A Unified Framework for Generalized Symmetric Contractions and Economic Dynamics via Fractional Differential Equations

Abstract

This study has developed a unified framework for modeling economic growth through Caputo fractional differential equations. The framework has established the existence and uniqueness of solutions by employing a generalized fixed-point approach. In particular, the analysis has introduced and utilized new classes of symmetric operators, including symmetric Lipschitz-type mappings, symmetric Kannan-type contractions, and symmetric Chatterjea-type contractions. These mappings are based on a refined symmetric Lipschitz condition that enables the examination of the behavior of their iterative sequences. The study has focused on several forms of symmetric contractions defined on metric spaces endowed with a binary relation, providing a setting that generalizes and unifies various existing fixed-point theorems. This framework has extended classical results by Goebel and Sims, Goebel and Japon-Pineda, and others. Finally, to illustrate the practical significance of the theoretical findings, the developed results have been applied to demonstrate the existence of solutions for fractional models of economic growth and a related Fredholm integral equation.

1. Introduction

The fundamental Banach fixed-point result [1], introduced in 1922 in Banach’s thesis, remains a cornerstone in the theory of metric fixed points. According to this result, any Banach contraction mapping $\mathcal{T}$ on a complete metric space $(\mathcal{X}, d)$, possesses a unique fixed point ${\mathfrak{x}}^{*}\in \mathcal{X}$, i.e., $\mathcal{T}{\mathfrak{x}}^{*}={\mathfrak{x}}^{*}$. Specifically, a mapping $\mathcal{T}$ is known as a Banach contraction if for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$, the distance between $\mathcal{T}\mathfrak{x}$ and $\mathcal{T}\mathfrak{y}$ satisfies the condition

$$d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\le \mu ·d(\mathfrak{x}, \mathfrak{y})$$

(1)

for $\mu \in [0, 1)$. This principle, known for its reliability and implementation, has become an indispensable tool for proving the presence of solutions to a wide variety of problems in mathematics and the applied sciences, including integral and differential equations across fields such as physics, engineering, biology, social sciences, etc.

In the years following its introduction, Banach’s contraction theorem inspired numerous extensions and generalizations. Prominent contributions from Edelstein [2], Rakotch [3], Boyd and Wong [4], Meir and Keeler [5], Reich [6], Matkowski [7], and others have expanded the scope of the theorem, exploring different types of contractions, the role of completeness, and various related fixed point results. Notably, the theorem’s applicability has been broadened with new contractive conditions and extensions to broader domains, as discussed in the works of Nadler [8] and others. For further developments on metric fixed point theory and its applications, the recent research by Joshi and Bose [9], Almezel et al. [10], and Hammad and Manuel [11,12] offers valuable insights. These extensions not only retain the original power of the Banach contraction principle but also explore its use in solving more complex problems across diverse scientific areas.

Various modifications of the Lipschitz condition have been proposed, based on different distance measures. These include six specific distances, as follows: $d(\mathfrak{x}, \mathfrak{y})$, $d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})$, $d(\mathfrak{x}, \mathcal{T}\mathfrak{x})$, $d(\mathfrak{y}, \mathcal{T}\mathfrak{y})$, $d(\mathfrak{x}, \mathcal{T}\mathfrak{y})$, and $d(\mathfrak{y}, \mathcal{T}\mathfrak{x})$ for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$. In its classical form, the Lipschitz condition typically involves an inequality between two of these distances: $(\mathfrak{x}, \mathfrak{y})$ and $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})$, which ensures that an operator is continuous or uniformly continuous. However, several authors have explored alternative inequalities that involve some or all of these six distances. While these alternative conditions offer insight, they do not always guarantee the continuity of the mapping in question.

In addition, Kannan [13] and Chatterjea [14] contractions have been studied extensively due to their importance in proving the presence of fixed points. An operator $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is classified as a Kannan contraction if we find a constant $\mu \in [0, {\displaystyle \frac{1}{2}})$ such that

$$d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\le \mu \left[d(\mathfrak{x}, \mathcal{T}\mathfrak{x})+d(\mathfrak{y}, \mathcal{T}\mathfrak{y})\right]\mathrm{for}\mathrm{all}\mathfrak{x}, \mathfrak{y}\in \mathcal{X}.$$

(2)

Similarly, an operator $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is referred to as a Chatterjea contraction if we find a constant $\mu \in [0, {\displaystyle \frac{1}{2}})$ for which

$$d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\le \mu \left[d(\mathfrak{x}, \mathcal{T}\mathfrak{y})+d(\mathfrak{y}, \mathcal{T}\mathfrak{x})\right]\mathrm{for}\mathrm{all}\mathfrak{x}, \mathfrak{y}\in \mathcal{X}.$$

(3)

Alam and Imdad [15] gave a weaker form of the Banach contraction condition, applying it only to point pairs connected by a specific binary relation, rather than throughout the whole space. This viewpoint inspired substantial research on fixed point theory in the framework of binary relations, commonly known as relational metric spaces. In particular, Din and his collaborators (see [16,17]) explored extensions of Perov’s classical fixed point theorems within this context.

Goebel and Pineda [18] and Goebel and Sims [19] introduced a new class of mappings termed $(\tau , \beta )$-Lipschitz mappings. Consider $\tau =({\tau }_1, {\tau }_2, \dots , {\tau }_k)$, where each ${\tau }_j\in \mathbb{R}$, with the conditions ${\sum }_{j=1}^k{\tau }_j=1$ and ${\tau }_j\ge 0$ for all ${\tau }_j$, and ${\tau }_1, {\tau }_k>0$. We then refer to $\tau$ as a multi-index, and the number k is called the length of the multi-index $\tau$. For further information on $(\tau , \beta )$-Lipschitz mappings, see the works of Goebel and Japon-Pineda [18], where they introduce a new type of nonexpansiveness that generalizes traditional Lipschitz conditions, offering a broader framework for studying fixed-point theory. Their approach is further expanded by Goebel and Sims [19], who explore mean Lipschitzian mappings, which offer insights into the behavior of mappings under more relaxed Lipschitz conditions. Additionally, the work of Asem et al. [20] on $(\tau , k)$-cyclic contractions delves into the relationships between such mappings and the existence of fixed points, contributing to a deeper understanding of cyclic iterations and their convergence properties. These contributions collectively highlight the versatility and applicability of $(\tau , \beta )$-Lipschitz mappings in modern fixed point theory.

Motivated by the concept of $(\tau , \beta )$-Lipschitz mappings, $(\tau , \beta )$-nonexpansive mappings with multi-indices [18,19], and utilizing the technique introduced by Alam and Imdad [15], we initiated the development of $(\tau , \beta , \mathcal{R})$-type symmetric nonexpansive mappings with multi-indices. We also define the concept of $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contractions and $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contractions with some multi-indices, and establish related fixed point results. Finally, we provide several detailed examples to verify our findings and present an existence condition to show the presence of a solution for an integral equation. We also study the existence and uniqueness of solutions for economic growth models with fractional differential equations using generalized fixed-point techniques.

2. Preliminaries

We now introduce the following consolidated definitions and notations that will be used throughout the paper.

Definition 1

([15,21,22,23,24,25,26,27]). Let $(\mathcal{X}, d)$ be a metric space, $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ a mapping, and $\mathcal{R}\subseteq {\mathcal{X}}^2$ a nonempty binary relation. Then,

1. 

${\mathcal{R}}^{-1}=\{(\mathfrak{x}, \mathfrak{y}):(\mathfrak{y}, \mathfrak{x})\in \mathcal{R}\}$, ${\mathcal{R}}^{\#}=\mathcal{R}\cup {\Delta }_{\mathcal{X}}$, and ${\mathcal{R}}^s=\mathcal{R}\cup {\mathcal{R}}^{-1}$ denote the inverse, reflexive, and symmetric closure relations of the binary relation $\mathcal{R}$, respectively.

2. 

Elements $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$ are $\mathcal{R}$-comparable if $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$ or $(\mathfrak{y}, \mathfrak{x})\in \mathcal{R}$, denoted $[\mathfrak{x}, \mathfrak{y}]\in \mathcal{R}$. It is also important to note that $[\mathfrak{x}, \mathfrak{y}]\in \mathcal{R}$ if and only if $[\mathfrak{x}, \mathfrak{y}]\in {\mathcal{R}}^s$.

3. 

A sequence $\left\{{\mathfrak{x}}_j\right\}\subset \mathcal{X}$ is $\mathcal{R}$-preserving if $({\mathfrak{x}}_j, {\mathfrak{x}}_{j+1})\in \mathcal{R}$ for all j.

4. 

$\mathcal{R}$ is $\mathcal{T}$-closed if $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}\Rightarrow (\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\in \mathcal{R}$; in this case, $\mathcal{R}$ and ${\mathcal{R}}^s$ remain ${\mathcal{T}}^j$-closed.

5. 

$(\mathcal{X}, d)$ is $\mathcal{R}$-complete if every $\mathcal{R}$-preserving Cauchy sequence converges.

6. 

$\mathcal{T}$ is $\mathcal{R}$-continuous if convergence of a $\mathcal{R}$-preserving sequence $\left\{{\mathfrak{x}}_j\right\}\to \mathfrak{x}$ implies $\mathcal{T}{\mathfrak{x}}_j\to \mathcal{T}\mathfrak{x}$.

7. 

$\mathcal{R}$ is d-self-closed if every convergent $\mathcal{R}$-preserving sequence has a subsequence $\left\{{\mathfrak{x}}_{j_l}\right\}$ with $[{\mathfrak{x}}_{j_l}, \mathfrak{x}]\in \mathcal{R}$.

8. 

A subset $\mathcal{B}\subseteq \mathcal{X}$ is $\mathcal{R}$-directed if for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{B}$ there exists $\mathfrak{q}\in \mathcal{X}$ such that $(\mathfrak{x}, \mathfrak{q}),$$(\mathfrak{y}, \mathfrak{q})\in \mathcal{R}$.

9. 

A path of length k in $\mathcal{R}$ from $\mathfrak{x}$ to $\mathfrak{y}$ is a sequence $\{{\mathfrak{q}}_0, \dots , {\mathfrak{q}}_k\}\subseteq \mathcal{X}$ with ${\mathfrak{q}}_0=\mathfrak{x}$, ${\mathfrak{q}}_k=\mathfrak{y}$, and $({\mathfrak{q}}_i, {\mathfrak{q}}_{i+1})\in \mathcal{R}$ for all $0\le i<k$.

10. 

Denote $\mathrm{Fix}\left(\mathcal{T}\right)=\{\mathfrak{x}\in \mathcal{X}:\mathcal{T}(\mathfrak{x})=\mathfrak{x}\}$, ${\mathcal{R}}_{\mathcal{T}}=\{\mathfrak{x}\in \mathcal{X}:(\mathfrak{x}, \mathcal{T}\left(\mathfrak{x}\right))\in \mathcal{R}\}$, and $\mathrm{Paths}(\mathfrak{x}, \mathfrak{y};\mathcal{R})$ as the set of all $\mathcal{R}$-paths connecting $\mathfrak{x}$ and $\mathfrak{y}$.

11. 

If $\mathrm{Paths}(\mathfrak{x}, \mathfrak{y};\mathcal{R})$ is non-empty for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$, then $\mathcal{X}$ is called $\mathcal{R}$-connected.

Furthermore, Alam et al. [23] extended the findings of [15] by introducing even more relaxed conditions. This allowed them to prove the presence and uniqueness of fixed points for self-mappings defined with respect to binary relations, as further supported by [28].

Definition 2

([23]). The self-mapping $\mathcal{T}$ is $\mathcal{R}$-preserving contraction if there exists $\mu \in [0, 1)$ such that

$$forall\mathfrak{x}, \mathfrak{y}\in \mathcal{X}with(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}\Rightarrow d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\le \mu d(\mathfrak{x}, \mathfrak{y}).$$

(4)

Theorem 1

([23]). Let $(\mathcal{X}, d)$ be a metric space with a binary relation $\mathcal{R}$ defined on $\mathcal{X}$, and let $\mathcal{T}$ be a self-mapping on $\mathcal{X}$. Assume further that the following assumptions are satisfied:

1. 

$(\mathcal{X}, d)$ is $\mathcal{R}$-complete;

2. 

$\mathcal{R}$ is $\mathcal{T}$-closed;

3. 

either $\mathcal{T}$ is $\mathcal{R}$-continuous or $\mathcal{R}$ is d-self-closed;

4. 

${\mathcal{R}}_{\mathcal{T}}$ is non-empty;

5. 

$\mathcal{T}$ is an $\mathcal{R}$-preserving contraction.

Then, $\mathcal{T}$ admits at least one fixed point. Furthermore, if the image $\mathcal{T}\left(\mathcal{X}\right)$ is ${\mathcal{R}}^s$-connected, the fixed point of $\mathcal{T}$ is unique.

Example 1.

Let $\mathcal{X}=\mathbb{R}$ be the set of real numbers equipped with the usual metric given by $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$. Define the binary relation $\mathcal{R}$ over $\mathcal{X}$ as $\mathcal{R}:=\left\{\right(\mathfrak{x}, \mathfrak{y}\left)\right|\mathfrak{x}\in [0, 1]\&\mathfrak{y}\in [1, 2]\}$ and mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ by $\mathcal{T}\left(\mathfrak{x}\right)=sin\left(\mathfrak{x}\right)$. Then $\mathcal{T}$ is a $\mathcal{R}$-preserving contraction. In fact, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have

$$d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})={\displaystyle \frac{1}{9}}|sin\mathfrak{x}-sin\mathfrak{y}|\le {\displaystyle \frac{1}{9}}|\mathfrak{x}-\mathfrak{y}|={\displaystyle \frac{1}{9}}d(\mathfrak{x}, \mathfrak{y}).$$

All other assumptions of Theorem 1 are fulfilled. Therefore, the operator $\mathcal{T}$ admits a unique fixed point, namely ${\mathfrak{x}}^{*}=\left\{0\right\}$.

Goebel and Pineda [18], as well as Goebel and Sims [19], introduced a new class of mappings, referred to as $(\tau , \beta )$-Lipschitz mappings, defined as follows:

Definition 3

([18,19]). Let us consider a metric space $(\mathcal{X}, d)$ and let $\mathcal{A}$ be a nonempty subset of $\mathcal{X}$. A mapping $\mathcal{T}:\mathcal{A}\to \mathcal{A}$ is known as a $(\tau , \beta )$-Lipschitzian with constant $\mu \in [0, \infty )$ if one finds a multi-index $\tau =({\tau }_1, {\tau }_2, \dots , {\tau }_k)$ of length k, and a parameter $\beta \in [1, \infty )$, such that

$$\sum _{j=1}^k{\tau }_j·{\left[d({\mathcal{T}}^j\mathfrak{x}, {\mathcal{T}}^j\mathfrak{y})\right]}^{\beta }\le \mu {\left[d(\mathfrak{x}, \mathfrak{y})\right]}^{\beta }$$

(5)

for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{A}.$

The smallest $\mu$ that satisfies the $(\tau , \beta )$-Lipschitz condition is called the $(\tau , \beta )$-Lipschitz constant. The mapping $\mathcal{T}$ is termed a $(\tau , \beta )$-contraction if $\mu <1$, and $(\tau , \beta )$-nonexpansive if $\mu =1$. Similarly, when $\beta =1$, we call $\mathcal{T}$ a $\tau$-contraction or $\tau$-nonexpansive if Equation (5) holds with $\mu <1$ or $\mu =1$, respectively. If $\tau$ is not explicitly specified with $\beta =1$ in Equation (5), then $\mathcal{T}$ is referred to as a mean Lipschitzian mapping.

