Research article

A tripled coincidence point technique for solving integral equations via an upper class of type II

  • The goal of this paper is to obtain some tripled coincidence point results for generalized contraction mappings in the setting of JS-metric spaces endowed with a partial order. Furthermore, illustrative examples to support the theoretical results and the application are obtained. Finally, some theoretical results are applied to discuss the existence of a solution for a system of non-homogeneous and homogeneous integral equations as applications.

    Citation: Hasanen A. Hammad, Hassen Aydi, Aiman Mukheimer. A tripled coincidence point technique for solving integral equations via an upper class of type II[J]. AIMS Mathematics, 2023, 8(4): 9795-9819. doi: 10.3934/math.2023494

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  • The goal of this paper is to obtain some tripled coincidence point results for generalized contraction mappings in the setting of JS-metric spaces endowed with a partial order. Furthermore, illustrative examples to support the theoretical results and the application are obtained. Finally, some theoretical results are applied to discuss the existence of a solution for a system of non-homogeneous and homogeneous integral equations as applications.



    The fixed point theory is one of the most powerful and productive tools from the nonlinear analysis and it can be considered to be the kernel of the nonlinear analysis since 1960. It has been classified into two major areas: metric fixed point theory and topological fixed point theory. The fixed point theory finds its roots with the method of successive approximations to prove the existence of solutions of differential equations introduced independently by Liouville [1] in 1837 and Picard [2] in 1890. While, officially, it was launched at the start of the 20th century as an important part of functional analysis. The best known result from the fixed point theory is Banach Contraction Principle [3] (1922), which can be considered to be the beginning of this theory. In a metric space, setting it can be briefly stated as follows: Every contraction on a complete metric space admits a unique fixed point. Furthermore, it provides a fixed point approximation algorithm as the limit of an iterated sequence. This result has been extended and generalized last years in various directions. Among these extensions, we cite the main result of Jleli and Samet [4], where they initiated a new generalized metric, called as JS-metric. It is given as follows:

    Definition 1.1. [4] Let ξ:×[0,+) be a mapping justifying, for each ϑ,ϖ,

    (i) if ξ(ϑ,ϖ)=0, then ϑ=ϖ;

    (ii) ξ(ϑ,ϖ)=ξ(ϖ,ϑ);

    (iii) there is Z>0 so that

    limiξ(ϑi,ϑ)ξ(ϑ,ϖ)Zlim supiξ(ϑi,ϖ).

    Then ξ is a JS-metric on and the pair (,ξ) is called a JS-metric space.

    Definition 1.2. Let (,ξ) be a JS-metric space and {xn} be a sequence in (,ξ).

    (i) {xn} is said to be ξ-convergent to x in if limnξ(xn,x)=0.

    (ii) {xn} is said to be ξ-Cauchy sequence if limn,mξ(xn,xm)=0.

    (iii) (,ξ) is said to be ξ-complete if every ξ-Cauchy sequence in is ξ-convergent to some point x.

    After the work of Jleli and Samet [4], there is an immense literature in fixed point theory and its applications in this setting. For more details, see [5,6,7].

    In paper [8], the notions of mixed-monotone functions and coupled fixed points were initiated and studied. Under partially ordered metric spaces (POMSs) and abstract spaces, some main results in this direction have been driven, for broadening, see [4,8,11,12,13,14].

    In 2011, Berinde and Borcut [15], introduced the definition of mixed monotone property and the definition of tripled fixed point for a mapping T:3=:×× and established tripled fixed point theorems for contractive type mappings having that property in partially ordered metric spaces. Later, Borcut [16] and Berinde and Borcut [17] have introduced the notion of a tripled coincidence point for a pair of nonlinear contractive mappings T:3 and f:. Subsequently, Aydi et al. [18] have proved some new tripled fixed point theorems in abstract metric spaces. Further works dealing in this direction have been appeared, see [19,20,21].

    Combining the notion of triangular αadmissible property and the concept of upper class of type II, the aim of this paper is to establish some tripled coincidence point results for a pair of generalized contraction type mappings Ξ:3 and r: in the context of JS-metric spaces equipped with a partial order. The obtained results are supported by some concrete examples. At the end, we ensure the existence of a solution for a system of non-homogeneous and homogeneous integral equations.

    We start this section with the following concepts.

    Consider a nonempty set . Assume that Ξ:3 and r: are two mappings. We say that r commutes with Ξ if

    rΞ(ϑ,ϖ,ϰ)=Ξ(rϑ,rϖ,rϰ), ϑ,ϖ,ϰ.

    According to [4], for a partial order , define ={(ϑ,ϖ)2:ϑϖ}. Then Ξ has the r monotone property, if for each ϑ,ϖ,ϰ,

    ϑ1,ϑ2, (rϑ1,rϑ2) implies (Ξ(ϑ1,ϖ,ϰ),Ξ(ϑ2,ϖ,ϰ)),ϖ1,ϖ2, (rϖ1,rϖ2) implies (Ξ(ϑ,ϖ1,ϰ),Ξ(ϑ,ϖ2,ϰ))

    and

    ϰ1,ϰ2, (rϰ1,rϰ2) implies (Ξ(ϑ,ϖ,ϰ1),Ξ(ϑ,ϖ,ϰ2)).

    After that, we generalize the notion of triangular αadmissible property as follows:

    Definition 2.1. Assume that Ξ:3, r: and α:3×3[0,+] so that the postulates below hold:

    (i) If α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))1, then

    α((Ξ(ϑ,ϖ,ϰ),Ξ(ϖ,ϰ,ϑ),Ξ(ϰ,ϑ,ϖ)),(Ξ(˜ϑ,˜ϖ,˜ϰ),Ξ(˜ϖ,˜ϰ,˜ϑ),Ξ(˜ϰ,˜ϑ,˜ϖ)))1.

    (ii) If α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))1 and

    α((r˜ϑ,r˜ϖ,r˜ϰ),(Ξ(˜ϑ,˜ϖ,˜ϰ),Ξ(˜ϖ,˜ϰ,˜ϑ),Ξ(˜ϰ,˜ϑ,˜ϖ)))1,

    then

    α((rϑ,rϖ,rϰ),(Ξ(˜ϑ,˜ϖ,˜ϰ),Ξ(˜ϖ,˜ϰ,˜ϑ),Ξ(˜ϰ,˜ϑ,˜ϖ)))1.

    Here, we say that Ξ and r is a generalized triangular α admissible (αGTA, for short).

    Next, we extend the concept of upper class of type II [11] as follows.

    Definition 2.2. Let :[0,+]3R{+} and :[0,+]2R{+} be two given mappings. We say that the pair (,) is an upper class of type II (ΩII, for short) if for all ϑ,ϖ,ϰ,˜ϑ,˜ϖ,˜ϰ[0,+], the postulates below hold:

    (i) (1,1,ϖ)(ϑ,ϖ,ϰ), whenever 1ϖ,ϰ;

    (ii) (˜ϑ,˜ϖ)(1,˜ϖ), whenever ˜ϑ1;

    (iii) ϰ˜ϑ˜ϖ, whenever (1,1,ϰ)(˜ϑ,˜ϖ).

    Example 2.3. Let ,:[0,+]2R{+} be a function described as follows:

    (1) (ϑ,ϖ,ϰ)={(ϰ+z)ϑϖ,if ϑ,ϖ,ϰ[0,+),+,otherwise,   (˜ϑ,˜ϖ)={˜ϑ˜ϖ+z,if ˜ϑ,˜ϖ[0,+),+,otherwise.

