A new common coupled fixed point theorem in generalized metric space and applications to integral equations

Abstract

In the present paper, we prove a common coupled fixed point theorem in the setting of a generalized metric space in the sense of Mustafa and Sims. Our results improve and extend the corresponding results of Shatanawi. We also present an application to integral equations.

1 Introduction and preliminaries

The study of fixed points of mappings satisfying certain contractive conditions has been in the center of rigorous research activity. For a survey of common fixed point theory in metric and cone metric spaces, we refer the reader to [1–9]. In 2006, Bhaskar and Lakshmikantham [10] initiated the study of a coupled fixed point in ordered metric spaces and applied their results to prove the existence and uniqueness of solutions for a periodic boundary value problem. For more works in coupled and coincidence point theorems, we refer the reader to [11–13].

Some authors generalized the concept of metric spaces in different ways. Mustafa and Sims [14] introduced the notion of G-metric space, in which the real number is assigned to every triplet of an arbitrary set as a generalization of the notion of metric spaces. Based on the notion of G-metric spaces, many authors (for example, [15–33]) obtained some fixed point and common fixed point theorems for mappings satisfying various contractive conditions. Fixed point problems have also been considered in partially ordered G-metric spaces [34–39].

The purpose of this paper is to obtain some common coupled coincidence point theorems in G-metric spaces satisfying some contractive conditions.

The following definitions and results will be needed in the sequel.

Definition 1.1 [14]

Let X be a nonempty set, and let $G:X\times X\times X\to R^{+}$ be a function satisfying the following axioms:

• (G1) $G(x,y,z)=0$ if $x=y=z$;

• (G2) $0<G(x,x,y)$ for all $x,y\in X$ with $x\ne y$;

• (G3) $G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $z\ne y$;

• (G4) $G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots$ (symmetry in all three variables);

• (G5) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$ (rectangle inequality),

then the function G is called a generalized metric, or more specifically, a G-metric on X, and the pair $(X,G)$ is called a G-metric space.

Definition 1.2 [14]

Let $(X,G)$ be a G-metric space, and let $\{x_n\}$ be a sequence of points in X, a point x in X is said to be the limit of the sequence $\{x_n\}$ if ${lim}_{m,n\to \mathrm{\infty }}G(x,x_n,x_m)=0$, and one says that the sequence $\{x_n\}$ is G-convergent to x.

Thus, if $x_n\to x$ in a G-metric space $(X,G)$, then for any $ϵ>0$, there exists $N\in \mathbb{N}$ such that $G(x,x_n,x_m)<ϵ$ for all $n,m\ge N$.

Proposition 1.3 [14]

Let $(X,G)$ be a G-metric space, then the following are equivalent:

• (1) $\{x_n\}$ is G-convergent to x.

• (2) $G(x_n,x_n,x)\to 0$ as $n\to \mathrm{\infty }$.

• (3) $G(x_n,x,x)\to 0$ as $n\to \mathrm{\infty }$.

• (4) $G(x_n,x_m,x)\to 0$ as $n,m\to \mathrm{\infty }$.

Definition 1.4 [14]

Let $(X,G)$ be a G-metric space. A sequence $\{x_n\}$ is called G-Cauchy sequence if for each $ϵ>0$, there exists a positive integer $N\in \mathbb{N}$ such that $G(x_n,x_m,x_l)<ϵ$ for all $n,m,l\ge N$; i.e., if $G(x_n,x_m,x_l)\to 0$ as $n,m,l\to \mathrm{\infty }$.

Definition 1.5 [14]

A G-metric space $(X,G)$ is said to be G-complete if every G-Cauchy sequence in $(X,G)$ is G-convergent in X.

Proposition 1.6 [14]

Let $(X,G)$ be a G-metric space, then the following are equivalent:

• (1) The sequence $\{x_n\}$ is G-Cauchy.

• (2) For every $ϵ>0$, there exists $k\in \mathbb{N}$ such that $G(x_n,x_m,x_m)<ϵ$ for all $n,m\ge k$.

Proposition 1.7 [14]

Let $(X,G)$ be a G-metric space. Then the function $G(x,y,z)$ is jointly continuous in all three of its variables.

Definition 1.8 [14]

Let $(X,G)$ and $(X^{\prime },G^{\prime })$ be G-metric space, and let $f:(X,G)\to (X^{\prime },G^{\prime })$ be a function. Then f is said to be G-continuous at a point $a\in X$ if and only if for every $ϵ>0$, there is $\delta >0$ such that $x,y\in X$ and $G(a,x,y)<\delta$ implies that $G^{\prime }(f(a),f(x),f(y))<ϵ$. A function f is G-continuous at X if and only if it is G-continuous at all $a\in X$.

Proposition 1.9 [14]

Let $(X,G)$ and $(X^{\prime },G^{\prime })$ be G-metric spaces, then a function $f:X\to X^{\prime }$ is G-continuous at a point $x\in X$ if and only if it is G-sequentially continuous at x; that is, whenever $(x_n)$ is G-convergent to x, $(f(x_n))$ is G-convergent to $f(x)$.

Proposition 1.10 [14]

Let $(X,G)$ be a G-metric space. Then for any x, y, z, a in X, it follows that

• (i) if $G(x,y,z)=0$, then $x=y=z$;

• (ii) $G(x,y,z)\le G(x,x,y)+G(x,x,z)$;

• (iii) $G(x,y,y)\le 2G(y,x,x)$;

• (iv) $G(x,y,z)\le G(x,a,z)+G(a,y,z)$;

• (v) $G(x,y,z)\le \frac{2}{3}(G(x,y,a)+G(x,a,z)+G(a,y,z))$;

• (vi) $G(x,y,z)\le G(x,a,a)+G(y,a,a)+G(z,a,a)$.

Definition 1.11 [10]

An element $(x,y)\in X\times X$ is called a coupled fixed point of a mapping $F:X\times X\to X$ if $F(x,y)=x$ and $F(y,x)=y$.

Definition 1.12 [11]

An element $(x,y)\in X\times X$ is called a coupled coincidence point of the mappings $F:X\times X\to X$ and $g:X\to X$ if $F(x,y)=gx$ and $F(y,x)=gy$.

Definition 1.13 [11]

Let X be a nonempty set. Then we say that the mappings $F:X\times X\to X$ and $g:X\to X$ are commutative if $gF(x,y)=F(gx,gy)$.

2 Main results

We start our work by proving the following crucial lemma.

