376 Β© 2025 The Author(s). Published by College of Education for Pure Science (Ibn Al-Haitham), University of Baghdad. This is an open-access article distributed under the terms of the Creative Commons Attribution 4.0 International License Fixed Point Results and Application of Cyclic Contractive Maps in b- Metric Spaces Abbas Karim Nahi1* and Salwa Salman Abed2 1,2Department of Mathematics, College of Education for Pure Science (Ibn Al-Haitham), University of Baghdad, Baghdad, Iraq. *Corresponding Author. Received:10 November 2023 Accepted:16 January 2024 Published: 20 April 2025 Abstract One of the generalizations of the usual metric function is the b-metric, which provides researchers with a broader field for deriving numerous results and applications related to fixed point theory. The aim of this paper is to develop three new fixed point principles in the complete b-metric space (Β£,𝜌) when 𝜌 is a continuous function in two variables. Here there are three directions to prove the existence and uniqueness of fixed points. First, we derive a result in terms of Branciari’s theorem by combining integral contractive conditions with the notion of a cyclic map. Second, we apply the notion of cyclic representation to maps satisfying general weak conditions, including a changing distance function, to simulate the content of Boyd and Wong's theorem. Using this result, an application to the existence and uniqueness of the solution of an integral equation is given. Finally, an implicit relation with a changing distance function is used to construct a cyclic contractive map. Some examples are also presented to analyze and illustrate the main results. Keywords: Alternating distance functions, Complete b-metric spaces, Contractive conditions, Cyclic representation, Fixed points. 1. Introduction In 2003, Cyclic contraction was first introduced by Kirk et al. (1), who introduced results dealing with mappings of the type 𝑓: £𝑖 β†’ £𝑖+1, 𝑖 = 1, 2, Β· Β· Β·, 𝑝 + 1, with £𝑖 = £𝑖+1, where the contractive hypotheses are restricted to pairs (π‘Ž, 𝑏) ∈ £𝑖 Γ— £𝑖+1. Extensions of the Banach’s theorem and an extension of the Caristi theorem were proved. In addition, results related to non-expansive mappings in a Banach space were included. Several authors have contributed to research on fixed points in various cases for cyclic contractions (2- 9). Backhtin (10) presented a definition of b-metric by replacing the triangle inequality as the following Definition1.1: Let Β£ be a nonempty set and π‘ž β‰₯ 1. A function 𝜌 ∢ Β£ Γ— Β£ β†’ 𝑅+ is said to be a b-metric on Β£ if (10): 1. 𝜌(π‘Ž, 𝑏) = 0 if and only if π‘Ž = 𝑏; doi.org/10.30526/38.2.3828 https://creativecommons.org/licenses/by/4.0/ https://creativecommons.org/licenses/by/4.0/ mailto:https://orcid.org/0009-0006-3158-1348 mailto:https://orcid.org/0000-0002-0581-253X mailto:salwa.s.a@ihcoedu.uobaghdad.edu.iq mailto:abbas.nahi2103m@ihcoedu.uobaghdad.edu.iq IHJPAS. 2025,38(2) 377 2. 𝜌(π‘Ž, 𝑏) = 𝜌(𝑏, π‘Ž) for all π‘Ž, 𝑏 ∈ Β£; 3. 𝜌(π‘Ž, 𝑏) ≀ π‘ž(𝜌(π‘Ž, 𝑐) + 𝜌(𝑐, 𝑏)) for all π‘Ž, 𝑏, 𝑐 ∈ Β£. The pair (Β£, 𝜌) is called a b-metric space. Definition 1.2: Let (Β£, 𝜌) be a b-metric space, π‘Ž ∈ Β£ and (π‘Žπ‘›) be a sequence in Β£. Then 1. (π‘Žπ‘›) converges to π‘Ž if and only if lim π‘›β†’βˆž 𝜌(π‘Žπ‘›, π‘Ž) = 0. We denote this by π‘™π‘–π‘š π‘›β†’βˆž π‘Žπ‘› = π‘Ž or π‘Žπ‘› β†’ π‘Ž( as 𝑛 β†’ ∞). 2. (π‘Žπ‘›) is Cauchy if and only if lim 𝑛,π‘šβ†’βˆž 𝜌(π‘Žπ‘›, π‘Žπ‘š) = 0. 3. (Β£, 𝜌) is complete if and only if every Cauchy sequence in Β£ is convergent. Remark 1.3: In a b-metric space (Β£, 𝜌), the following assertions hold(11,12): 1. A convergent sequence has a unique limit. 2. Each convergent sequence is Cauchy. 3. In general, a b-metric is not continuous. As in the usual metric space )1(, we reform the following definition: Definition1.4: Let {£𝑖}𝑖=1 𝑛 be a nonempty subsets of a b-metric space (Β£, 𝜌), Β£ = ⋃ £𝑖 𝑛 𝑖=1 and 𝑓: Β£ β†’ Β£ such that 1) 𝑓(Β£1) βŠ‚ Β£2, … , 𝑓(Β£π‘›βˆ’1) βŠ‚ £𝑛, 𝑓(£𝑛) βŠ‚ Β£1 for 1 ≀ 𝑖 ≀ 𝑛; 2) βˆƒπ‘˜ ∈ (0,1) such that 𝜌(π‘“π‘Ž, 𝑓𝑏) ≀ π‘˜ 𝜌(π‘Ž, 𝑏) βˆ€π‘Ž ∈ £𝑖 , 𝑏 ∈ £𝑖+1 for 1 ≀ 𝑖 ≀ 𝑛 Then 𝑓 is the cyclic contraction map and Β£ is cyclic representation w.r.t., 𝑓. In the field of fixed point theory for cyclicity, see (13-20) Example1.5: Let Β£ = [βˆ’1,1], 𝜌(π‘Ž, 𝑏) = |π‘Ž βˆ’ 𝑏|2 is b-metric with 𝑠 = 2 and Β£1 = [βˆ’1,0], Β£2 = [0,1], Β£3 = [βˆ’1,0], Β£4 = [0,1], Β£5 = [βˆ’1,0], Β£6 = [0,1]. So, Β£ = ⋃ £𝑖 6 𝑖=1 . Define 𝑓:⋃ £𝑖 6 𝑖=1 β†’ ⋃ £𝑖 6 𝑖=1 such that π‘“π‘Ž = βˆ’ π‘Ž 2+π‘Ž , βˆ€π‘Ž ∈ ⋃ £𝑖 6 𝑖=1 . Here Β£ is cyclic representation w.r.t.,𝑓. (𝑓(Β£1) βŠ‚ Β£2, 𝑓(Β£2) βŠ‚ Β£3, 𝑓(Β£3) βŠ‚ Β£4, and 𝑓(Β£4) βŠ‚ Β£5, 𝑓(Β£5) βŠ‚ Β£6, 𝑓(Β£6) βŠ‚ Β£1) and 𝑓is cyclic contraction with constant 0 < π‘˜ = 1 2 , where π‘Ž ∈ £𝑖 𝑏 ∈ £𝑖+1. Definition 1.6: A point π‘Ž is called a fixed point of a map 𝑓: Β£ β†’ Β£ if 𝑓(π‘Ž) = π‘Ž (21). An example, π‘Ž = 0 is a fixed point of 𝑓 in the previous example. This work includes four main fixed point theorems based on different cyclic contractive conditions. Here, (Β£, ρ) denote to complete b-metric space where ρ is continuous. 2. Materials and Methods In this section, there are three axes; the first one is Fixed point for cyclic contractive maps with integral condition Not that, A Lebesgue-integrable function Β₯: [0,1) β†’ [0,1) is called summable if ∫Β₯(π‘Ÿ)π‘‘π‘Ÿ < ∞ (22). Theorem 2.1: Let (Β£, 𝜌) be b-metric space, π‘˜ ∈ (0,1), and let 𝑓: Β£ β†’ Β£ be a map such that for each π‘Ž, 𝑏 ∈ Β£ ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Ž, 𝑓𝑏) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ)π‘‘π‘Ÿ 𝜌(π‘Ž, 𝑏) 0 , βˆ€π‘Ž ∈ £𝑖, 𝑏 ∈ £𝑖+1 (1) where, Β₯ is summable on each compact subset of [0,∞), nonnegative and for any πœ€ > 0, ∫ Β₯(π‘Ÿ) Ξ΅ 0 π‘‘π‘Ÿ > 0. Then βˆƒ! 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 , moreover, βˆ€π‘Ž ∈ Β£, lim π‘šβ†’βˆž π‘“π‘š(π‘Ž) = 𝑐. Proof: Let π‘Ž0 ∈ Β£ = ⋃ £𝑖 𝑛 𝑖=1 and consider π‘Žπ‘š+1 = π‘“π‘Žπ‘š for each π‘š ∈ 𝑁⋃{0}, so for any π‘š ∈ 𝑁⋃{0}, βˆƒπ‘–π‘š ∈ {1,2, … , 𝑛} such that π‘Žπ‘š ∈ Β£π‘–π‘š and π‘Žπ‘š+1 ∈ Β£π‘–π‘š+1. If π‘Žπ‘š0 = π‘Žπ‘š0+1 for some π‘š0 then, since π‘Žπ‘š0+1 = π‘“π‘Žπ‘š0 = π‘Žπ‘š0 this mean π‘Žπ‘š0 is a fixed point IHJPAS. 2025,38(2) 378 of 𝑓. Thus, assume that π‘Žπ‘š β‰  π‘Žπ‘š+1 for all π‘š ∈ 𝑁⋃{0}. So, ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘š, π‘“π‘Žπ‘š+1) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) 0 π‘‘π‘Ÿ. (2) By repeating the inequality (2) m times, it follows directly ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘š, π‘“π‘Žπ‘š+1) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) 0 π‘‘π‘Ÿ = π‘˜π‘š ∫ Β₯(π‘Ÿ) 𝜌(π‘Ž0, π‘“π‘Ž0) 0 π‘‘π‘Ÿ. As consequence, since π‘˜ ∈ (0,1), obtaining a monotone decreasing sequence (∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘š, π‘“π‘Žπ‘š+1) 0 π‘‘π‘Ÿ) = (∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘š+1, π‘Žπ‘š+2) 0 π‘‘π‘Ÿ) which has lower bound is 0. We have 𝜌(π‘“π‘Žπ‘š, π‘“π‘Žπ‘š+1) β†’ 0 asπ‘š β†’ ∞. By properties of real sequence, (∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘š, π‘“π‘Žπ‘š+1) 0 π‘‘π‘Ÿ) convergences πœ€ β‰₯ 0 such that lim π‘šβ†’βˆž ∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘š+1,π‘Žπ‘š+2) 0 π‘‘π‘Ÿ = πœ€. Suppose that πœ€ > 0, it is enough to assume lim π‘šβ†’βˆž 𝑠𝑒𝑝 𝜌(π‘Žπ‘š+1, π‘Žπ‘š+2) = πœ€ > 0. Then there exists a π‘’πœ€πœ–π‘ and a sequence (π‘“π‘Žπ‘šπ‘’ )𝑒β‰₯π‘’πœ€ such that 𝜌(π‘“π‘Žπ‘šπ‘’ , π‘“π‘Žπ‘šπ‘’+1) β†’ πœ€ > 0 as 𝑒 β†’ ∞ and 𝜌(π‘“π‘Žπ‘šπ‘’ , π‘“π‘Žπ‘šπ‘’+1) β‰₯ πœ€ 2 . For each 𝑒 β‰₯ π‘’πœ€, by πœ€ 2 ≀ lim π‘šβ†’βˆž 𝜌(π‘“π‘Žπ‘šπ‘’ , π‘“π‘Žπ‘šπ‘’+1) = lim π‘šβ†’βˆž sup 𝜌(π‘“π‘Žπ‘šπ‘’ , π‘“π‘Žπ‘šπ‘’+1) = πœ€ > 0, this is true only if πœ€ = 0. The next step is proving that for each π‘Ž0 ∈ Β£, (π‘Žπ‘š) is a Cauchy sequence. For 𝑗 > 𝑛 define £𝑗 = £𝑖 if 𝑗 = 𝑖 mod 𝑛. Claim I: for all πœ€ > 0 there exist π‘š ∈ 𝑁 such that for all 𝑗, 𝑖 β‰₯ π‘š, 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) then 𝜌(π‘Žπ‘—, π‘Žπ‘–) < πœ€. Suppose that there exists πœ€ > 0 such that for each π‘š ∈ 𝑁, one can find 𝑗 > 𝑖 > π‘š with 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) satisfying 𝜌(π‘Žπ‘—, π‘Žπ‘–) β‰₯ πœ€. Clearly, 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) < πœ€. Now, take π‘š β‰₯ 2(mod 𝑛). Then, corresponding to 𝑖 β‰₯ π‘š use can choose 𝑗 in such a way that it is the smallest integer with 𝑗 > 𝑖 satisfying 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) and 𝜌(π‘Žπ‘— , π‘Žπ‘–) β‰₯ πœ€. Therefore, 𝜌(π‘Žπ‘—βˆ’π‘›, π‘Žπ‘–) ≀ πœ€. By triangular inequality πœ€ ≀ 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ π‘ž(𝜌(π‘Žπ‘— , π‘Žπ‘–βˆ’π‘›) + βˆ‘ 𝜌(π‘Žπ‘–βˆ’π‘˜, π‘Žπ‘–βˆ’π‘˜+1) 𝑛 π‘˜=1 ) ≀ π‘ž(βˆ‘ 𝜌(π‘Žπ‘–βˆ’π‘˜, π‘Žπ‘–βˆ’π‘˜+1) 𝑛 π‘˜=1 ) + π‘ž πœ€ β†’ π‘ž πœ€ as 𝑛 β†’ ∞. Again, by triangular inequality πœ€ ≀ 