Existence Theorem for Integral and Functional Integral Equations with Discontinuous Kernels

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): 𝑥(𝑡)=𝑢(𝑡)+^𝑡₀𝑓(𝑡,𝜏,𝑥(𝜏))𝑑𝜏, (1.1) and the functional Volterra integral equations (for short FVIEs): 𝑥(𝑡)=𝑢(𝑡)+^𝑡₀𝑓(𝑡,𝜏,𝑥(𝜏),𝑥)𝑑𝜏, (1.2) where 𝑓 may be discontinuous with respect to all of their arguments. The special case of (1.1) 𝑥(𝑡)=^𝑡₀𝑘(𝑡−𝜏)𝑔(𝑥(𝜏))𝑑𝜏, (1.3) 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 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 [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 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 ℝ) 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 𝑓∶[0,1]×[0,1]×ℝ→ℝ satisfies conditions (𝐶1) and (𝐶3). Let 𝑥₁,𝑥₂∶[0,1]→ℝ be continuous and satisfy the inequality 𝑥₁(𝜏)<𝑥₂(𝜏) for all 𝜏∈[0,1]. Then the functions 𝜑(𝑡,𝜏)=inf_{𝑥𝑦∈1(𝜏),𝑥2(𝑡,𝜏)}𝑓𝑡,𝜏,𝑦,𝜓(𝑡,𝜏)=sup_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓𝑡,𝜏,𝑦, (1.4) 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 𝑥∈𝐶([0,1]).

Lemma 1.2. Suppose that 𝑓∶[0,1]×[0,1]×ℝ→ℝ satisfies conditions (𝐶1), (𝐶2), and (𝐶3). Let 𝑥₁,𝑥₂∶[0,1]→ℝ 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]×ℝ, 𝑓^∗(𝑡,𝜏,𝑥)=liminf_𝑦↓𝑥𝑓𝑡,𝜏,𝑦,𝑓_∗(𝑡,𝜏,𝑥)=limsup_𝑦↑𝑥𝑓𝑡,𝜏,𝑦. (1.5) The compositions 𝑓^∗(𝑡,⋅,𝑥(⋅)) and 𝑓_∗(𝑡,⋅,𝑥(⋅)) are Lebesgue measurable for all 𝑡∈[0,1] any continuous 𝑥∶[0,1]→ℝ, and, for almost all 𝜏∈[0,1], inf_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓_∗𝑡,𝜏,𝑦=inf_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓𝑡,𝜏,𝑦=inf_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓^∗𝑡,𝜏,𝑦,sup_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓_∗𝑡,𝜏,𝑦=sup_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓𝑡,𝜏,𝑦=sup_{𝑥𝑦∈1(𝜏),𝑥2(𝜏)}𝑓^∗𝑡,𝜏,𝑦. (1.6)

The outline of the work is as follows. In Section 2 we present our existence theorem for (1.1) in ℝ. 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]→ℝ and let 𝑓∶[0,1]×[0,1]×ℝ→ℝ be given. Suppose that (C1)–(C4) are fulfilled.(𝐶1)𝑢 is continuous.(𝐶2) For each (𝑡,𝑥)∈[0,1]×ℝ, the function 𝜏↦𝑓(𝑡,𝜏,𝑥) is Lebesgue measurable. For all (𝑡,𝑥)∈[0,1]×ℝ and for almost all 𝜏∈[0,1], 𝑓(𝑡,𝜏,𝑥)<𝑀(𝜏), (2.1) where 𝑀∶[0,1]→[0,∞] is a Lebesgue integrable function.(𝐶3) For each (𝑡,𝜏,𝑥)∈[0,1]×[0,1]×ℝ, limsup_y↑x𝑓𝑡,𝜏,𝑦≤𝑓(𝑡,𝜏,𝑥)=liminf_y↓x𝑓𝑡,𝜏,𝑦. (2.2)(𝐶4) Let ϝ={𝑦∈ℝ;|𝑦|≤|𝑢|+|∫¹₀𝑀(𝜏)𝑑𝜏|}, where, |𝑢|=max{|𝑢(𝑡)|;𝑡∈[0,1]}. For every 𝑦∈ϝ and all 𝑛∈𝑁, the functions 𝑡↦^𝑡₀sup_{𝑥−𝑦≤1/3𝑛}𝑓(𝑡,𝜏,𝑥)𝑑𝜏, (2.3) are equicontinuous and tend to zero as 𝑡↓0.

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 𝑥_𝑛, 𝑛∈𝑁.
