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Journal of Function Spaces and Applications
Volume 2013 (2013), Article ID 479049, 9 pages
http://dx.doi.org/10.1155/2013/479049
Research Article

Second-Order Impulsive Differential Equations with Functional Initial Conditions on Unbounded Intervals

Dipartimento di Matematica e Informatica, Università della Calabria, 87036 Arcavacata di Rende, Cosenza, Italy

Received 29 December 2012; Accepted 27 February 2013

Academic Editor: Feliz Minhós

Copyright © 2013 Luigi Muglia and Paolamaria Pietramala. 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

We present some results on the existence of solutions for second-order impulsive differential equations with deviating argument subject to functional initial conditions. Our results are based on Schaefer's fixed point theorem for completely continuous operators.

1. Introduction

Differential equations with impulses arise quite often in the study of different problems in particular are used as a model for evolutionary processes subject to a sudden rapid change of their state at certain moments. The theory of impulsive differential equations has become recently a quite active area of research. For an introduction to this theory we refer to the books [14], which also contain a variety of interesting examples and applications.

In this paper, we establish new results for the existence of solutions for the second-order impulsive differential equation with deviating argument subject to initial conditions: where is in the space ( not necessarily the Euclidean norm), , is a real matrix, , and are the left and right limits of in .

Such type of problem arises from a few of scientific applications as, for example, the problem of impulsive maneuver of a spacecraft (see [5] in finite-dimentional setting and [6] in infinite-dimentional setting). More recently, impulsive second-order differential equations or inclusions on compact intervals subject to nonhomogeneous conditions have been studied by several authors (see for example [714] and the references therein). The study of differential problems on unbounded interval has been done, for example, in [1418].

Differential equations with deviating argument are investigated, for example, in [6, 7, 9, 12, 14, 16, 17, 19, 20].

Various techniques are utilized in the above papers: Schauder's fixed point Theorem [6, 9, 16, 17, 19, 20], Leray-Schauder's nonlinear alternative [14, 18], Contractions Principle [8, 13], fixed point index theory [12], and Sadovskii's fixed point Theorem [7]. Moreover, in order to prove the compactness of involved operator, the Ascoli-Arzelà Theorem is often used (see [6, 9, 12, 14, 18]).

The methodology here is to write the problem as a perturbed integral equation, and we look for fixed points of an operator in a suitable functions space. For this purpose we want to utilize the Schaefer's fixed point theorem for completely continuous operators. In order to prove that is completely continuous, we make use of a variant of the compactness result in the Banach space of continuous bounded functions from a topological space into due to De Pascale, Lewicki and G. Marino. We have used similar techniques and tools in [16, 17, 20] to study first-order, impulsive, or nonimpulsive, differential or integrodifferential equations on unbounded intervals. In this paper we extend these ideas to second-order impulsive equations.

Moreover we prove that our method can be easily used also in the case of functional conditions that, to the best of the our knowledge, are not studied for this class of problems. In fact, we discuss in details the Problem in the case of the initial conditions because it sheds light on the techniques used, but the same approach may be applied to impulsive equations subject to a more general functional initial condition that covers a large number of cases, namely where is a bounded linear operator and is a completely continuous operator. This is done in the last section.

2. Notations and Preliminaries

Firstly, we recall definitions, notations, and useful facts regarding the cosine families (see [2123] for the detailed study of cosine and sine families).

Definition 1. A one-parameter family of bounded linear mappings on into itself is called a strongly continuous cosine family if and only if , for all , , where is the identity map, , for all . Moreover, the sine family is defined as
By definition one obtains that: is continuous in , and , for all , , for all , commute, for all , , for all , , for all .

Definition 2. The infinitesimal generator of a strongly continuous cosine family is the operator defined by and .
One obtains that, for and , and then So, by one has , for all .

Definition 3. A cosine family (resp., a sine family ) is uniformly bounded if there exists (resp., ) such that where denotes the norm in the space of the matrices.

Example 4. In , let be the family of bounded linear operators Let us observe that is a uniformly bounded cosine family and its infinitesimal generator is

We will utilize the Schaefer's fixed point theorem for completely continuous operators.

Theorem 5 (see [24]). Let be a normed linear space. Let be a completely continuous operator. Let Then either in unbounded or has a fixed point.

We use the condition for compactness in the Banach space of continuous bounded functions from a topological space into , endowed with the norm , due to De Pascale, Lewicki and G. Marino.

Theorem 6 (see [25]). Let be a continuous operator. Suppose that, for any bounded set , is a bounded set and there exist bounded functions , , such that, for all and for all , Then is a compact operator.

