Existence and Unique Coupled Solution in S_b-Metric Spaces by Rational Contraction with Application

Abstract

In this paper, we prove a unique common coupled fixed point theorem for two pairs of w-compatible mappings in $S_b$-metric spaces. We also furnish an example to support our main result.

1. Introduction

In 2012, Sedghi et al. [1] introduced the notion of S-metric space and proved several results. Some other authors also worked on this (e.g., [2,3,4,5,6]). On the other hand, the concept of b-metric space was introduced by Bakhtin [7] and Czerwik [8] (see also [9,10,11]).

Recently, Sedghi et al. [1] defined $S_b$-metric spaces using the concepts of S and b-metric spaces and proved common fixed point theorem for four maps in $S_b$-metric spaces (see also [12]).

Bhaskar and Lakshmikantham [13] introduced the notion of coupled fixed point and proved some coupled fixed point results as well.

The aim of this paper is to prove a unique common coupled fixed point theorem for four mappings in $S_b$-metric spaces. Throughout this paper, ${\mathbb{R}}^{+}$ and $\mathbb{N}$ denote the set of all non-negative real numbers and positive integers, respectively.

First, we recall some definitions, lemmas and examples.

Definition 1.

Ref. [4] Let X be a nonempty set. A S-metric on X is a function $S:X^3\to {\mathbb{R}}^{+}$ that satisfies the following conditions for each $x,y,z,a\in X$:

(S1) 

$S(x,y,z)=0$ if and only if $x=y=z$,

(S2) 

$S(x,y,z)\le S(x,x,a)+S(y,y,a)+S(z,z,a)$.

Then, the pair $(X,S)$ is called a S-metric space.

Definition 2.

([8]) Let X be a nonempty set and $s\ge 1$ be a given real number. A function $d:X\times X\to {\mathbb{R}}^{+}$ is called a b-metric if the following axioms are satisfied for all $x,y,z\in X$:

(b1) 

$d(x,y)=0$ if and only if $x=y$;

(b2) 

$d(x,y)=d(y,x)$; and

(b3) 

$d(x,y)\le s\left[d\right(x,z)+d(z,y\left)\right]$.

The pair $(X,d)$ is called a b b-metric space.

Definition 3.

([1]) Let X be a nonempty set and $b\ge 1$ be a given real number. Suppose that a mapping $S_b:X^3\to {\mathbb{R}}^{+}$ is a function satisfying the following properties:

(S_b1)

$S_b(x,y,z)=0$ if and only if $x=y=z$; and

(S_b2)

$S_b(x,y,z)\le b(S_b(x,x,a)+S_b(y,y,a)+S_b(z,z,a))$ for all $x,y,z,a\in X$.

Then, the function $S_b$ is called a $S_b$-metric on X and the pair $(X,S_b)$ is called a $S_b$-metric space.

Remark 1.

([1]) It should be noted that the class of $S_b$-metric spaces is effectively larger than that of S-metric spaces. Indeed, each S-metric space is a $S_b$-metric space with $b=1$.

The following example shows that a $S_b$-metric on X need not be a S-metric on X.

Example 1.

([1]) Let $(X,S_b)$ be a $S_b$-metric space and $S_b(x,y,z)=S^p(x,y,z)$, where $p>1$ is a real number. Note that $S_b$ is a $S_b$-metric with $b=2^{2(p-1)}$. In addition, $(X,S_b)$ is not necessarily a S-metric space.

Definition 4.

([1]) Let $(X,S_b)$ be a $S_b$-metric space. Then, for $x\in X$, $r>0$, we defined the open ball $B_{S_b}(x,r)$ and closed ball $B_{S_b}[x,r]$ with center x and radius r as follows, respectively:

$$\begin{array}{cc}\hfill B_{S_b}(x,r)& =\{y\in X:S_b(y,y,x)<r\},\hfill \\ \hfill B_{S_b}[x,r]& =\{y\in X:S_b(y,y,x)\le r\}.\hfill \end{array}$$

Lemma 1.

([1]) In a $S_b$-metric space, we have

$$S_b(x,x,y)\le b{S}_b(y,y,x)$$

and

$$S_b(y,y,x)\le b{S}_b(x,x,y).$$

Lemma 2.

([1]) In a $S_b$-metric space we have

$$S_b(x,x,z)\le 2b{S}_b(x,x,y)+b^2{S}_b(y,y,z).$$

Definition 5.

([1]) Let $(X,S_b)$ be a $S_b$-metric space. A sequence $\left\{x_n\right\}$ in X is said to be:

(1) 

$S_b$-Cauchy sequence if, for each $ϵ>0$, there exists $n_0\in \mathbb{N}$ such that $S_b(x_n,x_n,x_m)<ϵ$ for each $m,n\ge n_0$.

(2) 

$S_b$-convergent to a point $x\in X$ if, for each $ϵ>0$, there exists a positive integer $n_0$ such that $S_b(x_n,x_n,x)<ϵ$ or $S_b(x,,x,x_n)<ϵ$ for all $n\ge n_0$ and we denote that by $\underset{n\to \infty }{lim}{x}_n=x$.

Definition 6.

([1]) A $S_b$-metric space $(X,S_b)$ is called complete if every $S_b$-Cauchy sequence is $S_b$-convergent in X.

Lemma 3.

([14]) If $(X,S_b)$ is a $S_b$-metric space with $b\ge 1$ and $\left\{x_n\right\}$ is a $S_b$-convergent to x, then for all $y\in X$, we have

(i) 

${\displaystyle \frac{1}{2b}}{S}_b(y,x,x)\le \underset{n\to \infty }{lim}inf{S}_b(y,y,x_n)\le \underset{n\to \infty }{lim}sup{S}_b(y,y,x_n)\le 2b{S}_b(y,y,x)$; and

(ii) 

${\displaystyle \frac{1}{b^2}}{S}_b(x,x,y)\le \underset{n\to \infty }{lim}inf{S}_b(x_n,x_n,y)\le \underset{n\to \infty }{lim}sup{S}_b(x_n,x_n,y)\le b^2{S}_b(x,x,y)$.

In particular, if $x=y$, then we have $\underset{n\to \infty }{lim}{S}_b(x_n,x_n,y)=0.$

Definition 7.

([13]) Let X be a nonempty set. An element $(x,y)\in X\times X$ is called a coupled fixed point of a mapping $F:X\times X\to X$ if $x=F(x,y)$ and $y=F(y,x)$.

Definition 8.

([15]) Let X be a nonempty set. An element $(x,y)\in X\times X$ is called:

(i) 

a coupled coincident point of mappings $F:X\times X\to X$ and $f:X\to X$ if $f\left(x\right)=F(x,y)$ and $f\left(y\right)=F(y,x)$; and

(ii) 

a common coupled fixed point of mappings $F:X\times X\to X$ and $f:X\to X$ if $x=f\left(x\right)=F(x,y)$ and $y=f\left(y\right)=F(y,x)$.

Definition 9.

([16]) Let X be a nonempty set and $F:X\times X\to X$ and $f:X\to X$. The $\{F,f\}$ is said to be w-compatible pair if $f\left(F\right(x,y\left)\right)=F\left(f\right(x),f(y\left)\right)$ and $f\left(F\right(y,x\left)\right)=F\left(f\right(y),f(x\left)\right)$ whenever there exist $x,y\in X$ with $f\left(x\right)=F(x,y)$ and $f\left(y\right)=F(y,x)$.

