Open AccessArticle

A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions

Sumati Kumari Panda

^1,*

Asifa Tassaddiq

^2,*

and

Ravi P Agarwal

^3,4

Department of Mathematics, Basic Sciences and Humanities, GMR Institute of Technology, Rajam 532127, Andhra Pradesh, India

College of Computer and Information Sciences, Majmaah University, Majmaah 11952, Saudi Arabia

Department of Mathematics, Texas A&M University-Kingsville, Kingsville, TX 78363, USA

⁴

Distinguished University Professor of Mathematics, Florida Institute of Technology, Melbourne, FL 32901, USA

Authors to whom correspondence should be addressed.

Symmetry 2019, 11(2), 206; https://doi.org/10.3390/sym11020206

Submission received: 8 January 2019 / Revised: 23 January 2019 / Accepted: 28 January 2019 / Published: 12 February 2019

(This article belongs to the Special Issue Nonlinear, Convex, Nonsmooth, Functional Analysis in Symmetry)

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Abstract

In this article, we introduce and establish various approaches related to the F-contraction using new sorts of contractions, namely the extended

F_{B_{e}}

-contraction, the extended

F_{B_{e}}

-expanding contraction, and the extended generalized

F_{B_{e}}

-contraction. Thereafter, we propose a simple and efficient solution for non-linear integral equations using the fixed point technique in the setting of a

B_{e}

-metric space. Moreover, to address conceptual depth within this approach, we supply illustrative examples where necessary.

Keywords:

extended b-metric space; extended

ℱ

_Be-contraction; extended F_Be-expanding contraction; extended weak generalized F_Be-contraction; non-linear integral equation

MSC:

AMS Subject Classification (2000): 47H09; 47H10; 54E50; 54H40; 55M20

1. Introduction

Widely renowned for the “Fredholm Integral Equation”, Erik Ivar Fredholm [1], a mathematician and researcher par excellence, has provided research contributions on various aspects of integral equation theory.

Inspired by his great work, many fixed point researchers have focused their work on solving the Fredholm integral equation [2,3,4,5].

There was an amazing publication called F-contraction, which was one of the most influential publication in metric fixed point theory. It was introduced by a fellow named Wardkowski in 2012, and he brought this development to mathematical world with his idealistic touch [6]. It contained topological notions such as Cauchy, completeness, converges, and fixed point.

Definition 1.

Let

(X, d)

be a metric space. A mapping

H : X \to X

is said to be an F-contraction if there exists

τ > 0

such that for all

x, y \in X

d (H x, H y) > 0 \Rightarrow τ + F (d (H x, H y)) \leq F (d (x, y)) .

(1)

F-expanding mappings were introduced in 2017 by Gornicki [6] as below:

Let

(X, d)

be a metric space. A mapping

H : X \to X

is said to be F-expanding if there exists

τ > 0

such that for all

x, y \in X

d (x, y) > 0 \Rightarrow F (d (H x, H y)) \geq + F (d (x, y)) + τ

(2)

where

F : R^{+} \to R

is a mapping satisfying:

$(F 1)$: F is strictly increasing, i.e., for all $α, β \in R^{+}$ such that if $α < β$ then $F (α) < F (β);$
$(F 2)$: For each sequence of positive numbers ${α_{n}}_{n \in N}$ ,

$lim_{n \to \infty} α_{n} = 0 iff lim_{n \to \infty} F (α_{n}) = - \infty;$
$(F 3)$: There exists $k \in (0, 1)$ such that $lim_{α \to 0^{+}} α^{k} F (α) = 0 .$

We represent by

F

the set of all functions satisfying the conditions

(F 1)

–

(F 3)

. There is an effort, however, to convert fixed point theorems that are in the theory of topological fixed point theory into non-linear integral equations and differential equations. This effort is spearheaded major developments in related research areas ( see for more info References [6,7,8,9,10,11,12,13]).

Recently, a new kind of generalized metric space was introduced by Kamran et al. [14], as shown below, and named an extended b-metric space (simply,

B_{e}

-metric space).

Definition 2.

Let X be a non-empty set and

s : X \times X \to [1, \infty)

. A function

B_{e} : X \times X \to [0, \infty)

is called a

B_{e}

-metric if, for all

x, y, z \in X,

it satisfies:

(i): $B_{e} (x, y) = 0$ iff $x = y$ ;
(ii): $B_{e} (x, y) = B_{e} (y, x)$ ;
(iii): $B_{e} (x, y) \leq s (x, y) [B_{e} (x, z) + B_{e} (z, y)] .$

The pair

(X, B_{e})

is called a

B_{e}

-metric space.

Example 1.

Let

X = {- 1, 0, 1}

. Define the function

s : X \times X \to R^{+}

and

B_{e} : X \times X \to R^{+}

s (x, y) = 2 + x + y .

B_{e} (- 1, - 1) = B_{e} (0, 0) = B_{e} (1, 1) = 0; B_{e} (- 1, 0) = B_{e} (0, - 1) = 3; B_{e} (- 1, 1) = B_{e} (1, - 1) = 7;

B_{e} (0, 1) = B_{e} (1, 0) = 1 .

First, we prove that

B_{e}

is a

B_{e}

-metric space. It is clear that (i) and (ii) trivially hold. For (iii), we have

B_{e} (- 1, 0) = 3; s (- 1, 0) [B_{e} (- 1, 1) + B_{e} (1, 0)] = 8 .

Thus,

B_{e} (- 1, 0) \leq s (- 1, 0) [B_{e} (- 1, 1) + B_{e} (1, 0)] .

B_{e} (0, 1) = 1; s (0, 1) [B_{e} (0, - 1) + B_{e} (- 1, 1)] = 30 .

B_{e} (- 1, 1) = 7; s (- 1, 1) [B_{e} (- 1, 0) + B_{e} (0, 1)] = 8 .

Hence, for all

x, y, z \in X,

B_{e} (x, z) \leq s (x, z) [B_{e} (x, y) + B_{e} (y, z)] .

Hence,

(X, B_{e})

is a

B_{e}

-metric space.

Definition 3.

Let

(X, B_{e})

be a

B_{e}

-metric space and a sequence

{x_{n}}

in X is said to

(a): Converge to $x \in X$ iff if for every $ϵ > 0$ there exists $N = N (ϵ) \in N$ such that $B_{e} (x_{n}, x) < ϵ,$ for all $n \geq N .$ For this particular case, we write ${lim}_{n \to \infty} x_{n} = x .$
(b): Cauchy iff for every $ϵ > 0$ there exists $N = N (ϵ) \in N$ such that $B_{e} (x_{m}, x_{n}) < ϵ,$ for all $m, n \geq N .$

Definition 4.

B_{e}

-metric space

(X, B_{e})

is complete if every Cauchy sequence in X is convergent.

Observe that usually a b-metric is not a continuous functional. Analogously, the functional

B_{e}

-metric is also not necessarily a continuous function [15,16,17,18,19].

Within the past century, mathematical research has been increasingly drawn towards understanding the link between the Banach contraction principle and non-linear integral equations. The brief and chronological history of these two topics are explored through a developing conceptual model. Since then, many researchers have formulated and developed fixed point approaches of non-linear integral equations in many directions.

Motivated by the above facts, we establish fixed point theorems by using F-contractions in the context of an extended b-metric space since it was very hard to obtain fixed points via the Warkowski [15] approach, which gives a solutions for non-linear integral equations by using the fixed point technique.

2. An Extended $F_{B_{e}}$ -Contraction

Now, we introduce the following definition:

Definition 5.

