# Some applications via fixed point results in partially ordered $$S_{b}$$-metric spaces

## Abstract

In this paper we give some applications to integral equations as well as homotopy theory via fixed point theorems in partially ordered complete $$S_{b}$$-metric spaces by using generalized contractive conditions. We also furnish an example which supports our main result.

## Introduction

Banach contraction principle in metric spaces is one of the most important results in fixed theory and nonlinear analysis in general. Since 1922, when Stefan Banach [1] formulated the concept of contraction and posted a famous theorem, scientists around the world have published new results related to the generalization of a metric space or with contractive mappings (see [124]). Banach contraction principle is considered to be the initial result of the study of fixed point theory in metric spaces.

In the year 1989, Bakhtin introduced the concept of b-metric spaces as a generalization of metric spaces [6]. Later several authors proved so many results on b-metric spaces (see [1316]). Mustafa and Sims defined the concept of a generalized metric space which is called a G-metric space [12]. Sedghi, Shobe and Aliouche gave the notion of an S-metric space and proved some fixed point theorems for a self-mapping on a complete S-metric space [22]. Aghajani, Abbas and Roshan presented a new type of metric which is called $$G_{b}$$-metric and studied some properties of this metric [2].

Recently Sedghi et al. [20] defined $$S_{b}$$-metric spaces using the concept of S-metric spaces [22].

The aim of this paper is to prove some unique fixed point theorems for generalized contractive conditions in complete $$S_{b}$$-metric spaces. Also, we give applications to integral equations as well as homotopy theory. Throughout this paper $$R, R^{+}$$ and N denote the sets of all real numbers, non-negative real numbers and positive integers, respectively.

First we recall some definitions, lemmas and examples.

## Preliminaries

### Definition 2.1

[22]

Let X be a non-empty set. An S-metric on X is a function $$S:X^{3} \to[0,+\infty)$$ that satisfies the following conditions for each $$x,y,z,a \in X$$:

$$(S1)$$::

$$0 < S(x, y, z)$$ for all $$x,y,z \in X$$ with $$x \neq y \neq z \neq x$$,

$$(S2)$$::

$$S(x, y, z) = 0 \mbox{ if and only if } x = y = z$$,

$$(S3)$$::

$$S(x,y,z) \leq S(x ,x, a) + S(y, y, a) + S(z, z, a)$$ for all $$x,y,z,a \in X$$.

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

### Definition 2.2

[20]

Let X be a non-empty set and $$b \geq1$$ be a given real number. Suppose that a mapping $$S_{b}:X^{3} \to\mathcal{[}0,\infty )$$ is a function satisfying the following properties:

$$(S_{b} 1)$$ :

$$0 < S_{b}(x,y,z)$$ for all $$x,y,z \in X$$ with $$x \neq y \neq z \neq x$$,

$$(S_{b} 2)$$ :

$$S_{b}(x, y, z) = 0 \mbox{ if and only if } x = y = z$$,

$$(S_{b} 3)$$ :

$$S_{b}(x,y,z) \leq b(S_{b}(x, x, a) + S_{b}(y, y, a) + S_{b}(z, z, a))$$ for all $$x,y,z,a \in X$$.

Then the function $$S_{b}$$ is called an $$S_{b}$$-metric on X and the pair $$(X,S_{b})$$ is called an $$S_{b}$$-metric space.

### Remark 2.3

[20]

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 an $$S_{b}$$-metric space with $$b=1$$.

The following example shows that an $$S_{b}$$-metric on X need not be an S-metric on X.

### Example 2.4

[20]

Let $$(X,S)$$ be an S-metric space and $$S_{*}(x,y,z) = S(x,y,z)^{p}$$, where $$p>1$$ is a real number. Note that $$S_{*}$$ is an $$S_{b}$$-metric with $$b = 2^{2(p-1)}$$. Also, $$(X,S_{*})$$ is not necessarily an S-metric space.

### Definition 2.5

[20]

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

\begin{aligned} &B_{S_{b}} (x,r) = \bigl\{ y \in X: S_{b}(y,y,x) < r\bigr\} , \\ &B_{S_{b}} [x,r] = \bigl\{ y \in X: S_{b}(y,y,x) \le r\bigr\} . \end{aligned}

### Lemma 2.6

[20]

In an $$S_{b}$$-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 2.7

[20]

In an $$S_{b}$$-metric space, we have

$$S_{b}(x,x,z) \le2 b S_{b}(x,x,y)+b^{2} S_{b}(y,y,z).$$

### Definition 2.8

[20]

If $$(X,S_{b})$$ is an $$S_{b}$$-metric space, a sequence $$\{x_{n}\}$$ in X is said to be:

1. (1)

$$S_{b}$$-Cauchy sequence if, for each $$\epsilon> 0$$, there exists $$n_{0} \in\mathcal {N}$$ such that $$S_{b}(x_{n},x_{n},x_{m}) < \epsilon$$ for each $$m,n \geq n_{0}$$.

2. (2)

$$S_{b}$$-convergent to a point $$x \in X$$ if, for each $$\epsilon> 0$$, there exists a positive integer $$n_{0}$$ such that $$S_{b}(x_{n},x_{n},x) < \epsilon$$ or $$S_{b}(x,x,x_{n}) < \epsilon$$ for all $$n \geq n_{0}$$, and we denote $$\mathop{\lim} _{n \rightarrow\infty}x_{n} = x$$.

### Definition 2.9

[20]

An $$S_{b}$$-metric space $$(X,S_{b})$$ is called complete if every $$S_{b}$$-Cauchy sequence is $$S_{b}$$-convergent in X.

