# Existence and Uniqueness of Solution for Perturbed Nonautonomous Systems with Nonuniform Exponential Dichotomy

Authors: Yong-Hui Xia; Xiaoqing Yuan; Kit Ian Kou; Patricia J. Y. Wong
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Nonuniform exponential dichotomy has been investigated extensively. The essential condition of these previous results is based on the assumption that the nonlinear term satisfies|f(t,x)|≤μe−ε|t|. However, this condition is very restricted. There are few functions satisfying |f(t,x)|≤μe−ε|t|. In some sense, this assumption is not reasonable enough. More suitable assumption should be |f(t,x)|≤μ. To the best of the authors' knowledge, there is no paper considering the existence and uniqueness of solution to the perturbed nonautonomous system with a relatively conservative assumption |f(t,x)|≤μ. In this paper, we prove that if the nonlinear term is bounded, the perturbed nonautonomous system with nonuniform exponential dichotomy has a unique solution. The technique employed to prove Theorem 4 is the highlight of this paper.


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## Body

## 1. Introduction

The notion of exponential dichotomy, introduced by Perron in [1], plays an important role in the theory of differential equations and dynamical systems (also see [2–5]). It is well known that if linear system x˙(t)=A(t)x(t) admits an (uniform) exponential dichotomy, the nonlinear term f(t,x) is bounded and has a small Lipschitz constant, then the nonlinear system x˙(t)=A(t)x(t)+f(t,x) has a unique bounded solution (see [6]). However, many scholars argued that (uniform) exponential dichotomy restricted the behavior of dynamical systems. For this reason, we need a more general concept of hyperbolicity. Recently, Barreira and Valls [7, 8] have introduced the notion of nonuniform exponential dichotomy. General nonuniform exponential dichotomy has also been proposed (see [9–11]). Many properties of nonuniform exponential dichotomy have been extensively studied. For example, the topological conjugacies between linear and nonlinear perturbations were explored and some new Grobman-Hartman type theorems for nonuniform exponential dichotomy were established ([12, 13]). However, the essential condition of these results is based on the assumption that the nonlinear term satisfies |f(t,x)|≤μe-ε|t|. Under the same condition, Zhang et al. studied nonlinear perturbations of nonuniform exponential dichotomy on measure chains ([14]).However, the condition|f(t,x)|≤μe-ε|t| is very restricted. There are few functions satisfying |f(t,x)|≤μe-ε|t|. Thus, it is necessary to find a more conservative condition for the nonlinear term f(t,x). In this paper, our main objective is to explore the existence and uniqueness of solution to the perturbed nonautonomous system with a relatively conservative assumption |f(t,x)|≤μ. Finally, we prove that if |f(t,x)|≤μ, the perturbed nonautonomous system with nonuniform exponential dichotomy has a unique solution x(t) satisfying |x(t)|=O(eε|t|).The outline of this paper is arranged as follows. Next section is to state our main results. In Section3, we prove the main results.


## 2. Main Results

In this section, we will state our main theorems. First, we introduce the definition of nonuniform exponential dichotomy.Consider systems(1)x˙(t)=A(t)x(t),(2)x˙(t)=A(t)x(t)+f(t,x),
where t∈ℝ, x∈ℝn, A(t) is a continuous matrix function, and f:ℝ×ℝn→ℝn is a continuous function.LetTA(t,s) be the evolution operator satisfying x(t)=TA(t,s)x(s), t,s∈ℝ, where x(t) is a solution of (1).Definition 1.
Linear system (1) is said to admit a nonuniform exponential dichotomy if there exists a projection P(t)(P2=P) and constants α>0,K>0,ε≥0, such that
(3)|TA(t,s)P(s)|≤Ke-α(t-s)·eε|s|,t≥s,|TA(t,s)Q(s)|≤Keα(t-s)·eε|s|,t≤s,
where P(t)+Q(t)=Id(identity),TA(t,s)P(s)=P(t)TA(t,s),t,s∈ℝ.Remark 2.
Whenε≡0, system (1) is said to have an exponential dichotomy; and when ε≡0, α≡0, system (1) is said to have a uniform dichotomy.To present our main results, we give a theorem under the trivial condition|f(t,x)|≤μe-ε|t|.Theorem 3.
Suppose that linear system (1) admits a nonuniform exponential dichotomy. For t∈ℝ,x,x1,x2∈ℝn, if the nonlinear term f(t,x) satisfies (H~1)
|f(t,x)|≤μe-ε|t|,(H~2)
|f(t,x1)-f(t,x2)|≤re-ε|t||x1-x2|,(H~3)
4Kr<α,where μ,r,ε,α are all positive constants, then nonlinear system (2) has a unique bounded solution x~(t) satisfying
(4)x~(t)=∫-∞t‍TA(t,s)P(s)f(s,x~(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x~(s))ds.Discussion. One of the essential conditions of Theorem 3 is (H~1). However, this condition is very restricted. There are few functions satisfying |f(t,x)|≤μe-ε|t|. Thus, it is necessary to find a more conservative condition for the nonlinear term f(t,x). The main objective of this paper is to prove that the perturbed system has a unique solution under |f(t,x)|≤μ. But Theorem 3 cannot be valid yet. For this case, we have the following.Theorem 4.
Suppose that linear system (1) admits a nonuniform exponential dichotomy with the estimates (3). For t∈ℝ,x,x1,x2∈ℝn, if f(t,x) satisfies (H1)
|f(t,x)|≤μ,(H2)
|f(t,x1)-f(t,x2)|≤re-ε|t||x1-x2|,(H3)
4Kr<α-ε,where α-ε is a positive constant, then system (2) has a unique solution x(t) satisfying
(5)|x(t)|=O(eε|t|).Remark 5.
The method used to prove Theorem3 cannot be applied to this case. To see how to overcome the difficulty, one can refer to the main proof of Theorem 4. The technique employed to prove Theorem 4 is very skillful and interesting, which is the highlight of this paper.


## 3. Proofs of Main Results

In what follows, to prove Theorem3, a preliminary lemma is needed.Lemma 6 (see [15], Lemma 4).
If system (1) admits a nonuniform exponential dichotomy, then system (1) has no nontrivial bounded solutions; that is, x(t)=0 is the unique bounded solution of (1).