Lemma 2.1 Let $(X,G)$ be a G-metric space. Let $F_1,F_2,F_3:X\times X\to X$ and $g:X\to X$ be four mappings such that

$$\begin{array}{rcl}G(F_1(x,y),F_2(u,v),F_3(w,z))& \le & a_{1}G(gx,gu,gw)+a_{2}G(gy,gv,gz)+a_{3}G(gx,gu,gu)\\ +a_{4}G(gy,gv,gv)+a_{5}G(gu,gw,gw)+a_{6}G(gv,gz,gz)\\ +a_{7}G(gw,gx,gx)+a_{8}G(gz,gy,gy)\end{array}$$

(2.1)

for all $x,y,u,v,w,z\in X$, where $a_i\ge 0$, $i=1,2,\dots ,8$ and $a_1+a_2+a_3+a_4+a_7+a_8<1$. Suppose that $(x,y)$ is a common coupled coincidence point of the mappings pair $(F_1,g)$, $(F_2,g)$ and $(F_3,g)$. Then

$$F_1(x,y)=F_2(x,y)=F_3(x,y)=gx=gy=F_1(y,x)=F_2(y,x)=F_3(y,x).$$

Proof Since $(x,y)$ is a common coupled coincidence point of the mappings pair $(F_1,g)$, $(F_2,g)$ and $(F_3,g)$, we have $gx=F_1(x,y)=F_2(x,y)=F_3(x,y)$ and $gy=F_1(y,x)=F_2(y,x)=F_3(y,x)$. Assume that $gx\ne gy$. Then by (2.1), we get

$$\begin{array}{rcl}G(gx,gy,gy)& =& G(F_1(x,y),F_2(y,x),F_3(y,x))\\ \le & a_{1}G(gx,gy,gy)+a_{2}G(gy,gx,gx)+a_{3}G(gx,gy,gy)+a_{4}G(gy,gx,gx)\\ +a_{5}G(gy,gy,gy)+a_{6}G(gx,gx,gx)+a_{7}G(gy,gx,gx)+a_{8}G(gx,gy,gy)\\ =& (a_1+a_3+a_8)G(gx,gy,gy)+(a_2+a_4+a_7)G(gy,gx,gx).\end{array}$$

Also by (2.1), we have

$$\begin{array}{rcl}G(gy,gx,gx)& =& G(F_1(y,x),F_2(x,y),F_3(x,y))\\ \le & a_{1}G(gy,gx,gx)+a_{2}G(gx,gy,gy)+a_{3}G(gy,gx,gx)+a_{4}G(gx,gy,gy)\\ +a_{5}G(gx,gx,gx)+a_{6}G(gy,gy,gy)+a_{7}G(gx,gy,gy)+a_{8}G(gy,gx,gx)\\ =& (a_1+a_3+a_8)G(gy,gx,gx)+(a_2+a_4+a_7)G(gx,gy,gy).\end{array}$$

Therefore,

$$\begin{array}{c}G(gx,gy,gy)+G(gy,gx,gx)\hfill \\ \le (a_1+a_2+a_3+a_4+a_7+a_8)[G(gx,gy,gy)+G(gy,gx,gx)].\hfill \end{array}$$

Since $0\le a_1+a_2+a_3+a_4+a_7+a_8<1$, we get

$$G(gx,gy,gy)+G(gy,gx,gx)<G(gx,gy,gy)+G(gy,gx,gx),$$

which is a contradiction. So, $gx=gy$, and hence,

$$F_1(x,y)=F_2(x,y)=F_3(x,y)=gx=gy=F_1(y,x)=F_2(y,x)=F_3(y,x).$$

 □

Theorem 2.1 Let $(X,G)$ be a G-metric space. Let $F_1,F_2,F_3:X\times X\to X$ and $g:X\to X$ be four mappings such that

$$\begin{array}{rcl}G(F_1(x,y),F_2(u,v),F_3(w,z))& \le & a_{1}G(gx,gu,gw)+a_{2}G(gy,gv,gz)+a_{3}G(gx,gu,gu)\\ +a_{4}G(gy,gv,gv)+a_{5}G(gu,gw,gw)+a_{6}G(gv,gz,gz)\\ +a_{7}G(gw,gx,gx)+a_{8}G(gz,gy,gy)\end{array}$$

(2.2)

for all $x,y,u,v,w,z\in X$, where $a_i\ge 0$, $i=1,2,\dots ,8$ and $a_1+a_2+a_3+a_4+2a_5+2a_6+a_7+a_8<1$. Suppose that $F_1$, $F_2$, $F_3$ and g satisfy the following conditions:

• (i) $F_1(X\times X)\subseteq gX$, $F_2(X\times X)\subseteq gX$, $F_3(X\times X)\subseteq gX$;

• (ii) gX is G-complete;

• (iii) g is G-continuous and commutes with $F_1$, $F_2$, $F_3$.

Then there exist unique $x\in X$ such that

$$gx=F_1(x,x)=F_2(x,x)=F_3(x,x)=x.$$

Proof Let $x_0,y_0\in X$. Since $F_1(X\times X)\subseteq gX$, $F_2(X\times X)\subseteq gX$, $F_3(X\times X)\subseteq gX$, we can choose $x_1,x_2,x_3,y_1,y_2,y_3\in X$ such that $g{x}_1=F_1(x_0,y_0)$, $g{y}_1=F_1(y_0,x_0)$, $g{x}_2=F_2(x_1,y_1)$, $g{y}_2=F_2(y_1,x_1)$, $g{x}_3=F_3(x_2,y_2)$ and $g{y}_3=F_3(y_2,x_2)$. Combining this process, we can construct two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$\begin{array}{c}g{x}_{3n}=F_3(x_{3n-1},y_{3n-1}),g{y}_{3n}=F_3(y_{3n-1},x_{3n-1}),n=1,2,3,\dots ,\hfill \\ g{x}_{3n+1}=F_1(x_{3n},y_{3n}),g{y}_{3n+1}=F_1(y_{3n},x_{3n}),n=0,1,2,3,\dots ,\hfill \\ g{x}_{3n+2}=F_2(x_{3n+1},y_{3n+1}),g{y}_{3n+2}=F_2(y_{3n+1},x_{3n+1}),n=0,1,2,3,\dots .\hfill \end{array}$$

