AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 232314 10.1155/2012/232314 232314 Research Article Existence Theorem for Integral and Functional Integral Equations with Discontinuous Kernels Hassan Ezzat R. Galdi Giovanni 1 Department of Mathematics Faculty of Science King Abdulaziz University P.O. Box 80203 Jeddah 21589 Saudi Arabia kau.edu.sa 2012 12 7 2012 2012 04 03 2012 09 05 2012 09 05 2012 2012 Copyright © 2012 Ezzat R. Hassan. 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.

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): (1.1)x(t)=u(t)+0tf(t,τ,x(τ))dτ, and the functional Volterra integral equations (for short FVIEs): (1.2)x(t)=u(t)+0tf(t,τ,x(τ),x)dτ, where f may be discontinuous with respect to all of their arguments. The special case of (1.1) (1.3)x(t)=0tk(t-τ)g(x(τ))dτ, has been studied extensively under continuity and/or monotonicity . Meehan and O’Regan  established, by placing some monotonicity assumption on a nonlinear L1-Carathéodory kernel of the form k(t,s,x(s)), existence of a C[0,T] solution to (1.1). It is proven in  that, providing some type of discontinuous nonlinearities, (1.1) has extremal solutions. Dhage  proved under mixed Lipschitz, Carathéodory, and monotonicity conditions existence of extremal solutions of nonlinear discontinuous functional integral equations. Other remarkable work was done in .

The main objective in this paper is to emphasize that the kernel f 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 , and hence the proofs are omitted.

Lemma 1.1.

Suppose that f:[0,1]×[0,1]× satisfies conditions (C1) and (C3). Let x1,x2:[0,1] be continuous and satisfy the inequality x1(τ)<x2(τ) for all τ[0,1]. Then the functions (1.4)φ(t,τ)=infy(x1(τ),x2(t,τ))f(t,τ,y),ψ(t,τ)=supy(x1(τ),x2(τ))f(t,τ,y), are Lebesgue measurable for each fixed t[0,1]. In particular, for each t[0,1], f(t,·,x(·)) is Lebesgue measurable for each fixed xC([0,1]).

Lemma 1.2.

Suppose that f:[0,1]×[0,1]× satisfies conditions (C1), (C2), and (C3). Let x1,x2:[0,1] be continuous and satisfy the inequality x1(τ)<x2(τ) for all τ[0,1]. Let, for each (t,τ,x)[0,1]×[0,1]×, (1.5)f*(t,τ,x)=liminfyxf(t,τ,y),f*(t,τ,x)=limsupyxf(t,τ,y). The compositions f*(t,·,x(·)) and f*(t,·,x(·)) are Lebesgue measurable for all t[0,1] any continuous x:[0,1], and, for almost all τ[0,1], (1.6)infy(x1(τ),x2(τ))f*(t,τ,y)=infy(x1(τ),x2(τ))f(t,τ,y)=infy(x1(τ),x2(τ))f*(t,τ,y),supy(x1(τ),x2(τ))f*(t,τ,y)=supy(x1(τ),x2(τ))f(t,τ,y)=supy(x1(τ),x2(τ))f*(t,τ,y).

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 u:[0,1] and let f:[0,1]×[0,1]× be given. Suppose that (C1)–(C4) are fulfilled.

u is continuous.

For each (t,x)[0,1]×, the function τf(t,τ,x) is Lebesgue measurable. For all (t,x)[0,1]× and for almost all τ[0,1], (2.1)|f(t,τ,x)|<M(τ), where M:[0,1][0,] is a Lebesgue integrable function.

For each (t,τ,x)[0,1]×[0,1]×, (2.2)limsupyxf(t,τ,y)f(t,τ,x)=liminfyxf(t,τ,y).

Let Ϝ={y;|y||u|+|01M(τ)dτ|}, where, |u|=max{|u(t)|;t[0,1]}. For every yϜ and all nN, the functions (2.3)t0tsup|x-y|1/3nf(t,τ,x)dτ, are equicontinuous and tend to zero as t0.

Under the above assumptions VIE expressed by (1.1) has extremal solutions in the interval [0,1].

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 xn, nN.

