Fixed points for generalized contractions via rational expressions in partially ordered b-metric spaces and applications to integral equations

Abstract

The purpose of this paper is to give some fixed point results for mappings involving generalized $(\psi ,\varphi )$-contractions via rational expressions in the setup of partially ordered b-metric spaces. Our main results extend, generalize and enrich several well known comparable results in the recent literature. Moreover, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.

1 Introduction

Fixed point theory is one of the most powerful and productive tools from the nonlinear analysis and it can be considered the kernel of the nonlinear analysis. In this theory, contraction is one of the main tools to prove the existence and uniqueness of a fixed point. The Banach contraction principle [6], which is a useful tool in the study of many branches of mathematics and mathematical sciences, is one of the earlier and fundamental results in fixed point theory. Because of its importance in nonlinear analysis, a number of authors have improved, generalized and extended this basic result either by defining a new contractive mapping in the context of a complete metric space or by investigating the existing contractive mappings in various abstract spaces (see, e.g., [7, 13, 16, 25, 32]).

Ran and Reurings [31] extended the Banach contraction principle in partially ordered sets with some applications to linear and nonlinear matrix equations. While Nieto and Rodŕiguez-López [30] extended the result of Ran and Reurings and applied their main theorems to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions. For more details on fixed point results, their applications, comparison of different contractive conditions and related results in ordered metric spaces we refer the reader to [1, 3, 5, 8, 20, 21] and the references mentioned therein.

The concept of a b-metric space was introduced by Bakhtin in [5], and later extensively used by Czerwik in [17, 18]. Since then, several papers have been published on the fixed point theory of various classes of single-valued and multi-valued operators in b-metric spaces. For further works in this direction, we refer to [4, 9,10,11,12, 22, 33, 34]. Khmasi and Hussain [28] and Hussain and Shah [24] discussed KKM mappings and related results in b-metric and cone b-metric spaces.

In this paper, we establish some fixed point results for nonlinear generalized $(\psi ,\varphi )$-weakly contractive mappings of rational type in the framework of partially ordered b-metric spaces. Our main results extend and generalize the results in [14, 15] and [19] from the context of ordered metric spaces to the setting of ordered b-metric spaces. Furthermore, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.

2 Preliminaries

In 1975, Dass and Gupta [19] proved the following fixed point theorem satisfying a contractive condition of a rational type.

Theorem 2.1

(Dass and Gupta [19]) Let (X, d) be a complete metric space and $T:X\rightarrow X$ a mapping such that there exist $\alpha ,\beta \ge 0$ with $\alpha +\beta <1$ satisfying

$$\begin{aligned} d(Tx,Ty)\le \alpha \frac{d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta d(x,y), \end{aligned}$$

for all $x,y\in X$. Then T has a unique fixed point.

In 2013, Cabrera et al. [14] proved the above theorem in the context of partially ordered metric spaces.

Use of auxiliary functions to generalize the contractive conditions on maps have been a subject of interest in fixed point theory. Now, we define the following class of auxiliary functions which will be used densely in the sequel.

Let $\Psi$ denotes the class of all functions $\psi :[0,+\infty )\rightarrow [0,+\infty )$ with the following properties:

($\psi _{i}$):

$\psi$ is continuous and nondecreasing;

($\psi _{ii}$):

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

Also, let $\Phi$ denotes the class of all functions $\varphi :[0,+\infty )\rightarrow [0,+\infty )$ with the following properties:

($\varphi _{i}$):

$\varphi$ is lower semi-continuous;

($\varphi _{ii}$):

$\varphi (t)=0$ if and only if $t=0$.

The functions $\psi \in \Psi$ and $\varphi \in \Phi$ are called control functions.

So far, many authors have studied fixed point theorems, which are based on control functions. In 2015, Chandok et al. [15] established some fixed point theorems satisfying a generalized contraction mapping of rational type in metric spaces endowed with partial order by using some auxiliary functions.

Consistent with [18, 26, 35], the following definitions and results will be needed in the sequel.

Definition 2.2

([18]) Let X be a (nonempty) set and $s\ge 1$ be a given real number. A function $d:X\times X\rightarrow [\mathbf{0}, + \infty )$ is a b-metric if, for all $x,y,z\in X$, the following conditions hold:

$(b_1)$ :

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

$(b_2)$ :

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

$(b_3)$ :

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

In this case, the pair (X, d) is called a b-metric space.

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

The following example shows that in general a b-metric need not necessarily be a metric (see, also, [35, p. 264]).

Example 2.3

Let (X, d) be a metric space and $\rho (x,y)=(d(x,y))^p$, where $p>1$ is a real number. We show that $\rho$ is a b-metric with $s=2^{p-1}$.

Obviously, conditions $\mathrm (b_1)$ and $\mathrm (b_2)$ of Definition 2.2 are satisfied. If $1<p<+\infty$, then the convexity of the function $f(x)=x^p\,(x>0)$ implies

$$\begin{aligned} \bigg (\frac{a+b}{2}\bigg )^p\le \frac{1}{2}\big (a^p+b^p\big ), \end{aligned}$$

and hence, $(a+b)^p\le 2^{p-1}\big (a^p+b^p\big )$ holds.

Thus, for each $x,y,z\in X$, we obtain

$$\begin{aligned} \rho (x,y)&=\big (d(x,y)\big )^p\le \big (d(x,z)+d(z,y)\big )^p\le 2^{p-1}\big (\big (d(x,z)\big )^p+\big (d(z,y)\big )\big )\nonumber \\&=2^{p-1}\big (\rho (x,z)+\rho (z,y)\big ). \end{aligned}$$

So, condition $\mathrm (b_3)$ of Definition 2.2 is also satisfied and $\rho$ is a b-metric.

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 $\rho (x,y)=(x-y)^2$ is a b-metric on $\mathbb {R}$ with $s=2$, but is not a metric on $\mathbb {R}$.

Also, the following example of a b-metric space is given in [28].

Example 2.4

Let X be the set of Lebesgue measurable functions on [0, 1] such that

$$\begin{aligned} \int _0^1|f(x)|^2dx<+\infty . \end{aligned}$$

Define $D:X\times X\rightarrow [0,+\infty )$ by

$$\begin{aligned} D(f,g)=\int _0^1|f(x)-g(x)|^2dx. \end{aligned}$$

As

$$\begin{aligned} \bigg (\int _0^1|f(x)-g(x)|^2dx\bigg )^{\frac{1}{2}} \end{aligned}$$

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.

Before stating and proving our results, we need the following definitions and proposition in b-metric spaces.

Definition 2.5

Let X be a nonempty set. Then $(X,\preceq ,d)$ is called a partially ordered b-metric space if and only if d is a b-metric on a partially ordered set $(X,\preceq )$.

Definition 2.6

([10]) Let (X, d) be a b-metric space, and let $\{x_n\}$ be a sequence in X. Then one has the following:

• (a) The sequence $\{x_n\}$ in X is called b-convergent if and only if there exists $x\in X$ such that $d(x_n,x)\rightarrow 0$ as $n\rightarrow +\infty$. In this case, we write $\lim _{n\rightarrow +\infty }x_n=x$.

