Solvability of a Maximum Quadratic Integral Equation of Arbitrary Orders

Home , Quadratic integral

Advances in Dynamical Systems and Applications ISSN 0973-5321, Volume 11, Number 2, pp. 93–104 (2016) http://campus.mst.edu/adsa

Mohamed Abdalla Darwish King Abdulaziz University Sciences Faculty for Girls Department of Mathematics Jeddah, Saudi Arabia [email protected] Johnny Henderson Baylor University Department of Mathematics Waco, Texas 76798-7328, USA Johnny [email protected] Kishin Sadarangani Universidad de Las Palmas de Gran Canaria Departamento de Matematicas´ Campus de Taﬁra Baja, 35017 Las Palmas de Gran Canaria, Spain [email protected]

Abstract

We investigate a new quadratic integral equation of arbitrary orders with maximum and prove an existence result for it. We will use a ﬁxed point theorem due to Darbo as well as the monotonicity measure of noncompactness due to Banas´ and Olszowy to prove that our equation has at least one solution in C[0, 1] which is monotonic on [0, 1].

AMS Subject Classiﬁcations: 45G10, 47H09, 45M99. Keywords: Fractional, monotonic solutions, monotonicity measure of noncompactness, quadratic integral equation, Darbo theorem.

Received August 24, 2016; Accepted November 27, 2016 Communicated by Martin Bohner 94 M. A. Darwish, J. Henderson and K. Sadarangani

1 Introduction

In several papers, among them [1,11], the authors studied differential and integral equations with maximum. In [6–9] Darwish et al. studied fractional integral equations with supremum. Also, in [4, 5], Caballero et al. studied the Volterra quadratic integral equations with supremum. They showed that these equations have monotonic solutions in the space C[0, 1]. Darwish [7] generalized and extended the Caballero et al. [4] results to the case of quadratic fractional integral equations with supremum. In this paper we will study the fractional quadratic integral equation with maximum Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |y(τ)| y(t) = f(t) + 1−β ds, t ∈ J = [0, 1], 0 < β < 1, Γ(β) 0 (ϕ(t) − ϕ(s)) (1.1) where f, ϕ : J → R, T : C(J) → C(J), σ : J → J and κ : J × J → R+. By using the monotonicity measure of noncompactness due to Banas´ and Olszowy [3] as well as the Darbo ﬁxed point theorem, we prove the existence of monotonic solutions to (1.1) in C[0, 1]. Now, we assume that (E, k · k) is a real Banach space. We denote by B(x, r) the closed ball centred at x with radius r and Br ≡ B(θ, r), where θ is a zero element of E. We let X ⊂ E. The closure and convex closure of X are denoted by X and ConvX, respectively. The symbols X + Y and λY are using for the usual algebraic operators on sets and ME and NE stand for the families deﬁned by ME = {A ⊂ E : A 6= ∅,A is bounded} and NE = {B ⊂ ME : B is relatively compact}, respectively.

Deﬁnition 1.1 (See [2]). A function µ : ME → [0, +∞) is called a measure of noncompactness in E if the following conditions: ◦ 1 ∅= 6 {X ∈ ME : µ(X) = 0} = kerµ ⊂ NE, 2◦ if X ⊂ Y , then µ(X) ≤ µ(Y ), 3◦ µ(X) = µ(X) = µ(ConvX), 4◦ µ(λX + (1 − λ)Y ) ≤ λµ(X) + (1 − λ)µ(Y ), 0 ≤ λ ≤ 1 and

◦ 5 if (Xn) is a sequence of closed subsets of ME with Xn ⊃ Xn+1(n = 1, 2, 3, ...) ∞ and lim µ(Xn) = 0 then X∞ = ∩n=1Xn 6= ∅, n→∞ hold. We will establish our result in the Banach space C(J) of all defined, real and continuous functions on J ≡ [0, 1] with standard norm kyk = max{|y(τ)| : τ ∈ J}. Next, we define the measure of noncompactness related to monotonicity in C(J); see [2, 3]. Let ∅= 6 Y ⊂ C(J) be a bounded set. For y ∈ Y and ε ≥ 0, the modulus of continuity of the function y , denoted by ω(y, ε), is defined by ω(y, ε) = sup{|y(t) − y(s)| : t, s ∈ J, |t − s| ≤ ε}. A Maximum Quadratic Integral Equation 95

Moreover, we let ω(Y, ε) = sup{ω(y, ε): y ∈ Y } and

ω0(Y ) = lim ω(Y, ε). ε→0 Deﬁne d(y) = sup (|y(t) − y(s)| − [y(t) − y(s)]) t,s∈J, s≤t and d(Y ) = sup d(y). y∈Y Notice that all functions in Y are nondecreasing on J if and only if d(Y ) = 0.