Step  1.  Since u, being a continuous function on compact set, is uniformly continuous and EEMdτ, being absolutely continuous with respect to Lebesgue measure, is uniformly continuous on [0,1]; then for all nN there exists δn>0 such that (2.4)|u(s)-u(t)|+|stM(τ)dτ|13n+2, for all s,t[0,1], with |s-t|δn. Next, we take for nN subdivisions Dn(2.5)t0n<t1n<<tknn of [0,1] in such a way that 0=t0n<t1n<<tknn=1, Dn+1 is a refinement of Dn, that is, DnDn+1 and (2.6)tk+1n-tknδn,k=0,1,,kn-1. For any nN, gn:[0,1]×[0,1] and xn:[0,1] are recursively defined by, for s,t[0,t1n], (2.7)gn(t,τ)=sup|x-u(0)|2/3nf(t,τ,x),xn(t)=u(t)+0tgn(t,τ)dτ. Once gn,xn have already been defined on [0,tkn], with k<kn, they are defined in [tkn,tk+1n] by putting (2.8)gn(t,τ)=sup|x-xn(tkn)|2/3nf(t,τ,x),(2.9)xn(t)=xn(tkn)+u(t)-u(tkn)+tkntgn(t,τ)dτ. It follows, by Lemma 1.1, that functions gn are Lebesgue measurable; taking into account this together with (C2), xn is well defined. Moreover it is easy to see that for all t[0,1] and all nN(2.10)xn(t)=u(t)+0tgn(t,τ)dτ.

Step  2.  We claim that, for all nN,

xn+1xn,

if x:[0,1] is a continuous function, which serves as a dummy function, satisfying, x(0)u(0) and x(t)x(s)+u(t)-u(s)+stf(t,τ,x(τ))dτ, for s,t[0,1], then xxn in [0,1].

To prove these assertions we shall proceed inductively. Clearly, xn+1(0)=xn(0)=p(0). Let us suppose that xn+1(t)xn(t), for t[0,tkn], with some k<kn. Since DnDn+1, there exist i,j{0,1,kn+1}, i<j, such that tin+1=tkn and tjn+1=tk+1n. Let us suppose that xn+1(t)xn(t), for [0,tmn+1], with an m{i,i+1,,j-1}. At this point we have just two possibilities:

( P 11 ) xn+1(tmn+1)<xn(tmn+1)-2/3n+1,

( P 12 ) xn(tmn+1)-2/3n+1xn+1(tmn+1)xn(tmn+1).

If (P11) holds, since tm+1n+1-tmn+1δn+1, it follows, by (2.4) and (2.9), that for t[tmn+1,tm+1n+1](2.11)xn+1(t)=xn+1(tmn+1)+u(t)-u(tmn+1)+tmn+1tgn(t,τ)dτ<xn+1(tmn+1)+13n+2<xn(tmn+1)-23n+1+13n+2=xn(t)+[xn(tmn+1)-xn(t)]-53n+2<xn(t)+13n+2-53n+2=xn(t)+13n+1-53n+2<xn(t). Assume the validity of (P12); it follows, by (2.4) and (2.9), that (2.12)xn+1(tmn+1)-23n+1xn(tmn+1)-23n+1-23n+1=xn(tkn)+[xn(tmn+1)-xn(tkn)]-43n+1>xn(tkn)-13n+2-43n+1=xn(tkn)-13n+1-43n+1>xn(tkn)-23n. On the other hand we have (2.13)xn+1(tmn+1)+23n+1xn(tmn+1)+23n+1=xn(tkn)+[xn(tmn+1)-xn(tkn)]+23n+1<xn(tkn)+13n+2+23n+1=xn(tkn)+13n+1+23n+1<xn(tkn)+23n. We thus have (2.14)(xn+1(tmn+1)-23n+1,xn+1(tmn+1)+23n+1)(xn(tkn)-23n,xn(tkn)+23n), and hence, for each τ[tmn+1,tm+1n+1], (2.15)gn+1(t,τ)=sup|x-xn+1(tmn+1)|2/3nf(t,τ,x)sup|x-xn(tkn)|2/3nf(t,τ,x)=gn(t,τ). By (2.9), (2.16)xn(tmn+1)=xn(tkn)+u(tmn+1)-u(tkn)+tkntmn+1gn(t,τ)dτ. This in turn implies that, for all t[tmn+1,tm+1n+1], (2.17)xn+1(t)=xn+1(tmn+1)+u(t)-u(tmn+1)+tmn+1tgn+1(t,τ)dτxn(tmn+1)+u(t)-u(tmn+1)+tmn+1tgn(t,τ)dτ=xn(tin)+u(tmn+1)-u(tkn)+tkntmn+1gn(t,τ)dτ+u(t)-u(tmn+1)+tmn+1tgn(t,τ)dτ=xn(tkn)+u(t)-u(tkn)+tkntgn(t,τ)dτ=xn(t), which completes the proof of (i). Now we shall handle (ii) inductively too. Let us fix an arbitrary nN, by assumption, x(0)u(0)=xn(0). Let us suppose that x(t)xn(t), for t[0,tkn], with k<kn. At this point we have just two possibilities:

( P 21 ) x(tkn)<xn(tkn)-2/3n+1,

( P 22 ) xn(tkn)-2/3n+1x(tkn)xn(tkn).

