## Dynamical Aspects of Initial/Boundary Value Problems for Ordinary Differential Equations 2014

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# On Homoclinic Solutions for First-Order Superquadratic Hamiltonian Systems with Spectrum Point Zero

**Academic Editor:**Yongli Song

#### Abstract

The existence and multiplicity of homoclinic solutions for a class of first-order periodic Hamiltonian systems with spectrum point zero are obtained. The proof is based on two critical point theorems for strongly indefinite functionals. Some recent results are improved and extended.

#### 1. Introduction

Consider the following first-order Hamiltonian systems: where , is the standard symplectic matrix in and has the form with being a symmetric matrix-valued function and . A solution of is a homoclinic orbit if and as .

To continue the discussion, we need the following notation:

Homoclinic orbits of dynamical systems are important in applications for a number of reasons. They may be “organizing centers” for the dynamics in their neighborhood. From their existence one may, under certain conditions, infer the existence of chaos nearby or the bifurcation behavior of periodic orbits. As a special case of dynamical systems, Hamiltonian systems are very important in the study of gas dynamics, fluid mechanics, relativistic mechanics, and nuclear physics.

During the last decades, many authors are devoted to the study of homoclinic orbits for Hamiltonian systems via modern variational methods. For example, see [1–4] for the second-order systems and [5–16] for the first-order systems. To be precise, in 1999, Ding and Willem [15] studied the existence of homoclinic solutions for a class of the first-order periodic Hamiltonian systems with spectrum point zero under the well-known (AR) condition as follows: Later, under condition (4), Ding and Girardi [11] obtained that has infinitely many homoclinic orbits provided that is even in . However, there are many potentials satisfying superquadratic condition, not satisfying the (AR) condition (4).

Motivated by the above facts, in this paper, our aim is to study homoclinic solutions for under the conditions that zero is a continuous spectrum point and satisfies weak superquadratic conditions.

Set Given a matrix , we say that if and only if Also letting be the identity matrix in and , we denote the matrix by . Moreover, denote by and the spectrum and the continuous spectrum of the operator , respectively.

We make the following assumptions:(L_{1}) is -periodic on . and there exists such that .(H_{1}) is -periodic in , .(H_{2}) uniformly in as .(H_{3}) as uniformly in .(H_{4}) if , and there exist and such that if .

*Remark 1. *The following functions satisfy weak superquadratic conditions – but do not verify (4):
where and is -periodic in and , .

Now we only check .). It is easy to verify that are satisfied. However, following the discussion of Remark 1.2 in [17], let , where . Then for any , one has
That is, condition (4) is not satisfied for .

Observe that, due to the periodicity of and , if is a homoclinic orbit of , then so is for each , where . Two solutions and are said to be geometrically distinct if for all .

Now we state our main result.

Theorem 2. *Assume that and hold. Then has at least one homoclinic orbit. In addition, if is even in , then has infinitely many geometrically distinct homoclinic orbits.*

This paper is organized as follows. In Section 2, we formulate the variational setting and recall some critical point theorems required. In Section 3, we discuss linking structure and the Cerami condition of the functional. In Section 4, we prove Theorem 2.

*Notation. *Throughout the paper we will denote by various positive constants which may vary from line to line and are not essential to the problem.

#### 2. Preliminaries

In what follows by we denote the usual -norm and by the usual inner product of . A standard Floquet reduction argument shows that (see Proposition 2.2 in [15]).

Let be the spectral family of . We have , called the polar decomposition, where . By , has an orthogonal decomposition where .

Let be the linear space of the completion of under the norm Then is a Hilbert space under the inner product possesses an orthogonal decomposition where , the corresponding projections being denoted by .

By , it is easy to check where “~" means “equivalence." Therefore, can be embedded continuously into for any and compactly into for any .

