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

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.

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 𝑥̇(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 𝑥̇(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(𝑒^𝜀|𝑡|).

The outline of this paper is arranged as follows. Next section is to state our main results. In Section 3, 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 𝑥̇(𝑡)=𝐴(𝑡)𝑥(𝑡),𝑥̇(𝑡)=𝐴(𝑡)𝑥(𝑡)+𝑓(𝑡,𝑥), (1)(2) where 𝑡∈ℝ, 𝑥∈ℝ^𝑛, 𝐴(𝑡) is a continuous matrix function, and 𝑓∶ℝ×ℝ^𝑛→ℝ^𝑛 is a continuous function.

Let 𝑇_𝐴(𝑡,𝑠) be the evolution operator satisfying 𝑥(𝑡)=𝑇_𝐴(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 𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)≤𝐾𝑒^{−𝛼(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|𝑇,𝑡≥𝑠,_𝐴(𝑡,𝑠)𝑄(𝑠)≤𝐾𝑒^{𝛼(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|,𝑡≤𝑠, (3) where 𝑃(𝑡)+𝑄(𝑡)=Id(identity),𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)=𝑃(𝑡)𝑇_𝐴(𝑡,𝑠),𝑡,𝑠∈ℝ.

Remark 2. When 𝜀≡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∈ℝ^𝑛, if the nonlinear term 𝑓(𝑡,𝑥) satisfies (H₁)|𝑓(𝑡,𝑥)|≤𝜇𝑒^{−𝜀|𝑡|},(H₂)|𝑓(𝑡,𝑥₁)−𝑓(𝑡,𝑥₂)|≤𝑟𝑒^{−𝜀|𝑡|}|𝑥₁−𝑥₂|,(H₃)4𝐾𝑟<𝛼,where 𝜇,𝑟,𝜀,𝛼 are all positive constants, then nonlinear system (2) has a unique bounded solution 𝑥(𝑡) satisfying 𝑥(𝑡)=^𝑡_−∞𝑇_𝐴()𝑃(𝑠)𝑓(𝑥(𝑠−𝑡,𝑠𝑠,))𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (4)

Discussion. One of the essential conditions of Theorem 3 is (H₁). 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∈ℝ^𝑛, 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 |𝑥(𝑡)|=𝑂𝑒^𝜀|𝑡|. (5)

Remark 5. The method used to prove Theorem 3 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 Theorem 3, 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 Theorem 3

Let 𝐁={𝜑(𝑡)∣𝜑(𝑡) be continuous and |𝜑(𝑡)|≤2𝐾𝜇𝛼⁻¹}, for ∀𝜑∈𝐁, define a mapping 𝒯₁: 𝒯₁𝜑(𝑡)=^𝑡_−∞𝑇_𝐴()𝑃(𝑠)𝑓(𝑠)−𝑡,𝑠𝑠,𝜑𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑(𝑠)𝑑𝑠. (6) From (3) and (H₁)and(H₂), we have 𝒯₁𝜑(𝑡)≤^𝑡_−∞𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝜇𝑒^{−𝜀|𝑠|}+𝑑𝑠_𝑡+∞𝐾𝑒^{𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝜇𝑒^{−𝜀|𝑠|}𝑑𝑠=𝐾𝜇𝛼⁻¹+𝐾𝜇𝛼⁻¹=2𝐾𝜇𝛼⁻¹. (7) Therefore, 𝒯₁𝜑(𝑡)∈𝐁, which implies 𝒯₁ maps 𝐁 onto itself. On the other hand, 𝒯₁𝜑₁(𝑡)−𝒯₁𝜑₂(𝑡≤_𝑡−∞)𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑟𝑒^{−𝜀|𝑠|}𝜑₁(𝑠)−𝜑₂(𝑠+)𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑟𝑒^{−𝜀|𝑠|}𝜑₁(𝑠)−𝜑₂(𝑠)𝑑𝑠≤^𝑡_−∞𝐾𝑟𝑒^{−𝛼(𝑡−𝑠)}𝜑₁(𝑠)−𝜑₂(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑟𝑒^{𝛼(𝑡−𝑠)}𝜑₁(𝑠)−𝜑₂(𝑠)𝑑𝑠≤2𝐾𝑟𝛼⁻¹sup_𝑠≥0𝜑₁(𝑠)−𝜑₂(𝑠≤1)2sup_𝑠≥0𝜑₁(𝑠)−𝜑₂(𝑠). (8) Then 𝒯₁ is a contraction mapping. Therefore, in 𝐁, there exists a unique fixed point 𝜑₀(𝑡), such that 𝜑₀(𝑡)=𝒯₁𝜑₀(𝑡)=^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑₀(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑₀(𝑠)𝑑𝑠. (9) Differentiating the above equality, we see that 𝜑₀(𝑡) satisfies (2). Now we are going to show that the solution of system (2) satisfying (H₁),(H₂), and(H₃) is unique. Assume that system (2) has another bounded solution 𝜑^∗(𝑡) satisfying (H₁),(H₂), and(H₃); we have 𝜑^∗(𝑡)=𝑇_𝐴(𝑡,0)𝜑^∗(0)+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑇⁻¹(𝑠,𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠=𝑇_𝐴(𝑡,0)𝜑^∗(0)+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)+𝑑𝑠_𝑡0𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠=𝑇_𝐴(𝑡,0)𝜑^∗(0)+^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_0−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)+𝑑𝑠_0+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠=𝑇_𝐴(𝑡,0)𝜑^∗(0)−⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)+𝑑𝑠_𝑡−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠. (10) By calculating, we get ^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠≤2𝐾𝜇𝛼⁻¹. (11) As 𝜑^∗(𝑡) is bounded, we obtain 𝑇_𝐴(𝑡,0)𝜑^∗(0)−⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠, (12) is bounded. In addition, the formula above is the solution of system (1), so it is a bounded solution. From Lemma 6, we have 𝑇_𝐴(𝑡,0)𝜑^∗(0)−⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠=0. (13) Therefore, 𝜑^∗(𝑡)=^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝜑^∗(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝜑^∗(𝑠)𝑑𝑠. (14) From (3), (H₂), and (H₃), we have 𝜑₀(𝑡)−𝜑^∗(𝑡≤_𝑡−∞)𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝜑₀(𝑠)−𝜑^∗(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝜑₀(𝑠)−𝜑^∗(𝑠)𝑑𝑠=2𝐾𝑟𝛼⁻¹𝜑₀(𝑠)−𝜑^∗(𝑠=1)2sup_𝑡∈ℝ𝜑₀(𝑡)−𝜑^∗(𝑡). (15) 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 Theorem 4

