# Exponential Dichotomy of the Two-Dimensional Linear Differential System

Authors: Ni Hua; Wang Jing; Bai Min
Journal: Journal of Mathematics
 (2022)

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## 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.


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## Body

## 1. Introduction

Consider the second-order linear differential equation(1)d2xdt2+ptdxdt+qtx=0.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:(2)dxdt=y.Then Equation (1) becomes(3)dxdtdydt=01−qt−ptxy.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 system(4)dxdt=Atx,where At is a square matrix of order n, and At is continuous on ℝ; if there is a projection P and positive constants K≥1,α>0, then(5)XtPX−1s≤Kexp−αt−s,t≥s,XtI−PX−1s≤Kexpαt−s,s≥t,where Xt is a fundamental solution matrix of (4), I is the n-order unit matrix, then (4) is said to have exponential dichotomy on ℝ. 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(6)dxdtdydt=atbtctdtxy.Remark 1.
Ifat≡0,bt≡1, 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), at,bt,ct, and dt are ω-periodic continuous functions on R, ct and dt have continuous derivatives on ℝ, and assume that the following conditions hold true:(7)H1∫0ωat+dtdt≠0,H2ct≠0,H3dtc′tct+atdt−btct−d′t<0.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(8)dxdt=Atx+ft.Here,At is an ω-periodic continuous n-order square matrix function on ℝ, ft is an ω-periodic continuous n-dimensional vector function on ℝ, and the corresponding homogeneous linear system of Equation (8) is as follows:(9)dxdt=Atx.Lemma 1.
(see [7]). If At is an ω-periodic continuous n-order square matrix function on ℝ, ft is an ω-periodic n-dimensional continuous vector function on ℝ, and (9) has exponential dichotomy (5), then the nonhomogeneous linear periodic system (8) has a unique ω-periodic continuous solution, which can be expressed as follows:(10)xt=∫−∞tXtPX−1sfsds−∫t+∞XtI−PX−1sfsds,where mod(xt⊆modAt,ft, xt≤2K/αft, and Xt is a fundamental solution matrix of (9).Lemma 2.
(see [9]). Consider the following equation(11)dxdt=atx+bt,where at,bt are ω-periodic continuous functions on ℝ; if ∫0ωatdt≠0, then Equation (11) has a unique ω-periodic continuous solution ηt, then modη⊆modat,bt, and ηt can be written as follows:(12)ηt=∫−∞te∫staθdθbsds,∫0ωatdt<0,−∫t+∞e∫staθdθbsds.∫0ωatdt>0.Lemma 3.
(see [10]). Assume that ω-periodic sequence fnt is convergent uniformly on any compact set of ℝ,ftis an ω-periodic function, and modfn⊆modfn=1,2,…, then fnt is convergent uniformly onℝ.Lemma 4.
(see [11]). Assume that V is a metric space, ℂ is a convex closed set of V, and its boundary is ∂ℂ, and if T:ℂ⟶ℂ is a continuous compact mapping, such that T∂ℂ⊆ℂ, then T has at least a fixed point on ℂ.Consider one-dimensional periodic differential equation as follows:(13)dxdt=ft,x.Here,f:ℝ×I⟶ℝ is a continuous function, and ft+ω,x=ft,x,ω>0,I⊆ℝ.Lemma 5.
(see [12]).If ft,x has second-order continuous partial derivatives on x, and fxx″t,x≠0, then Equation (13) has at most two periodic continuous solutions.For the sake of convenience, assume thatft is an ω-periodic continuous function on ℝ, then we denote(14)fM=supt∈0,ωft,fL=inft∈0,ωft.The rest of the paper is arranged as follows. In section3, 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:(15)dxdt=atx2+btx+ct.The existence of two periodic solutions of Riccati equation is obtained by using the fixed point theorem.Theorem 2.
Consider Equation (15), at,bt, and ct are all ω-periodic continuous functions on ℝ, and assume that the following condition holds true:(16)H1atct<0.Remark 2.
Without loss of generality, assumeat<0,ct>0.
