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MODIFIED ORDERED S-METRIC SPACES AND SOME FIXED POINT THEOREMS
FOR CONTRACTIVE MAPPINGS WITH APPLICATION TO PERIODIC BOUNDARY
VALUE PROBLEMS
Preprint · July 2019
DOI: 10.13140/RG.2.2.29922.79042
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MODIFIED ORDERED S-METRIC SPACES AND SOME FIXED
POINT THEOREMS FOR CONTRACTIVE MAPPINGS WITH
APPLICATION TO PERIODIC BOUNDARY VALUE PROBLEMS.
Z. MUSTAFA1,∗ , R.J. SHAHKOOHI2 , V. PARVANEH3 , Z. KADELBURG4 AND M.M.M.
JARADAT5
Abstract. In this paper, we introduce the structure of modified S-metric spaces
as a generalization of both S-metric and Sb -metric spaces. Also, we present the noe
tions of S-contractive
mappings in the setup of ordered modified S-metric spaces
and investigate the existence of fixed point for such mappings under various contractive conditions. We provide examples to illustrate the results presented herein. An
application to periodic boundary value problems is presented.
1. Introduction and preliminaries
There is a large number of generalizations of Banach contraction principle via using
different forms of contractive conditions in various generalized metric spaces. Some
of such generalizations are obtained via contractive conditions expressed by rational
terms (see, e.g., [7, 9, 17, 19]).
Ran and Reurings initiated the studying of fixed point results on partially ordered
sets in [15]. Further, many researchers have focused on different contractive conditions
in complete metric spaces endowed with a partial order. For more details we refer the
reader to [12, 13].
Parvaneh introduced in [14] the concept of extended b-metric spaces as follows.
Definition 1.1. [14] Let X be a (nonempty) set. A function de : X × X → R+ is
a p-metric if there exists a strictly increasing continuous function Ω : [0, ∞) → [0, ∞)
with Ω−1 (t) ≤ t ≤ Ω(t) and Ω−1 (0) = 0 = Ω(0) such that for all x, y, z ∈ X, the
following conditions hold:
e y) = 0 iff x = y,
(de1 ) d(x,
e
e y) = d(y,
e x),
(d2 ) d(x,
e z) ≤ Ω(d(x,
e y) + d(y,
e z)).
(de3 ) d(x,
e is called a p-metric space, or an extended b-metric space.
In this case, the pair (X, d)
A b-metric [2] is a p-metric, with Ω(t) = st, for some fixed s ≥ 1, while a metric is a
p-metric, when Ω(t) = t. We have the following proposition.
e y) = ξ(d(x, y)) where
Proposition 1.2. [14] Let (X, d) be a metric space and let d(x,
ξ : [0, ∞) → [0, ∞) is a strictly increasing function with t ≤ ξ(t) and 0 = ξ(0). In this
case, de is a p-metric with Ω(t) = ξ(t).
The above proposition provides several examples of p-metric spaces.
Date: Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 47H10; Secondary 54H25.
Key words and phrases. Fixed point, complete metric space, ordered b-metric space.
1
2
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
e y) = ed(x,y) sec−1 (ed(x,y) ).
Example 1.3. Let (X, d) be a metric space and let d(x,
Then de is a p-metric with Ω(t) = et sec−1 (et ).
In [16], Sedghi et al. have introduced the notion of an S-metric space as follows.
Definition 1.4. [16] Let X be a nonempty set and S : X × X × X → R+ be a function
satisfying the following properties:
(S1) S(x, y, z) = 0 iff x = y = z;
(S2) S(x, y, z) ≤ S(x, x, a) + S(y, y, a) + S(z, z, a), for all x, y, z, a ∈ X (rectangle
inequality).
Then, the function S is called an S-metric on X and the pair (X, S) is called an
S-metric space.
Example 1.5. [16] Let R be the real line. Then S(x, y, z) = |x − y| + |x − z| for all
x, y, z ∈ R is an S-metric on R. This S-metric on R is called the usual S-metric on R.
Souayaha and Mlaiki in [18], motivated by the concepts of b-metric and S-metric, introduced the concept of Sb -metric spaces and then they presented some basic properties
of such spaces.
The following is the definition of modified S-metric spaces which is a proper generalization of the notions of S-metric spaces and Sb -metric spaces.