For more details of other generalized metric spaces as well as on some rational contraction, see [9,17,18,19].

Now, we give our main result.

2. Main Results

Let $\Phi$ denote the class of all functions $\varphi :[0,\infty )\to [0,\infty )$ such that $\varphi$ is a non-decreasing, continuous, $\varphi \left(t\right)<{\displaystyle \frac{t}{4b^4}}$ for all $t>0$ and $\varphi \left(0\right)=0$.

Theorem 1.

Let $(X,S_b)$ be a $S_b$-metric space. Suppose that $A,B:X\times X\to X$ and $P,Q:X\to X$ satisfy:

(1) 

$A(X\times X)\subseteq Q\left(X\right),B(X\times X)\subseteq P\left(X\right)$;

(2) 

$\{A,P\}$ and $\{B,Q\}$ are w-compatible pairs;

(3) 

one of $P\left(X\right)$ or $Q\left(X\right)$ is $S_b$-complete subspace of X; and

(4) 

$2b^5{S}_b(A(x,y),A(x,y),B(u,v))$

$$\le \varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(x\right),P\left(x\right),Q\left(u\right)),S_b(P\left(y\right),P\left(y\right),Q\left(v\right)),S_b(A(x,y),A(x,y),P\left(x\right)),\\ S_b(A(y,x),A(y,x),P\left(y\right)),S_b(B(u,v),B(u,v),Q\left(u\right)),S_b(B(v,u),B(v,u),Q\left(v\right)),\\ {\displaystyle \frac{S_b(A(x,y),A(x,y),Q\left(u\right))S_b(B(u,v),B(u,v),P\left(x\right))}{1+S_b(P\left(x\right),P\left(x\right),Q\left(u\right))}},\\ {\displaystyle \frac{S_b(A(y,x),A(y,x),Q\left(v\right))S_b(B(v,u),B(v,u),P\left(y\right))}{1+S_b(P\left(y\right),P\left(y\right),Q\left(v\right))}}\end{array}\right\}\right),$$

for all $x,y,u,v\in X$, $\varphi \in \Phi$.

Then, $A,B,P$ and Q have a unique common coupled fixed point in $X\times X$.

Proof of Theorem.

Let $x_0,y_0\in X$. From Equation (1), we can construct the sequences $\left\{x_n\right\}$, $\left\{y_n\right\}$, $\left\{z_n\right\}$ and $\left\{w_n\right\}$ such that

$$\begin{array}{c}A(x_{2n},y_{2n})=Q\left(x_{2n+1}\right)=z_{2n},\hfill \\ A(y_{2n},x_{2n})=Q\left(y_{2n+1}\right)=w_{2n},\hfill \\ B(x_{2n+1},y_{2n+1})=P\left(x_{2n+2}\right)=z_{2n+1},\hfill \\ B(y_{2n+1},x_{2n+1})=P\left(y_{2n+2}\right)=w_{2n+1},n=0,1,2,\cdots .\hfill \end{array}$$

Case (i). Suppose $z_{2m}=z_{2m+1}$ and $w_{2m}=w_{2m+1}$ for some m. Assume that $z_{2m+1}\ne z_{2m+2}$ or $w_{2m+1}\ne w_{2m+2}$.

From Equation (4), we have

$$\begin{array}{c}{S}_b(z_{2m+2},z_{2m+2},z_{2m+1})\hfill \\ \le 2b^5{S}_b(A(x_{2m+2},y_{2m+2}),A(x_{2m+2},y_{2m+2}),B(x_{2m+1},y_{2m+1}))\hfill \\ \le \varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(x_{2m+2}\right),P\left(x_{2m+2}\right),Q\left(x_{2m+1}\right)),S_b(P\left(y_{2m+2}\right),P\left(y_{2m+2}\right),Q\left(y_{2m+1}\right)),\\ S_b(A(x_{2m+2},y_{2m+2}),A(x_{2m+2},y_{2m+2}),P\left(x_{2m+2}\right)),\\ S_b(A(y_{2m+2},x_{2m+2}),A(y_{2m+2},x_{2m+2}),P\left(y_{2m+2}\right)),\\ S_b(B(x_{2m+1},y_{2m+1}),B(x_{2m+1},y_{2m+1}),Q\left(x_{2m+1}\right)),\\ S_b(B(y_{2m+1},x_{2m+1}),B(y_{2m+1},x_{2m+1}),Q\left(y_{2m+1}\right)),\\ {\displaystyle \frac{S_b(A(x_{2m+2},y_{2m+2}),A(x_{2m+2},y_{2m+2}),Q\left(x_{2m+1}\right))S_b(B(x_{2m+1},y_{2m+1}),B(x_{2m+1},y_{2m+1}),P\left(x_{2m+2}\right))}{1+S_b(P\left(x_{2m+2}\right),P\left(x_{2m+2}\right),Q\left(x_{2m+1}\right))}}\\ {\displaystyle \frac{S_b(A(y_{2m+2},x_{2m+2}),A(y_{2m+2},x_{2m+2}),Q\left(y_{2m+1}\right))S_b(B(y_{2m+1},x_{2m+1}),B(y_{2m+1},x_{2m+1}),P\left(y_{2m+2}\right))}{1+S_b(P\left(y_{2m+2}\right),P\left(y_{2m+2}\right),Q\left(y_{2m+1}\right))}}\end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2m+1},z_{2m+1},z_{2m}),S_b(w_{2m+1},w_{2m+1},w_{2m}),S_b(z_{2m+2},z_{2m+2},z_{2m+1}),\hfill \\ S_b(w_{2m+2},w_{2m+2},w_{2m+1}),S_b(z_{2m+1},z_{2m+1},z_{2m}),S_b(w_{2m+1},w_{2m+1},w_{2m}),\hfill \\ {\displaystyle \frac{S_b(z_{2m+2},z_{2m+2},z_{2m+1})S_b(z_{2m+1},z_{2m+1},z_{2m})}{1+S_b(z_{2m+1},z_{2m+1},z_{2m})}},{\displaystyle \frac{S_b(w_{2m+2},w_{2m+2},w_{2m+1})S_b(w_{2m+1},w_{2m+1},w_{2m})}{1+S_b(w_{2m+1},w_{2m+1},w_{2m})}}\hfill \end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}0,0,S_b(z_{2m+2},z_{2m+2},z_{2m+1}),S_b(w_{2m+2},w_{2m+2},w_{2m+1}),0,0,0,0\hfill \end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2m+2},z_{2m+2},z_{2m+1}),S_b(w_{2m+2},w_{2m+2},w_{2m+1})\hfill \end{array}\right\}\right).\hfill \end{array}$$

Similarly, we can prove that

$$S_b(w_{2m+2},w_{2m+2},w_{2m+1})\le \varphi \left(max\left\{S_b(z_{2m+2},z_{2m+2},z_{2m+1}),S_b(w_{2m+2},w_{2m+2},w_{2m+1})\right\}\right).$$

It follows that

$$\begin{array}{c}max\left\{S_b(z_{2m+2},z_{2m+2},z_{2m+1}),S_b(w_{2m+2},w_{2m+2},w_{2m+1})\right\}\hfill \\ \le \varphi \left(max\left\{S_b(z_{2m+2},z_{2m+2},z_{2m+1}),S_b(w_{2m+2},w_{2m+2},w_{2m+1})\right\}\right).\hfill \end{array}$$

Hence, $z_{2m+2}=z_{2m+1}$ and $w_{2m+2}=w_{2m+1}$.