### Lemma 2.10

[20]

If $$(X,S_{b})$$ is an $$S_{b}$$-metric space with $$b\geq1$$, and suppose that $$\{x_{n}\}$$ is $$S_{b}$$-convergent to x, then we have

$$\mathrm{(i)}\quad \frac{1}{2b}S_{b}(y,x,x) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b}(y,y,x_{n}) \le\mathop{\lim} _{n \rightarrow\infty} \sup S_{b}(y,y,x_{n}) \le2b S_{b}(y,y,x)$$

and

$$\mathrm{(ii)}\quad \frac{1}{b^{2}}S_{b}(x,x,y) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b}(x_{n},x_{n},y) \le\mathop{\lim} _{n \rightarrow\infty} \sup S_{b}(x_{n},x_{n},y) \le b^{2} S_{b}(x,x,y)$$

for all $$y \in X$$.

In particular, if $$x=y$$, then we have $$\mathop{\lim} _{n \rightarrow \infty} S_{b}(x_{n},x_{n},y) = 0$$.

Now we prove our main results.

## Results and discussions

### Definition 3.1

Let $$(X, S_{b}, \preceq)$$ be a partially ordered complete $$S_{b}$$-metric space which is said to be regular if every two elements of X are comparable,

$$\mbox{i.e., if } x, y \in X \Rightarrow\mbox{ either } x \preceq y \mbox{ or } y \preceq x.$$

### Definition 3.2

Let $$(X, S_{b}, \preceq)$$ be a partially ordered complete $$S_{b}$$-metric space which is also regular; let $$f: X \to X$$ be a mapping. We say that f satisfies $$(\psi, \phi)$$-contraction if there exist $$\psi, \phi : [0, \infty) \to[0, \infty)$$ such that

(3.2.1):

f is non-decreasing,

(3.2.2):

ψ is continuous, monotonically non-decreasing and ϕ is lower semi-continuous,

(3.2.3):

$$\psi(t) = 0 = \phi(t)$$ if and only if $$t= 0$$,

(3.2.4):

$$\psi (4b^{4} {S_{b} ( {fx, fx, fy} )} ) \le \psi ( { M_{f}^{i} ( {x,y} )} ) - \phi ( {M_{f}^{i} ( {x,y} )} )$$, $$\forall x, y \in X$$, $$x \preceq y$$, $$i = 3, 4, 5$$ and

\begin{aligned} &M_{f}^{5} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy),S_{b}(x, x,fy), S_{b}(y, y, fx) } \right \}, \\ &M_{f}^{4} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy), \frac{1}{4b^{4}} \bigl[S_{b}(x, x,fy)+S_{b}(y, y, fx) \bigr] } \right \}, \\ &M_{f}^{3} ( {x,y} ) = \max\biggl\{ S_{b}(x, x,y), \frac{1}{4b^{4}} \bigl[S_{b}(x, x,fx)+S_{b}(y, y, fy) \bigr],\\ &\phantom{M_{f}^{3} ( {x,y} ) =}\frac{1}{4b^{4}} \bigl[S_{b}(x, x,fy)+S_{b}(y, y, fx) \bigr] \biggr\} . \end{aligned}

### Definition 3.3

Suppose that $$(X, \preceq)$$ is a partially ordered set and f is a mapping of X into itself. We say that f is non-decreasing if for every $$x, y \in X$$,

$$x \preceq y \mbox{ implies that } f x \preceq f y.$$
(1)

### Theorem 3.4

Let $$(X,S_{b}, \preceq)$$ be an ordered complete $$S_{b}$$ metric space, which is also regular, and let $$f : X \to X$$ satisfy $$(\psi, \phi )$$-contraction with $$i = 5$$. If there exists $$x_{0} \in X$$ with $$x_{0} \preceq f x_{0}$$, then f has a unique fixed point in X.

### Proof

Since f is a mapping from X into X, there exists a sequence $$\{x_{n} \}$$ in X such that

$$x_{n+1} = f x_{n}, \quad n = 0, 1, 2, 3, \ldots.$$

Case (i): If $$x_{n} = x_{n+1}$$, then $$x_{n}$$ is a fixed point of f.

Case (ii): Suppose $$x_{n} \neq x_{n+1}\ \forall n$$.

Since $$x_{0} \preceq fx_{0} = x_{1}$$ and f is non-decreasing, it follows that

$$x_{0} \preceq f x_{0} \preceq f^{2} x_{0} \preceq f^{3} x_{0} \preceq\cdots \preceq f^{n} x_{0} \preceq f^{n+1} x_{0} \preceq\cdots.$$

Now

\begin{aligned} \psi \bigl(4b^{4} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr) &= \psi \bigl(4b^{4} S_{b} ( fx_{0}, fx_{0}, fx_{1} ) \bigr) \\ & \leq\psi \bigl(M_{f}^{5} ( x_{0},x_{1} ) \bigr) - \phi \bigl(M_{f}^{5} ( x_{0},x_{1} ) \bigr), \end{aligned}

where

\begin{aligned} M_{f}^{5} ( x_{0},x_{1} ) & = \max \left \{ \begin{matrix} S_{b} ( x_{0}, x_{0}, x_{1} ), S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} ( x_{1}, x_{1}, fx_{1} )\\ S_{b} ( x_{0}, x_{0}, f^{2}x_{0} ), S_{b} ( fx_{0}, fx_{0}, f x_{0} ) \end{matrix} \right \} \\ & = \max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} . \end{aligned}

Therefore

\begin{aligned} &\psi \bigl(4b^{4} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr) \\ &\quad \leq \psi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ) \\ &\qquad{} - \phi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ) \\ &\quad \leq \psi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ). \end{aligned}

By the definition of ψ, we have that

$$S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \max \left \{ \begin{matrix} {\frac{1}{4b^{4}} S_{b} ( x_{0}, x_{0}, fx_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )}\\ {\frac{1}{4b^{4}} S_{b} ( x_{0}, x_{0}, f^{2}x_{0} )} \end{matrix} \right \}.$$
(2)

But

\begin{aligned} \frac{1}{4b^{4}} S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr)&\leq \frac{1}{4b^{4}} \bigl[2b S_{b} ( x_{0}, x_{0}, fx_{0} ) + b^{2} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr] \\ &\leq \max \biggl\{ \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ) , \frac{1}{2b^{2}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \biggr\} . \end{aligned}

