New Contractive Mappings and Solutions to Boundary-Value Problems in Triple Controlled Metric Type Spaces

Abstract

In this study, we utilize the notion of triple controlled metric type space that preserves the symmetry property, which is a generalization of b-metric-type spaces, to prove new fixed-point results. We introduce ($\alpha$-$\mathcal{F}$)-contractive mappings and $\Theta$-contractive mappings on triple controlled metric type space settings. Then, we establish the existence and uniqueness of fixed-point results on complete triple controlled metric type space. Moreover, some examples and applications to boundary-value problems of the fourth-order differential equation are presented to display the usage of the obtained result.

1. Introduction

Primary fixed-point theory comprises the Banach contractive principle, which was proved in 1922 in a metric space setting [1]. Today, the fixed-point theory is a fast-growing and exciting field of mathematics with various applications in diverse areas of mathematics. A Banach contractive principle is a handy tool in nonlinear analysis, which led to the appearance of numerous expansions of the Banach Theorem in various directions. Among those directions are new types of metric spaces. As a result, several researchers have generalized the structure of the metric space; for instance, Bakhtin [2] proposed the b-metric space, which was then developed into extended b-metric spaces [3], controlled metric-type spaces [4], and double-controlled metric-type spaces [5]. On the other hand, Branciari [6] presented the idea of generalized metric space (or rectangular metric space) in which a weaker assumption called quadrilateral inequality $d(a,b)\le d(a,u)+d(u,v)+d(v,b),$ for all pairwise distinct points $a,b,u,v\in W,$ is used to substitute the property of the triangular inequality, and many fixed-point results were established on such spaces. Rectangular metric spaces can lack the Hausdorffness separation property (consult [7,8,9,10] for examples). As a result, the topological structure of the rectangular metric space is incompatible with the topological structure of ordinary metric spaces. Thus, it developed into an intriguing area and drew various researchers to work on a such metric. For more information on fixed-point theory in rectangular metric spaces, we refer the reader to [7,8,9,10,11,12]. The rectangular metric was extended into rectangular b-metric space [13,14]; controlled b-Branciari metric-type space [15]; and recently, triple controlled metric type space [16,17].

Numerous researchers have investigated the fixed-point theory in a variety of metric-type spaces under various contractive circumstances. For instance, Wardowski, in 2012, [18] introduced a new contractive mapping later denoted as $\mathcal{F}$-contractive, while Samet et al. [12] proposed the class of $\alpha$-admissible mappings in metric spaces. The new concept of $\alpha$-type $\mathcal{F}$-contractive mappings, which are essentially weaker than the class of $\mathcal{F}$-contractive mappings as in [18], was presented in 2016 by Gopal et al. [19]. Few authors have investigated fixed-point theorems for $(\alpha$-$\mathcal{F}$)-contractive on some complete metric spaces [19,20,21,22]; moreover, a new Wardowski-type fixed-point result was illustrated in [23]. The concept of $\Theta$-contractive was introduced by Jleli [24], and they established a generalization of the Banach fixed-point theorem in the situation of Branciari metric spaces.

In this article, we discuss two contractive mappings: the ($\alpha$-$\mathcal{F}$)-contractive mappings on triple controlled metric type space, and $\Theta$-contractive mappings. The existence and uniqueness of fixed-point results in a complete triple controlled metric type space are then established. We also provide some examples of our findings and provide a solution for a fourth-order differential-equation boundary-value problem. Finally, we make some suggestions for potential future research areas.

It deserves to be noted that by using various contraction mappings other than those we describe in this article, [16,17] have established fixed-point results in triple controlled metric type space.

2. Preliminaries

In 2000, Brainciari [6] introduced the notion of rectangular metric spaces as follows:

Definition 1.

Let X be a nonempty set, and let $d:X\times X\to [0,\infty )$ be a mapping such that for all $\widehat{x},\widehat{z}\in X$ and all distinct points $a,b\in X,$ it satisfies the following:

1. 

$d(\widehat{x},\widehat{z})=0⟺\widehat{x}=\widehat{z}$;

2. 

$d(\widehat{x},\widehat{z})=d(\widehat{z},\widehat{x})$, the symmetry condition;

3. 

$d(\widehat{x},\widehat{z})\le d(\widehat{x},a)+d(a,b)+d(b,\widehat{z}).$

The pair $(X,d)$ is called a rectangular metric space.

The rectangular metric space that had a topological structure that was incompatible with the topological structure of ordinary metric spaces became interesting and drew the attention of numerous researchers to work on such a metric [11]. It was extended into a rectangular b-metric space [13,14].

Definition 2.

Let X be a nonempty set, $s\ge 1$ be a given real number, and $d:X\times X\to [0,\infty )$ be a mapping, such that for all $\widehat{x},\widehat{z}\in X$ and all distinct points $a,b\in X$, each different from $\widehat{x},\widehat{z}$:

1. 

$d(\widehat{x},\widehat{z})=0⟺\widehat{x}=\widehat{z}$, for all $\widehat{x},\widehat{z}\in X$;

2. 

$d(\widehat{x},\widehat{z})=d(\widehat{z},\widehat{x}),$ a symmetry condition for all $\widehat{x},\widehat{z}\in X$;

3. 

$d(\widehat{x},\widehat{z})\le s[d(\widehat{x},a)+d(a,b)+d(b,\widehat{z})].$

The pair $(X,d)$ is called a b-rectangular metric space.

In 2020, a controlled b-Branciari metric-type space was introduced [15].

Definition 3.

Let X be a nonempty set, and let $\beta :X\times X\to [1,\infty )$ be a function. A mapping $d:X\times X\to [0,\infty )$ is called a controlled b-Branciari metric type if it satisfies the following:

1. 

$d(\widehat{x},\widehat{z})=0⟺\widehat{x}=\widehat{z}$, for all $\widehat{x},\widehat{z}\in X$;

2. 

$d(\widehat{x},\widehat{z})=d(\widehat{z},\widehat{x}),$ a symmetry condition for all $\widehat{x},\widehat{z}\in X$;

3. 

$d(\widehat{x},\widehat{z})\le \beta (\widehat{x},a)d(\widehat{x},a)+\beta (a,b)d(a,b)+\beta (b,\widehat{z})d(b,\widehat{z}).$

For all $\widehat{x},\widehat{z}\in X$, and for all distinct points $a,b\in X$, each different from $\widehat{x},\widehat{z}$. The pair $(X,d)$ is called a controlled b-Branciari metric-type space that preserves the symmetry condition.

The controlled b-Branciari metric-type space was expanded to a triple controlled metric type space as follows; for details, consult [16,17].

Definition 4

([16]). Let Z be a nonempty set and consider the functions $\beta ,\mu ,\gamma :Z\times Z\to [1,\infty )$. A mapping $d_{Tri}:Z\times Z\to [0,\infty )$ is called a triple controlled metric type if it satisfies the following:

($q1$) 

$d_{Tri}(z_1,z_2)=0$ if and only if $z_1=z_2,$ for all $z_1,z_2\in Z$;

($q2$) 

$d_{Tri}(z_1,z_2)=d_{Tri}(z_2,z_1),$ a symmetry condition for all $z_1,z_2\in Z$;

($q3$) 

$d_{Tri}(z_1,z_2)\le \beta (z_1,z_3)d_{Tri}(z_1,z_3)+\mu (z_3,z_4)d_{Tri}(z_3,z_4)+\gamma (z_4,z_2)d_{Tri}(z_4,z_2)$.

