Journal of Inequalities and Applications
Volume 2009 (2009), Article ID 369185, 11 pages
doi:10.1155/2009/369185
Research Article

Integrodifferential Inequality for Stability of Singularly Perturbed Impulsive Delay Integrodifferential Equations

Danhua He1 and Liguang Xu2

1Yangtze Center of Mathematics, Sichuan University, Chengdu 610064, China
2Department of Applied Mathematics, Zhejiang University of Technology, Hangzhou 310023, China

Received 24 February 2009; Accepted 11 June 2009

Academic Editor: Ondrej Dosly

Copyright © 2009 Danhua He and Liguang Xu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The exponential stability of singularly perturbed impulsive delay integrodifferential equations (SPIDIDEs) is concerned. By establishing an impulsive delay integrodifferential inequality (IDIDI), some sufficient conditions ensuring the exponentially stable of any solution of SPIDIDEs for sufficiently small 𝜀 > 0 are obtained. A numerical example shows the effectiveness of our theoretical results.

1. Introduction

Integrodifferential equations (IDEs) arise from many areas of science (from physics, biology, medicine, etc.), which have extensive scientific backgrounds and realistic mathematical models, and hence have been emerging as an important area of investigation in recent years, see [1–6]. Correspondingly, the stability of impulsive delay integrodifferential equations has been studied quite well, for example, [7–9]. However, besides delay and impulsive effects, singular perturbation likewise exists in a wide models for physiological processes or diseases [10]. And many good results on the stability of singularly perturbed delay differential equations have been reported, see, for example, [11–14]. Therefore, it is necessary to consider delay, impulse and singular perturbation on the stability of integrodifferential equations. However, to the best of our knowledge, there are no results on the problems of the exponential stability of solutions for SPIDIDEs due to some theoretical and technical difficulties. Based on this, this article is devoted to the discussion of this problem.

Applying differential inequalities, in [14–17], authors investigated the stability of impulsive differential equations. In [14], Zhu et al. established a delay differential inequality with impulsive initial conditions and derived some sufficient conditions ensuring the exponential stability of solutions for the singular perturbed impulsive delay differential equations (SPIDDEs). In this paper, we will improve the inequality established in [14] such that it is effective for SPIDIDEs. By establishing an IDIDI, some sufficient conditions ensuring the exponential stability of any solution of SPIDIDEs for sufficiently small 𝜀 > 0 are obtained. The results extend and improve the earlier publications, and which will be shown by the Remarks 3.2 and 3.5 provided later. An example is given to illustrate the theory.

2. Preliminaries

Throughout this letter, unless otherwise specified, let 𝑅 𝑛 be the space of 𝑛 -dimensional real column vectors and 𝑅 𝑚 × 𝑛 be the set of 𝑚 × 𝑛 real matrices. 𝒩 Δ = { 1 , 2 , … , 𝑛 } . For 𝐴 , 𝐵 ∈ 𝑅 𝑚 × 𝑛 or 𝐴 , 𝐵 ∈ 𝑅 𝑛 , 𝐴 ≥ 𝐵 ( 𝐴 ≤ 𝐵 , 𝐴 > 𝐵 , 𝐴 < 𝐵 ) means that each pair of corresponding elements of 𝐴 and 𝐵 satisfies the inequality “ ≥ ( ≤ , > , < )". Especially, 𝐴 is called a nonnegative matrix if 𝐴 ≥ 0 , and 𝑧 is called a positive vector if 𝑧 > 0 .

𝐶 [ 𝑋 , 𝑌 ] denotes the space of continuous mappings from the topological space 𝑋 to the topological space 𝑌 . In particular, let 𝐶 Δ = 𝐶 [ ( − ∞ , 0 ] , 𝑅 𝑛 ] denote the family of all bounded continuous 𝑅 𝑛 -valued functions 𝜙 defined on ( − ∞ , 0 ] with the norm ‖ 𝜙 ‖ = s u p − ∞ < 𝜃 ≤ 0 | 𝜙 ( 𝜃 ) | , where | ⋅ | is Euclidean norm of 𝑅 𝑛 .

𝑃 𝐶 [ 𝐼 , 𝑅 𝑛 ] Δ = { 𝜑 ∶ 𝐼 → 𝑅 𝑛 ∣ 𝜑 ( 𝑡 + ) = 𝜑 ( 𝑡 ) for 𝑡 ∈ 𝐼 , 𝜑 ( 𝑡 − ) exist for 𝑡 ∈ 𝐼 , 𝜑 ( 𝑡 − ) = 𝜑 ( 𝑡 ) for all but points 𝑡 𝑘 ∈ 𝐼 } , where 𝐼 ⊂ 𝑅 is an interval, 𝜑 ( 𝑡 + ) and 𝜑 ( 𝑡 − ) denote the left limit and right limit of scalar function 𝜑 ( 𝑡 ) , respectively. Especially, let 𝑃 𝐶 Δ = 𝑃 𝐶 [ ( − ∞ , 0 ] , 𝑅 𝑛 ] .