Now, we deﬁne the map µ on MC(J) as

µ(Y ) = d(Y ) + ω0(Y ).

Clearly, µ satisﬁes all conditions in Deﬁnition 3, and therefore, it is a measure of noncompactness in C(J) [3].

Deﬁnition 1.2. Let P : M → E be a continuous mapping, where ∅= 6 M ⊂ E. Suppose that P maps bounded sets onto bounded sets. Let Y be any bounded subset of M with µ(PY ) ≤ αµ(Y ), α ≥ 0, then P is called verify the Darbo condition with respect to a measure of noncompactness µ.

In the case α < 1, the operator P is said to be a contraction with respect to µ.

Theorem 1.3 (See [10]). Let ∅= 6 Ω ⊂ E be a closed, bounded and convex set. If P :Ω → Ω is a continuous contraction mapping with respect to µ, then P has a ﬁxed point in Ω.

We will need the following two lemmas in order to prove our results [4].

Lemma 1.4. Let r : J → J be a continuous function and y ∈ C(J). If, for t ∈ J,

(F y)(t) = max |y(τ)|, [0,σ(t)] then F y ∈ C(J).

Lemma 1.5. Let (yn) be a sequence in C(J) and y ∈ C(J). If (yn) converges to y ∈ C(J), then (F yn) converges uniformly to F y uniformly on J. 96 M. A. Darwish, J. Henderson and K. Sadarangani

2 Main Theorem

Let us consider the following assumptions:

(a1) f ∈ C(J). Moreover, f is nondecreasing and nonnegative on J.

(a2) The operator T : C(J) → C(J) is continuous and satisﬁes the Darbo condition with a constant c for the measure of noncompactness µ . Moreover, T y ≥ 0 if y ≥ 0.

(a3) There exist constants a, b ≥ 0 such that |(T y)(t)| ≤ a + bkyk ∀y ∈ C(J), t ∈ J.

1 (a4) The function ϕ : J → R is C (J) and nondecreasing.

(a5) The function κ : J ×J → R+ is continuous on J ×J and nondecreasing ∀t and s separately. Moreover, κ∗ = sup κ(t, s). (t,s)∈J×J

(a6) The function σ : J → J is nondecreasing and continuous on J.

(a7) ∃ r0 > 0 such that

κ∗r (a + br ) kfk + 0 0 (ϕ(1) − ϕ(0))β ≤ r (2.1) Γ(β + 1) 0

ck∗r and 0 < (ϕ(1) − ϕ(0))−β. Γ(β + 1) Now, we deﬁne two operators K and F on C(J) as follows

Z t 0 1 ϕ (s)κ(t, s) max[0,σ(s)] |y(τ)| (Ky)(t) = 1−β ds (2.2) Γ(β) 0 (ϕ(t) − ϕ(s)) and (Fy)(t) = f(t) + (T y)(t) · (Ky)(t), (2.3) respectively. Solving (1.1) is equivalent to ﬁnd a ﬁxed point of the operator F. Under the above assumptions, we will prove the following theorem.

Theorem 2.1. Assume the assumptions (a1) − (a7) are satisﬁed. Then (1.1) has at least one solution y ∈ C(J) which is nondecreasing on J.

Proof. First, we claim that the operator F transforms C(J) into itself. For this, it is sufﬁcient to show that if y ∈ C(J), then Ky ∈ C(J). Let y ∈ C(J) and t1, t2 ∈ J (t1 ≤ t2) such that |t2 − t1| ≤ ε for ﬁxed ε > 0, then we have

