Some Fixed-Point Theorems in b-Dislocated Metric Space and Applications

Abstract

In this article, we prove some fixed-point theorems in b-dislocated metric space. Thereafter, we propose a simple and efficient solution for a non-linear integral equation and non-linear fractional differential equations of Caputo type by using the technique of fixed point.

1. Introduction

The F-contraction is widely renowned, and has provided research contributions on various aspects in fixed-point theory for over 10 years.

Definition 1

([1]). Let $(X,d)$ be a metric space. A mapping $T:X\to X$ is said to be a F-contraction on $(X,d)$ if there exist F along with $C_1,C_2,C_3$ and $\tau >0$ such that

$$\begin{array}{}(\star )& d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F(d(x,y)).\end{array}$$

From above we have:

• every F-contraction is necessarily continuous;
• every F-contraction is need not be a Banach contraction.

Now we present a list of certain F-contractive conditions for a self-map on a metric space:

(E1)

$\frac{1}{2}d(x,Tx)<d(x,y)\Rightarrow \tau +F(d(Tx,Ty))\le F(d(x,y))$,

(E2)

$\frac{1}{2}d(x,Tx)<d(x,y)\Rightarrow \tau +F(d(Tx,Ty))\le F\left(d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)+d(y,Ty)}{4}\right)$,

(E3)

$d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F\left\{max\left(d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\right)\right\}$

(E4)

$d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F(d(x,y)+Ld(y,Tx))$ and

$d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F(d(x,y)+Ld(x,Ty))$

(E5)

$\varpi (d(x,y))+F(d(Tx,Ty))\le F(f_{1}d(x,y)+f_{2}d(x,Tx)+f_{3}d(y,Ty)+f_{4}d(x,Ty)+f_{5}d(y,Tx))$

(E6)

$\varpi (d(x,y))+F(d(Tx,Ty))\le F(d(x,y))$

for all $x,y\in X$, where $\tau >0$, $T:X\to X,F:{\mathbb{R}}_0^{+}\to \mathbb{R}$ and $\varpi :(0,\infty )\to (0,\infty ).$

Below is the list of necessary conditions which helps authors to establish the statements of the certain F-contractions and treated as main tool to obtain fixed points:

$(C_1)$. 

F is strictly increasing;

$(C_2)$. 

$\underset{n\to \infty }{lim}{t}_{1_n}=0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\underset{n\to \infty }{lim}F(t_{1_n})=-\infty ;$

$(C_3)$. 

There exists $k\in (0,1)$ such that $\underset{t_1\to 0^{+}}{lim}{t}_1^{k}F(t_1)=0;$

$(C_4)$. 

inf F = $-\infty$;

$(C_5)$. 

F is continuous on $(0,\infty )$;

$(C_6)$. 

${lim\; inf}_{u\to v^{+}}\varpi (u)>0$ for all $v\ge 0$;

$(C_7)$. 

$\underset{v\to 0^{+}}{lim}F(v)=-\infty .$

We present a list of F-contractions and we named these conditions with their respective authors:

1°.

Secelean type F-contraction [2] if T satisfies $(\star )$ along with $(C_1)$, $(C_4)$ and $(C_5)$;

2°.

Piri type F-contraction [3] if T satisfies $(E1)$ along with $(C_1)$, $(C_4)$ and $(C_5)$;

3°.

Karapınar type F-contraction [4] if T satisfies $(E2)$ along with $(C_1)$, $(C_4)$ and $(C_5)$;

4°.

Wardowski-I type F-contraction [5] if T satisfies $(E3)$ along with $(C_1)$, $(C_2)$ and $(C_3)$;

5°.

Minak type F-contraction [6] if T satisfies $(E3)$ along with $(C_1)$, $(C_2)$ and $(C_3)$;

6°.

Minak-I type F-contraction [6] if T satisfies $(E4)$ along with $(C_1)$, $(C_2)$ and $(C_3)$;

7°.

Vetro type F-contraction [7] if T satisfies $(E5)$ along with $(C_1)$, $(C_2),(C_3)$ and $(C_6)$;

8°.

Wardowski-II type F-contraction [8] if T satisfies $(E6)$ along with $(C_1)$, $(C_6)$ and $(C_7)$;

9°.

A. Lukacs and S. Kajanto type F-contraction [9] if T satisfies $(\star )$ along with $(C_1)$ and $(C_3)$;

If we consider the F-contraction in the theory of fixed point, we can see various developments in it. More particularly, the techniques used to obtain fixed points in [1] have attracted several authors. In this scenario, many authors imposed various restrictions to obtain the existence of a fixed point (see for example [10,11,12,13,14,15]).

On the other hand, there are many generalizations on the concept of metric spaces in the literature. In particular, Matthews [16] introduced the notion of dislocated metric space under the name ’metric domains’. Later, in 2000, Hitzler and Seda [17] renamed these spaces as ’dislocated metric spaces’ as below:

Definition 2

([17]). A dislocated metric on a nonempty set X is a function $d:X\times X\to [0,+\infty )$ such that for all $x,y,z\in X$. Then the following conditions hold good:

1.

$d(x,y)=0\Rightarrow x=y$;

2.

$d(x,y)=d(y,x)$;

3.

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

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

Very recently, Alghamdi et al. [18] introduced b-dislocated metric space as below.

Definition 3

([18]). A b-dislocated metric on a nonempty set X is a function $d:X\times X\to [0,+\infty )$ such that for all $x,y,z\in X$ and a constant $b\ge 1$ the following conditions hold good:

1.

$d(x,y)=0\Rightarrow x=y$;

2.

$d(x,y)=d(y,x)$;

3.

$d(x,y)\le k[d(x,z)+d(z,y)]$.

The pair $(X,d)$ is called a b-dislocated metric space. Moreover, if $k=1$ then b-dislocated metric space becomes dislocated metric space as in [16,17].

The structure properties of convergent, Cauchy and complete of b-dislocated metric space can be found in [18] as below:

Definition 4

([18]). Let $(X,d)$ be a b-dislocated metric space, and let $\left\{x_n\right\}$ be a sequence of points of X. Then $\left\{x_n\right\}$ called,

• Convergent to x if and only if $\underset{n\to +\infty }{lim}d(x,x_n)=0$
• Cauchy if and only if $\underset{m,n\to \infty }{lim}d(x_n,x_m)$ exists and is tends to finite.
• Complete if and only if every Cauchy sequence $\left\{x_n\right\}$ in X converges to $x\in X$ this gives

  $$\underset{m,n\to \infty }{lim}d(x_n,x_m)=0=\underset{n\to \infty }{lim}d(x_n,x).$$

The study of dislocated metric space and its generalizations was very interesting and expanded very extensively. For more info, the reader can refer to [19,20,21,22,23,24,25,26].

Motivated by Wardowski [1], we introduce the notions of an extended F-contraction and weak-generalized F-contraction. Furthermore, we establish some fixed-point results for given named contractions. Thereafter, we propose a simple and efficient solution for a non-linear integral equation by using the technique of fixed point in the setting of b-dislocated metric space.

2. Extended $\mathit{F}$-Contraction

Now we start this section by introducing below definition.

Definition 5.

Let $(X,d)$ be a b-dislocated metric space. A mapping $T:X\to X$ is said to be an Extended F-contraction if there exists $F\in \mathcal{F}$ and $\tau >0$ such that for all $x,y\in X$ with $d(Tx,Ty)>0,$

$$\frac{1}{2k}d(x,Tx)<d(x,y)\Rightarrow \tau +F(d(Tx,Ty))\le F(\mathcal{A}(x,y))$$

where,

$$\begin{array}{cc}\hfill \mathcal{A}(x,y)& =\frac{{\mathcal{A}}_1(k+1)}{k}d(x,y)+{\mathcal{A}}_2[d(x,Tx)+d(y,Ty)]+{\mathcal{A}}_3\left[\frac{d(x,Ty)+d(y,Tx)}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x,x)+d(y,y)}{4k}\right]\hfill \end{array}$$

here ${\mathcal{A}}_1,{\mathcal{A}}_2,{\mathcal{A}}_3,{\mathcal{A}}_4\ge 0$, $\frac{{\mathcal{A}}_1(k+1)}{k}+2({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)<1$ and $F:{\mathbb{R}}^{+}\to \mathbb{R}$ is a mapping satisfying the following conditions:

1.

F is an order embedding. i.e., for all $t_1,t_2\in {\mathbb{R}}^{+}$ we have $t_1\le t_2\Rightarrow F(t_1)\le F(t_2)$.

2.

F is sub-additive, i.e., for $t_1,t_2,s_1,s_2\in {\mathbb{R}}^{+}$ we have

$$F(s_1{t}_1+s_2{t}_2)\le s_{1}F(t_1)+s_{2}F(t_2)$$

3.