Let

(X, B_{e})

be a

B_{e}

-metric space. A mapping

H : X \to X

is said be an extended

F_{B_{e}}

-contraction if there exists

τ > 0

such that for all

x, y \in X

B_{e} (H x, H y) > 0 \Rightarrow τ + F_{B_{e}} (B_{e} (H x, H y)) \leq F_{B_{e}} (B_{e} (x, y)),

(3)

such that for each

x_{0} \in X, {lim}_{n, m \to \infty} s (x_{n}, x_{m}) < \frac{1}{k},

where

k \in (0, 1)

, here

x_{n} = H^{n} x_{0}; n = 1, 2, 3, \dots

and

F_{B_{e}} : R^{+} \to R

is a mapping satisfying:

$(F 1)$: $F_{B_{e}}$ is strictly increasing, i.e., for all $α, β \in R^{+}$ such that $α < β implies F_{B_{e}} (α) < F_{B_{e}} (β);$
$(F 2)$: For each sequence ${α_{n}}_{n \in N}$ of positive numbers $lim_{n \to \infty} α_{n} = 0$ iff $lim_{n \to \infty} F_{B_{e}} (α_{n}) = - \infty;$
$(F 3)$: There exists $k \in (0, 1)$ such that $lim_{α \to 0^{+}} α^{k} F_{B_{e}} (α) = 0 .$

We denote by

F_{B_{e}}

the set of all functions satisfying the conditions

(F 1)

–

(F 3)

Theorem 1.

Let

(X, B_{e})

be a complete

B_{e}

-metric space such that

B_{e}

is a continuous functional and let

H : X \to X

be an extended

F_{B_{e}}

-contraction, then

H

has a fixed point.

Proof.

In order to show that

H

has a fixed point, let

x_{0} \in X

be arbitrary and fixed. We define a sequence

{x_{n}}_{n \in N \in X}

, by

x_{0}, H x_{0} = x_{1}, x_{2} = H x_{1} = H (H x_{0}) = H^{2} (x_{0}) \dots x_{n} = H^{n} x_{0} \dots

Denote

γ_{n} = B_{e} (x_{n + 1}, x_{n}), n = 0, 1, 2, \dots .

If there exists

n_{0} \in N

for which

x_{n_{0} + 1} = x_{n_{0}}

then

H x_{n_{0}} = x_{n_{0}}

and the proof is finished. Suppose now that

x_{n + 1} \neq x_{n}

for every

n \in N

which yields

B_{e} (x_{n + 1}, x_{n}) > 0

, i.e.,

B_{e} (H x_{n}, H x_{n - 1}) > 0

. Thus, by using (3), the following holds for every

n \in N

\begin{matrix} F_{B_{e}} (γ_{n}) & \leq F_{B_{e}} (γ_{n - 1}) - τ \\ \leq F_{B_{e}} (γ_{n - 2}) - 2 τ \\ ⋮ \\ \leq F_{B_{e}} (γ_{0}) - n τ, \end{matrix}

(4)

which yields,

{lim}_{n \to \infty} F_{B_{e}} (γ_{n}) = - \infty

F 2

lim_{n \to \infty} γ_{n} = 0 .

(5)

From

F 3

, there exists

k \in (0, 1)

such that

lim_{n \to \infty} γ_{n}^{k} F_{B_{e}} (γ_{n}) = 0 .

(6)

By Equation (4), the following holds for all

n \in N

. Thus,

\begin{matrix} γ_{n}^{k} F_{B_{e}} (γ_{n}) - γ_{n}^{k} F_{B_{e}} (γ_{0}) & \leq γ_{n}^{k} (F_{B_{e}} (γ_{0}) - n τ) - γ_{n}^{k} F_{B_{e}} (γ_{0}) \\ = - γ_{n}^{k} n τ \\ \leq 0 . \end{matrix}

(7)

Letting

n \to \infty

in (7) and using (4) and (5), we obtain

lim_{n \to \infty} n γ_{n}^{k} = 0 .

(8)

Now, let us observe that from (8) there exists

n_{1} \in N

such that

n γ_{n}^{k} \leq 1

for all

n \geq n_{1} .

Consequently, we have

γ_{n} \leq \frac{1}{n^{\frac{1}{k}}} for all n \geq n_{1} .

(9)

In order to prove that

{x_{n}}_{n \in N}

is a Cauchy sequence, consider

m, n \in N

such that

m > n \geq n_{1} .

By triangle inequality,

\begin{matrix} B_{e} (x_{n}, x_{m}) & \leq s (x_{n}, x_{m}) [B_{e} (x_{n}, x_{n + 1}) + B_{e} (x_{n + 1}, x_{m})] \\ \leq s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) [B_{e} (x_{n + 1}, x_{n + 2}) + B_{e} (x_{n + 2}, x_{m})] \\ \leq s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) B_{e} (x_{n + 1}, x_{n + 2}) + \dots \\ + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) s (x_{n + 2}, x_{m}) \dots s (x_{m - 2}, x_{m}) s (x_{m - 1}, x_{m}) B_{e} (x_{m - 1}, x_{m}) \\ \leq s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) \\ + s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{n + 1}, x_{m}) B_{e} (x_{n + 1}, x_{n + 2}) + \dots \\ + s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{m - 1}, x_{m}) B_{e} (x_{m - 1}, x_{m}) . \end{matrix}

(10)

Note that this series

\sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m}) converges .

Since

\begin{matrix} \sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m}) & \leq \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}} \prod_{i = 1}^{n} s (x_{i}, x_{m}) \\ < \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}} . \frac{1}{k} \\ = \frac{1}{k} \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}}; which is convergent . \end{matrix}

(11)

Let

S = \sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m});

S_{n} = \sum_{j = 1}^{n} B_{e} (x_{j}, x_{j + 1}) \prod_{i = 1}^{j} s (x_{i}, x_{m}) .

Thus, for

m > n

, the above inequality implies

B_{e} (x_{n}, x_{m}) \leq S_{m - 1} - S_{n - 1} .

Letting

n \to \infty

, we conclude that

{x_{n}}

is a Cauchy sequence. Since X is complete, let

x_{n} \to ρ \in X .

Case 1.

H

is continuous, we have

\begin{matrix} B_{e} (H ρ, ρ) & = lim_{n \to \infty} B_{e} (H x_{n}, x_{n}) \\ = lim_{n \to \infty} B_{e} (x_{n + 1}, x_{n}) \\ = 0 . \end{matrix}

Thus,

H ρ = ρ

. Thus

ρ

is a fixed point of

H

Case 2.

F_{B_{e}}

is continuous, in this case, we consider two following subcases:

Case 2.1.: For each $n \in N$ , there exists $i_{n} \in N$ such that $x_{i_{n}} = H ρ$ and $i_{n} > i_{n - 1}$ where $i = 0 .$ Then, we have $ρ = lim_{n \to \infty} x_{i_{n}} = lim_{n \to \infty} H ρ = H ρ$ , which yields that $ρ$ is a fixed point of $H$ .
Case 2.2.: There exists $n_{0} \in N$ such that $x_{n + 1} \neq H ρ$ for al $n \geq n_{0}$ . That is $B_{e} (H x_{n}, H ρ) > 0$ for all $n \geq n_{0}$ .

It follows from (3) that

\begin{matrix} τ + F_{B_{e}} (B_{e} (x_{n + 1}, H ρ)) & = τ + F_{B_{e}} (B_{e} (H x_{n}, H ρ)) \\ \leq F_{B_{e}} (B_{e} (x_{n}, ρ)) . \end{matrix}

Since

F_{B_{e}}

is continuous, taking the limit as

n \to \infty

, then we obtain

τ + F_{B_{e}} (B_{e} (ρ, H ρ)) \leq F_{B_{e}} (B_{e} (ρ, ρ))

\Rightarrow F_{B_{e}} (B_{e} (ρ, ρ)) \leq F_{B_{e}} (B_{e} (ρ, ρ)) - τ,

which is a contradiction due to

F 1

. Therefore,

B_{e} (ρ, H ρ) = 0

. Hence,

ρ

is a fixed point of

H

Thus, from above two cases, we can conclude that

H

has a fixed point

ρ

. Hence,

H ρ = ρ .

