Solution of Nonlinear Integral Equation via Fixed Point of Cyclic \(\alpha _{L}^{ \psi }\)-Rational Contraction Mappings in Metric-Like Spaces

Abstract

In this paper, we introduce the notions of \(\alpha _{L}^{\psi }\)-rational contractive and cyclic \(\alpha _{L}^{\psi }\)- rational contractive mappings and establish the existence and uniqueness of fixed points for such mappings in complete metric-like spaces (dislocated metric spaces). The results presented here substantially generalize and extend several comparable results in the existing literature. As an application, we prove new fixed point results for \(\psi L\)-graphic and cyclic \(\psi L\)-graphic rational contractive mappings. Moreover, some examples and an application to integral equation are presented here to illustrate the usability of the obtained results.

Introduction and Preliminaries

Let (Xd) be a metric and \(S:X\rightarrow X\) be a mapping. The mapping S is called a Banach contraction if there exists \(\lambda \in [0,1)\) such that

$$\begin{aligned} d(Sx,Sy)\le \lambda d(x,y) \end{aligned}$$
(1)

for all \(x,y\in X.\) According to the Banach contraction principle (contraction principle) every Banach contraction on a complete metric space has a unique fixed point, i.e., there exists a point \(x\in X\) such that \( Sx=x. \) This principle appeared in Banach’s thesis (Banach 1922), where it was used in proving the existence and uniqueness of solution of integral equations. Later, several generalizations and applications of this principle appeared in this line of research (Agarwal et al. 2012; Hussain et al. 2013; Gopal et al. 2014; Liu et al. 2011; Pathak et al. 2007; Radenović et al. 2016; Shahzad et al. 2015; Suzuki 2008, 2009; Salimi et al. 2013).

Notice that, the contractive condition (1) is satisfied for all \(x,y\in X\) which forces the mapping S to be continuous, and so in case of discontinuous mappings the Banach contraction principle is not applicable. In view of the applicability of contraction principle this is the major draw-back of this principle. Several authors attempted to overcome this drawback, see, e.g., Ran and Reuring (2004), Kirk et al. (2003), Shatanawi and Postolache (2013) and the recent result of Salimi et al. (2013).

Salimi et al. (2013) defined the notion of \(\alpha \)-admissible mappings as follows:

Definition 1.1

A self-mapping S on a nonempty set X is called an \(\alpha \)-admissible mapping if

$$\begin{aligned} \alpha (x,y)\ge 1\Rightarrow \alpha (Sx,Sy)\ge 1 \end{aligned}$$

for all \(x,y\in X\) and \(\alpha :X\times X\rightarrow [0,+\infty )\) be a function.

By using this concept, they proved some fixed point results.

Theorem 1.1

Let \(S:X\rightarrow X\) be an \(\alpha \)-admissible mapping on a complete metric space X. Assume that the following statements are satisfied:

(i) for all \(x,y\in X,\) we have

$$\begin{aligned} \alpha (x,y)d(Sx,Sy)\le \psi (d(x,y)) \end{aligned}$$
(2)

where \(\psi :[0,+\infty )\rightarrow [0,+\infty )\) is a nondecreasing function such that

$$\begin{aligned} \sum \limits _{n=1}^{+\infty }\psi ^{n}(t)<+\infty ,\text { }\forall t>0; \end{aligned}$$

(ii) there exists \(x_{\circ }\in X\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

(iii) either S is continuous or for any sequence \(\{x_{n}\}\) in X with \( \alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow +\infty ,\) we have \(\alpha (x_{n},x)\ge 1.\) Then S has a fixed point.

If we take \(\alpha (x,y)=1\) in (2) for all \(x,y\in X,\) then S reduces into a Banach contraction.

Notice that, in Theorem 1.1 the mapping S is a weaker version of Banach contraction in the sense that S need not satisfy the contractive condition (1) (see Example 2.3 and Example 2.4 of Salimi et al. 2013).

On the other hand, Matthews (1994), introduced the concept spaces with nonzero self-distance of points, Aage and Salunke (2008), generalized it in dislocated and dislocated quasi-metric spaces. In Amini-Harandi (2012) the notion of a metric-like space was presented, which is an important extensions of the spaces defined in Aage and Salunke (2008) and Amini-Harandi (2012).

Now, we recall some basic concepts, notations and known results of metric-like spaces which will be used in the sequel.

Definition 1.2

(Amini-Harandi 2012) Let X be a nonempty set, a mapping \(\omega :X\times X\rightarrow {\mathbb {R}} ^{+}=[0,\infty )\) is said to be a metric-like mapping on X if for all \( x,y,z\in X\) the following three conditions hold true:

\((\omega _{1})\) \(\omega (x,y)=0\) \(\Rightarrow x=y;\)

\((\omega _{2})\) \(\omega (x,y)=\omega (y,x);\)

\((\omega _{3})\) \(\omega (x,z)\le \omega (x,y)+\omega (y,z).\)

Then the pair \((X,\omega )\) is called a metric-like space.

A metric-like mapping on X satisfies all of the conditions of a metric except that \(\omega (x,x)\) may be positive for \(x\in X\).

Definition 1.3

(Amini-Harandi 2012) Let \((X,\omega )\) be a metric-like space and \(\{x_{n}\}\) be a sequence of point on X. A point \(x\in X\) is called the limit of the sequence \(\{x_{n}\}\) if \(\lim _{n\rightarrow \infty }\omega (x,x_{n})=\omega (x,x),\) and we say that the sequence \(\{x_{n}\}\) is convergent to x and denote it by \(x_{n}\rightarrow x\) as \(n\rightarrow \infty .\)

Definition 1.4

(Amini-Harandi 2012) Let \((X,\omega )\) be a metric-like space.

  1. (i)

    if \(\lim _{m,n\rightarrow \infty }\omega (x_{m},x_{n})\) exists and is finite, then a sequence \(\{x_{n}\}\) is said to be a \(\omega -\)Cauchy sequence.

  2. (ii)

    if every \(\omega -\)Cauchy sequence \(\{x_{n}\}\) in X converges to \( x\in X,\) with \(\lim _{m,n\rightarrow \infty }\omega (x_{m},x_{n})=\omega (x,x)=\lim _{n\rightarrow \infty }\omega (x_{n},x),\) then the space \((X,\omega )\) is said to be complete.

Definition 1.5

(Hitzler 2001) Let \((X,\omega )\) be a metric-like space and V be a subset of X. We say that V is a \(\omega \)-open subset of X,  if for all \(x\in X\) there exists \(r>0\) such that \(B_{\omega }(x,r)\subseteq V.\) Also \(U\subseteq X\) is a \(\omega -\)closed subset of X if \(X\backslash V\) is a \(\omega \)-open subset of X.

Lemma 1.1

(Karapinar and Salimi 2013) Let \((X,\omega )\) be a metric-like space and U be a \( \omega -\)closed subset of X. Let \(\{x_{n}\}\) be a sequence of U such that \(x_{n}\rightarrow x\) as \(n\rightarrow \infty .\) Then \(x\in U.\)

Lemma 1.2

(Karapinar and Salimi 2013) Let \((X,\omega )\) be a metric-like space and \( \{x_{n}\}\) be a sequence of X such that \(x_{n}\rightarrow x\) as \( n\rightarrow \infty \) and \(\omega (x,x)=0.\) Then \(\lim _{n\rightarrow \infty }\omega (x_{n},y)=\omega (x,y)\) for all \(y\in X.\)

Lemma 1.3

(Karapinar and Salimi 2013) Let \((X,\omega )\) be a metric-like space. Then,

  1. (i)

    if \(\omega (x,y)=0,\) then \(\omega (x,x)=\omega (y,y)=0;\)

  2. (ii)

    if \(\{x_{n}\}\) be a sequence such that \(\lim _{n\rightarrow \infty }\omega (x_{n},x_{n+1})=0,\) then

    $$\begin{aligned} \lim _{n\rightarrow \infty }\omega (x_{n},x_{n})=\lim _{n\rightarrow \infty }\omega (x_{n+1},x_{n+1})=0; \end{aligned}$$
  3. (iii)

    if \(x\ne y\), then \(\omega (x,y)>0;\)

  4. (iv)

    \(\omega (x,x)\le \frac{2}{n}\sum \nolimits _{i=1}^{n}\omega (x,x_{i})\) holds for all \(x,x_{i}\in X\) where \(1\le i\le n.\)

Lemma 1.4

(Shukla et al. 2013) Let \((X,\omega )\) be a metric-like space and \( \{x_{n}\}\) be a sequence of X such that

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

If \(\lim _{n,m\rightarrow \infty }\omega (x_{n},x_{m})\ne 0,\) then there exist \(\varepsilon >0\) and two sequences \(\{m(k)\}\) and \(\{n(k)\}\) such that \(n(k)>m(k)>k\) and the following tend to \(\varepsilon \) when \(k\rightarrow \infty \):

$$\begin{aligned} \{\omega (x_{n(k)},x_{m(k)})\},\text { }\{\omega (x_{n(k)+1},x_{m(k)})\}, \text { }\{\omega (x_{n(k)},x_{m(k)-1})\},\text { }\{\omega (x_{n(k)+1},x_{m(k)-1})\}. \nonumber \\ \end{aligned}$$
(3)

Remark 1.1

If the condition of the above lemma is satisfied, then the sequences \(\{\omega (x_{n(k)+s},x_{m(k)})\}\) and \(\{\omega (x_{n(k)+s},x_{m(k)+1})\}\) also converge to \(\varepsilon \) when \( k\rightarrow \infty ,\) where \(s\in {\mathbb {N}} .\)

Proof

for \(s\in {\mathbb {N}} ,\) we have

$$\begin{aligned}&\omega (x_{n(k)+s},x_{m(k)})\le \omega (x_{n(k)+s},x_{n(k)+s-1})+\cdots \\&\quad +\,\omega (x_{n(k)+2},x_{n(k)+1})+\omega (x_{n(k)+1},x_{m(k)}) \end{aligned}$$

and

$$\begin{aligned}&\omega (x_{n(k)+1},x_{m(k)})\le \omega (x_{n(k)+1},x_{n(k)+2})+\cdots \\&\quad +\,\omega (x_{n(k)+s-1},x_{n(k)+s})+\omega (x_{n(k)+s},x_{m(k)}). \end{aligned}$$

By Lemma 1.4 we obtain \(\lim _{k\rightarrow \infty }\omega (x_{n(k)+s},x_{m(k)})=\varepsilon .\) Similarly, we can show that the sequence \(\{\omega (x_{n(k)+s},x_{m(k)+1})\}\) also converge to \(\varepsilon \) when \(k\rightarrow \infty .\)\(\square \)

In this paper, the notions of \(\alpha _{L}^{\psi }\)-rational contractive and cyclic \(\alpha _{L}^{\psi }\)-rational contractive mappings are given and we establish some fixed point results for such mappings in complete metric-like spaces. As an application, we prove new fixed point results for \( \psi L\)-graphic and cyclic \(\psi L\)-graphic rational contractive mappings and obtained fixed point theorems to certain type of integral equation by means of metric-like spaces.

Cyclic \(\alpha _{L}^{\psi }\)-Rational Contractive Mappings

In this section, we present some new fixed point theorems for a cyclic \(\alpha _{L}^{\psi }\)-rational contraction self-mapping on complete metric-like spaces. Before stating our main results, we need to give the following definition.