To prove Theorem 4, a standard method is to employ a linear transformation 𝑥=𝑒^𝜀|𝑡|𝑦. However, 𝑥=𝑒^𝜀|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 𝑥̇(𝑡)=𝐵(𝑡)𝑣(𝑡)+𝐹(𝑡,𝑣), (16) where 𝑢∈ℝ^𝑛, 𝐵(𝑡) is a continuous matrix function, and 𝐹∶ℝ×ℝ^𝑛→ℝ^𝑛 is a continuous function.

Lemma 7. Suppose that system 𝑣(̇𝑡)=𝐵(𝑡)𝑣(𝑡) admits a nonuniform exponential dichotomy; that is, its evolution operator 𝑇_𝐵(𝑡,𝑠) satisfies 𝑇_𝐵(𝑡,𝑠)𝑃(𝑠)≤𝐾𝑒^{−𝜆(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|𝑇,𝑡≥𝑠,_𝐵(𝑡,𝑠)𝑄(𝑠)≤𝐾𝑒^{𝜆(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|,𝑡≤𝑠, (17) where 𝜆 is a positive constant. In addition, 𝐹𝑡,𝑣₁−𝐹𝑡,𝑣₂≤𝑟𝑒^{−𝜀|𝑡|}𝑣₁−𝑣₂,4𝐾𝑟<𝜆. (18) If |𝐹(𝑡,𝑣)|≤𝜇𝑒^{−𝜀|𝑡|} for 𝑡≥0, then for any 𝑎∈ℝ^𝑛, 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)in    ℝ⁺,𝑣⁺(𝑡) satisfies integral equation 𝑣⁺(𝑡)=𝑇_𝐵(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐵(𝑡,𝑠)𝑃(𝑠)𝐹𝑠,𝑣⁺(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐵(𝑡,𝑠)𝑄(𝑠)𝐹𝑠,𝑣⁺(𝑠)𝑑𝑠. (19)
If |𝐹(𝑡,𝑣)|≤𝜇𝑒^{−𝜀|𝑡|} for 𝑡≤0, then for any 𝑎∈ℝ^𝑛, 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)in   ℝ⁻,𝑣⁻(𝑡) satisfies integral equation 𝑣⁻(𝑡)=𝑇_𝐵(𝑡,0)𝑄(0)𝑎+^𝑡_−∞𝑇_𝐵(𝑡,𝑠)𝑃(𝑠)𝐹(𝑠,𝑣⁻(𝑠−))𝑑𝑠_0𝑡𝑇_𝐵(𝑡,𝑠)𝑄(𝑠)𝐹(𝑠,𝑣⁻(𝑠))𝑑𝑠. (20)

