Fixed points of $(\psi ,\varphi ,\theta )$-contractive mappings in partially ordered b-metric spaces and application to quadratic integral equations

Abstract

We prove some coupled coincidence and coupled common fixed point theorems for mappings satisfying $(\psi ,\varphi ,\theta )$-contractive conditions in partially ordered complete b-metric spaces. The obtained results extend and improve many existing results from the literature. As an application, we prove the existence of a unique solution to a class of nonlinear quadratic integral equations.

MSC:47H10, 54H25.

1 Introduction and preliminaries

In [1, 2], Czerwik introduced the notion of a b-metric space, which is a generalization of the usual metric space, and generalized the Banach contraction principle in the context of complete b-metric spaces. After that, many authors have carried out further studies on b-metric spaces and their topological properties (see, e.g., [1–14]). In this paper, some coupled coincidence and coupled common fixed point theorems for mappings satisfying $(\psi ,\varphi ,\theta )$-contractive conditions in partially ordered complete b-metric spaces are proved. Also, we apply our results to study the existence of a unique solution to a large class of nonlinear quadratic integral equations. There are many papers in the literature concerning coupled fixed points introduced by Bhaskar and Lakshmikantham [15] and their applications in the existence and uniqueness of solutions for boundary value problems. A number of articles on this topic have been dedicated to the improvement and generalization; see [16–20] and references therein. Also, to see some results on common fixed points for generalized contraction mappings, we refer the reader to [21–23]. For the sake of convenience, some definitions and notations are recalled from [1, 3, 24] and [25].

Definition 1.1 [1]

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

• (i) $d(x,y)=0$ iff $x=y$,

• (ii) $d(x,y)=d(y,x)$,

• (iii) $d(x,y)\le s[d(x,z)+d(z,y)]$.

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

It should be noted that the class of b-metric spaces is effectively larger than that of metric spaces since a b-metric is a metric when $s=1$.

The following example shows that, in general, a b-metric need not necessarily be a metric (see also [14]).

Example 1.2 [3]

Let $(X,d)$ be a metric space and $\rho (x,y)={(d(x,y))}^p$, where $p>1$ is a real number. Then ρ is a b-metric with $s=2^{p-1}$. However, if $(X,d)$ is a metric space, then $(X,\rho )$ is not necessarily a metric space. For example, if $X=\mathbb{R}$ is the set of real numbers and $d(x,y)=|x-y|$ is the usual Euclidean metric, then $\rho (x,y)={(x-y)}^s$ is a b-metric on ℝ with $s=2$, but is not a metric on ℝ.

Also, the following example of a b-metric space is given in [26].

Example 1.3 [26]

Let X be the set of Lebesgue measurable functions on $[0,1]$ such that ${\int }_0^1{|f(x)|}^{2}dx<\mathrm{\infty }$. Define $D:X\times X⟶[0,\mathrm{\infty })$ by $D(f,g)={\int }_0^1{|f(x)-g(x)|}^{2}dx$. As ${({\int }_0^1{|f(x)-g(x)|}^{2}dx)}^{\frac{1}{2}}$ is a metric on X, then, from the previous example, D is a b-metric on X, with $s=2$.

Khamsi [27] also showed that each cone metric space over a normal cone has a b-metric structure.

Since, in general, a b-metric is not continuous, we need the following simple lemma about the b-convergent sequences in the proof of our main result.

Lemma 1.4 [3]

Let $(X,d)$ be a b-metric space with $s\ge 1$, and suppose that $\{x_n\}$ and $\{y_n\}$ are b-convergent to x, y, respectively. Then we have

$$\frac{1}{s^2}d(x,y)\le lim\hspace{0.17em}infd(x_n,y_n)\le lim\hspace{0.17em}supd(x_n,y_n)\le s^{2}d(x,y).$$

In particular, if $x=y$, then we have $limd(x_n,y_n)=0$. Moreover, for each $z\in X$, we have

$$\frac{1}{s}d(x,z)\le lim\hspace{0.17em}infd(x_n,z)\le lim\hspace{0.17em}supd(x_n,z)\le sd(x,z).$$

In [25], Lakshmikantham and Ćirić introduced the concept of mixed g-monotone property as follows.

Definition 1.5 [25]

Let $(X,\le )$ be a partially ordered set and $F:X\times X⟶X$ and $g:X⟶X$. We say F has the mixed g-monotone property if F is non-decreasing g-monotone in its first argument and is non-increasing g-monotone in its second argument, that is, for any $x,y\in X$,

$$x_1,x_2\in X,g{x}_1\le g{x}_2⟹F(x_1,y)\le F(x_2,y)$$

and

$$y_1,y_2\in X,g{y}_1\le g{y}_2⟹F(x,y_1)\ge F(x,y_2).$$

Note that if g is an identity mapping, then F is said to have the mixed monotone property (see also [15]).

Definition 1.6 [25]

An element $(x,y)\in X\times X$ is called a coupled coincidence point of a mapping $F:X\times X⟶X$ and a mapping $g:X⟶X$ if

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

Similarly, note that if g is an identity mapping, then $(x,y)$ is called a coupled fixed point of the mapping F (see also [15]).

Definition 1.7 [24]

An element $x\in X$ is called a common fixed point of a mapping $F:X\times X⟶X$ and $g:X⟶X$ if

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

(1.1)

Definition 1.8 [25]

Let X be a nonempty set and $F:X\times X⟶X$ and $g:X⟶X$. One says that F and g are commutative if for all $x,y\in X$,

$$F(gx,gy)=g(F(x,y)).$$

Definition 1.9 [28]

The mappings F and g, where $F:X\times X⟶X$ and $g:X⟶X$, are said to be compatible if

$$\underset{n⟶\mathrm{\infty }}{lim}d(g(F(x_n,y_n)),F(g{x}_n,g{y}_n))=0$$

and

$$\underset{n⟶\mathrm{\infty }}{lim}d(g(F(y_n,x_n)),F(g{y}_n,g{x}_n))=0,$$

whenever $\{x_n\}$ and $\{y_n\}$ are sequences in X such that ${lim}_{n⟶\mathrm{\infty }}F(x_n,y_n)={lim}_{n⟶\mathrm{\infty }}g{x}_n=x$ and ${lim}_{n⟶\mathrm{\infty }}F(y_n,x_n)={lim}_{n⟶\mathrm{\infty }}g{y}_n=y$ for all $x,y\in X$.

2 Main results

Throughout the paper, let Ψ be a family of all functions $\psi :[0,\mathrm{\infty })⟶[0,\mathrm{\infty })$ satisfying the following conditions:

• (a) ψ is continuous,

• (b) ψ non-decreasing,

• (c) $\psi (t)=0$ if and only if $t=0$.