For all $z_1,z_2\in Z$ and for all distinct points $z_3,z_4\in Z,$ each distinct from $z_1$ and $z_2$. The pair $(Z,d_{Tri})$ is called a triple controlled metric type space (in short it will be denoted by $\mathcal{TCMTS}$).

Remark 1.

In Definition 4, if we have only two functions $\beta ,\mu :Z\times Z\to [1,\infty )$ satisfying $\left(q1\right)$ and $\left(q2\right)$, while $\left(q3\right)$ is written as $d_{Tri}(z_1,z_2)\le \beta (z_1,z_3)d_{Tri}(z_1,z_3)+\mu (z_3,z_2)d_{Tri}(z_3,z_2)$, for all $z_1,z_2\in Z,$ and $z_3\in Z$ distinct from $z_1$ and $z_2$. Then, the pair $(Z,d_{Tri})$ is referred to as a double controlled metric type space. The example below illustrates that $\mathcal{TCMTS}$ is a generalization of double controlled metric type space and controlled b-Branciari metric-type space.

Example 1

([16]). Let $A=X\cup Y$, where $X=\{\frac{1}{n}:n\in \mathbb{N}\}$ and Y is the set of all positive integers. Define the symmetric mapping $d_{Tri}:A\times A\to [0,\infty )$ by

$$d_{Tri}(x,y)=\left\{\begin{array}{cc}0\hfill & iffx=y,\hfill \\ x+10\hfill & ifx\in X,y\in \{6,9\},orx\in \{6,9\},y\in X,\hfill \\ 2\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Let $\beta ,\mu ,\gamma :A\times A\to [1,\infty )$ be defined as,

$$\beta (x,y)=\left\{\begin{array}{cc}\frac{1}{x}\hfill & \mathit{if}x\in X,y\in Y,\hfill \\ 1\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

$$\mu (x,y)=\left\{\begin{array}{cc}x+y\hfill & \mathit{if}x\in X,y\in Y,\hfill \\ 2\hfill & \mathit{otherwise},\hfill \end{array}\right.$$

and

$$\gamma (x,y)=\left\{\begin{array}{cc}x+1\hfill & \mathit{if}\mathit{both}x,y\in X,\mathit{or}x,y\in Y,\hfill \\ \frac{3}{2}\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

One can easily show that $d_{Tri}$ is a $\mathcal{TCMTS}$. Observe that

$$d_{Tri}(\frac{1}{3},9)=\frac{31}{3}>\beta (\frac{1}{3},4)d_{Tri}(\frac{1}{3},4)+\mu (4,9)d_{Tri}(4,9)=10.$$

Hence, the $\mathcal{TCMTS}$ is not a double controlled metric type space; furthermore,

$$d_{Tri}(\frac{1}{3},9)=\frac{31}{3}>\beta (\frac{1}{3},4)d_{Tri}(\frac{1}{3},4)+\beta (4,1)d_{Tri}(4,1)+\beta (1,9)d_{Tri}(1,9)=10.$$

Thus, the $\mathcal{TCMTS}$ is not a controlled b-Branciari metric-type space.

The concept of convergent and Cauchy sequences in $\mathcal{TCMTS}$; completeness; and open balls are mentioned below.

Definition 5.

Let $(Z,d_{Tri})$ be a $\mathcal{TCMTS}$, and let $\left\{z_n\right\}$ be any sequence in Z.

(1) 

Let $z\in Z$ and $\epsilon >0$. The open ball $B(z,\epsilon )$ is defined as

$$B(z,\epsilon )=\{w\in Z,d_{Tri}(z,w)<\epsilon \}.$$

(2) 

The mapping $T:Z\to Z$ is said to be continuous at $z\in Z,$ if for all $\epsilon >0$, $\delta >0$ exists such that $T\left(B\right(z,\delta \left)\right)\subseteq B(Tz,\epsilon )$.

(3) 

A sequence $\left\{z_n\right\}$ in Z is said to converge to some w in $Z,$ if for every $\epsilon >0$, there exists $N=N\left(\epsilon \right)\in \mathbb{N},$ such that $d_{Tri}(z_n,w)<\epsilon$ for all $n\ge N.$ In this case, we write ${lim}_{n\to \infty }{z}_n=w.$ Thus, the continuty of the mapping T at z can be defined in terms of sequences, if $z_n\to z$, then $T{z}_n\to Tz$ as $n\to \infty$.

(4) 

We say $\left\{z_n\right\}$ is a Cauchy sequence, if for every $\epsilon >0$ there exists $N=N\left(ϵ\right)\in \mathbb{N}$ such that $d_{Tri}(z_m,z_n)<\epsilon$ for all $m,n\ge N.$

(5) 

The space $(Z,d_{Tri})$ is called complete if every Cauchy sequence in Z is convergent.

Remark 2.

Since the b-rectangular metric space topology is different than the standard metric space topology, the example below was constructed in [25], see also [13,26]. It illustrates some features in b-rectangular metric space that are not present in the standard metric space.

Example 2.

Let $A=\{0,2\}$, $B=\{\frac{1}{n}:n\in \mathbb{N}\}$, and $X=A\cup B$. Define $\rho :X\times X\to [0,\infty )$ as follows:

$$\rho (x,y)=\left\{\begin{array}{cc}0,\hfill & x=y,\hfill \\ 1\hfill & x\ne y,\mathit{and}\{x,y\}\subset A\mathit{or}\{x,y\}\subset B,\hfill \\ y,\hfill & x\in A,y\in B,\hfill \\ x,\hfill & x\in B,y\in A.\hfill \end{array}\right.$$

Let $d(x,y)=\rho {(x,y)}^2$ and $s=3$; then, $(X,d)$ is a b-rectangular metric space. Observe the sequence $\left\{\frac{1}{n}\right\}$ converges to both 0 and 2. For details, consult [13,25].

The example shows that a sequence in b-rectangular metric space can have two limits. However, there is a special situation, where this is not possible, and this will be useful in proving our main theorems regarding the fixed point. The following lemma is a variant of a lemma in [27]; see also [13,26] for details. We need it in the sequel.

Lemma 1.

Let $(X,d)$ be a b-rectangular metric space, and let $\left\{x_n\right\}$ be a Cauchy sequence in X such that $x_m\ne x_n$ whenever $m\ne n$. Then, $\left\{x_n\right\}$ can converge to at most one point.

This article discusses two different kinds of contractive mappings: ($\alpha$-$\mathcal{F}$)-contractive mappings and $\Theta$-contractive mappings on triple controlled metric type space, to establish fixed-point results. The pertinent preliminaries and definitions are therefore supplied.

The class of $\alpha$-admissible mappings was first introduced by Samet et al. [12]; for further information, see [28].

Definition 6.

Let Z be a nonempty set, $T:Z⟶Z$ be a given mapping, and $\alpha :Z\times Z\to [0,\infty )$. We say T is an α-admissible if whenever $\alpha (\widehat{x},\widehat{y})\ge 1$ implies $\alpha (T\widehat{x},T\widehat{y})\ge 1,$ for all $\widehat{x},\widehat{y}\in Z$.