Proof. We prove the existence of 𝑣⁺(𝑡) by successive approximation method. For any 𝑎∈ℝ^𝑛, let 𝑣⁺₀(𝑡)=𝑇_𝐵(𝑡,0)𝑃(0)𝑎. We define 𝑣⁺_𝑚(𝑡),𝑣⁺_𝑚+1(𝑡) recursively as follows: 𝑣⁺_𝑚+1(𝑡)=𝑇_𝐵(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐵(𝑡,𝑠)𝑃(𝑠)𝐹𝑠,𝑣⁺_𝑚(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐵(𝑡,𝑠)𝑄(𝑠)𝐹𝑠,𝑣⁺_𝑚(𝑠)𝑑𝑠. (21) From (17) and |𝐹(𝑡,𝑣)|≤𝜇𝑒^{−𝜀|𝑡|}, for 𝑡≥0, we have 𝑣⁺_𝑚+1(𝑡)≤𝐾𝑒^−𝜆𝑡|𝑎|+^𝑡₀𝐾𝑒^{−𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝜇𝑒^−𝜀𝑠+𝑑𝑠_𝑡+∞𝐾𝑒^{𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝜇𝑒^−𝜀𝑠𝑑𝑠=𝐾𝑒^−𝜆𝑡|𝑎|+𝐾𝜇𝜆⁻¹1−𝑒^−𝜆𝑡+𝐾𝜇𝜆⁻¹≤𝐾|𝑎|+2𝐾𝜇𝜆⁻¹. (22) For any bounded function 𝑣(𝑡) defined on ℝ⁺, denote ‖𝑣‖=sup_𝑡∈ℝ|𝑣(𝑡)|; then it follows from (18) and (21) that 𝑣⁺_𝑚+1(𝑡)−𝑣⁺_𝑚(𝑡≤_𝑡0)𝐾𝑒^{−𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝑟𝑒^−𝜀𝑠𝑣⁺_𝑚(𝑠)−𝑣⁺_𝑚−1(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝑟𝑒^−𝜀𝑠𝑣⁺_𝑚(𝑠)−𝑣⁺_𝑚−1(𝑠)𝑑𝑠≤2𝐾𝑟𝜆⁻¹𝑣⁺_𝑚(𝑠)−𝑣⁺_𝑚−1(𝑠)≤2𝐾𝑟𝜆⁻¹𝑣⁺_𝑚−𝑣⁺_𝑚−1≤12𝑣⁺_𝑚−𝑣⁺_𝑚−1, (23) which implies that ‖𝑣⁺_𝑚+1−𝑣⁺_𝑚‖≤(1/2)‖𝑣⁺_𝑚−𝑣⁺_𝑚−1‖. Hence, the series ∑^∞_𝑚=0(𝑣⁺_𝑚+1(𝑡)−𝑣⁺_𝑚(𝑡)) converges uniformly on ℝ⁺. It means that the series {𝑣⁺_𝑚(𝑡)} converges uniformly to a limit 𝑣⁺(𝑡) on ℝ⁺.
From (21), for any fixed 𝑡, let 𝑚→∞, we have 𝑣⁺(𝑡)=𝑇_𝐵(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐵(𝑡,𝑠)𝑃(𝑠)𝐹𝑠,𝑣⁺(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐵(𝑡,𝑠)𝑄(𝑠)𝐹𝑠,𝑣⁺(𝑠)𝑑𝑠. (24) 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 𝑣⁺(𝑡)−𝑣⁺(𝑡≤_𝑡0)𝐾𝑒^{−𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝑟𝑒^−𝜀𝑠𝑣⁺(𝑠)−𝑣⁺(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{𝜆(𝑡−𝑠)}𝑒^𝜀𝑠⋅𝑟𝑒^−𝜀𝑠𝑣⁺(𝑠)−𝑣⁺(𝑠)𝑑𝑠≤2𝐾𝑟𝜆⁻¹𝑣⁺−𝑣⁺≤12𝑣⁺−𝑣⁺, (25) which implies ‖𝑣⁺−𝑣⁺‖≤(1/2)‖𝑣⁺−𝑣⁺‖. Hence, 𝑣⁺(𝑡)≡𝑣⁺(𝑡).
The proof of the existence and uniqueness of 𝑣⁻(𝑡) is similar to that of 𝑣⁺(𝑡). The proof of Lemma 7 is complete.

Lemma 8. Suppose that 𝛼>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: 𝑣(𝑡)≤𝐶𝑒^−𝛼𝑡+𝐿^𝑡₀𝑒^{−𝛼(𝑡−𝑠)}𝑣(𝑠)𝑑𝑠+𝑀_𝑡^+∞𝑒^{𝛿(𝑡−𝑠)}𝑣(𝑠)(),𝑣(𝑡)𝑑𝑠,𝑡≥0≤𝐶𝑒^𝛼𝑡+𝐿⁰_𝑡𝑒^{𝛼(𝑡−𝑠)}𝑣(𝑠)𝑑𝑠+𝑀^𝑡_−∞𝑒^{−𝛿(𝑡−𝑠)}𝑣(𝑠)𝑑𝑠,(𝑡≤0). (26) In addition, if 𝛾=𝐿/𝛼+𝑀/𝛿<1, then for 𝑡≥0 or 𝑡≤0, one has 𝑣(𝑡)≤1−𝛾⁻¹𝐶𝑒^{−[𝛼−(1−𝛾)−1𝐿]|𝑡|}. (27)

Proof. The proof is straightforward by Lemma 6.2 of Chapter 3 in [6].