For 𝑥 ∈ 𝑅 𝑛 and 𝜑 ∈ 𝐶 or 𝜑 ∈ 𝑃 𝐶 , we define [ 𝑥 ] + | | 𝑥 = ( 1 | | | | 𝑥 , … , 𝑛 | | ) 𝑇 , [ 𝐴 ] + = | | 𝑎 𝑖 𝑗 | | 𝑛 × 𝑛 , [ 𝜑 ( 𝑡 ) ] 𝜏 = 𝜑 1 ( 𝑡 ) 𝜏 , … , [ 𝜑 𝑛 ( 𝑡 ) ] 𝜏 𝑇 , [ 𝜑 ( 𝑡 ) ] + 𝜏 = [ ] 𝜑 ( 𝑡 ) + 𝜏 , [ 𝜑 𝑖 ( 𝑡 ) ] 𝜏 = s u p − 𝜏 ≤ 𝑠 ≤ 0 𝜑 𝑖 𝐷 ( 𝑡 + 𝑠 ) , 𝑖 ∈ 𝒩 , + 𝜑 ( 𝑡 ) = l i m 𝑠 → 0 + 𝜑 s u p ( 𝑡 + 𝑠 ) − 𝜑 ( 𝑡 ) 𝑠 . ( 2 . 1 )

In this paper, we consider a class of SPIDIDEs described by 𝜀 ̇ 𝑥 ( 𝑡 ) = 𝐴 ( 𝑡 ) 𝑥 ( 𝑡 ) + 𝑓 ( 𝑡 , 𝑥 ( 𝑡 − 𝜏 ( 𝑡 ) ) ) + 𝑡 − ∞ 𝑅 ( 𝑡 − 𝑠 ) 𝐺 ( 𝑥 ( 𝑠 ) ) 𝑑 𝑠 , 𝑡 ≥ 𝑡 0 , 𝑡 ≠ 𝑡 𝑘 , 𝑥 𝑡 𝑘 = 𝐽 𝑘 𝑡 𝑘 𝑡 , 𝑥 − 𝑘 , 𝑘 = 1 , 2 , … , ( 2 . 2 ) with the initial conditions 𝑥 𝑡 0 ] + 𝜃 = 𝜙 ( 𝜃 ) ∈ 𝑃 𝐶 , 𝜃 ∈ ( − ∞ , 0 , ( 2 . 3 ) where 0 ≤ 𝜏 ( 𝑡 ) ≤ 𝜏 , 𝑥 ( 𝑡 ) = ( 𝑥 1 ( 𝑡 ) , … , 𝑥 𝑛 ( 𝑡 ) ) 𝑇 ∈ 𝑃 𝐶 [ 𝑅 , 𝑅 𝑛 ] , 𝐴 ( 𝑡 ) = ( 𝑎 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ∈ 𝑃 𝐶 [ 𝑅 , 𝑅 𝑛 × 𝑛 ] , 𝐽 𝑘 ∈ 𝐶 [ 𝑅 × 𝑅 𝑛 , 𝑅 𝑛 ] , 𝑅 ( 𝑡 ) = ( 𝑟 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ∈ 𝑃 𝐶 [ 𝑅 + , 𝑅 𝑛 × 𝑛 ] , 𝜀 ∈ ( 0 , 𝜀 0 ] is a small parameter, and 𝑡 1 < 𝑡 2 < ⋯ is a strictly increasing sequence such that l i m 𝑘 → ∞ 𝑡 𝑘 = ∞ .

Definition 2.1. The solution of (2.2) is said to be exponentially stable for sufficiently small 𝜀 if there exist finite constant vectors 𝐾 > 0 and 𝜎 > 0 , which are independent of 𝜀 ∈ ( 0 , 𝜀 0 ] for some 𝜀 0 , and a constant 𝜆 > 0 such that [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝐾 𝑒 − 𝜆 ( 𝑡 − 𝑡 0 ) for 𝑡 ≥ 𝑡 0 and for any initial perturbation satisfying s u p 𝑠 ∈ ( − ∞ , 0 ] [ 𝜙 ( 𝑠 ) − 𝜑 ( 𝑠 ) ] + < 𝜎 . Here 𝑦 ( 𝑡 ) is the solution of (2.2) corresponding to the initial condition 𝜑 .

3. Main Results

In order to prove the main result in this paper, we first need the following technique lemma.