Definition 1.6. Let X be a nonempty set and Ω : [0, ∞) → [0, ∞) be a strictly
increasing continuous function such that Ω−1 (t) ≤ t ≤ Ω(t) for all t > 0 and Ω−1 (0) =
0 = Ω(0). Suppose that a mapping Se : X × X × X → R+ satisfies:
e S(x,
e y, z) = 0 iff x = y = z,
(S1)
e S(x,
e y, z) ≤ Ω[S(x,
e x, a) + S(y,
e y, a) + S(z,
e z, a)] for all x, y, z, a ∈ X (rectangle
(S2)
inequality).
e
e is called a modified S-metric
Then S is called a modified S-metric and the pair (X, S)
space.
e
e y, y) = S(y,
e x, x), for all x, y ∈ X.
Remark 1.7. In an S-metric
space, we have S(x,
e
Each S-metric space is an S-metric
space with Ω(t) = t and every Sb -metric space
e
with parameter s ≥ 1 is an S-metric space with Ω(t) = st.
Proposition 1.8. Let (X, S) be an Sb -metric space with coefficient s ≥ 1 and let
e y, z) = ξ(S(x, y, z)) where ξ : [0, ∞) → [0, ∞) is a strictly increasing function with
S(x,
t ≤ ξ(t) for all t > 0 and ξ(0) = 0. Then Se is a modified S-metric with Ω(t) = ξ(st).
Proof. For all x, y, z, a ∈ X,
e y, z) = ξ(S(x, y, z)) ≤ ξ(sS(x, x, a) + sS(y, y, a) + sS(z, z, a))
S(x,
≤ ξ(sξ(S(x, x, a)) + sξ(S(y, y, a) + sξ(S(z, z, a)))
e x, a) + sS(y,
e y, a) + sS(z,
e z, a))
= ξ(sS(x,
e x, a) + S(y,
e y, a) + S(z,
e z, a)).
= Ω(S(x,
So, Se is a modified S-metric with Ω(t) = ξ(st).
The above proposition provides several examples of modified S-metric spaces.
Example 1.9. Let (X, S) be an Sb -metric space with coefficient s ≥ 1. Then:
e y, z) = eS(x,y,z) sec−1 (eS(x,y,z) ) is an S-metric
e
1. S(x,
with Ω(t) = est sec−1 (est ).
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
3
e y, z) = [S(x, y, z) + 1] sec−1 ([S(x, y, z) + 1]) is an S-metric
e
2. S(x,
with Ω(t) =
−1
[st + 1] sec ([st + 1]).
e y, z) = eS(x,y,z) tan−1 (eS(x,y,z) − 1) is an S-metric
e
3. S(x,
with Ω(t) = est tan−1 (est −
1).
e y, z) = S(x, y, z) cosh(S(x, y, z)) is an S-metric
e
4. S(x,
with Ω(t) = st cosh(st).
S(x,y,z)
e y, z) = e
e
5. S(x,
ln(1 + S(x, y, z)) is an S-metric
with Ω(t) = est ln(1 + st).
e y, z) = S(x, y, z) + ln(1 + S(x, y, z)) is an S-metric
e
6. S(x,
with Ω(t) = st + ln(1 + st).
e
Definition 1.10. Let X be an S-metric
space. A sequence {xn } in X is said to be:
e
(1) S-Cauchy if, for each ε > 0, there exists a positive integer n0 such that for all
e m , xn , xn ) < ε;
m, n ≥ n0 , S(x
e
(2) S-convergent to a point x ∈ X if, for each ε > 0, there exists a positive integer
e n , xn , x) < ε.
n0 such that for all n ≥ n0 , S(x
e
e
e
e
(3) An S-metric space X is called S-complete,
if every S-Cauchy
sequence is Sconvergent in X.
e
e with a super-additive function Ω,
Proposition 1.11. (1) Every S-metric
space (X, S)
f
defines a p-metric dS , where
e
e
df
S (x, y) = S(x, y, y) = S(y, x, x),
e = 1 Ω(3t). Moreover,
for all x, y ∈ X with the function Ω
2
e if and only if xn → x in (X, df
(2) xn → x in (X, S)
S );
e if and only if {xn } is a Cauchy sequence in
(3) {xn } is a Cauchy sequence in (X, S)
(X, df
S ).