From (2) we have that

\begin{aligned} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \max \biggl\{ \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ) , \frac{1}{2b^{2}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \biggr\} . \end{aligned}

If $$\frac{1}{2b^{2}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )$$ is maximum, we get a contradiction. Hence

$$S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ).$$
(3)

Also

\begin{aligned} \psi \bigl(4b^{4} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr) &= \psi \bigl(4b^{4} S_{b} ( fx_{1}, fx_{1}, fx_{2} ) \bigr) \\ & \leq\psi \bigl(M_{f}^{5} ( x_{1},x_{2} ) \bigr) - \phi \bigl(M_{f}^{4} ( x_{1},x_{2} ) \bigr), \end{aligned}

where

\begin{aligned} M_{f}^{5} ( x_{1},x_{2} ) & = \max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{2} x_{0} ) \end{matrix} \right \} \\ & = \max \left \{ S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr), S_{b} \bigl( fx_{0}, fx_{0}, f^{3}x_{0} \bigr) \right \} . \end{aligned}

Therefore

\begin{aligned} &\psi \bigl(4b^{4} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr)\\ &\quad\leq \psi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ) \\ &\qquad{} - \phi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ) \\ &\quad\leq \psi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ). \end{aligned}

By the definition of ψ, we have that

$$S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq \max \left \{ \begin{matrix} {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} )} \end{matrix} \right \}.$$
(4)

But

\begin{aligned} &\frac{1}{4b^{4}} S_{b} \bigl( fx_{0}, fx_{0}, f^{3}x_{0} \bigr)\\ &\quad\leq \frac {1}{4b^{4}} \bigl[2b S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) + b^{2} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr] \\ &\quad\leq \max \biggl\{ \frac{1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) , \frac{1}{2b^{2}} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \biggr\} . \end{aligned}

From (4) we have that

\begin{aligned} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq \max \biggl\{ \frac {1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) , \frac{1}{2 b^{2}} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \biggr\} . \end{aligned}

If $$\frac{1}{2b^{2}} S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )$$ is maximum, we get a contradiction. Hence

\begin{aligned} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq& \frac{1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \\ \le&\frac{1}{(b^{3})^{2}} S_{b} ( x_{0}, x_{0}, fx_{0} ) . \end{aligned}

Continuing this process, we can conclude that

\begin{aligned} S_{b} \bigl( f^{n}x_{0}, f^{n}x_{0}, f^{n+1}x_{0} \bigr) &\leq \frac {1}{(b^{3})^{n}} S_{b} ( x_{0}, x_{0}, f x_{0} ) \\ &\to 0 \quad\mbox{as } n \to\infty. \end{aligned}

That is,

\begin{aligned} \mathop{\lim} _{n \rightarrow\infty} S_{b} \bigl( f^{n}x_{0}, f^{n}x_{0}, f^{n+1}x_{0} \bigr) = 0. \end{aligned}
(5)

Now we prove that $$\{ f^{n} x_{0} \}$$ is an $$S_{b}$$-Cauchy sequence in $$(X, S_{b})$$. On the contrary, we suppose that $$\{ f^{n} x_{0} \}$$ is not $$S_{b}$$-Cauchy. Then there exist $$\epsilon> 0$$ and monotonically increasing sequences of natural numbers $$\{m_{k}\}$$ and $$\{ n_{k}\}$$ such that $$n_{k} > m_{k}$$.

$$S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr) \geq\epsilon$$
(6)

and

$$S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k} - 1}x_{0} \bigr) < \epsilon .$$
(7)

From (6) and (7), we have

\begin{aligned} \epsilon\le{}& S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr) \\ \le{}& 2b S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} \bigr) \\ &{}+ b^{2} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr). \end{aligned}

So that

\begin{aligned} 4 b^{2} \epsilon\le{}& 8 b^{3} S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} \bigr)\\ &{} + 4b^{4} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr). \end{aligned}

Letting $$k \to\infty$$ and applying ψ on both sides, we have that

\begin{aligned} \psi \bigl(4 b^{2} \epsilon \bigr) &\le \mathop{\lim} _{k \rightarrow \infty} \psi \bigl(4b^{4} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr) \bigr) \\ &= \mathop{\lim} _{k \rightarrow\infty} \psi \bigl(4b^{4} S_{b} ( fx_{m_{k}}, fx_{m_{k}}, fx_{n_{k}-1} ) \bigr) \\ &\leq \mathop{\lim} _{k \rightarrow\infty} \psi \bigl( M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \bigr) - \mathop{\lim} _{k \rightarrow \infty} \phi \bigl( M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \bigr), \end{aligned}
(8)

where

\begin{aligned} & \mathop{\lim} _{k \rightarrow\infty} M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \\ &\quad= \mathop{\lim} _{k \rightarrow\infty} \max \left \{ \begin{matrix} {S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}-1}x_{0} ), S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} ) }\\ {S_{b} ( f^{n_{k}-1}x_{0} , f^{n_{k}-1}x_{0} , f^{n_{k}}x_{0} )}, S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} ) \\ S_{b} ( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} ) \end{matrix} \right \} \\ &\quad < \mathop{\lim} _{k \rightarrow\infty} \max \left \{ {\epsilon, 0, 0, } S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr), S_{b} \bigl( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} \bigr) \right \}. \end{aligned}

But

\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr)&\le \mathop{\lim} _{k \rightarrow \infty} \left [ \begin{matrix} 2b S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}-1}x_{0} )\\ {}+ b^{2} S_{b} ( f^{n_{k}-1}x_{0}, f^{n_{k}-1}x_{0}, f^{n_{k}}x_{0} ) \end{matrix} \right ] < 2 b \epsilon. \end{aligned}

Also

\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} S_{b} \bigl( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} \bigr) &\le \mathop {\lim} _{k \rightarrow\infty} \left [ \begin{matrix} 2b S_{b} ( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}}x_{0} )\\ {}+ b^{2}S_{b} ( f^{m_{k}}x_{0},f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} ) \end{matrix} \right ] < 2 b^{2} \epsilon. \end{aligned}