Lemma 3.1. Assume that 0 ≤ 𝑢 ( 𝑡 ) = ( 𝑢 1 ( 𝑡 ) , … , 𝑢 𝑛 ( 𝑡 ) ) 𝑇 ∈ 𝑅 𝑛 , 𝑡 ≥ 𝑡 0 satisfy 𝐷 + 𝑢 ( 𝑡 ) ≤ 𝑃 ( 𝑡 ) 𝑢 ( 𝑡 ) + 𝑄 ( 𝑡 ) [ 𝑢 ( 𝑡 ) ] 𝜏 + ∞ 0 𝑊 ( 𝑠 ) 𝑢 ( 𝑡 − 𝑠 ) 𝑑 𝑠 , 𝑡 ≥ 𝑡 0 , 𝑢 𝑡 0 ] , + 𝜃 = 𝜑 ( 𝜃 ) ∈ 𝑃 𝐶 , 𝜃 ∈ ( − ∞ , 0 ( 3 . 1 ) where 𝑃 ( 𝑡 ) = ( 𝑝 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ≥ 0 for 𝑡 ≥ 𝑡 0 and 𝑖 ≠ 𝑗 , 𝑄 ( 𝑡 ) = ( 𝑞 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ≥ 0 for 𝑡 ≥ 𝑡 0 , 𝑊 ( 𝑠 ) = ( 𝑤 𝑖 𝑗 ( 𝑠 ) ) 𝑛 × 𝑛 ≥ 0 .
If there exist a positive constant 𝜉 and a positive vector 𝑧 = ( 𝑧 1 , … , 𝑧 𝑛 ) 𝑇 ∈ 𝑅 𝑛 and two positive diagonal matrices 𝐿 = d i a g { 𝐿 1 , … , 𝐿 𝑛 } , 𝐻 = d i a g { ℎ 1 , … , ℎ 𝑛 } with 0 < ℎ 𝑖 < 1 such that 𝑄 ( 𝑡 ) + 𝐻 𝑃 ( 𝑡 ) + 𝐿 + ∞ 0 𝑊 ( 𝑠 ) 𝑒 𝜉 𝑠 𝑑 𝑠 𝑧 < 0 , 𝑡 ≥ 𝑡 0 . ( 3 . 2 ) Then one has 𝑢 ( 𝑡 ) ≤ 𝑧 𝑒 − 𝜆 ( 𝑡 − 𝑡 0 ) , 𝑡 ≥ 𝑡 0 , ( 3 . 3 ) where the positive constant 𝜆 is defined as 0 < 𝜆 < 𝜆 0 = m i n 1 ≤ 𝑖 ≤ 𝑛 i n f 𝑡 ≥ 𝑡 0 𝜆 𝑖 ( 𝑡 ) ∶ 𝜆 𝑖 ( 𝑡 ) 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) + 𝑞 𝑖 𝑗 ( 𝑡 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝜏 + ∞ 0 𝑤 𝑖 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝑠 𝑧 𝑑 𝑠 𝑗 = 0 , ( 3 . 4 ) for the given 𝑧 .