    (2) (ϑ,ϖ,ϰ)={(ϑ+z)ϰϖ,if ϑ,ϖ,ϰ[0,+),+,otherwise,   (˜ϑ,˜ϖ)={(1+z)˜ϑ˜ϖ,if ˜ϑ,˜ϖ[0,+),+,otherwise.

    (3) (ϑ,ϖ,ϰ)={ϰ,if ϑ,ϖ,ϰ[0,+),+,otherwise,   (˜ϑ,˜ϖ)={˜ϑ˜ϖ,if ˜ϑ,˜ϖ,˜ϰ[0,+),+,otherwise.

    (4) (ϑ,ϖ,ϰ)={ϑiϖjϰkif ϑ,ϖ,ϰ[0,+),+,otherwise,   (˜ϑ,˜ϖ)={(˜ϑ˜ϖ)k,if ˜ϑ,˜ϖ[0,+),+,otherwise,

    for z>1 and i,j,kN. Then the pair (,) is ΩII.

    Here, the triplet (,ξ,) is a complete POJSM-space and ˜Φ consists of all ϕ:[0,+]3[0,1) satisfying:

    (ϕ1) For any ϑ,ϖ[0,+], ϕ(ϑ,ϖ)=ϕ(ϖ,ϑ).

    (ϕ2) For {ϑi}, {ϖi}[0,+], limiϕ(ϑi,ϖi,ϰi)=1limiϑi=limiϖi=limiϰi=0.

    Also, we define the mappings Ξ:3 and r: so that the following properties hold:

    (a) Ξ(3)r();

    (b) Ξ is r monotone;

    (c) r commutes with Ξ and is ξcontinuous.

    Now, our first theorem becomes valid for presentation, which generalizes the results of [7].

    Theorem 2.4. Assume that the postulates below are fulfilled:

    (i) For any ΩII of the pair (,), there exist α:3×3[0,+] and ϕΦ fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the below inequality is verified:

    (1,α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ)),ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ)))(ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)),((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))),

    where

    ((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))=max{ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ),ξ(rϑ,Ξ(ϑ,ϖ,ϰ)),ξ(rϖ,Ξ(ϖ,ϰ,ϑ)),ξ(rϰ,Ξ(ϰ,ϑ,ϖ)),ξ(r˜ϑ,Ξ(r˜ϑ,r˜ϖ,r˜ϰ)),ξ(r˜ϖ,Ξ(r˜ϖ,r˜ϰ,r˜ϑ)),ξ(rϰ,Ξ(r˜ϰ,r˜ϑ,r˜ϖ))};

    (ii) Ξ and r is αGTA, and there exist ϑ0,ϖ0,ϰ0,

    (rϑ0,Ξ(ϑ0,ϖ0,ϰ0)),(rϖ0,Ξ(ϖ0,ϰ0,ϑ0)),(rϰ0,Ξ(ϰ0,ϑ0,ϖ0))

    so that

    α((rϑ0,rϖ0,rϰ0),(Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϑ0,ϖ0,ϰ0)))1,α((rϖ0,rϰ0,rϑ0),(Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϰ0,ϑ0,ϖ0),Ξ(ϖ0,ϰ0,ϑ0)))1

    and

    α((rϰ0,rϑ0,rϖ0),(Ξ(ϰ0,ϑ0,ϖ0),Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϰ0,ϑ0,ϖ0)))1;

    (iii) If limiξ(rϑi,rϑi+1)=0, limiξ(rϖi,rϖi+1)=0 and limiξ(rϰi,rϰi+1)=0, then

    sup{ξ(rϑ0,rϑi),ξ(rϖ0,rϖi),ξ(rϰ0,rϰi)}<,

    where {ϑi}, {ϖi} and {ϰi} are sequences in ; (iv) Ξ is ξcontinuous.

    Then Ξ and r have a tripled coincidence point (TCP) in .

    Proof. Assume that ϑ0,ϖ0,ϰ0 justifying assertion (ii). Because Ξ(3)r(), we can select ϑ1,ϖ1,ϰ1 so that rϑ1=Ξ(ϑ0,ϖ0,ϰ0), rϖ1=Ξ(ϖ0,ϰ0,ϑ0) and rϰ1=Ξ(ϰ0,ϑ0,ϖ0). Analogously, rϑ2=Ξ(ϑ1,ϖ1,ϰ1), rϖ2=Ξ(ϖ1,ϰ1,ϑ1) and rϰ2=Ξ(ϰ1,ϑ1,ϖ1). In the same scenario, {ϑi}, {ϖi} and {ϰi} are obtained with

    rϑi+1=Ξ(ϑi,ϖi,ϰi), rϖi+1=Ξ(ϖi,ϰi,ϑi) and rϰi+1=Ξ(ϰi,ϑi,ϖi).

    For some natural number i0, if rϑi0+1=rϑi0, rϖi0+1=rϖi0 and rϰi0+1=rϰi0, then Ξ and r have a TCP. So, for some positive integer i, assume that

    rϑi+1rϑi, rϖi+1rϖi and rϰi+1rϰi.

    Based on assumption (ii), we have

    (rϑ0,rϑ1),(rϖ0,rϖ1) and (rϰ0,rϰ1).

    Because Ξ is r monotone,

    (Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϑ1,ϖ1,ϰ1)), (Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϖ1,ϰ1,ϑ1)),(Ξ(ϰ0,ϑ0,ϖ0),Ξ(ϰ1,ϑ1,ϖ1)),

    that is,

    (rϑ1,rϑ2),(rϖ1,rϖ2) and (rϰ1,rϰ2).

    Repeating the same approach, we get

    (rϑi,rϑi+1),(rϖi,rϖi+1) and (rϰi,rϰi+1), iN.

    From the transitivity of , one can obtain

    (rϑi,rϑi+m),(rϖi,rϖi+m) and (rϰi,rϰi+m), i,mN.

    Again, applying the postulate (ii), we have

    α((rϑ0,rϖ0,rϰ0),(rϑ1,rϖ1,rϰ1))=α((rϑ0,rϖ0,rϰ0),(Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϰ0,ϑ0,ϖ0)))1.

    Since Ξ and r are αGTA,

    α((rϑ1,rϖ1,rϰ1),(rϑ2,rϖ2,rϰ2))=α((Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϰ0,ϑ0,ϖ0)),(Ξ(ϑ1,ϖ1,ϰ1),Ξ(ϖ1,ϰ1,ϑ1),Ξ(ϰ1,ϑ1,ϖ1))).

    By induction, one can deduce that

    α((rϑi,rϖi,rϰi),(rϑi+1,rϖi+1,rϰi+1))1, iN.

    Analogously, we can obtain

    α((rϖi,rϰi,rϑi),(rϖi+1,rϰi+1,rϑi+1))1, iN,

    and

    α((rϰi,rϑi,rϖi),(rϰi+1,rϑi+1,rϖi+1))1, iN.

    Since Ξ and r are αGTA,

    α((rϑi,rϖi,rϰi),(rϑi+m,rϖi+m,rϰi+m))1,α((rϖi,rϰi,rϑi),(rϖi+m,rϰi+m,rϑi+m))1,α((rϰi,rϑi,rϖi),(rϰi+m,rϑi+m,rϖi+m))1, i,mN.