• (b) The sequence $\{x_n\}$ is said to be b-Cauchy sequence if and only if $d(x_n,x_m)\rightarrow 0$ as $n,m\rightarrow +\infty$.

• (c) (X, d) is called b-complete if every b-Cauchy sequence $\{x_n\}$ in X be b-converges.

Proposition 2.7

(See, Remark 2.1 in [10]) In a b-metric space (X, d) the following assertions hold:

$\mathrm (a)$ :

A b-convergent sequence has a unique limit.

$\mathrm (b)$ :

Each b-convergent sequence is b-Cauchy.

$\mathrm (c)$ :

In general, a b-metric is not continuous.

It should be noted that in general, a b-metric function d(x, y) for $s>1$ need not be jointly continuous in both variables (see Example 2 in [23]).

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 2.8

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

$$\begin{aligned} \frac{1}{s^2}d(x,y)\le \liminf _{n\rightarrow +\infty }d(x_n,y_n)\le \limsup _{n\rightarrow +\infty }d(x_n,y_n)\le s^2d(x,y). \end{aligned}$$

In particular, if $x=y$, then $\lim _{n\rightarrow +\infty }d(x_n,y_n)=0$. Moreover, for each $z\in X$, we have

$$\begin{aligned} \frac{1}{s}d(x,z)\le \liminf _{n\rightarrow +\infty }d(x_n,z)\le \limsup _{n\rightarrow +\infty }d(x_n,z)\le sd(x,z). \end{aligned}$$

3 Main results

Now, we are ready to state and prove our main results.

Theorem 3.1

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a nondecreasing, with respect to $\preceq$, continuous mapping such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} \psi \big (sd(Tx,Ty)\big )\le \psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ), \end{aligned}$$

(3.1)

where $\psi \in \Psi$, $\varphi \in \Phi$,

$$\begin{aligned} M(x,y)=\max \bigg \{d(x,y),\frac{d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)},\frac{d(y,Tx)[1+d(x,Ty)]}{1+d(x,y)}\bigg \} \end{aligned}$$

and

$$\begin{aligned} N(x,y)=\max \bigg \{d(x,y),\frac{d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}\bigg \}. \end{aligned}$$

If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

If $Tx_0=x_0$, then there is nothing to prove. Suppose that $x_0\prec Tx_0$. Then, we define a sequence $\{x_n\}$ in X such that

$$\begin{aligned} x_{n+1}=Tx_n,\qquad \text {for all }n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

(3.2)

Since $x_0\prec Tx_0=x_1$ and T is nondecreasing, we have $x_1=Tx_0\preceq x_2=Tx_1$. Again, as $x_1\preceq x_2$ and T is nondecreasing, we have $x_2=Tx_1\preceq x_3=Tx_2$. By induction on n, we conclude that

$$\begin{aligned} x_0\prec x_1\preceq x_2\preceq \cdots \preceq x_n\preceq x_{n+1}\preceq \cdots . \end{aligned}$$

(3.3)

If $x_n=x_{n+1}$ for some $n\in \mathbb {N}$, then from (3.2), $x_n=x_{n+1}=Tx_n$, that is, $x_n$ is a fixed point of T and the proof is finished. So, we may assume that $x_n\ne x_{n+1}$, that is, $d(x_n,x_{n+1})\ne 0$ for all $n\in \mathbb {N}$.

We complete the proof in the following steps.

Step 1 We will prove that

$$\begin{aligned} \lim _{n\rightarrow +\infty }d(x_n,x_{n+1})=0. \end{aligned}$$

(3.4)

Since $x_{n-1}\prec x_n$ for all $n\in \mathbb {N}$, by applying (3.1), we have

$$\begin{aligned} \psi \big (d(x_n,x_{n+1})\big )&\le \psi \big (sd(x_n,x_{n+1})\big )\nonumber \nonumber \\&=\psi \big (sd(Tx_{n-1},Tx_n)\big )\nonumber \nonumber \\&\le \psi \big (M(x_{n-1},x_n)\big )-\varphi \big (N(x_{n-1},x_n)\big ), \end{aligned}$$

(3.5)

where

$$\begin{aligned} M(x_{n-1},x_n)=&\max \bigg \{d(x_{n-1},x_n),\frac{d(x_n,Tx_n)\big [1+d(x_{n-1},Tx_{n-1})\big ]}{1+d(x_{n-1},x_n)},\nonumber \\&\frac{d(x_n,Tx_{n-1})\big [1+d(x_{n-1},Tx_n)\big ]}{1+d(x_{n-1},x_n)}\bigg \}\nonumber \nonumber \\ =&\max \bigg \{d(x_{n-1},x_n),\frac{d(x_n,x_{n+1})\big [1+d(x_{n-1},x_n)\big ]}{1+d(x_{n-1},x_n)},\nonumber \\&\frac{d(x_n,x_n)\big [1+d(x_{n-1},x_{n+1})\big ]}{1+d(x_{n-1},x_n)}\bigg \}\nonumber \nonumber \\ =&\max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \} \end{aligned}$$

(3.6)

and

$$\begin{aligned} N(x_{n-1},x_n)&=\max \bigg \{d(x_{n-1},x_n),\frac{d(x_n,Tx_n)\big [1+d(x_{n-1},Tx_{n-1})\big ]}{1+d(x_{n-1},x_n)}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(x_{n-1},x_n),\frac{d(x_n,x_{n+1})\big [1+d(x_{n-1},x_n)\big ]}{1+d(x_{n-1},x_n)}\bigg \}\nonumber \nonumber \\&=\max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \}. \end{aligned}$$

(3.7)

By applying the inequality (3.5) and using (3.6) and (3.7) and the properties of $\psi$ and $\varphi$, we get

$$\begin{aligned} \psi \big (d(x_n,x_{n+1})\big )\le&\psi \big (\max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \}\big )\nonumber \\&-\varphi \big (\max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \}\big ). \end{aligned}$$

(3.8)

Now, if

$$\begin{aligned} \max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \}=d(x_n,x_{n+1}), \end{aligned}$$

then by (3.8) and the property of $\varphi$, we have

$$\begin{aligned} \psi \big (d(x_n,x_{n+1})\big )&\le \psi \big (d(x_n,x_{n+1})\big )-\varphi \big (d(x_n,x_{n+1})\big )\nonumber \\&<\psi \big (d(x_n,x_{n+1})\big ), \end{aligned}$$

which gives a contradiction. Thus,

$$\begin{aligned} \max \big \{d(x_{n-1},x_n),d(x_n,x_{n+1})\big \}=d(x_{n-1},x_n). \end{aligned}$$

Therefore, the inequality (3.8) yields that

$$\begin{aligned} \psi \big (d(x_n,x_{n+1})\big )\le \psi \big (d(x_{n-1},x_n)\big )-\varphi \big (d(x_{n-1},x_n)\big )<\psi \big (d(x_{n-1},x_n)\big ). \end{aligned}$$

(3.9)

Since $\psi$ is nondecreasing, we have $d(x_n,x_{n+1})<d(x_{n-1},x_n)$. Hence, $\{d(x_n,x_{n+1})\,:\,n\in \mathbb {N}\cup \{0\}\}$ is a monotone decreasing sequence of nonnegative numbers which is bounded below. So, there exists $r\ge 0$ such that

$$\begin{aligned} \lim _{n\rightarrow +\infty }d(x_n,x_{n+1})=r. \end{aligned}$$

Now, we shall show that $r=0$. Suppose, on the contrary, that $r>0$. By taking the upper limit on both sides of (3.9) and using the properties of $\psi$ and $\varphi$, we obtain

$$\begin{aligned} \psi (r)\le \psi (r)-\varphi (r)<\psi (r), \end{aligned}$$

which is a contradiction. Hence, we conclude that $r=0$.