For every sequence ${\left\{t_{1_n}\right\}}_{n\in \mathbb{N}}$ of positive numbers

$$\underset{n\to \infty }{lim}{t}_{1_n}=0\mathit{if}\mathit{and}\mathit{only}\mathit{if}\underset{n\to \infty }{lim}F(t_{1_n})=-\infty ;$$

Let $\mathcal{F}$ denote the family of all functions $F:{\mathbb{R}}^{+}\to \mathbb{R}$ which satisfy conditions $(1),(2)$ and $(3).$

Under this new scenario, we will prove below theorem.

Theorem 1.

Let $(X,d)$ be a complete b-dislocated metric space such that d is a continuous function. If T is an Extended F-contraction then T has a unique fixed point.

Proof. 

Now, take $x\in X$ and build a sequence $\left\{x_n\right\}$ as follows:

$$x_n=T{x}_{n-1}=T^{n}x,\mathrm{for}\mathrm{all}n\in \mathbb{N}\mathrm{where}{x}_0=x.$$

If there exists $n_{\star }\in \mathbb{N}$ such that $d(x_{n_{\star }},T{x}_{n_{\star }})=0$ then $x_{\star }=x_{n_{\star }}$ turn into a fixed point which completes the proof. As a result, we assume for every $n\in \mathbb{N},$

$$0<d(x_n,T{x}_n).$$

(1)

Therefore, from (1), we have

$$\frac{1}{2k}d(x_n,T{x}_n)<d(x_n,T{x}_n)=d(x_n,x_{n+1}),\mathrm{for}\mathrm{all}n\in \mathbb{N}$$

(2)

Since T is Extended F-contraction,

$$\begin{array}{cc}\hfill \tau +F(d(T{x}_n,T{x}_{n+1}))& \le F(\mathcal{A}(x_n,x_{n+1}))\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_n,x_{n+1})+{\mathcal{A}}_2[d(x_n,T{x}_n)+d(x_{n+1},T{x}_{n+1})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_n,T{x}_{n+1})+d(x_{n+1},T{x}_n)}{3k}\right]+{\mathcal{A}}_4\left[\frac{d(x_n,x_n)+d(x_{n+1},x_{n+1})}{4k}\right]\}\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_n,x_{n+1})+{\mathcal{A}}_2[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_n,x_{n+2})+d(x_{n+1},x_{n+1})}{3k}\right]+{\mathcal{A}}_4\left[\frac{d(x_n,x_n)+d(x_{n+1},x_{n+1})}{4k}\right]\}\hfill \end{array}$$

(3)

From triangle inequality, we have,

$$d(x_n,x_{n+2})\le k[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})]$$

(4)

and

$$\begin{array}{cc}\hfill d(x_{n+1},x_{n+1})& \le 2kd(x_{n+2},x_{n+1})\hfill \\ & \le 2k[d(x_{n+1},x_{n+2})+d(x_n,x_{n+1})]\hfill \end{array}$$

(5)

By using $(4)$ and $(5)$,

$$\frac{d(x_n,x_{n+2})+d(x_{n+1},x_{n+1})}{3k}\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2}).$$

(6)

Also,

$$\begin{array}{cc}\hfill d(x_n,x_n)& \le 2kd(x_n,x_{n+1})\hfill \\ & \le 2k[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})]\hfill \end{array}$$

(7)

From $(5)$ and $(7)$,

$$\frac{d(x_n,x_n)+d(x_{n+1},x_{n+1})}{4k}\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2}).$$

(8)

From $(3),(7)$ and $(8)$ we get

$$\begin{array}{cc}\hfill \tau +F(d(T{x}_n,T{x}_{n+1}))& \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_n,x_{n+1})+{\mathcal{A}}_2[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})]\hfill \\ & +{\mathcal{A}}_3\left[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})\right]\hfill \\ & +{\mathcal{A}}_4\left[d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})\right]\}\hfill \end{array}$$

(9)

Since $F\in \mathcal{F}$,

$$\begin{array}{cc}\hfill \tau +F(d(x_{n+1},x_{n+2}))& \le \frac{{\mathcal{A}}_1(k+1)}{k}F(d(x_n,x_{n+1}))+{\mathcal{A}}_{2}F(d(x_n,x_{n+1}))+{\mathcal{A}}_{2}F(d(x_{n+1},x_{n+2}))\hfill \\ & +{\mathcal{A}}_{3}F(d(x_n,x_{n+1}))+{\mathcal{A}}_{3}F(d(x_{n+1},x_{n+2}))\hfill \\ & +{\mathcal{A}}_{4}F(d(x_n,x_{n+1}))+{\mathcal{A}}_{4}F(d(x_{n+1},x_{n+2}))\hfill \\ & \le \left[\frac{{\mathcal{A}}_1(k+1)}{k}+{\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4\right]F(d(x_n,x_{n+1}))+({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)F(d(x_{n+1},x_{n+2}))\hfill \end{array}$$

(10)

$$\Rightarrow (1-{\mathcal{A}}_2-{\mathcal{A}}_3-{\mathcal{A}}_4)F(d(x_{n+1},x_{n+2}))\le \left[\frac{{\mathcal{A}}_1(k+1)}{k}+{\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4\right]F(d(x_n,x_{n+1})-\tau$$

$$\Rightarrow F(d(x_{n+1},x_{n+2}))\le \frac{\left[\frac{{\mathcal{A}}_1(k+1)}{k}+{\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4\right]}{(1-{\mathcal{A}}_2-{\mathcal{A}}_3-{\mathcal{A}}_4)}F(d(x_n,x_{n+1})-\frac{\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}$$

$$F(d(x_{n+1},x_{n+2}))\le {\rm Y}F(d(x_n,x_{n+1})-\frac{\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}$$

(11)

where ${\rm Y}=\frac{\left[\frac{{\mathcal{A}}_1(k+1)}{k}+{\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4\right]}{(1-{\mathcal{A}}_2-{\mathcal{A}}_3-{\mathcal{A}}_4)}<1$.

Thus,

$$F(d(x_{n+1},x_{n+2}))\le F(d(x_n,x_{n+1})-\frac{\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}$$

(12)

If we continue same scenario, we get,

$$\begin{array}{cc}\hfill F(d(x_n,T{x}_n))& =F(d(x_n,x_{n+1}))\hfill \\ & \le F(d(x_{n-1},x_n))-\frac{\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}\hfill \\ & \le F(d(x_{n-2},x_{n-1}))-\frac{2\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}\hfill \\ & \le F(d(x_{n-3},x_{n-2}))-\frac{3\tau }{1-({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)}\hfill \\ & ⋮\hfill \\ & \le F(d(x_0,x_1))-n\tau \hfill \\ & \le F(d(x_0,T{x}_0))-n\tau \hfill \end{array}$$

(13)

From (13), we obtain,

$$\underset{n\to \infty }{lim}F(d(x_n,T{x}_n))=-\infty$$

(14)

From (15) and from the hypothesis, we have

$$\underset{n\to \infty }{lim}F(d(x_n,T{x}_n))=\underset{n\to \infty }{lim}F(d(x_n,x_{n+1}))=0.$$

(15)

Now, we claim that

$$\underset{n\to \infty }{lim}F(d(x_n,x_m))=0.$$

(16)

i.e., $\left\{x_n\right\}$ is a Cauchy sequence. Let us suppose that $\left\{x_n\right\}$ is not a Cauchy sequence, then there exist $ϵ>0$ and sequence ${\left\{a(n)\right\}}_{n=1}^{\infty }$ and ${\left\{b(n)\right\}}_{n=1}^{\infty }$ of natural numbers such that $a(n)>b(n)>n$, $d(x_{a(n)},x_{b(n)})\ge ϵ$,

$$d(x_{a(n)-1},x_{b(n)})<\frac{ϵ}{k},\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

(17)

From the triangle inequality, we get,

$$\begin{array}{cc}\hfill ϵ& \le d(x_{a(n)},x_{b(n)})\le k[d(x_{a(n)},x_{a(n)-1})+d(x_{a(n)-1},x_{b(n)})]\hfill \\ & \le kd(x_{a(n)},x_{a(n)-1})+ϵ\hfill \\ & =kd(x_{a(n)-1},T{x}_{a(n)-1})+ϵ\hfill \end{array}$$

(18)

From (15) and (18) and the sandwich theorem, we get,

$$\underset{n\to \infty }{lim}d(x_{a(n)},x_{b(n)})=ϵ,\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

(19)

From Extended F-contraction, we have

$$\begin{array}{cc}\hfill \tau +F(d(T{x}_{a(n)},T{x}_{b(n)}))& \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_{a(n)},x_{b(n)})+{\mathcal{A}}_2[d(x_{a(n)},T{x}_{a(n)})+d(x_{b(n)},T{x}_{b(n)})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_{a(n)},T{x}_{b(n)})+d(x_{b(n)},T{x}_{a(n)})}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x_{a(n)},x_{a(n)})+d(x_{b(n)},x_{b(n)})}{4k}\right]\}\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_{a(n)},x_{b(n)})+{\mathcal{A}}_2[d(x_{a(n)},x_{a(n)+1})+d(x_{b(n)},x_{b(n)+1})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_{a(n)},x_{b(n)+1})+d(x_{b(n)},x_{a(n)+1})}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x_{a(n)},x_{a(n)})+d(x_{b(n)},x_{b(n)})}{4k}\right]\}\hfill \end{array}$$

(20)

Letting $n\to \infty$,

$$\begin{array}{cc}\hfill \tau +F(ϵ)& \le F\left[\frac{{\mathcal{A}}_1(k+1)}{k}ϵ+{\mathcal{A}}_3\left(\frac{k(ϵ)+k(ϵ)}{3k}\right)\right]\hfill \\ & \le F\left(\frac{{\mathcal{A}}_1(k+1)}{k}ϵ+\frac{2{\mathcal{A}}_3ϵ}{3}\right),\hfill \end{array}$$

(21)

which is a contradiction.