Therefore

\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \le& \max \left \{ {\epsilon, 2 b \epsilon, 2 b^{2} \epsilon} \right \} \\ =& 2 b^{2} \epsilon. \end{aligned}

From (8), by the definition of ψ, we have that

$$4 b ^{2} \epsilon\le 2 b^{2} \epsilon,$$

which is a contradiction. Hence $$\{ f^{n} x_{0} \}$$ is an $$S_{b}$$-Cauchy sequence in complete regular $$S_{b}$$-metric spaces $$(X, S_{b}, \preceq)$$. By the completeness of $$(X, S_{b})$$, it follows that the sequence $$\{ f^{n} x_{0} \}$$ converges to α in $$( X, S_{b} )$$. Thus

$$\mathop{\lim} _{k \rightarrow\infty} f^{n} x_{0} = \alpha= \mathop{\lim } _{k \rightarrow\infty} f^{n+1} x_{0}.$$

Since $$x_{n}, \alpha\in X$$ and X is regular, it follows that either $$x_{n} \preceq\alpha$$ or $$\alpha\preceq x_{n}$$.

Now we have to prove that α is a fixed point of f.

Suppose $$f\alpha\neq\alpha$$, by Lemma (2.10), we have that

$$\frac{1}{2b} S_{b}(f\alpha, f\alpha, \alpha) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b} (f\alpha, f\alpha, f^{n+1}x_{0} ).$$

Now from (3.2.4) and applying ψ on both sides, we have that

\begin{aligned} \psi \bigl(2b^{3} S_{b}(f\alpha, f\alpha, \alpha) \bigr) \le{}& \mathop {\lim} _{n \rightarrow\infty} \inf \psi \bigl(4 b^{4} S_{b} \bigl(f\alpha, f\alpha, f^{n+1}x_{0} \bigr) \bigr) \\\le{}& \mathop{\lim} _{n \rightarrow\infty} \inf \psi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr) - \mathop{\lim} _{n \rightarrow\infty} \inf \phi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr). \end{aligned}
(9)

Here

\begin{aligned} \mathop{\lim} _{n \rightarrow\infty} \inf M_{f}^{5} ( \alpha, x_{n} ) &= \mathop{\lim} _{n \rightarrow\infty} \inf\max \left \{ \begin{matrix} {S_{b} (\alpha, \alpha, x_{n} ), S_{b} (\alpha, \alpha, f\alpha ), S_{b} (x_{n}, x_{n}, fx_{n} ) }\\ {S_{b} (\alpha, \alpha, fx_{n} ), S_{b} (x_{n}, x_{n}, f\alpha ) } \end{matrix} \right \} \\ &\le \mathop{\lim} _{n \rightarrow\infty} \sup\max \left \{ {0, S_{b} ( \alpha, \alpha, f\alpha ), 0, 0, S_{b} (x_{n}, x_{n}, f\alpha ) } \right \} \\ & \le\max \left \{ S_{b} (\alpha, \alpha, f\alpha ), b^{2} S_{b} (\alpha, \alpha, f\alpha ) \right \} \\ &\le b^{3} S_{b} (f\alpha, f\alpha, \alpha ). \end{aligned}

Hence from (9) we have that

\begin{aligned} \psi \bigl(2 b^{3} S_{b}(f\alpha, f\alpha, \alpha) \bigr) & \le \psi \bigl(b^{3}S_{b} (\alpha, \alpha, f\alpha ) \bigr) - \mathop {\lim} _{n \rightarrow\infty} \inf \phi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr) \\ &\le \psi \bigl(b^{3} S_{b} (f\alpha, f\alpha, \alpha ) \bigr) , \end{aligned}

which is a contradiction. So that α is a fixed point of f.

Suppose that $$\alpha^{\ast}$$ is another fixed point of f such that $$\alpha\neq\alpha^{\ast}$$.

Consider

\begin{aligned} \psi \bigl( 4 b^{4} S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr) \bigr) \le{}& \psi \bigl( M_{f}^{5} \bigl(\alpha, \alpha^{\ast} \bigr) \bigr) - \phi \bigl( M_{f}^{5} \bigl( \alpha, \alpha^{\ast} \bigr) \bigr) \\ = {}& \psi \bigl(\max \bigl\{ S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr), S_{b} \bigl(\alpha^{\ast}, \alpha^{\ast}, \alpha \bigr) \bigr\} \bigr) \\ &{}- \phi \bigl(\max \bigl\{ S_{b} \bigl(\alpha, \alpha, \alpha^{\ast } \bigr), S_{b} \bigl(\alpha^{\ast}, \alpha^{\ast}, \alpha \bigr) \bigr\} \bigr) \\ \le{} & \psi \bigl(b S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr) \bigr), \end{aligned}

Hence α is a unique fixed point of f in $$(X, S_{b} )$$. □

### Example 3.5

Let $$X = [0, 1]$$ and $$S : X \times X \times X \to\mathbb {R}^{+}$$ by $$S_{b}(x,y,z) = ( \vert y+z-2x \vert + \vert y-z \vert )^{2}$$ and by $$a \preceq b \iff a\le b$$, then $$(X, S_{b} , \preceq )$$ is a complete ordered $$S_{b}$$-metric space with $$b = 4$$. Define $$f: X \rightarrow X$$ by $$f(x) = \frac{x}{32\sqrt{2}}$$. Also define $$\psi,\phi:\mathbb{R}^{+} \to\mathbb{R}^{+}$$ by $$\psi(t) = t$$ and $$\phi(t) = \frac{t}{2}$$.

\begin{aligned} \psi \bigl(4 b^{4} S_{b}(fx, fx, fy) \bigr) &= 4 b^{4} \bigl( \vert fx + fy - 2 fx \vert + \vert fx - fy \vert \bigr)^{2} \\ &= 4 b^{4} \biggl(2 \biggl\vert \frac{x}{32\sqrt{2}} - \frac{y}{32\sqrt {2}} \biggr\vert \biggr)^{2} \\ &= \frac{4 b^{4}}{8 b^{4}}S_{b}(x,x, y) \\ &\le \frac{1}{2}M_{f}^{5}(x, y) \\ &\le \psi \bigl(M_{f}^{5}(x, y) \bigr) - \phi \bigl(M_{f}^{5}(x, y) \bigr), \end{aligned}

where

$$M_{f}^{5} ( {x,y} ) = \max \left \{ S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy),S_{b}(x, x,fy), S_{b}(y, y, fx) \right \}.$$

Hence, all the conditions of Theorem 3.4 are satisfied and 0 is a unique fixed point of f.