Step  1.  Since 𝑢, being a continuous function on compact set, is uniformly continuous and 𝐸↦∫_𝐸𝑀𝑑𝜏, being absolutely continuous with respect to Lebesgue measure, is uniformly continuous on [0,1]; then for all 𝑛∈𝑁 there exists 𝛿_𝑛>0 such that |𝑢(𝑠)−𝑢(𝑡)|+^𝑡_𝑠𝑀(𝜏)≤1𝑑𝜏3^𝑛+2, (2.4) for all 𝑠,𝑡∈[0,1], with |𝑠−𝑡|≤𝛿_𝑛. Next, we take for 𝑛∈𝑁 subdivisions 𝐷_𝑛𝑡^𝑛₀<𝑡^𝑛₁<⋯<𝑡^𝑛_𝑘𝑛 (2.5) of [0,1] in such a way that 0=𝑡^𝑛₀<𝑡^𝑛₁<⋯<𝑡^𝑛_𝑘𝑛=1, 𝐷_𝑛+1 is a refinement of 𝐷_𝑛, that is, 𝐷_𝑛⊂𝐷_𝑛+1 and 𝑡^𝑛_𝑘+1−𝑡^𝑛_𝑘≤𝛿_𝑛,𝑘=0,1,…,𝑘_𝑛−1. (2.6) For any 𝑛∈𝑁, 𝑔_𝑛∶[0,1]×[0,1]→ℝ and 𝑥_𝑛∶[0,1]→ℝ are recursively defined by, for 𝑠,𝑡∈[0,𝑡^𝑛₁], 𝑔_𝑛(𝑡,𝜏)=sup_{|(0𝑥−𝑢)|≤2/3𝑛}𝑓(),𝑥𝑡,𝜏,𝑥_𝑛(𝑡)=𝑢(𝑡)+^𝑡₀𝑔_𝑛(𝑡,𝜏)𝑑𝜏. (2.7) Once 𝑔_𝑛,𝑥_𝑛 have already been defined on [0,𝑡^𝑛_𝑘], with 𝑘<𝑘_𝑛, they are defined in [𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1] by putting 𝑔_𝑛(𝑡,𝜏)=sup_{𝑥−𝑥𝑛𝑡𝑛𝑘≤2/3𝑛}𝑓(),𝑥𝑡,𝜏,𝑥_𝑛(𝑡)=𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑔_𝑛(𝑡,𝜏)𝑑𝜏. (2.8)(2.9) It follows, by Lemma 1.1, that functions 𝑔_𝑛 are Lebesgue measurable; taking into account this together with (𝐶2), 𝑥_𝑛 is well defined. Moreover it is easy to see that for all 𝑡∈[0,1] and all 𝑛∈𝑁𝑥_𝑛(𝑡)=𝑢(𝑡)+^𝑡₀𝑔_𝑛(𝑡,𝜏)𝑑𝜏. (2.10)
Step  2.  We claim that, for all 𝑛∈𝑁,(i)𝑥_𝑛+1≤𝑥_𝑛,(ii)if 𝑥∶[0,1]→ℝ is a continuous function, which serves as a dummy function, satisfying, 𝑥(0)≤𝑢(0) and 𝑥(𝑡)≤𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+∫^𝑡_𝑠𝑓(𝑡,𝜏,𝑥(𝜏))𝑑𝜏, for 𝑠,𝑡∈[0,1], then 𝑥≤𝑥_𝑛 in [0,1].
To prove these assertions we shall proceed inductively. Clearly, 𝑥_𝑛+1(0)=𝑥_𝑛(0)=𝑝(0). Let us suppose that 𝑥_𝑛+1(𝑡)≤𝑥_𝑛(𝑡), for 𝑡∈[0,𝑡^𝑛_𝑘], with some 𝑘<𝑘_𝑛. Since 𝐷_𝑛⊂𝐷_𝑛+1, there exist 𝑖,𝑗∈{0,1…,𝑘_𝑛+1}, 𝑖<𝑗, such that 𝑡_𝑖^𝑛+1=𝑡^𝑛_𝑘 and 𝑡_𝑗^𝑛+1=𝑡^𝑛_𝑘+1. Let us suppose that 𝑥_𝑛+1(𝑡)≤𝑥_𝑛(𝑡), for [0,𝑡_𝑚^𝑛+1], with an 𝑚∈{𝑖,𝑖+1,…,𝑗−1}. At this point we have just two possibilities:
(𝑃₁₁)   𝑥_𝑛+1(𝑡_𝑚^𝑛+1)<𝑥_𝑛(𝑡_𝑚^𝑛+1)−2/3^𝑛+1,
(𝑃₁₂)   𝑥_𝑛(𝑡_𝑚^𝑛+1)−2/3^𝑛+1≤𝑥_𝑛+1(𝑡_𝑚^𝑛+1)≤𝑥_𝑛(𝑡_𝑚^𝑛+1).