If $g{x}_{3n}=g{x}_{3n+1}$, then $gx=F_1(x,y)$, where $x=x_{3n}$, $y=y_{3n}$. If $g{x}_{3n+1}=g{x}_{3n+2}$, then $gx=F_2(x,y)$, where $x=x_{3n+1}$, $y=y_{3n+1}$. If $g{x}_{3n+2}=g{x}_{3n+3}$, then $gx=F_3(x,y)$, where $x=x_{3n+2}$, $y=y_{3n+2}$. On the other hand, if $g{y}_{3n}=g{y}_{3n+1}$, then $gy=F_1(y,x)$, where $y=y_{3n}$, $x=x_{3n}$. If $g{y}_{3n+1}=g{y}_{3n+2}$, then $gy=F_2(y,x)$, where $y=y_{3n+1}$, $x=x_{3n+1}$. If $g{y}_{3n+2}=g{y}_{3n+3}$, then $gy=F_3(y,x)$, where $y=y_{3n+2}$, $x=x_{3n+2}$. Without loss of generality, we can assume that $g{x}_n\ne g{x}_{n+1}$ and $g{y}_n\ne g{y}_{n+1}$, for all $n=0,1,2,\dots$ .

By (2.2) and (G3), we have

$$\begin{array}{rcl}G(g{x}_{3n},g{x}_{3n+1},g{x}_{3n+2})& =& G(F_3(x_{3n-1},y_{3n-1}),F_1(x_{3n},y_{3n}),F_2(x_{3n+1},y_{3n+1}))\\ =& G(F_1(x_{3n},y_{3n}),F_2(x_{3n+1},y_{3n+1}),F_3(x_{3n-1},y_{3n-1}))\\ \le & a_{1}G(g{x}_{3n},g{x}_{3n+1},g{x}_{3n-1})+a_{2}G(g{y}_{3n},g{y}_{3n+1},g{y}_{3n-1})\\ +a_{3}G(g{x}_{3n},g{x}_{3n+1},g{x}_{3n+1})+a_{4}G(g{y}_{3n},g{y}_{3n+1},g{y}_{3n+1})\\ +a_{5}G(g{x}_{3n+1},g{x}_{3n-1},g{x}_{3n-1})+a_{6}G(g{y}_{3n+1},g{y}_{3n-1},g{y}_{3n-1})\\ +a_{7}G(g{x}_{3n-1},g{x}_{3n},g{x}_{3n})+a_{8}G(g{y}_{3n-1},g{y}_{3n},g{y}_{3n})\\ \le & (a_1+a_3+a_5+a_7)G(g{x}_{3n-1},g{x}_{3n},g{x}_{3n+1})\\ +(a_2+a_4+a_6+a_8)G(g{y}_{3n-1},g{y}_{3n},g{y}_{3n+1}).\end{array}$$

(2.3)

Similarly, we have

$$\begin{array}{rcl}G(g{y}_{3n},g{y}_{3n+1},g{y}_{3n+2})& \le & (a_1+a_3+a_5+a_7)G(g{y}_{3n-1},g{y}_{3n},g{y}_{3n+1})\\ +(a_2+a_4+a_6+a_8)G(g{x}_{3n-1},g{x}_{3n},g{x}_{3n+1}).\end{array}$$

(2.4)

By combining (2.3) and (2.4), we get

$$\begin{array}{c}G(g{x}_{3n},g{x}_{3n+1},g{x}_{3n+2})+G(g{y}_{3n},g{y}_{3n+1},g{y}_{3n+2})\hfill \\ \le \left(\sum _{i=1}^8{a}_i\right)[G(g{x}_{3n-1},g{x}_{3n},g{x}_{3n+1})+G(g{y}_{3n-1},g{y}_{3n},g{y}_{3n+1})].\hfill \end{array}$$

(2.5)

In the same way, we can show that

$$\begin{array}{c}G(g{x}_{3n-1},g{x}_{3n},g{x}_{3n+1})+G(g{y}_{3n-1},g{y}_{3n},g{y}_{3n+1})\hfill \\ \le \left(\sum _{i=1}^8{a}_i\right)[G(g{x}_{3n-2},g{x}_{3n-1},g{x}_{3n})+G(g{y}_{3n-2},g{y}_{3n-1},g{y}_{3n})]\hfill \end{array}$$

(2.6)

and

$$\begin{array}{c}G(g{x}_{3n-2},g{x}_{3n-1},g{x}_{3n})+G(g{y}_{3n-2},g{y}_{3n-1},g{y}_{3n})\hfill \\ \le \left(\sum _{i=1}^8{a}_i\right)[G(g{x}_{3n-3},g{x}_{3n-2},g{x}_{3n-1})+G(g{y}_{3n-3},g{y}_{3n-2},g{y}_{3n-1})].\hfill \end{array}$$

(2.7)

It follows from (2.5), (2.6) and (2.7) that for all $n\in \mathbb{N}$, we have

$$\begin{array}{c}G(g{x}_n,g{x}_{n+1},g{x}_{n+2})+G(g{y}_n,g{y}_{n+1},g{y}_{n+2})\hfill \\ \le \left(\sum _{i=1}^8{a}_i\right)[G(g{x}_{n-1},g{x}_n,g{x}_{n+1})+G(g{y}_{n-1},g{y}_n,g{y}_{n+1})]\hfill \\ =k[G(g{x}_{n-1},g{x}_n,g{x}_{n+1})+G(g{y}_{n-1},g{y}_n,g{y}_{n+1})]\hfill \\ \le k^2[G(g{x}_{n-2},g{x}_{n-1},g{x}_n)+G(g{y}_{n-2},g{y}_{n-1},g{y}_n)]\hfill \\ ⋮\hfill \\ \le k^n[G(g{x}_0,g{x}_1,g{x}_2)+G(g{y}_0,g{y}_1,g{y}_2)].\hfill \end{array}$$

(2.8)

Where $k={\sum }_{i=1}^8{a}_i\in [0,1)$. From (G3), we have $G(g{x}_n,g{x}_{n+1},g{x}_{n+1})\le G(g{x}_n,g{x}_{n+1},g{x}_{n+2})$ and $G(g{y}_n,g{y}_{n+1},g{y}_{n+1})\le G(g{y}_n,g{y}_{n+1},g{y}_{n+2})$. Hence, by the (G3) and (2.8), we get

$$\begin{array}{c}G(g{x}_n,g{x}_{n+1},g{x}_{n+1})+G(g{y}_n,g{y}_{n+1},g{y}_{n+1})\hfill \\ \le G(g{x}_n,g{x}_{n+1},g{x}_{n+2})+G(g{y}_n,g{y}_{n+1},g{y}_{n+2})\hfill \\ \le k^n[G(g{x}_0,g{x}_1,g{x}_2)+G(g{y}_0,g{y}_1,g{y}_2)].\hfill \end{array}$$