### Theorem 3.6

Let $$(X,S_{b}, \preceq)$$ be an ordered complete $$S_{b}$$ metric space, and let $$f : X \to X$$ satisfy $$(\psi, \phi)$$-contraction with $$i = 3\textit{ or }4$$. If there exists $$x_{0} \in X$$ with $$x_{0} \preceq f x_{0}$$, then f has a unique fixed point in X.

### Proof

Follows along similar lines of Theorem 3.4 if we take $$M_{f}^{3} ( {x,y} )$$ or $$M_{f}^{4} ( {x,y} )$$ in place of $$M_{f}^{5} ( {x,y} )$$ in Theorem 3.4. □

### Theorem 3.7

Let $$(X,S_{b}, \preceq)$$ be an ordered complete $$S_{b}$$ metric space, and let $$f : X \to X$$ satisfy

$$4 b^{4} S_{b} ( {fx, fx, fy} ) \le M_{f}^{i} ( {x,y} ) - \varphi \bigl( {M_{f}^{i} ( {x,y} )} \bigr),$$

where $$\varphi:[0, \infty) \to[0, \infty)$$ and i= 3 or 4 or 5. If there exists $$x_{0} \in X$$ with $$x_{0} \preceq f x_{0}$$, then f has a unique fixed point in X.

### Proof

The proof follows from Theorems 3.4 and 3.6 by taking $$\psi(t) = t$$ and $$\phi(t) = \varphi(t)$$. □

### Theorem 3.8

Let $$(X,S_{b}, \preceq)$$ be an ordered complete $$S_{b}$$ metric space, and let $$f : X \to X$$ satisfy

$$S_{b} ( {fx, fx, fy} ) \le\lambda M_{f}^{i} ( {x,y} ),$$

where $$\lambda\in [ {0, \frac{1}{4b^{4}}} )$$ and $$i = 3, 4, 5$$. If there exists $$x_{0} \in X$$ with $$x_{0} \preceq f x_{0}$$, then f has a unique fixed point in X.

### Application to integral equations

In this section, we study the existence of a unique solution to an initial value problem as an application to Theorem 3.4.

### Theorem 3.9

Consider the initial value problem

$$x^{1} (t) = T \bigl(t, x(t) \bigr),\quad t \in I= [0, 1], x(0) = x_{0},$$
(10)

where $$T: I \times [ {\frac{{x_{0} }}{4},\infty} ) \to [ {\frac{{x_{0} }}{4},\infty} )$$ and $$x_{0} \in\mathbb{R}$$. Then there exists a unique solution in $$C (I, [ {\frac{{x_{0} }}{4},\infty} ) )$$ for initial value problem (10).

### Proof

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

\begin{aligned} x(t) = x_{0} +3 b^{2} \int_{0}^{t} {T \bigl(s, x(s) \bigr)} \,ds . \end{aligned}

Let $$X =C (I, [ {\frac{{x_{0} }}{4},\infty} ) )$$ and $$S_{b}(x,y,z) = ( \vert y+z-2x \vert + \vert y-z \vert )^{2} for x, y \in X$$. Define $$\psi, \phi: [0, \infty) \to[0, \infty)$$ by $$\psi(t) = t$$, $$\phi(t)= \frac{5t}{9}$$. Define $$f: X \to X$$ by

$$f(x) (t) = \frac{x_{0}}{3b^{2}} + \int_{0}^{t} {T \bigl(s, x(s) \bigr)} \,ds .$$
(11)

Now

\begin{aligned} &\psi \bigl(4 b^{4} S_{b}\bigl(fx(t), fx(t), fy(t)\bigr) \bigr) \\ &\quad = 4 b^{4} \bigl\{ { \bigl\vert fx(t)+ fy(t)-2fx(t) \bigr\vert + \bigl\vert fx(t)- fy(t) \bigr\vert } \bigr\} ^{2} \\ &\quad = 16 b^{4} \bigl\vert fx(t)- fy(t) \bigr\vert ^{2} \\ &\quad = \frac{16 b^{4}}{9b^{4}} \biggl\vert x_{0} + 3 b^{2} \int_{0}^{t} {T\bigl(s, x(s)\bigr)} \,ds - y_{0} - 3 b^{2} \int_{0}^{t} {T\bigl(s, y(s)\bigr)} \,ds \biggr\vert ^{2} \\ & \quad= \frac{16}{9} \bigl\vert x(t) - y(t) \bigr\vert ^{2} \\ &\quad = \frac{4}{9} S(x,x,y) \\ & \quad\le\frac{4}{9} M_{f}^{5}(x, y) \\ &\quad= \psi \bigl(M_{f}^{5}(x, y) \bigr) - \phi \bigl(M_{f}^{5}(x, y) \bigr) , \end{aligned}

where

$$M_{f}^{5} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy), S_{b}(x, x,fy), S_{b}(y, y, fx) } \right \}.$$

It follows from Theorem 3.4 that f has a unique fixed point in X. □

### Application to homotopy

In this section, we study the existence of a unique solution to homotopy theory.

