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Research article Special Issues

Fixed point results for generalized contractions in controlled metric spaces with applications

  • The purpose of this article is to establish some common fixed point results for generalized contractions including some precise control functions of two variables in the setting of controlled metric spaces. As consequences of our leading results, we derive common fixed point and fixed point results for contractions with control functions of one variable and constants. We also discuss controlled metric spaces endowed with a graph and obtain some common fixed point results in this newly introduced space. As an application of our leading result, we examine the solution of a Fredholm type integral equation.

    Citation: Zhenhua Ma, Jamshaid Ahmad, Abdullah Eqal Al-Mazrooei. Fixed point results for generalized contractions in controlled metric spaces with applications[J]. AIMS Mathematics, 2023, 8(1): 529-549. doi: 10.3934/math.2023025

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  • The purpose of this article is to establish some common fixed point results for generalized contractions including some precise control functions of two variables in the setting of controlled metric spaces. As consequences of our leading results, we derive common fixed point and fixed point results for contractions with control functions of one variable and constants. We also discuss controlled metric spaces endowed with a graph and obtain some common fixed point results in this newly introduced space. As an application of our leading result, we examine the solution of a Fredholm type integral equation.



    The theory of fixed points has worldwide applications in distinct fields of science and Engineering [1,2,3]. M. Frechet is a principal researcher in this theory who defined the notion of metric space in 1906. Metric space methods have been employed for decades in numerous applications, for example in internet search engines, protein classification and image classification. The best applied fixed point result is the Banach contraction principle that has been extended by either changing the contractive condition or by functioning on a further generalized metric spaces [4,5,6,7,8]. In current years, some novel types of generalized metric spaces were presented and numerous spaces established as hybrids of the foregoing varieties were examined such as rectangular metric space, b -metric space, bvs-metric space, extended b-metric space. The key in creating these spaces is to generalize or extend the third axiom of metric space that is, its triangle property. In 1993, Czerwik [9] introduced the notion of b-metric space which extends the metric space by enhancing the triangle equality metric axiom by placing a constant s1 multiplied to the right-hand side, is one of the great applied extensions for metric spaces. Later on, Kamran et al. [10] gave a new kind of extended b-metric spaces by putting a function σ(ξ,ϱ) on the place of constant s and this function depends on the parameters used on left-hand side of the triangle inequality. Recently, Mlaiki et al. [11] replaced the constant s by a function σ(ξ,ϱ) which act separately on each term in the right-hand side of the triangle inequality and defined controlled metric space. They established Banach contraction principle in the background of this newly introduced space. Subsequently, Lateef [18] obtained Fisher type fixed point result and generalized the leading result of Mlaiki et al. [11]. For more characteristics, we assign the researchers to [12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33]. In this article, we utilize the notion of controlled metric space to establish common fixed point theorems for rational contractive mappings dealing with some precise control functions of two variables in the background of controlled metric space. As an outcome of our pioneering theorems, we derive common fixed point and fixed point theorems for contractive mappings including control functions of one variable and constants. In this way, we generalize the main result of Lateef [18] as well as the leading theorem of Mlaiki et al. [11]. We also discuss controlled metric spaces equipped with a graph and obtain some common fixed point results in this newly introduced space.

    In 1993, Czerwik [9] introduced the concept of b-metric space (b-MS) in this way.

    Definition 1. ([9])Let U be a non empty set and s1. A function :U×U [0,) is said to be b-metric if following conditions hold:

    (1) (ξ,ϱ)0 and (ξ,ϱ)=0 if and only if ξ=ϱ;

    (2) (ξ,ϱ)=(ϱ,ξ);

    (3)(ξ,ω)s[(ξ,ϱ)+(ϱ,ω)];

    for all ξ,ϱ,ωU. The pair (U,) is said to be a b-metric space (b-MS).

    In 2017, Kamran et al. [10] gave the notion of extended b-metric space (EbMS) as follows:

    Definition 2. ([10]) Let U be a non empty set and σ:U×U[1,). A function :U×U [0,) is said to be extended b-metric if following conditions hold:

    (i) (ξ,ϱ)0 and (ξ,ϱ)=0 if and only if ξ=ϱ;

    (ii) (ξ,ϱ)=(ϱ,ξ);

    (iii) (ξ,ϱ)σ(ξ,ϱ)[(ξ,ω)+(ϱ,ω)];

    for all ξ,ϱ,ωU, then (U,) is called an extended b-metric space (EbMS).

    In 2018, a contemporary extended b-metric space was initiated by Mlaiki et al. [11] which is known as controlled metric space as follows:

    Definition 3. ([11]) Let U be a non empty set and σ:U×U[1,). A function :U×U [0,) is said to be controlled metric if following conditions hold:

    (i) (ξ,ϱ)=0 if and only if ξ=ϱ;

    (ii) (ξ,ϱ)=(ϱ,ξ);

    (iii) (ξ,ϱ)σ(ξ,ω)(ξ,ω)+σ(ϱ,ω)(ϱ,ω);

    for all ξ,ϱ,ωU, then (U,,σ) is said to be a controlled metric space (CMS).

