New Generalization of Metric-Type Spaces—Strong Controlled

Abstract

In this manuscript, we establish a new type of metric space that is called controlled strong metric spaces by introducing a controlled function to the triangle inequality as follows: $\wp (s,r)\le \wp (s,z)+\eta (z,r)\wp (z,r),$ and keeping the symmetry condition that is $\wp (s,r)=\wp (r,s)\mathrm{for}\mathrm{all}r,s$. We demonstrate the existence of the fixed point of self-mapping and its uniqueness in such spaces that satisfy linear and nonlinear contractions. Moreover, we provide three applications of results to polynomial equations of high degree, systems of linear equations, along with fractional differential equations.

1. Introduction

In this century, theories based on fixed points are proving to be an important tool for studying non-linear phenomena. Particularly, the fixed-point technique has been applied to various fields, such as biology, chemistry, economics and engineering, the theory of games, computer science, physics, and the physics of fluid and elasticity. The Banach principle is a fundamental result in fixed points, which claims all contractions in metric spaces have fixed points, which are unique. It has been referred to us that the generalization of Banach’s principle has been extensively studied by many researchers in various methods; see the Refs. [1,2,3,4,5]. In this paper, we focus on the b-metric space (bMS) and its generalization. Bakhtin [6] and Czerwik [7] made a step forward in the generalization of (MS) by introducing (bMS). It can be obtained by altering the triangle inequality of a (MS). Definitely, every (MS) is a (bMS) when the constant coefficient equals one. However, it is not the same as the reverse. Later on, many authors obtained various fixed-point results pertaining to b-metric spaces; see the Refs. [8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24]. Concerning our special interest in the current study, we make reference to papers that include some main generalizations of b-metric spaces, such as a strong b-metric space (SbMS) (see the Ref. [25]), extended (bMS) (see the Refs. [26,27,28]), (CMTS) and double (CMTS) (see the Ref. [29]). Subsequently, we generalized the (SbM) to (SCbMS). It is crystal clear that every (SbMS) is a controlled strong b-metric space (SCbMS) when the function $\vartheta$ is equal to a constant. In the main results section, we prove the existence of the fixed point of self-mapping on a complete (SCMTS) and its uniqueness. Each presented theorem gives a sufficient condition to pass convergence via the ratio test. In closing, we present different applications of the results to the polynomial equations, the system of linear equations, along with fractional differential equations.

2. Preliminary Assertions

First, we remind the reader of the definition of strong $b-$metric spaces.

Definition 1 

([30]). Given a set that contains at least one element ϝ and $t\ge 1$.

The function $\Delta :ϝ\times ϝ\to [0,\infty )$ is called a strong $b-$ metric space if

1.

$\Delta (p,q)=0$ if and only if $p=q$;

2.

$\Delta (p,q)=\Delta (q,p)$;

3.

$\Delta (p,q)\le \Delta (p,l)+t\Delta (l,q)$,

for all $p,q,l\in ϝ$. The pair $(ϝ,\Delta )$ is called an (SbMS).

Next, we demonstrate the definition of a controlled strong $b-$metric spaces (CSbMS).

Definition 2.

Let $ϝ$ be a set that contains at least one element and $\zeta :ϝ\times ϝ\to [1,\infty )$. The function $\Delta :ϝ\times ϝ\to [0,\infty )$ is called a (CSbMS) if

1.

$\Delta (p,q)=0$ if, and only if $p=q$;

2.

$\Delta (p,q)=\Delta (q,p)$;

3.

$\Delta (p,q)\le \Delta (p,l)+\zeta (l,q)\Delta (l,q)$,

for all $p,q,l\in ϝ$. The pair $(ϝ,\Delta )$ is called a controlled strong $b-$metric space (CSbMS).

We observe that every (SbMS) is a (CSbMS), and we just take $\zeta (l,q)=t$; however, the reverse is not necessarily true. The following example illustrates the observation by giving a (CSbMS) that is not a strong $b-$metric space. Now, we present an example of a (CSbMS) that is not an (SbMS).

Example 1.