Continuing in this process, we can conclude that $z_{2m+k}=z_{2m}$ and $w_{2m+k}=w_{2m}$ for all $k\ge 0$.

It follows that $\left\{z_m\right\}$ and $\left\{w_m\right\}$ are Cauchy sequences.

Case (ii). Assume that $z_{2n}\ne z_{2n+1}$ or $w_{2n}\ne w_{2n+1}$ for all n.

Put $S_n=max\left\{S_b(z_{n+1},z_{n+1},z_n),S_b(w_{n+1},w_{n+1},w_n)\right\}$.

From Equation (4), we have

$$\begin{array}{c}{S}_b(z_{2n+2},z_{2n+2},z_{2n+1})\hfill \\ \le 2b^5{S}_b(A(x_{2n+2},y_{2n+2}),A(x_{2n+2},y_{2n+2}),B(x_{2n+1},y_{2n+1}))\hfill \\ \le \varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(x_{2n+2}\right),P\left(x_{2n+2}\right),Q\left(x_{2n+1}\right)),S_b(P\left(y_{2n+2}\right),P\left(y_{2n+2}\right),Q\left(y_{2n+1}\right)),\\ S_b(A(x_{2n+2},y_{2n+2}),A(x_{2n+2},y_{2n+2}),P\left(x_{2n+2}\right)),\\ S_b(A(y_{2n+2},x_{2n+2}),A(y_{2n+2},x_{2n+2}),P\left(y_{2n+2}\right)),\\ S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),Q\left(x_{2n+1}\right)),\\ S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),Q\left(y_{2n+1}\right)),\\ {\displaystyle \frac{S_b(A(x_{2n+2},y_{2n+2}),A(x_{2n+2},y_{2n+2}),Q\left(x_{2n+1}\right))S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),P\left(x_{2n+2}\right))}{1+S_b(P\left(x_{2n+2}\right),P\left(x_{2n+2}\right),Q\left(x_{2n+1}\right))}},\\ {\displaystyle \frac{S_b(A(y_{2n+2},x_{2n+2}),A(y_{2n+2},x_{2n+2}),Q\left(y_{2n+1}\right))S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),P\left(y_{2n+2}\right))}{1+S_b(P\left(y_{2n+2}\right),P\left(y_{2n+2}\right),Q\left(y_{2n+1}\right))}}\end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n}),S_b(z_{2n+2},z_{2n+2},z_{2n+1}),\hfill \\ S_b(w_{2n+2},w_{2n+2},w_{2n+1}),S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n}),\hfill \\ {\displaystyle \frac{S_b(z_{2n+2},z_{2n+2},z_{2n})S_b(z_{2n+1},z_{2n+1},z_{2n+1})}{1+S_b(z_{2n+1},z_{2n+1},z_{2n})}},{\displaystyle \frac{S_b(w_{2n+2},w_{2n+2},w_{2n})S_b(w_{2n+1},w_{2n+1},w_{2n+1})}{1+S_b(w_{2n+1},w_{2n+1},w_{2n})}}\hfill \end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(z_{2n+2},z_{2n+2},z_{2n+1}),\hfill \\ S_b(w_{2n+1},w_{2n+1},w_{2n}),S_b(w_{2n+2},w_{2n+2},w_{2n+1})\hfill \end{array}\right\}\right)\hfill \\ =\varphi \left(max\left\{\begin{array}{c}{S}_{2n+1},S_{2n}\hfill \end{array}\right\}\right).\hfill \end{array}$$

Similarly, we can prove

$$S_b(w_{2n+2},w_{2n+2},w_{2n+1})\le \varphi \left(max\left\{S_{2n+1},S_{2n}\right\}\right).$$

Thus,

$$S_{2n+1}\le \varphi (max\{S_{2n},S_{2n+1}\}).$$

If $S_{2n+1}$ is maximum, then we get contradiction so that $S_{2n}$ is maximum.

Thus,

$$S_{2n+1}\le \varphi \left(S_{2n}\right)<S_{2n}.$$

(1)

Similarly, we can conclude that $S_{2n}<S_{2n-1}$.

It is clear that $\left\{S_n\right\}$ is a non-increasing sequence of non-negative real numbers and must converge to a real number, say $r\ge 0$.

Suppose $r>0$. Letting $n\to \infty$, in Equation (1), we have $r\le \varphi \left(r\right)\le r$.

It is a contradiction. Hence, $r=0$.

Thus,

$$\underset{n\to \infty }{lim}{S}_b(z_{n+1},z_{n+1},z_n)=0$$

(2)

and

$$\underset{n\to \infty }{lim}{S}_b(w_{n+1},w_{n+1},w_n)=0.$$

(3)

Now, we prove that $\left\{z_{2n}\right\}$ and $\left\{w_{2n}\right\}$ are Cauchy sequences in $(X,S_b)$. On teh contrary, we suppose that $\left\{z_{2n}\right\}$ or $\left\{w_{2n}\right\}$ is not Cauchy. Then, there exist $ϵ>0$ and monotonically increasing sequence of natural numbers $\left\{2m_k\right\}$ and $\left\{2n_k\right\}$ such that for $n_k>m_k$

$$max\{S_b(z_{2m_k},z_{2m_k},z_{2n_k}),S_b(w_{2m_k},w_{2m_k},w_{2n_k})\}\ge ϵ$$

(4)

and

$$max\{S_b(z_{2m_k},z_{2m_k},z_{2n_{k-2}}),S_b(w_{2m_k},w_{2m_k},w_{2n_{k-2}})\}<ϵ.$$

(5)

From Equations (4) and (5), we have

$$\begin{array}{cc}\hfill ϵ& \le max\{S_b(z_{2m_k},z_{2m_k},z_{2n_k}),S_b(w_{2m_k},w_{2m_k},w_{2n_k})\}\hfill \\ & \le 2bmax\{S_b(z_{2m_k},z_{2m_k},z_{2m_k+2}),S_b(w_{2m_k},w_{2m_k},w_{2m_k+2})\}\hfill \\ & +bmax\{S_b(z_{2n_k},z_{2n_k},z_{2m_k+2}),S_b(w_{2n_k+2},w_{2n_k},w_{2m_k+2})\}\hfill \\ & \le 2b\left(2bmax\{S_b(z_{2m_k},z_{2m_k},z_{2m_k+1}),S_b(w_{2m_k},w_{2m_k},w_{2m_k+1})\}\right)\hfill \\ & +2b\left(bmax\{S_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2m_k+1})\}\right)\hfill \\ & +b\left(2bmax\{S_b(z_{2n_k},z_{2n_k},z_{2n_k+1}),S_b(w_{2n_k},w_{2n_k},w_{2n_k+1})\}\right)\hfill \\ & +b\left(bmax\{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k+1})\}\right)\hfill \\ & \le 4b^{3}max\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ & +2b^{2}max\{S_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2m_k+1})\}\hfill \\ & +2b^{3}max\{S_b(z_{2n_k+1},z_{2n_k+1},z_{2n_k}),S_b(w_{2n_k+1},w_{2n_k+1},w_{2n_k})\}\hfill \\ & +b^{2}max\{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k+1})\}.\hfill \end{array}$$