In order to prove uniqueness, first, let us observe that

H

has at most one fixed point. Indeed, if

x_{1}, x_{2} \in X

H x_{1} = x_{1} \neq x_{2} = H x_{2},

then

B_{e} (x_{1}, x_{2}) > 0

, i.e.,

B_{e} (H x_{1}, H x_{2}) > 0

. From (3), we get

τ \leq F_{B_{e}} (B_{e} (H x_{1}, H x_{2})) \leq F_{B_{e}} (B_{e} (x_{1}, x_{2})),

\Rightarrow τ \leq F_{B_{e}} (B_{e} (x_{1}, x_{2})) - F_{B_{e}} (B_{e} (x_{1}, x_{2})) = 0,

which is a contradiction. Hence,

H

has a unique fixed point. □

Example 2.

Let

X = {\frac{1}{5^{n - 1}}; n \in N} \cup {0}

. Define

B_{e} : X \times X \to R^{+}

B_{e} (x, y) = {(x - y)}^{2}

and

s : X \times X \to [1, \infty)

s (x, y) = 1 + x + y

. Then,

(X, B_{e})

is a complete

B_{e}

-metric space.

Define

H : X \to X

H (x) = \{\begin{matrix} {\frac{1}{5^{2 n}}}, if x \in {\frac{1}{5^{2 n - 1}}; n \in N} \\ 0, if x = 0 . \end{matrix}

Define the function

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = \ln α

for all

α \in R^{+}

and

τ > 0 .

Case 1.: Let $x = \frac{1}{5^{2 n - 1}}, y = \frac{1}{5^{2 m - 1}}$ , for $m > n \geq 1$ .
Consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) & = F (B_{e} (H \frac{1}{5^{2 n - 1}}, H \frac{1}{5^{2 m - 1}})) \\ = F_{B_{e}} (B_{e} (\frac{1}{5^{2 n}}, \frac{1}{5^{2 m}})) \\ = F_{B_{e}} ({(\frac{1}{5^{2 n}} - \frac{1}{5^{2 m}})}^{2}) \\ = F_{B_{e}} ({(\frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}})}^{2}) \\ = \ln {(\frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}})}^{2} \\ = 2 \ln (\frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}}) . \end{matrix}$

$\begin{matrix} F_{B_{e}} (B_{e} (x, y)) & = F (B_{e} (\frac{1}{5^{2 n - 1}}, \frac{1}{5^{2 m - 1}})) \\ = F_{B_{e}} ({(\frac{1}{5^{2 n - 1}} - \frac{1}{5^{2 m - 1}})}^{2}) \\ = F_{B_{e}} ({(\frac{5^{2 m - 1} - 5^{2 n - 1}}{5^{2 n + 2 m - 2}})}^{2}) \\ = \ln {(\frac{5^{2 m - 1} - 5^{2 n - 1}}{5^{2 n + 2 m - 2}})}^{2} \\ = 2 \ln (\frac{5^{2 m - 1} - 5^{2 n - 1}}{5^{2 n + 2 m - 2}}) . \end{matrix}$

Consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) - F_{B_{e}} (B_{e} (x, y)) & = 2 (\ln \frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}} - \ln \frac{5^{2 m - 1} - 5^{2 n - 1}}{5^{2 n + 2 m - 2}}) \\ = 2 (\ln (\frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}} \times \frac{5^{2 n + 2 m - 2}}{5^{2 m - 1} - 5^{2 n - 1}})) \\ = 2 (\ln (\frac{5^{2 m} - 5^{2 n}}{5^{2 n + 2 m}} \times \frac{5^{2 n + 2 m} . 5^{- 2}}{5^{- 1} (5^{2 m} - 5^{2 n})})) \\ = 2 (\ln (\frac{1}{5})) \\ < - 3 . \end{matrix}$

Thus, $H$ is an extended $F_{B_{e}}$ contraction for $τ = 3$ .
Case 2.: Let $x = \frac{1}{5^{2 n - 1}}; y = 0 .$
$H x = \frac{1}{5^{2 n}}; H y = 0 .$
Consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) & = F_{B_{e}} (B_{e} (\frac{1}{5^{2 n}}, 0)) \\ = F_{B_{e}} (\frac{1}{5^{4 n}}) \\ = \ln (\frac{1}{5^{4 n}}) . \end{matrix}$

Now consider

$\begin{matrix} F_{B_{e}} (B_{e} (x, y)) & = F_{B_{e}} (B_{e} (\frac{1}{5^{2 n - 1}}, 0)) \\ = F_{B_{e}} (\frac{1}{5^{4 n - 2}}) \\ = \ln (\frac{1}{5^{4 n - 2}}) . \end{matrix}$

Now take

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) - F_{B_{e}} (B_{e} (x, y)) & = \ln (\frac{1}{5^{4 n}}) - \ln (\frac{1}{5^{4 n - 2}}) \\ = \ln (\frac{5^{4 n - 2}}{5^{4 n}}) \\ = \ln (\frac{1}{5^{2}}) \\ = \ln (\frac{1}{25}) \\ < - 3 . \end{matrix}$

For $τ = 3, H$ satisfied all the conditions of the above theorem and 0 is the unique fixed point.

Similarly, for

x = 0

and

\frac{1}{5^{2 n - 1}}

, the same proof follows as above. Hence, all the conditions of the above theorem are satisfied for all the cases and 0 is the unique fixed point.

Example 3.

Let

X = {- 1, 0, 1} .

Define the function

s : X \times X \to [1, \infty)

s (x, y) = 2 + x + y

and

B_{e} : X \times X \to R^{+}

as:

B_{e} (- 1, - 1) = B_{e} (0, 0) = B_{e} (1, 1) = 0;

B_{e} (- 1, 0) = B_{e} (0, - 1) = 3;

B_{e} (- 1, 1) = B_{e} (1, - 1) = 7;

B_{e} (0, 1) = B_{e} (1, 0) = 1 .

It is clear that

(X, B_{e})

is a complete

B_{e}

-metric space.

Let

H : X \to X

given by

H 0 = 0 = H 1, H (- 1) = 1 .

Define

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = - \frac{1}{α} + α

and

τ \in (0, 2]

Case 1.: Let $x = 0$ . Now, $B_{e} (H 0, H 1) = B_{e} (H 0, H 0) = B_{e} (0, 0) = 0$ . Therefore, we only need to consider $y = - 1$ .
Now, $B_{e} (H 0, H (- 1)) = B_{e} (0, 1) = 1$ and $B_{e} (0, - 1) = 3 .$
Consider

$\begin{matrix} τ + F_{B_{e}} (B_{e} (H 0, H (- 1))) & = τ - \frac{1}{B_{e} (H 0, H (- 1))} + (B_{e} (H 0, H (- 1)) \\ = τ - 1 + 1 \\ = τ \end{matrix}$

$\begin{matrix} F_{B_{e}} (B_{e} (0, - 1)) & = - \frac{1}{B_{e} (0, - 1)} + B_{e} (0, - 1) \\ = - \frac{1}{3} + 3 \\ = 3 - \frac{1}{3} \\ = \frac{8}{3} . \end{matrix}$

Clearly for $τ \in (0, \frac{1}{2})$ ,

$τ + F_{B_{e}} (B_{e} (H 0, H (- 1))) \leq F_{B_{e}} (B_{e} (0, - 1)) .$
Case 2.: Let $x = 1$ . Now, $B_{e} (H 1, H 1) = B_{e} (H 1, H 0) = 0$ . Therefore, we only need to consider $y = - 1$ .
Now, $B_{e} (H 1, H (- 1)) = B_{e} (0, 1) = 1$ and $B_{e} (1, - 1) = 7 .$
Consider

$\begin{matrix} τ + F_{B_{e}} (B_{e} (H 1, H (- 1))) & = τ - \frac{1}{B_{e} (H 1, H (- 1))} + (B_{e} (H 1, H (- 1)) \\ = τ - 1 + 1 \\ = τ . \end{matrix}$

$\begin{matrix} F_{B_{e}} (B_{e} (1, - 1)) & = - \frac{1}{B_{e} (1, - 1)} + B_{e} (1, - 1) \\ = - \frac{1}{7} + 7 \\ = 7 - \frac{1}{7} \\ = \frac{48}{7} . \end{matrix}$

Clearly for $τ \in (0, \frac{1}{2})$ ,

$τ + F_{B_{e}} (B_{e} (H 1, H (- 1))) \leq F_{B_{e}} (B_{e} (1, - 1)) .$

For

x = - 1

, the proof is similar as above cases. Hence, all the conditions of the Theorem 1 are satisfied and 0 is the unique fixed point. Thus, the above examples illustrate the above theorem.

