Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 879084, 7 pages http://dx.doi.org/10.1155/2013/879084 Research Article Common Fixed Points for Weak π-Contractive Mappings in Ordered Metric Spaces with Applications Sumit Chandok1 and Simona Dinu2 1 Department of Mathematics, Khalsa College of Engineering & Technology (Punjab Technical University), Ranjit Avenue, Amritsar 143001, India 2 Faculty of Applied Sciences, University Politehnica of Bucharest, 313 Splaiul Independentei, 060042 Bucharest, Romania Correspondence should be addressed to Simona Dinu; [email protected] Received 7 July 2013; Accepted 29 August 2013 Academic Editor: Lishan Liu Copyright © 2013 S. Chandok and S. Dinu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We obtain some new common fixed point theorems satisfying a weak contractive condition in the framework of partially ordered metric spaces. The main result generalizes and extends some known results given by some authors in the literature. 1. Introduction and Preliminaries Fixed point and common fixed point theorems for different types of nonlinear contractive mappings have been investigated extensively by various researchers (see [1β41]). Fixed point problems involving weak contractions and the mappings satisfying weak contractive type inequalities have been studied by many authors (see [10β20] and references cited therein). Recently, many researchers have obtained fixed point, common fixed point, coupled fixed point, and coupled common fixed point results in partially ordered metric spaces (see [3, 6β8, 10β12, 29, 30, 32, 36]) and other spaces (see [5, 15, 31, 35, 38, 40, 41]). Let (π, β€) be a partially ordered set and π, π two selfmappings on π. A pair (π, π) of self-mappings of π is said to be weakly increasing [4] if ππ₯ β€ πππ₯ and ππ₯ β€ πππ₯ for any π₯ β π. An ordered pair (π, π) is said to be partially weakly increasing if ππ₯ β€ πππ₯ for all π₯ β π. Note that a pair (π, π) is weakly increasing if and only if the ordered pairs (π, π) and (π, π) are partially weakly increasing. Example 1 (see [3]). Let π = [0, 1] be endowed with usual ordering and π, π : π β π two mappings given by ππ₯ = π₯2 and ππ₯ = βπ₯. Clearly, the pair (π, π) is partially weakly increasing. But ππ₯ = βπ₯ β° π₯ = πππ₯ for any π₯ β (0, 1) implies that the pair (π, π) is not partially weakly increasing. Let (π, β€) be a partially ordered set. A mapping π : π β π is called a weak annihilator of a mapping π : π β π if πππ₯ β€ π₯ for all π₯ β π. Example 2 (see [3]). Let π = [0, 1] be endowed with usual ordering and π, π : π β π be two mappings given by ππ₯ = π₯2 and ππ₯ = π₯3 . It is clear that πππ₯ = π₯6 β€ π₯ for π₯ β π implies that π is a weak annihilator of π. Let (π, β€) be a partially ordered set. A mapping π is called a dominating if π₯ β€ ππ₯ for any π₯ β π. Example 3 (see [3]). Let π = [0, 1] be endowed with usual ordering and π : π β π a mapping defined by ππ₯ = π₯1/3 , since π₯ β€ π₯1/3 = ππ₯ for π₯ β π implies that π is a dominating mapping. A subset π of a partially ordered set π is said to be well ordered if every two elements of π are comparable. Let π be a nonempty subset of a metric space (π, π). Let π and π be mappings from a metric space (π, π) into itself. A point π₯ β π is a common fixed (resp., coincidence ) point of π and π if π₯ = ππ₯ = ππ₯ (resp., Sx = Tx). The set of fixed points 2 Abstract and Applied Analysis (resp., coincidence points) of π and π is denoted by πΉ(π, π) (resp., πΆ(π, π)). In 1986, Jungck [24] introduced the more generalized commuting mappings in metric spaces, called compatible mappings, which also are more general than the concept of weakly commuting mappings (that is, the mappings π, π : π β π are said to be weakly commuting if π(πππ₯, πππ₯) β€ π(ππ₯, ππ₯) for all π₯ β π) introduced by Sessa [34] as follows. Definition 4. Let π and π be mappings from a metric space (π, π) into itself. The mappings π and π are said to be compatible if lim π (πππ₯π , πππ₯π ) = 0, (1) πββ whenever {π₯π } is a sequence in π such that limπ β β ππ₯π = limπ β β ππ₯π = π§ for some π§ β π. In general, commuting mappings are weakly commuting and weakly commuting mappings are compatible, but the converses are not necessarily true and some examples can be found in [24β26]. In [27], Jungck and Rhoades introduced the concept of weakly compatible mappings and proved some common fixed point theorems for these mappings. Definition 5. The mappings π and π are said to be weakly compatible if they commute at coincidence points of π and π. In Djoudi and Nisse [21], we can find an example to show that there exists weakly compatible mappings which are not compatible mappings in metric spaces. Let Ξ¨ denote the set of all functions π : [0, β)5 β [0, β) such that (a) π is continuous; (b) π is strictly increasing in all the variables; π (π‘, π‘, π‘, 2π‘, 0) < π‘, π (0, 0, π‘, π‘, 0) < π‘, π (0, π‘, 0, 0, π‘) < π‘, (2) π (π‘, 0, 0, π‘, π‘) < π‘. It is easy to verify that the following functions are from the class Ξ¨, see [18]: π‘ π‘ π (π‘1 , π‘2 , π‘3 , π‘4 , π‘5 ) = π max {π‘1 , π‘2 , π‘3 , 4 , 5 } , 2 2 for π β (0, 1) ; π (π‘1 , π‘2 , π‘3 , π‘4 , π‘5 ) = π max {π‘1 , π‘2 , π‘3 , π‘4 + π‘5 }, 2 π (ππ₯, ππ¦) β€ π (π (π₯, π¦) , π (π₯, ππ₯) , π (π¦, ππ¦) , π (π₯, ππ¦) , π (π¦, ππ₯)) , (4) for π₯ β₯ π¦. Recently, Chen introduced π-contractive mappings. The purpose of this paper is to extend the results of Chen for four mappings, in the framework of ordered metric spaces. 2. Main Results Now, we give the main results in this paper. Theorem 7. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π, π, π, and π are selfmappings on π, the pairs (π, π) and (π, π) are partially weakly increasing with π(π) β π(π) and π(π) β π(π), and the dominating mappings π and π are weak annihilators of π and π, respectively. Further, suppose that for any two comparable elements π₯, π¦ β π, and π β Ξ¨, π (ππ₯, ππ¦) β€ π (π (ππ₯, ππ¦) , π (ππ₯, ππ₯) , π (ππ¦, ππ¦) , π (ππ₯, ππ¦) , π (ππ¦, ππ₯)) , (5) holds. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and either (a) π and π are compatible, π or π is continuous, and π, π are weakly compatible or (b) π and π are compatible, π or π is continuous, and π, π are weakly compatible, then π, π, π, and π have a common fixed point in π. Moreover, the set of common fixed points of π, π, π, and π is well ordered if and only if π, π, π, and π have one and only one common fixed point in π. (c) for all π‘ β [0, β) \ {0}, π (π‘, π‘, π‘, 0, 2π‘) < π‘, is a metric space. The mapping π : π β π is said to be a πcontractive mapping, if