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ π‘ž 𝜌(π‘Žπ‘— , π‘Žπ‘—+1) + π‘žπœŒ(π‘Žπ‘—+1, π‘Žπ‘–+1) + π‘žπœŒ(π‘Žπ‘–+1, π‘Žπ‘–) β†’ π‘ž Ξ΅ as 𝑗, 𝑖 β†’ ∞, we get ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘—+1,π‘“π‘Žπ‘–+1) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘—+1,π‘Žπ‘–+1) 0 π‘‘π‘Ÿ. (3) Letting 𝑗, 𝑖 β†’ ∞ implies to ∫ Β₯(π‘Ÿ) π‘žπœ€ 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) π‘žπœ€ 0 π‘‘π‘Ÿ, which is a contradiction. Therefore, the (Claim I) is proved. Now, to prove (π‘Žπ‘š) is Cauchy sequence in (Β£, 𝜌). Fix Ξ΅ > 0. By the claim, βˆƒπ‘š0 such that if 𝑗, 𝑖 β‰₯ π‘š0 with 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ Ξ΅ 𝑛 . Since lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = 0, βˆƒπ‘š1 ∈ 𝑁 such that 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ πœ€ 𝑛 , βˆ€π‘š β‰₯ π‘š1 Suppose 𝑐, 𝑣 β‰₯ π‘šπ‘Žπ‘₯{π‘š0, π‘š1} and 𝑐 > 𝑣. Then there exists β„Ž ∈ {1,2, … . . , 𝑛} such that 𝑣 βˆ’ 𝑐 ≑ β„Ž(mod 𝑛). Therefore, 𝑣 βˆ’ 𝑐 + π‘Ÿ ≑ 1(mod 𝑛) for π‘Ÿ = 𝑛 βˆ’ β„Ž + 1. So, getting 𝜌(π‘Žπ‘, π‘Žπ‘£) ≀ π‘ž(𝜌(π‘Žπ‘, π‘Žπ‘£+π‘Ÿ) + 𝜌(π‘Žπ‘£+π‘Ÿ , π‘Žπ‘£)). ≀ π‘ž 𝜌(π‘Žπ‘, π‘Žπ‘£+π‘Ÿ) + π‘ž 2 𝜌(π‘Žπ‘£+π‘Ÿ , π‘Žπ‘£+π‘Ÿβˆ’1) + π‘ž 3𝜌(π‘Žπ‘£+π‘Ÿβˆ’1, π‘Žπ‘£+π‘Ÿβˆ’2) + β‹―+ π‘ž π‘ŸπœŒ(π‘Žπ‘£+1, π‘Žπ‘£) By 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ Ξ΅ 𝑛 and 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ πœ€ 𝑛 and from the last inequality, 𝜌(π‘Žπ‘, π‘Žπ‘£) ≀ π‘ž πœ€ 𝑛 + π‘ž2 πœ€ 𝑛 + π‘ž3 πœ€ 𝑛 +β‹―+ π‘žπ‘Ÿ πœ€ 𝑛 ≀ π‘ž πœ€ 𝑛 ( 1 1βˆ’π‘ž ) β†’ 0, as 𝑛 β†’ ∞ This proves that (π‘Žπ‘š) is the Cauchy sequence. The completeness of (Β£, 𝜌) implies to exist 𝑐 ∈ IHJPAS. 2025,38(2) 379 Β£ such that lim π‘šβ†’βˆž π‘Žπ‘š = 𝑐. To prove 𝑐 is a fixed point for 𝑓. Since Β£ = ⋃ £𝑖 𝑛 𝑖=1 is a cyclic representation of Β£ w.r.t., 𝑓, the sequence (π‘Žπ‘š) has infinite terms in each Β£π‘–π‘šfor π‘–π‘š ∈ {1,2, . . . , 𝑛}. Closeness of Β£π‘–π‘š for π‘–π‘š ∈ {1,2, . . . , 𝑛} implies to 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Suppose that 𝑐 ∈ £𝑖 and 𝑓𝑐 ∈ £𝑖+1. Since (Β£, 𝜌) is complete, there exists appoint 𝑐 ∈ Β£ = ⋃ £𝑖 𝑛 𝑖=1 such that 𝑐 = lim π‘šβ†’βˆž π‘“π‘Žπ‘š 0 < 𝜌(𝑐, 𝑓𝑐) ≀ π‘ž 𝜌(𝑐, π‘Žπ‘š+1) + π‘ž 𝜌(π‘“π‘Žπ‘š, 𝑓𝑐) β†’ 0 as π‘š β†’ ∞. Indeed both 𝜌(𝑐, π‘“π‘Žπ‘šπ‘Ÿ ) and 𝜌(π‘“π‘Žπ‘šπ‘Ÿ , 𝑓𝑐) converge to 0 as π‘š β†’ ∞, for the first one it is obvious, while for the second one we have ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘š,𝑓𝑐) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) 𝜌(π‘Žπ‘š,𝑐) 0 π‘‘π‘Ÿ β†’ 0 as π‘š β†’ ∞. Now, if 𝜌(π‘“π‘Žπ‘š, 𝑓𝑐) does not converge to 0 as π‘š β†’ ∞, then there exists a subsequence (π‘Žπ‘šπ‘Ÿ )π‘Ÿβˆˆπ‘ of (π‘Žπ‘š) with π‘Žπ‘šπ‘Ÿ ∈ Β£π‘–βˆ’1 such that 𝜌(π‘“π‘Žπ‘šπ‘Ÿ , 𝑓𝑐) β‰₯ πœ€ for a certain πœ€ > 0, we have the following contradiction 0 < ∫ Β₯(π‘Ÿ) πœ€ 0 π‘‘π‘Ÿ ≀ ∫ Β₯(π‘Ÿ) 𝜌(π‘“π‘Žπ‘šπ‘Ÿ ,𝑓𝑐) 0 π‘‘π‘Ÿ β†’ 0 as π‘Ÿ β†’ ∞. This means that 𝜌(𝑐, 𝑓𝑐) ≀ 0 thus, 𝜌(𝑐, 𝑓𝑐) = 0, 𝑐 is the fixed point of 𝑓. For the uniqueness of fixed point 𝑐. Assume a fixed point 𝑀 of 𝑓 differs from 𝑐, i.e., 𝑓𝑀 = 𝑀. The cyclic character of 𝑓 and 𝑐, 𝑀 ∈ Β£ = ⋃ £𝑖 𝑛 𝑖=1 are fixed points of 𝑓 implying that 𝑐, 𝑀 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . By (1), we obtain ∫ Β₯(π‘Ÿ) 𝜌(𝑐,𝑀) 0 π‘‘π‘Ÿ = ∫ Β₯(π‘Ÿ) 𝜌(𝑓𝑐,𝑓𝑀) 0 π‘‘π‘Ÿ ≀ π‘˜ ∫ Β₯(π‘Ÿ) 𝜌(𝑐,𝑀) 0 π‘‘π‘Ÿ, which is contradiction, consequently, 𝑐 = 𝑀 for each π‘Ž ∈ Β£, lim π‘šβ†’βˆž π‘“π‘šπ‘Ž = 𝑐. Example 2.2 :Let Β£ = [βˆ’1, 1] and 𝜌(π‘Ž, 𝑏) = |π‘Ž βˆ’ 𝑏|2 is b-metric with π‘ž = 2. Suppose Β£1 = [βˆ’1, 0], Β£2 = [0, 1] and Β£ = ⋃ £𝑖 2 𝑖=1 . Define 𝑓:⋃ £𝑖 2 𝑖=1 β†’ ⋃ £𝑖 2 𝑖=1 such that (π‘Ž) = βˆ’π‘Ž 2 βˆ€π‘Ž. So, 𝑓(Β£1) βŠ‚ Β£2, 𝑓(Β£2) βŠ‚ Β£1 and 𝑓 is contraction of integral type with constant π‘˜ = 1 2 ∈ (0, 1) and Β₯(π‘Ÿ) = π‘Ÿ 3 , for π‘Ž ∈ Β£1, 𝑏 ∈ Β£2 ∫ π‘Ÿ 3 𝜌(π‘“π‘Ž,𝑓𝑏) 0 π‘‘π‘Ÿ = ∫ π‘Ÿ 3 1 2 |π‘βˆ’π‘Ž|2 0 π‘‘π‘Ÿ ≀ 1 2 ∫ π‘Ÿ 3 π‘‘π‘Ÿ 𝜌(π‘Ž,𝑏) 0 . Hence 𝑓 satisfies the hypothesis of (Theorem 2.1), which has a unique fixed point at 0. Fixed point for general cyclic(βˆ… βˆ’ πœ“) weak contractive maps Recall the following two definitions: Definition 2.3: Let πœ“ is function where ψ: [0,∞) β†’ [0,∞) is called altering distance function if satisfies(23). 