### 3.1. Proof of Theorem3

LetB={φ(t)∣φ(t) be continuous and |φ(t)|≤2Kμα-1}, for ∀φ∈B, define a mapping 𝒯1:
(6)𝒯1φ(t)=∫-∞t‍TA(t,s)P(s)f(s,φ(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ(s))ds.
From (3) and (H~1)and(H~2), we have
(7)|𝒯1φ(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·μe-ε|s|ds+∫t+∞‍Keα(t-s)eε|s|·μe-ε|s|ds=Kμα-1+Kμα-1=2Kμα-1.
Therefore, 𝒯1φ(t)∈B, which implies 𝒯1 maps B onto itself. On the other hand,
(8)|𝒯1φ1(t)-𝒯1φ2(t)|≤∫-∞t‍TA(t,s)P(s)re-ε|s||φ1(s)-φ2(s)|ds+∫t+∞‍TA(t,s)Q(s)re-ε|s||φ1(s)-φ2(s)|ds≤∫-∞t‍Kre-α(t-s)|φ1(s)-φ2(s)|ds+∫t+∞‍Kreα(t-s)|φ1(s)-φ2(s)|ds≤2Krα-1sup⁡s≥0|φ1(s)-φ2(s)|≤12sup⁡s≥0|φ1(s)-φ2(s)|.
Then 𝒯1 is a contraction mapping. Therefore, in B, there exists a unique fixed point φ0(t), such that
(9)φ0(t)=𝒯1φ0(t)=∫-∞t‍TA(t,s)P(s)f(s,φ0(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ0(s))ds.
Differentiating the above equality, we see that φ0(t) satisfies (2). Now we are going to show that the solution of system (2) satisfying (H~1),(H~2), and(H~3) is unique. Assume that system (2) has another bounded solution φ*(t) satisfying (H~1),(H~2), and(H~3); we have
(10)φ*(t)=TA(t,0)φ*(0)+∫0t‍TA(t,s)T-1(s,s)f(s,φ*(s))ds=TA(t,0)φ*(0)+∫0t‍TA(t,s)P(s)f(s,φ*(s))ds+∫0t‍TA(t,s)Q(s)f(s,φ*(s))ds=TA(t,0)φ*(0)+∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫-∞0‍TA(t,s)P(s)f(s,φ*(s))ds+∫0+∞‍TA(t,s)Q(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]φ*(0)-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))ds-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds).
By calculating, we get
(11)∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds≤2Kμα-1.
As φ*(t) is bounded, we obtain
(12)TA(t,0)[φ*(0)-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))dsvvvvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)],
is bounded. In addition, the formula above is the solution of system (1), so it is a bounded solution. From Lemma 6, we have
(13)TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]φ*(0)vvvvvvv-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))dsvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]=0.
Therefore,
(14)φ*(t)=∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds.
From (3), (H~2), and (H~3), we have
(15)|φ0(t)-φ*(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·re-ε|s||φ0(s)-φ*(s)|ds+∫t+∞‍Keα(t-s)eε|s|·re-ε|s||φ0(s)-φ*(s)|ds=2Krα-1|φ0(s)-φ*(s)|=12sup⁡t∈ℝ|φ0(t)-φ*(t)|.
That is, sup⁡t∈ℝ|φ0(t)-φ*(t)|≤(1/2)sup⁡t∈ℝ|φ0(t)-φ*(t)|, which implies φ0(t)=φ*(t). Then the uniqueness is proved. The proof of Theorem 3 is complete.


