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Study on Pata Econtractions
Advances in Difference Equations volume 2020, Article number: 539 (2020)
Abstract
In this paper, we introduce the notion of an α–ζ̃––Pata contraction that combines wellknown concepts, such as the Pata contraction, the Econtraction and the simulation function. Existence and uniqueness of a fixed point of such mappings are investigated in the setting of a complete metric space. An example is stated to indicate the validity of the observed result. At the end, we give an application on the solution of nonlinear fractional differential equations.
1 Introduction
In 2015, Khojasteh et al. [1] initiated the concept of simulation functions.
Definition 1.1
([1])
A mapping \(\zeta :[0,\infty )\times {}[ 0,\infty )\rightarrow \mathbb{R}\) is called a simulation function if the following conditions hold:
 \((\zeta _{1})\):

for all ;
 \((\zeta _{2})\):

if , are sequences in \((0,\infty )\) such that , then
(1.1)
We denote by \(\mathcal{Z}\) the family of all above simulation functions.
Let be a metric space and \(\alpha :\mathcal{X}\times \mathcal{X}\rightarrow [0,\infty )\) be a function. A mapping is called αorbital admissible if the following condition holds:
for all \(\nu \in \mathcal{X}\). Moreover, an αorbital admissible mapping is called triangular αorbital admissible if for all \(\nu , \omega \in \mathcal{X}\), we have
Definition 1.2
A set \(\mathcal{X}\) is said to be regular with respect to a given function \(\alpha :\mathcal{X}\times \mathcal{X} \to [0,\infty )\) if for each sequence \(\{\nu _{n}\}\) in \(\mathcal{X}\) such that \(\alpha (\nu _{n},\nu _{n+1})\geq 1\) for all n and \(\nu _{n} \rightarrow \nu \in \mathcal{X}\) as \(n\rightarrow \infty \), then \(\alpha (\nu _{n},\nu )\geq 1\) for all n.
The notion of αadmissible \(\mathcal{Z}\)contractions with respect to a given simulation function was merged and used by Karapinar in [2]. Using this new type of contractive mappings, he investigated the existence and uniqueness of a fixed point in standard metric spaces.
Definition 1.3
([2])
Let T be a selfmapping defined on a metric space \((\mathcal{X}, d)\). If there exist a function \(\zeta \in \mathcal{Z}\) and \(\alpha :\mathcal{X}\times \mathcal{X} \to [0,\infty )\) such that
then we say that T is an αadmissible \(\mathcal{Z}\)contraction with respect to ζ.
Theorem 1.4
([2])
Let \((\mathcal{X},d)\)be a complete metric space and let \(T:\mathcal{X}\rightarrow \mathcal{X}\)be an αadmissible \(\mathcal{Z}\)contraction with respect to ζ. Suppose that:

(a)
T is triangular αorbital admissible;

(b)
there exists \(\nu _{0}\in \mathcal{X}\)such that \(\alpha (\nu _{0}, T\nu _{0})\geq 1\);