3. An Extended $F_{B_{e}}$ -Expanding Contraction

We start this section by introducing following definition:

Definition 6.

Let

(X, B_{e})

be a

B_{e}

-metric space. A mapping

H : X \to X

is said to be an extended expanding if

\forall x, y \in X B_{e} (H x, H y) \geq κ B_{e} (x, y); w h e r e κ > 1 .

Theorem 2.

Let

(X, B_{e})

be a complete

B_{e}

-metric space such that

B_{e}

is a continuous functional. Let

H : X \to X

be surjective and extended expanding. Then,

H

is bijective and has a unique fixed point.

Proof.

First, we will prove that

H

is bijective. For this, we need to prove

H

is injective.

Let

x, y \in X

with

x \neq y

. From the definition of extended expanding,

B_{e} (H x, H y) \geq κ B_{e} (x, y) > 0 .

which yields

H x \neq H y

. Hence,

H

is bijective.

Since

H

is bijective,

H

has an inverse on its range. Note that

H^{- 1}

is a Banach contraction in the setting of an

B_{e}

-metric space. In addition, since

\frac{1}{κ} < 1,

we can conclude that

H^{- 1}

has a unique fixed point by using Theorem 3 of Kamran et al. [13]. This completes the proof of the theorem. □

Theorem 3.

Let

(X, B_{e})

be a complete

B_{e}

-metric space such that

B_{e}

is a continuous functional. If

H : X \to X

is surjective then there exists a mapping

H^{*} : X \to X

such that

H \circ H^{*}

is the identity map on X.

The proof is omitted as it is easy to prove.

Now, we define a new definition.

Definition 7.

Let

(X, B_{e})

be a complete

B_{e}

-metric space. A mapping

H

is said to be extended F-expanding if there exists

F \in F^{*}

and

τ > 0

such that for all

x, y \in X

B_{e} (x, y) > 0 \Rightarrow F_{B_{e}} (B_{e} (H x, H y)) \geq F_{B_{e}} (B_{e} (x, y)) + τ

(12)

where

F_{B_{e}} : (0, + \infty) \to R

is a mapping satisfying:

$(F 1)$: $F_{B_{e}}$ is strictly increasing, i.e., for all $α, β \in R^{+}$ such that if $α < β$ then $F_{B_{e}} (α) < F_{B_{e}} (β);$
$(F 2)$: For each sequence ${α_{n}} \subset (0, + \infty)$ , then

$lim_{n \to \infty} α_{n} = 0 \Leftrightarrow lim_{n \to \infty} F_{B_{e}} (α_{n}) = - \infty;$
$(F 3)$: There exists $k \in (0, 1)$ such that $lim_{α \to 0^{+}} α^{k} F_{B_{e}} (α) = 0 .$

We represent by

F^{*}

the set of all functions satisfying the conditions

(F 1)

–

(F 3)

Theorem 4.

Let

(X, B_{e})

be a complete

B_{e}

-metric space such that

B_{e}

is a continuous functional. Let

H : X \to X

be surjective and extended F-expanding. Then,

H

has a unique fixed point.

Proof.

From Theorem 3, there exists a mapping

H^{*} : X \to X

such that

H \circ H^{*}

is the identity mapping on X.

Let

x, y \in X

be arbitrary points such that

x \neq y

, and let

η = H^{*} x

and

ξ = H^{*} y

(obviously

η \neq ξ

) which yields

B_{e} (η, ξ) > 0 .

From the definition of extended F-expanding, we get

F_{B_{e}} (B_{e} (H η, H ξ)) \geq F_{B_{e}} (B_{e} (η, ξ)) + τ .

Since

H η = H (H^{*} x) = x

and

H ξ = H (H^{*} y) = y

, then

F_{B_{e}} (B_{e} (x, y)) \geq F_{B_{e}} (B_{e} (H^{*} x, H^{*} y)) + τ .

Therefore,

H^{*} : X \to X

is an extended F-contraction. By Theorem 1,

H^{*}

has a unique fixed point

δ \in X .

Now consider

\begin{matrix} H δ & = H (H^{*} δ) \\ = δ \end{matrix}

(13)

Hence,

δ

is also a fixed point of

H

In order to get uniqueness, let us suppose that

H

has at most two fixed points. If

δ_{1}, δ_{2} \in X

and

H δ_{1} = δ_{1} \neq δ_{2} = H δ_{2}

, then

B_{e} (δ_{1}, δ_{2}) > 0

which yields

F_{B_{e}} (B_{e} (H δ_{1}, H δ_{2})) \geq F_{B_{e}} (B_{e} (δ_{1} δ_{2})) + τ

0 = F_{B_{e}} (B_{e} (H δ_{1}, H δ_{2})) - F_{B_{e}} (B_{e} (δ_{1} δ_{2})) \geq τ > 0,

which is a contradiction. Thus,

δ_{1} = δ_{2} .

Therefore, the fixed point of

H

is unique. □

Remark 1.

H

is not surjective, the above theorem is false.

For example, let

X = [0, 1]

. Define

B_{e} (x, y) : X \times X \to R^{+}

and

s : X \times X \to [1, \infty)

B_{e} (x, y) = {(x - y)}^{2}, s (x, y) = x + y + 1

Then,

B_{e}

is a complete

B_{e}

-metric space on

X .

Define

H : X \to X

H x = 2 x + 1

for all

x \in X

. Then,

\begin{matrix} B_{e} (H x, H y) & = B_{e} (2 x + 1, 2 y + 1) \\ = {(2 x - 2 y)}^{2} \\ = 4 {(x - y)}^{2} \\ > B_{e} (x, y) . \end{matrix}

(14)

Thus,

H

satisfies all the conditions of the theorem but

H

has no fixed point.

s (x, y) = 1

, then the above theorem will reduce to Theorem 2.1 of Jaroslaw Gornicki [7]. Thus, we can conclude that our theorem is a standard generalization of Theorem 2.1 of Jaroslaw Gornicki [7].

4. An Extended Generalized $F_{B_{e}}$ -Contraction

Definition 8.

Let

(X, B_{e})

be a

B_{e}

-metric space. A map

H : X \to X

is said to be an extended generalized

F_{B_{e}}

-contraction on

(X, B_{e})

if there exists

F \in F^{*}

and

τ > 0

such that for all

x, y \in X

satisfying

B_{e} (H x, H y) > 0,

the following holds:

τ + F_{B_{e}} (B_{e} (H x, H y)) \leq F_{B_{e}} (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\}),

and for each

x_{0} \in X

{lim}_{n, m \to \infty} s (x_{n}, x_{m}) < \frac{1}{k}

, where

k \in (0, 1)

. Here

x_{n} = H^{n} x_{0}; n = 1, 2, 3, \dots .