### Theorem 3.10

Let $$(X, S_{b})$$ be a complete $$S_{b}$$-metric space, U be an open subset of X and be a closed subset of X such that $$U \subseteq\overline{U}$$. Suppose that $$H : \overline{U} \times[0, 1] \to X$$ is an operator such that the following conditions are satisfied:

1. (i)

$$x \neq H(x, \lambda)$$ for each $$x \in\partial{U}$$ and $$\lambda \in[0, 1]$$ (here ∂U denotes the boundary of U in X),

2. (ii)

$$\psi(4b^{4} S_{b}(H(x, \lambda),H(x, \lambda), H(y, \lambda) )) \leq\psi( S_{b}(x, x, y)) - \phi( S_{b}(x, x, y))$$ $$\forall x, y \in\overline{U}$$ and $$\lambda\in[0, 1]$$, where $$\psi :[0,\infty) \to[0,\infty)$$ is continuous, non-decreasing and $$\phi :[0,\infty) \to[0,\infty)$$ is lower semi-continuous with $$\phi(t)>0$$ for $$t>0$$,

3. (iii)

there exists $$M\geq0$$ such that

$$S_{b} \bigl(H(x, \lambda),H(x, \lambda), H(x, \mu) \bigr) \leq M \vert \lambda - \mu \vert$$

for every $$x \in\overline{U}$$ and $$\lambda, \mu\in[0, 1]$$.

Then $$H(\cdot, 0)$$ has a fixed point if and only if $$H(\cdot, 1)$$ has a fixed point.

### Proof

Consider the set

$$A = \bigl\{ \lambda\in[0, 1] : x = H(x, \lambda) \mbox{ for }\mbox{ some }x \in U \bigr\} .$$

Since $$H(\cdot, 0)$$ has a fixed point in U, we have that $$0 \in A$$. So that A is a non-empty set.

We will show that A is both open and closed in $$[0, 1]$$, and so, by the connectedness, we have that $$A = [0, 1]$$. As a result, $$H(\cdot, 1)$$ has a fixed point in U. First we show that A is closed in $$[0, 1]$$. To see this, let $$\{ { \lambda_{n}} \}_{n = 1}^{\infty }\subseteq A$$ with $$\lambda_{n} \to\lambda\in[0, 1]$$ as $$n \to\infty$$.

We must show that $$\lambda\in A$$. Since $$\lambda_{n} \in A$$ for $$n = 1, 2, 3, \ldots$$ , there exists $$x_{n} \in U$$ with $$x_{n} = H(x_{n}, \lambda _{n})$$.

Consider

\begin{aligned} S_{b}(x_{n}, x_{n}, x_{n + 1}) ={}& S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n+1}) \bigr) \\ \leq{}&2b S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda _{n})\bigr) \\ &{} + b^{2}S_{b}\bigl(H(x_{n+1}, \lambda_{n}), H(x_{n+1},\lambda_{n}), H(x_{n+1}, \lambda_{n+1})\bigr) \\ \leq{}& S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n})\bigr) + M \vert \lambda_{n} - \lambda_{n+1} \vert . \end{aligned}

Letting $$n \to\infty$$, we get

$$\mathop{\lim} _{n \to\infty} S_{b}(x_{n}, x_{n}, x_{n + 1}) \leq \mathop{\lim} _{n \to\infty} S_{b} \bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda _{n}), H(x_{n+1}, \lambda_{n}) \bigr) + 0.$$

Since ψ is continuous and non-decreasing, we obtain

\begin{aligned} \mathop{\lim} _{n \to\infty} \psi\bigl(4b^{4} S_{b}(x_{n}, x_{n}, x_{n + 1})\bigr) &\leq\mathop{\lim} _{n \to\infty}\psi \bigl( 4b^{4} S_{b}\bigl(H(x_{n}, \lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n})\bigr)\bigr) \\ &\leq \mathop{\lim} _{n \to\infty} \bigl[ {\psi\bigl( S_{b}(x_{n}, x_{n}, x_{n+1})\bigr) - \phi\bigl( S_{b}(x_{n}, x_{n}, x_{n+1})\bigr)} \bigr]. \end{aligned}

By the definition of ψ, it follows that

$$\mathop{\lim} _{n \to\infty} \bigl(4 b^{4} -1 \bigr)S_{b} (x_{n}, x_{n}, x_{n+1} ) \leq0.$$

So that

$$\mathop{\lim} _{n \to\infty} S_{b}(x_{n}, x_{n}, x_{n+1}) = 0 .$$
(12)

Now we prove that $$\{x_{n}\}$$ is an $$S_{b}$$-Cauchy sequence in $$(X, d_{p})$$. On the contrary, suppose that $$\{x_{n} \}$$ is not $$S_{b}$$-Cauchy.

There exists $$\epsilon> 0$$ and monotone increasing sequences of natural numbers $$\{ m_{k}\}$$ and $$\{ n_{k}\}$$ such that $$n_{k} > m_{k}$$,

$$S_{b}(x_{m_{k}},x_{m_{k}}, x_{n_{k}}) \geq \epsilon$$
(13)

and

$$S_{b}(x_{m_{k}}, x_{m_{k}}, x_{n_{k}-1}) < \epsilon.$$
(14)

From (13) and (14), we obtain

\begin{aligned} \epsilon&\leq S_{b}(x_{m_{k}}, x_{m_{k}}, x_{n_{k}}) \\ &\leq2 b S_{b}(x_{m_{k}},x_{m_{k}}, x_{m_{k} + 1}) + b^{2} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}). \end{aligned}

Letting $$k \to\infty$$ and applying ψ on both sides, we have that

$$\psi \bigl(2b^{2} \epsilon \bigr) \leq \mathop{\lim} _{n \to\infty } \psi \bigl(4b^{4} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \bigr).$$
(15)