Example 3

([12]). Let $Z=[0,\infty )$. Consider the mappings $T:Z⟶Z$, and $\alpha :Z\times Z\to [0,\infty )$ defined by $T\left(\widehat{z}\right)=\sqrt{\widehat{z}},$ for all $\widehat{z}\in Z$; and $\alpha (\widehat{z},\widehat{w})=e^{\widehat{z}-\widehat{w}}$ for $\widehat{z}\ge \widehat{w}$, and $\alpha (\widehat{z},\widehat{w})=0$ otherwise. Then, T is α-admissible.

Wardowski [18] introduced a new type of contraction called $\mathcal{F}$-contraction and established some new related fixed-point theorems in the context of complete metric spaces.

Definition 7.

Let $\mathcal{F}$ be the family of all functions $F:(0,\infty )\to (-\infty ,\infty )$ satisfying the following:

($F1$) 

F is a strictly increasing function.

($F2$) 

For each sequence $\left\{s_n\right\}$ of postive real numbers, this holds;

$$\underset{n\to \infty }{lim}{s}_n=0\iff \underset{n\to \infty }{lim}F\left(s_n\right)=-\infty .$$

($F3$) 

There exists $k\in (0,1)\ni {lim}_{s\to 0^{+}}{s}^{k}F\left(s\right)=0$.

Example 4.

Let $L\left(s\right)=\frac{-1}{\sqrt{s}}$, and $T\left(s\right)=ln\left(s\right)$, for $s>0$. Then, clearly both $L\left(s\right)$ and $T\left(s\right)$ satisfy the conditions $\left(F1\right),\left(F2\right)$, and $\left(F3\right)$. Hence, they belong to $\mathcal{F}$; for more details, consult [18].

For the $\Theta$-contractive mappings, the following definition will be required; for more information, consult [24].

Definition 8

([24]). Let Θ be the set of all functions $\theta :(0,\infty )\to (1,\infty )$ satisfying the following conditions:

($\theta 1$) 

θ is nondecreasing.

($\theta 2$) 

For each sequence $\left\{t_m\right\}$ of positive real numbers, this holds;

$$\underset{m\to \infty }{lim}{t}_m=0^{+}\iff \underset{m\to \infty }{lim}\theta \left(t_m\right)=1.$$

($\theta 3$) 

There exists k, with $0<k<1$, and an $M\in (0,\infty ]$, such that this holds

$$\underset{t\to 0^{+}}{lim}\frac{\theta \left(t\right)-1}{t^k}=M.$$

3. Main Results

This section consists of two subsections, each of which discusses the fixed point outcomes for a specific contractive mapping on $\mathcal{TCMTS}$.

3.1. The ($\alpha$-$\mathcal{F}$)-Contractive Mappings and Fixed-Point Results

Definition 9.

Let $(Z,d_{Tri})$ be a $\mathcal{TCMTS},$ where Z is a nonempty set. A self mapping $T:Z\to Z$ is said to be an $\mathcal{F}$-contractive if there exists a function $F\in$$\mathcal{F}$ and a constant $\tau >0$ such that this holds:

$$d_{Tri}(Tx,Ty)>0\Rightarrow \tau +F\left(d_{Tri}(Tx,Ty)\right)\le F\left(d_{Tri}(x,y)\right),\mathit{for}\mathit{all}x,y\in Z.$$

(1)

On $\mathcal{TCMTS}$, the ($\alpha$-$\mathcal{F}$)-contractive mappings are defined as follows.

Definition 10.

Let $(Z,d_{Tri})$ be a $\mathcal{TCMTS}$, where Z is a nonempty set. A self-mapping $T:Z\to Z$ is said to be an (α-$\mathcal{F}$)-contractive mapping, if there exists a mapping $\alpha :Z\times Z\to [0,\infty )$, $F\in$$\mathcal{F}$, and a constant $\tau >0$, such that this holds;

$$\begin{array}{c}\hfill \tau +\alpha (x,y)F\left(d_{Tri}(Tx,Ty)\right)\le F\left(d_{Tri}(x,y)\right),\end{array}$$

(2)

for all $x,y\in Z$, with $d_{Tri}(Tx,Ty)>0.$

The first main fixed-point result is shown below.

Theorem 1.

Let $(Z,d_{Tri})$ be a complete $\mathcal{TCMTS}$, and let $T:Z\to Z$ be an (α-$\mathcal{F}$)-contractive mapping, such that the following holds:

(1) 

T is α-admissible.

(2) 

There exists $z_0\in Z$, such that $\alpha (z_0,T{z}_0)\ge 1$.

(3) 

T is continuous,

(4) 

For $z_0\in Z$, define the sequence $\left\{z_n\right\}$ by $z_n=T^n{z}_0$, and assume these hold

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta (z_{2i+2},z_{2i+3})}{\beta (z_{2i},z_{2i+1})}\gamma (z_{2i+2},z_m)<1,$$

(3)

and

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\mu (z_{2i+3},z_{2i+4})}{\mu (z_{2i+1},z_{2i+2})}\gamma (z_{2i+2},z_m)<1.$$

(4)

In addition, for each $z\in Z$

$$\underset{n\to \infty }{lim}\beta (z,z_n),\underset{n\to \infty }{lim}\mu (z_n,z)\mathit{and}\underset{n\to \infty }{lim}\gamma (z_n,z)\mathit{exists}\mathit{and}\mathit{finite}.$$

(5)

Then, T has a fixed point. Moreover, if for any two fixed points of T in Z, say ξ, and η with $\alpha (\xi ,\eta )\ge 1$, then T has a unique fixed point in $Z.$

Proof. 

Choose $z_0\in Z$ such that $\alpha (z_0,T{z}_0)\ge 1$. Define the sequence $\left\{z_n\right\}$ by letting $T{z}_0=z_1,T^2{z}_0=T{z}_1=z_2$. Thus, for any $n\in \mathbb{N}$,

$$T^n{z}_0=T^{n-1}{z}_1=\cdots =T{z}_{n-1}=z_n.$$

Note that if n exists such that $z_n=z_{n+1}$, then we are done and $z_n$ is the fixed point of T. Therefore, we may assume that $z_n\ne z_{n+1}$ for all $n\ge 0$.

The map T being $\alpha$-admissible, we deduce $\alpha (z_n,z_{n+1})\ge 1$, for all $n\in \mathbb{N}$. As T is ($\alpha$-$\mathcal{F}$)-contractive mapping, we have

$$\begin{array}{ccc}\hfill \tau +F\left(d_{Tri}(z_n,z_{n+1})\right)& =& \tau +F\left(d_{Tri}(T{z}_{n-1},z_{n+1})\right).\hfill \\ & \le & \tau +\alpha (z_n,z_{n+1})F\left(d_{Tri}(T{z}_{n-1},T{z}_n)\right).\hfill \\ & \le & F\left(d_{Tri}(z_{n-1},z_n)\right),\hfill \end{array}$$

which implies

$$\begin{array}{ccc}\hfill F\left(d_{Tri}(z_n,z_{n+1})\right)& \le & F\left(d_{Tri}(z_{n-1},z_n)\right)-\tau .\hfill \\ \hfill & \le & F\left(d_{Tri}(z_{n-2},z_{n-1})\right)-2\tau .\hfill \\ \hfill & \le & \cdots \le F\left(d_{Tri}(z_0,z_1)\right)-n\tau .\hfill \end{array}$$

(6)

