EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6133 ISSN 1307-5543 – ejpam.com Published by New York Business Global A Fixed Point Result in a DCMS Setting and a Fredholm Integral Equation Hassen Aydi1,2,∗, Hamdi Hammouda3, Saber Mansour4 1 Institut Supérieur d’Informatique et des Techniques de Communication, Université de Sousse, H. Sousse 4000, Tunisia 2 Department of Mathematics and Applied Mathematics, Sefako Makgatho Health Sciences University, Ga-Rankuwa, South Africa. 3 Université de Monastir, Institut Préparatoire des Études d’Ingénieurs de Monastir, Monastir, Tunisia 4 Department of Mathematics, Umm Al-Qura University, Faculty of Sciences, P.O. Box 14035, Holy Makkah 21955, Saudi Arabia Abstract. In this present work, we prove a fixed point result for generalized contraction mappings in the setting of a double control metric space (DCMS) by using α-orbital admissibility. The uniqueness of the fixed point is established by adding further hypotheses. The presented result is supported by a concrete example. Moreover, we ensure the existence of a solution of a Fredholm type integral equation via a fixed point technique. Some consequences are also presented to make effective the obtained results. 2020 Mathematics Subject Classifications: 74H10, 34A08, 26A33, 34B15 Key Words and Phrases: Contraction, Fixed point, Double controlled metric space, α-admissible, Fredholm integral equation 1. Introduction The fixed point result due to Banach [1] is considered as the most essential theorem in fixed point theory. It asserts that a contraction mapping on a complete metric space admits a unique fixed point. In 1989, an interesting extension of the metric space was explored by Bakhtin [2] and Czerwik [3] by initiating the concept of b-metric spaces. Later, in 2017, this setting was generalized to extended b-metric spaces initiated by Kiran et al. [4], where the triangular inequality is extended via a controlled function. On the other hand, using two control functions ϖ, ϵ : ℧ × ℧ −→ [1,∞), the notion of a double controlled metric space [5] (DCMS) was considred by Abdeljawad et al. [5]. Many related ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6133 Email addresses: hassen.aydi@isima.rnu.tn (H. Aydi), 7amdi7ammouda@gmail.com (H. Hammouda), samansour@uqu.edu.sa (S. Mansour) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 2 of 18 works in this direction appeared, see as examples [6–9]. These generalizations led to study several fixed point results for contractions arising in many real applications for several problems in nonlinear analysis. Integral equations appear naturally in several branches of and engineering and science. Particularly, Fredholm integral equations appeared widely in various scientific areas, like computational mathematics, physics, continuum mechanics, medicine, acoustics and approximation theory. There are several analytical and numerical methods to solve Fredholm integral equations. In 2018, Karapinar et al. [10] solved a Fredholm integral equation given as follows: ℏ(ς) = ∫ b a ℵ(ς, υ, ℏ(υ))dυ + ζ(ς), (1) by using a fixed point method in the context of extended b-metric spaces. Several works in literature dealing with the existence of a solution of a Fredholm type integhral equations arise. For more details, see the papers [11–14]. In this work, by generalizing the equation (1), the following Fredholm functional integral equation is studied: ℏ(t) = ∫ b a K(t, r, ℏ(r), ℏ(g(r)), ℏ(a), ℏ(b))dr + f(t). (2) Under appropriate conditions on functions K, g and f , we aim to resolve the equation (2) via a fixed point technique. Namely, we give some fixed point results in a DCMS via orbital α-admissibility. We also present some illustrated concrete examples. At the end, we solve a Fredhom type integral equation in order to show that our required conditions are applicable. 