    Next, we want to prove limiξ(rϑi,rϑi+1)=0, limiξ(rϖi,rϖi+1)=0 and limiξ(rϰi,rϰi+1)=0. For this, we use the opposite technique, suppose that either limiξ(rϑi,rϑi+1)0, or limiξ(rϖi,rϖi+1)0 or limiξ(rϰi,rϰi+1)0. Then there is ϵ>0 for which we have a subsequence {is} so that sis and

    max{ξ(rϑis,rϑis+1),ξ(rϖis,rϖis+1),ξ(rϰis,rϰis+1),ξ(rϰis,rϰis+1)}ϵ.

    Consider

    (1,1,ξ(rϑis,rϑis+1))=(1,1,ξ(Ξ(ϑis1,ϖis1,ϰis1),Ξ(ϑis,ϖis,ϰis)))(1,α((ϑis1,ϖis1,ϰis1),(ϑis,ϖis,ϰis)),ξ(Ξ(ϑis1,ϖis1,ϰis1),Ξ(ϑis,ϖis,ϰis)))(ϕ(ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1),ξ(rϰis1,rϰis)),((rϑis1,rϑis),(rϖis1,rϖis1),(rϰis1,rϰis))). (2.1)

    Again,

    (1,1,ξ(rϖis,rϖis+1))=(1,1,ξ(Ξ(ϖis1,ϰis1,ϑis1),Ξ(ϖis,ϰis,ϑis)))(1,α((ϖis1,ϰis1,ϑis1),(ϖis,ϰis,ϑis)),ξ(Ξ(ϖis1,ϰis1,ϑis1),Ξ(ϖis,ϰis,ϑis)))(ϕ(ξ(rϖis1,rϖis1),ξ(rϰis1,rϰis),ξ(rϑis1,rϑis)),((rϖis1,rϖis1),(rϰis1,rϰis),(rϑis1,rϑis))), (2.2)

    and

    (1,1,ξ(rϰis,rϰis+1))=(1,1,ξ(Ξ(ϰis1,ϑis1,ϖis1),Ξ(ϰis,ϑis,ϖis)))(1,α((ϰis1,ϑis1,ϖis1),(ϰis,ϑis,ϖis)),ξ(Ξ(ϰis1,ϑis1,ϖis1),Ξ(ϰis,ϑis,ϖis)))(ϕ(ξ(rϰis1,rϰis),ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1)),((rϰis1,rϰis),(rϑis1,rϑis),(rϖis1,rϖis1))). (2.3)

    The inequalities (2.1)–(2.3) lead to

    ξ(rϑis,rϑis+1)ϕ(ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1),ξ(rϰis1,rϰis))×((rϑis1,rϑis),(rϖis1,rϖis1),(rϰis1,rϰis)), (2.4)
    ξ(rϖis,rϖis+1)ϕ(ξ(rϖis1,rϖis1),ξ(rϰis1,rϰis),ξ(rϑis1,rϑis))×((rϖis1,rϖis1),(rϰis1,rϰis),(rϑis1,rϑis)), (2.5)

    and

    ξ(rϰis,rϰis+1)ϕ(ξ(rϰis1,rϰis),ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1))×((rϰis1,rϰis),(rϑis1,rϑis),(rϖis1,rϖis1)). (2.6)

    Because ϕ(ϑ,ϖ,ϰ)[0,1) for any ϑ,ϖ,ϰ[0,+], we get

    ((rϑis1,rϑis),(rϖis1,rϖis1),(rϰis1,rϰis))=((rϖis1,rϖis1),(rϰis1,rϰis),(rϑis1,rϑis))=((rϰis1,rϰis),(rϑis1,rϑis),(rϖis1,rϖis1))=max{ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1),ξ(rϰis1,rϰis)}. (2.7)

    It follows from (2.4)–(2.7) that max{ξ(rϑis,rϑis+1),ξ(rϖis,rϖis+1),ξ(rϰis,rϰis+1)}

    max{ξ(rϑis,rϑis+1),ξ(rϖis,rϖis+1),ξ(rϰis,rϰis+1)}ϕ(ξ(rϰis1,rϰis),ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1))×max{ξ(rϰis1,rϰis),ξ(rϑis1,rϑis),ξ(rϖis1,rϖis1)}.

    Using this concept, we can write

    max{ξ(rϑis,rϑis+1),ξ(rϖis,rϖis+1),ξ(rϰis,rϰis+1)}isv=1ϕ(ξ(rϰisv,rϰis+1v),ξ(rϑisv,rϑis+1v),ξ(rϖisv,rϖis+1v))×max{ξ(rϰ0,rϰ1),ξ(rϑ0,rϑ1),ξ(rϖ0,rϖ1)}.

    Select vs so that

    ϕ(ξ(rϰisvs,rϰis+1vs),ξ(rϑisvs,rϑis+1vs),ξ(rϖisvs,rϖis+1vs))=max1vis{ϕ(ξ(rϰisv,rϰis+1v),ξ(rϑisv,rϑis+1v),ξ(rϖisv,rϖis+1v))}.

    Consider

    =lim sups(ϕ(ξ(rϰisvs,rϰis+1vs),ξ(rϑisvs,rϑis+1vs),ξ(rϖisvs,rϖis+1vs))).

    If <1, then

    limsmax{ξ(rϑis,rϑis+1),ξ(rϖis,rϖis+1),ξ(rϰis,rϰis+1)}=0.

    This contradicts the hypothesis. If =1, for suitability, we assume that

    limsϕ(ξ(rϰisvs,rϰis+1vs),ξ(rϑisvs,rϑis+1vs),ξ(rϖisvs,rϖis+1vs))=1.

    Because ϕ˜Φ,

    limsξ(rϰisvs,rϰis+1vs)=0, limsξ(rϑisvs,rϑis+1vs)=0 and limsξ(rϖisvs,rϖis+1vs)=0.

    That is, there is an s0N so that

    ξ(rϰis0vs0,rϰis0+1vs0)<ϵ3, ξ(rϑis0vs0,rϑis0+1vs0)<ϵ3and ξ(rϖis0vs0,rϖis0+1vs0)<ϵ3.

    Thus, we get

    ϵmax{ξ(rϑis0,rϑis0+1),ξ(rϖis0,rϖis0+1),ξ(rϰis0,rϰis0+1)}vs0j=1ϕ(ξ(rϰis0j,rϰis0+1j),ξ(rϑis0j,rϑis0+1j),ξ(rϖis0j,rϖis0+1j))×max{ξ(rϰis0vs0,rϰis0+1vs0),ξ(rϑis0vs0,rϑis0+1vs0),ξ(rϖis0vs0,rϖis0+1vs0)}<ϵ3,

    a contradiction, hence, we have

    limiξ(rϑi,rϑi+1)=0, limiξ(rϖi,rϖi+1)=0 and limiξ(rϰi,rϰi+1)=0. (2.8)

    Now, we claim that {rϑi}, {rϖi} and {rϰi} are ξCauchy sequences. Suppose that {rϑi}, {rϖi} and {rϰi} are not ξCauchy sequences, so for each sN with subsequences is,jss such that

    max{ξ(rϑis,rϑis+js),ξ(rϖis,rϖis+js),ξ(rϰis,rϰis+js)}ϵ for some ϵ>0.

