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Research article

The ordered implicit relations and related fixed point problems in the cone b-metric spaces

  • Received: 09 September 2021 Revised: 01 December 2021 Accepted: 02 December 2021 Published: 04 January 2022
  • MSC : 47H09, 47H10, 54H25

  • In this paper, we introduce an ordered implicit relation. We present some examples for the illustration of the ordered implicit relation. We investigate conditions for the existence of the fixed points of an implicit contraction. We obtain some fixed point theorems in the cone b-metric spaces and hence answer a fixed-point problem. We present several examples and consequences to explain the obtained theorems. We solve an homotopy problem and show existence of solution to a Urysohn Integral Equation as applications of the obtained fixed point theorem.

    Citation: Anam Arif, Muhammad Nazam, Aftab Hussain, Mujahid Abbas. The ordered implicit relations and related fixed point problems in the cone b-metric spaces[J]. AIMS Mathematics, 2022, 7(4): 5199-5219. doi: 10.3934/math.2022290

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  • In this paper, we introduce an ordered implicit relation. We present some examples for the illustration of the ordered implicit relation. We investigate conditions for the existence of the fixed points of an implicit contraction. We obtain some fixed point theorems in the cone b-metric spaces and hence answer a fixed-point problem. We present several examples and consequences to explain the obtained theorems. We solve an homotopy problem and show existence of solution to a Urysohn Integral Equation as applications of the obtained fixed point theorem.



    The Banach Contraction Principle [7] for ordered metric spaces was given by Ran and Reuring [39]. Subsequently, Nieto and Rodriguez-Lopez [34], Ó Regan and Petrusel [40] and Agarwal et al. [1] extended this work. Popa [36] introduced an implicit relation and proved a fixed-point theorem for the self-mappings on the complete metric spaces. Altun and Simsek [2] presented a generalization of the results in [1,34,36,39] by using implicit relation in ordered metric space as follow:

    Theorem 1.1. [2] Let (X,d,) be a partially ordered metric space. Suppose S:XX is a non-decreasing mapping such that for all x,yX with xy

    T(d(Sx,Sy),d(x,y),d(x,Sx),d(y,Sy),d(x,Sy),d(y,Sx))0, (1.1)

    where T:[0,)6(,). Also suppose that either S is continuous or (X,d,) is regular, if there exists an element x0X with x0S(x0), then S admits a fixed point.

    We can obtain several contractive conditions from (1.1), for example, defining T:[0,)6(,) by

    T(x1,x2,x3,x4,x5,x6)=x1ψ(max{x2,x3,x4,12(x5+x6)}),

    we have the main result presented in [1]. Similarly, if we choose

    T(x1,x2,x3,x4,x5,x6)=x1kx2;k[0,1)

    in (1.1), we have the main result presented in [39]. Thus, different definitions of T:[0,)6(,) produce different contractive conditions. Moreover, the investigation of fixed points of implicit contractions was done by Popa [37,38], Beg et al. [8,9], Berinde et al. [11,12] and Sedghi [43].

    The graphical metric spaces [20], partial metric spaces [31], dualistic partial metric spaces [32], b-metric spaces, multivalued contractions [27,33], αψ-contractions [42], F-contractions [25], (ψ,ϕ)-contractions [35] are extensively being used in metric fixed point theory. Motivated by the above discoveries, Huang and Zhang [16] presented the idea of cone metric by replacing the set of positive real numbers with ordered Banach space, and utilized this idea to universalize Banach contraction principle. Huang and Zhang [16] worked with the concept of normal cone, however, Rezapour et al. [41] neglected normality of the cone and improved the various theorems presented by Huang. The idea of cone b-metric space [17] was directly influenced by b-metric space [14]. Huang and Xu [18] applied the cone b-metric axioms to prove some fixed point theorems. For detail readings about cone metric, b-metric and cone b-metric, we suggest [3,13,14,21-23,26,29].

    We observe that the implicit relation defined in Popa [37,38], can be generalized to vector spaces. So, in this research paper, motivated by Beg et al. [8,9], Berinde et al. [11,12] and Sedghi [43], we define an ordered implicit relation in a cone b-metric space and contribute a fixed point problem. We answer the proposed problem subject to monotone mappings satisfying an implicit contraction. We solve an homotopy problem and show existence of solution to a Urysohn Integral Equation as applications of the obtained fixed point theorem. The obtained fixed point theorems are independent of the observations presented by Ercan [15]. These observations apply on linear contractions and in this paper, we considered nonlinear contractions (implicit relation involving the contraction mappings). So, the obtained results are real generalizations and could not be followed from known ones in literature.

    In this section, we recall cone, cone metric space and some related properties. Let E represent the real Banach space.

    Definition 2.1. [19] The set E is called a cone if and only if the following axioms hold:

    (1) is closed, non empty and {0E};

    (2) αv+βw, for all v,w and α,βR such that α,β0;

    (3) ()={0E}.

    The partial order with respect to is defined as follows:

    vwwvforallv,wE.

    vw serves as vw but vw, and vw represents that wv(interior of ).

    Definition 2.2. [19] The cone E is normal if for all v,w, there exists S>0 such that,

    0EvwimpliesvSw.