Hence $\underset{m,n\to \infty }{lim}d(x_n,x_m)=0.$ Thus, the sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ is a Cauchy in X.

Since $(X,d)$ is a Complete b-dislocated metric space, there exists $\omega \in X$ such that

$$d(\omega ,\omega )=\underset{n\to \infty }{lim}d(x_n,\omega )=\underset{m,n\to \infty }{lim}d(x_n,x_m)=0.$$

(22)

From (22)

$$d(x_{n+1},T\omega )\le k[d(x_{n+1},\omega )+d(\omega ,T\omega )]$$

letting $n\to \infty ,$

$$\underset{n\to \infty }{lim}d(x_{n+1},T\omega )=k{b}_d(\omega ,T\omega )$$

(23)

We now prove that for every $n\in N$

$$\begin{array}{cc}\hfill \frac{1}{2k}d(x_n,T{x}_n)& <d(x_n,\omega )\mathrm{or}\hfill \\ \hfill \frac{1}{2k}d(T{x}_n,T^2{x}_n)& <d(T{x}_n,\omega )\forall n\in \mathbb{N}\hfill \end{array}$$

(24)

Arguing by contradiction, we assume that there exists $e\in \mathbb{N}$ such that

$$\begin{array}{cc}\hfill \frac{1}{2k}d(x_e,T{x}_e)& \ge d(x_e,\omega )\mathrm{and}\hfill \\ \hfill \frac{1}{2k}d(T{x}_e,T^2{x}_e)& \ge d(T{x}_e,\omega )\hfill \end{array}$$

(25)

Now from (12),

$$d(T{x}_e,T^2{x}_e)<d(x_e,T{x}_e)$$

(26)

From (25) and (26),

$$\begin{array}{cc}\hfill d(x_e,T{x}_e)& \le k[d(x_e,\omega )+d(\omega ,T{x}_e)]\hfill \\ & \le k\left[\frac{1}{2k}d(x_e,T{x}_e)+\frac{1}{2k}d(T{x}_e,T^2{x}_e)\right]\hfill \\ & <k\left[\frac{1}{2k}d(x_e,T{x}_e)+\frac{1}{2k}d(x_e,T{x}_e)\right]\hfill \\ & =d(x_e,T{x}_e),\hfill \end{array}$$

(27)

which is a contradiction. Hence (24) holds.

Suppose $\frac{1}{2k}d(x_n,T{x}_n)<d(x_n,\omega )$ is satisfied and $d(\omega ,T\omega )>0$. From (23) there exists $N_1\in \mathbb{N}$ such that $d(T{x}_n,T\omega )=d(x_{n+1},T\omega )>0$ for $n\ge N_1.$ Then from Extended F-contraction (with $n\ge N_1$), we have,

$$\begin{array}{cc}\hfill \tau +F(d(x_{n+1},T\omega ))& =\tau +F(d(T{x}_n,T\omega ))\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_n,\omega )+{\mathcal{A}}_2[d(x_n,T{x}_n)+d(\omega ,T\omega )]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_n,T\omega )+d(\omega ,T{x}_n)}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x_n,x_n)+d(\omega ,\omega )}{4k}\right]\}\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_n,\omega )+{\mathcal{A}}_2[d(x_n,x_{n+1})+d(\omega ,T\omega )]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_n,T\omega )+d(\omega ,x_{n+1})}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x_n,x_n)+d(\omega ,\omega )}{4k}\right]\}\hfill \end{array}$$

(28)

From (23) there exists $N_k\in \mathbb{N}$ with $(N_k\ge N_1)$ such that for all $n\ge N_k$.

$$\begin{array}{cc}\hfill \tau +F(d(x_{n+1},T\omega ))& \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_{n+1},\omega )+{\mathcal{A}}_{2}d(\omega ,T\omega )\hfill \\ & +{\mathcal{A}}_3\left[k\frac{d(x_{n+1},T\omega )+d(\omega ,x_{n+1})}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[2s\frac{d(\omega ,T\omega )}{4k}\right]\}\hfill \\ & \le F\left\{{\mathcal{A}}_2[d(\omega ,T\omega )+\frac{{\mathcal{A}}_3}{3}d(x_{n+1},T\omega )+{\mathcal{A}}_4\frac{d(\omega ,T\omega )}{2}\right\},\hfill \end{array}$$

(29)

which implies, $(1-\frac{{\mathcal{A}}_3}{3})F(d(x_{n+1},T\omega ))\le ({\mathcal{A}}_3+\frac{{\mathcal{A}}_4}{2})F(d(\omega ,T\omega ))-\tau$

Letting $n\to \infty$,

$$\begin{array}{cc}\hfill F(d(\omega ,T\omega ))& \le \frac{{\mathcal{A}}_2+\frac{{\mathcal{A}}_4}{2}}{1-\frac{{\mathcal{A}}_3}{3}}F(d(\omega ,T\omega ))-\frac{\tau }{1-\frac{{\mathcal{A}}_3}{3}}\hfill \\ & <\frac{{\mathcal{A}}_2+\frac{{\mathcal{A}}_4}{2}}{1-\frac{{\mathcal{A}}_3}{3}}F(d(\omega ,T\omega ))\hfill \\ & <F(d(\omega ,T\omega ))\forall n\ge N_k,\hfill \end{array}$$

(30)

which is a contradiction.

Now suppose $\frac{1}{2k}d(T{x}_n,T^2{x}_n)<d(T{x}_n,\omega )$ of (24) is true with $d(\omega ,T\omega )>0$. Noted from (23), there exists $N_p\in \mathbb{N}$ such that $d(T(T{x}_n),T\omega )=d(x_{n+2},T\omega )>0$ for $n\ge N_p$. Then from our assumption with $n\ge N_p$, we have

$$\begin{array}{cc}\hfill \tau +F(d(x_{n+2},T\omega ))& =\tau +F(d(T^2{x}_n,T\omega ))\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(T{x}_n,\omega )+{\mathcal{A}}_2[d(T{x}_n,T^2{x}_n)+d(\omega ,T\omega )]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(T{x}_n,T\omega )+d(\omega ,T^2{x}_n)}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(T{x}_n,T{x}_n)+d(\omega ,\omega )}{4k}\right]\}\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(x_{n+1},\omega )+{\mathcal{A}}_2[d(x_{n+1},x_{n+2})+d(\omega ,T\omega )]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(x_{n+1},T\omega )+d(\omega ,x_{n+2})}{3k}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x_{n+1},x_{n+1})+d(\omega ,\omega )}{4k}\right]\}\hfill \\ & \le F({\mathcal{A}}_{2}d(\omega ,T\omega )+\frac{{\mathcal{A}}_3}{3k}k{b}_d(\omega ,T\omega ))\hfill \\ & \le {\mathcal{A}}_{2}F(d(\omega ,T\omega ))+\frac{{\mathcal{A}}_3}{3}F(d(\omega ,T\omega ))\hfill \end{array}$$

(31)

Letting $n\to \infty ,$

$$\tau +F(d(\omega ,T\omega )\le \left[{\mathcal{A}}_2+\frac{{\mathcal{A}}_3}{3}\right]F(d(\omega ,T\omega )$$

$$\begin{array}{cc}\hfill F(d(\omega ,T\omega )& \le \left[{\mathcal{A}}_2+\frac{{\mathcal{A}}_3}{3}\right]F(d(\omega ,T\omega ))-\tau \hfill \\ & <\left[{\mathcal{A}}_2+\frac{{\mathcal{A}}_3}{3}\right]F(d(\omega ,T\omega ))\hfill \\ & <F(d(\omega ,T\omega )),\hfill \end{array}$$

(32)

which is a contradiction.

