Int. J. Anal. Appl. (2023), 21:18 On Certain Fixed Point Theorems in Sb-Metric Spaces With Applications N. Mangapathi1,3,∗, B. Srinuvasa Rao2, K.R.K. Rao3, M.I. Pasha1,3 1Department of Mathematics, B V Raju Institute of Technology, Narsapur, Medak-502313, Telangana, India 2Department of Mathematics, Dr.B.R.Ambedkar University, Srikakulam, Etcherla-532410, Andhra Pradesh, India 3Department of Mathematics, Gitam School of Science, GITAM Deemed to be University, Hyderabad, Rudraram-502329, Telangana, India ∗Corresponding author: nmp.maths@gmail.com Abstract. In this paper, we introduce the notion of generalized (α,φ,ψ)- Geraghty contractive type mappings in the setup of Sb-metric spaces and α-orbital admissible mappings with respect to φ. Furthermore, the fixed-point theorems for such mappings in complete Sb-metric spaces are proven without assuming the subadditivity of ψ. Some examples are provided for supporting of our main results. Also, we gave an application to integral equations as well as Homotopy. 1. Introduction The Banach contraction principle [1] is one of the most significant findings in fixed point theory since it has applications in many areas of mathematics and mathematical sciences. By combining the ideas of S and b-metric spaces, Sedghi et al. [2] created Sb-metric spaces and established common fixed point outcomes in Sb- metric spaces. In order to improve, numerous authors developed numerous findings in Sb-metric spaces (see e.g. [3], [4], [5], [6], [7], [8]). One of the more intriguing findings is Geraghty’s [9] generalisation of the Banach contraction theorem. Multiple researchers have since studied this type of research in different metric spaces (see e.g [10], [11], [12], [13], [14], [15], [16], [17], [18]). Received: Dec. 9, 2022. 2020 Mathematics Subject Classification. 54H25, 47H10, 54E50. Key words and phrases. (α,φ,ψ)-Geraghty type contraction mappings; Sb-metric spaces; α-orbital admissible map- pings with respect to φ and completeness. https://doi.org/10.28924/2291-8639-21-2023-18 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-18 2 Int. J. Anal. Appl. (2023), 21:18 Triangular α-admissible mappings are a novel idea that Karapinar et al. [15] introduced to study fixed points for such mappings in metric spaces. Three new concepts triangular α-orbital admissible, α -orbital admissible and α-orbital attractive mappings were developed by Popescu [17] in 2014. The idea of triangular α-orbital admissible mappings with respect to η was first suggested in 2016 by Chuadchawna et al. [19]. The concept of generalised α−η−ψ-Geraghty contractive type mappings and α-orbital attractive mappings with regard to is introduced by Farajzadeh et al. in 2018 [20] in the framework of partial b-metric spaces, which will be effectively applied for establishing our key findings. In the context of Sb-metric spaces and α-orbital admissible mappings with regard to φ, this paper aims to demonstrate unique fixed point theorems for generalised (α,φ,ψ)-Geraghty contractive type mapping. Additionally, we may provide relevant applications for homotopy, integral equations, and appropriate examples. We first review some fundamental findings. 