EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 4, 2020, 995-1015 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Solving a class of integral equations via contractive mapping with rational type Abdullah1, M. Sarwar1, Z. Mustafa2, M.M.M. Jaradat2,∗ 1 Department of Mathematics, University of Malakand, Chakdara Dir(L), Pakistan. 2 Department of Mathematics, Statistics and Physics,Qatar University, Doha-Qatar. Abstract. In this paper, using rational type contractive conditions, the existence and uniqueness of common coupled fixed point theorem in the set up of Gb-metric spaces is studied. The derived result cover and generalize some well-known comparable results in the existing literature. Then we use the derived results to prove the existence and uniqueness solution for some classes of integral equations. Further more, an example of such type of integral equation is presented. 2020 Mathematics Subject Classifications: 47H10, 54H25 Key Words and Phrases: Integral equations, Complete Gb-metric space, Cauchy sequence, common coupled fixed point. 1. Introduction and Preliminaries The notion of metric space has been generalized in different directions. In particular, Mustafa and Sims [18] introduced a new generalization of metric space known as G-metric space. In fact, they assigned a non-negative real number to every triplet of elements of a metric space M and studied fixed point results. Further, in the setting of G-metric space, Mustafa et al. [20] investigated some fixed point results for mappings via rational type contractions. Abbas and Rhoades [2] opened the study of finding common fixed points in G-metric spaces. Shatanawi [15] studied applications to integral equations via fixed point results for two weakly increasing mappings f and g with respect to partial ordering relation � in G- metric spaces. After that several fixed point results were proved in these spaces. Some of these works are noted in [ [8],[3],[19],[22],[23] ] . Recently Aghajani et al. [1] introduced the concept of Gb-metric spaces by combining the definition of G-metric and b-metric spaces and studied a common fixed result for six mappings. Jamal Rezaei et al. [13] obtained common fixed point results for three maps in discontinuous Gb-metric spaces. Sedghi et al. [14] derived coupled fixed point theorems ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3831 Email addresses: abdullahibad474@gmail.com (Abdullah), sarwarswati@gmail.com (M. Sarwar), zead@qu.edu.qa (Z. Mustafa), mmjst4@qu.edu.qa (M.M.M. Jaradat) https://www.ejpam.com 995 c© 2020 EJPAM All rights reserved. M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 996 in Gb-metric spaces. Khomdram et al. [4] obtained a coupled fixed point theorem in Gb-metric space using rational type contractive conditions. In addition Mustafa et al. [21] worked on applications to a system of integral equations with the help of tripled coincidence point in ordered Gb-metric spaces. Guo and Lakshmikantham [5] gave the concept of coupled fixed point of non-linear operator with applications. Lakshmikantham et al. [16] introduced Ćirić type mappings in the framework of Gb−metric spaces. Sintunavarat et al. [17] used the nonlinear contraction to established a coupled fixed point results in complete metric spaces without the mixed monotone property. Using generalized contractive condition, Nashie et al. [6] proved the existence and uniqueness of a common coupled fixed point theorem for a pair of mappings in complete cone metric space. Radenović [10] presented some remarks on some recent coupled coincidence point results in symmetric G-metric spaces and reduce some coupled coincidence point results to more general forms. For more details on coupled fixed point results we refer the reader to [[12], [9],[10], [11], [7]]. In the current work we will obtain a common triple fixed point in Gb-metric space using contraction with rational type. Definition 1. [18] Let S 6= ∅ and G : S3 → R+ satisfies : G1) G(p, q, r) = 0 if p = q = r; G2) G(p, p, q) > 0 for all p, q ∈ S with p 6= q; G3) G(p, q, q) ≤ G(p, q, r) for all p, q, r ∈ S with q 6= r; G4) G(p, q, r) = G(q, r, p) = G(r, p, q) = · · · (symmetry in all three