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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 957696, 7 pages
http://dx.doi.org/10.1155/2013/957696
Research Article

Asymptotic Behavior of Solutions to a Vector Integral Equation with Deviating Arguments

Departamento de Análisis Matemático, Facultad de Ciencias, Universidad de Málaga, 29071 Málaga, Spain

Received 31 July 2013; Accepted 5 November 2013

Academic Editor: István Györi

Copyright © 2013 Cristóbal González and Antonio Jiménez-Melado. 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

In this paper, we propose the study of an integral equation, with deviating arguments, of the type in the context of Banach spaces, with the intention of giving sufficient conditions that ensure the existence of solutions with the same asymptotic behavior at as . A similar equation, but requiring a little less restrictive hypotheses, is In the case of , its solutions with asymptotic behavior given by yield solutions of the second order nonlinear abstract differential equation with the same asymptotic behavior at as .

1. Introduction

From the pioneering work of Atkinson [1], and subsequent works found in the literature (see, e.g., [210] for recent papers on the subject), we consider the following differential problem, with deviating arguments: with the task of finding solutions with the same behavior at as . Solutions with this prescription are given by the solutions of the following integral equation: which, by writing = , is of the type

The purpose of this note is to provide conditions that ensure the existence of solutions to the above integral equation, whose asymptotic behavior at is the same as that of , thus giving a procedure to show existence of solutions with prescribed asymptotic behavior of differential equation of the type (1). Our wish is to also work out this integral equation in the setting of Banach spaces.

Denote by the set of nonnegative real numbers. Assume that is a finite set of continuous mappings from to , that is a Banach space (with norm ), and also that is a continuous mapping from to . Finally, assume that is a given continuous mapping with certain regularity and integrability conditions, to be specified later. In order to give a better aspect to our equation, define, for each continuous , the mapping given by

Then, our equation becomes which, by writing , , is transformed into

A bit more of notation and preliminary results are needed. As customary, denotes the open ball in centered at with radius . The closure in of any set is written , and its closed convex hull, . The space of continuous -valued functions defined on is denoted by , while the space of bounded ones is . The latter is a Banach space when endowed with the sup norm , (i.e., for , ).

The Schauder fixed point theorem states that any continuous operator defined on a nonempty, bounded, closed and convex subset of a Banach space has necessarily a fixed point, provided that is a relatively compact subset of . We will also be needing a well-known version of the Arzelà-Ascoli theorem which, in the case that occupies us, is as follows: if a family is equicontinuous at each , and each section is relatively compact in , then each sequence contains a subsequence that converges uniformly on compact subsets of to a given -valued function .

Let us also say a word about vector integrals. For a brief introduction see, for example, [11] or [12]. If is a bounded vector function, the Riemann sum of associated to a finite partition of , (with norm ), and to a selection , , is . Without specifying the type of limit alluded, we say that is Riemann integrable on if there exists the limit of its Riemann sums as the norm of the partitions of tends to , in which case, this limit (which is unique) is called the integral of over and is denoted by . That is, being a little bit sloppy with the notation, we have

Observe that whenever is integrable on , then each Riemann sum associated to is times a linear convex combination of elements of . Therefore, the integral of over , being a limit of Riemann sums, is a multiple of an element of the closed convex hull of , that is,

2. Existence of Solutions

We begin this section enumerating the conditions that will be basic for our results on existence of solutions of   . Take , , and as above and, in analogy with what would happen with continuous mappings on finite dimensional spaces, assume that

The following conditions have already been motivated in previous work [7]. Recall that is continuous, but not necessarily bounded.

The result on existence of solutions to the integral equation is the following.

Theorem 1. Under hypotheses , , and , the integral equation has a solution asymptotically equal to as .

Remark 2. This theorem represents a generalization of the one presented in the work [7] in two aspects. First, we have made the jump to deal with integral equations in the setting of infinite dimensional spaces. And second, we have included deviating arguments in the equation.