### 3.2. Proof of Theorem4

To prove Theorem4, a standard method is to employ a linear transformation x=eε|t|y. However, x=eε|t|y is not differentiable at t=0. Thus, we cannot use such transformation directly. We have to discuss by dividing into two pieces t≥0 and t≤0.Consider system(16)x˙(t)=B(t)v(t)+F(t,v),
where u∈ℝn, B(t) is a continuous matrix function, and F:ℝ×ℝn→ℝn is a continuous function.Lemma 7.
Suppose that systemv˙(t)=B(t)v(t) admits a nonuniform exponential dichotomy; that is, its evolution operator TB(t,s) satisfies
(17)|TB(t,s)P(s)|≤Ke-λ(t-s)·eε|s|,t≥s,|TB(t,s)Q(s)|≤Keλ(t-s)·eε|s|,t≤s,
where λ is a positive constant. In addition,
(18)|F(t,v1)-F(t,v2)|≤re-ε|t||v1-v2|,4Kr<λ.
If |F(t,v)|≤μe-ε|t| for t≥0, then for any a∈ℝn, system (16) has a unique solution v+(t) satisfying the following:(i)
|v+(t)|<+∞, for t≥0;(ii)
P(0)v+(0)=P(0)a;(iii)
inℝ+,v+(t) satisfies integral equation
(19)v+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,v+(s))ds.
If|F(t,v)|≤μe-ε|t| for t≤0, then for any a∈ℝn, system (16) has a unique solution v-(t) satisfying the following:(i)
|v-(t)|<+∞, for t≤0;(ii)
Q(0)v-(0)=Q(0)a;(iii)
inℝ-,v-(t) satisfies integral equation
(20)v-(t)=TB(t,0)Q(0)a+∫-∞t‍TB(t,s)P(s)F(s,v-(s))ds-∫t0‍TB(t,s)Q(s)F(s,v-(s))ds.Proof.
We prove the existence ofv+(t) by successive approximation method. For any a∈ℝn, let v0+(t)=TB(t,0)P(0)a. We define vm+(t),vm+1+(t) recursively as follows:
(21)vm+1+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,vm+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,vm+(s))ds.
From (17) and |F(t,v)|≤μe-ε|t|, for t≥0, we have
(22)|vm+1+(t)|≤Ke-λt|a|+∫0t‍Ke-λ(t-s)eεs·μe-εsds+∫t+∞‍Keλ(t-s)eεs·μe-εsds=Ke-λt|a|+Kμλ-1(1-e-λt)+Kμλ-1≤K|a|+2Kμλ-1.
For any bounded function v(t) defined on ℝ+, denote ∥v∥=sup⁡t∈ℝ|v(t)|; then it follows from (18) and (21) that
(23)|vm+1+(t)-vm+(t)|≤∫0t‍Ke-λ(t-s)eεs·re-εs|vm+(s)-vm-1+(s)|ds+∫t+∞‍Keλ(t-s)eεs·re-εs|vm+(s)-vm-1+(s)|ds≤2Krλ-1|vm+(s)-vm-1+(s)|≤2Krλ-1∥vm+-vm-1+∥≤12∥vm+-vm-1+∥,
which implies that ∥vm+1+-vm+∥≤(1/2)∥vm+-vm-1+∥. Hence, the series ∑m=0∞‍(vm+1+(t)-vm+(t)) converges uniformly on ℝ+. It means that the series {vm+(t)} converges uniformly to a limit v+(t) on ℝ+.
From (21), for any fixed t, let m→∞, we have
(24)v+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,v+(s))ds.
Differentiating the above equality, we see that v+(t) satisfies the system (16).
From (22) and (24), we know that |v+(t)|<∞ for t≥0, and it is easy to demonstrate that P(0)v+(0)=P(0)a. Now we are going to show the uniqueness of v+(t). If there is another bounded function v~+(t) satisfying (i), (ii), and (iii) on ℝ+, in view of (iii) and (18) we have
(25)|v~+(t)-v+(t)|≤∫0t‍Ke-λ(t-s)eεs·re-εs|v~+(s)-v+(s)|ds+∫t+∞‍Keλ(t-s)eεs·re-εs|v~+(s)-v+(s)|ds≤2Krλ-1∥v~+-v+∥≤12∥v~+-v+∥,
which implies ∥v~+-v+∥≤(1/2)∥v~+-v+∥. Hence, v~+(t)≡v+(t).
The proof of the existence and uniqueness ofv-(t) is similar to that of v+(t). The proof of Lemma 7 is complete.Lemma 8.
Suppose thatα>0,δ>0,C,L, and M are nonnegative constants and that v(t) is a nonnegative bounded continuous function which satisfies two of the following inequalities:
(26)v(t)≤Ce-αt+L∫0t‍e-α(t-s)v(s)ds+M∫t+∞‍eδ(t-s)v(s)ds,(t≥0),v(t)≤Ceαt+L∫t0‍eα(t-s)v(s)ds+M∫-∞t‍e-δ(t-s)v(s)ds,(t≤0).
In addition, if γ=L/α+M/δ<1, then for t≥0 or t≤0, one has
(27)v(t)≤(1-γ)-1Ce-[α-(1-γ)-1L]|t|.Proof.
The proof is straightforward by Lemma 6.2 of Chapter 3 in [6].Lemma 9.
For anya∈ℝn, system (2) has a unique solution x+(t) with the following properties:(i)
|x+(t)e-εt|<+∞, for t≥0;(ii)
P(0)x+(0)=P(0)a;(iii)
x+(t) on ℝ+ satisfies integral equation
(28)x+(t)=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,x+(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x+(s))ds.Similarly, for any a∈ℝn, system (2) also has a unique solution x-(t) with the following properties:(i)
|x-(t)eεt|<+∞, for t≤0;(ii)
Q(0)x-(0)=Q(0)a;(iii)
x-(t) on ℝ- satisfies integral equation
(29)x-(t)=TA(t,0)Q(0)a+∫-∞t‍TA(t,s)P(s)f(s,x-(s))ds-∫t0‍TA(t,s)Q(s)f(s,x-(s))ds.Proof.
We firstly prove the existence and uniqueness ofx+(t). Let v=xe-εt, then system (2) becomes
(30)v˙(t)=(A(t)-εI)v(t)+e-εtf(t,veεt).
Let F(t,v)=e-εtf(t,veεt). From (H1) and (H2), we have
(31)|F(t,v)|≤e-εtμ=μe-ε|t|,fort≥0;|F(t,v1)-F(t,v2)|=e-εt|f(t,v1eεt)-f(t,v2eεt)|≤e-εtre-ε|t||v1eεt-v2eεt|≤re-ε|t||v1-v2|.
LetTC(t,s) be the evolution operator of the linear system v˙(t)=(A(t)-εt)v(t). Since x(t)=TA(t,s)x(s),v=e-εtx, we have TC(t,s)=e-ε(t-s)TA(t,s). Hence, from (3), we obtain
(32)|TC(t,s)P(s)|≤Ke-(α-ε)(t-s)·eε|s|,fort≥s,|TC(t,s)Q(s)|≤Ke(α-ε)(t-s)·eε|s|,fort≤s.
Since 4Kr<α-ε, system (30) satisfies all conditions of Lemma 7. Therefore, for any a∈ℝn, system (30) has a unique solution v+(t) with the following properties:(i)
|v+(t)|<+∞, for t≥0;(ii)
P(0)v+(0)=P(0)a;(iii)
v+(t) on ℝ+ satisfies integral equation
(33)v+(t)=TC(t,0)P(0)a+∫0t‍TC(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TC(t,s)Q(s)F(s,v+(s))ds=e-εtTA(t,0)P(0)a+∫0t‍e-ε(t-s)TA(t,s)P(s)·e-εsf(s,v+(s)eεs)ds-∫t+∞‍e-ε(t-s)TA(t,s)Q(s)vvvvv·e-εsf(s,v+(s)eεs)ds.Hence,
(34)v+(t)eεt=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,v+(s)eεs)ds-∫t+∞‍TA(t,s)Q(s)f(s,v+(s)eεs)ds.
Let x+(t)=v+(t)eεt, we have
(35)x+(t)=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,x+(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x+(s))ds.
Then x+(t) is the solution of system (2) and it satisfies all conditions of Lemma 9.
The proof for the existence and uniqueness ofx-(t) is similar to that of x+(t), so we omit it. This completes the proof of Lemma 9.Lemma 10.
If|TA(t,0)P(0)a|≤Meε|t|, then a=0.Proof.
Ifa≠0, then P(0)a≠0 or Q(0)a≠0. Without loss of generality, we assume P(0)a≠0, then we have
(36)|TA(t,0)P(0)a|=|TA(t,0)P(0)TA-1(s,0)TA(s,0)P(0)a|=|TA(t,0)TA(0,s)P(s)TA(s,0)P(0)a|=|TA(t,s)P(s)TA(s,0)P(0)a|≤|TA(t,s)P(s)|·|TA(s,0)P(0)a|.
Since |TA(t,s)P(s)|≤Ke-α(t-s)eε|s| for t≥s,
(37)|TA(t,0)P(0)a|≤Ke-α(t-s)eε|s|·|TA(s,0)P(0)a|,
which implies
(38)|TA(s,0)P(0)a|≥|TA(t,0)P(0)a|Ke-α(t-s)eε|s|.
Taking t=0, we obtain
(39)|TA(s,0)P(0)a|≥|P(0)a|Keαse-εs=K-1e-(α-ε)s|P(0)a|(s≤0).
Therefore, when s≤0,
(40)|TA(s,0)P(0)a|eε|s|≥K-1e-αs|P(0)a|⟶+∞ass⟶-∞.
On the other hand, when s≤0, from (3), we have
(41)|TA(s,0)Q(0)a|≤Keαseε|0|=Keαs,
hence,
(42)|TA(s,0)Q(0)a|eε|s|≤Ke(α+ε)s.
It follows from (40) and (42) that
(43)|TA(s,0)a|eε|s|=|TA(s,0)(P(s)+Q(s))a|eε|s|≥|TA(s,0)P(s)a|eε|s|-|TA(s,0)Q(s)a|eε|s|≥|TA(s,0)P(s)a|eε|s|-Ke(α+ε)s.
