides , that is, , operator is not invertible, the “small divisor” appears. Therefore, the removing of a “small set” (in Lebesgue measure sense) is needed; that is, we require . Then, we construct If dose not divide , operator is invertible. Then we can also construct as the same form.
It is easy to verify that is the solution of (46) and satisfies condition (22). This completes the proof.

Remark 6. In fact, to obtain th step approximation , we need to solvewhere By the method in Lemma 3, we can construct th step solution for (48) as

Now, in order to prove the convergence of our algorithm, we need the following KAM estimates.

Lemma 7 (KAM estimates). Assume that . Then there exist and such that, for any and any , the following estimates hold:

Proof. We first estimate the case that divides . It follows from (30) thatBy the definition of and (52), we haveBy (53) and the definition in (42), we getBy (52), (53), and (55), there exist and such that For the case of not dividing , 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. For and , we set Then, we have the following result about the convergence of Nash–Moser algorithm.

Lemma 8. Assume that and . Then, defined in (40) is a solution of (26); that is, , .

Proof. We claim that, for ,In fact, if (58) holds, then by the decay of , we obtain that In the following, we will prove (58) by induction. Firstly, we check (58) for the case of . Let and . By (51), we have which implies that , so, (58) holds for .
Let . Assume that (58) holds true for . Now we will prove that it also holds for . Let and . Note that . By (51), we get which shows that . Hence, our claim holds. This completes the proof.

Remark 9. In this lemma, we do not care whether divides or not. Because the convergence of Nash–Moser algorithm is the same.

Lemma 10 (uniqueness). Assume that and are solutions of (26) in the domain . Then, ; that is, the solution of (26) is unique.

Proof. Let and . ThenwhereNote that . Therefore, by (62), we have It follows from condition (22) thatNote that , so we have that Next we will estimate (66). By (63) and the similar estimates in (53), for , we haveIf we take and . Then, it follows from (68) thatwhere and .
By (69), we have which shows thatThere exists such that Then, it follows from (71) thatNote our assumption and . Therefore, by (73), we obtain which implies that 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. Let , , , and for some . Assume that “initial approximate solution” and , , , and . Then, (26) possesses solutions and . Moreover, if is also the solution of (26) and satisfies , then, ; that is, the solution of (26) is unique.

Proof. This result is the conclusion of Lemmas 514. Let and be defined in Lemma 7. We choose such that . 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 on in the case of . 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 , and sufficient small , there is a unique traveling wave solution for (4). However, it is difficult to find the initial approximation solution which must make the error function satisfying .

Now, we will use Theorem 11 to prove our main result.

Proof of Theorem 1. Let , . We choose the initial approximation solution Let and (). Then, the error function defined in (42) is given by Here, we require that be sufficiently small so that .
It follows from Theorem 11 that our result holds. This completes the proof.

Remark 13. By the proof of Theorem 1, we can see that our result also holds for the case of . It suffices to take and .

2.2. The Case of

We now focus on the proof of Theorem 2 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 waveform satisfies the following equation:Hence, as in [11], we investigate the traveling wave of the typewhere the wave velocity and satisfies Inserting (79) into (78), we get

Define the operator as

Then, (81) can be written as

We will also use the idea of Newton scheme to obtain the solution of (83). Firstly, we need to give some notations:where denotes the th step approximation solution.

Next, the spectrum analysis of operator is essential.

Lemma 14. Fix . The “diagonal” operator (on Fourier spaces) satisfies the following:
(1) , where (2) Let and . The operator is bounded and invertible, and maps onto , If divides , that is, , thenwhere If does not divide , thenwhere Furthermore,

Proof. The idea of this proof is similar to the proof of Lemma 3. Here, we only need to verify (88), (90), and (92).
Note that , , and Hence, in the case that divides , in the case that does not divide , where This completes the proof.

Lemma 15. Fix any and . Assume that . Then, for any , one obtains the “first step approximation”:

Proof. DefineThen we have For getting , we need to solve the following equation: By condition (22), we can construct “the first approximation solution”: This completes the proof.

Remark 16. In fact, we can construct the th step approximation solution as by solving the following equation:

Lemma 17 (KAM estimates). Assume that . Then, there exist and such that, for any and any , the following estimates hold:

Proof. The proof is the same as Lemma 7, so we omitted it.

Lemma 18. Assume that and . Then, is a solution of (83); that is, , . Furthermore, in the domain (83) admits a unique solution .

Proof. This proof is also similar to Lemmas 8 and 10, so we omitted it.

Based on Lemma 18, we show the following Nash–Moser theorem for the conservative lattice systems.

Theorem 19. Let , , , and for some . Assume that “initial approximate solution” and , , , and . Then, (83) possesses solutions and . Moreover, the solution of (83) is unique in the domain .

Proof of Theorem 2. Let , . We choose the initial approximation solution Let and (). Then, the error function defined in (84) is given by It follows from Theorem 19 that our result holds. This completes the proof.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The authors express their thanks to Dr. Weiping Yan for his suggestion and giving this problem to them.