(2.9)

Therefore, for all $n,m\in \mathbb{N}$, $n<m$, by (G5) and (2.9), we have

$$\begin{array}{c}G(g{x}_n,g{x}_m,g{x}_m)+G(g{y}_n,g{y}_m,g{y}_m)\hfill \\ \le [G(g{x}_n,g{x}_{n+1},g{x}_{n+1})+G(g{y}_n,g{y}_{n+1},g{y}_{n+1})]\hfill \\ +[G(g{x}_{n+1},g{x}_{n+2},g{x}_{n+2})+G(g{y}_{n+1},g{y}_{n+2},g{y}_{n+2})]\hfill \\ +\cdots +[G(g{x}_{m-1},g{x}_m,g{x}_m)+G(g{y}_{m-1},g{y}_m,g{y}_m)]\hfill \\ \le (k^n+k^{n+1}+\cdots +k^{m-1})[G(g{x}_0,g{x}_1,g{x}_2)+G(g{y}_0,g{y}_1,g{y}_2)]\hfill \\ \le \frac{k^n}{1-k}[G(g{x}_0,g{x}_1,g{x}_2)+G(g{y}_0,g{y}_1,g{y}_2)]\to 0\text{as }n,m\to \mathrm{\infty }.\hfill \end{array}$$

(2.10)

Which implies that

$$G(g{x}_n,g{x}_m,g{x}_m)\to 0\text{and}G(g{y}_n,g{y}_m,g{y}_m)\to 0\text{as }n,m\to \mathrm{\infty }.$$

Thus, $\{g{x}_n\}$ and $\{g{y}_n\}$ are all G-Cauchy in gX. Since gX is G-complete, we get that $\{g{x}_n\}$ and $\{g{y}_n\}$ are G-convergent to some $x\in gX$ and $y\in gX$, respectively. Since g is G-continuous, we have $\{gg{x}_n\}$ is G-convergent to gx and $\{gg{y}_n\}$ is G-convergent to gy. That is,

$$gg{x}_n\to gx\text{and}gg{y}_n\to gy\text{as }n\to \mathrm{\infty }.$$

(2.11)

Also, since g commutes with $F_1$, $F_2$ and $F_3$, respectively, we have

$$\begin{array}{c}gg{x}_{3n}=g{F}_3(x_{3n-1},y_{3n-1})=F_3(g{x}_{3n-1},g{y}_{3n-1}),\hfill \\ gg{y}_{3n}=g{F}_3(y_{3n-1},x_{3n-1})=F_3(g{y}_{3n-1},g{x}_{3n-1}),\hfill \\ gg{x}_{3n+1}=g{F}_1(x_{3n},y_{3n})=F_1(g{x}_{3n},g{y}_{3n}),\hfill \\ gg{y}_{3n+1}=g{F}_1(y_{3n},x_{3n})=F_1(g{y}_{3n},g{x}_{3n}),\hfill \\ gg{x}_{3n+2}=g{F}_2(x_{3n+1},y_{3n+1})=F_2(g{x}_{3n+1},g{y}_{3n+1}),\hfill \\ gg{y}_{3n+2}=g{F}_2(y_{3n+1},x_{3n+1})=F_2(g{y}_{3n+1},g{x}_{3n+1}).\hfill \end{array}$$

Thus, from condition (2.2), we have

$$\begin{array}{c}G(gg{x}_{3n},gg{x}_{3n+1},F_2(x,y))\hfill \\ =G(F_1(g{x}_{3n},g{y}_{3n}),F_2(x,y),F_3(g{x}_{3n-1},g{y}_{3n-1}))\hfill \\ \le a_{1}G(gg{x}_{3n},gx,gg{x}_{3n-1})+a_{2}G(gg{y}_{3n},gy,gg{y}_{3n-1})+a_{3}G(gg{x}_{3n},gx,gx)\hfill \\ +a_{4}G(gg{y}_{3n},gy,gy)+a_{5}G(gx,gg{x}_{3n-1},gg{x}_{3n-1})+a_{6}G(gy,gg{y}_{3n-1},gg{y}_{3n-1})\hfill \\ +a_{7}G(gg{x}_{3n-1},gg{x}_{3n},gg{x}_{3n})+a_{8}G(gg{y}_{3n-1},gg{y}_{3n},gg{y}_{3n}).\hfill \end{array}$$

Letting $n\to \mathrm{\infty }$, using (2.11) and the fact that G is continuous on its variables, we get that

$$G(gx,gx,F_2(x,y))=0.$$

Hence, $gx=F_2(x,y)$. Similarly, we may show that $gy=F_2(y,x)$. Also for the same reason, we may show that $gx=F_1(x,y)$, $gy=F_1(y,x)$, $gx=F_3(x,y)$ and $gy=F_3(y,x)$. Therefore, $(x,y)$ is a common coupled coincidence point of the pair $(F_1,g)$, $(F_2,g)$ and $(F_3,g)$. By Lemma 2.1, we obtain

$$gx=F_1(x,y)=F_2(x,y)=F_3(x,y)=F_1(y,x)=F_2(y,x)=F_3(y,x)=gy.$$

(2.12)

Since the sequences $\{g{x}_{3n-1}\}$, $\{g{x}_{3n}\}$ and $\{g{x}_{3n+1}\}$ are all a subsequence of $\{g{x}_n\}$, then they are all G-convergent to x. Similarly, we may show that $\{g{y}_{3n-1}\}$, $\{g{y}_{3n}\}$ and $\{g{y}_{3n+1}\}$ are all G-convergent to y. From (2.2), we have

$$\begin{array}{rcl}G(g{x}_{3n},gx,gx)& =& G(F_1(x,y),F_2(x,y),F_3(x_{3n-1},y_{3n-1}))\\ \le & a_{1}G(gx,gx,g{x}_{3n-1})+a_{2}G(gy,gy,g{y}_{3n-1})+a_{3}G(gx,gx,gx)\\ +a_{4}G(gy,gy,gy)+a_{5}G(gx,g{x}_{3n-1},g{x}_{3n-1})+a_{6}G(gy,g{y}_{3n-1},g{y}_{3n-1})\\ +a_{7}G(g{x}_{3n-1},gx,gx)+a_{8}G(g{y}_{3n-1},gy,gy).\end{array}$$