Then Equation (15) has exactly one positive, one negative, and two ω-periodic continuous solutions as follows:(1)
One negativeω-periodic continuous solution is γ1t, and then(17)−b+b2−4ac2aL≤γ1t≤−b+b2−4ac2aM.(2)
Another positiveω-periodic continuous solution is γ2t, and then(18)−b−b2−4ac2aL≤γ2t≤−b−b2−4ac2aM.Remark 3.
In Ref. [16], the author got a proposition as follows:Proposition 1.
(see [16]). Consider Equation (15), at,bt, and ct are all ω-periodic continuous functions on ℝ, and assume that the following conditions hold true:(19)H1atct<0,
then Equation (15) has exactly one positive, one negative, and two ω-periodic continuous solutions γ1t,γ2t, and then(20)γ1t<0<γ2t.The proof of Theorem1 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.
ByH1, Equation (15) can be turned into(21)dxdt=atx+bt−b2t−4atct2atx+bt+b2t−4atct2at.
Denote(22)λ1t=−bt+b2t−4atct2at,λ2t=−bt−b2t−4atct2at.
It follows fromH1 that(23)λ1L≤λ1t≤λ1M<0<λ2L≤λ2t≤λ2M.
It follows from (22) and Equation (21) that(24)dxdt=atx−λ1tx−λ2t.
Now, we divide the proof into three steps.(1)
We prove the existence of the periodic solutionγ1t of Equation (15).Assume that(25)S=ϕt∈Cℝ,ℝϕt+ω=ϕt.∀ϕt,φt∈S, then the distance is defined as follows:(26)ρϕ,φ=supt∈0,ωϕt−φt.Thus,S,ρ is a complete metric space.Take a convex closed set ofS as follows:(27)B1=ϕt∈Sλ1L≤ϕt≤λ1M,modϕ⊆moda,b,c.∀ϕt∈B1, then consider the following equation:(28)dxdt=atx−λ1tϕt−λ2t=atϕt−λ2tx−atϕt−λ2tλ1t.It follows fromat<0 (23) and (27) that(29)0<aMλ1M−λ2L≤atϕt−λ2t≤aLλ1L−λ2M.Hence, we have(30)∫0ωatϕt−λ2tdt>0.Sinceat,ϕt,λ1t, and λ2t are ω-periodic continuous functions, it follows that(31)atϕt−λ2t,atϕt−λ2tλ1t.areω-periodic continuous functions, by (30), according to Lemma 2, (24) has a unique ω-periodic continuous solution as follows:(32)ηt=∫t+∞e∫staθϕθ−λ2θdθasϕs−λ2sλ1sds.and(33)modη⊆modatϕt−λ2t,atϕt−λ2tλ1t.It follows from (22) and (27) that(34)modatϕt−λ2t⊆moda,b,c,modatϕt−λ2tλ1t⊆moda,b,c.Hence, we have(35)modη⊆moda,b,c.It follows from (27), (29), and (32) that(36)ηt≥λ1L∫t+∞e∫staθϕθ−λ2θdθasϕs−λ2sds=−λ1L∫t+∞e∫staθϕθ−λ2θdθd∫staθϕθ−λ2θdθ=−λ1Le∫staθϕθ−λ2θdθt+∞=−λ1Le∫+∞taθϕθ−λ2θdθ−1≥−λ1Le∫+∞taLλ1L−λ2Mdθ−1=λ1L,ηt≤λ1M∫t+∞e∫staθϕθ−λ2θdθasϕs−λ2sds=−λ1M∫t+∞e∫staθϕθ−λ2θdθd∫staθϕθ−λ2θdθ=−λ1Me∫staθϕθ−λ2θdθt+∞=−λ1Me∫+∞taθϕθ−λ2θdθ−1≤−λ1Me∫+∞taMλ1M−λ2Ldθ−1=λ1M.Hence,ηt∈B1.We define a mapping as follows:(37)Tϕt=∫t+∞e∫staθϕθ−λ2θdθasϕs−λ2sλ1sds.Thus, if∀ϕt∈B1, then