3. The Integral Problem

From now on, we assume that the functions and the matrix have the following properties. The matrix is the infinitesimal generator of a strongly continuous cosine family , uniformly bounded by . Suppose moreover that the corresponding sine family is uniformly bounded by a constant . is a continuous function such that there exist a continuous integrable function and a continuous nondecreasing function for which is a continuous increasing function such that , for all .The function belongs to . and , are continuous functions such that there are for which

We will work in the Banach space endowed with the supremum norm .

We define for : For any let be the function defined on by In the next propositions we show the properties of the operator useful for us.

Proposition 7. The operator maps into and . Moreover Tx has derivative in , , and .

Proof. The continuity of , , and guarantees that is continuous in , and there exists for . Moreover, since is integrable, for , one has so . We need to show that, for , there exist and . Note that(i) . Indeed, using (u5), for (ii)It results that . Indeed let . Then Analogously one can see that in such a way that .Since, by (u1), , a similar proof permits to verify that By the above steps, it follows that .
Since there exists if , to conclude our proof we show that there exists , is finite, and . We also observe that, for and , by (u6), Thus, following the same idea, when , we have Moreover, for Finally, for while and these permit to obtain that there exist and and that one has .

Proposition 8. The fixed points of are solutions of the problem .

Proof. In fact, and . Moreover, by the hypotheses on and on the sine and cosine families, one obtains that is derivable in , and it results in the following: So one can conclude that that is, satisfies the differential equation in problem . This combined with the results in Proposition 7 yields the thesis.

In order to prove that is a completely continuous operator, we make use of the following compactness criterion in .

Proposition 9 (see [16]). Let be a continuous operator. Suppose that, for any bounded set , is a bounded set, and there exist bounded functions , , such that, for all and, for all Then is a compact operator.

Proof. The Banach space is isometric to the Banach space: where . Of course, is closed in . So the thesis follows from Theorem 6.

Proposition 10. is a completely continuous operator.

Proof. First we prove that is a continuous operator. Let a sequence in such that in . We prove that . By the continuity of it results that, for fixed , Moreover, by the monotonicity of , we have where . Therefore is dominated by an integrable function that does not depend on . Then, from the uniform boundedness of the sine family and by the dominated convergence theorem, we have The uniform boundedness of the cosine family and the continuity of and imply that, as , Thus Now, let Thanks to Proposition 9, it is enough to show that is a bounded set and that it is possible to control the oscillations of each function in by means of a finite number of bounded functions. The boundedness of follows by the inequalities: To control the oscillations of we should need to distinguish three cases: , , and .
Here we study the case only because the proofs of the other cases are similar.
Let and . Then Now, by property , one obtains that so we have Using we obtain also that To control the oscillations of the operators and , first we note that if , and we have nothing to prove because . The boundedness of and solves also the case since we have Thus it remains to prove only the case . We have that In a similar way we can show that So, defining where , and we obtain that for all

4. Main Result

Theorem 11. Assume that the hypotheses hold. Then the problem has at least one solution.

Proof. Our problem can be reduced, by Propositions 7 and 8, to find a fixed point for the operator . Proposition 10 assures that is a completely continuous operator.
To apply the Schaefer's fixed point theorem, it remains to prove that the set is bounded. The proof is based on an idea in [19].
Let with . First of all, since , for we have We consider now that . Thus we have Consider the function defined by Observe that is not necessarily continuous in but and exist, for all .
For , we have Moreover, being increasing and for , one has so, taking the supremum over in the inequality (52), we obtain that Denoting by the right-hand side of the last inequality, we have that the function is continuous: and for .
Moreover, since is nondecreasing, for , This implies that and so, for any , Since is a continuous function for all , we have This, together with condition (13), permits us to conclude that is bounded by a constant depending on the functions and only.
Summarizing, implies that .

5. Functional Initial Conditions

Let us consider the problem The modularity of the operator used in the proof of our Theorem 11 permits to prove a result on the existence of solutions for Problem modifying few parts of the above proof.

For , we define the function as We suppose that is a bounded linear operator for which if are such that , then . is a completely continuous operator such that for a certain constant . For , .

Theorem 12. Assume that the hypotheses hold. Then the problem has at least one solution.

Proof. Let us consider the operator defined on as Following the proof of Theorem 11, it is necessary to study only the part of the operator arising from the functional initial condition. First, for , one has and, by the continuity of , as These, following Proposition 7, are enough to prove that and = .
Moreover, it is not difficult to verify that for , and this is enought to prove that so that one has From the hypotheses on one has so, it follows from that
Moreover, and so, using Proposition 8, the fixed points of are solutions of . It remains to prove that we can control the oscillations of by a finite number of bounded functions.
We need to control only the oscillations of . If (the other case are similar), we have So, by defining we have
Hence is a completely continuous operator. Following the same proof of Theorem 11 one can see that the set is bounded, so has fixed points that are solutions of . This complete the proof.

Remark 13. The reader is referred to [5, 9, 13, 19] for some examples and applications.

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