    Example 1. Let U={0,1,2}. Define the mapping :U×U[0,) by

    (0,0)=(1,1)=(2,2)=0

    and

    (0,1)=(1,0)=1,
    (0,2)=(2,0)=12,
    (1,2)=(2,1)=25.

    Define the symmetric control function σ:U×U[1,) by

    σ(0,0)=σ(1,1)=σ(2,2)=σ(0,2)=1,
    σ(1,2)=54,σ(0,1)=1110.

    Then (U,,σ) is CMS.

    Theorem 1. ([11]) Let (U,σ,) be a complete CMS and :UU such that

    (ξ,ϱ)τ((ξ,ϱ))

    for all ξ,ϱU, where τ[0,1). For ξ0U, take ξȷ=ȷξ0. Assume that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1τ.

    In addition, assume that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Lateef [12] obtained the following result in a complete CMS as follows:

    Theorem 2. ([12]) Let (U,σ,) be a complete CMS and :UU be such that

    (ξ,ϱ)τ((ξ,ξ)+(ϱ,ϱ))

    for all ξ,ϱU, where τ(0,12). For ξ0U, take ξȷ=ȷξ0. Suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ(ξi,ξi+1)<1τ.

    In addition, assume that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Later on, Lateef [18] established Fisher type fixed point result in this way.

    Theorem 3. Let (U,σ,) be a complete CMS, :UU and there exist τ1,τ2[0,1) with τ=τ1+τ2<1 such that

    (ξ,ϱ)τ1(ξ,ϱ)+τ2(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ)

    for all ξ,ϱU. For ξ0U, take ξȷ=ȷξ0. Suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1τ.

    Moreover, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    We give our leading result in this way.

    Theorem 4. Let (U,σ,) be a complete CMS, 1,2:UU and there exist the mappings τ1,τ2:U×U[0,1) such that

    (i) τ1(21ξ,ϱ)τ1(ξ,ϱ) and τ1(ξ,12ϱ)τ1(ξ,ϱ);

    (ii) τ2(21ξ,ϱ)τ2(ξ,ϱ) and τ2(ξ,12ϱ)τ2(ξ,ϱ);

    (iii) τ1(ξ,ϱ)+τ2(ξ,ϱ)<1;

    (iv)

    (1ξ,2ϱ)τ1(ξ,ϱ)(ξ,ϱ)+τ2(ξ,ϱ)(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ), (3.1)

    for all ξ,ϱU. For ξ0U, a sequence {ξȷ}ȷ0 is defined as ξ2ȷ+1=1ξ2ȷ and ξ2ȷ+2=2ξ2ȷ+1 for each ȷ0. Assume that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1τ, (3.2)

    where τ1(ξ0,ξ1)1τ2(ξ0,ξ1)=τ. In addition, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that 1ξ2ξ=ξ.

    Proof. Let ξ0 U. We construct {ξȷ} in U by ξ2ȷ+1=1ξ2ȷ and ξ2ȷ+2=2ξ2ȷ+1 for each ȷ0. From assumption and (3.1) we get

    (ξ2ȷ+1,ξ2ȷ+2)=(1ξ2ȷ,2ξ2ȷ+1)τ1(ξ2ȷ,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ2ȷ,ξ2ȷ+1)(ξ2ȷ,1ξ2ȷ)(ξ2ȷ+1,2ξ2ȷ+1)1+(ξ2ȷ,ξ2ȷ+1)=τ1(21ξ2ȷ2,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(21ξ2ȷ2,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)(ξ2ȷ+1,ξ2ȷ+2)1+(ξ2ȷ,ξ2ȷ+1)τ1(ξ2ȷ2,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ2ȷ2,ξ2ȷ+1)(ξ2ȷ+1,ξ2ȷ+2)=τ1(21ξ2ȷ4,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(21ξ2ȷ4,ξ2ȷ+1)(ξ2ȷ+1,ξ2ȷ+2)τ1(ξ2ȷ4,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ2ȷ4,ξ2ȷ+1)(ξ2ȷ+1,ξ2ȷ+2)τ1(ξ0,ξ2ȷ+1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ0,ξ2ȷ+1)(ξ2ȷ+1,ξ2ȷ+2)=τ1(ξ0,12ξ2ȷ1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ0,12ξ2ȷ1)(ξ2ȷ+1,ξ2ȷ+2)τ1(ξ0,ξ2ȷ1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ0,ξ2ȷ1)(ξ2ȷ+1,ξ2ȷ+2)τ1(ξ0,ξ1)(ξ2ȷ,ξ2ȷ+1)+τ2(ξ0,ξ1)(ξ2ȷ+1,ξ2ȷ+2).

    This implies that

    (ξ2ȷ+1,ξ2ȷ+2)(τ1(ξ0,ξ1)1τ2(ξ0,ξ1))(ξ2ȷ,ξ2ȷ+1).