If (𝑃₁₁) holds, since 𝑡^𝑛+1_𝑚+1−𝑡_𝑚^𝑛+1≤𝛿_𝑛+1, it follows, by (2.4) and (2.9), that for 𝑡∈[𝑡_𝑚^𝑛+1,𝑡^𝑛+1_𝑚+1]𝑥_𝑛+1(𝑡)=𝑥_𝑛+1𝑡_𝑚^𝑛+1+𝑢(𝑡)−𝑢𝑡_𝑚^𝑛+1+^𝑡_{𝑡𝑚𝑛+1}𝑔_𝑛(𝑡,𝜏)𝑑𝜏<𝑥_𝑛+1𝑡_𝑚^𝑛+1+13^𝑛+2<𝑥_𝑛𝑡_𝑚^𝑛+1−23^𝑛+1+13^𝑛+2=𝑥_𝑛(𝑡)+𝑥_𝑛𝑡_𝑚^𝑛+1−𝑥_𝑛(𝑡−5)3^𝑛+2<𝑥_𝑛(𝑡)+13^𝑛+2−53^𝑛+2=𝑥_𝑛(𝑡)+13^𝑛+1−53^𝑛+2<𝑥_𝑛(𝑡). (2.11) Assume the validity of (𝑃₁₂); it follows, by (2.4) and (2.9), that 𝑥_𝑛+1𝑡_𝑚^𝑛+1−23^𝑛+1≥𝑥_𝑛𝑡_𝑚^𝑛+1−23^𝑛+1−23^𝑛+1=𝑥_𝑛𝑡^𝑛_𝑘+𝑥_𝑛𝑡_𝑚^𝑛+1−𝑥_𝑛𝑡^𝑛_𝑘−43^𝑛+1>𝑥_𝑛𝑡^𝑛_𝑘−13^𝑛+2−43^𝑛+1=𝑥_𝑛𝑡^𝑛_𝑘−13^𝑛+1−43^𝑛+1>𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛. (2.12) On the other hand we have 𝑥_𝑛+1𝑡_𝑚^𝑛+1+23^𝑛+1≤𝑥_𝑛𝑡_𝑚^𝑛+1+23^𝑛+1=𝑥_𝑛𝑡^𝑛_𝑘+𝑥_𝑛𝑡_𝑚^𝑛+1−𝑥_𝑛𝑡^𝑛_𝑘+23^𝑛+1<𝑥_𝑛𝑡^𝑛_𝑘+13^𝑛+2+23^𝑛+1=𝑥_𝑛𝑡^𝑛_𝑘+13^𝑛+1+23^𝑛+1<𝑥_𝑛𝑡^𝑛_𝑘+23^𝑛. (2.13) We thus have 𝑥_𝑛+1𝑡_𝑚^𝑛+1−23^𝑛+1,𝑥_𝑛+1𝑡_𝑚^𝑛+1+23^𝑛+1⊂𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛,𝑥_𝑛𝑡^𝑛_𝑘+23^𝑛, (2.14) and hence, for each 𝜏∈[𝑡_𝑚^𝑛+1,𝑡^𝑛+1_𝑚+1], 𝑔_𝑛+1(𝑡,𝜏)=sup_{𝑥−𝑥𝑛+1𝑡𝑚𝑛+1≤2/3𝑛}𝑓(𝑡,𝜏,𝑥)≤sup_{𝑥−𝑥𝑛𝑡𝑛𝑘≤2/3𝑛}𝑓(𝑡,𝜏,𝑥)=𝑔_𝑛(𝑡,𝜏). (2.15) By (2.9), 𝑥_𝑛𝑡_𝑚^𝑛+1=𝑥_𝑛𝑡^𝑛_𝑘+𝑢𝑡_𝑚^𝑛+1−𝑢𝑡^𝑛_𝑘+^{𝑡𝑚𝑛+1}_𝑡𝑛𝑘𝑔_𝑛(𝑡,𝜏)𝑑𝜏. (2.16) This in turn implies that, for all 𝑡∈[𝑡_𝑚^𝑛+1,𝑡^𝑛+1_𝑚+1], 𝑥_𝑛+1(𝑡)=𝑥_𝑛+1𝑡_𝑚^𝑛+1+𝑢(𝑡)−𝑢𝑡_𝑚^𝑛+1+^𝑡_{𝑡𝑚𝑛+1}𝑔_𝑛+1(𝑡,𝜏)𝑑𝜏≤𝑥_𝑛𝑡_𝑚^𝑛+1+𝑢(𝑡)−𝑢𝑡_𝑚^𝑛+1+^𝑡_{𝑡𝑚𝑛+1}𝑔_𝑛(𝑡,𝜏)𝑑𝜏=𝑥_𝑛𝑡^𝑛_𝑖+𝑢𝑡_𝑚^𝑛+1−𝑢𝑡^𝑛_𝑘+^{𝑡𝑚𝑛+1}_𝑡𝑛𝑘𝑔_𝑛(𝑡,𝜏)𝑑𝜏+𝑢(𝑡)−𝑢𝑡_𝑚^𝑛+1+^𝑡_{𝑡𝑚𝑛+1}𝑔_𝑛(𝑡,𝜏)𝑑𝜏=𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑔_𝑛(𝑡,𝜏)𝑑𝜏=𝑥_𝑛(𝑡), (2.17) which completes the proof of (i). Now we shall handle (ii) inductively too. Let us fix an arbitrary 𝑛∈𝑁, by assumption, 𝑥(0)≤𝑢(0)=𝑥_𝑛(0). Let us suppose that 𝑥(𝑡)≤𝑥_𝑛(𝑡), for 𝑡∈[0,𝑡^𝑛_𝑘], with 𝑘<𝑘_𝑛. At this point we have just two possibilities:
(𝑃₂₁) 𝑥(𝑡^𝑛_𝑘)<𝑥_𝑛(𝑡^𝑛_𝑘)−2/3^𝑛+1,
(𝑃₂₂) 𝑥_𝑛(𝑡^𝑛_𝑘)−2/3^𝑛+1≤𝑥(𝑡^𝑛_𝑘)≤𝑥_𝑛(𝑡^𝑛_𝑘).
Suppose that we are in (𝑃₂₁); since 𝑡^𝑛_𝑘+1−𝑡^𝑛_𝑘≤𝛿_𝑛, 𝑘=0,1,2,…,𝑘_𝑛−1, it follows, by (2.4) and (2.9); and (ii), that 𝑥(𝑡)≤𝑥𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑓((𝜏𝑡𝑡,𝜏,𝑥))𝑑𝜏<𝑥^𝑛_𝑘+13^𝑛+2<𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛+1+13^𝑛+2=𝑥_𝑛(𝑡)+𝑥_𝑛𝑡^𝑛_𝑘−𝑥_𝑛(𝑡−5)3^𝑛+2<𝑥_𝑛(𝑡)+13^𝑛+2−53^𝑛+2<𝑥_𝑛(𝑡), (2.18) for 𝑡∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1].
If we are in (𝑃₂₂), it follows, by (2.4) and (2.9), that, for 𝑡∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1], 𝑥(𝑡)=𝑥𝑡^𝑛_𝑘+𝑥(𝑡)−𝑥𝑡^𝑛_𝑘≥𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛+1+𝑥(𝑡)−𝑥𝑡^𝑛_𝑘>𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛+1−13^𝑛+2>𝑥_𝑛𝑡^𝑛_𝑘−23^𝑛. (2.19) By (ii) and (2.9), 𝑥(𝑡)≤𝑥𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑓(𝑡,𝜏,𝑥(𝜏))𝑑𝜏≤𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑓(𝑡,𝜏,𝑥(𝜏))𝑑𝜏<𝑥_𝑛𝑡^𝑛_𝑘+13^𝑛+2<𝑥_𝑛𝑡^𝑛_𝑘+23^𝑛. (2.20) We thus have for all 𝜏∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1], 𝑥(𝜏)∈(𝑥_𝑛(𝑡^𝑛_𝑘)−2/3^𝑛,𝑥_𝑛(𝑡^𝑛_𝑘)+2/3^𝑛), and hence 𝑓(𝑡,𝜏,𝑥(𝜏))≤sup_{𝑥−𝑥𝑛𝑡𝑛𝑘≤2/3𝑛}𝑓(𝑡,𝜏,𝑥)=𝑔_𝑛(𝑡,𝜏), (2.21) for all 𝜏∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1], so it follows, by (2.4) and (2.9), that 𝑥(𝑡)≤𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑓(𝑡,𝜏,𝑥(𝜏))𝑑𝜏≤𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘𝑔_𝑛(𝑡,𝜏)𝑑𝜏=𝑥_𝑛(𝑡), (2.22) for 𝑡∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1], which completes the proof of the assertion (ii).