    Let be a partial order in any ordinary set X and be a partial order in cone E. If XE then and would be considered as identical.

    Definition 2.3. [19] Let Xϕ be a set, and the mapping dc:X×XE satisfies the following axioms:

    (dc1) dc(x,y)0E, for all x,yX and dc(x,y)=0E if and only if x=y;

    (dc2) dc(x,y)=dc(y,x);

    (dc3) dc(x,ξ)dc(x,y)+dc(y,ξ), for all x,y,ξX.

    Then dc is known as a cone metric on X, and (X,dc) is called a cone metric space.

    Example 2.4. [6] Let X=R, E=R2, and ={(x,y)E:x,y0}R2. Define the mapping dc:X×XE by

    dc(x,y)=(xy,αxy),

    where α0 is to be taken as a constant. Then dc defines a cone metric on X.

    Proposition 2.5. [6] Suppose that (X,dc) is a cone metric space, with cone . Then for w,ζ,ξE, we have

    (1) If wβw and β[0,1), then w=0E.

    (2) If 0Ewζ for each 0Eζ, then w=0E.

    (3) If wζ and ζξ, then wξ.

    Definition 2.6. [19] Let X be a non-empty set and x,y,υX. The mapping c:X×XE satisfying the following axioms:

    (cb1) 0Ec(x,y) and c(x,y)=0E if and only if x=y;

    (cb2) c(x,y)=c(y,x);

    (cb3) c(x,υ)s[c(x,ξ)+c(ξ,υ)] for s1,

    is known as a b-cone metric on X, and (X,c) is called a cone b-metric space.

    Example 2.7. [6] Let E=R2, ={(x,y)E:x,y0}R2, X={1,2,3,4}. Define the mapping c:X×XE by

    c(x,y)={(|xy|1,|xy|1)ifxy0Eifx=y,

    and the partial order on E by

    vwifandonlyifvwforallv,wE.

    Then (X,c) is a cone b-metric space for s=65. We note that c(1,2)c(1,4)+c(4,2), this shows that c is not a cone metric.

    Remark 2.8. Every cone metric space is a cone b-metric space, but converse is not true in general as seen in Example 2.7.

    Definition 2.9. [10] Let E be a real Banach space and (X,c) be a cone b-metric space. Then for every ϵE with 0Eϵ, we have the following information.

    (1) A sequence {xn} is said to be a Cauchy sequence, if there exists a natural number KN so that c(xn,xm)ϵ for all n,mK.

    (2) A sequence {xn} is said to be a convergent sequence (converging to xX), if there exists KN so that c(xn,x)ϵ for all nK.

    (3) A cone b-metric space (X,c) is complete if each Cauchy sequence converges in X.

    Let (E,.) be a real Banach space and B(E,E) be a space of all bounded linear operators S:EE such that S1<1 and .1 is taken as usual norm in B(E,E).

    Following the implicit relations presented in [4,5,8,9,11,12], we define a new ordered implicit relation as follows:

    Definition 3.1. Let (E,.) be a real Banach space. The relation L:E6E is said to be an ordered implicit relation, if it is continuous on E6 and satisfies the following axioms:

    (L1) v1υ1, v5υ5 and v6υ6

    implies L(υ1,v2,v3,v4,υ5,υ6)L(v1,v2,v3,v4,v5,v6);

    (L2) if L(v1,v2,v2,v1,α[v1+v2],0E)0E

    or,

    if L(v1,v2,v1,v2,0E,α[v1+v2])0E, then there exists SB(E,E) such that v1S(v2) (for all v1,v2E) and α1;

    (L3) L(αv,0E,0E,v,αv,0E)0E whenever v>0 and α1.

    Let G={L:E6E|LsatisfiesconditionsL1,L2,L3}.

    Example 3.2. Let be a partial order with respect to cone as defined in Section 2 and (E,.), be a real Banach space. For viE(i=1to6), α>2 and γ>1, we define the relation L:E6E by

    L(v1,v2,v3,v4,v5,v6)=αv1{v5+γv6}.

    Then LG. Indeed

    (L1): Let v1γ2, v5γ5 and v6γ6, then γ5v5 and γ6v6, we show that L(v1,v2,v3,v4,v5,v6)L(γ1,v2,v3,v4,γ5,γ6). Consider,

    L(v1,v2,v3,v4,v5,v6)L(γ1,v2,v3,v4,γ5,γ6)=αγ1{v5+γv6}αγ1+{γ5+γγ6}=γ5v5+γ(γ6v6).

    Thus, L(γ1,v2,v3,v4,γ5,γ6)L(v1,v2,v3,v4,v5,v6).

    (L2): Let v1,v2,v3E be such that 0Ev1,0Ev2. If L(v1,v2,v2,v1,s[v1+v2],0E)0E then, we have αv1+s{v1+v2}.

    So,sv2(αs)v1.implies (3.1)
    v2(αs)sv1. (3.2)

    If v1=0E, then v2 (by (3.1)). Thus, there exists T:EE defined by T(v2)=ηv2 (where η=s) such that T∥=s>1 (not possible). Now if v10E, then, (3.2) implies v2s(αs)v1. So for α2>s1, there exists T:EE defined by T(v2)=ηv2 (η=s(αs) is a scalar) such that v1T(v2), for α>2.