variables); G5) G(p, q, r) ≤ G(p, t, t) +G(t, q, r) for all p, q, r, t ∈ S. Then G is called generalized metric on S, and the pair (S,G) is called G-metric space. Definition 2. [1] Let Y 6= ∅ and s ≥ 1 be a given real number. Suppose that a mapping Gb : Y × Y × Y → R+ satisfies: (Gb1) Gb(a, b, c) = 0 if a = b = c, (Gb2) Gb(a, a, b) > 0 for all a, b ∈ Y with a 6= b, (Gb3) Gb(a, b, b) ≤ Gb(a, b, c) for all a, b, c ∈ Y with b 6= c, (Gb4) Gb(a, b, c) = Gb(p{a, b, c}) where p is a permutation of a, b, c (symmetry) (Gb5) Gb(a, b, c) ≤ s(Gb(a, u, u) +Gb(u, b, c)) for all a, b, c, u ∈ Y . Then Gb is called Gb-metric on Y , and the pair (Y,Gb) is called Gb-metric space. If s = 1, then Gb reduce to G-metric space. Gb is the generalization of G-metric, evidently every G-metric is a Gb-metric. Remark 1. Every G-metric space (X,G) defines a Gb-metric space with s = 2p−1 by Gb(x, y, z) = (G(x, y, z))p, where p > 1 is a real number. Further every Gb-metric is a G-metric when s = 1, but, in general, not every Gb-metric is a G-metric, for example, let X = R and the metric Gb is defined by Gb(x, y, z) = max { |x− y|2, |y − z|2, |z − x|2 } , for all x, y, z ∈ R. Then Gb is a Gb-metric on R with s = 22−1 = 2, but it is not a G-metric on R. M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 997 Proposition 1. [1] Let X be a Gb-metric space, then for each x, y, z, a ∈ X the following holds: (1) if Gb(x, y, z) = 0 then x = y = z. (2) G(x, y, z) ≤ s(Gb(x, x, y) +Gb(x, x, z)). (3) Gb(x, y, y) ≤ 2sGb(y, x, x). (4) Gb(x, y, z) ≤ s(G(x, a, z) +G(a, y, z)). Definition 3. [1] Let X ′ be a Gb-metric space. A sequence {x′n} in X is said to be: (1) Gb-Cauchy sequence if, for each ε > 0 there exists n0 ∈ Z+ such that for all m,n, l ≥ n0, Gb(xn, xm, xl) < ε. (2) Gb-convergent to a point x ∈ X if for each ε > 0, there exists n0 ∈ Z+ such that, for all m,n ≥ n0, Gb(xn, xm, x) < ε. Proposition 2. [1] Let X be a Gb-metric space, Then the following are equivalent: (1) the sequence xn is Gb-Cauchy. (2) for any ε > 0, there exists n0 ∈ N such that G(xn, xm, xm) < ε, for all m,n ≥ n0. Proposition 3. [1] The following are equivalent in (X,Gb) metric spaces: (A1) {xn} is Gb-convergent to x. (A2) Gb(xn, xn, x)→ 0 as n→ +∞. (A3) Gb(xnx, x)→ 0 as n→ +∞. Definition 4. [1] A Gb-metric space X is called Gb-complete if every Gb-Cauchy sequence is Gb-convergent in X. Definition 5. [5] Let M ′ be a metric space and let F : M × M → M be a function. An element (m,n) ∈ M × M is said to be a coupled fixed point of the mapping F if F (m,n) = m,F (n,m) = n. 2. Main Result Throughout the paper we will use the following generalized contraction. Assume that (X,Gb) is generalized Gb-metric space. The mappings F,H, T : X ×X → X are said to satisfy the generalized contraction if for a, b, u, v, x, y ∈ X Gb ( F (a, b), H(u, v), T (x, y) ) ≤ N (a, b, u, v, x, y), where N (a, b, u, v, x, y) = α1(a, u) · Gb(a, u, x) +Gb(b, v, y) 2 +α2(a, u) Gb ( F (a, b), H(u, v), T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) +α3(a, u) Gb ( F (a, b), H(u, v), T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 998 +α4(a, u) Gb ( a, a, F (a, b) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α5(u, x) Gb ( a, a, F (a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α6(u, x) Gb ( u, u,H(u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α7(u, x) Gb ( u, u,H(u, v) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α8(u, x) Gb ( x, x, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α9(x, a) Gb ( x, x, T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α10(x, a) Gb ( a, u, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α11(x, a) Gb ( u, x, F (a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α12(x, a) Gb ( x, a,H(u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) , (1) Ω1(a, u, x, b, v, y) = [1 + Gb(a, u, x) + Gb(b, v, y)], Ωs(a, u, x, b, v, y) = sΩ1(a, u, x, b, v, y) and αi : X ×X → [0, 1), i = 1, 2, 3, · · · , 12. such that 0 ≤ s[α1(x, y)+α4(x, y)+α5(x, y)+α10(x, y)+α12(x, y)]+[α2(x, y)+α3(x, y)+ 11∑ i=6 αi(x, y)] < 1. (2) Theorem 1. Let (X,Gb) be a complete Gb-metric space and let F,H, T : X × X → X be mappings satisfying the above generalized contraction(1) and the following conditions