Step  3. It follows, by (C4), that the constructed bounded nonincreasing sequence 𝑥_𝑛, 𝑛∈𝑁 is uniformly convergence, and hence let us set 𝑥⁺(𝑡)=lim_𝑛→∞𝑥_𝑛(𝑡),𝑡∈[0,1]. (2.23) By (2.4) and (2.9), we have|𝑥_𝑛(𝜏)−𝑥_𝑛(𝑡^𝑛_𝑘)|≤1/3^𝑛+2, for 𝜏∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1]. Thus, 𝑔_𝑛(𝑡,𝜏)=sup_{𝑥−𝑥𝑛𝑡𝑛𝑘≤2/3𝑛}𝑓(𝑡,𝜏,𝑥)≥𝑓𝑡,𝜏,𝑥_𝑛(𝜏), (2.24) for 𝜏∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1]. Therefore, by formula just before Step  2, for 𝑡∈[0,1], and 𝑛∈𝑁𝑥_𝑛(𝑡)=𝑢(𝑡)+^𝑡₀𝑔_𝑛(𝑡,𝜏)𝑑𝜏≥𝑢(𝑡)+^𝑡₀𝑓𝑡,𝜏,𝑥_𝑛(𝜏)𝑑𝜏. (2.25) Applying Fatou Lemma and taking into account the condition (𝐶3) and in view of Lemma 1.2, we obtain 𝑥⁺(𝑡)=lim_𝑛→∞𝑥_𝑛(𝑡)=liminf_𝑛→∞𝑥_𝑛(𝑡)≥𝑢(𝑡)+^𝑡₀liminf_𝑛→∞𝑓𝑡,𝜏,𝑥_𝑛(𝜏)(𝑡)+𝑑𝜏=𝑢^𝑡₀𝑓𝑡,𝜏,𝑥⁺(𝜏)𝑑𝜏, (2.26) for each 𝑡∈[0,1]. Let us observe that limsup_𝑛→∞𝑔_𝑛(𝑡,𝜏)≤𝑓𝑡,𝜏,𝑥⁺(𝜏), (2.27) almost everywhere in [0,1]. Indeed, let us fix an 𝑛∈𝑁. For any 𝜏∈[0,1] there exists a 𝑘=0,…,𝑘_𝑛 such that 𝜏∈[𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1]. Since 𝑔_𝑛(𝑡,𝜏)=sup_{𝑥−𝑥𝑛𝑡𝑛𝑘≤2/3𝑛}𝑓(𝑡,𝜏,𝑥), (2.28) there exists an 𝑥_𝑛, with |𝑥_𝑛−𝑥_𝑛(𝑡^𝑛_𝑘)|≤2/3^𝑛 such that 𝑔_𝑛()−1𝑡,𝜏3^𝑛≤𝑓𝑡,𝜏,𝑥_𝑛≤𝑔_𝑛(𝑡,𝜏). (2.29) Whence, limsup_𝑛→∞𝑔_𝑛(𝑡,𝜏)=limsup_𝑛→∞𝑓𝑡,𝜏,𝑥_𝑛.𝑥_𝑛−𝑥_𝑛(𝜏)≤𝑥_𝑛−𝑥_𝑛𝑡^𝑛_𝑘+𝑥_𝑛𝑡^𝑛_𝑘−𝑥_𝑛(𝜏≤4)3^𝑛. (2.30)(2.31) We thus have lim_𝑛→∞𝑥_𝑛=lim_𝑛→∞𝑥_𝑛(𝜏)=𝑥⁺(𝜏), (2.32) which together with (𝐶3) and (2.30) implies that limsup_𝑛→∞𝑔_𝑛(𝑡,𝜏)≤𝑓𝑡,𝜏,𝑥⁺(𝜏), (2.33) almost everywhere in [0,1]. Applying Fatou lemma once again we obtain 𝑥⁺(𝑡)=lim_𝑛→∞𝑥_𝑛(𝑡)=limsup_𝑛→∞𝑥_𝑛(𝑡)≤𝑢(𝑡)+^𝑡₀limsup_𝑛→∞𝑔_𝑛(𝑡,𝜏)𝑑𝜏≤𝑢(𝑡)+^𝑡₀𝑓𝑡,𝜏,𝑥⁺(𝜏)𝑑𝜏, (2.34) which together with (2.26) means that 𝑥⁺ is a solution (1.1).