    (L3): Let vE be such that v>0 and consider, 0EL(sv,0E,0E,v,sv,0E) then (αs1)v, which hold whenever v>0.

    Example 3.3. Let L1,L2,L3:E6E be defined by

    (i) L1(v1,v2,v3,v4,v5,v6)=αv1v2;α<1.

    (ii) L2(v1,v2,v3,v4,v5,v6)=v3α{v3+v4}(1α)βv5;α<12,βR.

    (iii) L3(v1,v2,v3,v4,v5,v6)=v1+v5α{v2+v4};α>12.

    Then Li defines an ordered implicit relation for each i=1,2,3.

    In the next section, we employ this implicit relation in association with a few other conditions to construct an iterative sequence and hence to answer the following fixed-point problem:

    "find p(X,c) such that f(p)=p" where TB(E,E), I:EE an identity operator, LG and f:XX satisfies (3.3), for all comparable x,yXands1.

    (IT)(c(x,f(x)))sc(x,y)implies
    L(c(f(x),f(y)),c(x,y),c(x,f(x)),c(y,f(y)),c(x,f(y)),c(y,f(x)))0E. (3.3)

    The following assertion is essential in the sequel.

    Remark 3.4. If SB(E,E), then Neumann series I+S+S2++Sn+ converges whenever S1<1 and diverge if S1>1. Also for S1<1 there exists μ>0 so that S1<μ<1 and Sn1μn<1.

    Popa[36] applied implicit type contractive conditions on a self-mapping to establish some fixed point results. Altun and Simsek [2] extended the research work in [36] to partially ordered metric spaces. Nazam et al. [30] have further generalized the results given in [2] by using the concept of the cone metric space [19]. In this section, we shall address the proposed fixed-point problem that generalizes the results in [2,30,36]. For that purpose, we have the following theorem.

    Theorem 4.1. Let (X,c) be a complete cone b-metric space with E as a cone and f:XX. If TB(E,E), I:EE is an identity operator and LG such that, for all comparable elements x,yX and s1, we have

    (IT)(c(x,f(x)))sc(x,y)implies
    L(c(f(x),f(y)),c(x,y),c(x,f(x)),c(y,f(y)),c(x,f(y)),c(y,f(x)))0E. (4.1)

    If,

    (1) There exists x0X, so that x0f(x0);

    (2) For all x,yX, xy implies f(x)f(y);

    (3) The sequence {xn} satisfies xn1xn and xnp, then xnp for all nN.

    Then, there exists a point pX such that p=f(p).

    Proof. Suppose that x0X be an arbitrary point such that x0f(x0). We construct a sequence {xn} by f(xn1)=xn taking x0 as an initial guess. Since, x0x1, by assumption (2) we have x1x2, x2x3 xn1xn. Since, x0x1, by (4.1) we have

    (IT)(c(x0,f(x0)))=(IT)(c(x0,x1))sc(x0,x1)implies
    L(c(f(x0),f(x1)),c(x0,x1),c(x0,f(x0)),c(x1,f(x1)),c(x0,f(x1)),0E)0E,

    that is,

    L(c(x1,x2),c(x0,x1),c(x0,x1),c(x1,x2),c(x0,x2),0E)0E. (4.2)

    By triangle property of the cone b-metric, we have

    c(x0,x2)s[c(x0,x1)+c(x1,x2)].

    Rewriting (4.2) and using condition (L1), we get:

    L(c(x1,x2),c(x0,x1),c(x0,x1),c(x1,x2),s[c(x0,x1)+c(x1,x2)],0E)0E.

    By using (L2), there exists KB(E,E) with K1<1 such that

    c(x1,x2)K(c(x0,x1)).

    Again since, x1x2, by (4.1) we obtain

    (IT)(c(x1,f(x1)))=(IT)(c(x1,x2))sc(x1,x2)implies
    L(c(f(x1),f(x2)),c(x1,x2),c(x1,f(x1)),c(x2,f(x2)),c(x1,f(x2)),c(x2,f(x1)))0E,

    that is,

    L(c(x2,x3),c(x1,x2),c(x1,x2),c(x2,x3),c(x1,x3),0E)0E.

    By (cb3), we have

    c(x1,x3)s[c(x1,x2)+c(x2,x3)]

    using (L1) we get

    L(c(x2,x3),c(x1,x2),c(x1,x2),c(x2,x3),s[c(x1,x2)+c(x2,x3)],0E)0E.

    By (L2) there exists KB(E,E) with K1<1 such that

    c(x2,x3)K(c(x1,x2))K2(c(x0,x1)).

    Now keeping in view the above pattern and with the relation xnxn+1 n1, we can construct a sequence {xn} so that xn+1=f(xn) and

    (IT)(c(xn1,f(xn1)))=(IT)(c(xn1,xn))sc(xn1,xn)

    implies

    c(xn,xn+1)K(c(xn1,xn))K2(c(xn2,xn1))Kn(c(x0,x1)).

