Initial value problems for first order impulsive integro-differential equations of Volterra type in Banach spaces

In this paper, we use a new method and combining the partial ordering method to study the existence of the solutions for the first order nonlinear impulsive integro-differential equations of Volterra type on finite interval in Banach spaces and for the first order nonlinear impulsive integro-differential equations of Volterra type on infinite interval with infinite number impulsive times in Banach spaces. By introducing an interim space and using progressive estimation method, some restrictive conditions on impulsive terms, used before, such as, prior estimation, noncompactness measure estimations are deleted. ∗The first author is supported by natural science foundation of the EDJP and JSQLGC, China. 10 Initial value problems


Introduction
Impulsive differential equations arise naturally from a wide variety of applications, such as spacecraft control, inspection processes in operations research, drug administration, and threshold theory in biology, see [1], [8] and [10].Initial value problems for first order impulsive integro-differential equations of Volterra type in Banach Spaces has been studied in [5], [6] and [9].However, to obtain the solutions, some restricted conditions on impulsive terms, such as, prior estimation, noncompactness measure estimation, are needed in [5], [6] and [9].In this paper, we use a new method and combining the partial ordering method to study the existence of the solutions for the first order nonlinear impulsive integro-differential equations of Volterra type on infinite interval with infinite number impulsive times in Banach spaces.By introducing an interim space and using progressive estimation method, the restrictive conditions on impulsive terms, such as, prior estimation, noncompactness measure estimations are deleted.Our results and methods are different from the ones in [5], [6] and [9].
Let E be a Banach space, P ⊂ E be a closed convex cone in E .The partial order ≤ in E is induced by P , that is x ≤ y , if and only if y−x ∈ P .P is said to be normal if there exists a positive constant K , such that θ ≤ x ≤ y implies x ≤ K y , where θ denotes the zero element of E , and the smallest K is called the normal constant of P .P is said to be regular if x 1 ≤ x 2 ≤ • • • ≤ y implies x n − x → 0 as n → ∞ for some x ∈ E. For details on cone theory, see [6].
In this paper, we consider the following initial value problems for the first order nonlinear impulsive integro-differential equations of Volterra type on infinite interval with infinite number impulsive times in Banach spaces. (1) where u(t + k ) and u(t − k ) denote the right and left limits of u(t) at t = t k , respectively.
For fixed we introduce the interim spaces as follows: (2) where