Letting $n\to \infty$ in (6), and with $\tau >0$, we have

$$\underset{n\to \infty }{lim}F\left(d_{Tri}(z_n,z_{n+1})\right)=-\infty .$$

(7)

Since $F\in$$\mathcal{F}$, thus by (F2) of Definition 7, it follows that ${lim}_{n\to \infty }{d}_{Tri}(z_n,z_{n+1})=0$, and by (F3), there exists $k\in (0,1)$ such that

$$\underset{n\to \infty }{lim}{\left(d_{Tri}(z_n,z_{n+1})\right)}^{k}F\left(d_{Tri}(z_n,z_{n+1})\right)=0.$$

(8)

From (6), we obtain

$$F\left(d_{Tri}(z_n,z_{n+1})\right)-F\left(d_{Tri}(z_0,z_1)\right)\le -n\tau .$$

This implies for any n

$$\begin{array}{c}{\left(d_{Tri}(z_n,z_{n+1})\right)}^{k}F\left(d_{Tri}(z_n,z_{n+1})\right)-{\left(d_{Tri}(z_n,z_{n+1})\right)}^{k}F\left(d_{Tri}(z_0,z_1)\right)\\ \le -n\tau {\left(d_{Tri}(z_n,z_{n+1})\right)}^k\le 0.\end{array}$$

(9)

Taking the limit as n tends to infinity in (9), we have

$$\underset{n\to \infty }{lim}n({\left(d_{Tri}(z_n,z_{n+1})\right)}^k=0.$$

(10)

Therefore, ${lim}_{n\to \infty }{n}^{1/k}(d_{Tri}(z_n,z_{n+1})=0$, so some $n_0\in \mathbb{N}$ exists, such that

$$d_{Tri}(z_n,z_{n+1})\le \frac{1}{n^{1/k}},\mathrm{for}\mathrm{all}n\ge n_0.$$

(11)

To demonstrate that the sequence $\left\{z_n\right\}$ is a Cauchy sequence, we take into account two cases. Consequently, for any $m,n\in \mathbb{N}$, the following will hold.

Case 1. Let $p=2m+1$ be an odd number, with $m\ge 1$. Then, using the triangular inequality of $\mathcal{TCMTS}$, we have

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m+1})& \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\gamma (z_{n+2},z_{n+2m+1})d_{Tri}(z_{n+2},z_{n+2m+1}).\hfill \\ \hfill & \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\gamma (z_{n+2},z_m)[\beta (z_{n+2},z_{n+3})d_{Tri}(z_{n+2},z_{n+3})+\mu (z_{n+3},z_{n+4})d_{Tri}(z_{n+3},z_{n+4})\hfill \\ \hfill & +\gamma (z_{n+4},z_{n+2m+1})d_{Tri}(z_{n+4},z_{n+2m+1})].\hfill \\ \hfill ⋮\end{array}$$

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m+1})& \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-2}{2}}[\beta (z_{2i},z_{2i+1})d_{Tri}(z_{2i},z_{2i+1})+\mu (z_{2i+1},z_{2i+2})d_{Tri}(z_{2i+1},z_{2i+2})]\hfill \\ \hfill & \left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)+\prod _{i=\frac{n}{2}+1}^{\frac{n+2m}{2}}\gamma (z_{2i},z_{n+2m+1})d_{Tri}(z_{n+2m},z_{n+2m+1}).\hfill \\ \hfill & \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-2}{2}}\beta (z_{2i},z_{2i+1})d_{Tri}(z_{2i},z_{2i+1})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)\hfill \\ \hfill & +\prod _{i=\frac{n}{2}+1}^{\frac{n+2m}{2}}\gamma (z_{2i},z_{n+2m+1})d_{Tri}(z_{n+2m},z_{n+2m+1})\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-2}{2}}\mu (z_{2i+1},z_{2i+2})d_{Tri}(z_{2i+1},z_{2i+2})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)\hfill \end{array}$$

The last inequality can be written as

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m+1})& \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m}{2}}\beta (z_{2i},z_{2i+1})d_{Tri}(z_{2i},z_{2i+1})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-2}{2}}\mu (z_{2i+1},z_{2i+2})d_{Tri}(z_{2i+1},z_{2i+2})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right).\hfill \end{array}$$

When (11) is used in the above inequality, it implies that $d_{Tri}(z_{2i},z_{2i+1})\le \frac{1}{{\left(2i\right)}^{1/k}}$; we obtain

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m+1})& \le \beta (z_n,z_{n+1})\frac{1}{{\left(n\right)}^{1/k}}+\mu (z_{n+1},z_{n+2})\frac{1}{{(n+1)}^{1/k}}\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m}{2}}\beta (z_{2i},z_{2i+1})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)\frac{1}{{\left(2i\right)}^{1/k}}\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-2}{2}}\mu (z_{2i+1},z_{2i+2})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m+1})\right)\frac{1}{{(2i+1)}^{1/k}}.\hfill \end{array}$$

We write it as follows:

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m+1})& \le \beta (z_n,z_{n+1})\frac{1}{{\left(n\right)}^{1/k}}+\mu (z_{n+1},z_{n+2})\frac{1}{{(n+1)}^{1/k}}\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & +[{\Delta }_{n+2m/2}-{\Delta }_{n/2}]+[{\Omega }_{n+2m-2/2}-{\Omega }_{n/2}].\hfill \end{array}$$

(13)

where

$${\Delta }_p=\sum _{i=1}^p\left(\prod _{j=1}^i\gamma (z_{2j},z_{n+2m+1})\right)\beta (z_{2i},z_{2i+1})\frac{1}{{\left(2i\right)}^{1/k}},$$

and

$${\Omega }_q=\sum _{i=1}^q\left(\prod _{j=1}^i\gamma (z_{2j},z_{n+2m+1})\right)\mu (z_{2i+1},z_{2i+2})\frac{1}{{(2i+1)}^{1/k}}.$$

By using the ratio test and applying (3), we obtain ${lim}_{n,m\to \infty }[{\Delta }_{n+2m/2}-{\Delta }_{n/2}]=0$. Similarly, using (4) and the ratio test, we conclude that ${lim}_{n,m\to \infty }[{\Omega }_{n+2m-2/2}-{\Omega }_{n/2}]=0$. Moreover, (5) implies that ${lim}_{n\to \infty }\beta (z_n,z_{n+1})\left(\frac{1}{n^{1/k}}\right)=0$, and ${lim}_{n\to \infty }\mu (z_{n+1},z_{n+2})\frac{1}{{(n+1)}^{1/k}}=0$, as $k\in (0,1)$. Hence,

$$\underset{n,m\to \infty }{lim}{d}_{Tri}(z_n,z_{n+2m+1})=0.$$

(14)

Case 2: Let $p=2m$ be an even number, with $m\ge 1$. Then, using the triangular inequality of $\mathcal{TCMTS}$, we derive:

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m})& \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\gamma (z_{n+2},z_{n+2m})d_{Tri}(z_{n+2},z_{n+2m}).\hfill \\ \hfill ⋮\end{array}$$