2. Preliminaries Definition 2.1. ( [2]) Let P be a non-empty set and b ≥ 1 be given real number. Suppose that a mapping Sb : P3 → [0,∞) be a function satisfying the following properties : (Sb1) 0 < Sb(σ,ς,τ) for all σ,ς,τ ∈ P with σ 6= ς 6= τ 6= σ, (Sb2) Sb(σ,ς,τ) = 0 ⇔ σ = ς = τ, (Sb3) Sb(σ,ς,τ) ≤ b(Sb(σ,σ,a) + Sb(ς,ς,a) + Sb(τ,τ,a)) for all σ,ς,τ,a ∈ P. The function Sb is then referred to as a Sb-metric on P, and the pair (P,Sb) is referred to as a Sb-metric space. Remark 2.1. ( [2]) It should be noted that the Sb-metric space class is effectively larger than the S-metric space class. Each S-metric space is, in fact, a Sb-metric space with b = 1. Example 2.1. ( [2]) Let (P,S) be S-metric space and S∗(σ,ς,τ) = S(σ,ς,τ)p, where p > 1 is a real number.Then obviously, S∗ is a Sb-metric with b = 22(p−1) but (P,S∗) is not necessarily a S-metric space. Definition 2.2. ( [2]) Let (P,Sb) be a Sb-metric space. Then, for σ ∈ P, λ > 0 we defined the open ball BSb (σ,λ) and closed ball BSb [σ,λ] with center σ and radius λ as follows respectively: BSb (σ,λ) = {ς ∈ P : Sb(ς,ς,σ) < λ} and BSb [σ,λ] = {ς ∈ P : Sb(ς,ς,σ) ≤ λ}. Lemma 2.1. ( [2]) In the Sb-metric space, we have Sb(µ,µ,ν) ≤ 2bSb(µ,µ,ξ) + b2Sb(ξ,ξ,ν). Definition 2.3. ( [2]) Let {σn} be a sequence in Sb-metric space (P,Sb) is said to be: (1) Sb-Cauchy sequence if, for each � > 0, there exists n0 ∈N such that Sb(σn,σn,σm) < � for each m,n ≥ n0. Int. J. Anal. Appl. (2023), 21:18 3 (2) Sb-convergent to a point σ ∈ P if, for each � > 0, there exists a positive integer n0 such that Sb(σn,σn,σ) < � or Sb(σ,,σ,σn) < � for all n ≥ n0 and we denote by lim n→∞ σn = σ. (3) (P,Sb) is Sb-complete if every Sb-Cauchy sequence is Sb-convergent in P. Lemma 2.2. ( [2]) If (P,Sb) be a Sb-metric space with b ≥ 1 and suppose that {σn} is a Sb-convergent to σ, then we have (i) 1 2b Sb(ς,ς,σ) ≤ lim n→∞ inf Sb(ς,ς,σn) ≤ lim n→∞ sup Sb(ς,ς,σn) ≤ 2bSb(ς,ς,σ) (ii) 1 b2 Sb(σ,σ,ς) ≤ lim n→∞ inf Sb(σn,σn,ς) ≤ lim n→∞ sup Sb(σn,σn,ς) ≤ b2Sb(σ,σ,ς) for all ς ∈ P. In particular, if σ = ς, then we have lim n→∞ Sb(σn,σn,ς) = 0. We should always consider the following factors in order to obtain our results. 3. Main Results We say F be the class of all functions β : [0,∞) → [0, 1) satisfying the following condition: lim n→∞ β(tn) = 1 implies lim n→∞ tn = 0 Definition 3.1. Let G : P → P be a self mapping defined on non-empty set P and α,φ : P × P × P → R+ be two functions. We say that G is an α- admissible mapping with respect to φ, if α(σ,σ,ς) ≥ φ(σ,σ,ς) implies α(Gσ,Gσ,Gς) ≥ φ(Gσ,Gσ,Gς) for all σ,ς ∈ P. We say that G is an α- admissible mapping if for all σ,ς ∈ P, α(σ,σ,ς) ≥ 1 implies α(Gσ,Gσ,Gς) ≥ 1. Definition 3.2. Let P be a non-empty set. G : P → P be a self mapping and α : P×P×P →R+. We say that G is a triangular α- admissible mapping, if (a) G is an α- admissible mapping; (b) α(σ,σ,ς) ≥ 1 and α(ς,ς,τ) ≥ 1 implies α(σ,σ,τ) ≥ 1 for all σ,ς,τ ∈ P. Definition 3.3. Let P be a non-empty set. G : P → P be a self mapping and α,φ : P×P×P →R+ be two functions. We say that G is an α- orbital admissible mapping with respect to φ, if α(σ,σ,Gσ) ≥ φ(σ,σ,Gσ) implies α(Gσ,Gσ,G2σ) ≥ φ(Gσ,Gσ,G2σ) for all σ ∈ P. Definition 3.4. Let G : P → P be a self mapping defined on nonempty