Let \(\Psi \) denote the class of all function \(\psi :[0,\infty )\rightarrow [0,\infty ),\) satisfying the following conditions:

  1. (i)

    \(\psi \) non-decreasing and continuous;

  2. (ii)

    \(\lim _{n\rightarrow \infty }\psi ^{n}(t)=0\) for all \(t>0.\)

Definition 2.1

Let \((X,\omega )\) be a metric-like space, \(q\in {\mathbb {N}} ,\)\(B_{1},B_{2},\ldots ,B_{q}\) be \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. We say that S is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping if

  1. (a)
    $$\begin{aligned} S(B_{j})\subseteq B_{j+1},\quad j=1,2,\ldots ,q,\quad \text { where }B_{q+1}=B_{1}. \end{aligned}$$
    (4)
  2. (b)

    for any \(x\in B_{i}\) and \(y\in B_{i+1},\)\(i=1,2,\ldots ,q\) where \( B_{q+1}=B_{1}\) and \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\) we get

    $$\begin{aligned} \psi (\omega (Sx,Sy))\le \psi (M_{\omega }(x,y))-LM_{\omega }(x,y), \end{aligned}$$
    (5)

    where \(\psi \in \Psi ,\)\(0<L<1\) and

    $$\begin{aligned}&M_{\omega }(x,y)\\&\quad =\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{ \omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{ \omega (x,Sy)+\omega (y,Sx)}{4}\right\} . \end{aligned}$$

If we take \(X=B_{i},\)\(i=1,2,\ldots ,q,\) in Definition 2.1, then we say that S is \(\alpha _{ L}^{ \psi }\)-rational contractive mapping. We denote the set of all fixed points of S by Fix(S),  that is, \( Fix(S)=\{x\in X:Sx=x\}.\)

Example 2.1

Let \(X= {\mathbb {R}} \) be equipped with the metric-like mapping \(\omega (x,y)=\max \{|x|,|y|\}\) for all \(x,y\in X.\) Suppose \(B_{1}=[-\frac{\pi }{2},0]\) and \(B_{2}=[0,\frac{\pi }{2}]\) and \(Y=B_{1}\cup B_{2}\). Define \(S:X\rightarrow X\) and \(\alpha :X\times X\rightarrow [0,\infty )\) by

$$\begin{aligned} Sx=\left\{ \begin{array}{lll} -\frac{x}{16}\left| \cos (\frac{1}{x})\right| , &{} &{} \text {if }x\in [-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}] \\ 0, &{} &{} \text {if }x=0 \end{array} \right. \quad \text { and } \quad \alpha (x,y)=\left\{ \begin{array}{lll} 1, &{} &{} \text {if }x,y\in [-\frac{\pi }{2},\frac{\pi }{2}] \\ 0, &{} &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Also, define \(\psi :[0,\infty )\rightarrow [0,\infty )\) by \(\psi (t)= \frac{1}{2}t\) and \(L=\frac{15}{32}<1.\) Clearly, \(S(B_{1})\subseteq B_{2}\) and \(S(B_{2})\subseteq B_{1}.\) Let, \(x\in B_{1}\), \(y\in B_{2}\) and \(\alpha (x,Sx)\alpha (y,Sy)\ge 1.\) Now, if \(x\notin [-\frac{\pi }{2},\frac{ \pi }{2}]\) or \(y\notin [-\frac{\pi }{2},\frac{\pi }{2}],\) then \( \alpha (x,Sx)=0\) or \(\alpha (y,Sy)=0.\) That is \(\alpha (x,Sx)\alpha (y,Sy)=0<1\), which is a contradiction. Hence \(x\in B_{1}\), \(y\in B_{2}\) and \( x,y\in [-\frac{\pi }{2},\frac{\pi }{2}].\) This implies that \(x\in [-\frac{\pi }{2},0]\) and \(y\in [0,\frac{\pi }{2}].\) We consider the following cases:

Case 1\(x=0\) and \(y=0\) with \((x,y)\in B_{1}\times B_{2}\) or \( (x,y)\in B_{2}\times B_{1}.\) (trivial case).

Case 2 \(x\ne 0\) and \(y=0\) with \((x,y)\in B_{1}\times B_{2}\) or \( (x,y)\in B_{2}\times B_{1}.\) Then

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \frac{1}{2}\max \left\{ \left| - \frac{x}{16}\left| \cos \left( \frac{1}{x}\right) \right| \right| ,0\right\} =\frac{1}{32}\left| x\right| \left| \cos \left( \frac{1}{x}\right) \right| \\\le & {} \frac{1}{32}\left| x\right| =\frac{1}{32}\omega (x,0)=\psi (M_{\omega }(x,0))-LM_{\omega }(x,0). \end{aligned}$$

Case 3 \(x=0\) and \(y\ne 0\) with \((x,y)\in B_{1}\times B_{2}\) or \( (x,y)\in B_{2}\times B_{1}.\) Then

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \frac{1}{2}\max \left\{ 0,\left| - \frac{y}{16}\left| \cos \left( \frac{1}{y}\right) \right| \right| \right\} =\frac{1}{32}\left| y\right| \left| \cos \left( \frac{1}{y}\right) \right| \\\le & {} \frac{1}{32}\left| y\right| =\frac{1}{32}\omega (0,y)=\psi (M_{\omega }(0,y))-LM_{\omega }(0,y). \end{aligned}$$

Case 4 \(x\ne 0\) and \(y\ne 0\) with \((x,y)\in B_{1}\times B_{2}\) or \((x,y)\in B_{2}\times B_{1}.\) Then

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \frac{1}{2}\max \left\{ \left| - \frac{x}{16}\left| \cos \left( \frac{1}{x}\right) \right| \right| ,\left| -\frac{y}{16}\left| \cos \left( \frac{1}{y} \right) \right| \right| \right\} \\= & {} \frac{1}{32}\max \left\{ \left| x\right| \left| \cos \left( \frac{1}{x}\right) \right| ,\left| y\right| \left| \cos \left( \frac{1}{y} \right) \right| \right\} \\\le & {} \frac{1}{32}\max \left\{ \left| x\right| ,\left| y\right| \right\} =\frac{1}{32}\omega (x,y)=\psi (M_{\omega }(x,y))-LM_{\omega }(x,y). \end{aligned}$$

By the above cases, we conclude that, S is a cyclic \(\alpha _{{ L}}^{ { \psi }}\)-rational contractive mapping.

Remark 2.1

If \(S:X\rightarrow X\) is a cyclic \(\alpha _{{ L}}^{ { \psi }}\)-rational contractive mapping, \(x\in Fix(S)\) and \(\alpha (x,x)\ge 1,\) then \(\omega (x,x)=0.\)

Proof

Let \(\omega (x,x)>0,\) then \(M_{\omega }(x,x)=\omega (x,x)\) and by (5), we can write

$$\begin{aligned} \psi (\omega (x,x))=\psi (\omega (Sx,Sx))\le \psi (\omega (x,x))-L\omega (x,x). \end{aligned}$$

This contradiction shows that \(\omega (x,x)=0.\)\(\square \)

Definition 2.2

(Hussain et al. 2014) Let \((X,\omega )\) be a metric-like space and \( \alpha :X\times X\rightarrow [0,\infty ).\) We say that an \(\alpha -\)admissible mapping \( S:X\rightarrow X\) is \(\alpha -\)continuous on \((X,\omega ),\) if

$$\begin{aligned} x_{n}\rightarrow x\text { as }n\rightarrow \infty ,\text { }\alpha (x_{n},x_{n+1})\ge 1\Longrightarrow Sx_{n}\rightarrow Sx \quad \text { for all } n\in {\mathbb {N}}. \end{aligned}$$

Example 2.2

Let \(X=[0,\infty )\) and \(\omega (x,y)=x+y\) be a metric-like on X. Consider an \(\alpha \)-admissible mapping \(S:X\rightarrow X\) and \(\alpha :X\times X\rightarrow X\) are defined by

$$\begin{aligned} Sx=\left\{ \begin{array}{l} x^{3},\qquad \qquad \qquad \text {if }x\in [0,\frac{1}{3}] \\ \cos \frac{\pi }{2}x+3,\quad \text { if }x>\frac{1}{3} \end{array} \right. \quad \quad \alpha (x,y)=\left\{ \begin{array}{l} 2,\quad \text {if }x,y\in [0,\frac{1}{3}] \\ \frac{1}{5}, \quad \text {if }x>\frac{1}{3} \end{array} \right. . \end{aligned}$$

Clearly, S is not continuous. Let \(x_{\circ }=0\in X\) so, \(Sx_{\circ }=0\) and \(\alpha (x_{\circ },Sx_{\circ })=\alpha (0,0)=2>1\). Let \( \{x_{n}\}\subset [0,\frac{1}{3}]\) for all \(n\in {\mathbb {N}} .\) If \(x_{n}\rightarrow x\) as \(n\rightarrow \infty ,\) then, we can get \(x\in [0,\frac{1}{3}]\) and \(\alpha (x_{n},Sx_{n+1})=2\ge 1.\) Hence

$$\begin{aligned} \lim _{n\rightarrow \infty }Sx_{n}=\lim _{n\rightarrow \infty }x_{n}^{3}=x^{3}=Sx. \end{aligned}$$

That is, S is \(\alpha \)-continuous on \((X,\omega ).\)

Now, we want to state and prove our main result of this section.

Theorem 2.1

Let \((X,\omega )\) be a complete metric-like space, q be a positive integer, \(B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Assume that \(S:Y\rightarrow Y\) is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping satisfying the following conditions:

  1. (i)

    S is an \(\alpha \)-admissible mapping;

  2. (ii)

    there exists \(x_{\circ }\in Y\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  3. (iii)

    \((\spadesuit _{1})\) either S is \(\alpha \)-continuous, or; \((\spadesuit _{2})\) for any sequence \(\{x_{n}\}\) in X with \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1.\) Then, S has a fixed point \(x\in \cap _{i=1}^{q}B_{i}.\) Moreover, if

  4. (iv)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}\).

