#### Abstract

Existence of extremal solutions of nonlinear discontinuous integral equations of Volterra type is proved. This result is extended herein to functional Volterra integral equations (FVIEs) and to a system of discontinuous VIEs as well.

#### 1. Introduction

In this work the existence of extremal solutions of nonlinear discontinuous integral as well as functional integral equations is proved by weakening all forms of Caratheodory’s condition. We consider the nonlinear Volterra integral equation (for short VIEs): and the functional Volterra integral equations (for short FVIEs): where may be discontinuous with respect to all of their arguments. The special case of (1.1) has been studied extensively under continuity and/or monotonicity [1–4]. Meehan and O’Regan [5] established, by placing some monotonicity assumption on a nonlinear -Carathéodory kernel of the form , existence of a solution to (1.1). It is proven in [6] that, providing some type of discontinuous nonlinearities, (1.1) has extremal solutions. Dhage [7] proved under mixed Lipschitz, Carathéodory, and monotonicity conditions existence of extremal solutions of nonlinear discontinuous functional integral equations. Other remarkable work was done in [8–11].

The main objective in this paper is to emphasize that the kernel is not required to be neither continuous nor monotonic in any of its arguments to establish an existence of extremal solutions for (1.1) (in ) which generalizes in some aspects some of the previously mentioned works. A monotonicity type condition with respect to the functional term is needed to establish existence of extremal solutions to (1.2). We base the proof of the main result on, among other tools, the following lemmas which could analogously be proved as Lemma 1.1 and Lemma 1.2, see [12], and hence the proofs are omitted.

Lemma 1.1. *Suppose that satisfies conditions and . Let be continuous and satisfy the inequality for all . Then the functions
**
are Lebesgue measurable for each fixed . In particular, for each , is Lebesgue measurable for each fixed .*

Lemma 1.2. *Suppose that satisfies conditions , , and . Let be continuous and satisfy the inequality for all . Let, for each ,
**
The compositions and are Lebesgue measurable for all any continuous , and, for almost all , *

The outline of the work is as follows. In Section 2 we present our existence theorem for (1.1) in . In Sections 3 and 4 generalizations of this established existence theorem for functional Volterra integral equation as well as for system of nonlinear Volterra integral equations are presented. Comparison with the literature is provided throughout the paper.

#### 2. Volterra Integral Equations

Theorem 2.1. *Let and let be given. Suppose that (C1)–(C4) are fulfilled.*)* is continuous.*()* For each , the function is Lebesgue measurable. For all and for almost all ,
where is a Lebesgue integrable function.*()* For each ,
*()* Let , where, . For every and all , the functions
are equicontinuous and tend to zero as .*

Under the above assumptions VIE expressed by (1.1) has extremal solutions in the interval .

*Proof. *We will prove the existence of a maximal solution the proof of the existence of a minimal solution is analogous and hence is omitted. The pattern of the proof consists of four steps. Similarly as it was done in [13, 14] we define the maximal solution as the limit of an appropriate sequence of approximations , .*Step 1. * Since , being a continuous function on compact set, is uniformly continuous and , being absolutely continuous with respect to Lebesgue measure, is uniformly continuous on ; then for all there exists such that
for all , with . Next, we take for subdivisions
of in such a way that , is a refinement of , that is, and
For any , and are recursively defined by, for ,
Once have already been defined on , with , they are defined in by putting
It follows, by Lemma 1.1, that functions are Lebesgue measurable; taking into account this together with , is well defined. Moreover it is easy to see that for all and all *Step 2*. We claim that, for all ,(i),(ii)if is a continuous function, which serves as a dummy function, satisfying, and , for , then in .

To prove these assertions we shall proceed inductively. Clearly, . Let us suppose that , for , with some . Since , there exist , , such that and . Let us suppose that , for , with an . At this point we have just two possibilities:

,

.

If () holds, since , it follows, by (2.4) and (2.9), that for
Assume the validity of (); it follows, by (2.4) and (2.9), that
On the other hand we have
We thus have
and hence, for each ,
By (2.9),
This in turn implies that, for all ,
which completes the proof of . Now we shall handle inductively too. Let us fix an arbitrary , by assumption, . Let us suppose that , for , with . At this point we have just two possibilities:

,

.

Suppose that we are in ; since , , it follows, by (2.4) and (2.9); and , that
for .