    Consider

    (1,1,ξ(rϑis,rϑis+js))=(1,1,ξ(Ξ(ϑis1,ϖis1,ϰis1),Ξ(ϑis+js1,ϖis+js1,ϰis+js)))(1,α((ϑis1,ϖis1,ϰis1),(ϑis+js1,ϖis+js1,ϰis+js1)),ξ(Ξ(ϑis1,ϖis1,ϰis1),Ξ(ϑis+js1,ϖis+js1,ϰis+js1)))(ϕ(ξ(rϑis1,rϑis+js1),ξ(rϖis1,rϖis+js1),ξ(rϰis1,rϰis+js1)),((rϑis1,rϑis+js1),(rϖis1,rϖis+js1),(rϰis1,rϰis+js1))), (2.9)
    (1,1,ξ(rϖis,rϖis+js))=(1,1,ξ(Ξ(ϖis1,ϰis1,ϑis1),Ξ(ϖis+js1,ϰis+js,ϑis+js1)))(1,α((ϖis1,ϰis1,ϑis1),(ϖis+js1,ϰis+js1,ϑis+js1)),ξ(Ξ(ϖis1,ϰis1,ϑis1),Ξ(ϖis+js1,ϰis+js1,ϑis+js1)))(ϕ(ξ(rϖis+js1,rϖis1),ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1)),((rϖis1,rϖis+js1),(rϰis1,rϰis+js1),(rϑis1,rϑis+js1))), (2.10)

    and

    (1,1,ξ(rϰis,rϰis+js))=(1,1,ξ(Ξ(ϰis1,ϑis1,ϖis1),Ξ(ϰis+js,ϑis+js1),ϖis+js1))(1,α((ϰis1,ϑis1,ϖis1),(ϰis+js1,ϑis+js1,ϖis+js1)),ξ(Ξ(ϰis1,ϑis1,ϖis1),Ξ(ϰis+js1,ϑis+js1,ϖis+js1)))(ϕ(ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1),ξ(rϖis+js1,rϖis1)),((rϰis1,rϰis+js1),(rϑis1,rϑis+js1),(rϖis1,rϖis+js1))). (2.11)

    Based on (2.9)–(2.11) and the above properties, one can obtain

    ξ(rϑis,rϑis+js)ϕ(ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1),ξ(rϖis+js1,rϖis1))×((rϰis1,rϰis+js1),(rϑis1,rϑis+js1),(rϖis1,rϖis+js1)), (2.12)
    ξ(rϖis,rϖis+js)ϕ(ξ(rϖis+js1,rϖis1),ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1))×((rϖis1,rϖis+js1),(rϰis1,rϰis+js1),(rϑis1,rϑis+js1)), (2.13)

    and

    ξ(rϰis,rϰis+js)ϕ(ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1),ξ(rϖis+js1,rϖis1))×((rϰis1,rϰis+js1),(rϑis1,rϑis+js1),(rϖis1,rϖis+js1)), (2.14)

    respectively. Using (2.8), we get

    ((rϰis1,rϰis+js1),(rϑis1,rϑis+js1),(rϖis1,rϖis+js1))=((rϖis1,rϖis+js1),(rϰis1,rϰis+js1),(rϑis1,rϑis+js1))=((rϰis1,rϰis+js1),(rϑis1,rϑis+js1),(rϖis1,rϖis+js1))=max{ξ(rϰis1,rϰis+js1),ξ(rϑis1,rϑis+js1),ξ(rϖis1,rϖis+js1). (2.15)

    From (2.12)–(2.15), we have

    max{ξ(rϑis,rϑis+js),ξ(rϖis,rϖis+js),ξ(rϰis,rϰis+js)}ϕ(ξ(rϰis+js1,rϰis1),ξ(rϑis1,rϑis+js1),ξ(rϖis+js1,rϖis1))×max{ξ(rϰis1,rϰis+js1),ξ(rϑis1,rϑis+js1),ξ(rϖis1,rϖis+js1).

    Thus,

    max{ξ(rϑis,rϑis+js),ξ(rϖis,rϖis+js),ξ(rϰis,rϰis+js)}isv=1ϕ(ξ(rϰisv,rϰis+jsv),ξ(rϑisv,rϑis+jsv),ξ(rϖisv,rϖis+jsv))×max{ξ(rϰ0,rϰjs),ξ(rϑ0,rϑjs),ξ(rϖ0,rϖjs)}.

    Select vs so that

    ϕ(ξ(rϰisvs,rϰis+jsvs),ξ(rϑisvs,rϑis+jsvs),ξ(rϖisvs,rϖis+jsvs))=max1vis{ϕ(ξ(rϰisv,rϰis+jsv),ξ(rϑisv,rϑis+jsv),ξ(rϖisv,rϖis+jsv))}.

    Define

    =lim sups(ϕ(ξ(rϰisvs,rϰis+jsvs),ξ(rϑisvs,rϑis+jsvs),ξ(rϖisvs,rϖis+jsvs))).

    If <1, then

    limsmax{ξ(rϑis,rϑis+js),ξ(rϖis,rϖis+js),ξ(rϰis,rϰis+js)}=0.

    This is impossible to happen because of our hypothesis.

    If =1, for convenience, consider

    limsϕ(ξ(rϰisvs,rϰis+jsvs),ξ(rϑisvs,rϑis+jsvs),ξ(rϖisvs,rϖis+jsvs))=1.

    Because ϕ˜Φ,

    limsξ(rϰisvs,rϰis+jsvs)=0, limsξ(rϑisvs,rϑis+jsvs)=0 and limsξ(rϖisvs,rϖis+jsvs)=0.

    That is, there is an s0N so that

    ξ(rϰis0vs0,rϰis0+js0vs0)<ϵ3, ξ(rϑis0vs0,rϑis0+js0vs0)<ϵ3and ξ(rϖis0vs0,rϖis0+js0vs0)<ϵ3.

    Thus, we get

    ϵmax{ξ(rϑis0,rϑis0+js0),ξ(rϖis0,rϖis0+js0),ξ(rϰis0,rϰis0+js0)}vs0j=1ϕ(ξ(rϰis0j,rϰis0+js0j),ξ(rϑis0j,rϑis0+js0j),ξ(rϖis0j,rϖis0+js0j))×max{ξ(rϰis0vs0,rϰis0+js0vs0),ξ(rϑis0vs0,rϑis0+js0vs0),ξ(rϖis0vs0,rϖis0+js0vs0)}<ϵ3.

    This is a contradiction, therefore, {rϑi}, {rϖi} and {rϰi} are ξCauchy sequences. The completeness of (,ξ) implies that for some u,u,u,

    limiξ(Ξ(ϑi,ϖi,ϰi),u)=limiξ(rϑi,u)=0,limiξ(Ξ(ϖi,ϰi,ϑi),u)=limiξ(rϖi,u)=0and limiξ(Ξ(ϰi,ϑi,ϖi),u)=limiξ(rϰi,u)=0.

    Since r is continuous, we have

    limiξ(r(Ξ(ϑi,ϖi,ϰi)),ru)=0,limiξ(r(Ξ(ϖi,ϰi,ϑi)),ru)=0and limiξ(r(Ξ(ϰi,ϑi,ϖi)),ru)=0.

    Also, the continuity of Ξ leads to

    limiξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u))=0,limiξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u))=0and limiξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))=0.

    By alteration between Ξ and r, we have that Ξ(u,u,u)=ru, Ξ(u,u,u)=ru and Ξ(u,u,u)=ru, therefore Ξ and r have a TCP.

    Corollary 2.5. The results of Theorem 2.4 are still true if we replace the stipulation (i) with one of the hypotheses below: (note ((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)) is defined in the above theorem):

    (1) There exist ϕΦ and z>1, fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the following inequality is satisfied:

    (ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))+z)α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))+z.

