The Generalized Iterated Function System and Common Attractors of Generalized Hutchinson Operators in Dislocated Metric Spaces

Abstract

In this paper, we present the generalized iterated function system for the construction of common fractals of generalized contractive mappings in the setup of dislocated metric spaces. The well-posedness of attractors’ problems of rational contraction maps in the framework of dislocated metric spaces is also established. Moreover, the generalized collage theorem is also obtained in dislocated metric spaces.

1. Introduction

Metric fixed-point theory serves as an essential tool for solving problems arising in various branches of mathematical analysis, for instance, split feasibility problems, variational inequality problems, nonlinear optimization problems, equilibrium problems, complementary problems, selection and matching problems, and problems of proving the existence of a solution of integral and differential equations, adaptive control systems, fractal image decoding, the convergence of recurrent networks, and many more. In particular, it has deep roots in nonlinear functional analysis.

Hitzler and Seda [1] introduced the notion of dislocated metric space and proved a fixed-point result as an interesting generalization of the Banach contraction principle. Then, several useful results of fixed-point theory were established in [2,3,4,5] by introducing the notion of contractions and related fixed point results in dislocated metric spaces. Rasham et al. [6] presented some multivalued fixed-point results in dislocated metric space with having some applications.

In this paper, by using the concept of the Hausdorff dislocated metric and generalized contraction mappings, we establish the existence of common attractors of generalized rational contractive Hutchinson operators in the framework of dislocated metric spaces. The sections of the paper are organized as follows. Section 2 discusses the basic concepts of the dislocated metric space, dislocated Hausdorff metric space, and generalized contraction, which are required for this research work. Then, the generalized rational contractive Hutchinson operators, generalized iterated function system, and common attractors of generalized iterated function system are defined in Section 3. Further, the existence and some properties of common attractors of generalized rational contractive Hutchinson operators in dislocated metric spaces are proved in Section 4. Then, the well-posedness of attractors’ problems for generalized rational contractive Hutchinson operators in the framework of dislocated metric spaces is proved in Section 5. The application of obtained results to solve a pair of functional equations arising in the dynamic programming is in Section 6. Finally, the obtained results are concluded in Section 7.

2. Preliminary Results

In the preliminary part, we recall some basic theory of dislocated metric space and generalized contraction that is required for the proposed research work.

Throughout this work, a set of real numbers shall be represented by $\mathbb{R}$, a set of non-negative real numbers by ${\mathbb{R}}_{+}$, a set of a-tuples of real numbers by ${\mathbb{R}}^a$, and a set of natural numbers by $\mathbb{N}$. First, we review some key concepts.

Definition 1

([1]). Let X be a nonempty set. A function $\delta :X\times X\to {\mathbb{R}}_{+}$ is said to be a dislocated metric (or metric-like) on X if for any $x,y,z\in X,$ the following conditions hold:

(i)

$\delta (x,y)=0$ implies that $x=y;$

(ii)

$\delta (x,y)=\delta (y,x);$

(iii)

$\delta (x,z)\le \delta (x,y)+\delta (y,z).$

Then, δ is called a dislocated metric and the pair $(X,\delta )$ is called a dislocated metric space.

Example 1.

We take $X=\{a,b,c\}\subseteq \mathbb{R}$ and consider the dislocated metric $\delta :X\times X\to {\mathbb{R}}_{+}$ defined as

$$\begin{array}{cc}\hfill & \delta (a,a)=0,\delta (b,b)=1,\delta (c,c)=\frac{2}{3},\hfill \\ \hfill & \delta (a,b)=\delta (b,a)=\frac{9}{10},\delta (b,c)=\delta (c,b)=\frac{4}{5},\hfill \\ \hfill & \delta (a,c)=\delta (c,a)=\frac{7}{10}.\hfill \end{array}$$

Since $\delta (b,b)\ne 0,$ δ is not a metric, and since $\delta (b,b)\ge \delta (a,b),$ δ is not a partial metric defined in [3].

Example 2.

Let $X={\mathbb{R}}_{+}$ and $a,b\in {\mathbb{R}}_{+}.$ Consider $\delta :X\times X\to {\mathbb{R}}_{+}$ defined as

$$\delta (x,y)=a\left|x-y\right|+bmax\{x,y\}.$$

If we take $a=1$ and $b=0$, then δ is a metric on X. If we take $a=0$ and $b=1$, then δ is a partial metric on X. If we take $a=2$ and $b=4$, then δ is a dislocated metric on X, which is neither a metric nor a partial metric on $X.$

Definition 2

(Open Ball). Let $(X,\delta )$ be a dislocated metric space and $\epsilon \in {\mathbb{R}}_{+}.$ We define the open ball as follows:

$$B_{\epsilon }\left(x\right)=\{y\in X:|\delta (x,y)-\delta (x,x)|<\epsilon \}.$$

The topology ${\tau }_{\delta }$ on $(X,\delta )$ is as follows:

$${\tau }_{\delta }=\{U\subseteq X:\forall u\in U,\exists \epsilon \in {\mathbb{R}}_{+}\mathrm{such}\mathrm{that}{B}_{\epsilon }\left(x\right)\subseteq U\}.$$

A sequence $\left\{u_n\right\}$ in $(X,\delta )$ is said to converge to u in X if and only if $\underset{n\to \infty }{lim}\delta (u_n,u)$ = $\delta (u,u).$

A limit point of every convergence sequence in dislocated metric space $(X,\delta )$ is unique [7].

Definition 3

([1]). Let $(X,\delta )$ be a dislocated metric space.

(i)

A sequence $\left\{x_n\right\}$ in X is said to be a Cauchy sequence if $\underset{n,m\to \infty }{lim}\delta (x_n,x_m)$ exists and is finite.

(ii)

$(X,\delta )$ is said to be complete if every Cauchy sequence $\left\{x_n\right\}$ in X converges with respect to ${\tau }_{\delta }$ to a point $x\in X$ such that $\underset{n\to \infty }{lim}\delta (x_n,x)=\delta (x,x)=\underset{n\to \infty }{lim}\delta (x_n,x_m).$

Definition 4

([7]). Let $(X_1,{\delta }_1)$ and $(X_2,{\delta }_2)$ be two dislocated metric spaces. A function $\mathfrak{f}:X_1\to X_2$ is said to be continuous if for each sequence $\left\{u_n\right\}$, which converges to $u_0$ in $X_1$, the sequence $\left\{\mathfrak{f}\left(u_n\right)\right\}$ converges to $\mathfrak{f}\left(u_0\right)$ in $X_2$.

A subset Y in dislocated metric space $(X,\delta )$ is said to be bounded if and only if the set $\left\{\delta \right(x,y):x,y\in Y\}$ is bounded above.

Definition 5

([3]). Let $\overline{C}$ be a closure of C with respect to dislocated metric $\delta .$ Then,

$$c\in \overline{C}⟺B_{\delta }(c,\epsilon )\cap C\ne \varnothing \mathrm{for}\mathrm{all}\epsilon >0.$$

Set C in dislocated metric space is closed if and only if $\overline{C}=C.$

Definition 6.

Let $(X,\delta )$ be a dislocated metric space. Subset K of X is said to be compact if and only if every open cover of K (by open sets in M) has a finite subcover. If M itself has this property, then we say that M is a compact dislocated metric space.

Theorem 1.

Let $(X,\delta )$ be a dislocated metric space, and let K be a compact subset of X. Then, K is a closed and bounded subset of X.

Proof. 

Let K be a compact subset of a dislocated metric space $(X,\delta ).$ To show that K is closed, we show that the complement, $O=X\setminus K$, is open. Let $z\in O$. We need to find $\epsilon >0$ such that $B\epsilon \left(z\right)\subseteq O$. Now, for any $x\in K$, let ${\epsilon }_x=d(x,z)$. Since $z\notin K$, ${\epsilon }_x>0$. The collection of open sets $\{B{\epsilon }_x\left(x\right):x\in K\}$ is an open cover of K (since any $x\in K$ is covered by $B{\epsilon }_x\left(x\right)$). Since K is compact, there is a finite subcover of this cover, that is, there is a finite set $x_1,\dots ,x_n$ such that the corresponding open balls already cover K. Let $\epsilon =\frac{1}{2}min\{{\epsilon }_{x_1},\dots ,{\epsilon }_{x_n}\}$. The claim is that $B_{\epsilon }\left(z\right)\subseteq O$. To show this, let $y\in B_{\epsilon }\left(z\right)$. We want to show $y\in O$, that is, $y\notin K$. Consider $x_i$, where $1\le i\le n$. By the triangle inequality, $\delta (x_i,y)+\delta (y,z)\ge \delta (x_i,z)={\epsilon }_{x_i}\ge 2\epsilon$. So, $\delta (x_i,y)\ge 2\epsilon -\delta (y,z)>2\epsilon -\epsilon =\epsilon$. (The last inequality follows because $\delta (y,z)<\epsilon$). Then, since $\delta (x_i,y)>\epsilon$, y is not in the open ball of radius ${\epsilon }_{x_i}$ about $x_i$. Since the open balls $B{\epsilon }_{x_i}\left(x_i\right)$ cover K, we have that $y\notin K$.

Now, to prove that K is bounded, let x be an element of K, and consider the collection of open balls of integral radius, $\{B_i\left(x\right):i=1,2,\dots \}$. Since every element of K has some finite distance from x, this collection is an open cover of K. Since K is compact, it has a finite subcover $\{B_{i_1}\left(x\right),\dots ,B_{i_n}\left(x\right)\}$, where we can assume $i_1<\dots <i_n$. But since $B_{i_1}\left(x\right)\subseteq \dots \subseteq B_{i_n}\left(x\right)$, this means that $B_{i_n}$ by itself already covers K. Then, for $y,z\in K$, y and z are in $B_{x_n}\left(x\right)$, and $\delta (y,z)\le \delta (y,x)+\delta (x,z)\le i_n+i_n$. It follows that $2i_n$ is an upper bound for $\left\{\delta \right(y,z):y,z\in K\}$. Thus, K is bounded. □

Definition 7.

A dislocated metric space $(X,\delta )$ is sequentially compact if every sequence has a convergent subsequence.

Theorem 2.

A dislocated metric space $(X,\delta )$ is compact if and only if it is sequentially compact.

Proof. 

Suppose that X is compact. Let $\left\{F_n\right\}$ be a decreasing sequence of closed nonempty subsets of X, and let $G_n=F_n^c$.