Tϕt∈B1; hence, T:B1⟶B1.Now, we prove that the mappingT is a compact mapping.Consider any sequenceϕnt⊆B1n=1,2,…, then it is given as(38)λ1L≤ϕnt≤λ1M,modϕn⊆moda,b,c.n=1,2,….On the other hand,Tϕnt=xϕnt satisfies(39)dxϕntdt=atϕnt−λ2txϕnt−atϕnt−λ2tλ1t.Thus, we have(40)dxϕntdt≤2aLλ2M−λ1Lλ1L,modxϕnt⊆moda,b,c.Hence,dxϕnt/dt is uniformly bounded; therefore, xϕnt is uniformly bounded and equicontinuous on ℝ. By the theorem of Ascoli–Arzela, for any sequence xϕnt⊆B1, there exists a subsequence (also denoted by xϕnt) such that xϕnt which is convergent uniformly on any compact set of ℝ. From (40), combined with Lemma 3, xϕnt is convergent uniformly on ℝ; that is, T is relatively compact on B1.Next, we prove thatT is a continuous mapping.Assumeϕnt⊆B1,ϕt∈B1, and then(41)ϕnt⟶ϕt.n⟶∞.It follows from (37) that(42)Tϕnt−Tϕt=∫t+∞e∫staθϕnθ−λ2θdθasϕns−λ2sλ1sds−∫t+∞e∫staθϕθ−λ2θdθasϕs−λ2sλ1sds=∫t+∞e∫staθϕnθ−λ2θdθasϕns−ϕsλ1sds+∫t+∞e∫staθϕnθ−λ2θdθ−e∫staθϕθ−λ2θdθasϕs−λ2sλ1sds=∫t+∞e∫staθϕnθ−λ2θdθasϕns−ϕsλ1sds+∫t+∞eξ∫staθϕnθ−ϕθdθasϕs−λ2sλ1sds.Here,ξ is between ∫staθϕnθ−λ2θdθ and ∫staθϕθ−λ2θdθ; thus, ξ is between aLλ1L−λ2Mt−s and aMλ1M−λ2Lt−s; hence, we have(43)Tϕnt−Tϕt≤∫t+∞eaMλ1M−λ2Lt−saMλ1Mds+∫t+∞eaMλ1M−λ2Lt−ss−taM2λ2M−λ1Lλ1Mdsρϕn,ϕ=aMλ1MaMλ1M−λ2L+aM2λ2M−λ1Lλ1MaMλ1M−λ2L2ρϕn,ϕ=aLλ1LaMλ1M−λ2L−aL2λ2M−λ1Lλ1LaMλ1M−λ2L2ρϕn,ϕ.By (41) and the above inequality, we get(44)Tϕnt⟶Tϕt,n⟶∞.Therefore,T is continuous. From equation (37), T∂B1⊆B1. According to Lemma 4, T has at least a fixed point on B1, and the fixed point is the ω-periodic continuous solution γ1t of Equation (15), and then we get(45)λ1L≤γ1t≤λ1M<0.(2)
We prove the existence of the periodic solutionγ2t of Equation (15).Assume that(46)S=ϕt∈Cℝ,ℝ|ϕt+ω=ϕt.When∀ϕt,φt∈S, the distance is defined as follows:(47)ρϕ,φ=supt∈0,ω|ϕt−φt|Thus,S,ρ is a complete metric space.Take a convex closed set ofS as follows:(48)B2=ϕt∈S|λ2L≤ϕt≤λ2M,modϕ⊆moda,b,c.∀ϕt∈B2, then consider the following equation:(49)dxdt=atϕt−λ1tx−λ2t=atϕt−λ1tx−atϕt−λ1tλ2t.It follows fromat<0 and (23) and (51) that(50)aLλ2M−λ1L≤atϕt−λ1t≤aMλ2L−λ1M<0.Hence, we have(51)∫0ωatϕt−λ1tdt<0.Sinceat,ϕt,λ1t, and λ2t are ω-periodic continuous functions, we get(52)atϕt−λ1t,atϕt−λ1tλ2t.From (54), according to Lemma 2, we get unique ω-periodic continuous solution as follows:(53)ηt=−∫−∞te∫staθϕθ−λ1θdθasϕs−λ1sλ2sds.and(54)modη⊆modatϕt−λ1t,atϕt−λ1tλ2t.It follows from (22) and (51) that(55)modatϕt−λ1t⊆moda,b,c,modatϕt−λ1tλ2t⊆moda,b,c.Hence, we