Example 2

([29]). Let $\mathcal{X}=[0, \infty )$ be equipped with the standard metric $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$. Define a self-mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ by $\mathcal{T}\left(\mathfrak{x}\right)=\mathfrak{x}+a$, where $a>0$. Then, for $\beta \ge 1$ and $\tau =({\tau }_1, {\tau }_2)\in {(0, 1)}^2$, the mapping satisfies the following:

$${\tau }_1{\left|\mathcal{T}\mathfrak{x}-\mathcal{T}\mathfrak{y}\right|}^{\beta }+{\tau }_2{\left|{\mathcal{T}}^2\mathfrak{x}-{\mathcal{T}}^2\mathfrak{y}\right|}^{\beta }={\left|\mathfrak{x}-\mathfrak{y}\right|}^{\beta }$$

for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$. This shows that a $\mathcal{T}$ is $(\tau , \beta )$-nonexpansive with multi-index τ of length $k=2$.

3. Main Results

In this section, we introduced the concept of $(\tau , \beta , \mathcal{R})$-type symmetric nonexpansive mappings with multi-indices. We also defined the notions of $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contractions and $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contractions for certain multi-indices, and established corresponding fixed point results. Lastly, we presented several detailed examples to illustrate and validate our findings.

Definition 4.

A mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is referred to as a $(\tau , \beta , \mathcal{R})$-type symmetric Lipschitzian if, for some multi-index τ of length k and $\beta \in [1, \infty )$, there exists a constant $\mu \ge 0$ such that

$$\sum _{j=1}^k{\tau }_j{\left[d\left({\mathcal{T}}^j\mathfrak{x}, {\mathcal{T}}^j\mathfrak{y}\right)\right]}^{\beta }\le \mu {\left[d\left(\mathfrak{x}, \mathfrak{y}\right)\right]}^{\beta }$$

(6)

for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$ with $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$.

The smallest $\mu$ satisfying the $(\tau , \beta , \mathcal{R})$-type symmetric Lipschitz condition is termed the corresponding Lipschitz constant. The mapping $\mathcal{T}$ is called a $(\tau , \beta , \mathcal{R})$-type contraction if $\mu <1$, and non-expansive if $\mu =1$. When $\beta =1$ and $\tau$ is unspecified, $\mathcal{T}$ is referred to as a mean-theoretic order $\mathcal{R}$-contraction if $\mu <1$ or an $\mathcal{R}$-non-expansive mapping if $\mu =1$. If $\mathcal{R}={\mathcal{X}}^2$, then a $(\tau , \beta , \mathcal{R})$-type contraction reduces to a $(\tau , \beta )$-contraction. Moreover, if $\beta =1$ in (6), the mapping $\mathcal{T}$ is referred to as a $(\tau , \mathcal{R})$-type contraction.

Definition 5.

A mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is said to be a $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contraction if, for some multi-index τ of length k and $\beta \in [1, \infty )$, there exists a constant $\mu \in [0, {\displaystyle \frac{1}{2}})$ such that

$$\sum _{j=1}^k{\tau }_j{\left[d\left({\mathcal{T}}^j\mathfrak{x}, {\mathcal{T}}^j\mathfrak{y}\right)\right]}^{\beta }\le \mu \left[{\left[d\left(\mathfrak{x}, \mathcal{T}\mathfrak{x}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{y}, \mathcal{T}\mathfrak{y}\right)\right]}^{\beta }\right]$$

(7)

for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$ with $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$.

Definition 6.

A mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is termed a $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contraction if, for some multi-index τ of length k and $\beta \in [1, \infty )$, there exists a constant $\mu \in [0, {\displaystyle \frac{1}{2}})$ such that

$$\sum _{j=1}^k{\tau }_j{\left[d\left({\mathcal{T}}^j\mathfrak{x}, {\mathcal{T}}^j\mathfrak{y}\right)\right]}^{\beta }\le \mu \left[{\left[d\left(\mathfrak{x}, \mathcal{T}\mathfrak{y}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{y}, \mathcal{T}\mathfrak{x}\right)\right]}^{\beta }\right]$$

(8)

for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$ with $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$.

For simplicity, we focus on the multi-index $\tau$ of length $k=2$ in this section. This simplification aids in better illustrating the concepts, although the method can be generalized to indices of higher dimensions.

Example 3.

Consider $\mathcal{X}=[0, 1]$ equipped with the standard metric $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$. Define the binary relation $\mathcal{R}$ and self-mapping $\mathcal{T}$ as given below:

$$\mathcal{R}=\left\{(\mathfrak{x}, \mathfrak{y})\in \mathcal{X}\times \mathcal{X}:\mathfrak{x}\in \left[{\displaystyle \frac{5}{8}}, 1\right]\&\mathfrak{y}\in \left[{\displaystyle \frac{1}{2}}, 1\right]\right\}$$

and

$$\mathcal{T}\mathfrak{x}={\displaystyle \frac{1+{\mathfrak{x}}^2}{2}}\mathrm{for}\mathrm{all}\mathfrak{x}\in \mathcal{X}.$$

Additionally, ${\mathcal{T}}^2\mathfrak{x}={\displaystyle \frac{5+2{\mathfrak{x}}^2+{\mathfrak{x}}^4}{8}}$ for all $\mathfrak{x}\in \mathcal{X}$. Then for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have

$$|\mathcal{T}\mathfrak{x}-\mathcal{T}\mathfrak{y}|={\displaystyle \frac{1}{2}}\left|{\mathfrak{x}}^2-{\mathfrak{y}}^2\right|={\displaystyle \frac{\mathfrak{x}+\mathfrak{y}}{2}}|\mathfrak{x}-\mathfrak{y}|\le |\mathfrak{x}-\mathfrak{y}|.$$

Also, we have the following:

$$\left|{\mathcal{T}}^2\mathfrak{x}-{\mathcal{T}}^2\mathfrak{y}\right|\le {\displaystyle \frac{1}{4}}\left|{\mathfrak{x}}^2-{\mathfrak{y}}^2\right|+{\displaystyle \frac{1}{8}}\left|{\mathfrak{x}}^4-{\mathfrak{y}}^4\right|\le |\mathfrak{x}-\mathfrak{y}|.$$

So, for any $\beta \ge 1$, we get

$${\tau }_1{\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y}\right)\right]}^{\beta }\le {\left[d(\mathfrak{x}, \mathfrak{y})\right]}^{\beta }$$

for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$. It is evident that $\mathcal{T}$ constitutes a $(\tau , \beta , \mathcal{R})$-type symmetric nonexpansive mapping, where the multi-index τ has a length of $k=2$.

For additional examples of $(\tau , \beta , \mathcal{R})$-type symmetric contraction mappings, as well as $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contractions and $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contractions, we direct the reader to Examples 4, 5, and 7 for further clarification and detailed illustrations.

Theorem 2.

Consider $(\mathcal{X}, d)$ as a metric space, equipped with a binary relation $\mathcal{R}$ and a self-mapping $\mathcal{T}$ on $\mathcal{X}$. Suppose that the following assumptions are fulfilled:

1. 

$(\mathcal{X}, d)$ is $\mathcal{R}$-complete;

2. 

$\mathcal{R}$ is $\mathcal{T}$-closed;

3. 

either $\mathcal{T}$ is $\mathcal{R}$-continuous or $\mathcal{R}$ is d-self-closed;

4. 

${\mathcal{R}}_{\mathcal{T}}$ is non-empty;

5. 

$\mathcal{T}$ is a $(\tau , \beta , \mathcal{R})$-type symmetric contraction with constant $\mu <\delta =min\{{\tau }_1, {\tau }_2\}$.

Then, $\mathcal{T}$ has a fixed point.

Proof. 

In lieu of assumption (4), there exists a ${\mathfrak{x}}_0\in {\mathcal{R}}_{\mathcal{T}}$, that is, $\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\in \mathcal{R}$. Define a Picard sequence $\left\{{\mathfrak{x}}_n\right\}$ at the initial guess ${\mathfrak{x}}_0$, that is, ${\mathfrak{x}}_{n+1}=\mathcal{T}{\mathfrak{x}}_n$, for all $n\in \mathbb{N}$. As $\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\in \mathcal{R}$, using assumption (2), we obtain

$$({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0), (\mathcal{T}{\mathfrak{x}}_0, \mathcal{T}\mathcal{T}{\mathfrak{x}}_0), (\mathcal{T}\mathcal{T}{\mathfrak{x}}_0, \mathcal{T}\mathcal{T}\mathcal{T}{\mathfrak{x}}_0), \dots \in \mathcal{R},$$

which further implies that

$$\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)\in \mathcal{R}, \mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

(9)

This indicates that the Picard sequence $\left\{{\mathfrak{x}}_n\right\}$ maintains the relation $\mathcal{R}$ throughout its iterations, making it a $\mathcal{R}$-preserving sequence. If there exists some $n^{*}\in \mathbb{N}$ such that ${\mathfrak{x}}_{n^{*}+1}={\mathfrak{x}}_{n^{*}}$, then $\mathcal{T}{\mathfrak{x}}_{n^{*}}={\mathfrak{x}}_{n^{*}}$, i.e., ${\mathfrak{x}}_{n^{*}}\in Fix\left(\mathcal{T}\right)$. From this point onward in the proof, we proceed under the assumption that ${\mathfrak{x}}_{n+1}\ne {\mathfrak{x}}_n$, for all $n\ge 0$. Setting $\xi =max\left\{d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right), d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right\}$, we have $d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\le \xi$ and $d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\le \xi$.

Now, we discuss the following cases.

Case (i): When $2\mu <{\tau }_1$ and setting ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (4), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\right]}^{\beta }\le \mu {\xi }^{\beta }\hfill \\ \hfill \Rightarrow {\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{{\tau }_1}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }.\hfill \end{array}$$

It follows that:

$$d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\le \phi \xi .$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (4), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }=\mu {\left(\phi \xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{\delta }}{\left(\phi \xi \right)}^{\beta }={\left({\phi }^2\xi \right)}^{\beta }.\hfill \end{array}$$

Therefore, $d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\le {\phi }^2\xi$. And from (4), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }=\mu {\left({\phi }^2\xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\left({\phi }^3\xi \right)}^{\beta }.\hfill \end{array}$$

Therefore, $d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\le {\phi }^3\xi$. Additionally, we have the following:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\phi }^4\xi .\hfill \end{array}$$

Similarly, we obtain the following:

$$d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\le {\phi }^n\xi , \mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Hence, we have $d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\to 0$ as $n\to \infty$. We now aim to demonstrate that the sequence $\left\{{\mathfrak{x}}_n\right\}$ forms a Cauchy sequence in the space $\mathcal{X}$. To this end, consider any $n_0\ge 0$, and let $n\ge n_0$, $p\ge 1$; then we get

$$\begin{array}{cc}\hfill d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+p}{\mathfrak{x}}_0\right)\le & d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)+d\left({\mathcal{T}}^{n+1}{\mathfrak{x}}_0, {\mathcal{T}}^{n+2}{\mathfrak{x}}_0\right)\hfill \\ & +\dots +d\left({\mathcal{T}}^{n+p-1}{\mathfrak{x}}_0, {\mathcal{T}}^{n+p}{\mathfrak{x}}_0\right)\hfill \\ \hfill \le & {\phi }^n\left(1+\phi +{\phi }^2+\dots +{\phi }^{p-1}\right)\xi \hfill \\ \hfill \le & {\displaystyle \frac{{\phi }^n}{1-\phi }}\xi .\hfill \end{array}$$

Taking limit as $n\to \infty$, we obtain $d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+p}{\mathfrak{x}}_0\right)\to 0$, which means that $\left\{{\mathfrak{x}}_n\right\}$ is a $\mathcal{R}$-preserving Cauchy sequence in $\mathcal{X}$.

Case (ii): When $\mu <{\tau }_2$ and setting ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (4), we have the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\right]}^{\beta }=\mu \xi \hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{{\tau }_2}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }={\left(\phi \xi \right)}^{\beta }.\hfill \end{array}$$

Therefore, we obtain the following:

$$d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\le \phi \xi .$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (4), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }=\mu \xi \hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)& \le \phi \xi .\hfill \end{array}$$

And

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)& \le {\phi }^2\xi .\hfill \end{array}$$

Additionally, we have the following:

$$d\left({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0\right)\le {\phi }^2\xi .$$

Moreover, we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0\right)& \le {\phi }^3\xi .\hfill \end{array}$$

Further, we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^7{\mathfrak{x}}_0, {\mathcal{T}}^8{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^7{\mathfrak{x}}_0, {\mathcal{T}}^8{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu {\left[d\left({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^7{\mathfrak{x}}_0, {\mathcal{T}}^8{\mathfrak{x}}_0\right)& \le {\phi }^3\xi .\hfill \end{array}$$

Proceeding with the same steps as previously outlined, we arrive at the following:

$$d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0\right)\le {\phi }^l\xi ,$$

whenever $m=2l$ or $m=2l+1$, it shows that $d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0\right)\to 0$ as $m\to +\infty$. To demonstrate that the sequence $\left\{{\mathfrak{x}}_n\right\}$ is Cauchy in $\mathcal{X}$, consider the two possible subcases by selecting positive integers m and n such that $m<n=m+q$.