However, since may belong to a spectrum of , then may not be equivalent to -norm on . Therefore, in the following we use the spectrum family of to separate into two segments. That is, for any , set and . Let , where denotes the closure of the set in . Similar to , since the spectrum of restricted to is bounded away from , thus, one has However, is not complete with respect to the norm ; thus it is reasonable to introduce a new norm. Define and let be the completion of under the norm .

Lemma 3 (see [16], Lemma 2.1). * and is embedded compactly in and continuously in for all .*

Let denote the completion of with respect to the norm . Since is continuously embedded in for all , by (15), is a closed subspace of . Note that and is orthogonal to with respect to . Then has the following decomposition: Let be the completion of under the norm . We have the following lemma.

Lemma 4. *Suppose that (L _{1}) is satisfied. Then has the direct sum decomposition
*

*and is embedded continuously in and compactly in for any .*

*Proof. *By (13), (15), and (17) and Lemma 3, and are closed, and using the decomposition of , it is easy to check that , and so (18) holds. Using the same facts above and Lemma 3, one can obtain easily the desired conclusion on embedding.

It is easy to verify that is a Hilbert space under the inner product , and is its induced norm. From now on, we consider the space as our working space. Clearly, and all norms , , and are equivalent to . It is not difficult to check that is uniformly convex, so is reflexive. Set From Lemma 4 and (25) in Section 3, it follows that is defined on the Banach space and belongs to , and Consider the functional for . Then and a standard argument shows that critical points of are homoclinic orbits of (see [15]).

In order to study the critical points of , we now recall some abstract critical point theory developed recently in [18].

Let be a Banach space with direct sum decomposition and the corresponding projections onto , respectively. For a functional we write , , and . Recall that is said to be weakly sequentially lower semicontinuous if for any in one has . is said to be weakly sequentially continuous if for each . A sequence is said to be a -sequence if and . is said to satisfy the -condition if any -sequence has a convergent subsequence. A set is said to be a -attractor if for any and any -sequence there is such that for . Given an interval , is said to be a -attractor if it is a -attractor for all .

From now on, we assume that is separable and reflexive and fix a countable dense subset . For each there exists a seminorm on defined by We denote by the induced topology. Let denote the weak*-topology on .

Suppose the following.For any , is -closed, and is continuous.For any , there exists such that for all .There exists with , where .There exists an increasing sequence of finite-dimensional subsequences and a sequence of positive numbers such that, letting and , and .For any interval , there exists a -attractor with bounded and .

Now we state two critical point theorems which will be used later.

Theorem 5. *Let be satisfied and suppose that there are and with such that , where . Then has a -sequence with .*

Theorem 6. *Assume is even with and let be satisfied. Then possesses an unbounded sequence of positive critical values.*

#### 3. Linking Structure and the -Sequence

We now study the linking structure and the -sequence of . Observe that if holds, then ; hence for . Remark that (23) and - imply that, given , for any , there is such that for all .

Lemma 7. *Assume that and hold. Then there exists such that , where .*

*Proof. *Choose such that (25) holds for any . This yields
for all ; together with the equivalence of and on , the lemma follows from the form of .

Set . implies that . Choose . Then we can take a number such that Since , the subspace is infinite dimensional, where denotes the spectral family of . Note that and Let be a sequence with For each , take an element with and define . Then is an increasing sequence of finite dimensional subsequence of . Set .

Lemma 8. *Suppose that and are satisfied and is given by Lemma 7. Then , and there exists a sequence such that , where .*

*Proof. *It suffices to show that as in . If not, we assume that, for some sequence with , there exists such that for all . Define ; we have . Passing to subsequence , , in for and with respect to and . Then, by we have
We claim that . Indeed, if not, it follows from (30) that . Also .

By , there exists such that
For , let
Thus, by (31) and (32), one has
From (33) and and together with it follows that . Therefore, , which contradicts . So .

By (27)-(28) and the fact that , one has
Then there exists a finite interval such that
Note that
where denotes the Lebesgue measure of . Combining this with (30) and (35), one has
which is a contradiction.