Proof

Let \(x_{\circ }\) be an arbitrary point of Y such that \( \alpha (x_{\circ },Sx_{\circ })\ge 1\). Then there exists some \(i_{\circ }\) such that \(x_{\circ }\in B_{i_{\circ }}\). Now \(S(B_{i_{\circ }})\subseteq B_{i_{\circ }+1}\) implies that \(Sx_{\circ }\in B_{i_{\circ }+1}\). Thus there exists \(x_{1}\) in \(B_{i_{\circ }+1}\) such that \(Sx_{\circ }=x_{1}\). Similarly, \(Sx_{n}=x_{n+1}\), where \(x_{n}\in B_{i_{n}}\). Hence for \(n\ge 0\) , there exists \(i_{n}\in \{1,2,\ldots ,q\}\) such that \(x_{n}\in B_{i_{n}}\) and \( Sx_{n}=x_{n+1}\). On the other hand, S is an \(\alpha \)-admissible mapping, hence we get

$$\begin{aligned} \alpha (x_{1},Sx_{1})=\alpha (Sx_{\circ },S(Sx_{\circ }))\ge 1. \end{aligned}$$

Again, since S is an \(\alpha -\)admissible mapping, it follows that

$$\begin{aligned} \alpha (x_{2},Sx_{2})=\alpha (Sx_{1},S(Sx_{1}))\ge 1. \end{aligned}$$

Continuing this process, we have

$$\begin{aligned} \alpha (x_{n},Sx_{n})\ge 1\text { }\quad \forall n\ge 0. \end{aligned}$$

and so

$$\begin{aligned} \alpha (x_{n},Sx_{n})\alpha (x_{n-1},Sx_{n-1})\ge 1 \quad \text { for all }n\in {\mathbb {N}}. \end{aligned}$$
(6)

In the case when \(x_{n_{\circ }}=x_{n_{\circ }+1}\) for some \(n_{\circ }=0,1,2,\ldots ,\) it is clear that \(x_{n_{\circ }}\) is a fixed point of S. Now assume that \(x_{n}\ne x_{n+1}\) for all n. Hence, by Lemma 1.3 (iii) we have \(\omega (x_{n-1},x_{n})>0\) for all n. We shall show that the sequence \( \{\omega _{n}=\omega (x_{n},x_{n+1})\}\) is non-increasing. Assume that there exists some \(n_{\circ }\in {\mathbb {N}} \) such that

$$\begin{aligned} \omega (x_{n_{\circ }-1},x_{n_{\circ }})\le \omega (x_{n_{\circ }},x_{n_{\circ }+1}). \end{aligned}$$

Hence

$$\begin{aligned} \psi (\omega (x_{n_{\circ }-1},x_{n_{\circ }}))\le \psi (\omega (x_{n_{\circ }},x_{n_{\circ }+1})). \end{aligned}$$
(7)

By (5) and (6), we have

$$\begin{aligned} \psi (\omega (x_{n},x_{n+1}))=\psi (\omega (Sx_{n-1},Sx_{n}))\le \psi (M_{\omega }(x_{n-1},x_{n}))-LM_{\omega }(x_{n-1},x_{n}), \end{aligned}$$
(8)

where

$$\begin{aligned} M_{\omega }(x_{n-1},x_{n})= & {} \max \left\{ \begin{array}{c} \omega (x_{n-1},x_{n}),\frac{\omega (x_{n-1},Sx_{n-1})\omega (x_{n},Sx_{n})}{ \omega (x_{n-1},x_{n})},\frac{\omega (x_{n},Sx_{n})(\omega (x_{n-1},Sx_{n-1})+1)}{1+\omega (x_{n-1},x_{n})}, \\ \frac{\omega (x_{n-1},Sx_{n})+\omega (x_{n},Sx_{n-1})}{4} \end{array} \right\} \\\le & {} \max \left\{ \begin{array}{c} \omega (x_{n-1},x_{n}),\frac{\omega (x_{n-1},x_{n})\omega (x_{n},x_{n+1})}{ \omega (x_{n-1},x_{n})},\frac{\omega (x_{n},x_{n+1})(\omega (x_{n-1},x_{n})+1)}{1+\omega (x_{n-1},x_{n})}, \\ \frac{\omega (x_{n-1},x_{n+1})+\omega (x_{n},x_{n})}{4} \end{array} \right\} \\= & {} \max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1}),\frac{\omega (x_{n-1},x_{n+1})+\omega (x_{n},x_{n})}{4}\right\} . \end{aligned}$$

On the other hand, from Lemma 1.3 (iv), we get

$$\begin{aligned} \omega (x_{n},x_{n})\le \omega (x_{n},x_{n-1})+\omega (x_{n},x_{n+1}), \end{aligned}$$

and by \((\omega _{3})\), we obtain that

$$\begin{aligned} \omega (x_{n-1},x_{n+1})\le \omega (x_{n-1},x_{n})+\omega (x_{n},x_{n+1}). \end{aligned}$$

That is

$$\begin{aligned} M_{\omega }(x_{n-1},x_{n})\le & {} \max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1}),\frac{\omega (x_{n},x_{n+1})+\omega (x_{n},x_{n-1})}{2} \right\} \\= & {} \max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1})\right\} . \end{aligned}$$

Therefore from (8), we get

$$\begin{aligned} \psi (\omega (x_{n},x_{n+1}))\le & {} \psi (\max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1})\right\} )\\&-L\max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1})\right\} . \end{aligned}$$

Now, if \(\max \left\{ \omega (x_{n-1},x_{n}),\omega (x_{n},x_{n+1})\right\} =\omega (x_{n},x_{n+1}),\) then

$$\begin{aligned} \psi (\omega (x_{n},x_{n+1}))\le \psi (\omega (x_{n},x_{n+1}))-L\omega (x_{n},x_{n+1}), \end{aligned}$$

which is a contradiction. Hence,

$$\begin{aligned} \psi (\omega (x_{n},x_{n+1}))\le \psi (\omega (x_{n-1},x_{n}))-L\omega (x_{n-1},x_{n}). \end{aligned}$$
(9)

By taking \(x=x_{n_{\circ }-1}\) and \(y=y_{n_{\circ }}\) in (9) and using (7) for all \(n\in {\mathbb {N}} ,\) we deduce

$$\begin{aligned} \psi (\omega (x_{n_{\circ }-1},x_{n_{\circ }}))\le \psi (\omega (x_{n_{\circ }-1},x_{n_{\circ }}))-L\omega (x_{n_{\circ }-1},x_{n_{\circ }}), \end{aligned}$$

which is a contradiction. Thus \(\omega _{n}<\omega _{n-1}\) holds for all \( n\in {\mathbb {N}} \) and there exists \(r\ge 0\) such that \(\lim _{n\rightarrow \infty }\omega _{n}=r.\) We will show that \(r=0\) by method of reductio ad absurdum. For this, let \(r>0.\) By (9), together with the properties of \(\psi ,\) we have

$$\begin{aligned} \psi (r)=\lim _{n\rightarrow \infty }\psi (\omega _{n})\le \lim _{n\rightarrow \infty }[\psi (\omega _{n-1})-L\omega _{n-1}]\le \psi (r)-Lr. \end{aligned}$$

This is a contradiction again. Hence

$$\begin{aligned} \lim _{n\rightarrow \infty }\omega _{n}=\lim _{n\rightarrow \infty }\omega (x_{n},x_{n+1})=0. \end{aligned}$$
(10)

Now, we shall show that \(\lim _{n,m\rightarrow \infty }\omega (x_{n},x_{m})=0\) . Let if possible, \(\lim _{n,m\rightarrow \infty }\omega (x_{n},x_{m})\ne 0\) then by Remark 1.1, there exist \(\varepsilon >0\) and two sequences \(\{m(k)\}\) and \(\{n(k)\}\) such that \(n(k)>m(k)>k\) and

$$\begin{aligned} \lim _{k\rightarrow \infty }\omega (x_{n(k)+s},x_{m(k)})=\varepsilon ,\text { }\quad s\in {\mathbb {N}} . \end{aligned}$$

Let \(s,t\in {\mathbb {N}} \) such that \(s=m(k)+tq+1-n(k)\), then we have \(x_{m(k)}\in B_{i}\) and \( x_{n(k)+s}\in B_{i+1}\) for some \(i\in \{1,2,\ldots ,q\}.\) Therefore, using the contractive condition (5) with \(x=x_{m(k)}\), \(y=x_{n(k)+s},\) we get

$$\begin{aligned} \psi ( \omega (x_{m(k)+1},x_{n(k)+s+1})) \le \psi ( M_{\omega }(x_{m(k)},x_{n(k)+s})) -LM_{\omega }(x_{m(k)},x_{n(k)+s}),\nonumber \\ \end{aligned}$$
(11)

where

$$\begin{aligned}&M_{\omega }(x_{m(k)},x_{n(k)+s})\\&\quad =\max \left\{ \begin{array}{c} \omega (x_{m(k)},x_{n(k)+s}),\frac{\omega (x_{m(k)},Sx_{m(k)})\omega (x_{n(k)+s},Sx_{n(k)+s})}{\omega (x_{m(k)},x_{n(k)+s})}, \\ \frac{\omega (x_{n(k)+s},Sx_{n(k)+s})(\omega (x_{m(k)},Sx_{m(k)})+1)}{ 1+\omega (x_{m(k)},x_{n(k)+s})},\frac{\omega (x_{m(k)},Sx_{n(k)+s})+\omega (x_{n(k)+s},Sx_{m(k)})}{4} \end{array} \right\} \\&\quad \le \max \left\{ \begin{array}{c} \omega (x_{m(k)},x_{n(k)+s}),\frac{\omega (x_{m(k)},x_{m(k)+1})\omega (x_{n(k)+s},x_{n(k)+s+1})}{\omega (x_{m(k)},x_{n(k)+s})}, \\ \frac{\omega (x_{n(k)+s},x_{n(k)+s+1})(\omega (x_{m(k)},x_{m(k)+1})+1)}{ 1+\omega (x_{m(k)},x_{n(k)+s})},\frac{\omega (x_{m(k)},x_{n(k)+s+1})+\omega (x_{n(k)+s},x_{m(k)+1})}{4} \end{array} \right\} \\&\quad \rightarrow \max \left\{ \varepsilon ,0,0,\frac{\epsilon }{2}\right\} =\varepsilon \quad \text { as }\quad k\rightarrow \infty . \end{aligned}$$

Taking the limit as \(k\rightarrow \infty \) in (11), we can write

$$\begin{aligned} \psi \left( \varepsilon \right) \le \psi \left( \varepsilon \right) -L\varepsilon . \end{aligned}$$

This contradiction proves that \(\lim _{n,m\rightarrow \infty }\omega (x_{n},x_{m})=0\), that is, the sequence \(\{x_{n}\}\) is a \(\omega \)-Cauchy sequence. Since Y is \(\omega \)-closed in \((X,\omega )\), there exists \(x\in Y=\cup _{i=1}^{q}B_{i}\) such that \(\lim _{n\rightarrow \infty }x_{n}=x\) in \( (Y,\omega )\), equivalently,

$$\begin{aligned} \omega (x,x)=\lim _{n\rightarrow \infty }\omega (x,x_{n})=\lim _{n,m\rightarrow \infty }\omega (x_{n},x_{m})=0. \end{aligned}$$
(12)

At first, assume that (\(\spadesuit _{1}\)) of (iii) holds, that is, S is \( \alpha \)-continuous. Then it is obvious that,

$$\begin{aligned} x=\lim _{n\rightarrow \infty }x_{n+1}=\lim _{n\rightarrow \infty }Sx_{n}=Sx. \end{aligned}$$
(13)