Remark 2.

1.: Every $F$ -contraction is an extended generalized $F_{B_{e}}$ -contraction.
2.: Let $H$ be an extended generalized $F_{B_{e}}$ -contraction and from the definition of extended generalized $F_{B_{e}}$ -contractions we have for all $x, y \in X, H x \neq H y$ , which gives $B_{e} (H x, H y) > 0 .$ Thus,

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) & < τ + F_{B_{e}} (B_{e} (H x, H y)) \\ \leq F_{B_{e}} (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\}) . \end{matrix}$

(15)

Then, by $(F 1)$ , we get

$B_{e} (H x, H y) \leq max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\};$

$\forall_{x, y \in X,} H x \neq H y .$

Counter example for Remark: The following example shows that the inverse implication of the remark does not hold. Let

X = [0, \infty)

define

B_{e} : X \times X \to R

B_{e} (x, y) = {(x - y)}^{2}

and

s : X \times X \to [1, \infty)

s (x, y) = x + y + 1

. Then,

B_{e}

is an

B_{e}

-metric. Define

H : X \to X

H x = \{\begin{matrix} 0, if 0 \leq x < 1 \\ \frac{1}{4}, if x \geq 1 . \end{matrix}

Clearly

H

is not continuous.

Thus,

H

is not an

F

-contraction. For

x \in [0, 1)

and

y = 1

we have

B_{e} (H x, H 1) = B_{e} (0, \frac{1}{4}) = \frac{1}{16} > 0

and

\begin{matrix} max \{B_{e} (x, 1), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (1, H 1)}{1 + B_{e} (1, H 1)}, \frac{B_{e} (x, H x) + B_{e} (1, H 1)}{2}\} & \geq B_{e} (1, H 1) \\ = B_{e} (1, \frac{1}{4}) \\ = \frac{9}{16} \\ > \frac{1}{16} \\ = B_{e} (H x, H 1) . \end{matrix}

(16)

Define the function

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = \ln α, \forall α \in R^{+} & τ > 0 .

Then consider

\begin{matrix} F_{B_{e}} (B_{e} (H x, H 1)) - F_{B_{e}} (B_{e} (1, H 1)) & = F_{B_{e}} (\frac{1}{16}) - F_{B_{e}} (\frac{9}{16}) \\ = \ln (\frac{1}{16}) - \ln (\frac{9}{16}) \\ = \ln ((\frac{1}{16}) \times (\frac{16}{9})) \\ = \ln \frac{1}{9} \\ < - 2 . \end{matrix}

(17)

Thus,

H

is an extended generalized

F_{B_{e}}

-contraction for

τ = 2

Theorem 5.

Let

(X, B_{e})

be a

B_{e}

-metric space such that

B_{e}

is a continuous functional and

H : X \to X

be an extended generalized

F_{B_{e}}

-contraction. Then,

H

has a unique fixed point.

Proof.

Let

x \in X

be arbitrary and fixed. We define

x_{n + 1} = H x_{n}; \forall n \in N \cup {0},

where

x_{0} = x .

If there exists

n_{0} \in N \cup {0}

such that

x_{n_{0} + 1} = x_{n_{0}},

then

H x_{n_{0}} = x_{n_{0}} .

This concludes that

x_{n_{0}}

is a fixed point of

H

Let us suppose that

x_{n + 1} \neq x_{n}

for all

n \in N \cup {0} .

Which gives

B_{e} (x_{n + 1}, x_{n}) > 0

. It follows from extended generalized

F_{B_{e}}

-contraction that for each

n \in N .

\begin{matrix} F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) & = F_{B_{e}} (B_{e} (H x_{n}, H x_{n - 1})) \\ \leq F_{B_{e}} (max {B_{e} (x_{n}, x_{n - 1}), \frac{B_{e} (x_{n}, x_{n + 1})}{1 + B_{e} (x_{n}, x_{n + 1})}, \frac{B_{e} (x_{n - 1}, x_{n})}{1 + B_{e} (x_{n - 1}, x_{n})}, \\ \frac{B_{e} (x_{n}, x_{n + 1}) + B_{e} (x_{n - 1}, x_{n})}{2}}) - τ \\ \leq F_{B_{e}} (max {B_{e} (x_{n}, x_{n - 1}), B_{e} (x_{n}, x_{n + 1}), B_{e} (x_{n - 1}, x_{n}), \\ \frac{B_{e} (x_{n}, x_{n + 1}) + B_{e} (x_{n - 1}, x_{n})}{2}}) - τ \\ \leq F_{B_{e}} (max \{B_{e} (x_{n}, x_{n - 1}), B_{e} (x_{n}, x_{n + 1})\}) - τ . \end{matrix}

(18)

B_{e} (x_{n + 1}, x_{n}) = B_{e} (x_{n}, x_{n + 1})

then

F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) \leq F_{B_{e}} (B_{e} (x_{n}, x_{n + 1})) - τ

, which is a contradiction due to

F 1

Thus,

F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) \leq F_{B_{e}} (B_{e} (x_{n}, x_{n - 1})) - τ; \forall n \in N \cup {0} .

(19)

Similarly,

F_{B_{e}} (B_{e} (x_{n}, x_{n - 1})) \leq F_{B_{e}} (B_{e} (x_{n - 1}, x_{n - 2})) - τ; \forall n \in N \cup {0} .

(20)

By using (20)&(21), we have

F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) \leq F_{B_{e}} (B_{e} (x_{n - 1}, x_{n - 2})) - 2 τ .

(21)

By repeating same scenario, we get

F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) \leq F_{B_{e}} (B_{e} (x_{1}, x_{0})) - n τ; \forall n \in N \cup {0} .

(22)

Taking the limit as

n \to \infty

in (23), we get

lim_{n \to \infty} F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) = - \infty .

(23)

By using

(F 2)

, we get

lim_{n \to \infty} B_{e} (x_{n + 1}, x_{n}) = 0 .

(24)

From

(F 3)

, there exists

k \in (0, 1)

such that

lim_{n \to \infty} ({(B_{e} (x_{n + 1}, x_{n}))}^{k} F_{B_{e}} (B_{e} (x_{n + 1}, x_{n}))) = 0 .

(25)

Now consider

\begin{matrix} {(B_{e} (x_{n + 1}, x_{n}))}^{k} (F_{B_{e}} (B_{e} (x_{n + 1}, x_{n})) - F_{B_{e}} (B_{e} (x_{1}, x_{0}))) & \leq - {(B_{e} (x_{n + 1}, x_{n}))}^{k} n τ \\ \leq 0; \forall n \in N . \end{matrix}

(26)

By using (25)&(26) and taking the limit as

n \to \infty

in (27), we get

lim_{n \to \infty} (n {(B_{e} (x_{n + 1}, x_{n}))}^{k}) = 0 .

(27)

Then, there exists

n_{1} \in N

such that

n {(B_{e} (x_{n + 1}, x_{n}))}^{k} \leq 1; \forall n \geq n_{1},

which yields

B_{e} (x_{n + 1}, x_{n}) \leq \frac{1}{n^{\frac{1}{k}}}; \forall n \geq n_{1} .

(28)

In order to prove that

{x_{n}}

is a Cauchy sequence, consider

m, n \in N

such that

m > n \geq n_{1} .