have(56)modη⊆moda,b,c.It follows from (51), (53), and (56) that(57)ηt≥−λ2L∫−∞te∫staθϕθ−λ1θdθasϕs−λ1sds=λ2L∫−∞te∫staθϕθ−λ1θdθd∫staθϕθ−λ1θdθ=λ2Le∫staθϕθ−λ1θdθ−∞t=λ2L1−e∫−∞taθϕθ−λ1θdθ≥λ2L1−e∫−∞taMλ2L−λ1Mdθ=λ2L,ηt≤−λ2M∫−∞te∫staθϕθ−λ1θdθasϕs−λ1sds,=λ2M∫−∞te∫staθϕθ−λ1θdθd∫staθϕθ−λ1θdθ=λ2Me∫staθϕθ−λ1θdθ−∞t=λ2M1−e∫−∞taθϕθ−λ1θdθ≤λ2M1−e∫−∞taLλ2M−λ1Ldθ=λ2M.Hence,ηt∈B2.We define a mapping as follows:(58)Tϕt=−∫−∞te∫staθϕθ−λ1θdθasϕs−λ1sλ2sds.Thus, if∀ϕt∈B2, then Tϕt∈B2; hence, T:B2⟶B2.Now, we prove that the mappingT is a compact mapping.Consider any sequenceϕnt⊆B2n=1,2,…, then it follows that(59)λ2L≤ϕnt≤λ2M,modϕn⊆moda,b,c.n=1,2,….On the other hand,Tϕnt=xϕnt satisfies(60)dxϕntdt=atϕnt−λ1txϕnt−atϕnt−λ1tλ2t.Thus, we have(61)dxϕntdt≤−2aLλ2M−λ1Lλ2M,modxϕnt⊆moda,b,c.Hence,ℝ e dxϕnt/dt is uniformly bounded; therefore, xϕnt is uniformly bounded and equicontinuous on ℝ. By the theorem of Ascoli–Arzela, for any sequence, xϕnt⊆B2, then there exists a subsequence (also denoted by xϕnt) such that xϕnt is convergent uniformly on any compact set . From (40), combined with Lemma 3, xϕnt is convergent uniformly on ℝ; that is, T is relatively compact on B2.Next, we prove thatT is a continuous mapping.Assume thatϕnt⊆B2,ϕt∈B2, and then(62)ϕnt⟶ϕt.n⟶∞.It follows from (61) that(63)Tϕnt−Tϕt=∫−∞te∫staθϕnθ−λ1θdθasϕns−λ1sλ2sds−∫−∞te∫staθϕθ−λ1θdθasϕs−λ1sλ2sds=∫−∞te∫staθϕnθ−λ1θdθasϕns−ϕsλ2sds+∫−∞te∫staθϕnθ−λ1θdθ−e∫staθϕθ−λ1θdθasϕs−λ1sλ2sds=∫−∞te∫staθϕnθ−λ1θdθasϕns−ϕsλ2sds+∫−∞teξ∫staθϕnθ−ϕθdθasϕs−λ1sλ2sds,Here,ξ is between ∫staθϕnθ−λ1θdθ and ∫staθϕθ−λ1θdθ; thus, ξ is between aLλ2M−λ1Lt−s and aMλ2L−λ1Mt−s; hence, we have(64)Tϕnt−Tϕnt≤∫−∞teaMλ2L−λ1Mt−s|a|M|λ2|Mds+∫−∞teaMλ2L−λ1Mt−st−s|a|M2λ2M−λ1L|λ2|Mdsρϕn,ϕ=|a|M|λ2|M−aMλ2L−λ1M+|a|M2λ2M−λ1L|λ2|MaMλ2L−λ1M2ρϕn,ϕ=aLλ2MaMλ2L−λ1M+aL2λ2M−λ1Lλ2MaMλ2L−λ1M2ρϕn,ϕ.From (65) and the above inequality, we get(65)Tϕnt⟶Tϕt,n⟶∞.Therefore,T is continuous. From Equation (61), T∂B2⊆B2. According to Lemma 4, T has at least a fixed point on B2, then the fixed point is the ω-periodic continuous solution γ2t of Equation (15), and then(66)0<λ2L≤γ2t≤λ2M.(3)
We prove Equation (15) has exactly two periodic solutions γ1t and γ2t.Let(67)ft,x=atx2+btx+ct,then(68)fxx″t,x=2at<0.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: γ1t,γ2t; thus, it follows that Equation (15) has exactly two periodic solutions γ1t,γ2t, and then(69)λ1L≤γ1t≤λ1M<0,0<λ2L≤γ2t≤λ2M.