set P and α,φ : P × P × P →R+. We say that G is a triangular α-orbital admissible mapping with respect to φ, if (a) G is an α-orbital admissible mapping with respect to φ; (b) α(σ,σ,ς) ≥ φ(σ,σ,ς) and α(ς,ς,Gς) ≥ φ(ς,ς,Gς) imply α(σ,σ,Gς) ≥ φ(σ,σ,Gς) for all σ,ς ∈ P. 4 Int. J. Anal. Appl. (2023), 21:18 Definition 3.5. Let Ω = {Γ/Γ : [0,∞) → [0,∞)} be a family of functions that satisfy the following properties; (i) Γ is a continuously nondecreasing map; (ii) Γ(t) = 0 if and only if t = 0; (iii) Γ(t) is subadditive, Γ(p + q) ≤ Γ(p) + Γ(q). Definition 3.6. Let (P,Sb) be an Sb-metric space, a mapping G : P → P is said to be a generalized (α,φ, Γ)-Geraghty contractive type mapping if there exist Γ ∈ Ω, α,φ : P × P × P → [0,∞) and β ∈ F such that α(σ,σ,ς) ≥ φ(σ,σ,ς) implies Γ ( ( 1 + b3 2 )Sb(Gσ,Gσ,Gς) ) ≤ β ( Γ(MG b (σ,σ,ς)) ) Γ(MG b (σ,σ,ς)) (3.1) where MG b (σ,σ,ς) = max { Sb (σ,σ,ς) ,Sb (σ,σ,Gσ) , Sb (ς,ς,Gς) , Sb(σ,σ,Gς)+Sb(ς,ς,Gσ) 4b3 } ∀ σ,ς ∈ P Lemma 3.1. Let G : P → P be a triangular α-orbital admissible mapping with respect to φ. Assume that there exists σ1 ∈ P such that α(σ1,Gσ1) ≥ φ(σ1,Gσ1). Define a sequence {σn} by σn+1 = Gσn. Then we have α(σn,σm) ≥ φ(σn,σm) for all m,n ∈ N with n < m. Proof. Since α(σ1,σ1,Gσ1) ≥ φ(σ1,σ1,Gσ1) and G is α-orbital admissible with respect to φ, we obtain that α(σ2,σ2,σ3) = α(Gσ1,Gσ1,G(Gσ1)) ≥ φ(Gσ1,Gσ1,G(Gσ1)) = φ(σ2,σ2,σ3). By continuing the process as above, we have α(σn,σn,σn+1) ≥ Γ(σn,σn,σn+1) for all n ∈ N. Suppose that α(σn,σn,σm) ≥ φ(σn,σn,σm) (3.2) and we will prove that α(σn,σn,σm+1) ≥ φ(σn,σn,σm+1), where m > n. Since α(σm,σm,σm+1) ≥ φ(σm,σm,σm+1), we obtain that α(σm,σm,Gσm) = α(σm,σm,σm+1) ≥ φ(σm,σm,σm+1) = φ(σm,σm,Gσm). (3.3) By (3.2), (3.3) and triangular α-orbital admissibility of G, we have α(σn,σn,Gσm) ≥ φ(σn,σn,Gσm). This implies that α(σn,σn,σm+1) ≥ φ(σn,σn,σm+1). Hence, α(σn,σm) ≥ φ(σn,σm) for all m,n ∈ N with n < m. � Theorem 3.1. Let (P,Sb) be a complete Sb-metric space with coefficient b ≥ 1. Let G : P → P be an be a generalized (α,φ, Γ)-Geraghty contractive type mapping. Assume the following conditions are hold: (i) G is a triangular α-orbital admissible mapping w.r.t φ; (ii) there exists σ1 ∈ P such that α(σ1,σ1,Gσ1) ≥ φ(σ1,σ1,Gσ1); Int. J. Anal. Appl. (2023), 21:18 5 (iii) if {σn} is a Sb-convergent sequence to ν in P and α(σn,σn,σn+1) ≥ φ(σn,σn,σn+1) for each n ∈ N then α(ν,ν,ν) ≥ φ(ν,ν,ν); (iv) G is continuous. Then G has a fixed point. Proof. Let σ1 ∈ P such that α(σ1,σ1,Gσ1) ≥ φ(σ1,σ1,Gσ1). Define the sequence {σn} in P by σn+1 = Tσn for all n ∈ N. By Lemma(2.1), we get that α(σn,σn,σn+1) ≥ φ(σn,σn,σn+1) for all n ∈ N. (3.4) If σn = σn+1 for some n ∈ N, then σn is a fixed point of G. Assume that σn 6= σn+1 for all n ∈ N. The sequence {Sb(σn,σn,σn+1)} is first shown to be non-increasing and to trend to 0 as n →∞. By using (3.4), for each n ∈ N, we have Γ ( ( 1 + b3 2 )Sb(σn+1,σn+1,σn+2) ) = Γ ( ( 1 + b3 2 )Sb(Gσn,Gσn,Gσn+1) ) ≤ β ( Γ(MG b (σn,σn,σn+1)) ) Γ(MG b (σn,σn,σn+1)) < Γ(MG b (σn,σn,σn+1)) (3.5) where, MG b (σn,σn,σn+1) = max { Sb (σn,σn,σn+1) ,Sb (σn,σn,Gσn) , Sb (σn+1,σn+1,Gσn+1) , Sb(σn,σn,Gσn+1)+Sb(σn+1,σn+1,Gσn) 4b3 } = max { Sb (σn,σn,σn+1) ,Sb (σn,σn,σn+1) , Sb (σn+1,σn+1,σn+2) , Sb(σn,σn,σn+2)+Sb(σn+1,σn+1,σn+1) 4b3 } = max { Sb (σn,σn,σn+1) ,Sb (σn,σn,σn+1) , Sb (σn+1,σn+1,σn+2) , 2bSb(σn,σn,σn+1)+b 2Sb(σn+1,σn+1,σn+2) 4b3 } = max { Sb (σn,σn,σn+1) ,Sb (σn+1,σn+1,σn+2) } . If max { Sb (σn,σn,σn+1) ,Sb (σn+1,σn+1,σn+2) } = Sb (σn+1,σn+1,σn+2) then Γ ( ( 1+b 3 2 )Sb(σn+1,σn+1,σn+2) ) < Γ(Sb (σn+1,σn+1,σn+2)) implies that ( 1+b 3 2 )Sb(σn+1,σn+1,σn+2) < Sb (σn+1,σn+1,σn+2) which is contradiction. Hance, max { Sb (σn,σn,σn+1) ,Sb (σn+1,σn+1,σn+2) } = Sb (σn,σn,σn+1). It follows that 0 < Sb(σn+1,σn+1,σn+2) ≤ Sb(σn,σn,σn+1). Hence the sequence {Sb(σn,σn,σn+1)} is nonnegative non-increasing and bounded below. Thus there exist some ξ ≥ 0 such that lim n→∞ Sb (σn,σn,σn+1) = ξ. Suppose that ξ > 0. 6 Int. J. Anal. Appl. (2023), 21:18 By using (3.5), we have Γ (Sb(σn+1,σn+1,σn+2)) Γ (Sb(σn,σn,σn+1)) ≤ Γ ( ( 1+b 3 2 )Sb(σn+1,σn+1,σn+2) ) Γ (Sb(σn,σn,σn+1)) ≤ β ( Γ(MG b (σn,σn,σn+1)) ) < 1, for all n ∈ N. On letting n →∞ in above inequality, we have lim n→∞ β ( Γ(MG b (σn,σn,σn+1)) ) = 1. Since β ∈ F, we have lim n→∞ Γ(MG b (σn,σn,σn+1)) = 0 and so ξ = lim n→∞ Sb (σn,σn,σn+1) = 0. We now demonstrate that the (P,Sb) Cauchy sequence is {σn}. On the other hand, we assume that {σn} is not Cauchy. In this case, a monotonically rising sequence of the natural numbers {mk} and {nk} exists, where nk > mk. Sb ( σmk,σmk,σnk ) ≥ � (3.6) and Sb ( σmk,σmk,σnk−1 ) < �. (3.7) From (3.6) and (3.7), we have � ≤ Sb ( σmk,σmk,σnk ) ≤ 2bSb ( σmk,σmk,σmk +1 ) + b2Sb ( σmk +1,σmk +1,σnk ) . So that ( 1+b 3 2b2 )� ≤ ( 1+b 3 b )Sb ( σmk,σmk,σmk +1 ) + ( 1+b 3 2 )Sb ( σmk +1,σmk +1,σnk ) . We obtain that by applying Γ on both sides and letting k →∞. Γ(( 1 + b3 2b2 )�) ≤ lim k→∞ Γ ( ( 1 + b3 2 )Sb ( σmk +1,σmk +1,σnk )) ≤ lim k→∞ Γ ( ( 1 + b3 2 )Sb ( Gσmk,Gσmk,Gσnk−1 )) . (3.8) By applying the triangular inequality, we get that Sb ( σmk +1,σmk +1,σnk ) ≤ 2bSb ( σmk +1,σmk +1,σmk ) + b2Sb ( σmk,σmk,σnk ) . ≤ 2bSb ( σmk +1,σmk +1,σmk ) + 2b3Sb ( σmk,σmk,σnk−1 ) +b3Sb ( σnk,σnk,σnk−1 ) ≤ 2bSb ( σmk +1,σmk +1,σmk ) + 2b3� + b3Sb ( σnk,σnk,σnk−1 ) . In the above inequality, we get that by taking the limit as k →∞. lim k→∞ Sb ( σmk +1,σmk +1,σnk ) ≤ 2b3�. (3.9) Int. J. Anal. Appl. (2023), 21:18 7 We obtain (3.5) because G is a triangular α-orbital admissible mapping with respect to φ. and α(σmk,σmk,σnk−1) ≥ φ(σmk,σmk,σnk−1). By using (3.1), we have Γ ( ( 1 + b3 2 )Sb(σmk +1,σmk +1,σnk ) ) ≤ β ( Γ(MGb (σmk,σmk,σnk−1)) ) Γ(MGb (σmk,σmk,σnk−1)) (3.10) where, MG b (σmk,σmk,σnk−1) = max   Sb ( σmk,σmk,σnk−1 ) ,Sb ( σmk,σmk,Gσmk ) , Sb ( σnk−1,σnk−1,Gσnk−1 ) , Sb(σmk ,σmk ,Gσnk−1)+Sb(σnk−1,σnk−1,Gσmk ) 4b3   ≤ max   Sb ( σmk,σmk,σnk−1 ) ,Sb ( σmk,σmk,σmk +1 ) ,Sb ( σnk−1,σnk−1,σnk ) , 2bSb(σmk ,σmk ,σnk−1)+b 2Sb(σnk−1,σnk−1,σnk )+Sb(σnk−1,σnk−1,σmk +1) 4b3   ≤ max   Sb ( σmk,σmk,σnk−1 ) ,Sb ( σmk,σmk,σmk +1 ) ,Sb ( σnk−1,σnk−1,σnk ) , 2bSb(σmk ,σmk ,σnk−1)+b 2Sb(σnk−1,σnk−1,σnk ) 4b3 + 2bSb(σnk−1,σnk−1,σnk )+bSb(σmk +1,σmk +1,σnk ) 4b3   Using (3.7) and (3.9) and treating the limit of the inequality above as k →∞, this results. lim k→∞ MG b (σmk,σmk,σnk−1) ≤ lim k→∞ max   Sb ( σmk,σmk,σnk−1 ) ,Sb ( σmk,σmk,σmk +1 ) ,Sb ( σnk−1,σnk−1,σnk ) , 2bSb(σmk ,σmk ,σnk−1)+b 2Sb(σnk−1,σnk−1,σnk ) 4b3 + 2bSb(σnk−1,σnk−1,σnk )+bSb(σmk +1,σmk +1,σnk ) 4b3   ≤ max { �, 0, 0,�( 1+b 3 2b2 ) } = �( 1 + b3 2b2 ) (3.11) By taking the limit in (3.10) as k →∞ and using (3.8) and (3.11), we have Γ(( 1 + b3 2b2 )�) ≤ Γ ( lim k→∞ ( 1 + b3 2 )Sb ( σmk +1,σmk +1,σnk )) . ≤ β ( Γ( lim k→∞ MG b (σmk,σmk,σnk−1)) ) Γ( lim k→∞ MG b (σmk,σmk,σnk−1)) ≤ β ( Γ( lim k→∞ MG b (σmk,σmk,σnk−1)) ) Γ(�( 1 + b3 2b2 )) This implies that Γ(( 1+b 3 2b2 )�) Γ(( 1+b 3 2b2 )�) ≤ β ( Γ ( lim k→∞ MG b (σmk,σmk,σnk−1) )) . Since β ∈ F, we have lim n→∞ β ( Γ( lim k→∞ MG b (σmk,σmk,σnk−1)) ) = 1. 8 Int. J. Anal. Appl. (2023), 21:18 It follows that Γ( lim k→∞ MG b (σmk,σmk,σnk−1)) = 0. By using (3.10) we obtain lim n→∞ Sb(σmk +1,σmk +1,σnk ) = 0. which contradicts to(3.9).In the Sb-metric space (P,Sb), the sequence {σn} is a Sb-Cauchy sequence. The sequence {σn} → ν ∈ (P,Sb) emerges from the completeness of (P,Sb). We begin by presuming that G is continuous. Therefore, we have ν = lim n→∞ σn+1 = lim n→∞ Gσn = G lim n→∞ σn = Gν. Since {σn} is a Sb-convergent sequence to ν in P and α(ν,ν,ν) ≥ φ(ν,ν,ν). Then to prove ν = Gν. Suppose that ν 6= Gν. From (3.1), we have Γ (Sb(ν,ν,Gν)) ≤ Γ ( ( 1 + b3 2 )Sb(ν,ν,Gν) ) = Γ ( ( 1 + b3 2 ) lim n→∞ Sb(σn+1,σn+1,Gσn+1) ) = Γ ( ( 1 + b3 2 ) lim n→∞ Sb(Gσn,Gσn,Gσn+1) ) ≤ β ( Γ( lim n→∞ MG b (σn,σn,σn+1)) ) Γ( lim n→∞ MG b (σn,σn,σn+1)) (3.12) where lim n→∞ MG b (σn,σn,σn+1) = lim n→∞ max   Sb (σn,σn,σn+1) ,Sb (σn,σn,Gσn) , Sb (σn+1,σn+1,Gσn+1) , Sb(σn,σn,Gσn+1)+Sb(σn+1,σn+1,Gσn) 4b3   = max { Sb (ν,ν,ν) ,Sb (ν,ν,Gν) , Sb (ν,ν,Gν) , Sb(ν,ν,Gν) 2b3 } = Sb (ν,ν,Gν) . By taking limit as n →∞ in (3.12), we have Γ (Sb(ν,ν,Gν)) ≤ β ( Γ( lim n→∞ MG b (σn,σn,σn+1)) ) Γ( lim n→∞ MG b (σn,σn,σn+1)) ≤ β (Γ(Sb (ν,ν,Gν))) Γ(Sb (ν,ν,Gν)) we can deduce that Γ(Sb(ν,ν,Gν)) Γ(Sb(ν,ν,Gν)) ≤ β (Γ(Sb (ν,ν,Gν))) We obtain that lim n→∞ β (Γ(Sb (ν,ν,Gν))) = 1. Therefore, Sb (ν,ν,Gν) = 0 implies Gν = ν. and thus ν is a fixed point of G. Assume further that ν and ν∗ are two fixed points of G such that ν 6= ν∗. Consider Γ (Sb(ν,ν,ν ∗)) ≤ Γ ( ( 1 + b3 2 )Sb(Gν,Gν,Gν∗) ) ≤ β ( Γ(MG b (ν,ν,ν∗)) ) Γ(MG b (ν,ν,ν∗)) (3.13) where MG b (ν,ν,ν∗) = max { Sb (ν,ν,ν ∗) ,Sb (ν,ν,Gν) , Sb (ν ∗,ν∗,Gν∗) , Sb(ν,ν,Gν ∗)+Sb(ν ∗,ν∗,Gν) 4b3 } ≤ max { Sb (ν,ν,ν ∗) ,Sb (ν,ν,ν) , Sb (ν ∗,ν∗,ν∗) , Sb(ν,ν,ν ∗)+2bSb(ν ∗,ν∗,ν∗)+Sb(ν,ν,ν ∗) 4b3 } = Sb (ν,ν,ν ∗) Int. J. Anal. Appl. (2023), 21:18 9 Using by (3.13), we have Γ (Sb(ν,ν,ν∗)) ≤ β (Γ(Sb (ν,ν,ν∗))) Γ(Sb (ν,ν,ν∗)). we can deduce that Γ(Sb(ν,ν,ν ∗)) Γ(Sb(ν,ν,ν ∗)) ≤ β (Γ(Sb (ν,ν,ν∗))). We obtain that lim n→∞ β (Γ(Sb (ν,ν,ν ∗))) = 1. Therefore, Sb (ν,ν,ν∗) = 0 implies ν = ν∗. Consequently, ν is a unique fixed point of G. � Example 3.1. Let Sb : P3 → R+ be defined as Sb(µ,ν,ξ) = (|ν + ξ − 2µ| + |ν − ξ|)2 where P = [0,∞). It is obvious that (P,Sb) is a complete with b = 4. Define G : P → P by G(µ) = µ43 and Γ : [0,∞) → [0,∞) and β : [0,∞) → [0, 1) as Γ(t) = t, β(t) =   e − 4 5 3969 t 1+ 4 5 3969 t , t ∈ (0,∞) 0, t = 0 also define α,φ : P×P×P →R+ α(µ,µ,ν) = { 4, (µ,µ,ν) ∈ [0, 1] 0, Otherwise φ(µ,µ,ν) = { 1, (µ,µ,ν) ∈ [0, 1] 0, Otherwise Let α(µ,µ,Gµ) ≥ φ(µ,µ,Gµ). Thus µ,Gµ ∈ [0, 1] and so G2µ = G(Gµ) ∈ [0, 1] which implies that α(Gµ,Gµ,G2µ) ≥ (Gµ,Gµ,G2µ) that