Let $ϝ=[1,\infty )$ and define $\Delta (q,p)=max\{|q-p|,2|q-p|-1\}\mathit{for}\mathit{every}q,p\in [1,\infty )$ and $\zeta (q,p)=q+p+2.$

• if $p=q$ then $|q-p|=0\Rightarrow \Delta (q,p)=0.$ Conversely, if $\Delta (q,p)=0$ then $|q-p|\le \Delta (q,p)=0\Rightarrow q=p.$
•  

  $$\begin{array}{cc}\hfill \Delta (q,p)& =max\{|q-p|,2|q-p|-1\}\hfill \\ & =max\{|p-q|,2|p-q|-1\}\hfill \\ & =\Delta (p,q).\hfill \end{array}$$
•  

  $$\begin{array}{cc}\hfill |q-p|& \le |q-l|+|l-p|\hfill \\ & \le \Delta (q,l)+\Delta (l,p)\hfill \\ & \le \Delta (q,l)+\zeta (q,p)\Delta (l,p).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{cc}\hfill 2|q-p|-1& \le 2|q-l|+2|l-p|-1\hfill \\ & \le \Delta (q,l)+2\Delta (l,p)\hfill \\ & \le \Delta (q,l)+\zeta (q,p)\Delta (l,p).\hfill \end{array}$$

Hence, all the conditions of Definition 2 are satisfied. Thus, $(ϝ,\Delta )$ is a (CSbMS).

Example 2.

Let $ϝ=\{0,1,2\}$. Assume that the function Δ is given as follows:

$$\Delta (0,0)=\Delta (1,1)=\Delta (2,2)=0$$

and

$$\Delta (0,1)=\Delta (1,0)=\frac{1}{2},\Delta (0,2)=\Delta (2,0)=6,\Delta (1,2)=\Delta (2,1)=5.$$

Consider $\zeta :{ϝ}^2\to [1,\infty )$ that is symmetric and specified as below:

$$\zeta (0,0)=\zeta (1,1)=\zeta (2,2)=\zeta (0,2)=1,\zeta (1,2)=\frac{5}{4},\zeta (0,1)=\frac{11}{10}.$$

It is quite easy to show that $(ϝ,\Delta )$ is a (CSbMS).

Next, we present some topological properties of (CSbMS).

Definition 3.

Let $(ϝ,\Delta )$ be a (CSbMS) by one or two functions and ${\left\{x_n\right\}}_{n\ge 0}$ be a sequence in ϝ.

• (1) This affirms that the sequence $\left\{x_n\right\}$ converges to arbitrary x that belongs to $ϝ,$ if ∀ $\xi >0$, ∃$N=N\left(\xi \right)\in \mathbb{N}$ such that $\Delta (x_n,x)<\xi$ for every $n\ge N.$ Here, the ${lim}_{n\to \infty }{x}_n=x.$
• (2) The sequence $\left\{x_n\right\}$ is said to be Cauchy if ∀$\xi >0$, ∃$N=N\left(\xi \right)\in \mathbb{N}$ where $\Delta (x_m,x_n)<\xi$ for every $m,n\ge N.$
• (3) The (CSbMS) $(ϝ,\Delta )$ is said to be complete if for all Cauchy sequences, it is convergent.

Definition 4.

Consider $(ϝ,\Delta )$ to be a (CSbMS) by either one function or two functions. Take $x\in ϝ$ along with $\xi >0$.

• (i) An open ball $B(x,\xi )$ is

  $$B(x,\xi )=\{y\in F,\Delta (x,y)<\xi \}.$$
• (ii) The mapping $T:ϝ\to ϝ$ is called continuous at $x\in ϝ$ if ∀$\xi >0$, ∃$\delta >0$, satisfying $T\left(B\right(x,\delta \left)\right)\subseteq B(Tx,\xi )$.

Obviously, if T is continuous at x in the (CSbMS) $(ϝ,\Delta )$, then $x_n\to x$ implies that $T{x}_n\to Tx$ as $n\to \infty$.

3. Main Results

Currently, we are prepared to investigate the main result pertaining to the Banach contraction principle in (CSbMS).

Theorem 1.

Let $(ϝ,\Delta )$ be a complete (CSbMS) by the function $\omega :ϝ\times ϝ\to [1,\infty )$. Suppose that $\mathsf{\Gamma }:ϝ\to ϝ$ holds

$$\Delta (\mathsf{\Gamma }p,\mathsf{\Gamma }q)\le t\Delta (p,q),$$

(1)

for all $p,q\in ϝ$, where $t\in (0,1)$. For $p_0\in ϝ$, choose $p_n={\mathsf{\Gamma }}^n{p}_0$. Suppose that

$$\omega (p_{i+1},p_m)<\frac{1}{t}.$$

(2)

Additionally, for all $p\in ϝ$, assume that

$$\underset{n\to \infty }{lim}\omega (p,p_n)existandisfiniteaswell.$$

(3)

Hence, Γ has a point that is fixed and unique.

Proof. 