    (2) There exist ϕΦ and z>1, fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the following inequality is satisfied:

    (α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))+z)ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))(1+z)ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    (3) There exists ϕΦ, justifying, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the following inequality holds:

    ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    (4) There exists ϕΦ, satisfying, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the inequality below is obtained:

    {α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))}j{ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))}k{ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))}k{((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))}k,

    for all positive integers j and k.

    Proof. The proof follows immediately from Example 2.3 by applying the following in Theorem 2.4.

    (1) Use the value of and from Example 2.3(1).

    (2) Use the value of and from Example 2.3(2).

    (3) Use the value of and from Example 2.3(3).

    (4) Use the value of and from Example 2.3(4).

    Remark 2.6 ● In a bmetric and a bmetric-like space, Corollary 2.5 holds.

    ● The stipulation (3) of Corollary 2.5 generalizes Theorem 3.2 of [13].

    ● The stipulation (4) of Corollary 2.5 extends and unifies Theorem 3.1 of [7] when j=k=1.

    Example 2.7. Consider =[0,+]. Define ξ(ϑ,ϖ)=max{ϑ,ϖ}, ϑ,ϖ. Describe the mappings Ξ:3, r: and α:3×3[0,+] by

    Ξ(ϑ,ϖ,ϰ)={ϑ+ϖ+ϰ3,if ϑ,ϖ,ϰ[0,+),+,otherwise,  rϑ={2ϑ,if ϑ[0,+),+,otherwise, 

    and

    α((ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ))={6,if ϑϖϰ and ˜ϑ˜ϖ˜ϰ,0,otherwise, 

    respectively. Assume that ϑ˜ϑ, ϖ˜ϖ and ϰ˜ϰ. Then

    ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))=max{ϑ+ϖ+ϰ3,˜ϑ+˜ϖ+˜ϰ3}13max{2˜ϑ,2˜ϖ,2˜ϰ}=ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    Hence, Stipulation (3) of Corollary 2.5 holds for ϕ(m,z)=13, where m,z[0,+]. In an easy way, it can be shown that all the hypotheses of Theorem 2.4 are fulfilled, and then there is a TCP for the mappings Ξ and r. Notice that, here ξ is not a metric on because, if we take ϑϖϰ˜ϰ, and ϑ˜ϖ˜ϑ˜ϰ, we have

    α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))=6max{ϑ+ϖ+ϰ3,˜ϑ+˜ϖ+˜ϰ3}6(˜ϑ+˜ϖ+˜ϰ3)=2˜ϑ+2˜ϖ+2˜ϰ>ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)).2˜ϰ=ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ))((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    Therefore, in the sense of TCPs, Theorem 3.1 in [13] and one of the hypotheses of Theorems 3.1 and 3.2 in [7] fails in studying the existence of a TCP for r and Ξ.

    Let Φ represent the class of mappings ϕ:[0,+][0,1) so that

    ϕ(ϑi)1 implies ϑi0 for all ϑi[0,+].

    The second part of this section is neglected the continuity condition for the mapping Ξ as follows:

    Theorem 2.8. Let the following postulates be satisfied:

    (hi) For any ΩII of the pair (,), there exist α:3×3[0,+] and ϕΦ fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the below inequality is justified:

    (1,α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ)),ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ)))(ϕ(Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))),Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))),

    where

    Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))=max{ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ),ξ(rϑ,Ξ(ϑ,ϖ,ϰ)),ξ(rϖ,Ξ(ϖ,ϰ,ϑ)),ξ(rϰ,Ξ(ϰ,ϑ,ϖ)),ξ(r˜ϑ,Ξ(r˜ϑ,r˜ϖ,r˜ϰ)),ξ(r˜ϖ,Ξ(r˜ϖ,r˜ϰ,r˜ϑ)),ξ(rϰ,Ξ(r˜ϰ,r˜ϑ,r˜ϖ))};

    (hii) Ξ and r is αGTA and there exist ϑ0,ϖ0,ϰ0,

    (rϑ0,Ξ(ϑ0,ϖ0,ϰ0)),(rϖ0,Ξ(ϖ0,ϰ0,ϑ0)),(rϰ0,Ξ(ϰ0,ϑ0,ϖ0))

    so that

    α((rϑ0,rϖ0,rϰ0),(Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϑ0,ϖ0,ϰ0)))1,α((rϖ0,rϰ0,rϑ0),(Ξ(ϖ0,ϰ0,ϑ0),Ξ(ϰ0,ϑ0,ϖ0),Ξ(ϖ0,ϰ0,ϑ0)))1

    and

    α((rϰ0,rϑ0,rϖ0),(Ξ(ϰ0,ϑ0,ϖ0),Ξ(ϑ0,ϖ0,ϰ0),Ξ(ϰ0,ϑ0,ϖ0)))1;

    (hiii) If limiξ(rϑi,rϑi+1)=0, limiξ(rϖi,rϖi+1)=0 and limiξ(rϰi,rϰi+1)=0, then

    sup{ξ(rϑ0,rϑi),ξ(rϖ0,rϖi),ξ(rϰ0,rϰi)}<,

    where {ϑi}, {ϖi} and {ϰi} are sequences in ; (hiv) If {ϑi}, {ϖi} and {ϰi} are sequences in with (rϑi,rϑi+1), (rϖi,rϖi+1), (rϰi,rϰi+1),

    α((rϑi,rϖi,rϰi),(rϑi+1,rϖi+1,rϰi+1))1,α((rϖi,rϰi,rϑi),(rϖi+1,rϰi+1,rϑi+1))1,α((rϰi,rϑi,rϖi),(rϰi+1,rϑi+1,rϖi+1))1,iN,

    and limiξ(rϑi,u)=0, limiξ(rϖi,u)=0, limiξ(rϰi,u)=0, for each u,u,u, then (rϑi,ru), (rϖi,ru), (rϰi,u),

    α((rϑi,rϖi,rϰi),(ru,ru,ru))1,α((rϖi,rϰi,rϑi),(ru,ru,ru))1,α((rϰi,rϑi,rϖi),(ru,ru,ru))1,iN;

    (hv) There is (0,1] so that

    ξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)),ξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u))andξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u)).

    Then Ξ and r have a TCP.

    Proof. Based on Theorem 2.4, the sequences {rϑi}, {rϖi} and {rϰi} are obtained. Furthermore, replacing ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)) with ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)) in Theorem 2.4, where ϑ,˜ϑ,ϖ,˜ϖ,ϰ,˜ϰ, we conclude that the three sequences are Cauchy in (,ξ). The completness of this space leads to for some u,u,u,

    limiξ(Ξ(rϑi,rϖi,rϰi),u)=limiξ(rϑi,u)=0,limiξ(Ξ(rϖi,rϰi,rϑi),u)=limiξ(rϖi,u)=0and limiξ(Ξ(rϰi,rϑi,rϖi),u)=limiξ(rϰi,u)=0.

    Since r is continuous,

    limiξ(rΞ(rϑi,rϖi,rϰi),ru)=limiξ(rrϑi,ru)=0,limiξ(rΞ(rϖi,rϰi,rϑi),ru)=limiξ(rϖi,ru)=0and limiξ(rΞ(rϰi,rϑi,rϖi),ru)=limiξ(rϰi,ru)=0.