Suppose that we are in (P21); since tk+1n-tknδn, k=0,1,2,,kn-1, it follows, by (2.4) and (2.9); and (ii), that (2.18)x(t)x(tkn)+u(t)-u(tkn)+tkntf(t,τ,x(τ))dτ<x(tkn)+13n+2<xn(tkn)-23n+1+13n+2=xn(t)+[xn(tkn)-xn(t)]-53n+2<xn(t)+13n+2-53n+2<xn(t), for t[tkn,tk+1n].

If we are in (P22), it follows, by (2.4) and (2.9), that, for t[tkn,tk+1n], (2.19)x(t)=x(tkn)+[x(t)-x(tkn)]xn(tkn)-23n+1+[x(t)-x(tkn)]>xn(tkn)-23n+1-13n+2>xn(tkn)-23n. By (ii) and (2.9), (2.20)x(t)x(tkn)+u(t)-u(tkn)+tkntf(t,τ,x(τ))dτxn(tkn)+u(t)-u(tkn)+tkntf(t,τ,x(τ))dτ<xn(tkn)+13n+2<xn(tkn)+23n. We thus have for all τ[tkn,tk+1n], x(τ)(xn(tkn)-2/3n,xn(tkn)+2/3n), and hence (2.21)f(t,τ,x(τ))sup|x-xn(tkn)|2/3nf(t,τ,x)=gn(t,τ), for all τ[tkn,tk+1n], so it follows, by (2.4) and (2.9), that (2.22)x(t)xn(tkn)+u(t)-u(tkn)+tkntf(t,τ,x(τ))dτxn(tkn)+u(t)-u(tkn)+tkntgn(t,τ)dτ=xn(t), for t[tkn,tk+1n], which completes the proof of the assertion (ii).

Step  3. It follows, by (C4), that the constructed bounded nonincreasing sequence xn, nN is uniformly convergence, and hence let us set (2.23)x+(t)=limnxn(t),t[0,1]. By (2.4) and (2.9), we have|xn(τ)-xn(tkn)|1/3n+2, for τ[tkn,tk+1n]. Thus, (2.24)gn(t,τ)=sup|x-xn(tkn)|2/3nf(t,τ,x)f(t,τ,xn(τ)), for τ[tkn,tk+1n]. Therefore, by formula just before Step  2, for t[0,1], and nN(2.25)xn(t)=u(t)+0tgn(t,τ)dτu(t)+0tf(t,τ,xn(τ))dτ. Applying Fatou Lemma and taking into account the condition (C3) and in view of Lemma 1.2, we obtain (2.26)x+(t)=limnxn(t)=liminfnxn(t)u(t)+0tliminfnf(t,τ,xn(τ))dτ=u(t)+0tf(t,τ,x+(τ))dτ, for each t[0,1]. Let us observe that (2.27)limsupngn(t,τ)f(t,τ,x+(τ)), almost everywhere in [0,1]. Indeed, let us fix an nN. For any τ[0,1] there exists a k=0,,kn such that τ[tkn,tk+1n]. Since (2.28)gn(t,τ)=sup|x-xn(tkn)|2/3nf(t,τ,x), there exists an xn^, with |xn^-xn(tkn)|2/3n such that (2.29)gn(t,τ)-13nf(t,τ,xn^)gn(t,τ). Whence, (2.30)limsupngn(t,τ)=limsupnf(t,τ,xn^).(2.31)|xn^-xn(τ)||xn^-xn(tkn)|+|xn(tkn)-xn(τ)|43n. We thus have (2.32)limnxn^=limnxn(τ)=x+(τ), which together with (C3) and (2.30) implies that (2.33)limsupngn(t,τ)f(t,τ,x+(τ)), almost everywhere in [0,1]. Applying Fatou lemma once again we obtain (2.34)x+(t)=limnxn(t)=limsupnxn(t)u(t)+0tlimsupngn(t,τ)dτu(t)+0tf(t,τ,x+(τ))dτ, which together with (2.26) means that x+ is a solution (1.1).