2. Preliminaries Definition 1. [5] Consider a nonempty set ℧. A function ζ : ℧×℧ −→ [0,∞) is termed as a DCM with controlled functions ϖ, ϵ : ℧× ℧ −→ [1,∞) if for all η, y, ג ∈ ℧, we have (µ1): ζ (η, y) = 0 ⇐⇒ η = y; (µ2): ζ(η, y) = ζ(y, η); (µ3): ζ(η, y) ≤ ϖ(η, ,ζ(η(ג (ג + ϵ(ג, y)ζ(ג, y). Here, (℧, ζ) is termed as a DCMS. Example 1. Let ℧ = [0,∞]. Given ζ : ℧× ℧ −→ [0,∞) as ζ(η, ι) =  0 if η = ι η η+1 if η ̸= 0 and ι = 0 ι ι+1 if ι ̸= 0 and η = 0 η + ι if 0 ̸= η ̸= ι ̸= 0. Take ϖ, ϵ : ℧× ℧ −→ [1,∞) as ϖ(η, ι) = ϵ(η, ι) = 2η + 2ι+ 2. Here, (℧, ζ) is a DCMS. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 3 of 18 Definition 2. [5] Let {s̃n} be a sequence in a DCMS (℧, ζ). Then, (i) {s̃ȷ} is called a convergent sequence, if, for any ϵ > 0, there is an integer ȷ0 = ȷ0(ϵ) so that ζ(s̃ȷ, η) < ϵ, for all ȷ ≥ ȷ0. One writes lim n→∞ s̃ȷ = η; (ii) {s̃ȷ} is named Cauchy if lim ȷ,m→∞ ζ(s̃ȷ, s̃m) = 0; (iii) (℧, ζ) is called complete if each Cauchy sequence converges in ℧. Definition 3. Let ⊤ be a self-mapping on a DCMS (℧, ζ). For s̃0 ∈ ℧, the set O(s̃0,⊤) = { s̃0,⊤s̃0,⊤2s̃0,⊤3s̃0, · · · } is named as an orbit of ⊤ at s̃0. ⊤ is said to be orbitally continuous at ℘ ∈ ℧ if lim k→∞ ⊤ks̃0 = ℘ yields that lim k→∞ ⊤⊤ks̃0 = ⊤℘. Also, when each Cauchy sequence hav- ing the form {⊤ks̃0}k converges in ℧, then (℧, ζ) is orbitally complete. In the sequel, Fix(⊤) = {℘ ∈ ℧ /⊤℘ = ℘}. Remark 1. The continuity yields the orbital continuity. Definition 4. [15] Let α : ℧×℧ −→ R be a function (℧ is a nonempty set). ⊤ : ℧ −→ ℧ is termed an α-admissible mapping, if for every ı, ℓ ∈ ℧, α(ı, ℓ) ≥ 1 implies α(⊤ı,⊤ℓ) ≥ 1. ⊤ is called α-orbitally admissible if, for each ı ∈ ℧, α(ı,⊤ı) ≥ 1 leads to α(⊤ı,⊤2ı) ≥ 1. Definition 5. ψ : [0,+∞) −→ [0,+∞) is termed as a comparison function, if it is nondecreasing and lim ȷ→∞ ψȷ(τ) = 0 for each τ > 0. Here, ψȷ is the nth iteration of ψ. Ψ is denoted the set of all comparison functions. FP is the abbreviation of a fixed point. Lemma 1. For each ψ ∈ Ψ, we have ψ(0) = 0 and ψ(Θ) < Θ for any Θ > 0. The next result is needful in the sequel. Proposition 1. Let (℧, ζ) be a DCMS with two controlled functions ϖ, ϵ. Let {s̃n} be a convergent sequence in ℧ so that lim n→∞ ϵ(γ, s̃n) and lim n→∞ ϖ(s̃n, β) exist and are finite for any γ, β ∈ ℧, then such a convergent sequence possesses a unique limit. Proof. Suppose {s̃ℓ} ∈ ℧ converges to σ and ς in ℧. We have lim ℓ→∞ ζ(s̃ℓ, σ) = 0 and lim ℓ→∞ ζ(s̃ℓ, ς) = 0. By the triangle inequality in the DCMS, we obtain ζ(σ, ς) ≤ ϖ(σ, s̃ℓ)ζ(σ, s̃ℓ) + ϵ(s̃ℓ, ς)ζ(s̃ℓ, ς). Taking ℓ −→ 0, we get ζ(σ, ς) ≤ 0. Thus, ζ(σ, ς) = 0, which further implies that σ = ς. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 4 of 18 3. Main results Our first theorem is stated as follows: Theorem 1. Let ⊤ be a self-mapping on a orbitally complete DCMS (℧, ζ) with controlled functions ϖ, ϵ : ℧×℧ −→ [1,+∞[. Suppose there are two functions α : ℧×℧ −→ [0,+∞), and ψ ∈ Ψ so that α(x, y)ϖ(x, y)ϵ(x, y)ζ(⊤x,⊤y) ≤ ψ(R(x, y)) ∀x, y ∈ ℧, (3) where R(x, y) = max { ζ(x, y), ζ(x,⊤x), ζ(y,⊤y), ζ(x,⊤x)[ϖ(x, y)ϵ(x, y) + ζ(y,⊤y)] ϖ(x, y)ϵ(x, y) + ζ(x, y) , ζ(y,⊤y)[ϖ(x, y)ϵ(x, y) + ζ(x,⊤x)] ϖ(x, y)ϵ(x, y) + ζ(x, y) } . Assume that: (i) ⊤ is α-orbitally admissible; (ii) there is s̃0 ∈ ℧ satisfying α(s̃0,⊤s̃0) ≥ 1; (iii) supm≥1 lim i→∞ ϵ(s̃i+1, s̃m)ϖ(s̃i+1, s̃i+2)ψ i+1(ζ(s̃0, s̃1)) ϖ(s̃i, s̃i+1)ψi(ζ(s̃0, s̃1)) < 1, where s̃i = ⊤i(s̃0); (iv) ⊤ is orbitally continuous on ℧; (v) lim n→∞ ϵ(s̃n, x) and lim n→∞ ϖ(s̃n, x) exist and are finite. Therefore, ⊤ possesses a FP in ℧. If in addition, we have (U) : x, x∗ ∈ Fix(⊤) implies α(x, x∗) ≥ 1, then such a FP is unique. Proof. By (ii), define a sequence {s̃ℓ} in ℧ such that s̃ℓ+1 = ⊤s̃ℓ = ⊤n+1s̃0, for all ℓ ∈ N. When s̃ℓ = s̃ℓ+1 for some ℓ ∈ N, then s̃ℓ is a FP of ⊤. Now, suppose that s̃ℓ ̸= s̃ℓ+1, for any ℓ ∈ N. Due to (i), α(s̃0, s̃1) = α(s̃0,⊤s̃0) ≥ 1 implies that α(s̃1, s̃2) = α(⊤s̃0,⊤s̃1) ≥ 