In line with the postulate with the theorem, consider the sequence $\{p_n={\mathsf{\Gamma }}^n{p}_0\}$ in $ϝ$. By using (1), we obtain

$$\Delta (p_n,p_{n+1})\le t^n\Delta (p_0,p_1)\mathrm{for} \mathrm{every}nin[0,\infty ).$$

(4)

For $n<m$ where $n,m$ are two integers, we undergo

$$\begin{array}{cc}\hfill \Delta (p_n,p_m)& \le \Delta (p_n,p_{n+1})+\omega (p_{n+1},p_m)\Delta (p_{n+1},p_m)\hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\omega (p_{n+1},p_m)\Delta (p_{n+1},p_{n+2})+\omega (p_{n+1},p_m)\omega (p_{n+2},p_m)\Delta (p_{n+2},p_m)\hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\omega (p_{n+1},p_m)\Delta (p_{n+1},p_{n+2})+\omega (p_{n+1},p_m)\omega (p_{n+2},p_m)\Delta (p_{n+2},p_{n+3})\hfill \\ \hfill & +\omega (p_{n+1},p_m)\omega (p_{n+2},p_m)\omega (p_{n+3},p_m)\Delta (p_{n+3},p_m)\hfill \\ \hfill & \le \cdots \hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\omega (p_j,p_m)\right)\Delta (p_i,p_{i+1})\hfill \\ \hfill & +\prod _{t=n+1}^{m-1}\omega (p_k,p_m)\Delta (p_{m-1},p_m)\hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\omega (p_j,p_m)\right)\Delta (p_i,p_{i+1})\hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\omega (p_j,p_m)\right)t^i\Delta (p_0,p_1)\hfill \\ \hfill & \le \Delta (p_n,p_{n+1})+\sum _{i=0}^{m-1}\left(\prod _{j=0}^i\omega (p_j,p_m)\right)t^i\Delta (p_0,p_1)-\sum _{i=0}^n\left(\prod _{j=0}^i\omega (p_j,p_m)\right)t^i\Delta (p_0,p_1)\hfill \end{array}$$

We used $\omega (p,q)\ge 1$. Let

$$S_p=\sum _{i=0}^p\left(\prod _{j=0}^i\omega (p_j,p_m)\right)t^i\Delta (p_0,p_1).$$

Hence, we reach

$$\Delta (p_n,p_m)\le \Delta (p_n,p_{n+1})+(S_{m-1}-S_n)].$$

(5)

Using the ratio test on the term

$U_i=\left({\prod }_{j=0}^i\omega (p_j,p_m)\right)t^i\Delta (p_0,p_1)$

$$\begin{array}{cc}& \frac{U_{i+1}}{U_i}=\frac{{\prod }_{j=0}^{i+1}w\left(p_j,p_m\right)t^{i+1}\Delta \left(p_0,p_1\right)}{{\prod }_{j=0}^i\omega \left(p_j,p_m\right)t^i\Delta \left(p_0,p_i\right)}\hfill \\ & \mathrm{But}\frac{{\prod }_{j=0}^{i+1}w\left(p_j,P_m\right)}{{\prod }_{j=0}^{i}w\left(p_j,P_m\right)}=w\left(P_{i+1},P_m\right)t\hfill \\ & \mathrm{then}\frac{U_{i+1}}{U_i}=\omega \left(p_{i+1},P_m\right)t\hfill \\ & \mathrm{using}\mathrm{Equation}(2),\mathrm{we}\mathrm{deduce}\mathrm{that}\frac{U_{i+1}}{U_i}<\frac{1}{t}\times t<1\hfill \end{array}$$

proves that this term is convergent. Utilizing this result and Equation (2) leads to the existence of the limit of the real number sequence $\left\{S_n\right\}$. Accordingly, $\left\{S_n\right\}$ is Cauchy.

Assuming the $n,m$ approach to ∞ in (5), it follows that

$$\underset{n,m\to \infty }{lim}\Delta (p_n,p_m)=0.$$

(6)

Then, the mentioned sequence $\left\{p_n\right\}$ is said to be Cauchy. Because $(ϝ,\Delta )$ is a complete (CSbMS), there exists some r in $ϝ$ such that

$$\underset{n\to \infty }{lim}\Delta (p_n,r)=0.$$

We argue that $\mathsf{\Gamma }r=r$. By using the definition of (SCMTS) and applying the triangular inequality, it yields

$$\Delta (r,p_{n+1})\le \Delta (r,p_n)+\Delta (p_n,p_{n+1})\Delta (p_n,p_{n+1}).$$

Based on (3) and (6), we obtain the following:

$$\underset{n\to \infty }{lim}\Delta (r,p_{n+1})=0.$$

(7)

By the use of (1), we obtain

$$\begin{array}{cc}\hfill \Delta (r,\mathsf{\Gamma }r)& \le \Delta (r,p_{n+1})+\omega (p_{n+1},\mathsf{\Gamma }r)\Delta (p_{n+1},\mathsf{\Gamma }r)\hfill \\ \hfill & \le \Delta (r,p_{n+1})+t\omega (p_{n+1},\mathsf{\Gamma }r)\Delta (p_n,r).\hfill \end{array}$$

