Int. J. Anal. Appl. (2023), 21:13 Some Results on Conditionally Sequential Absorbing Maps in Multiplicative Metric Space T. Thirupathi1,∗, V. Srinivas2 1Department of Mathematics, Sreenidhi Institute of Science and Technology, Hyderabad, Telangana, India 2Department of Mathematics, Osmania University, Hyderabad, Telangana, India ∗Corresponding author: thotathirupathi1986@gmail.com Abstract. This paper aims to prove two general fixed point theorems in multiplicative metric space (MMS) by using reciprocally continuous mappings and conditionally sequential absorbing mappings. Further our outcomes are validated by discussing two appropriate examples. 1. Introduction One of the most exciting areas of contemporary mathematics is fixed point theory, which is also interesting topic of the analysis. Further this topic has became a platform due to its wide applications in pure and applied mathematics. In this connection S. Young Cho et al [1] proved a common fixed point theorem over a complete metric space. Later,many researchers generated results in diffeent spaces. In this process Monika Verma et al [2] generalized [1] for multiplicative metric space.Furthermore some results can be witnessed like [3], [4], [5], [6], [7] [8] and [9] in MMS. Using the conditions conditionally sequential absorption and reciprocally continuous mappings, the goal of this research is to derive two common fixed point theorems for MMS. Further two suitable examples are discussed to validate our theorems. 2. Preliminaries Definition 2.1 Let X be a non empty set and d : X × X → R+ then (X,d) is said to be MMS if satisfying the following conditions: Received: Nov. 28, 2022. 2010 Mathematics Subject Classification. 47H10, 54H25. Key words and phrases. multiplicative metric space; fixed point; reciprocally continuous mapping; conditionally se- quential absorbing mapping. https://doi.org/10.28924/2291-8639-21-2023-13 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-13 2 Int. J. Anal. Appl. (2023), 21:13 (i) d(ζ,η)≥ 1 for all ζ,η ∈ X and d(ζ,η)=1 if and only if x = y (ii) d(ζ,η)= d(ζ,η) for all ζ,η ∈ X (iii) d(ζ,η)≤ d(ζ,β).d(β,η) for all ζ,η,β ∈ X (multiplicative triangle inequality). Then (X,d) is called MMS. The pair of mapping (G,J) of a MMS (X,d) is said to be Definition 2.2 Compatible if lim j→∞ d(GJηj,JGηj) = 1, whenever ηj is a sequence in X such that Gηj = Jηj = ζ for some ζ ∈ X. Definition 2.3 Weakly compatible if Gζ = Jζ for some ζ ∈ X such that IJζ = JIζ. Definition 2.4 If there is a coincidence point where the mappings commute then it is said to be Occasionally Weakly Compatible (OWC). Example 2.4.1 Let (X,d) be a MMS and ∀η,ζ ∈ X we have (.