Proof. Let π₯0 β π be an arbitrary point. Since π(π) β π(π) and π(π) β π(π), we can construct the sequences {π₯π } and {π¦π } in π such that π¦2πβ1 = ππ₯2πβ2 = ππ₯2πβ1 , π¦2π = ππ₯2πβ1 = ππ₯2π , for each π β₯ 1. By assumptions, we have π₯2πβ2 β€ ππ₯2πβ2 = ππ₯2πβ1 β€ πππ₯2πβ1 β€ π₯2πβ1 , (3) for π β (0, 1) . Definition 6 (see [18]). Let (π, β€) be a partially ordered set and suppose that there exists a metric π in π such that (π, π) (6) π₯2πβ1 β€ ππ₯2πβ1 = ππ₯2π β€ πππ₯2π β€ π₯2π , (7) for each π β₯ 1. Thus, for each π β₯ 1, we have π₯π β€ π₯π+1 . Without loss of generality, we assume that π¦2π =ΜΈ π¦2π+1 for each π β₯ 1. Now, we claim that for all π β N, we have π (π¦π+1 , π¦π+2 ) < π (π¦π , π¦π+1 ) . (8) Abstract and Applied Analysis 3 Suppose to the contrary that π(π¦2π , π¦2π+1 ) β€ π(π¦2π+1 , π¦2π+2 ) for some π β N. Since π¦2π and π¦2π+1 are comparable, from (5), we have Again, π (π¦π(π)β1 , π¦π(π)β1 ) β€ π (π¦π(π)β1 , π¦π(π) ) + π (π¦π(π) , π¦π(π) ) π (π¦2π+1 , π¦2π+2 ) + π (π¦π(π) , π¦π(π)β1 ) , = π (ππ₯2π , ππ₯2π+1 ) π (π¦π(π) , π¦π(π) ) β€ π (π¦π(π) , π¦π(π)β1 ) + π (π¦π(π)β1 , π¦π(π)β1 ) β€ π (π (ππ₯2π , ππ₯2π+1 ) , π (ππ₯2π , ππ₯2π ) , π (ππ₯2π+1 , ππ₯2π+1 ) , + π (π¦π(π)β1 , π¦π(π) ) . (14) π (ππ₯2π , ππ₯2π+1 ) , π (ππ₯2π+1 , ππ₯2π )) Letting π β β in the above inequalities and using (10) and (13), we have = π (π (π¦2π , π¦2π+1 ) , π (π¦2π , π¦2π+1 ) , π (π¦2π+1 , π¦2π+2 ) , π (π¦2π , π¦2π+2 ) , π (π¦2π+1 , π¦2π+1 )) lim π (π¦π(π)β1 , π¦π(π)β1 ) = π. πββ = π (π (π¦2π , π¦2π+1 ) , π (π¦2π , π¦2π+1 ) , Again, π (π¦2π+1 , π¦2π+2 ) , π (π¦2π , π¦2π+2 ) , 0) π (π¦π(π)β1 , π¦π(π) ) β€ π (π¦π(π)β1 , π¦π(π)β1 ) + π (π¦π(π)β1 , π¦π(π) ) , β€ π (π (π¦2π , π¦2π+1 ) , π (π¦2π , π¦2π+1 ) , π (π¦2π+1 , π¦2π+2 ) , π (π¦π(π)β1 , π¦π(π)β1 ) β€ π (π¦π(π)β1 , π¦π(π) ) + π (π¦π(π)β1 , π¦π(π) ) . (16) π (π¦2π , π¦2π+1 ) + π (π¦2π+1 , π¦2π+2 ) , 0) β€ π (π (π¦2π+1 , π¦2π+2 ) , π (π¦2π+1 , π¦2π+2 ) , Letting π β β in the above inequalities and using (10) and (15), we have π (π¦2π+1 , π¦2π+2 ) , 2π (π¦2π+1 , π¦2π+2 ) , 0) < π (π¦2π+1 , π¦2π+2 ) , (9) which is a contradiction. Hence π(π¦2π+1 , π¦2π+2 ) β€ π(π¦2π , π¦2π+1 ) for each π β₯ 1. Similarly, we can prove that π(π¦2π+1 , π¦2π ) β€ π(π¦2π , π¦2πβ1 ) for each π β₯ 1. Therefore, we can conclude that (8) holds. Let us denote ππ = π(π¦π+1 , π¦π ). Then, from (8), ππ is a nonincreasing sequence and bounded below. Thus, it must converge to some π β₯ 0. If π > 0, then by the above inequalities, we have π β€ ππ+1 β€ π(ππ , ππ , ππ , 2ππ , 0). Taking the limit, as π β β, we have π β€ π β€ π(π, π, π, 2π, 0) < π, which is a contradiction. Hence, π (π¦π+1 , π¦π ) σ³¨β 0. π (π¦π(π) , π¦π(π)β1 ) < π. (11) From (11), it follows that (12) (18) πββ Also, again from (10), (15), and the inequality π (π¦π(π)β1 , π¦π(π)+1 ) β π (π¦π(π)β1 , π¦π(π) ) β€ π (π¦π(π) , π¦π(π)+1 ) , (19) it follows that lim π (π¦π(π)β1 , π¦π(π)+1 ) = π. πββ (20) Now, we have π (π¦π(π) , π¦π(π)+1 ) β€ π (π (π¦π(π)β1 , π¦π(π) ) , π (π¦π(π)β1 , π¦π(π) ) , π (π¦π(π) , π¦π(π)+1 ) , π (π¦π(π)β1 , π¦π(π)+1 ) , (21) Letting π β β, we get < π, Letting π β β and using (10), we have lim π (π¦π(π) , π¦π(π) ) = π. lim π (π¦π(π)β1 , π¦π(π) ) = π. Similarly, we have π β€ π (π, 0, 0, π, π) < π + π (π¦π(π)β1 , π¦π(π) ) . πββ (17) π (π¦π(π) , π¦π(π) )) . π