We denote by Φ the set of all functions $\varphi :[0,\mathrm{\infty })⟶[0,\mathrm{\infty })$ satisfying the following conditions:

• (a) ϕ is lower semi-continuous,

• (b) $\varphi (t)=0$ if and only if $t=0$,

and Θ the set of all continuous functions $\theta :[0,\mathrm{\infty })⟶[0,\mathrm{\infty })$ with $\theta (t)=0$ if and only if $t=0$.

Let $(X,d,\le )$ be a partially ordered b-metric space, and let $T:X\times X⟶X$ and $g:X⟶X$ be two mappings. Set

$$\begin{array}{rcl}{M}_{s,T,g}(x,y,u,v)& =& max\{d(gx,gu),d(gy,gv),d(gx,T(x,y)),\\ \frac{1}{2s}d(gu,T(u,v)),d(gy,T(y,x)),\frac{1}{2s}d(gv,T(v,u)),\\ \frac{d(gx,T(u,v))+d(gu,T(x,y))}{2s},\frac{d(gy,T(v,u))+d(gv,T(y,x))}{2s}\}\end{array}$$

and

$$N_{T,g}(x,y,u,v)=min\{d(gx,T(x,y)),d(gu,T(u,v)),d(gu,T(x,y)),d(gx,T(u,v))\}.$$

Now, we introduce the following definition.

Definition 2.1 Let $(X,d,\le )$ be a partially ordered b-metric space and $\psi \in \mathrm{\Psi }$, $\varphi \in \mathrm{\Phi }$ and $\theta \in \mathrm{\Theta }$. We say that $T:X\times X⟶X$ is an almost generalized $(\psi ,\varphi ,\theta )$-contractive mapping with respect to $g:X⟶X$ if there exists $L\ge 0$ such that

$$\begin{array}{rl}\psi \left(s^{3}d(T(x,y),T(u,v))\right)\le & \psi (M_{s,T,g}(x,y,u,v))\\ -\varphi (M_{s,T,g}(x,y,u,v))+L\theta (N_{T,g}(x,y,u,v))\end{array}$$

(2.1)

for all $x,y,u,v\in X$ with $gx\le gu$ and $gy\ge gv$.

Now, we establish some results for the existence of a coupled coincidence point and a coupled common fixed point of mappings satisfying almost generalized $(\psi ,\varphi ,\theta )$-contractive condition in the setup of partially ordered b-metric spaces. The first result in this paper is the following coupled coincidence theorem.

Theorem 2.2 Suppose that $(X,d,\le )$ is a partially ordered complete b-metric space. Let $T:X\times X⟶X$ be an almost generalized $(\psi ,\varphi ,\theta )$-contractive mapping with respect to $g:X⟶X$, and T and g are continuous such that T has the mixed g-monotone property and commutes with g. Also, suppose $T(X\times X)\subseteq g(X)$. If there exists $(x_0,y_0)\in X\times X$ such that $g{x}_0\le T(x_0,y_0)$ and $g{y}_0\ge T(y_0,x_0)$, then T and g have coupled coincidence point in X.

Proof By the given assumptions, there exists $(x_0,y_0)\in X\times X$ such that $g{x}_0\le T(x_0,y_0)$ and $g{y}_0\ge T(y_0,x_0)$. Since $T(X\times X)\subseteq g(X)$, we can define $(x_1,y_1)\in X\times X$ such that $g{x}_1=T(x_0,y_0)$ and $g{y}_1=T(y_0,x_0)$, then $g{x}_0\le T(x_0,y_0)=g{x}_1$ and $g{y}_0\ge T(y_0,x_0)=g{y}_1$. Also, there exists $(x_2,y_2)\in X\times X$ such that $g{x}_2=T(x_1,y_1)$ and $g{y}_2=T(y_1,x_1)$. Since T has the mixed g-monotone property, we have

$$g{x}_1=T(x_0,y_0)\le T(x_0,y_1)\le T(x_1,y_1)=g{x}_2$$

and

$$g{y}_2=T(y_1,x_1)\le T(y_0,x_1)\le T(y_0,x_0)=g{y}_1.$$

Continuing in this way, we construct two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$g{x}_{n+1}=T(x_n,y_n)\text{and}g{y}_{n+1}=T(y_n,x_n)\text{for all }n=0,1,2,\dots$$

(2.2)

for which

$$\begin{array}{r}g{x}_0\le g{x}_1\le g{x}_2\le \cdots \le g{x}_n\le g{x}_{n+1}\le \cdots ,\\ g{y}_0\ge g{y}_1\ge g{y}_2\ge \cdots \ge g{y}_n\ge g{y}_{n+1}\ge \cdots .\end{array}$$

(2.3)

From (2.2) and (2.3) and inequality (2.1) with $(x,y)=(x_n,y_n)$ and $(u,v)=(x_{n+1},y_{n+1})$, we obtain

$$\begin{array}{rl}\psi (d(g{x}_{n+1},g{x}_{n+2}))\le & \psi (s^{3}d(g{x}_{n+1},g{x}_{n+2}))=\psi \left(s^{3}d(T(x_n,y_n),T(x_{n+1},y_{n+1}))\right)\\ \le & \psi (M_{s,T,g}(x_n,y_n,x_{n+1},y_{n+1}))-\varphi (M_{s,T,g}(x_n,y_n,x_{n+1},y_{n+1}))\\ +L\theta (N_{T,g}(x_n,y_n,x_{n+1},y_{n+1})),\end{array}$$

(2.4)

where

$$\begin{array}{rl}{M}_{s,T,g}(x_n,y_n,x_{n+1},y_{n+1})=& max\{d(g{x}_n,g{x}_{n+1}),d(g{y}_n,g{y}_{n+1}),d(g{x}_n,T(x_n,y_n)),\\ \frac{1}{2s}d(g{x}_{n+1},T(x_{n+1},y_{n+1})),d(g{y}_n,T(y_n,x_n)),\\ \frac{1}{2s}d(g{y}_{n+1},T(y_{n+1},x_{n+1})),\\ \frac{d(g{x}_n,T(x_{n+1},y_{n+1}))+d(g{x}_{n+1},T(x_n,y_n))}{2s},\\ \frac{d(g{y}_n,T(y_{n+1},x_{n+1}))+d(g{y}_{n+1},T(y_n,x_n))}{2s}\}\\ =& max\{d(g{x}_n,g{x}_{n+1}),d(g{y}_n,g{y}_{n+1}),\\ \frac{1}{2s}d(g{x}_{n+1},g{x}_{n+2}),\frac{1}{2s}d(g{y}_{n+1},g{y}_{n+2}),\\ \frac{d(g{x}_n,g{x}_{n+2})}{2s},\frac{d(g{y}_n,g{y}_{n+2})}{2s}\}\end{array}$$

and

$$\begin{array}{rl}{N}_{T,g}(x_n,y_n,x_{n+1},y_{n+1})=& min\{d(g{x}_n,T(x_n,y_n)),d(g{x}_{n+1},T(x_{n+1},y_{n+1})),\\ d(g{x}_{n+1},T(x_n,y_n)),d(g{x}_{n+1},T(x_{n+1},y_{n+1}))\}=0.\end{array}$$