This is the end of the proof of Theorem2.


## 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),at,bt,ct, and dt are ω-periodic continuous functions on ℝ, at and bt have continuous derivatives on ℝ, then assume that the following conditions hold true:(70)H1bt≠0,H2atb′tbt+atdt−btct−a′t<0.Then, Equation (6) has exponential dichotomy.Proof.
From Equation (6), the following equation can be obtained:(71)d2xdt2−at+dt+b′tbtdxdt+atb′tbt+atdt−btct−a′ttx=0,where dx/dt=atx+bty.
Let(72)x=eu,then Equation (72) becomes(73)d2udt2+dudt2−at+dt+b′tbtdudt+atb′tbt+atdt−btct−a′t=0.
Let(74)z=dudt,then Equation (73) becomes(75)dzdt=−z2+at+dt+b′tbtz−atbtbt+atdt−btct−a′t.
This is Riccati’s equation. FromH1,H2, Equation (75) satisfies all the conditions of Theorem 2. According to Theorem 2, Equation (75) has exactly one negative, one positive, and two ω-periodic continuous solutions γ1t<0 and γ2t>0.
It follows from Equation (74) that Equation (73) has two continuous solutions:(76)u1t=u10+∫0tγ1sds,u2t=u20+∫0tγ2sds.
Here,u10≠u20.
It follows from Equation (72) that Equation (71) has two continuous solutions:(77)Φ1t=eu1t=eu10+∫0tγ1sds=eu10e∫0tγ1sds1=k1e∫0tγ1sds,Φ2t=eu2t=eu20+∫0tγ2sds=eu20e∫0tγ2sds=k2e∫0tγ2sds.
Here,k1=eu10,k2=eu20.
Morveover, we get(78)Φ2tΦ1t=eu20e∫0tγ2sdseu10e∫0tγ1sds=e(u20−u10e∫0tγ2s−γ1sds≠C.
Thus,Φ1t,Φ2t are linearly independent.
Since(79)dxdt=atx+bty,
then Equation (6) has two sets of linearly independent solutions as follows:(80)xy=k1e∫0tγ1θdθk1γ1t−ate∫0tγ1θdθbtk2e∫0tγ2θdθk2γ2t−ate∫0tγ2θdθbt.
Taking the fundamental solution matrixXt of Equation (6), we get as follows:(81)Xt=k1e∫0tγ1θdθk2e∫0tγ2θdθk1γ1t−ate∫0tγ1θdθbtk2γ2t−ate∫0tγ2θdθbt.
Thus, we have(82)X−1s=γ2s−ase−∫0sγ1θdθ/k1−bse−∫0sγ1θdθ/k1as−γ1se−∫0sγ2θdθ/k2bse−∫0sγ2θdθ/k2γ2s−γ1s.
Take projectionP, we get as follows:(83)P=1000,
and then it follows(84)XtPX−1s=γ2s−as−bsγ1t−atγ2s−as/btat−γ1tbs/bte∫stγ1θdθγ2s−γ1s,XtI−PX−1s=as−γ1sbsγ2t−atas−γ1s/btγ2t−atbs/bte∫stγ2θdθγ2s−γ1s.
Sinceat,bt,γ1t,γ2t are ω-periodic continuous functions on ℝ, it follows that there exist K≥1,α>0 such that the following inequalities hold true:(85)XtPX−1s≤Ke−αt−s,t≥s,XtI−PX−1s≤Keαt−s,s≥t.
Thus, Equation (6) has exponential dichotomy, and the dichotomy constants are K,α.
This is the end of the proof of Theorem3.
If we turn Equation (6) into a second-order linear differential equation about y, we can get the following:Theorem 4.
Consider Equation (6), and at,bt,ct, and dt are ω-periodic continuous functions on ℝ, and ct and dt have continuous derivatives on ℝ, then assume that the following conditions hold true:(86)H1ct≠0,H2dtc′tct+atdt−btct−d′t<0,
then Equation (6) has exponential dichotomy.Proof.