(6)

Now, from Equation (4), we have

$$\begin{array}{c}2b^5{S}_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k+1})\hfill \\ \hspace{1em}\hspace{1em}\le \varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k}),S_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1}),\hfill \\ S_b(w_{2m_k+2},w_{2m_k+2},w_{2m_k+1}),S_b(z_{2n_k+1},z_{2n_k+1},z_{2n_k}),S_b(w_{2n_k+1},w_{2n_k+1},w_{2n_k}),\hfill \\ {\displaystyle \frac{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k})S_b(z_{2n_k+1},z_{2n_k+1},z_{2m_k+1})}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}},\hfill \\ {\displaystyle \frac{S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k})S_b(w_{2n_k+1},w_{2n_k+1},w_{2m_k+1})}{1+S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})}}\hfill \end{array}\right\}\right).\hfill \end{array}$$

Similarly,

$$\begin{array}{c}2b^5{S}_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k+1})\hfill \\ \hfill \hspace{1em}\le \varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k}),S_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1}),\hfill \\ S_b(w_{2m_k+2},w_{2m_k+2},w_{2m_k+1}),S_b(z_{2n_k+1},z_{2n_k+1},z_{2n_k}),S_b(w_{2n_k+1},w_{2n_k+1},w_{2n_k}),\hfill \\ {\displaystyle \frac{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k})S_b(z_{2n_k+1},z_{2n_k+1},z_{2m_k+1})}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}},\hfill \\ {\displaystyle \frac{S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k})S_b(w_{2n_k+1},w_{2n_k+1},w_{2m_k+1})}{1+S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})}}\hfill \end{array}\right\}\right).\end{array}$$

Thus,

$$\begin{array}{c}\hfill 2b^{5}max\left\{\begin{array}{c}{S}_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k+1})\hfill \end{array}\right\}\end{array}$$

$$\begin{array}{c}\le \varphi \left(max\left\{\begin{array}{c}{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k}),\hfill \\ S_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2m_k+1}),\hfill \\ S_b(z_{2n_k+1},z_{2n_k+1},z_{2n_k}),S_b(w_{2n_k+1},w_{2n_k+1},w_{2n_k}),\hfill \\ {\displaystyle \frac{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k})S_b(z_{2n_k+1},z_{2n_k+1},z_{2m_k+1})}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}},\hfill \\ {\displaystyle \frac{S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k})S_b(w_{2n_k+1},w_{2n_k+1},w_{2m_k+1})}{1+S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})}}\hfill \end{array}\right\}\right).\hfill \end{array}$$

(7)

However,

$$\begin{array}{c}max\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})\}\hfill \\ \le 2bmax\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ +bmax\{S_b(z_{2n_k},z_{2n_k},z_{2m_k}),S_b(w_{2n_k},w_{2n_k},w_{2m_k})\}\hfill \\ \le 2bmax\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ +b^{2}max\{S_b(z_{2m_k},z_{2m_k},z_{2n_k}),S_b(w_{2m_k},w_{2m_k},w_{2n_k})\}\hfill \\ \le 2bmax\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ +b^2\left(2bmax\{S_b(z_{2m_k},z_{2m_k},z_{2n_{k-2}}),S_b(w_{2m_k},w_{2m_k},w_{2n_{k-2}})\}\right)\hfill \\ +b^2\left(bmax\{S_b(z_{2n_k},z_{2n_k},z_{2n_{k-2}}),S_b(w_{2n_k},w_{2n_k},w_{2n_{k-2}})\}\right)\hfill \\ <2bmax\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ +2b^3ϵ+b^3\left(2bmax\{S_b(z_{2n_k},z_{2n_k},z_{2n_k-1}),S_b(w_{2n_k},w_{2n_k},w_{2n_k-1})\}\right)\hfill \\ +b^3\left(bmax\{S_b(z_{2n_{k-2}},z_{2n_k-2},z_{2n_k-1}),S_b(w_{2n_k-2},w_{2n_k-2},w_{2n_k-1})\}\right)\hfill \\ \le 2bmax\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2m_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2m_k})\}\hfill \\ +2b^3ϵ+2b^{4}max\{S_b(z_{2n_k},z_{2n_k},z_{2n_k-1}),S_b(w_{2n_k},w_{2n_k},w_{2n_k-1})\}\hfill \\ +b^{5}max\{S_b(z_{2n_k-1},z_{2n_k-1},z_{2n_k-2}),S_b(w_{2n_k-1},w_{2n_k-1},w_{2n_k-2})\}.\hfill \end{array}$$

Letting $k\to \infty$, we have

$$\underset{k\to \infty }{lim}max\{S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k}),S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})\}\le 2b^3ϵ.$$

In addition,

$$\begin{array}{cc}\underset{k\to \infty }{lim}& {\displaystyle \frac{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k})S_b(z_{2n_k+1},z_{2n_k+1},z_{2m_k+1})}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}}\hfill \\ & \le \underset{k\to \infty }{lim}{\displaystyle \frac{1}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}}[2b{S}_b(z_{2m_k+2},z_{2m_k+2},z_{2m_k+1})+b{S}_b(z_{2n_k},z_{2n_k},z_{2m_k+1})]·\hfill \\ & [2b{S}_b(z_{2n_k+1},z_{2n_k+1},z_{2n_k})+b{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})]\hfill \\ & \le \underset{k\to \infty }{lim}{\displaystyle \frac{b^3{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}{1+S_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})}}\hfill \\ & \le \underset{k\to \infty }{lim}{b}^3{S}_b(z_{2m_k+1},z_{2m_k+1},z_{2n_k})\hfill \\ & \le 2b^6ϵ.\hfill \end{array}$$

Similarly,

$$\begin{array}{c}\hfill \underset{k\to \infty }{lim}{\displaystyle \frac{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k})S_b(w_{2n_k+1},w_{2n_k+1},w_{2m_k+1})}{1+S_b(w_{2m_k+1},w_{2m_k+1},w_{2n_k})}}\le 2b^6ϵ.\end{array}$$

Letting $k\to \infty$ in Equation (7), we have

$$\begin{array}{cc}\underset{k\to \infty }{lim}\hfill & max\{S_b(z_{2m_k+2},z_{2m_k+2},z_{2n_k+1}),S_b(w_{2m_k+2},w_{2m_k+2},w_{2n_k+1})\}\hfill \\ & \le {\displaystyle \frac{1}{2b^5}}\varphi \left(max\{2b^3ϵ,0,0,0,0,2b^6ϵ,2b^6ϵ\}\right)\hfill \\ & ={\displaystyle \frac{1}{2b^5}}\varphi \left(2b^6ϵ\right).\hfill \end{array}$$

(8)

Now, letting $n\to \infty$ in Equation (6), from Equations (2), (3) and (8), we have

$$ϵ\le 0+0+0+b^2{\displaystyle \frac{1}{2b^5}}\varphi \left(2b^6ϵ\right)<ϵ.$$

It is a contradiction. Hence, $\left\{z_{2n}\right\}$ and $\left\{w_{2n}\right\}$ are $S_b$-Cauchy sequences in $(X,S_b)$.