If ${\displaystyle \bigcup _{n=1}^{\infty }}{G}_n=X$, then $\{G_n:n\in \mathbb{N}\}$ is an open cover of X, so it has a finite subcover $\{G_{n_k}:k=1,\dots ,l\}$ since X is compact. Let $N=max\{n_k:k=1,\dots ,l\}$. Then, ${\displaystyle \bigcup _{n=1}^N}{G}_n=X$, so $F_N={\displaystyle \bigcap _{n=1}^N}{F}_n={\left({\displaystyle \bigcup _{n=1}^N}{G}_n\right)}^c=\varnothing$, contrary to our assumption that every $F_n$ is nonempty. It follows that ${\displaystyle \bigcup _{n=1}^{\infty }}{G}_n\ne X$, and then ${\displaystyle \bigcap _{n=1}^{\infty }}{F}_n={\left({\displaystyle \bigcup _{n=1}^{\infty }}{G}_n\right)}^c\ne \varnothing ,$ meaning that X has the finite intersection property for closed sets, so X is sequentially compact.

Conversely, suppose that X is sequentially compact. Let $\{G\alpha \subset X:\alpha \in I\}$ be an open cover of X. Then, there exists $\eta >0$ such that every ball $B_{\eta }\left(x\right)$ is contained in some $G\alpha$. Since X is sequentially compact, it is totally bounded, so there exists a finite collection of balls of radius $\eta ,$

$$\{B_{\eta }\left(x_i\right):i=1,\dots ,n\}$$

that covers X. Choose ${\eta }_i\in I$ such that $B_{\eta }\left(x_i\right)\subset G{\alpha }_i$. Then $\{G{\alpha }_i:i=1,\dots ,n\}$ is a finite subcover of X, so X is compact. □

Theorem 3.

Let f be a continuous self-map on compact set X in dislocated metric space $(X,\delta$) into itself. Then the range set $f\left(X\right)$ of f is also compact.

Proof. 

We are to show that, for any sequence $\left\{a_n\right\}$ in $X,$ the sequence $\left\{f{a}_n\right\}$ has a convergent subsequence with limit $\left\{f{a}_0\right\}$ for some $a_0$ in $X.$

Since the sequence $\left\{a_n\right\}$ in $X,$ we have a subsequence $\left\{a_{n_k}\right\}$ of $\left\{a_n\right\}$ such that $\left\{a_{n_k}\right\}$ converges to some $a_0$ in $X.$ By continuity of $f,$ we obtain that $\left\{f{a}_{n_k}\right\}$ converges to $\left\{f{a}_0\right\}.$ As the sequence $\left\{f{a}_n\right\}$ is in X has a convergent subsequence $\left\{f{a}_{n_k}\right\}$ with limit $\left\{f{a}_0\right\}$ for some $a_0$ in $X.$ Consequently, the range $f\left(X\right)$ of f is also compact. □

In the dislocated metric space $\left(X,\delta \right),$ we define the following sets:

$$\begin{array}{ccc}\hfill \mathcal{B}\left(X\right)& =& \{Y:Y\mathrm{is}\mathrm{nonempty}\mathrm{closed}\mathrm{and}\mathrm{bounded}\mathrm{subset}\mathrm{of}X\},\hfill \\ \hfill \mathcal{C}\left(X\right)& =& \{Y:Y\mathrm{is}\mathrm{nonempty}\mathrm{compact}\mathrm{subset}\mathrm{of}X\}.\hfill \end{array}$$

For $R,S\in \mathcal{B}\left(X\right)$ and $x\in X$, we define

$$\delta (x,R)=inf\{\delta (x,r),r\in R\},{\sigma }_{\delta }(R,S)=sup\{\delta (r,S):r\in R\}.$$

Definition 8

([3]). Let $(X,\delta )$ be a dislocated metric space. For $R,S\in \mathcal{B}\left(X\right),$ we define dislocated Hausdorff metric $H_{\delta }$ by

$$H_{\delta }(R,S)=max\{{\sigma }_{\delta }(R,S),{\sigma }_{\delta }(S,R)\}.$$

The pair $(X,H_{\delta })$ is called dislocated Hausdorff metric space.

Theorem 4

([3]). Let $(X,\delta )$ be a dislocated metric space. Then, for all $R,S,T\in \mathcal{B}\left(X\right),$ the following hold:

(H1) 

$H_{\delta }(R,R)={\sigma }_{\delta }(R,R)=sup\{\sigma (r,R):r\in R\};$

(H2) 

$H_{\delta }(R,S)=H_{\delta }(S,R);$

(H3) 

$H_{\delta }(R,S)=0⟹S=R;$

(H4) 

$H_{\delta }(R,T)\le H_{\delta }(R,S)+H_{\delta }(S,T).$

If $(X,\delta )$ is a complete dislocated metric space, then $\left(\mathcal{C}\left(X\right),H_{\delta }\right)$ is also a complete dislocated Hausdorff metric space.

Lemma 1.

Let $(X,\delta )$ be a dislocated metric space. For all $R,S,U,V\in \mathcal{B}\left(X\right)$, the following hold:

(i)

If $S\subseteq U,$ then $\underset{r\in R}{sup}\delta (r,U)\le \underset{r\in R}{sup}\delta (r,S).$

(ii)

$\underset{x\in R\cup S}{sup}\delta (x,U)=max\{\underset{r\in R}{sup}\delta (r,U),\underset{s\in S}{sup}\delta (s,U)\}.$

(iii)

$H_{\delta }(R\cup S,U\cup V)\le max\{H_{\delta }(R,U),H_{\delta }(S,V)\}.$

Proof. 

(i) Since $S\subseteq U,$ for all $r\in R,$ we have

$$\begin{array}{ccc}\hfill \delta (r,U)& =& inf\{\delta (r,u):u\in U\}\le inf\{\delta (r,s):s\in S\}=\delta \left(r,S\right),\hfill \end{array}$$

which implies that $\underset{r\in R}{sup}\delta (r,U)\le \underset{r\in R}{sup}\delta (r,S).$

(ii)

$$\begin{array}{ccc}\hfill \underset{x\in R\cup S}{sup}\delta \left(x,U\right)& =& sup\{\delta \left(x,U\right):x\in R\cup S\}\hfill \\ & =& max\{sup\{\delta \left(x,U\right):x\in R\},sup\{\delta \left(x,U\right):x\in S\}\}\hfill \\ & =& max\{\underset{r\in R}{sup}\delta \left(r,U\right),\underset{s\in S}{sup}\delta \left(s,U\right)\}.\hfill \end{array}$$

(iii) Note that

$$\begin{array}{cc}\hfill & \underset{x\in R\cup S}{sup}\delta (x,U\cup V)\hfill \\ \hfill & \le max\{\underset{r\in R}{sup}\delta (r,U\cup V),\underset{s\in S}{sup}\delta (s,U\cup V)\}(\mathrm{by}\mathrm{using}\left(\mathrm{ii}\right))\hfill \\ \hfill & \le max\{\underset{r\in R}{sup}\delta (r,U),\underset{s\in S}{sup}\delta (s,V)\}(\mathrm{by}\mathrm{using}\left(\mathrm{i}\right))\hfill \\ \hfill & \le max\left\{max\{\underset{r\in R}{sup}\delta (r,U),\underset{u\in U}{sup}\delta (u,R)\},max\{\underset{s\in S}{sup}\delta (s,V),\underset{v\in V}{sup}\delta (v,S)\}\right\}\hfill \\ \hfill & =max\left\{H_{\delta }\left(R,U\right),H_{\delta }\left(S,V\right)\right\}.\hfill \end{array}$$

In the similar way, we obtain that

$$\underset{y\in U\cup V}{sup}\delta (y,R\cup S)\le max\left\{H_{\delta }\left(R,U\right),H_{\delta }\left(S,V\right)\right\}.$$

Hence, it follows that

$$\begin{array}{ccc}\hfill H_{\delta }(R\cup S,U\cup V)& =& max\{\underset{x\in R\cup S}{sup}\delta (x,U\cup V),\underset{y\in U\cup V}{sup}\delta (y,R\cup S)\}\hfill \\ & \le & max\left\{H_{\delta }\left(R,U\right),H_{\delta }\left(S,V\right)\right\}.\hfill \end{array}$$

□

Definition 9.

Let $(X,\delta )$ be a dislocated metric space and $f,g:X\to X$ be two self-mappings. A pair of mappings $\left(f,g\right)$ is called a generalized contraction if for all $x,y\in X,$

$$\delta \left(fx,gy\right)\le \alpha \delta \left(x,y\right)$$

where $0\le \alpha <1$.

Theorem 5.

Let $(X,\delta )$ be a dislocated metric space and $f,g:X\to X$ be two continuous mappings. If the pair of mappings $\left(f,g\right)$ is generalized contraction with $0\le \alpha <1$. Then:

(i)

The elements in $\mathcal{C}\left(X\right)$ are mapped to elements in $\mathcal{C}\left(X\right)$ under f and g;

(ii)

For any $U\in \mathcal{C}\left(X\right)$, if

$$f\left(U\right)=\left\{f\right(u):u\in U\}\mathrm{and}g\left(U\right)=\left\{g\right(u):u\in U\},$$

then $f,g:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$. Also, the pair $\left(f,g\right)$ is a generalized contraction map on $(\mathcal{C}\left(X\right),H_{\delta })$.

Proof. 