From the above inequality, we know that |TA(s,0)a|/eε|s|→+∞ass→-∞, which contradicts the original condition |TA(s,0)a|/eε|s|≤M and it implies a=0. This ends the proof of Lemma 10.Proof of Theorem4.
For any solutionx(t) of system (2), it can be written as follows:
(44)x(t)=TA(t,0)x(0)+∫0t‍TA(t,s)TA-1(s,s)f(s,x(s))ds=TA(t,0)x(0)+∫0t‍TA(t,s)P(s)f(s,x(s))ds+∫0t‍TA(t,s)Q(s)f(s,x(s))ds=TA(t,0)x(0)+∫-∞t‍TA(t,s)P(s)f(s,x(s))ds-∫-∞0‍TA(t,s)P(s)f(s,x(s))ds+∫0+∞‍TA(t,s)Q(s)f(s,x(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,x(s))ds)]x(0)vvvvvvvvv.-(∫-∞0‍TA(0,s)P(s)f(s,x(s))dsvvvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,x(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,x(s))dsvvvvv.-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds).
Let ξ(t) be anyn-variable continuous function defined on ℝ. From (3) and (H1), we have
(45)|∫-∞t‍TA(t,s)P(s)f(s,ξ(s))ds|≤∫-∞t‍Ke-α(t-s)eε|s|μds=Kμe-αt∫-∞t‍eαseε|s|ds.
For t≥0,
(46)e-αt∫-∞t‍eαseε|s|ds=e-αt(∫-∞0‍eαse-εsds+∫0t‍eαseεsds)=e-αt(1α-ε+1α+ε(e(α+ε)t-1))≤1α+εeεt+1α-ε≤1α-εeε|t|+1α-ε;
for t≤0,
(47)e-αt∫-∞t‍eαseε|s|ds=e-αt∫-∞t‍eαse-εsds=1α-εe-εt≤1α-εeε|t|+1α-ε.
Hence, for any t∈ℝ, we have
(48)e-αt∫-∞t‍eαseε|s|ds≤1α-εeε|t|+1α-ε.
Therefore, for any t∈ℝ, we have
(49)|∫-∞t‍TA(t,s)P(s)f(s,ξ(s))ds|≤Kμα-εeε|t|+Kμα-ε.
By the same calculation, for any t∈ℝ, we have
(50)|∫t+∞‍TA(t,s)Q(s)f(s,ξ(s))ds|≤Kμα-εeε|t|+Kμα-ε.
In Lemma9, x+(t) and x-(t) are uniquely determined by a; we denote them by x+(t,a) and x-(t,a), respectively. Let ρ=(2Kμ)/(α-ε), denote by G the closed sphere on ℝn whose center is at the origin of the coordinate system and whose radius is ρ. For any a∈G, we define a mapping 𝒯2:G→ℝn as follows:
(51)𝒯2a=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a))ds-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a))ds.
It follows from (3) and (H1) that
(52)|𝒯2a|≤∫-∞0‍Keαseε|s|μds+∫0+∞‍Ke-αseε|s|μds=Kμα-ε+Kμα-ε=2Kμα-ε=ρ,
which implies that 𝒯2 maps G onto itself. Now we are going to show that 𝒯2 is continuous. For any a1,a2∈G, from (3) and (H2), we have
(53)|𝒯2a1-𝒯2a2|≤∫-∞0‍Keαseε|s|·re-ε|s||x-(s,a1)-x-(s,a2)|ds+∫0+∞‍Ke-αseε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds≤∫-∞0‍Kreαs|x-(s,a1)-x-(s,a2)|ds+∫0+∞‍Kre-αs|x+(s,a1)-x+(s,a2)|ds.
From (3) and the condition (iii) of Lemma 9, for t≥0, we have
(54)|x+(t,a1)-x+(t,a2)|≤TA(t,0)P(0)|a1-a2|+∫0t‍Ke-α(t-s)eε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds+∫t+∞‍Keα(t-s)eε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds≤Ke-αt|a1-a2|+∫0t‍Kre-α(t-s)|x+(s,a1)-x+(s,a2)|ds+∫t+∞‍Kreα(t-s)|x+(s,a1)-x+(s,a2)|ds.
Multiplying by e-εt on both sides of the above inequality, for t≥0, we get
(55)e-εt|x+(t,a1)-x+(t,a2)|≤Ke-(α+ε)t|a1-a2|+∫0t‍Kre-(α+ε)(t-s)(e-εs|x+(s,a1)-x+(s,a2)|)ds+∫t+∞‍Kre(α-ε)(t-s)(e-εs|x+(s,a1)-x+(s,a2)|)ds.
By Lemma 9, for t≥0, e-εt|x+(t,a1)-x+(t,a2)| is a bounded function. And it follows from Lemma 8 that
(56)e-εt|x+(t,a1)-x+(t,a2)|≤K|a1-a2|(1-γ)-1e-[(α+ε)-(1-γ)-1Kr]t,
where γ=Kr/(α+ε)+Kr/(α-ε). From (H3) and (α+ε)-1<(α-ε)-1, we get
(57)γ≤2Kr(α-ε)-1<12.
Therefore, for α-2Kr>0, we have
(58)e-εt|x+(t,a1)-x+(t,a2)|≤2K|a1-a2|e-[(α+ε)-2Kr]t(t≥0).
Hence,
(59)|x+(t,a1)-x+(t,a2)|≤2K|a1-a2|e-(α-2Kr)t(t≥0).
Similarly,
(60)|x-(t,a1)-x-(t,a2)|≤2K|a1-a2|e(α-2Kr)t(t≤0).
So from (53), it follows that
(61)|𝒯2a1-𝒯2a2|≤∫-∞0‍Kreαs·2K|a1-a2|e(α-2Kr)sds+∫0+∞‍Kre-αs·2K|a1-a2|e-(α-2Kr)sds≤∫-∞0‍2K2r|a1-a2|e2(α-Kr)sds+∫0+∞‍2K2r|a1-a2|e-2(α-Kr)sds=2K2rα-Kr|a1-a2|,
which show that 𝒯2 is a continuous mapping. By fixed point theorem, 𝒯2 has at least one fixed point on G. We denote this fixed point by a0, then
(62)a0=𝒯2a0=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))dsvvvvvvvvv-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a0))ds.
As P2(s)=P(s),P(t)TA(t,s)=TA(t,s)P(s),P(s)+Q(s)=Id, we obtain
(63)P(0)a0=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))ds,Q(0)a0=-∫0+∞‍TA(0,s)Q(s)f(s,x-(s,a0))ds.
From Lemma 9, we have
(64)x+(0,a0)=P(0)a0-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a0))ds,x-(0,a0)=Q(0)a0+∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))ds.
Hence,
(65)x+(0,a0)=x-(0,a0)=a0.
By the existence and uniqueness of the initial value problem, we conclude that x+(t,a0)=x-(t,a0). We can denote it by x0(t). Hence,
(66)x0(0)=a0=∫-∞0‍TA(0,s)P(s)f(s,x(s,a0))ds-∫0+∞‍TA(0,s)Q(s)f(s,x(s,a0))ds.
From the above equation, it follows from (44) that
(67)x0(t)=∫-∞t‍TA(t,s)P(s)f(s,x(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds.
From (49) and (50), we have
(68)|x0(t)|≤2Kμα-εeε|t|+2Kμα-ε,(-∞<t<+∞),
which implies that x0(t) satisfies (5); that is, x0(t)=O(eε|t|).
Now we are going to prove that the solution of (2) which satisfies (5) is unique. We assume that system (2) has another solution x0*(t) satisfying (5). From (44), x0*(t) can be written as
(69)x0*(t)=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)]x0*(0)vvvvvvvvv-(∫-∞0‍TA(0,s)P(s)f(s,x0*(s))dsvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,x0*(s))dsvvvvv.-∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds).
From (49) and (50), we get
(70)|∫-∞t‍TA(t,s)P(s)f(s,x0*(s))ds|≤Kμα-εeε|t|+Kμα-ε,|∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds|≤Kμα-εeε|t|+Kμα-ε.
It follows from |x0*(t)|=O(eε|t|) and Lemma 10 that
(71)x0*-(∫-∞0‍TA(0,s)P(s)f(s,x0*(s))ds-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)=0.
Therefore,
(72)x0*(t)=∫-∞t‍TA(t,s)P(s)f(s,x0*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds.
From (67), (72), and (H2), we have
(73)|x0(t)-x0*(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·re-ε|t||x0(s)-x0*(s)|ds+∫t+∞‍Keα(t-s)eε|t|·re-ε|t||x0(s)-x0*(s)|ds=∫-∞t‍Kre-α(t-s)|x0(s)-x0*(s)|ds+∫t+∞‍Kreα(t-s)|x0(s)-x0*(s)|ds.
Let L=sup⁡t∈ℝe-ε|t||x0(t)-x0*(t). Since x0(t) and x0*(t) satisfy (5), e-ε|s||x0(s)-x0*(s)| is bounded. Thus, for t≥0, we have
(74)e-ε|t||x0(t)-x0*(t)|=e-εt|x0(t)-x0*(t)|≤∫-∞t‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫t+∞‍Ke(α-ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds=∫-∞0‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫0t‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫t+∞‍Ke(α-ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds≤L(Krα+εe-(α-ε)t+Krα+ε(1-e-(α+ε)t)+Krα-ε)≤L(Krα+ε+Krα+ε+Krα-ε)≤L(3Krα-ε)≤34L(by(H3)).
Similarly, we can prove that
(75)e-ε|t||x0(t)-x0*(t)|≤34L,whent≤0.
Therefore,
(76)L=sup⁡t∈ℝe-ε|t||x0(t)-x0*(t)|≤34L,(-∞<t+∞).
That is, L≤(3/4)L, which implies L=0. Consequently, x0(t)=x0*(t). This completes the proof of Theorem 4.