Proof. Note that the result is trivial if 𝜏 = 0 . In the following, we assume that 𝜏 > 0 . Denote 𝐹 𝜆 𝑖 ( 𝑡 ) = 𝜆 𝑖 ( 𝑡 ) 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) + 𝑞 𝑖 𝑗 ( 𝑡 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝜏 + ∞ 0 𝑤 𝑖 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝑠 𝑧 𝑑 𝑠 𝑗 , 𝑡 ≥ 𝑡 0 , 𝑖 ∈ 𝒩 , ( 3 . 5 ) then for any given 𝑡 ≥ 𝑡 0 , we have 𝐹 ( 0 ) = 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) + 𝑞 𝑖 𝑗 ( 𝑡 ) + ∞ 0 𝑤 𝑖 𝑗 𝑧 ( 𝑠 ) 𝑑 𝑠 𝑗 ≤ 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) 𝑧 𝑗 − ℎ 𝑖 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) 𝑧 𝑗 = 1 − ℎ 𝑖 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) 𝑧 𝑗 ≤ − 1 − ℎ 𝑖 𝐿 𝑖 ℎ 𝑖 𝑧 𝑖 < 0 , ( 3 . 6 ) the first inequality and the second inequality are from (3.2), the last inequality is because 0 < ℎ 𝑖 < 1 , 𝐿 𝑖 > 0 , 𝑧 𝑖 > 0 , 𝑖 ∈ 𝒩 .
We also have l i m 𝜆 𝑖 ( 𝑡 ) → ∞ 𝐹 𝜆 𝑖 ( 𝑡 ) = ∞ , 𝐹 𝜆 𝑖 ( 𝑡 ) = 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑞 𝑖 𝑗 ( 𝑡 ) 𝜏 𝑒 𝜆 𝑖 ( 𝑡 ) 𝜏 + ∞ 0 𝑤 𝑖 𝑗 ( 𝑠 ) 𝑠 𝑒 𝜆 𝑖 ( 𝑡 ) 𝑠 𝑧 𝑑 𝑠 𝑗 > 0 . ( 3 . 7 ) So by (3.6) and (3.7), for any 𝑡 ≥ 𝑡 0 , there is a unique positive 𝜆 𝑖 ( 𝑡 ) such that 𝜆 𝑖 ( 𝑡 ) 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑝 𝑖 𝑗 ( 𝑡 ) + 𝑞 𝑖 𝑗 ( 𝑡 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝜏 + ∞ 0 𝑤 𝑖 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑖 ( 𝑡 ) 𝑠 𝑧 𝑑 𝑠 𝑗 = 0 , 𝑖 ∈ 𝒩 . ( 3 . 8 ) Therefore, from the definition of 𝜆 0 , one can know that 𝜆 0 ≥ 0 .
Next, we will show that 𝜆 0 ≠ 0 .
If this is not true, fix 𝑣 𝑖 satisfying 0 < ℎ 𝑖 < 𝑣 𝑖 < 1 and 1 − ℎ 𝑖 / 𝑣 𝑖 − ℎ 𝑖 > 0 , 𝑖 ∈ 𝒩 , there exist a 𝑡 ∗ ≥ 𝑡 0 and some integer 𝑙 such that 𝜆 𝑙 ( 𝑡 ∗ ) < 𝛿 , where 0 < 𝛿 < m i n { ( 1 − ℎ 𝑙 / 𝑣 𝑙 − ℎ 𝑙 ) ( 𝐿 𝑙 / ℎ 𝑙 ) , ( 𝑙 / 𝜏 ) l n ( 1 / 𝑣 𝑙 ) , 𝜉 } , such that 𝜆 𝑙 𝑡 ∗ 𝑧 𝑙 + 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) + 𝑞 𝑙 𝑗 ( 𝑡 ) 𝑒 𝜆 𝑙 ( 𝑡 ∗ ) 𝜏 + ∞ 0 𝑤 𝑖 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑙 ( 𝑡 ∗ ) 𝑠 𝑧 𝑑 𝑠 𝑗 = 0 . ( 3 . 9 )