Hence we conclude that $d(\omega ,T\omega )=0,$ which gives $\omega =T\omega$. Therefore, $\omega$ is a fixed point of T.

To prove uniqueness, let $\omega$ and ${\omega }^{*}$ are two fixed points of T such that $\omega \ne {\omega }^{*}$, which means,

$$\begin{array}{cc}\hfill \frac{1}{2k}d(\omega ,T{\omega }^{*})& <d(\omega ,T{\omega }^{*})\hfill \\ & =d(\omega ,{\omega }^{*})\hfill \end{array}$$

(33)

Thus, from, Extended F-contraction,

$$\begin{array}{cc}\hfill \tau +F(d(T\omega ,T{\omega }^{*}))& \le F(\mathcal{A}(\omega ,{\omega }^{*}))\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(\omega ,{\omega }^{*})+{\mathcal{A}}_2[d(\omega ,T\omega )+d({\omega }^{*},T{\omega }^{*})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(\omega ,T{\omega }^{*})+d({\omega }^{*},T\omega )}{3k}\right]+{\mathcal{A}}_4\left[\frac{d(\omega ,{\omega }^{*})+d({\omega }^{*},{\omega }^{*})}{4k}\right]\}\hfill \\ & \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(\omega ,{\omega }^{*})+{\mathcal{A}}_2[d(\omega ,\omega )+d({\omega }^{*},{\omega }^{*})]\hfill \\ & +{\mathcal{A}}_3\left[\frac{d(\omega ,{\omega }^{*})+d({\omega }^{*},\omega )}{3k}\right]\}\hfill \end{array}$$

(34)

From (22), $d(\omega ,\omega )=d({\omega }^{*},{\omega }^{*})=0.$ Therefore (34) becomes:

$$\begin{array}{cc}\hfill \tau +F(d(\omega ,{\omega }^{*}))& \le F\{\frac{{\mathcal{A}}_1(k+1)}{k}d(\omega ,{\omega }^{*})+2{\mathcal{A}}_3\frac{d(\omega ,{\omega }^{*})}{3k}\}\hfill \\ & \le \left(\frac{{\mathcal{A}}_1(k+1)}{k}+\frac{2{\mathcal{A}}_3}{3k}\right)F(d(\omega ,{\omega }^{*}))\hfill \\ & =\left(\frac{3{\mathcal{A}}_1(k+1)+2{\mathcal{A}}_{3}k}{k}\right)F(d(\omega ,{\omega }^{*}))\hfill \\ & <F(d(\omega ,{\omega }^{*}))\hfill \end{array}$$

(35)

Thus, $F(d(\omega ,{\omega }^{*}))<F(d(\omega ,{\omega }^{*}))-\tau$, this gives a contradiction. Hence $\omega ={\omega }^{*}.$ ☐

Example 1.

Consider $X=\{0,1,2\}$. Let $d:X\times X\to [0,\infty )$ be a mapping defined by

$$d(0,0)=d(1,1)=0,d(2,2)=2.5;$$

$$d(0,2)=d(2,0)=2,d(1,2)=d(2,1)=3;$$

$$d(0,1)=d(1,0)=1.5.$$

We can easily prove that $(X,d)$ is a b-dislocated metric space with $k=2$. Clearly $(X,d)$ is a complete b-dislocated metric space.

Let $T:X\to X$ be given by $T0=T1=0$ and $T2=1$. Suppose that $F(\nu )=\frac{-1}{\nu }+\nu$.

Case-1: Let $x=0$, now $d(T0,T0)=d(T0,T1)=d(0,0)=0.$

So, we need to consider $y=2.$ Then we have $0.25d(0,T0)<d(0,2)$.

Consider,

$$\begin{array}{cc}\hfill \mathcal{A}(0,2)& ={\mathcal{A}}_{1}1.5d(0,2)+{\mathcal{A}}_2[d(0,T0)+d(2,T2)]+{\mathcal{A}}_3\left[\frac{d(0,T2)+d(2,T0)}{6}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(0,0)+d(2,2)}{8}\right]\hfill \\ & ={\mathcal{A}}_{1}1.5\times 2+{\mathcal{A}}_2[d(0,0)+d(2,1)]+{\mathcal{A}}_3\left[\frac{d(0,1)+d(2,0)}{6}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(0,0)+d(2,2)}{8}\right]\hfill \\ & =3{\mathcal{A}}_1+{\mathcal{A}}_2[0+3]+{\mathcal{A}}_3\left[\frac{1.5+2}{6}\right]+{\mathcal{A}}_4\left[\frac{0+2.5}{8}\right]\hfill \\ & =3{\mathcal{A}}_1+3{\mathcal{A}}_2+0.58{\mathcal{A}}_3+0.31{\mathcal{A}}_4\hfill \end{array}$$

(36)

Let ${\mathcal{A}}_1=0.56,{\mathcal{A}}_2={\mathcal{A}}_3={\mathcal{A}}_4=0$ such that $1.5{\mathcal{A}}_1+2({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)<1$.

Now, $\mathcal{A}(0,2)=1.68.$

Consider,

$$\begin{array}{cc}\hfill F(d(T0,T2))-F(\mathcal{A}(0,2))& =F(d(0,1))-F(\mathcal{A}(0,2))\hfill \\ & =F(1.5)-F(1.68)\hfill \\ & =-0.66+1.5+0.59-1.68\hfill \\ & =-0.25\hfill \\ & =-\tau \hfill \end{array}$$

(37)

In this case, for $\tau =0.25,$T satisfies all the conditions of Theorem 1 and 0 is the unique fixed point.

Case-2: Let $x=1$, now $d(T0,T1)=d(T1,T1)=d(0,0)=0$. So, we must consider $y=2.$

$$d(1,T1)=d(1,0)=1.5,d(1,2)=3.$$

Therefore $\frac{1}{4}d(1,T1)<d(1,2)$.

$$\begin{array}{cc}\hfill \mathcal{A}(1,2)& ={\mathcal{A}}_{1}1.5d(1,2)+{\mathcal{A}}_2[d(1,T1)+d(2,T2)]+{\mathcal{A}}_3\left[\frac{d(1,T2)+d(2,T1)}{6}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(1,1)+d(2,2)}{8}\right]\hfill \\ & ={\mathcal{A}}_{1}1.5\times 3+{\mathcal{A}}_2[d(1,0)+d(2,1)]+{\mathcal{A}}_3\left[\frac{d(1,1)+d(2,0)}{6}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(1,1)+d(2,2)}{8}\right]\hfill \\ & =4.5{\mathcal{A}}_1+{\mathcal{A}}_2[1.5+3]+{\mathcal{A}}_3\left[\frac{0+2}{6}\right]+{\mathcal{A}}_4\left[\frac{0+2.5}{8}\right]\hfill \\ & =4.5{\mathcal{A}}_1+4.5{\mathcal{A}}_2+0.33{\mathcal{A}}_3+0.31{\mathcal{A}}_4\hfill \end{array}$$

(38)

Let ${\mathcal{A}}_1=0.56,{\mathcal{A}}_2={\mathcal{A}}_3={\mathcal{A}}_4=0$ such that $1.5{\mathcal{A}}_1+2({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)<1$.

Consider,

$$\begin{array}{cc}\hfill F(d(T1,T2))-F(\mathcal{A}(1,2))& =F(d(0,1))-F(\mathcal{A}(1,2))\hfill \\ & =F(1.5)-F(2.52)\hfill \\ & =-1.29\hfill \\ & =-\tau \hfill \end{array}$$

(39)

In this case, for $\tau =1.29,$T satisfies all the conditions of Theorem 1 and 0 is the unique fixed point.

Case-3: Let $x=2,$ now $d(T2,T2)=0$. So, we need only consider the case $y\in \{0,1\}.$

Subcase-I: If $y=0.$ Please note that $0.25d(2,T2)<d(2,0)$ as $0.75=0.25d(2,1)<d(2,0)=2$.

$$\tau +F(d(T2,T0))\le F(\mathcal{A}(2,0)),$$

follows as in Case-1.

Subcase-II: If $y=1$, note that $0.25d(2,T2)<d(2,1)$ as $0.75=0.25d(2,1)<d(2,1)=3$.

$$\tau +F(d(T2,T1))\le F(\mathcal{A}(2,1)),$$

follows as in Case-2.

Hence T is an extended F-contraction and it is clear that 0 is the fixed point of T.

If we take $k=1$, then the above theorem reduces to below corollary.

Corollary 1.