    Similarly,

    (ξ2ȷ+2,ξ2ȷ+3)=(2ξ2ȷ+1,1ξ2ȷ+2)=(1ξ2ȷ+2,2ξ2ȷ+1)τ1(ξ2ȷ+2,ξ2ȷ+1)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,ξ2ȷ+1)(ξ2ȷ+2,1ξ2ȷ+2)(ξ2ȷ+1,1ξ2ȷ+1)1+(ξ2ȷ+2,ξ2ȷ+1)=τ1(ξ2ȷ+2,12ξ2ȷ1)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,12ξ2ȷ1)(ξ2ȷ+2,ξ2ȷ+3)(ξ2ȷ+1,ξ2ȷ+2)1+(ξ2ȷ+2,ξ2ȷ+1)τ1(ξ2ȷ+2,ξ2ȷ1)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,ξ2ȷ1)(ξ2ȷ+2,ξ2ȷ+3)=τ1(ξ2ȷ+2,12ξ2ȷ3)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,12ξ2ȷ3)(ξ2ȷ+2,ξ2ȷ+3)τ1(ξ2ȷ+2,ξ2ȷ3)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,ξ2ȷ3)(ξ2ȷ+2,ξ2ȷ+3)τ1(ξ2ȷ+2,ξ0)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ+2,ξ0)(ξ2ȷ+2,ξ2ȷ+3)=τ1(21ξ2ȷ,ξ0)(ξ2ȷ+2,ξ2ȷ+1)+τ2(21ξ2ȷ,ξ0)(ξ2ȷ+2,ξ2ȷ+3)τ1(ξ2ȷ,ξ0)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ2ȷ,ξ0)(ξ2ȷ+2,ξ2ȷ+3)...τ1(ξ1,ξ0)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ1,ξ0)(ξ2ȷ+2,ξ2ȷ+3)=τ1(ξ0,ξ1)(ξ2ȷ+2,ξ2ȷ+1)+τ2(ξ0,ξ1)(ξ2ȷ+2,ξ2ȷ+3).

    This implies that

    (ξ2ȷ+2,ξ2ȷ+3)(τ1(ξ0,ξ1)1τ2(ξ0,ξ1))(ξ2ȷ+1,ξ2ȷ+2)=τ(ξ2ȷ+1,ξ2ȷ+2).

    By pursuing in this direction, we get

    (ξȷ,ξȷ+1)τ(ξȷ1,ξȷ)τ2(ξȷ2,ξȷ1)...τȷ(ξ0,ξ1).

    Thus

    (ξȷ,ξȷ+1)τȷ(ξ0,ξ1). (3.3)

    Now for all ȷ,mN with ȷ<m, we get

    (ξȷ,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)(ξȷ+1,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)σ(ξȷ+1,ξȷ+2)(ξȷ+1,ξȷ+2)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)(ξȷ+2,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)σ(ξȷ+1,ξȷ+2)(ξȷ+1,ξȷ+2)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)σ(ξȷ+2,ξȷ+3)(ξȷ+2,ξȷ+3)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)σ(ξȷ+3,ξm)(ξȷ+3,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)(ξi,ξi+1)+m1i=ȷ+1σ(ξi,ξm)(ξm1,ξm)

    which further implies that

    (ξȷ,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)(ξi,ξi+1)+(m1i=ȷ+1σ(ξi,ξm))σ(ξm1,ξm)(ξm1,ξm)σ(ξȷ,ξȷ+1)τȷ(ξ0,ξ1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1)+(m1i=ȷ+1σ(ξi,ξm))σ(ξm1,ξm)τm1(ξ0,ξ1)=σ(ξȷ,ξȷ+1)τȷ(ξ0,ξ1)+m1i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1).

    Thus

    (ξȷ,ξm)σ(ξȷ,ξȷ+1)τȷ(ξ0,ξ1)+m1i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1). (3.4)

    Let

    Ψl=li=0(ik=0σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1).

    From (3.4), we get

    (ξȷ,ξm)(ξ0,ξ1)[τȷσ(ξȷ,ξȷ+1)+(Ψm1Ψȷ)]. (3.5)

    Since σ(ξ,ϱ)1, and by employing the ratio test, limȷ+Ψȷ exists. Thus {Ψȷ} is Cauchy sequence. Lastly, letting ȷ,m+ in (3.5), we get that

    limȷ,m+(ξȷ,ξm)=0. (3.6)

    Hence, {ξȷ} is a Cauchy sequence in (U,,σ). As (U,,σ) is complete, so there exists ξU such that

    limȷ+(ξȷ,ξ)=0, (3.7)

    that is ξȷξ as ȷ+. Now, by (3.1) and assumption (iii), we have

    (ξ,1ξ)σ(ξ,ξ2ȷ+2)(ξ,ξ2ȷ+2)+σ(ξ2ȷ+2,1ξ)(ξ2ȷ+2,1ξ)=σ(ξ,ξ2ȷ+2)(ξ,ξ2ȷ+2)+σ(ξ2ȷ+2,1ξ)(2ξ2ȷ+1,1ξ)=σ(ξ,ξ2ȷ+2)(ξ,ξ2ȷ+2)+σ(ξ2ȷ+2,1ξ)(1ξ,2ξ2ȷ+1)=σ(ξ,ξ2ȷ+2)(ξ,ξ2ȷ+2)+σ(ξ2ȷ+2,1ξ)[τ1(ξ,ξ2ȷ+1)(ξ,ξ2ȷ+1)+τ2(ξ,ξ2ȷ+1)(ξ,1ξ)(ξ2ȷ+1,ξ2ȷ+2)1+(ξ,ξ2ȷ+1)]=σ(ξ,ξ2ȷ+2)(ξ,ξ2ȷ+2)+σ(ξ2ȷ+2,1ξ)[τ1(ξ,ξ2ȷ+1)(ξ,ξ2ȷ+1)+τ2(ξ,ξ2ȷ+1)(ξ,1ξ)(ξ2ȷ+1,ξ2ȷ+2)1+(ξ,ξ2ȷ+1)].