1. Then, α(s̃2, s̃3) = α(⊤s̃1,⊤s̃2) ≥ 1. Continuing this process, one gets α(s̃ℓ, s̃ℓ+1) ≥ 1, for any ℓ ∈ N. Letting x = s̃ℓ−1 and y = s̃ℓ in (3), we have ζ(s̃ℓ, s̃ℓ+1) = ζ(⊤s̃ℓ−1,⊤s̃ℓ) ≤ α(s̃ℓ−1, s̃ℓ)ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ)ζ(⊤s̃ℓ−1,⊤s̃ℓ) ≤ ψ(R(s̃ℓ−1, s̃ℓ)), (4) H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 5 of 18 where R(s̃ℓ−1, s̃ℓ) = max{ζ(s̃ℓ−1, s̃ℓ), ζ(s̃ℓ−1,⊤s̃ℓ−1), ζ(s̃ℓ,⊤s̃ℓ); ζ(s̃ℓ−1,⊤s̃ℓ−1)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ,⊤s̃ℓ)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) , ζ(s̃ℓ,⊤s̃ℓ)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,⊤s̃ℓ−1)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) } = max{ζ(s̃ℓ−1, s̃ℓ), ζ(s̃ℓ−1, s̃ℓ), ζ(s̃ℓ, s̃ℓ+1); ζ(s̃ℓ−1,s̃ℓ)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ,s̃ℓ+1)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) ; ζ(s̃ℓ,s̃ℓ+1)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) } = max{ζ(s̃ℓ−1, s̃ℓ), ζ(s̃ℓ, s̃ℓ+1), ζ(s̃ℓ−1,s̃ℓ)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ,s̃ℓ+1)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) }. (5) We should take the following: case 1 If R(s̃ℓ−1, s̃ℓ) = ζ(s̃ℓ, s̃ℓ+1), so using (4), one writes 0 < ζ(s̃ℓ, s̃ℓ+1) ≤ ψ(ζ(s̃ℓ, s̃ℓ+1)) < ζ(s̃ℓ, s̃ℓ+1). It is a contradiction. case 2 If R(s̃ℓ−1, s̃ℓ) = ζ(s̃ℓ−1,s̃ℓ)[ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ,s̃ℓ+1)] ϖ(s̃ℓ−1,s̃ℓ)ϵ(s̃ℓ−1,s̃ℓ)+ζ(s̃ℓ−1,s̃ℓ) , then by (5), we have max {ζ(s̃ℓ−1, s̃ℓ); ζ(s̃ℓ, s̃ℓ+1)} ≤ ζ(s̃ℓ−1, s̃ℓ)[ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ, s̃ℓ+1)] ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ−1, s̃ℓ) . We study two subcases: subcase 1 : If max {ζ(s̃ℓ−1, s̃ℓ); ζ(s̃ℓ, s̃ℓ+1)} = ζ(s̃ℓ−1, s̃ℓ), then ζ(s̃ℓ, s̃ℓ+1) < ζ(s̃ℓ−1, s̃ℓ) and ζ(s̃ℓ−1, s̃ℓ) ≤ ζ(s̃ℓ−1, s̃ℓ)[ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ, s̃ℓ+1)] ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ−1, s̃ℓ) . that is, ζ(s̃ℓ−1, s̃ℓ) ≤ ζ(s̃ℓ, s̃ℓ+1), which is a contradiction. subcase 2 : If max {ζ(s̃ℓ−1, s̃ℓ), ζ(s̃ℓ, s̃ℓ+1)} = ζ(s̃ℓ, s̃ℓ+1), then ζ(s̃ℓ, s̃ℓ+1) > ζ(s̃ℓ−1, s̃ℓ) and H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 6 of 18 ζ(s̃ℓ, s̃ℓ+1) ≤ ζ(s̃ℓ−1, s̃ℓ)[ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ, s̃ℓ+1)] ϖ(s̃ℓ−1, s̃ℓ)ϵ(s̃ℓ−1, s̃ℓ) + ζ(s̃ℓ−1, s̃ℓ) . That is, ζ(s̃ℓ−1, s̃ℓ) ≥ ζ(s̃ℓ, s̃ℓ+1), which is a contradiction. Thus, R(s̃ℓ−1, s̃ℓ) = ζ(s̃ℓ−1, s̃ℓ) and by (4), we get ζ(s̃ℓ, s̃ℓ+1) ≤ ψ(ζ(s̃ℓ−1, s̃ℓ)) ≤ ... ≤ ψℓ(ζ(s̃0, s̃1)). When ℓ −→ ∞, we find lim ℓ→∞ ζ(s̃ℓ, s̃ℓ+1) = 0. Now, we show that {s̃ℓ} is a Cauchy sequence in ℧. For m, ℓ ∈ N with m > ℓ, we have ζ(s̃ℓ, s̃m) ≤ ϖ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1) + ϵ(s̃ℓ+1, s̃m)ζ(s̃ℓ+1, s̃m) ≤ ϖ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1) + ϵ(s̃ℓ+1, s̃m)[ϖ(s̃ℓ+1, s̃ℓ+2)ζ(s̃ℓ+1, s̃ℓ+2)+ ϵ(s̃ℓ+2, s̃m)ζ(s̃ℓ+2, s̃m)] = ϖ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1) + ϵ(s̃ℓ+1, s̃m)ϖ(s̃ℓ+1, s̃ℓ+2)ζ(s̃ℓ+1, s̃ℓ+2) + ϵ(s̃ℓ+1, s̃m)ϵ(s̃ℓ+2, s̃m)ζ(s̃ℓ+2, s̃m) . . . ≤ ϖ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1) + m−2∑ i=ℓ+1  i∏ j=ℓ+1 ϵ(s̃j , s̃m) ϖ(s̃i, s̃i+1)ζ(s̃i, s̃i+1) +  m−1∏ j=ℓ+1 ϵ(s̃j , s̃m)  ζ(s̃m−1, s̃m). By using the fact that ϖ(σ, ς) ≥ 1 and ϵ(σ, ς) ≥ 1 for all σ, ς ∈ X, we deduce ζ(s̃ℓ, s̃m) ≤ ϖ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1) + m−1∑ i=n+1  i∏ j=n+1 ϵ(s̃j , s̃m) ϖ(s̃i, s̃i+1)ζ(s̃i, s̃i+1) ≤ m−1∑ i=n  i∏ j=n+1 ϵ(s̃j , s̃m) ϖ(s̃i, s̃i+1)ζ(s̃i, s̃i+1) ≤ m−1∑ i=n  i∏ j=n+1 ϵ(s̃j , s̃m) ϖ(s̃i, s̃i+1)ψ i(ζ(s̃0, s̃1)). H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 7 of 18 Choose ai =  i∏ j=n+1 ϵ(s̃j , s̃m) ϖ(s̃i, s̃i+1)ψ i(ζ(s̃0, s̃1)), and Ωp = ∑p i=1 ai. Then we have ζ(s̃ℓ, s̃m) ≤ Ωm−1 − Ωn−1. (6) Since lim i→∞ ai+1 ai ≤ sup m≥1 lim i→∞ ϵ(s̃i+1, s̃m)ϖ(s̃i+1, s̃i+2)ψ i+1(ζ(s̃0, s̃1)) ϖ(s̃i, s̃i+1)ψi(ζ(s̃0, s̃1)) , by condition (iii) one gets lim i→∞ ai+1 ai < 1, so we conclude that the real