Let $(X,d)$ be a dislocated metric space such that d is a continuous function. If T is an extended F-contraction, i.e., there exists $F\in \mathcal{F}$ and $\tau >0$ such that for all $x,y\in X$ with $d(Tx,Ty)>0,$

$$\frac{1}{2}d(x,Tx)<d(x,y)\Rightarrow \tau +F(d(Tx,Ty))\le F(\mathcal{A}(x,y))$$

where,

$$\begin{array}{cc}\hfill \mathcal{A}(x,y)& =2{\mathcal{A}}_{1}d(x,y)+{\mathcal{A}}_2[d(x,Tx)+d(y,Ty)]+{\mathcal{A}}_3\left[\frac{d(x,Ty)+d(y,Tx)}{3}\right]\hfill \\ & +{\mathcal{A}}_4\left[\frac{d(x,x)+d(y,y)}{4}\right]\hfill \end{array}$$

here ${\mathcal{A}}_1,{\mathcal{A}}_2,{\mathcal{A}}_3,{\mathcal{A}}_4\ge 0$, $2{\mathcal{A}}_1+2({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)<1$. Then T has a unique fixed point.

If we take ${\mathcal{A}}_2={\mathcal{A}}_3={\mathcal{A}}_4=0$ and ${\mathcal{A}}_1<\frac{1}{2}$ in above theorem, we get below corollary.

Corollary 2.

Let $(X,d)$ be a b-dislocated metric space. If T is an extended F-contraction, i.e., there exists $F\in \mathcal{F}$ and $\tau >0$ such that for all $x,y\in X$ with $d(Tx,Ty)>0,$ $\frac{1}{2}d(x,Tx)<d(x,y)\Rightarrow$ $\tau +F(d(Tx,Ty))\le F(\mathcal{A}(x,y))$, here, $\mathcal{A}(x,y)=2{\mathcal{A}}_{1}d(x,y))$. Then T has a unique fixed point.

3. Weak-Generalized $\mathit{F}$-Contraction

Definition 6.

Let $(X,d)$ be a b-dislocated metric space and $T:X\to X$ be a mapping. Then T is said to be a Weak-generalized F-contraction if F along with $C_1,C_2,C_3$ and there exists $\tau >0$ such that $\forall x,y\in X,$

$$d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F(\mathcal{S}(x,y))$$

(40)

where

$$\begin{array}{cc}\hfill \mathcal{S}(x,y)& =max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)+d(y,Ty)}{2},\hfill \\ & \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)},\frac{d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)},\hfill \\ & \frac{d^2(x,Tx)d(y,Ty)}{1+d^2(x,y)}\}\hfill \end{array}$$

(41)

Theorem 2.

Let $(X,d)$ be a b-dislocated metric space such that d is continuous functional. If T is a Weak-generalized F-contraction, then T has a unique fixed point.

Proof. 

Choose an arbitrary point $x_0\in X$ and define a sequence $\left\{x_n\right\}$ in X by $x_n=T{x}_{n-1}$ for $n\in \{1,2,3\dots \}$.

i.e., $x_1=T{x}_0,x_2=T{x}_1=T(T{x}_0)=T^2{x}_0,\dots x_n=T{x}_{n-1}=T^n{x}_0.$ If $d(x_{n_0},x_{n_0+1})=0$ for some $n_0\in \{0,1,2\dots \},$ then $x_{n_0}=x_{n_0+1}$, which yields $x_{n_0}=T{x}_{n_0}$, and so T has a fixed point. Then there is nothing left to prove, and our proof is complete.

Thus, let us assume $d(x_n,x_{n+1})>0$ for every $n\in \{0,1,2\dots \}.$ i.e., $d(T{x}_{n-1},T{x}_n)>0.$

Now using (36), we get,

$$\tau +F(k{b}_d(T{x}_{n-1},T{x}_n))\le F(\mathcal{S}(x_{n-1},x_n))$$

(42)

where

$$\begin{array}{cc}\hfill \mathcal{S}(x_{n-1},x_n)& =max\{d(x_{n-1},x_n),d(x_{n-1},T{x}_{n-1}),d(x_n,T{x}_n),\frac{d(x_{n-1},T{x}_{n-1})+d(x_n,T{x}_n)}{2},\hfill \\ & \frac{d(x_n,T{x}_n)[1+d(x_{n-1},T{x}_{n-1})]}{1+d(x_{n-1},x_n)},\frac{d(x_{n-1},T{x}_{n-1})d(x_n,T{x}_n)}{1+d(x_{n-1},x_n)},\hfill \\ & \frac{d^2(x_{n-1},T{x}_{n-1})d(x_n,T{x}_n)}{1+d^2(x_{n-1},x_n)}\}\hfill \\ & =max\{d(x_{n-1},x_n),d(x_{n-1},x_n),d(x_n,x_{n+1}),\frac{d(x_{n-1},x_n)+d(x_n,x_{n+1})}{2},\hfill \\ & \frac{d(x_n,x_{n+1})[1+d(x_{n-1},x_n)]}{1+d(x_{n-1},x_n)},\frac{d(x_{n-1},x_n)d(x_{n+1},x_n)}{1+d(x_{n-1},x_n)},\hfill \\ & \frac{d^2(x_{n-1},x_n)d(x_n,x_{n+1})}{1+d^2(x_{n-1},x_n)}\}\hfill \\ & \le max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}\hfill \end{array}$$

(43)

This yields,

$$\tau +F(k{B}_d(x_n,x_{n+1}))\le F(max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\})$$

(44)

If $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_n,x_{n+1})$ then $\tau +F(k{b}_d(x_n,x_{n+1}))\le F(d(x_n,x_{n+1}))$, which is a contradiction, since $\tau >0.$

Hence $max\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}=d(x_{n-1},x_n)$. Thus, $(40)$ reduces to $\tau +F(k{B}_d(x_n,x_{n+1}))\le F(d(x_{n-1},x_n))$. Therefore, we obtain,

$$\begin{array}{cc}\hfill F(k^{n}d(x_n,x_{n+1}))& \le F(k^{n-1}d(x_{n-1},x_n))-\tau \hfill \\ & \le F(k^{n-2}d(x_{n-2},x_{n-1}))-2\tau \hfill \\ & ⋮\hfill \\ & \le F(k^{0}d(x_0,x_1))-n\tau \hfill \end{array}$$

(45)

Applying $\underset{n\to \infty }{lim}$ in above, we get,

$$\underset{n\to \infty }{lim}F(k^{n}d(x_n,x_{n+1}))=-\infty .$$

Thus, by using $(F_2)$ of Wardowski, we get,

$$\underset{n\to \infty }{lim}{k}^{n}d(x_n,x_{n+1}))=0.$$

Thus, from $(F_3)$ there exists $\vartheta \in (0,1)$ such that

$$\underset{n\to \infty }{lim}{\left[k^{n}d(x_n,x_{n+1})\right]}^{\vartheta }F(d(x_n,x_{n+1}))=0.$$

By (41), the following holds good for all $n\in \{1,2,3\dots \}$,

$$\begin{array}{cc}\hfill {\left[k^{n}d(x_n,x_{n+1})\right]}^{\vartheta }F(d(x_n,x_{n+1}))-{\left[k^{n}d(x_n,x_{n+1})\right]}^{\vartheta }F(d(x_0,x_1))& \le -{\left[k^{n}d(x_n,x_{n+1})\right]}^{\vartheta }n\tau \hfill \\ & \le 0\hfill \end{array}$$

(46)

Letting $n\to \infty$ in (42), we get

$$\underset{n\to \infty }{lim}n{\left[k^{n}d(x_n,x_{n+1})\right]}^{\vartheta }=0.$$

(47)

From (43), there exists $N_1\in \{1,2,3\dots \}$ such that $n{\left[d(x_n,x_{n+1})\right]}^{\vartheta }\le 1$ for all $n\ge N_1.$ Therefore, we have for all $n\ge N_1$.

$$k^{n}d(x_n,x_{n+1})\le \frac{1}{n^{\frac{1}{\vartheta }}}$$

(48)

Now to prove that $\left\{x_n\right\}$ is a Cauchy sequence, consider $m,n\in N$ such that $m>n\ge N_1.$ Using the triangle inequality and from (44), we have,

$$\begin{array}{cc}\hfill d(x_n,x_m)& \le k[d(x_n,x_{n+1})+d(x_{n+1},x_m)]\hfill \\ & \le kd(x_n,x_{n+1})+k\left[k(d(x_{n+1},x_{n+2})+d(x_{n+2},x_m))\right]\hfill \\ & \le k{b}_d(x_n,x_{n+1})+k^{2}d(x_{n+1},x_{n+2})+k^{2}d(x_{n+2},x_m)\hfill \\ & ⋮\hfill \\ & \le k{b}_d(x_n,x_{n+1})+k^{2}d(x_{n+1},x_{n+2})+k^{3}d(x_{n+2},x_{n+3})+...+k^{m-1}d(x_{m-1},x_m)\hfill \end{array}$$

(49)

Thus,

$$\begin{array}{cc}\hfill d(x_n,x_m)& \le \sum _{e=n}^{m-1}{k}^{e}d(x_e,x_{e+1})\hfill \\ & \le \sum _{e=n}^{\infty }{k}^{e}d(x_e,x_{e+1})\hfill \\ & \le \sum _{e=n}^{\infty }\frac{1}{e^{\frac{1}{\vartheta }}}\hfill \end{array}$$

(50)

which is convergent, passing to limit $n\to \infty ,$ then we get $\underset{n\to \infty }{lim}d(x_n,x_m)=0.$ This yields that $\left\{x_n\right\}$ is a Cauchy sequence in $(X,d)$.