    For ,nN and s1, consider

    c(xn+,xn)s[c(xn+,xn+1)+c(xn+1,xn)]sc(xn+,xn+1+s2[c(xn+1,xn+2)+c(xn+2,xn)]sc(xn+,xn+1)+s2c(xn+1,xn+2)+......+s(1)c(xn+1,xn)sKn+1(c(x0,x1))+s2Kn+2(c(x0,x1))+......+s(1)Kn(c(x0,x1))=sKn+((sK1)11)sK(c(x0,x1))+s(1)Kn(c(x0,x1))sKn+1sK(c(x0,x1))+s(1)Kn(c(x0,x1)).

    Since K1<1, so, Kn0E as n (Remark 3.4). Hence, limnc(xn,xm)=0E, this shows that {xn} is a Cauchy sequence in X. Since, (X,c) is a complete cone b-metric space, so there exists pX so that xnp for large n, alternately, for a given 0ϵ, there is a natural number N2 so that

    c(xn,p)ϵforallnN2.

    Now since xn1xn and xnp, by the assumption (3), we have xnp for all nN. We claim that,

    (IT)(c(xn,f(xn)))sc(xn,p).

    Suppose on the contrary that

    (IT)(c(xn,f(xn)))sc(xn,p)and
    (IT)(c(xn+1,f(xn+1)))sc(xn+1,p)fornN.

    By (cb3) and (4.1), we get

    c(xn,f(xn))s[c(xn,p)+c(p,xn+1)]s[1s(IT)(c(xn,f(xn)))+1s(IT)c(xn+1,f(xn+1)))][(IT)(c(xn,f(xn)))+(IT)T(c(xn,f(xn)))](IT)(I+T)(c(xn,f(xn)))(IT2)(c(xn,f(xn))).

    Thus,

    T2(c(xn,f(xn)))0E,

    which leads to a contradiction. So, for each n1 and s1, we get

    (IT)(c(xn,f(xn)))sc(xn,p),

    thus, by (4.1), we have

    L(c(f(xn),f(p)),c(xn,p),c(xn,f(xn)),c(p,f(p)),c(xn,f(p)),c(p,f(xn)))0E. (4.3)

    We need to show that c(p,f(p))=0. If c(p,f(p))>0, then we have the following information.

    c(f(xn),f(p))s[c(f(xn),p)+c(p,f(p))]limnc(f(xn),f(p))slimn[c(xn+1,p)+c(p,f(p))]=sc(p,f(p)),and
    c(xn,f(p))s[c(xn,p)+c(p,f(p))]limnc(xn,f(p))slimn[c(xn,p)+c(p,f(p))]limnc(xn,f(p))sc(p,f(p)).

    In view of the condition (L1) and (4.3), we have

    L(s(c(p,f(p))),0E,0E,c(p,f(p)),s(c(p,f(p))),0E)0E.

    This contradicts the condition (L3). Thus, c(p,f(p))=0. So, c(p,f(p))=0E, and hence p=f(p).

    Remark 4.2. If the operator L:E6E is defined by

    L(x1,x2,x3,x4,x5,x6)=x1ψ(max{x2,x3,x4,12(x5+x6)}),forallxiE

    where, ψ:EE is a non-decreasing operator satisfying limnψn(v)=0E. Then Theorem (4.1), generalizes the corresponding result in [1]. If we define L:E6E by

    L(x1,x2,x3,x4,x5,x6)=x1kx2;k[0,1)

    in Theorem (4.1), we obtain a generalization of the corresponding result in [39]. Thus, different definitions of L:E6E produce different ordered contractive conditions. Moreover, Theorem (4.1) generalizes the results in Popa [37,38], Beg et al. [8,9], Berinde et al. [11,12].

    The following theorem is for decreasing self-mappings.

    Theorem 4.3. Let (X,c) be a complete cone b-metric space with E as a cone and f:XX. Let TB(E,E), I:EE be an identity operator, LG so that, for all comparable elements x,yX and s1

    (IT)(c(x,f(x)))sc(x,y)implies
    L(c(f(x),f(y)),c(x,y),c(x,f(x)),c(y,f(y)),c(x,f(y)),c(k,f(x)))0E. (4.4)

    If,

    (1) There exists x0X, so that f(x0)x0;

    (2) For all x,yX, xy implies f(y)f(x);

    (3) The sequence {xn} satisfies xn1xn and xnx, then xnx for all nN.

    Then, there exists a point xX such that x=f(x).

    Proof. Let x0X be arbitrary satisfying assumption (1). We define a sequence {xn} by xn=f(xn1) for all n. As x1=f(x0)x0 and by condition (2) x1=f(x0)f(x1)=x2 and repeated implementation of hypothesis (2) leads to have xnxn1. Since, x=x1x0, by (4.4), we get

    (IT)(c(f(x0),x0)=(IT)(c(x1,x0))sc(x1,x0)implies
    L(c(f(x1),f(x0)),c(x1,x0),c(x1,f(x1)),c(x0,f(x0)),c(x1,f(x0)),c(x0,f(x1)))0E
    L(c(x1,x2),c(x0,x1),c(x1,x2),c(x0,x1),0,c(x0,x2))0E.

    By (cb3), we have

    c(x0,x2)s[c(x0,x1)+c(x1,x2)]

    and then using L1, we obtain

    L(c(x1,x2),c(x0,x1),c(x1,x2),c(x0,x1),0E,s[c(x0,x1)+c(x1,x2)]0E.