    By postulates (i) and (ii), we get

    (1,1,ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)))(1,α((rrϑi,rrϖi,rrϰi),(ru,ru,ru)),ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)))(ϕ(Q((rrϑi,ru),(rrϖi,ru),(rrϰi,ru))),Q((rrϑi,ru),(rrϖi,ru),(rrϰi,ru))),
    (1,1,ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u)))(1,α((rrϖi,rrϰi,rrϑi),(ru,ru,ru)),ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u)))(ϕ(Q((rrϖi,ru),(rrϰi,ru),(rrϑi,ru))),Q((rrϖi,ru),(rrϰi,ru),(rrϑi,ru)))

    and

    (1,1,ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u)))(1,α((rrϰi,rrϑi,rrϖi),(ru,ru,ru)),ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u)))(ϕ(Q((rrϰi,ru),(rrϑi,ru),(rrϖi,ru)))Q((rrϰi,ru),(rrϑi,ru),(rrϖi,ru))),

    where

    Q((rrϑi,ru),(rrϖi,ru),(rrϰi,ru))=Q((rrϖi,ru),(rrϰi,ru),(rrϑi,ru))=Q((rrϰi,ru),(rrϑi,ru),(rrϖi,ru))=max{ξ(rrϑi,ru),ξ(rrϖi,ru),ξ(rrϰi,ru),ξ(rrϑi,Ξ(rϑi,rϖi,rϰi)),ξ(rrϖi,Ξ(rϖi,rϰi,rϑi)),ξ(rrϰi,Ξ(rϰi,rϑi,rϖi)),ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u))}. (2.16)

    Accordingly,

    ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u))ϕ(Q((rrϖi,ru),(rrϰi,ru),(rrϑi,ru)))Q((rrϖi,ru),(rrϰi,ru),(rrϑi,ru)), (2.17)
    ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))ϕ(Q((rrϰi,ru),(rrϑi,ru),(rrϖi,ru)))Q((rrϰi,ru),(rrϑi,ru),(rrϖi,ru)) (2.18)

    and

    ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u))ϕ(Q((rrϑi,ru),(rrϖi,ru),(rrϰi,ru)))Q((rrϑi,ru),(rrϖi,ru),(rrϰi,ru)). (2.19)

    Assume that ruΞ(u,u,u) or ruΞ(u,u,u) or ruΞ(u,u,u), that is,

    W=max{ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u))}>0.

    Applying hypothesis (hv), there is (0,1] so that

    ξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u))W,ξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u))W

    and

    ξ(ru,Ξ(u,u,u))lim supiξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))W.

    Hence

    W=max{ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u)),ξ(ru,Ξ(u,u,u))}lim supimax{ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)),ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u)),ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))}W.

    Because 11, we have

    W1Wlim supimax{ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)),ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u)),ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))}W,

    this implies that

    W=lim supimax{ξ(Ξ(rϑi,rϖi,rϰi),Ξ(u,u,u)),ξ(Ξ(rϖi,rϰi,rϑi),Ξ(u,u,u)),ξ(Ξ(rϰi,rϑi,rϖi),Ξ(u,u,u))}.

    Then there exists a subsequence max{ξ(Ξ(rϑis,rϖis,rϰis),Ξ(u,u,u)),ξ(Ξ(rϖis,rϰis,rϑis),Ξ(u,u,u)),ξ(Ξ(rϰis,rϑis,rϖis),Ξ(u,u,u))} so that

    limsmax{ξ(Ξ(rϑis,rϖis,rϰis),Ξ(u,u,u)),ξ(Ξ(rϖis,rϰis,rϑis),Ξ(u,u,u)),ξ(Ξ(rϰis,rϑis,rϖis),Ξ(u,u,u))}=W.

    Passing i in (2.16), we obtain that

    limsQ((rrϑi,ru),(rrϖi,ru),(rrϰi,ru))=W. (2.20)

    Using (2.17)–(2.19), one can write

    max{ξ(Ξ(rϑis,rϖis,rϰis),Ξ(u,u,u)),ξ(Ξ(rϖis,rϰis,rϑis),Ξ(u,u,u)),ξ(Ξ(rϰis,rϑis,rϖis),Ξ(u,u,u))}Q((rrϑis,ru),(rrϖis,ru),(rrϰis,ru))ϕ(Q((rrϑis,ru),(rrϖis,ru),(rrϰis,ru))).

    Putting s on both sides, we get

    limsϕ(Q((rrϑis,ru),(rrϖis,ru),(rrϰis,ru)))=1.

    Thus, limsQ((rrϑis,ru),(rrϖis,ru),(rrϰis,ru))=0. This contradicts (2.20). Hence ru=Ξ(u,u,u) and ru=Ξ(u,u,u) and ru=Ξ(u,u,u). Therefore, the element(u,u,u) is a TCP of r and Ξ.

    Corollary 2.9. The statements of Theorem 2.8 are still valid if we replace the stipulation (hi) with one of the assumptions below: (note Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)) is defined in Theorem 2.8):

    (1) There exist ϕΦ and z>1, fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the following inequality is obtained:

    (ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))+z)α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))ϕ(Q(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))+z.

    (2) There exist ϕΦ and z>1, fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the following inequality is verified:

    (α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))+z)ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))(1+z)ϕ(Q(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    (3) There is ϕΦ, fulfilling, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ) so that

    ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))ϕ(Q(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    (4) There is ϕΦ, satisfying, for any (rϑ,r˜ϑ), (rϖ,r˜ϖ) and (rϰ,r˜ϰ), the inequality below is obtained:

    {α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))}j{ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))}k{Q(ϕ(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))}k{Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))}k,

    for all positive integers j and k.

    Proof. The proof follows immediately from Example 2.3 by applying the following in Theorem 2.8:

    (1) Use the value of and from Example 2.3(1).

    (2) Use the value of and from Example 2.3(2).

    (3) Use the value of and from Example 2.3(3).

    (4) Use the value of and from Example 2.3(4).

    Example 2.10. Consider =[0,+]. Define ξ(ϑ,ϖ)=|ϑ|+|ϖ|, for all ϑ,ϖ. Define Ξ:3, r: and α:3×3[0,+] by

    Ξ(ϑ,ϖ,ϰ)={|ϑϖ|+|ϖϰ|+|ϰϑ|8,if ϑ,ϖ,ϰ[0,+),+,otherwise,  r(ϑ)={4ϑ,if ϑ[0,+),+,otherwise, 

    and

    α((ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ))={1,if ϑϖϰ and~ ϑ˜ϖ˜ϰ,0,otherwise, 

    respectively. Suppose that ϑ˜ϑ, ϖ˜ϖ and ϰ˜ϰ. When z>1, we have

    (α((rϑ,rϖ,rϰ),(r˜ϑ,r˜ϖ,r˜ϰ))+z)ξ(Ξ(ϑ,ϖ,ϰ),Ξ(˜ϑ,˜ϖ,˜ϰ))×1(1+z)(||ϑϖ|+|ϖϰ|+|ϰϑ|6|+||˜ϑ˜ϖ|+|˜ϖ˜ϰ|+|˜ϰ˜ϑ|6|)(1+z)14(|ϑ|+|ϖ|+|ϰ|+|˜ϑ|+|˜ϖ|+|˜ϰ|)(1+z)34max{|4ϑ|+|4˜ϑ|,|4ϖ|+|4˜ϖ|,|4ϰ|+|4˜ϰ|}=(1+z)ϕ(Q(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ))=(1+z)ϕ(Q(ξ(rϑ,r˜ϑ),ξ(rϖ,r˜ϖ),ξ(rϰ,r˜ϰ)))Q((rϑ,r˜ϑ),(rϖ,r˜ϖ),(rϰ,r˜ϰ)).