Step 2 We shall prove that $\{x_n\}$ is a b-Cauchy sequence in X. Suppose, to the contrary, that $\{x_{n}\}$ is not a b-Cauchy sequence. Then there exists $\varepsilon >0$ for which we can find two subsequences of positive integers $\{x_{m_i}\}$ and $\{x_{n_i}\}$ of $\{x_n\}$ such that $n_i$ is the smallest index for which

$$\begin{aligned} n_i>m_i>i,\qquad d(x_{m_i},x_{n_i})\ge \varepsilon . \end{aligned}$$

(3.10)

This means that

$$\begin{aligned} d(x_{m_i},x_{n_i-1})<\varepsilon . \end{aligned}$$

(3.11)

From (3.10) and by using the triangular inequality, we get

$$\begin{aligned} \varepsilon&\le d(x_{m_i},x_{n_i})\nonumber \\&\le s d(x_{m_i},x_{m_i-1})+sd(x_{m_i-1},x_{n_i})\nonumber \\&\le sd(x_{m_i},x_{m_i-1})+s^2d(x_{m_i-1},x_{n_i-1})+s^2d(x_{n_i-1},x_{n_i}). \end{aligned}$$

By taking the upper limit as $i\rightarrow +\infty$ in the above inequality and using (3.4), we obtain

$$\begin{aligned} \frac{\varepsilon }{s^2}\le \limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1}). \end{aligned}$$

On the other hand, we have

$$\begin{aligned} d(x_{m_i-1},x_{n_i-1})\le sd(x_{m_i-1},x_{m_i})+sd(x_{m_i},x_{n_i-1}). \end{aligned}$$

By using (3.4) and (3.11) and taking the upper limit as $i\rightarrow +\infty$ in the above inequality, we get

$$\begin{aligned} \limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1})\le s\varepsilon . \end{aligned}$$

So, we have

$$\begin{aligned} \frac{\varepsilon }{s^2}\le \limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1})\le s\varepsilon . \end{aligned}$$

(3.12)

Further, from

$$\begin{aligned} d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{n_i-1})+sd(x_{n_i-1},x_{n_i}) \end{aligned}$$

and by using (3.4), (3.10) and (3.11), we get

$$\begin{aligned} \frac{\varepsilon }{s}\le \limsup _{i\rightarrow +\infty }d(x_{n_i-1},x_{m_i})\le \varepsilon . \end{aligned}$$

(3.13)

Moreover, from

$$\begin{aligned} \varepsilon \le d(x_{m_i},x_{n_i})\le sd(x_{m_i},x_{m_i-1})+sd(x_{m_i-1},x_{n_i}) \end{aligned}$$

and

$$\begin{aligned} d(x_{m_i-1},x_{n_i})\le sd(x_{m_i-1},x_{n_i-1})+sd(x_{n_i-1},x_{n_i}), \end{aligned}$$

and taking into account (3.4) and (3.12), we get

$$\begin{aligned} \frac{\varepsilon }{s}\le \limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i})\le s^2\varepsilon . \end{aligned}$$

(3.14)

Since $x_{m_i-1}\preceq x_{n_i-1}$, by applying the inequality (3.1), we have

$$\begin{aligned} \psi \big (sd(x_{m_i},x_{n_i})\big )&=\psi \big (sd(Tx_{m_i-1},Tx_{n_i-1})\big )\nonumber \nonumber \\&\le \psi \big (M(x_{m_i-1},x_{n_i-1})\big )-\varphi \big (N(x_{m_i-1},x_{n_i-1})\big ), \end{aligned}$$

(3.15)

where

$$\begin{aligned} M(x_{m_i-1},x_{n_i-1})&=\max \bigg \{d(x_{m_i-1},x_{n_i-1}),\frac{d(x_{n_i-1},Tx_{n_i-1})\big [1+d(x_{m_i-1},Tx_{m_i-1})\big ]}{1+d(x_{m_i-1},x_{n_i-1})},\nonumber \nonumber \\&\quad \frac{d(x_{n_i-1},Tx_{m_i-1})\big [1+d(x_{m_i-1},Tx_{n_i-1})\big ]}{1+d(x_{m_i-1},x_{n_i-1})}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(x_{m_i-1},x_{n_i-1}),\frac{d(x_{n_i-1},x_{n_i})\big [1+d(x_{m_i-1},x_{m_i})\big ]}{1+d(x_{m_i-1},x_{n_i-1})},\nonumber \nonumber \\&\quad \frac{d(x_{n_i-1},x_{m_i})\big [1+d(x_{m_i-1},x_{n_i})\big ]}{1+d(x_{m_i-1},x_{n_i-1})}\bigg \} \end{aligned}$$

(3.16)

and

$$\begin{aligned} N(x_{m_i-1},x_{n_i-1})&=\max \bigg \{d(x_{m_i-1},x_{n_i-1}),\frac{d(x_{n_i-1},Tx_{n_i-1})\big [1+d(x_{m_i-1},Tx_{m_i-1})\big ]}{1+d(x_{m_i-1},x_{n_i-1})}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(x_{m_i-1},x_{n_i-1}),\frac{d(x_{n_i-1},x_{n_i})\big [1+d(x_{m_i-1},x_{m_i})\big ]}{1+d(x_{m_i-1},x_{n_i-1})}\bigg \}. \end{aligned}$$

(3.17)

By taking the upper limit as $i\rightarrow +\infty$ in (3.16) and (3.17) and using (3.12), (3.13) and (3.14), we get

$$\begin{aligned} \limsup _{i\rightarrow +\infty }M(x_{m_i-1},x_{n_i-1})&=\max \bigg \{\limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1}),0,\nonumber \nonumber \\&\quad \frac{\limsup _{i\rightarrow +\infty }d(x_{n_i-1},x_{m_i})\big [1+\limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i})\big ]}{1+\limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1})}\bigg \}\nonumber \nonumber \\&\le \max \bigg \{s\varepsilon ,\frac{\varepsilon (1+s^2\varepsilon )}{1+\frac{\varepsilon }{s^2}}\bigg \}=s\varepsilon \end{aligned}$$

(3.18)

and

$$\begin{aligned} \frac{\varepsilon }{s^2}\le \limsup _{i\rightarrow +\infty }N(x_{m_i-1},x_{n_i-1})=\max \Bigg \{\limsup _{i\rightarrow +\infty }d(x_{m_i-1},x_{n_i-1}),0\Bigg \}\le s\varepsilon . \end{aligned}$$