Letting $n\to \mathrm{\infty }$, and using the fact that G is continuous on its variables, we get that

$$G(x,gx,gx)\le (a_1+a_7)G(gx,gx,x)+(a_2+a_8)G(gy,gy,y)+a_{5}G(gx,x,x)+a_{6}G(gy,y,y).$$

Similarly, we may show that

$$G(y,gy,gy)\le (a_1+a_7)G(gy,gy,y)+(a_2+a_8)G(gx,gx,x)+a_{5}G(gy,y,y)+a_{6}G(gx,x,x).$$

Thus, using the Proposition 1.10(iii), we have

$$\begin{array}{rcl}G(x,gx,gx)+G(y,gy,gy)& \le & (a_1+a_2+a_7+a_8)[G(gx,gx,x)+G(gy,gy,y)]\\ +(a_5+a_6)[G(gx,x,x)+G(gy,y,y)]\\ \le & (a_1+a_2+2a_5+2a_6+a_7+a_8)[G(gx,gx,x)+G(gy,gy,y)].\end{array}$$

Since $0\le a_1+a_2+a_3+a_4+2a_5+2a_6+a_7+a_8<1$, so the last inequality happens only if $G(x,gx,gx)=0$ and $G(y,gy,gy)=0$. Hence, $x=gx$ and $y=gy$. From (2.12), we have $x=gx=gy=y$, thus, we get

$$gx=F_1(x,x)=F_2(x,x)=F_3(x,x)=x.$$

To prove the uniqueness, let $z\in X$ with $z\ne x$ such that

$$z=gz=F_1(z,z)=F_2(z,z)=F_3(z,z).$$

Again using condition (2.2) and Proposition 1.10(iii), we have

$$\begin{array}{rcl}G(z,z,x)& =& G(F_1(z,z),F_2(z,z),F_3(x,x))\\ \le & a_{1}G(gz,gz,gx)+a_{2}G(gz,gz,gx)+a_{3}G(gz,gz,gz)+a_{4}G(gz,gz,gz)\\ +a_{5}G(gz,gx,gx)+a_{6}G(gz,gx,gx)+a_{7}G(gx,gz,gz)+a_{8}G(gx,gz,gz)\\ \le & (a_1+a_2+2a_5+2a_6+a_7+a_8)G(z,z,x).\end{array}$$

Since $0\le a_1+a_2+a_3+a_4+2a_5+2a_6+a_7+a_8<1$, we get $G(z,z,x)<G(z,z,x)$, which is a contradiction. Thus, $F_1$, $F_2$, $F_3$ and g have a unique common fixed point. □

Remark 2.1 Theorem 2.1 extends and improves Theorem 3.2 of Shatanawi [26].

The following corollary can be obtained from Theorem 2.1 immediately.

Corollary 2.1 Let $(X,G)$ be a G-metric space. Let $F_1,F_2,F_3:X\times X\to X$ and $g:X\to X$ be mappings such that

$$G(F_1(x,y),F_2(u,v),F_3(w,z))\le a_{1}G(gx,gu,gw)+a_{2}G(gy,gv,gz)$$

(2.13)

for all $x,y,u,v,w,z\in X$, where $a_i\ge 0$, $i=1,2$ and $a_1+a_2<1$. Suppose that $F_1$, $F_2$, $F_3$ and g satisfy the following conditions:

• (1) $F_1(X\times X)\subseteq gX$, $F_2(X\times X)\subseteq gX$, $F_2(X\times X)\subseteq gX$;

• (2) gX is G-complete;

• (3) g is G-continuous and commutes with $F_1$, $F_2$, $F_3$.

Then there exist unique $x\in X$ such that

$$gx=F_1(x,x)=F_2(x,x)=F_3(x,x)=x.$$

Remark 2.2 If $F_1(x,y)=F_2(x,y)=F_3(x,y)$ and $a_1=a_2=k$, then Corollary 2.1 is reduced to Theorem 3.2 of Shatanawi [26].

Now, we give an example to support Corollary 2.1.

Example 2.1 Let $X=[0,1]$. Define $G:X\times X\times X\to {\mathbb{R}}^{+}$ by

$$G(x,y,z)=|x-y|+|y-z|+|z-x|$$

for all $x,y,z\in X$. Then $(X,G)$ is a complete G-metric space. Define a map

$$F_1,F_2,F_3:X\times X\to X$$

by

$$F_1(x,y)=F_2(x,y)=F_3(x,y)=\frac{x+y}{8}$$

for all $x,y\in X$. Also, define $g:X\to X$ by $gx=\frac{x}{2}$ for $x\in X$. Then $F(X\times X)\subseteq gX$. Through calculation, we have

$$\begin{array}{c}G(F_1(x,y),F_2(u,v),F_3(w,z))\hfill \\ \le G(\frac{x+y}{8},\frac{u+v}{8},\frac{w+z}{8})\hfill \\ =\frac{1}{8}(|x-u+y-v|+|u-w+v-z|+|w-x+z-y|)\hfill \\ \le \frac{1}{8}(|x-u|+|y-v|+|u-w|+|v-z|+|w-x|+|z-y|)\hfill \\ =\frac{1}{4}(G(gx,gu,gw)+G(gy,gv,gz)).\hfill \end{array}$$

Then the mappings $F_1$, $F_2$, $F_3$ and g are satisfying condition (2.13) of Corollary 2.1 with $a_1=a_2=\frac{1}{4}$. So that all the conditions of Corollary 2.1 are satisfied. By Corollary 2.4, $F_1$, $F_2$, $F_3$ and g have a unique common fixed point. Moreover, 0 is the unique common fixed point for all of the mappings $F_1$, $F_2$, $F_3$ and g.

If $a_1=a_2=0$, then Theorem 2.1 is reduced to the following.

