The Existence of Periodic Solutions for Non-Autonomous Differential Delay Equations via Minimax Methods

Abstract

By using variational methods directly, we establish the existence of periodic solutions for a class of nonautonomous differential delay equations which are superlinear both at zero and at infinity.

1. Introduction and Main Result

Many equations arising in nonlinear population growth models [1], communication systems [2], and even in ecology [3] can be written as the following differential delay equation:

(1.1)

where is odd and is parameter. Since Jone's work in [4], there has been a great deal of research on problems of existence, multiplicity, stability, bifurcation, uniqueness, density of periodic solutions to (1.1) by applying various approaches. See [2, 4–23]. But most of those results concern scalar equations (1.1) and generally slowly oscillating periodic solutions. A periodic solution of (1.1) is called a "slowly oscillating periodic solution" if there exist numbers and such that for , for and for all .

In a recent paper [17], Guo and Yu applied variational methods directly to study the following vector equation:

(1.2)

where is odd and is a given constant. By using the pseudo index theory in [24], they established the existence and multiplicity of periodic solutions of (1.2) with satisfying the following asymptotically linear conditions both at zero and at infinity:

(1.3)

where and are symmetric constant matrices. Before Guo and Yu's work, many authors generally first use the reduction technique introduced by Kaplan and Yorke in [7] to reduce the search for periodic solutions of (1.2) with and its similar ones to the problem of finding periodic solutions for a related system of ordinary differential equations. Then variational method was applied to study the related systems and the existence of periodic solutions of the equations is obtained.

The previous papers concern mainly autonomous differential delay equations. In this paper, we use minimax methods directly to study the following nonautonomous differential-delay equation:

(1.4)

where is odd with respect to and satisfies the following superlinear conditions both at zero and at infinity

(1.5)

When (1.2) satisfies (1.3), we can apply the twist condition between the zero and at infinity for to establish the existence of periodic solutions of (1.2). Under the superlinear conditions (1.5), there is no twist condition for , which brings difficulty to the study of the existence of periodic solutions of (1.4). But we can use minimax methods to consider the problem without twist condition for .

Throughout this paper, we assume that the following conditions hold.

• (H1) is odd with respect to and -periodic with respect to .

• (H2) write . There exist constants and such that

  

  (1.6)

  with and .

• (H3) there exist constants , and such that

  

  (1.7)

  with and .

Then our main result can be read as follows.

Theorem 1.1.

Suppose that satisfies (1.5) and the conditions hold. Then (1.4) possesses a nontrivial -periodic solution.

Remark 1.2.

We shall use a minimax theorem in critical point theory in [25] to prove our main result. The ideas come from [25–27]. Theorem 1.1 will be proved in Section 2.

2. Proof of the Main Result

First of all in this section, we introduce a minimax theorem which will be used in our discussion. Let be a Hilbert space with . Let be the projections of onto and , respectively.

Write

(2.1)

where is compact.

Definition 2.1.

Let and be boundary. One calls and link if whenever and for all , then .

Definition 2.2.

A functional satisfies condition, if every sequence that , and being bounded, possesses a convergent subsequence.

Then [25, Theorem ] can be stated as follows.

Theorem 2 A.

Let be a real Hilbert space with , and inner product . Suppose satisfies condition,

, where and is bounded and selfadjoint, ,

is compact, and

there exists a subspace and sets , and constants such that

and ,

is bounded and ,

and link.

Then possesses a critical value .

Let

(2.2)

Then and , where denotes the gradient of with respect to . We have the following lemma.

Lemma 2.3.

Under the conditions of Theorem 1.1, the function satisfies the following.

• (i) is 2-periodic with respect to and for all ,

• (ii)
  

  (2.3)
  

  (2.4)

• (iii) There exist constants , and such that for all with and , , and

  

  (2.5)
  

  (2.6)

  where denotes the inner product in .

Proof.

The definition of implies (i) directly. We prove case (ii) and case (iii).

Case (ii). Let

(2.7)

Then and or is equivalent to or , respectively.

From (1.5) and L'Hospital rules, we have (2.3) by a direct computation.

Case (iii). By (H2), we have a constant such that for with .

Now we prove for with , that is,

(2.8)

Firstly, it follows from that .

Now we show . Let , . By , , that is, . Then

(2.9)

By reducing method, we have

(2.10)

Thus, the inequality for holds.

Take and . Then (2.5) and (2.6) hold with and .

Below we will construct a variational functional of (1.4) defined on a suitable Hilbert space such that finding -periodic solutions of (1.4) is equivalent to seeking critical points of the functional.

Firstly, we make the change of variable

(2.11)

Then (1.4) can be changed to

(2.12)

where is -periodic with respect to . Therefore we only seek -periodic solution of (2.12) which corresponds to the -periodic solution of (1.4).