On the other hand, we assume that (\(\spadesuit _{2}\)) of (iii) holds and \(\omega (x,Sx)>0.\) Then, we have \(\alpha (x,Sx)\ge 1\) and so

$$\begin{aligned} \alpha (x,Sx)\alpha (x_{n(k)},Sx_{n(k)})\ge 1. \end{aligned}$$

In the following, we prove that x is a fixed point of S. Since \( \lim _{n\rightarrow \infty }x_{n}=x\) and \(S(B_{j})\subseteq B_{j+1}\), \( j=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1},\) the sequence \(\{x_{n}\}\) has infinitely many terms in each \(B_{i}\) for \(i\in \{1,2,\ldots ,q\}.\) Suppose that \(x\in B_{i}.\) Then \(Sx\in B_{i+1}\) and we take a subsequence \(\{x_{n(k)}\}\) of \( \{x_{n}\}\) with \(x_{n(k)}\in B_{i-1}\) (by the above mentioned comment this subsequence exists). By using the contractive condition, we have

$$\begin{aligned}&\psi \left( \omega (Sx,Sx_{n(k)})\right) \\&\quad \le \psi \left( \max \left\{ \begin{array}{c} \omega (x,x_{n(k)}),\frac{\omega (x,Sx)\omega (x_{n(k)},Sx_{n(k)})}{\omega (x,x_{n(k)})},\frac{\omega (x_{n(k)},Sx_{n(k)})(\omega (x,Sx)+1)}{1+\omega (x,x_{n(k)})}, \\ \frac{\omega (x,Sx_{n(k)})+\omega (x_{n(k)},Sx)}{4} \end{array} \right\} \right) \\&\quad -L\max \left\{ \begin{array}{c} \omega (x,x_{n(k)}),\frac{\omega (x,Sx)\omega (x_{n(k)},Sx_{n(k)})}{\omega (x,x_{n(k)})},\frac{\omega (x_{n(k)},Sx_{n(k)})(\omega (x,Sx)+1)}{1+\omega (x,x_{n(k)})}, \\ \frac{\omega (x,Sx_{n(k)})+\omega (x_{n(k)},Sx)}{4} \end{array} \right\} , \end{aligned}$$

passing to limit as \(k\rightarrow \infty \) and using (12), (13), we get

$$\begin{aligned} \psi \left( \omega (Sx,x)\right) \le \psi \left( \omega (Sx,x)\right) -L\omega (Sx,x). \end{aligned}$$

This contradiction shows that \(\omega (x,Sx)=0\), that is, \(Sx=x\). Therefore, x is a fixed point of S. The cyclic character of S and the fact that \( x\in X\) is a fixed point of S, imply that \(x\in \bigcap \nolimits _{i=1}^{q}B_{i}.\)

At last, we shall prove the uniqueness of the fixed point, suppose that \( x,y\in \bigcap \nolimits _{i=1}^{q}B_{i}\) are two distinct fixed points of S and (iv) holds. Then, we have \(\omega (x,y)>0\), \(\omega (x,x)\ge 1\), \( \omega (y,y)\ge 1\) and then by (5), we have

$$\begin{aligned} \psi (\omega (Sx,Sy))\le & {} \psi \left( \max \left\{ \begin{array}{c} \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)}, \\ \frac{\omega (x,Sy)+\omega (y,Sx)}{4} \end{array} \right\} \right) \\&-L\max \left\{ \begin{array}{c} \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)}, \\ \frac{\omega (x,Sy)+\omega (y,Sx)}{4} \end{array} \right\} . \end{aligned}$$

Using Remark 2.1 we have \(\omega (x,x)=\omega (y,y)=0\), therefore it follows from the above inequality that

$$\begin{aligned} \psi (\omega (x,y))\le \psi (\omega (x,y))-L\omega (x,y), \end{aligned}$$

which is a contradiction. Hence, \(\omega (x,y)=0\), that is, \(x=y.\) The proof is completed. \(\square \)

Next, we give the following examples to illustrate the usefulness of Theorem 2.1.

Example 2.3

Let \(X= {\mathbb {R}} \) be equipped with the metric-like mapping \(\omega (x,y)=max\{|x|,|y|\}\) for all \(x,y\in X.\) Suppose \(B_{1}=(-\infty ,0]\) and \(B_{2}=[0,\infty )\) and \( Y=B_{1}\cup B_{2}\). Define \(S:Y\rightarrow Y\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) by

$$\begin{aligned} Sx=\left\{ \begin{array}{lll} -3x, &{} &{} \text {if }\,x\in (-\infty ,-1) \\ \frac{-x}{5}, &{} &{} \text {if }\,x\in [-1,0] \\ \frac{-x^{3}}{4}, &{} &{} \text {if }\,x\in [0,1] \\ -6x, &{} &{} x\in (1,\infty ) \end{array} \right. \text { and } \quad \alpha (x,y){=}\left\{ \begin{array}{lll} x^{2}+y^{2}+2, &{} &{} \text {if }\,x,y\in [-1,1] \\ 0, &{} &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Also, define \(\psi :[0,\infty )\rightarrow [0,\infty )\) by \(\psi (t)= \frac{1}{2}t\) and \(L=\frac{3}{8}<1.\) Clearly, \(S(B_{1})\subseteq B_{2}\) and \( S(B_{2})\subseteq B_{1}.\)

Let \(x\in B_{1}\), \(y\in B_{2}\) and \(\alpha (x,Sx)\alpha (y,Sy)\ge 1.\) Now, if \(x\notin [-1,1]\) or \(y\notin [-1,1],\) then \(\alpha (x,Sx)=0\) or \(\alpha (y,Sy)=0.\) That is, \(\alpha (x,Sx)\alpha (y,Sy)=0<1,\) which is a contradiction. Hence \(x\in B_{1}\), \(y\in B_{2}\) and \(x,y\in [-1,1].\) This implies that \(x\in [-1,0]\) and \(y\in [0,1].\) Then,

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \frac{1}{2}\max \left\{ \left| \frac{-x}{5}\right| ,\left| \frac{-y^{3}}{4}\right| \right\} = \frac{1}{2}\max \left\{ \frac{x}{5},\frac{y^{3}}{4}\right\} \\\le & {} \frac{1}{2}\max \left\{ \frac{x}{4},\frac{y}{4}\right\} \le \frac{1}{8}\max \left\{ \left| x\right| ,\left| y\right| \right\} \\= & {} \frac{1}{8}\omega (x,y)=\psi (M_{\omega }(x,y))-LM_{\omega }(x,y). \end{aligned}$$

Then, S is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping. Clearly, \(\alpha (0,S0)\ge 1\) and so the condition (ii) of Theorem 2.1 is satisfied. If \(\alpha (x,y)\ge 1\), then \(x,y\in [-1,1]\) which implies that \(\alpha (Sx,Sy)\ge 1\), that is, S is an \(\alpha \)-admissible mapping. Let \(\{x_{n}\}\) be a sequence in X such that, \(\alpha (x_{n},Sx_{n})\ge 1\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty \). Then, we must have \(x_{n}\in [-1,1]\) and so, \( x\in [-1,1]\), that is, \(\alpha (x,Sx)\ge 1\). Hence, all the conditions of Theorem 2.1 hold and S has a fixed point \(x=0\in B_{1}\cap B_{2}.\)

Example 2.4

Consider \(X=\{0,1,2\}.\) Let \(\omega :X\times X\rightarrow [0,\infty )\) be a mapping defined by

$$\begin{aligned} \omega (0,0)= & {} \omega (1,1)=0,\quad \omega (2,2)=\frac{5}{2}, \\ \omega (0,2)= & {} \omega (2,0)=2,\quad \omega (1,2)=\omega (2,1)=3, \quad \omega (0,1)=\omega (1,0)=\frac{3}{2}. \end{aligned}$$

It is clear that \((X,\omega )\) is a complete metric-like space. Suppose that \(B_{1}=\{0,1\}\) and \(B_{2}=\{1,2\}\) and \(Y=B_{1}\cup B_{2}\). Define \( S:Y\rightarrow Y\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) by

$$\begin{aligned} Sx=\left\{ \begin{array}{lll} 2, &{} &{} \text {if }\,x=0 \\ 1, &{} &{} \text {if }\,x=1 \\ 0, &{} &{} \text {if }\,x=2 \end{array} \right. \quad \text { and } \quad \alpha (x,y)=\left\{ \begin{array}{lll} x^{2}+3, &{} &{} \text {if }\,x,y\in \{1,2\} \\ 0, &{} &{} \text {if }\,x,y\in \{0\}\text {.} \end{array} \right. \end{aligned}$$

for all \(x,y\in Y.\) Then \(S(B_{1})\subseteq B_{2}\) and \(S(B_{2})\subseteq B_{1}.\) Thus \(x\in B_{1}\), \(y\in B_{2}.\) Now, if \(x\in \{0\}\) or \(y\in \{0\},\) then \(\alpha (x,Sx)=0\) or \(\alpha (y,Sy)=0.\) That is, \(\alpha (x,Sx)\alpha (y,Sy)=0<1,\) which is contradiction. Hence \(x\in \{1\}\) or \( y\in \{1,2\}.\)

Also, define \(\psi :[0,\infty )\rightarrow [0,\infty )\) by \(\psi (t)= \frac{1}{2}t.\) Next, our process of the proof is divided into two steps as follows:

Step 1 Assume that \(x=2\), \(y=1.\) On the one hand, \(Sx=0\) and \( Sy=1. \) Then

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \psi \left( \omega (0,1)\right) =\frac{3 }{4}, \\= & {} \frac{3}{2}-\frac{3}{4}=\psi (M_{\omega }(x,y))-LM_{\omega }(x,y),\text { with }L=\frac{1}{4}<1. \end{aligned}$$

Step 2 Assume that \(x=2\), \(y=2.\) On the one hand, \(Sx=0\) and \( Sy=0. \) Then

$$\begin{aligned} \psi \left( \omega (Sx,Sy)\right)= & {} \psi \left( \omega (0,0)\right) =0, \\= & {} \frac{5}{4}-\frac{5}{4}=\psi (M_{\omega }(x,y))-LM_{\omega }(x,y),\text { with }L=\frac{1}{2}<1. \end{aligned}$$

Then, S is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping. Clearly, \(\alpha (0,S0)=\alpha (0,2)=3>1\) and so the condition (ii) of Theorem 2.1 is satisfied. If \(\alpha (x,y)\ge 1\), then \( x,y\in \{1,2\}\) which implies that \(\alpha (Sx,Sy)\ge 1\), that is, S is an \(\alpha \)-admissible mapping. Let \(\{x_{n}\}\) be a sequence in X such that, \(\alpha (x_{n},Sx_{n})\ge 1\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty \). Then, we must have \(x_{n}\in \{1,2\}\) and so, \(x\in \{1,2\}\), that is, \(\alpha (x,Sx)\ge 1\). Hence, all the conditions of Theorem 2.1 hold and S has a unique fixed point \(x=1\in B_{1}\cap B_{2}.\)

Definition 2.3

Let (Xd) be a metric space, \(q\in {\mathbb {N}} ,\)\(B_{1},B_{2},\ldots ,B_{q}\) closed nonempty subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. We say that S is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping if

  1. (a)
    $$\begin{aligned} S(B_{j})\subseteq B_{j+1},\ j=1,2,\ldots ,q,\quad \text { where }B_{q+1}=B_{1}. \end{aligned}$$
  2. (b)

    for any \(x\in B_{i}\) and \(y\in B_{i+1},\)\(i=1,2,\ldots ,q\) where \( B_{q+1}=B_{1}\) and \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\) we get

    $$\begin{aligned} \psi (d(Sx,Sy))\le \psi (M_{d}(x,y))-LM_{d}(x,y), \end{aligned}$$

    where \(\psi \in \Psi ,\)\(0<L<1\) and

    $$\begin{aligned}&M_{d}(x,y)\\&\quad =\max \left\{ d(x,y),\frac{d(x,Sx)d(y,Sy)}{d(x,y)},\frac{ d(y,Sy)(d(x,Sx)+1)}{1+d(x,y)},\frac{d(x,Sy)+d(y,Sx)}{4}\right\} . \end{aligned}$$

Now, we have the following result in metric spaces. Since a metric space must be a metric-like space, we omit the proof of the following corollary in here.