As a special case we have the following.

Lemma 9. *Suppose that and are satisfied and is given by Lemma 7. Then, letting with , there exists such that , where .*

Lemma 10. *Suppose that are satisfied. Then any -sequence is bounded in .*

*Proof. *Let be such that
Then, for ,
Suppose to the contrary that is unbounded. Setting , then and for all . Passing to subsequence, in , in for , and for a.e., .

Note that
From (40), we obtain
Set, for ,
By and , for all and as .

Let
Since depends periodically on and if , one has and
where is defined in (32). It follows from (32) and (39) that
Using (45) we obtain
as uniformly in , and for any fixed ,
as . It follows from (46) that, for any ,
as uniformly in .

Let . By there is such that
for all . Consequently,
for all .

Set . By , (48), and Hölder inequality, we can take large enough such that
for all . Note that there is independent of such that for . By (47) there is such that
for all . By (50)–(52), one has
On the other hand, by (48), take large enough such that
Similar to (47), one has
Thus (54) and (55) imply that
By (53) and (56), one has
which contradicts (41). The proof is complete.

Let be a -sequence of ; by Lemma 10, it is bounded, up to a subsequence; we may assume in , in for , and a.e. on . Plainly, is a critical point of . Set .

Lemma 11. *Under the assumptions of Theorem 2, one has, as ,*(1)*;*(2)*.*

*Proof. *The proof of this lemma is similar to the one of Lemma 5.7 in [9] (or see [10]), so we omit it here.

Let be the set of nontrivial critical points of .

Lemma 12. *Under the assumptions of Theorem 2, the following two conclusions hold:*(1)*;*(2)*.*

*Proof. *(1) For any , there holds
together with (25), which implies that
where . Choose small enough; one has
for each . Therefore,
(2) Suppose to the contrary that there exists a sequence such that .

Then
By (1), . Clearly is a -sequence of and hence is bounded by Lemma 10. Using (63) and the discussion in the proof of Lemma 10, we see that, for any and , and as . Therefore, it follows from (24) and (62) that for any
which contradicts (61). The proof is complete.

In the following lemma we discuss further the -sequence. Let denote the integer part of . The following lemma is standard by using Lemmas 11 and 12 (see [1, 19]).

Lemma 13. *Under the assumptions of Theorem 2, let be a -sequence of . Then either*(i)* (and hence ), or*(ii)* and there exist a positive integer , , and sequence , , such that, after extraction of a subsequence of ,
* *and for ,
* *as .*

#### 4. Proof of Theorem 2

In order to apply the abstract Theorems 5 and 6 to , we choose and . is separable and reflexive and let be a countable dense subset of . First we have the following.

Lemma 14. * satisfies .*

*Proof. *We first show that is -closed for every . Consider a sequence in which -converges to and write and . Observe that converges to in norm topology. Since , there exists such that
Therefore, is bounded. So is bounded. Now we claim that is bounded. Assume by contradiction that is unbounded. Clearly, is also unbounded. Setting , then . Passing to subsequence, in , in for . Note that
Therefore, is bounded and . Similar to Lemma 8, we can obtain (33). From (33) and and together with it follows that . Since , we obtain , which contradicts the fact that . So is bounded. Therefore, is bounded in , and we have . Therefore, . It is easy to show that is weakly sequentially lower semicontinuous. Thus, from the form of it follows that . So and is -closed.

Next we show that is continuous. It is sufficient to show that has the same property. Let in . Then in for . It is obvious that
as . Now using the density of in one can obtain the desired conclusion.

Lemma 15. * satisfies .*

*Proof. *We assume by contradiction that, for some , there exists a sequence with and . The form of implies that
Since and , it follows that as . Hence . Set . Then and . Therefore, in . By (70) we have
Thus . By (31)-(32) and (71), one has