Subcase (a): For $m=2l$, where $l, q\ge 1$, we have

$$\begin{array}{cc}\hfill d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+q}{\mathfrak{x}}_0\right)=& d\left({\mathcal{T}}^{2l}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q}{\mathfrak{x}}_0\right)\hfill \\ \hfill \le & d\left({\mathcal{T}}^{2l}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+1}{\mathfrak{x}}_0\right)+d\left({\mathcal{T}}^{2l+1}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+2}{\mathfrak{x}}_0\right)\hfill \\ & +d\left({\mathcal{T}}^{2l+2}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+3}{\mathfrak{x}}_0\right)+\dots +d\left({\mathcal{T}}^{2l+q-2}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q-1}{\mathfrak{x}}_0\right)\hfill \\ & +d\left({\mathcal{T}}^{2l+q-1}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q}{\mathfrak{x}}_0\right)\hfill \\ \hfill \le & {\phi }^l\xi +{\phi }^l\xi +{\phi }^{l+1}\xi +{\phi }^{l+1}\xi +\dots \hfill \\ \hfill \le & 2{\phi }^l\left(1+\phi +{\phi }^2+\dots \right)\xi \hfill \\ \hfill \le & 2{\phi }^l{\displaystyle \frac{1}{1-\phi }}\xi .\hfill \end{array}$$

Subcase (b): Similarly, for $m=2l+1$ with $l, q\ge 1$, we obtain the following:

$$\begin{array}{cc}\hfill d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+q}{\mathfrak{x}}_0\right)=& d\left({\mathcal{T}}^{2l+1}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q+1}{\mathfrak{x}}_0\right)\hfill \\ \hfill \le & d\left({\mathcal{T}}^{2l+1}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+2}{\mathfrak{x}}_0\right)+d\left({\mathcal{T}}^{2l+2}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+3}{\mathfrak{x}}_0\right)\hfill \\ & +d\left({\mathcal{T}}^{2l+3}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+4}{\mathfrak{x}}_0\right)+\dots +d\left({\mathcal{T}}^{2l+q-1}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q}{\mathfrak{x}}_0\right)+\hfill \\ & d\left({\mathcal{T}}^{2l+q}{\mathfrak{x}}_0, {\mathcal{T}}^{2l+q+1}{\mathfrak{x}}_0\right)\hfill \\ \hfill \le & {\phi }^l\xi +{\phi }^{l+1}\xi +{\phi }^{l+1}\xi +{\phi }^{l+2}\xi +\dots \hfill \\ \hfill \le & 2{\phi }^l\left(1+\phi +{\phi }^2+\dots \right)\xi \hfill \\ \hfill \le & 2{\phi }^l{\displaystyle \frac{1}{1-\phi }}\xi .\hfill \end{array}$$

Letting $l\to \infty$ in all sub-cases and noting that $\phi <1$, it follows that

$$d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0\right)\to 0.$$

Therefore, $\left\{{\mathfrak{x}}_n\right\}$ forms a $\mathcal{R}$-preserving Cauchy sequence in $\mathcal{X}$. Consequently, in all cases considered, the sequence retains the $\mathcal{R}$-preserving property. Given that $\mathcal{X}$ is $\mathcal{R}$-complete, there exists a point $\mathfrak{q}\in \mathcal{X}$ such that ${\mathfrak{x}}_n={\mathcal{T}}^n{\mathfrak{x}}_0\to \mathfrak{q}$ as $n\to +\infty$.

Finally, to establish that $\mathfrak{q}$ is a fixed point of $\mathcal{T}$, assumption (3) is employed. Assuming $\mathcal{T}$ is $\mathcal{R}$-continuous, and given that the sequence $\left\{{\mathfrak{x}}_k\right\}$ is $\mathcal{R}$-preserving with ${\mathfrak{x}}_k\to \mathfrak{q}$, the $\mathcal{R}$-continuity of $\mathcal{T}$ ensures that ${\mathfrak{x}}_{k+1}=\mathcal{T}\left({\mathfrak{x}}_k\right)\to \mathcal{T}\left(\mathfrak{q}\right)$. By the uniqueness of the limit, we conclude that $\mathcal{T}\left(\mathfrak{q}\right)=\mathfrak{q}$. Thus, $\mathfrak{q}$ is a fixed point of $\mathcal{T}$.

Alternatively, suppose that $\mathcal{R}$ is d-self-closed. Since the sequence $\left\{{\mathfrak{x}}_n\right\}$ is preserved under $\mathcal{R}$ and converges to $\mathfrak{q}$, ∃$\left\{{\mathfrak{x}}_{n_k}\right\}$ from $\left\{{\mathfrak{x}}_n\right\}$ for which, for all $n\in \mathbb{N}$, the pair $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]$ lies within $\mathcal{R}$. By invoking assumption (5) and Definition 1(1), we deduce that $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]\in \mathcal{R}$ and ${\mathfrak{x}}_{n_k}\to \mathfrak{q}$, which leads to the following conclusion:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d({\mathfrak{x}}_{n_{k+1}}, \mathcal{T}\mathfrak{q})\right]}^{\beta }+{\tau }_2{\left[d({\mathfrak{x}}_{n_{k+2}}, {\mathcal{T}}^2\mathfrak{q})\right]}^{\beta }& ={\tau }_1{\left[d(\mathcal{T}{\mathfrak{x}}_{n_k}, \mathcal{T}\mathfrak{q})\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^2{\mathfrak{x}}_{n_k}, {\mathcal{T}}^2\mathfrak{q})\right]}^{\beta }\hfill \\ & \le \mu {\left[d({\mathfrak{x}}_{n_k}, \mathfrak{q})\right]}^{\beta }.\hfill \end{array}$$

Taking the limit as $n\to \infty$, we obtain

$${\tau }_1{\left[d(\mathfrak{q}, \mathcal{T}\mathfrak{q})\right]}^{\beta }+{\tau }_2{\left[d(\mathfrak{q}, {\mathcal{T}}^2\mathfrak{q})\right]}^{\beta }\le 0.$$

This implies that $d(\mathfrak{q}, \mathcal{T}\mathfrak{q})=0$ and $d(\mathfrak{q}, {\mathcal{T}}^2\mathfrak{q})=0$. Therefore, $\mathcal{T}\mathfrak{q}=\mathfrak{q}$, and hence, $Fix\left(\mathcal{T}\right)\ne ⌀$. □

To ensure that the fixed point is unique, we impose stronger conditions on the contraction mapping as well as the associated relation. These supplementary constraints prevent the existence of multiple distinct fixed points, thereby guaranteeing that the fixed point set contains exactly one element, that is, it is a singleton.

Theorem 3.

If all the hypotheses outlined in Theorem 2 hold, and in addition, for any pair of elements $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$, there exists at least one path connecting them within the relation $\mathcal{R}$, that is,

$$\mathrm{Paths}(\mathfrak{x}, \mathfrak{y};\mathcal{R})\ne ⌀,$$

then the self-mapping $\mathcal{T}$ possesses exactly one fixed point.

Proof. 

The existence part of the result follows directly from Theorem 2. We now proceed to prove the uniqueness part. Suppose, for contradiction, that the operator $\mathcal{T}$ has two distinct fixed points, denoted by ${\mathfrak{x}}^{*}$ and ${\mathfrak{y}}^{*}$. Then, there exists a finite sequence (or path) $\{{\mathfrak{q}}_0, {\mathfrak{q}}_1, \dots , {\mathfrak{q}}_m\}\subseteq \mathcal{X}$ such that

$${\mathfrak{q}}_0={\mathfrak{x}}^{*}, {\mathfrak{q}}_m={\mathfrak{y}}^{*}, [{\mathfrak{q}}_i, {\mathfrak{q}}_{i+1}]\in \mathcal{R}\mathrm{for}\mathrm{each}i(0\le i\le m-1).$$

Since $\mathcal{R}$ is $\mathcal{T}$-closed, so by Definition 1(4), we obtain

$$[{\mathcal{T}}^k{\mathfrak{q}}_j, {\mathcal{T}}^k{\mathfrak{q}}_{j+1}]\in \mathcal{R}, \mathrm{for}\mathrm{all}k\in \mathbb{N}\cup \left\{0\right\}.$$

(10)

Utilizing the triangle inequality in conjunction with the contraction-type condition for all $k\in \mathbb{N}$, we get

$$\begin{array}{cc}\hfill {\tau }_1^k\left[d({\mathfrak{x}}^{*}, {\mathfrak{y}}^{*})\right]& ={\tau }_1^k\left[d({\mathcal{T}}^k{\mathfrak{x}}^{*}, {\mathcal{T}}^k{\mathfrak{y}}^{*})\right]\hfill \\ & \le {\tau }_1^k\sum _{j=0}^{m-1}\left[d({\mathcal{T}}^k{\mathfrak{q}}_j, {\mathcal{T}}^k{\mathfrak{q}}_{j+1})\right]\hfill \\ & \le {\tau }_1^k\sum _{j=0}^{m-1}{\left[d({\mathcal{T}}^k{\mathfrak{q}}_j, {\mathcal{T}}^k{\mathfrak{q}}_{j+1})\right]}^{\beta }+{\tau }_2\sum _{j=0}^{m-1}{\left[d({\mathcal{T}}^{k+1}{\mathfrak{q}}_j, {\mathcal{T}}^{k+1}{\mathfrak{q}}_{j+1})\right]}^{\beta }\hfill \\ & ={\tau }_1^{k-1}\sum _{j=0}^{m-1}\left\{{\tau }_1{\left[d({\mathcal{T}}^k{\mathfrak{q}}_j, {\mathcal{T}}^k{\mathfrak{q}}_{j+1})\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^{k+1}{\mathfrak{q}}_j, {\mathcal{T}}^{k+1}{\mathfrak{q}}_{j+1})\right]}^{\beta }\right\}\hfill \\ & \le {\tau }_1^{k-1}\sum _{j=0}^{m-1}\left\{\mu {\left[d\left({\mathcal{T}}^{k-1}{\mathfrak{q}}_j, {\mathcal{T}}^{k-1}{\mathfrak{q}}_{j+1}\right)\right]}^{\beta }\right\}\hfill \\ & =\mu {\tau }_1^{k-1}\sum _{j=0}^{m-1}{\left[d\left({\mathcal{T}}^{k-1}{\mathfrak{q}}_j, {\mathcal{T}}^{k-1}{\mathfrak{q}}_{j+1}\right)\right]}^{\beta }\hfill \\ & \le \mu {\tau }_1^{k-2}\sum _{j=0}^{m-1}\left\{{\tau }_1{\left[d\left({\mathcal{T}}^{k-1}{\mathfrak{q}}_j, {\mathcal{T}}^{k-1}{\mathfrak{q}}_{j+1}\right)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^k{\mathfrak{q}}_j, {\mathcal{T}}^k{\mathfrak{q}}_{j+1})\right]}^{\beta }\right\}\hfill \\ & \le \mu {\tau }_1^{k-2}\sum _{j=0}^{m-1}\mu {\left[d\left({\mathcal{T}}^{k-1}{\mathfrak{q}}_j, {\mathcal{T}}^{k-1}{\mathfrak{q}}_{j+1}\right)\right]}^{\beta }\hfill \\ & ={\mu }^2{\tau }_1^{k-2}\sum _{j=0}^{m-1}{\left[d\left({\mathcal{T}}^{k-1}{\mathfrak{q}}_j, {\mathcal{T}}^{k-1}{\mathfrak{q}}_{j+1}\right)\right]}^{\beta }\hfill \\ & ⋮\hfill \\ & \le {\mu }^k\sum _{j=0}^{m-1}{\left[d\left({\mathfrak{q}}_j, {\mathfrak{q}}_{j+1}\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow \left[d({\mathfrak{x}}^{*}, {\mathfrak{y}}^{*})\right]& \le {\left({\displaystyle \frac{\mu }{{\tau }_1}}\right)}^k·\sum _{j=0}^{m-1}{\left[d\left({\mathfrak{q}}_j, {\mathfrak{q}}_{j+1}\right)\right]}^{\beta }⟶0\mathrm{as}k\to \infty .\hfill \end{array}$$

Thus, we conclude that ${\mathfrak{x}}^{*}={\mathfrak{y}}^{*}$, which leads to a contradiction. Therefore, the set $Fix\left(\mathcal{T}\right)$ must be a singleton set. □

The following example is presented to illustrate the above result.

Example 4.

Let $\mathcal{X}=[0, 1]$ and define a metric d on $\mathcal{X}$ by $d(\mathfrak{x}, \mathfrak{y}):={|\mathfrak{x}-\mathfrak{y}|}^{\frac{1}{2}}$ for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$. Then $(\mathcal{X}, d)$ is a complete metric space. Consider the binary relation

$$\mathcal{R}:=\left\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}\in [0, \frac{1}{2}], \mathfrak{y}\in [0, 1]\right\}\cup \{(\mathfrak{y}, \mathfrak{x})\in {\mathcal{X}}^2:\mathfrak{y}\in [0, \frac{1}{2}], \mathfrak{x}\in [0, 1]\},$$

and define the mapping $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ by $\mathcal{T}\mathfrak{x}={\displaystyle \frac{1-{\mathfrak{x}}^2}{4}}$ for all $\mathfrak{x}\in \mathcal{X}.$ Then, we obtain ${\mathcal{T}}^2\mathfrak{x}={\displaystyle \frac{15+2{\mathfrak{x}}^2-{\mathfrak{x}}^4}{64}}$ for all $\mathfrak{x}\in \mathcal{X}.$ For all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have

$${\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^2=|\mathcal{T}\mathfrak{x}-\mathcal{T}\mathfrak{y}|={\displaystyle \frac{1}{4}}|{\mathfrak{x}}^2-{\mathfrak{y}}^2|\le {\displaystyle \frac{1}{2}}|\mathfrak{x}-\mathfrak{y}|={\displaystyle \frac{1}{2}}{\left[d(\mathfrak{x}, \mathfrak{y})\right]}^2,$$

and

$${\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^2=|{\mathcal{T}}^2\mathfrak{x}-{\mathcal{T}}^2\mathfrak{y}|\le {\displaystyle \frac{5}{16}}|\mathfrak{x}-\mathfrak{y}|={\displaystyle \frac{5}{16}}{\left[d(\mathfrak{x}, \mathfrak{y})\right]}^2.$$

We consider the multi-index $\tau =({\tau }_1, {\tau }_2)$, where ${\tau }_1={\tau }_2={\displaystyle \frac{1}{2}}$. Then, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we get

$${\tau }_1{\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^2+{\tau }_2{\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^2\le \mu {\left[d(\mathfrak{x}, \mathfrak{y})\right]}^2,$$

where

$$\mu =\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{16}}\right)={\displaystyle \frac{13}{16}}<\sigma :=min\{{\tau }_1, {\tau }_2\}={\displaystyle \frac{1}{2}}.$$

Moreover, there exists $0\in \mathcal{X}$ such that $(0, \mathcal{T}0)\in \mathcal{R}$, $\mathcal{R}$ is $\mathcal{T}$-closed and T is continuous. Hence, all the assumptions of Theorem 2 are satisfied, and therefore $Fix\left(\mathcal{T}\right)\ne ⌀.$

Furthermore, for any $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$, we have $\mathrm{Paths}(\mathfrak{x}, \mathfrak{y};\mathcal{R})\ne ⌀.$ Consequently, by Theorem 3, $Fix\left(\mathcal{T}\right)=\{\sqrt{5}-2\},$ which is a singleton set.

By considering the binary relation $\mathcal{R}$ on $\mathcal{X}$ as $\mathcal{R}={\mathcal{X}}^2$, we can deduce the main result of Khan et al. [29] as follows.

Corollary 1

([29]). Consider $(\mathcal{X}, d)$ as a complete metric space, and let $\mathcal{T}$ be a self-mapping on $\mathcal{X}$ that satisfies the conditions of a $(\tau , \beta )$-type contraction, where τ is a multi-index of length $k=2$ and $\mu +{\tau }_1<1$. In this case, the set of fixed points $Fix\left(\mathcal{T}\right)$ contains exactly one element.