|(Ky)(t2) − (Ky)(t1)| A Maximum Quadratic Integral Equation 97

Z t2 0 1 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| = 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t1) − ϕ(s)) Z t2 0 1 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| ≤ 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t2) − ϕ(s)) Z t2 0 1 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t2) − ϕ(s)) Z t1 0 1 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t1) − ϕ(s)) Z t2 0 1 ϕ (s)|κ(t2, s) − κ(t1, s)| max[0,σ(s)] |y(τ)| ≤ 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 Z t1 1 ϕ (s)|κ(t1, s)| max[0,σ(s)] |y(τ)| 1 + 1−β ds + |κ(t1, s)| Γ(β) t1 (ϕ(t2) − ϕ(s)) Γ(β) 0 0 β−1 β−1 ×ϕ (s) |(ϕ(t2) − ϕ(s)) − (ϕ(t1) − ϕ(s)) | max |y(τ)| ds [0,σ(s)] kyk Z t2 ϕ0(s) κ∗kyk ≤ ωκ(ε, .) 1−β ds + Γ(β) 0 (ϕ(t2) − ϕ(s)) Γ(β) Z t1 0 β−1 β−1 × ϕ (s)[(ϕ(t1) − ϕ(s)) − (ϕ(t2) − ϕ(s)) ]ds 0 Z t2 ϕ0(s) + 1−β ds t1 (ϕ(t2) − ϕ(s)) kyk κ∗kyk = ω (ε, .)(ϕ(t ) − ϕ(0))β + Γ(β + 1) κ 2 Γ(β + 1) β β β × (ϕ(t1) − ϕ(0)) − (ϕ(t2) − ϕ(0)) + 2(ϕ(t2) − ϕ(t1)) kyk 2κ∗kyk ≤ ω (ε, .)(ϕ(t ) − ϕ(0))β + (ϕ(t ) − ϕ(t ))β Γ(β + 1) κ 2 Γ(β + 1) 2 1 kyk 2κ∗kyk ≤ ω (ε, .)(ϕ(1) − ϕ(0))β + [ω(ϕ, ε)]β, (2.4) Γ(β + 1) κ Γ(β + 1) where we used ωκ(ε, .) = sup |κ(t, s) − κ(τ, s)| t, τ∈J, |t−τ|≤ε 98 M. A. Darwish, J. Henderson and K. Sadarangani

and the fact that ϕ(t1) − ϕ(0) ≤ ϕ(t2) − ϕ(0). Notice that, since the function κ is uniformly continuous on J ×J and the function ϕ is continuous on J, then when ε → 0, we have that ωκ(ε, .) → 0 and ω(ϕ, ε) → 0. Therefore, Ky ∈ C(J) and consequently, Fy ∈ C(J). Now, for t ∈ J, we have

Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |y(τ)| |(Fy)(t)| ≤ f(t) + 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) a + bkyk Z t ϕ0(s) ≤ kfk + κ(t, s) 1−β max |y(τ)|ds Γ(β) 0 (ϕ(t) − ϕ(s)) [0,σ(s)] (a + bkyk)κ∗kyk ≤ kfk + (ϕ(t) − ϕ(0))β. Γ(β + 1)

Hence (a + bkyk)κ∗kyk kFyk ≤ kfk + (ϕ(1) − ϕ(0))β. Γ(β + 1)

By assumption (a7), if kyk ≤ r0, we get

(a + br )κ∗r kFyk ≤ kfk + 0 0 (ϕ(1) − ϕ(0))β Γ(β + 1) ≤ r0.

Therefore, F maps Br0 into itself. + Next, we consider the operator F on the set Br0 = {y ∈ Br0 : y(t) ≥ 0, ∀t ∈ J}. It + is clear that Br0 6= ∅ is closed, convex and bounded. By these facts and our assumptions, + we obtain F maps Br0 into itself. + In what follows, we will show that F is continuous on Br0 . For this, let (yn) be a + sequence in Br0 such that yn → y and we will show that Fyn → Fy. We have, for t ∈ J,

|(Fyn)(t) − (Fy)(t)| Z t 0 (T yn)(t) ϕ (s)κ(t, s) max[0,σ(s)] |yn(τ)| = 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 (T yn)(t) ϕ (s)κ(t, s) max[0,σ(s)] |yn(τ)| ≤ 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |yn(τ)| − 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |yn(τ)| + 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) A Maximum Quadratic Integral Equation 99

Z t 0 (T y)(t) ϕ (s)κ(t, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 |(T yn)(t) − (T y)(t)| ϕ (s)|κ(t, s)| max[0,σ(s)] |yn(τ)| ≤ 1−β ds Γ(β) 0 (ϕ(t) − ϕ(s)) Z t 0 |(T y)(t)| ϕ (s)|κ(t, s)| + 1−β max |yn(τ)| − max |y(τ)| ds. Γ(β) 0 (ϕ(t) − ϕ(s)) [0,σ(s)] [0,σ(s)] By applying Lemma 1.5, we get

