Fixed point results for single and setvalued αηψcontractive mappings
 Nawab Hussain^{1},
 Peyman Salimi^{2} and
 Abdul Latif^{1}Email author
https://doi.org/10.1186/168718122013212
© Hussain et al.; licensee Springer 2013
Received: 21 May 2013
Accepted: 23 July 2013
Published: 8 August 2013
Abstract
Samet et al. (Nonlinear Anal. 75:21542165, 2012) introduced αψcontractive mappings and proved some fixed point results for these mappings. More recently Salimi et al. (Fixed Point Theory Appl. 2013:151, 2013) modified the notion of αψcontractive mappings and established certain fixed point theorems. Here, we continue to utilize these modified notions for singlevalued Geraghty and MeirKeelertype contractions, as well as multivalued contractive mappings. Presented theorems provide main results of Hussain et al. (J. Inequal. Appl. 2013:114, 2013), Karapinar et al. (Fixed Point Theory Appl. 2013:34, 2013) and Asl et al. (Fixed Point Theory Appl. 2012:212, 2012) as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results.
MSC:46N40, 47H10, 54H25, 46T99.
Keywords
Dedication
Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday
1 Introduction and preliminaries
In metric fixed point theory, the contractive conditions on underlying functions play an important role for finding solution of fixed point problems. Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions by several mathematicians (see [1–21]). In 2012, Samet et al. [15] introduced the concepts of αψcontractive and αadmissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Afterwards, Karapinar and Samet [13] generalized these notions to obtain fixed point results. More recently, Salimi et al. [14] modified the notions of αψcontractive and αadmissible mappings and established fixed point theorems, which are proper generalizations of the recent results in [13, 15]. Here, we continue to utilize these modified notions for singlevalued Geraghty and MeirKeelertype contractions, as well as multivalued contractive mappings. Presented theorems provide main results of Hussain et al. [9], Karapinar et al. [11] and Asl et al. [12] as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results.
Denote with Ψ the family of nondecreasing functions $\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ such that ${\sum}_{n=1}^{\mathrm{\infty}}{\psi}^{n}(t)<+\mathrm{\infty}$ for all $t>0$, where ${\psi}^{n}$ is the n th iterate of ψ.
The following lemma is obvious.
Lemma 1.1 If $\psi \in \mathrm{\Psi}$, then $\psi (t)<t$ for all $t>0$.
Samet et al. [15] defined the notion of αadmissible mappings as follows.
Theorem 1.1 [15]
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},T{x}_{0})\ge 1$;
 (ii)
either T is continuous or for any sequence $\{{x}_{n}\}$ in X with $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$.
Then T has a fixed point.
Very recently Salimi et al. [14] modified the notions of αadmissible and αψcontractive mappings as follows.
Definition 1.2 [14]
Note that if we take $\eta (x,y)=1$, then this definition reduces to Definition 1.1. Also, if we take $\alpha (x,y)=1$, then we say that T is ηsubadmissible mapping.
The following result properly contains Theorem 1.1 and Theorems 2.3 and 2.4 of [13].
Theorem 1.2 [14]
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0})$;
 (ii)
either T is continuous or for any sequence $\{{x}_{n}\}$ in X with $\alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1})$ for all $n\in \mathbb{N}\cup \{0\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, we have $\alpha ({x}_{n},x)\ge \eta ({x}_{n},x)$ for all $n\in \mathbb{N}\cup \{0\}$.
Then T has a fixed point.
2 Modified αηGeraghty type contractions
Our first main result of this section is concerning αηGeraghtytype [4] contractions.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1})$ for all n, then $\alpha (x,fx)\ge \eta (x,fx)$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge \eta ({x}_{0},f{x}_{0})$, then f has a fixed point.
Letting $n\to \mathrm{\infty}$ in the inequality above, we get $d(fz,z)=0$, that is, $z=fz$. □
If in Theorem 2.1 we take, $\eta (x,y)=1$, then we have the following corollary.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},f{x}_{n})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
Further, if in Theorem 2.1 we take $\alpha (x,y)=1$, then we have the following corollary.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\eta ({x}_{n},{x}_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.
If there exists ${x}_{0}\in X$ such that $\eta ({x}_{0},f{x}_{0})\le 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\eta ({x}_{n},{x}_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.
If there exists ${x}_{0}\in X$ such that $\eta ({x}_{0},f{x}_{0})\le 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\eta ({x}_{n},{x}_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.
If there exists ${x}_{0}\in X$ such that $\eta ({x}_{0},f{x}_{0})\le 1$, then f has a fixed point.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\eta ({x}_{n},f{x}_{n})\le 1$ for all n, then $\eta (x,fx)\le 1$.
If there exists ${x}_{0}\in X$ such that $\eta ({x}_{0},f{x}_{0})\le 1$, then f has a fixed point.
From Corollary 2.1, we can deduce the following corollary.
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
Also, from the corollary above, we can deduce the following corollaries.