(i) Since f is continuous mapping and the image of a compact subset under $f:X\to X$ is compact, that is,

$$U\in \mathcal{C}\left(X\right)\mathrm{implies}f\left(U\right)\in \mathcal{C}\left(X\right).$$

Similarly, we have

$$U\in \mathcal{C}\left(X\right)\mathrm{implies}g\left(U\right)\in \mathcal{C}\left(X\right).$$

(ii) Let $A_1,A_2\in \mathcal{C}\left(X\right)$. Since for $f,g:X\to X,$ the pair of mappings $\left(f,g\right)$ is a generalized contraction, we obtain that for all $u,v\in X$,

$$\delta \left(fu,gv\right)\le \alpha \delta \left(u,v\right)$$

where $0\le \alpha <1$. Thus, we have

$$\delta \left(fu,g\left(V\right)\right)=\underset{v\in V}{inf}\delta \left(fu,gv\right)\le \underset{v\in V}{inf}\alpha \delta \left(u,v\right)=\alpha \delta \left(u,V\right).$$

Also,

$$\delta \left(fv,g\left(U\right)\right)=\underset{u\in U}{inf}\delta \left(fv,gu\right)\le \underset{u\in U}{inf}\alpha \delta \left(v,u\right)=\alpha \delta \left(v,U\right).$$

Now,

$$\begin{array}{ccc}\hfill H_{\delta }\left(f\left(U\right),g\left(V\right)\right)& =& max\{\underset{u\in U}{sup}\delta (fu,f\left(V\right)),\underset{v\in V}{sup}\delta (fv,f\left(U\right))\},\hfill \\ & \le & max\{\underset{u\in U}{sup}\alpha \delta (u,V),\underset{v\in V}{sup}\alpha \delta (v,U)\},\hfill \\ & =& max\{\alpha \left(\underset{u\in U}{sup}\delta (u,V)\right),\alpha \left(\underset{v\in B}{sup}\delta (v,U)\right)\},\hfill \\ & =& \alpha max\{\underset{u\in U}{sup}\delta (u,V),\underset{v\in V}{sup}\delta (v,U)\},\hfill \\ & =& \alpha H_{\delta }\left(U,V\right).\hfill \end{array}$$

Consequently,

$$H_{\delta }\left(f\left(U\right),g\left(V\right)\right)\le \alpha H_{\delta }\left(U,V\right).$$

Hence, the pair $\left(f,g\right)$ is a generalized contraction map on $(\mathcal{CB}\left(X\right),H_{\delta })$. □

3. Generalized Iterated Function System

The iterated function system is a very useful tool for generating fractals by a finite family of contractive and generalized contractive self-mappings in complete metric spaces. Recently, some useful results appeared in [8,9,10,11,12]. Pasupathi et al. [13] developed new iterated function systems consisting of cyclic contractions and discussed some special properties of the Hutchinson operator associated with cyclic iterated function systems. Recently, Thangaraj and Easwaramoorthy [14] obtained some interesting results and consequences of the controlled Fisher iterated function system and controlled Fisher fractals. In addition, they established a collage theorem on controlled Fisher fractals. The newly developing IFS and fractal set in the controlled metric space can provide the novel directions in the fractal theory. Thangaraj et al. [15] constructed a controlled Kannan iterated function system with Kannan contraction maps in a controlled metric space and developed a new kind of invariant set called a controlled Kannan attractor.

In this section, we construct a fractal set of generalized iterated function systems, a certain finite collection of mappings defined in the setup of dislocated metric space. We also define Hutchinson operators with the help of a finite collection of generalized rational contraction mappings on a dislocated metric space. We start with the following result.

Proposition 1.

Let $(X,\delta )$ be a dislocated metric space. Suppose that the mappings $f_n,g_n$: $X\to X$ for $n=1,\dots ,N$ are satisfying

$$\delta \left(f_{n}x,g_{n}y\right)\le {\alpha }_n\delta \left(x,y\right)\mathrm{for}\mathrm{all}x,y\in X,$$

where $0\le {\alpha }_n<1$ for each $n\in \left\{1,\dots ,N\right\}.$ Then, the mappings $T,S:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ defined as

$$\begin{array}{c}\hfill T\left(U\right)=f_1\left(U\right)\cup \dots \cup f_N\left(U\right)=\bigcup _{n=1}^N{f}_n\left(U\right),\mathrm{for}\mathrm{each}U\in \mathcal{C}\left(X\right),\\ \hfill S\left(U\right)=g_1\left(U\right)\cup \dots \cup g_N\left(U\right)=\bigcup _{n=1}^N{g}_n\left(U\right),\mathrm{for}\mathrm{each}U\in \mathcal{C}\left(X\right)\end{array}$$

also satisfy

$$H_{\delta }\left(TU,SV\right)\le {\alpha }_{*}{H}_{\delta }\left(U,V\right)\mathrm{for}\mathrm{all}U,V\in \mathcal{C}\left(X\right),$$

where ${\alpha }_{*}=max\{{\alpha }_i:i\in \{1,\dots ,N\}\},$ that is, the pair $\left(T,S\right)$ is a generalized contraction on $\mathcal{C}\left(X\right)$.

Proof. 

We will prove the result for $N=2$. Let $f_1,f_2,g_1,g_2:X\to X$ be two contractions. For $A_1,A_2\in \mathcal{C}\left(X\right)$ and using Lemma 1(iii), we have

$$\begin{array}{ccc}\hfill H_{\delta }(T\left(A_1\right),S\left(A_2\right))& =& H_{\delta }(f_1\left(A_1\right)\cup f_2\left(A_1\right),g_1\left(A_2\right)\cup g_2\left(A_2\right))\hfill \\ & \le & max\left\{H_{\delta }(f_1\left(A_1\right),g_1\left(A_2\right))\right\},H_{\delta }(f_2\left(A_1\right),f_2\left(A_2\right))\}\hfill \\ & \le & max\{{\alpha }_1{H}_{\delta }(A_1,A_2),{\alpha }_2{H}_{\delta }(A_1,A_2)\}\hfill \\ & =& {\alpha }_{*}{H}_{\delta }(A_1,A_2).\hfill \end{array}$$

□

Definition 10.

Let $(X,\delta )$ be a dislocated metric space and $T,S:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ be two mappings. A pair of mappings $\left(T,S\right)$ is called a generalized rational contractive, if there exists an $\alpha \in [0,1)$ such that

$$H_{\delta }(T\left(U\right),S\left(V\right))\le \alpha M_{T,S}(U,V),$$

where

$$\begin{array}{ccc}\hfill M_{T,S}(U,V)& =& max\{H_{\delta }(U,V),H_{\delta }(U,T\left(U\right)),H_{\delta }(V,S\left(V\right)),\hfill \\ & & \frac{H_{\delta }(U,S\left(V\right))+H_{\delta }(V,T\left(U\right))}{4},{\displaystyle \frac{H_{\delta }(U,V)(1+H_{\delta }(U,T\left(U\right)))}{1+H_{\delta }\left(U,V\right)}},\hfill \\ & & \frac{H_{\delta }(V,S\left(V\right))(1+H_{\delta }(U,T\left(U\right)))}{1+H_{\delta }(U,V)},\frac{H_{\delta }(U,S\left(V\right))(1+H_{\delta }(U,T\left(U\right)))}{2(1+H_{\delta }(U,V))}\}.\hfill \end{array}$$

The above defined operator $(T,S)$ is also known as a generalized rational contractive Hutchinson operator, which is the extension of Hutchinson operator given in [16]. Moreover, if $(T,S)$ defined in Proposition 1 is a generalized contraction, then it is a trivially generalized rational contraction, and so $(T,S)$ is a generalized rational contractive Hutchinson operator.

Definition 11.

Let X be a dislocated metric space. If $f_n,g_n:X\to X$, $n=1,\dots ,N$ are generalized contraction mappings, then $(X;f_1,\dots ,f_N;g_1,\dots ,g_N)$ is called a generalized iterated function system (GIFS).

Definition 12.

A nonempty closed and bounded subset U of X is called a common attractor of T and $S$ generated by GIFS if

(i)

$T\left(U\right)=S\left(U\right)=U$;

(ii)

There exists an open subset V of X such that $U\subseteq V$ and $\underset{k\to \infty }{lim}{T}^k\left(B\right)=\underset{k\to \infty }{lim}{S}^k\left(B\right)=U$ for any closed and bounded subset B of V, where the limit is applied with respect to the dislocated Hausdorff metric.

The largest open set V satisfying (ii) is called a basin of attraction.

4. Main Results

By using the concept of Hausdorff dislocated metrics and generalized rational contractive Hutchinson operators, we establish the existence of unique common attractors of such mappings satisfying more general contractive conditions than those in [2,4,16,17,18] in the framework of complete dislocated metric spaces. We start with the following Theorem.

Theorem 6.

Let $(X,\delta )$ be a complete dislocated metric space, with $(X;f_1,\dots ,f_N;g_1,\dots ,g_N)$ being a given generalized iterated function system. Suppose that the pair of self-mappings $(T,S)$ defined by

$$\begin{array}{c}\hfill T\left(B\right)=f_1\left(B\right)\cup \dots \cup f_N\left(B\right)=\bigcup _{n=1}^N{f}_n\left(B\right),\mathrm{for}\mathrm{each}B\in \mathcal{C}\left(X\right),\\ \hfill S\left(B\right)=g_1\left(B\right)\cup \dots \cup g_N\left(B\right)=\bigcup _{n=1}^N{g}_n\left(B\right),\mathrm{for}\mathrm{each}B\in \mathcal{C}\left(X\right)\end{array}$$

is generalized rational contractive. Then, T and S have a unique common attractor $A\in \mathcal{C}\left(X\right),$

$$A=T\left(A\right)=\bigcup _{n=1}^N{f}_n\left(A\right)=S\left(A\right)=\bigcup _{n=1}^N{g}_n\left(A\right).$$

Furthermore, for the initial set $B_0\in \mathcal{C}\left(X\right)$, the iterative sequence of compact sets defined as

$$B_{2n+1}=T\left(B_{2n}\right)\mathrm{and}{B}_{2n+2}=S\left(B_{2n+1}\right)\mathrm{for}n=0,1,2,\dots$$

converges to the common attractor of T and S.

Proof. 

Let $B_0$ be any arbitrary element in $\mathcal{C}\left(X\right).$ Define $B_1=T\left(B_0\right),B_2=S\left(B_1\right)$ and the sequence $\left\{B_n\right\}$, where $B_{2n+1}=T\left(B_{2n}\right)$ and $B_{2n+2}=S\left(B_{2n+1}\right)$ for $n=0,1,2,\dots$.