Corollary 2.2 Let $(X,G)$ be a G-metric space. Let $F_1,F_2,F_3:X\times X\to X$ and $g:X\to X$ be four mappings such that

$$\begin{array}{rcl}G(F_1(x,y),F_2(u,v),F_3(w,z))& \le & c_{1}G(gx,gu,gu)+c_{2}G(gy,gv,gv)\\ +c_{3}G(gu,gw,gw)+c_{4}G(gv,gz,gz)\\ +c_{5}G(gw,gx,gx)+c_{6}G(gz,gy,gy)\end{array}$$

(2.14)

for all $x,y,u,v,w,z\in X$, where $c_i\ge 0$, $i=1,2,\dots ,6$ and $c_1+c_2+2c_3+2c_4+c_5+c_6<1$. Suppose that $F_1$, $F_2$, $F_3$ and g satisfy the following conditions:

• (i) $F_1(X\times X)\subseteq gX$, $F_2(X\times X)\subseteq gX$, $F_3(X\times X)\subseteq gX$;

• (ii) gX is G-complete;

• (iii) g is G-continuous and commutes with $F_1$, $F_2$, $F_3$.

Then there exist unique $x\in X$ such that

$$gx=F_1(x,x)=F_2(x,x)=F_3(x,x)=x.$$

If we take $F_1(x,y)=F_2(x,y)=F_3(x,y)$ in Corollary 2.2, then the following corollary is obtained.

Corollary 2.3 Let $(X,G)$ be a G-metric space. Let $F:X\times X\to X$ and $g:X\to X$ be four mappings such that

$$\begin{array}{rcl}G(F(x,y),F(u,v),F(w,z))& \le & c_{1}G(gx,gu,gu)+c_{2}G(gy,gv,gv)\\ +c_{3}G(gu,gw,gw)+c_{4}G(gv,gz,gz)\\ +c_{5}G(gw,gx,gx)+c_{6}G(gz,gy,gy)\end{array}$$

(2.15)

for all $x,y,u,v,w,z\in X$, where $c_i\ge 0$, $i=1,2,\dots ,6$ and $c_1+c_2+2c_3+2c_4+c_5+c_6<1$. Suppose that F and g satisfy the following conditions:

• (i) $F(X\times X)\subseteq gX$;

• (ii) gX is G-complete;

• (iii) g is G-continuous and commutes with F.

Then there exist unique $x\in X$ such that

$$gx=F(x,x)=x.$$

Now, we give an example to support Corollary 2.3.

Example 2.2 Let $X=[0,1]$. Define $G:X\times X\times X\to {\mathbb{R}}^{+}$ by

$$G(x,y,z)=|x-y|+|y-z|+|z-x|$$

for all $x,y,z\in X$. Then $(X,G)$ is a complete G-metric space. Define a map $F:X\times X\to X$ by

$$F(x,y)=\frac{xy}{8}$$

for all $x,y\in X$. Also, define $g:X\to X$ by $gx=x$ for $x\in X$. Then $F(X\times X)\subseteq gX$. Through calculation, we have

$$\begin{array}{c}G(F(x,y),F(u,v),F(w,z))\hfill \\ =\frac{1}{8}(|xy-uv|+|uv-wz|+|wz-xy|)\hfill \\ \le \frac{1}{8}(|y||x-u|+|u||y-v|+|v||u-w|+|w||v-z|+|z||w-x|+|x||z-y|)\hfill \\ \le \frac{1}{8}(|x-u|+|y-v|+|u-w|+|v-z|+|w-x|+|z-y|)\hfill \\ =\frac{1}{16}(G(gx,gu,gu)+G(gy,gv,gv)+G(gu,gw,gw)+G(gv,gz,gz)\hfill \\ +G(gw,gx,gx)+c_{6}G(gz,gy,gy)).\hfill \end{array}$$

Then the mappings $F_1$, $F_2$, $F_3$ and g are satisfying condition (2.15) of Corollary 2.3 with $c_1=c_2=c_3=c_4=c_5=c_6=\frac{1}{16}$. So that all the conditions of Corollary 2.3 are satisfied. By Corollary 2.3, F and g have a unique common fixed point. Moreover, 0 is the unique common fixed point for all of the mappings F and g.

If we take $F_1(x,y)=F_2(x,y)=F_3(x,y)$ in Theorem 2.1, then the following corollary is obtained.

Corollary 2.4 Let $(X,G)$ be a G-metric space. Let $F:X\times X\to X$ and $g:X\to X$ be mappings such that

$$\begin{array}{c}G(F(x,y),F(u,v),F(w,z))\hfill \\ \le a_{1}G(gx,gu,gw)+a_{2}G(gy,gv,gz)+a_{3}G(gx,gu,gu)\hfill \\ +a_{4}G(gy,gv,gv)+a_{5}G(gu,gw,gw)+a_{6}G(gv,gz,gz)\hfill \\ +a_{7}G(gw,gx,gx)+a_{8}G(gz,gy,gy)\hfill \end{array}$$

(2.16)

for all $x,y,u,v,w,z\in X$, where $a_i\ge 0$, $i=1,2,\dots ,8$ and $a_1+a_2+a_3+a_4+2a_5+2a_6+a_7+a_8<1$. Suppose that F and g satisfy the following conditions:

• (1) $F(X\times X)\subseteq gX$;

• (2) gX is G-complete;

• (3) g is G-continuous and commutes with F.

Then there exist unique $x\in X$ such that $gx=F(x,x)=x$.

Now, we introduce an example to support Corollary 2.4.

Example 2.3 Let $X=[-1,1]$. Define $G:X\times X\times X\to {\mathbb{R}}^{+}$ by

$$G(x,y,z)=|x-y|+|y-z|+|z-x|$$

for all $x,y,z\in X$. Then $(X,G)$ is a complete G-metric space. Define a map

$$F:X\times X\to X$$

by

$$F(x,y)=\frac{1}{16}{x}^2+\frac{1}{16}{y}^2-1$$

for all $x,y\in X$. Also, define $g:X\to X$ by $gx=x$ for $x\in X$.

Clearly, we can get $F(X\times X)=[-1,-\frac{7}{8}]\subseteq gX$, and g is G-continuous and commutes with F.