β€ π (π¦π(π) , π¦π(π) ) β€ π (π¦π(π) , π¦π(π)β1 ) + π (π¦π(π)β1 , π¦π(π) ) lim π (π¦π(π)β1 , π¦π(π) ) = π. πββ (10) Now, we show that {π¦π } is a Cauchy sequence. Suppose that {π¦π } is not a Cauchy sequence. Then, there exists π > 0 for which we can find two sequences of natural numbers {π(π)} and {π(π)} with π(π) > π(π) > π such that π (π¦π(π) , π¦π(π) ) β₯ π, (15) (13) (22) which is a contradiction. Thus {π¦π } is a Cauchy sequence. Since π is a complete metric space, there exists π§ β π such 4 Abstract and Applied Analysis that π¦π β π§. Therefore, we have lim π¦ π β β 2π+1 Now, π₯2π β€ ππ₯2π and ππ₯2π β π§ implies π₯2π β€ π§. Thus, from (5), we obtain = lim ππ₯2π+1 = lim ππ₯2π = π§, πββ πββ lim π¦2π+2 = lim ππ₯2π+2 = lim ππ₯2π+1 = π§. πββ πββ (23) πββ Assume that π is continuous. Since π and π are compatible, we have lim πππ₯2π+2 = lim πππ₯2π+2 = ππ§. πββ πββ Letting π β β, we get π (πππ₯2π , ππ₯2π+1 ) π (π§, ππ§) β€ π (π (π§, ππ§) , 0, 0, π (π§, ππ§) , π (π§, ππ§)) β€ π (π (π2 π₯2π , ππ₯2π+1 ) , π (π2 π₯2π , πππ₯2π ) , < π (π§, ππ§) , Letting π β β, we get < π (ππ§, π§) , (26) which implies that ππ§ = π§. Now, it follows that π₯2π+1 β€ ππ₯2π+1 and ππ₯2π+1 β π§, π₯2π+1 β€ π§. From (5), we have π (ππ§, ππ₯2π+1 ) β€ π (π (ππ§, ππ₯2π+1 ) , π (ππ§, ππ§) , π (ππ₯2π+1 , ππ₯2π+1 ) , π (ππ§, ππ₯2π+1 ) , π (ππ₯2π+1 , ππ§)) . (27) < π (ππ§, π§) , π (π’, V) = π (ππ’, πV) β€ π (π (ππ’, πV) , π (ππ’, ππ’) , π (πV, πV) , π (ππ’, πV) , π (πV, ππ’)) (32) < π (π’, V) . (28) which implies that ππ§ = π§. Since π(π) β π(π), there exists π€ β π such that ππ§ = π§ = ππ§ = ππ€. Suppose that ππ€ =ΜΈ ππ€. Since π§ β€ ππ§ = ππ€ β€ πππ€ β€ π€ implies π§ β€ π€, from (5), we obtain π (ππ€, ππ€) = π (ππ§, ππ€) β€ π (π (ππ§, ππ€) , π (ππ§, ππ§) , π (ππ€, ππ€) , π (ππ§, ππ€) , π (ππ€, ππ§)) which implies that ππ§ = π§. Therefore, we have ππ§ = ππ§ = ππ§ = ππ§ = π§. If π is continuous, then, following the similar arguments, also we get the result. Similarly, the result follows when (b) holds. Now, suppose that the set of common fixed points of π, π, π, and π is well ordered. We claim that common fixed points of π, π, π, and π are unique. Assume that ππ’ = ππ’ = ππ’ = ππ’ = π’ and πV = πV = πV = πV = V, but π’ =ΜΈ V. Then, from (5), we have = π (π (π’, V) , 0, 0, π (π’, V) , π (V, π’)) Letting π β β, we get π (ππ§, π§) β€ π (0, π (ππ§, π§) , 0, 0, π (π§, ππ§)) (31) (25) π (ππ₯2π+1 , πππ₯2π ) ) . π (ππ§, π§) β€ π (π (ππ§, π§) , 0, 0, π (ππ§, π§) , π (π§, ππ§)) π (ππ§, ππ§) , π (ππ₯2π , ππ§) , π (ππ§, ππ₯2π )) . (30) (24) Also, π₯2π+1 β€ ππ₯2π+1 = ππ₯2π+2 . Now, we have π (ππ₯2π+1 , ππ₯2π+1 ) , π (π2 π₯2π , ππ₯2π+1 ) , π (ππ₯2π , ππ§) β€ π (π (ππ₯2π , ππ§) , π (ππ₯2π , ππ₯2π ) , (29) = π2 (0, 0, π (ππ€, ππ€) , π (ππ€, ππ€) , 0) < π (ππ€, ππ€) , which implies that ππ€ = ππ€. Since π and π are weakly compatible, ππ§ = πππ§ = πππ€ = πππ€ = πππ§ = ππ§. Thus, π§ is a coincidence point of π and π. This implies that π(π’, V) = 0, and hence π’ = V. Conversely, if π, π, π, and π have only one common fixed point, then the set of common fixed point of π, π, π, and π being singleton is well ordered. This completes the proof. Example 8. Consider π = [0, 1] βͺ {2, 3, 4, . . .