κ∗r (ϕ(1) − ϕ(0))βkT y − T yk kFy − Fyk ≤ 0 n n Γ(β + 1) κ∗(a + br )(ϕ(1) − ϕ(0))βky − yk + 0 n . (2.5) Γ(β + 1)

By the continuity of T , ∃n1 ∈ N such that εΓ(β + 1) kT yn − T yk ≤ ∗ β , ∀n ≥ n1. 2κ r0(ϕ(1) − ϕ(0))

Also, ∃n2 ∈ N such that εΓ(β + 1) kyn − yk ≤ ∗ β , ∀n ≥ n2. 2κ (a + br0)(ϕ(1) − ϕ(0))

Now, take n ≥ max{n1, n2}, then (2.5) gives us that

kFyn − Fyk ≤ ε.

+ This shows that F is continuous in Br0 . + Next, let Y ⊂ Br0 be a nonempty set. Let us choose y ∈ Y and t1, t2 ∈ J with |t2 − t1| ≤ ε for ﬁxed ε > 0. Since no generality will loss, we will assume that t2 ≥ t1. Then, by using our assumptions and (2.4), we obtain

|(Fy)(t2) − (Fy)(t1)|

≤ |f(t2) − f(t1)| + |(T y)(t2)(Ky)(t2) − (T y)(t2)(Ky)(t1)|

+ |(T y)(t2)(Ky)(t1) − (T y)(t1)(Ky)(t1)|

≤ ω(f, ε) + |(T y)(t2)| |(Ky)(t2) − (Ky)(t1)| + |(T y)(t2) − (T y)(t1)| |(Ky)(t1)| (a + bkyk)kyk ≤ ω(f, ε) + ω (ε, .)(ϕ(1) − ϕ(0))β + 2κ∗(ω(ϕ, ε))β Γ(β + 1) κ ω(T y, ε)kykκ∗(ϕ(t ) − ϕ(0))β + 1 Γ(β + 1) r (a + br ) ≤ ω(f, ε) + 0 0 ω (ε, .)(ϕ(1) − ϕ(0))β + 2κ∗(ω(ϕ, ε))β Γ(β + 1) κ 100 M. A. Darwish, J. Henderson and K. Sadarangani

κ∗r (ϕ(1) − ϕ(0))β + 0 ω(T y, ε). Γ(β + 1) Hence, r (a + br ) ω(Fy, ε) ≤ ω(f, ε) + 0 0 ω (ε, .)(ϕ(1) − ϕ(0))β + 2κ∗(ω(ϕ, ε))β Γ(β + 1) κ κ∗r (ϕ(1) − ϕ(0))β + 0 ω(T y, ε). Γ(β + 1)

Consequently, r (a + br ) ω(FY, ε) ≤ ω(f, ε) + 0 0 ω (ε, .)(ϕ(1) − ϕ(0))β + 2κ∗(ω(ϕ, ε))β Γ(β + 1) κ κ∗r (ϕ(1) − ϕ(0))β + 0 ω(T Y, ε). Γ(β + 1)

The uniform continuity of the function κ on J × J and the continuity of the functions f and ϕ on J, implies the last inequality becomes

κ∗r (ϕ(1) − ϕ(0))β ω (FY ) ≤ 0 ω (TY ). (2.6) 0 Γ(β + 1) 0

In the next step, ﬁx arbitrary y ∈ Y and t1, t2 ∈ J with t2 > t1. Then, by our assumptions, we have

|(Fy)(t2) − (Fy)(t1)| − [(Fy)(t2) − (Fy)(t1)] Z t2 0 (T y)(t2) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| = f(t2) + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 (T y)(t1) ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| −f(t1) − 1−β ds Γ(β) 0 (ϕ(t1) − ϕ(s)) Z t2 0 (T y)(t2) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| − f(t2) + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 (T y)(t1) ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| −f(t1) − 1−β ds Γ(β) 0 (ϕ(t1) − ϕ(s)) ≤ {|f(t2) − f(t1)| − [f(t2) − f(t1)]} Z t2 0 (T y)(t2) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 (T y)(t1) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 (T y)(t1) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) A Maximum Quadratic Integral Equation 101