    By (L2), there exists KB(E,E) with K1<1 such that

    c(x1,x2)K(c(x0,x1)).

    Again since, x=x1x2, by (4.4)

    (IT)(c(x1,f(x1))=(IT)(c(x1,x2))sc(x1,x2)implies
    L(c(f(x1),f(x2)),c(x1,x2),c(x1,f(x1)),c(x2,f(x2)),c(x1,f(x2)),c(x2,f(x1)))0E
    L(c(x2,x3),c(x1,x2),c(x1,x2),c(x2,x3),c(x1,x3),0E)0E.

    By (cb3), (L1) and (L2), we get

    c(x2,x3)T(c(x1,x2))T2(c(x0,x1)).

    By following same steps, we can construct a sequence {xn} such that

    c(xn,xn+1)K(c(xn1,xn))K2(c(xn2,xn1))Kn(c(x0,x1)).

    Hence copying the arguments for the proof of Theorem 4.1, we get x=f(x).

    The following theorem encapsulate the statements of Theorem 4.1 and Theorem 4.3.

    Theorem 4.4. Let (X,c) be a complete cone b-metric space with E as a cone and f:XX be monotone mapping. Let TB(E,E), I:EE be an identity operator. If there exists LG so that, for all comparable x,yX and s1

    (IT)(c(x,f(x)))sc(x,y)implies
    L(c(f(x),f(y)),c(x,y),c(x,f(x)),c(y,f(y)),c(x,f(y)),c(k,f(x)))0E, (4.5)

    and,

    (1) There exists x0X, so that x0f(x0) or f(x0)x0,

    (2) The sequence {xn} satisfies xn1xn and xna, then xna for all nN.

    Then, there exists a point aX such that a=f(a).

    Proof. Proof is obvious.

    Remark 4.5. We can get a unique fixed point in Theorem 4.1, Theorem 4.3 and Theorem 4.4 by taking an additional condition, "for each pair x,yX, we have either an upper bound or lower bound." (2). The cone is assumed as non-normal.

    We illustrate above theorems with the help of the following examples.

    Example 5.1. Let E=C1R[0,1], and ζ=ζ+ˊζ,={ζE:ζ(t)>0,t[0,1]}. For each K1, take ζ=x and y=x2K. By definition ζ=1 and y=2k+1. Clearly ζy, and Kζ y. Hence is a non-normal cone. Define the operator T:EE by

    (Tζ)(t)=12t0ζ(s)ds.Thus,Tislinearandboundedasshownbelow.
    (T(aζ+by))(t)=12t0(aζ+by)(s)ds=a2t0ζ(s)ds+b2t0y(s)ds,and
    Tnζζn+12(n+1)!foreachn1.

    So

    (Tnζ)12(n+1)!.
    (Tnζ)ζn2(n)!12(n)!forn1.
    (Tnζ)=(Tnζ)+(Tnζ)12(n+1)!+12(n)!forn1
    (Tnζ)=0whennn1forn1N.

    Hence TB(E,E). Let X={1,2,3} and f:XX defined by f(1)=f(2)=1 and f(3)=2, then f is increasing with respect to usual order. Define the mapping c by

    c(x1,x2)={0ifx1=x2ζ3ifx1,x2{1,2};ζEζ8otherwise.
    ζ3=c(1,2)c(1,3)+c(3,2)=ζ8+ζ8,

    since triangular inequality does not hold, so c is not a cone metric space, but one can check that c is a cone b-metric space for s=43. Now, for x=1 and y=2 (xy), we have

    c(x,y)=ζ3=c(y,f(x))=c(y,f(y))
    c(x,f(x))=0E=c(f(x),f(y))=c(x,f(y)).

    For x=2 and y=3

    c(x,y)=ζ8,c(x,f(x))=ζ3=c(f(x),f(y))
    c(x,f(y))=0E,c(y,f(x)=ζ8=c(y,f(y)).

    For α>2 and γ>163, define

    L(c(f(x),f(y)),c(x,y),c(x,f(x)),c(y,f(y)),c(x,f(y)),c(y,f(x)))=αc(f(x),f(y))[c(x,f(y))+γc(y,f(x)].

    Clearly

    (IT)c(x,f(x))sc(x,y)implies
    αc(f(x),f(y))c(x,f(y))+γc(y,f(x).

    Thus, by Theorem 4.1, f has a fixed point which is given by f(1)=1.

    Remark 5.2. Since, c is not a cone metric, this shows that Theorem 4.1 does not hold in a cone metric space. The Example 5.1 also endorses the choice of cone b-metric space for this paper.

    In the following, we have some consequences of the main results given above.

    Corollary 5.3. Let (X,c) be a complete cone b-metric space with E as a cone and f:XX. If TB(E,E), I:EE an identity operator and there exist LG so that, for all comparable elements x,yX and s1

    (IT)(c(x,f(x)))sc(x,y)implies
    c(f(x),f(y))T(c(x,y)), (5.1)

    and,

    (1) xX so that x0f(x0) or f(x0)x0;

    (2) For all x,yX, xy implies f(x)f(y) or f(y)f(x);

    (3) The sequence {xn} satisfies xn1xn and xnb, then xnb for all nN.

    Then, there exists a point bX such that b=f(b).