From Equation (6), the following equation can be obtained:(87)d2ydt2−at+dt+c′tctdydt+dtc′tct+atdt−btct−d′ty=0,where(88)dydt=ctx+dty,
Let(89)y=eu,
then Equation (89) becomes(90)d2udt2+dudt2−at+dt+c′tctdudt+dtc′tct+atdt−btct−d′t=0.
Let(91)z=dudt,
then Equation (87) becomes(92)dzdt=−z2+at+dt+c′tctz−dtctct+atdt−btct−d′t,
and this is Riccati’s equation. FromH1,H2, Equation (92) satisfies all the conditions of Theorem 2. According to Theorem 2, Equation (92) has exactly one negative, one positive, and two ω-periodic continuous solutions γ1t<0 and γ2t>0.
It follows from Equation (92) two continuous solutions as follows:(93)u1t=u10+∫0tγ1sds,
and(94)u2t=u20+∫0tγ2sds.
Here,u10≠u20.
It follows from Equation (89) that Equation (87) has two continuous solutions as follows:(95)Φ1t=eu1t=eu10e∫0tγ1sds=k1e∫0tγ1sds,
and(96)Φ2t=eμ2t=eu20e∫0tγ2sds=k2e∫0tγ2sds.
Here,k1=eu10,k2=eu20.
Moreover,(97)Φ2tΦ1t=k2e∫0tγ2sdsk1e∫0tγ1sds=k2k1e∫0tγ2s−γ1sds≠C.
Thus,Φ1t,Φ2t are linearly independent.
Since(98)dydt=ctx+dty,
Then Equation (6) has two sets of linearly independent solutions given as follows:(99)xy=k1γ1t−dte∫0tγ1θdθctk1e∫0tγ1θdθk2γ2t−dte∫0tγ2θdθctk2e∫0tγ2θdθ.
Take the fundamental solution matrixYt of Equation (6) as follows:(100)Yt=k1γ1t−dte∫0tγ1θdθctk2γ2t−dte∫0tγ2θdθctk1e∫0tγ1θdθk2e∫0tγ2θdθ.
Thus, we have(101)Y−1s=cse−∫0sγ1θdθ/k1ds−γ2se−∫0sγ1θdθ/k1−cse−∫0sγ2θdθ/k2γ1s−dse−∫0sγ2θdθ/k2γ1s−γ2s.
Take projectionP as(102)P=1000,
Then it follows(103)YtPY−1s=γ1t−dtcs/ctγ1t−dtds−γ2s/ctcsds−γ2se∫stγ1θdθγ1s−γ2s,YtI−PY−1s=d(t)−γ2tcs/c(t)γ2t−dtγ1s−ds/ct−csγ1s−dse∫stγ2θdθγ1s−γ2s.
Sinceat,bt,γ1t,γ2t are ω-periodic continuous functions on ℝ, it follows that there exist L≥1,β>0 such that the following inequalities hold true:(104)YtPY−1s≤Le−βt−s,t≥s,YtI−PY−1s≤Leβt−s,s≥t.
Thus, Equation (6) has exponential dichotomy, and the dichotomy constants are L,β.
This is the end of the proof of Theorem4.
Next, we will calculate the characteristic exponents of system (6). We first give the definition of the characteristic exponent of the linear system.Definition 1.
(see[17]). Consider the following linear differential system:(105)dxdt=Atx,where At is a square matrix of order n, and At continuous on t0,+∞,x=xt is a nonzero solution of (105). Define(106)χxt=limt⟶+∞¯1tlnxt,where χxt is called the characteristic exponent of x=xt.
According the proof of Theorem4, we get as follows.Theorem 5.
Consider Equation (6), and at,bt,ct and dt are ω-periodic continuous functions on ℝ, and ct and dt have continuous derivatives on ℝ, assume that the following conditions hold true:(107)H1ct≠0,H2dtc′tct+atdt−btct−d′t<0.Then the characteristic exponents of system Equation (6) must be positive and negative.Proof.