Since $(X,d)$ is a complete b-dislocated metric space, the sequence $\left\{x_n\right\}$ converges to some point $\varrho \in X$, that is $\underset{n\to \infty }{lim}{x}_n=\varrho$, and $d(\varrho ,\varrho )=0$. As $d(\varrho ,\varrho )=\underset{n\to \infty }{lim}d(x_n,\varrho )=\underset{n,m\to \infty }{lim}d(x_n,x_m)=0$.

Case-I: If T is continuous. Then we have,

$$\varrho =\underset{n\to \infty }{lim}{x}_{n+1}=\underset{n\to \infty }{lim}T{x}_n=T(\underset{n\to \infty }{lim}{x}_n)=T\varrho$$

and so, $\varrho$ is a fixed point of T.

Case-II: If F is continuous. In this case, we will prove that $\varrho =T\varrho .$ Let us assume that $\varrho \ne T\varrho .$

In this case, there exists an $n_0\in \mathbb{N}$ and a subsequence $\left\{x_{n_e}\right\}$ of $\left\{x_n\right\}$ such that $d(T{x}_{n_e},T\varrho )>0$ for all $n_e\ge n_0.$

Otherwise, if not, there exist $n_1\in \mathbb{N}$ such that $x_n=T\varrho$ for all $n\ge n_1,$ which yields that $x_n\to T\varrho .$ This is a contradiction, since $\varrho \ne T\varrho .$ Being $d(T{x}_{n_e},T\varrho )>0$ for all $n_e\ge n_0,$ then from hypothesis, we have,

$$\begin{array}{cc}\hfill \tau +F(d(x_{n_{e+1}},T\varrho ))& =\tau +F(d(T{x}_{n_{e+1}},T\varrho ))\hfill \\ & \le F(\mathcal{A}(x_{n_e},\varrho ))\hfill \end{array}$$

(51)

where,

$$\begin{array}{cc}\hfill \mathcal{A}(x_{n_e},\varrho )& =max\{d(x_{n_e},\varrho ),d(x_{n_e},T{x}_{n_e}),d(\varrho ,T\varrho ),\frac{d(x_{n_e},T{x}_{n_e})+d(\varrho ,T\varrho )}{2},\hfill \\ & \frac{d(\varrho ,T\varrho )[1+d(x_{n_e},T{x}_{n_e})]}{1+d(x_{n_e},\varrho )},\frac{d(x_{n_e},T{x}_{n_e})d(\varrho ,T\varrho )}{1+d(x_{n_e},\varrho )},\hfill \\ & \frac{d^2(x_{n_e},T{x}_{n_e})d(\varrho ,T\varrho )}{1+d^2(x_{n_e},\varrho )}\}\hfill \end{array}$$

(52)

From (47),

$$\begin{array}{cc}\hfill \tau +F(d(x_{n_e+1},T\varrho ))& =max\{d(x_{n_e},\varrho ),d(x_{n_e},x_{n_e+1}),d(\varrho ,T\varrho ),\frac{d(x_{n_e},x_{n_e+1})+d(\varrho ,T\varrho )}{2},\hfill \\ & \frac{d(\varrho ,T\varrho )[1+d(x_{n_e},x_{n_e+1})]}{1+d(x_{n_e},\varrho )},\frac{d(x_{n_e},x_{n_e+1})d(\varrho ,T\varrho )}{1+d(x_{n_e},\varrho )},\hfill \\ & \frac{d^2(x_{n_e},x_{n_e+1})d(\varrho ,T\varrho )}{1+d^2(x_{n_e},\varrho )}\}\hfill \end{array}$$

(53)

Taking the limit $n_e\to \infty$ and using the continuity of F, we get,

$$\begin{array}{cc}\hfill \tau +F(d(\varrho ,T\varrho ))& \le max\left\{d(\varrho ,\varrho ),d(\varrho ,T\varrho ),\frac{d(\varrho ,\varrho )+d(\varrho ,T\varrho )}{2}\right\}\hfill \\ & \le max\{d(\varrho ,\varrho ),d(\varrho ,T\varrho )\}\hfill \end{array}$$

(54)

Case-A: If $max\{d(\varrho ,\varrho ),d(\varrho ,T\varrho )\}=d(\varrho ,T\varrho )$.

Then from (50), we get,

$$\tau +F(d(\varrho ,T\varrho ))\le d(\varrho ,T\varrho ),$$

which is a contradiction as $\tau >0.$

Case-B: If $max\{d(\varrho ,\varrho ),d(\varrho ,T\varrho )\}=d(\varrho ,\varrho )$.

Then from (50), we get,

$$\tau +F(d(\varrho ,T\varrho ))\le d(\varrho ,\varrho ),$$

$$\Rightarrow \tau +F(d(\varrho ,T\varrho ))\le F(0),$$

which is a contradiction.

Thus, in the both cases, we get the contradiction due to $d(x_{n_e+1},T\varrho )>0.$ Therefore, $x_{n_e+1}=T\varrho ,$ which implies $T{x}_{n_e}=T\varrho .$ Thus, $T\varrho =\varrho .$ This gives, $\varrho$ is a fixed point of T. To prove uniqueness, let $\varrho$ and ${\varrho }^{*}$ are two fixed points of T, i.e. $T\varrho =\varrho$ and $T{\varrho }^{*}={\varrho }^{*}.$

Let us assume that $\varrho \ne {\varrho }^{*}$.

$$\Rightarrow d(\varrho ,{\varrho }^{*})>0$$

$$\Rightarrow d(T\varrho ,T{\varrho }^{*})>0$$

From hypothesis, we get,

$$\tau +F(d(T\varrho ,T{\varrho }^{*}))\le F(\mathcal{A}(\varrho ,{\varrho }^{*}))$$

(55)

where,

$$\begin{array}{cc}\hfill \mathcal{A}(\varrho ,{\varrho }^{*})& =max\{d(\varrho ,{\varrho }^{*}),d(\varrho ,T\varrho ),d({\varrho }^{*},T{\varrho }^{*}),\frac{d(\varrho ,T\varrho )+d({\varrho }^{*},T{\varrho }^{*})}{2},\hfill \\ & \frac{d({\varrho }^{*},T{\varrho }^{*})[1+d(\varrho ,T\varrho )]}{1+d(\varrho ,{\varrho }^{*})},\frac{d(\varrho ,T\varrho )d({\varrho }^{*},T{\varrho }^{*})}{1+d(\varrho ,{\varrho }^{*})},\hfill \\ & \frac{d^2(\varrho ,T\varrho )d({\varrho }^{*},T{\varrho }^{*})}{1+d^2(\varrho ,{\varrho }^{*})}\}\hfill \\ & =max\{d(\varrho ,{\varrho }^{*}),d(\varrho ,\varrho ),d({\varrho }^{*},{\varrho }^{*}),\frac{d(\varrho ,\varrho )+d({\varrho }^{*},{\varrho }^{*})}{2},\hfill \\ & \frac{d({\varrho }^{*},{\varrho }^{*})[1+d(\varrho ,\varrho )]}{1+d(\varrho ,{\varrho }^{*})},\frac{d(\varrho ,\varrho )d({\varrho }^{*},{\varrho }^{*})}{1+d(\varrho ,{\varrho }^{*})},\hfill \\ & \frac{d^2(\varrho ,\varrho )d({\varrho }^{*},{\varrho }^{*})}{1+d^2(\varrho ,{\varrho }^{*})}\}\hfill \\ & =max\left\{d(\varrho ,{\varrho }^{*})\right\}\hfill \\ & =d(\varrho ,{\varrho }^{*})\hfill \end{array}$$

(56)

From hypothesis, we get,

$$\tau +F(d(T\varrho ,T{\varrho }^{*}))\le F(d(\varrho ,{\varrho }^{*}))$$

$$\Rightarrow \tau +F(d(\varrho ,{\varrho }^{*}))\le F(d(\varrho ,{\varrho }^{*})),$$

which is a contradiction. Hence $\varrho ={\varrho }^{*}$, this completes the proof of the theorem. ☐

Corollary 3.