(c)
T is continuous.
Then there is \(\nu _{*}\in \mathcal{X}\)such that \(T\nu _{*}=\nu _{*}\).
Remark 1.5
The continuity condition in Theorem 1.4 can be replaced by the “regularity” condition, which is considered in Definition 1.2.
We will consider the following set of functions:
and we denote
Several interesting extensions and generalizations of the Banach contraction principle [3] appeared in the literature. For instance, see [4–10]. Among these generalizations, we cite the paper of Pata [11]. Since then, much work appeared in the same direction; see [12–15].
Theorem 1.6
([11])
Let be a complete metric space and let \(\Lambda \geq 0\), \(\lambda \geq 1\), \(\beta \in [0,\lambda ]\)be fixed constants. The mapping has a fixed point in \(\mathcal{X}\)if the inequality
is satisfied for every \(\varepsilon \in [0,1]\)and .
Definition 1.7
Let be a metric space. We say that is a Pata type Zamfirescu mapping if for all \(\nu , \omega \in \mathcal{X}\), \(\psi \in \mathcal{Z}\) and for every \(\varepsilon \in [0,1]\), , it satisfies the following inequality:
where
and \(\Lambda \geq 0\), \(\lambda \geq 1\) and \(\beta \in [0,\lambda ]\) are constants.
Theorem 1.8
([16])
Let be a complete metric space and let be a Pata type Zamfirescu mapping. Then has a unique fixed point in \(\mathcal{X}\).
We state the following useful known lemma.
Lemma 1.9
Let be a complete metric space and be a sequence in \(\mathcal{X}\)such that . If the sequence is not Cauchy, then there exist and subsequences and of such that
and
In this paper, we combine the concepts of simulation functions and αadmissibility to give a generalized Pata type fixed point result. At the end, we present an application on fractional calculus.
2 Main results
We denote by \(\tilde{\mathcal{Z}}\) the set of all functions \(\tilde{\zeta }:[0,\infty )\times {}[ 0,\infty )\rightarrow \mathbb{R}\) satisfying the following condition:
 \((\tilde{\zeta }_{1})\):

for all .
Definition 2.1
Let be a metric space and \(\phi \in \Phi \). Let \(\Lambda \geq 0\), \(\lambda \geq 1\) and \(\beta \in [0,\lambda ]\) be fixed constants. A triangular αorbital admissible mapping is called an αζ̃ Pata contraction if there exists a function \(\tilde{\zeta }\in \tilde{\mathcal{Z}}\) such that, for every \(\varepsilon \in [0,1]\), the following condition is satisfied:
for all \(\nu ,\omega \in \mathcal{X}\), where
and
Remark 2.2
It is clear that any Pata type Zamfirescu mapping is also an αζ̃ Pata mapping. Indeed, letting \(\alpha (\nu , \omega )=1\) and , the inequality (2.1) becomes
Moreover, note that for all \(\nu , \omega \in \mathcal{X}\).
Theorem 2.3
Every α–ζ̃––Pata contraction on a complete metric space possesses a fixed point if
 (i):

there exists such that ;
 (ii):

is triangular αorbital admissible;
 (iii):

either is continuous, or the set \(\mathcal{X}\)is regular.
If in addition we assume that the following condition is satisfied:
 (iv):