Proof of Theorem 1. First observe that it suffices to find a solution to the integral equation in , and this will be achieved by proving the existence of a fixed point in of the operator:
We proceed to check that the conditions of the Schauder fixed point Theorem are fulfilled. First observe that is a nonempty, bounded, closed, and convex subset of the Banach space . Also, by , for all and all . So provided for all .
In fact, we will prove four assertions. (a) is uniformly equicontinuous on . This will give the desired continuity of for each . (b) is uniformly continuous on , hence it will be continuous on . (c) is relatively compact in . Thus, (a) and (c) will imply, by the Arzelà-Ascoli theorem, that each sequence in has a subsequence which converges uniformly on each compact subset of to a given function in (actually in ). Finally, in order to obtain that is relatively compact in , we will use the “funnel” structure of to prove (d), that any sequence in which converges uniformly on each compact subset of to a given function in must indeed converge uniformly to that function in all of . With all these assertions, the Schauder fixed point theorem can be applied to conclude the existence of a fixed point of , as we want.
Start fixing an arbitrary once for all. In what follows, we will build up different objects indexed by this , ( , , , ), knowing that even if for each assertion we have to start taking an arbitrary , the objects will vary accordingly but not the way to obtain them.
Since as , there exists such that
We start proving (d), as it is quite independent of the rest of assertions. Assume that a sequence converges uniformly on each compact subset of to a function , and let us show that indeed converges uniformly to in all of . For the above (so for any ), find the corresponding to satisfy (8). Since , , and all belong to , then
In particular,
Now, since converges uniformly to in , there exists such that
This tells us that if , then , proving that the convergence of to is uniform on , and thus (d) is proven.
Next, continue building up other objects associated with the arbitrary fixed above. Observe that, by (8) and ,
Also, since as , there exists such that so, by ,
The continuity of and the 's, , and the uniform bound for functions in (given by the bound of ) imply that there exists a bounded set , depending on , , the 's, and , but not on , such that
Now, observe that is bounded in , so by ,
By the uniform continuity of on , there exists such that, whenever and with , , and , , then
Notice that if with and, without loss of generality, , then several cases are possible. If , then, as , , and so, by (12),
If, on the other hand, , then, by (15), for all , and, by (17) and (14), we have, for any , This proves (a), the uniform equicontinuity of over .
Now notice that if with and , then, again, two cases are possible. If , then, by (12), while for , using (15), (17), (14), and (13), This proves that , showing (b), that is uniformly continuous on .
For the compactness of in , it suffices to show that is totally bounded [13, page 298]; that is, for the given (so for any ) there exists a finite covering of with balls of radii not bigger than . Observe first that, by (12), for all and all , that is,
Now, in order to control the elements of , observe that each of these can be decomposed as the sum of a “head” and a “tail,”
The “head” can be approximated by Riemann sums, which, in turn, are nothing else but times a convex linear combination of elements of , that is, the “head” is an element of . By (16), has compact closure in , so by Mazur's theorem [14], is compact, and therefore it can be covered with a finite number of balls, say , of radii not bigger than . This yields a finite covering of with balls of radii not bigger than , precisely the collection . On the other hand, by (14), the “tail” of each of the above integrals is bounded by , so they are elements of . All this can be summarized as follows: that is, can be covered with a finite collection of balls of radii smaller than , because each is readily seen to be a ball of radius not bigger than .
At the end, by (22) and (24), we have that is, we have given a finite covering of with balls of radii not bigger than . With this we conclude (c) and, with all four assertions proved, the theorem too.

Coming back to the integral equation (2) underlying the differential equation (1), we just need to adapt the hypotheses presented above to the function , to obtain a corresponding result on existence of solutions to (2) with asymptotic behavior given by . Notice that these hypotheses, , , and , are natural generalizations from the 1-dimensional case. However, the proof in the abstract setting has shown many more properties for the operator that needed in order to show the existence of a fixed point. This, somehow, is telling us that the hypotheses could be weakened. For the moment, we content ourselves noticing that, in (2), the corresponding hypothesis of uniform continuity on bounded sets is “not needed,” because it will be “consequence” of the other hypotheses, and a little trick of changing the domain of definition of the operator. Thus, for our next result, we will be using the following hypothesis: Also, instead of considering just the integral equation (2), we generalize a little bit to a convolution type integral equation, with a kernel satisfying the right properties for us. Observe that, when , and ,

That is, satisfies a Lipschitz condition on the first variable, independent of the second. Actually, much less is needed, just continuity of suffices.