In addition,

$$\begin{array}{c}max\{S_b(z_{2n+1},z_{2n+1},z_{2m+1}),S_b(w_{2n+1},w_{2n+1},w_{2m+1})\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le 2bmax\{S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n})\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+bmax\{S_b(z_{2m+1},z_{2m+1},z_{2n}),S_b(w_{2m+1},w_{2m+1},w_{2n})\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\le 2bmax\{S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n})\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}+2b^{2}max\{S_b(z_{2m+1},z_{2m+1},z_{2m}),S_b(w_{2m+1},w_{2m+1},w_{2m})\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+b^{2}max\{S_b(z_{2n},z_{2n},z_{2m}),S_b(w_{2n},w_{2n},w_{2m})\}.\hfill \end{array}$$

Since $\left\{z_{2n}\right\}$ and $\left\{w_{2n}\right\}$ are $S_b$-Cauchy sequences, from Equations (2) and (3), it follows that $\left\{z_{2n+1}\right\}$ and $\left\{w_{2n+1}\right\}$ are also $S_b$-Cauchy sequences in $(X,S_b)$. Hence, $\left\{z_n\right\}$ and $\left\{w_n\right\}$ are $S_b$-Cauchy sequences in $(X,S_b)$.

Suppose $P\left(X\right)$ is a $S_b$-complete subspace of $(X,S_b)$. Then, the sequences $\left\{z_n\right\}$ and $\left\{w_n\right\}$ converge to α and β in $P\left(X\right)$. Thus, there exist a and b in $P\left(X\right)$ such that

$$\underset{n\to \infty }{lim}{z}_n=\alpha =P\left(a\right)and\underset{n\to \infty }{lim}{w}_n=\beta =P\left(b\right).$$

(9)

Now, we have to prove that $A(a,b)=\alpha$ and $A(b,a)=\beta$. On the contrary, suppose that $A(a,b)\ne \alpha$ or $A(b,a)\ne \beta$.

From Equation (4) and Lemma 3, we obtain that

$$\begin{array}{c}{\displaystyle \frac{1}{2b}}{S}_b(A(a,b),A(a,b),\alpha )\hfill \\ \hspace{1em} \le \underset{n\to \infty }{lim}inf2b^5{S}_b(A(a,b),A(a,b),B(x_{2n+1},y_{2n+1}))\hfill \\ \hspace{1em} \le \underset{n\to \infty }{lim}inf\varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(a\right),P\left(a\right),Q\left(x_{2n+1}\right)),S_b(P\left(b\right),P\left(b\right),Q\left(y_{2n+1}\right)),\\ S_b(A(a,b),A(a,b),P\left(a\right)),S_b(A(b,a),A(b,a),P\left(b\right)),\\ S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),Q\left(x_{2n+1}\right)),\\ S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),Q\left(y_{2n+1}\right)),\\ {\displaystyle \frac{S_b(A(a,b),A(a,b),Q\left(x_{2n+1}\right))S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),P\left(a\right))}{1+S_b(P\left(a\right),P\left(a\right),Q\left(x_{2n+1}\right))}},\\ {\displaystyle \frac{S_b(A(b,a),A(b,a),Q\left(y_{2n+1}\right))S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),P\left(b\right))}{1+S_b(P\left(b\right),P\left(b\right),Q\left(y_{2n+1}\right))}}\end{array}\right\}\right)\hfill \\ \hspace{1em} \le \underset{n\to \infty }{lim}inf\varphi \left(max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,z_{2n}),S_b(\beta ,\beta ,w_{2n}),\\ S_b(A(a,b),A(a,b),\alpha ),S_b(A(b,a),A(b,a),\beta ),\\ S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n}),\\ {\displaystyle \frac{S_b(A(a,b),A(a,b),Q\left(x_{2n+1}\right))S_b(z_{2n+1},z_{2n+1},\alpha )}{1+S_b(\alpha ,\alpha ,Q\left(x_{2n+1}\right))}},\\ {\displaystyle \frac{S_b(A(b,a),A(b,a),w_{2n})S_b(w_{2n+1},w_{2n+1},\beta )}{1+S_b(\beta ,\beta ,Q\left(y_{2n+1}\right))}}\end{array}\right\}\right)\hfill \\ \hspace{1em} =\varphi \left(max\left\{\begin{array}{c}0,0,S_b(A(a,b),A(a,b),\alpha ),S_b(A(b,a),A(b,a),\beta ),0,0,0,0\end{array}\right\}\right)\hfill \\ \hspace{1em} =\varphi \left(max\left\{\begin{array}{c}{S}_b(A(a,b),A(a,b),\alpha ),S_b(A(b,a),A(b,a),\beta )\end{array}\right\}\right).\hfill \end{array}$$

Similarly,

$${\displaystyle \frac{1}{2b}}{S}_b(A(b,a),A(b,a),\beta )\le \varphi \left(max\left\{\begin{array}{c}{S}_b(A(a,b),A(a,b),\alpha ),S_b(A(b,a),A(b,a),\beta )\end{array}\right\}\right).$$

Thus,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2b}}max\left\{\begin{array}{c}{S}_b(A(a,b),A(a,b),\alpha ),\\ S_b(A(b,a),A(b,a),\beta )\end{array}\right\}\le \varphi \left(max\left\{\begin{array}{c}{S}_b(A(a,b),A(a,b),\alpha ),\\ S_b(A(b,a),A(b,a),\beta )\end{array}\right\}\right).\end{array}$$

By the definition of ϕ, it follows that $A(a,b)=\alpha =P\left(a\right)$ and $A(b,a)=\beta =P\left(b\right)$. Since $(A,P)$ is w-compatible pair, we have that $A(\alpha ,\beta )=P\left(\alpha \right)$ and $A(\beta ,\alpha )=P\left(\beta \right)$.