Then, we have 0 = 𝜆 𝑙 𝑡 ∗ 𝑧 𝑙 + 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) + 𝑞 𝑙 𝑗 ( 𝑡 ) 𝑒 𝜆 𝑙 ( 𝑡 ∗ ) 𝜏 + ∞ 0 𝑤 𝑙 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑙 ( 𝑡 ∗ ) 𝑠 𝑧 𝑑 𝑠 𝑗 < 𝛿 𝑧 𝑙 + 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) + 𝑞 𝑙 𝑗 ( 𝑡 ) 𝑒 𝛿 𝜏 𝑧 𝑗 + 𝑛 𝑗 = 1 ∞ 0 𝑤 𝑙 𝑗 ( 𝑠 ) 𝑧 𝑗 𝑒 𝛿 𝑠 𝑑 𝑠 < 𝛿 𝑧 𝑙 + 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 1 ( 𝑡 ) + 𝑣 𝑙 𝑞 𝑙 𝑗 𝑧 ( 𝑡 ) 𝑗 + 𝑛 𝑗 = 1 ∞ 0 𝑤 𝑙 𝑗 ( 𝑠 ) 𝑧 𝑗 𝑒 𝜉 𝑠 𝑑 𝑠 ≤ 𝛿 𝑧 𝑙 + 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) 𝑧 𝑗 − ℎ 𝑙 𝑣 𝑙 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) 𝑧 𝑗 − ℎ 𝑙 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) 𝑧 𝑗 = 𝛿 𝑧 𝑙 + ℎ 1 − 𝑙 𝑣 𝑙 − ℎ 𝑙 𝑛 𝑗 = 1 𝑝 𝑙 𝑗 ( 𝑡 ) 𝑧 𝑗 ≤ 𝛿 𝑧 𝑙 − ℎ 1 − 𝑙 𝑣 𝑙 − ℎ 𝑙 𝐿 𝑙 ℎ 𝑙 𝑧 𝑙 < 0 , ( 3 . 1 0 ) this contradiction shows that 𝜆 0 > 0 , so there at least exists a positive constant 𝜆 0 such that 0 < 𝜆 < 𝜆 0 , that is, the definition of 𝜆 for (3.3) is reasonable.
Since 𝜑 ( 𝑡 ) ∈ 𝑃 𝐶 is bounded, we always can choose a sufficiently large 𝑧 > 0 such that 𝑢 ( 𝑡 ) ≤ 𝑧 𝑒 − 𝜆 ( 𝑡 − 𝑡 0 ) , − ∞ < 𝑡 ≤ 𝑡 0 . ( 3 . 1 1 ) In order to prove (3.3), we first prove for any given 𝑘 > 1 , 𝑢 𝑖 ( 𝑡 ) < 𝑘 𝑧 𝑖 𝑒 − 𝜆 ( 𝑡 − 𝑡 0 ) ≡ 𝑣 𝑖 ( 𝑡 ) , 𝑡 ≥ 𝑡 0 , 𝑖 ∈ 𝒩 . ( 3 . 1 2 ) If (3.12) is not true, then by continuity of 𝑢 ( 𝑡 ) , there must exist some integer 𝑚 and ̂ 𝑡 > 𝑡 0 such that 𝑢 𝑚 ̂ 𝑡 = 𝑣 𝑚 ̂ 𝑡 , 𝐷 + 𝑢 𝑚 ̂ 𝑡 ≥ 𝑣 𝑚 ̂ 𝑡 𝑢 , ( 3 . 1 3 ) 𝑖 ( 𝑡 ) ≤ 𝑣 𝑖 ̂ ( 𝑡 ) , − ∞ < 𝑡 ≤ 𝑡 , 𝑖 ∈ 𝒩 . ( 3 . 1 4 ) So, by (3.1), the equality of (3.13), (3.14) and 𝑝 𝑖 𝑗 ( 𝑡 ) ≥ 0 and 𝑖 ≠ 𝑗 , 𝑞 𝑖 𝑗 ( 𝑡 ) ≥ 0 , for 𝑡 ≥ 𝑡 0 , and the definition of 𝜆 , we derive that 𝐷 + 𝑢 𝑚 ̂ 𝑡 ≤ 𝑛 𝑗 = 1 𝑝 𝑚 𝑗 ̂ 𝑡 𝑢 𝑗 ̂ 𝑡 + 𝑞 𝑚 𝑗 ̂ 𝑡 𝑢 𝑗 ̂ + 𝑡 − 𝜏 𝑛 𝑗 = 1 ∞ 0 𝑤 𝑚 𝑗 ( 𝑠 ) 𝑢 𝑗 ̂ ≤ 𝑡 − 𝑠 𝑑 𝑠 𝑛 𝑗 = 1 𝑝 𝑚 𝑗 ̂ 𝑡 𝑘 𝑧 𝑗 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑡 0 ) + 𝑞 𝑚 𝑗 ̂ 𝑡 𝑘 𝑧 𝑗 𝑒 − 𝜆 ( ̂ 𝑡 − 𝜏 − 𝑡 0 ) + 𝑛 𝑗 = 1 ∞ 0 𝑤 𝑚 𝑗 ( 𝑠 ) 𝑘 𝑧 𝑗 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑠 − 𝑡 0 ) = 𝑑 𝑠 𝑛 𝑗 = 1 𝑝 𝑚 𝑗 ̂ 𝑡 + 𝑞 𝑚 𝑗 ̂ 𝑡 𝑒 𝜆 𝜏 + ∞ 0 𝑤 𝑚 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑠 𝑑 𝑠 𝑘 𝑧 𝑗 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑡 0 ) < 𝑛 𝑗 = 1 𝑝 𝑚 𝑗 ̂ 𝑡 + 𝑞 𝑚 𝑗 ̂ 𝑡 𝑒 𝜆 𝑚 ( ̂ 𝑡 ) 𝜏 + ∞ 0 𝑤 𝑚 𝑗 ( 𝑠 ) 𝑒 𝜆 𝑚 ( ̂ 𝑡 ) 𝜏 𝑑 𝑠 𝑘 𝑧 𝑗 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑡 0 ) = − 𝜆 𝑚 ̂ 𝑡 𝑧 𝑚 𝑘 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑡 0 ) < − 𝜆 𝑧 𝑚 𝑘 𝑒 − 𝜆 ( ̂ 𝑡 − 𝑡 0 ) = 𝑣 𝑚 ̂ 𝑡 , ( 3 . 1 5 ) which contradicts the inequality in (3.13), and so (3.12) holds for all 𝑡 ≥ 𝑡 0 . Letting 𝑘 → 1 , then (3.3) holds, and the proof is completed.