1. πœ“ is monotone increasing and lower semi-continuous; 2. πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0. As in usual metric spaces )4(, below, we reform many concepts in a b-metric space Definition 2.4: Let (Β£, 𝜌) be a b-metric space, 𝑛 a positive integer Β£1, Β£2, …, £𝑛 nonempty closed subsets of Β£ and Β£ = ⋃ £𝑖 𝑛 𝑖=1 . An operator 𝑓 ∢ Β£ β†’ Β£ is said to be a cyclic weak (βˆ… βˆ’ πœ“)-contraction if 1) Β£ = ⋃ £𝑖 𝑛 𝑖=1 is a cyclic representation of Β£ w.r.t., 𝑓. 2) βˆ…(𝜌(π‘“π‘Ž, 𝑓𝑏)) ≀ βˆ…(𝜌(π‘Ž, 𝑏)) βˆ’ πœ“(𝜌(π‘Ž, 𝑏)), for any π‘Ž ∈ £𝑖, 𝑏 ∈ £𝑖+1, 𝑖 = 1, 2, … , 𝑛, where £𝑛+1 = Β£1 and βˆ…,πœ“: [0,∞) β†’ [0,∞) is a non-decreasing and continuous function satisfying βˆ…(π‘Ÿ) > 0, πœ“(π‘Ÿ) > 0 for π‘Ÿ ∈ (0,∞) and βˆ…(0) = 0, πœ“(0) = 0. Theorem 2.5: Let 𝑓 be a self-map of (Β£, 𝜌) satisfies βˆ…(𝜌(π‘“π‘Ž, 𝑓𝑏)) ≀ βˆ…(𝑀(π‘Ž, 𝑏)) βˆ’ πœ“(𝑁(π‘Ž, 𝑏)), βˆ€π‘Ž ∈ £𝑖, 𝑏 ∈ £𝑖+1 (4) where IHJPAS. 2025,38(2) 380 𝑀(π‘Ž, 𝑏) = 𝑑 max{𝜌(π‘Ž, 𝑏), 𝜌(π‘Ž, π‘“π‘Ž), 𝜌(𝑏, 𝑓𝑏), 𝜌(π‘Ž, 𝑓𝑏), 𝜌(𝑏, π‘“π‘Ž)}, and 𝑁(π‘Ž, 𝑏) = 𝑑 min{𝜌(π‘Ž, 𝑏), 𝜌(π‘Ž, π‘“π‘Ž), 𝜌(𝑏, 𝑓𝑏), 𝜌(π‘Ž, 𝑓𝑏), 𝜌(𝑏, π‘“π‘Ž)}, where 𝑑 ∈ (0,1), and πœ“, βˆ…: [0,∞) β†’ [0,∞) are altering distance functions. Then βˆƒ 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 , 𝑐 is a unique fixed point of 𝑓. Proof: Let π‘Ž0 ∈ ⋃ £𝑖 𝑛 𝑖=1 and consider π‘Žπ‘š+1 = π‘“π‘Žπ‘š for each π‘š ∈ 𝑁⋃{0}, so for any π‘š ∈ 𝑁⋃{0}, βˆƒπ‘–π‘š ∈ {1, 2, … , 𝑛} such that π‘Žπ‘š ∈ Β£π‘–π‘š and π‘Žπ‘š+1 ∈ Β£π‘–π‘š+1. If π‘Žπ‘š0 = π‘Žπ‘š0+1 for some π‘š0 then, since π‘Žπ‘š0+1 = π‘“π‘š0π‘Ž0 = π‘Žπ‘š0 this means π‘Žπ‘š0 is fixed point of 𝑓. Thus, assume that π‘Žπ‘š β‰  π‘Žπ‘š+1 for all π‘š ∈ 𝑁⋃{0}. getting βˆ…(𝜌(π‘Žπ‘š, π‘Žπ‘š+1)) ≀ βˆ…(𝑀(π‘Žπ‘šβˆ’1, π‘Žπ‘š)) βˆ’ πœ“(𝑁(π‘Žπ‘šβˆ’1, π‘Žπ‘š)), (5) where 𝑀(π‘Žπ‘šβˆ’1, π‘Žπ‘š) = 𝑑 max {𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š), 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š+1) , 𝜌(π‘Žπ‘š, π‘Žπ‘š)}, and 𝑁(π‘Žπ‘šβˆ’1, π‘Žπ‘š) = 𝑑min {𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š), 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š+1), 0} = 0. By triangular inequality, so 𝑀(π‘Žπ‘šβˆ’1, π‘Žπ‘š) = 𝑑 max {𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š), 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), π‘ž 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š) + π‘ž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), 0}, if 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š) < 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), then 𝑀(π‘Žπ‘šβˆ’1, π‘Žπ‘š) ≀ 𝑑 2 π‘ž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1). And by (5), we obtain βˆ…(𝜌(π‘Žπ‘š, π‘Žπ‘š+1)) ≀ βˆ…(𝑑 2 π‘ž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1)). Since βˆ… is altering distance function, subsequently 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ 𝑑 2 π‘ž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1), this not true for all 𝑑 ∈ (0,1), as result 𝑀(π‘Žπ‘šβˆ’1, π‘Žπ‘š) = 𝑑 2 π‘ž 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š) and 𝑁(π‘Žπ‘šβˆ’1, π‘Žπ‘š) = 0. (6) Now put (6) in (5) βˆ…(𝜌(π‘Žπ‘š, π‘Žπ‘š+1)) ≀ βˆ…(𝑑 2 π‘ž 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š)). (7) Since βˆ… is altering distance function, as a result 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ 𝑑 2 π‘ž 𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š). Thus for all π‘š ∈ 𝑁⋃{0} we have a monotone decreasing sequence (𝜌(π‘Žπ‘š, π‘Žπ‘š+1)) = (𝜌(π‘“π‘Žπ‘šβˆ’1, π‘“π‘Žπ‘š)). By properties of real sequence, (𝜌(π‘Žπ‘š, π‘Žπ‘š+1)) there exist πœ€ β‰₯ 0 such that lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = πœ€. On letting π‘š β†’ ∞ in (7), obtaining βˆ…(πœ€) ≀ βˆ…(𝑑 2 π‘ž πœ€). Assume that πœ€ β‰  0. So since βˆ… is altering distance function, getting πœ€ ≀ 𝑑 2 π‘ž πœ€ < Ξ΅, which a contradiction. Hence lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = 0. For 𝑗 > 𝑛 define £𝑗 = £𝑖 if 𝑗 = 𝑖 mod 𝑛. Claim I: for all πœ€ > 0 there exist π‘š ∈ 𝑁 such that for all 𝑗, 𝑖 β‰₯ π‘š, 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) then 𝜌(π‘Žπ‘—, π‘Žπ‘–) < πœ€. Suppose that there exists πœ€ > 0 such that for each π‘š ∈ 𝑁 we can find 𝑗 > 𝑖 > π‘š with 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) satisfying 𝜌(π‘Žπ‘–, π‘Žπ‘—) β‰₯ πœ€. Now, take π‘š β‰₯ 2(mod 𝑛). Then, corresponding to 𝑖 β‰₯ π‘š use can choose 𝑗 in such a way that it is the smallest integer with 𝑗 > 𝑖 satisfying 𝑗 βˆ’ 𝑖 ≑ 1(mod 𝑛) and 𝜌(π‘Žπ‘— , π‘Žπ‘–) β‰₯ πœ€. Therefore, 𝜌(π‘Žπ‘—βˆ’π‘›, π‘Žπ‘–) ≀ πœ€. By triangular inequality πœ€ ≀ 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ π‘ž 𝜌(π‘Žπ‘— , π‘Žπ‘—βˆ’π‘›) + π‘ž βˆ‘ 𝜌(π‘Žπ‘–βˆ’π‘˜, π‘Žπ‘–βˆ’π‘˜+1) 𝑛 π‘˜=1 ≀ π‘ž βˆ‘ 𝜌(π‘Žπ‘–βˆ’π‘˜, π‘Žπ‘–βˆ’π‘˜+1) 𝑛 π‘˜=1 + π‘žπœ€. Taking 𝑛 β†’ ∞ and since lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = 0, we obtain lim 𝑖,π‘—β†’βˆž 𝜌(π‘Žπ‘— , π‘Žπ‘–) = π‘žπœ€. Again, by triangular inequality πœ€ ≀ 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ 2π‘žπœŒ(π‘Žπ‘—+1, π‘Žπ‘—) + π‘žπœŒ(π‘Žπ‘— , π‘Žπ‘–) + 2π‘žπœŒ(π‘Žπ‘–+1, π‘Žπ‘–). Letting 𝑗, 𝑖 β†’ ∞ and since lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = 0, so lim 𝑖,π‘—β†’βˆž 𝜌(π‘Žπ‘—+1, π‘Žπ‘–+1) = π‘žπœ€. IHJPAS. 