    Taking ȷ+ and utilizing (3.7), we get a contradiction to the fact that (ξ,1ξ)>0. Thus (ξ,1ξ)=0. It implies that ξ=1ξ. Likewise, we can prove that ξ=2ξ.Thus, ξ is a common fixed point of 1 and 2. In due course, we prove that ξ is unique. Suppose that there exists another point ξ/U such that ξ/=1ξ/=2ξ/. It follows from

    (ξ,ξ/)=(1ξ,2ξ/)τ1(ξ,ξ/)(ξ,ξ/)+τ2(ξ,ξ/)(ξ,1ξ)(ϱ,2ξ/)1+(ξ,ξ/)=τ1(ξ,ξ/)(ξ,ξ/).

    Since τ1(ξ,ξ/)[0,1), so we have (ξ,ξ/)=0. Thus, we get ξ=ξ/, which shows that ξ is unique.

    Example 2. Let U=[0,1]. Now we define :U×U[0,) by

    (ξ,ϱ)=(ξ+ϱ)2,

    where σ(ξ,ϱ)=2+ξ+ϱ, for all ξ,ϱU. Now we define 1,2:UU by

    1ξ=ξ3 and 2ξ=ξ4,

    for ξR. Choose τ1,τ2:U×U[0,1) by

    τ1(ξ,ϱ)=16+ξ+ϱ144 and τ2(ξ,ϱ)=15+ξ+ϱ144.

    Then evidently,

    τ1(ξ,ϱ)+τ2(ξ,ϱ)<1.

    Now

    τ1(21ξ,ϱ)=19+ξ1726+ϱ14416+ξ+ϱ144=τ1(ξ,ϱ),

    and

    τ1(ξ,12ϱ)=19+ξ144+ϱ172616+ξ+ϱ144=τ1(ξ,ϱ),

    also,

    τ2(21ξ,ϱ)=546+ξ1726+ϱ14415+ξ+ϱ144=τ1(ξ,ϱ),

    and

    τ1(ξ,12ϱ)=546+ξ144+ϱ172615+ξ+ϱ144=τ1(ξ,12ϱ).

    Take ξ0=0, so (3.2) is satisfied. Let ξ,ϱU. Then

    (1ξ,2ϱ)=(4ξ+3ϱ)2144(4ξ+4ϱ)214416+ξ+ϱ144(ξ+ϱ)2+15+ξ+ϱ144(5ξ4)2(6ϱ4)21+(ξ+ϱ)2=τ1(ξ,ϱ)(ξ,ς)+τ2(ξ,ϱ)(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ).

    Thus all the assumption of Theorem 4 are satisfied and there exists ξ=0 U such that 1ξ2ξ=ξ.

    By setting 1 = 2= in Theorem 4, we derive the following result.

    Corollary 1. Let (U,σ,) be a complete CMS, :UU and there exist the mappings τ1,τ2:U×U[0,1) such that

    (i) τ1(ξ,ϱ)τ1(ξ,ϱ) and τ1(ξ,ϱ)τ1(ξ,ϱ);

    (ii) τ2(ξ,ϱ)τ2(ξ,ϱ) and τ2(ξ,ϱ)τ2(ξ,ϱ);

    (iii) τ1(ξ,ϱ)+τ2(ξ,ϱ)<1;

    (iv)

    (ξ,ϱ)τ1(ξ,ϱ)(ξ,ϱ)+τ2(ξ,ϱ)(ξ,ξ)(ϱ,ϱ)1+(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, a sequence {ξȷ}ȷ0 is generated as ξȷ+1=ξȷ for each ȷ0. Assume that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1τ,

    where τ1(ξ0,ξ1)1τ2(ξ0,ξ1)=τ. Additionally, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Theorem 5. Let (U,σ,) be a complete CMS, 1,2:UU and there exist the mappings β1,β2:U[0,1) such that

    (i) β1(1ξ)β1(ξ) and β2(1ξ)β2(ξ);

    (ii) β1(2ξ)β1(ξ) and β2(2ξ)β2(ξ);

    (iii) (β1+β2)(ξ)<1;

    (iv)

    (1ξ,2ϱ)β1(ξ)(ξ,ϱ)+β2(ξ)(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, a sequence {ξȷ}ȷ0 is generated as ξ2ȷ+1=1ξ2ȷ and ξ2ȷ+2=2ξ2ȷ+1 for each ȷ0. Suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1β,

    where β1(ξ0)1β2(ξ0)=β. In addition, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that 1ξ2ξ=ξ.