Remark 3.2. If 𝑊 ( 𝑠 ) = ( 𝑤 𝑖 𝑗 ( 𝑠 ) ) 𝑛 × 𝑛 = 0 in Lemma 3.1, then we get [14, Lemma  1].

Theorem 3.3. Assume that 𝐴 ( 𝑡 ) = ( 𝑎 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ≥ 0 for 𝑡 ≥ 𝑡 0 and 𝑖 ≠ 𝑗 , further suppose the following ( 𝐻 1 ) For any 𝑥 , 𝑦 ∈ 𝑅 𝑛 , there exist nonnegative matrices 𝑈 ( 𝑡 ) = ( 𝑢 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 and 𝐵 = ( 𝑏 𝑖 𝑗 ) 𝑛 × 𝑛 , 𝑡 ≥ 𝑡 0 , such that [ 𝑓 ( 𝑡 , 𝑥 ) − 𝑓 ( 𝑡 , 𝑦 ) ] + ≤ 𝑈 ( 𝑡 ) [ 𝑥 − 𝑦 ] + , 𝑡 ≥ 𝑡 0 , [ 𝐺 ( 𝑥 ) − 𝐺 ( 𝑦 ) ] + ≤ 𝐵 [ 𝑥 − 𝑦 ] + , 𝑡 ≥ 𝑡 0 . ( 3 . 1 6 ) ( 𝐻 2 ) For any 𝑥 , 𝑦 ∈ 𝑅 𝑛 ,there exist nonnegative constant matrices 𝑀 𝑘 such that [ 𝐽 𝑘 ( 𝑡 , 𝑥 ) − 𝐽 𝑘 ( 𝑡 , 𝑦 ) ] + ≤ 𝑀 𝑘 [ 𝑥 − 𝑦 ] + , 𝑡 ≥ 𝑡 0 . ( 3 . 1 7 ) ( 𝐻 3 ) There exist a positive constant 𝜉 and a positive vector 𝑧 = ( 𝑧 1 , 𝑧 2 , … , 𝑧 𝑛 ) 𝑇 ∈ 𝑅 𝑛 and two positive diagonal matrices 𝑉 = d i a g { 𝑣 1 , … , 𝑣 𝑛 } , 𝑆 = d i a g { 𝑠 1 , 𝑠 2 , … , 𝑠 𝑛 } , with 0 < 𝑠 𝑖 < 1 , 𝑖 ∈ 𝒩 such that 𝑈 ( 𝑡 ) + 𝑆 𝐴 ( 𝑡 ) + 𝑉 + ∞ 0 𝑁 ( 𝑠 ) 𝑒 𝜉 𝑠 𝑑 𝑠 𝑧 < 0 , 𝑡 ≥ 𝑡 0 , ( 3 . 1 8 ) where 𝑁 ( 𝑠 ) = ( 𝑛 𝑖 𝑗 ( 𝑠 ) ) 𝑛 × 𝑛 ≥ 0 , 𝑛 𝑖 𝑗 ( 𝑠 ) = | 𝑟 𝑖 𝑗 ( 𝑠 ) | 𝑏 𝑖 𝑗 . ( 𝐻 4 ) There exists a positive constant 𝜂 satisfying l n 𝜂 𝑘 𝑡 𝑘 − 𝑡 𝑘 − 1 ≤ 𝜂 < 𝜆 ( 𝜀 ) , 𝑘 = 1 , 2 , … , ( 3 . 1 9 ) where 𝜂 𝑘 satisfy 𝜂 𝑘 ≥ 1 , 𝜂 𝑘 𝑧 ≥ 𝑀 𝑘 𝑧 , ( 3 . 2 0 ) and 𝜆 ( 𝜀 ) is defined as 0 < 𝜆 ( 𝜀 ) < 𝜆 0 ( 𝜀 ) = m i n 1 ≤ 𝑖 ≤ 𝑛 i n f 𝑡 ≥ 𝑡 0 𝜆 𝑖 ( 𝑡 , 𝜀 ) ∶ 𝜆 𝑖 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑎 𝑖 𝑗 ( 𝑡 ) 𝜀 + 𝑢 𝑖 𝑗 ( 𝑡 ) 𝜀 𝑒 𝜆 𝑖 𝜏 + ∞ 0 𝑛 𝑖 𝑗 ( 𝑠 ) 𝜀 𝑒 𝜆 𝑖 𝑠 𝑧 𝑑 𝑠 𝑗 . = 0 ( 3 . 2 1 ) for the given 𝑧 .
Then there exists a small 𝜀 0 > 0 such that the solution of (2.2) is exponentially stable for sufficiently small 𝜀 ∈ ( 0 , 𝜀 0 ] .