for all ,
then such a fixed point of is unique.
Proof
Let be a point such that . On account of the assumption that is a triangular αorbital admissible mapping, we derive that
and iteratively we find
Moreover, by (2.4) together with (1.3), we have
Again, iteratively, one writes
Starting from this point , we build an iterative sequence where for \(n=1,2,3, \ldots \). We can presume that any two consequent terms of this sequence are distinct. Indeed, if, on the contrary, there exists \(i_{0} \in \mathbb{N}\) such that
then is a fixed point. To avoid this, we will assume in the following that for all \(n\in \mathbb{N}\)
We mention that (2.4) can be rewritten as
respectively,
for any \(n \in \mathbb{N}\). In the sequel, we will denote for all \(\nu \in X\).
Since is an αζ̃Pata contraction, we have
Thus, taking into account \((\tilde{\zeta }_{1})\), together with (2.6) we get
where
and
Denoting by , we have
Thus, (2.8) becomes
We claim that the sequence \(\{ \gamma _{n} \} \) is nonincreasing. Indeed, if we suppose the contrary that, for some p, \(\gamma _{p}<\gamma _{p+1}\), and so \(\max \{ \gamma _{p}, \gamma _{p+1} \} =\gamma _{p+1}\), then we have \(\vert \gamma _{p}\gamma _{p+1} \vert =\gamma _{p+1}\gamma _{p}\).
Consequently, from (2.9), we get, for such an integer p,
The above inequality is true for all \(\varepsilon \in [0,1]\). In particular, for \(\varepsilon =0\), we get \(\gamma _{p+1}\leq \gamma _{p+1}\), which clearly is a contradiction. In this case, we find that the sequence \(\{ \gamma _{n} \} \) is nonincreasing. So we can find a nonnegative real number γ such that
We claim that \(\gamma =0\). In order to prove this, we have to show that the sequence \(\{ \kappa _{n} \} \) is bounded, where . Since the sequence is nonincreasing, we have
By the triangle inequality, we get
On account of (2.5), regarding that is an αζ̃Pata contraction, we have
Taking into account (2.7), this is equivalent to
Using (2.12) and the above inequality, we get
Moreover, since \(\beta \leq \lambda \), we have
Now, supposing that the sequence \(\{ \kappa _{n} \} \) is not bounded, there exists a subsequence \(\{ \kappa _{n_{l}} \} \) of \(\{ \kappa _{n} \} \) such that \(\kappa _{n_{l}}\rightarrow \infty \) as \(l\rightarrow \infty \). In this case, letting \(\varepsilon =\varepsilon _{l}=\frac{1+3\kappa _{1}}{\kappa _{n_{l}}} ( \in [0,1])\), the above inequality yields
This is a contradiction. Thus, we conclude that our presumption is false and then the sequence \(\{ \kappa _{n} \} \) is bounded. Furthermore, there exists such that for all \(n\in \mathbb{N}\).
Let us go back now and prove that \(\gamma =0\) (where \(\gamma =\lim_{n\rightarrow \infty }\gamma _{n}\)). In view of (2.10) and the fact that the sequence \(\{ \gamma _{n} \} \) is nonincreasing, one writes
Recall that
Taking into account that is an αζ̃\({E_{*}}\) contraction, keeping in mind (2.6) and using \((\tilde{\zeta }_{1})\), we have
We have
Letting \(n \to \infty \) in the previous inequality, we obtain
which is equivalent to
When \(\varepsilon \rightarrow 0\), we get \(\gamma \leq 0\). Therefore,
As a next step, we claim that is a Cauchy sequence. On the contrary, assuming that the sequence is not Cauchy, it follows from Lemma 1.9 that there exist and subsequences and such that (1.7) and (1.8) hold. Replacing and in (2.1), we have
where
The triangular αorbital admissibility of shows that . Thus,
Letting \(l\rightarrow \infty \) and taking into account (2.14) and Lemma 1.9, we have
At the same time, one writes
Denoting by and , by Lemma 1.9, it follows that
Thus, passing to the limit as \(l\rightarrow \infty \) in (2.16), we get
Furthermore,
i.e.,
That is, . Therefore, is a Cauchy sequence in the complete metric space. For this reason, there exists \(\nu ^{*}\in \mathcal{X}\) such that , as \(n\rightarrow \infty \).
Furthermore, in the case that is a continuous mapping, we get , that is, is a fixed point of .
Now, suppose that \(\mathcal{X}\) is regular. From (2.1), one writes
Using the regularity of \(\mathcal{X}\) and \((\tilde{\zeta }_{1})\), we get
where
and
Taking into account the boundedness of the sequence \(\{ \kappa _{n} \} \), we have
On the other hand,
Letting \(n\rightarrow \infty \) in the inequality (2.19), we find
which is equivalent to
Obviously, we obtain for \(\varepsilon =0\) that , so . Thus, \(\nu ^{*}\) is a fixed point of . Finally, to prove the uniqueness of the fixed point, we suppose that there exist two fixed points such that \(\nu ^{*}\neq \omega ^{*}\). We have
Taking into account \((\mathit{iv})\), we obtain
which leads to
In the limit \(\varepsilon \rightarrow 0\), we get , that is, \(\nu ^{*}=\omega ^{*}\), which is a contradiction. Therefore, the fixed point of is unique. □
In the following, we present an example that supports our statement, that is, Theorem 2.3 is a generalization of Theorem 1.8.
Example 2.4
Take \(\mathcal{X}=A \times A\), where \(A=[0,11]\) and is the usual distance. Define the mapping by
where . For \(\nu _{1}=(11,0)\) and \(\nu _{2}=(2,0)\), we have
and
Thus,
so that the inequality (1.6) does not hold for \(\varepsilon =0\). That is, is not a Pata type Zamfirescu mapping.
Consider the function \(\alpha :\mathcal{X}\times \mathcal{X} \to [0,\infty )\) given as
Since the assumptions \((i)\)–\((\mathit{iv})\) are obviously satisfied, we have to prove that is an αζ̃ Pata contraction. Take \(\alpha =\beta =1\), \(\Lambda =6\) and the functions \(\Psi (t)=\frac{t}{2}\), .
For \(\nu , \omega \in B\), we have , so that (2.1) holds.
For \(\nu =(11,0)\) and \(\omega =(2,0)\) we have
Due to the way the function α was defined, we omit the other cases.
3 An application on a fractional boundary value problem
In this section, we ensure the existence of a solution of a nonlinear fractional differential equation (for more related details, see [17–23]). Denote by \(\mathcal{X}= C[0,1]\) the set of all continuous functions defined on \([0,1]\). We endow \(\mathcal{X}\) with the metric given as
Consider the fractional differential equation
with boundary conditions
Here, \({}^{c}D^{\mu }\) corresponds for the Caputo fractional derivative of order μ, given as
where \(n1<\mu <n\) and \(n=[\mu ]+1\), and \(I^{\mu }f\) is the Riemann–Liouville fractional integral of order μ of a continuous function f, defined by
In [24], it is showed that the problem (3.1) and (3.2) can be written in the following integral form:
Theorem 3.1
Assume that