    Proof. Define τ1,τ2:U×U[0,1) by

    τ1(ξ,ϱ)=β1(ξ)andτ2(ξ,ϱ)=β2(ξ),

    for all ξ,ϱU. Then for all ξ,ϱU,

    (i) τ1(21ξ,ϱ)=β1(21ξ)β1(1ξ)β1(ξ)=τ1(ξ,ϱ) and τ1(ξ,12ϱ)=β1(ξ)=β1(ξ,ϱ);

    (ii) τ2(21ξ,ϱ)=β2(21ξ)β2(1ξ)β2(ξ)=τ2(ξ,ϱ) and τ2(ξ,12ϱ)=β2(ξ)=β2(ξ,ϱ);

    (iii) τ1(ξ,ϱ)+τ2(ξ,ϱ)=β1(ξ)+β2(ξ)<1;

    (iv)

    (1ξ,2ϱ)β1(ξ)(ξ,ϱ)+β2(ξ)(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ)=τ1(ξ,ϱ)(ξ,ϱ)+τ2(ξ,ϱ)(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ).

    By Theorem 4, 1 and 2 have a unique common fixed point.

    Corollary 2. Let (U,σ,) be a complete CMS, :UU and there exist the mappings β1,β2:U[0,1) such that

    (i) β1(ξ)β1(ξ) and β2(ξ)β2(ξ);

    (ii) (β1+β2)(ξ)<1;

    (iii)

    (ξ,ϱ)β1(ξ)(ξ,ϱ)+β2(ξ)(ξ,ξ)(ϱ,ϱ)1+(ξ,ϱ), (4.1)

    for all ξ,ϱU. For ξ0U, we set β1(ξ0)1β2(ξ0)=β. Assume that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1β,

    where ξȷ+1=ξȷ for each ȷ0. Furthermore, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Proof. Taking 1= 2= in the Theorem 5.

    Theorem 6. Let (U,σ,) be a complete CMS and :UU. Let there exist β1,β2:U[0,1) such that

    (i) β1(ȷξ)β1(ξ) and β2(ȷξ)β2(ξ);

    (ii) (β1+β2)(ξ)<1;

    (iii)

    (ȷξ,ȷϱ)β1(ξ)(ξ,ϱ)+β2(ξ)(ξ,ȷξ)(ϱ,ȷϱ)1+(ξ,ϱ),

    for all ξ,ϱU and for some ȷN. For ξ0U, we set β1(ξ0)1β2(ξ0)=β. Suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1β,

    where ξȷ+1=ξȷ for each ȷ0. In addition, suppose that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Proof. By result 2, we get that ȷ ξ=ξ. Now, as

    ȷ(ξ)=(ȷξ)=ξ,

    so, ξ is a fixed point of ȷ. Thus ξ = ξ. Since the fixed point of ȷ is unique, so ξ is also a fixed point of .

    Corollary 3. Let (U,σ,) be a complete CMS, 1, 2:UU and there exist γ1,γ2[0,1) with γ1+γ2<1 such that

    (1ξ,2ϱ)γ1(ξ,ϱ)+γ2(ξ,1ξ)(ϱ,2ϱ)1+(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, a sequence {ξȷ}ȷ0 is generated as ξ2ȷ+1=1ξ2ȷ and ξ2ȷ+2=2ξ2ȷ+1 for each ȷ0. Assume that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1γ,

    where γ11γ2=γ. In addition, assume that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that 1ξ2ξ=ξ.

    Proof. Taking γ1() = γ1 and γ2() = γ2 in Theorem 5.

    Corollary 4. Let (U,σ,) be a complete CMS, :UU and there exist γ1,γ2[0,1) with γ1+γ2<1 such that

    (ξ,ϱ)γ1(ξ,ϱ)+γ2(ξ,ξ)(ϱ,ϱ)1+(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, we set γ11γ2=γ. Suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ(ξi,ξi+1)<1γ,

    where ξȷ+1=ξȷ, for all ȷ0. Furthermore, assume that limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) are finite and exist, then there exists a unique point ξU such that ξ=ξ.

    Proof. Taking 1= 2= in Theorem 3.

    Corollary 5. Let (U,σ,) be a complete CMS, :UU and there exist γ1,γ2[0,1) with γ1+γ2<1 such that

    (ȷξ,ȷϱ)γ1(ξ,ϱ)+γ2(ξ,ȷξ)(ϱ,ȷϱ)1+(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, with γ11γ2=γ, suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ (ξi,ξi+1)<1γ,

    where ξȷ+1=ξȷ, for each ȷ0. Furthermore, assume that, limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finite, then there exists a unique point ξU such that ξ=ξ.

    Proof. Setting γ1()=γ1 and γ2()=γ2 in Theorem 6.

    Corollary 6. Let (U,σ,) be a complete CMS, :UU and there exists γ1[0,1) such that

    (ξ,ϱ)γ1(ξ,ϱ),

    for all ξ,ϱU. For ξ0U, with γ11γ1=γ, suppose that

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ(ξi,ξi+1)<1γ,

    where ξȷ+1=ξȷ for each ȷ0. Moreover, assume that, limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and are finitet, then there exists a unique point ξU such that ξ=ξ.