η,ζ)= e |η−ζ|. Now the self mappings G,J are defined on X = [0,∞) and given below G(η)= η+1 2 and J(η)= η 2+1 2 for all η ∈ X. From above η =0,1 are coincidence points for the mappings G,J. At η =0 G(0)= J(0)= 1 2 , GJ(0)= G(1 2 )= 3 4 , JG(0)= J(1 2 )= 5 8 . Therefore GJ(0) 6= JG(0). And also GJ(1)= JG(1)=1. Resulting that the maps G, J are OWC but not weakly compatible. Definition 2.5 Conditionally sequentially absorbing if whenever a sequence (ζj) satisfying {(ζj): lim j→∞ Gζj = lim j→∞ Jζj} 6= ∅ then there exists another sequence (ηj) in X with lim j→∞ Gηj= lim j→∞ Jηj = u for some u ∈ X such that lim j→∞ d(Gηj,GJηj)=1 and lim j→∞ d(Jηj,JGηj)=1. Example 2.5.1 Lt (X,d) be an MMS and ∀η,ζ ∈ X we have d(η,ζ)= e|η−ζ|. Now the self mappings G,J are defined on X = [0,∞) and given below G(η)= { sinη if 0≤ η < π 2 2η2 if 0π 2 ≤ η ≤ π; J(η)= { cosη if 0≤ η < π 2 πη if 0π 2 ≤ η ≤ π; From above η =0, π 2 are coincidence points for the mappings G,J. At η =0 G(0)= J(0)=0, GJ(0)= G(0)=0, JG(0)= J(0)=0. Int. J. Anal. Appl. (2023), 21:13 3 Therefore GJ(0)= JG(0). And also GJ(π 2 )= G(π 2 2 )= π 4 2 , JG(π 2 )= G(π2)=2π4, Therefore GJ(π 2 ) 6= JG(π 2 ). Resulting that the maps G, J are not weakly compatible. Let (pj)= √ 4 j , for all j ≥ 1. Then lim j→∞ Gpj = lim j→∞ G( √ 4 j )= lim j→∞ sin( √ 4 j )=0 (2.1) and lim j→∞ Jpj = lim j→∞ J( √ 4 j )= lim j→∞ 1−cos( √ 4 j )=1−1=0. (2.2) From (2.1) and (2.2), we get lim j→∞ Gpj = lim j→∞ Jpj (2.3) From (2.3) implies {(pj) : lim j→∞ Gpj = lim j→∞ Jpj} 6= ∅. Then ∃ another sequence qj = π2 + 5 j , for all j ≥ 1. lim j→∞ Gqj = lim j→∞ G( π 2 + 5 j )= lim j→∞ =2( π 2 + 5 j )2 = π2 2 (2.4) and lim j→∞ Jqj = lim j→∞ J( π 2 + 5 j )= lim j→∞ π( π 2 + 5 j )= π2 2 . (2.5) From (2.4) and (2.5), we get lim j→∞ Gpj = lim j→∞ Jpj = π2 2 . (2.6) Now lim j→∞ GJ(qj)= GJ( π 2 + 5 j ) = lim j→∞ G(π(π 2 + 5 j )) = lim j→∞ G(π 2 2 + 5π j )= π 3 2 and lim j→∞ JG(qj)= JG( π 2 + 5 j ) = lim j→∞ J(2((π 2 + 5 j )2)) =π 3 2 . Therefore lim j→∞ d(Gqj,GJqj)=1 and lim j→∞ d(Jqj,JGqj)=1. Hence the pair(G,J) is conditionally sequentially absorbing but not weakly compatible. Definition 2.6 Reciprocally continuous whenever (ηj) is a sequence in X such that lim j→∞ Gηj = lim j→∞ Jηj = ζ for some ζ ∈ X such that lim j→∞ d(Gζ,GJηj)=1 and lim j→∞ d(Jζ,JGηj)=1. Example 2.5.1 Lt (X,d) be a Multiplicative metric space and ∀η,ζ ∈ X we have d(η,ζ)= e|η−ζ|. Now the self mappings G,J are defined on X = [0,∞) and