## 3.1. Proof of Theorem3

LetB={φ(t)∣φ(t) be continuous and |φ(t)|≤2Kμα-1}, for ∀φ∈B, define a mapping 𝒯1:
(6)𝒯1φ(t)=∫-∞t‍TA(t,s)P(s)f(s,φ(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ(s))ds.
From (3) and (H~1)and(H~2), we have
(7)|𝒯1φ(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·μe-ε|s|ds+∫t+∞‍Keα(t-s)eε|s|·μe-ε|s|ds=Kμα-1+Kμα-1=2Kμα-1.
Therefore, 𝒯1φ(t)∈B, which implies 𝒯1 maps B onto itself. On the other hand,
(8)|𝒯1φ1(t)-𝒯1φ2(t)|≤∫-∞t‍TA(t,s)P(s)re-ε|s||φ1(s)-φ2(s)|ds+∫t+∞‍TA(t,s)Q(s)re-ε|s||φ1(s)-φ2(s)|ds≤∫-∞t‍Kre-α(t-s)|φ1(s)-φ2(s)|ds+∫t+∞‍Kreα(t-s)|φ1(s)-φ2(s)|ds≤2Krα-1sup⁡s≥0|φ1(s)-φ2(s)|≤12sup⁡s≥0|φ1(s)-φ2(s)|.
Then 𝒯1 is a contraction mapping. Therefore, in B, there exists a unique fixed point φ0(t), such that
(9)φ0(t)=𝒯1φ0(t)=∫-∞t‍TA(t,s)P(s)f(s,φ0(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ0(s))ds.
Differentiating the above equality, we see that φ0(t) satisfies (2). Now we are going to show that the solution of system (2) satisfying (H~1),(H~2), and(H~3) is unique. Assume that system (2) has another bounded solution φ*(t) satisfying (H~1),(H~2), and(H~3); we have
(10)φ*(t)=TA(t,0)φ*(0)+∫0t‍TA(t,s)T-1(s,s)f(s,φ*(s))ds=TA(t,0)φ*(0)+∫0t‍TA(t,s)P(s)f(s,φ*(s))ds+∫0t‍TA(t,s)Q(s)f(s,φ*(s))ds=TA(t,0)φ*(0)+∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫-∞0‍TA(t,s)P(s)f(s,φ*(s))ds+∫0+∞‍TA(t,s)Q(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]φ*(0)-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))ds-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds).
By calculating, we get
(11)∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds≤2Kμα-1.
As φ*(t) is bounded, we obtain
(12)TA(t,0)[φ*(0)-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))dsvvvvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)],
is bounded. In addition, the formula above is the solution of system (1), so it is a bounded solution. From Lemma 6, we have
(13)TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]φ*(0)vvvvvvv-(∫-∞0‍TA(0,s)P(s)f(s,φ*(s))dsvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,φ*(s))ds)]=0.
Therefore,
(14)φ*(t)=∫-∞t‍TA(t,s)P(s)f(s,φ*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,φ*(s))ds.
From (3), (H~2), and (H~3), we have
(15)|φ0(t)-φ*(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·re-ε|s||φ0(s)-φ*(s)|ds+∫t+∞‍Keα(t-s)eε|s|·re-ε|s||φ0(s)-φ*(s)|ds=2Krα-1|φ0(s)-φ*(s)|=12sup⁡t∈ℝ|φ0(t)-φ*(t)|.
That is, sup⁡t∈ℝ|φ0(t)-φ*(t)|≤(1/2)sup⁡t∈ℝ|φ0(t)-φ*(t)|, which implies φ0(t)=φ*(t). Then the uniqueness is proved. The proof of Theorem 3 is complete.