Now, as the pair $\left(T,S\right)$ is a generalized rational contraction, for all $n\in \{0,1,2,\dots \},$

$$\begin{array}{ccc}\hfill H_{\delta }(B_{2n+1},B_{2n+2})& =& H_{\delta }(T\left(B_{2n}\right),S\left(B_{2n+1}\right))\hfill \\ & \le & \alpha M_{T,S}(B_{2n},B_{2n+1})\hfill \\ & =& \alpha max\{H_{\delta }(B_{2n},B_{2n+1}),H_{\delta }(B_{2n},T\left(B_{2n}\right)),H_{\delta }(B_{2n+1},S\left(B_{2n+1}\right)),\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(B_{2n+1}\right))+H_{\delta }(B_{2n+1},T\left(B_{2n}\right))}{4},\hfill \\ & & \frac{H_{\delta }\left(B_{2n},B_{2n+1}\right)(1+H_{\delta }\left(B_{2n},T\left(B_{2n}\right)\right))}{1+H_{\delta }(B_{2n},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n+1},S\left(B_{2n+1}\right)(1+H_{\delta }\left(B_{2n},T\left(B_{2n}\right)\right))}{1+H_{\delta }(B_{2n},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(B_{2n+1}\right))(1+H_{\delta }(B_{2n},T\left(B_{2n}\right)))}{2(1+H_{\delta }(B_{2n},B_{2n+1}))}\}\hfill \\ & =& \alpha max\{H_{\delta }(B_{2n},B_{2n+1}),H_{\delta }(B_{2n},B_{2n+1}),H_{\delta }(B_{2n+1},B_{2n+2}),\hfill \\ & & \frac{H_{\delta }(B_{2n},B_{2n+2})+H_{\delta }(B_{2n+1},B_{2n+1})}{4},\hfill \\ & & \frac{H_{\delta }\left(B_{2n},B_{2n+1}\right)(1+H_{\delta }\left(B_{2n},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n+1},B_{2n+2})(1+H_{\delta }\left(B_{2n},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n},B_{2n+2})(1+H_{\delta }(B_{2n},B_{2n+1}))}{2(1+H_{\delta }(B_{2n},B_{2n+1}))}\}\hfill \\ & \le & \alpha max\{H_{\delta }(B_{2n},B_{2n+1}),H_{\delta }(B_{2n+1},B_{2n+2})\}\hfill \\ & =& \alpha H_{\delta }\left(B_{2n},B_{2n+1}\right).\hfill \end{array}$$

Similarly, we have

$$\begin{array}{cc}\hfill & H_{\delta }(B_{2n+2},B_{2n+3})=H_{\delta }(T\left(B_{2n+1}\right),S\left(B_{2n+2}\right))=H_{\delta }(T\left(B_{2n+1}\right),S\left(B_{2n+2}\right))\hfill \\ \hfill & \le \alpha M_{T,S}(B_{2n+1},B_{2n+2})\hfill \\ \hfill & =\alpha max\{H_{\delta }(B_{2n+1},B_{2n+2}),H_{\delta }(B_{2n+1},T\left(B_{2n+1}\right)),H_{\delta }(B_{2n+2},S\left(B_{2n+2}\right)),\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+1},S\left(B_{2n+2}\right))+H_{\delta }(B_{2n+2},T\left(B_{2n+1}\right))}{4},\hfill \\ \hfill & \frac{H_{\delta }\left(B_{2n+1},B_{2n+2}\right)(1+H_{\delta }\left(B_{2n+1},T\left(B_{2n+1}\right)\right))}{1+H_{\delta }(B_{2n+1},B_{2n+2})},\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+2},S\left(B_{2n+2}\right)(1+H_{\delta }\left(B_{2n+1},T\left(B_{2n+1}\right)\right))}{1+H_{\delta }(B_{2n+1},B_{2n+2})},\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+1},S\left(B_{2n+2}\right))(1+H_{\delta }(B_{2n+1},T\left(B_{2n+1}\right)))}{2(1+H_{\delta }(B_{2n+1},B_{2n+2}))}\}\hfill \\ \hfill & =\alpha max\{H_{\delta }(B_{2n+1},B_{2n+2}),H_{\delta }(B_{2n+1},B_{2n+2}),H_{\delta }(B_{2n+2},B_{2n+3}),\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+1},B_{2n+3})+H_{\delta }(B_{2n+2},B_{2n+2})}{4},\hfill \\ \hfill & \frac{H_{\delta }\left(B_{2n+1},B_{2n+2}\right)(1+H_{\delta }\left(B_{2n+1},B_{2n+2}\right))}{1+H_{\delta }(B_{2n+1},B_{2n+2})},\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+2},B_{2n+3})(1+H_{\delta }\left(B_{2n+1},B_{2n+2}\right))}{1+H_{\delta }(B_{2n+1},B_{2n+2})},\hfill \\ \hfill & \frac{H_{\delta }(B_{2n+1},B_{2n+3})(1+H_{\delta }(B_{2n+1},B_{2n+2}))}{2(1+H_{\delta }(B_{2n+1},B_{2n+2}))}\}\hfill \\ \hfill & \le \alpha max\{H_{\delta }(B_{2n+1},B_{2n+2}),H_{\delta }(B_{2n+2},B_{2n+3})\}\hfill \\ \hfill & =\alpha H_{\delta }(B_{2n+1},B_{2n+2}).\hfill \end{array}$$

Thus, $H_{\delta }(B_{2n+2},B_{2n+3})\le \alpha H_{\delta }(B_{2n+1},B_{2n+2})$ for all $n\in \{0,1,2,\dots \}.$ Continuing this way, we obtain

$$H_{\delta }(B_n,B_{n+1})\le \alpha H_{\delta }(B_{n-1},B_n)\le \dots \le {\alpha }^k{H}_{\delta }(B_{n-k},B_{n-k+1})\le \dots \le {\alpha }^n{H}_{\delta }(B_0,B_1).$$

(1)

Now for $m,n\in \mathbb{N}$ with $m>n,$ we have

$$\begin{array}{ccc}\hfill H_{\delta }(B_n,B_m)& \le & H_{\delta }(B_n,B_{n+1})+\dots +H_{\delta }(B_{n+j},B_{n+j+1})+\dots +H_{\delta }(B_{m-1},B_m)\hfill \\ & \le & {\alpha }^n{H}_{\delta }(B_0,B_1)+\dots +{\alpha }^{n+j}{H}_{\delta }(B_0,B_1)+\dots +{\alpha }^{m-1}{H}_{\delta }(B_0,B_1)\hfill \\ & =& ({\alpha }^n+\dots +{\alpha }^{n+j}+\dots +{\alpha }^{m-1})H_{\delta }(B_0,B_1)\hfill \\ & \le & {\alpha }^n(1+\alpha +{\alpha }^2+\dots )H_{\delta }(B_0,B_1)\hfill \\ & =& \frac{{\alpha }^n}{1-\alpha }{H}_{\delta }(B_0,B_1).\hfill \end{array}$$

It follows that by taking the limits as $n\to \infty$, we have $H_{\delta }(B_n,B_m)\to 0,$ that is, the sequence $\left\{B_n\right\}$ is a Cauchy. Since $(C\left(X\right),H_{\delta })$ is complete dislocated metric space, there exists a set $A^{*}\in C\left(X\right)$ such that $B_n$$\to A^{*}$ as $n\to \infty ,$ that is, $\underset{n\to \infty }{lim}{H}_{\delta }(B_n,A^{*})=\underset{n\to \infty }{lim}{H}_{\delta }(B_n,B_{n+1})=H_{\delta }(A^{*},A^{*}).$ It follows from (1) that $H_{\delta }(B_n,B_{n+1})\le {\alpha }^n{H}_{\delta }(B_0,B_1)$, and taking the limit as $n\to \infty$ yields $\underset{n\to \infty }{lim}{H}_{\delta }(B_n,B_{n+1})=0$, and due to uniqueness of the limit, $H_{\delta }(A^{*},A^{*})=0.$

Next, we will show that $A^{*}$ is the common attractor of T and S. Now,

$$\begin{array}{ccc}\hfill H_{\delta }(A^{*},S\left(A^{*}\right))& \le & H_{\delta }(A^{*},T\left(B_{2n}\right))+H_{\delta }(T\left(B_{2n}\right),S\left(A^{*}\right))\hfill \\ & \le & H_{\delta }(A^{*},B_{2n+1})+\alpha M_{T,S}\left(B_{2n},A^{*}\right),\hfill \end{array}$$

(2)

where

$$\begin{array}{ccc}\hfill M_{T,S}(B_{2n},A^{*})& =& max\{H_{\delta }(B_{2n},A^{*}),H_{\delta }(B_{2n},T\left(B_{2n}\right)),H_{\delta }(A^{*},S\left(A^{*}\right))\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))+H_{\delta }(A^{*},T\left(B_{2n}\right))}{4},\hfill \\ & & \frac{H_{\delta }\left(B_{2n},A^{*}\right)(1+H_{\delta }\left(B_{2n},T\left(B_{2n}\right)\right))}{1+H_{\delta }(B_{2n},A^{*})},\hfill \\ & & \frac{H_{\delta }(A^{*},S\left(A^{*}\right)(1+H_{\delta }\left(B_{2n},T\left(B_{2n}\right)\right))}{1+H_{\delta }(B_{2n},A^{*})},\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},T\left(B_{2n}\right)))}{2(1+H_{\delta }(B_{2n},A^{*}))}\}\hfill \\ & =& max\{H_{\delta }(B_{2n},A^{*}),H_{\delta }(B_{2n},B_{2n+1}),H_{\delta }(A^{*},S\left(A^{*}\right)),\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))+H_{\delta }(A^{*},B_{2n+1})}{4},\hfill \\ & & \frac{H_{\delta }\left(B_{2n},A^{*}\right)(1+H_{\delta }\left(A^{*},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},A^{*})},\hfill \\ & & \frac{H_{\delta }(A^{*},S\left(A^{*}\right))(1+H_{\delta }\left(B_{2n},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},A^{*})},\hfill \\ & & \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},B_{2n+1}))}{2(1+H_{\delta }(B_{2n},A^{*}))}\}.\hfill \end{array}$$

We have the following cases.

(i)

In case $M_{T,S}\left(B_{2n},A^{*}\right)=H_{\delta }\left(B_{2n},A^{*}\right),$ then by (2), we obtain

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha H_{\delta }\left(B_{2n},A^{*}\right)$$

and taking limit as $n\to \infty$ yields

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{H}_{\delta }(A^{*},S\left(A^{*}\right))& \le & \underset{n\to \infty }{lim}{H}_{\delta }(A^{*},B_{2n+1})+\alpha \underset{n\to \infty }{lim}{H}_{\delta }\left(B_{2n},A^{*}\right)=0,\hfill \end{array}$$

and we obtain that $A^{*}=S\left(A^{*}\right)$.

(ii)

In case $M_{T,S}\left(B_{2n},A^{*}\right)=H_{\delta }\left(B_{2n},B_{2n+1}\right),$ then by (2), we have

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha H\left(B_{2n},B_{2n+1}\right)$$

and taking limit as $n\to \infty$ yields

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{H}_{\delta }(A^{*},S\left(A^{*}\right))& \le & \underset{n\to \infty }{lim}{H}_{\delta }(A^{*},B_{2n+1})+\alpha \underset{n\to \infty }{lim}{H}_{\delta }\left(B_{2n},B_{2n+1}\right)=0,\hfill \end{array}$$

which implies $A^{*}=S\left(A^{*}\right)$.

(iii)

In case $M_{T,S}\left(B_{2n},A^{*}\right)=H_{\delta }\left(A^{*},S\left(A^{*}\right)\right),$ by (2), we obtain

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha H_{\delta }\left(A^{*},S\left(A^{*}\right)\right)$$

and taking limit as $n\to \infty$ yields

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le \alpha H_{\delta }\left(A^{*},S\left(A^{*}\right)\right),$$

which implies

$$(1-\alpha )H_{\delta }\left(A^{*},S\left(A^{*}\right)\right)\le 0$$

and since $1-\alpha >0,$ we obtain $A^{*}=S\left(A^{*}\right)$.