    Proof. Define L as in Example 3.3 (i), and the operator T:EE by T(v)=αv for all vE and 0α<1, then TB(E,E and application of Theorem 4.4 provide the proof.

    Corollary 5.4. Let (X,c) be a complete cone b-metric space with E as a cone and f:XX. If TB(E,E), I:EE an identity operator and there exist LG so that, for any comparable x,yX and s1

    (IT)(c(x,f(x)))sc(x,y)implies
    c(f(x),f(y))1sT(c(x,y)), (5.2)

    and,

    (1) xX so that x0f(x0) or f(x0)x0;

    (2) For all x,yX, xy implies f(x)f(y) or f(y)f(x);

    (3) The sequence {xn} satisfies xn1xn and xnr, then xnr for all nN.

    Then, there exists a point rX such that r=f(r).

    Proof. Define the operator T:EE by T(v)=v for all vE and following the proof of Corollary 5.3, we receive the result.

    The following examples illustrates Corollary 5.3 and Corollary 5.4.

    Example 5.5. Let E=(R,) be a real Banach space. Define ={xR:x0}, then, it is a cone in E. Let X={12,23,34,45}, define f:XX so that, f(23)=f(34)=45 and f(12)=f(45)=12, then f is deceasing with respect to usual order. Let TB(E,E) be defined by T(x)=x2. Define the mapping c by

    c(x1,x2)={0.3if(x1,x2)=(12,23)0.1if(x1,x2)=(12,45)0.01if(x1,x2)=(12,34)0.8if(x1,x2)=(34,45),(x1,x2)=(23,34)or(x1,x2)=(23,45)x1x2otherwise.

    Observe that:

    0.8=c(23,34)c(23,12)+c(12,34)=0.3+0.01=0.31
    0.8=c(34,45)c(34,12)+c(12,45)=0.01+0.1=0.11.

    So (cb3) holds for s=10, and c is a cone b-metric, but not a cone metric, as (dc3) does not hold. Consider x=12 and y=23. Then

    c(f(x),f(y))=c(x,f(y))=0.1,c(x,f(x))=0
    c(x,y)=c(y,f(x))=0.3,c(y,f(y))=0.8
    (IT)c(x,f(x))=0.

    Now take x=23 and y=34. Then

    c(f(x),f(y))=0,c(x,y)=c(x,f(x))=0.8
    (IT)c(x,f(x))=0.4,sc(x,f(x))=10(0.8)=8.

    For x=34 and y=45. Then

    c(f(x),f(y))=0.1,c(x,y)=c(x,f(x))=0.8
    (IT)c(x,f(x))=0.4,sc(x,f(x))=10(0.8)=8.

    Thus, for all x,yX such that xy, we have

    (IT)(c(x,f(x)))sc(x,y)impliesc(f(x),f(y))T(c(x,y)).

    Hence, Corollary 5.3 holds for all comparable x,yX. Notice that 12 is a fixed point of f.

    Remark 5.6. Since, c is not a cone metric, this shows that Corollary 5.3 does not hold in a cone metric space. The Example 5.5 also endorses the choice of cone b-metric space for this paper.

    Example 5.7. Consider E=(R,) be a real Banach space. Define ={xR:x0}, then, it is a cone in E. Take X={0,1,2}, and f:XX, f(0)=f(2)=0, f(1)=2. Define the mapping c by

    c(x1,x2)={0ifx1=x23ifx1,x2{1,2}10ifx1,x2{0,1}0.5otherwise.

    Notice that:

    10=c(0,1)c(0,2)+c(2,1)=0.5+3=3.5.

    For s=3, c is a cone b-metric space, but not cone metric space, as (dc3) does not hold. Define T(x)=x2 Consider x=0, y=1. Then

    c(x,y)=10,c(f(x),f(y))=0.5,c(x,f(x))=0
    Tc(x,y)=102=5,Tsc(x,y)=53=1.6667.

    Take x=1, y=2. Then

    c(x,y)=3,c(f(x),f(y))=0.5,c(x,f(x))=3
    Tc(x,y)=32=1.5,Tsc(x,y)=1.53=0.5
    (IT)c(x,f(x))=0.25.

    Clearly for all x,yX,

    (IT)c(x,f(x))sc(x,y)

    implies

    c(f(x),f(y))1sT(c(x,y)).

    So for all values of x,yX, the Corollary 5.4 holds. Here 0 is a fixed point of f.

    This section consists of a homotopy theorem as an application of Corollary 5.4.

    Theorem 6.1. Let (E,.) be a real Banach space with E taken as a cone and (X,c) be a complete cone b-metric space with open set UX. Suppose that TB(E,E) such that T1<1 with T(). If the mapping h:¯U×[0,1]X admits conditions of Corollary 5.4 in the first variable and

    (1) xh(x,θ) for each xU (U represents the boundary of U in X);

    (2) There exists M0 so that

    c(h(x,μ1),h(x,μ2))M|μ1μ2|

    for some x¯U and μ1,μ2[0,1];

    (3) For any xU there is yX such that c(x,y)r, then xy, here r represents radius of U.

    Then, whenever h(,0) possesses a fixed point in U, h(,1) also possesses a fixed point in U.

    Proof. Let

    B={t[0,1]|x=h(x,t);forxU}.