By following the same procedures as in case 1, we obtain

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m})& \le \beta (z_n,z_{n+1})d_{Tri}(z_n,z_{n+1})+\mu (z_{n+1},z_{n+2})d_{Tri}(z_{n+1},z_{n+2})\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-1}{2}}\beta (z_{2i},z_{2i+1})d_{Tri}(z_{2i},z_{2i+1})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m})\right)\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-3}{2}}\mu (z_{2i+1},z_{2i+2})d_{Tri}(z_{2i+1},z_{2i+2})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m})\right).\hfill \end{array}$$

Using the fact that $d_{Tri}(z_{2i},z_{2i+1})\le \frac{1}{{\left(2i\right)}^{1/k}}$, the above inequalities can be written as

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m})& \le \beta (z_n,z_{n+1})\frac{1}{{\left(n\right)}^{1/k}}+\mu (z_{n+1},z_{n+2})\frac{1}{{(n+1)}^{1/k}}\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-1}{2}}\beta (z_{2i},z_{2i+1})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m})\right)\frac{1}{{\left(2i\right)}^{1/k}}\hfill \\ \hfill & +\sum _{i=\frac{n}{2}+1}^{\frac{n+2m-3}{2}}\mu (z_{2i+1},z_{2i+2})\left(\prod _{j=\frac{n}{2}+1}^i\gamma (z_{2j},z_{n+2m})\right)\frac{1}{{(2i+1)}^{1/k}}.\hfill \end{array}$$

Writing it as in (13)

$$\begin{array}{cc}\hfill d_{Tri}(z_n,z_{n+2m})& \le \beta (z_n,z_{n+1})\frac{1}{{\left(n\right)}^{1/k}}+\mu (z_{n+1},z_{n+2})\frac{1}{{(n+1)}^{1/k}}\hfill \\ \hfill & +[{\Delta }_{n+2m/2}-{\Delta }_{n/2}]+[{\Omega }_{n+2m-2/2}-{\Omega }_{n/2}].\hfill \end{array}$$

where

$${\Delta }_p=\sum _{i=1}^p\left(\prod _{j=1}^i\gamma (z_{2j},z_{n+2m})\right)\beta (z_{2i},z_{2i})\frac{1}{{\left(2i\right)}^{1/k}},$$

and

$${\Omega }_q=\sum _{i=1}^q\left(\prod _{j=1}^i\gamma (z_{2j},z_{n+2m})\right)\mu (z_{2i+1},z_{2i+2})\frac{1}{{(2i+1)}^{1/k}}.$$

By applying (3)–(5), employing the ratio test, and then going through the same process as before, we arrive at the conclusion that

$$\underset{n,m\to \infty }{lim}{d}_{Tri}(z_n,z_{n+2m})=0.$$

(15)

As a result, $\left\{z_n\right\}$ is a Cauchy sequence in a complete $\mathcal{TCMTS}$$(Z,d_{Tri})$; therefore, it converges to some $\widehat{z}\in Z$.

In the subsequent paragraph, we demonstrate that $\widehat{z}$ is a fixed point of the mapping T, i.e., $T\widehat{z}=\widehat{z}$. We have ${lim}_{n\to \infty }{d}_{Tri}(T{z}_n,T\widehat{z})=0$ because T is continuous and ${lim}_{n\to \infty }{d}_{Tri}(z_n,\widehat{z})=0.$ This provides

$$d_{Tri}(\widehat{z},T\widehat{z})=\underset{n\to \infty }{lim}{d}_{Tri}(z_{n+1},T\widehat{z})=\underset{n\to \infty }{lim}{d}_{Tri}(T{z}_n,T\widehat{z})=0,$$

(16)

hence $T\widehat{z}=\widehat{z}$.

To prove the uniqueness of the fixed point, assume that there are two fixed points, $\widehat{a}$, and $\widehat{b}$, such that $\widehat{a}\ne \widehat{b}$ and $\alpha (\widehat{a},\widehat{b})\ge 1$. $T\widehat{a}=\widehat{a}\ne \widehat{b}=T\widehat{b}$ implies that $d_{Tri}(T\widehat{a},T\widehat{b})>0$. Since T is ($\alpha$-$\mathcal{F}$)-contractive mapping, by utilizing (2), we have

$$\begin{array}{ccc}\hfill \tau +F(d_{Tri}(T\widehat{a},T\widehat{b}))& \le & \tau +\alpha (\widehat{a},\widehat{b})F(d_{Tri}(T\widehat{a},T\widehat{b})).\hfill \\ & \le & F(d_{Tri}(\widehat{a},\widehat{b}))=F(d_{Tri}(T\widehat{a},T\widehat{b})).\hfill \end{array}$$

This shows that $\tau \le 0,$ leading to a contradiction. As a result, $\widehat{a}=\widehat{b}$ and the fixed point is unique. □

Next, we state a corollary to our primary theorem, [21].

Corollary 1.

Let $(Z,d_{Tri})$ be a complete $\mathcal{TCMTS}$, and let $T:Z\to Z$ be a continuous mapping satisfying:

$$\begin{array}{c}\hfill \tau +F\left(d_{Tri}(T\xi ,T\eta )\right)\le F\left(d_{Tri}(\xi ,\eta )\right),\mathit{for}\mathit{all}\xi ,\eta \in Z.\end{array}$$

(17)

Let ${\xi }_0\in Z$, consider the sequence ${\xi }_n=T^n{\xi }_0$. Suppose

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta ({\xi }_{2i+2},{\xi }_{2i+3})}{\beta ({\xi }_{2i},{\xi }_{2i+1})}\gamma ({\xi }_{2i+2},{\xi }_m)<1,$$

(18)

and

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\mu ({\xi }_{2i+3},{\xi }_{2i+4})}{\mu ({\xi }_{2i+1},{\xi }_{2i+2})}\gamma ({\xi }_{2i+2},{\xi }_m)<1.$$

(19)

In addition, for each $\xi \in Z$, assume this holds

$$\underset{n\to \infty }{lim}\beta (\xi ,{\xi }_n),\underset{n\to \infty }{lim}\mu ({\xi }_n,\xi )\mathit{and}\underset{n\to \infty }{lim}\gamma ({\xi }_n,\xi )\mathit{exists}\mathit{and}\mathit{finite}.$$

(20)

Then, T has a unique fixed point.

Proof. 

Let the map $\alpha :Z\times Z\to [0,\infty )$ be defined by $\alpha (\xi ,\eta )=1$, for all $\xi ,\eta \in Z$. Repeat the proof of Theorem 1 by considering the defined α. □

Letting $\beta =\mu =\gamma$, in Theorem 1, we obtain the following corollary, (see [21]).

Corollary 2.

Let $(Z,d_{\beta })$ be a complete $\mathcal{TCMTS}$, and let $T:Z\to Z$ be an (α-$\mathcal{F}$)-contractive mapping, such that the following holds:

(1) 

T is α-admissible.

(2) 

There exists $z_0\in Z$ such that $\alpha (z_0,T{z}_0)\ge 1$.

(3) 

T is continuous.

(4) 

For $z_0\in Z$, define the sequence $\left\{z_n\right\}$ by $z_n=T^n{z}_0$, and assume

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta (z_{2i+2},z_{2i+3})}{\beta (z_{2i},z_{2i+1})}\beta (z_{2i+2},z_m)<1,$$

(21)

and

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta (z_{2i+3},z_{2i+4})}{\beta (z_{2i+1},z_{2i+2})}\beta (z_{2i+2},z_m)<1.$$

(22)

In addition, for each $z\in Z$

$$\underset{n\to \infty }{lim}\beta (z,z_n),\underset{n\to \infty }{lim}\beta (z_n,z)\mathit{exists}\mathit{and}\mathit{finite}.$$

(23)

Then, the mapping T has a unique fixed point.

