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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 698234, 5 pages

http://dx.doi.org/10.1155/2013/698234

## On the Existence of the Solutions for Some Nonlinear Volterra Integral Equations

^{1}İnönü Üniversitesi, Eğitim Fakültesi, A-Blok, 44280 Malatya, Turkey^{2}Nevşehir Üniversitesi, Fen Edebiyat Fakültesi, Matematik Bölümü, 50300 Nevşehir, Turkey^{3}Malatya Fen Lisesi, 44110 Malatya, Turkey

Received 2 April 2013; Revised 4 June 2013; Accepted 8 June 2013

Academic Editor: Aref Jeribi

Copyright © 2013 İsmet Özdemir et al. 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 a theorem which gives sufficient conditions for existence of at least one solution for some nonlinear functional integral equations in the space of continuous functions on the interval . To do this, we will use Darbo's fixed-point theorem associated with the measure of noncompactness. We give also an example satisfying the conditions of our main theorem but not satisfying the conditions described by Maleknejad et al. (2009).

#### 1. Introduction

As it is known, nonlinear integral equations constitute an important branch of nonlinear analysis. Particularly integral equations are often used in the characterization of several problems of engineering, mechanics, physics, economics, and so on. Some authors have given some results for solvability of some functional integral equations such as Mureşan in [1], Banaś and Sadarangani in [2], and Djebali and Hammache in [3]. The following equation has been considered in [4]: for . Maleknejad et al. in [5] studied the existence of solutions of the following equation: under the following conditions.(K1) is continuous and there exist nonnegative constants and such that for all and .(K2) is continuous and satisfies the sublinearity condition, so there exist constants and such that for all and .(K3) for and .(K4) for .

In this paper, we consider the following nonlinear functional integral equation: which is more general than the equation given in [5].

In Section 2, we present some definitions and preliminary results about the concept of measure of noncompactness. In Section 3, we give our main results concerning the existence of solutions of the integral equation (5) by applying Darbo's fixed-point theorem associated with the measure of noncompactness defined by Banaś and Goebel [6] and, finally, we establish an example to show that these results are applicable.

#### 2. Definitions and Auxiliary Facts

In this section, we give some definitions and results which will be needed next section. Let be an infinite Banach space with zero element . We write to denote the closed ball centered at with radius and, especially, we write in case of . We write , to denote the closure and closed convex hull of , respectively. Moreover, let indicate the family of all nonempty bounded subsets of and indicate the subfamily of all relatively compact sets. Finally, the standard algebraic operations on sets are denoted by and , respectively [2].

We use the following definition of the measure of noncompactness, given in [6].

*Definition 1. *A mapping is said to be a measure of noncompactness in if it satisfies the following conditions. (1)The family is nonempty and .(2).(3).(4) for .(5)If is a sequence of closed sets from such that () and if , then the intersection set is nonempty.

Theorem 2 (see [6]). *Let be a nonempty, closed, bounded, and convex subset of the Banach space and let be a continuous mapping. Assume that there exists a constant such that
**
for any nonempty subset of . Then, has a fixed point in the set . *

As it is known, the family of all real-valued and continuous functions defined on the interval forms a Banach space with the standard norm

Let be a fixed subset of . For and , we denote by the modulus continuity of defined by Furthermore, let and be defined by The authors have shown in [6] that the previous function is a measure of noncompactness in the space .

#### 3. The Main Result

First of all, we write to denote the interval throughout this section. We study the functional integral equation (5) under the following conditions.

(a) , and are continuous.

*Remark 3. *Note that assumption (a) implies that there exists positive constant such that
for all .

(b) Functions are continuous and there exist positive constants and such that for all and .

*Remark 4. *Note that assumption (b) implies that there exist positive constants and such that
for all .

(c) is continuous and there exist positive constants , , and such that for all , and .

(d) The inequality holds.

Theorem 5. *Under assumptions (a)–(d), there exists at least one such that (5) has at least one solution which belongs to . *

*Proof. *We define the continuous function such that
where is the constant given in assumption (c). Then and by assumption (d). The continuity of guarantees that there exists the number such that and . Now, we will prove that (5) has at least one solution which belongs to . Note that we will use Theorem 2 as our main tool. We define operator by
for any . Using the conditions of Theorem 5, we infer that is continuous on . For any ,
This result shows that . Now we will prove that operator is continuous. To do this, consider and any such that . Then, taking into account the equality
we have by (18) that the inequality
holds, where
On the other hand, since function is uniformly continuous on , we infer that as . Hence, the previous estimate (19) proves that operator is continuous on ball . Now, we will show that operator satisfies (6) with respect to measure of noncompactness given by (9). To do this, we choose a fixed arbitrary . Let us take and with , for any nonempty subset of . Since
we get
from (21), where
for such that and . Also,
for such that , and . Thus, by using the previous estimate (22), we can write
We have that , , and as since functions , , and are uniformly continuous on . Similarly, we get , , and as since functions , , and are uniformly continuous on , , and , respectively. Hence, we obtain that
Thus, since from (23) and from condition (14), we derive that operator is a contraction on ball with respect to measure of noncompactness . Therefore, from Theorem 2 we get that has at least one fixed point in . Consequently, nonlinear functional integral equation (5) has at least one continuous solution in . This completes the proof.

*Example 6. *Consider the following nonlinear functional integral equation in :
Put
It is easy to prove that the assumptions of Theorem 5 hold. Therefore, Theorem 5 guarantees that (27) has at least one solution . Since there exist no constants and such that
for all and , the results presented in [5] are inapplicable to integral equation (27) with

#### Acknowledgments

The authors would like to thank the editor Professor Aref Jeribi and the referees for their valuable comments and suggestions that improved this paper.

#### References

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*Seminar on Fixed Point Theory Cluj-Napoca*, vol. 3, pp. 297–303, 2002. View at Zentralblatt MATH · View at MathSciNet - J. Banaś and K. Sadarangani, “Solutions of some functional-integral equations in Banach algebra,”
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