    So, the condition (2) of Corollary 2.9 is fulfilled for ϕ(m)=34, where m[0,+]. In an easy way, it can be shown that all assumptions of Theorem 2.8 are satisfied, therefore Ξ and r have a TCP.

    This section serves as the basis and core of our paper in which the existence of a solution to a system of integral equations is discussed. This system takes the following form:

    {ϑ(b)=Λ(b)+101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc,ϖ(b)=Λ(b)+101(b,c)1(c,ϖ(c))dc102(b,c)2(c,ϰ(c))dc103(b,c)3(c,ϑ(c))dc,ϰ(b)=Λ(b)+101(b,c)1(c,ϰ(c))dc102(b,c)2(c,ϑ(c))dc103(b,c)3(c,ϖ(c))dc, (3.1)

    where b[0,1]. Consider =C[0,1] equipped with ξ(ϑ,ϖ)=supb[0,1]|ϑ(b)ϖ(b)|, for each ϑ,ϖ. Clearly, (,ξ,) is a complete POJSM-space. Based on the theoretical results presented in the upper section of the paper, we are able to present theorems related to the existence of a solution to system (3.1) as follows:

    Theorem 3.1. Consider the integral equations (3.1) via the following postulates:

    (i) The functions Λ:[0,1]R, s:[0,1]×[0,1]R+ and s:[0,1]×RR+ are continuous, where s=1,2,3;

    (ii) For ϑ1,ϑ2, if ϑ1ϑ2, then 1(b,ϑ1(b))1(b,ϑ2(b)), 2(b,ϑ1(b))2(b,ϑ2(b)) and 3(b,ϑ1(b))3(b,ϑ2(b));

    (iii) There exist ˆA, ˆB and ˆC in R+ so that 3ˆAˆBˆC<1, 10s(b,c)dcˆA, 10s(b,c)s(c,ϑ(c))dcˆC and

    |s(b,ϑ(b)s(b,˜ϑ(b)|ˆB|ϑ(b)˜ϑ(b)|,

    where ϑ,˜ϑ, b[0,1] and s=1,2,3;

    (iv) There exist ϑ0,ϖ0,ϰ0 such that ϑ0Ξ(ϑ0,ϖ0,ϰ0), ϖ0Ξ(ϖ0,ϰ0,ϑ0) and ϰ0Ξ(ϰ0,ϑ0,ϖ0);

    (v) If {ϑi},{ϖi},{ϰi} so that limi|ϑiϑi+1|=0, limi|ϖiϖi+1|=0 and limi|ϰiϰi+1|=0, then

    sup{|ϑ0ϑi|,|ϖ0ϖi|,|ϰ0ϰi|:i1}<,

    where ϑ0,ϖ0,ϰ0.

    Then Problem (3.1) possesses a solution.

    Proof. Define the mappings Ξ:3 and r: by

    Ξ(ϑ,ϖ,ϰ)(b)=Λ(b)+101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc

    and rϑ=ϑ for ϑ,ϖ,ϰ. Clearly Ξ is ξcontinuous Ξ(3)r() and r is ξcontinuous and commutes with Ξ. After that, we claim that Ξ is r monotone. Suppose that ϑ1,ϑ2,ϖ,ϰ, ϑ1ϑ2. Based on postulate (ii), we get

    Ξ(ϑ1,ϖ,ϰ)(b)=Λ(b)+101(b,c)1(c,ϑ1(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dcΛ(b)+101(b,c)1(c,ϑ2(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc=Ξ(ϑ2,ϖ,ϰ)(b),
    Ξ(ϖ,ϰ,ϑ1)(b)=Λ(b)+101(b,c)1(c,ϖ(c))dc102(b,c)2(c,ϰ(c))dc103(b,c)3(c,ϑ1(c))dc,Λ(b)+101(b,c)1(c,ϖ(c))dc102(b,c)2(c,ϰ(c))dc103(b,c)3(c,ϑ2(c))dc,=Ξ(ϖ,ϰ,ϑ2)(b).

    Similarly, one can show that Ξ(ϰ,ϑ1,ϖ)(b)Ξ(ϰ,ϑ2,ϖ)(b). Hence, Ξ has the r monotone property.

    Now, for (ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ)3, describe α:3×3[0,+] by α((ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ))=1. Obviously, Ξ and r is αGTA. Also, assumptions (ii) and (iii) of Theorem 2.4 follow immediately from postulates (iv) and (v), respectively. In order to be able to validate Theorem 2.4, it remains to fulfill condition (i), which can be replaced by condition (3) of Corollary 2.5. Therefore, let ϑ,ϖ,ϰ,˜ϑ,˜ϖ,˜ϰ. If ϑ˜ϑ, ϖ˜ϖ and ϰ˜ϰ, then using postulate (iii), we have

    |Ξ(ϑ,ϖ,ϰ)(b)Ξ(˜ϑ,˜ϖ,˜ϰ)(b)|=|101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc101(b,c)1(c,˜ϑ(c))dc102(b,c)2(c,˜ϖ(c))dc103(b,c)3(c,˜ϰ(c))dc||101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)[3(c,ϰ(c))3(c,˜ϰ(c))]dc|+|101(b,c)1(c,ϑ(c))dc102(b,c)[2(c,ϖ(c))2(c,˜ϖ(c))]dc103(b,c)3(c,ϰ(c))dc|+|101(b,c)[1(c,ϑ(c))1(c,˜ϑ(c))]dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc|ˆAˆBˆCsupb[0,1]|ϰ(b)˜ϰ(b)|+ˆAˆBˆCsupb[0,1]|ϖ(b)˜ϖ(b)|+ˆAˆBˆCsupb[0,1]|ϑ(b)˜ϑ(b)|3ˆAˆBˆCmax{supb[0,1]|ϰ(b)˜ϰ(b)|,supb[0,1]|ϖ(b)˜ϖ(b)|,supb[0,1]|ϑ(b)˜ϑ(b)|}.

    Consider ϕ=3ˆAˆBˆC<1, then ϕ˜Φ. Hence the condition (3) of Corollary 2.5 is fulfilled. Therefore, a tripled fixed point of Ξ exists. Thus, there is a solution to the problem (3.1).

    It should be noted that in the above theorem, JS-metric ξ is a usual metric, therefore, the results seem easy to obtain, for this reason, we define JS-metric ξ:2[0,] by

    ξ(ϑ,ϖ)=supb[0,1](|ϑ(b)|+|ϖ(b)|).

    Clearly, under this distance, JS-metric ξ is not a metric. As a result, (,ξ,) is complete partially ordered. Moreover, we can guarantee the existence of the solution to system (3.1) only if it is homogeneous.

    Theorem 3.2. From the integral equations (3.1), let Ξ:3 be described as

    Ξ(ϑ,ϖ,ϰ)(b)=101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc,

    where b[0,1]. Assume also,

    (1) the functions s:[0,1]×[0,1]R+ and s:[0,1]×RR+ are continuous, where s=1,2,3;

    (2) for ϑ1,ϑ2, if ϑ1ϑ2, then 1(b,ϑ1(b))1(b,ϑ2(b)), 2(b,ϑ1(b))2(b,ϑ2(b)) and 3(b,ϑ1(b))3(b,ϑ2(b));

    (3) there exist ˆA, ˆB and ˆC in R+ so that 3ˆAˆBˆC<1, 10s(b,c)dcˆA, 10s(b,c)s(c,ϑ(c))dcˆC and

    |s(b,ϑ(b)|+|s(b,˜ϑ(b)|ˆB(|ϑ(b)|+|˜ϑ(b)|),

    where ϑ,˜ϑ, b[0,1] and s=1,2,3;

    (4) there exist ϑ0,ϖ0,ϰ0 so that ϑ0Ξ(ϑ0,ϖ0,ϰ0), ϖ0Ξ(ϖ0,ϰ0,ϑ0) and ϰ0Ξ(ϰ0,ϑ0,ϖ0).