Similarly, we can obtain

$$\begin{aligned} \frac{\varepsilon }{s^2}\le \liminf _{i\rightarrow +\infty }N(x_{m_i-1},x_{n_i-1})\le s\varepsilon . \end{aligned}$$

(3.19)

Now, taking the upper limit as $i\rightarrow +\infty$ in (3.15) and using (3.10), (3.18) and (3.19), we obtain

$$\begin{aligned} \psi (s\varepsilon )&\le \psi \Bigg (s\limsup _{i\rightarrow +\infty }d(x_{m_i},x_{n_i})\Bigg )\nonumber \\&\le \psi \Bigg (\limsup _{i\rightarrow +\infty }M(x_{m_i-1},x_{n_i-1})\Bigg )-\liminf _{i\rightarrow +\infty }\varphi \Bigg (N(x_{m_i-1},x_{n_i-1})\Bigg )\nonumber \\&\le \psi (s\varepsilon )-\varphi \Bigg (\liminf _{i\rightarrow +\infty }N(x_{m_i-1},x_{n_i-1})\Bigg ), \end{aligned}$$

which further implies that

$$\begin{aligned} \varphi \Bigg (\liminf _{i\rightarrow +\infty }N(x_{m_i-1},x_{n_i-1})\Bigg )=0, \end{aligned}$$

and so $\liminf _{i\rightarrow +\infty }N(x_{m_i-1},x_{n_i-1})=0$, which contradicts (3.19).

Thus, we have proved that $\{x_n\}$ is a b-Cauchy sequence in the b-metric space (X, d). Since (X, d) is b-complete, there exists $u\in X$ such that $\lim _{n\rightarrow +\infty }x_n=u$ and

$$\begin{aligned} \lim _{n\rightarrow +\infty }x_{n+1}=\lim _{n\rightarrow +\infty }fx_n=u. \end{aligned}$$

By using the triangular inequality, we have

$$\begin{aligned} d(u,fu)\le sd(u,fx_n)+sd(fx_n,fu). \end{aligned}$$

Now, by taking the upper limit as $n\rightarrow +\infty$ in the above inequality and using the continuity of f, we obtain

$$\begin{aligned} d(u,fu)\le s\limsup _{n\rightarrow +\infty }d(u,fx_n)+s\limsup _{n\rightarrow +\infty }d(fx_n,fu)=0. \end{aligned}$$

So, we have $fu=u$. Thus, u is a fixed point of f.

We will show now that the continuity of T in Theorem 3.1 is not necessary and can be replaced by another assumption.

Theorem 3.2

Under the same hypotheses of Theorem 3.1, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Proof

Following similar arguments as those given in the proof of Theorem 3.1, we construct a sequence $\{x_n\}$ in X such that $x_{n+1}=Tx_n$ for all $n\in \mathbb {N}\cup \{0\}$. Then, we have $x_0\prec x_1\preceq x_2\preceq \cdots \preceq x_n\preceq x_{n+1}\preceq \cdots$, that is, $\{x_n\}$ is a nondecreasing sequence. Also, this sequence converges to u for some $u\in X$. By using the assumption on X, we have $x_n\preceq u$ for all $n\in \mathbb {N}$. Now, by applying (3.1), we have

$$\begin{aligned} \psi \big (sd(x_{n+1},Tu)\big )&=\psi \big (sd(Tx_n,Tu)\big )\nonumber \nonumber \\&\le \psi \big (M(x_n,u)\big )-\varphi \big (N(x_n,u)\big ), \end{aligned}$$

(3.20)

where

$$\begin{aligned} M(x_n,u)&=\max \bigg \{d(x_n,u),\frac{d(u,Tu)\big [1+d(x_n,Tx_n)\big ]}{1+d(x_n,u)},\frac{d(u,Tx_n)\big [1+d(x_n,Tu)\big ]}{1+d(x_n,u)}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(x_n,u),\frac{d(u,Tu)\big [1+d(x_n,x_{n+1})\big ]}{1+d(x_n,u)},\frac{d(u,x_{n+1})\big [1+d(x_n,Tu)\big ]}{1+d(x_n,u)}\bigg \} \end{aligned}$$

(3.21)

and

$$\begin{aligned} N(x_n,u)&=\max \bigg \{d(x_n,u),\frac{d(u,Tu)\big [1+d(x_n,Tx_n)\big ]}{1+d(x_n,u)}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(x_n,u),\frac{d(u,Tu)\big [1+d(x_n,x_{n+1})\big ]}{1+d(x_n,u)}\bigg \}. \end{aligned}$$

(3.22)

Letting $n\rightarrow +\infty$ in (3.21) and (3.22) and using Lemma 2.8, we get

$$\begin{aligned} \limsup _{n\rightarrow +\infty }M(x_n,u)=d(u,Tu) \end{aligned}$$

(3.23)

and

$$\begin{aligned} \limsup _{n\rightarrow +\infty }N(x_n,u)=d(u,Tu). \end{aligned}$$

(3.24)

Similarly, we can obtain

$$\begin{aligned} \liminf _{n\rightarrow +\infty }N(x_n,u)=d(u,Tu). \end{aligned}$$

(3.25)

Again, taking the upper limit as $n\rightarrow +\infty$ in (3.20) and using Lemma 2.8 and (3.23), we obtain

$$\begin{aligned} \psi \big (d(u,Tu)\big )&=\psi \Bigg (s\frac{1}{s}d(u,Tu)\Bigg )\le \psi \Bigg (s\limsup _{n\rightarrow +\infty }d(x_{n+1},Tu)\Bigg )\nonumber \\&\le \psi \Bigg (\limsup _{n\rightarrow +\infty }M(x_n,u)\Bigg )-\liminf _{n\rightarrow +\infty }\varphi \Bigg (N(x_n,u)\Bigg )\nonumber \\&\le \psi \big (d(u,Tu)\big )-\varphi \Bigg (\liminf _{n\rightarrow +\infty }N(x_n,u)\Bigg ), \end{aligned}$$

which yields that $\varphi \big (\liminf _{n\rightarrow +\infty }N(x_n,u)\big )\le 0$ or equivalently $\liminf _{n\rightarrow +\infty }N(x_n,u)=0$. Thus, from (3.25), we get $u=Tu$ and hence u is a fixed point of T. $\square$

Now, we shall prove the uniqueness of the fixed point.

Theorem 3.3

In addition to the hypotheses of Theorem 3.1 (or Theorem 3.2), suppose that for every $x,y\in X$, there exists $u\in X$ such that $u\preceq x$ and $u\preceq y$. Then T has a unique fixed point.

Proof

It follows from Theorem 3.1 (or Theorem 3.2) that the set of fixed points of T is nonempty. We shall show that if $x^*$ and $y^*$ are two fixed points of T, that is, if $x^*=Tx^*$ and $y^*=Ty^*$, then $x^*=y^*$.