Since

$$\begin{array}{r}\frac{d(g{x}_n,g{x}_{n+2})}{2s}\le \frac{d(g{x}_n,g{x}_{n+1})+d(g{x}_{n+1},g{x}_{n+2})}{2}\le max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2})\}\end{array}$$

and

$$\begin{array}{r}\frac{d(g{y}_n,g{y}_{n+2})}{2s}\le \frac{d(g{y}_n,g{y}_{n+1})+d(g{y}_{n+1},g{y}_{n+2})}{2}\le max\{d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\},\end{array}$$

then we get

$$\begin{array}{r}{M}_{s,T,g}(x_n,y_n,x_{n+1},y_{n+1})\le max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),\\ \phantom{M_{s,T,g}(x_n,y_n,x_{n+1},y_{n+1})\le }d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\},\\ N_{T,g}(x_n,y_n,x_{n+1},y_{n+1})=0.\end{array}$$

(2.5)

By (2.4) and (2.5), we have

$$\begin{array}{r}\psi (d(g{x}_{n+1},g{x}_{n+2}))\\ \le \psi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\})\\ -\varphi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}).\end{array}$$

(2.6)

Similarly, we can show that

$$\begin{array}{r}\psi (d(g{y}_{n+1},g{y}_{n+2}))\\ \le \psi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\})\\ -\varphi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}).\end{array}$$

(2.7)

Now, denote

$${\delta }_n=max\{d(g{x}_n,g{x}_{n+1}),d(g{y}_n,g{y}_{n+1})\}.$$

(2.8)

Combining (2.6), (2.7) and the fact that $max\{\psi (a),\psi (b)\}=\psi (max\{a,b\})$ for $a,b\in [0,+\mathrm{\infty })$, we have

$$\psi ({\delta }_{n+1})=max\{\psi (d(g{x}_{n+1},g{x}_{n+2})),\psi (d(g{y}_{n+1},g{y}_{n+2}))\}.$$

(2.9)

So, using (2.6), (2.7), (2.8) together with (2.9), we obtain

$$\begin{array}{r}\psi ({\delta }_{n+1})\\ \le \psi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\})\\ -\varphi (max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}).\end{array}$$

(2.10)

Now we prove that for all $n\in \mathbb{N}$,

$$\begin{array}{r}max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}\\ ={\delta }_n\text{and}{\delta }_{n+1}\le {\delta }_n.\end{array}$$

(2.11)

For this purpose, consider the following three cases.

Case 1. If $max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}={\delta }_n$, then by (2.10) we have

$$\psi ({\delta }_{n+1})\le \psi ({\delta }_n)-\varphi ({\delta }_n)<\psi ({\delta }_n),$$

(2.12)

so (2.11) obviously holds.

Case 2. If $max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}=d(g{x}_{n+1},g{x}_{n+2})>0$, then by (2.6) we have

$$\psi (d(g{x}_{n+1},g{x}_{n+2}))\le \psi (d(g{x}_{n+1},g{x}_{n+2}))-\varphi (d(g{x}_{n+1},g{x}_{n+2}))<\psi (d(g{x}_{n+1},g{x}_{n+2})),$$

which is a contradiction.

Case 3. If $max\{d(g{x}_n,g{x}_{n+1}),d(g{x}_{n+1},g{x}_{n+2}),d(g{y}_n,g{y}_{n+1}),d(g{y}_{n+1},g{y}_{n+2})\}=d(g{y}_{n+1},g{y}_{n+2})>0$, then from (2.7) we have

$$\begin{array}{r}\psi (d(g{y}_{n+1},g{y}_{n+2}))\le \psi (d(g{y}_{n+1},g{y}_{n+2}))-\varphi (d(g{y}_{n+1},g{y}_{n+2}))<\psi (d(g{y}_{n+1},g{y}_{n+2})),\end{array}$$

which is again a contradiction.

Thus, in all the cases, (2.11) holds for each $n\in \mathbb{N}$. It follows that the sequence $\{{\delta }_n\}$ is a monotone decreasing sequence of nonnegative real numbers and, consequently, there exists $\delta \ge 0$ such that

$$\underset{n⟶\mathrm{\infty }}{lim}{\delta }_n=\delta .$$

(2.13)

We show that $\delta =0$. Suppose, on the contrary, that $\delta >0$. Taking the limit as $n⟶\mathrm{\infty }$ in (2.12) and using the properties of the function ϕ, we get

$$\psi (\delta )\le \psi (\delta )-\varphi (\delta )<\psi (\delta ),$$

which is a contradiction. Therefore $\delta =0$, that is,

$$\underset{n⟶\mathrm{\infty }}{lim}{\delta }_n=\underset{n⟶\mathrm{\infty }}{lim}max\{d(g{x}_n,g{x}_{n+1}),d(g{y}_n,g{y}_{n+1})\}=0,$$

which implies that

$$\underset{n⟶\mathrm{\infty }}{lim}d(g{x}_n,g{x}_{n+1})=0\text{and}\underset{n⟶\mathrm{\infty }}{lim}d(g{y}_n,g{y}_{n+1})=0.$$

(2.14)

Now, we claim that

$$\underset{n,m⟶\mathrm{\infty }}{lim}max\{d(g{x}_n,g{x}_m),d(g{y}_n,g{y}_m)\}=0.$$

(2.15)

Assume, on the contrary, that there exist $ϵ>0$ and subsequences $\{g{x}_{m(k)}\}$, $\{g{x}_{n(k)}\}$ of $\{g{x}_n\}$ and $\{g{y}_{m(k)}\}$, $\{g{y}_{n(k)}\}$ of $\{g{y}_n\}$ with $m(k)>n(k)\ge k$ such that

$$max\{d(g{x}_{n(k)},g{x}_{m(k)}),d(g{y}_{n(k)},g{y}_{m(k)})\}\ge ϵ.$$

(2.16)

Additionally, corresponding to $n(k)$, we may choose $m(k)$ such that it is the smallest integer satisfying (2.16) and $m(k)>n(k)\ge k$. Thus,

$$max\{d(g{x}_{n(k)},g{x}_{m(k)-1}),d(g{y}_{n(k)},g{y}_{m(k)-1})\}<ϵ.$$

(2.17)