## 3.2. Proof of Theorem4

To prove Theorem4, a standard method is to employ a linear transformation x=eε|t|y. However, x=eε|t|y is not differentiable at t=0. Thus, we cannot use such transformation directly. We have to discuss by dividing into two pieces t≥0 and t≤0.Consider system(16)x˙(t)=B(t)v(t)+F(t,v),
where u∈ℝn, B(t) is a continuous matrix function, and F:ℝ×ℝn→ℝn is a continuous function.Lemma 7.
Suppose that systemv˙(t)=B(t)v(t) admits a nonuniform exponential dichotomy; that is, its evolution operator TB(t,s) satisfies
(17)|TB(t,s)P(s)|≤Ke-λ(t-s)·eε|s|,t≥s,|TB(t,s)Q(s)|≤Keλ(t-s)·eε|s|,t≤s,
where λ is a positive constant. In addition,
(18)|F(t,v1)-F(t,v2)|≤re-ε|t||v1-v2|,4Kr<λ.
If |F(t,v)|≤μe-ε|t| for t≥0, then for any a∈ℝn, system (16) has a unique solution v+(t) satisfying the following:(i)
|v+(t)|<+∞, for t≥0;(ii)
P(0)v+(0)=P(0)a;(iii)
inℝ+,v+(t) satisfies integral equation
(19)v+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,v+(s))ds.
If|F(t,v)|≤μe-ε|t| for t≤0, then for any a∈ℝn, system (16) has a unique solution v-(t) satisfying the following:(i)
|v-(t)|<+∞, for t≤0;(ii)
Q(0)v-(0)=Q(0)a;(iii)
inℝ-,v-(t) satisfies integral equation
(20)v-(t)=TB(t,0)Q(0)a+∫-∞t‍TB(t,s)P(s)F(s,v-(s))ds-∫t0‍TB(t,s)Q(s)F(s,v-(s))ds.Proof.
We prove the existence ofv+(t) by successive approximation method. For any a∈ℝn, let v0+(t)=TB(t,0)P(0)a. We define vm+(t),vm+1+(t) recursively as follows:
(21)vm+1+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,vm+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,vm+(s))ds.
From (17) and |F(t,v)|≤μe-ε|t|, for t≥0, we have
(22)|vm+1+(t)|≤Ke-λt|a|+∫0t‍Ke-λ(t-s)eεs·μe-εsds+∫t+∞‍Keλ(t-s)eεs·μe-εsds=Ke-λt|a|+Kμλ-1(1-e-λt)+Kμλ-1≤K|a|+2Kμλ-1.
For any bounded function v(t) defined on ℝ+, denote ∥v∥=sup⁡t∈ℝ|v(t)|; then it follows from (18) and (21) that
(23)|vm+1+(t)-vm+(t)|≤∫0t‍Ke-λ(t-s)eεs·re-εs|vm+(s)-vm-1+(s)|ds+∫t+∞‍Keλ(t-s)eεs·re-εs|vm+(s)-vm-1+(s)|ds≤2Krλ-1|vm+(s)-vm-1+(s)|≤2Krλ-1∥vm+-vm-1+∥≤12∥vm+-vm-1+∥,
which implies that ∥vm+1+-vm+∥≤(1/2)∥vm+-vm-1+∥. Hence, the series ∑m=0∞‍(vm+1+(t)-vm+(t)) converges uniformly on ℝ+. It means that the series {vm+(t)} converges uniformly to a limit v+(t) on ℝ+.
From (21), for any fixed t, let m→∞, we have
(24)v+(t)=TB(t,0)P(0)a+∫0t‍TB(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TB(t,s)Q(s)F(s,v+(s))ds.
Differentiating the above equality, we see that v+(t) satisfies the system (16).
From (22) and (24), we know that |v+(t)|<∞ for t≥0, and it is easy to demonstrate that P(0)v+(0)=P(0)a. Now we are going to show the uniqueness of v+(t). If there is another bounded function v~+(t) satisfying (i), (ii), and (iii) on ℝ+, in view of (iii) and (18) we have
(25)|v~+(t)-v+(t)|≤∫0t‍Ke-λ(t-s)eεs·re-εs|v~+(s)-v+(s)|ds+∫t+∞‍Keλ(t-s)eεs·re-εs|v~+(s)-v+(s)|ds≤2Krλ-1∥v~+-v+∥≤12∥v~+-v+∥,
which implies ∥v~+-v+∥≤(1/2)∥v~+-v+∥. Hence, v~+(t)≡v+(t).
The proof of the existence and uniqueness ofv-(t) is similar to that of v+(t). The proof of Lemma 7 is complete.Lemma 8.
Suppose thatα>0,δ>0,C,L, and M are nonnegative constants and that v(t) is a nonnegative bounded continuous function which satisfies two of the following inequalities:
(26)v(t)≤Ce-αt+L∫0t‍e-α(t-s)v(s)ds+M∫t+∞‍eδ(t-s)v(s)ds,(t≥0),v(t)≤Ceαt+L∫t0‍eα(t-s)v(s)ds+M∫-∞t‍e-δ(t-s)v(s)ds,(t≤0).
In addition, if γ=L/α+M/δ<1, then for t≥0 or t≤0, one has
(27)v(t)≤(1-γ)-1Ce-[α-(1-γ)-1L]|t|.Proof.
The proof is straightforward by Lemma 6.2 of Chapter 3 in [6].Lemma 9.
For anya∈ℝn, system (2) has a unique solution x+(t) with the following properties:(i)
|x+(t)e-εt|<+∞, for t≥0;(ii)
P(0)x+(0)=P(0)a;(iii)
x+(t) on ℝ+ satisfies integral equation
(28)x+(t)=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,x+(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x+(s))ds.Similarly, for any a∈ℝn, system (2) also has a unique solution x-(t) with the following properties:(i)
|x-(t)eεt|<+∞, for t≤0;(ii)
Q(0)x-(0)=Q(0)a;(iii)
x-(t) on ℝ- satisfies integral equation
(29)x-(t)=TA(t,0)Q(0)a+∫-∞t‍TA(t,s)P(s)f(s,x-(s))ds-∫t0‍TA(t,s)Q(s)f(s,x-(s))ds.Proof.
We firstly prove the existence and uniqueness ofx+(t). Let v=xe-εt, then system (2) becomes
(30)v˙(t)=(A(t)-εI)v(t)+e-εtf(t,veεt).
Let F(t,v)=e-εtf(t,veεt). From (H1) and (H2), we have