Let $(X,d)$ be a dislocated metric space such that d is continuous functional. If T is a Weak-generalized F-contraction, i.e., there exist F along with $C_1,C_2,C_3$ and $\tau >0$ such that $\forall x,y\in X,$

$$d(Tx,Ty)>0\Rightarrow \tau +F(d(Tx,Ty))\le F(\mathcal{O}(x,y))$$

where

$$\begin{array}{cc}\hfill \mathcal{O}(x,y)& =max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)+d(y,Ty)}{2},\hfill \\ & \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)},\frac{d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)},\hfill \\ & \frac{d^2(x,Tx)d(y,Ty)}{1+d^2(x,y)}\}.\hfill \end{array}$$

Then T has a unique fixed point.

4. Application to Non-linear Integral Equation

Now we study the existence of unique solutions for non-linear integral equations as an application using the Theorem 1.

Let us consider the non-linear integral equation:

$$\Xi (i)={\int }_0^t\Lambda (i,j)\Gamma (j,\Xi (j))dj$$

(57)

for all $i,j\in [0,T]$, where the unknown function $\Xi (i)$ takes real values.

Let $X=\mathbb{C}([0,T])$ be the space of all real continuous functions defined on $[0,T]$. We endow X with the d-metric $d:X\times X\to \mathbb{R}$ by

$$d(\Xi ,\Theta )=su{p}_{t\in [0,T]}(|\Xi (i)+{\Theta (i)|)}^2$$

for all $\Xi ,\Theta \in X.$

Clearly $(X,d,2)$ is a complete b-dislocated metric space.

Let for all $j\in [0,T]$, $\Gamma (j,.)$ be a decreasing function, that is:

$$x,y\in \mathbb{R},x\ge y\Rightarrow \Gamma (j,x)\le \Gamma (j,y)$$

Define a mapping $T:X\to X$ by

$$T\Xi (i)={\int }_0^i\Lambda (i,j)\Gamma (j,\Xi (j))dj;\mathrm{for}\mathrm{all}i\in [0,T].$$

Furthermore, we assume the following conditions:

N1:

$\Lambda (i,j)\le e^{\frac{j\tau }{2}}$ where $\tau >0$ and $j\in [0,T]$

N2:

For all $j\in [0,T]$ and $\Xi (j),\Theta (j)\in X$

$$\begin{array}{cc}\hfill |\Gamma (j,\Xi (j))|+|\Gamma (j,\Theta (j))|& \le [\frac{3{\mathcal{A}}_1}{2}{(\Xi (j)+\Theta (j))}^2+{\mathcal{A}}_2[{(\Xi (j)+T\Xi (j))}^2+{(\Theta (j)+T\Theta (j))}^2]\hfill \\ & +{\mathcal{A}}_3\left(\frac{{(\Xi (j)+T\Xi (j))}^2+{(\Theta (j)+T\Theta (j))}^2}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{{(2\Xi (j))}^2+{(2\Theta (j))}^2}{8}\right){]}^{\frac{1}{2}}\frac{\tau }{2}{e}^{\frac{-\tau }{2}}\hfill \end{array}$$

where ${\mathcal{A}}_1,{\mathcal{A}}_2,{\mathcal{A}}_3,{\mathcal{A}}_4\ge 0$ and $\frac{3{\mathcal{A}}_1}{2}+2({\mathcal{A}}_2+{\mathcal{A}}_3+{\mathcal{A}}_4)<1.$

For $\Xi \in X$, we define a norm ${\left|\right|\Xi \left|\right|}_{\tau }=su{p}_{i\in [0,T]}|\Xi (i)|e^{-\tau i}$. It is easy to check that ${\left|\right|.\left|\right|}_{\tau }$ is equivalent to the supremum norm $\left|\right|.\left|\right|$ in X, and X is endowed with the d-metric d defined by

$$\begin{array}{cc}\hfill d(\Xi ,\Theta )& =\left|\right|\Xi +{\Theta \left|\right|}_{\tau }\hfill \\ & =su{p}_{i\in [0,T]}\{(|\Xi (i)|+{|\Theta (i)|)}^2{e}^{-i\tau }\};\mathrm{for}\Xi ,\Theta \in X.\hfill \end{array}$$

(58)

Then $(X,d)$ is a complete b-dislocated metric space.

Now we will prove the existence of unique solution of the non-linear integral Equation (53).

Theorem 3.

Let $(X,d)$ be a b-dislocated metric space as defined above with assuming the above conditions $N1$ and $N2$. Then the non-linear integral Equation (53) have a unique solution.

Proof: Let $\Xi ,\Theta \in X$ and $i\in [0,T]$ then consider,

$$\begin{array}{cc}\hfill [|T\Xi (i)|+{|T\Theta (i)|]}^2& ={\left(|{\int }_0^i\Lambda (i,j)\Gamma (j,\Xi (j))dj|+|{\int }_0^i\Lambda (i,j)\Gamma (j,\Theta (j))dj|\right)}^2\hfill \\ & ={\left({\int }_0^i\Lambda (i,j)|\Gamma (j,\Xi (j))|dj+{\int }_0^i\Lambda (i,j)|\Gamma (j,\Theta (j))|dj\right)}^2\hfill \\ & ={\left({\int }_0^i\Lambda (i,j)(|\Gamma (j,\Xi (j))|+|\Gamma (j,\Theta (j))|)dj\right)}^2\hfill \\ & \le ({\int }_0^i[\frac{3{\mathcal{A}}_1}{2}{(\Xi (j)+\Theta (j))}^2+{\mathcal{A}}_2[{(\Xi (j)+T\Xi (j))}^2+{(\Theta (j)+T\Theta (j))}^2]\hfill \\ & +{\mathcal{A}}_3\left(\frac{{(\Xi (j)+T\Xi (j))}^2+{(\Theta (j)+T\Theta (j))}^2}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{{(2\Xi (j))}^2+{(2\Theta (j))}^2}{8}\right){]}^{\frac{1}{2}}\frac{\tau }{2}{e}^{\frac{-\tau }{2}}dj{)}^2\hfill \\ & \le ({\int }_0^i\Lambda (i,j)[\frac{3{\mathcal{A}}_1}{2}(|\Xi (j)|+{|\Theta (j)|)}^2+{\mathcal{A}}_2[(|\Xi (j)|+{|T\Xi (j)|)}^2+(|\Theta (j)|+{|T\Theta (j)|)}^2]\hfill \\ & +{\mathcal{A}}_3\left(\frac{(|\Xi (j)|+{|T\Xi (j)|)}^2+(|\Theta (j)|+{|T\Theta (j)|)}^2}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{(2|\Xi (j)|+{2|\Theta (j)|)}^2}{8}\right){]}^{\frac{1}{2}}\frac{\tau }{2}{e}^{\frac{-\tau }{2}}dj{)}^2\hfill \\ & \le ({\int }_0^i\Lambda (i,j)e^{j\tau }(e^{-j\tau }[\frac{3{\mathcal{A}}_1}{2}(|\Xi (j)|+{|\Theta (j)|)}^2+{\mathcal{A}}_2[(|\Xi (j)|+{|T\Xi (j)|)}^2+(|\Theta (j)|+{|T\Theta (j)|)}^2]\hfill \\ & +{\mathcal{A}}_3\left(\frac{(|\Xi (j)|+{|T\Xi (j)|)}^2+(|\Theta (j)|+{|T\Theta (j)|)}^2}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{(2|\Xi (j)|+{2|\Theta (j)|)}^2}{8}\right){]}^{\frac{1}{2}}\frac{\tau }{2}{e}^{\frac{-\tau }{2}}dj){)}^2\hfill \\ & \le ({\int }_0^i\Lambda (i,j)e^{j\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right){]}^{\frac{1}{2}}\frac{\tau }{2}{e}^{\frac{-\tau }{2}}dj){)}^2\hfill \\ & \le \frac{{\tau }^2}{4}{e}^{-\tau }({\int }_0^i\Lambda (i,j)[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right){]}^{\frac{1}{2}}dj{)}^2\hfill \\ & =\frac{{\tau }^2}{4}{e}^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]{\left({\int }_0^i\Lambda (i,j)dj\right)}^2\hfill \end{array}$$

(59)

$$\begin{array}{cc}\hfill \left[\right|T\Xi (i)|+{|T\Theta (i)|]}^2& \le \frac{{\tau }^2}{4}{e}^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]{\left({\int }_0^i{e}^{\frac{j\tau }{2}}dj\right)}^2\hfill \\ & =\frac{{\tau }^2}{4}{e}^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]{\left({\left(\frac{e^{\frac{j\tau }{2}}}{\frac{\tau }{2}}\right)}_0^i\right)}^2\hfill \\ & <\frac{{\tau }^2}{4}{e}^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]{\left(e^{\frac{i\tau }{2}}\times \frac{2}{\tau }\right)}^2\hfill \\ & =\frac{{\tau }^2}{4}{e}^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]\left(e^{i\tau }\frac{4}{{\tau }^2}\right)\hfill \\ & =e^{-\tau (1-i)}[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]\hfill \end{array}$$