1.
\(f:[0,1]\times \mathbb{R}\rightarrow \mathbb{R}\)is continuous;

2.
for all \(\rho ,\omega \in \mathcal{X}\), we have
$$ \bigl\vert f\bigl(s,\rho (s)\bigr)f\bigl(s,\omega (s)\bigr) \bigr\vert \leq \frac{\varepsilon ^{2}}{4}\Gamma ( \mu +1) \bigl\vert \rho (s)\omega (s) \bigr\vert , $$(3.6)for each \(s\in [0,1]\), where \(\varepsilon \in [0,1]\).
Then the problem 3.1and 3.2possesses a unique solution.
Proof
Consider the functional
Note that a solution of (3.5) is also a fixed point of T. We mention that T is well posed. For all \(\rho ,\omega \in \mathcal{X}\) and \(s\in [0,1]\), we have
where B is the beta function. Consequently, one has
where \(\psi (\varepsilon )=\varepsilon \), \(\beta =\lambda =1\) and \(\Lambda =2\). Applying Theorem 2.3, the functional T admits a unique fixed point, that is, the problem (3.1) and (3.2) possesses a unique solution. □
4 Conclusion and remarks
Our results merged from and generalized several existing results in the related literature. First of all, as underlined in Remark 2.2, the main result of [16] is a consequence of our given theorem. On the other hand, by choosing the auxiliary functions in a proper way, we may state a long list of corollaries. More precisely, by choosing the mapping α in a proper way, we can get the analogue of our result in the setting of partially ordered metric spaces, or in the setup of cyclic mappings. Note that, if we take \(\alpha (x,y)=1\) for all x, y, we get the standard fixed point theorems in the context of complete metric spaces; see [25–29]. In addition, by choosing the appropriate simulation function, one can get several more results; see [30–35].
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The authors thank the anonymous referees for their remarkable comments, suggestions, and ideas, which helped to improve this paper. The authors also thanks to their universities.
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Karapinar, E., Fulga, A. & Aydi, H. Study on Pata Econtractions. Adv Differ Equ 2020, 539 (2020). https://doi.org/10.1186/s13662020029924
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DOI: https://doi.org/10.1186/s13662020029924