Step  4.  Let x:[0,1] be a solution of (1.1). Clearly, x is continuous and satisfies the conditions of the assertion (ii), so, xxn, for any nN. Since xnx+ as n, then xx+. This shows that x+ is a maximal solution of (1.1).

We proceed similarly to prove the existence of minimal solution; we first define recursively the functions hn(t,τ) and xn(t), by setting (2.35)hn(t,τ)=inf|x-xn(tkn)|2/3nf(t,τ,x),τ[0,1],xn(t)=xn(tkn)+u(t)-u(tkn)+tknthn(t,τ)dτ,t[tkn,tk+1n]. Taking x:[0,1] to be a continuous function, which serves as a dummy function, satisfying x(0)u(0) and x(t)x(s)+u(t)-u(s)+stf(t,τ,x(τ))dτ, for s,t[0,1], and just following the previous steps one can show that (1.1) has a minimal solution x- in [0,1]. This completes the proof.

It is interesting to point out that “=” in condition (C3) could not be replaced with “” as it was done in . Probably the reason consists in the fact that the composition f(t,·,x(·)) is no longer measurable for any continuous function x:[0,1]; the following example illustrates this fact.

Example 2.2.

Let S[0,1] be any non-Lebesgue measurable subset. Define f:[0,1]×[0,1]× by (2.36)f(t,τ,x)={1ifx>τ,1ifx=τ,τS,0otherwise. It is easy to see that conditions (C1) and (C2) are satisfied. Since f is nondecreasing in x, we have, for all (t,τ,x)[0,1]×[0,1]×, (2.37)limsupyxf(t,τ,y)f(t,τ,x)liminfyxf(t,τ,y). However the composition f(t,·,x(·)) is not Lebesgue measurable if x(τ)=τ, for τ[0,1].

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.

M : [ 0,1 ] [ 0 , ] is a Lebesgue integrable function, CM([0,1];) is the set of all continuous functions x:[0,1] satisfying |x(t)-x(s)||u(t)-u(s)|+|stM(τ)dτ| for all s,t[0,1]. For a fixed φCM([0,1]) let fφ:[0,1]×[0,1]× be the function defined by fφ(t,τ,x)=f(t,τ,x,φ), and let Γφ={xCM([0,1];)x(0)u(0),x(t)x(s)+u(t)-u(s)+stfφ(t,τ,x(τ))dτ,fors,t[0,1]}.

Theorem 3.1.

Let Υ denote the set of all f:[0,1]×[0,1]××CM([0,1];) satisfying the following conditions (F1)–(F5).

u:[0,1] is continuous,

For each (t,x)[0,1]× and φCM([0,1];),τf(t,τ,x,φ) is Lebesgue measurable, and for almost all τ[0,1], (3.1)|f(t,τ,x,φ)|<M(τ).

For each (t,τ,x,φ)[0,1]×[0,1]××CM([0,1];), (3.2)limsupyxf(t,τ,y,φ)f(t,τ,x,φ)=liminfyxf(t,τ,y,φ).

For each (t,τ,x)[0,1]×[0,1]×, f(t,τ,x,φ)f(t,τ,x,ψ) whenever φ,ψCM([0,1];) with φψ.

Let Ϝ={y;|y||u|+|01M(τ)dτ|}, where, |u|=max{|u(t)|;t[0,1]}. Let φCM([0,1];) be fixed, for every yϜ and all nN; the functions (3.3)t0tsup|x-y|1/3nfφ(t,τ,x)dτ,

are equicontinuous and tend to zero as t0.

Under the previous assumptions FVIE expressed by (1.2) has extremal solutions in the interval [0,1].