Proof. 

Consider a multi-index $\tau =({\tau }_1, {\tau }_2)$ of length $k=2$, where ${\tau }_1+{\tau }_2=1$. Therefore, we can express ${\tau }_2$ as $1-{\tau }_1$. The condition $\mu +{\tau }_1<1$ implies that

$${\displaystyle \frac{\mu }{1-{\tau }_1}}={\displaystyle \frac{\mu }{{\tau }_2}}\le {\displaystyle \frac{\mu }{\delta }}<1.$$

From this, we deduce that $\mu <\delta$, where $\delta =min\{{\tau }_1, {\tau }_2\}$. Furthermore, all the other conditions of Theorem 3 are met, and thus, $Fix\left(\mathcal{T}\right)$ contains exactly one element. □

Next, we state and prove the fixed point result for a $(\tau , \beta , \mathcal{R})$-type Kannan contraction mapping, which is as follows:

Theorem 4.

Consider $(\mathcal{X}, d)$ as a metric space, where $\mathcal{R}$ represents a binary relation on $\mathcal{X}$, and $\mathcal{T}$ is a self-mapping of $\mathcal{X}$. Assume that the following conditions hold:

1. 

$(\mathcal{X}, d)$ is $\mathcal{R}$-complete;

2. 

$\mathcal{R}$ is $\mathcal{T}$-closed;

3. 

either $\mathcal{T}$ is $\mathcal{R}$-continuous or $\mathcal{R}$ is d-self-closed;

4. 

${\mathcal{R}}_{\mathcal{T}}$ is non-empty;

5. 

$\mathcal{T}$ is a $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contraction with constant $\mu <\delta ={\displaystyle \frac{1}{2}}min\{{\tau }_1, {\tau }_2\}$.

Then, $\mathcal{T}$ has a fixed point.

Proof. 

Using the same steps as in Theorem 2, we can define a Picard sequence $\left\{{\mathfrak{x}}_n\right\}$ such that $({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})\in \mathcal{R}$ and ${\mathfrak{x}}_n\ne {\mathfrak{x}}_{n+1}$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Next, assume $\xi =max\left\{d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right), d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right\}$, then we have $d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\le \xi$ and $d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\le \xi$. Now, we show that the sequence $\left\{d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)\right\}=\left\{d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\right\}$ is a monotone decreasing sequence of positive real numbers. As

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_n, \mathcal{T}{\mathfrak{x}}_{n+1}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_n, {\mathcal{T}}^2{\mathfrak{x}}_{n+1}\right)\right]}^{\beta }& \le \mu \left[{\left[d\left({\mathfrak{x}}_n, \mathcal{T}{\mathfrak{x}}_n\right)\right]}^{\beta }+{\left[d\left({\mathfrak{x}}_{n+1}, \mathcal{T}{\mathfrak{x}}_{n+1}\right)\right]}^{\beta }\right]\hfill \\ \hfill \Rightarrow {\tau }_1{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathfrak{x}}_{n+2}, {\mathfrak{x}}_{n+3}\right)\right]}^{\beta }& \le \mu \left[{\left[d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)\right]}^{\beta }+{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }\right].\hfill \end{array}$$

Therefore, we obtain

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathfrak{x}}_{n+2}, {\mathfrak{x}}_{n+3}\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)\right]}^{\beta }+{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }\right].\hfill \end{array}$$

If for some $n\in \mathbb{N}$, we have $d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)<d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)$, then

$${\tau }_1{\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }\le 2\mu {\left[d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\right]}^{\beta }\Rightarrow 1\le {\displaystyle \frac{2\mu }{{\tau }_1}}={\displaystyle \frac{\mu }{\frac{{\tau }_1}{2}}}\le {\displaystyle \frac{\mu }{\delta }}.$$

This leads to a contradiction, and hence we have $d\left({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2}\right)\le d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Thus, $\left\{d\left({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1}\right)\right\}=\left\{d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\right\}$ is a monotone decreasing sequence of positive real numbers. Setting:

$$\xi =d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\ge \dots \ge {\left[d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\right]}^{\beta }\ge {\left[d\left({\mathcal{T}}^{n+1}{\mathfrak{x}}_0, {\mathcal{T}}^{n+2}{\mathfrak{x}}_0\right)\right]}^{\beta }\ge \dots .$$

Now, as in Theorem 2, the following cases arise:

Case (i): If $2\mu <{\tau }_1$ and then define ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (7), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\le 2\mu \xi \hfill \\ \hfill \Rightarrow {\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{2\mu }{{\tau }_1}}{\xi }^{\beta }={\displaystyle \frac{\mu }{\frac{{\tau }_1}{2}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }={\left(\phi \xi \right)}^{\beta }\Rightarrow d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\le \phi \xi .\hfill \end{array}$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (7), we get

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\hfill \\ & \le 2\mu {\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }=2\mu {\left(\phi \xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)& \le {\displaystyle \frac{\mu }{\frac{{\tau }_1}{2}}}{\left(\phi \xi \right)}^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\left(\phi \xi \right)}^{\beta }={\phi }^2\xi .\hfill \end{array}$$

Additionally, from (7), we have

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\hfill \\ & \le 2\mu {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }=2\mu {\left({\phi }^2\xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)& \le {\phi }^3\xi .\hfill \end{array}$$

Similarly, we obtain the following:

$$d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\le {\phi }^4\xi .$$

In general, we obtain the following:

$$d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\le {\phi }^n\xi , \mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Therefore, $d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\to 0$ as $n\to +\infty$.

Case (ii): When $2\mu <{\tau }_2$ and setting ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (7), we have the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\le 2\mu \xi \hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{2\mu }{{\tau }_2}}{\xi }^{\beta }={\displaystyle \frac{\mu }{\frac{{\tau }_2}{2}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }={\left(\phi \xi \right)}^{\beta }.\hfill \end{array}$$

Therefore, we obtain the following:

$$d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\le \phi \xi .$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (5), we obtain

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\le 2\mu {\xi }^{\beta }\hfill \\ \hfill \Rightarrow {\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{\frac{{\tau }_2}{2}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }\Rightarrow d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\le \phi \xi \hfill \end{array}$$

and also

$$\begin{array}{cc}\hfill {\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0\right)\right]}^{\beta }+{\left[d\left({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0\right)\right]}^{\beta }\right]\le 2\mu {\xi }^{\beta }\hfill \\ \hfill \Rightarrow d\left({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0\right)& \le {\phi }^2\xi .\hfill \end{array}$$

Additionally, we get

$$d\left({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0\right)\le {\phi }^2\xi .$$

Further, we obtain the following: $d\left({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0\right)\le {\phi }^3\xi ,$ and $d\left({\mathcal{T}}^7{\mathfrak{x}}_0, {\mathcal{T}}^8{\mathfrak{x}}_0\right)\le {\phi }^3\xi .$ Thus, in general, it follows that

$$d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0\right)\le {\phi }^l\xi ,$$

whenever $m=2l$ or $m=2l+1$. Therefore, $d\left({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0\right)\to 0$ as $m\to \infty$.

Following the same reasoning as in Theorem 2, it can be established that, in all cases, the sequence $\left\{{\mathfrak{x}}_n\right\}$ is a $\mathcal{R}$-preserving Cauchy sequence in $\mathcal{X}$. Given that $\mathcal{X}$ is $\mathcal{R}$-complete, there exists a point $\mathfrak{q}\in \mathcal{X}$ such that ${\mathfrak{x}}_n={\mathcal{T}}^n{\mathfrak{x}}_0$ converges to $\mathfrak{q}$ as $n\to \infty$.

Using assumption (3), we show that $\mathfrak{q}$ is a fixed point of $\mathcal{T}$. Assume that $\mathcal{T}$ is $\mathcal{R}$-continuous. Since the sequence $\left\{{\mathfrak{x}}_k\right\}$ is an $\mathcal{R}$-preserving sequence and converges to $\mathfrak{q}$, the $\mathcal{R}$-continuity of $\mathcal{T}$ implies that ${\mathfrak{x}}_{k+1}=\mathcal{T}\left({\mathfrak{x}}_k\right)$ converges to $\mathcal{T}\left(\mathfrak{q}\right)$. Due to the uniqueness of the limit, it follows that $\mathcal{T}\left(\mathfrak{q}\right)=\mathfrak{q}$, which confirms that $\mathfrak{q}$ is a fixed point of $\mathcal{T}$.

Alternatively, assume that $\mathcal{R}$ is a d-self-closed binary relation. Given that the sequence $\left\{{\mathfrak{x}}_n\right\}$ is preserved by $\mathcal{R}$ and converges to $\mathfrak{q}$, there exists a subsequence $\left\{{\mathfrak{x}}_{n_k}\right\}$ from $\left\{{\mathfrak{x}}_n\right\}$ such that, for every $n\in \mathbb{N}$, the pair $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]$ is contained within $\mathcal{R}$. By invoking assumption (5) and Definition 1(1), we deduce that $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]\in \mathcal{R}$ and ${\mathfrak{x}}_{n_k}\to \mathfrak{q}$, which leads to the following conclusion:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathfrak{x}}_{n_{k+1}}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathfrak{x}}_{n_{k+1}}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathfrak{x}}_{n_{k+2}}, {\mathcal{T}}^2\mathfrak{q}\right)\right]}^{\beta }\hfill \\ & ={\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_{n_k}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_{n_k}, {\mathcal{T}}^2\mathfrak{q}\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathfrak{x}}_{n_k}, \mathcal{T}{\mathfrak{x}}_{n_k}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\right]\hfill \\ & =\mu \left[{\left[d\left({\mathfrak{x}}_{n_k}, {\mathfrak{x}}_{n_k+1}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\right].\hfill \end{array}$$

Taking the limit as $n\to \infty$, we obtain

$$\begin{array}{cc}\hfill {\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{{\tau }_1}}{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow \left(1-{\displaystyle \frac{\mu }{\delta }}\right){\left[d(\mathfrak{q}, \mathcal{T}\mathfrak{q})\right]}^{\beta }& \le 0.\hfill \end{array}$$

This implies that $\left(1-{\displaystyle \frac{\mu }{\delta }}\right){\left[d(\mathfrak{q}, \mathcal{T}\mathfrak{q})\right]}^{\beta }\le 0$, and consequently, $d(\mathfrak{q}, \mathcal{T}\mathfrak{q})=0$. Therefore, $Fix\left(\mathcal{T}\right)\ne ⌀$. □

To establish the uniqueness of the fixed point for a $(\tau , \beta , \mathcal{R})$-type Kannan contraction mapping, we additionally assume the following condition:

Theorem 5.

If all the conditions specified in Theorem 4 are satisfied, and for every pair of elements $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$, they are comparable with respect to the binary relation $\mathcal{R}$, then $Fix\left(\mathcal{T}\right)$ must be a singleton.

Proof. 

Let $\mathfrak{q}, {\mathfrak{q}}^{*}\in Fix\left(\mathcal{T}\right)$. Then $[\mathfrak{q}, {\mathfrak{q}}^{*}]\in \mathcal{R}$ and so using Definition 1(1) and inequality (5), we get

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left(\mathcal{T}\mathfrak{q}, \mathcal{T}{\mathfrak{q}}^{*}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2\mathfrak{q}, {\mathcal{T}}^2{\mathfrak{q}}^{*}\right)\right]}^{\beta }& \le \mu \left[{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\left[d\left({\mathfrak{q}}^{*}, \mathcal{T}{\mathfrak{q}}^{*}\right)\right]}^{\beta }\right]\hfill \\ \hfill \Rightarrow \left({\tau }_1+{\tau }_2\right){\left[d\left(\mathfrak{q}, {\mathfrak{q}}^{*}\right)\right]}^{\beta }& \le 0.\hfill \end{array}$$

This leads to the conclusion that ${\tau }_1+{\tau }_2=0$, which is a contradiction. Therefore, $Fix\left(\mathcal{T}\right)$ must be a singleton. □

Example 5.

Let $\mathcal{X}=[0, 1]$ with the usual metric $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$ and $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is defined by

$$\mathcal{T}\mathfrak{x}=\left\{\begin{array}{cc}{\displaystyle \frac{1}{5}}, \hfill & 0\le \mathfrak{x}\le {\displaystyle \frac{1}{2}}, \hfill \\ {\displaystyle \frac{1}{5}}(1-\mathfrak{x}), \hfill & {\displaystyle \frac{1}{2}}<\mathfrak{x}\le 1\hfill \end{array}\right.$$

and construct a binary relation $\mathcal{R}:=\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}, \mathfrak{y}\in [0, {\displaystyle \frac{1}{2}}]\}$. Notice that there exists ${\displaystyle \frac{1}{2}}\in \mathcal{X}$ such that $({\displaystyle \frac{1}{2}}, \mathcal{T}{\displaystyle \frac{1}{2}})=({\displaystyle \frac{1}{2}}, {\displaystyle \frac{1}{5}})\in \mathcal{R}$, also whenever $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, then $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})=({\displaystyle \frac{1}{5}}, {\displaystyle \frac{1}{5}})\in \mathcal{R}$ showing that the binary relation $\mathcal{R}$ is $\mathcal{T}$-closed.

Now for any $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$ and taking $\beta =5$, ${\tau }_1=1$, and ${\tau }_2={\displaystyle \frac{1}{2}}$, we get

$$|\mathcal{T}\mathfrak{x}-\mathcal{T}\mathfrak{y}|=\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}\right|=0$$

as well as

$$|{\mathcal{T}}^2\mathfrak{x}-{\mathcal{T}}^2\mathfrak{y}|=\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}\right|=0.$$

Therefore, we further obtain

$${\tau }_1{\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^5+{\tau }_2{\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^5=0.$$

Hence, the following inequality holds

$${\tau }_1{\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^{\beta }\le \mu \left[{\left[d\left(\mathfrak{x}, \mathcal{T}\mathfrak{x}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{y}, \mathcal{T}\mathfrak{y}\right)\right]}^{\beta }\right]$$

for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$ with $\beta =5$, ${\tau }_1={\displaystyle \frac{1}{2}}$, ${\tau }_2={\displaystyle \frac{1}{2}}$ and $\mu ={\displaystyle \frac{1}{5}}$. This shows that $\mathcal{T}$ is a $(\tau , 5, \mathcal{R})$-type symmetric Kannan contraction mapping of multi-index τ of length $k=2$ with $\mu ={\displaystyle \frac{1}{5}}<\delta ={\displaystyle \frac{1}{2}}min({\tau }_1, {\tau }_2)={\displaystyle \frac{1}{4}}$. Therefore, all the conditions of Theorem 4 are satisfied, which leads to the conclusion that $Fix\left(\mathcal{T}\right)\ne ⌀$.

In the following example, we modify the binary relation, and its impact can be clearly observed. This change leads to a different structure in the mapping behavior and fixed-point analysis.

Example 6.