Using the triangle inequality in a b-metric space and (2.16) and (2.17), we obtain that

$$\begin{array}{rl}ϵ& \le d(g{x}_{m(k)},g{x}_{n(k)})\le sd(g{x}_{m(k)},g{x}_{m(k)-1})+sd(g{x}_{m(k)-1},g{x}_{n(k)})\\ <sd(g{x}_{m(k)},g{x}_{m(k)-1})+sϵ.\end{array}$$

Taking the upper limit as $k⟶\mathrm{\infty }$ and using (2.14), we obtain

$$ϵ\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)})\le sϵ.$$

(2.18)

Similarly, we obtain

$$ϵ\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{n(k)},g{y}_{m(k)})\le sϵ.$$

(2.19)

Also,

$$\begin{array}{rl}ϵ& \le d(g{x}_{n(k)},g{x}_{m(k)})\le sd(g{x}_{n(k)},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)})\\ \le s^{2}d(g{x}_{n(k)},g{x}_{m(k)})+s^{2}d(g{x}_{m(k)},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)})\\ \le s^{2}d(g{x}_{n(k)},g{x}_{m(k)})+(s^2+s)d(g{x}_{m(k)},g{x}_{m(k)+1}).\end{array}$$

So, from (2.14) and (2.18), we have

$$\frac{ϵ}{s}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)+1})\le s^2ϵ.$$

(2.20)

Similarly, we obtain

$$\frac{ϵ}{s}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{n(k)},g{y}_{m(k)+1})\le s^2ϵ.$$

(2.21)

Also,

$$\begin{array}{rl}ϵ& \le d(g{x}_{m(k)},g{x}_{n(k)})\le sd(g{x}_{m(k)},g{x}_{n(k)+1})+sd(g{x}_{n(k)+1},g{x}_{n(k)})\\ \le s^{2}d(g{x}_{m(k)},g{x}_{n(k)})+s^{2}d(g{x}_{n(k)},g{x}_{n(k)+1})+sd(g{x}_{n(k)+1},g{x}_{n(k)})\\ \le s^{2}d(g{x}_{m(k)},g{x}_{n(k)})+(s^2+s)d(g{x}_{n(k)},g{x}_{n(k)+1}).\end{array}$$

So, from (2.14) and (2.18), we have

$$\frac{ϵ}{s}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{m(k)},g{x}_{n(k)+1})\le s^2ϵ.$$

(2.22)

In a similar way, we obtain

$$\frac{ϵ}{s}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{m(k)},g{y}_{n(k)+1})\le s^2ϵ.$$

(2.23)

Also,

$$d(g{x}_{n(k)+1},g{x}_{m(k)})\le sd(g{x}_{n(k)+1},g{x}_{m(k)+1})+sd(g{x}_{m(k)+1},g{x}_{m(k)}).$$

So, from (2.14) and (2.22), we have

$$\frac{ϵ}{s^2}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)+1},g{x}_{m(k)+1}).$$

(2.24)

Similarly, we obtain

$$\frac{ϵ}{s^2}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{n(k)+1},g{y}_{m(k)+1}).$$

(2.25)

Linking (2.14), (2.18), (2.19), (2.20), (2.21), (2.22) together with (2.23), we get

$$\begin{array}{rl}\frac{ϵ}{s^2}=& min\{ϵ,ϵ,\frac{\frac{ϵ}{s}+\frac{ϵ}{s}}{2s},\frac{\frac{ϵ}{s}+\frac{ϵ}{s}}{2s}\}\\ \le & max\{\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)},g{x}_{m(k)}),\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{n(k)},g{y}_{m(k)}),\\ \frac{{lim\hspace{0.17em}sup}_{k⟶\mathrm{\infty }}d(g{x}_{n(k)},g{x}_{m(k)+1})+{lim\hspace{0.17em}sup}_{k⟶\mathrm{\infty }}d(g{x}_{m(k)},g{x}_{n(k)+1})}{2s},\\ \frac{{lim\hspace{0.17em}sup}_{k⟶\mathrm{\infty }}d(g{y}_{n(k)},g{y}_{m(k)+1})+{lim\hspace{0.17em}sup}_{k⟶\mathrm{\infty }}d(g{y}_{m(k)},g{y}_{n(k)+1})}{2s}\}\\ \le & max\{sϵ,sϵ,\frac{s^2ϵ+s^2ϵ}{2s},\frac{s^2ϵ+s^2ϵ}{2s}\}=sϵ.\end{array}$$

So,

$$\frac{ϵ}{s^2}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})\le ϵs.$$

(2.26)

Similarly, we have

$$\frac{ϵ}{s^2}\le \underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})\le ϵs$$

(2.27)

and

$$\underset{k⟶\mathrm{\infty }}{lim}{N}_{T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})=0.$$

(2.28)

Since $m(k)>n(k)$, from (2.2) we have

$$g{x}_{n(k)}\le g{x}_{m(k)},g{y}_{n(k)}\ge g{y}_{m(k)}.$$

Thus,

$$\begin{array}{r}\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))=\psi \left(s^{3}d(T(x_{n(k)},y_{n(k)}),T(x_{m(k)},y_{m(k)}))\right)\\ \phantom{\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))}\le \psi (M_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ \phantom{\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))=}-\varphi (M_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ \phantom{\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1}))=}+L\theta (N_{T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})),\\ \psi (s^{3}d(g{y}_{n(k)+1},g{y}_{m(k)+1}))=\psi \left(s^{3}d(T(y_{n(k)},x_{n(k)}),T(y_{m(k)},x_{m(k)}))\right)\\ \phantom{\psi (s^{3}d(g{y}_{n(k)+1},g{y}_{m(k)+1}))}\le \psi (M_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ \phantom{\psi (s^{3}d(g{y}_{n(k)+1},g{y}_{m(k)+1}))=}-\varphi (M_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ \phantom{\psi (s^{3}d(g{y}_{n(k)+1},g{y}_{m(k)+1}))=}+L\theta (N_{T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})).\end{array}$$

Since ψ is a non-decreasing function, we have

$$\begin{array}{r}max\{\psi (s^{3}d(g{x}_{n(k)+1},g{x}_{m(k)+1})),\psi (s^{3}d(g{y}_{n(k)+1},g{y}_{m(k)+1}))\}\\ =\psi (s^{3}max\{d(g{x}_{n(k)+1},g{x}_{m(k)+1}),d(g{y}_{n(k)+1},g{y}_{m(k)+1})\}).\end{array}$$