Step  4.  Let 𝑥∶[0,1]→ℝ be a solution of (1.1). Clearly, 𝑥 is continuous and satisfies the conditions of the assertion (ii), so, 𝑥≤𝑥_𝑛, for any 𝑛∈𝑁. Since 𝑥_𝑛→𝑥⁺ as 𝑛→∞, then 𝑥≤𝑥⁺. 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 ℎ_𝑛(𝑡,𝜏) and 𝑥_𝑛(𝑡), by setting ℎ_𝑛(𝑡,𝜏)=inf_{|𝑥−𝑥𝑛(𝑡𝑛𝑘)|≤2/3𝑛}𝑓()[],𝑥𝑡,𝜏,𝑥,𝜏∈0,1_𝑛(𝑡)=𝑥_𝑛𝑡^𝑛_𝑘+𝑢(𝑡)−𝑢𝑡^𝑛_𝑘+^𝑡_𝑡𝑛𝑘ℎ_𝑛(𝑡,𝜏)𝑑𝜏,𝑡∈𝑡^𝑛_𝑘,𝑡^𝑛_𝑘+1. (2.35) Taking 𝑥∶[0,1]→ℝ 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)+∫^𝑡_𝑠𝑓(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 [12]. Probably the reason consists in the fact that the composition 𝑓(𝑡,⋅,𝑥(⋅)) is no longer measurable for any continuous function 𝑥∶[0,1]→ℝ; 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]×ℝ→ℝ by 𝑓(𝑡,𝜏,𝑥)=1if𝑥>𝜏,1if𝑥=𝜏,𝜏∈𝑆,0otherwise. (2.36) 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]×ℝ, limsup_𝑦↑𝑥𝑓𝑡,𝜏,𝑦≤𝑓(𝑡,𝜏,𝑥)≤liminf_𝑦↓𝑥𝑓𝑡,𝜏,𝑦. (2.37) 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, 𝐶_𝑀([0,1];ℝ) is the set of all continuous functions 𝑥∶[0,1]→ℝ satisfying |𝑥(𝑡)−𝑥(𝑠)|≥|𝑢(𝑡)−𝑢(𝑠)|+|∫^𝑡_𝑠𝑀(𝜏)𝑑𝜏| for all 𝑠,𝑡∈[0,1]. For a fixed 𝜑∈𝐶_𝑀([0,1]) let 𝑓_𝜑∶[0,1]×[0,1]×ℝ→ℝ be the function defined by 𝑓_𝜑(𝑡,𝜏,𝑥)=𝑓(𝑡,𝜏,𝑥,𝜑), and let Γ_𝜑={𝑥∈𝐶_𝑀([0,1];ℝ)∣𝑥(0)≥𝑢(0),𝑥(𝑡)≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+∫^𝑡_𝑠𝑓_𝜑(𝑡,𝜏,𝑥(𝜏))𝑑𝜏,for𝑠,𝑡∈[0,1]}.

Theorem 3.1. Let Υ denote the set of all 𝑓∶[0,1]×[0,1]×ℝ×𝐶_𝑀([0,1];ℝ)→ℝ satisfying the following conditions (F1)–(F5).(𝐹1)𝑢∶[0,1]→ℝ is continuous,(𝐹2) For each (𝑡,𝑥)∈[0,1]×ℝ and 𝜑∈𝐶_𝑀([0,1];ℝ),𝜏↦𝑓(𝑡,𝜏,𝑥,𝜑) is Lebesgue measurable, and for almost all 𝜏∈[0,1], 𝑓𝑡,𝜏,𝑥,𝜑<𝑀(𝜏). (3.1)(𝐹3)   For each (𝑡,𝜏,𝑥,𝜑)∈[0,1]×[0,1]×ℝ×𝐶_𝑀([0,1];ℝ), limsup_𝑦↑𝑥𝑓𝑡,𝜏,𝑦,𝜑≤𝑓𝑡,𝜏,𝑥,𝜑=liminf_𝑦↓𝑥𝑓𝑡,𝜏,𝑦,𝜑. (3.2)(𝐹4) For each (𝑡,𝜏,𝑥)∈[0,1]×[0,1]×ℝ, 𝑓(𝑡,𝜏,𝑥,𝜑)≤𝑓(𝑡,𝜏,𝑥,𝜓) whenever 𝜑,𝜓∈𝐶_𝑀([0,1];ℝ) with 𝜑≤𝜓. (𝐹5) Let ϝ={𝑦∈ℝ;|𝑦|≤|𝑢|+|∫¹₀𝑀(𝜏)𝑑𝜏|}, where, |𝑢|=max{|𝑢(𝑡)|;𝑡∈[0,1]}. Let 𝜑∈𝐶_𝑀([0,1];ℝ) be fixed, for every 𝑦∈ϝ and all 𝑛∈𝑁; the functions 𝑡↦^𝑡₀sup_{𝑥−𝑦≤1/3𝑛}𝑓_𝜑(𝑡,𝜏,𝑥)𝑑𝜏, (3.3)are equicontinuous and tend to zero as 𝑡↓0.