By the assumption, there exists $u_0\in X$ such that $u_0\preceq x^*$ and $u_0\preceq y^*$. Then, similarly as in the proof of Theorem 3.1, we define the sequence $\{u_n\}$ in X as follows:

$$\begin{aligned} u_{n+1}=Tu_n=T^{n+1}u_0\quad \text {for all }n\in \mathbb {N}\cup \{0\}. \end{aligned}$$

(3.26)

Monotonicity of T implies that

$$\begin{aligned} T^{n}u_0=u_n\preceq x^*=T^nx^*\quad \text {and}\quad T^{n}u_0=u_n\preceq y^*=T^ny^*. \end{aligned}$$

By induction, we get $z_n\preceq x^*$.

If there exists a positive integer m such that $x^*=u_m$, then $x^*=Tx^*=Tu_n=u_{n+1}$ for all $n\ge m$. Hence, $u_n\rightarrow x^*$ as $n\rightarrow +\infty$.

Now, we suppose that $x^*\ne u_n$ for all $n\in \mathbb {N}\cup \{0\}$. So, $u_n\prec x^*$ for all $n\in \mathbb {N}\cup \{0\}$. Then, $d(u_n,x^*)\ne 0$ for all $n\in \mathbb {N}\cup \{0\}$.

Since $u_n\prec x^*$ for all $n\in \mathbb {N}\cup \{0\}$, by applying (3.1), we have

$$\begin{aligned} \psi \big (d(u_{n+1},x^*)\big )&\le \psi \big (sd(u_{n+1},x^*)\big )\nonumber \nonumber \\&=\psi \big (sd(Tu_n,Tx^*)\big )\nonumber \nonumber \\&\le \psi \big (M(u_n,x^*)\big )-\psi \big (N(u_n,x^*)\big ), \end{aligned}$$

(3.27)

where

$$\begin{aligned} M(u_n,x^*)&=\max \bigg \{d(u_n,x^*),\frac{d(x^*,Tx^*)\big [1+d(u_n,Tu_n)\big ]}{1+d(u_n,x^*)}, \frac{d(x^*,Tu_n)\big [1+d(u_n,Tx^*)\big ]}{1+d(u_n,x^*)}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(u_n,x^*),\frac{d(x^*,x^*)\big [1+d(u_n,u_{n+1})\big ]}{1+d(u_n,x^*)}, \frac{d(x^*,u_{n+1})\big [1+d(u_n,x^*)\big ]}{1+d(u_n,x^*)}\bigg \}\nonumber \nonumber \\&=\max \big \{d(u_n,x^*),d(x^*,u_{n+1})\big \} \end{aligned}$$

(3.28)

and

$$\begin{aligned} N(u_n,x^*)&=\max \bigg \{d(u_n,x^*),\frac{d(x^*,Tx^*)\big [1+d(u_n,Tu_n)\big ]}{1+d(u_n,x^*)}\bigg \}\nonumber \nonumber \\&=\max \bigg \{d(u_n,x^*),\frac{d(x^*,x^*)\big [1+d(u_n,u_{n+1})\big ]}{1+d(u_n,x^*)}\bigg \}\nonumber \nonumber \\&=d(u_n,x^*). \end{aligned}$$

(3.29)

By using (3.27) and taking into account (3.28) and (3.29), we get

$$\begin{aligned} \psi \big (d(u_{n+1},x^*)\big )\le \psi \big (\max \big \{d(u_n,x^*),d(x^*,u_{n+1})\big \}\big )-\varphi \big (d(u_n,x^*)\big ). \end{aligned}$$

(3.30)

Now, if $\max \big \{d(u_n,x^*),d(x^*,u_{n+1})\big \}=d(u_{n+1},x^*)$, then from the above inequality and the property of $\varphi$, we have

$$\begin{aligned} \psi \big (d(u_{n+1},x^*)\big )\le \psi \big (d(u_{n+1},x^*)\big )-\varphi \big (d(u_n,x^*)\big )<\psi \big (d(u_{n+1},x^*)\big ), \end{aligned}$$

which is a contradiction. Hence,

$$\begin{aligned} \max \big \{d(u_n,x^*),d(x^*,u_{n+1})\big \}=d(u_n,x^*) \end{aligned}$$

and from (3.30), we obtain

$$\begin{aligned} \psi \big (d(u_{n+1},x^*)\big )&\le \psi \big (d(u_n,x^*)\big )-\varphi \big (d(u_n,x^*)\big )\nonumber \nonumber \\&<\psi \big (d(u_n,x^*)\big ). \end{aligned}$$

(3.31)

Since $\psi$ is nondecreasing, we have $d(u_{n+1},x^*)<d(u_n,x^*)$, that is, $\big \{d(u_n,x^*)\,:\,n\in \mathbb {N}\cup \{0\}\big \}$ is a decreasing sequence of positive real numbers which is bounded below. Therefore, there exists $r\ge 0$ such that

$$\begin{aligned} \lim _{n\rightarrow +\infty }d(z_n,x^*)=r. \end{aligned}$$

We shall show that $r=0$. Suppose, on the contrary, that $r>0$. By taking the upper limit as $n\rightarrow +\infty$ in (3.31) and using the properties of $\psi$ and $\varphi$, we get

$$\begin{aligned} \psi (r)\le \psi (r)-\varphi (r)<\psi (r), \end{aligned}$$

which is a contradiction. Thus, we conclude that $r=0$. Hence, $d(z_n,x^*)\rightarrow 0$ as $n\rightarrow +\infty$, that is,

$$\begin{aligned} z_n\rightarrow x^*\quad \text {as}\quad n\rightarrow +\infty . \end{aligned}$$

(3.32)

By using a similar argument, we can prove that

$$\begin{aligned} z_n\rightarrow y^*\quad \text {as}\quad n\rightarrow +\infty . \end{aligned}$$

(3.33)

Finally, from (3.32) and (3.33), the uniqueness of the limit gives us $x^*=y^*$. Hence, T has a unique fixed point. $\square$

Corollary 3.4

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a nondecreasing, with respect to $\preceq$, continuous mapping such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} \psi \big (sd(Tx,Ty)\big )\le \psi \big (M(x,y)\big )-\varphi \big (M(x,y)\big ), \end{aligned}$$

(3.34)

where $\psi \in \Psi$, $\varphi \in \Phi$ and

$$\begin{aligned} M(x,y)=\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)},\frac{d(y,Tx)\big [1+d(x,Ty)\big ]}{1+d(x,y)}\bigg \}. \end{aligned}$$

If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

By noting that

$$\begin{aligned}&\psi \bigg (\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\bigg \}\bigg )\nonumber \\&\quad \le \psi \bigg (\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)},\frac{d(y,Tx)\big [1+d(x,Ty)\big ]}{1+d(x,y)}\bigg \}\bigg ), \end{aligned}$$

the proof follows from Theorem 3.1.

Corollary 3.5

Under the same hypotheses of Corollary 3.4, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Corollary 3.6

In addition to the hypotheses of Corollary 3.4 (or Corollary 3.5), suppose that for every $x,y\in X$, there exists $u\in X$ such that $u\preceq x$ and $u\preceq y$. Then T has a unique fixed point.

Corollary 3.7

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a nondecreasing, with respect to $\preceq$, continuous mapping such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} \psi \big (sd(Tx,Ty)\big )\le \psi \big (N(x,y)\big )-\varphi \big (N(x,y)\big ), \end{aligned}$$

(3.35)

where $\psi \in \Psi$, $\varphi \in \Phi$ and

$$\begin{aligned} N(x,y)=\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\bigg \}. \end{aligned}$$

If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

Since the inequality (3.35) implies the inequality (3.1), by Theorem 3.1, we have the required proof.