Lemma 9. For any 𝑎∈ℝ^𝑛, 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 𝑥⁺(𝑡)=𝑇_𝐴(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑥⁺(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁺(𝑠)𝑑𝑠. (28)Similarly, for any 𝑎∈ℝ^𝑛, system (2) also has a unique solution 𝑥⁻(𝑡) with the following properties:(i)|𝑥⁻(𝑡)𝑒^𝜀𝑡|<+∞, for 𝑡≤0;(ii)𝑄(0)𝑥⁻(0)=𝑄(0)𝑎;(iii)𝑥⁻(𝑡) on ℝ⁻ satisfies integral equation 𝑥⁻(𝑡)=𝑇_𝐴(𝑡,0)𝑄(0)𝑎+^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓(𝑠,𝑥⁻(𝑠−))𝑑𝑠_0𝑡𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥⁻(𝑠))𝑑𝑠. (29)

Proof. We firstly prove the existence and uniqueness of 𝑥⁺(𝑡). Let 𝑣=𝑥𝑒^−𝜀𝑡, then system (2) becomes 𝑣̇(𝑡)=(𝐴(𝑡)−𝜀𝐼)𝑣(𝑡)+𝑒^−𝜀𝑡𝑓𝑡,𝑣𝑒^𝜀𝑡. (30) Let 𝐹(𝑡,𝑣)=𝑒^−𝜀𝑡𝑓(𝑡,𝑣𝑒^𝜀𝑡). From (H₁) and (H₂), we have |𝐹(𝑡,𝑣)|≤𝑒^−𝜀𝑡𝜇=𝜇𝑒^{−𝜀|𝑡|}𝐹,for𝑡≥0;𝑡,𝑣₁−𝐹𝑡,𝑣₂=𝑒^−𝜀𝑡𝑓𝑡,𝑣₁𝑒^𝜀𝑡−𝑓𝑡,𝑣₂𝑒^𝜀𝑡≤𝑒^−𝜀𝑡𝑟𝑒^{−𝜀|𝑡|}𝑣₁𝑒^𝜀𝑡−𝑣₂𝑒^𝜀𝑡≤𝑟𝑒^{−𝜀|𝑡|}𝑣₁−𝑣₂. (31)
Let 𝑇_𝐶(𝑡,𝑠) be the evolution operator of the linear system 𝑣(̇𝑡)=(𝐴(𝑡)−𝜀𝑡)𝑣(𝑡). Since 𝑥(𝑡)=𝑇_𝐴(𝑡,𝑠)𝑥(𝑠),𝑣=𝑒^−𝜀𝑡𝑥, we have 𝑇_𝐶(𝑡,𝑠)=𝑒^{−𝜀(𝑡−𝑠)}𝑇_𝐴(𝑡,𝑠). Hence, from (3), we obtain 𝑇_𝐶(𝑡,𝑠)𝑃(𝑠)≤𝐾𝑒^{−(𝛼−𝜀)(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|𝑇,for𝑡≥𝑠,_𝐶(𝑡,𝑠)𝑄(𝑠)≤𝐾𝑒^{(𝛼−𝜀)(𝑡−𝑠)}⋅𝑒^𝜀|𝑠|,for𝑡≤𝑠. (32) Since 4𝐾𝑟<𝛼−𝜀, system (30) satisfies all conditions of Lemma 7. Therefore, for any 𝑎∈ℝ^𝑛, 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 𝑣⁺(𝑡)=𝑇_𝐶(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐶(𝑡,𝑠)𝑃(𝑠)𝐹𝑠,𝑣⁺(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐶(𝑡,𝑠)𝑄(𝑠)𝐹𝑠,𝑣⁺(𝑠)𝑑𝑠=𝑒^−𝜀𝑡𝑇_𝐴()𝑃(0)𝑎+𝑡,0_𝑡0𝑒^{−𝜀(𝑡−𝑠)}𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)⋅𝑒^−𝜀𝑠𝑓𝑠,𝑣⁺(𝑠)𝑒^𝜀𝑠−𝑑𝑠_𝑡+∞𝑒^{−𝜀(𝑡−𝑠)}𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)⋅𝑒^−𝜀𝑠𝑓𝑠,𝑣⁺(𝑠)𝑒^𝜀𝑠𝑑𝑠. (33)Hence, 𝑣⁺(𝑡)𝑒^𝜀𝑡=𝑇_𝐴(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑣⁺(𝑠)𝑒^𝜀𝑠−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑣⁺(𝑠)𝑒^𝜀𝑠𝑑𝑠. (34) Let 𝑥⁺(𝑡)=𝑣⁺(𝑡)𝑒^𝜀𝑡, we have 𝑥⁺(𝑡)=𝑇_𝐴(𝑡,0)𝑃(0)𝑎+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑥⁺(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁺(𝑠)𝑑𝑠. (35) Then 𝑥⁺(𝑡) is the solution of system (2) and it satisfies all conditions of Lemma 9.
The proof for the existence and uniqueness of 𝑥⁻(𝑡) is similar to that of 𝑥⁺(𝑡), so we omit it. This completes the proof of Lemma 9.

Lemma 10. If |𝑇_𝐴(𝑡,0)𝑃(0)𝑎|≤𝑀𝑒^𝜀|𝑡|, then 𝑎=0.