(31)|F(t,v)|≤e-εtμ=μe-ε|t|,fort≥0;|F(t,v1)-F(t,v2)|=e-εt|f(t,v1eεt)-f(t,v2eεt)|≤e-εtre-ε|t||v1eεt-v2eεt|≤re-ε|t||v1-v2|.
LetTC(t,s) be the evolution operator of the linear system v˙(t)=(A(t)-εt)v(t). Since x(t)=TA(t,s)x(s),v=e-εtx, we have TC(t,s)=e-ε(t-s)TA(t,s). Hence, from (3), we obtain
(32)|TC(t,s)P(s)|≤Ke-(α-ε)(t-s)·eε|s|,fort≥s,|TC(t,s)Q(s)|≤Ke(α-ε)(t-s)·eε|s|,fort≤s.
Since 4Kr<α-ε, system (30) satisfies all conditions of Lemma 7. Therefore, for any a∈ℝn, system (30) has a unique solution v+(t) with the following properties:(i)
|v+(t)|<+∞, for t≥0;(ii)
P(0)v+(0)=P(0)a;(iii)
v+(t) on ℝ+ satisfies integral equation
(33)v+(t)=TC(t,0)P(0)a+∫0t‍TC(t,s)P(s)F(s,v+(s))ds-∫t+∞‍TC(t,s)Q(s)F(s,v+(s))ds=e-εtTA(t,0)P(0)a+∫0t‍e-ε(t-s)TA(t,s)P(s)·e-εsf(s,v+(s)eεs)ds-∫t+∞‍e-ε(t-s)TA(t,s)Q(s)vvvvv·e-εsf(s,v+(s)eεs)ds.Hence,
(34)v+(t)eεt=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,v+(s)eεs)ds-∫t+∞‍TA(t,s)Q(s)f(s,v+(s)eεs)ds.
Let x+(t)=v+(t)eεt, we have
(35)x+(t)=TA(t,0)P(0)a+∫0t‍TA(t,s)P(s)f(s,x+(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x+(s))ds.
Then x+(t) is the solution of system (2) and it satisfies all conditions of Lemma 9.
The proof for the existence and uniqueness ofx-(t) is similar to that of x+(t), so we omit it. This completes the proof of Lemma 9.Lemma 10.
If|TA(t,0)P(0)a|≤Meε|t|, then a=0.Proof.
Ifa≠0, then P(0)a≠0 or Q(0)a≠0. Without loss of generality, we assume P(0)a≠0, then we have
(36)|TA(t,0)P(0)a|=|TA(t,0)P(0)TA-1(s,0)TA(s,0)P(0)a|=|TA(t,0)TA(0,s)P(s)TA(s,0)P(0)a|=|TA(t,s)P(s)TA(s,0)P(0)a|≤|TA(t,s)P(s)|·|TA(s,0)P(0)a|.
Since |TA(t,s)P(s)|≤Ke-α(t-s)eε|s| for t≥s,
(37)|TA(t,0)P(0)a|≤Ke-α(t-s)eε|s|·|TA(s,0)P(0)a|,
which implies
(38)|TA(s,0)P(0)a|≥|TA(t,0)P(0)a|Ke-α(t-s)eε|s|.
Taking t=0, we obtain
(39)|TA(s,0)P(0)a|≥|P(0)a|Keαse-εs=K-1e-(α-ε)s|P(0)a|(s≤0).
Therefore, when s≤0,
(40)|TA(s,0)P(0)a|eε|s|≥K-1e-αs|P(0)a|⟶+∞ass⟶-∞.
On the other hand, when s≤0, from (3), we have
(41)|TA(s,0)Q(0)a|≤Keαseε|0|=Keαs,
hence,
(42)|TA(s,0)Q(0)a|eε|s|≤Ke(α+ε)s.
It follows from (40) and (42) that
(43)|TA(s,0)a|eε|s|=|TA(s,0)(P(s)+Q(s))a|eε|s|≥|TA(s,0)P(s)a|eε|s|-|TA(s,0)Q(s)a|eε|s|≥|TA(s,0)P(s)a|eε|s|-Ke(α+ε)s.
From the above inequality, we know that |TA(s,0)a|/eε|s|→+∞ass→-∞, which contradicts the original condition |TA(s,0)a|/eε|s|≤M and it implies a=0. This ends the proof of Lemma 10.Proof of Theorem4.
For any solutionx(t) of system (2), it can be written as follows:
(44)x(t)=TA(t,0)x(0)+∫0t‍TA(t,s)TA-1(s,s)f(s,x(s))ds=TA(t,0)x(0)+∫0t‍TA(t,s)P(s)f(s,x(s))ds+∫0t‍TA(t,s)Q(s)f(s,x(s))ds=TA(t,0)x(0)+∫-∞t‍TA(t,s)P(s)f(s,x(s))ds-∫-∞0‍TA(t,s)P(s)f(s,x(s))ds+∫0+∞‍TA(t,s)Q(s)f(s,x(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,x(s))ds)]x(0)vvvvvvvvv.-(∫-∞0‍TA(0,s)P(s)f(s,x(s))dsvvvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,x(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,x(s))dsvvvvv.-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds).
Let ξ(t) be anyn-variable continuous function defined on ℝ. From (3) and (H1), we have
(45)|∫-∞t‍TA(t,s)P(s)f(s,ξ(s))ds|≤∫-∞t‍Ke-α(t-s)eε|s|μds=Kμe-αt∫-∞t‍eαseε|s|ds.
For t≥0,
(46)e-αt∫-∞t‍eαseε|s|ds=e-αt(∫-∞0‍eαse-εsds+∫0t‍eαseεsds)=e-αt(1α-ε+1α+ε(e(α+ε)t-1))≤1α+εeεt+1α-ε≤1α-εeε|t|+1α-ε;
for t≤0,
(47)e-αt∫-∞t‍eαseε|s|ds=e-αt∫-∞t‍eαse-εsds=1α-εe-εt≤1α-εeε|t|+1α-ε.
Hence, for any t∈ℝ, we have
(48)e-αt∫-∞t‍eαseε|s|ds≤1α-εeε|t|+1α-ε.
Therefore, for any t∈ℝ, we have
(49)|∫-∞t‍TA(t,s)P(s)f(s,ξ(s))ds|≤Kμα-εeε|t|+Kμα-ε.
By the same calculation, for any t∈ℝ, we have
(50)|∫t+∞‍TA(t,s)Q(s)f(s,ξ(s))ds|≤Kμα-εeε|t|+Kμα-ε.
In Lemma9, x+(t) and x-(t) are uniquely determined by a; we denote them by x+(t,a) and x-(t,a), respectively. Let ρ=(2Kμ)/(α-ε), denote by G the closed sphere on ℝn whose center is at the origin of the coordinate system and whose radius is ρ. For any a∈G, we define a mapping 𝒯2:G→ℝn as follows:
(51)𝒯2a=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a))ds-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a))ds.
It follows from (3) and (H1) that
(52)|𝒯2a|≤∫-∞0‍Keαseε|s|μds+∫0+∞‍Ke-αseε|s|μds=Kμα-ε+Kμα-ε=2Kμα-ε=ρ,
which implies that 𝒯2 maps G onto itself. Now we are going to show that 𝒯2 is continuous. For any a1,a2∈G, from (3) and (H2), we have
(53)|𝒯2a1-𝒯2a2|≤∫-∞0‍Keαseε|s|·re-ε|s||x-(s,a1)-x-(s,a2)|ds+∫0+∞‍Ke-αseε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds≤∫-∞0‍Kreαs|x-(s,a1)-x-(s,a2)|ds+∫0+∞‍Kre-αs|x+(s,a1)-x+(s,a2)|ds.