(iv)

In case $M_{T,S}\left(B_{2n},A^{*}\right)={\displaystyle \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))+H_{\delta }(A^{*},B_{2n+1})}{4}},$ we obtain

$$\begin{array}{c}{H}_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\frac{\alpha }{4}(H_{\delta }(B_{2n},S\left(A^{*}\right))+H_{\delta }(A^{*},B_{2n+1}))\hfill \\ \hfill \le H_{\delta }(A^{*},B_{2n+1})+\frac{\alpha }{4}(H_{\delta }(B_{2n},A^{*})+H_{\delta }(A^{*},S\left(A^{*}\right))+H_{\delta }(A^{*},B_{2n+1}))\end{array}$$

and taking limit as $n\to \infty$ yields

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le \frac{\alpha }{4}{H}_{\delta }\left(A^{*},S\left(A^{*}\right)\right),$$

which implies

$$(1-\frac{\alpha }{4})H_{\delta }\left(A^{*},S\left(A^{*}\right)\right)\le 0,$$

and since $1-{\displaystyle \frac{\alpha }{4}}>0,$ we obtain $A^{*}=S\left(A^{*}\right)$.

(v)

In case $M_{T,S}\left(B_{2n},A^{*}\right)={\displaystyle \frac{H_{\delta }\left(B_{2n},A^{*}\right)(1+H_{\delta }\left(A^{*},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},A^{*})}},$ then we have

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha \left(\frac{H_{\delta }\left(B_{2n},A^{*}\right)(1+H_{\delta }\left(A^{*},B_{2n+1}\right))}{1+H_{\delta }(B_{2n},A^{*})}\right)$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},S\left(A^{*}\right))\le 0,$ which gives $A^{*}=S\left(A^{*}\right)$.

(vi)

When $M_{T,S}\left(B_{2n},A^{*}\right)={\displaystyle \frac{H_{\delta }(A^{*},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},B_{2n+1}))}{1+H_{\delta }(B_{2n},A^{*})}},$ we obtain

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha \left(\frac{H_{\delta }(A^{*},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},B_{2n+1}))}{1+H_{\delta }(B_{2n},A^{*})}\right)$$

and taking limit as $n\to \infty$ yields

$$H_{\delta }(A^{*},S\left(A^{*}\right))\le \alpha H_{\delta }(A^{*},S\left(A^{*}\right)),$$

since $1-\alpha >0,$ we obtain $A^{*}=S\left(A^{*}\right)$.

(vii)

Finally, if $M_{T,S}\left(B_{2n},A^{*}\right)={\displaystyle \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},B_{2n+1}))}{2(1+H_{\delta }(B_{2n},A^{*}))}},$ then

$$\begin{array}{c}{H}_{\delta }(A^{*},S\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+1})+\alpha \left({\displaystyle \frac{H_{\delta }(B_{2n},S\left(A^{*}\right))(1+H_{\delta }(B_{2n},B_{2n+1}))}{2(1+H_{\delta }(B_{2n},A^{*}))}}\right)\hfill \\ \hfill \le H_{\delta }(A^{*},B_{2n+1})+\alpha \left({\displaystyle \frac{(H_{\delta }(B_{2n},A^{*})+H_{\delta }(A^{*},S\left(A^{*}\right)))(1+H_{\delta }(B_{2n},B_{2n+1}))}{2(1+H_{\delta }(B_{2n},A^{*}))}}\right)\end{array}$$

and taking limit as $n\to \infty$ yields $H_{\delta }\left(A^{*},S\left(A^{*}\right)\right)\le \alpha H_{\delta }(A^{*},S\left(A^{*}\right))$ and so $A^{*}=S\left(A^{*}\right)$.

Thus, from all cases, we obtain that $A^{*}=S\left(A^{*}\right).$ Again, we have

$$\begin{array}{ccc}\hfill H_{\delta }(A^{*},T\left(A^{*}\right))& \le & H_{\delta }(A^{*},S\left(B_{2n+1}\right))+H_{\delta }(T\left(A^{*}\right),S\left(B_{2n+1}\right))\hfill \\ & \le & H_{\delta }(A^{*},B_{2n+2})+\alpha M_{T,S}\left(A^{*},B_{2n+1}\right),\hfill \end{array}$$

(3)

where

$$\begin{array}{ccc}\hfill M_{T,S}(A^{*},B_{2n+1})& =& \alpha max\{H_{\delta }(A^{*},B_{2n+1}),H_{\delta }(A^{*},T\left(A^{*}\right)),H_{\delta }(B_{2n+1},S\left(B_{2n+1}\right)),\hfill \\ & & \frac{H_{\delta }(A^{*},S\left(B_{2n+1}\right))+H_{\delta }(B_{2n+1},T\left(A^{*}\right))}{4},\hfill \\ & & \frac{H_{\delta }\left(A^{*},B_{2n+1}\right)(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n+1},S\left(B_{2n+1}\right)(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(A^{*},S\left(B_{2n+1}\right))(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{2(1+H_{\delta }(A^{*},B_{2n+1}))}\}\hfill \\ & =& \alpha max\{H_{\delta }(A^{*},B_{2n+1}),H_{\delta }(A^{*},T\left(A^{*}\right)),H_{\delta }(B_{2n+1},B_{2n+2}),\hfill \\ & & \frac{(H_{\delta }(A^{*},B_{2n+2})+H_{\delta }(B_{2n+1},T\left(A^{*}\right)))}{4},\hfill \\ & & \frac{H_{\delta }\left(A^{*},B_{2n+1}\right)(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n+1},B_{2n+2})(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})},\hfill \\ & & \frac{H_{\delta }(B_{2n+2},A^{*})(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{2(1+H_{\delta }(A^{*},B_{2n+1}))}\}.\hfill \end{array}$$

Now, we have again the following cases.

(i)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)=H_{\delta }(A^{*},B_{2n+1}),$ then by (3), we have

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha H_{\delta }(A^{*},B_{2n+1})$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},T\left(A^{*}\right))=0,$ which implies $A^{*}=T\left(A^{*}\right)$.

(ii)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)=H_{\delta }(A^{*},T\left(A^{*}\right)),$ then by (3), we obtain

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha H_{\delta }(A^{*},T\left(A^{*}\right))$$

and taking limit as $n\to \infty$ yields

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le 0+\alpha H_{\delta }(A^{*},T\left(A^{*}\right)),$$

that is, $(1-\alpha )H_{\delta }(A^{*},T\left(A^{*}\right))\le 0$, and since $(1-\alpha )>0$, we obtain $A^{*}=T\left(A^{*}\right).$

(iii)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)=H_{\delta }\left(B_{2n+1},B_{2n+2}\right),$ then by (3), we obtain that

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha H_{\delta }\left(B_{2n+1},B_{2n+2}\right)$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},T\left(A^{*}\right))=0,$ that is, $A^{*}=T\left(A^{*}\right)$.

(iv)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)={\displaystyle \frac{H_{\delta }(A^{*},B_{2n+2})+H_{\delta }(B_{2n+1},T\left(A^{*}\right))}{4}},$ then

$$\begin{array}{c}{H}_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\frac{\alpha }{4}(H_{\delta }(A^{*},B_{2n+2})+H_{\delta }(B_{2n+1},T\left(A^{*}\right)))\hfill \\ \hfill \le H_{\delta }(A^{*},B_{2n+2})+\frac{\alpha }{4}(H_{\delta }(A^{*},B_{2n+2})+H_{\delta }(B_{2n+1},A^{*})+H_{\delta }(A^{*},T\left(A^{*}\right)))\end{array}$$

and taking limit as $n\to \infty$ yields

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le 0+\frac{\alpha }{4}{H}_{\delta }(A^{*},T\left(A^{*}\right)),$$

that is,

$$(1-\frac{\alpha }{4})H_{\delta }(A^{*},T\left(A^{*}\right))\le 0$$

and since $1-{\displaystyle \frac{\alpha }{4}}>0$, we obtain $A^{*}=T\left(A^{*}\right).$

(v)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)={\displaystyle \frac{H_{\delta }\left(A^{*},B_{2n+1}\right)(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})}},$ we have

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha {\displaystyle \frac{H_{\delta }\left(A^{*},B_{2n+1}\right)(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right))}{1+H_{\delta }(A^{*},B_{2n+1})}}$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},T\left(A^{*}\right))=0,$ that is, $A^{*}=T\left(A^{*}\right)$.

(vi)

If $M_{T,S}\left(B_{2n+1},A^{*}\right)={\displaystyle \frac{H_{\delta }(B_{2n+1},B_{2n+2})(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right)}{1+H_{\delta }(A^{*},B_{2n+1})}},$ then

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha \frac{H_{\delta }(B_{2n+1},B_{2n+2})(1+H_{\delta }\left(A^{*},T\left(A^{*}\right)\right)}{1+H_{\delta }(A^{*},B_{2n+1})}$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},T\left(A^{*}\right))=0$, that is, $A^{*}=T\left(A^{*}\right).$

(vii)

Finally if $M_{T,S}\left(B_{2n+1},A^{*}\right)={\displaystyle \frac{H_{\delta }(B_{2n+2},A^{*})(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{2(1+H_{\delta }(A^{*},B_{2n+1}))}},$ then

$$H_{\delta }(A^{*},T\left(A^{*}\right))\le H_{\delta }(A^{*},B_{2n+2})+\alpha {\displaystyle \frac{H_{\delta }(B_{2n+2},A^{*})(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{2(1+H_{\delta }(A^{*},B_{2n+1}))}}$$

and taking limit as $n\to \infty$ yields $H_{\delta }(A^{*},T\left(A^{*}\right))=0$, we obtain $A^{*}=T\left(A^{*}\right).$

Thus, $A^{*}$ is the common attractor of T and S. Finally, to prove that the common attractor of T and S is unique, we suppose that $A^{*}$ and $B^{*}$ are the two common attractors of T and S in $\mathcal{C}\left(X\right)$. Since the pair $(T,S)$ are generalized rational contractive mappings, we obtain that

$$H_{\delta }(A^{*},B^{*})=H_{\delta }(T\left(A^{*}\right),S\left(B^{*}\right))\le \alpha M_{T,S}\left(A^{*},B^{*}\right),$$

where

$$\begin{array}{cc}\hfill & M_{T,S}\left(A^{*},B^{*}\right)=max\{H_{\delta }\left(A^{*},B^{*}\right),H_{\delta }(A^{*},T\left(A^{*}\right)),H_{\delta }(B^{*},S\left(B^{*}\right)),\hfill \\ \hfill & \frac{H_{\delta }(A^{*},S\left(B^{*}\right))+H_{\delta }(B^{*},T\left(A^{*}\right))}{4},{\displaystyle \frac{H_{\delta }(A^{*},B^{*})(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{1+H_{\delta }\left(A^{*},B^{*}\right)}},\hfill \\ \hfill & \frac{H_{\delta }(B^{*},S\left(B^{*}\right))(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{1+H_{\delta }\left(A^{*},B^{*}\right)},\frac{H_{\delta }(A^{*},S\left(B^{*}\right))(1+H_{\delta }(A^{*},T\left(A^{*}\right)))}{2(1+H_{\delta }\left(A^{*},B^{*}\right))}\}\hfill \\ \hfill & =max\{H_{\delta }\left(A^{*},B^{*}\right),\frac{H_{\delta }(A^{*},B^{*})+H_{\delta }(B^{*},A^{*})}{4},\frac{H_{\delta }(A^{*},B^{*})}{2(1+H_{\delta }\left(A^{*},B^{*}\right))}\}\hfill \\ \hfill & =H_{\delta }\left(A^{*},B^{*}\right),\hfill \end{array}$$

that is, $H_{\delta }(A^{*},B^{*})\le \alpha H_{\delta }\left(A^{*},B^{*}\right),$ which implies $A^{*}=B^{*}$. Hence, T and S have a unique common attractor $A^{*}$ in $\mathcal{C}\left(X\right)$. □

Remark 1.

