#### Abstract

This paper deals with a class of two-dimensional linear differential systems with periodic coefficients. A sufficient condition for the exponential dichotomy of the linear system is given by the fixed point theorem and variable substitution, and some new results are obtained.

#### 1. Introduction

Consider the second-order linear differential equation

Equation (1) is often encountered in engineering, elasticity, electricity, and various oscillation problems.

Because of its importance, scholars have never stopped studying Equation (1) [1–4], and some scholars used transformation as follows:

Then Equation (1) becomes

Huang [5] used Lyapunov’s direct method to study the stability of the ordinary solution of Equation (3) and thereby obtained some sufficient conditions for the stability of the ordinary solution of Equation (1).

Exponential dichotomy is an important property of linear differential equation. The theory of exponential dichotomy is a generalization of the concept of hyperbolic rate of linear autonomous equation in linear nonautonomous equation and plays an important role in the analysis of nonautonomous equation. The theory of exponential dichotomy of linear differential equations can be traced back to Perron’s study of the stability of linear differential equations and the existence of bounded solutions of nonlinear differential equations. In 1934, Li established the theory of exponential dichotomy on the linear difference equation. The theory of exponential dichotomy has been playing an important role in the qualitative and stability research of differential and difference equations and has been applied to various research fields of mathematics with its rich theoretical ideas and complex mathematical skills [6].

Consider the homogeneous linear differential systemwhere is a square matrix of order , and is continuous on ; if there is a projection and positive constants , thenwhere is a fundamental solution matrix of (4), is the -order unit matrix, then (4) is said to have exponential dichotomy on . The exponential dichotomy of linear system on the whole axis is a powerful tool in the stability theory and also a very useful tool in the study of (almost) periodic differential systems [7].

Consider the following two-dimensional linear differential system

*Remark 1. *If , then Equation (6) becomes Equation (3). So, without losing generality, we just need to discuss Equation (6).

Shi [8] established some conclusions on the sign of characteristic exponents directly in terms of the coefficients. Theorem 1 in Ref. [8] is as follows:

Theorem 1. *(see [8]). Consider Equation (6), , and are -periodic continuous functions on , and have continuous derivatives on , and assume that the following conditions hold true:*

Then the characteristic exponents of system (6) must be positive and negative*.*

In this paper, we are devoted to obtaining the sufficient conditions for the exponential dichotomy of system (6) and generalized Theorem 1 in [8].

#### 2. Some Lemmas and Abbreviations

Consider the following equation

Here, is an -periodic continuous -order square matrix function on , is an -periodic continuous -dimensional vector function on , and the corresponding homogeneous linear system of Equation (8) is as follows:

Lemma 1. *(see [7]). If is an -periodic continuous -order square matrix function on , is an -periodic -dimensional continuous vector function on , and (9) has exponential dichotomy (5), then the nonhomogeneous linear periodic system (8) has a unique -periodic continuous solution, which can be expressed as follows:where , , and is a fundamental solution matrix of (9).*

Lemma 2. *(see [9]). Consider the following equationwhere are -periodic continuous functions on ; if , then Equation (11) has a unique -periodic continuous solution , then , and can be written as follows:*

Lemma 3. *(see [10]). Assume that -periodic sequence is convergent uniformly on any compact set of , is an -periodic function, and , then is convergent uniformly on.*

Lemma 4. *(see [11]). Assume that is a metric space, is a convex closed set of , and its boundary is , and if is a continuous compact mapping, such that , then has at least a fixed point on .*

Consider one-dimensional periodic differential equation as follows:

Here, is a continuous function, and .

Lemma 5. *(see [12]).If has second-order continuous partial derivatives on , and , then Equation (13) has at most two periodic continuous solutions .*

For the sake of convenience, assume that is an -periodic continuous function on , then we denote

The rest of the paper is arranged as follows. In section 3, we discuss the existence of two periodic solutions of Riccati’s equation. In section 4, we discuss the exponential dichotomy of two-dimensional linear differential system. In section 5, we consider the nonhomogeneous second-order linear differential equation and get some results about the existence of the unique periodic solution on the equation. Finally, we draw a conclusion of this paper.

#### 3. Two Periodic Solutions of Riccati’s Equation

Topological degree theory and fixed point theorem are often used by scholars to find (almost) periodic solutions of differential equations. Stimulated by the works of [13–15], in this section, we consider the existence of two periodic solutions of the following Riccati’s equation:

The existence of two periodic solutions of Riccati equation is obtained by using the fixed point theorem.

Theorem 2. *Consider Equation (15), , and are all -periodic continuous functions on , and assume that the following condition holds true:*

*Remark 2. *Without loss of generality, assume *.*

Then Equation (15) has exactly one positive, one negative, and two -periodic continuous solutions as follows:(1)One negative -periodic continuous solution is , and then(2)Another positive -periodic continuous solution is , and then

*Remark 3. *In Ref. [16], the author got a proposition as follows:

Proposition 1. *(see [16]). Consider Equation (15), , and are all -periodic continuous functions on , and assume that the following conditions hold true:**then Equation (15) has exactly one positive, one negative, and two -periodic continuous solutions , and then*

The proof of Theorem 1 in this paper is different from that of Theorem 1 in [16], and from Equation (17), (18), and (20), the ranges of periodic solutions obtained by us is more accurate. It can be seen that Theorem 1 in this paper is an extension of Theorem 1 in paper [16].