given below G(η)= { πcosη if 0≤ η < π 2 η2 if 0π 2 ≤ η ≤ π; 4 Int. J. Anal. Appl. (2023), 21:13 J(η)= { πsecη if 0≤ η < π 2 πη if 0π 2 ≤ η ≤ π; From above η =0,π are coincidence points for the mappings G,J. From above at η =0 G(0)= J(0)= π, GJ(0)= G(π)= π3, JG(0)= J(π)= π3. Therefore GJ(0)= JG(0). And also GJ(π)= G(π3)= π7, JG(π)= J(π3)= π9. Therefore GJ(π) 6= JG(π) Resulting that the maps G, J are not weakly compatible. Let (rj)= π − 3j3 for all j ≥ 1. Then lim j→∞ Grj = lim j→∞ G(π − 3 j3 )= lim j→∞ (π − 3 j3 )2 = π2 (2.7) and lim j→∞ Jrj = lim j→∞ J(π − 3 j3 )= lim j→∞ π(π − 3 j3 )= π2. (2.8) From (2.7) and (2.8), we get lim j→∞ Grj = lim j→∞ Jrj (2.9) Now lim j→∞ GJ(rj)= GJ(π − 3j3) = limj→∞ G(π(π − 3 j3 )) = π3 and lim j→∞ JG(rj)= JG(π − 3j3) = limj→∞ J(π(π − 3 j3 )) =π4. Therefore lim j→∞ d(G(π2),GJrj)=1 and lim j→∞ d(J(π2),JGrj)=1. Hence the maps G, J are reciprocally continuous. In [1], The following Theorem was established. Theorem 2.7 Assume that (X,d) is an MMS which is complete and the mappings B, S, A, and T are defined on X such that (B1) B(X)⊆ S(X) and A(X)⊆ T(X) (B2) d(Au,Bv)≤ (max{d(Au,Su),d(Bv,Tv), √ [d(Au,Tv).d(Bv,Su)],d(Su,Tv)})p. (max{d(Au,Su),d(Bv,Tv)})q.(max{d(Au,Tv),d(Bv,Su)})r for all u,v ∈ X, where 0 < h = p+q +2r < 1 (p,q and r are non-ve real numbers). (B3) Among the subspaces AX or BX or SX or TX is complete (B4) both the pairs (A,S) and (B,T) are weakly compatible. Then the four maps A,B,S and T above share a common single fixed point. Now we generalize the above Theorem 2.7 as below. Int. J. Anal. Appl. (2023), 21:13 5 3. Main Result Theorem 3.1 Assume that (X,d) is an MMS which is complete and the mappings A, B, S, and T are defined on X such that (D1) A(X)⊆ T(X) and B(X)⊆ S(X) (D2) d(Au,Bv)≤ (max{d(Au,Su),d(Bv,Tv), √ [d(Au,Tv).d(Bv,Su)],d(Su,Tv)})p. (max{d(Au,Su),d(Bv,Tv)})q.(max{d(Au,Tv),d(Bv,Su)})r for all u,v ∈ X, where h = p+q +2r and o < h < 1 (p,qandrarenon−verealnumbers). (D3) the pairs (A,S) reciprocally continuous and conditionally sequentially absorbing and (B,T) is occasionally weakly compatible. Then the four mappings share a single fixed point which is common in X. Proof: By (D1), there is a point here u0 ∈ X such that Au0 = Tu1 = y1. For this point u1 ∈ X there exists a point u2 in X such that Bu1 = Su2 = y2 and so on. Similarly, we can inductively define Bu2j−1 = Su2j = y2j;Au2j = Tu2j+1 = y2j+1 for n =0,1,2, ... We can now show that the sequence {vj} is a Cauchy in X. Put u = u2j and v = u2j+1 in (D2) then d(v2j+1,v2n+2)= d(Au2j,Bu2j+1)≤ (max{d(Au2j,Su2j),d(Bu2j+1,Tu2j+1), √ [d(Au2j,Tu2j+1).d(Bu2j+1,Su2j)],d(Su2j,Tu2j+1)})p. (max{d(Au2j,Su2j),d(Bu2j+1,Tu2j+1)})q.