(60)

which implies,

$$\begin{array}{cc}\hfill (|T\Xi (i)|+{|T\Theta (i)|)}^2& \le e^{-\tau (1-i)}[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]\hfill \end{array}$$

(61)

$$\begin{array}{cc}\hfill \Rightarrow (|T\Xi (i)|+{|T\Theta (i)|)}^2{e}^{-i\tau }& \le e^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]\hfill \end{array}$$

(62)

this gives,

$$\begin{array}{cc}\hfill d(T\Xi ,T\Theta )& \le e^{-\tau }[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)]\hfill \end{array}$$

(63)

Applying logarithms on both sides, we get,

$$\begin{array}{cc}\hfill ln\left(d\right(T\Xi ,T\Theta \left)\right)& \le ln\left(e^{-\tau }\right[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)\left]\right)\hfill \\ & \le ln{e}^{-\tau }+ln\left(\right[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)\left]\right)\hfill \\ & \le -\tau +ln\left(\right[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)\left]\right)\hfill \end{array}$$

(64)

$$\begin{array}{cc}\hfill \tau +ln\left(d\right(T\Xi ,T\Theta \left)\right)& \le ln\left(\right[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)\left]\right)\hfill \end{array}$$

(65)

Define $F:{\mathbb{R}}^{+}\to \mathbb{R}$ by $F(\alpha )=ln(\alpha ),\alpha >0$. Then from (61), we get

$$\begin{array}{cc}\hfill \tau +F\left(d\right(T\Xi ,T\Theta \left)\right)& \le F\left(\right[\frac{3{\mathcal{A}}_1}{2}d(\Xi ,\Theta )+{\mathcal{A}}_2[d(\Xi ,T\Xi )+d(\Theta ,T\Theta )]\hfill \\ & +{\mathcal{A}}_3\left(\frac{d(\Xi ,T\Theta )+d(\Theta ,T\Xi )}{6}\right)\hfill \\ & +{\mathcal{A}}_4\left(\frac{d(\Xi ,\Xi )+d(\Theta ,\Theta )}{8}\right)\left]\right)\hfill \end{array}$$

(66)

Thus, T is an Extended F-contraction and satisfied all the conditions of the Theorem 1. Thus, T has a unique fixed point which is the unique solution of the non-linear integral equation.

5. Application to Fractional Calculus

In this section, we will start with the existence of solution for the non-linear fractional differential equation.

$$C_{D^{\varpi }}(x(\kappa ))=\Lambda (\kappa ,x(\kappa )),(0<\kappa <1,1<\varpi \le 2)$$

(67)

via the integral boundary conditions

$$x(0)=0,x(1)={\int }_0^{\xi }x(s)ds,(0<\xi <1)$$

(68)

where $C_{D^{\varpi }}$ denotes the Caputo fractional derivative of order $\varpi$ and $\Lambda :[0,1]\times \mathbb{R}\to \mathbb{R}$ is a continuous function. Let the space $X=\mathbb{C}([0,1],\mathbb{R})$ be the set of real continuous functions on $[0,1]$. We confer X with the dislocated metric

$$D_{\mathcal{M}(\infty )}(x,y)=su{p}_{\kappa \in [0,1]}(|x(\kappa )|+\left|y(\kappa )\right|),\mathrm{for}\mathrm{all}x,y\in x.$$

Then $(X,D_{\mathcal{M}(\infty )})$ is a complete dislocated metric space. It is very well known that, for a continuous function $\Theta :[0,\infty ]\to \mathbb{R}$, the Caputo derivative of fractional order $\varpi$ is defined as:

$$C_{D^{\varpi }}(\Theta (\kappa ))=\frac{1}{\Gamma (n-\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{n-\varpi -1}{\Theta }^n(s)ds;n-1<\varpi <n,n=\left[\varpi \right]+1,$$

where $\left[\varpi \right]$ is represented as integer part of the real number $\varpi$ and $\Gamma$ is a gamma function.

Now, we consider the following conditions:

${\mathcal{C}}_1$:

For $u,v\in \mathbb{R},\tau >0$ and ${\mathcal{A}}_1<\frac{1}{2}$

$$(|\Lambda (\kappa ,u)|+|\Lambda (\kappa ,v)|)\le e^{-\tau }2{\mathcal{A}}_1\frac{\Gamma (\varpi +1)}{5}(|u|+\left|v\right|)$$

${\mathcal{C}}_2$:

Define $T:X\to X$ by

$$\begin{array}{cc}\hfill Tx(\kappa )& =\frac{1}{\Gamma \varpi }{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\Lambda (s,x(s))ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\Lambda (s,x(s))ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\Lambda (\ell ,x(\ell ))d\ell )ds;\mathrm{for}\kappa \in [0,1].\hfill \end{array}$$

Theorem 4.

If ${\mathcal{C}}_1\&{\mathcal{C}}_2$ are satisfied then the problem (67) has at least one solution.

Proof. 

It is very well known that $x\in X$ is a solution of (67) if and only if $x\in X$ is a solution of the integral equation.

$$\begin{array}{cc}\hfill x(\kappa )& =\frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\Lambda (s,x(s))ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\Lambda (s,\varpi (s))ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\Lambda (\ell ,x(\ell ))d\ell )ds;\mathrm{for}\kappa \in [0,1].\hfill \end{array}$$

Then, the problem (67) is equivalent to find $x^{\star }\in X$ which is a fixed point of $T.$

Now, let $x,y\in X$ for all $t\in [0,1]$.

Consider,

$$\begin{array}{cc}\hfill |Tx(\kappa )|+|Ty(\kappa )|& =|\frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\Lambda (s,x(s))ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\Lambda (s,x(s))ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\Lambda (\ell ,x(\ell ))d\ell )ds|\hfill \\ & +|\frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\Lambda (s,y(s))ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\Lambda (s,y(s))ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\Lambda (\ell ,y(\ell ))d\ell )ds|\hfill \\ & =|\frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}[\Lambda (s,x(s))+\Lambda (s,y(s))]ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}[\Lambda (s,x(s)+\Lambda (s,y(s))]ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}[\Lambda (\ell ,x(\ell ))+\Lambda (\ell ,y(\ell ))]d\ell )ds|\hfill \\ & \le \frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\left[\right|\Lambda (s,x(s))+\Lambda (s,y(s))\left|\right]ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\left[\right|\Lambda (s,x(s)+\Lambda (s,y(s))|]ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\left[\right|\Lambda (\ell ,x(\ell ))+\Lambda (\ell ,y(\ell ))\left|\right]d\ell )ds\hfill \\ & \le \frac{1}{\Gamma (\varpi )}{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}\left[\right|\Lambda (s,x(s))|+|\Lambda (s,y(s))|]ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}\left[\right|\Lambda (s,x(s)|+|\Lambda (s,y(s))|]ds\hfill \\ & +\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}\left[\right|\Lambda (\ell ,x(\ell ))|+|\Lambda (\ell ,y(\ell ))|]d\ell )ds\hfill \\ & \le \frac{\Gamma \varpi +1}{5}\left[\right|x(s)|+\left|y(s)\right|]e^{-\tau }sup\{\frac{1}{\Gamma \varpi }{\int }_0^{\kappa }{(\kappa -s)}^{\varpi -1}ds\hfill \\ & -\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^1{(1-s)}^{\varpi -1}ds+\frac{2\kappa }{(2-{\xi }^2)\Gamma (\varpi )}{\int }_0^{\xi }({\int }_0^s{(s-m)}^{\varpi -1}d\ell )ds\}\hfill \\ & \le e^{-\tau }2{\mathcal{A}}_1\left|x(s)\right|+\left|y(s)\right|\hfill \end{array}$$

By taking supremum on both sides, we get,

$$D_{\mathcal{M}(\infty )}(Tx,Ty)\le e^{-\tau }2{\mathcal{A}}_1{D}_{\mathcal{M}(\infty )}(x,y).$$

Taking logarithms on both sides, $ln(D_{\mathcal{M}(\infty )}(Tx,Ty))\le -\tau +ln(2{\mathcal{A}}_1{D}_{\mathcal{M}(\infty )}(x,y)),$ which implies, $F(D_{\mathcal{M}(\infty )}(Tx,Ty))+\tau \le F(2{\mathcal{A}}_1{D}_{\mathcal{M}(\infty )}(x,y)).$

By taking $F(\alpha )=ln\alpha$ for $\alpha >0$ and ${\mathcal{A}}_1<\frac{1}{2},{\mathcal{A}}_2,{\mathcal{A}}_3,{\mathcal{A}}_4=0$, all the conditions of Corollary 2 satisfied. Hence T has a unique fixed point. i.e., $T{x}^{\star }=x^{\star }.$ Hence $x^{\star }$ is a solution of (67). ☐