Employing (3) and (7), we attain the limit $\Delta (r,\mathsf{\Gamma }r)=0$, which implies that $\mathsf{\Gamma }r=r$. Assume that s belongs to $ϝ$, by which $\mathsf{\Gamma }s=s\mathrm{and}r\ne s.$ It leads to

$$0<\Delta (r,s)=\Delta (\mathsf{\Gamma }r,\mathsf{\Gamma }s)\le t\Delta (r,s).$$

Thus, the extremities of the inequality contradict our assumption, and $r=s.$ Consequently, $\mathsf{\Gamma }$ contains a point r that is unique and fixed. □

In the posterior Theorem, we inserted a control function defined by Matkowski [31] to the non-linear contraction that leads to a fixed point.

Theorem 2.

Assume that $(ϝ,g)$ is a complete (CSbMS) via the function $\omega (p,q)$. Suppose that $\mathsf{\Gamma }:ϝ\to ϝ$ fulfills for all $p,q\in ϝ$

$$g(\mathsf{\Gamma }p,\mathsf{\Gamma }q)\le \psi (\Delta (p,q\left)\right),\Delta (p,q)=max\left\{g\right(p,q),g(p,\mathsf{\Gamma }p),g(q,\mathsf{\Gamma }q\left)\right\},$$

(8)

where $\psi :[0,\infty )\to [0,\infty )$ and $\psi \left(b\right)\ge \psi \left(a\right)$ for all b $>a$, is continuous and fulfills ${\displaystyle \underset{i\to \infty }{lim}{\psi }^i\left(t\right)=0,wheret>0}$. Further, suppose that for whatever $r_0\in ϝ$, we obtain

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\omega (p_{i+1},p_m){\psi }^{i+1}(\Delta (p_0,p_1))}{{\psi }^i(\Delta (p_0,p_1))}<1,$$

(9)

where $r_n={\mathsf{\Gamma }}^n{r}_0,n=0,1,...$. If the considered mapping Γ is continuous, then ∃ a point that is fixed as well as unique of Γ (say ζ) such by which for every $r\in ϝ$, we obtain ${\mathsf{\Gamma }}^{n}r\to \zeta$.

Proof. 

Take $\left\{r_n\right\}$ and $r_0$ to be the same as in the representation of this theorem. If $r_m=r_{m+1}=\mathsf{\Gamma }{r}_m$ for any arbitrary m, then it can be easily seen that the fixed point is $r_m$. Meanwhile, assume that $r_{n+1}\ne r_n$ for every n. Employing the condition (8),

$$g(r_n,r_{n+1})=g(\mathsf{\Gamma }{r}_{n-1},\mathsf{\Gamma }{r}_n)\le \psi (\Delta (r_{n-1},r_n)),$$

(10)

where clearly, $\Delta (r_{n-1},r_n)=max\{g(r_{n-1},r_n),g(r_n,r_{n+1})\}$. If for any arbitrary n, we approve that $\Delta (r_{n-1},r_n)=g(r_n,r_{n+1})$, then using (10) in addition to $\psi \left(k\right)<k$, for every positive t we obtain

$$0<g(r_n,r_{n+1})\le \psi \left(g(r_n,r_{n+1})\right)<g(r_n,r_{n+1}),$$

(11)

which clearly yields a contradiction. Subsequently, for every n it should be expressed as $\Delta (r_{n-1},r_n)=g(r_{n-1},r_n)$. By conclusion, it means that $0<g(r_n,r_{n+1})\le \psi \left(g(r_{n-1},r_n)\right)$. If we continue inductively, we conclude that for every $n\in [0,\infty )$, we obtain

$$0<g(r_n,r_{n+1})\le {\psi }^n\left(g(r_0,r_1)\right).$$

Using the supposition of $\psi$, we deduce that the

$$\underset{n\to \infty }{lim}g(r_n,r_{n+1})=0.$$

To prove that $\left\{r_n\right\}$ is Cauchy, we follow the same procedure we used in proving the Theorem 1. For every $m>n$, we obtain

$$g(p_n,p_m)\le g(p_n,p_{n+1})+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\omega (p_j,p_m)\right){\psi }^i\left(g(p_0,p_1)\right).$$

(12)

By using the ratio test and applying it to the second addend on the right-hand side of (12) along with the condition (9), it verifies that $\left\{r_n\right\}$ is Cauchy-similar to the proof of Theorem 1. With reference to $(ϝ,\Delta )$ being complete, ∃$\zeta \in ϝ$ satisfies ${lim}_{n\to \infty }g(r_n,\zeta )=0$. Thus, since $\mathsf{\Gamma }$ is continuous, we deduce that