    Then Problem (3.1) has a solution provided that the integral equations are homogeneous.

    Proof. Assume that α((ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ))=1 for any (ϑ,ϖ,ϰ),(˜ϑ,˜ϖ,˜ϰ)3. Repeating the same arguments used in Theorem 3.1, under this JS-{metric }ξ and considering r is the identity mapping, we have the ξcontinuities of r and Ξ and all assumption except condition (i) of Theorem 2.4 are valid.

    So, our task is to prove the condition (i) of Theorem 2.4. Again, this condition can be replaced by assumption (3) of Corollary 2.5. Let ϑ,ϖ,ϰ,˜ϑ,˜ϖ,˜ϰ. If ϑ˜ϑ, ϖ˜ϖ and ϰ˜ϰ, then by (3), we get

    |Ξ(ϑ,ϖ,ϰ)(b)|+|Ξ(˜ϑ,˜ϖ,˜ϰ)(b)|=|101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc|
    +|101(b,c)1(c,˜ϑ(c))dc102(b,c)2(c,˜ϖ(c))dc103(b,c)3(c,˜ϰ(c))dc||101(b,c)1(c,ϑ(c))dc102(b,c)2(c,ϖ(c))dc103(b,c)[3(c,ϰ(c))+3(c,˜ϰ(c))]dc|+|101(b,c)1(c,ϑ(c))dc102(b,c)[2(c,ϖ(c))+2(c,˜ϖ(c))]dc103(b,c)3(c,ϰ(c))dc|
    +|101(b,c)[1(c,ϑ(c))+1(c,˜ϑ(c))]dc102(b,c)2(c,ϖ(c))dc103(b,c)3(c,ϰ(c))dc|ˆAˆBˆCsupb[0,1](|ϰ(b)|+|˜ϰ(b)|)+ˆAˆBˆCsupb[0,1](|ϖ(b)|+|˜ϖ(b)|)+ˆAˆBˆCsupb[0,1](|ϑ(b)|+|˜ϑ(b)|)
    3ˆAˆBˆCmax{supb[0,1](|ϰ(b)|+|˜ϰ(b)|),supb[0,1](|ϖ(b)|+|˜ϖ(b)|),supb[0,1](|ϑ(b)|+|˜ϑ(b)|)}.

    Analogously, if we take ϕ=3ˆAˆBˆC<1, then ϕ˜Φ. Hence, the assumption (3) of Corollary 2.5 is satisfied. Hence, Ξ possesses a tripled fixed point, which is a unique solution for the problem (3.1) under the condition of homogeneity for these equations.

    The following example supports Theorem 3.2:

    Example 3.3. Consider the problem below:

    {ϑ(b)=10c21+b4.11+c3.|ϑ(c)|1+|ϑ(c)|dc10ceb2.c21+c4.|ϖ(c)|2+|ϖ(c)|dc10c3eb2.c31+c7.|ϰ(c)|3+|ϰ(c)|dcϖ(b)=10c21+b4.11+c3.|ϖ(c)|1+|ϖ(c)|dc10ceb2.c21+c4.|ϰ(c)|2+|ϰ(c)|dc10c3eb2.c31+c7.|ϑ(c)|3+|ϑ(c)|dcϰ(b)=10c21+b4.11+c3.|ϰ(c)|1+|ϰ(c)|dc10ceb2.c21+c4.|ϑ(c)|2+|ϑ(c)|dc10c3eb2.c31+c7.|ϖ(c)|3+|ϖ(c)|dc, (3.2)

    where b[0,1]. By comparing this system with system (3.1) in the homogeneous case, we can write

    1(b,c)=c21+b4,1(c,ϑ(c))=11+c3.|ϑ(c)|1+|ϑ(c)|,2(b,c)=c2eb2,2(c,ϑ(c))=c1+c4.|ϑ(c)|2+|ϑ(c)|,3(b,c)=c2eb3,3(c,ϑ(c))=c41+c7.|ϑ(c)|3+|ϑ(c)|,

    for b,c[0,1]. It is easy to see that s and s are continuous, and s(c,ϑ)0 for s=1,2,3. Furthermore, s(b,ϑ(b))s(b,ϖ(b)) whenever ϑϖ for all s=1,2,3. Moreover, for non-positive-valued ϑ0,ϖ0,ϰ0, the assumption (4) of Theorem 3.2 holds.

    Now, consider

    |1(b,ϑ(b))|+|1(b,˜ϑ(b))|=11+b3(|ϑ(b)|1+|ϑ(b)|+|˜ϑ(b)|1+|˜ϑ(b)|)12(|ϑ(b)|+|˜ϑ(b)|),
    |2(b,ϖ(b))|+|2(b,˜ϖ(b))|=b1+b4(|ϖ(b)|2+|ϖ(b)|+|˜ϖ(b)|2+|˜ϖ(b)|)12(|ϖ(b)|+|˜ϖ(b)|),

    and

    |3(b,ϰ(b))|+|3(b,˜ϰ(b))|=b41+b7(|ϰ(b)|3+|ϰ(b)|+|˜ϰ(b)|3+|˜ϰ(b)|)12(|ϰ(b)|+|˜ϰ(b)|),

    so, assume that ˆB=12. Also, one can write

    101(b,c)1(c,ϑ(c))dc10c21+b4.11+c3dc=11+b4.13ln(2)13ln(2),102(b,c)2(c,ϑ(c))dc10c2eb2.c1+c4dc=eb2.14ln(2)14ln(2),102(b,c)2(c,ϑ(c))dc10c2eb3.c41+c7dc=eb3.17ln(2)17ln(2).

    Putting ˆC=13ln(2), we have 10s(b,c)s(c,ϑ(c))dcˆC for s=1,2,3. Moreover,

    101(b,c)dc=10c21+b4dc=13(1+b4)13,102(b,c)dc=10c2eb2dc=13eb213,103(b,c)dc=10c2eb3dc=13eb313,

    select ˆA=13, then we have 10s(b,c)dcˆA. Consequently, 3ˆAˆBˆC=16ln(2)<1. Therefore, all hypotheses of Theorem 3.2 are fulfilled. This implies that Problem (3.2) owns a solution in C[0,1].

    The existence of a tripled coincidence point of a generalized contraction type mapping, which is considered in JS-metric spaces endowed with a partial order, was investigated in this article. Furthermore, the theoretical results have been supported by illustrative examples. Ultimately, as an application, the existence of a solution for a system of non-homogeneous and homogeneous integral equations is provided. Moreover, a numerical example of the system is derived. Along with the works presented in [4,24,25] as future works, we launch the following two inquiries: What would the results look like if the JS-metric space was replaced by a Modular space? What if we used the variational principle?

    The authors would like to acknowledge the support of Prince Sultan University for paying the Article Processing Charges (APC) of this publication through the Theoretical and Applied Sciences Lab.

    The authors declare that they have no competing interests concerning the publication of this article.



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