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 φCM([0,1];) let us consider the nonfunctional Volterra integral equation N-FVIE: (3.4)x(t)=u(t)+0tfφ(t,τ,x(τ))dτ. Obviously, the function fφ(t,τ,x(τ)) satisfies the hypotheses of Theorem 2.1; we thus conclude that N-FVIE (3.4) has a maximal solution which is given by (3.5)xφ(t)=infxΓφx(t). Let ={φCM([0,1];)xφφ}. Since -u(t)-stM(τ)dτ belongs to , then the set is not empty. Define (3.6)x¯(t)=infφφ(t). Given φ and xΓφ, it follows, by (F4), that (3.7)x(t)x(s)+u(t)-u(s)+stf(t,τ,x(τ),φ)dτx(s)+u(t)-u(s)+stf(t,τ,x(τ),x¯). We thus have xΓx¯, ΓφΓx¯ and xφxx¯. Since φ is arbitrary, then φxφxx¯, and hence (3.8)x¯xx¯. On the other hand, for xΓφ we have (3.9)x(t)x(s)+u(t)-u(s)+stf(t,τ,x(τ),φ)dτx(s)+u(t)-u(s)+stf(t,τ,x(τ),x¯)x(s)+u(t)-u(s)+stf(t,τ,x(τ),xx¯). Thus xΓφΓx¯Γxx¯. Consequently, xφxx¯xxx¯ which implies that xx¯, and thus x¯xx¯ which together with (3.8) implies that x¯=xx¯. Since every solution of (1.2) belongs to , then x¯ 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 n makes the condition, (4.1)limsupyxf(t,τ,y)f(t,τ,x)=liminfyxf(t,τ,y), 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 u=(u1,,un):[0,1]n and f=(f1,,fn):[0,1]×[0,1]×nn satisfying the conditions (B1)(B4) below

f(t,·,x(·))is Lebesgue measurable for any continuous x:[0,1]n,

for each i=1,,n and Lebesgue almost all t[0,1], fi is nondecreasing in xk, k=1,,i-1,i+1,,n and for all x=(x1,,xn)n, (4.2)limsupyxifi(t,τ,x1,,xi-1,y,xi+1,,xn)fi(t,τ,x1,,xi-1,xi,xi+1,,xn)=limyxifi(t,τ,x1,,xi-1,y,xi+1,,xn),

for each xn and Lebesgue almost all t[0,1], f(t,τ,x)M(τ), where x=max{|xi|;i=1,,n} and M:[0,1][0,] is a Lebesgue integrable function,

let Ϝ={yn;yu+|01M(τ)dτ|}, for every yϜ and all kN, the functions (4.3)t0tsup|x-y|1/3kf(t,τ,x)dτ, are equicontinuous and tend to zero as t0, where supf and f are interpreted componentwise.

Under the above assumptions VIE expressed by (1.1) (in n) has extremal solutions in the interval [0,1].

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 x,yn, we write xy if xiyi, for each i=1,,n. Let us denote by X the set of all x=(x1,,xn):[0,1]n satisfying the following conditions (4.4)xi(t)ui(t)+0tfi(t,τ,x1(τ),,xn(τ))dτ,i=1,2,,n,x(t)-x(s)u(t)-u(s)+|stM(τ)dτ|,s,t[0,1]. For every i=1,2,,n, we let (4.5)xi+(t)=sup(x1,,xn)Xxi(t),t[0,1]. It follows, by monotonicity, that for every xX, for each i=1,,n, and for all t[0,1], (4.6)xi(t)ui(t)+0tfi(t,τ,x1+(τ),,xi-1+(τ),xi(τ),xi+1+(τ),,xn+(τ))dτ. 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; (ym), ym=(ym1,,ymn):[0,1]n such that, for each i=1,,n(4.7)ymi(0)=ui(0),ymi(t)=ui(t)+0tgmi(τ)dτ,y(t)-y(s)u(t)-u(s)+|stM(τ)dτ|,s,t[0,1], where (4.8)gmi(τ)=sup|x-ymi(tkn)|2/3mfi(t,τ,x1+(τ),,xi-1+(τ),x,xi+1+(τ),,xn+(τ)). Let, for each i=1,,n, (4.9)limmymi(τ)=yi+(τ),τ[0,1]. Clearly xi+(t)yi+(t). Regarding fi as a function only of yi+ 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 i=1,,n, and for all t[0,1], (4.10)yi+(t)=ui(t)+0tfi(t,τ,x1+(τ),,xi-1+(τ),yi+(τ),xi+1+(τ),,xn+(τ))dτ. By monotonicity (4.11)yi+(t)ui(t)+0tfi(t,τ,y1+(τ),,yi-1+(τ),yi+(τ),yi+1+(τ),,yn+(τ))dτ,i=1,,n. We thus have y+=(y1+,,yn+)X, so, y+=x+, which together with (4.10) implies that y+ is a solution of (1.1) on [0,1]. Proceeding analogously as Step  4 of the proof of Theorem 2.1, one can show that y+ is a maximal solution of (1) on [0,1].

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.

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