Corollary 2.1

Let (Xd) be a complete metric space, q be a positive integer, \(B_{1},B_{2},\ldots ,B_{q}\) be nonempty closed subsets of X\( Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Assume that \(S:Y\rightarrow Y\) is a cyclic \(\alpha _{{ L} }^{{ \psi }}\)-rational contractive mapping satisfying the following conditions:

  1. (i)

    S is an \(\alpha \)-admissible mapping;

  2. (ii)

    there exists \(x_{\circ }\in Y\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  3. (iii)

    \((\spadesuit _{1})\) either S is \(\alpha \)-continuous, or; \((\spadesuit _{2})\) for any sequence \(\{x_{n}\}\) in X with \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1\). Then S has a fixed point \(x\in \cap _{i=1}^{q}B_{i}.\) Moreover, if

  4. (iv)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then the fixed point of S is a unique.

If in Theorem 2.1, we take \(\psi (t)=t\) and \(L=(1-r),\) where \(0<r<1\), then we can get the following corollary.

Corollary 2.2

Let \((X,\omega )\) be a complete metric like-space, q be a positive integer, \(B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Assume that \(S:Y\rightarrow Y\) be an operator such that:

  1. (i)

    \(S(B_{j})\subseteq B_{j+1},\)\(j=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1};\)

  2. (ii)

    S is an \(\alpha \)-admissible mapping;

  3. (iii)

    there exists \(x_{\circ }\in Y\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  4. (iv)

    \((\spadesuit _{1})\)S is \(\alpha \)-continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1;\)

  5. (v)

    there exists \(r\in (0,1)\) such that \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\Rightarrow \)

    $$\begin{aligned}&\omega (Sx,Sy)\\&\quad \le r\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)}, \frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} , \end{aligned}$$

    for any \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1}.\) Then S has a fixed point \(x\in \cap _{i=1}^{q}B_{i}.\) Moreover, if

  6. (vi)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}\).

If we put \(X=B_{i},\)\(i=1,2,\ldots ,q,\) in Theorem 2.1, we obtain the following result.

Theorem 2.2

Let \((X,\omega )\) be a complete metric-like space and \( \alpha :X\times X\rightarrow [0,\infty )\) be a mapping. Assume that \( S:X\rightarrow X\) is an \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping satisfying the following conditions:

  1. (i)

    S is an \(\alpha \)-admissible mapping;

  2. (ii)

    there exists \(x_{\circ }\in X\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  3. (iii)

    \((\spadesuit _{1})\) either S is \(\alpha \)-continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1.\) Then S has a fixed point. Moreover, if

  4. (iv)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then S has a unique fixed point.

We give the following example to illustrate the usefulness of Theorem 2.2.

Example 2.5

Let \(X= {\mathbb {R}} ^{+}\) be equipped with the metric-like mapping \(\omega (x,y)=\max \{x,y\}\) for all \(x,y\in X\). Let \(S:X\rightarrow X\) and \(\alpha :X\times X\rightarrow [0,\infty )\) be defined by

$$\begin{aligned} Sx=\left\{ \begin{array}{lll} \frac{1}{4}x^{3}, &{} &{} \text {if }\,0\le x<\frac{1}{4} \\ \frac{1}{8}x^{2}, &{} &{} \text {if }\,\frac{1}{4}\le x\le 1 \\ \frac{1}{24}x, &{} &{} \text {if }\,1<x\le 3 \\ 3x^{3}+2, &{} &{} \text {if }\,x>3 \end{array} \right. \quad \text { and }\quad \alpha (x,y)=\left\{ \begin{array}{lll} 6, &{} &{} \text {if }\, x,y\in [0,3] \\ 0, &{} &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Also, define \(\psi :[0,\infty )\rightarrow [0,\infty )\) by \(\psi (t)= \frac{1}{2}t\) and \(L=\frac{3}{8}<1.\) Let \(\alpha (x,Sx)\alpha (y,Sy)\ge 1,\) then \(x,y\in [0,3].\) Now we discuss the following cases:

\(\blacklozenge \) Let \(0\le x,y<\frac{1}{4},\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x^{3}}{4},\frac{y^{3}}{4} \right\} \le \frac{1}{8}\max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y). \end{aligned}$$

\(\blacklozenge \) Let \(\frac{1}{4}\le x,y\le 1,\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x^{2}}{8},\frac{y^{2}}{8} \right\} =\frac{1}{16}\max \left\{ x^{2},y^{2}\right\} \le \frac{1}{8}\max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y). \end{aligned}$$

\(\blacklozenge \) Let \(1<x,y\le 3,\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x}{24},\frac{y}{24} \right\} \le \frac{1}{8}\max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y). \end{aligned}$$

\(\blacklozenge \) Let \(0\le x<\frac{1}{4}\) and \(\frac{1}{4}\le y<1,\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x^{3}}{4},\frac{y^{2}}{8} \right\} =\frac{1}{8}\max \left\{ x^{3},\frac{y^{2}}{2}\right\} \le \frac{1 }{8}\max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y). \end{aligned}$$

\(\blacklozenge \) Let \(0\le x<\frac{1}{4}\) and \(1<y\le 3,\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x^{3}}{4},\frac{y}{24} \right\} =\frac{1}{8}\max \left\{ x^{3},\frac{y}{6}\right\} \le \frac{1}{8} \max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y). \end{aligned}$$

\(\blacklozenge \) Let \(\frac{1}{4}\le x\le 1\) and \(1<y\le 3,\) then

$$\begin{aligned} \psi (\omega (Sx,Sy))=\frac{1}{2}\max \left\{ \frac{x^{2}}{8},\frac{y}{24} \right\} =\frac{1}{16}\max \left\{ x^{2},\frac{y}{3}\right\} \le \frac{1}{8} \max \left\{ x,y\right\} =\frac{1}{8}\omega (x,y) \end{aligned}$$

and so

$$\begin{aligned} \psi (\omega (Sx,Sy))\le \frac{1}{8}\omega (x,y)=\psi (M_{\omega }(x,y))-LM_{\omega }(x,y). \end{aligned}$$

Then S is an \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping. As in Example 2.3, we can show that the conditions (i), (ii) and (iv) of Theorem 2.2 hold true and 0 is a unique fixed point of S.

Corollary 2.3

Let \((X,\omega )\) be a complete metric like-space, q be a positive integer, \(B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Assume that \(S:Y\rightarrow Y\) be an operator such that:

  1. (i)

    \(S(B_{j})\subseteq B_{j+1},\)\(j=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1};\)

  2. (ii)

    S is an \(\alpha \)-admissible mapping;

  3. (iii)

    there exists \(x_{\circ }\in Y\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  4. (iv)

    (\(\spadesuit _{1}\)) S is \(\alpha \)-continuous, or; (\(\spadesuit _{2}\)) if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1;\)

  5. (v)

    there exists \(r\in (0,1)\) such that \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\Rightarrow \)

    $$\begin{aligned} {\large e}^{\omega (Sx,Sy)}\le r{\large e}^{\max \left\{ \omega (x,y),\frac{ \omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} }, \end{aligned}$$

    for any \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2,\ldots ,q,\) where \(e:[0,\infty )\rightarrow [0,\infty )\) is a Lebesgue-integrable mapping satisfying \(e^{\epsilon }>0\) for \(\epsilon >0.\) Then S has a fixed point \(x\in \cap _{i=1}^{q}B_{i}.\) Moreover, if

  6. (vi)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}\).

If in the above corollary, we take \(X=B_{i},\)\(i=1,2,\ldots ,q\), then the following result holds.

Corollary 2.4

Let \((X,\omega )\) be a complete metric like-space and \(\alpha :X\times X\rightarrow [0,\infty )\) be a mapping. Assume that \( S:X\rightarrow X\) is a self-mapping such that:

  1. (i)

    S is an \(\alpha \)-admissible mapping;

  2. (ii)

    there exists \(x_{\circ }\in X\) such that \(\alpha (x_{\circ },Sx_{\circ })\ge 1;\)

  3. (iii)

    \((\spadesuit _{1})\)S is \(\alpha \)-continuous, or; (\(\spadesuit _{2}\)) if \(\{x_{n}\}\) is a sequence in X such that \(\alpha (x_{n},x_{n+1})\ge 1\) for all \(n\ge 0\) and \(x_{n}\rightarrow x\) as \( n\rightarrow \infty ,\) then \(\alpha (x,Sx)\ge 1;\)

  4. (iv)

    there exists \(r\in (0,1)\) such that \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\Rightarrow \)

    $$\begin{aligned} {\large e}^{\omega (Sx,Sy)}\le r{\large e}^{\max \left\{ \omega (x,y),\frac{ \omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} } \end{aligned}$$

    for any \(x,y\in X\) where \(e:[0,\infty )\rightarrow [0,\infty )\) is a Lebesgue-integrable mapping satisfying \(e^{\epsilon }>0\) for \(\epsilon >0.\) Then S has a fixed point. Moreover, if

  5. (v)

    for all \(x\in Fix(S)\) we have \(\alpha (x,x)\ge 1.\) Then S has a unique fixed point.

Fixed Point Results for Cyclic \({\psi L}\)-Graphic Rational Contractions

In line with Jachymski (2008), let \((X,\omega )\) be a metric-like space and \(\mathfrak {R}\) denote the diagonal of the Cartesian product \(X\times X\). Consider a directed graph G such that the set V(G) of its vertices coincides with X,  and the set E(G) of its edges contains all loops, i.e., \(E(G)\supseteq \mathfrak {R}.\) We assume that G has no parallel edges, so we can identify G with the pair (V(G), E(G)). Moreover, we may treat G as a weighted graph (see (Jachymski 2008)) by assigning to each edge the distance between its vertices.

By \(G^{-1}\) we denote the conversion of a graph G, that is, the graph obtained from G by reversing the direction of edges. Thus, we have

$$\begin{aligned} E(G^{-1})=\{(x,y)\in X\times X:(y,x)\in E(G)\}. \end{aligned}$$

The letter \(\breve{G}\) denotes the undirected graph obtained from G by ignoring the direction of edges. Actually, it will be more convenient for us to treat \(\breve{G}\) as a directed graph for which the set of its edges is symmetric. Under this convention

$$\begin{aligned} E(\breve{G})=E(G)\cup E(G^{-1}). \end{aligned}$$

If x and y are vertices in a graph G, then a path in G from x to y of length N (\(N\in {\mathbb {N}} \)) is a sequence \(\{x_{i}\}_{i}^{N}=0\) of \(N+1\) vertices such that \(x_{\circ }=x\), \(x_{N}=y\) and \((x_{n-1},x_{n})\in E(G)\) for \(i=1,\ldots ,N\). A graph G is connected if there is a path between any two vertices. G is weakly connected if \(\breve{G}\) is connected.

Recently, some results have appeared providing sufficient conditions for a self mapping of X to be a Picard operator when (Xd) is endowed with a graph. The first result in this direction was given by Jachymski (2008).