Proof. 

In Theorem 1, let $\mu =\beta =\gamma$ and repeat the proof. □

3.2. The $\Theta$-Contractive Mappings and Fixed-Point Results

With $\Theta$ as in Definition 8, we present the notion of $\Theta$-contractive mapping on $\mathcal{TCMTS}$.

Definition 11.

Let $(Z,d_{Tri})$ be a $\mathcal{TCMTS}$, where Z is a nonempty set. Let $T:Z\to Z$ be a self-mapping. Then, T is said to be Θ-contractive mapping if there exists a function $\theta \in \Theta$ and an $r\in (0,1)$, such that the following holds:

$$z,w\in Z,d_{Tri}(Tz,Tw)\ne 0\Rightarrow \theta \left(d_{Tri}(Tz,Tw)\right)\le {\left[\theta \left(d_{Tri}(z,w)\right)\right]}^r.$$

(24)

We state and prove our second main result on the fixed-point theorem, which is inspired by [24].

Theorem 2.

Let $(Z,d_{Tri})$ be a complete $\mathcal{TCMTS}$, where Z is a nonempty set. Let $T:Z\to Z$ be a Θ-contractive mapping, such that for any $z_0\in Z$, define the sequence $\left\{z_n\right\}$ by $z_n=T^n{z}_0$, and assume these hold:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta (z_{2i+2},z_{2i+3})}{\beta (z_{2i},z_{2i+1})}\gamma (z_{2i+2},z_m)<1,$$

(25)

and

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\mu (z_{2i+3},z_{2i+4})}{\mu (z_{2i+1},z_{2i+2})}\gamma (z_{2i+2},z_m)<1.$$

(26)

In addition, for each $z\in Z$

$$\underset{n\to \infty }{lim}\beta (z,z_n),\underset{n\to \infty }{lim}\mu (z_n,z),\mathit{and}\underset{n\to \infty }{lim}\gamma (z_n,z)\mathit{exists}\mathit{and}\mathit{finite}.$$

(27)

Then, T has a unique fixed point in $Z.$

Proof. 

Let $z_0$ be any arbitrary point in Z. Construct a sequence $\left\{z_n\right\}$ using the iteration as follows: $T{z}_0=z_1,T{z}_1=z_2$, thus $T^n{z}_0=z_n,$ for all $n\in N$.

If for some $m\in \mathbb{N}$, $T^m{z}_0=T^{m+1}{z}_0$, then this implies that $T^m{z}_0$ is a fixed point of the mapping T. Thus, without loss of generality, we may assume that $z_n\ne z_{n+1}$, i.e., $d_{Tri}(T^n{z}_0,T^{n+1}{z}_0)>0,$ for all $n\in \mathbb{N}$.

Utilizing (24) and applying it recursively, we obtain

$$\begin{array}{cc}\hfill \theta \left(d_{Tri}(z_n,z_{n+1})\right)& =\theta \left(d_{Tri}(T{z}_{n-1},T{z}_n)\right)\hfill \\ \hfill & \le {\left[\theta \left(d_{Tri}(z_{n-1},z_n)\right)\right]}^r\hfill \\ \hfill & \le {\left[\theta \left(d_{Tri}(z_{n-2},z_{n-1})\right)\right]}^{r^2}\hfill \\ \hfill & \le {\left[\theta \left(d_{Tri}(z_{n-3},z_{n-2})\right)\right]}^{r^3}\le \cdots \hfill & \hfill \le {\left[\theta \left(d_{Tri}(z_0,z_1)\right)\right]}^{r^n}.\end{array}$$

(28)

Therefore, as $\theta \left(t\right)>1$, we have

$$1<\theta \left(d_{Tri}(z_n,z_{n+1})\right)\le {\left[\theta \left(d_{Tri}(z_0,z_1)\right)\right]}^{r^n}.$$

(29)

Since $0<r<1$, letting n tends to infinity in (29); we deduce

$$\underset{n\to \infty }{lim}\theta \left(d_{Tri}(z_n,z_{n+1})\right)=1.$$

Employing property $\left(\theta 2\right)$, we obtain

$$\underset{n\to \infty }{lim}{d}_{Tri}(z_n,z_{n+1})=0.$$

(30)

In a similar method, one can show that

$$\underset{n\to \infty }{lim}{d}_{Tri}(z_n,z_{n+2})=0.$$

(31)

By $\left(\theta 3\right)$, there exists $k\in (0,1)$ and $M\in (0,\infty ]$ such that

$$\underset{n\to \infty }{lim}\frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{{\left[d_{Tri}(z_n,z_{n+1})\right]}^k}=M.$$

(32)

Case 1: Let $0<M<\infty$, and define $L=\frac{M}{2}$, from Equation (32), there exists some $n_0\in N$, such that for all $n\ge n_0$ we obtain

$$\left|\frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{{\left[d_{Tri}(z_n,z_{n+1})\right]}^k}-M\right|\le L,$$

which implies that

$$L=M-L\le \frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{{\left[d_{Tri}(z_n,z_{n+1})\right]}^k},\forall n\ge n_0.$$

Hence, for all $n\ge n_0$, we have

$$n{\left[d_{Tri}(z_n,z_{n+1})\right]}^k\le n\left[\frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{L}\right].$$

By employing (29), we obtain;

$$n{\left[d_{Tri}(z_n,z_{n+1})\right]}^k\le n\left[\frac{{\left[\theta \left(d_{Tri}(z_0,z_1)\right)\right]}^{r^n}-1}{L}\right].$$

Letting $n\to \infty$, in the above inequality, we obtain

$$\underset{n\to \infty }{lim}n{\left[d_{Tri}(z_n,z_{n+1})\right]}^k=0.$$

(33)

Case 2: $M=\infty$, in this case let $L>0$ be any arbitrary number. Thus, by the definition of the limit, we can find some $n_1\in N$ such that

$$\frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{{\left[d_{Tri}(z_n,z_{n+1})\right]}^k}\ge L,\mathrm{for}\mathrm{all}n\ge n_1,$$

which gives

$$n{\left[d_{Tri}(z_n,z_{n+1})\right]}^k\le n\left[\frac{\theta \left(d_{Tri}(z_n,z_{n+1})\right)-1}{L}\right].$$

Again employing (29) in the above inequality and then letting $n\to \infty$, we obtain

$$\underset{n\to \infty }{lim}n{\left[d_{Tri}(z_n,z_{n+1})\right]}^k=0.$$

(34)

Thus, from Equations (33) and (34) we deduce that for any $M\in (0,\infty ]$ and $0<k<1$, there exists some $\widehat{N}\in N$, where $\tilde{N}=max\{n_0,n_1\}$ such that

$$d_{Tri}(z_n,z_{n+1})\le \frac{1}{n^{1/k}},\forall n\ge \widehat{N},$$

(35)

holds, (35) is comparable to (11). Following the steps of the proof of Theorem 1, one can easily show that for any $n,m\in \mathbb{N}$, and with $m\ge n$, then

$$\underset{n,m\to \infty }{lim}{d}_{Tri}(z_n,z_m)=0.$$

(36)

As a result, $\left\{z_n\right\}$ is a Cauchy sequence in a complete $\mathcal{TCMTS}$$(Z,d_{Tri})$, and therefore it converges to some $\widehat{z}\in Z$, i.e. ${lim}_{n\to \infty }{d}_{Tri}(z_n,\widehat{z})=0$.