    Define the partial order in E by uv uv for all u,vE. Clearly 0B, since h(.,0) possesses a fixed point in U. So Bϕ, In consideration of c(x,h(x,θ))=c(x,y), (IT)(c(x,h(x,θ)))sc(x,y) for all xy and s1, by Corollary 5.4, we get

    c(h(x,θ),h(y,θ))1sT(c(x,y)).

    Firstly, we prove that B is closed in [0,1]. For this, let {θn}n=1B with θnθ[0,1] as n. It is necessary to prove that θB. Since, θnB for nN, there exists xnU with xn=h(xn,θn). Since, h(,θ) is monotone, so, for n,mN, we have xmxn. Since for s1

    (IT)(c(xn,h(xm,θm)))=(IT)(c(xn,xm))sc(xn,xm),

    we have

    c(h(xn,θm),h(xm,θm))1sT(c(xnxm)),

    and

    c(xn,xm)=c(h(xn,θn),h(xm,θm))s[c(h(xn,θn),h(xn,θm))+c(h(xn,θm),h(xm,θm))]c(xn,xm)sM|θnθm|+ssT(c(xn,xm))c(xn,xm)sM1T[|θnθm|].

    As {θn}n=1 is a Cauchy sequence in [0,1], we have

    limn,mc(xn,xm)=0E.

    So {xn} is a Cauchy sequence in X. As X is a complete cone b-metric space, so we have x¯U such that limnc(xn,x)ϵ. Hence xnx for all nN. By triangle property, we have

    c(x,h(x,θ))s[c(x,xn)+c(xn,h(x,θ))]sc(x,xn)+s2[c(xn,h(xn,θ))+c(h(xn,θ),h(x,θ))]sc(x,xn)+s2[c(h(xn,θn),h(xn,θ))+c(h(xn,θ),h(x,θ))]c(x,h(x,θ))sc(x,xn)+s2M|θnθ|+s2sT(c(xn,x)).

    Thus, c(x,h(x,θ))=0 as n. So θB, hence B is closed in [0,1]. Now we show that B is open in [0,1]. Let θ1B, so, there exists x1U such that h(θ1,x1)=x1. As U is open, we have r>0 so that B(x1,r)U. Consider

    l=c(x1,U)=inf{c(x1,ξ):ξU}.

    Then r=l>0. Given ϵ>0 such that ω<(1T)lsM for s1. Let θ(θ1ω,θ1+ω). Then

    x¯B(x1,r)={xX:c(x,x1)r},sothatxx1.

    Consider

    c(h(x,θ),x1)=c(h(x,θ),h(x1,θ1)s[c(h(x,θ),h(x,θ1)+c(h(x,θ1),h(x1,θ1)]c(h(x,θ),x1)sM|θ1θ|+ssT(c(x1,x))sMω+Tl=sMω+Tl<l.

    Thus, for each θ(θ1ω,θ1+ω), h(,θ):¯B(x,r)¯B(x,r) has a fixed point in ¯U by applying Corollary 5.4. Hence θB for any θ(θ1ω,θ1+ω) and so B is open in [0,1]. Thus, B is open as well as closed in [0,1] and by connectedness, B=[0,1]. Hence h(,1) has a fixed point in U.

    In this section, we use the homotopy to describe the process of aging of human body. The aging process is considered by choosing suitable values for the time parameters t of the homotopy a(t,x). The values of the parameters t and x in the function a(t,x) are adjusted to control the process of the aging. For example, if t[0,1], and there is a homotopy a(t,x) from f(x) to g(x) such that a(0,x)=f(x) and a(1,x)=g(x) then the body is only one year old. Thus, if t[l,n], where n>l, and there is a continuous function a(t,x) called homotopy from one function from f(x) to w(x) satisfying the condition a(l,x)=f(x) and a(n,x)=w(x), then the body is described as being n years old. It is observed that for the interval t[l,n], the supremum a(n,x)=y(x) is the actual age of the body. Topologically the infant is equal to the adult since the infant continuously grows into the adult. The study found an algebraic way of relating homotopy to the process of aging of human body. The compact connected human body with boundary is assumed to be topologically equivalent to a cylinder X=S×I, where S is a circle and I=[0,α]. The initial state of the body X=S×I is the topological shape of the infant. The aging process, called homotopy, is the family of continuous functions a(t,x) on the interval I=[0,α]. It is an increasing sequence of the function a(t,x) of the body X. The homotopy relates the topological shape of the infant to the topological shape of the adult. For the human body X, let xX and tI define the growth of the body and the age of the body respectively. Since the final age of the human body is not known let t= represent the final age of the body such that t[θ,] denotes the age interval of the body from t=θ to t=. The time t= is the age threshold value of the human body. The aging process for all t[0,] is the family or the sequence of the functions a(t,x) such that a(0,x)=f(x) and a(,x)=y(x).

    Theorem 7.1. Let X=S×I be a cylinder. Let a(t,x) be a homotopy related to the process of aging of human body. Then, whenever a(0,x) possesses a fixed point in X, Then, a(,x) also possesses a fixed point in X.

    Proof. Since, the human body is compact connected and bounded, so, human body is topologically equivalent to a cylinder X=S×I. It is known that continuous reshaping of a cylinder possesses many invariant points. Thus, whenever a(0,x) possesses a fixed point in X, Then, a(,x) possesses a fixed point in X.