} with usual ordering and σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π₯ β π¦σ΅¨σ΅¨σ΅¨ if π₯, π¦ β [0, 1] , π₯ =ΜΈ π¦; { { { {π₯ + π¦ if at least one of π₯ or π¦ β [0, 1] , π (π₯, π¦) = { { π₯ =ΜΈ π¦; { { if π₯ = π¦. {0 (33) Then (π, β€, π) is a complete partially ordered metric space. Abstract and Applied Analysis 5 Let π, π, π, and π be self-mappings on π defined as 0 { { { { { 1 { { { { 2 { { { π (π₯) = { { 1 { { { { { { { { {π₯ { { 0 { { { { { 1 { { { π (π₯) = { 2 { { { { { { {π₯ { if π₯ = 0; 1 if π₯ β (0, ] ; 2 then π, π, and π have a common fixed point in π. Moreover, the set of common fixed points of π, π, and π is well ordered if and only if π, π, and π have one and only one common fixed point in π. 1 if π₯ β ( , 1] ; 2 if π₯ β {2, 3, 4, . . .} ; if π₯ = 0; 1 if π₯ β (0, ] ; 2 1 if π₯ β ( , 1] βͺ {2, 3, 4, . . .} ; 2 (34) π (ππ₯, ππ¦) , π (ππ¦, ππ₯)) (a) π, π are compatible, π or π is continuous, and π, π are weakly compatible or (b) π, π are compatible, π or π is continuous, and π, π are weakly compatible, 1 if π₯ β€ ; 2 1 if π₯ β ( , 1] ; 2 then π, π, and π have a common fixed point in π. Moreover, the set of common fixed points of π, π, and π is well ordered if and only if π, π, and π have one and only one common fixed point in π. if π₯ β {2, 3, 4, . . .} . π‘ +π‘ 6 max {π‘1 , π‘2 , π‘3 , 4 5 } . 7 2 (35) Note that π, π, π, and π satisfy all the conditions given in Theorem 7. Moreover, 0 is a common fixed point of π, π, π, and π. If π = π, then we have the following result. Corollary 9. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π, π, and π are self-mappings on π, the pairs (π, π) and (π, π) are partially weakly increasing with π(π) β π(π) and π(π) β π(π), and the dominating mapping π is a weak annihilator of π and π. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, π (ππ₯, ππ¦) β€ π (π (ππ₯, ππ¦) , π (ππ₯, ππ₯) , π (ππ¦, ππ¦) , π (ππ₯, ππ¦) , π (ππ¦, ππ₯)) (37) holds. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and either Define function π : [0, β)5 β [0, β) by the formula π (π‘1 , π‘2 , π‘3 , π‘4 , π‘5 ) = Corollary 10. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π, π, and π are selfmappings on π, the pairs (π, π) and (π, π) are partially weakly increasing with π(π) β π(π) and π(π) β π(π), and the dominating mappings π and π are weak annihilators of π. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, π (ππ₯, ππ¦) β€ π (π (ππ₯, ππ¦) , π (ππ₯, ππ₯) , π (ππ¦, ππ¦) , 1 0 if π₯ β€ ; { { { 2 { { { 1 {1 if π₯ β ( , 1] ; π (π₯) = { 2 2 { { { { { {π₯ β 1 if π₯ β {2, 3, 4, . . .} ; { 0 { { { { { { { π (π₯) = {2π₯ β 1 { { { { { {π₯ { (a) π, π are compatible, π or π is continuous, and π, π are weakly compatible or (b) π, π are compatible, π or π is continuous, and π, π are weakly compatible, Corollary 11. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π and π are self-mappings on π, the pair (π, π) is partially weakly increasing with π(π) β π(π), and the dominating mapping π is a weak annihilator of π. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, π (ππ₯, ππ¦) β€ π (π (ππ₯, ππ¦) , π (ππ₯, ππ₯) , π (ππ¦, ππ¦) , π (ππ₯, ππ¦) , π (ππ¦, ππ₯)) (38) holds. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and, further, π, π are compatible, π or π is continuous, and π, π are weakly compatible, then π and π have a common fixed point in π. Moreover, the set of common fixed points of π and π is well ordered if and only if π and π have one and only one common fixed point in π. 3. Applications (36) holds. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and either The aim of the section is to apply our new results to mappings involving contractions of integral type. For this purpose, denote by Ξ the set of functions π : [0, β) β [0, β) satisfying the following hypotheses: 6 Abstract and Applied Analysis (h1) π is a Lebesgue-integrable mapping on each compact of [0, β); π (h2) for any π > 0, we have β«0 π(π‘) > 0. Corollary 12. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π, π, π, and π are selfmappings on π, the pairs (π, π) and (π, π) are partially weakly increasing with π(π) β π(π) and π(π) β π(π), and the dominating mappings π and π are weak annihilators of π and π, respectively. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, π(ππ₯,ππ¦) β« 0 π(π(ππ₯,ππ¦),π(ππ₯,ππ₯),π(ππ¦,ππ¦),π(ππ₯,ππ¦),π(ππ¦,ππ₯)) πΌ (π ) ππ (39) holds, where πΌ β Ξ. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and either (a) π, π are compatible, π or π is continuous, and π, π are weakly compatible or (b) π, π are compatible, π or π is continuous, and π, π are weakly compatible, then π, π, π, and π have a common fixed point in π. Moreover, the set of common fixed points of π, π, π, and π is well ordered if and only if π, π, π, and π have one and only one common fixed point in π. Corollary 13. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π, π, and π are self-mappings on π, the pairs (π, π) and (π, π) are partially weakly increasing with π(π) β π(π) and π(π) β π(π), and the dominating mappings π and π are weak annihilators of π. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, 0 0 πΌ (π ) ππ β€β« π(π(ππ₯,ππ¦),π(ππ₯,ππ₯),π(ππ¦,ππ¦),π(ππ₯,ππ¦),π(ππ¦,ππ₯)) πΌ (π ) ππ 0 π(ππ₯,ππ¦) π(ππ₯,ππ¦) β« 0 β€β« β« Corollary 14. Let (π, β€) be a partially ordered set, and suppose that there exists a metric π on π such that (π, π) is a complete metric space. Suppose that π and π are self-mappings on π, the pair (π, π) is a partially weakly increasing with π(π) β π(π), and the dominating mapping π is a weak annihilator of π. Further, suppose that there exists the function π β Ξ¨ such that, for any two comparable elements π₯, π¦ β π, πΌ (π ) ππ π(π(ππ₯,ππ¦),π(ππ₯,ππ₯),π(ππ¦,ππ¦),π(ππ₯,ππ¦),π(ππ¦,ππ₯)) β€β« 0 πΌ (π ) ππ (40) holds, where πΌ β Ξ. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and either (a) π, π are compatible, π or π is continuous, and π, π are weakly compatible or (b) π, π are compatible, π or π is continuous, and π, π are weakly compatible, then π, π, and π have a common fixed point in π. Moreover, the set of common fixed points of π, π, and π is well ordered if and only if π, π, and π have one and only one common fixed point in π. πΌ (π ) ππ (41) holds, where πΌ β Ξ. If, for a nondecreasing sequence {π₯π } with π₯π β€ π¦π for all π β₯ 1, π¦π β π’ implies that π₯π β€ π’ and, further, π, π are compatible, π or π is continuous, and π, π are weakly compatible, then π and π have a common fixed point in π. Moreover, the set of common fixed points of π and π is well ordered if and only if π and π have one and only one common fixed point in π. 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