Definition 3.1

(Jachymski 2008) We say that a mapping \(S:X\rightarrow X\) is a Banach G-contraction or simply G-contraction if S preserves edges of G, i.e.,

$$\begin{aligned} \forall x,y\in X:(x,y)\in E(G)\Rightarrow (S(x),S(y))\in E(G) \end{aligned}$$

and S decreases weights of edges of G as for all \(x,y\in X,\) there exists \(c\in (0,1),\) such that

$$\begin{aligned} (x,y)\in E(G)\Rightarrow d(S(x),S(y))\le cd(x,y). \end{aligned}$$

Definition 3.2

(Türkoğlu et al. 1999) A mapping \(S:X\rightarrow X\) is called orbitally continuous, if given \(x\in X\) and any sequence \(\{k_{n}\}\) of positive integers,

$$\begin{aligned} S^{k_{n}}x\rightarrow y\in X\text { as }n\rightarrow \infty \text { implies } S(S^{k_{n}}x)\rightarrow Sy\,\,\text { as }\,\,n\rightarrow \infty . \end{aligned}$$

Definition 3.3

(Jachymski 2008) A mapping \(S:X\rightarrow X\) is called G -continuous, if given \(x\in X\) and sequence \(\{x_{n}\}\) for all \(n\in {\mathbb {N}} ,\)

$$\begin{aligned} x_{n}\rightarrow x\text { as }n\rightarrow \infty \text { and }(x_{n},x_{n+1})\in E(G)\,\,\text { imply }\,\,Sx_{n}\rightarrow Sx. \end{aligned}$$

Definition 3.4

(Jachymski 2008) A mapping \(S:X\rightarrow X\) is called orbitally G-continuous, if given \(x,y\in X\) and any sequence \(\{k_{n}\}\) of positive integers for all \(n\in {\mathbb {N}} ,\)

$$\begin{aligned} S^{k_{n}}x\rightarrow y\in X\text { and }(S^{k_{n}}x,S^{k_{n}+1}x)\in E(G) \,\,\text { implies }\,\,S(S^{k_{n}}x)\rightarrow Sy\text { as }n\rightarrow \infty . \end{aligned}$$

Now, we present our main result of this section.

Definition 3.5

Let \((X,\omega )\) be a metric-like space endowed with a graph G. Let q be positive integer,  \(B_{1},B_{2},\ldots ,B_{q}\) be \( \omega \)-closed subsets of X and \(Y=\cup _{i=1}^{q}B_{i}\). We say that S is a cyclic \(\psi L\)-graphic rational contractive mapping if

\(\diamond \)\(S(B_{j})\subseteq B_{j+1},\ j=1,2,\ldots ,q,\) where \( B_{q+1}=B_{1};\)

\(\diamond \) for all \(x,y\in X:(x,y)\in E(G)\Rightarrow \)\((Sx,Sy)\in E(G);\)

\(\diamond \) for and \(x\in B_{i},y\in B_{i+1},i=1,2,\ldots ,q\) where \( B_{q+1}=B_{1}\) and \((x,Sx),(y,Sy)\in E(G),\) we get

$$\begin{aligned} \psi (\omega (Sx,Sy))\le \psi (M_{\omega }(x,y))-LM_{\omega }(x,y), \end{aligned}$$

where \(\psi \in \Psi ,\)\(L>0\) and

$$\begin{aligned}&M_{\omega }(x,y)\\&\quad =\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{ \omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{ \omega (x,Sy)+\omega (y,Sx)}{4}\right\} . \end{aligned}$$

Notice that, if we take \(X=B_{i},\)\(i=1,2,\ldots ,q,\) in Definition 3.5, we say that S is \(\psi L\)-graphic rational contraction.

Theorem 3.1

Let \((X,\omega )\) be a complete metric-like space endowed with a graph G. Let q be a positive integer, \( B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(S:Y\rightarrow Y\) is a cyclic \(\psi L\)-graphic rational contractive mapping satisfying the following conditions:

  1. (i)

    there exists \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G);\)

  2. (ii)

    \((\spadesuit _{1})\) either S is orbitally G-continuous, or; \((\spadesuit _{2})\) for any sequence \(\{x_{n}\}\) in X with \( x_{n}\rightarrow x\) as \(n\rightarrow \infty \) and \((x_{n},x_{n+1})\in E(G).\) Then, S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}.\)

Proof

Define, \(\alpha :Y\times Y\rightarrow [0,+\infty )\) by

$$\begin{aligned} \alpha (x,y)=\left\{ \begin{array}{lll} 1, &{} &{} \text {if }\,(x,y)\in E(G) \\ 0, &{} &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

At first, we prove that S is \(\alpha \)-admissible. If \(\alpha (x,y)\ge 1\), then \((x,y)\in E(G)\). As S is a cyclic \(\psi L\)-graphic rational contraction mapping, we have \((Sx,Sy)\in E(G).\) That is, \(\alpha (Sx,Sy)\ge 1\). So S is an \(\alpha \)-admissible mapping. Let S be G-continuous on \( (X,\omega )\). That is,

$$\begin{aligned} x_{n}\rightarrow x\,\,\text { as }\,\,n\rightarrow \infty \,\, \text { and } \,\,(x_{n},x_{n+1})\in E(G)\,\,\text { for all }\,\,n\in {\mathbb {N}} \,\,\text { imply }\,\,Sx_{n}\rightarrow Sx. \end{aligned}$$

This implies

$$\begin{aligned} x_{n}\rightarrow x\,\,\text { as }\,\,n\rightarrow \infty \,\,\text { and }\,\,\alpha (x_{n},x_{n+1})\ge 1\,\,\text { for all }\,\,n\in {\mathbb {N}} \,\,\text { imply }\,\,Sx_{n}\rightarrow Sx, \end{aligned}$$

which implies that S is \(\alpha \)-continuous on \((X,\omega )\). From (i) there exists \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G).\) That is, \(\alpha (x_{\circ },Sx_{\circ })\ge 1\).

Let \(x\in B_{i}\) and \(y\in B_{i+1}\) where \(\alpha (x,Sx)\alpha (y,Sy)\ge 1.\) Then, \(x\in B_{i}\) and \(y\in B_{i+1}\) where \((x,Sx)\in E(G)\) and \((y,Sy)\in E(G).\)

Now, since S is a cyclic \(\psi L\)-graphic rational contraction, we have

$$\begin{aligned} \psi (\omega (Sx,Sy))\le \psi (M_{\omega }(x,y))-LM_{\omega }(x,y). \end{aligned}$$

That is, S is a cyclic \(\alpha _{L}^{\psi }\)-rational contractive mapping. Let \(\{x_{n}\}\subseteq X\) be a sequence such that \(x_{n}\rightarrow x\) as \( n\rightarrow \infty \) and \(\alpha (x_{n+1},x_{n})\ge 1\). Therefore, \( (x_{n+1},x_{n})\in E(G)\) and then from (ii), we get \((x,Sx)\in E(G).\) That is, \(\alpha (x,Sx)\ge 1\). Further, sine \(E(G)\supseteq \)\(\mathfrak {R}\), we have \( \alpha (x,x)=1\) for all \(x\in X.\) Thus, all the conditions of Theorem 2.1 are satisfied and S has a unique fixed point in \(Y=\cap _{i=1}^{q}B_{i}.\) This completes the proof. \(\square \)

If in Theorem 3.1, we put \(\psi (t)=t\) and \(L=(1-r)\) where \(0<r<1,\) then we deduce the following corollary.

Corollary 3.1

Let \((X,\omega )\) be a complete metric-like space endowed with a graph G. Let q be a positive integer, \( B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Let \(S:Y\rightarrow Y\) is an operator such that:

  1. (i)

    \(S(B_{j})\subseteq B_{j+1},\)\(j=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1};\)

  2. (ii)

    for all \(x,y\in Y,\)\((x,y)\in E(G)\Rightarrow (Sx,Sy)\in E(G);\)

  3. (iii)

    there exists \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G);\)

  4. (iv)

    \((\spadesuit _{1})\)S is orbitally G-continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \( (x_{n},x_{n+1})\in E(G)\) for all \(n\in {\mathbb {N}} \) and \(x_{n}\rightarrow x\) as \(n\rightarrow \infty ,\) then \((x,Sx)\in E(G);\)

  5. (v)

    there exists \(r\in (0,1)\) such that \((x,Sx)(y,Sy)\in E(G)\Rightarrow \)

    $$\begin{aligned}&\omega (Sx,Sy)\\&\quad \le r\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{\omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)}, \frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} , \end{aligned}$$

    for any \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1}.\) Then S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}.\)

If in Theorem 3.1, we take \(B_{i}=X,\) then, we have the following theorem.

Theorem 3.2

Let \((X,\omega )\) be a complete metric-like space endowed with a graph G. Assume that \(S:X\rightarrow X\) is a cyclic \(\psi L \)-graphic rational contractive mapping satisfying the following conditions:

  1. (i)

    for all \(x,y\in X,\)\((x,y)\in E(G)\Rightarrow (Sx,Sy)\in E(G);\)

  2. (ii)

    there exists an element \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G);\)

  3. (iii)

    \((\spadesuit _{1})\)S is orbitally \(G-\)continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \( (x_{n},x_{n+1})\in E(G)\) for all \(n\in {\mathbb {N}} \) and \(x_{n}\rightarrow x\) as \(n\rightarrow \infty ,\) then \((x,Sx)\in E(G).\) Then S has a unique fixed point.

By Corollary 3.1, we get the following corollary.

Corollary 3.2

Let \((X,\omega )\) be a complete metric-like space endowed with a graph G. Let q be a positive integer, \( B_{1},B_{2},\ldots ,B_{q}\) be nonempty \(\omega \)-closed subsets of X\(Y=\cup _{i=1}^{q}B_{i}\) and \(\alpha :Y\times Y\rightarrow [0,\infty )\) be a mapping. Let \(S:Y\rightarrow Y\) is an operator such that:

  1. (i)

    \(S(B_{j})\subseteq B_{j+1},\)\(j=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1};\)

  2. (ii)

    for all \(x,y\in Y,\)\((x,y)\in E(G)\Rightarrow (Sx,Sy)\in E(G);\)

  3. (iii)

    there exists \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G);\)

  4. (iv)

    \((\spadesuit _{1})\)S is orbitally G-continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \( (x_{n},x_{n+1})\in E(G)\) for all \(n\in {\mathbb {N}} \) and \(x_{n}\rightarrow x\) as \(n\rightarrow \infty ,\) then \((x,Sx)\in E(G);\)

  5. (v)

    there exists \(r\in (0,1)\) such that \((x,Sx)(y,Sy)\in E(G)\Rightarrow \)

    $$\begin{aligned} \int \limits _{0}^{\omega (Sx,Sy)}\varphi (t)dt\le r\int \limits _{0}^{\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{ \omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} }\varphi (t)dt, \end{aligned}$$

    for any \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1}\) and \(\varphi :[0,\infty )\rightarrow [0,\infty )\) is a Lebesgue-integrable mapping satisfying \(\int \nolimits _{0}^{\varepsilon }\varphi (t)dt>0\) for \(\epsilon >0.\) Then S has a unique fixed point \(x\in \cap _{i=1}^{q}B_{i}.\)

By Corollary 3.2, we get the following result.