satisfied ∀i, i = 1, · · · , 12: (i) αi(F (x1, y1), y) ≤ αi(x1, y) and αi(x, F (x1, y1)) ≤ αi(x, x1)); (ii) αi(H(x1, y1), y) ≤ αi(x1, y) and αi(x,H(x1, y1)) ≤ αi(x, x1)); (iii) αi(T (x1, y1), y) ≤ αi(x1, y) and αi(x, T (x1, y1)) ≤ αi(x, x1)). Then F,H, T have a unique common coupled fixed point in X. Proof. Define the sequences {xn} and {yn} in X by the rule: x3k+1 = F (x3k, y3k), (3) y3k+1 = F (y3k, x3k), . x3k+2 = H(x3k+1, y3k+1), (4) y3k+2 = H(yk+1, x3k+1), x3k+3 = T (x3k+2, y3k+2) (5) y3k+3 = T (yk+2, x3k+2), where k = 0, 1, 2, 3, · · · and x0, y0 to be arbitrary in X. Now, consider Gb(x3k+1, x3k+2, x3k+3) = Gb(F (x3k, y3k), H(x3k+1, y3k+1), T (x3k+2, y3k+2)) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 999 ≤ α1(x3k, x3k+1). Gb(x3k, x3k+1, x3k+2) +Gb(y3k, y3k+1, y3k+2) 2 +α2(x3k, x3k+1) · Gb ( F (x3k, y3k), H(x3k+1, y3k+1), T (x3k+2, y3k+2) ) Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α3(x3k, x3k+1) · Gb ( F (x3k, y3k), H(x3k+1, y3k+1), T (x3k+2, y3k+2) ) Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α4(x3k, x3k+1) · Gb(x3k, x3k, F (x3k, y3k))Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α5(x3k+1, x3k+2) · Gb(x3k, x3k, F (x3k, y3k))Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α6(x3k+1, x3k+2) · Gb(x3k+1, x3k+1, H(x3k+1, y3k+1))Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α7(x3k+1, x3k+2) · Gb(x3k+1, x3k+1, H(x3k+1, y3k+1))Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α8(x3k+1, x3k+2) · Gb(x3k+2, x3k+2, T (x3k+2, y3k+2))Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α9(x3k+2, x3k) · Gb(x3k+2, x3k+2, T (x3k+2, y3k+2))Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α10(x3k+2, x3k) · Gb(x3k, x3k+1, T (x3k+2, y3k+2))Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α11(x3k+2, x3k) · Gb(x3k+1, x3k+2, F (x3k, y3k))Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α12(x3k+2, x3k) · Gb(x3k+2, x3k, H(x3k+1, y3k+1))Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) Now, with the help of condition (i), we have αi(x3k, x3k+1) = αi(x3k, F (x3k, y3k)) ≤ αi(x3k, x3k) = αi(x3k, F (x3k−1, y3k−1)) ≤ αi(x3k, x3k−1) = αi(F (x3k−1, y3k−1), x3k−1) ≤ αi(x3k−1, x3k−1) ≤ · · · ≤ αi(x0, x0) for all i = 1, 2, 3, 4. Similarly by using (ii) and (iii) one can show that αi(x3k+1, x3k+2) ≤ αi(x0, x0) for all i = 5, 6, 7, 8. αi(x3k+2, x3k) ≤ αi(x0, x0) for all i = 9, 10, 11, 12. So by using the above and (3) and (4) one has Gb(x3k+1, x3k+2, x3k+3) ≤ α1(x0, x0). Gb(x3k, x3k+1, x3k+2) +Gb(y3k, y3k+1, y3k+2) 2 M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1000 +α2(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α3(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α4(x0, x0) · Gb(x3k, x3k, x3k+1)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α5(x0, x0) · Gb(x3k, x3k, x3k+1)Gb(y3k+1, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α6(x0, x0) · Gb(x3k+1, x3k+1, x3k+2)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α7(x0, x0) · Gb(x3k+1, x3k+1, x3k+2)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α8(x0, x0) · Gb(x3k+2, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α9(x0, x0) · Gb(x3k+2, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α10(x0, x0) · Gb(x3k, x3k+1, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α11(x0, x0) · Gb(x3k+1, x3k+2, x3k+1)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α12(x0, x0) · Gb(x3k+2, x3k, x3k+2)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) Now using (Gb3), (Gb4) and (Gb5) of Definition 2 we have Gb(x3k+1, x3k+2, x3k+3 ≤ α1(x0, x0). Gb(x3k, x3k+1, x3k+2) +Gb(y3k, y3k+1, y3k+2) 2 +α2(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α3(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α4(x0, x0) · Gb(x3k, x3k+1, x3k+2)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α5(x0, x0) · Gb(x3k, x3k+1, x3k+2)Gb(y3k+1, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α6(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α7(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1001 +α8(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α9(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α10(x0, x0) · Gb(x3k, x3k+1, x3k+2)Gb(x3k, x3k+1, x3k+2) Ω1(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α10(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(x3k, x3k+1, x3k+2) Ω1(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α11(x0, x0) · Gb(x3k+1, x3k+2, x3k+3)Gb(y3k, y3k+1, y3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) +α12(x0, x0) · Gb(x3k, x3k+1, x3k+2)Gb(x3k, x3k+1, x3k+2) Ωs(x3k, x3k+1, x3k+2, y3k, y3k+1, y3k+2) Which implies that Gb(x3k+1, x3k+2, x3k+3) ≤ α1(x0, x0). Gb(x3k, x3k+1, x3k+2) +Gb(y3k, y3k+1, y3k+2) 2 +α2(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) + α3(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) +α4(x0, x0) ·Gb(x3k, x3k+1, x3k+2) + α5(x0, x0) ·Gb(x3k, x3k+1, x3k+2) +α6(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) + α7(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) +α8(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) + α9(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) +α10(x0, x0) ·Gb(x3k, x3k+1, x3k+2) + α10(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) +α11(x0, x0) ·Gb(x3k+1, x3k+2, x3k+3) + α12(x0, x0) ·Gb(x3k, x3k+1, x3k+2). Which further implies that ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) ·Gb(x3k+1, x3k+2, x3k+3) ≤ (α1(x0, x0) 2 + α4(x0, x0) + α5(x0, x0) + α10(x0, x0) + α12(x0, x0) ) ·Gb(x3k, x3k+1, x3k+2) + α1(x0, x0) 2 ·Gb(y3k, y3k+1, y3k+2). Which gives Gb(x3k+1, x3k+2, x3k+3) (6) ≤ ( α1(x0,x0) 2 + α4(x0, x0) + α5(x0, x0) + α10(x0, x0) + α12(x0, x0) ) ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) .Gb(x3k, x3k+1, x3k+2) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1002 + α1(x0, x0) 2 ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) ·Gb(y3k, y3k+1, y3k+2). Similarly for the sequence {yn}, we have Gb(y3k+1, y3k+2, y3k+3) (7) ≤ ( α1(x0,x0) 2 + α4(x0, x0) + α5(x0, x0) + α10(x0, x0) + α12(x0, x0) ) ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) .Gb(y3k, y3k+1, y3k+2) + α1(x0, x0) 2 ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) ·Gb(x3k, x3k+1, x3k+2). Adding inequalities (6), and (7) we get Gb(x3k+1, x3k+2, x3k+3) +Gb(y3k+1, y3k+2, y3k+3) ≤ h. ( Gb(x3k, x3k+1, x3k+2) +Gb(y3k, y3k+1, y3k+2) ) , where 0 ≤ h = ( α1(x0, x0) + α4(x0, x0) + α5(x0, x0) + α10(x0, x0) + α12(x0, x0) ) ( 1− α2(x0, x0)− α3(x0, x0)− 11∑ i=6 αi(x0, x0) ) . By (2) we have 0 < h < 1 s . (8) Similarly, Gb(x3k+2, x3k+3, x3k+4) +Gb(y3k+2, y3k+3, y3k+4) ≤ h · ( Gb(x3k+1, x3k+2, x3k+3) +Gb(y3k+1, y3k+2, y3k+3) ) . Continuing this way, we have Gb(xn, xn+1, xn+2) +Gb(yn, yn+1, yn+2) ≤ h · (Gb(xn−1, xn, xn+1) +Gb(yn−1, yn, yn+1)) ≤ h2 · (Gb(xn−2, xn−1, xn) +Gb(yn−2, yn−1, yn)) ≤ h3 · (Gb(xn−3, xn−2, xn−1) +Gb(yn−3, yn−2, yn−1)) ≤ · · · ≤ hn+1 · (Gb(x0, x1, x2) +Gb(y0, y1, y2)). (9) Let Gb(xn, xn+1, xn+2) +Gb(yn, yn+1, yn+2) = Ψn. M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1003 Then the pattern (9) can be written as Ψn ≤ h.Ψn−1 ≤ h2.Ψn−2 ≤ · · · ≤ hn ·Ψ0. (10) Now, we prove that sequences {xn} and {yn} are Gb−cauchy. By (8) we have 0 ≤ sh < 1. Let j > k. Then Gb(xk, xj , xj) +Gb(yk, yj , yj) ≤ s[Gb(xk, xk+1, xk+1) +Gb(xk+1, xj , xj) +Gb(yk, yk+1, yk+1) +Gb(yk+1, yj , yj)] ≤ s[Gb(xk, xk+1, xk+1) +Gb(yk, yk+1, yk+1)] + s2[Gb(xk+1, xk+2, xk+2) +Gb(yk+1, yk+2, yk+2)] +s3[Gb(xk+2, xk+3, xk+3) +Gb(yk+2, yk+3, yk+3] + · · · +sj−k[Gb(xj−1, xj , xj) +Gb(yj−1, yj , yj ] ≤ shkΨ0 + s2hk+1Ψ0 + s3hk+2Ψ0 + · · ·+ sj−khj−1Ψ0 ≤ shk[1 + sh+ (sh)2 + (sh)3 + · · · ]Ψ0 = shk 1− sh Ψ0.→ 0 as k → +∞. Hence Gb(xk, xj , xj)+Gb(yk, yj , yj)→ 0 as k → +∞, which shows that {xn} and {yn} are Gb−cauchy sequences in X by Proposition 2. But due to the completeness of Gb-metric spaces, we have xn → x and yn → y as n→ +∞, for x, y ∈ X. Now, we show that (x, y) is common coupled fixed points of F,H and T. Suppose that x 6= F (x, y) and Gb(x, F (x, y), F (x, y)) > 0. Thus from definition of Gb-metric we have Gb(x, F (x, y), F (x, y)) ≤ Gb(x, x3k+2, F (x, y)) ≤ s · [ Gb(x, x3k+3, x3k+3) +Gb(x3k+3, x3k+2, F (x, y)) ] = s ·Gb(x, x3k+3, x3k+3) + s ·Gb ( T (x3k+2, y3k+2), H(x3k+1, y3k+1), F (x, y) ) ≤ s ·Gb(x, x3k+3, x3k+3) + sα1(x3k+2, x3k+1). Gb(x3k+2, x3k+1, x) +Gb(y3k+2, y3k+1, y) 2 +s.