By using (29) and the triangle inequality, we get

\begin{matrix} B_{e} (x_{n}, x_{m}) & \leq s (x_{n}, x_{m}) [B_{e} (x_{n}, x_{n + 1}) + B_{e} (x_{n + 1}, x_{m})] \\ \leq s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) [B_{e} (x_{n + 1}, x_{n + 2}) + B_{e} (x_{n + 2}, x_{m})] \\ \leq s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) B_{e} (x_{n + 1}, x_{n + 2}) + \dots \\ + s (x_{n}, x_{m}) s (x_{n + 1}, x_{m}) s (x_{n + 2}, x_{m}) \dots s (x_{m - 2}, x_{m}) s (x_{m - 1}, x_{m}) B_{e} (x_{m - 1}, x_{m}) \\ \leq s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{n}, x_{m}) B_{e} (x_{n}, x_{n + 1}) \\ + s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{n + 1}, x_{m}) B_{e} (x_{n + 1}, x_{n + 2}) + \dots \\ + s (x_{1}, x_{m}) s (x_{2}, x_{m}) \dots s (x_{m - 1}, x_{m}) B_{e} (x_{m - 1}, x_{m}) . \end{matrix}

(29)

Note that this series

\sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m}) converges .

Since

\begin{matrix} \sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m}) & \leq \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}} \prod_{i = 1}^{n} s (x_{i}, x_{m}) \\ < \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}} . \frac{1}{k} \\ = \frac{1}{k} \sum_{n = 1}^{\infty} \frac{1}{n^{\frac{1}{k}}}; which is convergent . \end{matrix}

(30)

Let

S = \sum_{n = 1}^{\infty} B_{e} (x_{n}, x_{n + 1}) \prod_{i = 1}^{n} s (x_{i}, x_{m});

S_{n} = \sum_{j = 1}^{n} B_{e} (x_{j}, x_{j + 1}) \prod_{i = 1}^{j} s (x_{i}, x_{m}) .

Thus, for

m > n

above inequality implies

B_{e} (x_{n}, x_{m}) \leq S_{m - 1} - S_{n - 1} .

Letting

n \to \infty

, we conclude that

{x_{n}}

is a Cauchy sequence. Hence, there exists

κ \in X

such that

{x_{n}} \to κ .

We shall prove that κ is a fixed point of

H

by two following cases:

Case 1.

H

is continuous, we have

\begin{matrix} B_{e} (κ, H κ) & = lim_{n \to \infty} B_{e} (x_{n}, H x_{n}) \\ = lim_{n \to \infty} B_{e} (x_{n}, x_{n + 1}) \\ = 0 . \end{matrix}

(31)

This proves that κ is a fixed point of

H

Case 2.

F_{B_{e}}

is continuous. In this case, we consider two following sub-cases:

Case 2.1.: For each $n \in N,$ there exists $i_{n} \in N$ such that $x_{i_{n} + 1} = H κ$ and $i_{n} > i_{n - 1}$ where $i_{0} = 1 .$ Then, we have

$κ = lim_{n \to \infty} x_{i_{n} + 1} = lim_{n \to \infty} H κ = H κ .$

This proves that κ is a fixed point of $H .$
Case 2.2.: There exists $n_{0} \in N$ such that $x_{n + 1} \neq H κ; \forall n \geq n_{0} .$

$i . e, B_{e} (H x_{n}, H κ) > 0; \forall n \geq n_{0} .$

It follows from extended generalized

F_{B_{e}}

-contraction and

F 1

\begin{matrix} τ + F_{B_{e}} (B_{e} (x_{n + 1}, H κ)) & = τ + F_{B_{e}} (B_{e} (H x_{n}, H κ)) \\ \leq F_{B_{e}} (max {B_{e} (x_{n}, κ), \frac{B_{e} (x_{n}, H x_{n})}{1 + B_{e} (x_{n}, H x_{n})}, \frac{B_{e} (κ, H κ)}{1 + B_{e} (κ, H κ)}, \\ \frac{B_{e} (x_{n}, H x_{n}) + B_{e} (κ, H κ)}{2}}) \\ \leq F_{B_{e}} (max {B_{e} (x_{n}, κ), B_{e} (x_{n}, x_{n + 1}), B_{e} (κ, H κ), \\ \frac{B_{e} (x_{n}, H x_{n}) + B_{e} (κ, H κ)}{2}}) \\ \leq F_{B_{e}} (max {B_{e} (x_{n}, κ), B_{e} (x_{n}, x_{n + 1}), B_{e} (κ, H κ), \\ \frac{B_{e} (x_{n}, x_{n + 1}) + B_{e} (κ, H κ)}{2}}) . \end{matrix}

(32)

B_{e} (κ, H κ) > 0

then

lim_{n \to \infty} B_{e} (x_{n}, κ) = lim_{n \to \infty} B_{e} (κ, x_{n + 1}) = 0 .

Then, there exists

n_{1} \in N

such that for all

n \geq n_{1},

we have

max \{B_{e} (x_{n}, κ), B_{e} (x_{n}, x_{n + 1}), B_{e} (κ, H κ), \frac{B_{e} (x_{n}, x_{n + 1}) + B_{e} (κ, H κ)}{2}\} = B_{e} (κ, H κ) .

From (33), we get

τ + F_{B_{e}} (B_{e} (x_{n + 1}, H κ)) \leq F_{B_{e}} (B_{e} (κ, H κ)); \forall n \geq max {n_{0}, n_{1}} .

(33)

Since

F_{B_{e}}

is continuous, taking the limit as

n \to \infty

(34)

, we obtain

τ + F_{B_{e}} (B_{e} (κ, H κ)) \leq F_{B_{e}} (B_{e} (κ, H κ)),

which is a contradiction. Hence,

B_{e} (κ, H κ) = 0 .

Therefore, κ is a fixed point of

H

By the above two cases,

H

has a fixed point κ.

To prove uniqueness, let

κ, κ^{*}

be two fixed points of

H

, such that

κ \neq κ^{*} .

Thus,

B_{e} (κ, κ^{*}) > 0

which implies

B_{e} (H κ, H κ^{*}) > 0

From extended generalized F-contraction,

\begin{matrix} τ + F_{B_{e}} (B_{e} (κ, κ^{*})) & = τ + F_{B_{e}} (B_{e} (H κ, H κ^{*})) \\ \leq F_{B_{e}} (max {B_{e} (κ, κ^{*}), \frac{B_{e} (κ, H κ)}{1 + B_{e} (κ, H κ)}, \frac{B_{e} (κ^{*}, H κ^{*})}{1 + B_{e} (κ^{*}, H κ^{*})}, \\ \frac{B_{e} (κ, H κ) + B_{e} (κ^{*}, H κ^{*})}{2}}) \\ = F_{B_{e}} (max \{B_{e} (κ, κ^{*}), \frac{B_{e} (κ, κ)}{1 + B_{e} (κ, κ)}, \frac{B_{e} (κ^{*}, κ^{*})}{1 + B_{e} (κ^{*}, κ^{*})}\}) \\ = F_{B_{e}} (B_{e} (κ, κ^{*})) . \end{matrix}

(34)

which implies,

τ \leq F_{B_{e}} (B_{e} (κ, κ^{*})) - F_{B_{e}} (B_{e} (κ, κ^{*})) = 0

. This is a contradiction.

Thus,

B_{e} (κ, κ^{*}) = 0,

which yields

κ = κ^{*}

. Hence, the fixed point of

H

is unique. □

Example 4.

Let

X = \{\frac{1}{2^{n - 1}}; n \in N\} \cup {0}

. Define

B_{e} : X \times X \to R^{+}

B_{e} (x, y) = {(x - y)}^{2}

and

s : X \times X \to [1, \infty)

s (x, y) = x + y + 1

. Then,

B_{e}

is a complete

B_{e}

-metric on X.