Corollary 3.3

Let \((X,\omega )\) be a complete metric-like space endowed with a graph G. Assume that \(S:X\rightarrow X\) is a self-mapping such that:

  1. (i)

    for all \(x,y\in X,\)\((x,y)\in E(G)\Rightarrow (Sx,Sy)\in E(G);\)

  2. (ii)

    there exists \(x_{\circ }\in X\) such that \((x_{\circ },Sx_{\circ })\in E(G);\)

  3. (iii)

    \((\spadesuit _{1})\)S is orbitally G-continuous, or; \((\spadesuit _{2})\) if \(\{x_{n}\}\) is a sequence in X such that \( (x_{n},x_{n+1})\in E(G)\) for all \(n\in {\mathbb {N}} \) and \(x_{n}\rightarrow x\) as \(n\rightarrow \infty ,\) then \((x,Sx)\in E(G);\)

  4. (iv)

    there exists \(r\in [0,1)\) such that \((x,Sx)(y,Sy)\in E(G)\Rightarrow \)

    $$\begin{aligned} \int \limits _{0}^{\omega (Sx,Sy)}\varphi (t)dt\le r\int \limits _{0}^{\max \left\{ \omega (x,y),\frac{\omega (x,Sx)\omega (y,Sy)}{\omega (x,y)},\frac{ \omega (y,Sy)(\omega (x,Sx)+1)}{1+\omega (x,y)},\frac{\omega (x,Sy)+\omega (y,Sx)}{4}\right\} }\varphi (t)dt, \end{aligned}$$

    for any \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2,\ldots ,q,\) where \(B_{q+1}=B_{1}\) and \(\varphi :[0,\infty )\rightarrow [0,\infty )\) is a Lebesgue-integrable mapping satisfying \(\int \nolimits _{0}^{\varepsilon }\varphi (t)dt>0\) for \(\epsilon >0.\) Then S has a unique fixed point.

Application

In fact, this section is the main result of our paper. Namely, we will apply the previous theoretical results to show the existence of solutions for the following integral equation.

$$\begin{aligned} x(t)=h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds,\quad \text { }t\in [0,1], \end{aligned}$$
(14)

where \(h:[0,1]\rightarrow {\mathbb {R}} ,\)\(k:[0,1]\times [0,1]\rightarrow [0,\infty )\) and \( f:[0,1]\times {\mathbb {R}} \rightarrow {\mathbb {R}} \) are three continuous functions.

Let \(X=C([0,1], {\mathbb {R}} )\) be the set of real continuous functions on [0, 1]. Take the metric-like space

$$\begin{aligned} \omega _{\infty }(x,y)=\sup _{t\in [0,1]}\{\left| \left| x(t)\right| -\left| y(t)\right| \right| \}\quad \text { for all } x,y\in X. \end{aligned}$$

Then \((X,\omega _{\infty })\) is a complete metric-like space.

Suppose that \(\pi :X\times X\rightarrow {\mathbb {R}} \) is a function with the following properties:

  • \(\pi (x,y)\ge 0\Longrightarrow \)\(\pi (Sx,Sy)\ge 0,\)

  • there exists \(x_{\circ }\in X\) such that \(\pi (x_{\circ }Sx_{\circ })\ge 0,\)

  • if \(\{x_{n}\}\) is a sequence in X such that \(\pi (x_{n},x_{n+1})\ge 0\) for all \(n\in {\mathbb {N}} \) and \(x_{n}\rightarrow x\) as \(n\rightarrow \infty \), then \(\pi (x,Sx)\ge 0\) , where

    $$\begin{aligned} Sx(t)=h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds,\quad \text { for all }t\in [0,1]. \end{aligned}$$

Let \((a,b)\in X\times X\), \((a_{\circ },b_{\circ })\in {\mathbb {R}} ^{2}\) such that

$$\begin{aligned} a_{\circ }\le a(t)\le b(t)\le b_{\circ },\quad \text { for all }t\in [0,1]. \end{aligned}$$
(15)

Assume that for all \(t\in [0,1]\), we get

$$\begin{aligned} a(t)\le h(t)+\int \limits _{0}^{1}k(t,s)f(s,b(s))ds \end{aligned}$$
(16)

and

$$\begin{aligned} b(t)\ge h(t)+\int \limits _{0}^{1}k(t,s)f(s,a(s))ds. \end{aligned}$$
(17)

Let for all \(s\in [0,1]\), f(s, .) be a decreasing function, that is,

$$\begin{aligned} x,y\in {\mathbb {R}} ,\text { }x\ge y\Rightarrow f(s,x)\le f(s,y). \end{aligned}$$
(18)

Assume that

$$\begin{aligned} \sup _{t\in [0,1]}\int \limits _{0}^{1}k(t,s)ds\le 1. \end{aligned}$$
(19)

Also, suppose that for all \(s\in [0,1],\)\(x,y\in {\mathbb {R}} \) with (\(x\le b_{\circ }\) and \(y\ge a_{\circ }\)) or (\(x\ge a_{\circ }\) and \(y\le b_{\circ }\)) and \(\pi (y,Sy)\ge 1\) and \(\pi (x,Sx)\ge 1\), we have

$$\begin{aligned}&\left| \left| f(s,x)\right| -\left| f(s,y)\right| \right| \nonumber \\&\quad \le r\max \left\{ \begin{array}{c} \left| \left| x\right| -\left| y\right| \right| , \frac{\left| \left| x\right| -\left| Sx\right| \right| \left| \left| y\right| -\left| Sy\right| \right| }{\left| \left| x\right| -\left| y\right| \right| },\frac{\left| \left| y\right| -\left| Sy\right| \right| (\left| \left| x\right| -\left| Sx\right| \right| +1)}{1+\left| \left| x\right| -\left| y\right| \right| }, \\ \frac{\left| \left| x\right| -\left| Sy\right| \right| +\left| \left| y\right| -\left| Sx\right| \right| }{4} \end{array} \right\} , \end{aligned}$$
(20)

where \(0<r<1.\)

Now, we state and prove our main theorem.

Theorem 4.1

Under conditions (15)–(20), the integral equation (14) has a solution \(\{x\in X:a\le x(t)\le b\) for all \(t\in [0,1]\}.\)

Proof

Define closed subsets of X, \(B_{1}(t)\) and \(B_{2}(t)\) for all \(t\in [0,1]\) by

$$\begin{aligned} B_{1}=\{x\in X:x\le b\} \end{aligned}$$

and

$$\begin{aligned} B_{2}=\{x\in X:x\ge a\}. \end{aligned}$$

Consider the operator \(S:X\rightarrow X\) defined by

$$\begin{aligned} Sx(t)=h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds,\quad \text { for all }t\in [0,1], \end{aligned}$$

Let us prove that S is a cyclic mapping, i.e.,

$$\begin{aligned} S(B_{1})\subseteq B_{2}\quad \text { and }\quad S(B_{2})\subseteq B_{1}. \end{aligned}$$
(21)

Suppose that \(x\in B_{1},\) that is

$$\begin{aligned} x(s)\le b(s),\quad \text { for all }s\in [0,1]. \end{aligned}$$

Applying condition (18), since \(k(t,s)\ge 0\) and f(sx(s)) non-increasing for all \(s,t\in [0,1],\) we can write

$$\begin{aligned} k(t,s)f(s,x(s))\ge k(t,s)f(s,b(s))\quad \text { for all }\,s,t\in [0,1]. \end{aligned}$$

Integrate with respect to s, yields

$$\begin{aligned} \int \limits _{0}^{1}k(t,s)f(s,x(s))ds\ge \int \limits _{0}^{1}k(t,s)f(s,b(s))ds, \end{aligned}$$

adding h(t) to both sides of the above inequality and using (16), we get

$$\begin{aligned} h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds\ge h(t)+\int \limits _{0}^{1}k(t,s)f(s,b(s))ds\ge a(t), \end{aligned}$$

for all \(t\in [0,1].\) Then we have \(Sx\in B_{2}.\)

Similarly, \(x\in B_{2},\) that is

$$\begin{aligned} x(s)\ge a(s),\quad \text { for all }s\in [0,1]. \end{aligned}$$

By condition (18), since \(k(t,s)\ge 0\) for all \(s,t\in [0,1],\) one can write

$$\begin{aligned} k(t,s)f(s,x(s))\le k(t,s)f(s,a(s))\quad \text { for all }s,t\in [0,1]. \end{aligned}$$

Taking the integral and using condition (17), it follows that

$$\begin{aligned} h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds\le h(t)+\int \limits _{0}^{1}k(t,s)f(s,a(s))ds\le b(t), \end{aligned}$$

for all \(t\in [0,1].\) Then we have \(Sx\in B_{2}.\) Therefore, (21) holds.

Now, let \((x,y)\in B_{1}\times B_{2}\), that is, for all \(t\in [0,1],\)

$$\begin{aligned} x(t)\le b(t),\,y(t)\ge a(t). \end{aligned}$$

This implies by condition (15) that for all \(t\in [0,1],\)

$$\begin{aligned} x(t)\le b_{\circ },\, y(t)\ge a_{\circ }. \end{aligned}$$

Let, \(x\in B_{1}\) and \(y\in B_{2}\) where \(\pi (y,Sy)\ge 0\) and \(\pi (x,Sx)\ge 0\). Then from (20), we have for all \(t\in [0,1],\)

$$\begin{aligned} \left| \left| Sx\right| -\left| Sy\right| \right|= & {} \left| \left| h(t)+\int \limits _{0}^{1}k(t,s)f(s,x(s))ds\right| -\left| h(t)+\int \limits _{0}^{1}k(t,s)f(s,y(s))ds\right| \right| \\\le & {} \int \limits _{0}^{1}k(t,s)\left( \left| \left| f(s,x(s))-f(s,y(s))\right| \right| \right) ds \\\le & {} r\int \limits _{0}^{1}k(t,s)\left( \max \left\{ \begin{array}{c} \left| \left| x\right| -\left| y\right| \right| , \frac{\left| \left| x\right| -\left| Sx\right| \right| \left| \left| y\right| -\left| Sy\right| \right| }{\left| \left| x\right| -\left| y\right| \right| },\frac{\left| \left| y\right| -\left| Sy\right| \right| (\left| \left| x\right| -\left| Sx\right| \right| +1)}{1+\left| \left| x\right| -\left| y\right| \right| }, \\ \frac{\left| \left| x\right| -\left| Sy\right| \right| +\left| \left| y\right| -\left| Sx\right| \right| }{4} \end{array} \right\} \right) ds \\\le & {} r\left[ \int \limits _{0}^{1}k(t,s)ds\right] \left( \max \left\{ \begin{array}{c} \omega _{\infty }(x,y),\frac{\omega _{\infty }(x,Sx)\omega _{\infty }(y,Sy)}{ \omega _{\infty }(x,y)},\frac{\omega _{\infty }(y,Sy)(\omega _{\infty }(x,Sx)+1)}{1+\omega _{\infty }(x,y)}, \\ \frac{\omega _{\infty }(x,Sy)+\omega _{\infty }(y,Sx)}{4} \end{array} \right\} \right) \\\le & {} r\max \left\{ \begin{array}{c} \omega _{\infty }(x,y),\frac{\omega _{\infty }(x,Sx)\omega _{\infty }(y,Sy)}{ \omega _{\infty }(x,y)},\frac{\omega _{\infty }(y,Sy)(\omega _{\infty }(x,Sx)+1)}{1+\omega _{\infty }(x,y)}, \\ \frac{\omega _{\infty }(x,Sy)+\omega _{\infty }(y,Sx)}{4} \end{array} \right\} , \end{aligned}$$

which implies

$$\begin{aligned} \omega _{\infty }(Sx,Sy)\le r\max \left\{ \begin{array}{c} \omega _{\infty }(x,y),\frac{\omega _{\infty }(x,Sx)\omega _{\infty }(y,Sy)}{ \omega _{\infty }(x,y)},\frac{\omega _{\infty }(y,Sy)(\omega _{\infty }(x,Sx)+1)}{1+\omega _{\infty }(x,y)}, \\ \frac{\omega _{\infty }(x,Sy)+\omega _{\infty }(y,Sx)}{4} \end{array} \right\} . \end{aligned}$$

By a similar way, we can show that the above inequality holds if \((x,y)\in B_{2}\times B_{1}\) where \(\pi (y,Sy)\ge 0\) and \(\pi (x,Sx)\ge 0.\)

Now, if we define \(\alpha :X\times X\rightarrow [0,\infty )\) by

$$\begin{aligned} \alpha (x,y)=\left\{ \begin{array}{lll} 1, &{} &{} \text {if }\,\pi (x,y)\ge 0 \\ 0, &{} &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

Then \(x\in B_{i},\)\(y\in B_{i+1},\)\(i=1,2\) and \(\alpha (x,Sx)\alpha (y,Sy)\ge 1\Rightarrow \)

$$\begin{aligned} \omega _{\infty }(Sx,Sy)\le r\max \left\{ \begin{array}{c} \omega _{\infty }(x,y),\frac{\omega _{\infty }(x,Sx)\omega _{\infty }(y,Sy)}{ \omega _{\infty }(x,y)},\frac{\omega _{\infty }(y,Sy)(\omega _{\infty }(x,Sx)+1)}{1+\omega _{\infty }(x,y)}, \\ \frac{\omega _{\infty }(x,Sy)+\omega _{\infty }(y,Sx)}{4} \end{array} \right\} . \end{aligned}$$

That is, S is a cyclic \(\alpha _{{ L}}^{{ \psi }}\)-rational contractive mapping. Hence, all the conditions of Corollary 2.2 hold and S has a fixed point z in

$$\begin{aligned} B_{1}\cap B_{2}=\{x\in C([0,1], {\mathbb {R}} ):a\le x(t)\le b,\quad \text { for all }t\in [0,1]\}. \end{aligned}$$

That is, \(z\in B_{1}\cap B_{2}\) is the solution to (14). \(\square \)

The following example support Theorem 4.1.