(max{d(Au2j,Tu2j+1),d(Bu2j+1,Su2j)})r d(v2j+1,v2n+2)≤ (max{d(v2j+1,v2j),d(v2j+2,v2j+1), √ [d(v2j+1,v2j+1).d(v2j+2,v2j)],d(v2j,v2j+1)})p. (max{d(v2j+1,v2j),d(v2j+2,v2j+1)})q.(max{d(v2j+1,v2j+1),d(v2j+2,v2j)})r d(v2j+1,v2n+2)≤ (max{d(v2j+1,v2j),d(v2j+2,v2j+1), √ [d(v2j+1,v2j+1).d(v2j+1,v2j).d(v2j+1,v2j+2)],d(v2j,v2j+1)})p. (max{d(v2j+1,v2j),d(v2j+2,v2j+1)})q.(max{d(v2j+1,v2j+1),d(v2j+1,v2j).d(v2j+1,v2j+2)})r In the above equation, if d(v2j+2,v2j+1) > d(v2j+1,v2j) for some +ve integer j, then we have d(v2j+1,v2j+2)≤ d(v2j+1,v2j+2)h, where o < h = p+q +2r < 1, a contradiction. Therefore we have d(v2j+2,v2j+1)≤ d(v2j,v2j+1)h. Likewise, we have d(vj,v2j+1)≤ (d(vj−1,vj)h)≤ (d(vj−2,vj −1)h 2 ≤ .... ≤ (d(v0,v1))h n . Let l, j lnN such that l > j, we get d(vl,vj)≤ d(vl,vl−1)....d(vj+1,vj) ≤ (d(v1,v0))h l−1+....hj ≤ (d(v1,v0)) hj 1−h → 1 as l, j →∞. As a result, the sequence {vj} is a Cauchy. 6 Int. J. Anal. Appl. (2023), 21:13 By the completeness of X ∃ w ∈ X such that vj → w as j →∞. Accordingly, the sequences Au2j,Su2j → z,Tu2j+1,Bu2j+1 → z (3.1) as j →∞. Use the notion L{A,S}= {(uj) : lim j→∞ Auj = lim j→∞ Suj}. By (D3) the pair of mapping (A,S) is conditionally sequentially absorbing from (3.1) L{A,S} 6= ∅⇒∃(vj) such that lim j→∞ Avj = lim j→∞ Svj = ψ (3.2) =⇒ d(Avj,ASvj)=1andd(Svj,SAvj)=1 (3.3) By the reciprocally continuous of the pair (A,S) implies whenever lim j→∞ Avj = lim j→∞ Svj = ψ (3.4) =⇒ d(Aψ,ASvj)=1andd(Sψ,SAvj)=1. (3.5) Using (3.2) and (3.5) in (3.3), we get Aψ = Sψ = ψ. Since Aψ is an element in A(X) by (D1) there exists ϕ such that ψ = Sψ = Aψ = Tϕ. (3.6) Claim Bϕ = Tϕ. Putting u = ψ , v = ϕ in (D2) d(Aψ,Bϕ)≤ (max{d(Aψ,Sψ),d(Bϕ,Tϕ), √ [d(Aψ,Tϕ).d(Bϕ,Sψ)],d(Sψ,Tϕ)})p. (max{d(Aψ,Sψ),d(Bϕ,Tϕ)})q.(max{d(Aψ,Tϕ),d(Bϕ,Sψ)})r. Letting n →∞ we get, d(Tϕ,Bϕ)≤ (max{d(ψ,ψ),d(Bϕ,Tϕ), √ [d(ψ,ψ).d(Bϕ,Tϕ)],d(ψ,ψ)})p. (max{d(ψ,ψ),d(Bϕ,Tϕ)})q.(max{d(ψ,ψ),d(Bϕ,Tϕ)})r d(Tϕ,Bϕ)≤ d(Tϕ,Bϕ)p+q+r, which is a contradiction. Hence Tϕ = Bϕ. Which gives ψ = Sψ = Aψ = Tϕ = Bϕ. (3.7) From (D3) we have the pair (B,T) is occasionally weakly compatible which gives BTϕ = TBϕ implies that Bψ = Tψ from (3.7). Int. J. Anal. Appl. (2023), 21:13 7 Claim ψ = Bψ. Putting u = v = ψ in (D2) d(ψ,Bψ)≤ (max{d(ψ,ψ),d(Tψ,Tψ), √ [d(ψ,Bψ).d(Bψ,ψ)],d(ψ,Bψ)})p. (max{d(ψ,ψ),d(Tψ,Tψ)})q.