Let $\mathcal{X}=[0, 1]$ with the usual metric $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$ and $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ be given as

$$\mathcal{T}\mathfrak{x}=\left\{\begin{array}{cc}{\displaystyle \frac{1}{5}}, \hfill & 0\le \mathfrak{x}\le {\displaystyle \frac{1}{2}}, \hfill \\ {\displaystyle \frac{1}{5}}(1-\mathfrak{x}), \hfill & {\displaystyle \frac{1}{2}}<\mathfrak{x}\le 1\hfill \end{array}\right..$$

We construct a binary relation $\mathcal{R}={\mathcal{R}}_1\cup {\mathcal{R}}_1^{-1}$, where ${\mathcal{R}}_1:=\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}\in [0, {\displaystyle \frac{1}{2}}]\&\mathfrak{y}\in [{\displaystyle \frac{1}{5}}, 1]\}$. Notice that there exists ${\displaystyle \frac{1}{2}}\in \mathcal{X}$ such that $({\displaystyle \frac{1}{2}}, \mathcal{T}{\displaystyle \frac{1}{2}})=({\displaystyle \frac{1}{2}}, {\displaystyle \frac{1}{5}})\in \mathcal{R}$, also whenever $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, then either $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\in {\mathcal{R}}_1$ or $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\in {\mathcal{R}}_1^{-1}$ which implies $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\in \mathcal{R}$, and it shows that the binary relation $\mathcal{R}$ is $\mathcal{T}$-closed.

Now, for $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have two cases, either $(\mathfrak{x}, \mathfrak{y})\in {\mathcal{R}}_1$ or $(\mathfrak{x}, \mathfrak{y})\in {\mathcal{R}}_1^{-1}.$ When $(\mathfrak{x}, \mathfrak{y})\in {\mathcal{R}}_1$, we discuss more subcases:

Subcase (1):  If $\mathfrak{x}\in \left[0, {\displaystyle \frac{1}{2}}\right]$ and $\mathfrak{y}\in \left({\displaystyle \frac{1}{2}}, 1\right]$, then $|\mathcal{T}\mathfrak{x}-\mathcal{T}\mathfrak{y}|=\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}(1-\mathfrak{y})\right|=\left|{\displaystyle \frac{1}{5}}\mathfrak{y}\right|$ and $\left|{\mathcal{T}}^2\mathfrak{x}-{\mathcal{T}}^2\mathfrak{y}\right|=\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}\right|=0$. So by taking $\beta =1$, ${\tau }_1={\tau }_2={\displaystyle \frac{1}{2}}$, we get

$${\tau }_{1}d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})+{\tau }_{2}d\left({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y}\right)={\displaystyle \frac{1}{2}}\left(\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}(1-\mathfrak{y})\right|+\left|{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{5}}\right|\right)={\displaystyle \frac{1}{2}}\left|{\displaystyle \frac{1}{5}}\mathfrak{y}\right|={\displaystyle \frac{1}{5}}\left|{\displaystyle \frac{\mathfrak{y}}{2}}\right|.$$

It can also be noticed that

$${\displaystyle \frac{6\mathfrak{y}-1}{5}}-{\displaystyle \frac{\mathfrak{y}}{2}}={\displaystyle \frac{7\mathfrak{y}-2}{10}}>0, \mathrm{for}\mathrm{all}\mathfrak{y}\in \left({\displaystyle \frac{1}{2}}, 1\right].$$

Therefore, we have

$$\begin{array}{cc}\hfill {\tau }_{1}d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})+{\tau }_{2}d\left({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y}\right)& ={\displaystyle \frac{1}{5}}\left|{\displaystyle \frac{\mathfrak{y}}{2}}\right|\hfill \\ & \le {\displaystyle \frac{1}{5}}\left[|\mathfrak{x}-\mathcal{T}\mathfrak{x}|+\left|{\displaystyle \frac{6\mathfrak{y}-1}{5}}\right|\right]\hfill \\ & ={\displaystyle \frac{1}{5}}\left[|\mathfrak{x}-\mathcal{T}\mathfrak{x}|+\left|\mathfrak{y}-{\displaystyle \frac{1}{5}}(1-\mathfrak{y})\right|\right]\hfill \\ & =\mu \left[d(\mathfrak{x}, \mathcal{T}\mathfrak{x})+d(\mathfrak{y}, \mathcal{T}\mathfrak{y})\right]\hfill \end{array}$$

for all $\mathfrak{x}\in \left[0, {\displaystyle \frac{1}{2}}\right]$ and $\mathfrak{y}\in \left({\displaystyle \frac{1}{2}}, 1\right]$, where $\mu ={\displaystyle \frac{1}{5}}$.

Subcase (2): If $\mathfrak{x}\in \left[0, {\displaystyle \frac{1}{2}}\right]$ and $\mathfrak{y}\in [{\displaystyle \frac{1}{5}}, {\displaystyle \frac{1}{2}}]$, then the above inequality holds trivially due to the constant function.

Similarly, due to the symmetric property of metric d, above inequality, that is, (5) also holds when $(\mathfrak{x}, \mathfrak{y})\in {\mathcal{R}}_1^{-1}.$

Hence, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we get

$${\tau }_1{\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y}\right)\right]}^{\beta }\le \mu \left[{\left[d(\mathfrak{x}, \mathcal{T}\mathfrak{x})\right]}^{\beta }+{\left[d(\mathfrak{y}, \mathcal{T}\mathfrak{y})\right]}^{\beta }\right],$$

where $\beta =1, \tau =({\tau }_1, {\tau }_2)=({\displaystyle \frac{1}{2}}, {\displaystyle \frac{1}{2}})$ and $\mu ={\displaystyle \frac{1}{5}}$.

This demonstrates that $\mathcal{T}$ is a $(\tau , 1, \mathcal{R})$-type symmetric Kannan contraction mapping with a multi-index τ of length $k=2$, where $\mu ={\displaystyle \frac{1}{5}}<\delta ={\displaystyle \frac{1}{2}}min\left\{{\tau }_1, {\tau }_2\right\}={\displaystyle \frac{1}{4}}$. Therefore, all the conditions of Theorem 4 are satisfied, and thus, $Fix\left(\mathcal{T}\right)\ne ⌀$. Furthermore, it is evident that every pair of elements in $\mathcal{X}$ is comparable. Therefore, by Theorem 5, the set $Fix\left(\mathcal{T}\right)$ contains exactly one element.

By taking the binary relation $\mathcal{R}$ on $\mathcal{X}$ as $\mathcal{R}={\mathcal{X}}^2$ in Theorem 5, we can derive the following result.

Corollary 2.

Let $(\mathcal{X}, d)$ be a complete metric space, and $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ be a $(\tau , \beta )$-type Kannan contraction with constant $\mu <\delta ={\displaystyle \frac{1}{2}}min\{{\tau }_1, {\tau }_2\}$. Then, $\mathcal{T}$ has a unique fixed point.

Theorem 6.

Consider $(\mathcal{X}, d)$ as a metric space, where $\mathcal{R}$ is a binary relation defined on $\mathcal{X}$, and $\mathcal{T}$ is a self-mapping on $\mathcal{X}$. Assume that the following conditions are met:

1. 

$(\mathcal{X}, d)$ is $\mathcal{R}$-complete;

2. 

$\mathcal{R}$ is $\mathcal{T}$-closed;

3. 

either $\mathcal{T}$ is $\mathcal{R}$-continuous or $\mathcal{R}$ is d-self-closed;

4. 

${\mathcal{R}}_{\mathcal{T}}$ is non-empty;

5. 

$\mathcal{T}$ is a $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contraction with constant $\mu <\delta ={\displaystyle \frac{1}{2^{\mathfrak{p}}}}min\{{\tau }_1, {\tau }_2\}$.

Then, $\mathcal{T}$ has a fixed point.

Proof. 

Using the same steps as in Theorem 2, we can define a Picard sequence $\left\{{\mathfrak{x}}_n\right\}$ such that $({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})\in \mathcal{R}$ and ${\mathfrak{x}}_n\ne {\mathfrak{x}}_{n+1}$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Next, assume $\xi =max\left\{d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right), d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\right\}$, then we have $d\left({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0\right)\le \xi$ and $d\left(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0\right)\le \xi$. We now demonstrate that $\left\{d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})\right\}$ forms a monotone decreasing sequence of positive real numbers. From inequality (8), we get

$${\tau }_1{\left[d(\mathcal{T}{\mathfrak{x}}_n, \mathcal{T}{\mathfrak{x}}_{n+1})\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^2{\mathfrak{x}}_n, {\mathcal{T}}^2{\mathfrak{x}}_{n+1})\right]}^{\beta }\le \mu {\left[d({\mathfrak{x}}_n, \mathcal{T}{\mathfrak{x}}_{n+1})\right]}^{\beta }+d({\mathfrak{x}}_{n+1}, \mathcal{T}{\mathfrak{x}}_n){]}^{\beta }$$

$$\Rightarrow {\tau }_1{\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }+{\tau }_2{\left[d({\mathfrak{x}}_{n+2}, {\mathfrak{x}}_{n+3})\right]}^{\beta }\le \mu {\left[d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+2})\right]}^{\beta }.$$

We obtain

$$\begin{array}{cc}\hfill {\tau }_1{\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }& \le {\tau }_1{\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }+{\tau }_2{\left[d({\mathfrak{x}}_{n+2}, {\mathfrak{x}}_{n+3})\right]}^{\beta }\hfill \\ & \le \mu {\left[d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+2})\right]}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }& \le {\displaystyle \frac{\mu }{{\tau }_1}}{\left[d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})+d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }.\hfill \end{array}$$

If for some $n\in \mathbb{N}$, we have $d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})<d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})$, then

$${\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }\le {\displaystyle \frac{2^{\beta }\mu }{{\tau }_1}}{\left[d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\right]}^{\beta }\Rightarrow 1\le {\displaystyle \frac{2^{\beta }\mu }{{\tau }_1}}={\displaystyle \frac{\mu }{\frac{{\tau }_1}{2^{\beta }}}}\le {\displaystyle \frac{\mu }{\delta }}.$$

This leads to a contradiction, and therefore, we have $d({\mathfrak{x}}_{n+1}, {\mathfrak{x}}_{n+2})\le d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Consequently, $\left\{d({\mathfrak{x}}_n, {\mathfrak{x}}_{n+1})\right\}=\left\{d({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0)\right\}$ is a monotone decreasing sequence of positive real numbers. By setting

$$\xi =d({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0)\ge \dots \ge {\left[d({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0)\right]}^{\beta }\ge {\left[d({\mathcal{T}}^{n+1}{\mathfrak{x}}_0, {\mathcal{T}}^{n+2}{\mathfrak{x}}_0)\right]}^{\beta }\ge \dots .$$

Proceeding as in Theorem 2, we consider the following distinct cases.

Case (i): If $2^{\beta }\mu <{\tau }_1$ and then by setting ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (8), we get

$$\begin{array}{cc}\hfill {\tau }_1{\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }& \le {\tau }_1{\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d({\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }+{\left[d(\mathcal{T}{\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0)\right]}^{\beta }\right]\hfill \\ & \le \mu {\left[d({\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0)+d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }& \le {\displaystyle \frac{2^{\beta }\mu }{{\tau }_1}}{\xi }^{\beta }={\displaystyle \frac{\mu }{\frac{{\tau }_1}{2^{\beta }}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }={\left(\phi \xi \right)}^{\beta }\Rightarrow d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\le \phi \xi .\hfill \end{array}$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (8), we obtain

$$\begin{array}{cc}\hfill {\tau }_1{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }& \le {\tau }_1{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }+{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }\right]\hfill \\ & \le \mu {\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)+d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{\frac{{\tau }_1}{2^{\beta }}}}{\left(\phi \xi \right)}^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\left(\phi \xi \right)}^{\beta }={\left({\phi }^2\xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)& \le {\phi }^2\xi .\hfill \end{array}$$

Similarly, we obtain the following:

$$d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\le {\phi }^3\xi , \mathrm{and}d({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0)\le {\phi }^4\xi .$$

In general, we can conclude that

$$d({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0)\le {\phi }^n\xi , \mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Therefore, $d\left({\mathcal{T}}^n{\mathfrak{x}}_0, {\mathcal{T}}^{n+1}{\mathfrak{x}}_0\right)\to 0$ as $n\to +\infty$. By following the same approach as in Theorem 2, we can establish that $\left\{{\mathfrak{x}}_n\right\}$ forms a $\mathcal{R}$-preserving Cauchy sequence.

Case (ii): When $2^{\beta }\mu <{\tau }_2$ and setting ${\phi }^{\beta }={\displaystyle \frac{\mu }{\delta }}<1$. Taking $\mathfrak{x}={\mathfrak{x}}_0$ and $\mathfrak{y}=\mathcal{T}{\mathfrak{x}}_0$ in (8), we have the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }& \le {\tau }_1{\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d({\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }+{\left[d(\mathcal{T}{\mathfrak{x}}_0, \mathcal{T}{\mathfrak{x}}_0)\right]}^{\beta }\right]\hfill \\ \hfill \Rightarrow {\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }& \le {\displaystyle \frac{2^{\beta }\mu }{{\tau }_2}}{\xi }^{\beta }={\displaystyle \frac{\mu }{\frac{{\tau }_2}{2^{\beta }}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }={\left(\phi \xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)& \le \phi \xi .\hfill \end{array}$$

Again, taking $\mathfrak{x}=\mathcal{T}{\mathfrak{x}}_0$ and $\mathfrak{y}={\mathcal{T}}^2{\mathfrak{x}}_0$ in (8), we obtain the following:

$$\begin{array}{cc}\hfill {\tau }_2{\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }& \le {\tau }_1{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d(\mathcal{T}{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }+{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^2{\mathfrak{x}}_0)\right]}^{\beta }\right]\hfill \\ \hfill \Rightarrow {\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{\frac{{\tau }_2}{2^{\beta }}}}{\xi }^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\xi }^{\beta }\hfill \\ \hfill \Rightarrow d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)& \le \phi \xi \hfill \end{array}$$

and also, we get

$$\begin{array}{cc}\hfill {\tau }_2{\left[d({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0)\right]}^{\beta }& \le {\tau }_1{\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }+{\tau }_2{\left[d({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d({\mathcal{T}}^2{\mathfrak{x}}_0, {\mathcal{T}}^4{\mathfrak{x}}_0)\right]}^{\beta }+{\left[d({\mathcal{T}}^3{\mathfrak{x}}_0, {\mathcal{T}}^3{\mathfrak{x}}_0)\right]}^{\beta }\right]\hfill \\ & \le 2^{\beta }\mu {\left(\phi \xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow {\left[d({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{\frac{{\tau }_2}{2^{\beta }}}}{\left(\phi \xi \right)}^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\left(\phi \xi \right)}^{\beta }={\left({\phi }^2\xi \right)}^{\beta }\hfill \\ \hfill \Rightarrow d({\mathcal{T}}^4{\mathfrak{x}}_0, {\mathcal{T}}^5{\mathfrak{x}}_0)& \le {\phi }^2\xi .\hfill \end{array}$$

Similarly, we can get

$$d({\mathcal{T}}^5{\mathfrak{x}}_0, {\mathcal{T}}^6{\mathfrak{x}}_0)\le {\phi }^2\xi \mathrm{and}d({\mathcal{T}}^6{\mathfrak{x}}_0, {\mathcal{T}}^7{\mathfrak{x}}_0)\le {\phi }^3\xi .$$

Therefore, in general, we can conclude that

$$d({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0)\le {\phi }^l\xi ,$$

whenever $m=2l$ or $m=2l+1$, it follows that $d({\mathcal{T}}^m{\mathfrak{x}}_0, {\mathcal{T}}^{m+1}{\mathfrak{x}}_0)\to 0$ as $m\to \infty$.