According to the proof of Theorem4, by (95)–(97), and (99), Equation (6) has two linearly independent solutions as follows:(108)x1y1=k1γ1t−dte∫0tγ1θdθctk1e∫0tγ1θdθ,x2y2=k2γ2t−dte∫0tγ2θdθctk2e∫0tγ2θdθ,where γ1t<0,γ2t>0. Thus, we have(109)χx1y1=limt⟶+∞¯1tlnk1γ1t−dte∫0tγ1(θ)dθctk1e∫0tγ1(θ)dθ=limt⟶+∞¯1tlnk1γ1(t)−dte∫0tγ1θdθct2+k1e∫0tγ1θdθ212=12limt⟶+∞¯1tlnk12γ1t−dt2+c2tc2te2∫0tγ1θdθ=12limt⟶+∞¯1tlnk12γ1t−dt2+c2tc2t+2∫0tγ1θdθ=γ1M<0.
Similarly, we have(110)χx2y2=limt⟶+∞1tlnk2γ2t−dte∫0tγ2θdθctk2e∫0tγ2θdθ=γ2M>0.
Thus, the characteristic exponents of system (6) must be positive and negative.
This is the end of the proof of Theorem5.
Similar as Theorem5, we get as follows.Theorem 6.
Consider Equation (6), and at,bt,ct, and dt are ω-periodic continuous functions on ℝ, and ct and dt have continuous derivatives on ℝ, assume that the following conditions hold true:(111)H1bt≠0,H2atb′tbt+atdt−btct−a′t<0,
then the characteristic exponents of system (1.5) must be positive and negative.Remark 4.
Comparing Theorem5 of this paper with Theorem 1 of (88), it can be found that the conditions of Theorem 5 of this paper is weaker than those of Theorem 1 of (88). It can be seen that Theorem 5 of this paper is the generalization of Theorem 1 of (88).


## 5. Existence of a Unique Periodic Solution

Consider the following equation(112)xy′=atbtctdtxy+etft.By Theorem3 and Theorem 4, according to Lemma 1, we can easily get as follows.Theorem 7.
Consider Equation (114), and at,bt,ct,dt,et and ft are ω-periodic continuous functions on ℝ, and at and bt have continuous derivatives on ℝ, assume that the following conditions hold true:(113)H1bt≠0,H2atb′tbt+atdt−btct−a′t<0,
then Equation (114) has a unique ω-periodic continuous solution.Theorem 8.
Consider Equation (114), and at,bt,ct,dt,et and ft are ω-periodic continuous functions on ℝ, and ct and dt have continuous derivatives on ℝ, assume that the following conditions hold true:(114)H1ct≠0,H2dtc′tct+atdt−btct−d′t<0,
Then Equation (114) has a unique ω-periodic continuous solution.In Equation (114), if at≡0,bt≡1,et≡0,ct=−qt,dt=−pt, then (112) becomes the following equation:(115)xy′=01−qt−ptxy+0ft.Setx′=y, then (115) is equivalent to the following equation:(116)x″+ptx′+qtx=ft.So we can easily get what as follows.Theorem 9.
Consider Equation (116), and pt,qt, and ft are ω-periodic continuous functions on ℝ, and assume that the following condition holds true:(117)H1qt<0,
Then Equation (116) has a unique ω-periodic continuous solution.Further, ifpt≡0, then we can obtain what as follows.Theorem 10.
Consider the following Hill’s equation(118)d2xdt2+qtx=rt.qt,rt are both ω-periodic continuous functions on ℝ, and assume that the following condition holds true:(119)H1qt<0,Then equation (117) has a unique ω-periodic continuous solution.


## 6. Conclusions

In this paper, the exponential dichotomy of two-dimensional linear system is studied. To verify whether the system has exponential dichotomy, the key is to find out the fundamental solution matrix of the system, but it is not easy to find the fundamental solution matrix of the system. In this paper, a new method for judging the exponential dichotomy of two-dimensional linear system is provided; that is, the two-dimensional linear system is transformed into a second-order linear system by using variable substitution; Then, the second-order linear system is transformed into Riccati’s equation, and the existence of two linearly independent periodic solutions of Riccati’s equation is obtained by using the fixed point theorem; thus, the fundamental solution matrix of the two-dimensional linear system is obtained. Under certain conditions, the two-dimensional linear system has exponential dichotomy.This method to verify the exponential dichotomy of two-dimensional linear system in this paper is new and feasible. It will have certain application value.



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