sequence {Ωp} converges, then it is a Cauchy sequence in R. By (6), {s̃ℓ} is a Cauchy sequence in ℧. Since (℧, ζ) is orbitally complete, {s̃ℓ} converges to x∗ ∈ ℧. Next, we show that ⊤x∗ = x∗. The orbital continuity of ⊤ on ℧ leads to s̃ℓ+1 = ⊤s̃ℓ = ⊤(⊤ℓs̃0) −→ ⊤x∗ as ℓ→ +∞ and ζ(s̃ℓ+1,⊤x∗) −→ 0 as ℓ→ +∞. The triangle inequality of DCMS implies ζ(x∗, s̃ℓ+1) ≤ ϖ(x∗, s̃ℓ)ζ(x ∗, s̃ℓ) + ϵ(s̃ℓ, s̃ℓ+1)ζ(s̃ℓ, s̃ℓ+1). Using the condition (v) and the fact that lim ℓ→∞ ζ(s̃ℓ, s̃ℓ+1) = 0, one writes ζ(x∗, s̃ℓ+1) −→ℓ→∞ 0. Since we have a unique limit of a convergent sequence in the DCMS, one has x∗ = ⊤x∗. Thus, ⊤ admits a FP x∗ ∈ ℧, i.e., Fix(⊤) is nonempty. Now, we show its uniqueness. Let ϑ and θ be two distinct FPs of ⊤. Due to condition (U), we have α(ϑ, θ) = α(⊤ϑ,⊤θ) ≥ 1. Taking x = ϑ and y = θ in (3), we obtain ζ(ϑ, θ) = ζ(⊤ϑ,⊤θ) ≤ α(ϑ, θ)ϖ(ϑ, θ)ϵ(ϑ, θ)ζ(⊤ϑ,⊤θ) ≤ ψ(ℜ(ϑ, θ)) = max{ζ(ϑ, θ); ζ(ϑ,⊤ϑ); ζ(θ,⊤θ); ζ(ϑ,⊤ϑ)[ϖ(ϑ,θ)ϵ(ϑ,θ)+ζ(θ,⊤θ)] ϖ(ϑ,θ)ϵ(ϑ,θ)+ζ(ϑ,θ) ; ζ(θ,⊤θ)[ϖ(ϑ,θ)ϵ(ϑ,θ)+ζ(ϑ,⊤ϑ)] ϖ(ϑ,θ)ϵ(ϑ,θ)+ζ(ϑ,θ) } = ψ(ζ(ϑ, θ)) < ζ(ϑ, θ), which is a contradiction. Therefore, ⊤ possesses a unique FP in ℧. Example 2. Given the DCMS as in Example 1. Consider ⊤ : ℧ −→ ℧ as ⊤x = { x β+βx if x ∈ [0, 1] 0 otherwise, where β > 108. To prove that the mapping ⊤ is continuous, let {s̃ℓ} be a convergent sequence in ℧ to x. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 8 of 18 Then ζ(s̃ℓ, x) −→ 0 as ℓ→ ∞. We require the following cases: Case 1: s̃ℓ = x for all but finitely many ℓ, then ⊤s̃ℓ = ⊤x. Here, ζ(s̃ℓ, x) = 0 and ζ(⊤s̃ℓ,⊤x) = 0 for all but finitely many. So ζ(⊤s̃ℓ,⊤x) −→ 0 as ℓ→ ∞. Case 2: s̃ℓ ̸= 0 and x ̸= 0 for all but finitely many ℓ, then ⊤s̃ℓ ̸= 0 and ⊤x ̸= 0. We have ζ(s̃ℓ, x) = s̃ℓ + x→ 0 and ζ(⊤s̃ℓ,⊤x) = ⊤s̃ℓ +⊤x = s̃ℓ β + βs̃ℓ + x β + βx = 1 β s̃ℓ + x+ 2s̃ℓx (x+ 1)(s̃ℓ + 1) ≤ (s̃ℓ + x) + (s̃ℓ + x)2 β(x+ 1)(s̃ℓ + 1) −→ 0. Thus, {⊤s̃ℓ} converges to ⊤x in (℧, ζ). Case 3: If s̃ℓ ̸= 0 for all but finitely many ℓ and x = 0, we have ζ(s̃ℓ, 0) = s̃ℓ 1 + s̃ℓ −→ 0 =⇒ s̃ℓ −→ 0, then ζ(⊤s̃ℓ,⊤0) = ⊤s̃ℓ 1 +⊤s̃ℓ = s̃ℓ β + (β + 1)s̃ℓ −→ 0. Thus, {⊤s̃ℓ} converges to ⊤x in (℧, ζ). Case 4: If s̃ℓ = 0 for all but finitely many ℓ, then, when ζ(s̃ℓ, x) −→ 0 we necessarily have x = 0, Consequently ζ(⊤s̃ℓ,⊤x) −→ 0. In all the cases, if ζ(s̃ℓ, x) −→ 0 =⇒ ζ(⊤s̃ℓ,⊤x) −→ 0, so ⊤ is continuous. We have ⊤ℓx = x βℓ + (∑ℓ k=1 β k ) x . It is obvious that s̃ℓ = ⊤ℓx −→ 0 as ℓ −→ ∞ and so for each x ∈ X, lim ℓ→∞ ϖ(s̃ℓ, x) = lim ℓ→∞ ϵ(s̃ℓ, x) = 2 + 2x <∞. In addition, we define a mapping ⊤ : ℧× ℧ −→ [0,+∞[ as α(x, y) = { 1 if x, y ∈ [0, 1] 0 otherwise. Let s̃0 ∈ ℧ be a point with α(s̃0,⊤s̃0) ≥ 1, then s̃0 ∈ [0, 1] and α(⊤s̃0,⊤2s̃0) = α( s̃0 β+βs̃0 , s̃0 β2+(β+β2)s̃0 ) ≥ 1. Therefore, ⊤ is α-orbitally admissible. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 9 of 18 Let ψ(χ) = kχ, for χ > 0, where k = 36 β . Here, ψℓ(χ) = kℓχ. For all x, y ∈ ℧, we will show that α(x, y)ϖ(x, y)ϵ(x, y)ζ(⊤x,⊤y) ≤ ψ(ζ(x, y). For this, we consider the following cases: Case 1: x = y. we have 0 = α(x, y)ϖ(x, y)ϵ(x, y)ζ(⊤x,⊤y) ≤ ψ(ζ(x, y). Case 2: (x ̸= 0 and y = 0) or (y ̸= 0 and x = 0). Without generality, suppose that x ̸= 0 and y = 0. Here, we have α(x, 0)ϖ(x, 0)ϵ(x, 0)ζ(⊤x,⊤0) = (2 + 2x)2ζ( x β + βx , 0) = (2 + 2x)2 x β+βx x β+βx + 1 = (2 + 2x)2 x β + (β + 1)x ≤ 1 β (2 + 2x)2 x x+ 1 ≤ 16 β x x+ 1 ≤ k.ζ(x, y) = ψ(ζ(x, y)). Case 3: 0 ̸= x ̸= y ̸= 0. Here, we have 0 ̸= x β+βx ̸= y β+βy ̸= 0. One writes α(x, y)ϖ(x, y)ϵ(x, y)ζ(⊤x,⊤y) = (2 + 2x+ 2y)2ζ( x β + βx , y β + βy ) = (2 + 2x+ 2y)2 [ x β + βx + y β + βy ] ≤ 1 β (2x+ 2y + 2)2(x+ y) ≤ 36 β [x+ y] = k.