is G is α-orbital admissible with respect to φ. Now, let α(µ,µ,ν) ≥ φ(µ,µ,ν) and α(ν,ν,Gν) ≥ φ(ν,ν,Gν), we get that µ,ν,Gν ∈ [0, 1] and so α(µ,µ,Gν) ≥ φ(µ,µ,Gν). Therefore G is triangular α-orbital admissible with respect to φ. Let {µn} be a sequence such that {µn} is Sb-convergent to χ and α(µn,µn,µn+1) ≥ φ(µn,µn,µn+1) for all n ∈ N. Then {µn} ∈ [0, 1] for any n ∈ N and so χ ∈ [0, 1] which we have, α(χ,χ,χ) ≥ φ(χ,χ,χ) and obviously the function G is continuous. Following that, we show that G is a generalised (α,φ, Γ)-Geraghty contraction type mapping. Let µ,ν ∈ P with α(µ,µ,ν) ≥ φ(µ,µ,ν). Thus µ,ν ∈ [0, 1].We can assume without losing generality that 0 ≤ ν ≤ µ ≤ 1. Therefore, Sb(Gµ,Gµ,Gν) = (|Gµ + Gν − 2Gµ| + |Gµ−Gν|)2 = ( 2 ∣∣∣ µ 43 − ν 43 ∣∣∣)2 = 1 46 Sb(µ,µ,ν) and MG b (µ,µ,ν) = max { (2|µ−ν|)2, 3969 45 µ2, 3969 45 ν2 ( (|43µ−ν|)2+(|µ−43ν|)2 46b3 } = 3969 45 µ2 Since 65 2×46 ≤ 1 2e ≤ e −µ2 1+µ2 so that 65 2×46 3969 45 µ2 ≤ e −µ2 1+µ2 3969 45 µ2 Γ ( ( 1+b 3 2 )Sb(Gµ,Gµ,Gν) ) = Γ ( 65 2×46 Sb(µ,µ,ν) ) = 65 2×46 Sb(µ,µ,ν) ≤ 65 2×46 3969 45 µ2 ≤ e −µ2 1+µ2 3969 45 µ2 ≤ β ( Γ( 3969 45 µ2) ) Γ( 3969 45 µ2) ≤ β ( Γ(MG b (µ,µ,ν)) ) Γ(MG b (µ,µ,ν)) As a result, all of Theorem (3.1)’s requirements are satisfied, and 0 is the only fixed point of G. 10 Int. J. Anal. Appl. (2023), 21:18 4. Application to Integral Equations As an application of Theorem (3.1), we will look at the existence of a unique solution to an initial value problem in this section. Theorem 4.1. Consider the I. V. P. σ′(t) = G(t,σ(t)), t ∈ I = [0, 1], σ(0) = σ0 (4.1) where G : I ×R→R is a continuous function and σ0 ∈R. Let Γ : [0,∞) → [0,∞),β : [0,∞) → [0, 1) be a two functions defined as Γ(t) = t and β(t) = 1 3 . Also examine the following conditions, (i) If there exist a function α,φ : R3 →R such that there is an σ1 ∈ C(I), for all t ∈ I,we’ve α  σ1(t),σ1(t), t∫ 0 G(s,σ1(s))ds   ≥ φ  σ1(t),σ1(t), t∫ 0 G(s,σ1(s))ds   . (ii) ∀ t ∈ I, and ∀ x,y ∈ C(I), α (σ(t),σ(t),ς(t)) ≥ α (σ(t),σ(t),ς(t)) ⇒ α ( σ0 3b3 + t∫ 0 G(s,σ(s))ds, σ0 3b3 + t∫ 0 G(s,σ(s))ds, ς0 3b3 + t∫ 0 G(s,ς(s))ds ) ≥ φ ( σ0 3b3 + t∫ 0 G(s,σ(s))ds, σ0 3b3 + t∫ 0 G(s,σ(s))ds, ς0 3b3 + t∫ 0 G(s,ς(s))ds ) . (iii) for any point σ of a sequence {σn} of points in C(I) with α (σn,σn,σn+1) ≥ φ (σn,σn,σn+1), lim n→∞ inf α (σn,σn,σ) ≥ lim n→∞ inf φ (σn,σn,σ). Then (4.1) has unique solution . Proof. The integral equation of I. V. P. ( 4.1 ) is σ(t) = σ0 + 3( 1 + b3 2 ) t∫ 0 G(s,σ(s))ds. Let P = C (I) and Sb(σ,ς,τ) = (|ς + τ −2σ|+ |ς−τ|)2 for σ,ς,ς ∈ P. Then (P,Sb) is a complete, also define T : P → P by T (σ)(t) = 2σ0 3(1 + b3) + t∫ 0 G(s,σ(s))ds. (4.2) Now Γ ( ( 1 + b3 2 )Sb(Tσ(t),Tσ(t),Tς(t)) ) = ( 1 + b3 2 ){| Tσ(t) + Tς(t) − 2Tσ(t) | + | Tσ(t) −Tς(t) |}2 Int. J. Anal. Appl. (2023), 21:18 11 = 8(1+b3) 9(1+b3)2 ∣∣∣∣ σ0 + 3( 1+b32 ) t∫ 0 G(s,σ(s))ds − ς0 − 3( 1+b 3 2 ) t∫ 0 G(s,ς(s))ds ∣∣∣∣2 = 8 9(1+b3) | σ(t) − ς(t) |2 = 8 9(1+b3) Sb(σ,σ,ς) ≤ 13 Sb(σ,σ,ς) ≤ β (Γ(Sb(σ,σ,ς))) Γ(Sb(σ,σ,ς)) ≤ β ( Γ(MTb (σ,σ,ς)) ) Γ(MTb (σ,σ,ς)) Thus we have Γ ( ( 1+b 3 2 )Sb(Tσ(t),Tσ(t),Tς(t)) ) ≤ β ( Γ(MTb (σ,σ,ς)) ) Γ(MTb (σ,σ,ς))∀σ,ς ∈ P Let us define α : P×P×P → [0,∞) and φ : P×P×P → [0,∞) by α(σ,σ,ς) = { 6, σ,ς ∈ [0, 1] 0, σ,ς ∈ (1,∞) , φ(σ,σ,ς) = { 2, σ,ς ∈ [0, 1] 1, σ,ς ∈ (1,∞) Then obviously, T is triangular α-orbital admissible with respect to φ. Let σ,ς ∈ P, if α(σ,σ,ς) = 6 and φ(σ,σ,ς) = 2, then α (σ(t),σ(t),ς(t)) ≥ φ (σ(t),σ(t),ς(t)). From (ii) we have α (Tσ(t),Tσ(t),Tς(t)) ≥ φ (Tσ(t),Tσ(t),Tς(t)) and so α(Tσ,Tσ,Tς) = 6 and φ(Tσ,Tσ,Tς) = 2. Thus T is is triangular α-orbital admissible with respect to φ. From (i), there exist σ1,ς1 ∈ P such that α(σ1,σ1,Tσ1) = 6 and φ(ς1,ς1,Tς1) = 2. By (iii), we have that for any point σ of a sequence {σn} of points in C(I) with α (σn,σn,σn+1) = 6, lim n→∞ inf α (σn,σn,σ) = 6 and φ (σn,σn,σn+1) = 2, lim n→∞ inf φ (σn,σn,σ) = 2. Therefore, for all σ,ς ∈ P and t ∈ I, we have α (σ(t),σ(t),ς(t)) ≥ φ (σ(t),σ(t),ς(t)) =⇒ Γ ( ( 1+b 3 2 )Sb(Tσ(t),Tσ(t),Tς(t)) ) ≤ β ( Γ(MTb (σ,σ,ς)) ) Γ(MTb (σ,σ,ς)) Theorem (3.1) states that T has a unique solution in P. � 5. Application to Homotopy The existence of a unique homotopy theory solution is investigated in this section. Theorem 5.1. Let (P,Sb) be a complete Sb-metric space, U and U be a open and closed subset of P such that U ⊆ U . Suppose α,φ : P × P × P → [0,∞), Hb : U × [0, 1] → P is a triangular α-orbital admissible operator with respect to φ and β ∈ F satisfying the following conditions: (τ0) σ 6= Hb(σ,κ), for each σ ∈ ∂U and κ ∈ [0, 1] (Here ∂U is boundary of U in P) (τ1) for all σ,ς ∈U and κ ∈ [0, 1], α(σ,σ,Hb(σ,κ)) ≥ φ(σ,σ,Hb(σ,κ)) implies Γ ( ( 1 + b3 2 )Sb(Hb(σ,κ),Hb(σ,κ),Hb(ς,κ)) ) ≤ β (Γ(Sb(σ,σ,ς))) Γ(Sb(σ,σ,ς)) (τ2) ∃ Mb ≥ 0 3 Sb(Hb(σ,κ),Hb(σ,κ),Hb(σ,ζ)) ≤ Mb|κ−ζ| for every σ ∈Uand κ,ζ ∈ [0, 1]. Then Hb(., 0) has a fixed point ⇐⇒ Hb(., 1) has a fixed point. 12 Int. J. Anal. Appl. (2023), 21:18 Proof. Let A = {κ ∈ [0, 1] : σ = Hb(σ,κ) for some σ ∈ U}. We have that 0 ∈ A since Hb(., 0) has a fixed point in U. As a result, the set A is not empty. By demonstrating that A is both open and closed in [0, 1], we shall establish that A = [0, 1]. As a result, in U, Hb(, 1) has a fixed point.First, we demonstrate that A is a closed set in [0, 1]. Let {κn}∞n=1 ⊆ A with κn → κ ∈ [0, 1] as n → ∞. We have to demonstrate that κ ∈ A. Since κn ∈ A for n = 0, 1, 2, · · · . there exists σn ∈ U with σn+1 = Hb(σn,κn). Since Hb is a triangular α-orbital admissible operator with respect to φ. α(σ0,σ0,Hb(σ0,κ0)) ≥ φ(σ0,σ0,Hb(σ0,κ0)). We can deduce from Lemma (2.1) that α(σn,σn,σn+1) ≥ φ(σn,σn,σn+1) for all n ≥ 0 Consider, Sb(σn+1,σn+1,σn+2) = Sb (Hb(σn,κn),Hb(σn,κn),Hb(σn+1,κn+1)) ≤ 2bSb (Hb(σn,κn),Hb(σn,κn),Hb(σn+1,κn)) +b2Sb (Hb(σn+1,κn),Hb(σn+1,κn),Hb(σn+1,κn+1)) ≤ 2bSb (Hb(σn,κn),Hb(σn,κn),Hb(σn+1,κn)) + b2Mb|κn −κn+1| Letting n →∞, we get lim n→∞ Sb(σn+1,σn+1,σn+2) ≤ lim n→∞ 2bSb (Hb(σn,κn),Hb(σn,κn),Hb(σn+1,κn)) . We get Γ since it is continuous and non-decreasing. lim n→∞ Γ ( ( 1 + b3 4b )Sb(σn+1,σn+1,σn+2) ) = lim n→∞ Γ ( ( 1 + b3 2 )Sb (Hb(σn,κn),Hb(σn,κn),Hb(σn+1,κn)) ) ≤ lim n→∞ β (Γ(Sb(σn,σn,σn+1))) Γ(Sb(σn,σn,σn+1)) Therefore, lim n→∞ Γ ( ( 1+b 3 4b )Sb(σn+1,σn+1,σn+2) ) lim n→∞ Γ(Sb(σn,σn,σn+1)) ≤ lim n→∞ β (Γ(Sb(σn,σn,ςn+1))) < 1. In above inequality, we have lim n→∞ β (Γ(Sb(σn,σn,σn+1))) = 1. Since β ∈ F, we have lim n→∞ Γ(Sb(σn,σn,σn+1)) = 0 and so lim n→∞ Sb (σn,σn,σn+1) = 0. It is now time to demonstrate {σn}, a Sb-Cauchy sequence