In Theorem 6, let $\mathcal{S}\left(X\right)$ be the class of all singleton subsets of $X.$ Obviously, $\mathcal{S}\left(X\right)\subseteq C\left(X\right).$ Further, suppose that $f_m=f$ and $g_m=g$ for every $m,$ with $f=f_1$ and $g=g_1$. Then, the operators $T,S:\mathcal{S}\left(X\right)\to \mathcal{S}\left(X\right)$ will be, for all $A\in \mathcal{S}\left(X\right)$,

$$\begin{array}{c}\hfill T\left(A\right)=f\left(A\right),S\left(A\right)=g\left(A\right).\end{array}$$

Following the above setup, we obtain the following result.

Corollary 1.

Let $\left(\mathcal{C}\right(X),\delta )$ be a complete dislocated metric space and $f,g:X\to X$ be two mappings. Suppose that $T,S:\mathcal{S}\left(X\right)\to \mathcal{S}\left(X\right)$ are maps defined above in Remark 1 satisfying

$$H_{\delta }(T\left(U\right),S\left(V\right))\le \alpha M_{T,S}(U,V),$$

for all $U,V\in C\left(X\right)$, where $\alpha \in [0,1)$, and

$$\begin{array}{ccc}\hfill M_{T,S}(U,V)& =& max\{H_{\delta }(U,V),H_{\delta }(U,T\left(U\right)),H_{\delta }(V,S\left(V\right)),\hfill \\ & & \frac{H_{\delta }(U,S\left(V\right))+H_{\delta }(V,T\left(U\right))}{4},{\displaystyle \frac{H_{\delta }(U,V)(1+H_{\delta }(U,T\left(U\right)))}{1+H_{\delta }\left(U,V\right)}},\hfill \\ & & \frac{H_{\delta }(V,S\left(V\right))(1+H_{\delta }(U,T\left(U\right)))}{1+H_{\delta }(U,V)},\frac{H_{\delta }(U,S\left(V\right))(1+H_{\delta }(U,T\left(U\right)))}{2(1+H_{\delta }(U,V))}\}.\hfill \end{array}$$

Then, T and S have at most one attractor, that is, there exists a unique $U^{*}\in \mathcal{S}\left(X\right)$ such that

$$U^{*}=T\left(U^{*}\right)=S\left(U^{*}\right).$$

Furthermore, for any singleton set $B_0\in \mathcal{S}\left(X\right),$ the iterative sequence of compact sets defined as

$$B_{2n+1}=T\left(B_{2n}\right),B_{2n+2}=S\left(B_{2n+1}\right),n=0,1,2,\dots$$

converges to the common attractor of T and S.

Corollary 2.

Let $\left(\mathcal{C}\right(X),\delta )$ be a complete dislocated metric space and $(X:f_m,g_m;m=1,\dots ,N)$ be a generalized iterated function system, where each $\left(f_i,g_i\right)$ is a generalized contraction on X for $i=1,\dots ,N.$ Then, $T,S:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ defined in Theorem 6 has a unique attractor in $\mathcal{C}\left(X\right).$ Furthermore, for any set $B_0\in \mathcal{C}\left(X\right)$, the sequence of compact sets

$$B_{2n+1}=T\left(B_{2n}\right),B_{2n+2}=S\left(B_{2n+1}\right)\mathrm{for}n=0,1,2,\dots$$

converges to the common attractor of T and S.

Example 3.

Let $X=[0,10]$ and consider a complete dislocated metric $\delta :X\times X\to {\mathbb{R}}_{+}$ defined by

$$\delta (x,y)=x+y+4max\{x,y\}\mathrm{for}\mathrm{all}x,y\in X.$$

Define $f_1,f_2:X\to X$ as

$$\begin{array}{ccc}\hfill f_1\left(u\right)& =& \frac{3u}{8}\mathrm{for}\mathrm{all}u\in X,\hfill \\ \hfill f_2\left(u\right)& =& \frac{2u}{9}\mathrm{for}\mathrm{all}u\in X\hfill \end{array}$$

and $g_1,g_2:X\to X$ as

$$\begin{array}{ccc}\hfill g_1\left(u\right)& =& \frac{5u}{8}\mathrm{for}\mathrm{all}u\in X,\hfill \\ \hfill g_2\left(u\right)& =& \frac{4u}{9}\mathrm{for}\mathrm{all}u\in X.\hfill \end{array}$$

Now for $u,v\in X,$ we have

$$\begin{array}{ccc}\hfill \delta \left(f_1\left(u\right),g_1\left(v\right)\right)& =& \frac{3u}{8}+\frac{5v}{8}+4max\left\{\frac{3u}{8},\frac{5v}{8}\right\}\hfill \\ & =& \frac{1}{8}\left(3u+5v\right)+\frac{1}{2}max\left\{3u,5v\right\}\hfill \\ & \le & \frac{7}{8}(u+v+4max\{u,v\})\hfill \\ & =& {\lambda }_1\delta \left(u,v\right),\hfill \end{array}$$

where ${\lambda }_1={\displaystyle \frac{7}{8}}.$ Also for $u,v\in X,$ we have

$$\begin{array}{ccc}\hfill \delta \left(f_2\left(u\right),g_2\left(v\right)\right)& =& \frac{2u}{9}+\frac{4v}{9}+4max\left\{\frac{2u}{9},\frac{4v}{9}\right\}\hfill \\ & =& \frac{2}{9}\left(u+2v\right)+\frac{8}{9}max\left\{u,2v\right\}\hfill \\ & \le & \frac{8}{9}(u+v+4max\{u,v\})\hfill \\ & =& {\lambda }_2\delta \left(u,v\right),\hfill \end{array}$$

where ${\lambda }_2={\displaystyle \frac{8}{9}}.$

Consider the generalized iterated function system $\{X;f_m,g_m;m=1,2\},$ where each $(f_i,g_i)$ is a generalized contraction on X for $i=1,2$. The mappings $T,S:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ are given by

$$\begin{array}{cc}\hfill T\left(U\right)& =f_1\left(U\right)\cup f_2\left(U\right)\mathrm{for}\mathrm{all}U\in \mathcal{C}\left(X\right),\hfill \\ \hfill S\left(U\right)& =g_1\left(U\right)\cup g_2\left(U\right)\mathrm{for}\mathrm{all}U\in \mathcal{C}\left(X\right).\hfill \end{array}$$

By Proposition 1, for $\mathcal{U},\mathcal{V}$$\in \mathcal{C}\left(X\right)$, we have

$$H_{\delta }\left(T\left(\mathcal{U}\right),S\left(\mathcal{V}\right)\right)\le {\lambda }_{*}{H}_{\delta }\left(\mathcal{U},\mathcal{V}\right),$$

where ${\lambda }_{*}=max\left\{{\displaystyle \frac{7}{8}},{\displaystyle \frac{8}{9}}\right\}={\displaystyle \frac{8}{9}}.$ Thus, all conditions of Corollary 2 are satisfied. Moreover, for any set $B_0\in \mathcal{C}\left(X\right)$, the sequence of compact sets

$$B_{2n+1}=T\left(B_{2n}\right),B_{2n+2}=S\left(B_{2n+1}\right)\mathrm{for}n=0,1,2,\dots$$

converges to the common attractor of T and S.

Theorem 7

(Generalized Collage Theorem). Let $(X,\delta )$ be a dislocated metric space and

$\left\{X;f_1,\dots ,f_N;g_1,\dots ,g_N\right\}$ be a GIFS. Suppose that the pair of self-mappings $(T,S)$ defined by

$$\begin{array}{ccc}\hfill T\left(B\right)& =& f_1\left(B\right)\cup \dots \cup f_N\left(B\right)\mathrm{for}\mathrm{each}B\in \mathcal{C}\left(X\right),\hfill \\ \hfill S\left(B\right)& =& g_1\left(B\right)\cup \dots \cup g_N\left(B\right)\mathrm{for}\mathrm{each}B\in \mathcal{C}\left(X\right)\hfill \end{array}$$

is satisfying

$$H_{\delta }(T\left(A\right),S\left(A\right))\le \alpha H_{\delta }(A,B)\mathrm{for}A,B\in \mathcal{C}\left(X\right),$$

where $0\le \alpha <1.$ If for any $A\in \mathcal{C}\left(X\right)$ and some $\epsilon \ge 0$ either $H_{\delta }(A,T\left(A\right))\le \epsilon$ or $H_{\delta }(A,S\left(A\right))\le \epsilon$, then

$$H_{\delta }(A,U)\le \frac{\epsilon }{1-\alpha },$$

where $U\in C\left(X\right)$ is the common attractor of $T$ and S.

Proof. 