Define

H : X \to X

H (x) = \{\begin{matrix} \frac{1}{2^{n}}, if x \in \{\frac{1}{2^{n - 1}}; n \in N\}; \\ 0, if x \in X . \end{matrix}

Define the function

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = \ln α

for all

α \in R^{+}

and

τ > 0

Case 1.: For $m > n \geq 1$ . Let $x = \frac{1}{2^{n - 1}}$ and $y = \frac{1}{2^{m - 1}}$ .
Now take $n = 1$ and $m = 2$ .
Consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) & = F_{B_{e}} (B_{e} (H 1, H \frac{1}{2})) \\ = F_{B_{e}} (B_{e} (\frac{1}{2}, \frac{1}{4})) \\ = F_{B_{e}} (\frac{1}{16}) \\ = \ln \frac{1}{16} \\ = - 2.7725 . \end{matrix}$

Additionally,

$\begin{matrix} F_{B_{e}} (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\}) \\ = F_{B_{e}} (max \{B_{e} (1, \frac{1}{2}), \frac{B_{e} (1, \frac{1}{2})}{1 + B_{e} (1, \frac{1}{2})}, \frac{B_{e} (\frac{1}{2}, \frac{1}{2})}{1 + B_{e} (\frac{1}{2}, \frac{1}{2})}, \frac{B_{e} (1, \frac{1}{2}) + B_{e} (\frac{1}{2}, \frac{1}{4})}{2}\}) \\ = F_{B_{e}} (max \{\frac{1}{4}, \frac{1}{5}, \frac{1}{17}, \frac{5}{16}\}) \\ = F_{B_{e}} (\frac{5}{16}) \\ = \ln \frac{5}{16} \\ = - 1.1631 . \end{matrix}$

Consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) - F (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\}) \\ = - 2.7725 + 1.1631 \\ = - 1.6094 \\ < - 1 . \end{matrix}$

Thus, $H$ is an extended generalized $F_{B_{e}}$ -contraction for $τ = 1 .$
Case 2.: Let $x = \frac{1}{2}$ and $y = 0$ .
Consider $F_{B_{e}} (B_{e} (H x, H y)) = F_{B_{e}} (B_{e} (\frac{1}{4}, 0)) = F_{B_{e}} (\frac{1}{16}) = \ln \frac{1}{16} = - 2.77$ .
Now,

$\begin{matrix} F_{B_{e}} (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H y) + B_{e} (y, H x)}{2}\}) \\ = F_{B_{e}} (max \{B_{e} (\frac{1}{2}, 0), \frac{B_{e} (\frac{1}{2}, \frac{1}{4})}{1 + B_{e} (\frac{1}{2}, \frac{1}{4})}, \frac{B_{e} (0, 0)}{1 + B_{e} (0, 0)}, \frac{B_{e} (\frac{1}{2}, \frac{1}{4}) + B_{e} (0, 0)}{2}\}) \\ = F_{B_{e}} (max \{\frac{1}{4}, \frac{1}{17}, 0, \frac{1}{32}\}) \\ = F_{B_{e}} (\frac{1}{4}) \\ = \ln \frac{1}{4} \\ = - 1.38 . \end{matrix}$

Now consider

$\begin{matrix} F_{B_{e}} (B_{e} (H x, H y)) - F (max \{B_{e} (x, y), \frac{B_{e} (x, H x)}{1 + B_{e} (x, H x)}, \frac{B_{e} (y, H y)}{1 + B_{e} (y, H y)}, \frac{B_{e} (x, H x) + B_{e} (y, H y)}{2}\}) \\ = - 2.77 + 1.38 \\ = - 1.39 \\ < - 1 . \end{matrix}$

Thus, $H$ is an extended generalized $F_{B_{e}}$ -contraction for $τ = 1 .$

Hence, we can conclude that all the conditions of above theorem are satisfied in all cases and 0 is the unique fixed point.

5. Applications to Existence of Solutions of Non-linear Integral Equation

As applications, we use Theorem 1 and Theorem 5 to study the existence problem of unique solutions of non-linear integral equations.

Theorem 6.

Let X be the set of all continuous real valued functions defined on

[a, b]

i . e ., X = C ([a, b], R)

Define

B_{e} : X \times X \to R

B_{e} (U, V) = s u p {| U (t) - V (t) |}^{2}, t \in [a, b]

with

s (U, V) = | U (t) | + | V (t) | + 1

, where

s : X \times X \to [1, \infty)

Note that

(X, B_{e})

is a complete

B_{e}

-metric space.

Consider the Fredholm integral equation as

U (t) = \int_{a}^{b} H (t, p, U (p)) d p + F_{B_{e}} (t) \forall t, p \in [a, b],

(35)

where

F_{B_{e}} : [a, b] \to R

and

H : [a, b] \times [a, b] \times R \to R

are continuous functions.

Define

H : A \cup B \to A \cup B

H (U (t)) = \int_{a}^{b} H (t, p, U (p)) d p + F_{B_{e}} (t) \forall t, p \in [a, b]

; where

F_{B_{e}} : [a, b] \to R

and

H : [a, b] \times [a, b] \times R \to R

are continuous functions.

Further assume that the following condition holds:

| H (t, p, U (p)) - H (t, p, V (p)) | \leq e^{- \frac{τ}{2}} | U (p) - V (p) |

for each

t, p \in [a, b], U, V \in X

and

τ > 0

. Then, the integral Equation (35) has a solution. We will prove that the operator

H

satisfies the conditions of Theorem 1.

For any

U (t), V (t) \in X

. Consider

\begin{matrix} {| H U (t) - H V (t) |}^{2} & = {(\int_{a}^{b} | H (t, p, U (p)) - H (t, p, V (p)) |)}^{2} d p \\ \leq {(\int_{a}^{b} e^{- \frac{τ}{2}} | U (p) - V (p) |)}^{2} d p \\ \leq e^{- τ} {(\int_{a}^{b} | U (p) - V (p) |)}^{2} d p \\ \leq e^{- τ} B_{e} (U (t), V (t)), \end{matrix}

which implies

B_{e} (H U (t), H V (t)) \leq e^{- τ} B_{e} (U (t), V (t))

Applying logarithms on both sides, we get

\ln (B_{e} (H U (t), H V (t))) \leq \ln (e^{- τ} B_{e} (U (t), V (t)));

\Rightarrow \ln (B_{e} (H U (t), H V (t))) \leq \ln (e^{- τ}) + \ln (B_{e} (U (t), V (t)));

\Rightarrow \ln (B_{e} (H U (t), H V (t))) \leq - τ + \ln (B_{e} (U (t), V (t))) .

Thus,

τ + \ln (B_{e} (H U (t), H V (t))) \leq \ln (B_{e} (U (t), V (t))) .

(36)

Let us define

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = \ln (α), α > 0 .

Then, from (36), we get

τ + F_{B_{e}} (B_{e} (H U (t), H V (t))) \leq F (B_{e} (U (t), V (t))) .

Thus all the conditions of the Theorem 1 are satisfied. Thus, the operator

H

has a unique fixed point. Hence, the Fredholm integral equation has a solution.

Theorem 7.

Let us consider the non-linear integral equation.

U (t) = F_{B_{e}} (t) + \int_{0}^{t} k (t, p) g (p, U (p)) d p,

(37)

where the unknown function

U (t)

takes real values.

Let

X = C ([0, β])

be the space of all real continuous functions defined on

[0, β] .

Define

B_{e} : X \times X \to R

B_{e} (U, V) = {max}_{t \in [0, β]} {| U (t) - V (t) |}^{2}

and

s : X \times X \to [1, \infty)

s (U, V) = | U (t) | + | V (t) | + 1

Clearly,

(X, B_{e})

is a complete

B_{e}

-metric space.

Define a mapping

H : X \to X

H U (t) = F_{B_{e}} (t) + \int_{0}^{t} k (t, p) g (p, U (p)) d p; \forall t \in [0, β] .