Z t1 0 (T y)(t1) ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t1) − ϕ(s)) Z t2 0 (T y)(t2) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 (T y)(t1) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t2 0 (T y)(t1) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 (T x)(t1) ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds Γ(β) 0 (ϕ(t1) − ϕ(s)) {|(T y)(t ) − (T y)(t )| − [(T y)(t ) − (T y)(t )]} ≤ 2 1 2 1 Γ(β) Z t2 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| × 1−β ds 0 (ϕ(t2) − ϕ(s)) Z t2 0 (T y)(t1) ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| + 1−β ds Γ(β) 0 (ϕ(t2) − ϕ(s)) Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t1) − ϕ(s)) Z t2 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| − 1−β ds 0 (ϕ(t2) − ϕ(s)) Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| − 1−β ds . (2.7) 0 (ϕ(t1) − ϕ(s)) But Z t2 0 Z t1 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| 1−β ds − 1−β ds 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t1) − ϕ(s)) Z t2 0 Z t2 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| = 1−β ds − 1−β ds 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t2) − ϕ(s)) Z t2 0 Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds − 1−β ds 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t2) − ϕ(s)) Z t1 0 Z t1 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds − 1−β ds 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t1) − ϕ(s)) Z t2 0 ϕ (s)(κ(t2, s) − κ(t1, s)) max[0,σ(s)] |y(τ)| = 1−β ds 0 (ϕ(t2) − ϕ(s)) Z t2 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds t1 (ϕ(t2) − ϕ(s)) Z t1 β−1 β−1 + κ(t1, s)[(ϕ(t2) − ϕ(s)) − (ϕ(t1) − ϕ(s)) ] max |y(τ)|ds. 0 [0,σ(s)] 102 M. A. Darwish, J. Henderson and K. Sadarangani

Since κ(t2, s) ≥ κ(t1, s) (κ(t, s) is nondecreasing with respect to t), we have

Z t2 0 ϕ (s)(κ(t2, s) − κ(t1, s)) max[0,σ(s)] |y(τ)| 1−β ds ≥ 0 (2.8) 0 (ϕ(t2) − ϕ(s)) 1 1 and, since 1−β ≥ 1−β for s ∈ [0, t1) then (ϕ(t2) − ϕ(s)) (ϕ(t1) − ϕ(s))

Z t1 0 β−1 β−1 ϕ (s)κ(t1, s)[(ϕ(t2) − ϕ(s)) − (ϕ(t1) − ϕ(s)) ] max |y(τ)|ds 0 [0,σ(s)] Z t2 0 ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| + 1−β ds t1 (ϕ(t2) − ϕ(s)) Z t1 0 β−1 β−1 ≥ ϕ (s)κ(t1, t1)[(ϕ(t2) − ϕ(s)) − (ϕ(t1) − ϕ(s)) ] max |y(τ)|ds 0 [0,σ(t1)] Z t2 0 ϕ (s)κ(t1, t1) max[0,σ(t1)] |y(τ)| + 1−β ds t1 (ϕ(t2) − ϕ(s)) Z t2 ϕ0(s)ds Z t1 ϕ0(s)ds = κ(t1, t1) max |y(τ)| 1−β − 1−β [0,σ(t1)] 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t1) − ϕ(s)) β β (ϕ(t2) − ϕ(0)) − (ϕ(t1) − ϕ(0)) = κ(t1, t1) max |y(τ)| β [0,σ(t1)] ≥ 0. (2.9) Finally, (2.8) and (2.9) imply that

Z t2 0 Z t1 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| ϕ (s)κ(t1, s) max[0,σ(s)] |y(τ)| 1−β ds − 1−β ds ≥ 0. 0 (ϕ(t2) − ϕ(s)) 0 (ϕ(t1) − ϕ(s)) The above inequality and (2.7) leads us to

|(Fy)(t2) − (Fy)(t1)| − [(Fy)(t2) − (Fy)(t1)] |(T y)(t ) − (T y)(t )| − [(T y)(t ) − (T y)(t )] ≤ 2 1 2 1 Γ(β) Z t2 0 ϕ (s)κ(t2, s) max[0,σ(s)] |y(τ)| × 1−β ds 0 (ϕ(t2) − ϕ(s)) κ∗r (ϕ(1) − ϕ(0))β ≤ 0 d(T y). Γ(β + 1) Thus, κ∗r (ϕ(1) − ϕ(0))β d(Fy) ≤ 0 d(T y) Γ(β + 1) and therefore, κ∗r (ϕ(1) − ϕ(0))β d(FY ) ≤ 0 d(TY ). (2.10) Γ(β + 1) A Maximum Quadratic Integral Equation 103