Corollary 2.10 (Theorem 4 of [9])
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
Corollary 2.11 (Theorem 6 of [9])
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
Corollary 2.12 (Theorem 8 of [9])
 (a)
f is continuous, or
 (b)
if $\{{x}_{n}\}$ is a sequence in X such that ${x}_{n}\to x$, $\alpha ({x}_{n},f{x}_{n})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.
If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
We prove that Corollary 2.9 can be applied to f, but Corollaries 2.10, 2.11 and 2.12 (Theorem 4, 6 and 8 of [9]) cannot be applied to f.
Clearly, $(X,d)$ is a complete metric space. We show that f is an αadmissible mapping. Let $x,y\in X$ with $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. On the other hand, for all $x\in [0,1]$, we have $fx\le 1$. It follows that $\alpha (fx,fy)\ge 1$. Hence, the assertion holds. Also, $\alpha (0,f0)\ge 1$. Now, if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\{{x}_{n}\}\subset [0,1]$, and hence $x\in [0,1]$. This implies that $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}$.
then the conditions of Corollary 2.1 hold, and f has a fixed point.
That is, Corollary 2.10 (Theorem 4 of [9]) cannot be applied for this example.
That is, Corollary 2.11 (Theorem 6 of [9]) cannot be applied for this example.
That is, Corollary 2.12 (Theorem 8 of [9]) cannot be applied for this example.
3 Modified αψMeirKeeler contractive mappings
Recently, Karapinar et al. [11] introduced the notion of a triangular αadmissible mapping as follows.
Definition 3.1 [11]
 (T1)
$\alpha (x,y)\ge 1$ implies that $\alpha (fx,fy)\ge 1$, $x,y\in X$;
 (T2)
$\{\begin{array}{l}\alpha (x,z)\ge 1,\\ \alpha (z,y)\ge 1\end{array}$ implies that $\alpha (x,y)\ge 1$.
Lemma 3.1 [11]
Denote with Ψ the family of nondecreasing functions $\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ continuous at $t=0$ such that

$\psi (t)=0$ if and only if $t=0$,

$\psi (t+s)\le \psi (t)+\psi (s)$.
Definition 3.2 [11]
Let $(X,d)$ be a metric space, and let $\psi \in \mathrm{\Psi}$. Suppose that $f:X\to X$ is a triangular αadmissible mapping satisfying the following condition:
for all $x,y\in X$. Then f is called an αψMeirKeeler contractive mapping.
Now, we modify Definition 3.2 as follows.
Definition 3.3 Let $(X,d)$ be a metric space, and let $\psi \in \mathrm{\Psi}$. Suppose that $f:X\to X$ is a triangular αadmissible mapping satisfying the following condition:
for all $x,y\in X$ with $\alpha (x,y)\ge 1$. Then f is called a modified αψMeirKeeler contractive mapping.
Theorem 3.1 Let $(X,d)$ be a complete metric space. Suppose that f is a continuous modified αψMeirKeeler contractive mapping, and that there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
and hence (3.6) holds.
that is, f has a fixed point. □
Corollary 3.1 (Theorem 10 of [11])
Let $(X,d)$ be a complete metric space. Suppose that f is a continuous αψMeirKeeler contractive mapping, and that there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
That is, conditions of Theorem 3.1 hold, and f has a fixed point. □
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$,
 (ii)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\alpha ({x}_{n},x)\ge 1$ for all n.
Then f has a fixed point.
By taking limit as $n\to +\mathrm{\infty}$, in the inequality above, we get $\psi (d(fz,z))\le 0$, that is, $d(fz,z)=0$. Hence $fz=z$. □
Corollary 3.2 (Theorem 11 of [11])
 (i)
there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$,
 (ii)
if $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all n, and ${x}_{n}\to x$ as $n\to +\mathrm{\infty}$, then $\alpha ({x}_{n},x)\ge 1$ for all n.
Then f has a fixed point.
and $\psi (t)=\frac{1}{4}t$. Clearly, $(X,d)$ is a complete metric space. We show that f is a triangular αadmissible mapping. Let $x,y\in X$, if $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. On the other hand, for all $x,y\in [0,1]$, we have $fx\le 1$ and $fy\le 1$. It follows that $\alpha (fx,fy)\ge 1$. Also, if $\alpha (x,z)\ge 1$ and $\alpha (z,y)\ge 1$, then $x,y,z\in [0,1]$, and hence, $\alpha (x,y)\ge 1$. Thus the assertion holds by the same arguments. Notice that $\alpha (0,f0)\ge 1$.
That is, Corollary 3.2 (Theorem 11 of [11]) cannot be applied for this example.
Denote with ${\mathrm{\Psi}}_{\mathrm{st}}$ the family of strictly nondecreasing functions ${\psi}_{\mathrm{st}}:[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ continuous at $t=0$ such that

${\psi}_{\mathrm{st}}(t)=0$ if and only if $t=0$,

${\psi}_{\mathrm{st}}(t+s)\le {\psi}_{\mathrm{st}}(t)+{\psi}_{\mathrm{st}}(s)$.