Proof. Let x, y, z ∈ X. Then we have
e
e
2df
S (x, y) = S(x, y, y) + S(y, x, x)
e x, z) + 2S(z,
e y, y)) + Ω(S(y,
e y, z) + 2S(z,
e x, x))
≤ Ω(S(x,
e x, z) + 2S(z,
e y, y) + S(y,
e y, z) + 2S(z,
e x, x)]
≤ Ω[S(x,
e x, z) + 3S(z,
e y, y)]
≤ Ω[3S(x,
f
= Ω[3df
S (x, z) + 3dS (z, y)],
which yields that
f
f
2df
S (x, y) ≤ Ω[3dS (x, z) + 3dS (z, y)].
e
e y, z) with nontrivial function Ω need not be
In general, an S-metric
mapping S(x,
jointly continuous in all its variables (see [10]). Thus, in some proofs we will need the
e
following simple lemma about the S-convergent
sequences.
e be an S-metric
e
Lemma 1.12. Let (X, S)
space.
e
1. Suppose that {xn } and {yn } are S-convergent to x and y, respectively. Then we
have
e y, y)]]
Ω−1 [ 12 Ω−1 [S(x,
e n , yn , yn ) ≤ lim sup S(x
e n , yn , yn ) ≤ Ω[2Ω[S(x,
e y, y)].
≤ lim inf S(x
n−→∞
2
n−→∞
e n , yn , yn ) = 0.
In particular, if x = y, then we have lim S(x
n→∞
4
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
e
2. Suppose that {xn } is S-convergent
to x and z ∈ X is arbitrary. Then we have
e z, z)]
Ω−1 [S(x,
e n , z, z) ≤ lim sup S(x
e n , z, z) ≤ Ω[2S(x,
e z, z)].
≤ lim inf S(x
n−→∞
2
n−→∞
e
Proof. 1. Using the rectangle inequality in the S-metric
space it is easy to see that
e x, xn ) + 2S(y,
e y, xn )
e y, y) ≤ Ω S(x,
S(x,
e x, xn ) + 2Ω 2S(y,
e y, yn ) + S(y
e n , xn , xn )
≤ Ω S(x,
and
e n , yn , x)
e n , xn , x) + 2S(y
e n , yn , yn ) ≤ Ω S(x
S(x
e n , yn , y) + S(y,
e x, x) .
e n , xn , x) + 2Ω 2S(y
≤ Ω S(x
Taking the lower limit as n → ∞ in the first inequality and the upper limit as n → ∞
in the second inequality we obtain the desired result.
2. Using the rectangle inequality we see that
e x, xn ) + 2S(z,
e z, xn )
e z, z) ≤ Ω S(x,
S(x,
and
e n , z, z) ≤ Ω S(x
e n , xn , x) + 2S(z,
e z, x) .
S(x
Let B denote the class of all real functions β : [0, ∞) → [0, 1) satisfying the condition
β(tn ) → 1 implies that tn → 0, as n → ∞.
In order to generalize the Banach contraction principle, Geraghty proved in 1973 the
following result.
Theorem 1.13. [5] Let (X, d) be a complete metric space and let f : X → X be a
self-map. Suppose that there exists β ∈ B such that
d(f x, f y) ≤ β(d(x, y))d(x, y)
holds for all x, y ∈ X. Then f has a unique fixed point z ∈ X and for each x ∈ X the
Picard sequence {f n x} converges to z.
In 2010, Amini-Harandi and Emami [1] characterized the result of Geraghty in the
setting of a partially ordered complete metric space. In [4], some fixed point theorems
for mappings satisfying Geraghty-type contractive conditions were proved in various
generalized metric spaces. Also, Zabihi and Razani [19] and Shahkoohi and Razani [17]
obtained some fixed point results for rational Geraghty contractions in b-metric spaces.
Motivated by [9], in this paper we present some fixed point theorems for various
contractive mappings in tripled partially ordered modified S-metric spaces. Our results
extend some existing results in the literature. Examples are provided to show the
usefulness of the results. In the last section, an application is given to a first-order
boundary value problem for differential equations.
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
5
2. Main results
e be an S-metric
e
2.1. Fixed point results using Geraghty contractions. Let (X, S)
space with function Ω and let BΩ denote the class of all functions β : [0, ∞) →
[0, Ω−1 (1)) satisfying the following condition:
lim sup β(tn ) = Ω−1 (1) implies that tn → 0, as n → ∞.
n→∞
e y, z) = e|x−y|+|y−z| − 1 for all x, y, z ∈ R, with
Example 2.1. (1) Let X = R and S(x,
t
Ω(t) = e − 1. Then, by β(t) = (ln 2)e−t for t > 0 and β(0) ∈ [0, ln 2), a function β
belonging to BΩ is given.