Corollary 3.8

Under the same hypotheses of Corollary 3.7, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Corollary 3.9

In addition to the hypotheses of Corollary 3.7 (or Corollary 3.8), suppose that for every $x,y\in X$, there exists $u\in X$ such that $u\preceq x$ and $u\preceq y$. Then T has a unique fixed point.

Corollary 3.10

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a nondecreasing, with respect to $\preceq$, continuous mapping. Suppose that there exists $k\in [0,\frac{1}{s})$ such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} d(Tx,Ty)\le \frac{k}{s}\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\bigg \}. \end{aligned}$$

If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

The result follows from Corollary 3.7 by taking $\psi (t)=t$ and $\varphi (t)=(1-k)t$ for all $t\in [0,+\infty )$, where $k\in [0,\frac{1}{s})$.

Corollary 3.11

Under the same hypotheses of Corollary 3.10, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Corollary 3.12

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a nondecreasing, with respect to $\preceq$, continuous mapping. Suppose that there exist $\alpha ,\beta \ge 0$ with $\alpha +\beta <\frac{1}{s}$ such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} sd(Tx,Ty)\le \alpha d(x,y)+\beta \frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}. \end{aligned}$$

(3.36)

If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

Since

$$\begin{aligned}&\alpha d(x,y)+\beta \frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\nonumber \\&\quad \le (\alpha +\beta )\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\bigg \}, \end{aligned}$$

it follows by (3.36) that

$$\begin{aligned} sd(Tx,Ty)\le k\max \bigg \{d(x,y),\frac{d(y,Ty)\big [1+d(x,Tx)\big ]}{1+d(x,y)}\bigg \}, \end{aligned}$$

where $k=\alpha +\beta$. Hence, all the conditions of Corollary 3.10 hold and T has a fixed point.

Corollary 3.13

Under the same hypotheses of Corollary 3.12, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Other consequences of our results are the following for the mappings involving contractions of integral type.

Denote by $\Lambda$ the set of functions $\gamma :[0,+\infty )\rightarrow [0,+\infty )$ satisfying the following hypotheses:

• (1) $\gamma$ is a Lebesgue-integrable mapping on each compact subset of $[0,+\infty )$;

• (2) for every $\varepsilon >0$, we have $\int _{0}^{\varepsilon }\gamma (t)dt>0$.

Corollary 3.14

Let $(X,\preceq )$ be a partially ordered set and suppose that there exists a b-metric space d on X with parameter $s>1$ such that (X, d) is a b-complete b-metric space. Let $T:X\rightarrow X$ be a continuous and nondecreasing mapping with respect to $\preceq$ such that for all $x,y\in X$ with $x\preceq y$,

$$\begin{aligned} \int _{0}^{\psi (sd(Tx,Ty))}\gamma (t)dt\le \int _{0}^{\psi (M(x,y))}\gamma (t)dt-\int _{0}^{\varphi (N(x,y))}\delta (t)dt, \end{aligned}$$

(3.37)

where $\gamma \in \Lambda$ and M(x, y), N(x, y) and the conditions upon $(\psi ,\varphi )$ are the same as in Theorem 3.1. If there exists $x_0\in X$ such that $x_0\preceq Tx_0$, then T has a fixed point.

Proof

Consider the function

$$\begin{aligned} \Gamma _1(x)=\int _0^x\gamma (t)dt\qquad \text {and}\qquad \Gamma _2(x)=\int _0^x\delta (t)dt. \end{aligned}$$

Then, (3.37) becomes

$$\begin{aligned} \Gamma _1\big (\psi \big (sd(Tx,Ty)\big )\big )\le \Gamma _1\big (\psi \big (M(x,y)\big )\big )-\Gamma _2\big (\varphi \big (N(x,y)\big )\big ). \end{aligned}$$

By taking $\psi _0=\Gamma _1\circ \psi$ and $\varphi _0=\Gamma _2\circ \varphi$, it is easy to verify that $\psi _0$ and $\varphi _0$ are control functions. Now, by applying Theorem 3.1, the result follows.

Corollary 3.15

Under the same hypotheses of Corollary 3.14, without the continuity assumption on T, assume that whenever $\{x_n\}$ is a nondecreasing sequence in X such that $x_n\rightarrow x\in X$, one has $x_n\preceq x$ for all $n\in \mathbb {N}$. Then T has a fixed point in X.

Remark 3.16

Since a b-metric is a metric when $s=1$, so our results can be viewed as the generalization and extension of corresponding results in [3, 14, 15, 19] and several other comparable results.

4 Some examples

In this section, we present some examples which illustrate our results.

The following example is an illustration of the existence of the fixed point in case d is continuous in each variable.

Example 4.1

Let $X=\{1,2,3,4\}$ with the usual order “$\preceq$”. Define a b-metric d on X by

$$\begin{aligned} \begin{aligned}&d(1,1)=d(2,2)=d(3,3)=d(4,4)=0,\\&d(1,2)=d(2,1)=d(1,3)=d(3,1)=d(2,3)=d(3,2)=1,\\&d(1,4)=d(4,1)=d(2,4)=d(4,2)=4,\\&d(3,4)=d(4,3)=10. \end{aligned} \end{aligned}$$

Then (X, d) is a b-complete b-metric space with $s=2$. Consider $T:X\rightarrow X$ given as

$$\begin{aligned} T1=T2=T3=1,\qquad T4=2. \end{aligned}$$

Take $\psi (t)=t$ and $\varphi (t)=\frac{t}{2}$ for all $t\in [0,+\infty )$. Note that $\psi \in \Psi$ and $\varphi \in \Phi$. We consider the following two cases.

Case 1 Let $x,y\in \{1,2,3\}$ and $x>y$. Then, we obtain

$$\begin{aligned} \psi \big (2d(Tx,Ty)\big )=0\le \psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ). \end{aligned}$$

Therefore in this case (3.1) is satisfied.

Case 2 Let $x=4$ and $y\in \{1,2,3\}$. Then, we have $d(Tx,Ty)=d(2,1)=1$, $M(4,3)=N(4,3)=10$ and $M(4,y)=N(4,y)=4$ if $y\in \{1,2\}$. Thus, we obtain

$$\begin{aligned} \psi \big (2d(Tx,Ty)\big )=2\le M(x,y)-\frac{N(x,y)}{2}=\psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ). \end{aligned}$$

Therefore in this case (3.1) is satisfied.

Hence, T, $\psi$ and $\varphi$ satisfy all the hypotheses of Theorems 3.1 and 3.2. Thus, T has a fixed point which is $u=1$.

Remark 4.2

In the above example, it is easy to see that d is not a metric on X. This proves that [15, Theorem 3.1] and [15, Theorem 3.2] are not applicable to Example 3.14.

The next example is an illustration of the existence of the fixed point in case d is a noncontinuous b-metric.