Proof. By a similar argument with (3.4), one can know that the 𝜆 ( 𝜀 ) defined by (3.21) is reasonable. For any 𝜙 , 𝜑 ∈ 𝑃 𝐶 , let 𝑥 ( 𝑡 ) , 𝑦 ( 𝑡 ) be two solutions of (2.2) through ( 𝑡 0 , 𝜙 ) , ( 𝑡 0 , 𝜑 ) , respectively. Since 𝜙 , 𝜑 ∈ 𝑃 𝐶 are bounded, we can always choose a positive vector 𝑧 such that [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝑧 𝑒 − 𝜆 ( 𝑡 − 𝑡 0 ) , 𝑡 ∈ − ∞ , 𝑡 0 . ( 3 . 2 2 ) Calculating the upper right derivative 𝐷 + [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + along the solution of (2.2), by condition ( 𝐻 1 ) , we have 𝐷 + [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + = S g n ( 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ) ( 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ) ≤ S g n ( 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ) 𝐴 ( 𝑡 ) 𝜀 1 ( 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ) + 𝜀 [ 𝑓 ( 𝑡 , 𝑥 ( 𝑡 − 𝜏 ( 𝑡 ) ) ) − 𝑓 ( 𝑡 , 𝑦 ( 𝑡 − 𝜏 ( 𝑡 ) ) ) ] + + 1 𝜀 𝑡 − ∞ 𝑅 ( 𝑡 − 𝑠 ) [ 𝐺 ( 𝑥 ( 𝑠 ) ) − 𝐺 ( 𝑦 ( 𝑠 ) ) ] + ≤ 𝑑 𝑠 𝐴 ( 𝑡 ) 𝜀 [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + + 𝑈 ( 𝑡 ) 𝜀 [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + 𝜏 + 1 𝜀 ∞ 0 [ 𝑅 ( 𝑠 ) ] + 𝐵 [ 𝑥 ( 𝑡 − 𝑠 ) − 𝑦 ( 𝑡 − 𝑠 ) ] + 𝑑 𝑠 . ( 3 . 2 3 ) From condition ( 𝐻 3 ) , we have 𝑈 ( 𝑡 ) 𝜀 + 𝑆 𝐴 ( 𝑡 ) 𝜀 + 𝑉 𝜀 + ∞ 0 𝑁 ( 𝑠 ) 𝜀 𝑒 𝜉 𝑠 𝑑 𝑠 𝑧 < 0 , 𝑡 ≥ 𝑡 0 . ( 3 . 2 4 ) Therefore, (3.23) and (3.24) imply that all the assumptions of Lemma 3.1 are true. So we have [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝑧 𝑒 − 𝜆 ( 𝜀 ) ( 𝑡 − 𝑡 0 ) 𝑡 , 𝑡 ∈ 0 , 𝑡 1 , ( 3 . 2 5 ) where 𝜆 ( 𝜀 ) is determined by (3.21) and the positive constant vector 𝑧 is determined by (3.18).
Using the discrete part of (2.2), condition ( 𝐻 2 ) , (3.20) and (3.25), we can obtain that 𝑡 [ 𝑥 1 𝑡 − 𝑦 1 ] + = [ 𝐽 1 𝑡 1 𝑡 , 𝑥 − 1 − 𝐽 1 𝑡 1 𝑡 , 𝑦 − 1 ] + ≤ 𝑀 1 𝑡 [ 𝑥 − 1 𝑡 − 𝑦 − 1 ] + ≤ 𝑀 1 𝑧 𝑒 − 𝜆 ( 𝜀 ) ( 𝑡 1 − 𝑡 0 ) ≤ 𝜂 1 𝑧 𝑒 − 𝜆 ( 𝜀 ) ( 𝑡 1 − 𝑡 0 ) , ( 3 . 2 6 ) and so, we have [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝜂 1 𝑧 𝑒 − 𝜆 ( 𝜀 ) 𝑡 − 𝑡 0 , 𝑡 ∈ − ∞ , 𝑡 1 . ( 3 . 2 7 ) By a similar argument with (3.25), we can use (3.27) derive that [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝜂 1 𝑧 𝑒 − 𝜆 ( 𝜀 ) 𝑡 − 𝑡 0 𝑡 , 𝑡 ∈ 1 , 𝑡 2 . ( 3 . 2 8 ) Therefore, by simple induction, we have [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝜂 1 ⋯ 𝜂 𝑘 − 1 𝑧 𝑒 − 𝜆 ( 𝜀 ) 𝑡 − 𝑡 0 𝑡 , 𝑡 ∈ 𝑘 − 1 , 𝑡 𝑘 , 𝑘 = 1 , 2 , … . ( 3 . 2 9 ) From (3.19) and (3.29), we obtain [ 𝑥 ( 𝑡 ) − 𝑦 ( 𝑡 ) ] + ≤ 𝑧 𝑒 − ( 𝜆 ( 𝜀 ) − 𝜂 ) 𝑡 − 𝑡 0 𝑡 , 𝑡 ∈ 𝑘 − 1 , 𝑡 𝑘 , 𝑘 = 1 , 2 , … , ∀ 𝜀 > 0 . ( 3 . 3 0 ) For any 𝑡 ≥ 𝑡 0 , let 𝜆 𝑖 ( 𝑡 , 𝜀 ) be defined as the unique positive zero of 𝜆 𝑖 𝑧 𝑖 + 𝑛 𝑗 = 1 𝑎 𝑖 𝑗 ( 𝑡 ) 𝜀 + 𝑢 𝑖 𝑗 ( 𝑡 ) 𝜀 𝑒 𝜆 𝑖 𝜏 + ∞ 0 𝑛 𝑖 𝑗 ( 𝑠 ) 𝜀 𝑒 𝜆 𝑖 𝑠 𝑧 𝑑 𝑠 𝑗 = 0 . ( 3 . 3 1 ) Differentiate both sides of (3.31) with respect to the variable 𝜀 , we have 𝑑 𝜆 𝑑 𝜀 𝑖 ( 𝑡 , 𝜀 ) = − 𝜆 𝑖 𝑧 𝑖 𝜀 𝑧 𝑖 + ∑ 𝑛 𝑗 = 1 𝑢 𝑖 𝑗 ( 𝑡 ) 𝜏 𝑒 𝜆 𝑖 𝜏 𝑧 𝑗 + ∑ 𝑛 𝑗 = 1 ∫ ∞ 0 𝑛 𝑖 𝑗 ( 𝑠 ) 𝑠 𝑧 𝑗 𝑒 𝜆 𝑖 𝑠 𝑑 𝑠 < 0 , ( 3 . 3 2 ) so 𝜆 𝑖 ( 𝑡 , 𝜀 ) is monotonically decreasing with respect to the variable 𝜀 , which implies that 𝜆 0 ( 𝜀 ) is also monotonically decreasing with respect to the variable 𝜀 . So we can choose the 𝜆 ( 𝜀 ) in (3.21) satisfying the same monotonicity with 𝜆 0 ( 𝜀 ) , for example, 𝜆 ( 𝜀 ) = 𝜆 0 ( 𝜀 ) − 𝛿 , where 0 < 𝛿 < 𝜆 0 ( 𝜀 ) − 𝜆 ( 𝜀 ) . Hence we can deduce that there exists a small 𝜀 0 > 0 such that the solution of (2.2) is exponentially stable for sufficiently small 𝜀 ∈ ( 0 , 𝜀 0 ] . The proof is completed.

Remark 3.4. Suppose that 𝑁 ( 𝑠 ) = ( 𝑛 𝑖 𝑗 ( 𝑠 ) ) 𝑛 × 𝑛 = 0 in Theorem 3.3, then we can easily get [14, Theorem  1]. In fact, “ 𝜂 𝑘 Δ = m a x { ‖ 𝑀 𝑘 ‖ , 1 } ” of condition ( 𝐻 4 ) in [14, Theorem  1] ensure that the above (3.20) holds.