in (P,Sb). On the other hand, suppose {σn} is not a Sb-Cauchy se- quence. There is a monotone increasing sequence with � > 0 and Natural numbers with the property that {mk} and {nk} such that nk > mk. Sb ( σmk,σmk,σnk ) ≥ � (5.1) and Sb ( σmk,σmk,σnk−1 ) < �. (5.2) Int. J. Anal. Appl. (2023), 21:18 13 From (5.1) and (5.2), we have � ≤ Sb ( σmk,σmk,σnk ) ≤ 2bSb ( σmk,σmk,σmk +1 ) + b2Sb ( σmk +1,σmk +1,σnk ) . So that ( 1+b 3 2b2 )� ≤ ( 1+b 3 b )Sb ( σmk,σmk,σmk +1 ) + ( 1+b 3 2 )Sb ( σmk +1,σmk +1,σnk ) . We get that by setting k →∞ and Γ is applied on both sides, Γ(( 1 + b3 2b2 )�) ≤ lim k→∞ Γ ( ( 1 + b3 2 )Sb ( σmk +1,σmk +1,σnk )) ≤ lim n→∞ β ( Γ(Sb(σmk,σmk,σnk−1)) ) Γ(Sb(σmk,σmk,σnk−1)) ≤ lim n→∞ β ( Γ(Sb(σmk,σmk,σnk−1)) ) Γ(( 1 + b3 2b2 )�). That is 1 ≤ lim n→∞ β ( Γ(Sb(σmk,σmk,σnk−1)) ) ⇒ lim n→∞ β ( Γ(Sb(σmk,σmk,σnk−1)) ) = 1 . This leads to the result lim n→∞ Sb(σmk,σmk,σnk−1) = 0. and hence, lim n→∞ Sb(σmk +1,σmk +1,σnk ) = 0.It contradicts itself. In the Sb-metric space (P,Sb), the sequence {σn} is a Sb-Cauchy sequence.The sequence {σn} → ν ∈ (P,Sb) emerges from the completeness of (P,Sb). lim n→∞ σn+1 = ν = lim n→∞ σn. Since {σn} is a Sb-convergent sequence to ν in X and α(ν,ν,ν) ≥ φ(ν,ν,ν). Then to prove ν = Hb(ν,κ). Now Γ ( 1 2b Sb(Hb(ν,κ),Hb(ν,κ),ν) ) ≤ lim n→∞ inf Γ (Sb(Hb(ν,κ),Hb(ν,κ),Hb(σn,κ))) ≤ lim n→∞ inf Γ ( ( 1 + b3 2 )Sb(Hb(ν,κ),Hb(ν,κ),Hb(σn,κ)) ) ≤ lim n→∞ β (Γ(Sb(ν,ν,σn))) Γ(Sb(ν,ν,σn)). So that Γ ( 1 2b Sb(Hb(ν,κ),Hb(ν,κ),ν) ) lim n→∞ Γ(Sb(ν,ν,σn)) ≤ lim n→∞ β (Γ(Sb(ν,ν,σn))) That is 1 ≤ lim n→∞ β (Γ(Sb(ν,ν,σn))) implies lim n→∞ β (Γ(Sb(ν,ν,σn))) = 1. As a result, we obtain lim n→∞ Γ(Sb(ν,ν,σn)) = 0 and hence Sb(Hb(ν,κ),Hb(ν,κ),ν) = 0. Thus, it follows ν = Hb(ν,κ). Thus κ ∈ A. Clearly, [0, 1] closes A. Let κ0 ∈ A.Consequently, there is σ0 ∈ U such that σ0 = Hb(σ0,κ0). Due to the fact that Uis open, r > 0 exists such that BSb (σ0, r) ⊆ U. Choose κ ∈ (κ0 − �,κ0 + �) such that |κ−κ0| ≤ 1Mn < �. Then, for Bp(σ0, r) = {σ ∈ P : Sb(σ,σ,σ0) ≤ r + b2Sb(σ0,σ0,σ0)}. Now Sb (Hb(σ,κ),Hb(σ,κ),σ0)) = Sb (Hb(σ,κ),Hb(σ,κ),Hb(σ0,κ0)) ≤ 2bSb (Hb(σ,κ),Hb(σ,κ),Hb(σ,κ0)) +b2Sb (Hb(σ,κ0),Hb(σ,κ0),Hb(σ0,κ0)) ≤ 2bM|κ−κ0| + b2Sb (Hb(σ,κ0),Hb(σ,κ0),Hb(σ0,κ0)) 14 Int. J. Anal. Appl. (2023), 21:18 Upon letting n →∞ and applying Γ to both sides, Γ (Sb (Hb(σ,κ),Hb(σ,κ),σ0))) ≤ Γ ( b2Sb (Hb(σ,κ0),Hb(σ,κ0),Hb(σ0,κ0)) ) ≤ Γ ( ( 1 + b3 2 )Sb (Hb(σ,κ0),Hb(σ,κ0),Hb(σ0,κ0)) ) ≤ β (Γ(Sb(σ,σ,σ0))) Γ(Sb(σ,σ,σ0)) ≤ Γ(Sb(σ,σ,σ0)). Therefore, Sb (Hb(σ,κ),Hb(σ,κ),σ0)) ≤ Sb(σ,σ,σ0) ≤ r + b2Sb(σ0,σ0,σ0). Thus for each fixed κ ∈ (κ0 −�,κ0 + �), Hb(.; κ) : Bp(σ0, r) → Bp(σ0, r). Thus, Theorem (5.1)’s criteria are met in full. Consequently, it may be said that Hb(.; κ) has a fixed point in U. 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Sci. 10 (2018), 455–478. https://doi.org/10.1186/s13663-015-0313-6 https://doi.org/10.1186/s13663-015-0313-6 https://doi.org/10.1186/1687-1812-2013-329 https://www.jstor.org/stable/24896705 https://doi.org/10.1186/1687-1812-2014-190 https://doi.org/10.1186/1687-1812-2013-151 https://doi.org/10.22436/jnsa.009.02.13 1. Introduction 2. Preliminaries 3. Main Results 4. Application to Integral Equations 5. Application to Homotopy References