From Equation (4) and Lemma 3, we have

$$\begin{array}{c}{\displaystyle \frac{1}{2b}}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha )\hfill \\ \hspace{1em}\le \underset{n\to \infty }{lim}sup2b^5{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B(x_{2n+1},y_{2n+1}))\hfill \\ \hspace{1em}\le \underset{n\to \infty }{lim}sup\varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(x_{2n+1}\right)),S_b(P\left(\beta \right),P\left(\beta \right),Q\left(y_{2n+1}\right)),\\ S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),P\left(\alpha \right)),\\ S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),P\left(\beta \right)),\\ S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),Q\left(x_{2n+1}\right)),\\ S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),Q\left(y_{2n+1}\right)),\\ {\displaystyle \frac{S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),Q\left(x_{2n+1}\right))S_b(B(x_{2n+1},y_{2n+1}),B(x_{2n+1},y_{2n+1}),P\left(\alpha \right))}{1+S_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(x_{2n+1}\right))}},\\ {\displaystyle \frac{S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),Q\left(y_{2n+1}\right))S_b(B(y_{2n+1},x_{2n+1}),B(y_{2n+1},x_{2n+1}),P\left(\beta \right))}{1+S_b(P\left(\beta \right),P\left(\beta \right),Q\left(y_{2n+1}\right))}}\end{array}\right\}\right)\hfill \\ \hspace{1em}=\underset{n\to \infty }{lim}sup\varphi \left(max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),z_{2n}),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),w_{2n}),\\ 0,0,S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n}),\\ {\displaystyle \frac{S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),z_{2n})S_b(z_{2n+1},z_{2n+1},A(\alpha ,\beta ))}{1+S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),z_{2n})}},\\ {\displaystyle \frac{S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),w_{2n})S_b(w_{2n+1},w_{2n+1},A(\beta ,\alpha ))}{1+S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),w_{2n})}}\end{array}\right\}\right)\hfill \\ \hspace{1em}\le \underset{n\to \infty }{lim}sup\varphi \left(max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),z_{2n}),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),w_{2n}),\\ S_b(z_{2n+1},z_{2n+1},z_{2n}),S_b(w_{2n+1},w_{2n+1},w_{2n}),\\ S_b(z_{2n+1},z_{2n+1},A(\alpha ,\beta )),S_b(w_{2n+1},w_{2n+1},A(\beta ,\alpha ))\end{array}\right\}\right)\hfill \\ \hspace{1em}\le \varphi \left(max\left\{\begin{array}{c}2b{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha ),2b{S}_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta ),\\ 0,0,b^2{S}_b(\alpha ,\alpha ,A(\alpha ,\beta )),b^2{S}_b(\beta ,\beta ,A(\beta ,\alpha ))\end{array}\right\}\right)\hfill \\ \hspace{1em}\le \varphi \left(2b^{2}max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha ),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta )\end{array}\right\}\right).\hfill \end{array}$$

Similarly,

$${\displaystyle \frac{1}{2b}}{S}_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta )\le \varphi \left(2b^{2}max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha ),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta )\end{array}\right\}\right).$$

Thus,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2b}}max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha ),\\ S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta )\end{array}\right\}\le \varphi \left(2b^{2}max\left\{\begin{array}{c}{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),\alpha ),\\ S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),\beta )\end{array}\right\}\right).\end{array}$$

By the definition of ϕ, it follows that $A(\alpha ,\beta )=\alpha =P\left(\alpha \right)$ and $A(\beta ,\alpha )=\beta =P\left(\beta \right)$.

Therefore, $(\alpha ,\beta )$ is a common coupled fixed point of A and P.

Since $A(X\times X)\subseteq Q\left(X\right)$, there exist x and y in X such that $A(\alpha ,\beta )=\alpha =Q\left(x\right)$ and $A(\beta ,\alpha )=\beta =Q\left(y\right)$.

From Equation (4), we have

$$\begin{array}{cc}\hfill S_b(\alpha ,\alpha ,B(x,y))& =S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B(x,y))\hfill \\ & \le 2b^5{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B(x,y))\hfill \\ & \le \varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(x\right)),S_b(P\left(\beta \right),P\left(\beta \right),Q\left(y\right)),\\ S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),P\left(\alpha \right)),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),P\left(\beta \right)),\\ S_b(B(x,y),B(x,y),Q\left(x\right)),S_b(B(y,x),B(y,x),Q\left(y\right)),\\ {\displaystyle \frac{S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),Q\left(x\right))S_b(B(x,y),B(x,y),P\left(\alpha \right))}{1+S_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(x\right))}},\\ {\displaystyle \frac{S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),Q\left(y\right))S_b(B(y,x),B(y,x),P\left(\beta \right))}{1+S_b(P\left(\beta \right),P\left(\beta \right),Q\left(y\right))}}\end{array}\right\}\right)\hfill \\ & =\varphi \left(max\left\{\begin{array}{c}0,0,0,0,S_b(B(x,y),B(x,y),\alpha ),S_b(B(y,x),B(y,x),\beta ),0,0\end{array}\right\}\right)\hfill \\ & \le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(x,y)),S_b(\beta ,\beta ,B(y,x))\end{array}\right\}\right).\hfill \end{array}$$

Similarly,

$$S_b(\beta ,\beta ,B(y,x))\le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(x,y)),S_b(\beta ,\beta ,B(y,x))\end{array}\right\}\right).$$

Thus,

$$\begin{array}{c}\hfill max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(x,y)),S_b(\beta ,\beta ,B(y,x))\end{array}\right\}\le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(x,y)),S_b(\beta ,\beta ,B(y,x))\end{array}\right\}\right).\end{array}$$

It follows that $B(x,y)=\alpha =Q\left(x\right)$ and $B(y,x)=\beta =Q\left(y\right)$.

Since $(B,Q)$ is w-compatible pair, we have $B(\alpha ,\beta )=Q\left(\alpha \right)$ and $B(\beta ,\alpha )=Q\left(\beta \right)$.

From Equation (4), we have

$$\begin{array}{cc}\hfill S_b(\alpha ,\alpha ,B(\alpha ,\beta ))& =S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B(\alpha ,\beta ))\hfill \\ & \le 2b^5{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B(\alpha ,\beta ))\hfill \\ & \le \varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(\alpha \right)),S_b(P\left(\beta \right),P\left(\beta \right),Q\left(\beta \right)),\\ S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),P\left(\alpha \right)),S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),P\left(\beta \right)),\\ S_b(B(\alpha ,\beta ),B(\alpha ,\beta ),Q\left(\alpha \right)),S_b(B(\beta ,\alpha ),B(\beta ,\alpha ),Q\left(\beta \right)),\\ {\displaystyle \frac{S_b(A(\alpha ,\beta ),A(\alpha ,\beta ),Q\left(\alpha \right))S_b(B(\alpha ,\beta ),B(\alpha ,\beta ),P\left(\alpha \right))}{1+S_b(P\left(\alpha \right),P\left(\alpha \right),Q\left(\alpha \right))}},\\ {\displaystyle \frac{S_b(A(\beta ,\alpha ),A(\beta ,\alpha ),Q\left(\beta \right))S_b(B(\beta ,\alpha ),B(\beta ,\alpha ),P\left(\beta \right))}{1+S_b(P\left(\beta \right),P\left(\beta \right),Q\left(\beta \right))}}\end{array}\right\}\right)\hfill \\ & =\varphi \left(max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(\alpha ,\beta )),S_b(\beta ,\beta ,B(\beta ,\alpha )),\\ S_b(B(\alpha ,\beta ),B(\alpha ,\beta ),\alpha ),S_b(B(\beta ,\alpha ),B(\beta ,\alpha ),\beta )\end{array}\right\}\right)\hfill \\ & \le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(\alpha ,\beta )),S_b(\beta ,\beta ,B(\beta ,\alpha ))\end{array}\right\}\right).\hfill \end{array}$$

Similarly,

$$S_b(\beta ,\beta ,B(\beta ,\alpha ))\le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(\alpha ,\beta )),S_b(\beta ,\beta ,B(\beta ,\alpha ))\end{array}\right\}\right).$$

Thus,

$$\begin{array}{c}\hfill max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(\alpha ,\beta )),S_b(\beta ,\beta ,B(\beta ,\alpha ))\end{array}\right\}\le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,B(\alpha ,\beta )),S_b(\beta ,\beta ,B(\beta ,\alpha ))\end{array}\right\}\right).\end{array}$$

It follows that $B(\alpha ,\beta )=\alpha =Q\left(\alpha \right)$ and $B(\beta ,\alpha )=\beta =Q\left(\beta \right)$.