    Let (U,σ,) be a CMS and G be a directed graph. Let us represent by G1, the graph achieved from G by changing the direction of E(G). Hence,

    E(G1)={(ξ,ϱ)U×U:(ϱ,ξ)E(G)}.

    Definition 4. An element ξ U is claimed to be common fixed point of (1,2), if 1(ξ)=2(ξ)=ξ. We shall represent by CFix(1,2), the set of all common fixed points of (1,2), i.e.

    CFix(1,2)={ξU:1(ξ)=2(ξ)=ξ}.

    Definition 5. Suppose that 1,2 :UU are two mappings on complete CMS (U,σ,) equipped with a directed graph G. Then (1,2) is said to be a G-orbital cyclic pair, if for any ξ U

    (ξ,1ξ)E(G)(1ξ,2(1ξ))E(G),
    (ξ,2ξ)E(G)(2ξ,1(2ξ))E(G).

    Let us consider the following sets

    U1={ξU:(ξ,1ξ)E(G)},U2={ξU:(ξ,2ξ)E(G)}.

    Remark 1. If the pair (1,2) be a G -orbital-cyclic pair, then U1U2.

    Proof. Let ξ0U1. Then (ξ0,1ξ0)E(G)(1ξ0,2(1ξ0))E(G). If we represent by ξ1=1ξ0, then we get that (ξ1,2(ξ1))E(G), thus U2.

    Theorem 7. Let (U,σ,) be a complete CMS equipped with a directed graph G and 1,2 :UU is G-orbital cyclic pair. Assume that there exists τ1[0,1) such that

    (i) U1;

    (ii) for all ξU1 and ϱU2,

    (1ξ,2ϱ)τ1max{(ξ,ϱ),(ξ,1ξ),(ϱ,2ϱ)}; (5.1)

    (iii) for all (ξȷ)ȷNU, one has (ξȷ,ξȷ+1)E(G),

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ(ξi,ξi+1)<1τ, (5.2)

    where τ=τ11τ1;

    (iv) 1 and 2 are continuous, or for all (ξȷ)ȷNU, with ξȷξ as ȷ+, and (ξȷ,ξȷ+1)E(G) for ȷN, we have ξU1U2. In these conditions, CFix(12);

    (v) for all ξU, we have limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and finite;

    (vi) if (ξ,ξ/)CFix(1,2) implies ξU1 and ξ/U2, then the pair (1,2) has a unique common fixed point.

    Proof. Let ξ0U1. Thus (ξ0,1ξ0)E(G). As the pair (1,2) is G-orbital cyclic, we get (1ξ0,21ξ0)E(G). Construct ξ1 by ξ1=1ξ0, we have (ξ1,2ξ1)E(G) and from here (2ξ1,12ξ1)E(G). Denoting by ξ2=2ξ1, we have (ξ2,1ξ2)E(G). Continuing along these lines, we generate a sequence (ξȷ)ȷN with ξ2ȷ=2ξ2ȷ1 and ξ2ȷ+1=1ξ2ȷ, such that (ξ2ȷ,ξ2ȷ+1)E(G). We assume that ξȷξȷ+1. If, there exists ȷ0N, such that ξȷ0=ξȷ0+1, then in the view of the fact that ΔE(G), (ξȷ0,ξȷ0+1)E(G) and thus ξ=ξȷ0 is a fixed point of 1. Now to manifest that ξ CFix(1,2), we shall discuss these two cases for ȷ0. If ȷ0 is even, then ȷ0=2ȷ. Then, ξ2ȷ=ξ2ȷ+1=1ξ2ȷ and thus, ξ2ȷ is a fixed point of 1. Assume that ξ2ȷ=ξ2ȷ+1=1ξ2ȷ but (1ξ2ȷ,2ξ2ȷ+1)>0, and let ξ=ξ2ȷU1 and ϱ=ξ2ȷ+1U2. So

    0<(ξ2ȷ+1,ξ2ȷ+2)=(1ξ2ȷ,2ξ2ȷ+1)τ1max{(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ,1ξ2ȷ),(ξ2ȷ+1,2ξ2ȷ+1)}=τ1max{(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ+1,ξ2ȷ+2)}=τ1(ξ2ȷ+1,ξ2ȷ+2)

    that is contradiction because τ1<1. Hence ξ2ȷ is a fixed point of 2 too. Likewise if ȷ0 is odd number, then there exists ξU such that 1ξ2ξ=ξ. So we assume that ξȷξȷ+1 for all ȷN. Now we shall show that (ξȷ)ȷN is Cauchy sequence. We have these two possible cases to discuss:

    Case 1. ξ=ξ2ȷU1 and ϱ=ξ2ȷ+1U2.