Taking the upper limit as $k⟶\mathrm{\infty }$ and using (2.25) and (2.26), we get

$$\begin{array}{rl}\psi (sϵ)\le & \psi (s^{3}max\{\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{x}_{n(k)+1},g{x}_{m(k)+1}),\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}d(g{y}_{n(k)+1},g{y}_{m(k)+1})\})\\ \le & \psi (\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ -\varphi (\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ +L\theta (\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}{N}_{T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))\\ \le & \psi (sϵ)-\varphi (\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})),\end{array}$$

which implies that

$$\varphi (\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)}))=0,$$

so

$$\underset{k⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{M}_{s,T,g}(x_{n(k)},y_{n(k)},x_{m(k)},y_{m(k)})=0,$$

a contradiction to (2.27). Therefore, (2.15) holds and we have

$$\underset{n,m⟶\mathrm{\infty }}{lim}d(g{x}_n,g{x}_m)=0\text{and}\underset{n,m⟶\mathrm{\infty }}{lim}d(g{y}_n,g{y}_m)=0.$$

Since X is a complete b-metric space, there exist $x,y\in X$ such that

$$\underset{n⟶\mathrm{\infty }}{lim}g{x}_{n+1}=x\text{and}\underset{n⟶\mathrm{\infty }}{lim}g{y}_{n+1}=y.$$

(2.29)

From the commutativity of T and g, we have

$$g(g{x}_{n+1})=g(T(x_n,y_n))=T(g{x}_n,g{y}_n),g(g{y}_{n+1})=g(T(y_n,x_n))=T(g{y}_n,g{x}_n).$$

(2.30)

Now, we shall show that

$$gx=T(x,y)\text{and}gy=T(y,x).$$

Letting $n⟶\mathrm{\infty }$ in (2.30), from the continuity of T and g, we get

$$\begin{array}{r}gx=\underset{n⟶\mathrm{\infty }}{lim}g(g{x}_{n+1})=\underset{n⟶\mathrm{\infty }}{lim}T(g{x}_n,g{y}_n)=T(\underset{n⟶\mathrm{\infty }}{lim}g{x}_n,\underset{n⟶\mathrm{\infty }}{lim}g{y}_n)=T(x,y),\\ gy=\underset{n⟶\mathrm{\infty }}{lim}g(g{y}_{n+1})=\underset{n⟶\mathrm{\infty }}{lim}T(g{y}_n,g{x}_n)=T(\underset{n⟶\mathrm{\infty }}{lim}g{y}_n,\underset{n⟶\mathrm{\infty }}{lim}g{x}_n)=T(y,x).\end{array}$$

This implies that $(x,y)$ is a coupled coincidence point of T and g. This completes the proof. □

Corollary 2.3 Let $(X,d,\le )$ be a partially ordered complete b-metric space, and let $T:X\times X⟶X$ be a continuous mapping such that T has the mixed monotone property. Suppose that there exist $\psi \in \mathrm{\Psi }$, $\varphi \in \mathrm{\Phi }$, $\theta \in \mathrm{\Theta }$ and $L\ge 0$ such that

$$\psi \left(s^{3}d(T(x,y),T(u,v))\right)\le \psi (M_s(x,y,u,v))-\varphi (M_s(x,y,u,v))+L\theta (N(x,y,u,v)),$$

where

$$\begin{array}{rl}{M}_s(x,y,u,v)=& max\{d(x,u),d(y,v),d(x,T(x,y)),\\ \frac{1}{2s}d(u,T(u,v)),d(y,T(y,x)),\frac{1}{2s}d(v,T(v,u)),\\ \frac{d(x,T(u,v))+d(u,T(x,y))}{2s},\frac{d(y,T(v,u))+d(v,T(y,x))}{2s}\}\end{array}$$

and

$$N(x,y,u,v)=min\{d(x,T(x,y)),d(u,T(u,v)),d(u,T(x,y)),d(x,T(u,v))\}$$

for all $x,y,u,v\in X$ with $x\le u$ and $y\ge v$. If there exists $(x_0,y_0)\in X\times X$ such that $x_0\le T(x_0,y_0)$ and $y_0\ge T(y_0,x_0)$, then T has a coupled fixed point in X.

Proof Take $g=I_X$ and apply Theorem 2.2. □

The following result is the immediate consequence of Corollary 2.3.

Corollary 2.4 Let $(X,d,\le )$ be a partially ordered complete b-metric space. Let $T:X\times X⟶X$ be a continuous mapping such that T has the mixed monotone property. Suppose that there exists $\varphi \in \mathrm{\Phi }$ such that

$$d(T(x,y),T(u,v))\le \frac{1}{s^3}{M}_s(x,y,u,v)-\frac{1}{s^3}\varphi (M_s(x,y,u,v)),$$

(2.31)

where

$$\begin{array}{rl}{M}_s(x,y,u,v)=& max\{d(x,u),d(y,v),d(x,T(x,y)),\\ \frac{1}{2s}d(u,T(u,v)),d(y,T(y,x)),\frac{1}{2s}d(v,T(v,u)),\\ \frac{d(x,T(u,v))+d(u,T(x,y))}{2s},\frac{d(y,T(v,u))+d(v,T(y,x))}{2s}\}\end{array}$$

for all $x,y,u,v\in X$ with $x\le u$ and $y\ge v$. If there exists $(x_0,y_0)\in X\times X$ such that $x_0\le T(x_0,y_0)$ and $y_0\ge T(y_0,x_0)$, then T has a coupled fixed point in X.

3 Uniqueness of a common fixed point

In this section we shall provide some sufficient conditions under which T and g have a unique common fixed point. Note that if $(X,\le )$ is a partially ordered set, then we endow the product $X\times X$ with the following partial order relation, for all $(x,y),(z,t)\in X\times X$,

$$(x,y)\le (z,t)⟺x\le z,y\ge t.$$

From Theorem 2.2, it follows that the set $C(T,g)$ of coupled coincidences is nonempty.