It follows from Corollary 2 that U is the common attractor of mappings $T,S$: $C\left(X\right)\to C\left(X\right),$ that is, $U=T\left(U\right)=S\left(U\right).$ Also, for any $B_0$ in $\mathcal{C}\left(X\right),$ we define a sequence $\left\{B_n\right\}$ as $B_{2n+1}=T\left(B_{2n}\right)$ and $B_{2n+2}=S\left(B_{2n+1}\right)$ for $n=0,1,2,\dots$. We have

$$\underset{n\to \infty }{lim}{H}_{\delta }\left(T\left(B_{2n}\right),U\right)=\underset{n\to \infty }{lim}{H}_{\delta }\left(S\left(B_{2n+1}\right),U\right)=0.$$

Now, as the pair $\left(T,S\right)$ is a generalized contraction, we have

$$H_{\delta }\left(B_n,B_{n+1}\right)\le {\alpha }^n{H}_{\delta }\left(B_0,B_1\right).$$

Assume that $H_{\delta }(A,T\left(A\right))\le \epsilon$ for any $A\in \mathcal{C}\left(X\right).$ Now, we have

$$H_{\delta }(A,U)\le H_{\delta }(A,T\left(A\right))+H_{\delta }(T\left(A\right),S\left(U\right))\le \epsilon +\alpha H_{\delta }(A,U),$$

which further implies that

$$H_{\delta }(A,U)\le \frac{\epsilon }{1-\alpha }.$$

Similarly, if we assume that $H_{\delta }(A,S\left(A\right))\le \epsilon$ for any $A\in \mathcal{C}\left(X\right).$ Then, we have

$$H_{\delta }(A,U)\le H_{\delta }(A,S\left(A\right))+H_{\delta }(S\left(A\right),T\left(U\right))\le \epsilon +\alpha H_{\delta }(A,U),$$

which gives $H_{\delta }(A,U)\le \frac{\epsilon }{1-\alpha }.$ □

5. Well-Posedness of Common Attractors

In this section, we will define the well-posedness of attractor-based problems of rational contraction maps in the framework of dislocated metric spaces.

Definition 13.

Let $(X,\delta )$ be a dislocated metric space. An attractor-based problem of mapping $T:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ is called well-posed if T has a unique attractor $A^{*}\in \mathcal{C}\left(X\right)$ and for any sequence $\left\{A_n\right\}$ in $\mathcal{C}\left(X\right)$ such that $\underset{n\to \infty }{lim}{H}_{\delta }(T\left(A_n\right),A_n)=0$ implies that $\underset{n\to \infty }{lim}{A}_n=A^{*}.$

Definition 14.

Let $(X,\delta )$ be a dislocated metric space. A common attractor-based problem of mappings $T,S:\mathcal{C}\left(X\right)\to \mathcal{C}\left(X\right)$ is called well-posed if T and S have a unique common attractor $A^{*}\in \mathcal{C}\left(X\right)$ and for any sequence $\left\{A_n\right\}$ in $\mathcal{C}\left(X\right)$ such that $\underset{n\to \infty }{lim}{H}_{\delta }(T\left(A_n\right),A_n)=0$ and $\underset{n\to \infty }{lim}{H}_{\delta }(S\left(A_n\right),A_n)=0$ implies that $\underset{n\to \infty }{lim}{A}_n=A^{*}.$

Theorem 8.

Let $(X,\delta )$ be a complete dislocated metric space and T and S be self-maps on $\mathcal{C}\left(X\right)$ as in Theorem 6. Then, the common attractor-based problem of T and S is well-posed.

Proof. 

It follows from Theorem 6 that maps T and S have a unique common attractor, say, $B^{*}$ and $H_{\delta }\left(B^{*},B^{*}\right)=0.$ Let $\left\{B_n\right\}$ be the sequence in $\mathcal{C}\left(X\right)$ such that $\underset{n\to \infty }{lim}{H}_{\delta }(T\left(B_n\right),B_n)=0$ and $\underset{n\to \infty }{lim}{H}_{\delta }(S\left(B_n\right),B_n)=0.$ We want to show that $B^{*}=\underset{n\to \infty }{lim}{B}_n$ for every integer $n.$ As the pair of $\left(S,T\right)$ is generalized rational contractive operators, so that

$$\begin{array}{ccc}\hfill H_{\delta }(B_n,B^{*})& =& H_{\delta }(B^{*},B_n)\hfill \\ & \le & H_{\delta }(B^{*},T\left(B_n\right))+H_{\delta }(T\left(B_n\right),B_n)\hfill \\ & =& H_{\delta }(S\left(B^{*}\right),T\left(B_n\right))+H_{\delta }(T\left(B_n\right),B_n)\hfill \\ & \le & \alpha M_{T,S}(B^{*},B_n)+H_{\delta }(T\left(B_n\right),B_n),\hfill \end{array}$$

(4)

where

$$\begin{array}{cc}\hfill & M_{T,S}(B^{*},B_n)=max\{H_{\delta }(B^{*},B_n),H_{\delta }(B^{*},T\left(B^{*}\right)),H_{\delta }(B_n,S\left(B_n\right)),\hfill \\ \hfill & \frac{H_{\delta }(B^{*},S\left(B_n\right))+H_{\delta }(B_n,T\left(B^{*}\right))}{4},{\displaystyle \frac{H_{\delta }(B^{*},B_n)(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{1+H_{\delta }(B^{*},B_n)}},\hfill \\ \hfill & \frac{H_{\delta }(B_n,S\left(B_n\right))(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{1+H_{\delta }(B^{*},B_n)},\frac{H_{\delta }(B^{*},S\left(B_n\right))(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{2(1+H_{\delta }(B^{*},B_n))}\}.\hfill \end{array}$$

Then, the following cases arise:

(i)

If $M_{T,S}(B^{*},B_n)=H_{\delta }(B^{*},B_n),$ then

$$H_{\delta }(B_n,B^{*})\le \alpha H_{\delta }(B^{*},B_n)+H_{\delta }(T\left(B_n\right),B_n),$$

which implies

$$H_{\delta }(B^{*},B_n)\le \frac{1}{1-\alpha }{H}_{\delta }(T\left(B_n\right),B_n)$$

and taking limit as $n\to \infty$ yields

$$\underset{n\to \infty }{lim}{H}_{\delta }(B^{*},B_n)\le 0,$$

that is, $\underset{n\to \infty }{lim}{B}_n=B^{*}.$

(ii)

In case $M_{T,S}(B^{*},B_n)=H_{\delta }(B^{*},T\left(B^{*}\right)),$ then

$$\begin{array}{ccc}\hfill H_{\delta }(B_n,B^{*})& \le & \alpha H_{\delta }(B^{*},T\left(B^{*}\right))+H_{\delta }(T\left(B_n\right),B_n)\hfill \\ & =& H_{\delta }(T\left(B_n\right),B_n)\hfill \end{array}$$

and taking limit as $n\to \infty$ yields

$$\underset{n\to \infty }{lim}H(B^{*},B_n)=0,$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}.$

(iii)

If $M_{T,S}(B^{*},B_n)=H_{\delta }(B_n,S\left(B_n\right)),$ we obtain

$$H_{\delta }(B_n,B^{*})\le \alpha H_{\delta }(B_n,S\left(B_n\right))+H_{\delta }(T\left(B_n\right),B_n),$$

and taking limit as $n\to \infty$ yields

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{H}_{\delta }(B_n,B^{*})& \le & \underset{n\to \infty }{lim}(\alpha H_{\delta }(B_n,S\left(B_n\right))+H_{\delta }(T\left(B_n\right),B_n))=0,\hfill \end{array}$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}.$

(iv)

If $M_{T,S}(B^{*},B_n)={\displaystyle \frac{H_{\delta }(B^{*},S\left(B_n\right))+H_{\delta }(B_n,T\left(B^{*}\right))}{4}}$, then

$$\begin{array}{c}{H}_{\delta }(B_n,B^{*})\le \frac{\alpha }{4}(H_{\delta }(B^{*},S\left(B_n\right))+H_{\delta }(B_n,B^{*}))+H_{\delta }(T\left(B_n\right),B_n)\hfill \\ \hfill \le \frac{\alpha }{4}(H_{\delta }(B_n,S\left(B_n\right))+2H_{\delta }(B_n,B^{*}))+H_{\delta }(T\left(B_n\right),B_n),\end{array}$$

that is,

$$H_{\delta }(B_n,B^{*})\le \frac{\alpha }{4\left(1-\alpha \right)}{H}_{\delta }(B_n,S\left(B_n\right))+\frac{1}{1-\alpha }{H}_{\delta }(T\left(B_n\right),B_n)$$

and taking limit as $n\to \infty$ yields

$$\underset{n\to \infty }{lim}{H}_{\delta }(B_n,B^{*})=0,$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}.$

(v)

If $M_{T,S}(B^{*},B_n)={\displaystyle \frac{H_{\delta }(B^{*},B_n)(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{1+H_{\delta }(B^{*},B_n)}},$ then

$$H_{\delta }(B_n,B^{*})\le \alpha \left({\displaystyle \frac{H_{\delta }(B^{*},B_n)(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{1+H_{\delta }(B^{*},B_n)}}\right)+H_{\delta }(T\left(B_n\right),B_n)$$

and taking limit $n\to \infty$ yields

$$\underset{n\to \infty }{lim}{H}_{\delta }(B_n,B^{*})\le 0,$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}.$

(vi)

If $M_{T,S}(B^{*},B_n)={\displaystyle \frac{H_{\delta }(B_n,S\left(B_n\right))(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{1+H_{\delta }(B^{*},B_n)}}$, then

$$H_{\delta }(B_n,B^{*})\le \alpha \frac{H_{\delta }(B_n,S\left(B_n\right))}{1+H_{\delta }(B^{*},B_n)}+H_{\delta }(T\left(B_n\right),B_n)$$

and taking limit as $n\to \infty$ yields

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{H}_{\delta }(B_n,B^{*})& \le & \underset{n\to \infty }{lim}(\alpha \frac{H_{\delta }(B_n,S\left(B_n\right))}{1+H_{\delta }(B^{*},B_n)}+H_{\delta }(T\left(B_n\right),B_n))\hfill \\ & =& 0,\hfill \end{array}$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}$.

(vii)

Finally, if $M_{T,S}(B^{*},B_n)={\displaystyle \frac{H_{\delta }(B^{*},S\left(B_n\right))(1+H_{\delta }(B^{*},T\left(B^{*}\right)))}{2(1+H_{\delta }(B^{*},B_n))}}$, then

$$\begin{array}{ccc}\hfill H_{\delta }(B_n,B^{*})& \le & \alpha \left({\displaystyle \frac{H_{\delta }(B^{*},S\left(B_n\right))}{2(1+H_{\delta }(B^{*},B_n))}}\right)+H_{\delta }(T\left(B_n\right),B_n)\hfill \\ & \le & \alpha \left({\displaystyle \frac{H_{\delta }(B^{*},B_n)+H_{\delta }(B_n,S\left(B_n\right))}{2(1+H_{\delta }(B^{*},B_n))}}\right)+H_{\delta }(T\left(B_n\right),B_n),\hfill \end{array}$$

which implies that

$$H_{\delta }(B_n,B^{*})\le \frac{1}{1-\alpha }(\alpha H_{\delta }(B_n,S\left(B_n\right))+H_{\delta }(T\left(B_n\right),B_n))$$

and taking limit $n\to \infty$ yields

$$\underset{n\to \infty }{lim}{H}_{\delta }(B_n,B^{*})\le 0,$$

which implies that $\underset{n\to \infty }{lim}{B}_n=B^{*}$.