2025,38(2) 381 Since π‘Žπ‘–, π‘Žπ‘— belong to different sets £𝑖 and £𝑖+1, and using (2.4) βˆ…(𝜌(π‘“π‘Žπ‘— , π‘“π‘Žπ‘–)) ≀ βˆ… (𝑀(π‘Žπ‘— , π‘Žπ‘–)) βˆ’ πœ“ (𝑁(π‘Žπ‘— , π‘Žπ‘–)), (8) where 𝑀(π‘Žπ‘—, π‘Žπ‘–) = 𝑑 max {𝜌(π‘Žπ‘— , π‘Žπ‘–), 𝜌(π‘Žπ‘— , π‘Žπ‘—+1), 𝜌(π‘Žπ‘–, π‘Žπ‘–+1), 𝜌(π‘Žπ‘— , π‘Žπ‘–+1), 𝜌(π‘Žπ‘–, π‘Žπ‘—+1)} 𝑁(π‘Žπ‘—, π‘Žπ‘–) = 𝑑 min {𝜌(π‘Žπ‘— , π‘Žπ‘–), 𝜌(π‘Žπ‘—, π‘Žπ‘—+1), 𝜌(π‘Žπ‘–, π‘Žπ‘–+1), 𝜌(π‘Žπ‘—, π‘Žπ‘–+1), 𝜌(π‘Žπ‘–, π‘Žπ‘—+1)}. Since lim 𝑖,π‘—β†’βˆž 𝜌(π‘Žπ‘— , π‘Žπ‘–) = π‘žπœ€, lim 𝑖,π‘—β†’βˆž 𝜌(π‘Žπ‘—+1, π‘Žπ‘–+1) = π‘žπœ€, and by the triangle inequality 𝜌(π‘Žπ‘–+1, π‘Žπ‘—+1) ≀ π‘žπœŒ(π‘Žπ‘–+1, π‘Žπ‘–) + π‘žπœŒ(π‘Žπ‘–, π‘Žπ‘—) + π‘žπœŒ(π‘Žπ‘— , π‘Žπ‘—+1) β†’ π‘žπœ€ as 𝑖, 𝑗 β†’ ∞. Again by the triangle inequality 𝜌(π‘Žπ‘—, π‘Žπ‘–+1) ≀ π‘žπœŒ(π‘Žπ‘— , π‘Žπ‘—+1) + π‘žπœŒ(π‘Žπ‘—+1, π‘Žπ‘–+1) β†’ π‘žπœ€ as 𝑖, 𝑗 β†’ ∞. Again by the triangle inequality 𝜌(π‘Žπ‘–, π‘Žπ‘—+1) ≀ π‘žπœŒ(π‘Žπ‘–, π‘Žπ‘–+1) + π‘žπœŒ(π‘Žπ‘–+1, π‘Žπ‘—+1) β†’ π‘žπœ€ as 𝑖, 𝑗 β†’ ∞. Letting 𝑖, 𝑗 β†’ ∞, we have 𝑀(π‘Žπ‘— , π‘Žπ‘–) = 𝑑 max {π‘žπœ€, 0,0, π‘žπœ€, π‘žπœ€} β†’ π‘‘π‘žπœ€ and 𝑁(π‘Žπ‘— , π‘Žπ‘–) β†’ 0 and by the inequality (8), we have βˆ…(πœ€) ≀ βˆ…(π‘‘π‘žπœ€) βˆ’ πœ“(0). Since βˆ… is altering distance function and 𝑑 ∈ (0,1), then πœ€ ≀ π‘‘π‘žπœ€ which a contradiction. Therefore, the claim (I) is held. Now, we will prove (π‘Žπ‘š) is a Cauchy sequence in (Β£, 𝜌). Fix Ξ΅ > 0. By the claim, βˆƒπ‘š0 such that if 𝑗, 𝑖 β‰₯ π‘š0 with 𝑗 βˆ’ 𝑖 ≑ 1(mod𝑛) such that 𝜌(π‘Žπ‘—, π‘Žπ‘–) ≀ Ξ΅ 2 . Since lim π‘šβ†’βˆž 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) = 0, also βˆƒπ‘š1 ∈ 𝑁 such that𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ πœ€ 2𝑛 , βˆ€π‘š β‰₯ π‘š1. Suppose 𝑐, 𝑣 β‰₯ max{π‘š0,π‘š1} and 𝑐 > 𝑣. Then there exists β„Ž ∈ {1, 2, … . . , 𝑛} such that 𝑣 βˆ’ 𝑐 ≑ β„Ž(mod 𝑛). Therefore, 𝑣 βˆ’ 𝑐 + π‘Ÿ ≑ 1(mod 𝑛) for π‘Ÿ = 𝑛 βˆ’ β„Ž + 1. So, we have 𝜌(π‘Žπ‘, π‘Žπ‘£) ≀ π‘ž 𝜌(π‘Žπ‘, π‘Žπ‘£+π‘Ÿ) + π‘ž 2 𝜌(π‘Žπ‘£+π‘Ÿ , π‘Žπ‘£+π‘Ÿβˆ’1) + π‘ž 3𝜌(π‘Žπ‘£+π‘Ÿβˆ’1, π‘Žπ‘£+π‘Ÿβˆ’2) + +π‘žπ‘ŸπœŒ(π‘Žπ‘£+1, π‘Žπ‘£) (9) By 𝜌(π‘Žπ‘— , π‘Žπ‘–) ≀ Ξ΅ 2 and 𝜌(π‘Žπ‘š, π‘Žπ‘š+1) ≀ πœ€ 2𝑛 and from (9), 𝜌(π‘Žπ‘, π‘Žπ‘£) ≀ π‘ž πœ€ 𝑛 ( 1 1βˆ’π‘ž ) β†’ 0, as 𝑛 β†’ ∞ This proves that (π‘Žπ‘š) is a Cauchy sequence. The completeness of (Β£, 𝜌) implies to exists 𝑐 ∈ Β£ such that lim π‘šβ†’βˆž π‘Žπ‘š = 𝑐. Now, to prove 𝑐 is a fixed point for 𝑓. Since Β£ = ⋃ £𝑖 𝑛 𝑖=1 is a cyclic representation of Β£ w.r.t., 𝑓, the sequence (π‘Žπ‘š) has infinite terms in each Β£π‘–π‘š for π‘–π‘š ∈ {1, 2, . . . , 𝑛}. Closeness of Β£π‘–π‘š for π‘–π‘š ∈ {1, 2, . . . , 𝑛} implies to 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Suppose that 𝑐 ∈ £𝑖 and 𝑓𝑐 ∈ £𝑖+1 and take a subsequence (π‘Žπ‘šπ‘Ÿ )π‘Ÿβˆˆπ‘ of (π‘Žπ‘š) with π‘Žπ‘šπ‘Ÿ ∈ Β£π‘–βˆ’1 and take π‘Ž = 𝑐, 𝑏 = π‘Žπ‘šπ‘Ÿ in (4) βˆ…(𝜌(𝑓𝑐, π‘“π‘Žπ‘šπ‘Ÿ )) ≀ βˆ…(𝑀(π‘Žπ‘šπ‘Ÿ , 𝑐)) βˆ’ πœ“ (𝑁(π‘Žπ‘šπ‘Ÿ , 𝑐)) (10) where 𝑀(π‘Žπ‘šπ‘Ÿ , 𝑐) = 𝑑 max {𝜌(π‘Žπ‘šπ‘Ÿ , 𝑐), 𝜌(π‘Žπ‘šπ‘Ÿ , π‘“π‘Žπ‘šπ‘Ÿ ), 𝜌(𝑐, 𝑓𝑐), 𝜌(π‘Žπ‘šπ‘Ÿ , 𝑓𝑐), 𝜌(𝑐, π‘“π‘Žπ‘šπ‘Ÿ )}, 𝑁(π‘Žπ‘šπ‘Ÿ , 𝑐) = 𝑑 min {𝜌(π‘Žπ‘šπ‘Ÿ , 𝑐), 𝜌(π‘Žπ‘šπ‘Ÿ , π‘“π‘Žπ‘šπ‘Ÿ ), 𝜌(𝑐, 𝑓𝑐), 𝜌(π‘Žπ‘šπ‘Ÿ , 𝑓𝑐), 𝜌(𝑐, π‘“π‘Žπ‘šπ‘Ÿ )}. Taking π‘Ÿ β†’ ∞, hence 𝑀(𝑐, π‘Žπ‘šπ‘Ÿ ) = 𝑑 𝜌(𝑐, 𝑓𝑐). And (𝑐, π‘Žπ‘šπ‘Ÿ ) = 0, using (10) subsequently βˆ…(𝜌(𝑓𝑐, 𝑐)) ≀ βˆ…(𝑑 𝜌(𝑐, 𝑓𝑐)) βˆ’ πœ“(0) ≀ βˆ…(𝑑 𝜌(𝑐, 𝑓𝑐)). Since βˆ… is altering distance function, then 𝜌(𝑓𝑐, 𝑐) < 𝑑 𝜌(𝑐, 𝑓𝑐), which a contradiction because 𝑑 ∈ (0,1), hence 𝑓𝑐 = 𝑐. Thus 𝑐 is a fixed point of 𝑓. For the uniqueness, suppose that there are two distinct points 𝑐, 𝑀 with 𝑐 and 𝑀 fixed points of 𝑓. The cyclic character of 𝑓 and the fact that 𝑐, 𝑀 ∈ Β£ = ⋃ £𝑖 𝑛 𝑖=1 are fixed points of 𝑓 imply IHJPAS. 2025,38(2) 382 that 𝑐, 𝑀 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Using (4), we can obtain βˆ…(𝜌(𝑓𝑐, 𝑓𝑀)) ≀ βˆ…(𝑀(𝑐, 𝑀)) βˆ’ πœ“(𝑁(𝑐, 𝑀)), Where, 𝑀(𝑐,𝑀) = 𝑑 𝜌(𝑐, 𝑀) and 𝑁(𝑐, 𝑀)} = 0. Hence βˆ…(𝜌(𝑐, 𝑀)) = βˆ…(𝜌(𝑓𝑐, 𝑓𝑀)) ≀ βˆ…(𝑑 𝜌(𝑐, 𝑀)). And since βˆ… is altering distance function, we obtain 𝜌(𝑐, 𝑀) ≀ π‘ž 𝜌(𝑐, 𝑀) which a is contradiction since it is not true for all 𝑑 ∈ (0,1), then 𝑐 = 𝑀. Hence 𝑓 has a unique fixed point in Β£ and 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Example 2.6: Let Β£ = {6, 7, 8, 9, 10} with 𝜌: Β£ Γ— Β£ β†’ [0,∞) defined by 𝜌(π‘Ž, 𝑏) = { 0, 𝑖𝑓 π‘Ž = 𝑏 6, 𝑖𝑓 π‘Ž β‰  𝑏, π‘Ž, 𝑏 ∈ {6,7,8,9} 17, 𝑖𝑓 π‘Ž, 𝑏 ∈ {9,10} and π‘Ž β‰  𝑏 40, π‘–π‘“π‘Ž ∈ {6,7,8} and 𝑏 = 10 (or 𝑏 ∈ {6,7,8} and π‘Ž = 10) . Since all Cauchy sequences in Β£ are constant. Therefore, are convergent. Then (Β£, 𝜌) is complete b-metric space with π‘ž = 2. And Β£1 = {6,8,10} and Β£2 = {6,7,9}, Β£ = ⋃ £𝑖 2 𝑖=1 . Define 𝑓:⋃ £𝑖 2 𝑖=1 β†’ ⋃ £𝑖 2 𝑖=1 such that 𝑓(π‘Ž) = 6 and 𝑖𝑓 π‘Ž ∈ {6,7,8,9} and 𝑓(10) = 8. So, (Β£1) βŠ‚ Β£2,𝑓(Β£2) βŠ‚ Β£1, for π‘Ž ∈ Β£1, 𝑏 ∈ Β£2, and take 𝑑 = 1 4 . Let βˆ…,πœ“: [0,∞) β†’ [0,∞) such that βˆ…(π‘Ÿ) = π‘Ÿ 4 and πœ“(π‘Ÿ) = π‘Ÿ 2 . Then βˆ… and πœ“ are altering distance functions. It is easy to check condition (4) holds with fixed point π‘Ž = 6. An application for solving integral equations. Consider the integral equation (24,25) 𝑀(𝑑) = ∫ 𝑄(𝑑, π‘Ÿ)𝑇(π‘Ÿ, 𝑀(π‘Ÿ)) 𝐽 0 π‘‘π‘Ÿ for all 𝑑 ∈ [0, 𝐽], (11) where 𝐽 > 0, 𝑇: [0, 𝐽] Γ— 𝑅 β†’ 𝑅 and 𝑄: [0, 𝐽] Γ— [0, 𝐽] β†’ [0,∞) are continuous functions. In this section, we look for a nonnegative solution to (11) in Β£ = 𝐢([0, 𝐽], 𝑅) by (Theorem 2.5). Let Β£ = 𝐢[0, 𝐽] be the set of real valued continuous functions on [0, 𝐽], where [0, 𝐽] is a closed and bounded interval in 𝑅. For 𝑝 > 1, define 𝜌: [0, 𝐽] Γ— [0, 𝐽] β†’ 𝑅 by 𝜌(𝑀, 𝑣) = max π‘‘πœ–[0,𝐽] |𝑀(𝑑) βˆ’ 𝑣(𝑑)|𝑝, for all 𝑀, 𝑣 ∈ Β£. Therefore, (Β£ ,𝜌)is a complete b-metric space with π‘ž = 2π‘βˆ’1. Let 𝛼, 𝛽 ∈ Β£ and 𝛼0, 𝛽0 ∈ 𝑅 such that 𝛼0 ≀ Ξ±(𝑑) ≀ 𝛽(𝑑) ≀ 𝛽0, βˆ€ 𝑑 ∈ [0, 𝐽]. (12) Suppose that for all 𝑑 ∈ [0, 𝐽], we have 𝛼(𝑑) ≀ ∫ 𝑄(𝑑, π‘Ÿ)𝑇(π‘Ÿ, 𝛽(π‘Ÿ)) 𝐽 0 π‘‘π‘Ÿ, (13) and 𝛽(𝑑) β‰₯ ∫ 𝑄(𝑑, π‘Ÿ)𝑇(π‘Ÿ, 𝛼(π‘Ÿ)) 𝐽 0 π‘‘π‘Ÿ. (14) We suppose that βˆ€ π‘Ÿ ∈ [0, 𝐽], 𝑇(π‘Ÿ, . )be a decreasing function, that π‘Ž, 𝑏 ∈ 𝑅, π‘Ž β‰₯ 𝑏 then 𝑇(π‘Ÿ, π‘Ž) ≀ 𝑇(π‘Ÿ, 𝑏). (15) Assume that π‘˜ > 0 is such that π‘˜(max π‘‘βˆˆ[0,𝐽] ∫ 𝑄(𝑑, π‘Ÿ) 𝐽 0 π‘‘π‘Ÿ) < 1. (16) Define a map 𝑓: Β£ β†’ Β£ by 𝑓𝑀(𝑑) = ∫ 𝑄(𝑑, π‘Ÿ)𝑇(π‘Ÿ, 𝑀(π‘Ÿ)) 𝐽 0 π‘‘π‘Ÿ, for all 𝑑 ∈ [0, 𝐽]. Suppose that βˆ€π‘Ÿ ∈ [0, 𝐽] and π‘Ž, 𝑏 ∈ Β£ with (π‘Ž(π‘Ÿ) ≀ 𝛼0 and 𝑏(π‘Ÿ) ≀ 𝛽0) or vice versa, 0 ≀ [𝑇(π‘ž, π‘Ž(π‘Ÿ)) βˆ’ 𝑇(π‘Ÿ, 𝑏(π‘Ÿ))] ≀ π‘˜ max {|π‘Ž(π‘Ÿ) βˆ’ 𝑏(π‘Ÿ)|𝑝, 0, |𝑏(π‘Ÿ) βˆ’ 𝑓𝑏(π‘Ÿ)|𝑝, |π‘Ž(π‘Ÿ) βˆ’ 𝑓𝑏(π‘Ÿ)|𝑝, |𝑏(π‘Ÿ) βˆ’ π‘Ž(π‘Ÿ)|𝑝}) 1 𝑝 (17) Theorem 2.7: Under the assumptions (12)-(17), the integral equation (11) has a solution in the set {𝑀 ∈ 𝐢([0, 𝐽]): 𝛼 ≀ 𝑀 ≀ 𝛽}. IHJPAS. 2025,38(2) 383 Proof : We omit the details because the proof steps are classic with some minor differences due to the specificity of the b-metric space. Fixed points by implicit conditions The following list of implicit functions under various conditions (26). Let Ξ© be the set of all real continuous functions 𝑀:𝑅+ 6 β†’ 𝑅, satisfying the following conditions: 𝑀1) is non-increasing in variables π‘Ÿ2, π‘Ÿ3, π‘Ÿ4, π‘Ÿ5, π‘Ÿ6 𝑀2) there exists a right continuous function 𝐴: [0,∞) β†’ [0,∞), 𝐴(0) = 0,𝐴(π‘Ÿ) < π‘Ÿ, for π‘Ÿ > 0 such that for 𝑐 β‰₯ 0 𝑀(𝑐, 𝑒, 𝑐, 𝑒, 0, 𝑐 + 𝑒) ≀ 0 or 𝑀(𝑐, 𝑒, 0,0, 𝑒, 𝑒) ≀ 0, implies 𝑐 ≀ 𝐴(𝑒). 𝑀3) 𝑀(𝑐, 0, 𝑐, 0,0, 𝑐) > 0 and 𝑀(𝑐, 𝑐, 0,0, 𝑐, 𝑐) > 0 βˆ€ 𝑐 > 0. Also, let 𝛹 ≔ the set of functions πœ“: [0,∞) β†’ [0,∞) such that: i) πœ“ is monotone increasing and continuous; ii) πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0; iii) πœ“ is subadditive, i.e., βˆ€π‘Ÿ1, π‘Ÿ2 ∈ [0,+∞),πœ“(π‘Ÿ1 + π‘Ÿ2) = πœ“(π‘Ÿ1) + πœ“(π‘Ÿ2). Lemma 2.8: Let 𝐴: [0,∞) β†’ [0,∞) be a right continuous function such that 𝐴(π‘Ÿ) < π‘Ÿ, for π‘Ÿ > 0. (27( . Then lim π‘šβ†’βˆž π΄π‘š(π‘Ÿ) = 0, where π΄π‘šβ‰”π‘š times repeated composition of 𝐴. Theorem 2.9: If 𝑀 ∈ 𝛺 exists and Β£ = ⋃ £𝑖 𝑛 𝑖=1 is a cyclic representation of Β£ w.r.t., 𝑓: Β£ β†’ Β£. If for any (π‘Ž, 𝑏) ∈ £𝑖 Γ— £𝑖+1, 𝑖 = 1,2, … , 𝑛 𝑀(πœ“(𝜌(π‘“π‘Ž, 𝑓𝑏)), πœ“(𝜌(π‘Ž, 𝑏)), πœ“(𝜌(π‘Ž, π‘“π‘Ž)), πœ“(𝜌(𝑏, 𝑓𝑏)), πœ“(𝜌(π‘Ž, 𝑓𝑏)), πœ“(𝜌(𝑏, π‘“π‘Ž))) ≀ 0 (18), and πœ“ ∈ 𝛹. βˆƒ! 