By applying the same method as in Theorem 2, it can be proven that, in all scenarios, the sequence $\left\{{\mathfrak{x}}_n\right\}$ forms a $\mathcal{R}$-preserving Cauchy sequence within $\mathcal{X}$. Given that $\mathcal{X}$ is $\mathcal{R}$-complete, there exists a point $\mathfrak{q}\in \mathcal{X}$ such that ${\mathfrak{x}}_n={\mathcal{T}}^n{\mathfrak{x}}_0\to \mathfrak{q}$ as $n\to +\infty$.

In conclusion, using assumption (3), we show that $\mathfrak{q}$ is a fixed point of $\mathcal{T}$. Assume that $\mathcal{T}$ is $\mathcal{R}$-continuous. Since the sequence $\left\{{\mathfrak{x}}_k\right\}$ is $\mathcal{R}$-preserving and converges to $\mathfrak{q}$, the $\mathcal{R}$-continuity of $\mathcal{T}$ ensures that ${\mathfrak{x}}_{k+1}=\mathcal{T}\left({\mathfrak{x}}_k\right)$ converges to $\mathcal{T}\left(\mathfrak{q}\right)$. Due to the uniqueness of the limit, we deduce that $\mathcal{T}\left(\mathfrak{q}\right)=\mathfrak{q}$, which means that $\mathfrak{q}$ is a fixed point of $\mathcal{T}$.

One the other hand, let us assume that $\mathcal{R}$ is d-self-closed. Since the sequence $\left\{{\mathfrak{x}}_n\right\}$ is preserved under $\mathcal{R}$ and converges to $\mathfrak{q}$, there exists a subsequence $\left\{{\mathfrak{x}}_{n_k}\right\}$ from $\left\{{\mathfrak{x}}_n\right\}$ such that for every $n\in \mathbb{N}$, the pair $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]$ lies within $\mathcal{R}$. By invoking assumption (5) and Definition 1(1), we deduce that $[{\mathfrak{x}}_{n_k}, \mathfrak{q}]\in \mathcal{R}$ and ${\mathfrak{x}}_{n_k}\to \mathfrak{q}$, which leads to the following conclusion:

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left({\mathfrak{x}}_{n_{k+1}}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }& \le {\tau }_1{\left[d\left({\mathfrak{x}}_{n_{k+1}}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathfrak{x}}_{n_{k+2}}, {\mathcal{T}}^2\mathfrak{q}\right)\right]}^{\beta }\hfill \\ & ={\tau }_1{\left[d\left(\mathcal{T}{\mathfrak{x}}_{n_k}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2{\mathfrak{x}}_{n_k}, {\mathcal{T}}^2\mathfrak{q}\right)\right]}^{\beta }\hfill \\ & \le \mu \left[{\left[d\left({\mathfrak{x}}_{n_k}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{q}, \mathcal{T}{\mathfrak{x}}_{n_k}\right)\right]}^{\beta }\right]\hfill \\ & =\mu \left[{\left[d\left({\mathfrak{x}}_{n_k}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }+{\left[d\left(\mathfrak{q}, {\mathfrak{x}}_{n_{k+1}}\right)\right]}^{\beta }\right].\hfill \end{array}$$

Taking the limit as $n\to \infty$, we obtain

$$\begin{array}{cc}\hfill {\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }& \le {\displaystyle \frac{\mu }{{\tau }_1}}{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\le {\displaystyle \frac{\mu }{\delta }}{\left[d\left(\mathfrak{q}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\hfill \\ \hfill \Rightarrow \left(1-{\displaystyle \frac{\mu }{\delta }}\right){\left[d(\mathfrak{q}, \mathcal{T}\mathfrak{q})\right]}^{\beta }& \le 0.\hfill \end{array}$$

This leads to $\left(1-{\displaystyle \frac{\mu }{\delta }}\right){\left[d(\mathfrak{q}, \mathcal{T}\mathfrak{q})\right]}^{\beta }\le 0$, and consequently, $d(\mathfrak{q}, \mathcal{T}\mathfrak{q})=0$. Therefore, $Fix\left(\mathcal{T}\right)\ne ⌀$. □

To establish the uniqueness of the fixed point for a $(\tau , \beta , \mathcal{R})$-type Chatterjea contraction mapping, we additionally assume the following condition:

Theorem 7.

If all the conditions of Theorem 6 are satisfied with ${\tau }_1-2\mu <1$, and every pair of elements $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$ is comparable under the relation $\mathcal{R}$, then the set of fixed points of $\mathcal{T}$, contains exactly one element.

Proof. 

Let $\mathfrak{q}, {\mathfrak{q}}^{*}\in Fix\left(\mathcal{T}\right)$. Then $[\mathfrak{q}, {\mathfrak{q}}^{*}]\in \mathcal{R}$ and so using Definition 1(1) and inequality (6), we get

$$\begin{array}{cc}\hfill {\tau }_1{\left[d\left(\mathcal{T}\mathfrak{q}, \mathcal{T}{\mathfrak{q}}^{*}\right)\right]}^{\beta }\le & {\tau }_1{\left[d\left(\mathcal{T}\mathfrak{q}, \mathcal{T}{\mathfrak{q}}^{*}\right)\right]}^{\beta }+{\tau }_2{\left[d\left({\mathcal{T}}^2\mathfrak{q}, {\mathcal{T}}^2{\mathfrak{q}}^{*}\right)\right]}^{\beta }\hfill \\ \hfill \le & \mu \left[{\left[d\left(\mathfrak{q}, \mathcal{T}{\mathfrak{q}}^{*}\right)\right]}^{\beta }+{\left[d\left({\mathfrak{q}}^{*}, \mathcal{T}\mathfrak{q}\right)\right]}^{\beta }\right]\hfill \\ \hfill \Rightarrow \left({\tau }_1-2\mu \right){\left[d\left(\mathfrak{q}, {\mathfrak{q}}^{*}\right)\right]}^{\beta }\le & 0.\hfill \end{array}$$

This implies that $d\left(\mathfrak{q}, {\mathfrak{q}}^{*}\right)=0$. Consequently, $Fix\left(\mathcal{T}\right)$ must consist of a single point. □

By considering the binary relation $\mathcal{R}$ on $\mathcal{X}$ as $\mathcal{R}={\mathcal{X}}^2$ in Theorem 7, we can deduce the following result.

Corollary 3.

Let $(\mathcal{X}, d)$ be a complete metric space, and $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ be a $(\tau , \beta )$-type Chatterjea contraction mapping with constant $\mu <\delta ={\displaystyle \frac{1}{2^{\mathfrak{p}}}}min\{{\tau }_1, {\tau }_2\}$. Then, $\mathcal{T}$ has a unique fixed point.

Example 7.

Let $\mathcal{X}=[0, 1]$ with the usual metric $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|$ and binary relation $\mathcal{R}=\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}\in [0, {\displaystyle \frac{1}{2}}]\&\mathfrak{y}\in [0, 1]\}$. Define $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ as $\mathcal{T}\left(\mathfrak{x}\right)={\displaystyle \frac{{\mathfrak{x}}^2}{5}}$, for all $\mathfrak{x}\in \mathcal{X}$. Then, we obtain ${\mathcal{T}}^2\left(\mathfrak{x}\right)={\displaystyle \frac{{\mathfrak{x}}^4}{125}}$.

Note that there exists ${\displaystyle \frac{1}{2}}\in \mathcal{X}$ such that $\left({\displaystyle \frac{1}{2}}, \mathcal{T}{\displaystyle \frac{1}{2}}\right)=\left({\displaystyle \frac{1}{2}}, {\displaystyle \frac{1}{20}}\right)\in \mathcal{R}$. Moreover, whenever $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, it follows that $(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\in \mathcal{R}$ as $\mathcal{T}\left([0, {\displaystyle \frac{1}{2}}]\right)\subseteq [0, {\displaystyle \frac{1}{2}}]$ and $\mathcal{T}\left([0, 1]\right)\subseteq [0, 1]$. This demonstrates that the binary relation $\mathcal{R}$ is $\mathcal{T}$-closed.

Next, for every $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have

$$|\mathcal{T}\left(\mathfrak{x}\right)-\mathcal{T}\left(\mathfrak{y}\right)|={\displaystyle \frac{1}{5}}|{\mathfrak{x}}^2-{\mathfrak{y}}^2|={\displaystyle \frac{\mathfrak{x}+\mathfrak{y}}{5}}|\mathfrak{x}-\mathfrak{y}|\le {\displaystyle \frac{3}{10}}|\mathfrak{x}-\mathfrak{y}|$$

and

$$\begin{array}{cc}\hfill |{\mathcal{T}}^2\left(\mathfrak{x}\right)-{\mathcal{T}}^2\left(\mathfrak{y}\right)|& =\left|{\displaystyle \frac{{\mathfrak{x}}^4}{125}}-{\displaystyle \frac{{\mathfrak{y}}^4}{125}}\right|\hfill \\ & ={\displaystyle \frac{({\mathfrak{x}}^2+{\mathfrak{y}}^2)(\mathfrak{x}+\mathfrak{y})}{125}}|\mathfrak{x}-\mathfrak{y}|\hfill \\ & \le {\displaystyle \frac{3}{200}}|\mathfrak{x}-\mathfrak{y}|.\hfill \end{array}$$

It can be also seen that for $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, that is $\mathfrak{x}\in [0, {\displaystyle \frac{1}{2}}]$ and $\mathfrak{y}\in [0, 1]$, we have ${\displaystyle \frac{1}{5}}(\mathfrak{x}+\mathfrak{y}+5)\ge 1$ with

$$\begin{array}{cc}\hfill |\mathfrak{x}-\mathfrak{y}|& \le \left|{\displaystyle \frac{1}{5}}(\mathfrak{x}+\mathfrak{y}+5)(\mathfrak{x}-\mathfrak{y})\right|\hfill \\ & =\left|{\displaystyle \frac{1}{5}}({\mathfrak{x}}^2-{\mathfrak{y}}^2)+(\mathfrak{x}-\mathfrak{y})\right|\hfill \\ & =\left|\left({\displaystyle \frac{1}{5}}{\mathfrak{x}}^2-\mathfrak{y}\right)+\left(\mathfrak{x}-{\displaystyle \frac{1}{5}}{\mathfrak{y}}^2\right)\right|\hfill \\ & \le {\left|{\displaystyle \frac{1}{5}}{\mathfrak{x}}^2-\mathfrak{y}\right|}^2+\left|\mathfrak{x}-{\displaystyle \frac{1}{5}}{\mathfrak{y}}^2\right|.\hfill \end{array}$$

By taking $\beta =1$ and $\tau =({\tau }_1, {\tau }_2)=({\displaystyle \frac{1}{2}}, {\displaystyle \frac{1}{2}})$, we get

$$\begin{array}{cc}\hfill {\tau }_1{|\mathcal{T}\left(\mathfrak{x}\right)-\mathcal{T}\left(\mathfrak{y}\right)|}^{\beta }+{\tau }_2{|{\mathcal{T}}^2\left(\mathfrak{x}\right)-{\mathcal{T}}^2\left(\mathfrak{y}\right)|}^{\beta }& \le {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{3}{10}}+{\displaystyle \frac{3}{200}}\right)|\mathfrak{x}-\mathfrak{y}|\hfill \\ & \le {\displaystyle \frac{63}{400}}\left[{|\mathfrak{x}-\mathcal{T}\left(\mathfrak{y}\right)|}^{\beta }+{|\mathfrak{y}-\mathcal{T}\left(\mathfrak{x}\right)|}^{\beta }\right].\hfill \end{array}$$

Hence, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we conclude that inequality (8) holds with $\mu ={\displaystyle \frac{63}{400}}=0.1575<\delta ={\displaystyle \frac{1}{2}}min\{{\tau }_1, {\tau }_2\}={\displaystyle \frac{1}{4}}=0.25$.

This demonstrates that $\mathcal{T}$ is a $(\tau , 1, \mathcal{R})$-type symmetric Chatterjea contraction mapping with a multi-index τ of length $k=2$. Therefore, all the conditions of Theorem 6 are satisfied, and thus, $Fix\left(\mathcal{T}\right)\ne ⌀$. Furthermore, it is evident that every pair of elements in $\mathcal{X}$ is comparable. Therefore, by Theorem 7, the set $Fix\left(\mathcal{T}\right)$ contains exactly one element.

4. Application to Nonlinear Fredholm Integral Equations

In this section, we utilize the results obtained earlier to establish the existence of a solution for a nonlinear Fredholm integral equation.

Let $\mathcal{X}:=\{f:[r, \mathfrak{e}]\to \mathbb{R}\mid f\mathrm{is}\mathrm{continuous}\},$ and d be a metric on $\mathcal{X}$ defined by $d(\mathfrak{x}, \mathfrak{y})=|\mathfrak{x}-\mathfrak{y}|={sup}_{t\in [r, \mathfrak{e}]}|\mathfrak{x}\left(t\right)-\mathfrak{y}\left(t\right)|$ for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$. Then, $(\mathcal{X}, d)$ is a complete metric space. We now consider the following nonlinear Fredholm integral equation

$$\mathfrak{x}\left(t\right)=f\left(t\right)+\beta {\int }_r^{\mathfrak{e}}\nabla (t, h, \mathfrak{x}\left(h\right))dh$$

(11)

where $t, h\in [r, \mathfrak{e}]$ and $\beta$ is a constant. Assume that $\nabla :[r, \mathfrak{e}]\times [r, \mathfrak{e}]\times \mathcal{X}\to \mathbb{R}$ and $f:[r, \mathfrak{e}]\to \mathbb{R}$ are continuous functions, where $\nabla (t, h, \mathfrak{x}(h\left)\right)$ is the kernel of the equation and f is a given function.

We also define a binary relation $\mathcal{R}$ over $\mathcal{X}$ as $\mathcal{R}:=\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}\left(t\right)\le \mathfrak{y}\left(t\right)$ for all $t\in [r, \mathfrak{e}]\}.$

Theorem 8.