Next, we demonstrate that $\widehat{z}$ is a fixed point of the mapping T, i.e., $T\widehat{z}=\widehat{z}$. As T fulfills property (24), this follows

$$ln\left(\theta \left(d_{Tri}(T{z}_n,T\widehat{z})\right)\right)\le r\left(ln\left(\theta \left(d_{Tri}(z_n,\widehat{z})\right)\right)\right)\le ln\left(\theta \left(d_{Tri}(z_n,\widehat{z})\right)\right),$$

(37)

which implies by $\left(\theta 1\right)$ that $d_{Tri}(T{z}_n,T\widehat{z})\le d_{Tri}(z_n,\widehat{z})$; hence, T is continuous. This gives ${lim}_{n\to \infty }{d}_{Tri}(T{z}_n,T\widehat{z})=0$. Therefore, from

$$d_{Tri}(\widehat{z},T\widehat{z})=\underset{n\to \infty }{lim}{d}_{Tri}(z_{n+1},T\widehat{z})=\underset{n\to \infty }{lim}{d}_{Tri}(T{z}_n,T\widehat{z})=0,$$

(38)

one can deduce that $T\widehat{z}=\widehat{z}$.

To prove the uniqueness of the fixed point, assume that T has two fixed points $\widehat{x},\widehat{y}$ such that $\widehat{x}\ne \widehat{y},$

$$\begin{array}{ccc}\hfill \theta \left(d_{Tri}(\widehat{x},\widehat{y})\right)& =& \theta \left(d_{Tri}(T\widehat{x},T\widehat{y})\right)\hfill \\ & \le & {\left[\theta \left(d_{Tri}(\widehat{x},\widehat{y})\right)\right]}^r\hfill \\ & <& \theta \left(d_{Tri}(\widehat{x},\widehat{y})\right).\hfill \end{array}$$

which is a contradiction; hence, $\widehat{x}=\widehat{y}$, proving that T has a unique fixed point. □

Example 5.

Let $Z=[0,4]$. Define the mapping $d_{Tri}:Z\times Z\to [0,\infty )$ by $d_{Tri}(x,y)={|x-y|}^4$. Consider the functions $\beta ,\mu ,\gamma :Z\times Z\to [1,\infty )$ defined by $\beta (x,y)=1+x+y$, $\mu (x,y)=y+1$, and

$$\gamma (x,y)=\left\{\begin{array}{cc}x+y\hfill & \mathit{if}0\le x<1\hfill \\ x\hfill & \mathit{if}x\ge 1.\hfill \end{array}\right.$$

One can verify that $(Z,d_{Tri})$ is a complete $\mathcal{TCMTS}$.

Consider the contractive mapping $T:Z\to Z$ defined by $T\left(x\right)=\frac{x}{4}$, and let $\theta :(0,\infty )\to (1,\infty )$, be $\theta \left(t\right)=e^{\sqrt{t}}$. Then, $\theta \in \Theta$. To form the sequence $\left\{z_n\right\}$, start with $z_0=1$; then, using iteration $T{z}_{n-1}=z_n$, we obtain $z_1=T{z}_0=T\left(1\right)=\frac{1}{4},z_2=T\left(\frac{1}{4}\right)=\frac{1}{4^2}$, thus $z_n=T^n\left(1\right)=\frac{1}{4^n}$ for all $n\in \mathbb{N}$. As

$$\frac{\beta (z_{2i+2},z_{2i+3})}{\beta (z_{2i},z_{2i+1})}\gamma (z_{2i+2},z_m)=\frac{(\frac{1}{4^{2i+2}}+\frac{1}{4^{2i+3}}+1)}{(\frac{1}{4^{2i}}+\frac{1}{4^{2i+1}}+1)}(\frac{1}{4^{2i+2}}+\frac{1}{4^m}).$$

Hence

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\beta (z_{2i+2},z_{2i+3})}{\beta (z_{2i},z_{2i+1})}\gamma (z_{2i+2},z_m)<1.$$

(39)

Similarly,

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\mu (z_{2i+3},z_{2i+4})}{\mu (z_{2i+1},z_{2i+2})}\gamma (z_{2i+2},z_m)=\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{(\frac{1}{4^{2i+4}}+1)}{(\frac{1}{4^{2i+2}}+1)}(\frac{1}{4^{2i+2}}+\frac{1}{4^m})<1.$$

(40)

Additionally, one can easily show

$$\underset{n\to \infty }{lim}\beta (z,z_n)=\underset{n\to \infty }{lim}(z+\frac{1}{4^n}+1)is finte,$$

and

$$\underset{n\to \infty }{lim}\mu (z_n,z)=\underset{n\to \infty }{lim}z,and,\underset{n\to \infty }{lim}\gamma (z_n,z)=\underset{n\to \infty }{lim}\frac{1}{4^n}are all finite.$$

To explore if T is Θ-contractive mapping with $r=1/2$, we investigate if $\theta \left(d_{Tri}(Tz,Tw)\right)\le {\left[\theta \left(d_{Tri}(z,w)\right)\right]}^{1/2}$. Take any $z,w\in Z$, such that $d_{Tri}(Tz,Tw)\ne 0.$ As $d_{Tri}(Tz,Tw)={|\frac{z}{4}-\frac{w}{4}|}^4$, this results in

$$\theta \left(d_{Tri}(Tz,Tw)\right)=e^{\sqrt{|\frac{z}{4}-\frac{w}{4}{|}^4}}=e^{\frac{1}{16}{|z-w|}^2}\le {[e^{\sqrt{{|z-w|}^4}}]}^{1/2}$$

Therefore, all of the conditions of Thereom 2 are satisfied, so T has a unique fixed point $z=0$ in Z.

4. Application to Fourth-Order Differential Equation

In this section, we discuss the application of our main Theorem 2 in solving the following boundary-value problem of a fourth-order differential equation:

$$\left\{\begin{array}{c}{f}^{″″}\left(t\right)=K(t,f\left(t\right),f{}^{\prime }\left(t\right),f^{{}^{″}}\left(t\right),f^{{}^{‴}}\left(t\right)),t\in [0,1],\hfill \\ f\left(0\right)=f{}^{\prime }\left(0\right)=f^{{}^{″}}\left(1\right)=f^{{}^{‴}}\left(1\right)=0,\hfill \end{array}\right.$$

(41)

where $K:[0,1]\times R^4⟶R$ is a function. Let $Z=C\left(\right[0,1],\mathbb{R})$ be the space of all continuous real valued functions defined on the interval $[0,1]$. Let the mapping $d_{Tri}:Z\times Z⟶[0,\infty )$, be defined by

$$d_{Tri}(f,g)=ma{x}_{t\in [0,1]}{|f\left(t\right)-g\left(t\right)|}^2.$$

(42)