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{r}_{n+1}& =\underset{n\to \infty }{lim}\mathsf{\Gamma }{r}_n\hfill \\ \hfill & =\mathsf{\Gamma }\underset{n\to \infty }{lim}{r}_n\hfill \\ \hfill & =\mathsf{\Gamma }\zeta \hfill \end{array}$$

and $\zeta$ is a fixed point of $\mathsf{\Gamma }.$ In order to verify that the fixed point is unique, let z satisfy the definition $\mathsf{\Gamma }z=z$ where $\zeta \ne$z. By (8), we reach

$$0<g(\zeta ,z)=g(\mathsf{\Gamma }\zeta ,\mathsf{\Gamma }z)\le \psi (\Delta (\zeta ,z\left)\right)=\psi \left(g\right(\zeta ,z\left)\right)<g(\zeta ,z),$$

that clearly yields a contradiction. □

Remark 1.

With reference to Theorem 2, if we consider how $\psi \left(r\right)=tr,$t lies in the interval $(0,1)$, then the (8) constraint will take the following form:

$$g(\mathsf{\Gamma }p,\mathsf{\Gamma }q)\le tmax\left\{g\right(p,q),g(p,\mathsf{\Gamma }p),g(q,\mathsf{\Gamma }q\left)\right\}.$$

(13)

In relation to [32] Kannan’s result of fixed points, the below theorem articulates the outcome.

Theorem 3.

Consider $(ϝ,g)$ that is a complete (CSbMS) by the function $\omega :ϝ\times ϝ\to [1,\infty )$. Let $\mathsf{\Gamma }:ϝ\to ϝ$ to be a Kannan mapping defined as follows:

$$g(\mathsf{\Gamma }p,\mathsf{\Gamma }q)\le b\left[g\right(p,\mathsf{\Gamma }p)+g(q,\mathsf{\Gamma }q\left)\right],$$

(14)

for all $p,q\in ϝ$, where $b\in (0,{\displaystyle \frac{1}{2}})$. For $r_0\in ϝ$, take $r_n={\mathsf{\Gamma }}^n{r}_0$. Suppose that

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\omega (p_{i+1},p_m)<\frac{1-b}{b}.$$

(15)

For each $r\in ϝ$, assume that

$$\underset{n\to \infty }{lim}\omega (r,r_n)exists,isfiniteand\underset{n\to \infty }{lim}\omega (r_n,r)<\frac{1}{b}.$$

(16)

Hence, Γ has a point that is fixed and unique.

Proof. 

Consider $\{r_n=\mathsf{\Gamma }{r}_{n-1}\}$ that belongs to $ϝ$ and satisfies the given conditions (15) and (16) in the Theorem. Using (14), we attain

$$\begin{array}{ccc}\hfill g(r_n,r_{n+1})& =& g(\mathsf{\Gamma }{r}_{n-1},\mathsf{\Gamma }{r}_n)\hfill \\ & \le & b[g(r_{n-1},\mathsf{\Gamma }{r}_{n-1})+g(r_n,\mathsf{\Gamma }{r}_n)]\hfill \\ & =& b[g(r_{n-1},r_n)+g(r_n,r_{n+1})].\hfill \end{array}$$

Then $g(r_n,r_{n+1})\le {\displaystyle \frac{b}{1-b}}g(r_{n-1},r_n)$. By induction, we obtain

$$g(r_n,r_{n+1})\le {\left({\displaystyle \frac{b}{1-b}}\right)}^{n}g(r_1,r_0),\forall n\ge 0.$$

(17)

Currently, our purpose is to prove that the sequence $\left\{r_n\right\}$ is Cauchy. Utilizing the triangular inequality, for every $n,m\in \mathbb{N}$, we obtain

$$g(r_n,r_m)\le g(r_n,r_{n+1})+\omega (r_{n+1},r_m)g(r_{n+1},r_m).$$

Using similar proof as the Theorem 1, we obtain

$$\begin{array}{cc}\hfill g(r_n,r_m)& \le g(r_n,r_{n+1})+\sum _{i=0}^{m-1}\left(\prod _{j=0}^i\omega (r_j,r_m)\right)g(r_i,r_{i+1})\hfill \\ \hfill & \le {\left({\displaystyle \frac{b}{1-b}}\right)}^{n}g(r_0,r_1)+\sum _{i=0}^{m-1}\left(\prod _{j=0}^i\omega (r_j,r_m)\right){\left({\displaystyle \frac{b}{1-b}}\right)}^{i}g(r_0,r_1).\hfill \end{array}$$