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 𝜑∈𝐶_𝑀([0,1];ℝ) let us consider the nonfunctional Volterra integral equation N-FVIE: 𝑥(𝑡)=𝑢(𝑡)+^𝑡₀𝑓_𝜑(𝑡,𝜏,𝑥(𝜏))𝑑𝜏. (3.4) Obviously, the function 𝑓_𝜑(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 𝑥_𝜑(𝑡)=inf_{𝑥∈Γ𝜑}𝑥(𝑡). (3.5) Let ℶ={𝜑∈𝐶_𝑀([0,1];ℝ)∣𝑥_𝜑≤𝜑}. Since −𝑢(𝑡)−∫^𝑡_𝑠𝑀(𝜏)𝑑𝜏 belongs to ℶ, then the set ℶ is not empty. Define 𝑥(𝑡)=inf_𝜑∈ℶ𝜑(𝑡). (3.6) Given 𝜑∈ℶ and 𝑥∈Γ_𝜑, it follows, by (F4), that 𝑥(𝑡)≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+^𝑡_𝑠𝑓𝑡,𝜏,𝑥(𝜏),𝜑𝑑𝜏≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+^𝑡_𝑠𝑓(𝑡,𝜏,𝑥(𝜏),𝑥). (3.7) We thus have 𝑥∈Γ_𝑥, Γ_𝜑⊆Γ_𝑥 and 𝑥_𝜑≥𝑥_𝑥. Since 𝜑∈ℶ is arbitrary, then 𝜑≥𝑥_𝜑≥𝑥_𝑥, and hence 𝑥≥𝑥_𝑥. (3.8) On the other hand, for 𝑥∈Γ_𝜑 we have 𝑥(𝑡)≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+^𝑡_𝑠𝑓𝑡,𝜏,𝑥(𝜏),𝜑𝑑𝜏≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+^𝑡_𝑠𝑓(𝑡,𝜏,𝑥(𝜏),𝑥)≥𝑥(𝑠)+𝑢(𝑡)−𝑢(𝑠)+^𝑡_𝑠𝑓𝑡,𝜏,𝑥(𝜏),𝑥_𝑥. (3.9) Thus 𝑥∈Γ_𝜑⊆Γ_𝑥⊆Γ_𝑥𝑥. Consequently, 𝑥_𝜑≥𝑥_𝑥≥𝑥_𝑥𝑥 which implies that 𝑥_𝑥∈ℶ, and thus 𝑥≤𝑥_𝑥 which together with (3.8) implies that 𝑥=𝑥_𝑥. Since every solution of (1.2) belongs to ℶ, 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 ℝ^𝑛 makes the condition, limsup_𝑦↑𝑥𝑓𝑡,𝜏,𝑦≤𝑓(𝑡,𝜏,𝑥)=liminf_𝑦↓𝑥𝑓𝑡,𝜏,𝑦, (4.1) 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,…,𝑢_𝑛)∶[0,1]→ℝ^𝑛 and 𝑓=(𝑓₁,…,𝑓_𝑛)∶[0,1]×[0,1]×ℝ^𝑛→ℝ^𝑛 satisfying the conditions (𝐵1)-(𝐵4) below(𝐵1)𝑓(𝑡,⋅,𝑥(⋅))is Lebesgue measurable for any continuous 𝑥∶[0,1]→ℝ^𝑛,(𝐵2) for each 𝑖=1,…,𝑛 and Lebesgue almost all 𝑡∈[0,1], 𝑓_𝑖 is nondecreasing in 𝑥_𝑘, 𝑘=1,…,𝑖−1,𝑖+1,…,𝑛 and for all 𝑥=(𝑥₁,…,𝑥_𝑛)∈ℝ^𝑛, limsup_{𝑦↑𝑥𝑖}𝑓_𝑖𝑡,𝜏,𝑥₁,…,𝑥_𝑖−1,𝑦,𝑥_𝑖+1,…,𝑥_𝑛≤𝑓_𝑖𝑡,𝜏,𝑥₁,…,𝑥_𝑖−1,𝑥_𝑖,𝑥_𝑖+1,…,𝑥_𝑛=lim_{𝑦↓𝑥𝑖}𝑓_𝑖𝑡,𝜏,𝑥₁,…,𝑥_𝑖−1,𝑦,𝑥_𝑖+1,…,𝑥_𝑛, (4.2)(𝐵3) for each 𝑥∈ℝ^𝑛 and Lebesgue almost all 𝑡∈[0,1], ‖𝑓(𝑡,𝜏,𝑥)‖≤𝑀(𝜏), where ‖𝑥‖=max{|𝑥_𝑖|;𝑖=1,…,𝑛} and 𝑀∶[0,1]→[0,∞] is a Lebesgue integrable function,(𝐵4) let ϝ={𝑦∈ℝ^𝑛;‖𝑦‖≤‖𝑢‖+|∫¹₀𝑀(𝜏)𝑑𝜏|}, for every 𝑦∈ϝ and all 𝑘∈𝑁, the functions 𝑡↦^𝑡₀sup_{𝑥−𝑦≤1/3𝑘}𝑓(𝑡,𝜏,𝑥)𝑑𝜏, (4.3) are equicontinuous and tend to zero as 𝑡↓0, where sup𝑓 and ∫𝑓 are interpreted componentwise.