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 , 𝑐 is a unique fixed point. Moreover, π‘™π‘–π‘š π‘šβ†’βˆž π‘“π‘š(π‘Ž) = 𝑐, for any π‘Ž ∈ Β£. Proof: Let π‘Ž0 ∈ ⋃ £𝑖 𝑛 𝑖=1 and π‘Žπ‘š define by π‘Žπ‘š+1 = π‘“π‘Žπ‘š. So for π‘š β‰₯ 0, βˆƒπ‘–π‘š ∈ {1,2, … 𝑛} such that π‘Žπ‘šβˆ’1 ∈ Β£π‘–π‘š and π‘Žπ‘š ∈ Β£π‘–π‘š+1. If π‘Žπ‘š0 = π‘Žπ‘š0βˆ’1for some π‘š0, π‘Žπ‘š0 = π‘“π‘Žπ‘š0βˆ’1 = π‘Žπ‘š0βˆ’1 then π‘Žπ‘š0 is fixed point of 𝑓. Thus, suppose that π‘Žπ‘š β‰  π‘Žπ‘šβˆ’1, for all π‘š ∈ 𝑁⋃{0}. By using (2.18) therefore 𝑀(πœ“(𝜌(π‘Žπ‘š+1, π‘Žπ‘š)), πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘šβˆ’1)), πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘š+1)), πœ“(𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š)), πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘š)), πœ“(𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š+1))) ≀ 0. And since πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0, also by using triangle inequality and since πœ“ is subadditive, therefore 𝑀(πœ“(𝜌(π‘Žπ‘š+1, π‘Žπ‘š)), πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘šβˆ’1)), πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘š+1)), πœ“(𝜌(π‘Žπ‘šβˆ’1, π‘Žπ‘š)),0, πœ“(π‘žπœŒ(π‘Žπ‘šβˆ’1, π‘Žπ‘š)) + πœ“(π‘žπœŒ(π‘Žπ‘š, π‘Žπ‘š+1))) ≀ 0. And from 𝑀2, there exists a right continuous function A: [0,∞) β†’ [0,∞), A(0) = 0, A(π‘Ÿ) < π‘Ÿ, for π‘Ÿ > 0, such that for all π‘š ∈ 𝑁⋃{0} πœ“(𝜌(π‘Žπ‘š+1, π‘Žπ‘š) ≀ 𝐴( πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘šβˆ’1))). If we use this procedure, we get πœ“(𝜌(π‘Žπ‘š+1, π‘Žπ‘š) ≀ 𝐴( πœ“(𝜌(π‘Žπ‘š, π‘Žπ‘šβˆ’1))) ≀ β‹― ≀ Aπ‘š (πœ“(𝜌(π‘Ž1, π‘Ž0)). (19) And by (Lemma 8) and continuity of πœ“, subsequently π‘™π‘–π‘š π‘šβ†’βˆž πœ“( 𝜌(π‘Žπ‘š+1, π‘Žπ‘š)) = 0 = πœ“( π‘™π‘–π‘š π‘šβ†’βˆž 𝜌(π‘Žπ‘š+1, π‘Žπ‘š)). Since πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0, therefore π‘™π‘–π‘š π‘šβ†’βˆž 𝜌(π‘Žπ‘š+1, π‘Žπ‘š) = 0. (20) To prove for each π‘Ž0 ∈ Β£, (π‘Žπ‘š) is a Cauchy sequence. Assume it is false. Then we can find a πœ€ > 0 and {π‘π‘Ÿ}, {π‘‘π‘Ÿ}, π‘‘π‘Ÿ > π‘π‘Ÿ β‰₯ π‘Ÿ where {π‘π‘Ÿ}, {π‘‘π‘Ÿ} two subsequences of integers with πœ“(𝜌(π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) β‰₯ πœ€ for 𝑛 ∈ {1,2, … }. (21) We also assume IHJPAS. 2025,38(2) 384 πœ“(𝜌(π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿβˆ’1)) < πœ€. (22( By selecting π‘‘π‘Ÿ to be the least number surpassing π‘π‘Ÿ for which inequality (21) holds, now by (19) and (21), (22), and since πœ“ is subadditive, getting πœ€ ≀ πœ“(𝜌(π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿβˆ’1)) ≀ πœ“(π‘žπœŒ(π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿβˆ’1)) + πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿβˆ’1, π‘Žπ‘π‘Ÿ)) ≀ π‘žπœ€ + π΄π‘‘π‘Ÿβˆ’1πœ“(π‘žπœŒ(π‘Ž0, π‘Ž1)). (23) And so lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘‘π‘Ÿβˆ’1, π‘Žπ‘π‘Ÿ)) = π‘žπœ€. (24) On the other hand, βˆ€π‘Ÿ, βˆƒπ‘–π‘Ÿ ∈ {1,2, … , 𝑛} such that π‘‘π‘Ÿ βˆ’ π‘π‘Ÿ + π‘–π‘Ÿ ≑ 1(mod 𝑛). Then π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ(for π‘Ÿ large enough, π‘π‘Ÿ > π‘–π‘Ÿ) and π‘Žπ‘‘π‘Ÿbelong to different sets £𝑖 and £𝑖+1 for 𝑖 ∈ {1,2, … , 𝑛}. By the triangle inequality, also πœ“ is subadditive, obtaining πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿ)) ≀ πœ“(π‘žπœŒ( π‘Žπ‘‘π‘Ÿ , π‘Žπ‘‘π‘Ÿβˆ’π‘–π‘Ÿ) + π‘žπœŒ(π‘Žπ‘‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘π‘Ÿ)) ≀ πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘‘π‘Ÿβˆ’π‘–π‘Ÿ)) + πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘π‘Ÿ)). Now taking π‘Ÿ β†’ ∞, by (19) and from (24), as a results πœ“(𝜌(π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) ≀ 𝐴 π‘‘π‘Ÿβˆ’π‘–π‘Ÿπœ“(π‘žπœŒ(π‘Ž0, π‘Ž1)) + π‘žπœ€. And so, lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿ)) = π‘žπœ€. (25) By using (20), so Lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘‘π‘Ÿ+1, π‘Žπ‘‘π‘Ÿ)) = 0, Lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1, π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ)) = 0. (26) And by using the triangle inequality, then πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) ≀ πœ“(π‘žπœŒ(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘π‘Ÿ)) + πœ“(π‘žπœŒ( π‘Žπ‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ)). Letting π‘Ÿ β†’ ∞ in the last inequality and using (2.19) and (2.25), consequently πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) ≀ π΄π‘π‘Ÿβˆ’π‘–π‘Ÿπœ“(π‘žπœŒ(π‘Ž0, π‘Ž1)) + π‘žπœ€ lim π‘Ÿβ†’βˆž 𝜌 (𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) = π‘žπœ€. (27) Again, by using the triangle inequality, obtaining πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘žπ‘Ÿ+1)) ≀ πœ“(π‘žπœŒ( π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ)) + πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ+1)). Letting π‘Ÿ β†’ ∞ in the last inequality and using (26) and (27), therefore lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ+1)) = π‘žπœ€. (28) And in the same way, as a results πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1)) ≀ πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ)) + πœ“(π‘žπœŒ(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1)). Letting π‘Ÿ β†’ ∞ and using (27) and (26), getting lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1)) = π‘žπœ€. (29) Again, by using the triangle inequality, therefore πœ“(𝜌( π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1, π‘Žπ‘‘π‘Ÿ+1)) ≀ πœ“(π‘žπœŒ( π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1, π‘Žπ‘‘π‘Ÿ)) + πœ“(π‘žπœŒ(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ+1)). Letting π‘Ÿ β†’ ∞ in the last inequality and using (2.26) and (2.29), having Lim π‘Ÿβ†’βˆž πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1, π‘Žπ‘‘π‘Ÿ+1)) = π‘žπœ€. (30) Using (2.18) for π‘Ž = π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ and 𝑏 = π‘Žπ‘‘π‘Ÿ , subsequently 𝑀(πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1, π‘Žπ‘‘π‘Ÿ+1)), πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘‘π‘Ÿ)), πœ“(𝜌(π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1)), IHJPAS. 2025,38(2) 385 πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘‘π‘Ÿ+1)), πœ“(𝜌(π‘Žπ‘‘π‘Ÿ+1, π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ)), πœ“(𝜌(π‘Žπ‘‘π‘Ÿ , π‘Žπ‘π‘Ÿβˆ’π‘–π‘Ÿ+1))) ≀ 0. Letting π‘Ÿ β†’ ∞ , and using (30), (27), (26), (28), (29), then, by continuity of 𝑀 and πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0, 𝑀(π‘žπœ€, π‘žπœ€, 0,0, π‘žπœ€, π‘žπœ€) ≀ 0, a contradiction with 𝑀3. Thus, (π‘Žπ‘š) is Cauchy sequence in (Β£, 𝜌). Now to prove that 𝑐 is fixed point of 𝑓. In fact π‘“π‘Žπ‘š β†’ 𝑐 and since Β£ = ⋃ £𝑖 𝑛 𝑖=1 is cyclic representation of Β£ w.r.t., 𝑓, the sequence (π‘Žπ‘š) has infinite terms in each Β£π‘–π‘šfor π‘–π‘š ∈ {1,2, … , 𝑛}. Considering that Β£π‘–π‘š is closed for π‘–π‘š ∈ {1,2, … , 𝑛} we have 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Suppose that 𝑐 ∈ £𝑖 and 𝑓𝑐 ∈ £𝑖+1, and take a subsequence (π‘Žπ‘šπ‘Ÿ ) π‘Ÿβˆˆπ‘ of (π‘Žπ‘š) with π‘Žπ‘šπ‘Ÿ ∈ Β£π‘–βˆ’1, using (2.18), take π‘Ž = 𝑐 and 𝑏 = π‘Žπ‘šπ‘Ÿ , as result 𝑀(πœ“(𝜌(𝑓𝑐, π‘“π‘Žπ‘šπ‘Ÿ )), πœ“(𝜌(𝑐, π‘Žπ‘šπ‘Ÿ )), πœ“(𝜌(𝑐, 𝑓𝑐)), πœ“(𝜌(π‘Žπ‘šπ‘Ÿ , π‘“π‘Žπ‘šπ‘Ÿ )), πœ“(𝜌(𝑐, π‘“π‘Žπ‘šπ‘Ÿ )), πœ“(𝜌(π‘Žπ‘šπ‘Ÿ , 𝑓𝑐))) ≀ 0. Taking π‘Ÿ β†’ ∞, hence 𝑀(πœ“(𝜌(𝑓𝑐, 𝑐)), πœ“(𝜌(𝑐, 𝑐)), πœ“(𝜌(𝑐, 𝑓𝑐)), πœ“(𝜌(𝑐, 𝑐)), πœ“(𝜌(𝑐, 𝑐)), πœ“(𝜌(𝑐, 𝑓𝑐))) ≀ 0, and πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0, then 𝑀(πœ“(𝜌(𝑓𝑐, 𝑐)),0, πœ“(𝜌(𝑐, 𝑓𝑐)), 0,0, πœ“(𝜌(𝑐, 𝑓𝑐))) ≀ 0, which is contradiction to 𝑀3. Thus πœ“(𝜌(𝑓𝑐, 𝑐)) = 0 then 𝜌(𝑓𝑐, 𝑐) = 0, then 𝑓𝑐 = 𝑐. We obtain 𝑐 is a fixed point of 𝑓. Suppose that there are two distinct points 𝑐, 𝑀 with 𝑐 and 𝑀 are two fixed points of 𝑓. The cyclic nature of 𝑓, also, the fact 𝑐, 𝑀 ∈ Β£ = ⋃ £𝑖 𝑛 𝑖=1 are fixed points of 𝑓 imply that 𝑐, 𝑀 ∈ β‹‚ £𝑖 𝑛 𝑖=1 and by (18), obtaining 𝑀(πœ“(𝜌(𝑓𝑐, 𝑓𝑀)),πœ“(𝜌(𝑐, 𝑀)), πœ“(𝜌(𝑐, 𝑓𝑐)), πœ“(𝜌(𝑀, 𝑓𝑀)),πœ“(𝜌(𝑐, 𝑓𝑀)),πœ“(𝜌(𝑀, 𝑓𝑐))) ≀ 0. Since 𝑐, 𝑀 are fixed points of 𝑓, πœ“(π‘Ÿ) = 0 if and only if π‘Ÿ = 0, we get 𝑀(πœ“(𝜌(𝑐, 𝑀)), πœ“(𝜌(𝑐, 𝑀)),0,0, πœ“(𝜌(𝑐, 𝑓𝑀)),πœ“(𝜌(𝑀, 𝑐))) ≀ 0, which is a contradiction to 𝑀3, then πœ“(𝜌(𝑐, 𝑀)) = 0, hence 𝜌(𝑐, 𝑀) = 0, that is, 𝑐 = 𝑀. Hence 𝑓 has a unique fixed point in Β£ and 𝑐 ∈ β‹‚ £𝑖 𝑛 𝑖=1 . Remark 2.10: It is worth noting, it is worth noting that we can obtain good results by including the concept of cyclicity in cases of )28,29(. 3. Discussion This work is classified within the field depending on the classification 2010 MSC: 47H09, 47H10. Our study of this topic is the first in Iraq (to the best of our knowledge) in the field of fixed points for cyclic maps, and it is taken from a master’s thesis by researcher Abbas Karim Nahi. The paper included new results in the field of integral contractions in the b-metric spaces of the cyclic type, as well as new generalizations of the results of other researchers in the case of the b-metric space. In the future, we would like to study the results in )30( in the case of cyclic maps. 4. Conclusions In this paper, new theorems were established to find fixed points. 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