(2) Another example of a function in BΩ may be given by β(t) = W (1)e−t for t > 0
e y, z) = (|x − y| + |y − z|)e|x−y|+|y−z| for all x, y, z ∈ R.
and β(0) ∈ [0, W (1)) where S(x,
Here, W is the Lambert W -function (see, e.g., [3]).
e be an ordered S-metric
e
Definition 2.2. Let (X, , S)
space. A mapping f : X → X
e
is called an S-Geraghty contraction if there exists β ∈ BΩ such that
e x, f y, f z)) ≤ β(M (x, y, z))M (x, y, z)
Ω2 (2S(f
(2.1)
for all mutually comparable elements x, y, z ∈ X, where
−1 [S(x,
e f x, f y)]
Ω
e f y, f z) .
e y, z),
, S(y,
M (x, y, z) = max S(x,
2
e
e is said to have the s.l.c. property, if whenever
An ordered S-metric
space (X, , S)
{xn } is an increasing sequence in X such that xn → u ∈ X, one has xn u for all
n ∈ N.
e be an ordered S-complete
e
e
Theorem 2.3. Let (X, , S)
S-metric
space. Let f : X → X
be an increasing mapping with respect to such that there exists an element x0 ∈ X
e
with x0 f x0 . Suppose that f is an S-Geraghty
contraction. If
(I) f is continuous, or
e has the s.l.c. property,
(II) (X, , S)
then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and
only if f has one and only one fixed point.
Proof. Put xn = f n x0 . Since x0 x1 and f is increasing, we obtain by induction that
e n , xn+1 , xn+1 ) = 0.
the sequence {xn } is increasing w.r.t. . We will show that lim S(x
n→∞
Without loss of generality, we may assume that xn 6= xn+1 , for all n ∈ N. Since xn and
xn+1 are comparable, then by (2.1) we have
e n , xn+1 , xn+1 ) = S(f
e xn−1 , f xn , f xn ) ≤ β(M (xn−1 , xn , xn ))M (xn−1 , xn , xn ), (2.2)
S(x
where
M (xn−1 , xn , xn )
o
n
e n−1 , f xn−1 , f xn )], S(x
e n , f xn , f xn )
e n−1 , xn , xn ), 1 Ω−1 [S(x
= max S(x
2
n
o
e n−1 , xn , xn ), 1 Ω−1 [S(x
e n−1 , xn , xn+1 )], S(x
e n , xn+1 , xn+1 )
= max S(x
2
n
o
e
e
e n−1 , xn , xn ), S(xn−1 , xn−1 , xn ) + S(xn+1 , xn+1 , xn ) , S(x
e n , xn+1 , xn+1 )
≤ max S(x
2
e
e
= max{S(xn−1 , xn , xn ), S(xn , xn+1 , xn+1 )}.
6
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
e n−1 , xn , xn ), S(x
e n , xn+1 , xn+1 )} = S(x
e n , xn+1 , xn+1 ), then from (2.2) we
If max{S(x
have
e n , xn+1 , xn+1 ) ≤ β(M (xn−1 , xn , xn ))S(x
e n , xn+1 , xn+1 )
S(x
e n , xn+1 , xn+1 )
< Ω−1 (1)S(x
e n , xn+1 , xn+1 ),
≤ S(x
which is a contradiction.
e n−1 , xn , xn ), S(x
e n , xn+1 , xn+1 )} = S(x
e n−1 , xn , xn ). So, from (2.2),
Hence, max{S(x
e n , xn+1 , xn+1 ) ≤ β(M (xn−1 , xn , xn ))S(x
e n−1 , xn , xn ) < S(x
e n−1 , xn , xn ).
S(x
(2.3)
e n , xn+1 , xn+1 )} is a decreasing sequence. Then there exists r ≥ 0 such
That is, {S(x
e
that lim S(xn , xn+1 , xn+1 ) = r. We will prove that r = 0. Suppose on the contrary,
n→∞
that r > 0. Then, letting n → ∞, from (2.3) we have
r ≤ lim β(M (xn−1 , xn , xn ))r ≤ Ω−1 (1)r,
n→∞
which implies that Ω−1 (1) ≤ 1 ≤ lim β(M (xn−1 , xn , xn )) ≤ Ω−1 (1). Now, as β ∈ BΩ
n→∞
we conclude that M (xn−1 , xn , xn ) → 0 which yields that r = 0, a contradiction. Hence,
the assumption that r > 0 is false. That is,
e n , xn+1 , xn+1 ) = lim S(x
e n , xn , xn+1 ) = 0.
lim S(x
n→∞
n→∞
(2.4)
e
Now, we prove that the sequence {xn } is an S-Cauchy
sequence. Suppose the contrary, i.e., there exists ε > 0 for which we can find two subsequences {xmi } and {xni }
of {xn } such that ni is the smallest index for which
e m , xn , xn ) ≥ ε.
ni > mi > i and S(x
i
i
i
(2.5)
e m , xn −1 , xn −1 ) < ε.