ζ(x, y) = ψ(ζ(x, y)). Moreover, there is s̃0 ∈ ℧ with α(s̃0,⊤s̃0) ≥ 1, then α(⊤s̃0,⊤2s̃0) ≥ 1. By induction, we obtain α(s̃ℓ, s̃ℓ+1) ≥ 1, where s̃ℓ = ⊤ℓs̃0 = s̃0 βℓ+( ∑ℓ k=1 β k)s̃0 = s̃0 βℓ+β βℓ−1 β−1 s̃0 , for each ℓ ∈ N. It is easy that s̃ℓ −→ 0 as ℓ −→ +∞. Thus, (℧, ζ) is an orbitally complete DCMS. Recall that lim i,m→+∞ ϖ(s̃i, s̃m) = lim i,m→+∞ ϵ(s̃i, s̃m) = 2 then we have sup m≥1 lim i→∞ ϵ(s̃i+1, s̃m)ϖ(s̃i+1, s̃i+2)ψ i+1(ζ(s̃0, s̃1)) ϖ(s̃i, s̃i+1)ψi(ζ(s̃0, s̃1)) = 2(2 + s̃0)k i+1ζ(s̃0, s̃1) 2.kiζ(s̃0, s̃1) ≤ 3k < 1, where s̃i = ⊤i(s̃0) and ψi(ζ(s̃0, s̃1)) = kiζ(s̃0, s̃1). H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 10 of 18 Proposition 2. If we replace the condition (U) by (W) : ∀µ, ν ∈ Fix(⊤), there is δ ∈ ℧ so that α(µ, δ) ≥ 1 and α(ν, δ) ≥ 1, then the map ⊤ has a unique FP in ℧. Proof. Assume there are two distinct FPs, say x ̸= y ∈ ℧. According to (W ), there is z ∈ ℧, so that α(x, z) ≥ 1, α(y, z) ≥ 1. The α-admissibility of ⊤ yields that α(x,⊤ℓz) ≥ 1; ;α(y,⊤ℓz) ≥ 1; ∀ℓ ∈ N. Consider the sequence {zℓ} ∈ ℧ defined as zℓ = ⊤ℓz. We have ζ(x, zℓ+1) = ζ(⊤x,⊤zℓ) ≤ α(x, zℓ)ζ(⊤x,⊤zℓ) ≤ ψ(R(x, zℓ)), where R(x, zℓ) = max{ζ(x, zℓ), ζ(x,⊤x), ζ(zℓ,⊤zℓ), ζ(x,⊤x)[ϖ(x,zℓ)ϵ(x,zℓ)+ζ(zℓ,⊤zℓ)] ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,zℓ) , ζ(zℓ,⊤zℓ)[ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,⊤x)] ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,zℓ) } = max { ζ(x, zℓ), ζ(zℓ, zℓ+1), ζ(zℓ,zℓ+1)ϖ(x,zℓ)ϵ(x,zℓ) ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,zℓ ) } . If R(x, zℓ) = ζ(zℓ,zℓ+1)ϖ(x,zℓ)ϵ(x,zℓ) ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,zℓ) for some ℓ, then ζ(x, zℓ+1) = ζ(⊤x,⊤zℓ) ≤ α(x, zℓ)ζ(⊤x,⊤zℓ) ≤ ψ(R(x, zℓ)) < R(x, zℓ) = ζ(zℓ,zℓ+1)ϖ(x,zℓ)ϵ(x,zℓ) ϖ(x,zℓ)ϵ(x,zℓ)+ζ(x,zℓ) . Therefore, ζ(zℓ, zℓ+1) < ζ(zℓ, zℓ+1)ϖ(x, zℓ)ϵ(x, zℓ) ϖ(x, zℓ)ϵ(x, zℓ) + ζ(x, zℓ) . That is, ζ(zℓ, zℓ+1)ϖ(x, zℓ)ϵ(x, zℓ) + ζ(zℓ, zℓ+1)ζ(x, zℓ) < ζ(zℓ, zℓ+1)ϖ(x, zℓ)ϵ(x, zℓ). Thus, ζ(zℓ, zℓ+1)ζ(x, zℓ) < 0, witch is a contradiction. We have R(x, zℓ) = max{ζ(x, zℓ), ζ(zℓ, zℓ+1)}. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 11 of 18 That is, ζ(zℓ, zℓ+1) ≤ ψℓ(ζ(z,⊤z)). Then lim ℓ→+∞ ζ(zℓ, zℓ+1) = 0. So for all δ > 0, there is ℓ0 ∈ N such that for all ℓ ≥ ℓ0, ζ(zℓ, zℓ+1) < δ. Now, we show that lim ℓ→+∞ ζ(x, zℓ) = 0. Suppose on the contrary that lim ℓ→+∞ ζ(x, zℓ) ̸= 0, then there exists a subsequence { zφ(ℓ) } of {zℓ} such that φ(ℓ) > n; ζ(x, zφ(ℓ)) ≥ δ. So for all ℓ ≥ ℓ0 we have φ(ℓ) > ℓ ≥ ℓ0 and then{ ζ(zφ(ℓ), zφ(ℓ)+1) < δ ζ(x, zφ(ℓ)) ≥ δ. Hence, for all ℓ ≥ ℓ0, R(x, zφ(ℓ)) = ζ(x, zφ(ℓ)). Also, we have for all ℓ ≥ ℓ0, ζ(x, zφ(ℓ)+1) = ζ(⊤x,⊤zφ(ℓ)) ≤ α(x, zφ(ℓ))ζ(⊤x,⊤zφ(ℓ)) ≤ ψ(ζ(x, zφ(ℓ))) . . . ≤ ψφ(ℓ)−ℓ0(ζ(x, zℓ0)). Letting ℓ −→ +∞ we deduce that lim ℓ→+∞ ζ(x, zφ(ℓ)+1) = 0. So for all ℓ ≥ ℓ0 and by triangle inequality, we obtain δ ≤ ζ(x, zφ(ℓ)) ≤ ϖ(x, zφ(ℓ)+1)ζ(x, zφ(ℓ)+1) + ϵ(zφ(ℓ)+1, zφ(ℓ))ζ(zφ(ℓ)+1, zφ(ℓ)). When ℓ −→ +∞, one finds δ ≤ lim ℓ→+∞ ζ(x, zφ(ℓ)) = 0. It is a contradiction. Therefore, lim ℓ→+∞ ζ(x, zℓ) = 0. Similarly, lim ℓ→+∞ ζ(y, zℓ) = 0. By uniqueness of the limit, we have x = y. 4. Applications Using Theorem 1, we will establish in this section an existence result of a solution of the following nonlinear Fredholm type functional integral equation x(t) = f(t) + λ ∫ τ2 τ1 K (t, r, x(r), x(g(r)), x(τ1), x(τ2)) dr, (7) H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 12 of 18 where τ1, τ2 ∈ R with τ1 < τ2, K ∈ C([τ1, τ2] × [τ1, τ2] × R4), g ∈ C([τ1, τ2] × [τ1, τ2]) and f ∈ C([τ1, τ2]) are given functions and x ∈ C[τ1, τ2] is an unknown function. Take ℧ = C([τ1, τ2]). Given ζ : ℧× ℧ −→ [0,+∞) as ζ(µ, ν) = sup t∈[τ1,τ2] |µ(t)− ν(t)|p , ∀µ, ν ∈ ℧. Then (℧, ζ) is a complete DCMS with the controlled functions ϖ(µ, ν) = 2p−1, ϵ(µ, ν) = 2p−1 + 1 1 + 1 1+∥µ∥∞ + 1 1 + 1 1+∥ν∥∞ , where p > 1 where, ∥u∥∞ = sup t∈[τ1,τ2] |u(t)| . Define the operator ⊤ : ℧ −→ ℧ by ⊤(x)(t) := f(t) + λ ∫ τ2 τ1 K (t, r, x(r), x(g(r)), x(τ1), x(τ2)) dr; ∀x ∈ ℧, t ∈ [τ1, τ2]. In what follows, we will establish the conditions so that the operator ⊤ has at least one FP. For this, define α : ℧× ℧ −→ [0,+∞) by α(µ, ν) = { 1 if µ(t) ≤ ν(t) 0 otherwise. Theorem 2. Assume the following conditions hold: (i) s̃0(t) ≤ f(t) + λ ∫ τ2 τ1 K (t, r, s̃0(r), s̃0(g(r)), s̃0(τ1), s̃0(τ2)) dr, ∀t ∈ [τ1, τ2]; (ii) For any x, y ∈ ℧ with x(r) ≤ y(r) for each r ∈ [τ1, τ2], |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))| ≤ γ(t,r) 2 1− 1 p ( 2p−1+ 1 1+ 1 1+∥x∥∞ + 1 1+ 1 1+∥y∥∞ ) 1 p [|x(r)− y(r)|p + |x(g(r))− y(g(r))|p + |x(τ1)− y(τ1)|p + |x(τ2)− y(τ2)|p] 1 p ; p > 1 (8) where t, r ∈ [τ1, τ2] and γ : [τ1, τ2]× [τ1, τ2] −→ R is continuous so that sup t∈[τ1,τ2] ∫ τ2 τ1 γp(t, r)dr < 1 2p+2 |λ|p (τ2 − 1)p−1 ; (9) (iii) For any s̃0 ∈ ℧, lim i,m→∞ ϵ(s̃i+1, s̃m) < 2p; (iv) K is non-decreasing. Then (7) has a unique solution in ℧. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 13 of 18 Proof. For s̃0 ∈ ℧, choose {s̃ℓ} in ℧ by s̃ℓ = ⊤ℓs̃0, ℓ ≥ 1. Consider s̃ℓ+1(t) = ⊤s̃ℓ(t) = f(t) + λ ∫ τ2 τ1 K (t, r, s̃ℓ(r), s̃ℓ(g(r)), s̃ℓ(τ1), s̃ℓ(τ2)) dr. The function K is non-decreasing in the last four arguments, and so α(ν, ς) ≥ 1 =⇒ α(⊤ν,⊤ς) ≥ 1. Hence ⊤ is α-admissible, and therefore ⊤ is α-orbitally admissible. Next, by condition (i), α(s̃0,⊤s̃0) ≥ 1. Consequently, it follows that α(s̃ℓ, s̃ℓ+1) ≥ 1 for each ℓ ∈ N. Let q > 1 so that 1 p + 1 q = 1. Using (8) and the Holder’s inequality, one has |⊤x(t)−⊤y(t)|p = ∣∣∣λ ∫ τ2 τ1 K (t, r, x(r), x(g(r)), x(τ1), x(τ2)) dr − λ ∫ τ2 τ1 K (t, r, y(r), y(g(r)), y(τ1), y(τ2)) dr ∣∣∣p ≤ (∫ τ2 τ1 |λ| |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))| dr )p ≤ (∫ τ2 τ1 |λ|q ) p q((∫ τ2 τ1 |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))|p dr ) 1 p )p = |λ|p (τ2 − τ1) p−1(∫ τ2 τ1 |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))|p dr ) ≤ |λ|p (τ2 − τ1) p−1 ∫ τ2 τ1 γp(t,r) 2p−1(2p−1+ 1 1+ 1 1+∥x∥∞ + 1 1+ 1 1+∥y∥∞ ) [|x(r)− y(r)|p + |x(g(r))− y(g(r))|p + |x(τ1)− y(τ1)|p + |x(τ2)− y(τ2)|p] dr ≤ |λ|p (τ2 − τ1) p−1 ∫ τ2 τ1 γp(t,r) 2p−1(2p−1+ 1 1+ 1 1+∥x∥∞ + 1 1+ 1 1+∥y∥∞ ) [4ζ(x, y)] dr ≤ 4 |λ|p (τ2 − τ1) p−1 ∫ τ2 τ1 γp(t,r) 2p−1(2p−1+ 1 1+ 1 1+∥x∥∞ + 1 1+ 1 1+∥y∥∞ ) [R(x, y)] dr ≤ 4|λ|p(τ2−τ1)p−1[R(x,y)] 2p−1(2p−1+ 1 1+ 1 1+∥x∥∞ + 1 1+ 1 1+∥y∥∞ ) sup t∈[τ1,τ2] (∫ τ2 τ1 γp(t, r)dr ) . From (9), it results that 2p−1 ( 2p−1 + 1 1 + 1 1+∥x∥∞ + 1 1 + 1 1+∥y∥∞ ) |⊤x(t)−⊤y(t)|p ≤ 1 2p (ℜ(x, y)) . Setting ψ(t) = 1 2p t, we obtain that α(x, y)ϖ(x, y)ϵ(x, y)ζ(⊤x,⊤y) ≤ ψ(ℜ(x, y)). Next, using the condition lim i,m→∞ ϵ(s̃i+1, s̃m) < 2p, H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 14 of 18 we deduce supm lim i→∞ ϵ(s̃i+1, s̃m)ϖ(s̃i+1, s̃i+2)ψ i+1(ζ(s̃0, s̃1)) ϖ(s̃i, s̃i+1)ψi(ζ(s̃0, s̃1)) = supm lim i→∞ ϵ(s̃i+1, s̃m) 1 2p(i+1) (ζ(s̃0, s̃1)) 1 2pi (ζ(s̃0, s̃1)) = supm lim i→∞ 1 2p ϵ(s̃i+1, s̃m) < 1. Thus, the condition (iii) in Theorem 1 holds. Since f, g andK are continuous, the operator ⊤ is continuous on ℧ and so ⊤ is orbitally continuous on ℧. The controlled functions ϖ; ϵ are defined by ϖ(µ, ν) = 2p−1; ϵ(µ, ν) = 2p−1 + 1 1 + 1 1+∥µ∥∞ + 1 1 + 1 1+∥ν∥∞ , where p > 1. Then lim ℓ→∞ ϖ(s̃ℓ, x) = 2p−1. On the other hand, { ⊤ℓx(t) } n converges because { ⊤ℓ } ℓ is a monotonic sequence and so it follows from Dini theorem that sup t∈[τ1,τ2] ∣∣∣⊤ℓx(t) ∣∣∣ converges and hence lim ℓ→∞ ϵ(s̃ℓ, x) exists and is finite. In ℧ = C([τ1, τ2]), if x ≤ y, then there is z ∈ ℧ so that x ≤ z and y ≤ z, and so (W) is satisfied. Thus, all the conditions of Theorem 1 are satisfied, and hence ⊤ possesses a unique FP in ℧. That is, the nonlinear Fredholm functional integral equation (7) has a unique solution. Example 3. Take the following nonlinear Fredholm functional integral equation: For t ∈ [0, 1] and λ = 1, x(t) = 1+ ∫ 1 0 1 256 ( |x(r)| (1 + |x(r)|) 2 + |x(r)| + ∣∣x( r2)∣∣ (1 + ∣∣x( r2)∣∣) 2 + ∣∣x( r2)∣∣ + tr |x(0)| (1 + |x(0)|) 2 + |x(0)| + tr |x(1)| (1 + |x(1)|) 2 + |x(1)| ) dr, (10) where K : [0, 1]× [0, 1]× R4 −→ R is defined by K(t, r, x(r), x( r 2 ), x(0), x(1)) = 1 256 ( |x(r)| (1 + |x(r)|) 2 + |x(r)| + ∣∣x( r2)∣∣ (1 + ∣∣x( r2)∣∣) 2 + ∣∣x( r2)∣∣ + tr |x(0)| (1 + |x(0)|) 2 + |x(0)| + tr |x(1)| (1 + |x(1)|) 2 + |x(1)| ) . Let g : [0, 1] −→ [0, 1] be given as g(r) = r 2 and f : [0, 1] −→ R be defined by f(t) = 1. These functions are continuous and x ∈ C([0, 1]) is the unknown function. In this case, we will use the space ℧ = C([0, 1]) endowed with the double controlled metric ζ : ℧× ℧ −→ [0,+∞[ defined by ζ(µ, ν) = sup t∈[0,1] |µ(t)− ν(t)|2 , ∀µ, ν ∈ ℧ H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 15 of 18 The space (℧, ζ) is a DCMS with controlled functions ϖ(µ, ν) = 2; ϵ(µ, ν) = 2 + 1 1 + 1 1+∥µ∥∞ + 1 1 + 1 1+∥ν∥∞ . Take the operator ⊤ : ℧ −→ ℧ defined for t ∈ [0, 1] and x ∈ ℧ as follows: ⊤x(t) = 1+ ∫ 1 0 1 256 ( |x(r)| (1 + |x(r)|) 2 + |x(r)| + ∣∣x( r2)∣∣ (1 + ∣∣x( r2)∣∣) 2 + ∣∣x( r2)∣∣ + tr |x(0)| (1 + |x(0)|) 2 + |x(0)| + tr |x(1)| (1 + |x(1)|) 2 + |x(1)| ) dr. In what follows, the conditions of Theorem 1 are checked. It is observed that the function K is non-decreasing. Thereafter, we will quote the following lemma that we will use in the sequel. Lemma 2. For all τ1, τ2 ≥ 0, we have the a+ b ≤ 2 ( a2 + b2 ) 1 2 . At this moment, for t, r ∈ [0, 1], x, y ∈ ℧ = C([0, 1]) with x(r) ≤ y(r) for all r ∈ [0, 1], we estimate the difference |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))| ≤ 1 256 ∣∣∣ |x(r)|(1+|x(r)|) 2+|x(r)| − |y(r)|(1+|y(r)|) 2+|y(r)| ∣∣∣+ 1 256 ∣∣∣∣ |x( r2 )|(1+|x( r2 )|)2+|x( r2 )| − |y( r2 )|(1+|y( r2 )|) 2+|y( r2 )| ∣∣∣∣ + 1 256 ∣∣∣ tr|x(0)|(1+|x(0)|) 2+|x(0)| − tr|y(0)|(1+|y(0)|) 2+|y(0)| ∣∣∣+ 1 256 ∣∣∣ tr|x(1)|(1+|x(1)|) 2+|x(1)| − tr|y(1)|(1+|y(1)|) 2+|y(1)| ∣∣∣ = 1 256 (2+2|x(r)|+2|y(r)|+|x(r)||y(r)|)||x(r)|−|y(r)|| (2+|x(r)|)(2+|y(r)|) + 1 256 (2+2|x( r2 )|+2|y( r2 )|+|x( r2 )||y( r2 )|)||x( r2 )|−|y( r2 )|| (2+|x( r2 )|)(2+|y( r2 )|) + 1 256 (2+2|x(0)|+2|y(0)|+|x(0)||y(0)|)||x(0)|−|y(0)|| (2+|x(0)|)(2+|y(0)|) + 1 256 (2+2|x(1)|+2|y(1)|+|x(1)||y(1)|)||x(1)|−|y(1)|| (2+|x(1)|)(2+|y(1)|) ≤ 1 256 2(1+|x(r)|)(1+|y(r)|)|x(r)−y(r)| (2+|x(r)|)(2+|y(r)|) + 1 256 2(1+|x( r2 )|)(1+|y( r2 )|)|x( r2 )−y( r 2 )| (2+|x( r2 )|)(2+|y( r2 )|) + 1 256 2(1+|x(0)|)(1+|y(0)|)|x(0)−y(0)| (2+|x(0)|)(2+|y(0)|) + 1 256 2(1+|x(1)|)(1+|y(1)|)|x(1)−y(1)| (2+|x(1)|)(2+|y(1)|) = 1 128 (1+|x(r)|)(1+|y(r)|)|x(r)−y(r)| (2+|x(r)|)(2+|y(r)|) + 1 128 (1+|x( r2 )|)(1+|y( r2 )|)|x( r2 )−y( r 2 )| (2+|x( r2 )|)(2+|y( r2 )|) + 1 128 (1+|x(0)|)(1+|y(0)|)|x(0)−y(0)| (2+|x(0)|)(2+|y(0)|) + 1 128 (1+|x(1)|)(1+|y(1)|)|x(1)−y(1)| (2+|x(1)|)(2+|y(1)|) = 1 64 1 2 ( 1+ 1 1+|x(r)| )( 1+ 1 1+|y(r)| ) |x(r)− y(r)|+ 1 64 1 2 ( 1+ 1 1+|x( r2 )| )( 1+ 1 1+|y( r2 )| ) ∣∣x( r2)− y( r2) ∣∣ 1 64 1 2 ( 1+ 1 1+|x(0)| )( 1+ 1 1+|y(0)| ) |x(0)− y(0)|+ 1 64 1 2 ( 1+ 1 1+|x(1)| )( 1+ 1 1+|y(1)| ) |x(1)− y(1)| ≤ 1 64 1 2+ 1 1+|x(r)|+ 1 1+|y(r)| |x(r)− y(r)|+ 1 64 1 2+ 1 1+|x( r2 )|+ 1 1+|y( r2 )| ∣∣x( r2)− y( r2) ∣∣ 1 64 1 2+ 1 1+|x(0)|+ 1 1+|y(0)| |x(0)− y(0)|+ 1 64 1 2+ 1 1+|x(1)|+ 1 1+|y(1)| |x(1)− y(1)| ≤ 1 64 1 2+ 1 1+∥x∥∞ + 1 1+∥y∥∞ |x(r)− y(r)|+ 1 64 1 2+ 1 1+∥x∥∞ + 1 1+∥y∥∞ ∣∣x( r2)− y( r2) ∣∣ 1 64 1 2+ 1 1+∥x∥∞ + 1 1+∥y∥∞ |x(0)− y(0)|+ 1 64 1 2+ 1 1+∥x∥∞ + 1 1+∥y∥∞ |x(1)− y(1)| = 1 64 1 2+ 1 1+∥x∥∞ + 1 1+∥y∥∞ [ |x(r)− y(r)|+ ∣∣x( r2)− y( r2) ∣∣+ |x(0)− y(0)|+ |x(1)− y(1)| ] . H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 16 of 18 So by lemma 2, one writes|x(r)− y(r)|+ ∣∣x( r2)− y( r2) ∣∣ ≤ 2 ( |x(r)− y(r)|2 + ∣∣x( r2)− y( r2) ∣∣2) 1 2 |x(0)− y(0)|+ |x(1)− y(1)| ≤ 2 ( |x(0)− y(0)|2 + |x(1)− y(1)|2 ) 1 2 . Thus, we have |K(t, r, x(r), x(g(r)), x(τ1), x(τ2))−K(t, r, y(r), y(g(r)), y(τ1), y(τ2))| ≤ 1 64 1 2 + 1 1+∥x∥∞ + 1 1+∥y∥∞ (2(|x(r)− y(r)|2 + ∣∣∣x(r 2 )− y( r 2 ) ∣∣∣2) 1 2 + 2(|x(0)− y(0)|2 + |x(1)− y(1)|2) 1 2 ) = 1 16 1 2(2 + 1 1+∥x∥∞ + 1 1+∥y∥∞ ) ((|x(r)− y(r)|2 + ∣∣∣x(r 2 )− y( r 2 ) ∣∣∣2) 1 2 + (|x(0)− y(0)|2 + |x(1)− y(1)|2) 1 2 ). We apply another time Lemma 2 to have a = ( |x(r)− y(r)|2 + ∣∣∣x(r 2 )− y( r 2 ) ∣∣∣2) 1 2 b = ( |x(0)− y(0)|2 + |x(1)− y(1)|2 ) 1 2 . That is, |K (t, r, x(r), x(g(r)), x(τ1), x(τ2))−K (t, r, y(r), y(g(r)), y(τ1), y(τ2))| ≤ 1 8 1 2(2 + 1 1+∥x∥∞ + 1 1+∥y∥∞ ) ( |x(r)− y(r)|2 + ∣∣∣x(r 2 )− y( r 2 ) ∣∣∣2 + |x(0)− y(0)|2 + |x(1)− y(1)|2 ) 1 2 ≤ 1 8 1 √ 2 √ 2 + 1 1+∥x∥∞ + 1 1+∥y∥∞ ( |x(r)− y(r)|2 + ∣∣∣x(r 2 )− y( r 2 ) ∣∣∣2 + |x(0)− y(0)|2 + |x(1)− y(1)|2 ) 1 2 . Thus, the second condition (ii) of Theorem 2 is satisfied with γ(t, r) = 1 8 . That is, sup t∈[0,1] ∫ 1 0 γ2(t, r)dr = 1 64 < 1 16 = 1 22+2 |1|p (1− 0)2−1 . Next, Using the properties of the functions K, g and f , it is observed that the hypothesis (i) is satisfied for s̃0 = f ∈ ℧ = C([0, 1]). On the other hand, by definition of the function ϵ the condition (iii) is satisfied. It remains to check that the operator K is non-decreasing. It is clear that the function σ : s 7−→ s(1+s) 2+s ; s ≥ 0 is increasing, so on the space of continuous functions on [0, 1] with value in [0,+∞[, we easily see that the operator is indeed non-decreasing with respect to the variable x. The conditions of Theorem 2 are fulfilled and it results that the integral equation (10) has a unique solution. H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 17 of 18 Acknowledgements The authors extend their appreciation to Umm Al-Qura University, Saudi Arabia for funding this research work through grant number: 25UQU4331214GSSR07. Funding This Research work was funded by Umm Al-Qura University, Saudi Arabia under grant number: 25UQU4331214GSSR07. References [1] S. Banach. Sur les opérations dans les ensembles abstraits et leur application aux Équations intégrales. Fundamenta Mathematicae, 3:133–181, 1922. [2] I. A. Bakhtin. The contraction mapping principle in quasi-metric spaces. Functional Analysis, Ulyanovsk State Pedagogical Institute, 298:26–37, 1989. [3] S. Czerwik. Contraction mappings in b-metric spaces. Acta Mathematica et Infor- matica Universitatis Ostraviensis, 1:5–11, 1983. [4] T. Kamran, M. Samreen, and Q. Ul Ain. Generalization of b-metric space and some fixed point theorems. Mathematics, 5(2):19, 2017. [5] T. Abdeljawad, N. Mlaiki, H. Aydi, and N. Souayah. Double controlled metric type spaces and some fixed point results. Mathematics, 6(12):320, 2018. [6] N. Mlaiki. Double controlled metric-like spaces. Journal of Inequalities and Applica- tions, 2020:189, 2020. [7] A. Tas. On double controlled metric like spaces and related fixed point theorems. Advances in Theory of Nonlinear Analysis and its Applications, 5(2):167–172, 2021. [8] H. Ahmad, M. Younis, and M. K. Kosal. Double controlled partial metric type spaces and convergence results. Journal of Mathematics, 2021:7008737, 2021. [9] S. Haque, A. Souayah, N. Mlaiki, and D. Rizk. Double controlled quasi metric like spaces. Symmetry, 14(618), 2022. [10] E. Karapinar, P. S. Kumari, and D. Lateef. A new approach to the solution of the fredholm integral equation via a fixed point on extended b-metric spaces. Symmetry, 10(512), 2018. [11] M. D. Rus. A note on the existence of positive solution of fredholm integral equations. Fixed Point Theory, 5:369–377, 2004. [12] M. I. Berenguer, M. V. F. Muñoz, A. I. G. Guillem, and M. R. Galan. Numerical treatment of fixed point applied to the nonlinear fredholm integral equation. Fixed Point Theory and Applications, pages 1–8, 2009. [13] H. K. Pathak, M. S. Khan, and R. Tiwari. Common fixed point theorem and its application to nonlinear integral equations. Computers and Mathematics with Appli- cations, 53:961–971, 2007. [14] A. Felhi and H. Aydi. New fixed point results for multi-valued maps via manageable H. Aydi, H. Hammouda, S. Mansour / Eur. J. Pure Appl. Math, 18 (3) (2025), 6133 18 of 18 functions and an application on a boundary value problem. U.P.B. Scientific Bulletin, Series A, 80(1):1–12, 2018. [15] B. Samet, C. Vetro, and P. Vetro. Fixed point theorems for α − ψ-contractive type mappings. Nonlinear Analysis: Theory, Methods and Applications, 75:2154–2165, 2012.