Theorem 3. Let , , , be as for Theorem 1, and let   be continuous. Under hypotheses , , and    where , the integral equation has a solution asymptotically equal to as .

Remark 4. Observe, as in [7], that if the kernel is the one we started with, , then hypothesis is redundant, because in that case, if , we always have, for , , and for , Consequently, from hypothesis , taking , we have as , and
In general, hypothesis is superfluous whenever there exist constants, and , such that

Proof of Theorem 3. We proceed as before. Consider initially the same nonempty, bounded, closed, and convex subset as before, as well as the same operator . The idea is to find a fixed point for using the Schauder fixed point theorem. For that, we start repeating the same scheme of four assertions established in the previous theorem. The two assertions that do not depend on the uniform continuity of on bounded sets are done the same way as before; hence we omit their proofs. These are (c), that is compact in , and (d), that uniform convergence on compact subsets of of a sequence in turns into actual uniform convergence on .
Let us now prove (a), that is uniformly equicontinuous on . This will finish showing that , and, by (c), (d), and the Arzelà-Ascoli theorem, that has compact closure in .
Let be fixed. Find, as before, , , and bounded subset of , so as to satisfy (8), (12), (13), (14), and (15). Now, observe that is bounded in , so by , is relatively compact in ; hence there exists such that
Notice now that is uniformly continuous on . So there is such that, whenever and are in with and , then
Now take with and, without loss of generality, assume that . Then again, several cases are possible. If , then, as , , and so, by (12),
If, on the other hand, , then, by (31), (30), and (14), we have, for any , This proves the uniform equicontinuity of over .
To finish the proof, we have to prove (b), that is uniformly continuous. It is here that we restrict the domain of definition of . Observe that is nonempty, closed and convex. Also, because and is closed and convex. More is true, since has compact closure, then, by Mazur's Theorem, is compact too. One more thing, leaves invariant :
With all of this, it suffices to prove that is continuous on this new -invariant set. Let us prove that, indeed, is uniformly continuous on . Let be given. Find that and as before, so as to satisfy (8), (12), (13), and (14). The continuity of and the 's, , and the compactness of tells us that is a compact subset of , giving us the opportunity to say that is uniformly continuous on and, consequently, that is uniformly continuous on . So there exists such that, for and with , , and , , then Now, it is a matter of repeating the same steps as was done in Theorem 1 to prove the uniform continuity of .
This concludes the proof of the theorem.

Remark 5. Instead of working with functions defined on , we could have worked with functions defined on any interval of the type , obtaining a result completely similar. Also, many times, one is just interested in giving partial solutions; that is, solutions defined not on the whole interval , but on some other interval with .

Next, we just mention an easy result on the existence of solutions of the underlying differential equation, just to illustrate the type of functions that could generate a condition like . This result is inspired from [7, Thm. 2] and [8, Thm. 1].

Theorem 6. Let be a continuous function mapping bounded sets into relatively compact ones, such that, where Let also be a set of continuous functions from to .
Then, for any , (1) has a solution with as .

Proof. Take , consider the corresponding integral equation (2), and define
Observe that, by (37), as . This allows us to consider the set which, adopting the notation used throughout the paper, gives by (36), for and ,
With this, Theorem 3 applies (hypothesis need not be verified by Remark 4) and then there exists , solution of (2) with as . Finally, it is just an exercise to check that is twice continuously differentiable and that satisfies the differential equation (1).

Conflict of Interests

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

Acknowledgments

Research partially supported by the Spanish (Grant MTM2011-25502) and regional Andalusian (Grants FQM210 and P09-FQM4468) Governments.

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