Definition 3.4 [11]
Let $(X,d)$ be a metric space, and let ${\psi}_{\mathrm{st}}\in {\mathrm{\Psi}}_{\mathrm{st}}$. Suppose that $f:X\to X$ is a triangular αadmissible mapping satisfying the following condition:
Then f is called a generalized α${\psi}_{\mathrm{st}}$MeirKeeler contractive mapping.
Definition 3.5 Let $(X,d)$ be a metric space, and let ${\psi}_{\mathrm{st}}\in {\mathrm{\Psi}}_{\mathrm{st}}$. Suppose that $f:X\to X$ is a triangular αadmissible mapping satisfying the following condition:
Then f is called a modified generalized α${\psi}_{\mathrm{st}}$MeirKeeler contractive mapping.
Proposition 3.1 Let $(X,d)$ be a metric space, and let $f:X\to X$ be a modified generalized α${\psi}_{\mathrm{st}}$MeirKeeler contractive mapping. If there exists ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then ${lim}_{n\to \mathrm{\infty}}d({f}^{n+1}{x}_{0},{f}^{n}{x}_{0})=0$.
□
Theorem 3.3 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an orbitally continuous modified generalized α${\psi}_{\mathrm{st}}$MeirKeeler contractive mapping. If there exist ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
which is a contradiction. We obtained that ${lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0$, and so, $\{{x}_{n}={f}^{n}{x}_{0}\}$ is a Cauchy sequence. Since X is complete, then there exists $z\in X$ such that ${f}^{n}{x}_{0}\to z$ as $n\to \mathrm{\infty}$. As f is orbitally continuous, so $z=fz$. □
Corollary 3.3 (Theorem 17 of [11])
Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an orbitally continuous generalized α${\psi}_{\mathrm{st}}$MeirKeeler contractive mapping. If there exist ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$, then f has a fixed point.
That is, Corollary 3.3 (Theorem 17 of [11]) cannot be applied for this example.
4 Modified αηcontractive multifunction
Recently, Asl et al. [12] introduced the following notion.
We generalize this concept as follows.
If we take $\eta (x,y)=1$ for all $x,y\in X$, then this definition reduces to Definition 4.1. In case $\alpha (x,y)=1$ for all $x,y\in X$, then T is called an ${\eta}_{\ast}$subadmissible mapping.
Notice that Ψ is the family of nondecreasing functions $\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ such that ${\sum}_{n=1}^{\mathrm{\infty}}{\psi}^{n}(t)<+\mathrm{\infty}$ for all $t>0$, where ${\psi}^{n}$ is the n th iterate of ψ.
As an application of our new concept, we develop now a fixed point result for a multifunction, which generalizes Theorem 1.1.
 (i)
there exist ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\alpha ({x}_{0},{x}_{1})\ge \eta ({x}_{0},{x}_{1})$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1})$ for all $n\in \mathbb{N}$, we have $\alpha ({x}_{n},x)\ge \eta ({x}_{n},x)$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
for all $n\in \mathbb{N}$. Taking limit as $n\to \mathrm{\infty}$ in the inequality above, we get $d(z,Tz)=0$, i.e., $z\in Tz$. □
If in Theorem 4.1 we take $\eta (x,y)=1$, we have the following corollary.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\alpha ({x}_{0},{x}_{1})\ge 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
If in Theorem 4.1 we take $\alpha (x,y)=1$, then we have the following result.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\eta ({x}_{0},{x}_{1})\le 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\eta ({x}_{n},{x}_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta ({x}_{n},x)\le 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
Corollary 4.3 (Theorem 2.1 and 2.3 of [12])
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\alpha ({x}_{0},{x}_{1})\ge 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
That is, conditions of Corollary 4.1 hold, and T has a fixed point. □
Similarly, we can deduce the following corollaries.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\alpha ({x}_{0},{x}_{1})\ge 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\alpha ({x}_{0},{x}_{1})\ge 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\alpha ({x}_{n},{x}_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha ({x}_{n},x)\ge 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\eta ({x}_{0},{x}_{1})\le 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\eta ({x}_{n},{x}_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta ({x}_{n},x)\le 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\eta ({x}_{0},{x}_{1})\le 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\eta ({x}_{n},{x}_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta ({x}_{n},x)\le 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
 (i)
there exists ${x}_{0}\in X$ and ${x}_{1}\in T{x}_{0}$ such that $\eta ({x}_{0},{x}_{1})\le 1$;
 (ii)
for a sequence $\{{x}_{n}\}\subset X$ converging to $x\in X$ and $\eta ({x}_{n},{x}_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta ({x}_{n},x)\le 1$ for all $n\in \mathbb{N}$.
Then T has a fixed point.
Declarations
Acknowledgements
This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first and third authors acknowledge with thanks DSR, KAU for the financial support.
Authors’ Affiliations
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