The three controlled functions $\beta ,\mu ,\gamma :Z\times Z\to [1,\infty ),$ are defined by

$$\beta (f,g)=\left\{\begin{array}{cc}2+ma{x}_{t\in [0,1]}{|f\left(t\right)-g\left(t\right)|}^2\hfill & \mathrm{if}f\ne g.\hfill \\ 1\hfill & \mathrm{if}f=g,\hfill \end{array}\right.$$

and

$$\mu (f,g)=\left\{\begin{array}{cc}1+ma{x}_{t\in [0,1]}{|f\left(t\right)-g\left(t\right)|}^2\hfill & \mathrm{if}f\ne g.\hfill \\ 1\hfill & \mathrm{if}f=g,\hfill \end{array}\right.$$

and $\gamma (f,g)=1.$

It is not hard to see that $(Z,d_{Tri})$ is a complete $\mathcal{TCMTS}$. The boundary-value problem of (41) can be written in integral form as

$$f\left(t\right)={\int }_0^1\mathcal{G}(t,s)K(s,f\left(s\right),f^{{}^{\prime }}\left(s\right))ds,f\in Z$$

(43)

where $\mathcal{G}(t,s):{[0,1]}^2⟶\mathbb{R}$ is Green’s function associated to (41) and given explicitly as

$$\mathcal{G}(t,s)=\left\{\begin{array}{cc}\frac{1}{6}{t}^2(3s-t)\hfill & \mathrm{for}0\le t\le s\le 1,\hfill \\ \frac{1}{6}{s}^2(3s-t)\hfill & \mathrm{for}0\le s\le t\le 1.\hfill \end{array}\right.$$

(44)

It follows from (43) that $\frac{1}{3}{t}^2{s}^2\le \mathcal{G}(t,s)\le \frac{1}{2}{t}^2$, and $\frac{1}{3}{t}^2{s}^2\le \mathcal{G}(t,s)\le \frac{1}{2}{s}^2$, for $t,s\in [0,1].$

Theorem 3.

Consider $(Z,d_{Tri})$ a complete $\mathcal{TCMTS}$ as defined above with norm as in (42), and assume the following conditions hold:

1. 

$K:[0,1]\times R^4⟶R$ is a continuous function.

2. 

There exists a $\tau \in [1,\infty ),$ such that the following holds for all $f,g\in Z$

$$|K(s,f,f^{{}^{\prime }})-K(s,g,g^{{}^{\prime }})|\le \sqrt{20}{e}^{\frac{-\tau }{2}}|f\left(s\right)-g\left(s\right)|$$

3. 

There exists $f_0\in Z$ such that for all $t\in [0,1]$, we have

$$T{f}_0\left(t\right)\le {\int }_0^1\mathcal{G}(t,s)K(s,f_0\left(s\right),f_0^{{}^{\prime }}\left(s\right))ds.$$

(45)

where $T:Z⟶Z$ is a mapping defined by

$$Tf\left(t\right)={\int }_0^1\mathcal{G}(t,s)K(s,f\left(s\right),f^{{}^{\prime }}\left(s\right))ds,f\in Z$$

(46)

Then, the boundary-value problem (41) has a unique solution.

Proof. 

Note that $f\in Z$ is a solution of (41) if and only if f is a solution of the integral equation $Tf\left(t\right)={\int }_0^1\mathcal{G}(t,s)K(s,f\left(s\right),f^{{}^{\prime }}\left(s\right))ds.$ Consider

$$\begin{array}{ccc}\hfill |Tf\left(t\right)-Tg\left(t\right){|}^2& =& |{\int }_0^1\mathcal{G}(t,s)K(s,f\left(s\right),f^{{}^{\prime }}\left(s\right))ds-{\int }_0^1\mathcal{G}(t,s)K(s,g\left(s\right),g^{{}^{\prime }}\left(s\right))ds{|}^2.\hfill \\ & \le & {\int }_0^1{\left(\mathcal{G}(t,s)\right)}^2|K(s,f\left(s\right),f^{{}^{\prime }}\left(s\right))-K(s,g\left(s\right),g^{{}^{\prime }}\left(s\right)){|}^{2}ds.\hfill \\ & \le & {\int }_0^1{\displaystyle \frac{1}{4}}{s}^{4}20e^{-\tau }|f\left(s\right)-g\left(s\right){|}^{2}ds\hfill \\ & \le & 20e^{-\tau }max|f-g|{\int }_0^1{\displaystyle \frac{1}{4}}{s}^{4}ds.\hfill \\ & \le & e^{-\tau }{d}_{Tri}(f,g).\hfill \end{array}$$

Which gives

$$d_{Tri}(Tf,Tg)\le e^{-\tau }{d}_{Tri}(f,g).$$

(47)

Let the map $\theta \left(t\right)=e^{\sqrt{t}}$, then $\theta \left(t\right)\in \Theta$. To show T is $\Theta$-contractive, note that by taking the square root of (47), we have

$$\sqrt{d_{Tri}(Tf,Tg)}\le \sqrt{e^{-\tau }{d}_{Tri}(f,g)}$$

Thus,

$$e^{\sqrt{d_{Tri}(Tf,Tg)}}\le {\left(e^{\sqrt{d_{Tri}(f,g)}}\right)}^r,\mathrm{where}r=\sqrt{e^{-\tau }}<1,\mathrm{since}\tau \ge 1.$$

Therefore, T is $\Theta$-contractive. One can easily see that $|T{f}_0{|}^2\le e^{-\tau }{\left|f_0\right|}^2$; thus, for any $n>1$, we have $|T^n{f}_0{|}^2\le {\left(e^{-\tau }\right)}^n{\left|f_0\right|}^2$. Moreover, one can deduce that

$$|T^{n+1}{f}_0-T^n{f}_0{|}^2\le e^{-\tau }|T^n{f}_0-T^{n-1}{f}_0{|}^2\le {\left(e^{-\tau }\right)}^2{|T^{n-1}{f}_0-T^{n-2}{f}_0|}^2$$

From this and the fact $\sqrt{e^{-\tau }}<1$, one can deduce that

$$\underset{m\ge 1}{sup}\underset{n\to \infty }{lim}\frac{\beta (T^{2n+2}{f}_0,T^{2n+3}{f}_0)}{\beta (T^{2n}{f}_0,T^{2n+1}{f}_0)}\gamma (T^{2n+2}{f}_0,T^m{f}_0)<1,$$

and

$$\underset{m\ge 1}{sup}\underset{n\to \infty }{lim}\frac{\mu (T^{2n+3}{f}_0,T^{2n+4}{f}_0)}{\mu (T^{2n+1}{f}_0,T^{2n+2}{f}_0)}\gamma (T^{2n+2}{f}_0,T^m{f}_0)<1.$$

Hence, the hypothesis of Theorem 2 is satisfied, which implies the boundary-value problem (41) has a unique solution. □

5. Conclusions

This article examined the concept of triple controlled metric type space, which was first introduced by Tasneem et al. [16]. Due to our investigation of ($\alpha$-$\mathcal{F}$)-contractive mappings and $\Theta$-contractive mappings in the class of triple controlled metric type space, we were able to generalize some prior findings and prove certain fixed-point theorems relevant to this context. Few researchers have recently studied the fixed-circle problem in metric spaces [29], s-metric spaces [30], and some generalized metric spaces by using various contractive mappings. For instance, in [31] a new fixed-circle theorem for self-mappings on an s-metric space was presented using Wardowski Type contractions, see also [32]. We suggest some potential research topics for the future, such as employing these contractive mappings on triple controlled metric type space to study the fixed-circle problem.