Furthermore, we assume the following conditions:

$g \in C ([0, β] \times (- \infty, - \infty))$ and $k \in C ([0, β] \times [0, β])$ such that $k (t, p) \geq 0 .$
$g (t, .) : (- \infty, + \infty) \to (- \infty, + \infty)$ is increasing for all $t \in [0, β] .$
There exists $τ \in [1, + \infty)$ such that for all ${U, V \in X, t \in [0, β], | g (t, U) - g (t, V) |}^{2} \leq τ e^{- τ} M (U, V),$
where, $M (U, V) = max \{{| U - V |}^{2}, \frac{{| U - H U |}^{2}}{{1 + | U - H U |}^{2}}, \frac{{| V - H V |}^{2}}{{1 + | V - H V |}^{2}}, \frac{{| U - H U |}^{2} + {| V - H V |}^{2}}{2}\} .$
${max}_{t, p \in [0, β]} {| k (t, s) |}^{2} \leq 1$ ; For $U \in X,$ we define a norm ${| | U | |}_{τ} = {max}_{t \in [0, β]} | U (t) | e^{- τ t},$ where $t \geq 1$ is chosen arbitrarily.

It is easy to check that

{∥ . ∥}_{τ}

is equivalent to the maximum norm

∥ . ∥

in X, and X be endowed with the

B_{e_{τ}}

defined by

\begin{matrix} B_{e_{τ}} (U, V) & = | | U - {V | |}_{τ} \\ = max_{t \in [0, β]} {{| U (t) - V (t) |}^{2} e^{- τ t}}; U, V \in X and e^{t τ} \geq 1 . \end{matrix}

(38)

Then,

(X, B_{e_{τ}})

is a complete

B_{e}

-metric space.

Now, we will prove that the non-linear integral Equation (37) has a unique solution. For any

U, V \in C ([0, β]), t \in [0, β]

we have

\begin{matrix} {| H U (t) - H V (t) |}^{2} & = | \int_{0}^{t} {k (t, p) [g (p, U (p)) - g (p, V (p))] d p |}^{2} \\ \leq \int_{0}^{t} {| k (t, p) |}^{2} {| | g (p, U (p)) - g (p, V (p)) |}^{2} d p \\ \leq \int_{0}^{t} {| | g (p, U (p)) - g (p, V (p)) |}^{2} d p \\ \leq \int_{0}^{t} τ e^{- τ} M (U (p), V (p)) d p \\ = τ e^{- τ} \int_{0}^{t} e^{p τ} max {{| U (p) - V (p) |}^{2} e^{- p τ}, \frac{{| U (p) - H U (p) |}^{2} e^{- 2 p τ}}{{1 + | U (p) - H U (p) |}^{2} e^{- p τ}}, \\ \frac{{| V (p) - H V (p) |}^{2} e^{- 2 p τ}}{{1 + | V (p) - H V (p) |}^{2} e^{- p τ}}, \\ \frac{{| U (p) - H U (p) |}^{2} + {| V (p) - H V (p) |}^{2}}{2} e^{- p τ}} d p \\ \leq τ e^{- τ} \int_{0}^{t} e^{s τ} max {B_{e_{τ}} (U, V), \frac{B_{e_{τ}} (U, H U)}{1 + B_{e_{τ}} (U, H U)}, \frac{B_{e_{τ}} (V, H V)}{1 + B_{e_{τ}} (V, H V)}, \\ \frac{B_{e_{τ}} (U, H U) + B_{e_{τ}} (V, H V)}{2}} d p \\ = τ e^{- τ} M (U, V) \int_{0}^{t} e^{s τ} d p \\ \leq τ e^{- τ} M (U, V) \frac{e^{t τ}}{τ} \\ \leq e^{- τ} M (U, V) e^{t τ} \\ \leq e^{- (1 - t) τ} M (U, V) \end{matrix}

(39)

which implies

{| H U (t) - H V (t) |}^{2} e^{- t τ} \leq e^{- τ} M (U, V),

which yields

\begin{matrix} B_{e_{τ}} (H U, H V) & = max_{t \in [0, β]} {{| H U (t) - H V (t) |}^{2} e^{- t τ}} \\ \leq e^{- τ} M (U, V) . \end{matrix}

(40)

Applying logarithms on both sides, we get

τ + \ln B_{e_{τ}} (H U, H V) \leq \ln M (U, V); \forall U, V \in X .

(41)

Define

F_{B_{e}} : R^{+} \to R

F_{B_{e}} (α) = \ln α, α > 0 .

Then, from

(41)

we get

τ + F_{B_{e}} (B_{e_{τ}} (H U, H V)) \leq F_{B_{e}} (M (U, V));

where

M (U, V) = max \{B_{e_{τ}} (U, V), \frac{B_{e_{τ}} (U, H U)}{1 + B_{e_{τ}} (U, H U)}, \frac{B_{e_{τ}} (V, H V)}{1 + B_{e_{τ}} (V, H V)}, \frac{B_{e_{τ}} (U, H U) + B_{e_{τ}} (V, H V)}{2}\} .

Thus,

H

is an extended generalized F-contraction. By Theorem 5,

H

has a unique fixed point. Hence, it is the unique solution of the non-linear integral equation.

6. Conclusions

The research topic of fixed point theory and applications, with an extended approach being the latest, has continued for decades.

An extended b-metric space was introduced in 2017 by Kamran et al. [14]. Since then, very few researchers established fixed point theorems using F-contractions in an extended b-metric space since it was very hard to obtain fixed points via the Warkowski [15] approach. In this article, we first introduce various topics called the extended

F_{B_{e}}

-contraction, the extended

F_{B_{e}}

-expanding contraction, and the extended generalized

F_{B_{e}}

-contraction. Thereafter, we presented various fixed point theorems related to F-contractions, which gives a solutions for a non-linear integral equation by using the fixed point technique. Our results are important as they open new research avenues for non-linear analysis and its applications.

Author Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

Funding

This research received no external funding.

Acknowledgments

The authors would like to thank the editor and referees for careful reading the original manuscript and giving comments which were useful for improving the manuscript.

Conflicts of Interest

The authors declare that they have no competing interests.

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Panda, S.K.; Tassaddiq, A.; Agarwal, R.P. A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions. Symmetry 2019, 11, 206. https://doi.org/10.3390/sym11020206

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Panda SK, Tassaddiq A, Agarwal RP. A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions. Symmetry. 2019; 11(2):206. https://doi.org/10.3390/sym11020206

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Panda, Sumati Kumari, Asifa Tassaddiq, and Ravi P Agarwal. 2019. "A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions" Symmetry 11, no. 2: 206. https://doi.org/10.3390/sym11020206

APA Style

Panda, S. K., Tassaddiq, A., & Agarwal, R. P. (2019). A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions. Symmetry, 11(2), 206. https://doi.org/10.3390/sym11020206

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A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions

Abstract

1. Introduction

2. An Extended $F_{B_{e}}$ -Contraction

3. An Extended $F_{B_{e}}$ -Expanding Contraction

4. An Extended Generalized $F_{B_{e}}$ -Contraction

5. Applications to Existence of Solutions of Non-linear Integral Equation

6. Conclusions

Author Contributions

Funding

Acknowledgments

Conflicts of Interest

References

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A New Approach to the Solution of Non-Linear Integral Equations via Various FBe-Contractions

Abstract

1. Introduction

2. An Extended F B e -Contraction

3. An Extended F B e -Expanding Contraction

4. An Extended Generalized F B e -Contraction

5. Applications to Existence of Solutions of Non-linear Integral Equation

6. Conclusions

Author Contributions

Funding

Acknowledgments

Conflicts of Interest

References

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A New Approach to the Solution of Non-Linear Integral Equations via Various F_Be-Contractions

2. An Extended $F_{B_{e}}$ -Contraction

3. An Extended $F_{B_{e}}$ -Expanding Contraction

4. An Extended Generalized $F_{B_{e}}$ -Contraction