(max{d(ψ,Bψ),d(Bψ,ψ)})r d(ψ,Bψ)≤ d(ψ,Bψ)p+r, a contradiction which impies ψ = Bψ. Therefore ψ = Sψ = Aψ = Tψ = Bψ. Which implies that ψ is the required common fixed point. For Uniqueness: Assume that ρ be the another fixed point then ρ = Sρ = Aρ = Tρ = Bρ. Putting u = ψ and v = ρ in (D2), we get d(Aψ,Bρ)≤ (max{d(Aψ,Sψ),d(Bρ,Tρ), √ [d(Aψ,Tρ).d(Bρ,Sψ)],d(Sψ,Tρ)})p. (max{d(Aψ,Sψ),d(Bρ,Tρ)})q.(max{d(Aψ,Tρ),d(Bρ,Sψ)})r d(ψ,ρ)≤ (max{d(ψψ),d(ρ,ρ), √ [d(ψ,ρ).d(ρ,ψ)],d(ψ,ρ)})p. (max{d(ψ,ψ),d(ρ,ρ)})q.(max{d(ψ,ρ),d(ρ,ψ)})r d(ψ,ρ)≤ d(ψ,ρ)p+q+r, a contradiction which implies ψ = ρ. This proves the uniqueness. Now we discuss an example. Example 3.2 Assume that (X,d) is an MMS space with d(u,v)= e|u−v| for all u,v ∈ X. A, B, S, and T are the self maps that are defined on X = [0,1] as follows: A(η)= { η2+1 2 if 0≤ η < 1 5 η if 1 5 ≤ η ≤ 1; S(η)= { η2+η+1 2 if 0≤ η < 1 5 η2 if 1 5 ≤ η ≤ 1; B(η)= { η2+4η+1 2 if 0≤ η < 1 5 1 5 if 1 5 ≤ η ≤ 1; T(η)= { η2+3η+1 2 if 0≤ η < 1 5 η if 1 5 ≤ η ≤ 1; Now A(X) = [1 2 ,0.52]∪ (1 5 ,1], S(X) = [1 2 ,0.9)∪{1 5 } , B(X) = [1 2 ,0.62]∪{1 5 } and T(X) = [1 2 ,0.52]∪ (1 5 ,1]. Clearly A(X)⊆ T(X) and B(X)⊆ S(X) so that (D1) is satisfied. For the pair of mappings (A,S) and (B,T), it is evident that 0 and 1 are coincidence points. At η =0 ⇒ A(0)= S(0)= 1 2 . 8 Int. J. Anal. Appl. (2023), 21:13 But AS(0)= A(1 2 )= 1 2 and SA(0)= S(1 2 )= 1 5 . Therefore AS(0) 6= SA(0). Also at η =0 ⇒ B(0)= T(0)= 1 2 . But BT(0)= B(1 2 )= 1 8 and TB(0)= T(1 2 )= 1 2 . Therefore BT(0) 6= TB(0). As a result, the mappings (A,S) and (B,T) are not weakly compatible. Take a sequence ηk = 3 2k for all k ≥ 0. Then lim k→∞ Aηk = A( 3 2k )= ( 3 2k )2+1 2 = 1 2 and lim n→∞ Sηk = S( 3 2k )= 1 2 . Implies lim k→∞ Aηk = lim k→∞ Sηk = 1 2 . Now lim k→∞ ASηk= lim k→∞ AS( 3 2k )= lim k→∞ A(1 2 + 9+6k 8k2 )= 1 2 and lim k→∞ SAηk= lim k→∞ SA( 3 2k2 + 1 2 )= (1 2 + 9 8k2 )2 = 1 4 . Therefore the pair (A,S) is non-compatible so that ∃ another sequence βk = 15 + 2 3k for all k ≥ 1. lim k→∞ Aβk = lim k→∞ Sβk = 1 5 . Also we have lim k→∞ ASβk= lim k→∞ AS(1 5 + 2 3m )= lim k→∞ A(1 5 )= 1 5 and lim k→∞ SAβk= lim k→∞ SA(1 5 + 2 3k )= lim k→∞ S(1 5 + 2 3k )= 1 5 . Thus from above lim k→∞ d(Aβk,ASβk)= d( 1 5 , 1 5 )= e| 1 5 −1 5 | =1 and lim k→∞ d(Sβk,SAβk)= d( 1 5 , 1 5 )= e| 1 5 −1 5 | =1. Further lim k→∞ d(ASβk,A( 1 5 ))= d(1 5 , 1 5 )= e| 1 5 −1 5 | =1. From the above we can conclude that the pairs (A,S) and (B,T) are non-compatible reciprocally continuous and conditionally sequential absorbing mappings. Also it is observed that A(1 5 )= S(1 5 )= B(1 5 )= T(1 5 )= 1 5 . It is found that the only common fixed point shared by the four self-maps is 1 5 . Now we prove another generalization of Theorem 2.7, as given below. Theorem 3.3 Assume that (X,d) is an MMS which is complete and the mappings A,B,S, and T are defined on X such that (E1) A(X)⊆ T(X) and B(X)⊆ S(X) (E2) d(Au,Bv)≤ (max{d(Au,Su),d(Bv,Tv), √ [d(Au,Tv).d(Bv,Su)],d(Su,Tv)})p. (max{d(Au,Su),d(Bv,Tv)})q.