Theorem 3.1 By adding to the hypotheses of Theorem  2.2, the condition: for every $(x,y)$ and $(z,t)$ in $X\times X$, there exists $(u,v)\in X\times X$ such that $(T(u,v),T(v,u))$ is comparable to $(T(x,y),T(y,x))$ and to $(T(z,t),T(t,z))$, then T and g have a unique coupled common fixed point; that is, there exists a unique $(x,y)\in X\times X$ such that

$$x=gx=T(x,y),y=gy=T(y,x).$$

Proof We know, from Theorem 2.2, that there exists at least a coupled coincidence point. Suppose that $(x,y)$ and $(z,t)$ are coupled coincidence points of T and g, that is, $T(x,y)=gx$, $T(y,x)=gy$, $T(z,t)=gz$ and $T(t,z)=gt$. We shall show that $gx=gz$ and $gy=gt$. By the assumptions, there exists $(u,v)\in X\times X$ such that $(T(u,v),T(v,u))$ is comparable to $(T(x,y),T(y,x))$ and to $(T(z,t),T(t,z))$. Without any restriction of the generality, we can assume that

$$(T(x,y),T(y,x))\le (T(u,v),T(v,u))\text{and}(T(z,t),T(t,z))\le (T(u,v),T(v,u)).$$

Put $u_0=u$, $v_0=v$ and choose $(u_1,v_1)\in X\times X$ such that

$$g{u}_1=T(u_0,v_0),g{v}_1=T(v_0,u_0).$$

For $n\ge 1$, continuing this process, we can construct sequences $\{g{u}_n\}$ and $\{g{v}_n\}$ such that

$$g{u}_{n+1}=T(u_n,v_n),g{v}_{n+1}=T(v_n,u_n)\text{for all }n.$$

Further, set $x_0=x$, $y_0=y$ and $z_0=z$, $t_0=t$ and in the same way define sequences $\{g{x}_n\}$, $\{g{y}_n\}$ and $\{g{z}_n\}$, $\{g{t}_n\}$. Then it is easy to see that

$$g{x}_n⟶T(x,y),g{y}_n⟶T(y,x)\text{and}g{z}_n⟶T(z,t),g{t}_n⟶T(t,z)$$

(3.1)

for all $n\ge 1$. Since $(T(x,y),T(y,x))=(gx,gy)=(g{x}_1,g{y}_1)$ is comparable to $(T(u,v),T(v,u))=(gu,gv)=(g{u}_1,g{v}_1)$, then it is easy to show $(gx,gy)\le (gu,gv)$. Recursively, we get that

$$(g{x}_n,g{y}_n)\le (g{u}_n,g{v}_n)\text{for all }n.$$

(3.2)

Thus from (2.1) we have

$$\begin{array}{rl}\psi (d(gx,g{u}_{n+1}))\le & \psi (s^{3}d(gx,g{u}_{n+1}))=\psi \left(s^{3}d(T(x,y),T(u_n,v_n))\right)\\ \le & \psi (M_{s,T,g}(x,y,u_n,v_n))-\varphi (M_{s,T,g}(x,y,u_n,v_n))\\ +L\theta (N_{T,g}(x,y,u_n,v_n)),\end{array}$$

where

$$\begin{array}{rl}{M}_{s,T,g}(x,y,u_n,v_n)=& max\{d(gx,g{u}_n),d(gy,g{v}_n),d(gx,T(x,y)),\\ \frac{1}{2s}d(g{u}_n,T(u_n,v_n)),d(gy,T(y,x)),\\ \frac{1}{2s}d(g{v}_n,T(v_n,u_n)),\frac{d(gx,T(u_n,v_n))+d(g{u}_n,T(x,y))}{2s},\\ \frac{d(gy,T(v_n,u_n))+d(g{v}_n,T(y,x))}{2s}\}\\ \le & max\{d(gx,g{u}_n),d(gy,g{v}_n),d(gy,g{v}_{n+1}),d(gx,g{u}_{n+1})\}.\end{array}$$

It is easy to show that

$$M_{s,T,g}(x,y,u_n,v_n)\le max\{d(gx,g{u}_n),d(gy,g{v}_n)\}$$

and

$$N_{T,g}(x,y,u_n,v_n)=0.$$

Hence,

$$\begin{array}{rl}\psi (d(gx,g{u}_{n+1}))\le & \psi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\})\\ -\varphi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\}).\end{array}$$

(3.3)

Similarly, one can prove that

$$\begin{array}{rl}\psi (d(gy,g{v}_{n+1}))\le & \psi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\})\\ -\varphi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\}).\end{array}$$

(3.4)

Combining (3.3), (3.4) and the fact that $max\{\psi (a),\psi (b)\}=\psi (max\{a,b\})$ for $a,b\in [0,+\mathrm{\infty })$, we have

$$\begin{array}{r}\psi (max\{d(gx,g{u}_{n+1}),d(gy,g{v}_{n+1})\})\\ =max\{\psi (d(gx,g{u}_{n+1})),\psi (d(gy,g{v}_{n+1}))\}\\ \le \psi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\})\varphi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\})\\ \le \psi (max\{d(gx,g{u}_n),d(gy,g{v}_n)\}).\end{array}$$

(3.5)

Using the non-decreasing property of ψ, we get that

$$max\{d(gx,g{u}_{n+1}),d(gy,g{v}_{n+1})\}\le max\{d(gx,g{u}_n),d(gy,g{v}_n)\}$$

implies that $max\{d(gx,g{u}_n),d(gy,g{v}_n)\}$ is a non-increasing sequence. Hence, there exists $r\ge 0$ such that

$$\underset{n⟶\mathrm{\infty }}{lim}max\{d(gx,g{u}_n),d(gy,g{v}_n)\}=r.$$

Passing the upper limit in (3.5) as $n⟶\mathrm{\infty }$, we obtain

$$\psi (r)\le \psi (r)-\varphi (r),$$

which implies that $\varphi (r)=0$, and then $r=0$. We deduce that

$$\underset{n⟶\mathrm{\infty }}{lim}max\{d(gx,g{u}_n),d(gy,g{v}_n)\}=0,$$

which concludes

$$\underset{n⟶\mathrm{\infty }}{lim}d(gx,g{u}_n)=\underset{n⟶\mathrm{\infty }}{lim}d(gy,g{v}_n)=0.$$

(3.6)

Similarly, one can prove that

$$\underset{n⟶\mathrm{\infty }}{lim}d(gz,g{u}_n)=\underset{n⟶\mathrm{\infty }}{lim}d(gt,g{v}_n)=0.$$

(3.7)

From (3.6) and (3.7), we have $gx=gz$ and $gy=gt$. Since $gx=T(x,y)$ and $gy=T(y,x)$, by the commutativity of T and g, we have

$$g(gx)=g(T(x,y))=T(gx,gy),g(gy)=g(T(y,x))=T(gy,gx).$$

(3.8)

Denote $gx=a$ and $gy=b$. Then from (3.8) we have

$$g(a)=T(a,b),g(b)=T(b,a).$$

(3.9)

Thus, $(a,b)$ is a coupled coincidence point. It follows that $ga=gz$ and $gb=gy$, that is,

$$g(a)=a,g(b)=b.$$

(3.10)

From (3.9) and (3.10), we obtain

$$a=g(a)=T(a,b),b=g(b)=T(b,a).$$

(3.11)

Therefore, $(a,b)$ is a coupled common fixed point of T and g. To prove the uniqueness of the point $(a,b)$, assume that $(c,d)$ is another coupled common fixed point of T and g. Then we have

$$c=gc=T(c,d),d=gd=T(d,c).$$

Since $(c,d)$ is a coupled coincidence point of T and g, we have $gc=gx=a$ and $gd=gy=b$. Thus $c=gc=ga=a$ and $d=gd=gb=b$, which is the desired result. □

Theorem 3.2 In addition to the hypotheses of Theorem  3.1, if $g{x}_0$ and $g{y}_0$ are comparable, then T and g have a unique common fixed point, that is, there exists $x\in X$ such that $x=gx=T(x,x)$.