If we are in , it follows, by (2.4) and (2.9), that, for ,
By and (2.9),
We thus have for all , , and hence
for all , so it follows, by (2.4) and (2.9), that
for , which completes the proof of the assertion .*Step 3.* It follows, by (*C*4), that the constructed bounded nonincreasing sequence , is uniformly convergence, and hence let us set
By (2.4) and (2.9), we have, for . Thus,
for . Therefore, by formula just before *Step 2*, for , and
Applying Fatou Lemma and taking into account the condition and in view of Lemma 1.2, we obtain
for each . Let us observe that
almost everywhere in . Indeed, let us fix an . For any there exists a such that . Since
there exists an , with such that
Whence,
We thus have
which together with and (2.30) implies that
almost everywhere in . Applying Fatou lemma once again we obtain
which together with (2.26) means that is a solution (1.1).*Step 4*. Let be a solution of (1.1). Clearly, is continuous and satisfies the conditions of the assertion , so, , for any . Since as , then . This shows that is a maximal solution of (1.1).

We proceed similarly to prove the existence of minimal solution; we first define recursively the functions and , by setting
Taking to be a continuous function, which serves as a dummy function, satisfying and , for , and just following the previous steps one can show that (1.1) has a minimal solution in . This completes the proof.

It is interesting to point out that “” in condition could not be replaced with “” as it was done in [12]. Probably the reason consists in the fact that the composition is no longer measurable for any continuous function ; the following example illustrates this fact.

*Example 2.2. *Let be any non-Lebesgue measurable subset. Define by
It is easy to see that conditions and are satisfied. Since is nondecreasing in , we have, for all ,
However the composition is not Lebesgue measurable if , for .

#### 3. Functional Volterra Integral Equations

Our main concern in this section is to extend result established herein (Theorem 2.1) to a functional Volterra integral equation in deriving existence of extremal solutions for a class of FVIEs (1.2).

*Notations.* is a Lebesgue integrable function, is the set of all continuous functions satisfying for all . For a fixed let be the function defined by , and let .

Theorem 3.1. *Let denote the set of all satisfying the following conditions (F1)–(F5).*)* is continuous,*()* For each and is Lebesgue measurable, and for almost all ,
*() * For each ,
*()* For each , whenever with . *()* Let , where, . Let be fixed, for every and all ; the functions
**are equicontinuous and tend to zero as .*

Under the previous assumptions FVIE expressed by (1.2) has extremal solutions in the interval .

*Proof. * Since the proofs of existence of maximal and minimal solutions are similar, we concentrate our attention on showing the existence of the minimal solution.

For a fixed let us consider the nonfunctional Volterra integral equation N-FVIE:
Obviously, the function satisfies the hypotheses of Theorem 2.1; we thus conclude that N-FVIE (3.4) has a maximal solution which is given by
Let . Since belongs to , then the set is not empty. Define
Given and , it follows, by (*F4*), that
We thus have , and . Since is arbitrary, then , and hence
On the other hand, for we have
Thus . Consequently, which implies that , and thus which together with (3.8) implies that . Since every solution of (1.2) belongs to , then is a minimal solution. This completes the proof.

#### 4. System of Volterra Integral Equations

The main obstacle to extending the results of the previous section for vector-valued functions is that the usual order in makes the condition, used for scalar functions, does not have a good equivalence for vector-valued functions. We now show how Theorem 2.1 may be exploited to derive existence of extremal solutions for a class of systems of discontinuous VIEs. The proof is based on a technique similar to that used for systems of differential and functional differential equations [12, 15].

Theorem 4.1. *Given and satisfying the conditions below*()*is Lebesgue measurable for any continuous ,*() *for each and Lebesgue almost all , is nondecreasing in , and for all ,
*()* for each and Lebesgue almost all , , where and is a Lebesgue integrable function,*()* let , for every and all , the functions
are equicontinuous and tend to zero as , where and are interpreted componentwise.*

Under the above assumptions VIE expressed by (1.1) (in ) has extremal solutions in the interval .

*Proof. *We shall only prove the existence of a maximal solution, since the same pattern could be followed to prove existence of a minimal solution. Note that, for , we write if , for each . Let us denote by the set of all satisfying the following conditions
For every , we let
It follows, by monotonicity, that for every , for each , and for all ,
Let us define a nonincreasing sequence, whose existence is guaranteed by hypotheses and for its recursively construction we follow arguments developed in *Step 1* of the proof of Theorem 2.1; , such that, for each
where
Let, for each ,
Clearly . Regarding as a function only of while the remaining variables are considered to be constant. It follows, by similar arguments used in the proof of Theorem 2.1, that for every , and for all ,
By monotonicity
We thus have , so, , which together with (4.10) implies that is a solution of (1.1) on . Proceeding analogously as *Step 4* of the proof of Theorem 2.1, one can show that is a maximal solution of (1) on .

#### Acknowledgments

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under Grant no. (017-3/430). The author, therefore, acknowledges with thanks DSR technical and financial support.