By the definition of the mappings of F and g, for all $x,y,z,u,v,w\in [-1,1]$, we have

$$\begin{array}{c}G(F(x,y),F(u,v),F(w,z))\hfill \\ \le G(\frac{1}{16}{x}^2+\frac{1}{16}{y}^2-1,\frac{1}{16}{u}^2+\frac{1}{16}{v}^2-1,\frac{1}{16}{w}^2+\frac{1}{16}{z}^2-1)\hfill \\ =\frac{1}{16}(|x^2-u^2+y^2-v^2|+|u^2-w^2+v^2-z^2|+|w^2-x^2+z^2-y^2|)\hfill \\ \le \frac{1}{16}(|x^2-u^2|+|y^2-v^2|+|u^2-w^2|+|v^2-z^2|+|w^2-x^2|+|z^2-y^2|)\hfill \\ \le \frac{1}{16}(2|x-u|+2|y-v|+2|u-w|+2|v-z|+2|w-x|+2|z-y|)\hfill \\ =\frac{1}{16}G(gx,gu,gu)+\frac{1}{16}G(gy,gv,gv)+\frac{1}{16}G(gu,gw,gw)\hfill \\ +\frac{1}{16}G(gv,gz,gz)+\frac{1}{16}G(gw,gx,gx)+\frac{1}{16}G(gz,gy,gy).\hfill \end{array}$$

Then the mappings F and g are satisfying condition (2.16) of Corollary 2.4 with $a_1=a_2=0$, $a_3=a_4=a_5=a_6=a_7=a_8=\frac{1}{16}$. So that all the conditions of Corollary 2.4 are satisfied. By Corollary 2.4, F and g have a unique common fixed point. Here $x=4-2\sqrt{6}$ is the unique common fixed point of mappings F and g; that is, $F(x,x)=gx=x$.

3 Application to integral equations

Throughout this section, we assume that $X=C[0,1]$ is the set of all continuous functions defined on $[0,1]$. Define $G:X\times X\times X\to {\mathbb{R}}^{+}$ by

$$G(x,y,z)=\underset{t\in [0,1]}{sup}|x(t)-y(t)|+\underset{t\in [0,1]}{sup}|y(t)-z(t)|+\underset{t\in [0,1]}{sup}|z(t)-x(t)|$$

for all $x,y,z\in X$. Then $(X,G)$ is a G-complete metric space.

Consider the following integral equations:

$$F_i(x,y)(t)={\int }_0^{1}k(t,s)(f_i(s,x(s))+g_i(s,y(s)))ds,t\in [0,1](i=1,2,3).$$

(3.1)

Next, we will analyze (3.1) under the following conditions:

• (i) $k:[0,1]\times [0,1]\to {\mathbb{R}}^{+}$ is continuous.

• (ii) $f_i,g_i:[0,1]\times \mathbb{R}\to \mathbb{R}$ ($i=1,2,3$) are continuous functions.

• (iii) There exist constants ${\lambda }_i,{\mu }_i>0$ ($i=1,2,3$) such that

  $$\left\{\begin{array}{c}|f_1(t,x)-f_2(t,y)|\le {\lambda }_1|x-y|,\hfill \\ |f_2(t,x)-f_3(t,y)|\le {\lambda }_2|x-y|,\hfill \\ |f_3(t,x)-f_1(t,y)|\le {\lambda }_3|x-y|\hfill \end{array}\text{and}\right\{\begin{array}{c}|g_1(t,x)-g_2(t,y)|\le {\mu }_1|x-y|,\hfill \\ |g_2(t,x)-g_3(t,y)|\le {\mu }_2|x-y|,\hfill \\ |g_3(t,x)-g_1(t,y)|\le {\mu }_3|x-y|\hfill \end{array}$$

for all $t\in [0,1]$ and $x,y\in \mathbb{R}$.

• (iv) ${\parallel k\parallel }_{\mathrm{\infty }}(max\{{\lambda }_1,{\mu }_1\}+2max\{{\lambda }_2,{\mu }_2\}+max\{{\lambda }_3,{\mu }_3\})<1$, where

  $${\parallel k\parallel }_{\mathrm{\infty }}=sup\{k(t,s):t,s\in [0,1]\}.$$

The aim of this section is to give an existence theorem for a solution of the above integral equations by using the obtained result given by Theorem 2.1.

Theorem 3.1 Under conditions (i)-(iv), integral equation (3.1) has a unique common solution in $C[0,1]$.

Proof First, we consider $F_i:X\times X\to X$ ($i=1,2,3$). By virtue of our assumptions, $F_i$ is well defined (this means that for $x,y\in X$ then $F_i(x,y)\in X$ ($i=1,2,3$)). Then we can get

$$\begin{array}{c}G(F_1(x,y),F_2(u,v),F_3(w,z))\hfill \\ =\underset{t\in [0,1]}{sup}|F_1(x,y)-F_2(u,v)|+\underset{t\in [0,1]}{sup}|F_2(u,v)-F_3(w,z)|+\underset{t\in [0,1]}{sup}|F_3(w,z)-F_1(x,y)|\hfill \\ =\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)(f_1(s,x(s))+g_1(s,y(s)))ds-{\int }_0^{1}k(t,s)(f_2(s,u(s))+g_2(s,v(s)))ds|\hfill \\ +\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)(f_2(s,u(s))+g_2(s,v(s)))ds-{\int }_0^{1}k(t,s)(f_3(s,w(s))+g_3(s,z(s)))ds|\hfill \\ +\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)(f_3(s,w(s))+g_3(s,z(s)))ds-{\int }_0^{1}k(t,s)(f_1(s,x(s))+g_1(s,y(s)))ds|\hfill \\ =\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)((f_1(s,x(s))-f_2(s,u(s)))+(g_1(s,y(s))-g_2(s,v(s))))ds|\hfill \\ +\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)((f_2(s,u(s))-f_3(s,w(s)))+(g_2(s,v(s))-g_3(s,z(s))))ds|\hfill \\ +\underset{t\in [0,1]}{sup}|{\int }_0^{1}k(t,s)((f_3(s,w(s))-f_1(s,x(s)))+(g_3(s,z(s))-g_1(s,y(s))))ds|\hfill \\ \le \underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)(|f_1(s,x(s))-f_2(s,u(s))|+|g_1(s,y(s))-g_2(s,v(s))|)ds\hfill \\ +\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)(|f_2(s,u(s))-f_3(s,w(s))|+|g_2(s,v(s))-g_3(s,z(s))|)ds\hfill \\ +\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)(|f_3(s,w(s))-f_1(s,x(s))|+|g_3(s,z(s))-g_1(s,y(s))|)ds.\hfill \end{array}$$

(3.2)