Proof Following the proof of Theorem 3.1, T and g have a unique coupled common fixed point $(x,y)$. We only have to show that $x=y$. Since $g{x}_0$ and $g{y}_0$ are comparable, we may assume that $g{x}_0\le g{y}_0$. By using the mathematical induction, one can show that

$$g{x}_n\le g{y}_n\text{for all }n\ge 0,$$

(3.12)

where $\{g{x}_n\}$ and $\{g{y}_n\}$ are defined by (2.2). From (2.29) and Lemma 1.4, we have

$$\begin{array}{rl}\psi (sd(x,y))=& \psi (s^3\frac{1}{s^2}d(x,y))\le \underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (s^{3}d(g{x}_{n+1},g{y}_{n+1}))\\ =& \underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}\psi \left(s^{3}d(T(x_n,y_n),T(y_n,x_n))\right)\\ \le & \underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}\psi (M_{s,T,g}(x_n,y_n,y_n,x_n))-\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (M_{s,T,g}(x_n,y_n,y_n,x_n))\\ +\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}L\theta (N_{T,g}(x_n,y_n,y_n,x_n))\\ \le & \psi (d(x,y))-\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (M_s(x_n,y_n,y_n,x_n))\\ <& \psi (d(x,y)),\end{array}$$

a contradiction. Therefore, $x=y$, that is, T and g have a common fixed point. □

Remark 3.3 Since a b-metric is a metric when $s=1$, from the results of Jachymski [29], the condition

$$\psi \left(d(F(x,y),F(u,v))\right)\le \psi (max\{d(gx,gu),d(gy,gv)\})-\varphi (max\{d(gx,gu),d(gy,gv)\})$$

is equivalent to

$$d(F(x,y),F(u,v))\le \phi (max\{d(gx,gu),d(gy,gv)\}),$$

where $\psi \in \mathrm{\Psi }$, $\varphi \in \mathrm{\Phi }$ and $\phi :[0,\mathrm{\infty })⟶[0,\mathrm{\infty })$ is continuous, $\phi (t)<t$ for all $t>0$ and $\phi (t)=0$ if and only if $t=0$. So, our results can be viewed as a generalization and extension of the corresponding results in [15, 25, 30–32] and several other comparable results.

4 Application to integral equations

Here, in this section, we wish to study the existence of a unique solution to a nonlinear quadratic integral equation, as an application to our coupled fixed point theorem. Consider the nonlinear quadratic integral equation

$$x(t)=h(t)+\lambda {\int }_0^1{k}_1(t,s)f_1(s,x(s))ds{\int }_0^1{k}_2(t,s)f_2(s,x(s))ds,t\in I=[0,1],\lambda \ge 0.$$

(4.1)

Let Γ denote the class of those functions $\gamma :[0,+\mathrm{\infty })⟶[0,+\mathrm{\infty })$ which satisfy the following conditions:

• (i) γ is non-decreasing and ${(\gamma (t))}^p\le \gamma (t^p)$ for all $p\ge 1$.

• (ii) There exists $\varphi \in \mathrm{\Phi }$ such that $\gamma (t)=t-\varphi (t)$ for all $t\in [0,+\mathrm{\infty })$.

For example, ${\gamma }_1(t)=kt$, where $0\le k<1$ and ${\gamma }_2(t)=\frac{t}{t+1}$ are in Γ.

We will analyze Eq. (4.1) under the following assumptions:

(a₁) $f_i:I\times \mathbb{R}⟶\mathbb{R}$ ($i=1,2$) are continuous functions, $f_i(t,x)\ge 0$ and there exist two functions $m_i\in L^1(I)$ such that $f_i(t,x)\le m_i(t)$ ($i=1,2$).

(a₂) $f_1(t,x)$ is monotone non-decreasing in x and $f_2(t,y)$ is monotone non-increasing in y for all $x,y\in \mathbb{R}$ and $t\in I$.

(a₃) $h:I⟶\mathbb{R}$ is a continuous function.

(a₄) $k_i:I\times I⟶\mathbb{R}$ ($i=1,2$) are continuous in $t\in I$ for every $s\in I$ and measurable in $s\in I$ for all $t\in I$ such that

$${\int }_0^1{k}_i(t,s)m_i(s)ds\le K,i=1,2,$$

and $k_i(t,x)\ge 0$.

(a₅) There exist constants $0\le L_i<1$ ($i=1,2$) and $\gamma \in \mathrm{\Gamma }$ such that for all $x,y\in \mathbb{R}$ and $x\ge y$,

$$|f_i(t,x)-f_i(t,y)|\le L_i\gamma (x-y)(i=1,2).$$

(a₆) There exist $\alpha ,\beta \in C(I)$ such that

$$\begin{array}{rl}\alpha (t)& \le h(t)+\lambda {\int }_0^1{k}_1(t,s)f_1(s,\alpha (s))ds{\int }_0^1{k}_2(t,s)f_2(s,\beta (s))ds\\ \le h(t)+\lambda {\int }_0^1{k}_1(t,s)f_1(s,\beta (s))ds{\int }_0^1{k}_2(t,s)f_2(s,\alpha (s))ds\le \beta (t).\end{array}$$

(a₇) $max\{L_1^p,L_2^p\}{\lambda }^p{K}^{2p}\le \frac{1}{2^{4p-3}}$.

Consider the space $X=C(I)$ of continuous functions defined on $I=[0,1]$ with the standard metric given by

$$\rho (x,y)=\underset{t\in I}{sup}|x(t)-y(t)|\text{for }x,y\in C(I).$$

This space can also be equipped with a partial order given by

$$x,y\in C(I),x\le y⟺x(t)\le y(t)\text{for any }t\in I.$$

Now, for $p\ge 1$, we define

$$d(x,y)={(\rho (x,y))}^p={\left(\underset{t\in I}{sup}|x(t)-y(t)|\right)}^p=\underset{t\in I}{sup}{|x(t)-y(t)|}^p\text{for }x,y\in C(I).$$

It is easy to see that $(X,d)$ is a complete b-metric space with $s=2^{p-1}$ [3].