Proof. If 𝑎≠0, then 𝑃(0)𝑎≠0 or 𝑄(0)𝑎≠0. Without loss of generality, we assume 𝑃(0)𝑎≠0, then we have 𝑇_𝐴(𝑡,0)𝑃(0)𝑎=𝑇_𝐴(𝑡,0)𝑃(0)𝑇_𝐴⁻¹(𝑠,0)𝑇_𝐴()𝑃(0)𝑎=𝑇𝑠,0_𝐴(𝑡,0)𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑇_𝐴()𝑃(0)𝑎=𝑇𝑠,0_𝐴(𝑡,𝑠)𝑃(𝑠)𝑇_𝐴()𝑃(0)𝑎≤𝑇𝑠,0_𝐴(𝑡,𝑠)𝑃(𝑠)⋅𝑇_𝐴(𝑠,0)𝑃(0)𝑎. (36) Since |𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)|≤𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠| for 𝑡≥𝑠, 𝑇_𝐴(𝑡,0)𝑃(0)𝑎≤𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑇_𝐴(𝑠,0)𝑃(0)𝑎, (37) which implies 𝑇_𝐴()𝑃(0)𝑎≥𝑇𝑠,0^𝐴(𝑡,0)𝑃(0)𝑎𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|. (38) Taking 𝑡=0, we obtain 𝑇_𝐴()𝑃(0)𝑎≥|𝑃(0)𝑎|𝑠,0𝐾𝑒^𝛼𝑠𝑒^−𝜀𝑠=𝐾⁻¹𝑒^{−(𝛼−𝜀)𝑠}|𝑃(0)𝑎|(𝑠≤0). (39) Therefore, when 𝑠≤0, 𝑇_𝐴(𝑠,0)𝑃(0)𝑎𝑒^𝜀|𝑠|≥𝐾⁻¹𝑒^−𝛼𝑠|𝑃(0)𝑎|⟶+∞as𝑠⟶−∞. (40) On the other hand, when 𝑠≤0, from (3), we have 𝑇_𝐴(𝑠,0)𝑄(0)𝑎≤𝐾𝑒^𝛼𝑠𝑒^𝜀|0|=𝐾𝑒^𝛼𝑠, (41) hence, 𝑇_𝐴(𝑠,0)𝑄(0)𝑎𝑒^𝜀|𝑠|≤𝐾𝑒^{(𝛼+𝜀)𝑠}. (42) It follows from (40) and (42) that 𝑇_𝐴(𝑠,0)𝑎𝑒^𝜀|𝑠|=𝑇_𝐴(𝑠,0)(𝑃(𝑠)+𝑄(𝑠))𝑎𝑒^𝜀|𝑠|≥𝑇_𝐴(𝑠,0)𝑃(𝑠)𝑎𝑒^𝜀|𝑠|−𝑇_𝐴(𝑠,0)𝑄(𝑠)𝑎𝑒^𝜀|𝑠|≥𝑇_𝐴(𝑠,0)𝑃(𝑠)𝑎𝑒^𝜀|𝑠|−𝐾𝑒^{(𝛼+𝜀)𝑠}. (43) From the above inequality, we know that |𝑇_𝐴(𝑠,0)𝑎|/𝑒^𝜀|𝑠|→+∞as𝑠→−∞, which contradicts the original condition |𝑇_𝐴(𝑠,0)𝑎|/𝑒^𝜀|𝑠|≤𝑀 and it implies 𝑎=0. This ends the proof of Lemma 10.