Consider $(\mathcal{X}, d)$ to be a metric space with binary relation $\mathcal{R}$ as defined above. Let $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ be a nonlinear operator, defined for all $t\in [r, \mathfrak{e}]$ by

$$\mathcal{T}\mathfrak{x}\left(t\right)=f\left(t\right)+\beta {\int }_r^{\mathfrak{e}}\nabla (t, h, \mathfrak{x}\left(h\right))dh$$

(12)

for all $\mathfrak{x}\in \mathcal{X}$. Assume that the following conditions are satisfied:

1. 

there exist $\mu , \tau \in (0, 1)$, and $\beta \ge 1$ such that $\mu <\delta =min\{\tau , 1-\tau \}$ and

$$\begin{array}{cc}\hfill {\tau |\nabla (t, h, \mathfrak{x}\left(h\right))-\nabla (t, h, \mathfrak{y}\left(h\right))|}^{\beta }+& {(1-\tau )|\nabla (t, h, \mathcal{T}\mathfrak{x}\left(h\right))-\nabla (t, h, \mathcal{T}\mathfrak{y}\left(h\right))|}^{\beta }\hfill \\ & \le {\eta (t, h)|\mathfrak{x}\left(h\right)-\mathfrak{y}\left(h\right)|}^{\beta }forall(\mathfrak{x}, \mathfrak{y})\in \mathcal{R};\hfill \end{array}$$

2. 

where $\eta :[r, \mathfrak{e}]\times [r, \mathfrak{e}]\to \mathbb{R}$ is a continuous function satisfying

$$\underset{t\in [r, \mathfrak{e}]}{sup}{\int }_r^{\mathfrak{e}}\eta (t, h)dh\le {\displaystyle \frac{\mu }{{\beta }^{\beta }{(\mathfrak{e}-r)}^{\beta -1}}}.$$

Then, the Fredholm integral Equation (11) has a unique solution.

Proof. 

It is evident that $\mathcal{X}$ is a complete space, and by applying the definition of the binary relation $\mathcal{R}$ and the nonlinear operator $\mathcal{T}$, it is straightforward to conclude that ${\mathcal{R}}_{\mathcal{T}}\ne ⌀$. Furthermore, the binary relation $\mathcal{R}$ is $\mathcal{T}$-closed.

Next, we show that $\mathcal{T}$ is a $(\tau , \beta , \mathcal{R})$-type symmetric contraction mapping of multi-index $\tau$ of length $k=2$ for $\beta \ge 1$. It can be shown that $\mathcal{T}$ satisfies Equation (8) for $\beta =1$. Next, we demonstrate that $\mathcal{T}$ also satisfies Equation (6) for $\beta >1$. Given that $\beta >1$, there exists $\mathrm{w}>1$ such that ${\displaystyle \frac{1}{\mathrm{w}}}+{\displaystyle \frac{1}{\beta }}=1$. Then, by applying Hölder’s inequality, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we obtain

$$\begin{array}{cc}& {\tau |\mathcal{T}\mathfrak{x}\left(t\right)-\mathcal{T}\mathfrak{y}\left(t\right)|}^{\beta }+(1-\tau ){\left|{\mathcal{T}}^2\mathfrak{x}\left(t\right)-{\mathcal{T}}^2\mathfrak{y}\left(t\right)\right|}^{\beta }\hfill \\ & =\tau {\beta }^{\beta }{\left|{\int }_r^{\mathfrak{e}}\nabla (t, h, \mathfrak{x}\left(h\right))dh-{\int }_r^{\mathfrak{e}}\nabla (t, h, \mathfrak{y}\left(h\right))dh\right|}^{\beta }\hfill \\ & +(1-\tau ){\beta }^{\beta }{\left|{\int }_r^{\mathfrak{e}}\nabla (t, h, \mathcal{T}\mathfrak{x}\left(h\right))dh-{\int }_r^{\mathfrak{e}}\nabla (t, h, \mathcal{T}\mathfrak{y}\left(h\right))dh\right|}^{\beta }\hfill \\ & \le \tau {\beta }^{\beta }{\left[{\int }_r^{\mathfrak{e}}\left|\nabla (t, h, \mathfrak{x}(h\left)\right)-\nabla (t, h, \mathfrak{y}(h\left)\right)\right|dh\right]}^{\beta }\hfill \\ & +(1-\tau ){\beta }^{\beta }{\left[{\int }_r^{\mathfrak{e}}\left|\nabla (t, h, \mathcal{T}\mathfrak{x}(h\left)\right)-\nabla (t, h, \mathcal{T}\mathfrak{y}(h\left)\right)\right|dh\right]}^{\beta }\hfill \\ & \le \tau {\beta }^{\beta }{\left[{\left\{{\int }_r^{\mathfrak{e}}{1}^{\mathrm{w}}dh\right\}}^{{\displaystyle \frac{1}{\mathrm{w}}}}{\left\{{\int }_r^{\mathfrak{e}}{\left|\nabla (t, h, \mathfrak{x}(h\left)\right)-\nabla (t, h, \mathfrak{y}(h\left)\right)\right|}^{\beta }dh\right\}}^{{\displaystyle \frac{1}{\beta }}}\right]}^{\beta }\hfill \\ & +(1-\tau ){\beta }^{\beta }{\left[{\left\{{\int }_r^{\mathfrak{e}}{1}^{\mathrm{w}}dh\right\}}^{{\displaystyle \frac{1}{\mathrm{w}}}}{\left\{{\int }_r^{\mathfrak{e}}{\left|\nabla (t, h, \mathcal{T}\mathfrak{x}(h\left)\right)-\nabla (t, h, \mathcal{T}\mathfrak{y}(h\left)\right)\right|}^{\beta }dh\right\}}^{{\displaystyle \frac{1}{\beta }}}\right]}^{\beta }\hfill \\ & \le \tau {\beta }^{\beta }{(\mathfrak{e}-r)}^{{\displaystyle \frac{\beta }{\mathrm{w}}}}{\int }_r^{\mathfrak{e}}{\left|\nabla (t, h, \mathfrak{x}(h\left)\right)-\nabla (t, h, \mathfrak{y}(h\left)\right)\right|}^{\beta }dh\hfill \\ & +(1-\tau ){\beta }^{\beta }{(\mathfrak{e}-r)}^{{\displaystyle \frac{\beta }{\mathrm{w}}}}{\int }_r^{\mathfrak{e}}{\left|\nabla (t, h, \mathcal{T}\mathfrak{x}(h\left)\right)-\nabla (t, h, \mathcal{T}\mathfrak{y}(h\left)\right)\right|}^{\beta }dh\hfill \\ & ={(\mathfrak{e}-r)}^{\beta -1}{\beta }^{\beta }\left[{\int }_r^{\mathfrak{e}}\left\{\tau {\left|\nabla (t, h, \mathfrak{x}(h\left)\right)-\nabla (t, h, \mathfrak{y}(h\left)\right)\right|}^{\beta }\right.\right.\hfill \\ & \left.\left.+(1-\tau ){\left|\nabla (t, h, \mathcal{T}\mathfrak{x}(h\left)\right)-\nabla (t, h, \mathcal{T}\mathfrak{y}(h\left)\right)\right|}^{\beta }\right\}dh\right]\hfill \\ & \le {(\mathfrak{e}-r)}^{\beta -1}{\beta }^{\beta }\left[{\int }_r^{\mathfrak{e}}\eta (t, h){|\mathfrak{x}\left(h\right)-\mathfrak{y}\left(h\right)|}^{\beta }dh\right].\hfill \end{array}$$

By taking the supremum on both sides of the inequality over the interval $[r, \mathfrak{e}]$, we get:

$$\begin{array}{cc}& {\tau d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})}^{\beta }+(1-\tau ){d(\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2{\mathfrak{y})}^{\beta }\hfill \\ \hfill =& \underset{t\in [r, \mathfrak{e}]}{sup}\left[{\tau |\mathcal{T}\mathfrak{x}\left(t\right)-\mathcal{T}\mathfrak{y}\left(t\right)|}^{\beta }+(1-\tau ){\left|{\mathcal{T}}^2\mathfrak{x}\left(t\right)-{\mathcal{T}}^2\mathfrak{y}\left(t\right)\right|}^{\beta }\right]\hfill \\ \hfill \le & {(\mathfrak{e}-r)}^{\beta -1}{\beta }^{\beta }\left(\underset{t\in [r, \mathfrak{e}]}{sup}{\int }_r^{\mathfrak{e}}\eta (t, \mathfrak{u})d\mathfrak{u}\right){\left[d(\mathfrak{x}, \mathfrak{y})\right]}^{\beta }\hfill \\ \hfill \le & \mu {\left[d(\mathfrak{x}, \mathfrak{y})\right]}^{\beta }\hfill \end{array}$$

where $\mu <\delta =min\{\tau , 1-\tau \}$. This demonstrates that $\mathcal{T}$ satisfies the inequality (6), meaning that $\mathcal{T}$ is a $(\tau , \beta , \mathcal{R})$-type symmetric contraction mapping with a multi-index $\tau$ of length $k=2$. Therefore, all the conditions of Theorems 2 and 3 are met, implying that the integral operator $\mathcal{T}$ defined by (11) possesses a unique solution in $\mathcal{X}$. □

Example 8.

Let $r=0$ and $\mathfrak{e}=1$, and consider the nonlinear Fredholm integral equation

$$\mathfrak{x}\left(t\right)=t^2+{\displaystyle \frac{1}{2}}{\int }_0^{1}th\mathfrak{x}\left(h\right)dhforallt\in [0, 1].$$

(13)

Let $\mathcal{X}=\{f:[0, 1]\to \mathbb{R}\mid f\mathit{is}\mathit{continuous}\},$ endowed with the metric $d(\mathfrak{x}, \mathfrak{y})={sup}_{t\in [0,1]}\left|\mathfrak{x}\right(t)-\mathfrak{y}(t\left)\right|$ for all $\mathfrak{x}, \mathfrak{y}\in \mathcal{X}$. Then $(\mathcal{X}, d)$ is a complete metric space. Define the kernel $\nabla (t, h, z)=thzforallt, h\in [0, 1], \&z\in \mathcal{X}$ together with $f\left(t\right)=t^{2}forallt\in [0, 1],$ and let $\beta =1$. The associated integral operator $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ is given by

$$\left(\mathcal{T}\mathfrak{x}\right)\left(t\right)=t^2+{\displaystyle \frac{1}{2}}{\int }_0^{1}th\mathfrak{x}\left(h\right)dhforallt\in [0, 1],$$

for all $\mathfrak{x}\in \mathcal{X}$. Define the binary relation:

$$\mathcal{R}=\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2:\mathfrak{x}\left(t\right)\le \mathfrak{y}\left(t\right)forallt\in [0, 1]\}.$$

We now verify that the kernel ∇ satisfies the contractive condition required in Theorem 8. Fix $\tau ={\displaystyle \frac{1}{2}}$ and $\beta =1$. For all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$ and for all $t, h\in [0, 1]$, we have

$$\begin{array}{cc}& \tau |\nabla (t, h, \mathfrak{x}\left(h\right))-\nabla (t, h, \mathfrak{y}\left(h\right)\left)\right|+(1-\tau )|\nabla (t, h, \mathcal{T}\mathfrak{x}\left(h\right))-\nabla (t, h, \mathcal{T}\mathfrak{y}\left(h\right)\left)\right|\hfill \\ & ={\displaystyle \frac{1}{2}}th|\mathfrak{x}\left(h\right)-\mathfrak{y}\left(h\right)|+{\displaystyle \frac{1}{2}}th|\mathcal{T}\mathfrak{x}\left(h\right)-\mathcal{T}\mathfrak{y}\left(h\right)|\hfill \\ & \le {\displaystyle \frac{1}{2}}th\left(\left|\mathfrak{x}\right(h)-\mathfrak{y}(h\left)\right|\right).\hfill \end{array}$$

Thus, condition (1) holds with $\eta (t, h)={\displaystyle \frac{1}{2}}th$ for all $t, h\in [0, 1]$. Moreover, for condition (2), we have

$$\underset{t\in [0,1]}{sup}{\int }_0^1\eta (t, h)dh=\underset{t\in [0,1]}{sup}{\displaystyle \frac{t}{4}}={\displaystyle \frac{1}{4}}.$$

Choose $\mu ={\displaystyle \frac{1}{4}}$. Then

$$\underset{t\in [0,1]}{sup}{\int }_0^1\eta (t, h)dh={\displaystyle \frac{1}{4}}\le {\displaystyle \frac{\mu }{{\beta }^{\beta }{(\mathfrak{e}-r)}^{\beta -1}}}={\displaystyle \frac{1}{4}},$$

and

$$\mu ={\displaystyle \frac{1}{4}}<\delta =min\{\tau , 1-\tau \}={\displaystyle \frac{1}{2}}.$$

Hence, all assumptions of Theorem 8 are satisfied; consequently, problem (13) has a unique solution.

5. Application to Economic Growth Modeling via Fractional Differential Equations

Fractional differential equations (FDEs) have emerged as powerful tools in modern applied mathematics, particularly due to their capability to incorporate memory and hereditary properties into dynamical systems. In economic contexts, such memory effects are crucial, as past investments, policy decisions, or shocks can influence current economic outcomes. Caputo-type derivatives, which enable the formulation of initial conditions in the same form as classical differential equations, are particularly suited to modeling economic growth trajectories over time.

We consider the following nonlinear fractional differential equation of Caputo type:

$${}^{C}D_t^e\left(q\left(t\right)\right)=\mathfrak{b}(t, q\left(t\right)),(0<t<1, 1<e\le 2),$$

(14)

with boundary conditions:

$$q\left(0\right)=0,{}^{RL}I_t^{e}q\left(1\right)=q^{\prime }\left(0\right),$$

(15)

where ${}^{C}D_t^e$ denotes the Caputo derivative of order e, defined as follows:

$${}^{C}D_t^e\mathfrak{b}\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }(k-e)}}{\int }_0^t{(t-\psi )}^{k-e-1}{\mathfrak{b}}^{\left(k\right)}\left(\psi \right)d\psi ,$$

(16)

with $k-1<e<k$, where $k=\left[e\right]+1$, the function $\mathfrak{b}:[0, 1]\times \mathbb{R}\to [0, +\infty )$ is continuous. The Riemann–Liouville fractional integral of a continuous function f is defined as follows:

$${}^{RL}I_t^{e}f\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}f\left(\psi \right)d\psi .$$

(17)

The economic variable $q\left(t\right)$ can represent the gross domestic product (GDP ) or any indicator reflecting economic performance. The nonlinear kernel $\mathfrak{b}(t, q(t\left)\right)$ embodies the influences of key economic drivers such as education, investment, and policy. The fractional order e captures memory-dependent dynamics, acknowledging that current economic states are shaped by past behavior.