    0<(ξ2ȷ+1,ξ2ȷ+2)=(1ξ2ȷ,2ξ2ȷ+1)τ1max{(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ,1ξ2ȷ),(ξ2ȷ+1,2ξ2ȷ+1)}=τ1max{(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ+1,ξ2ȷ+2)}=τ1max{(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ+1,ξ2ȷ+2)}τ1[(ξ2ȷ,ξ2ȷ+1)+(ξ2ȷ+1,ξ2ȷ+2)]

    that is

    (1τ1)(ξ2ȷ+1,ξ2ȷ+2)τ1(ξ2ȷ,ξ2ȷ+1),

    which implies

    (ξ2ȷ+1,ξ2ȷ+2)τ11τ1(ξ2ȷ,ξ2ȷ+1). (5.3)

    Case 2. ξ=ξ2ȷU1 and ϱ=ξ2ȷ1U2.

    0<(ξ2ȷ+1,ξ2ȷ)=(1ξ2ȷ,2ξ2ȷ1)τ1max{(ξ2ȷ,ξ2ȷ1),(ξ2ȷ,1ξ2ȷ),(ξ2ȷ1,2ξ2ȷ1)}=τ1max{(ξ2ȷ,ξ2ȷ1),(ξ2ȷ,ξ2ȷ+1),(ξ2ȷ1,ξ2ȷ)}=τ1max{(ξ2ȷ1,ξ2ȷ),(ξ2ȷ,ξ2ȷ+1)}τ1[(ξ2ȷ1,ξ2ȷ)+(ξ2ȷ,ξ2ȷ+1)]

    that is

    (1τ1)(ξ2ȷ+1,ξ2ȷ)τ1(ξ2ȷ,ξ2ȷ1),

    which implies

    (ξ2ȷ,ξ2ȷ+1)τ11τ1(ξ2ȷ1,ξ2ȷ). (5.4)

    Since τ=τ11τ1, so we have

    (ξȷ,ξȷ+1)τ(ξȷ1,ξȷ). (5.5)

    Thus, we have

    (ξȷ,ξȷ+1)τ(ξȷ1,ξȷ)τ2(ξȷ2,ξȷ1)τȷ(ξ0,ξ1).

    For all ȷ,mN(ȷ<m), we have

    (ξȷ,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)(ξȷ+1,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)σ(ξȷ+1,ξȷ+2)(ξȷ+1,ξȷ+2)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)(ξȷ+2,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+σ(ξȷ+1,ξm)σ(ξȷ+1,ξȷ+2)(ξȷ+1,ξȷ+2)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)σ(ξȷ+2,ξȷ+3)(ξȷ+2,ξȷ+3)+σ(ξȷ+1,ξm)σ(ξȷ+2,ξm)σ(ξȷ+3,ξm)(ξȷ+3,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)(ξi,ξi+1)+m1i=ȷ+1σ(ξi,ξm)(ξm1,ξm),

    which further implies that

    (ξȷ,ξm)σ(ξȷ,ξȷ+1)(ξȷ,ξȷ+1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)(ξi,ξi+1)+(m1i=ȷ+1σ(ξi,ξm))σ(ξm1,ξm)(ξm1,ξm)σ(ξȷ,ξȷ+1)τȷ(ξ0,ξ1)+m2i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1)+(m1i=ȷ+1σ(ξi,ξm))σ(ξm1,ξm)τm1(ξ0,ξ1)=σ(ξȷ,ξȷ+1)τȷ(ξ0,ξ1)+m1i=ȷ+1(ik=ȷ+1σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1). (5.6)

    Let

    Ψl=li=0(ik=0σ(ξk,ξm))σ(ξi,ξi+1)τi(ξ0,ξ1).

    From (5.6), we get

    (ξȷ,ξm)(ξ0,ξ1)[τȷσ(ξȷ,ξȷ+1)+(Ψm1Ψȷ)]. (5.7)

    Now as σ(ξ,ϱ)1, and by utilizing ratio test, limȷ+Ψȷ exists. Clearly, if we let ȷ,m+ in (5.7), we get that

    limȷ,m+(ξȷ,ξm)=0. (5.8)

    Hence, {ξȷ} is a Cauchy sequence in (U,). So there exists ξU such that

    limȷ+(ξȷ,ξ)=0. (5.9)

    that is ξȷξ as ȷ+. It is obvious that

    limȷ+ξ2ȷ=limȷ+ξ2ȷ+1=ξ. (5.10)

    As 1 and 2 are continuous, so we have

    ξ=limȷ+ξ2ȷ+1=limȷ+1(ξ2ȷ)=1(ξ),ξ=limȷ+ξ2ȷ+2=limȷ+2(ξ2ȷ+1)=2(ξ).

    Now letting ξ=ξU1 and ϱ=ξ2ȷ+2U2, we have

    0<(1ξ,ξ2ȷ+2)=(1ξ,2(ξ2ȷ+1))τ1max{(ξ,ξ2ȷ+1),(ξ,1ξ),(ξ2ȷ+1,2(ξ2ȷ+1)}=τ1max{(ξ,ξ2ȷ+1),(ξ,1ξ),(ξ2ȷ+1,ξ2ȷ+2)}.