We work in the Sobolev space . The simplest way to introduce this space seems as follows. Every function has a Fourier expansion:

(2.13)

where are -vectors. is the set of such functions that

(2.14)

With this norm , is a Hilbert space induced by the inner product defined by

(2.15)

where

We define a functional by

(2.16)

By Riesz representation theorem, H identifies with its dual space H*. Then we define an operator A:H→H*=H by extending the bilinear form:

(2.17)

It is not difficult to see that is a bounded linear operator on and .

Define a mapping as

(2.18)

Then the functional can be rewritten as

(2.19)

According to a standard argument in [24], one has for any ,

(2.20)

Moreover according to [28], is a compact operator defined by

(2.21)

Our aim is to reduce the existence of periodic solutions of (2.12) to the existence of critical points of . For this we introduce a shift operator defined by

(2.22)

It is easy to compute that is bounded and linear. Moreover is isometric, that is, and , where denotes the identity mapping on .

Write

(2.23)

Lemma 2.4.

Critical points of over are critical points of on , where is the restriction of over .

Proof.

Note that any is -periodic and is odd with respect to . It is enough for us to prove for any and being a critical point of in .

For any , we have

(2.24)

This yields , that is, .

Suppose that is a critical point of in . We only need to show that for any . Writing with and noting , one has

(2.25)

The proof is complete.

Remark 2.5.

By Lemma 2.4, we only need to find critical points of over . Therefore in the following will be assumed on .

For , yields that , where is in the Fourier expansion of . Thus . Moreover for any ,

(2.26)

Hence is self-adjoint on .

Let and denote the positive definite and negative definite subspace of in , respectively. Then . Letting , , we see that of Theorem A holds. Since is compact, of Theorem A holds. Now we establish of Theorem A by the following three lemmas.

Lemma 2.6.

Under the assumptions of Theorem 1.1, of holds for .

Proof.

From the assumptions of Theorem 1.1 and Lemma 2.3, one has

(2.27)

By (2.3), for any , there is a such that

(2.28)

Therefore, there is an such that

(2.29)

Since is compactly embedded in for all and by (2.29), we have

(2.30)

Consequently, for ,

(2.31)

Choose and so that . Then for any ,

(2.32)

Thus satisfies of with and .

Lemma 2.7.

Under the assumptions of Theorem 1.1, satisfies of .

Proof.

Set and let

(2.33)

where is free for the moment.

Let . Write

(2.34)

Case (1). If with , one has

(2.35)

Case (2). If , we have

(2.36)

That is

(2.37)

Denote . By appendix, there exists such that ,

(2.38)

Now for , set . By (2.4), for a constant , there is an such that

(2.39)

Choosing , for ,

(2.40)

For , we have

(2.41)

Henceforth, for any and , that is, . Then of holds.

Lemma 2.8.

and link.

Proof.

Suppose and for all . Then we claim that for each , there is a such that , that is,

(2.42)

where is a projection. Define

(2.43)

as follows:

(2.44)

It is easy to see that

(2.45)

However,

(2.46)

According to topological degree theory in [29], we have

(2.47)

since . Therefore and link.

Now it remains to verify that satisfies -condition.

Lemma 2.9.

Under the assumptions of Theorem 1.1, satisfies -condition.

Proof.

Suppose that

(2.48)

We first show that is bounded. If is not bounded, then by passing to a subsequence if necessary, let as .

By (2.4), there exists a constant such that as . By (2.5), one has

(2.49)

This yields

(2.50)

Write . By (2.6), there is a constant such that

(2.51)

Therefore,

(2.52)

This inequality and (2.50) imply that

(2.53)

as , since .

Denote . We have

(2.54)

where is a constant independent of .

By the above inequality, one has

(2.55)

as . This yields

(2.56)

Similarly, we have

(2.57)

Thus it follows from (2.56) and (2.57) that

(2.58)

which is a contradiction. Hence is bounded.

Below we show that has a convergent subsequence. Notice that and is compact. Since is bounded, we may suppose that

(2.59)

Since has continuous inverse in , it follows from

(2.60)

that

(2.61)

Henceforth has a convergent subsequence.

Now we are ready to prove Theorem 1.1.

Proof of Theorem 1.1.

It is obviously that Theorem 1.1 holds from Lemmas 2.3, 2.4, 2.6, 2.7, 2.8, and 2.9 and Theorem A.

Appendix

The purpose of this appendix is to prove the following lemma. The main idea of the proof comes from [26]. Lemma A.1. There exists such that, ,

(A.1)

Proof. If (A.1) is not true, , there exists such that

(A.2)

Write . Notice that dim and . In the sense of subsequence, we have

(A.3)

Then (2.37) implies that

(A.4)

Note that , in the sense of subsequence as . Thus in the sense of subsequence,

(A.5)

This means that in , that is,

(A.6)

By (A.4) we know that . Therefore, . Then there exist such that

(A.7)

Otherwise, for all , we must have

(A.8)

that is, , . We get a contradiction. Thus (A.7) holds. Let , , and . By (A.2), we have

(A.9)

Let be large enough such that and . Then we have

(A.10)

This implies that

(A.11)

This is a contradiction to (A.6). Therefore the lemma is true and (A.1) holds.