(max{d(Au,Tv),d(Bv,Su)})r for all u,v ∈ X, where h = p+q +2r and 0 < h < 1 (p,q and r are non-ve real numbers). (E3) The mappings for the pairs (A,S) and (B,T) are non-compatible reciprocally continuous and conditionally sequential absorbing mappings. Then the four mappings share a single fixed point which is common in X. Int. J. Anal. Appl. (2023), 21:13 9 Proof: By (E3) we have the pair (A,S) non-compatible =⇒ there is a sequence (uj) with lim j→∞ Auj = lim j→∞ Suj = ψ (3.8) for some ψ ∈ X. =⇒ limj→∞ d(ASuj,SAuj) not exist or limj→∞ d(ASuj,SAuj) 6=1. Considering that the pair (A,S) is conditionally sequentially absorbing from (3.8) we have L{A,S} 6= ∅=⇒∃(vj) such that lim j→∞ Avj = lim j→∞ Svj = ψ (say) =⇒ limj→∞ d(Avj,ASvj)=1 and limj→∞ d(Svj,SAvj)=1. Also from (E3) we have (A,S) is reciprocally continuous means whenever lim j→∞ Avj = lim j→∞ Svj = ψ. (3.9) =⇒ limj→∞ d(Aψ,ASvj)=1 and limj→∞ d(Sψ,SAvj)=1. Using the above equations, we get Aψ = Sψ = ψ. (3.10) Since the pair (B,T) is non compatible implies there is sequence (uj) with lim j→∞ Buj = lim j→∞ Tuj = ϕ (3.11) for some ϕ ∈ X. =⇒ limj→∞ d(BTuj,TBuj)1 not exist or limj→∞ d(BTuj,TBuj) 6=1. From (E3) the pair (B,T) is conditionally sequential absorbing from (3.11) L{B,T} 6= ∅=⇒∃(vj) such that lim j→∞ Bvj = lim j→∞ Tvj = β (say) =⇒ limj→∞ d(Bvj,BTvj)=1 and limj→∞ d(Tvj,TBvj)=1. Also the pair (B,T) is reciprocally continuous implies whenever lim j→∞ Bvj = lim j→∞ Tvj = β (say) =⇒ limj→∞ d(Bβ,BTvj)=1 and limj→∞ d(Tβ,TBvj)=1. Using the above equation, we get Bβ = Tβ = β. (3.12) Claim β = ψ. Assume that β 6= ψ. Putting u = ψ and v = β in (E2) d(Aψ,Bβ)≤ (max{d(Aψ,Sψ),d(Bβ,Tβ), √ [d(Aψ,Tβ).d(Bβ,Sψ)],d(Sψ,Tβ)})p. (max{d(Aψ,Sψ),d(Bβ,Tβ)})q.(max{d(Aψ,Tβ),d(Bβ,Sψ)})r. Letting as n →∞ 10 Int. J. Anal. Appl. (2023), 21:13 d(ψ,β)≤ (max{d(ψ,ψ),d(β,β), √ [d(ψ,β).d(β,ψ)],d(ψ,β)})p. (max{d(ψ,ψ),d(β,β)})q.(max{d(ψ,β),d(β,ψ)})r. d(ψ,β)≤ d(ψ,β)p+r. A contradiction hence ψ = β. Therefore Aψ = Sψ = Bψ = Tψ = ψ. Which implies that ψ is the required unique common fixed point. From (E2) uniqueness follows easily. Now we give an example to support Theorem 3.3. Example 3.4 Assume that (X,d) is a MMS space with d(u,v)= e|u−v| for all u,v ∈ X. A, B, S, and T are the self maps that are defined on X = [0,12] as follows: A(η)= { η4 if 0≤ η ≤ 1 2 if 1 < η ≤ 12; S(η)= { η2 if 0≤ η ≤ 1 2logη if 1 < η ≤ 12; B(η)= { η5 if 0≤ η ≤ 1 4 if 1 < η ≤ 12; T(η)= { η3 if 0≤ η ≤ 1 4logη if 1 < η ≤ 12. Now A(X)= [0,1]∪{2}, S(X)= [0,4.96], B(X)= [0,1]∪4 and T(X)= [0,5.45]. We have from above maps η = e and 1 are coincidence points for (A,S) and (B,T). At η = e , A(e)= S(e)=2 and B(e)= T(e)=4. AS(e)= A(2)=2,SA(e)= S(2)=2log2 and BT(e)= B(1)=1,TB(e)= T(4)=4log4. Clearly AS(e) 6= SA(e) and BT(e) 6= TB(e). As a result, the mappings (A,S) and (B,T) are not weakly comparable. Now take a sequence ηj = e + 3 4j , for all j ≥ 1. Then lim j→∞ Aηj = lim j→∞ A(e + 3 4j )= lim j→∞ 2=2 (3.13) and lim j→∞ Sηj = lim j→∞ S(e + 3 4j )= lim j→∞ 2log(e + 3 4j )=2. (3.14) Implies lim j→∞ Aηj = lim j→∞ Sηj = 2. Now lim j→∞ AS(e + 3 4j )= lim j→∞ A(2log(e + 3 4m ))= lim j→∞ 2=2 and lim j→∞ SA(e + 3 4j )= lim j→∞ S(2)= 2log2. Therefore the pair (A,S) is non-compatible. Further from (3.13) and (3.14) we get Int. J. Anal. Appl. (2023), 21:13 11 {(ηj) : lim j→∞ Aηj = lim j→∞ Sηj} 6= ∅. Further there exists another sequence βj =1− 54j for all j ≥ 1. lim j→∞ Aβj = lim j→∞ A(1− 5 4j )= lim j→∞ (1− 5 4j )4 =1 (3.15) and lim j→∞ Sβj = lim j→∞ S(1− 5 4j )= lim j→∞ (1− 5 4j )2 =1. (3.16) Now lim j→∞ AS(βj) = lim j→∞ AS(1− 5 4j )= lim j→∞ A(1− 5 4m )2= lim j→∞ A(1− 15 16j )= lim j→∞ (1− 5 4j )4= 1 and lim j→∞ SA(βj) = lim j→∞ SA(1− 5 4j )= lim j→∞ S(1− 5 4j )4= lim j→∞ (1− 5 4j )8= 1. Therefore lim j→∞ d(ASβj,Aβj) =d(1,1)=e|1−1| =1 and lim j→∞ d(SAβj,Sβj) =d(1,1)= e|1−1| =1. Further lim j→∞ d(ASβj,A(1)) =d(1,1)=e|1−1| =1 and lim j→∞ d(SAβj,S(1)) = d(1,1) e|1−1| =1. It follows that the pairs (A,S) and (B,T) have unique fixed point η = 1 and are non-compatible reciprocally continuous and conditionally sequentially absorbing mappings. Further the maps A,S,T and B are discontinuous at η =1.Moreover the pairs (A,S) and (B,T) are not weakly compatible and hence all the conditions of Theorem 3.3 are satisfied. 4. Conclusion In this paper we generalized Theorem 2.7 using (i) conditionally sequential absorbing, reciprocally continuous and OWC by removing weakly compatible mappings in Theorem 3.1. (ii) Further the weakly compatible mappings are replaced by non-compatible reciprocally continuous and conditionally sequential absorbing mappings in Theorem 3.3. Moreover the above two results are substantiated by two suitable examples. Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi- cation of this paper. References [1] S.Y. Cho, M.J. 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