By conditions (iii),

$$\{\begin{array}{c}|f_1(s,x(s))-f_2(s,u(s))|\le {\lambda }_1|x(s)-u(s)|,\hfill \\ |f_2(s,u(s))-f_3(s,w(s))|\le {\lambda }_2|u(s)-w(s)|,\hfill \\ |f_3(s,w(s))-f_1(s,x(s))|\le {\lambda }_3|w(s)-x(s)|\hfill \end{array}$$

and

$$\{\begin{array}{c}|g_1(s,y(s))-g_2(s,v(s))|\le {\mu }_1|y(s)-v(s)|,\hfill \\ |g_2(s,v(s))-g_3(s,z(s))|\le {\mu }_2|v(s)-z(s)|,\hfill \\ |g_3(s,z(s))-g_1(s,y(s))|\le {\mu }_3|z(s)-y(s)|.\hfill \end{array}$$

Taking these inequalities into (3.2), we obtain

$$\begin{array}{c}G(F_1(x,y),F_2(u,v),F_3(w,z))\hfill \\ \le \underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)({\lambda }_1|x(s)-u(s)|+{\mu }_1|y(s)-v(s)|)ds\hfill \\ +\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)({\lambda }_2|u(s)-w(s)|+{\mu }_2|v(s)-z(s)|)\hfill \\ +\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)({\lambda }_3|w(s)-x(s)|+{\mu }_3|z(s)-y(s)|)\hfill \\ \le max\{{\lambda }_1,{\mu }_1\}\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)\left(\right|x(s)-u(s)|+|y(s)-v(s)|)ds\hfill \\ +max\{{\lambda }_2,{\mu }_2\}\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)(|u(s)-w(s)|+|v(s)-z(s)|)ds\hfill \\ +max\{{\lambda }_3,{\mu }_3\}\underset{t\in [0,1]}{sup}{\int }_0^{1}k(t,s)(|w(s)-x(s)|+|z(s)-y(s)|)ds.\hfill \end{array}$$

(3.3)

Using the Cauchy-Schwartz inequality in (3.3), we get

$$\begin{array}{c}{\int }_0^{1}k(t,s)(|x(s)-u(s)|+|y(s)-v(s)|)ds\hfill \\ \le {({\int }_0^1{k}^2(t,s)ds)}^{\frac{1}{2}}{\left({\int }_0^1{(|x(s)-u(s)|+|y(s)-v(s)|)}^{2}ds\right)}^{\frac{1}{2}}\hfill \\ \le {\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|x(t)-u(t)|+\underset{t\in [0,1]}{sup}|y(t)-v(t)|).\hfill \end{array}$$

(3.4)

Similarly, we can obtain the following estimate

$$\begin{array}{c}{\int }_0^{1}k(t,s)(|u(s)-w(s)|+|v(s)-z(s)|)ds\hfill \\ \le {\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|u(t)-w(t)|+\underset{t\in [0,1]}{sup}|v(t)-z(t)|),\hfill \end{array}$$

(3.5)

$$\begin{array}{c}{\int }_0^{1}k(t,s)(|w(s)-x(s)|+|z(s)-y(s)|)ds\hfill \\ \le {\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|w(t)-x(t)|+\underset{t\in [0,1]}{sup}|z(t)-y(t)|).\hfill \end{array}$$

(3.6)

Substituting (3.4), (3.5) and (3.6) into (3.3), we obtain that

$$\begin{array}{c}G(F_1(x,y),F_2(u,v),F_3(w,z))\hfill \\ \le max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|x(t)-u(t)|+\underset{t\in [0,1]}{sup}|y(t)-v(t)|)\hfill \\ +max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|u(t)-w(t)|+\underset{t\in [0,1]}{sup}|v(t)-z(t)|)\hfill \\ +max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }}(\underset{t\in [0,1]}{sup}|w(t)-x(t)|+\underset{t\in [0,1]}{sup}|z(t)-y(t)|)\hfill \\ =\frac{1}{2}max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|x(t)-u(t)|\hfill \\ +\frac{1}{2}max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|y(t)-v(t)|\hfill \\ +\frac{1}{2}max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|u(t)-w(t)|\hfill \\ +\frac{1}{2}max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|v(t)-z(t)|\hfill \\ +\frac{1}{2}max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|w(t)-x(t)|\hfill \\ +\frac{1}{2}max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }}\cdot 2\underset{t\in [0,1]}{sup}|z(t)-y(t)|\hfill \\ =\frac{1}{2}max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }}G(x,u,u)+\frac{1}{2}max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }}G(y,v,v)\hfill \\ +\frac{1}{2}max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }}G(u,w,w)+\frac{1}{2}max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }}G(v,z,z)\hfill \\ +\frac{1}{2}max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }}G(w,x,x)+\frac{1}{2}max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }}G(z,y,y).\hfill \end{array}$$

(3.7)

Taking $gx=x$ for all $x\in X$, and

$$\begin{array}{c}{a}_1=a_2=0,a_3=a_4=\frac{1}{2}max\{{\lambda }_1,{\mu }_1\}{\parallel k\parallel }_{\mathrm{\infty }},\hfill \\ a_5=a_6=\frac{1}{2}max\{{\lambda }_2,{\mu }_2\}{\parallel k\parallel }_{\mathrm{\infty }},a_7=a_8=\frac{1}{2}max\{{\lambda }_3,{\mu }_3\}{\parallel k\parallel }_{\mathrm{\infty }},\hfill \end{array}$$

then inequality (3.7) becomes

$$\begin{array}{rcl}G(F_1(x,y),F_2(u,v),F_3(w,z))& \le & a_{1}G(gx,gu,gw)+a_{2}G(gy,gv,gz)+a_{3}G(gx,gu,gu)\\ +a_{4}G(gy,gv,gv)+a_{5}G(gu,gw,gw)+a_{6}G(gv,gz,gz)\\ +a_{7}G(gw,gx,gx)+a_{8}G(gz,gy,gy).\end{array}$$

(3.8)

By condition (iv), we know that

$$\begin{array}{c}{a}_1+a_2+a_3+a_4+2(a_5+a_6)+a_7+a_8\hfill \\ ={\parallel k\parallel }_{\mathrm{\infty }}(max\{{\lambda }_1,{\mu }_1\}+2max\{{\lambda }_2,{\mu }_2\}+max\{{\lambda }_3,{\mu }_3\})<1.\hfill \end{array}$$

This proves that the operator $F_i$ ($i=1,2,3$) and $g=I$ satisfy contractive condition (2.2) appearing in Theorem 2.1 with $g=I$. Therefore, $F_1$, $F_2$, $F_3$ have a unique common coupled fixed point, that is, $F_1(x,x)=F_2(x,x)=F_3(x,x)=x$, and so, $(x,x)$ is the unique solution of equation (3.1). □