But

\begin{aligned} &\mathop{\lim} _{n \to\infty}\psi \bigl( 4b^{4} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \bigr)\\ & \quad= \mathop{\lim} _{n \to\infty}\psi \bigl( S_{b}\bigl( 4b^{4} H(x_{m_{k}+1}, \lambda_{m_{k}+1}), H(x_{m_{k}+1},\lambda_{m_{k}+1}), H(x_{n_{k}}, \lambda_{n_{k}})\bigr)\bigr) \\ &\quad\leq \mathop{\lim} _{n \to\infty} \bigl[ {\psi\bigl( S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}})\bigr) - \phi\bigl( S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}})\bigr)} \bigr]. \end{aligned}

It follows that

$$\mathop{\lim} _{n \to\infty} \bigl( 4b^{4} -1 \bigr) S_{b} (x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \leq0.$$

Thus

$$\mathop{\lim} _{n \to\infty} S_{b} (x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) = 0.$$

Hence from (15) and the definition of ψ, we have that

$$\epsilon\le0,$$

Hence $$\{x_{n} \}$$ is an $$S_{b}$$-Cauchy sequence in $$(X, S_{b})$$ and, by the completeness of $$(X, S_{b})$$, there exists $$\alpha\in U$$ with

\begin{aligned} &\mathop{\lim} _{n \to\infty}x_{n} = \alpha= \mathop{\lim} _{n \to\infty} x_{n+1}, \\ &\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr) \leq\mathop{\lim} _{n \to\infty} \inf\psi \bigl(4b^{4} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), H(x_{n}, \lambda)\bigr) \bigr) \\ &\phantom{\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr)}\leq\mathop{\lim} _{n \to\infty} \inf\bigl[\psi\bigl( S_{b}(\alpha, \alpha, x_{n})\bigr) - \phi\bigl( S_{b}(\alpha, \alpha, x_{n})\bigr) \bigr] \\ &\phantom{\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr)}= 0. \end{aligned}
(16)

It follows that $$\alpha= H(\alpha, \lambda)$$.

Thus $$\lambda\in A$$. Hence A is closed in $$[0, 1]$$.

Let $$\lambda_{0} \in A$$. Then there exists $$x_{0} \in U$$ with $$x_{0} = H(x_{0}, \lambda_{0})$$.

Since U is open, there exists $$r > 0$$ such that $$B_{S_{b}}(x_{0}, r) \subseteq U$$.

Choose $$\lambda\in(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon)$$ such that $$\vert \lambda- \lambda_{0} \vert \leq\frac{1}{M^{n}} < \epsilon$$.

Then, for $$x \in\overline{B_{p} (x_{0}, r)} = \{x \in X / S_{b}(x, x, x_{0}) \leq r + b^{2} S_{b}(x_{0}, x_{0}, x_{0}) \}$$,

\begin{aligned} &S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0}\bigr)\\ &\quad= S_{b}\bigl(H(x,\lambda),H(x,\lambda ), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq2 b S_{b}\bigl(H(x,\lambda),H(x,\lambda), H(x, \lambda_{0})\bigr) + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq2 b M \vert \lambda- \lambda_{0} \vert + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq\frac{2 b}{M^{n-1}} + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr). \end{aligned}

Letting $$n \to\infty$$, we obtain

$$S_{b} \bigl(H(x, \lambda), H(x, \lambda), x_{0} \bigr) \leq b^{2} S_{b} \bigl(H(x, \lambda_{0}), H(x, \lambda_{0}), H(x_{0}, \lambda_{0}) \bigr).$$

Since ψ is continuous and non-decreasing, we have

\begin{aligned} \psi\bigl(S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0} \bigr)\bigr) &\leq\psi\bigl(4b^{2}S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0}\bigr)\bigr) \\ &\leq \psi\bigl( 4b^{4} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda _{0})\bigr)\bigr) \\ &\leq\psi\bigl( S_{b}(x, x, x_{0})\bigr) - \phi\bigl( S_{b}(x, x, x_{0})\bigr) \\ &\leq \psi\bigl( S_{b}(x, x, x_{0})\bigr). \end{aligned}

Since ψ is non-decreasing, we have

\begin{aligned} S_{b}\bigl(H(x, \lambda),H(x, \lambda), x_{0}\bigr) &\leq S_{b}(x, x, x_{0}) \\ &\leq r + b^{2} S_{b}(x_{0}, x_{0}, x_{0}). \end{aligned}

Thus, for each fixed $$\lambda\in(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon)$$, $$H(\cdot, \lambda): \overline{B_{p} (x_{0}, r)} \to\overline {B_{p} (x_{0}, r)}$$.

Since also (ii) holds and ψ is continuous and non-decreasing and ϕ is continuous with $$\phi(t)> 0$$ for $$t > 0$$, then all the conditions of Theorem (3.10) are satisfied.

Thus we deduce that $$H(\cdot, \lambda)$$ has a fixed point in . But this fixed point must be in U since (i) holds.

Thus $$\lambda\in A$$ for any $$\lambda\in(\lambda_{0} - \epsilon, \lambda _{0} + \epsilon)$$.

Hence $$(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon) \subseteq A$$ and therefore A is open in [0, 1].

For the reverse implication, we use the same strategy. □

### Corollary 3.11

Let $$(X, p)$$ be a complete partial metric space, U be an open subset of X and $$H : \overline{U} \times[0, 1] \to X$$ with the following properties:

1. (1)

$$x \neq H(x, t)$$ for each $$x \in\partial U$$ and each $$\lambda \in[0, 1]$$ (here ∂U denotes the boundary of U in X),

2. (2)

there exist $$x, y \in\overline{U}$$ and $$\lambda\in[0, 1], L \in [0, \frac{1}{4b^{4}} )$$ such that

$$S_{b} \bigl(H(x, \lambda),H(x, \lambda), H(y, \mu) \bigr) \leq L S_{b}(x, x, y),$$
3. (3)

there exists $$M \geq0$$ such that

$$S_{b} \bigl(H(x, \lambda), H(x, \lambda), H(x, \mu) \bigr) \leq M \vert \lambda- \mu \vert$$

for all $$x \in\overline{U}$$ and $$\lambda, \mu\in[0, 1]$$.