α2(x3k+2, x3k+1) · Gb ( T (x3k+2, y3k+2), H(x3k+1, y3k+1), F (x, y) ) )Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α3(x3k+2, x3k+1) · Gb ( T (x3k+2, y3k+2), H(x3k+1, y3k+1), F (x, y) ) )Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α4(x3k+2, x3k+1) · Gb(x3k+2, x3k+2, T (x3k+2, y3k+2))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α5(x3k+1, x) · Gb(x3k+2, x3k+2, T (x3k+2, y3k+2))Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α6(x3k+1, x) · Gb(x3k+1, x3k+1, H(x3k+1, y3k+1))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α7(x3k+1, x) · Gb(x3k+1, x3k+1, H(x3k+1, y3k+1))Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1004 +s.α8(x3k+1, x) · Gb(x, x, F (x, y))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α9(x, x3k+2) · Gb(x, x, F (x, y))Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α10(x, x3k+2) · Gb(x3k+2, x3k+1, F (x, y))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α11(x, x3k+2) · Gb(x3k+1, x, T (x3k+2, y3k+2))Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α12(x, x3k+2) · Gb(x, x3k+2, H(x3k+1, y3k+1))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) . Using (3) and (4), we have Gb(x, F (x, y), F (x, y)) ≤ Gb(x, x3k+2, F (x, y)) ≤ s ·Gb(x, x3k+3, x3k+3) + sα1(x3k+2, x3k+1). Gb(x3k+2, x3k+1, x) +Gb(y3k+2, y3k+1, y) 2 +s.α2(x3k+2, x3k+1) · Gb ( x3k+3, x3k+2, F (x, y) ) Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α3(x3k+2, x3k+1) · Gb ( x3k+3, x3k+2, F (x, y) ) Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α4(x3k+2, x3k+1) · Gb(x3k+2, x3k+2, x3k+3)Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α5(x3k+1, x) · Gb(x3k+2, x3k+2, x3k+3)Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α6(x3k+1, x) · Gb(x3k+1, x3k+1, x3k+2)Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α7(x3k+1, x) · Gb(x3k+1, x3k+1, x3k+2)Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α8(x3k+1, x) · Gb(x, x, F (x, y))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α9(x, x3k+2) · Gb(x, x, F (x, y))Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α10(x, x3k+2) · Gb(x3k+2, x3k+1, F (x, y))Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α11(x, x3k+2) · Gb(x3k+1, x, x3k+3)Gb(y3k+2, y3k+1, y) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) +s.α12(x, x3k+2) · Gb(x, x3k+2, x3k+2)Gb(x3k+2, x3k+1, x) Ωs(x3k+2, x3k+1, x, y3k+2, y3k+1, y) . Since {xn} and {yn} are Gb−convergent sequences converges to x and y respectively. M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1005 Therefore, by taking lim sup as k → +∞ of the above and using condition (i), we have Gb(x, F (x, y), F (x, y)) ≤ s ·Gb(x, x, x) + sα1(x0, x0). Gb(x, x, x) +Gb(y, y, y) 2 + s.α2(x0, x0) · lim supk→∞Gb ( x3k+3, x3k+2, F (x, y) ) Gb(x, x, x) Ωs(x, x, x, y, y, y) + s.α3(x0, x0) · lim supk→∞Gb ( x3k+3, x3k+2, F (x, y)Gb(y, y, y) Ωs(x, x, x, y, y, y) + s.α4(x0, x0) · Gb(x, x, x)Gb(x, x, x) Ωs(x, x, x, y, y, y) + s.α5(x0, x) · Gb(x, x, x)Gb(y, y, y) Ωs(x, x, x, y, y, y) + s.α6(x0, x) · Gb(x, x, x)Gb(x, x, x) Ωs(x, x, x, y, y, y) + s.α7(x0, x) · Gb(x, x, x)Gb(y, y, y) Ωs(x, x, x, y, y, y) + s.α8(x0, x) · Gb(x, x, F (x, y))Gb(x, x, x) Ωs(x, x, x, y, y, y) + s.α9(x, x0) · Gb(x, x, F (x, y))Gb(y, y, y) Ωs(x, x, x, y, y, y) + s.α10(x, x0) · lim supk→∞Gb ( x3k+2, x3k+1, F (x, y)Gb(x, x, x) Ωs(x, x, x, y, y, y) + s.α11(x, x0) · Gb(x, x, x)Gb(y, y, y) Ωs(x, x, x, y, y, y) + s.α12(x, x0) · Gb(x, x, x)Gb(x, x, x) Ωs(x, x, x, y, y, y) . SinceGb(x, x, x) = Gb(y, y, y) = Gb(z, z, z) = 0 for all x, y, z ∈ X. Therefore, Gb(x, F (x, y), F (x, y)) ≤ 0. Which is contradiction to our assumption that Gb(x, F (x, y), F (x, y)) > 0. Hence Gb(x, F (x, y), F (x, y)) = 0. Which implies that x = F (x, y) and similarly y = F (y, x). Thus (x, y) is a coupled fixed point of F. Similarly we can show that (x, y) is a coupled fixed point of H and T. Hence (x, y) is a common coupled fixed point of F , H and T. To show the uniqueness, suppose that F,H and T have two common coupled fixed points (x, y) and (u, v). Then from given condition one has Gb(x, u, u) = Gb(F (x, y), H(u, v), T (u, v)) ≤ α1(x, u) · Gb(x, u, u) +Gb(y, v, v) 2 + s.α2(x, u) · Gb ( F (x, y), H(u, v), T (u, v) ) )Gb(x, u, u) Ωs(x, u, u, y, v, v) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1006 + s.α3(x, u) · Gb ( F (x, y), H(u, v), T (u, v) ) Gb(y, v, v) Ωs(x, u, u, y, v, v) + s.α4(x, u) · Gb(x, x, F (x, y))Gb(x, u, u) Ωs(x, u, u, y, v, v) + s.