Remark 3.5. If 𝐽 𝑘 ( 𝑡 , 𝑥 ) = 𝑥 , 𝑡 ≥ 𝑡 0 , that is there have no impulses in (2.2), then by Theorem 3.3, we can obtain the following result.

Corollary 3.6. Assume that 𝐴 ( 𝑡 ) = ( 𝑎 𝑖 𝑗 ( 𝑡 ) ) 𝑛 × 𝑛 ≥ 0 for 𝑡 ≥ 𝑡 0 and 𝑖 ≠ 𝑗 , 𝑁 ( 𝑠 ) = ( 𝑛 𝑖 𝑗 ( 𝑠 ) ) 𝑛 × 𝑛 ≥ 0 , further suppose that ( 𝐻 1 ) and ( 𝐻 3 ) hold. Then there exists a small 𝜀 0 > 0 such that the solution of (2.2) is exponentially stable for sufficiently small 𝜀 ∈ ( 0 , 𝜀 0 ] .

Remark 3.7. From Lemma 3.1 and the proof of Theorem 3.3, it is obvious that the results obtained in this paper still hold for 𝜀 = 1 . So this type of exponential stability can obviously be applied to general impulsive delay integrodifferential equations.

Remark 3.8. When 𝜀 = 1 and 𝑟 𝑖 𝑗 ( 𝑠 ) = 0 , the global exponential stability criteria for (2.2) have been established in [18] by utilizing the Lyapunov functional method. However, the additional assumption that 𝑓 𝑗 is bounded is required in [18].

4. An Illustrative Example

In this section, we will give an example to illustrate the exponential stability of (2.2).

Example 4.1. Consider the following SPIDIDEs: 𝜀 ̇ 𝑥 1 ( 𝑡 ) = ( − 1 0 − s i n 𝑡 ) 𝑥 1 ( 𝑡 ) + ( 2 + s i n 𝑡 ) a r c t a n 𝑥 1 ( 𝑡 − 𝜏 ( 𝑡 ) ) + ( 1 + c o s 𝑡 ) a r c t a n 𝑥 2 ( 𝑡 − 𝜏 ( 𝑡 ) ) + 𝑡 − ∞ 𝑒 − ( 𝑡 − 𝑠 ) 𝑥 1 ( 𝑠 ) 𝑑 𝑠 , 𝑡 ≠ 𝑡 𝑘 , 𝜀 ̇ 𝑥 2 ( 𝑡 ) = ( − 8 + 2 c o s 𝑡 ) 𝑥 1 ( 𝑡 ) + s i n 2 𝑡 a r c t a n 𝑥 1 + ( 𝑡 − 𝜏 ( 𝑡 ) ) ( 1 − c o s 𝑡 ) a r c t a n 𝑥 2 ( 𝑡 − 𝜏 ( 𝑡 ) ) + 𝑡 − ∞ 𝑒 − 2 ( 𝑡 − 𝑠 ) 𝑥 2 ( 𝑠 ) 𝑑 𝑠 , 𝑡 ≠ 𝑡 𝑘 , 𝑥 1 𝑡 𝑘 = 𝛼 1 𝑘 𝑥 1 𝑡 − 𝑘 − 𝛽 1 𝑘 𝑥 2 𝑡 − 𝑘 , 𝑥 2 𝑡 𝑘 = 𝛽 2 𝑘 𝑥 1 𝑡 − 𝑘 + 𝛼 2 𝑘 𝑥 2 𝑡 − 𝑘 , ( 4 . 1 ) where 𝛼 𝑖 𝑘 , 𝛽 𝑖 𝑘 ≥ 0 are constants, 𝜏 ( 𝑡 ) = 𝑒 − 𝑡 ≤ 1 Δ = 𝜏 , 𝑡 ≥ 𝑡 0 , 𝑡 𝑘 = 𝑡 𝑘 − 1 + 3 𝑘 , 𝑘 = 1 , 2 , … .
We can easily find that conditions ( 𝐻 1 ) and ( 𝐻 2 ) are satisfied with 𝐴 ( 𝑡 ) = − 1 0 − s i n 𝑡 0 0 − 8 + 2 c o s 𝑡 , 𝑈 ( 𝑡 ) = 2 + s i n 𝑡 1 + c o s 𝑡 s i n 2 , 𝑒 𝑡 1 − c o s 𝑡 𝑁 ( 𝑠 ) = − 𝑠 0 0 𝑒 − 2 𝑠 , 𝑀 𝑘 = 𝛼 1 𝑘 𝛽 1 𝑘 𝛽 2 𝑘 𝛼 2 𝑘 . ( 4 . 2 ) So there exist 𝜉 = 0 . 5 , 𝑧 = ( 1 , 1 ) 𝑇 , 𝑉 = d i a g { 1 , 1 / 3 } and 𝑆 = d i a g { 0 . 8 , 0 . 5 } such that 𝑈 ( 𝑡 ) + 𝑆 𝐴 ( 𝑡 ) + 𝑉 + ∞ 0 𝑁 ( 𝑠 ) 𝑒 𝜉 𝑠 𝑑 𝑠 𝑧