Main Results
In this section, we discuss the existences of the solutions for the initial value problem (1) and need the following comparison result.
The proof of Lemma 1 is similar to the proof of Lemma 4.1.1 in [6] or refer to [3].
We give the following conditions for the next results: (H 1 ) There exist v 0 , w (H 2 ) There exist constants M ≥ 0 and N ≥ 0 such that (H 4 ) For any r > 0, and any i = 1, 2, 3, • • • , there exist constants c r,i ≥ 0 and c * r,i ≥ 0 such that where B r = {u ∈ E : u ≤ r} and α denotes the Kuratowski measure of noncompactness in E.
Theorem 2. Let P be normal and the conditions (H 1 ) ∼ (H 4 ) be satisfied.Assume that the inequality (5) holds.Then the initial value problem (1) , (H 2 ), (H 4 ) and by using Theorem 4.1.1 in [6] or Theorem 3.1.1 in [5] that initial value problem has a minimal solution and a maximal solution u 0 , u * 0 ∈ [v 0 , w 0 ] 0 , respectively.By (2) For any η ∈ [v 0 , w 0 ] 1 , we consider the initial value problem of linear integro-differential equation in E : (6) u ), where ( 7) It is easy to show that u ∈ P C[J 0,1 , u 0 , E] is a solution of the linear initial value problem (6) if and only if u ∈ P C[J 0,1 , u 0 , E] is a solution of the following integral equation: (8) can be written as follows: where Clearly, the integral equation ( 9) is a linear Volterra equation in E .Thus, by Theorem 1.4.2 in [5], it has a unique solution Therefore, u ∈ P C[J 0,1 , u 0 , E] is the unique solution of the initial value problems (6).Letting u = Aη, then the operator Then, from (6), we have and so, from the condition (H 1 ), (H 3 ), it follows that By the definition of P C[J 0,1 , u 0 , E], we have where u 1 = Aη 1 and u 2 = Aη 2 .From ( 6), it follows that and so, by using the condition (H 2 ), we can show that By the definition of P C[J 0,1 , u 0 , E], we have Therefore, by Lemma 1, it follows that p(t) ≤ θ for all t ∈ J 1 , that is, Aη 1 ≤ Aη 2 , for all t ∈ J 1 and so (ii) is proved. Let where v 1 0 = v 0 , w 1 0 = w 0 .By the conclusions (i) and (ii), we have (10) Then, by the normality of P, the set ).Since the condition (H 4 ) implies that the set f (J 1 , B r , B r ) is bounded for any r > 0 , So, there exists a constant c 0 > 0 such that (11) By the definition of v 1 n and (8), we have Then, from (11) and ( 12), all the functions in V is equicontinuous on J 1 and so by Theorem 1.2.2 in [6], the function [6] to (12), we have where On the other hand, by Corollary 1.2.1 in [6], we have Thus, from the condition (H 4 ), there exists constant c ≥ 0 and c * ≥ 0 such that Therefore, from (13)∼ (15), it follows that Then, from (16), we have and so z(t) ≤ z(t 1 ) = y(t 1 )e −τ t1 = 0 for all t ∈ J 1 .Thus z(t) ≡ 0 for all t ∈ J 1 , which implies that m(t) ≡ 0 for all t ∈ J 1 and, by Theorem n } is increasing, for any positive integer n, we have Then, {v 1 n } converges uniformly to v 1 on J 0,1 .Now, we have Further, from (11), it follows that (18) and so, from (17) and (18), by taking limits in both sides of (12), we get ( Hence we can easily show that v That is, v 1 (t) is a solution of the initial value problem (1) on J 0,1 .
Assume that we have get a solution v m−1 (t) of ( 1) on and where u k (t), is a uniformly limitation of the increasing sequence It is similar to the above proof of (10), we can get an increasing sequence Then, by the similar proof of (19), we have That is, v m (t) is a solution of the following By induction, we can get two sequences {v m (t From (20), ( 21) and ( 22) we can know that v(t) satisfies From the proof of Theorem 2 we can see that if ū0 is different from u * 0 on J 0 , then the initial value problem (1) has at least two solutions in [v 0 , w 0 ].

Remark 2. Condition (H 4
) is an improtant, most in use and general condition in nonolinear problems in abstract spaces.Concerening the checking and using of this kind of conditions, one can refer to [2], [4], [6] and [7].
Theorem 3. Let P be regular and the conditions (H 1 ) ∼ (H 3 ) be satisfied.Assume that the inequality (5) holds.Then the initial value problem (1) Proof.The proof is similar to that of Theorem 2. The only difference of the proof is that the conclusion m(t) = α(V (t)) = 0 for all t ∈ J 1 follows directly from the regularity of P and the fact that {v 1 n (t)} is an increasing sequence.This complete the proof.
From the proof of Theorem 2 we can get the following corollary.Corollary 4. If J = [0, a] and 0 < t 1 < • • • < t m < a.Let P be normal and the conditions (H 1 ) ∼ (H 4 ) be satisfied for i = 1, 2, • • •, m.Assume that the inequality (5) holds.Then the initial value problem (1)

Example
We consider the following initial value problem of infinite system for the scalar first order impulsive integro-differential equation: t 0 e −ts un(s)ds Evidently, u n (t) ≡ 0 (n = 1, 2, 3, • • • ) is not a solution of the initial value problem (23).

Conclusion.
The initial value problem (23) has at least a continuously differentiable solutions defined on [0, 1  2 ) with the norm be defined by u = sup n |u n | and Then P is a normal cone in E and (23) can be regarded as an initial value problem of form (1) in E, where Then we have and v 0 (t) ≤ w 0 (t) for all t ∈ J and so In addition, if 0 ≤ t ≤ Thus v 0 and w 0 satisfy the condition (H 1 ).On the other hand, if t and so, from (24), it follows that where we have used the following: Thus the condition (H 2 ) is satisfied, where M = 1 and N = 1 2 .It follows from (25) that (H 3 ) is satisfied, where L 1 = 1.Let f = g + h, where g = (g 1 , g 2 , Therefore, the conclusion of our example can not be obtained from the results of [5], [6] and [9].This also shows that our results are different from the results in [5], [6] and [9].