This completes the proof. □

6. Application

In this section, we are using obtained results to solve a pair of functional equations arising in the dynamic programming.

Let ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$ be two Banach spaces and $U\subseteq {\mathcal{B}}_1$ and $V\subseteq {\mathcal{B}}_2$. Consider the mappings

$$\omega :U\times V⟶U,r_1,r_2:U\times V⟶\mathbb{R},{\mathfrak{h}}_1,{\mathfrak{h}}_2:U\times V\times \mathbb{R}⟶\mathbb{R}.$$

Considering U and V as the state and decision spaces, respectively, the problem of dynamic programming reduces to the problem of solving the functional equations defined as

$$\begin{array}{cc}\hfill p_1\left(x\right)& =\underset{y\in V}{sup}\{r_1(x,y)+{\mathfrak{h}}_1(x,y,p_1\left(\omega (x,y)\right))\},\mathrm{for}x\in U,\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill p_2\left(x\right)& =\underset{y\in V}{sup}\{r_1(x,y)+{\mathfrak{h}}_2(x,y,p_2\left(\omega (x,y)\right))\},\mathrm{for}x\in U.\hfill \end{array}$$

(6)

Equations (5) and (6) can be reformulated as

$$\begin{array}{cc}\hfill p_1\left(x\right)& =\underset{y\in V}{sup}\{r_2(x,y)+{\mathfrak{h}}_1(x,y,p_1\left(\omega (x,y)\right))\}-\lambda ,\mathrm{for}x\in U,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill p_2\left(x\right)& =\underset{y\in V}{sup}\{r_2(x,y)+{\mathfrak{h}}_2(x,y,p_2\left(\omega (x,y)\right))\}-\lambda ,\mathrm{for}x\in U,\hfill \end{array}$$

(8)

where $\lambda$ is a positive constant.

Our aim is to study the existence of the bounded solution of (7) and (8) arising in dynamic programming in dislocated metric spaces.

Let $B\left(U\right)$ denote the set of all bounded real valued functions on U. For $u\left(t\right)\in B\left(U\right)$, define $∥u∥=\underset{t\in U}{sup}\left|u\left(t\right)\right|$. Then, $(B\left(U\right),∥·∥)$ is a Banach space. Now, consider

$${\delta }_B(u,v)=\underset{t\in U}{sup}(u\left(t\right)+v\left(t\right))+\lambda ,$$

where $u,v\in B\left(U\right)$. Then, ${\delta }_{{}_B}$ is a dislocated metric on $B\left(U\right)$.

Consider the following assumptions:

(C₁):

$r_1,$$r_2$, ${\mathfrak{h}}_1$ and ${\mathfrak{h}}_2$ are bounded and continuous.

(C₂):

For $x\in U$, $w\in B\left(U\right)$ and $\lambda >0$, take $T,S:B\left(U\right)\to B\left(U\right)$ as

$$\begin{array}{cc}\hfill Tw\left(x\right)& =\underset{y\in V}{sup}\{r_2(x,y)+{\mathfrak{h}}_1(x,y,w\left(\omega (x,y)\right))\}-\lambda ,\mathrm{for}x\in U,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill Sw\left(x\right)& =\underset{y\in V}{sup}\{r_2(x,y)+{\mathfrak{h}}_2(x,y,w\left(\omega (x,y)\right))\}-\lambda ,\mathrm{for}x\in U.\hfill \end{array}$$

(10)

Also, for any $x\in U,$$y\in V$, $u,v\in B\left(U\right)$ and $t\in U$, it follows that

$${\mathfrak{h}}_1(x,y,u\left(t\right))+{\mathfrak{h}}_2(x,y,v\left(t\right))+2r_2(x,y)\le \alpha M_{T,S}(u\left(t\right),v\left(t\right))-2\lambda ,$$

(11)

where

$$\begin{array}{cc}\hfill & M_{T,S}(u\left(t\right),v\left(t\right))=max\{{\delta }_{{}_B}(u\left(t\right),v\left(t\right)),{\delta }_{{}_B}(u\left(t\right),Tu\left(t\right)),{\delta }_{{}_B}(v\left(t\right),Sv\left(t\right)),\hfill \\ \hfill & \frac{{\delta }_{{}_B}(u\left(t\right),Sv\left(t\right))+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right))}{4},\frac{{\delta }_{{}_B}(u\left(t\right),v\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))},\hfill \\ \hfill & \frac{{\delta }_{{}_B}(v\left(t\right),Sv\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))},\frac{{\delta }_{{}_B}(u\left(t\right),Sv\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))}\}.\hfill \end{array}$$

Theorem 9.

Assume that the conditions (C${}_1$) and (C${}_2$) hold. Then, Equations (7) and (8) have a unique common solution in $B\left(U\right)$.

Proof. 

By (C${}_1$), $T,S:B\left(U\right)\to B\left(U\right)$. From (9) and (10) in (C${}_2$), it follows that for any $u,v\in B\left(U\right)$ and $\lambda >0$, and $x\in U$ and $y_1,y_2\in V$ such that

$$\begin{array}{c}\hfill Tu<r_2(x,y_1)+{\mathfrak{h}}_1(x,y_1,u\left(\omega (x,y_1)\right)),\end{array}$$

(12)

$$\begin{array}{c}\hfill Sv<r_2(x,y_2)+{\mathfrak{h}}_2(x,y_2,v\left(\omega (x,y_2)\right)),\end{array}$$

(13)

it holds that

$$\begin{array}{cc}\hfill Tu& \le r_2(x,y_2)+{\mathfrak{h}}_1(x,y_2,u\left(\omega (x,y_2)\right))+\lambda ,\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill Sv& \le r_2(x,y_1)+{\mathfrak{h}}_2(x,y_1,v\left(\omega (x,y_1)\right))+\lambda .\hfill \end{array}$$

(15)

From (12) and (15) together with (11), it follows that

$$\begin{array}{cc}Tu\left(t\right)+Sv\left(t\right)\hfill & <2r_2\left(x,y_1\right)+{\mathfrak{h}}_1(x,y_1,u\left(\omega (x,y_1)\right))+{\mathfrak{h}}_2(x,y_1,v\left(\omega (x,y_1)\right))+\lambda \hfill \\ \hfill & \le \alpha M_{T,S}(u\left(t\right),v\left(t\right))-\lambda .\hfill \end{array}$$

(16)

Also, (13) and (14) together with (11) imply

$$\begin{array}{cc}Tu\left(t\right)+Sv\left(t\right)\hfill & <2r_2\left(x,y_2\right)+{\mathfrak{h}}_2(x,y_2,u\left(\omega (x,y_2)\right))+{\mathfrak{h}}_1(x,y_2,v\left(\omega (x,y_2)\right))+\lambda \hfill \\ \hfill & \le \alpha M_{S,T}(u\left(t\right),v\left(t\right))-\lambda .\hfill \end{array}$$

(17)

From (16) and (17), we have

$$Tu\left(t\right)+Sv\left(t\right)+\lambda \le \alpha M_{S,T}(u\left(t\right),v\left(t\right)).$$

(18)

Inequality (18) implies

$${\delta }_B(Tu\left(t\right),Sv\left(t\right))\le \alpha M_{S,T}(u\left(t\right),v\left(t\right)),$$

(19)

where

$$\begin{array}{cc}\hfill & M_{T,S}(u\left(t\right),v\left(t\right))=max\{{\delta }_{{}_B}(u\left(t\right),v\left(t\right)),{\delta }_{{}_B}(u\left(t\right),Tu\left(t\right)),{\delta }_{{}_B}(v\left(t\right),Sv\left(t\right)),\hfill \\ \hfill & \frac{{\delta }_{{}_B}(u\left(t\right),Sv\left(t\right))+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right))}{4},\frac{{\delta }_{{}_B}(u\left(t\right),v\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))},\hfill \\ \hfill & \frac{{\delta }_{{}_B}(v\left(t\right),Sv\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))},\frac{{\delta }_{{}_B}(u\left(t\right),Sv\left(t\right))(1+{\delta }_{{}_B}(v\left(t\right),Tu\left(t\right)))}{1+{\delta }_{{}_B}(u\left(t\right),v\left(t\right))}\}.\hfill \end{array}$$

Thus all the conditions of Corollary 3.4 hold. Moreover, there exists a common attractor of T and S, that is, $u^{*}\in B\left(U\right)$, where $u^{*}\left(t\right)$ is a common solution of (7) and (8). □

It is known that the generalized rational contraction is reduced to a usual contraction, but the converse is not always true. Similarly, the generalized Ciric type contraction is one of the general cases of a contraction, which covers Kannan contraction and Chatterjea-type contraction. It is important to note that our generalized rational contraction that we used in this paper covers all these contractions such as Kannan contraction, Chatterjea-type contraction and generalized Ciric type contraction.

The proposed study is focused on the generation of fractals in dislocated metric space by using generalized rational contraction at the first time in the literature. The importance of this research work is to develop the fractal set in dislocated metric space through the generalized iterated function system based on generalized rational contractions. It is also demonstrated with the idea of constructing a new type of fractals that are common attractors generated by generalized iterated function system. It is believed that the proposed research work of common attractors with generalized rational contractive operators can be established for deriving fractals through admissible hybrid contractions [19] and generalized cyclic contractions [13] maps. The proposed theory will also lead to a new path for developing new kind of fractals and their consequences based on generalized contractions in the extraction of dislocated metric spaces.

7. Conclusions

In this paper, we investigated a method of generalized iterated function systems for common attractors based on finite family of generalized contractive self-mappings in the setup of dislocated metric spaces. The theories of dislocated metric spaces, dislocated Hausdorff metric space, and generalized contractions were explored. The generalized iterated function system is defined with the help of generalized contraction in dislocated metric spaces. Essentially, the generalized rational contractive Hutchinson operators have been constructed in a dislocated metric space to generate the fractals. The common attractors of the generalized iterated function system were proved in the dislocated metric space using generalized rational contractive Hutchinson operators. The generalized collage theorem based on the generalized iterated function system for common attractors is also established in a dislocated metric space. Moreover, the well-posedness of attractor-based problems of rational contraction maps in the framework of dislocated metric spaces is also presented.