α5(u, u) · Gb(x, x, F (x, y))Gb(y, v, v) Ωs(x, u, u, y, v, v) + s.α6(u, u) · Gb(u, u,H(u, v))Gb(x, u, u) Ωs(x, u, u, y, v, v) + s.α7(u, u) · Gb(u, u,H(u, v))Gb(y, v, v) Ωs(x, u, u, y, v, v) + s.α8(u, u) · Gb(u, u, T (u, v))Gb(x, u, u) Ωs(x, u, u, y, v, v) + s.α9(u, x) · Gb(u, u, T (u, v))Gb(y, v, v) Ωs(x, u, u, y, v, v) + s.α10(u, x) · Gb(x, u, T (u, v))Gb(x, u, u) Ωs(x, u, u, y, v, v) + s.α11(u, x) · Gb(u, u, F (x, y))Gb(y, v, v) Ωs(x, u, u, y, v, v) + s.α12(u, x) · Gb(u, x,H(u, v))Gb(x, u, u) Ωs(x, u, u, y, v, v) . Which implies that Gb(x, u, u) ≤ α1(x, u) · Gb(x, u, u) +Gb(y, v, v) 2 + α2(x, u) · Gb(x, u, u)Gb(x, u, u) Ωs(x, u, u, y, v, v) + α3(x, u) · Gb(x, u, u)Gb(y, v, v) Ωs(x, u, u, y, v, v) + α4(x, u) · Gb(x, x, x)Gb(x, u, u) Ωs(x, u, u, y, v, v) + α5(u, u) · Gb(x, x, x)Gb(y, v, v) Ωs(x, u, u, y, v, v) + α6(u, u) · Gb(u, u, u)Gb(x, u, u) Ωs(x, u, u, y, v, v) + α7(u, u) · Gb(u, u, u)Gb(y, v, v) Ωs(x, u, u, y, v, v) + α8(u, u) · Gb(u, u, u)Gb(x, u, u) Ωs(x, u, u, y, v, v) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1007 + α9(u, x) · Gb(u, u, u)Gb(y, v, v) Ωs(x, u, u, y, v, v) + α10(u, x) · Gb(x, u, u)Gb(x, u, u) Ωs(x, u, u, y, v, v) + α11(u, x) · Gb(u, u, x)Gb(y, v, v) Ωs(x, u, u, y, v, v) + α12(u, x) · Gb(u, x, u)Gb(x, u, u) Ωs(x, u, u, y, v, v) . Thus we have Gb(x, u, u) ≤ α1(x, u) · Gb(x, u, u) +Gb(y, v, v) 2 + α2(x, u) ·Gb(x, u, u) + α3(x, u) ·Gb(x, u, u) + α10(u, x) ·Gb(x, u, u) + α11(u, x) ·Gb(x, u, u) + α12(u, x) ·Gb(x, u, u) Further simplification gives( 1− α1(x, u) 2 − α2(x, u)− α3(x, u)− 12∑ i=10 αi(u, x) ) Gb(x, u, u) ≤ α1(x, u) 2 ·Gb(y, v, v). (11) Similarly( 1− α1(x, u) 2 − α2(x, u)− α3(x, u)− 12∑ i=10 αi(u, x) ) ·Gb(y, v, v) ≤ α1(x, u) 2 ·Gb(x, u, u). (12) Adding inequalities (11) and (12), we get( 1− α1(x, u) 2 − α2(x, u)− α3(x, u)− 12∑ i=10 αi(u, x) ) · [ Gb(x, u, u) +Gb(y, v, v) ] ≤ α1(x, u) 2 · [ Gb(x, u, u) +Gb(y, v, v) ] . Which implies that( 1− 3∑ i=1 αi(u, x)− 12∑ i=10 αi(u, x) ) · [ Gb(x, u, u) +Gb(y, v, v) ] ≤ 0, but 1− 3∑ i=1 αi(u, x)− 12∑ i=10 αi(u, x) > 0. Therefore, Gb(x, u, u) +Gb(y, v, v) = 0. M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1008 Thus we have Gb(x, u, u) = Gb(y, v, v) = 0 which implies x = u and y = v. Hence (x, y) is the unique common coupled fixed point of F,H and T. Now, we present some corollaries. Corollary 1. Let (X,Gb) be a complete Gb-metric space and let H,T : X × X → X be mappings satisfying the following condition, (i) αi(H(x1, y1), y) ≤ αi(x1, y), αi(x,H(x1, y1)) ≤ αi(x, x1)), αi(T (x1, y1), y) ≤ αi(x1, y) and αi(x, T (x1, y1)) ≤ αi(x, x1). (ii) Gb ( H(a, b), H(u, v), T (x, y) ) ≤ α1(a, u) · Gb(a, u, x) +Gb(b, v, y) 2 +α2(a, u) Gb ( H(a, b), H(u, v), T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) +α3(a, u) Gb ( H(a, b), H(u, v), T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α4(a, u) Gb ( a, a,H(a, b) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α5(u, x) Gb ( a, a,H(a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α6(u, x) Gb ( u, u,H(u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α7(u, x) Gb ( u, u,H(u, v) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α8(u, x) Gb ( x, x, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α9(x, a) Gb ( x, x, T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α10(x, a) Gb ( a, u, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α11(x, a) Gb ( u, x,H(a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α12(x, a) Gb ( x, a,H(u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) , where 0 ≤ s[α1(x, y) + α4(x, y) + α5(x, y) + α10(x, y) + α12(x, y)] + [α2(x, y) + α3(x, y) + 11∑ i=6 αi(x, y)] < 1. Then H,T have a unique common coupled fixed point in X. Proof. Taking F = H in Theorem 1 we get the required proof. Corollary 2. Let (X,Gb) be a complete Gb-metric space and let T : X2 → X be a map satisfying the following condition, (i) αi(T (x1, y1), y) ≤ αi(x1, y) and αi(x, T (x1, y1)) ≤ αi(x, x1)) (ii) Gb ( T (a, b), T (u, v), T (x, y) ) ≤ α1(a, u) · Gb(a, u, x) +Gb(b, v, y) 2 +α2(a, u) Gb ( T (a, b), T (u, v), T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1009 +α3(a, u) Gb ( T (a, b), T (u, v), T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α4(a, u) Gb ( a, a, T (a, b) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α5(u, x) Gb ( a, a, T (a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α6(u, x) Gb ( u, u, T (u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α7(u, x) Gb ( u, u, T (u, v) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α8(u, x) Gb ( x, x, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α9(x, a) Gb ( x, x, T (x, y) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α10(x, a) Gb ( a, u, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α11(x, a) Gb ( u, x, T (a, b) ) Gb(b, v, y) Ωs(a, u, x, b, v, y) +α12(x, a) Gb ( x, a, T (u, v) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) , where 0 ≤ s[α1(x, y) + α4(x, y) + α5(x, y) + α10(x, y) + α12(x, y)] + [α2(x, y) + α3(x, y) + 11∑ i=6 αi(x, y)] < 1. Then T has a unique coupled fixed point in X. Proof. Taking F = H = T in Theorem 1, we get the required proof. The proof of the following corollary follows word by word by using the same argument as in the proof of Theorem 1. Corollary 3. Let (X,Gb) be a complete Gb-metric space and let F,H, T : X ×X → X be mappings satisfying the above generalized contraction(1) and the following conditions. (i) αi(F (x1, y1), y) ≤ αi(x1, y) and αi(x, F (x1, y1)) ≤ αi(x, x1)); (ii) αi(H(x1, y1), y) ≤ αi(x1, y) and αi(x,H(x1, y1)) ≤ αi(x, x1)); (iii) αi(T (x1, y1), y) ≤ αi(x1, y) and αi(x, T (x1, y1)) ≤ αi(x, x1)); (iv) Gb ( F (a, b), H(u, v), T (x, y) ) ≤ N (a, b, u, v, x, y). where N (a, b, u, v, x, y) = α1(a, u) · Gb(a, u, x) +Gb(b, v, y) 2 +α2(a, u) Gb ( F (a, b), H(u, v), T (x, y) ) Gb(a, u, x) Ω1(a, u, x, b, v, y) +α3(a, u) Gb ( F (a, b), H(u, v), T (x, y) ) Gb(b, v, y) Ω1(a, u, x, b, v, y) +α4(a, u) Gb ( a, a, F (a, b) ) Gb(a, u, x) Ω1(a, u, x, b, v, y) + α5(u, x) Gb ( a, a, F (a, b) ) Gb(b, v, y) Ω1(a, u, x, b, v, y) M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1010 +α6(u, x) Gb ( u, u,H(u, v) ) Gb(a, u, x) Ω1(a, u, x, b, v, y) + α7(u, x) Gb ( u, u,H(u, v) ) Gb(b, v, y) Ω1(a, u, x, b, v, y) +α8(u, x) Gb ( x, x, T (x, y) ) Gb(a, u, x) Ω1(a, u, x, b, v, y) + α9(x, a) Gb ( x, x, T (x, y) ) Gb(b, v, y) Ω1(a, u, x, b, v, y) +α10(x, a) Gb ( a, u, T (x, y) ) Gb(a, u, x) Ωs(a, u, x, b, v, y) + α11(x, a) Gb ( u, x, F (a, b) ) Gb(b, v, y) Ω1(a, u, x, b, v, y) +α12(x, a) Gb ( x, a,H(u, v) ) Gb(a, u, x) Ω1(a, u, x, b, v, y) , (13) and αi : X × X → [0, 1), i = 1, 2, 3, · · · , 12. such that 0 ≤ s[α1(x, y) + α4(x, y) + α5(x, y) +α10(x, y) +α12(x, y)] + [α2(x, y) +α3(x, y) + 11∑ i=6 αi(x, y)] < 1. Then F,H, T have a unique common coupled fixed point in X. Remark 2. If we put αi(x, y) = αi for i = 1, 2, 3, · · · , 9 and αi(x, y) = 0 for i = 10, 11, 12 in condition (13) we get the Theorem 16 of Khomdram et.al [4] Remark 3. If we put αi(x, y) = αi for i = 1, 2, 3, · · · , 9 and αi(x, y) = 0 for i = 10, 11, 12 in Corollary 1 we get the Corollary 17 of Khomdram et.al [4] Example 1. Let X = [0,∞) with complete G-metric defined by G(x, y, z) = { 0, if x = y = z; max{x, y, z}, otherwise, and define the Gb metric by Gb(x, y, z) = (G(x, y, z))3. Then, (X,Gb) is a complete Gb-metric space with s = 3. Define the mappings F,H and T by F (x, y) = 2x+ y 160 , H(x, y) = x+ 3y 170 , and T (x, y) = 4x+ 5y 180 for all x, y, z ∈ X. We will use the following fact: For δ, β,≥ 0, then (δ + β)p ≤ 22p−2(δp + βp) (14) Now for a 6= u 6= x and b 6= v 6= y we have Gb(F (a, b), H(u, v), T (x, y)) = (max{2a+ b 160 , u+ 3v 170 , 4x+ 5y 180 })3 = max{(2a+ b 160 )3, ( u+ 3v 170 )3, ( 4x+ 5y 180 )3} = max{(2a+ b)3 (160)3 , (u+ 3v)3 (170)3 , (4x+ 5y)3 (180)3 } M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1011 ( by using (14) for p = 3) ≤ 16 max{8(a)3 + (b)3 (160)3 , (u)3 + 27(v)3 (170)3 , 64(x)3 + 125(y)3 (180)3 } ≤ 16 (160)3 max{8(a)3 + (b)3, (u)3 + 27(v)3, 64(x)3 + 125(y)3} ≤ 16 (160)3 max{125(a)3 + 125(b)3, 125(u)3 + 125(v)3, 125(x)3 + 125(y)3} ≤ 2000 (160)3 ( max{a3, u3, x3}+ max{b3, v3, y3} ) = 4000 (160)3 max{a3, u3, x3}+ max{b3, v3, y3} 2 = 4000 (160)3 (max{a, u, x})3 + (max{b, v, y})3 2 = 1 1024 Gb(a, u, x) +Gb(b, v, y) 2 ≤ N (a, b, u, v, x, y), where α1(x, y) = 1 1024 and αi(x, y) = 1 (1024)2i−1 for i = 2, 3, · · · , 12. Note that for s = 3 we have s[α1(x, y)+α4(x, y)+α5(x, y)+α10(x, y)+α12(x, y)]+[α2(x, y)+α3(x, y)+ 11∑ i=6 αi(x, y)] < 1. Moreover, it is clear that the conditions (i), (ii) and (iii) of Theorem 1 are satisfied. Hence, F,H, T satisfy all conditions of Theorem 1, and (x, y) = (0, 0) is the unique common coupled fixed point of F,H and T . 