Also, $X\times X=C(I)\times C(I)$ is a partially ordered set if we define the following order relation:

$$(x,y),(u,v)\in X\times X,(x,y)\le (u,v)⟺x\le u\text{and}y\ge v.$$

For any $x,y\in X$ and each $t\in I$, $max\{x(t),y(t)\}$ and $min\{x(t),y(t)\}$ belong to X and are upper and lower bounds of x, y, respectively. Therefore, for every $(x,y),(u,v)\in X\times X$, one can take $(max\{x,u\},min\{y,v\})\in X\times X$ which is comparable to $(x,y)$ and $(u,v)$. Now, we formulate the main result of this section.

Theorem 4.1 Under assumptions (a₁)-(a₇), Eq. (4.1) has a unique solution in $C(I)$.

Proof We consider the operator $T:X\times X⟶X$ defined by

$$T(x,y)(t)=h(t)+\lambda {\int }_0^1{k}_1(t,s)f_1(s,x(s))ds{\int }_0^1{k}_2(t,s)f_2(s,y(s))ds\text{for }t\in I.$$

By virtue of our assumptions, T is well defined (this means that if $x,y\in X$, then $T(x,y)\in X$). Firstly, we prove that T has the mixed monotone property. In fact, for $x_1\le x_2$ and $t\in I$, we have

$$\begin{array}{rl}T(x_1,y)(t)-T(x_2,y)(t)=& h(t)+\lambda {\int }_0^1{k}_1(t,s)f_1(s,x_1(s))ds{\int }_0^1{k}_2(t,s)f_2(s,y(s))ds\\ -h(t)-\lambda {\int }_0^1{k}_1(t,s)f_1(s,x_2(s))ds{\int }_0^1{k}_2(t,s)f_2(s,y(s))ds\\ =& \lambda {\int }_0^1{k}_1(t,s)[f_1(s,x_1(s))-f_1(s,x_2(s))]ds{\int }_0^1{k}_2(t,s)f_2(s,y(s))ds\\ \le & 0.\end{array}$$

Similarly, if $y_1\ge y_2$ and $t\in I$, then $T(x,y_1)(t)\le T(x,y_2)(t)$. Therefore, T has the mixed monotone property. Also, for $(x,y)\le (u,v)$, that is, $x\le u$ and $y\ge v$, we have

$$\begin{array}{r}|T(x,y)(t)-T(u,v)(t)|\\ \le |\lambda {\int }_0^1{k}_1(t,s)f_1(s,x(s))ds{\int }_0^1{k}_2(t,s)[f_2(s,y(s))-f_2(s,v(s))]ds\\ +\lambda {\int }_0^1{k}_2(t,s)f_2(s,v(s))ds{\int }_0^1{k}_1(t,s)[f_1(s,x(s))-f_1(s,u(s))]ds|\\ \le \lambda {\int }_0^1{k}_1(t,s)f_1(s,x(s))ds{\int }_0^1{k}_2(t,s)|f_2(s,y(s))-f_2(s,v(s))|ds\\ +\lambda {\int }_0^1{k}_2(t,s)f_2(s,v(s))ds{\int }_0^1{k}_1(t,s)|f_1(s,x(s))-f_1(s,u(s))|ds\\ \le \lambda {\int }_0^1{k}_1(t,s)m_1(s)ds{\int }_0^1{k}_2(t,s)L_2\gamma (y(s)-v(s))ds\\ +\lambda {\int }_0^1{k}_2(t,s)m_2(s)ds{\int }_0^1{k}_1(t,s)L_1\gamma (u(s)-x(s))ds.\end{array}$$

Since the function γ is non-decreasing and $x\le u$ and $y\ge v$, we have

$$\gamma (u(s)-x(s))\le \gamma \left(\underset{t\in I}{sup}|x(s)-u(s)|\right)=\gamma (\rho (x,u))$$

and

$$\gamma (y(s)-v(s))\le \gamma \left(\underset{t\in I}{sup}|y(s)-v(s)|\right)=\gamma (\rho (y,v)),$$

hence

$$\begin{array}{rl}|T(x,y)(t)-T(u,v)(t)|& \le \lambda K{\int }_0^1{k}_2(t,s)L_2\gamma (\rho (y,v))ds+\lambda K{\int }_0^1{k}_1(t,s)L_1\gamma (\rho (u,x))ds\\ \le \lambda K^{2}max\{L_1,L_2\}[\gamma (\rho (u,x))+\gamma (\rho (y,v))].\end{array}$$

Then we can obtain

$$\begin{array}{rl}d(T(x,y),T(u,v))& =\underset{t\in I}{sup}{|T(x,y)(t)-T(u,v)(t)|}^p\\ \le {\{\lambda K^{2}max\{L_1,L_2\}[\gamma (\rho (u,x))+\gamma (\rho (y,v))]\}}^p\\ ={\lambda }^p{K}^{2p}max\{L_1^p,L_2^p\}{[\gamma (\rho (u,x))+\gamma (\rho (y,v))]}^p,\end{array}$$

and using the fact that ${(a+b)}^p\le 2^{p-1}(a^p+b^p)$ for $a,b\in (0,+\mathrm{\infty })$ and $p>1$, we have

$$\begin{array}{rl}d(T(x,y),T(u,v))& \le 2^{p-1}{\lambda }^p{K}^{2p}max\{L_1^p,L_2^p\}[{\left(\gamma (\rho (u,x))\right)}^p+{\left(\gamma (\rho (y,v))\right)}^p]\\ \le 2^{p-1}{\lambda }^p{K}^{2p}max\{L_1^p,L_2^p\}[\gamma (d(u,x))+\gamma (d(y,v))]\\ \le 2^p{\lambda }^p{K}^{2p}max\{L_1^p,L_2^p\}\left[\gamma (M_s(x,y,u,v))\right]\\ \le 2^p{\lambda }^p{K}^{2p}max\{L_1^p,L_2^p\}[M_s(x,y,u,v)-\varphi (M_s(x,y,u,v))]\\ \le \frac{1}{2^{3p-3}}{M}_s(x,y,u,v)-\frac{1}{2^{3p-3}}\varphi (M_s(x,y,u,v)).\end{array}$$

This proves that the operator T satisfies the contractive condition (2.31) appearing in Corollary 2.4.

Finally, let α, β be the functions appearing in assumption (a₆); then, by (a₆), we get

$$\alpha \le T(\alpha ,\beta )\le T(\beta ,\alpha )\le \beta .$$

Theorem 3.1 gives us that T has a unique coupled fixed point $(x^{\ast },y^{\ast })\in X\times X$. Since $\alpha \le \beta$, Theorem 3.2 says that $x^{\ast }=y^{\ast }$ and this implies $x^{\ast }=T(x^{\ast },x^{\ast })$. So, $x^{\ast }\in C(I)$ is the unique solution of Eq. (4.1) and the proof is complete. □