Therefore, $(\alpha ,\beta )$ is a common coupled fixed point of $A,B,P$ and Q.

To prove uniqueness, let us take that $({\alpha }^1,{\beta }^1)$ is another common coupled fixed point of $A,B,P$ and Q.

From Equation (4), we have

$$\begin{array}{cc}\hfill S_b(\alpha ,\alpha ,{\alpha }^1)& \le 2b^5{S}_b(\alpha ,\alpha ,{\alpha }^1)\hfill \\ & =2b^5{S}_b(A(\alpha ,\beta ),A(\alpha ,\beta ),B({\alpha }^1,{\beta }^1))\hfill \\ & \le \varphi \left(max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,{\alpha }^1),S_b(\beta ,\beta ,{\beta }^1),S_b(\alpha ,\alpha ,\alpha ),\\ S_b(\beta ,\beta ,\beta ),S_b({\alpha }^1,{\alpha }^1,{\alpha }^1),S_b({\beta }^1,{\beta }^1,{\beta }^1),\\ {\displaystyle \frac{S_b(\alpha ,\alpha ,{\alpha }^1)S_b({\alpha }^1,{\alpha }^1,\alpha )}{1+S_b(\alpha ,\alpha ,{\alpha }^1)}},{\displaystyle \frac{S_b(\beta ,\beta ,{\beta }^1)S_b({\beta }^1,{\beta }^1,\beta )}{1+S_b(\beta ,\beta ,{\beta }^1)}}\end{array}\right\}\right)\hfill \\ & \le \varphi (bmax\{S_b(\alpha ,\alpha ,{\alpha }^1),S_b(\beta ,\beta ,{\beta }^1)\}).\hfill \end{array}$$

Similarly,

$$S_b(\beta ,\beta ,{\beta }^1)\le \varphi (max\{b{S}_b(\alpha ,\alpha ,{\alpha }^1),b{S}_b(\beta ,\beta ,{\beta }^1)\}).$$

Thus,

$$\begin{array}{c}\hfill max\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,{\alpha }^1),S_b(\beta ,\beta ,{\beta }^1)\end{array}\right\}\le \varphi \left(bmax\left\{\begin{array}{c}{S}_b(\alpha ,\alpha ,{\alpha }^1),S_b(\beta ,\beta ,{\beta }^1)\end{array}\right\}\right).\end{array}$$

It follows that $\alpha ={\alpha }^1$ and $\beta ={\beta }^1$. Hence, $(\alpha ,\beta )$ is a unique common coupled fixed point of $A,B,P$ and Q.

 □

Now, we give one example which support our main theoretical result.

Example 2.

Let $X=[0,1]$ and let $S_b:X\times X\times X\to {\mathbb{R}}^{+}$ be defined by $S_b(x,y,z)={\left(\right|y+z-2x|+|y-z\left|\right)}^2$. Then, $S_b$ is a $S_b$-metric space with $b=4$. Define $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ by $\varphi \left(t\right)={\displaystyle \frac{t}{4^6}}$, $A,B:X\times X\to X$ and $P,Q:X\to X$ by $A(x,y)={\displaystyle \frac{x^2+y^2}{4^8}},$ $B(x,y)={\displaystyle \frac{x^2+y^2}{4^9}},$ $P\left(x\right)={\displaystyle \frac{x^2}{4}}$ and $Q\left(x\right)={\displaystyle \frac{x^2}{16}}$, respectively. Then, we have

$$\begin{array}{c}\hspace{1em}2b^5{S}_b(A(x,y),A(x,y),B(u,v))\hfill \\ =2\left(4^5\right){\left(\right|A(x,y)+B(u,v)-2A(x,y)|+|A(x,y)-B(u,v)\left|\right)}^2\hfill \\ =2\left(4^5\right)\left(2\right|A(x,y)-{B(u,v)\left|\right)}^2\hfill \\ =2\left(4^6\right)\left(\right|A(x,y)-{B(u,v)\left|\right)}^2\hfill \\ =2\left(4^6\right){\left|{\displaystyle \frac{x^2+y^2}{4^8}}-{\displaystyle \frac{u^2+v^2}{4^9}}\right|}^2\hfill \\ =2\left(4^6\right){\left|{\displaystyle \frac{4x^2-u^2}{4^9}}+{\displaystyle \frac{4y^2-v^2}{4^9}}\right|}^2\hfill \\ ={\displaystyle \frac{2\left(4^6\right)}{{\left(4^6\right)}^2}}{\left({\displaystyle \frac{1}{4}}\left\{\left|{\displaystyle \frac{4x^2-u^2}{16}}\right|+\left|{\displaystyle \frac{4y^2-v^2}{16}}\right|\right\}\right)}^2\hfill \\ \le {\displaystyle \frac{2}{4\left(4^6\right)}}{\left({\displaystyle \frac{1}{2}}\left\{\left|{\displaystyle \frac{4x^2-u^2}{16}}\right|+\left|{\displaystyle \frac{4y^2-v^2}{16}}\right|\right\}\right)}^2\hfill \\ \le {\displaystyle \frac{2}{4^7}}{\left(max\left\{\left|{\displaystyle \frac{4x^2-u^2}{16}}\right|,\left|{\displaystyle \frac{4y^2-v^2}{16}}\right|\right\}\right)}^2\hfill \\ \le {\displaystyle \frac{2}{\left(4^7\right)4}}{\left(max\left\{4\left|{\displaystyle \frac{4x^2-u^2}{16}}\right|,4\left|{\displaystyle \frac{4y^2-v^2}{16}}\right|\right\}\right)}^2\hfill \\ ={\displaystyle \frac{1}{2\left(4^7\right)}}max\left\{S_b(P\left(x\right),P\left(x\right),Q\left(u\right)),S_b(P\left(y\right),P\left(y\right),Q\left(v\right))\right\}\hfill \\ \le {\displaystyle \frac{1}{2\left(4^7\right)}}max\left\{\begin{array}{c}{S}_b(P\left(x\right),P\left(x\right),Q\left(u\right)),S_b(P\left(y\right),P\left(y\right),Q\left(v\right)),S_b(A(x,y),A(x,y),P\left(x\right)),\\ S_b(A(y,x),A(y,x),P\left(y\right)),S_b(B(u,v),B(u,v),Q\left(u\right)),S_b(B(v,u),B(v,u),Q\left(v\right)),\\ {\displaystyle \frac{S_b(A(x,y),A(x,y),Q\left(u\right))S_b(B(u,v),B(u,v),P\left(x\right))}{1+S_b(P\left(x\right),P\left(x\right),Q\left(u\right))}},{\displaystyle \frac{S_b(A(y,x),A(y,x),Q\left(v\right))S_b(B(v,u),B(v,u),P\left(y\right))}{1+S_b(P\left(y\right),P\left(y\right),Q\left(v\right))}}\end{array}\right\}\hfill \\ <\varphi \left(max\left\{\begin{array}{c}{S}_b(P\left(x\right),P\left(x\right),Q\left(u\right)),S_b(P\left(y\right),P\left(y\right),Q\left(v\right)),S_b(A(x,y),A(x,y),P\left(x\right)),\\ S_b(A(y,x),A(y,x),P\left(y\right)),S_b(B(u,v),B(u,v),Q\left(u\right)),S_b(B(v,u),B(v,u),Q\left(v\right)),\\ {\displaystyle \frac{S_b(A(x,y),A(x,y),Q\left(u\right))S_b(B(u,v),B(u,v),P\left(x\right))}{1+S_b(P\left(x\right),P\left(x\right),Q\left(u\right))}},{\displaystyle \frac{S_b(A(y,x),A(y,x),Q\left(v\right))S_b(B(v,u),B(v,u),P\left(y\right))}{1+S_b(P\left(y\right),P\left(y\right),Q\left(v\right))}}\end{array}\right\}\right).\hfill \end{array}$$

It is clear that all conditions of Theorem 1 are satisfied and $(0,0)$ is a unique common coupled fixed point of $A,B,P$ and Q.