If $$H(\cdot, 0)$$ has a fixed point in U, then $$H(\cdot, 1)$$ has a fixed point in U.

### Proof

Proof follows by taking $$\psi(x) = x, \phi(x) = x - Lx \mbox{ with } L \in [0, \frac{1}{4b^{4}} )$$ in Theorem (3.10). □

## Conclusions

In this paper we conclude some applications to homotopy theory and integral equations by using fixed point theorems in partially ordered $$S_{b}$$-metric spaces.

## References

1. Banach, S: Theorie des Operations Lineaires. Manograic Mathematic Zne, Warsaw (1932)

2. Aghajani, A, Abbas, M, Roshan, JR: Common fixed point of generalized weak contractive mappings in partially ordered $$G_{b}$$-metric spaces. Filomat 28(6), 1087-1101 (2014)

3. Ansari, AH, Ege, O, Randenović, S: Some fixed point results on complex valued $$G_{b}$$-metric spaces. RACSAM (2017). doi:10.1007/s13398-017-0391-x

4. Anasari, AH, Vetro, P, Randenović, S: Integration of type pair $$(H, F)$$ upclass in fixed point result for $$GP_{(\Lambda,\Theta)}$$ contractive mappings. Filomat 31(8), 2211-2218 (2017)

5. Arshad, M, Kadelburg, Z, Randenović, S, Shoaib, A, Shukla, S: α-dominated mappings on dislocated metric spaces and fixed point results. Filomat 31(11), 3041-3056 (2017)

6. Bakhtin, IA: The contraction mapping principle in quasimetric spaces. Funct. Anal. Unianowsk Gos. Ped. Inst. 30, 26-37 (1989)

7. Czerwik, S: Contraction mapping in b-metric spaces. Acta Math. Inform. Univ. Ostrav. 1, 5-11 (1993)

8. Došenović, T, Randenović, S: Some critical remarks on the paper: “An essential remark on fixed point results on multiplicative metric spaces”. J. Adv. Math. Stud. 10(1), 20-24 (2017)

9. Jaradat, MMM, Mustafa, Z, Ansari, AH, Kumari, PS, Djekić, DD, Jaradat, HM: Some fixed point results for $$F_{\alpha- \omega,\varphi }$$-generalized cyclic contractions on metric-like space with applications to graphs and integral equations. J. Math. Anal. 8(1), 28-45 (2017)

10. Kishore, GNV, Rao, KPR, Hima Bindu, VML: Suzuki type unique common fixed point theorem in partial metric spaces by using (C): condition with rational expressions. Afr. Math. (2016). doi:10.1007/s13370-017-0484-x

11. Lakshmikantham, V, Lj, C: Coupled fixed point theorems for nonlinear contractions in partially ordered metric spaces. Nonlinear Anal. 70(12), 4341-4349 (2009)

12. Mustafa, Z, Sims, B: A new approach to generalized metric spaces. J. Nonlinear Convex Anal. 7(2), 289-297 (2006)

13. Mustafa, Z, Roshan, JR, Parvaneh, V, Kadelburg, Z: Some common fixed point results in ordered partial b-metric space. J. Inequal. Appl. 2013, 562 (2013)

14. Mustafa, Z, Roshan, JR, Parvaneh, V, Kadelburg, Z: Fixed point theorems for weakly T-Chatterjea and weakly T-Kannan contractions in b-metric spaces. J. Inequal. Appl. 2014, 46 (2014)

15. Mustafa, Z, Jaradat, M, Ansari, AH, Popovi, BZ, Jaradat, H: C-class functions with new approach on coincidence point results for generalized $$(\psi, \phi)$$-weakly contractions in ordered b-metric spaces. SpringerPlus 5, 802 (2016)

16. Mustafa, Z, Jaradata, MMM, Jaradat, HM: Some common fixed point results of graphs on b-metric space. J. Nonlinear Sci. Appl. 9(6), 4838-4851 (2016)

17. Mustafa, Z, Huang, H, Randenović, S: Some remarks on the paper “Some fixed point generalizations are not real generalizations”. J. Adv. Math. Stud. 9(1), 110-116 (2016)

18. Sedghi, S, Altun, I, Shobe, N, Salahshour, M: Some properties of S-metric space and fixed point results. Kyungpook Math. J. 54, 113-122 (2014)

19. Rohen, Y, Došenović, T, Randenović, S: A note on the paper “A fixed point theorems in $$S_{b}$$-metric spaces”. Filomat 31(11), 3335-3346 (2017)

20. Sedghi, S, Gholidahneh, A, Došenović, T, Esfahani, J, Randenović, S: Common fixed point of four maps in $$S_{b}$$-metric spaces. J. Linear Topol. Algebra 5(2), 93-104 (2016)

21. Sedghi, S, Rezaee, MM, Došenović, T, Sü, R: Common fixed point theorems for contractive mappings satisfying ϕ-maps in S-metric spaces. Acta Univ. Sapientiae Math. 8(2), 298-311 (2016)

22. Sedghi, S, Shobe, N, Aliouche, A: A generalization of fixed point theorem in S-metric spaces. Mat. Vesn. 64, 258-266 (2012)

23. Sedghi, S, Shobe, N, Došenović, T: Fixed point results in S-metric spaces. Nonlinear Funct. Anal. Appl. 20(1), 55-67 (2015)

24. Zhou, M, Liu, XL, Randenović, S: S-γ-φ-contractive type mappings in S-metric spaces. J. Nonlinear Sci. Appl. 10, 1613-1639 (2017)

## Acknowledgements

Authors are thankful to referees for their valuable suggestions.

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Correspondence to GNV Kishore.

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Kishore, G., Rao, K., Panthi, D. et al. Some applications via fixed point results in partially ordered $$S_{b}$$-metric spaces. Fixed Point Theory Appl 2017, 10 (2016). https://doi.org/10.1186/s13663-017-0603-2

• $$S_{b}$$-metric space
• $$S_{b}$$-completeness