- Research Article
- Open Access

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

- Rong Cheng
^{1, 2}Email author

**2009**:137084

https://doi.org/10.1155/2009/137084

© Rong Cheng. 2009

**Received:**9 April 2009**Accepted:**19 October 2009**Published:**26 October 2009

## 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.

## Keywords

- Periodic Solution
- Compact Operator
- Fourier Expansion
- Shift Operator
- Differential Delay Equation

## 1. Introduction and Main Result

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 .

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.

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

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)
- (H3)

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.

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 .

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)
- (iii)
where denotes the inner product in .

Proof.

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

*Case* (ii). Let

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
.

Firstly, it follows from that .

By reducing method, we have

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.

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

where

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

Define a mapping as

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

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

This yields , that is, .

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 .

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.

Thus satisfies of with and .

Lemma 2.7.

Under the assumptions of Theorem 1.1, satisfies of .

Proof.

where is free for the moment.

Let . Write

*Case* (1). If
with
, one has

*Case* (2). If
, we have

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

Lemma 2.8.

and link.

Proof.

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.

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

as , since .

Denote . We have

where is a constant independent of .

By the above inequality, one has

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

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.

## Declarations

### Acknowledgments

This work is supported by the Specialized Research Fund for the Doctoral Program of Higher Education for New Teachers and the Science Research Foundation of Nanjing University of Information Science and Technology (20070049).

## Authors’ Affiliations

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