Proof of Theorem 4. For any solution 𝑥(𝑡) of system (2), it can be written as follows: 𝑥(𝑡)=𝑇_𝐴(𝑡,0)𝑥(0)+^𝑡₀𝑇_𝐴(𝑡,𝑠)𝑇_𝐴⁻¹(𝑠,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠=𝑇_𝐴(𝑡,0)𝑥(0)+^𝑡₀𝑇_𝐴()𝑃(𝑠)𝑓((𝑠+𝑡,𝑠𝑠,𝑥))𝑑𝑠_𝑡0𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠=𝑇_𝐴(𝑡,0)𝑥(0)+^𝑡_−∞𝑇_𝐴()𝑃(𝑠)𝑓((𝑠−𝑡,𝑠𝑠,𝑥))𝑑𝑠_0−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠+₀^+∞𝑇_𝐴()𝑄(𝑠)𝑓((𝑠−𝑡,𝑠𝑠,𝑥))𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠=𝑇_𝐴(𝑥(0)−𝑡,0)_0−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠−₀^+∞𝑇_𝐴()𝑄(𝑠)𝑓((𝑠+0,𝑠𝑠,𝑥))𝑑𝑠_𝑡−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠−_𝑡^+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (44) Let 𝜉(𝑡) be any n-variable continuous function defined on ℝ. From (3) and (H₁), we have ^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓(𝑠,𝜉(𝑠))𝑑𝑠≤^𝑡_−∞𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|𝜇𝑑𝑠=𝐾𝜇𝑒^{−𝛼𝑡𝑡}_−∞𝑒^𝛼𝑠𝑒^𝜀|𝑠|𝑑𝑠. (45) For 𝑡≥0, 𝑒^{−𝛼𝑡𝑡}_−∞𝑒^𝛼𝑠𝑒^𝜀|𝑠|𝑑𝑠=𝑒^−𝛼𝑡0_−∞𝑒^𝛼𝑠𝑒^−𝜀𝑠𝑑𝑠+^𝑡₀𝑒^𝛼𝑠𝑒^𝜀𝑠𝑑𝑠=𝑒^−𝛼𝑡1+1𝛼−𝜀𝑒𝛼+𝜀^{(𝛼+𝜀)𝑡}≤1−1𝑒𝛼+𝜀^𝜀𝑡+1≤1𝛼−𝜀𝑒𝛼−𝜀^𝜀|𝑡|+1;𝛼−𝜀 (46) for 𝑡≤0, 𝑒^{−𝛼𝑡𝑡}_−∞𝑒^𝛼𝑠𝑒^𝜀|𝑠|𝑑𝑠=𝑒^{−𝛼𝑡𝑡}_−∞𝑒^𝛼𝑠𝑒^−𝜀𝑠=1𝑑𝑠𝑒𝛼−𝜀^−𝜀𝑡≤1𝑒𝛼−𝜀^𝜀|𝑡|+1.𝛼−𝜀 (47) Hence, for any 𝑡∈ℝ, we have 𝑒^{−𝛼𝑡𝑡}_−∞𝑒^𝛼𝑠𝑒^𝜀|𝑠|1𝑑𝑠≤𝑒𝛼−𝜀^𝜀|𝑡|+1.𝛼−𝜀 (48) Therefore, for any 𝑡∈ℝ, we have ^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓(𝑠,𝜉(𝑠))𝑑𝑠≤𝐾𝜇𝑒𝛼−𝜀^𝜀|𝑡|+𝐾𝜇.𝛼−𝜀 (49) By the same calculation, for any 𝑡∈ℝ, we have _𝑡^+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝜉(𝑠))𝑑𝑠≤𝐾𝜇𝑒𝛼−𝜀^𝜀|𝑡|+𝐾𝜇.𝛼−𝜀 (50)
In Lemma 9, 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 ℝ^𝑛 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→ℝ^𝑛 as follows: 𝒯₂𝑎=⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓(𝑠,𝑥⁻(−𝑠,𝑎))𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁺(𝑠,𝑎)𝑑𝑠. (51) It follows from (3) and (H₁) that 𝒯₂𝑎≤⁰_−∞𝐾𝑒^𝛼𝑠𝑒^𝜀|𝑠|𝜇𝑑𝑠+₀^+∞𝐾𝑒^−𝛼𝑠𝑒^𝜀|𝑠|=𝜇𝑑𝑠𝐾𝜇+𝛼−𝜀𝐾𝜇=𝛼−𝜀2𝐾𝜇𝛼−𝜀=𝜌, (52) 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 𝒯₂𝑎₁−𝒯₂𝑎₂≤⁰_−∞𝐾𝑒^𝛼𝑠𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝑥⁻𝑠,𝑎₁−𝑥⁻𝑠,𝑎₂+𝑑𝑠_0+∞𝐾𝑒^−𝛼𝑠𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂≤𝑑𝑠_0−∞𝐾𝑟𝑒^𝛼𝑠𝑥⁻𝑠,𝑎₁−𝑥⁻𝑠,𝑎₂+𝑑𝑠_0+∞𝐾𝑟𝑒^−𝛼𝑠𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂𝑑𝑠. (53) From (3) and the condition (iii) of Lemma 9, for 𝑡≥0, we have 