Since $0\le b<{\displaystyle \frac{1}{2}}$, we have ${\displaystyle \frac{b}{1-b}}\in [0,1)$. As we guarantee that the constant behaves as the constant t in Theorem 1, it grants us to prove that the sequence $\left\{r_n\right\}$ is Cauchy in the complete (SCMTS) $(ϝ,\Delta )$ by following the same method of proof in the mentioned Theorem. Thus, ∃r$\in ϝ$ as the limit of $\left\{r_n\right\}$ in $(ϝ,\Delta )$. Suppose that $\mathsf{\Gamma }r\ne r$. We obtain

$$\begin{array}{ccc}\hfill 0<g(r,\mathsf{\Gamma }r)& \le & g(r,r_{n+1})+\omega (r_{n+1},\mathsf{\Gamma }r)g(r_{n+1},\mathsf{\Gamma }r)\hfill \\ & \le & g(r,r_{n+1})+\omega (r_{n+1},\mathsf{\Gamma }r)[bg(r_n,r_{n+1})+bg(r,\mathsf{\Gamma }r)].\hfill \end{array}$$

(18)

Inserting the limit to both sides of (18) and utilizing the condition (16), we conclude that $0<g(r,\mathsf{\Gamma }r)<g(r,\mathsf{\Gamma }r)$, that leads to a contradiction. Then, $\mathsf{\Gamma }r=r$. Now, we have to prove that the fixed point r is unique. Assume that $\mathsf{\Gamma }$ has two fixed points r and s. Thus,

$$\begin{array}{ccc}\hfill g(r,s)=g(\mathsf{\Gamma }r,\mathsf{\Gamma }s)& \le & b\left[g\right(r,\mathsf{\Gamma }r)+g(s,\mathsf{\Gamma }s\left)\right]\hfill \\ & =& b\left[g\right(r,r)+g(s,s\left)\right]=0.\hfill \end{array}$$

Therefore, $s=r$ and we deduce that the mapping $\mathsf{\Gamma }$ has a unique fixed point. □

4. Application

4.1. Polynomial Equations

Finally, we support our Theorems by providing the below application.

Theorem 4.

Consider the following equation:

$$p^m+1=(m^4-1)p^{m+1}+m^{4}p,$$

(19)

where m represents any natural number that is greater than or equal to 3, $m\ge 3$. Then the equation has one solution that is real and unique.

Proof. 

To begin with, we should notice that if $|p|>1,$ the solution set of the equation $\left(3.1\right)$ would be $\varphi$. Thus, consider $ϝ=[-1,1]$, and along with that, for every $p,q\in ϝ$ take $\Delta (p,q)=|p-q|$ and $\eta (p,q)=max\{p,q\}+2.$ Without difficulty we can observe that $(ϝ,\Delta )$ is a complete (CSbMS). Now, let

$$\mathsf{\Gamma }p=\frac{p^m+1}{(m^4-1)p^m+m^4}.$$

Note that, because $m\ge 2,$ we conclude that $m^4\ge 6.$ Therefore,

$$\begin{array}{cc}\hfill \Delta (\mathsf{\Gamma }p,\mathsf{\Gamma }q)& =|\frac{p^m+1}{(m^4-1)p^m+m^4}-\frac{q^m+1}{(m^4-1)q^m+m^4}|\hfill \\ \hfill & =|\frac{p^m-q^m}{((m^4-1)p^m+m^4)((m^4-1)q^m+m^4)}|\hfill \\ \hfill & \le \frac{|p-q|}{m^4}\hfill \\ \hfill & \le \frac{|p-q|}{6}\hfill \\ \hfill & =\frac{1}{6}\Delta (p,q)\hfill \end{array}$$

Hence,

$$\Delta (\mathsf{\Gamma }p,\mathsf{\Gamma }q)\le t\Delta (p,q)\mathrm{where}t=\frac{1}{6}$$

Furthermore, every $p_0\in ϝ$$\mathsf{\Gamma }$ fulfills all the given conditions of Theorem 1. Consequently, $\mathsf{\Gamma }$ has a point that is unique and fixed in $ϝ,$ and for that reason, Equation (19) has the required real solution that is unique. □

4.2. Linear System of Equations

Consider the set $ϝ={\mathbb{R}}^n$ where $\mathbb{R}$ is the set of real numbers and n a positive integer. Now, consider the (CSbMS) $(ϝ,\Delta )$ defined by

$$\Delta (\delta ,\xi )=\underset{1\le i\le n}{max}|{\delta }_i-{\xi }_i|$$

for all $\delta =({\delta }_1,...,{\delta }_n),\xi =({\xi }_1,...,{\xi }_n)\in ϝ.$

Theorem 5.