Putting $A=B=P=Q$ in Theorem 1, we obtain the next important result on unique fixed point.

Theorem 2.

Let $(X,S_b)$ be a complete $S_b$-metric space. Suppose that $A:X\times X\to X$ satisfies condition

$$\begin{array}{c}2b^5{S}_b(A(x,y),A(x,y),A(u,v))\hfill \\ \hspace{1em}\le \varphi \left(max\left\{\begin{array}{c}{S}_b(x,x,u),S_b(y,y,v),S_b(A(x,y),A(x,y),x),\\ S_b(A(y,x),A(y,x),y),S_b(A(u,v),A(u,v),u),S_b(A(v,u),A(v,u),v),\\ {\displaystyle \frac{S_b(A(x,y),A(x,y),u)S_b(B(u,v),B(u,v),x)}{1+S_b(x,x,u)}},{\displaystyle \frac{S_b(A(y,x),A(y,x),v)S_b(B(v,u),B(v,u),y)}{1+S_b(y,y,v)}}\end{array}\right\}\right)\hfill \end{array}$$

for all $x,y,u,v\in X$, $\varphi \in \Phi$. Then, A has a unique coupled fixed point in $X\times X$.

3. Application

In this section, we study the existence of a unique solution to an initial value problem, as an application to Theorem 2. Consider the initial value problem:

$$x^1\left(t\right)=f(t,x\left(t\right),x\left(t\right)),t\in I=[0,1],x\left(0\right)=x_0$$

(10)

where $f:I\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\to \left[{\displaystyle \frac{x_0}{4}},\infty \right)$ and $x_0\in \mathbb{R}.$

Theorem 3.

Consider the initial value problem in Equation (10) with $f\in C\left(I\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\right)$ and

$$\begin{array}{c}\underset{0}{\overset{t}{\int }}f(s,x(s),y(s\left)\right)ds={\displaystyle \frac{1}{\sqrt{5}}}min\left\{\begin{array}{c}\underset{0}{\overset{t}{\int }}f(s,x(s),x(s\left)\right)ds,\underset{0}{\overset{t}{\int }}f(s,y(s),y(s\left)\right)ds\end{array}\right\}.\hfill \end{array}$$

Then, there exists a unique solution in $C\left(I\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\right)$ for the initial value problem in Equation (10).

Proof of Theorem.

The integral equation corresponding to the initial value problem in Equation (10) is

$$\begin{array}{c}x\left(t\right)=x_0+\underset{0}{\overset{t}{\int }}f(s,x(s),x(s\left)\right)ds.\end{array}$$

Let $X=C\left(I\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\times \left[{\displaystyle \frac{x_0}{4}},\infty \right)\right)$ and let $S(x,y,z)={\left(\right|y+z-2x|+|y-z\left|\right)}^{2}forx,y\in X$. Define $\varphi :[0,\infty )\to [0,\infty )$ by $\varphi \left(t\right)={\displaystyle \frac{4t}{5}}$ and $A:X\times X\to X$ by

$$\begin{array}{ccc}\hfill A(x,y)\left(t\right)& =& x_0+\underset{0}{\overset{t}{\int }}f(s,x(s),y(s\left)\right)ds.\hfill \end{array}$$

(11)

Now, we have

$$\begin{array}{c}S\left(A\left(x,y\right)\left(t\right),A\left(x,y\right)\left(t\right),A\left(u,v\right)\left(t\right)\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\begin{array}{c}={\left\{\left|A(x,y)\left(t\right)+A(u,v)\left(t\right)-2A(x,y)\left(t\right)\right|+\left|A(x,y)\left(t\right)-A(u,v)\left(t\right)\right|\right\}}^2\hfill \\ =4{\left|A(x,y)\left(t\right)-A(u,v)\left(t\right)\right|}^2\hfill \\ =4{\left|\underset{0}{\overset{t}{\int }}f(s,x(s),y(s\left)\right)ds-\underset{0}{\overset{t}{\int }}f(s,u(s),v(s\left)\right)ds\right|}^2\hfill \\ =4{\left|{\displaystyle \frac{1}{\sqrt{5}}}min\left\{\begin{array}{c}\underset{0}{\overset{t}{\int }}f(s,x(s),x(s\left)\right)ds,\\ \underset{0}{\overset{t}{\int }}f(s,y(s),y(s\left)\right)ds\end{array}\right\}-{\displaystyle \frac{1}{\sqrt{5}}}min\left\{\begin{array}{c}\underset{0}{\overset{t}{\int }}f(s,u(s),u(s\left)\right)ds,\\ \underset{0}{\overset{t}{\int }}f(s,v(s),v(s\left)\right)ds\end{array}\right\}\right|}^2\hfill \\ \le {\displaystyle \frac{4}{5}}{\left|max\left\{\begin{array}{c}\underset{0}{\overset{t}{\int }}f(s,x(s),x(s\left)\right)ds-\underset{0}{\overset{t}{\int }}f(s,u(s),u(s\left)\right)ds,\\ \underset{0}{\overset{t}{\int }}f(s,y(s),y(s\left)\right)ds-\underset{0}{\overset{t}{\int }}f(s,v(s),v(s\left)\right)ds\end{array}\right\}\right|}^2\hfill \\ ={\displaystyle \frac{4}{5}}max\left\{\begin{array}{c}{\left|\underset{0}{\overset{t}{\int }}f(s,x(s),x(s\left)\right)ds-\underset{0}{\overset{t}{\int }}f(s,u(s),u(s\left)\right)ds\right|}^2,\\ {\left|\underset{0}{\overset{t}{\int }}f(s,y(s),y(s\left)\right)ds-\underset{0}{\overset{t}{\int }}f(s,v(s),v(s\left)\right)ds\right|}^2\end{array}\right\}\hfill \\ ={\displaystyle \frac{1}{5}}max\left\{\begin{array}{c}4{\left|x\left(t\right)-u\left(t\right)\right|}^2,4{\left|y\left(t\right)-v\left(t\right)\right|}^2\end{array}\right\}\hfill \\ ={\displaystyle \frac{1}{5}}max\left\{S(x,x,u),S(y,y,v)\right\}\hfill \\ \le \varphi \left(M\right(x,u,y,v\left)\right).\hfill \end{array}\hfill \end{array}$$

Hence, from Theorem 2, we conclude that A has a unique coupled fixed point in X.  □