Define an operator $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ by:

$$\mathcal{T}\mathfrak{x}\left(t\right):={\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}\mathfrak{b}(\psi , \mathfrak{x}\left(\psi \right))d\psi +{\displaystyle \frac{2t}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^1\left({\int }_0^{\psi }{(\psi -l)}^{e-1}\mathfrak{b}(l, \mathfrak{x}\left(l\right))dl\right)d\psi .$$

(18)

Notice that the solution of the fractional differential Equation (14) is precisely the fixed point of the operator $\mathcal{T}$. That is, a function $q\in \mathcal{X}$ satisfies the boundary value problem if and only if it fulfills the operator equation:

$$\mathcal{T}q=q.$$

Hence, solving the fractional model reduces to identifying the fixed point of the operator $\mathcal{T}$ within the complete metric space $(\mathcal{X}, d)$.

Let $\mathcal{X}:=\{\psi :[0, 1]\to \mathbb{R}\mid \psi \mathrm{is}\mathrm{continuous}\}$ be equipped with the uniform norm $\parallel \varphi \parallel ={max}_{\psi \in [0,1]}\left|\varphi \left(\psi \right)\right|.$ Then $(\mathcal{X}, \parallel ·\parallel )$ is a complete metric space, where the metric is induced by the norm

$$d({\varphi }_1, {\varphi }_2)=\parallel {\varphi }_1-{\varphi }_2\parallel \mathrm{for}\mathrm{all}{\varphi }_1, {\varphi }_2\in \mathcal{X}.$$

Define a binary relation $\mathcal{R}\subseteq \mathcal{X}\times \mathcal{X}$ by the following:

$$\mathcal{R}:=\left\{(\mathfrak{x}, \mathfrak{y})\in {\mathcal{X}}^2\mid \mathfrak{x}\left(t\right)\le \mathfrak{y}\left(t\right), \mathrm{for}\mathrm{all}t\in [0, 1]\right\}.$$

Theorem 9.

Consider the nonlinear fractional differential Equation (14). Suppose the following conditions are satisfied:

1. 

For constants $\tau \in (0, 1), \beta \ge 1$, and a continuous function $\eta :{[0, 1]}^2\to {\mathbb{R}}_{+}$, the inequality

$$\begin{array}{c}\hfill \tau {\left|\mathfrak{b}(\psi , \mathfrak{x}(\psi \left)\right)-\mathfrak{b}(\psi , \mathfrak{y}(\psi \left)\right)\right|}^{\beta }+(1-\tau ){\left|\mathfrak{b}(\psi , \mathcal{T}\mathfrak{x}(\psi \left)\right)-\mathfrak{b}(\psi , \mathcal{T}\mathfrak{y}(\psi \left)\right)\right|}^{\beta }\\ \hfill \le \eta (t, \psi )\left({|\mathfrak{x}\left(\psi \right)-\mathfrak{y}\left(\psi \right)|}^{\beta }\right)\end{array}$$

holds for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$.

2. 

There exists $\mu >0$, such that

$$\underset{t\in [0,1]}{sup}{\int }_0^1\eta (t, \psi )d\psi \le {\displaystyle \frac{\mu }{{\left(\frac{2}{\mathrm{\Gamma }\left(e\right)}\right)}^{\beta }}}, \mu <min\{\tau , 1-\tau \}.$$

Then Equation (14) has a unique solution.

Proof. 

By taking the binary relation $\mathcal{R}$ and the self-mapping $\mathcal{T}$ as given above, it is clear that the binary relation $\mathcal{R}$ is $\mathcal{T}$-closed, $\mathcal{T}$ is continuous, and the set ${\mathcal{R}}_{\mathcal{T}}$ is also nonempty; that is, there exists some $\mathfrak{z}\in \mathcal{X}$ such that $(\mathfrak{z}, \mathcal{T}\mathfrak{z})\in \mathcal{R}$.

Next, we show that $\mathcal{T}$ is a $(\tau , 1-\tau , \mathcal{R})$-type symmetric contraction mapping, that is, for all $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$, we have

$$\tau {\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+(1-\tau ){\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^{\beta }\le \mu {\left(d(\mathfrak{x}, \mathfrak{y})\right)}^{\beta }.$$

We begin from the left-hand side. For any pair $(\mathfrak{x}, \mathfrak{y})\in \mathcal{R}$ and fixed $t\in [0, 1]$, define the following:

$$\begin{array}{ccc}\hfill I\left(t\right)& :=& {\tau |\mathcal{T}\mathfrak{x}\left(t\right)-\mathcal{T}\mathfrak{y}\left(t\right)|}^{\beta }+(1-\tau ){|{\mathcal{T}}^2\mathfrak{x}\left(t\right)-{\mathcal{T}}^2\mathfrak{y}\left(t\right)|}^{\beta }.\hfill \end{array}$$

From the definition of $\mathcal{T}$, we write the following:

$$\begin{array}{ccc}\hfill \mathcal{T}\mathfrak{x}\left(t\right)-\mathcal{T}\mathfrak{y}\left(t\right)& =& {\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}[\mathfrak{b}(\psi , \mathfrak{x}\left(\psi \right))-\mathfrak{b}(\psi , \mathfrak{y}\left(\psi \right))]d\psi \hfill \\ & & +{\displaystyle \frac{2t}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^1{\int }_0^{\psi }{(\psi -l)}^{e-1}[\mathfrak{b}(l, \mathfrak{x}\left(l\right))-\mathfrak{b}(l, \mathfrak{y}\left(l\right))]dld\psi .\hfill \end{array}$$

We now estimate each term using the triangle inequality to get

$$\begin{array}{ccc}\hfill \left|\mathcal{T}\mathfrak{x}\right(t)-\mathcal{T}\mathfrak{y}(t\left)\right|& \le & {\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}|\mathfrak{b}(\psi , \mathfrak{x}\left(\psi \right))-\mathfrak{b}(\psi , \mathfrak{y}\left(\psi \right))|d\psi \hfill \\ & & +{\displaystyle \frac{2t}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^1{\int }_0^{\psi }{(\psi -l)}^{e-1}|\mathfrak{b}(l, \mathfrak{x}\left(l\right))-\mathfrak{b}(l, \mathfrak{y}\left(l\right))|dld\psi .\hfill \end{array}$$

Now, applying Hölder’s inequality (with exponents $p=1$, $q=1$ for simplicity since all terms are nonnegative and the integral bounds are finite), we raise both sides to the power $\beta$ and apply ${(a+b)}^{\beta }\le 2^{\beta -1}(a^{\beta }+b^{\beta }), \mathrm{for}a, b\ge 0, \beta \ge 1.$ Thus,

$$\begin{array}{ccc}\hfill {|\mathcal{T}\mathfrak{x}\left(t\right)-\mathcal{T}\mathfrak{y}\left(t\right)|}^{\beta }& \le & 2^{\beta -1}\left[{\left({\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}|\mathfrak{b}(\psi , \mathfrak{x}\left(\psi \right))-\mathfrak{b}(\psi , \mathfrak{y}\left(\psi \right))|d\psi \right)}^{\beta }\right.\hfill \\ & & \left.+{\left({\displaystyle \frac{2t}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^1{\int }_0^{\psi }{(\psi -l)}^{e-1}|\mathfrak{b}(l, \mathfrak{x}\left(l\right))-\mathfrak{b}(l, \mathfrak{y}\left(l\right))|dld\psi \right)}^{\beta }\right].\hfill \end{array}$$

A similar estimate holds for $|{\mathcal{T}}^2\mathfrak{x}\left(t\right)-{\mathcal{T}}^2{\mathfrak{y}\left(t\right)|}^{\beta }$ with $\mathfrak{x}, \mathfrak{y}$ replaced by $\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y}$ inside $\mathfrak{b}$. Now, apply assumption (1) to each kernel difference:

$$\begin{array}{ccc}\hfill I\left(t\right)& \le & {\left({\displaystyle \frac{2}{\mathrm{\Gamma }\left(e\right)}}\right)}^{\beta }{\int }_0^1\eta (t, \psi )\left\{{|\mathfrak{x}\left(\psi \right)-\mathfrak{y}\left(\psi \right)|}^{\beta }\right\}d\psi .\hfill \end{array}$$

Taking the supremum over $t\in [0, 1]$ gives:

$$\begin{array}{ccc}& & \tau {\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+(1-\tau ){\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^{\beta }\hfill \\ & =& \underset{t\in [0,1]}{sup}I\left(t\right)\hfill \\ & \le & {\left({\displaystyle \frac{2}{\mathrm{\Gamma }\left(e\right)}}\right)}^{\beta }\underset{t\in [0,1]}{sup}{\int }_0^1\eta (t, \psi )\left\{{|\mathfrak{x}\left(\psi \right)-\mathfrak{y}\left(\psi \right)|}^{\beta }\right\}d\psi .\hfill \end{array}$$

Finally, by assumption (2), we obtain

$$\underset{t\in [0,1]}{sup}{\int }_0^1\eta (t, \psi )d\psi \le {\displaystyle \frac{\mu }{{\left(\frac{2}{\mathrm{\Gamma }\left(e\right)}\right)}^{\beta }}},$$

which yields

$$\tau {\left[d(\mathcal{T}\mathfrak{x}, \mathcal{T}\mathfrak{y})\right]}^{\beta }+(1-\tau ){\left[d({\mathcal{T}}^2\mathfrak{x}, {\mathcal{T}}^2\mathfrak{y})\right]}^{\beta }\le \mu {\left(\left[d\right(\mathfrak{x}, \mathfrak{y}\left)\right]\right)}^{\beta }.$$

Hence, all assumptions of Theorem 2 are satisfied. Therefore, the operator $\mathcal{T}$ admits a fixed point in $\mathcal{X}$. Moreover, since any two elements of the space $\mathcal{X}$ are $\mathcal{R}$-comparable, it follows from Theorem 3 that the set of fixed points is a singleton. Equivalently, the fractional differential Equation (14) has a unique solution. □

Example 9.

Consider the nonlinear fractional differential equation of Caputo type

$${}^{C}D_t^{e}q\left(t\right)=\mathfrak{b}(t, q\left(t\right))forallt\in (0, 1), 1<e\le 2,$$

(19)

subject to the boundary conditions $q\left(0\right)=0,$${}^{RL}I_t^{e}q\left(1\right)=q^{\prime }\left(0\right).$ Let the nonlinear term be defined by $\mathfrak{b}(t, q)={\displaystyle \frac{1}{4}}qforallt\in [0, 1], q\in \mathbb{R}.$ Clearly, the function $\mathfrak{b}:[0, 1]\times \mathbb{R}\to [0, \infty )$ is continuous. Let $\mathcal{X}=\{\psi :[0, 1]\to \mathbb{R}\mid \psi$ is continuous} be endowed with the usual metric and define the binary relation as $\mathcal{R}=\{(x, y)\in {\mathcal{X}}^2:x\left(t\right)\le y\left(t\right)forallt\in [0, 1]\}$. Define the operator $\mathcal{T}:\mathcal{X}\to \mathcal{X}$ by

$$\left(\mathcal{T}x\right)\left(t\right)={\displaystyle \frac{1}{\mathrm{\Gamma }\left(e\right)}}{\int }_0^t{(t-\psi )}^{e-1}{\displaystyle \frac{1}{4}}x\left(\psi \right)d\psi forallt\in [0, 1].$$

Fix the constants $\beta =1,$ and $\tau ={\displaystyle \frac{1}{2}}.$ Then, for any $(x, y)\in \mathcal{R}$ and $\psi \in [0, 1]$, we have

$$|\mathfrak{b}(\psi , x\left(\psi \right))-\mathfrak{b}(\psi , y\left(\psi \right))|={\displaystyle \frac{1}{4}}|x\left(\psi \right)-y\left(\psi \right)|\le {\displaystyle \frac{1}{4}}d(x, y).$$

Moreover, for all $(x, y)\in \mathcal{R}$ and $\psi \in [0, 1]$, we have

$$|\mathfrak{b}(\psi , \mathcal{T}x\left(\psi \right))-\mathfrak{b}(\psi , \mathcal{T}y\left(\psi \right))|\le {\displaystyle \frac{1}{4}}\parallel \mathcal{T}x-\mathcal{T}y\parallel \le {\displaystyle \frac{1}{4\mathrm{\Gamma }(e+1)}}d(x, y).$$

Hence,

$$\begin{array}{cc}& \tau \left|\mathfrak{b}\right(\psi , x\left(\psi \right))-\mathfrak{b}(\psi , y\left(\psi \right)\left)\right|+(1-\tau )\left|\mathfrak{b}\right(\psi , \mathcal{T}x\left(\psi \right))-\mathfrak{b}(\psi , \mathcal{T}y\left(\psi \right)\left)\right|\hfill \\ & \le \left({\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{32\mathrm{\Gamma }(e+1)}}\right)d(x, y).\hfill \end{array}$$

Define $\eta (t, \psi )={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{32\mathrm{\Gamma }(e+1)}}$$forallt, \psi \in [0, 1].$ Then η is continuous and ${sup}_{t\in [0,1]}{\int }_0^1\eta (t, \psi )$$d\psi ={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{32\mathrm{\Gamma }(e+1)}}.$ Choose $\mu ={\displaystyle \frac{1}{5}}$. Since ${\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{32\mathrm{\Gamma }(e+1)}}<{\displaystyle \frac{\mu }{\left(\frac{2}{\mathrm{\Gamma }\left(e\right)}\right)}},$ and $\mu <min\{\tau , 1-\tau \}={\displaystyle \frac{1}{2}},$ all the hypotheses of Theorem 9 are satisfied. Therefore, the operator $\mathcal{T}$ admits a unique fixed point in $\mathcal{X}$. Consequently, the nonlinear fractional boundary value problem (19) has a unique continuous solution on $[0, 1]$.

6. Conclusions and Future Directions

In this study, we explore the existence and uniqueness of solutions for fractional differential equations using a class of generalized symmetric operators. Our findings highlight that this approach provides a meaningful and effective way to analyze systems with memory-dependent dynamics. This work also introduces and analyzes generalized classes of mappings, including $(\tau , \beta , \mathcal{R})$-type symmetric Lipschitz mappings, $(\tau , \beta , \mathcal{R})$-type symmetric Kannan contractions, and $(\tau , \beta , \mathcal{R})$-type symmetric Chatterjea contractions, within the context of metric spaces connected with any binary relation. By enhancing the classical Lipschitz condition, we develop a comprehensive framework that broadens the scope of traditional fixed point results. Our approach also explores the impact of symmetric and comparable relations, providing new fixed point theorems and applying these findings to prove the existence of solutions to Fredholm integral equations. The provided examples demonstrate the practical applicability and importance of our results.

A numerical investigation of the proposed results would be a valuable extension of this work and could further enrich the theoretical analysis. Another promising direction for future research is the development of algorithms based on the newly defined contraction mappings, leading to convergence results via suitable iterative schemes. Moreover, it is interesting to consider interpolative-type contractive conditions involving Kannan, Chatterjea, or other interpolative contractions introduced by Karapınar, Aydi, and other authors [30,31,32], by incorporating them into the present relation-theoretic and multi-index setting. Such an investigation may yield new fixed-point results that unify interpolative contractions with the generalized contractive conditions developed in this work.