Consider the following system:

$$\left\{\begin{array}{c}{s}_{11}{\delta }_1+s_{12}{\delta }_2+s_{13}{\delta }_3+s_{1n}{\delta }_n=r_1\hfill \\ s_{21}{\delta }_1+s_{22}{\delta }_2+s_{23}{\delta }_3+s_{2n}{\delta }_n=r_2\hfill \\ ⋮\hfill \\ s_{n1}{\delta }_1+s_{n2}{\delta }_2+s_{n3}{\delta }_3+s_{nn}{\delta }_n=r_n\hfill \end{array}\right.$$

if ${\displaystyle \theta =\underset{1\le i\le n}{max}\left(\sum _{j=1,j\ne i}^n|s_{ij}|+|1+s_{ii}|\right)<1,}$ then the above linear system has a unique solution.

Proof. 

Consider the map $\sigma :ϝ\to ϝ$ defined by $\sigma \delta =(B+I_n)\delta -r$, where

$$B=\left(\begin{array}{cccc}{s}_{11}& s_{12}& \cdots & s_{1n}\\ s_{21}& s_{22}& \cdots & s_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ s_{n1}& s_{n2}& \cdots & s_{nn}\end{array}\right)$$

$\delta =({\delta }_1,{\delta }_2,\cdots ,{\delta }_n)\mathrm{and}\xi =({\xi }_1,{\xi }_2,\cdots ,{\xi }_n)\in {\mathbb{R}}^n$, $I_n$ is the identity matrix for $n\times n$ matrices and $r=(r_1,r_2,\cdots ,r_n)\in {\mathbb{C}}^n.$ Let us prove that $\Delta (\sigma \delta ,\sigma \xi )\le \theta \Delta (\delta ,\xi )$, $\forall \delta ,\xi \in {\mathbb{R}}^n$. We denote by

$$\tilde{B}=B+I_n=\left({\tilde{b}}_{ij}\right),i,j=1,...,n,$$

with ${\tilde{b}}_{ij}=\left\{\begin{array}{c}{s}_{ij},j\ne i\hfill \\ 1+s_{ii},j=i\hfill \end{array}\right.$ Hence,

$${\displaystyle \underset{1\le i\le n}{max}\sum _{j=1}^n|{\tilde{b}}_{ij}|=\underset{1\le i\le n}{max}\left(\sum _{j=1,j\ne i}^n|s_{ij}|+|1+s_{ii}|\right)=\theta <1.}$$

On the other hand, for all $i=1,...,n$, we have

$$\begin{array}{cc}\hfill {\left(\sigma \delta \right)}_i-{\left(\sigma \xi \right)}_i& {\displaystyle =\sum _{j=1}^n{\tilde{b}}_{ij}({\delta }_j-{\xi }_j),}\end{array}$$

(20)

Therefore, using (20) we obtain

$$\begin{array}{cc}\hfill \Delta (\sigma \delta ,\sigma \xi )& =\underset{1\le i\le n}{max}\left({|\left(\sigma \delta \right)}_i-{\left(\sigma \xi \right)}_i|\right)\hfill \\ \hfill & {\displaystyle \le \underset{1\le i\le n}{max}\left(\sum _{j=1}^n|{\tilde{b}}_{ij}||{\delta }_j-{\xi }_j|\right)}\hfill \\ \hfill & {\displaystyle \le \underset{1\le i\le n}{max}\sum _{j=1}^n|{\tilde{b}}_{ij}|\underset{1\le k\le n}{max}\left(|{\delta }_k-{\xi }_k|\right)}\hfill \\ \hfill & =\theta \Delta (\delta ,\xi ),\hfill \end{array}$$

Thus, all the hypotheses of Theorem 1 are satisfied. Hence, $\sigma$ has a unique fixed point. Therefore, the above linear system has a unique solution as desired. □

4.3. Fractional Differential Equation

In this section, we discuss the existence of a solution to the following problem:

$$\left(\mathcal{P}\right):\left\{\begin{array}{ccc}{D}^{\lambda }x\left(t\right)& =& f(t,x\left(t\right))=Fx\left(t\right)ift\in I_0=(0,T]\hfill \\ x\left(0\right)& =& x\left(T\right)=r\hfill \end{array}\right\}$$

where $T>0$ and $f:I\times \mathbb{R}⟶\mathbb{R}$ is a continuous function, and $I=[0,T]$ and $D^{\lambda }x$ denote a Riemann–Liouville fractional derivative of x with $\lambda \in (0,1)$.

Let $C_{1-\lambda }(I,\mathbb{R})=\{f\in C((0,T],\mathbb{R}):t^{1-\lambda }f\in C(I,\mathbb{R})\}$. We define the following weighted norm:

$${||f||}^{\ast }=\underset{t\in [0,T]}{max}{t}^{1-\lambda }|f\left(t\right)|.$$

Theorem 6.