Under the above assumptions VIE expressed by (1.1) (in ℝ^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,𝑦∈ℝ^𝑛, we write 𝑥≤𝑦 if 𝑥_𝑖≤𝑦_𝑖, for each 𝑖=1,…,𝑛. Let us denote by 𝑋 the set of all 𝑥=(𝑥₁,…,𝑥_𝑛)∶[0,1]→ℝ^𝑛 satisfying the following conditions 𝑥_𝑖(𝑡)≤𝑢_𝑖(𝑡)+^𝑡₀𝑓_𝑖𝑡,𝜏,𝑥₁(𝜏),…,𝑥_𝑛(𝜏)‖𝑥(𝑡)(𝑠≤‖𝑢(𝑡)(𝑠+𝑑𝜏,𝑖=1,2,…,𝑛,−𝑥)‖−𝑢)‖^𝑡_𝑠𝑀(𝜏)𝑑𝜏,𝑠,𝑡∈[0,1]. (4.4) For every 𝑖=1,2,…,𝑛, we let 𝑥⁺_𝑖(𝑡)=sup_{𝑥1,…,𝑥𝑛∈𝑋}𝑥_𝑖(𝑡),𝑡∈[0,1]. (4.5) It follows, by monotonicity, that for every 𝑥∈𝑋, for each 𝑖=1,…,𝑛, and for all 𝑡∈[0,1], 𝑥_𝑖(𝑡)≤𝑢_𝑖(𝑡)+^𝑡₀𝑓_𝑖𝑡,𝜏,𝑥⁺₁(𝜏),…,𝑥⁺_𝑖−1(𝜏),𝑥_𝑖(𝜏),𝑥⁺_𝑖+1(𝜏),…,𝑥⁺_𝑛(𝜏)𝑑𝜏. (4.6) 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; (𝑦_𝑚), 𝑦_𝑚=(𝑦¹_𝑚,…,𝑦^𝑛_𝑚)∶[0,1]→ℝ^𝑛 such that, for each 𝑖=1,…,𝑛𝑦^𝑖_𝑚(0)=𝑢_𝑖(0),𝑦^𝑖_𝑚(𝑡)=𝑢_𝑖(𝑡)+^𝑡₀𝑔^𝑖_𝑚(𝜏)𝑦(𝑡)(𝑠≤‖𝑢(𝑡)(𝑠+𝑑𝜏,−𝑦)−𝑢)‖^𝑡_𝑠𝑀(𝜏)𝑑𝜏,𝑠,𝑡∈[0,1], (4.7) where 𝑔^𝑖_𝑚(𝜏)=sup_{𝑥−𝑦𝑖𝑚𝑡𝑛𝑘≤2/3𝑚}𝑓_𝑖𝑡,𝜏,𝑥⁺₁(𝜏),…,𝑥⁺_𝑖−1(𝜏),𝑥,𝑥⁺_𝑖+1(𝜏),…,𝑥⁺_𝑛(𝜏). (4.8) Let, for each 𝑖=1,…,𝑛, lim_𝑚→∞𝑦^𝑖_𝑚(𝜏)=𝑦⁺_𝑖(𝜏),𝜏∈[0,1]. (4.9) Clearly 𝑥⁺_𝑖(𝑡)≤𝑦⁺_𝑖(𝑡). Regarding 𝑓_𝑖 as a function only of 𝑦⁺_𝑖 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], 𝑦⁺_𝑖(𝑡)=𝑢_𝑖(𝑡)+^𝑡₀𝑓_𝑖𝑡,𝜏,𝑥⁺₁(𝜏),…,𝑥⁺_𝑖−1(𝜏),𝑦⁺_𝑖(𝜏),𝑥⁺_𝑖+1(𝜏),…,𝑥⁺_𝑛(𝜏)𝑑𝜏. (4.10) By monotonicity 𝑦⁺_𝑖(𝑡)≤𝑢_𝑖(𝑡)+^𝑡₀𝑓_𝑖𝑡,𝜏,𝑦⁺₁(𝜏),…,𝑦⁺_𝑖−1(𝜏),𝑦⁺_𝑖(𝜏),𝑦⁺_𝑖+1(𝜏),…,𝑦⁺_𝑛(𝜏)𝑑𝜏,𝑖=1,…,𝑛. (4.11) We thus have 𝑦⁺=(𝑦⁺₁,…,𝑦⁺_𝑛)∈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.