𝑥⁺𝑡,𝑎₁−𝑥⁺𝑡,𝑎₂≤𝑇_𝐴(𝑡,0)𝑃(0)𝑎₁−𝑎₂+^𝑡₀𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂+𝑑𝑠_𝑡+∞𝐾𝑒^{𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑠|}𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂𝑑𝑠≤𝐾𝑒^−𝛼𝑡𝑎₁−𝑎₂+^𝑡₀𝐾𝑟𝑒^{−𝛼(𝑡−𝑠)}𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂+𝑑𝑠_𝑡+∞𝐾𝑟𝑒^{𝛼(𝑡−𝑠)}𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂𝑑𝑠. (54) Multiplying by 𝑒^−𝜀𝑡 on both sides of the above inequality, for 𝑡≥0, we get 𝑒^−𝜀𝑡𝑥⁺𝑡,𝑎₁−𝑥⁺𝑡,𝑎₂≤𝐾𝑒^{−(𝛼+𝜀)𝑡}𝑎₁−𝑎₂+^𝑡₀𝐾𝑟𝑒^{−(𝛼+𝜀)(𝑡−𝑠)}𝑒^−𝜀𝑠𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂+𝑑𝑠_𝑡+∞𝐾𝑟𝑒^{(𝛼−𝜀)(𝑡−𝑠)}𝑒^−𝜀𝑠𝑥⁺𝑠,𝑎₁−𝑥⁺𝑠,𝑎₂𝑑𝑠. (55) 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 𝑒^−𝜀𝑡𝑥⁺𝑡,𝑎₁−𝑥⁺𝑡,𝑎₂≤𝐾𝑎₁−𝑎₂1−𝛾⁻¹𝑒^{−[(𝛼+𝜀)−(1−𝛾)−1𝐾𝑟]𝑡}, (56) where 𝛾=𝐾𝑟/(𝛼+𝜀)+𝐾𝑟/(𝛼−𝜀). From (H₃) and (𝛼+𝜀)⁻¹<(𝛼−𝜀)⁻¹, we get 𝛾≤2𝐾𝑟(𝛼−𝜀)⁻¹<12. (57) Therefore, for 𝛼−2𝐾𝑟>0, we have 𝑒^−𝜀𝑡𝑥⁺𝑡,𝑎₁−𝑥⁺𝑡,𝑎₂≤2𝐾𝑎₁−𝑎₂𝑒^{−[(𝛼+𝜀)−2𝐾𝑟]𝑡}(𝑡≥0). (58) Hence, 𝑥⁺𝑡,𝑎₁−𝑥⁺𝑡,𝑎₂≤2𝐾𝑎₁−𝑎₂𝑒^{−(𝛼−2𝐾𝑟)𝑡}(𝑡≥0). (59) Similarly, 𝑥⁻𝑡,𝑎₁−𝑥⁻𝑡,𝑎₂≤2𝐾𝑎₁−𝑎₂𝑒^{(𝛼−2𝐾𝑟)𝑡}(𝑡≤0). (60) So from (53), it follows that 𝒯₂𝑎₁−𝒯₂𝑎₂≤⁰_−∞𝐾𝑟𝑒^𝛼𝑠⋅2𝐾𝑎₁−𝑎₂𝑒^{(𝛼−2𝐾𝑟)𝑠}+𝑑𝑠_0+∞𝐾𝑟𝑒^−𝛼𝑠⋅2𝐾𝑎₁−𝑎₂𝑒^{−(𝛼−2𝐾𝑟)𝑠}≤𝑑𝑠_0−∞2𝐾²𝑟𝑎₁−𝑎₂𝑒^{2(𝛼−𝐾𝑟)𝑠}+𝑑𝑠_0+∞2𝐾²𝑟𝑎₁−𝑎₂𝑒^{−2(𝛼−𝐾𝑟)𝑠}=𝑑𝑠2𝐾²𝑟𝑎𝛼−𝐾𝑟¹−𝑎₂, (61) 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 𝑎₀=𝒯₂𝑎₀=⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥⁻𝑠,𝑎₀−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁺𝑠,𝑎₀𝑑𝑠. (62) As 𝑃²(𝑠)=𝑃(𝑠),𝑃(𝑡)𝑇_𝐴(𝑡,𝑠)=𝑇_𝐴(𝑡,𝑠)𝑃(𝑠),𝑃(𝑠)+𝑄(𝑠)=Id, we obtain 𝑃(0)𝑎₀=⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥⁻𝑠,𝑎₀𝑄(0)𝑎𝑑𝑠,₀=−₀^+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁻𝑠,𝑎₀𝑑𝑠. (63) From Lemma 9, we have 𝑥⁺0,𝑎₀=𝑃(0)𝑎₀−₀^+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥⁺𝑠,𝑎₀𝑥𝑑𝑠,₋0,𝑎₀=𝑄(0)𝑎₀+⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥⁻𝑠,𝑎₀𝑑𝑠. (64) Hence, 𝑥⁺0,𝑎₀=𝑥⁻0,𝑎₀=𝑎₀. (65) 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, 𝑥₀(0)=𝑎₀=⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥𝑠,𝑎₀𝑑𝑠−₀^+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥𝑠,𝑎₀𝑑𝑠. (66) From the above equation, it follows from (44) that 𝑥₀(𝑡)=^𝑡_−∞𝑇_𝐴()𝑃(𝑠)𝑓((𝑠−𝑡,𝑠𝑠,𝑥))𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (67) From (49) and (50), we have 𝑥₀(𝑡)≤2𝐾𝜇𝑒𝛼−𝜀^𝜀|𝑡|+2𝐾𝜇,(),𝛼−𝜀−∞<𝑡<+∞ (68) which implies