    In this section, we will apply Theorem 4.1 for the existence of the unique solution to UIE:

    ()=f()+IRK1(,s,(s))ds. (8.1)

    This integral equation encapsulates both Volterra Integral Equation (VIE) and Fredholm Integral Equation (FIE), depending upon the region of integration (IR). If IR =(a,x) where a is fixed, then UIE is VIE and for IR =(a,b) where a,b are fixed, UIE is FIE. In the literature, one can find many approaches to find a unique solution to UIE (see [24,28,44] and references therein). We are interested to use a fixed-point technique for this purpose. The fixed-point technique is simple and elegant to show the existence of a unique solution to further mathematical models.

    Let IR be a set of finite measure and L2IR={|IR|(s)|2ds<}. Define the norm .:L2IR[0,) by

    2=IR|(s)|2ds,forall,ȷL2IR.

    An equivalent norm can be defined as follows:

    2,ν=sup{eνIRα(s)dsIR|(s)|2ds},forallL2IR,ν>1.

    Then E=(L2IR,.2,ν) is a Banach space. Let A={L2IR:(s)>0foralmosteverys} be a cone in E. The cone b-metric cν associated to norm .2,ν is given by cν(,ȷ)=ȷ22,ν for all ,ȷA. Define a partial order on E by

    aυifandonlyifa(s)υ(s)υ(s),foralla,υE.

    Then (E,,cν) is a complete cone b-metric space. Let

    (A1) The kernel K1:IR×IR×RR satisfies Carathéodory conditions and

    |K1(,s,(s))|w(,s)+e(,s)|(s)|;w,eL2(IR×IR),e(,s)>0.

    A2) The function f:IR[1,) is continuous and bounded on IR.

    (A3) There exists a positive constant C such that

    supIRIR|K1(,s)|dsC.

    (A4) For any 0L2IR, there is 1=R(0) such that 10 or 01.

    (A4) The sequence {n} satisfies n1n and np, then np for all nN.

    (A5) There exists a non-negative and measurable function q:IR×IRR such that

    α():=IRq2(,s)ds1ν

    and integrable over IR with

    |K1(,s,(s))K1(,s,ȷ(s))|q(,s)|(s)ȷ(s)|

    for all ,sIR and ,ȷE with ȷ.

    Theorem 8.1. Suppose that the mappings f and K1 mentioned above satisfy the conditions (A1)–(A5), then the UIE (8.1) has a unique solution.

    Proof. Define the mapping R:EE, in accordance with the above-mentioned notations, by

    (R)()=f()+IRK1(,s,(s))ds.

    The operator R is -preserving:

    Let ,ȷE with ȷ, then (s)ȷ(s)ȷ(s). Since, for almost every IR,

    (R)()=f()+IRK1(,s,(s))ds1,

    this implies that (R)()(Rȷ)()(Rȷ)(). Thus, (R)(Rȷ).

    Self-operator:

    The conditions (A1) and (A3) imply that R is continuous and compact mapping from A to A (see [24,Lemma 3]).

    By (A4), for any 0A there is 1=R(0) such that 10 or 01 and using the fact that R is -preserving, we have n=Rn(0)) with nn+1 or n+1n for all n0. We will check the contractive condition of Theorem 4.1 in the next lines. By (A5) and Holder inequality, we have

    |(R)()(Rȷ)()|2=|IRK1(,s,(s))dsIRK1(,s,ȷ(s))ds|2(IR|K1(,s,(s))K1(,s,ȷ(s))|ds)2(IRq(,s)|(s)ȷ(s)|ds)2IRq2(,s)dsIR|(s)ȷ(s)|2ds=α()IR|(s)ȷ(s)|2ds.

    This implies, by integrating with respect to ,

    IR|(R)()(Rȷ)()|2dIR(α()IR|(s)ȷ(s)|2ds)d=IR(α()eνIRα(s)dseνIRα(s)dsIR|(s)ȷ(s)|2ds)dȷ22,νIRα()eνIRα(s)dsd1νȷ22,νeνIRα(s)ds.

    Thus, we have

    eνIRα(s)dsIR|(R)()(Rȷ)()|2d1νȷ22,ν.

    This implies that

    RRȷ22,ν1νȷ22,ν.

    That is,

    cν(R,Rȷ)1νcν(,ȷ).

    Thus, defining L:E6E by

    L(x1,x2,x3,x4,x5,x6)=x1kx2;k[0,1),

    we have

    L(c(R,Rȷ),c(,ȷ),c(,R),c(ȷ,Rȷ),c(,Rȷ),c(ȷ,R))0E.

    Hence, by Theorem 4.1, the operator R has a unique fixed point. This means that the UIE (8.1) has a unique solution.

    The ordered implicit relation in relation with implicit contraction is useful to obtain fixed point theorems that unify many corresponding fixed point theorems. These results can be applied to show the existence of the solutions to DE's and FDE's. The ordered implicit relation can be generalized to orthogonal implicit relation. The study of fixed point theorems is valid in cone metric spaces and hence in the cone b-metric spaces for the nonlinear contractions.

    The authors declare that they have no competing interests.



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