    Letting ȷ+ and using (5.10), we can simply conclude that (ξ,1ξ)=0. This yields that ξ=1ξ. Similarly, suppose that ξ=ξ2ȷ+1U1 and ϱ=ξU2, we have

    0<(ξ2ȷ+2,2ξ)=(1(ξ2ȷ),2ξ)τ1max{(ξ2ȷ,ξ),(ξ2ȷ,1(ξ2ȷ)),(ξ,2ξ)}=τ1max{(ξ2ȷ,ξ),(ξ2ȷ,ξ2ȷ+1),(ξ,2ξ)}.

    Letting ȷ+ and using (5.10), we can simply conclude that (ξ,2ξ)=0. This yields that ξ=2ξ.

    Corollary 7. Let (U,σ,) be a complete CMS euipped with a directed graph G and :UU is a G-orbital-cyclic. Suppose that there exists τ1[0,1) such that

    (i) U;

    (ii) for all ξ,ϱU, we have

    (ξ,ϱ)τ1max{(ξ,ϱ),(ξ,ξ),(ϱ,ϱ)};

    (iii) for all (ξȷ)ȷNU one has (ξȷ,ξȷ+1)E(G),

    supm1limiσ(ξi+1,ξi+2)σ(ξi+1,ξm)σ(ξi,ξi+1)<1τ,

    where τ=τ11τ1;

    (iv) is continuous, or for each (ξȷ)ȷNU, with ξȷξ as ȷ+, and (ξȷ,ξȷ+1)E(G) for ȷN, we have ξU;

    (v) for all ξU, we have limȷ+σ(ξȷ,ξ) and limȷ+σ(ξ,ξȷ) exist and finite, then has a unique fixed point.

    Example 3. Let U={0,1,2,3,4}. Define :U×U[0,+) by

    (ξ,ϱ)=|ξϱ|2

    and :U×U[1,+) by

    σ(ξ,ϱ)=1+ξ+ϱ

    for all ξ,ϱU. Then (U,σ,) is complete CMS. Now define :UU by

    ξ=0,forξ{0,1},

    and

    ξ=1,forξ{2,3}.

    Also define G={(0,1),(0,2),(2,3),(0,0),(1,1),(2,2),(3,3)}, then G is directed graph. Then all assumptions of Corollary 3 are satisfied with τ1=13 and ξ=0 is the unique fixed point of .

    In this section, we investigate the solution of Fredholm-type integral equation

    ξ(t)=10K(t,s,ξ(t))ds, (6.1)

    for all t[0,1], where K(t,s,ξ(t)) is a continuous function from [0,1]×[0,1] into R. Let U=C([0,1],(,+)). Now we define :U×U[1, ) by

    (ξ,ϱ)=supt[0,1](|ξ(t)|+|ϱ(t)|2).

    Then (U,σ,) is a complete CMS with σ(ξ,ϱ)=2.

    Theorem 8. Assume that

    (a) |K(t,s,ξ(t))|+|K(t,s,ϱ(t))|τ1(supt[0,1]|ξ(t)|+|ϱ(t)|)(|ξ(t)|+|ϱ(t)|) for some τ1U[0,1);

    (b) K(t,s,10K(t,s,ξ(t))ds)<K(t,s,ξ(t));

    for all t,s[0,1]. Then the integral equation (6.1) has a unique solution.

    Proof. Define :UU by

    ξ(t)=10K(t,s,ξ(t))ds.

    Then

    (ξ,ϱ)=supt[0,1](|ξ(t)|+|ϱ(t)|2).

    Now

    (ξ(t),ϱ(t))=|ξ(t)|+|ϱ(t)|2=|10K(t,s,ξ(t))ds|+|10K(t,s,ϱ(t))ds|210|K(t,s,ξ(t))|ds+10|K(t,s,ϱ(t))|ds2=10(|K(t,s,ξ(t))|+|K(t,s,ϱ(t))|)ds210(τ1(supt[0,1]|ξ(t)|+|ϱ(t)|)(|ξ(t)|+|ϱ(t)|))ds2τ1(supt[0,1]|ξ(t)|+|ϱ(t)|)(ξ(t),ϱ(t)).

    Also we observe that

    σ(ξ,ϱ)=1τ1(supt[0,1]|ξ(t)|+|ϱ(t)|).

    Thus all the conditions of result 6 are satisfied. Hence Eq (6.1) has a unique solution.

    In the current work, we have utilized the notion of controlled metric space and proved common fixed point results of self mappings for generalized contractions involving control functions of two variables. We also established common fixed point results in controlled metric space equipped with a graph. We have derived common fixed point and fixed point results for contractions with control functions of one variable as consequences of our leading result. We also supply a non trivial example to support the obtained results As as application of our prime result, we have investigated the solution of Fredholm type integral equation.

    Some related generalizations of such contractions for the multivalued mappings :UCB(U) and for fuzzy mappings :UF(U) would be a special field for future work. A distinct way of future study would be to employ our results in the solution of fractional differential inclusions.

    First author acknowledges with thanks Natural Science Foundation of Hebei Province (Grant No. A2019404009) and The Major Project of Education Department of Hebei Province (No. ZD2021039) for financial support.

    The authors declare no conflicts of interests.



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