that 𝑥₀(𝑡) satisfies (5); that is, 𝑥₀(𝑡)=𝑂(𝑒^𝜀|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 𝑥^∗₀(𝑡) satisfying (5). From (44), 𝑥^∗₀(𝑡) can be written as 𝑥^∗₀(𝑡)=𝑇_𝐴(𝑡,0)𝑥^∗₀(0)−⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)+𝑑𝑠_𝑡−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)𝑑𝑠. (69) From (49) and (50), we get ^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)𝑑𝑠≤𝐾𝜇𝑒𝛼−𝜀^𝜀|𝑡|+𝐾𝜇,𝛼−𝜀_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)𝑑𝑠≤𝐾𝜇𝑒𝛼−𝜀^𝜀|𝑡|+𝐾𝜇.𝛼−𝜀 (70) It follows from |𝑥^∗₀(𝑡)|=𝑂(𝑒^𝜀|𝑡|) and Lemma 10 that 𝑥^∗₀−⁰_−∞𝑇_𝐴(0,𝑠)𝑃(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)−𝑑𝑠_0+∞𝑇_𝐴(0,𝑠)𝑄(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)𝑑𝑠=0. (71) Therefore, 𝑥^∗₀(𝑡)=^𝑡_−∞𝑇_𝐴(𝑡,𝑠)𝑃(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)−𝑑𝑠_𝑡+∞𝑇_𝐴(𝑡,𝑠)𝑄(𝑠)𝑓𝑠,𝑥^∗₀(𝑠)𝑑𝑠. (72) From (67), (72), and (H₂), we have 𝑥₀(𝑡)−𝑥^∗₀(𝑡≤_𝑡−∞)𝐾𝑒^{−𝛼(𝑡−𝑠)}𝑒^𝜀|𝑠|⋅𝑟𝑒^{−𝜀|𝑡|}𝑥₀(𝑠)−𝑥^∗₀(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{𝛼(𝑡−𝑠)}𝑒^𝜀|𝑡|⋅𝑟𝑒^{−𝜀|𝑡|}𝑥₀(𝑠)−𝑥^∗₀(𝑠=)𝑑𝑠_𝑡−∞𝐾𝑟𝑒^{−𝛼(𝑡−𝑠)}𝑥₀(𝑠)−𝑥^∗₀(𝑠)𝑑𝑠+_𝑡^+∞𝐾𝑟𝑒^{𝛼(𝑡−𝑠)}𝑥₀(𝑠)−𝑥^∗₀(𝑠)𝑑𝑠. (73) Let 𝐿=sup_𝑡∈ℝ𝑒^{−𝜀|𝑡|}|𝑥₀(𝑡)−𝑥^∗₀(𝑡). Since 𝑥₀(𝑡) and 𝑥^∗₀(𝑡) satisfy (5), 𝑒^{−𝜀|𝑠|}|𝑥₀(𝑠)−𝑥^∗₀(𝑠)| is bounded. Thus, for 𝑡≥0, we have 𝑒^{−𝜀|𝑡|}𝑥₀(𝑡)−𝑥^∗₀(𝑡)=𝑒^−𝜀𝑡𝑥₀(𝑡)−𝑥^∗₀(𝑡≤_𝑡−∞)𝐾𝑒^{−(𝛼+𝜀)(𝑡−𝑠)}𝑟𝑒^−𝜀𝑠𝑥₀(𝑠)−𝑥^∗₀(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{(𝛼−𝜀)(𝑡−𝑠)}𝑟𝑒^−𝜀𝑠𝑥₀(𝑠)−𝑥^∗₀(𝑠=)𝑑𝑠_0−∞𝐾𝑒^{−(𝛼+𝜀)(𝑡−𝑠)}𝑟𝑒^−𝜀𝑠𝑥₀(𝑠)−𝑥^∗₀(𝑠+)𝑑𝑠_𝑡0𝐾𝑒^{−(𝛼+𝜀)(𝑡−𝑠)}𝑟𝑒^−𝜀𝑠𝑥₀(𝑠)−𝑥^∗₀(𝑠+)𝑑𝑠_𝑡+∞𝐾𝑒^{(𝛼−𝜀)(𝑡−𝑠)}𝑟𝑒^−𝜀𝑠𝑥₀(𝑠)−𝑥^∗₀(𝑠)𝑑𝑠≤𝐿𝐾𝑟𝑒𝛼+𝜀^{−(𝛼−𝜀)𝑡}+𝐾𝑟𝛼+𝜀1−𝑒^{−(𝛼+𝜀)𝑡}+𝐾𝑟𝛼−𝜀≤𝐿𝐾𝑟+𝛼+𝜀𝐾𝑟+𝛼+𝜀𝐾𝑟𝛼−𝜀≤𝐿3𝐾𝑟≤3𝛼−𝜀4𝐿byH₃. (74) Similarly, we can prove that 𝑒^{−𝜀|𝑡|}𝑥₀(𝑡)−𝑥^∗₀(𝑡≤3)4𝐿,when𝑡≤0. (75) Therefore, 𝐿=sup_𝑡∈ℝ𝑒^{−𝜀|𝑡|}𝑥₀(𝑡)−𝑥^∗₀(𝑡≤3)4𝐿,(−∞<𝑡+∞). (76) That is, 𝐿≤(3/4)𝐿, which implies 𝐿=0. Consequently, 𝑥₀(𝑡)=𝑥^∗₀(𝑡). This completes the proof of Theorem 4.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Yong-Hui Xia was supported by the National Natural Science Foundation of China under Grant nos. 11271333 and 11171090. Kit Ian Kou was supported by the University of Macau nos. MYRG142 (Y3-L2)-FST11-KKI, MRG002/KKI/2013/FST, MYRG099 (Y1-L2)-FST13-KKI, MRG 012/KKI/2014/FST and the Macao Science and Technology Development Fund FDCT/094/2011/A, FDCT/099/2012/A3.