*Proof. *By , Equation (15) can be turned intoDenoteIt follows from thatIt follows from (22) and Equation (21) thatNow, we divide the proof into three steps.(1)We prove the existence of the periodic solution of Equation (15). Assume that , then the distance is defined as follows: Thus, is a complete metric space. Take a convex closed set of as follows: , then consider the following equation: It follows from (23) and (27) that Hence, we have Since , and are -periodic continuous functions, it follows that are -periodic continuous functions, by (30), according to Lemma 2, (24) has a unique -periodic continuous solution as follows: and It follows from (22) and (27) that Hence, we have It follows from (27), (29), and (32) that Hence, . We define a mapping as follows: Thus, if , then ; hence, . Now, we prove that the mapping is a compact mapping. Consider any sequence , then it is given as On the other hand, satisfies Thus, we have Hence, is uniformly bounded; therefore, is uniformly bounded and equicontinuous on . By the theorem of Ascoli–Arzela, for any sequence , there exists a subsequence (also denoted by ) such that which is convergent uniformly on any compact set of . From (40), combined with Lemma 3, is convergent uniformly on ; that is, is relatively compact on . Next, we prove that is a continuous mapping. Assume , and then It follows from (37) that Here, is between and ; thus, is between and ; hence, we have By (41) and the above inequality, we get Therefore, is continuous. From equation (37), . According to Lemma 4, has at least a fixed point on , and the fixed point is the -periodic continuous solution of Equation (15), and then we get(2)We prove the existence of the periodic solution of Equation (15). Assume that When , the distance is defined as follows: Thus, is a complete metric space. Take a convex closed set of as follows: , then consider the following equation: It follows from and (23) and (51) that Hence, we have Since , and are -periodic continuous functions, we get From (54), according to Lemma 2, we get unique -periodic continuous solution as follows: and It follows from (22) and (51) that Hence, we have It follows from (51), (53), and (56) that Hence, . We define a mapping as follows: Thus, if , then ; hence, . Now, we prove that the mapping is a compact mapping. Consider any sequence , then it follows that On the other hand, satisfies Thus, we have Hence, e is uniformly bounded; therefore, is uniformly bounded and equicontinuous on . By the theorem of Ascoli–Arzela, for any sequence, , then there exists a subsequence (also denoted by ) such that is convergent uniformly on any compact set . From (40), combined with Lemma 3, is convergent uniformly on ; that is, is relatively compact on . Next, we prove that is a continuous mapping. Assume that , and then It follows from (61) that Here, is between and ; thus, is between and ; hence, we have From (65) and the above inequality, we get Therefore, is continuous. From Equation (61), . According to Lemma 4, has at least a fixed point on , then the fixed point is the -periodic continuous solution of Equation (15), and then(3)We prove Equation (15) has exactly two periodic solutions and . Let then From Equation (68), according to Lemma 5, Equation (15) has at most two periodic continuous solutions, and we have known that Equation (15) has two periodic continuous solutions: ; thus, it follows that Equation (15) has exactly two periodic solutions , and thenThis is the end of the proof of Theorem 2.

#### 4. Exponential Dichotomy

In this section, we consider Equation (6) and get the sufficient condition for the exponential dichotomy of Equation (6).

Theorem 3. *Consider Equation (6),, and are -periodic continuous functions on , and have continuous derivatives on , then assume that the following conditions hold true:Then, Equation (6) has exponential dichotomy.*

*Proof. *From Equation (6), the following equation can be obtained:where .

Letthen Equation (72) becomesLetthen Equation (73) becomesThis is Riccati’s equation. From , Equation (75) satisfies all the conditions of Theorem 2. According to Theorem 2, Equation (75) has exactly one negative, one positive, and two -periodic continuous solutions and .

It follows from Equation (74) that Equation (73) has two continuous solutions:Here, .

It follows from Equation (72) that Equation (71) has two continuous solutions:Here, .

Morveover, we getThus, are linearly independent.

Sincethen Equation (6) has two sets of linearly independent solutions as follows:Taking the fundamental solution matrix of Equation (6), we get as follows:Thus, we haveTake projection , we get as follows:and then it followsSince are -periodic continuous functions on , it follows that there exist such that the following inequalities hold true:Thus, Equation (6) has exponential dichotomy, and the dichotomy constants are .

This is the end of the proof of Theorem 3.

If we turn Equation (6) into a second-order linear differential equation about , we can get the following:

Theorem 4. *Consider Equation (6), and , and are -periodic continuous functions on , and and have continuous derivatives on , then assume that the following conditions hold true:**then Equation (6) has exponential dichotomy.*

*Proof. *From Equation (6), the following equation can be obtained:whereLetthen Equation (89) becomesLetthen Equation (87) becomesand this is Riccati’s equation. From , Equation (92) satisfies all the conditions of Theorem 2. According to Theorem 2, Equation (92) has exactly one negative, one positive, and two -periodic continuous solutions and .

It follows from Equation (92) two continuous solutions as follows:andHere,

It follows from Equation (89) that Equation (87) has two continuous solutions as follows:and