Example 4.3

Let $X=\big \{0,1,\frac{1}{2},\frac{1}{3},\cdots ,\frac{1}{n},\cdots \big \}$ with the usual order “$\preceq$”. Define a b-metric d on X by

$$\begin{aligned} d(x,y)={\left\{ \begin{array}{ll} 0, &{}\text { if } x=y,\\ 1, &{}\text { if } x\ne y\in \{0,1\},\\ |x-y|, &{}\text { if } x,y\in \big \{0,\frac{1}{2n},\frac{1}{2m}\,:\,n\ne m\ge 1\big \},\\ 4, &{}\text { otherwise. } \end{array}\right. } \end{aligned}$$

Then (X, d) is a b-complete b-metric space with $s=\frac{8}{3}$ but d is a noncontinuous b-metric (see [29, Example 13]). Consider $T:X\rightarrow X$ defined by

$$\begin{aligned} T0=0,\qquad T\frac{1}{n}=\frac{1}{8n}\quad \text {for all }n\in \mathbb {N}. \end{aligned}$$

Take $\psi (t)=t$ and $\varphi (t)=\frac{2t}{3}$ for all $t\in [0,+\infty )$. Note that $\psi \in \Psi$ and $\varphi \in \Phi$. Then, for $x,y\in X$ with $y\prec x$, we consider the following two cases.

Case 1 Let $x=\frac{1}{n}$ with $n\in \mathbb {N}$ and $y=0$. Then, we have $d(Tx,Ty)=d(\frac{1}{8n},0)=\frac{1}{8n}$, $M(x,y)=N(x,y)=\frac{1}{n}$ or $M(x,y)=N(x,y)\in \{1,4\}$. Thus, we obtain

$$\begin{aligned} \psi \bigg (\frac{8}{3}d(Tx,Ty)\bigg )\le M(x,y)-\frac{2}{3}N(x,y)=\psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ). \end{aligned}$$

Hence in this case (3.1) is satisfied.

Case 2 Let $x=\frac{1}{n}$ and $y=\frac{1}{m}$ with $m>n\ge 1$. Then, we get

$$\begin{aligned} d(Tx,Ty)=d\bigg (\frac{1}{8n},\frac{1}{8m}\bigg )=\frac{1}{8}\bigg (\frac{1}{n}-\frac{1}{m}\bigg ), \end{aligned}$$

$$\begin{aligned} M(x,y)=N(x,y)\ge \frac{1}{n}-\frac{1}{m}\quad \text {or}\quad M(x,y)=N(x,y)=4. \end{aligned}$$

Thus, we have

$$\begin{aligned} \psi \bigg (\frac{8}{3}d(Tx,Ty)\bigg )\le M(x,y)-\frac{2}{3}N(x,y)=\psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ). \end{aligned}$$

Hence in this case (3.1) is satisfied.

Therefore, T, $\psi$ and $\varphi$ satisfy all the hypotheses of Theorems 3.1. Thus, T has a fixed point which is $u=0$.

Remark 4.4

In the above example, it is easy to see that d is not a metric on X. This proves that [15, Theorem 3.1] and [15, Theorem 3.2] are not applicable to Example 4.3.

5 Application to integral equations

In this section, we apply Theorem 3.3 to study the existence and uniqueness of solutions for a class of nonlinear integral equations.

Consider the nonlinear integral equation

$$\begin{aligned} x(t)=g(t)+\int _a^b K(t,s,x(s))ds\quad \text { for all }t\in [a,b]. \end{aligned}$$

(5.1)

Denote by $X=C[a,b]$ the set of real continuous functions defined on [a, b]. We endow X with the b-metric

$$\begin{aligned} d(x,y)=\sup _{t\in [a,b]}|x(t)-y(t)|^p \end{aligned}$$

for all $x,y\in X$ and some $p\ge 1$. It is easy to see that (X, d) is a b-complete b-metric space with the parameter $s=2^{p-1}$. The space X can also be equipped with a partial order given by

$$\begin{aligned} x,y\in X,\quad x\preceq y \quad \Longleftrightarrow \quad x(t)\le y(t)\quad \text { for all }t\in [a,b]. \end{aligned}$$

Moreover, in [30], it is proved that $(X,\preceq )$ is regular.

We will analyze the integral Eq. (5.1) under the following assumptions:

• (i) $K:[a,b]\times [a,b]\times \mathbb {R}\rightarrow \mathbb {R}$ is a continuous function.

• (ii) K(t, s, x(s)) is integrable with respect to s on [a, b].

• (iii) $g:[a,b]\rightarrow \mathbb {R}$ is a continuous function.

• (iv) For all $s,t\in [a,b]$ and $x,y\in X$ with $x\succeq y$,

  $$\begin{aligned} 0\le K(t,s,x(s))-K(t,s,y(s))\le \xi (t,s)|x(s)-y(s)|, \end{aligned}$$

  where $\xi :[a,b]\times [a,b]\rightarrow [0,+\infty )$ is a continuous function satisfying

  $$\begin{aligned} \sup _{t\in [a,b]}\bigg (\int _a^b\xi ^p(t,s)ds\bigg )<\frac{1}{2^{p-1}(b-a)^{p-1}}. \end{aligned}$$

• (v) There exists $x_0\in X$ such that $x_0(t)\le g(t)+\int _a^b K(t,s,x_0(s))ds$ for all $t\in [a,b]$.

Now, we formulate the main result of this section.

Theorem 5.1

Under the assumptions (i)-(v), the Eq. (5.1) has a unique solution in C[a, b].

Proof

We consider the self-map $T:X\rightarrow X$ defined by

$$\begin{aligned} Tx(t)=g(t)+\int _a^b K(t,s,x(s))ds\quad \text { for all } t\in [a,b]. \end{aligned}$$

By virtue of our assumptions, T is well defined, that is, if $x\in X$, then $Tx\in X$. Notice that the existence of a solution to (5.1) is equivalent to the existence of a fixed point of T.

Now, let $x,y\in C[a,b]$ with $x\succeq y$. By applying the condition (iv), we have

$$\begin{aligned} Tx(t)-Ty(t)=\int _a^b\big [K(t,s,x(s))-K(t,s,y(s))\big ]ds\ge 0 \end{aligned}$$

for all $t\in [a,b]$. This implies that $Tx(t)\ge Ty(t)$ for all $t\in [a,b]$ and hence $Tx\succeq Ty$. Therefore, T is a nondecreasing function.