Finally, (2.6) and (2.10) give us that κ∗r (ϕ(1) − ϕ(0))β ω (FY ) + d(FY ) ≤ 0 (ω (FY ) + d(TY )) 0 Γ(β + 1) 0 or r κ∗(ϕ(1) − ϕ(0))β µ(FY ) ≤ 0 µ(TY ) Γ(β + 1) κ∗cr (ϕ(1) − ϕ(0))β ≤ 0 µ(Y ). Γ(β + 1) κ∗r c Since 0 < (ϕ(1) − ϕ(0))−β, F is a contraction operator with respect to µ. Γ(β + 1) Finally, by Theorem 1.3, F has at least one ﬁxed point, or equivalently, (1.1) has at least one nondecreasing solution in Br0 . This ﬁnishes our proof. Next, we present the following numerical example in order to illustrate our results. Example 2.2. Let us consider the following integral equation with maximum

√ Z t 2 2 y(t) t + s max[0,ln(s+1)] |y(τ)| y(t) = arctan t + √ p√ √ ds, t ∈ J. (2.11) 5Γ(1/2) 0 2 s + 1 t + 1 − s + 1

Notice that (2.11)√ is a particular case√ of (1.1), where f(t) = arctan t, (T y)(t) = y(t)/5, β = 1/2, ϕ(s) = s + 1, κ(t, s) = t2 + s2 and σ(t) = ln(t + 1). It is not difficult to see that assumptions (a1), (a2), (a3)√, (a4) , (a5) and (a6) are verified with kfk = π/4, c = 1/5, a = 0, b = 1/5 and κ∗ = 2. Now, the inequality (2.1) in assumption (a7) takes the expression √ p√ π 2 2 − 1 + r2 ≤ r 4 5Γ(3/2) 0 0 which is satisfied by r0 = 1. Moreover, √ ∗ cκ r0 2 ∼ −β 1 ∼ = = 0.32 < (ϕ(1) − ϕ(0)) = p√ = 1.56. Γ(β + 1) 5Γ(3/2) 2 − 1 Therefore, by Theorem 2.1, (2.11) has at least one continuous and nondecreasing solution which is located in the ball B1.

Acknowledgements

The ﬁrst author’s permanent address is Department of Mathematics, Faculty of Science, Damanhour University, Damanhour, Egypt. 104 M. A. Darwish, J. Henderson and K. Sadarangani

References

[1] D. Bainov and D. Mishev, Oscillation Theory for Neutral Differenial Equations with Delay, Hilger, Bristol, 1991.

[2] J. Banas´ and K. Goebel, Measures of Noncompactness in Banach Spaces, Lecture Notes in Pure and Applied Mathematics, 60, Marcel Dekker, New York, 1980.

[3] J. Banas´ and L. Olszowy, Measures of noncompactness related to monotonicity, Comment. Math. 41 (2001), 13–23.

[4] J. Caballero, B. Lopez´ and K. Sadarangani, On monotonic solutions of an integral equation of Volterra type with supremum, J. Math. Anal. Appl. 305 (2005), 304– 315.

[5] J. Caballero, B. Lopez,´ K. Sadarangani, Existence of nondecreasing and continuous solutions for a nonlinear integral equation with supremum in the kernel, Z. Anal. Anwend. 26 (2007) 195–205.

[6] J. Caballero, M. A. Darwish and K. Sadarangani, Solvability of a fractional hy- brid initial value problem with supremum by using measures of noncompactness in Banach algebras, Appl. Math. Comput. 224 (2013), 553–563.

[7] M. A. Darwish, On monotonic solutions of a singular quadratic integral equation with supremum, Dynam. Systems Appl. 17 (2008), 539–549.

[8] M. A. Darwish and K. Sadarangani, Nondecreasing solutions of a quadratic Abel equation with supremum in the kernel, Appl. Math. Comput. 219 (2013), 7830– 7836.

[9] M. A. Darwish and K. Sadarangani, On a quadratic integral equation with supremum involving Erdelyi-Kober´ fractional order, Math. Nachr. 288 (2015), 566–576.

[10] J. Dugundji and A. Granas, Fixed Point Theory, Monograﬁe Mathematyczne, PWN, Warsaw, 1982.

[11] S. G. Hristova and D. D. Bainov, Monontone–iterative techniques for V. Lak- shmikantham for a boundary value problem for systems of impulsive differential equations with ”supremum”, J. Math. Anal. Appl. 172 (1993), 339–352.