3. Application Let X = (C[a, b],R) denote the set of all continuous functions from [a, b] to R. In this section, we will use corollary 2 to show that there is a solution to the following integral equation: u(t) = ∫ b a H(t, κ)f(κ, u(κ))dκ; t ∈ [a, b], (15) where u(κ) ∈ X. Theorem 2. Consider equation (15) and suppose: (i) H : [a, b]× [a, b]→ [0,∞) is a continuous function, (ii) f : [a, b]×R→ R,is continuous, (iii) maxt,κ∈[a,b]H(t, κ) < α = α1 4(b−a) ,where α1 = 1 30 , M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1012 (iv) For all u(κ) ∈ X we have |f(κ, u(κ))| ≤ (|u(κ)|)3. Then equation (15) has a solution. Proof. Let X be as defined above. Define a mapping T : X ×X → X by T (u(t), v(t)) = ∫ b a H(t, κ)f(κ, u(κ) 2 + v(κ) 2 )dκ; t ∈ [a, b]. (16) For all u, v, w ∈ X define the Gb-metric on X by Gb(u, v, w) = ({ 0, if u = v = w, max{supκ∈[a,b] |u|, supκ∈[a,b] |v|, supκ∈[a,b] |w|}, otherwise. })3 (17) Clearly that (X,Gb) is a complete Gb-metric space with constant (s = 4). Now, for li(κ) ∈ X, where i = 1, 2, 3, 4, 5, 6 we have Gb(T (l1, l2), T (l3, l4), T (l5, l6)) = ({ 0, if T (l1, l2) = T (l3, l4) = T (l5, l6), max { supκ∈[a,b] |T (l1, l2)|, supκ∈[a,b] |T (l3, l4)|, supκ∈[a,b] |T (l5, l6)| } , otherwise. })3 (18) Now, sup κ∈[a,b] |T (l1, l2)| = sup κ∈[a,b] | ∫ b a H(t, κ)f(κ, l1(κ) 2 + l2(κ) 2 )dκ| (19) ≤ sup κ∈[a,b] ∫ b a |H(t, κ)||f(κ, l1(κ) 2 + l2(κ) 2 )|dκ ≤ sup κ∈[a,b] ∫ b a α|f(κ, l1(κ) 2 + l2(κ) 2 )|dκ ≤ sup κ∈[a,b] ∫ b a α(| l1(κ) 2 |+ | l2(κ) 2 |)3dκ ( by using (14) for p = 3) ≤ sup κ∈[a,b] ∫ b a 16α [ (| l1(κ) 2 |)3 + (| l2(κ) 2 |)3 ] dκ ≤ sup κ∈[a,b] ∫ b a 2α [ (|l1(κ)|)3 + (|l2(κ)|)3 ] dκ ≤ ∫ b a 2α [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] dκ ≤ 2α [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] ∫ b a dκ M.M.M. Jaradat et al. / Eur. J. Pure Appl. Math, 13 (4) (2020), 995-1015 1013 = 2α [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] |b− a| = α1 2 [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] . Similarly, one can show that sup κ∈[a,b] |T (l3, l4)| ≤ α1 2 [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] . (20) and sup κ∈[a,b] |T (l5, l6)| ≤ α1 2 [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] . (21) Hence, by (19), (20) and (21) we have max { supκ∈[a,b] |T (l1, l2)|, supκ∈[a,b] |T (l3, l4)|, supκ∈[a,b] |T (l5, l6)| } ≤ α1 2 [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] (22) Thus, Gb(T (l1, l2), T (l3, l4), T (l5, l6)) ≤ α1 2 [ Gb(l1, l3, l3) +Gb(l2, l4, l6) ] (23) Therefore, all conditions of corollary 2 are satisfied for αi = 1 30 for i = 1, 2, · · · , 12. As a result of corollary 2 the mapping T has a unique coupled fixed point in X which is a solution of (15). The following example illustrate the validity of Theorem 2. Example 2. The following integral equation has a solution in X = (C[0, 1],R). u(t) = ∫ 1 0 tκ 130 (e−κ)(u(κ))3dκ; t ∈ [0, 1]. (24) Proof. Let T : X ×X → X be defined as T (u(t), v(t)) = ∫ 1 0 tκ 130 (u(κ) 2 + v(κ) 2 )3 e−κdκ; t ∈ [0, 1]. By specifying H(t, κ) = tκ 130 , f(κ, t) = t3e−κ in Theorem 2 we get that: (i) The function H(t, κ) is continuous on [0, 1]× [0, 1], (ii) f(κ, t) = t3e−κ is continuous on [0, 1]×R, . (iii) maxt,κ∈[0,1]H(t, κ) = maxt,κ∈[0,1] tκ 130 = 1 130 < 1 120 = α1 4(1−0) , where α1 = 1 30 . (iv) also |f(κ, u(κ))| = |e−κ||(u(κ))3| ≤ (|u(κ)|)3. REFERENCES 1014 Therefore, all conditions of Theorem 2 are satisfied, hence the mapping T has a fixed point in X, which is a solution to equation (24), that is T (u(t), u(t)) = ∫ 1 0 tκ 130 ( u(κ) 2 + u(κ) 2 )3e−κdκ = ∫ 1 0 tκ 130 (u(κ))3e−κdκ; t ∈ [0, 1] = u(t). 4. Conclusions We have introduced rational type contractive conditions on three mappings to give the existence and uniqueness of common coupled fixed point theorem in the set up of Gb-metric spaces. The derived result cover and generalize some well-known comparable results in the existing literature. 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