Let $1\le q<+\infty$ with $\frac{1}{p}+\frac{1}{q}=1$. For all $x,y\in X$ with $x\succeq y$ and $t\in [a,b]$, from condition (iv), we have

$$\begin{aligned} 2^{p-1}\big |Tx(t)-Ty(t)\big |^p&\le 2^{p-1}\bigg (\int _a^b\big |K(t,s,x(s))-K(t,s,y(s))\big |ds\bigg )^p\nonumber \\&\le 2^{p-1}\bigg [\bigg (\int _a^b 1^q\bigg )^{\frac{1}{q}}\bigg (\int _a^b\big |K(t,s,x(s))-K(t,s,y(s))\big |^pds\bigg )^{\frac{1}{p}}\bigg ]^p\nonumber \\&\le 2^{p-1}(b-a)^{\frac{p}{q}}\bigg (\int _a^b\xi ^p(t,s)\big |x(s)-y(s)\big |^pds\bigg )\nonumber \\&\le 2^{p-1}(b-a)^{p-1}\bigg (\int _a^b\xi ^p(t,s)ds\bigg )d(x,y)\nonumber \\&\le 2^{p-1}(b-a)^{p-1}\bigg (\int _a^b\xi ^p(t,s)ds\bigg )M(x,y)\nonumber \\&\le \lambda M(x,y)\nonumber \\&=M(x,y)-(1-\lambda )M(x,y), \end{aligned}$$

where $\lambda =2^{p-1}(b-a)^{p-1}\big (\sup _{t\in [a,b]}\int _a^b\xi ^p(t,s)ds\big )\in [0,1)$. Since $M(x,y)\ge N(x,y)$ and $\varphi (t)=(1-\lambda )t$ is a nondecreasing function in $[0,+\infty )$, it follows that

$$\begin{aligned} 2^{p-1}\Bigg (\sup _{t\in [a,b]}\big |Tx(t)-Ty(t)\big |\Bigg )^p\le M(x,y)-(1-\lambda )N(x,y). \end{aligned}$$

Consider the functions $\psi ,\varphi :[0,+\infty )\rightarrow [0,+\infty )$ defined by

$$\begin{aligned} \psi (t)=t\quad \text {and}\quad \varphi (t)=(1-\lambda )t. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \psi \big (sd(Tx,Ty)\big )\le \psi \big (M(x,y)\big )-\varphi \big (N(x,y)\big ), \end{aligned}$$

for all $x,y\in X$ such that $x\succeq y$. This proves that T satisfies the contractive condition (3.1). From the assumption (v), there exists $x_0\in X$ such that $x_0\preceq Tx_0$. Put $z=\max \{x,y\}$ for all $x,y\in X$. Then, we have $z\in X$ and z is comparable to x and y.

From the above, all required assumptions of Theorem 3.3 are satisfied. Consequently, by Theorem 3.3, T has a unique fixed point $x^*\in C[a,b]$ and hence the integral equation (5.1) has a unique solution $x^*\in C[a,b]$. This completes the proof.

Example 5.2

Let $C[0,\frac{\pi }{2}]$ be the set of real continuous functions defined on $[0,\frac{\pi }{2}]$ with the b-metric

$$\begin{aligned} d(x,y)=\sup _{t\in [0,\frac{\pi }{2}]}|x(t)-y(t)|^2 \end{aligned}$$

for all $x,y\in C[0,\frac{\pi }{2}]$. Then $(C[0,\frac{\pi }{2}],d)$ is a b-complete b-metric space with the parameter $s=2$. We endow $C[0,\frac{\pi }{2}]$ with the partial order $\preceq$ given by

$$\begin{aligned} x\preceq y \quad \Longleftrightarrow \quad 0\le x(t)\le y(t)\le \frac{\pi }{2}\quad \text { for all }t\in \left[ 0,\frac{\pi }{2}\right] . \end{aligned}$$

Hence, $(X,\preceq )$ is a partially ordered set and regular.

Consider the nonlinear integral equation

$$\begin{aligned} x(t)=-\sqrt{\frac{96}{\pi ^7}}t^2+t+\sqrt{\frac{96}{\pi ^7}}\int _0^{\frac{\pi }{2}}t^2s\sin x(s)ds \quad \text { for all } t\in \left[ 0,\frac{\pi }{2}\right] . \end{aligned}$$

(5.2)

Observe that this equation is a special case of the equation (5.1) with

$$\begin{aligned} g(t)=-\sqrt{\frac{96}{\pi ^7}}t^2+t\qquad \text {and}\qquad K(t,s,x(s))=\sqrt{\frac{96}{\pi ^7}}t^2s\sin x(s). \end{aligned}$$

Put $Tx(t)=-\sqrt{\frac{96}{\pi ^7}}t^2+t+\sqrt{\frac{96}{\pi ^7}}\int _0^{\frac{\pi }{2}}t^2s\sin x(s)ds$, for all $x\in C[0,\frac{\pi }{2}]$ and all $s,t\in [0,\frac{\pi }{2}]$. Then

• (i) $K:[0,\frac{\pi }{2}]\times [0,\frac{\pi }{2}]\times \mathbb {R}\rightarrow \mathbb {R}$ is a continuous function.

• (ii) Since $x\in C[0,\frac{\pi }{2}]$, it follows that K(t, s, x(s)) is integrable with respect to s on $[0,\frac{\pi }{2}]$.

• (iii) g is continuous on $[0,\frac{\pi }{2}]$.

• (iv) For every $t\in [0,\frac{\pi }{2}]$ and the sequence $\{t_n\}\subset [0,\frac{\pi }{2}]$ with $\lim _{n\rightarrow +\infty }t_n=t$. Thus, for all $x\in C[0,\frac{\pi }{2}]$, we have

  $$\begin{aligned} |Tx(t_n)-Tx(t)|&\le |g(t_n)-g(t)|+\sqrt{\frac{96}{\pi ^7}}\int _0^{\frac{\pi }{2}}s\big |t_n^2-t^2\big ||\sin x(s)|ds\nonumber \\&\le |g(t_n)-g(t)|+\sqrt{\frac{3}{2\pi ^3}}\big |t_n^2-t^2\big |. \end{aligned}$$

  This implies that $\lim _{n\rightarrow +\infty }Tx(t_n)=Tx(t)$ and hence $Tx\in C[0,\frac{\pi }{2}]$ for all $x\in C[0,\frac{\pi }{2}]$.

• (v) For all $s,t\in [0,\frac{\pi }{2}]$ and $x,y\in C[0,\frac{\pi }{2}]$ with $x\succeq y$, we have $0\le y(s)\le x(s)\le \frac{\pi }{2}$. Therefore,

  $$\begin{aligned} 0\le K(t,s,x(s))-K(t,s,y(s))&=\sqrt{\frac{96}{\pi ^7}}t^2s|\sin x(s)-\sin y(s)|\nonumber \\&\le \xi (t,s)|x(s)-y(s)| \end{aligned}$$

  where $xi:[0,\frac{\pi }{2}]\times [0,\frac{\pi }{2}]\rightarrow [0,+\infty )$ defined by $\xi (t,s)=\sqrt{\frac{96}{\pi ^7}}t^2s$. It is easy to check that $\xi$ is a continuous function and

  $$\begin{aligned} \sup _{t\in [0,\frac{\pi }{2}]}\bigg (\int _0^{\frac{\pi }{2}}\xi ^2(t,s)ds\bigg )=\frac{1}{4}<\frac{1}{\pi }. \end{aligned}$$

• (vi) By choosing $x_0(t)=t$ for all $t\in [0,\frac{\pi }{2}]$, we have $Tx_0(t)=t$ for all $t\in [0,\frac{\pi }{2}]$. Therefore, $x_0(t)\le Tx_0(t)$ for all $t\in [0,\frac{\pi }{2}]$.

Consequently, all required assumptions of Theorem 4.1 are satisfied. Hence the integral equation (5.2) has a unique solution in $C[0,\frac{\pi }{2}]$.