From (3) and the condition (iii) of Lemma 9, for t≥0, we have
(54)|x+(t,a1)-x+(t,a2)|≤TA(t,0)P(0)|a1-a2|+∫0t‍Ke-α(t-s)eε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds+∫t+∞‍Keα(t-s)eε|s|·re-ε|s||x+(s,a1)-x+(s,a2)|ds≤Ke-αt|a1-a2|+∫0t‍Kre-α(t-s)|x+(s,a1)-x+(s,a2)|ds+∫t+∞‍Kreα(t-s)|x+(s,a1)-x+(s,a2)|ds.
Multiplying by e-εt on both sides of the above inequality, for t≥0, we get
(55)e-εt|x+(t,a1)-x+(t,a2)|≤Ke-(α+ε)t|a1-a2|+∫0t‍Kre-(α+ε)(t-s)(e-εs|x+(s,a1)-x+(s,a2)|)ds+∫t+∞‍Kre(α-ε)(t-s)(e-εs|x+(s,a1)-x+(s,a2)|)ds.
By Lemma 9, for t≥0, e-εt|x+(t,a1)-x+(t,a2)| is a bounded function. And it follows from Lemma 8 that
(56)e-εt|x+(t,a1)-x+(t,a2)|≤K|a1-a2|(1-γ)-1e-[(α+ε)-(1-γ)-1Kr]t,
where γ=Kr/(α+ε)+Kr/(α-ε). From (H3) and (α+ε)-1<(α-ε)-1, we get
(57)γ≤2Kr(α-ε)-1<12.
Therefore, for α-2Kr>0, we have
(58)e-εt|x+(t,a1)-x+(t,a2)|≤2K|a1-a2|e-[(α+ε)-2Kr]t(t≥0).
Hence,
(59)|x+(t,a1)-x+(t,a2)|≤2K|a1-a2|e-(α-2Kr)t(t≥0).
Similarly,
(60)|x-(t,a1)-x-(t,a2)|≤2K|a1-a2|e(α-2Kr)t(t≤0).
So from (53), it follows that
(61)|𝒯2a1-𝒯2a2|≤∫-∞0‍Kreαs·2K|a1-a2|e(α-2Kr)sds+∫0+∞‍Kre-αs·2K|a1-a2|e-(α-2Kr)sds≤∫-∞0‍2K2r|a1-a2|e2(α-Kr)sds+∫0+∞‍2K2r|a1-a2|e-2(α-Kr)sds=2K2rα-Kr|a1-a2|,
which show that 𝒯2 is a continuous mapping. By fixed point theorem, 𝒯2 has at least one fixed point on G. We denote this fixed point by a0, then
(62)a0=𝒯2a0=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))dsvvvvvvvvv-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a0))ds.
As P2(s)=P(s),P(t)TA(t,s)=TA(t,s)P(s),P(s)+Q(s)=Id, we obtain
(63)P(0)a0=∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))ds,Q(0)a0=-∫0+∞‍TA(0,s)Q(s)f(s,x-(s,a0))ds.
From Lemma 9, we have
(64)x+(0,a0)=P(0)a0-∫0+∞‍TA(0,s)Q(s)f(s,x+(s,a0))ds,x-(0,a0)=Q(0)a0+∫-∞0‍TA(0,s)P(s)f(s,x-(s,a0))ds.
Hence,
(65)x+(0,a0)=x-(0,a0)=a0.
By the existence and uniqueness of the initial value problem, we conclude that x+(t,a0)=x-(t,a0). We can denote it by x0(t). Hence,
(66)x0(0)=a0=∫-∞0‍TA(0,s)P(s)f(s,x(s,a0))ds-∫0+∞‍TA(0,s)Q(s)f(s,x(s,a0))ds.
From the above equation, it follows from (44) that
(67)x0(t)=∫-∞t‍TA(t,s)P(s)f(s,x(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x(s))ds.
From (49) and (50), we have
(68)|x0(t)|≤2Kμα-εeε|t|+2Kμα-ε,(-∞<t<+∞),
which implies that x0(t) satisfies (5); that is, x0(t)=O(eε|t|).
Now we are going to prove that the solution of (2) which satisfies (5) is unique. We assume that system (2) has another solution x0*(t) satisfying (5). From (44), x0*(t) can be written as
(69)x0*(t)=TA(t,0)[-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)]x0*(0)vvvvvvvvv-(∫-∞0‍TA(0,s)P(s)f(s,x0*(s))dsvvvvvvvvvvv.-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)]+(∫-∞t‍TA(t,s)P(s)f(s,x0*(s))dsvvvvv.-∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds).
From (49) and (50), we get
(70)|∫-∞t‍TA(t,s)P(s)f(s,x0*(s))ds|≤Kμα-εeε|t|+Kμα-ε,|∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds|≤Kμα-εeε|t|+Kμα-ε.
It follows from |x0*(t)|=O(eε|t|) and Lemma 10 that
(71)x0*-(∫-∞0‍TA(0,s)P(s)f(s,x0*(s))ds-∫0+∞‍TA(0,s)Q(s)f(s,x0*(s))ds)=0.
Therefore,
(72)x0*(t)=∫-∞t‍TA(t,s)P(s)f(s,x0*(s))ds-∫t+∞‍TA(t,s)Q(s)f(s,x0*(s))ds.
From (67), (72), and (H2), we have
(73)|x0(t)-x0*(t)|≤∫-∞t‍Ke-α(t-s)eε|s|·re-ε|t||x0(s)-x0*(s)|ds+∫t+∞‍Keα(t-s)eε|t|·re-ε|t||x0(s)-x0*(s)|ds=∫-∞t‍Kre-α(t-s)|x0(s)-x0*(s)|ds+∫t+∞‍Kreα(t-s)|x0(s)-x0*(s)|ds.
Let L=sup⁡t∈ℝe-ε|t||x0(t)-x0*(t). Since x0(t) and x0*(t) satisfy (5), e-ε|s||x0(s)-x0*(s)| is bounded. Thus, for t≥0, we have
(74)e-ε|t||x0(t)-x0*(t)|=e-εt|x0(t)-x0*(t)|≤∫-∞t‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫t+∞‍Ke(α-ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds=∫-∞0‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫0t‍Ke-(α+ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds+∫t+∞‍Ke(α-ε)(t-s)r[e-εs|x0(s)-x0*(s)|]ds≤L(Krα+εe-(α-ε)t+Krα+ε(1-e-(α+ε)t)+Krα-ε)≤L(Krα+ε+Krα+ε+Krα-ε)≤L(3Krα-ε)≤34L(by(H3)).
Similarly, we can prove that
(75)e-ε|t||x0(t)-x0*(t)|≤34L,whent≤0.
Therefore,
(76)L=sup⁡t∈ℝe-ε|t||x0(t)-x0*(t)|≤34L,(-∞<t+∞).
That is, L≤(3/4)L, which implies L=0. Consequently, x0(t)=x0*(t). This completes the proof of Theorem 4.



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*Source: 725098-2014-05-05.xml*