Example 4.1

Let \((X,\omega _{\infty })\) be a complete metric-like space under the same distance in this section and C(X, [0, 1]) be the set of real continuous functions on [0, 1]. Consider the following nonlinear integral equation:

$$\begin{aligned} x(t)=\frac{7}{8}t+\frac{1}{4}\int \limits _{0}^{1}t\left( \frac{s}{1+s}\right) x(s)ds,\quad \text {for all }s\in [0,1]. \end{aligned}$$
(22)

The integral equation (22) is a particular case of (14), where

$$\begin{aligned} h(t)=\frac{7}{8}t,\,k(t,s)=\frac{1}{2}ts\quad \text { and }\quad f(s,x(s))= \frac{x(s)}{2(1+s)}. \end{aligned}$$

Let \(a(t)=\frac{3}{4}t\) and \(b(t)=t^{3},\) for all \(t\in [0,1].\) It is clear that

$$\begin{aligned} a_{\circ }\le a(t)=\frac{3}{4}t\le t^{3}=b(t)\le b_{\circ }, \end{aligned}$$

where \((a,b)\in X\times X\) and \((a_{\circ },b_{\circ })\in {\mathbb {R}} ^{2}.\)

Also, for all \(t\in [0,1],\) we can write

$$\begin{aligned} a(t)=\frac{3}{4}t\le \frac{7}{8}t+\frac{1}{4}\int \limits _{0}^{1}t\left( \frac{s}{1+s}\right) b(s)ds=h(t)+\int \limits _{0}^{1}k(t,s)f(s,b(s))ds \end{aligned}$$

and

$$\begin{aligned} b(t)=t^{3}\ge \frac{7}{8}t+\frac{1}{4}\int \limits _{0}^{1}t\left( \frac{s}{ 1+s}\right) a(s)ds=h(t)+\int \limits _{0}^{1}k(t,s)f(s,a(s))ds. \end{aligned}$$

Again, for all \(s\in [0,1],\) we observe that the function f(sx(s)) is decreasing and

$$\begin{aligned} \sup _{t\in [0,1]}\int _{0}^{1}k(t,s)ds=\frac{1}{2}\sup _{t\in [0,1]}\int _{0}^{1}tsds\le 1. \end{aligned}$$

Also, suppose that for each \(s\in [0,1],\)\(x,y\in {\mathbb {R}} \) with (\(x\le b_{\circ }\) and \(y\ge a_{\circ }\)) or (\(x\ge a_{\circ }\) and \(y\le b_{\circ }\)) and \(\pi (y,Sy)\ge 1\), \(\pi (x,Sx)\ge 1\), we obtain that

$$\begin{aligned} \left| \left| f(s,x)\right| -\left| f(s,y)\right| \right|= & {} \frac{1}{2(1+s)}\left| \left| x(s)\right| -\left| y(s)\right| \right| \le \frac{1}{2}\left| \left| x(s)\right| -\left| y(s)\right| \right| \\\le & {} r\max \left\{ \begin{array}{c} \left| \left| x\right| -\left| y\right| \right| , \frac{\left| \left| x\right| -\left| Sx\right| \right| \left| \left| y\right| -\left| Sy\right| \right| }{\left| \left| x\right| -\left| y\right| \right| },\frac{\left| \left| y\right| -\left| Sy\right| \right| (\left| \left| x\right| -\left| Sx\right| \right| +1)}{1+\left| \left| x\right| -\left| y\right| \right| }, \\ \frac{\left| \left| x\right| -\left| Sy\right| \right| +\left| \left| y\right| -\left| Sx\right| \right| }{4} \end{array} \right\} \end{aligned}$$

where \(r=\frac{1}{2}<1.\) Therefore, all conditions (15)–(20) of Theorem 4.1 are satisfied. Hence the integral equation (22) has a solution \(\{x\in X:\)\( x(t)\in [\frac{3}{4}t,t^{3}]\}.\)

References

  1. Aage, C.T., Salunke, J.N.: The results on fixed points in dislocated and dislocated quasi-metric space. Appl. Math. Sci. 2(59), 2941–2948 (2008)

    MathSciNet  MATH  Google Scholar 

  2. Agarwal, R.P., Hussain, N., Taoudi, M.A.: Fixed point theorems in ordered Banach spaces and applications to nonlinear integral equations. Abstr. Appl. Anal. 2012, 245872 (2012)

  3. Amini-Harandi, A.: Metric-like spaces, partial metric spaces and fixed points. Fixed Point Theory Appl. 2012, 204 (2012)

    MathSciNet  Article  Google Scholar 

  4. Banach, S.: Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fund. Math. 3, 133–181 (1922)

    MathSciNet  Article  Google Scholar 

  5. Gopal, D., Abbas, M., Vetro, C.: Some new fixed point theorems in Menger PM-spaces with application to Volterra type integral equation. Appl. Math. Comput. 232, 955–967 (2014)

    MathSciNet  MATH  Google Scholar 

  6. Hitzler, P.: Generalized metrics and topology in logic programming semantics. Ph.D. thesis, School of Mathematics, Applied Mathematics and Statistics, National University Ireland, University college Cork (2001)

  7. Hussain, N., Al-Mezel, S., Salimi, P.: Fixed points for \(\psi \)-graphic contractions with application to integral equations. Abstr. Appl. Anal. 2013, 575869 (2013)

  8. Hussain, N., Kutbi, M.A., Salimi, P.: Fixed point theory in \(\alpha \)-complete metric spaces with applications. Abstr. Appl. Anal. 2014, 280817 (2014)

  9. Jachymski, J.: The contraction principle for mappings on a metric space with a graph. Proc. Am. Math. Soc. 1(136), 1359–1373 (2008)

    MathSciNet  MATH  Google Scholar 

  10. Karapinar, E., Salimi, P.: Dislocated metric space to metric spaces with some fixed point theorems. Fixed Point Theory Appl. 2013, 222 (2013)

    MathSciNet  Article  Google Scholar 

  11. Kirk, W.A., Srinavasan, P.S., Veeramani, P.: Fixed points for mapping satisfying cyclical contractive conditions. Fixed Point Theory 4, 79–89 (2003)

    MathSciNet  MATH  Google Scholar 

  12. Liu, Z., Li, X., Minkan, S., Cho, S.Y.: Fixed point theorems for mappings satisfying contractive condition of integral type and applications. Fixed Point Theory Appl. 2011, 64 (2011). https://doi.org/10.1186/1687-1812-2011-64

    MathSciNet  Article  Google Scholar 

  13. Matthews, S.G.: Partial metric topology, In: Proceedings of the 8th summer conference on general topology and applications, Annals of the New York Academy of Sciences, vol. 728, pp. 183–197 (1994)

    MathSciNet  Article  Google Scholar 

  14. Pathak, H.K., Khan, M.S., Tiwari, R.: A common fixed point theorem and its application to nonlinear integral equations. Comput. Math. Appl. 53, 961–971 (2007)

    MathSciNet  Article  Google Scholar 

  15. Radenović, S., Došenović, T., Lampert, T.A., Golubovíć, Z.: A note on some recent fixed point results for cyclic contractions in \(b\)-metric spaces and an application to integral equations. Appl. Math. Comput. 273, 155–164 (2016)

    MathSciNet  MATH  Google Scholar 

  16. Ran, A.C.M., Reuring, M.C.B.: A fixed point theorem in partially ordered sets and some applications to matrix equations. Proc. Am. Math. Soc. 132, 1435–1443 (2004)

    MathSciNet  Article  Google Scholar 

  17. Salimi, P., Vetro, C., Vetro, P.: Fixed point theorems for twisted \((\alpha,\beta )-\psi \)-contractive type mappings and applications. Filomat 27(4), 605–615 (2013)

    MathSciNet  Article  Google Scholar 

  18. Shahzad, N., Valero, O., Alghamdi, M.A., Alghamdi, M.A.: A fixed point theorem in partial quasi-metric spaces and an application to software engineering. Appl. Math. Comput. 268, 1292–1301 (2015)

    MathSciNet  MATH  Google Scholar 

  19. Shatanawi, W., Postolache, M.: Common fixed point results for mappings under nonlinear contraction of cyclic form in ordered metric spaces. Fixed Point Theory Appl. 2013, 60 (2013). https://doi.org/10.1186/1687-1812-2013-60

    MathSciNet  Article  MATH  Google Scholar 

  20. Shukla, S., Radenović, S., Rajić, V.Ć.: Some common fixed point theorems in 0-\(\sigma \)-complete metric-like spaces. Vietnam J. Math. 41, 341–352 (2013)

    MathSciNet  Article  Google Scholar 

  21. Suzuki, T.: A generalized Banach contraction principle that characterizes metric completeness. Proc. Am. Math. Soc. 136, 1861–1869 (2008)

    MathSciNet  Article  Google Scholar 

  22. Suzuki, T.: A new type of fixed point theorem in metric spaces. Nonlinear Anal. 71(11), 5313–5317 (2009)

    MathSciNet  Article  Google Scholar 

  23. Türkoğlu, D., Özer, O., Fisher, B.: Fixed point theorems for \(T\)-orbitally complete spaces. Mathematica 9, 211–218 (1999)

    MathSciNet  MATH  Google Scholar 

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Hammad, H.A., De la Sen, M. Solution of Nonlinear Integral Equation via Fixed Point of Cyclic \(\alpha _{L}^{ \psi }\)-Rational Contraction Mappings in Metric-Like Spaces. Bull Braz Math Soc, New Series 51, 81–105 (2020). https://doi.org/10.1007/s00574-019-00144-1

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Keywords

  • Cyclic contractive mapping
  • \(\alpha \)-Admissible
  • Nonlinear integral equations

Mathematics Subject Classification

  • 46N40
  • 47H10
  • 46T99