Let $\lambda \in (0,1)$, $f\in C(I\times I,\mathbb{R})$ be increasing and $0<\alpha <1$. In addition, we assume the following hypothesis:

$$|f(u_1\left(t\right),v_1\left(t\right))-f(u_2\left(t\right),v_2\left(t\right))|\le {\displaystyle \frac{\mathsf{\Gamma }\left(2\lambda \right)}{T^{2\lambda -1}}}\alpha |v_1-v_2|$$

Then the problem $\left(\mathcal{P}\right)$ has a unique solution.

Proof. 

Problem $\left(\mathcal{P}\right)$ is equivalent to the problem $\mathcal{M}x\left(t\right)=x\left(t\right)$ where

$$\mathcal{M}x\left(t\right)=r{t}^{\lambda -1}+{\displaystyle \frac{1}{\mathsf{\Gamma }\left(\lambda \right)}}{\int }_0^t{(t-s)}^{\lambda -1}Fx\left(s\right)ds.$$

Note that, showing that $\mathcal{M}$ has a fixed point is equivalent to showing that $\mathcal{P}$ has a unique solution. Now, suppose that $\mathcal{M}x\left(t\right)=x\left(t\right)$; by applying $D^{\lambda }$ we get $D^{\lambda }x\left(t\right)=Fx\left(t\right).$ Thus, we need to show that all the conditions of Theorem 1 are satisfied.

Note that $(A=C_{1-\lambda }(I,\mathbb{R}),\Delta )$ is a complete (CSbMS) if

$$\Delta (x,y)=\underset{[0,T]}{max}{t}^{1-\lambda }\left(|x(t)-y(t)|\right),x,y\in C_{1-\lambda }(I,\mathbb{R}).$$

Note that, $\mathcal{M}$ is increasing and that is due to the fact that is f is increasing.

Next, we show that $\mathcal{M}$ is a contractive. Let $x,y,z\in C_{1-\lambda }(J,\mathbb{R})$, $0<\lambda <1$.

$$\begin{array}{ccc}\hfill \Delta (\mathcal{M}x,\mathcal{M}y)& =& \underset{[0,T]}{max}{t}^{1-\lambda }\left(|\mathcal{M}x(t)-\mathcal{M}y(t)|\right)\hfill \\ & \le & {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\lambda \right)}}\underset{t\in [0,T]}{max}{t}^{1-\lambda }{\int }_0^t{(t-s)}^{\lambda -1}\left(|f(t,x\left(s\right))-f(t,y\left(s\right))|\right)ds.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill J(\mathcal{M}x,\mathcal{M}y)& \le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\lambda \right)}}\underset{t\in [0,T]}{max}{t}^{1-\lambda }{\int }_0^t{(t-s)}^{\lambda -1}{\displaystyle \frac{\mathsf{\Gamma }\left(2\lambda \right)}{T^{2\lambda -1}}}\left[\alpha |x(s)-y(s)|\right]ds\hfill \\ \hfill & \le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\lambda \right)}}\underset{t\in [0,T]}{max}{t}^{1-\lambda }{\int }_0^t{(t-s)}^{\lambda -1}{\displaystyle \frac{\mathsf{\Gamma }\left(2\lambda \right)}{T^{2\lambda -1}}}\left[\alpha ||x-{y||}^{\ast }{s}^{\lambda -1}\right]ds\hfill \\ \hfill & \le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\lambda \right)}}\underset{t\in [0,T]}{max}{t}^{1-\lambda }\alpha (||x-{y||}^{\ast }){\displaystyle \frac{\mathsf{\Gamma }\left(2\lambda \right)}{T^{2\lambda -1}}}{\int }_0^t{(t-s)}^{\lambda -1}{s}^{\lambda -1}ds.\hfill \end{array}$$

From the Riemann–Liouville fractional integral, we have

$${\int }_0^t{(t-s)}^{\lambda -1}{s}^{\lambda -1}ds={\displaystyle \frac{\mathsf{\Gamma }\left(\lambda \right)}{\mathsf{\Gamma }\left(2\lambda \right)}}{t}^{2\lambda -1}.$$

Therefore, we have

$$\Delta (\mathcal{M}x,\mathcal{M}y)\le \alpha \Delta (x,y).$$

Thus, by Theorem 1, we deduce that $\mathcal{M}$ has a unique fixed point which leads us to conclude that equation $\left(\mathcal{P}\right)$ has a unique solution as desired. □

5. Conclusions

In this manuscript, we introduced a new type of metric space, called the strong controlled metric space. We proved the existence and uniqueness of fixed points for self-mappings that satisfies linear and nonlinear contractions in such spaces. Moreover, we provided different applications of our results to polynomial equations, the system of linear equations, along with fractional differential equations. Our work generalizes many results in the literature and opens the door for many future studies.