S(x
i
i
i
(2.6)
This means that
From the rectangular inequality, we get
e m , xm +1 , xn )
S(x
i
i
i
e m , xm , xn −1 ) + S(x
e m +1 , xm +1 , xn −1 ) + S(x
e n , xn , xn −1 )]
≤ Ω[S(x
i
i
i
i
i
i
i
i
i
e m , xm , xn −1 ) + Ω[2S(x
e m +1 , xm +1 , xm ) + S(x
e n −1 , xn −1 , xm )]
≤ Ω[S(x
i
i
i
i
i
i
i
i
i
e n , xn , xn −1 )].
+ S(x
i
i
i
Taking the upper limit as i → ∞ and by (2.4), we get
e m , xm +1 , xn ) ≤ Ω(ε + Ω(ε)).
lim sup S(x
i
i
i
i→∞
From the rectangular inequality, we get
e m , xn , xn ) ≤ Ω[S(x
e m , xm , xm +1 ) + S(x
e n , xn , xm +1 ) + S(x
e n , xn , xm +1 )].
ε ≤ S(x
i
i
i
i
i
i
i
i
i
i
i
i
Taking the upper limit as i → ∞ and by (2.4), we get
1 −1
e m +1 , xn , xn ).
Ω (ε) ≤ lim sup S(x
i
i
i
2
i→∞
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
7
From the definition of M (x, y, z) and the above limits,
lim sup M (xmi , xni −1 , xni −1 )
i→∞
n
e m , xn −1 , xn −1 ), 1 Ω−1 [S(x
e m , f xm , f xn −1 )],
= lim sup max S(x
i
i
i
i
i
i
2
i→∞
o
e n −1 , f xn −1 , f xn −1 )
S(x
i
i
i
n
e m , xn −1 , xn −1 ), 1 Ω−1 [S(x
e m , xm +1 , xn )],
= lim sup max S(x
i
i
i
i
i
i
2
i→∞
e n −1 , xn , xn )}
S(x
i
i
i
≤ Ω(ε).
Now, since the sequence {xn } is increasing, we can apply (2.1) and the above inequalities
to get
1
−1
2
Ω(ε) = Ω 2 · Ω (ε)
2
2
e m +1 , xn , xn )]
≤ Ω [lim sup 2S(x
i
i
i
i→∞
≤ lim sup β(M (xmi , xni −1 , xni −1 )) lim sup M (xmi , xni −1 , xni −1 )
i→∞
i→∞
≤ Ω(ε) · lim sup β(M (xmi , xni −1 , xni −1 )),
which implies that
i→∞
−1
Ω (1) ≤
1 ≤ lim β(M (xmi , xni −1 , xni −1 )) ≤ Ω−1 (1). Now, as β ∈
n→∞
e m , xn −1 , xn −1 ) →
BΩ we conclude that M (xmi , xni −1 , xni −1 ) → 0 which yields that S(x
i
i
i
0. Consequently,
e m , xn , xn ) ≤ Ω[S(x
e m , xm , xn −1 ) + 2S(x
e n , xn , xn −1 )] → 0,
S(x
i
i
i
i
i
i
i
i
i
e
e
a contradiction with (2.5). Therefore, {xn } is an S-Cauchy
sequence. S-completeness
e
of X yields that {xn } S-converges
to a point u ∈ X.
We will prove that u is a fixed point of f . First, let f be continuous. Then we have
u = lim xn+1 = lim f xn = f u.
n→∞
n→∞
Now, let (II) hold. Using the assumption on X we have that xn u for all n ∈ N.
Now, by Lemma 1.12,
h Ω−1 [S(u,
e f u, f u)] i
e n+1 , f u, f u)]
≤ Ω2 [2 lim sup S(x
Ω2 2
2
n→∞
≤ lim sup β(M (xn , u, u)) lim sup M (xn , u, u),
n→∞
n→∞
where
e n , u, u), S(x
e n , xn+1 , f u), S(u,
e f u, f u)} = S(u,
e f u, f u).
lim M (xn , u, u) = lim max{S(x
n→∞
n→∞
e f u, f u) = 0, so u = f u.
Therefore, we deduce that S(u,
Finally, suppose that the set of fixed point of f is well ordered. Assume on the
contrary, that u and v are two fixed points of f such that u 6= v. Then by (2.1), we
have
e v, v) = S(f
e u, f v, f v) ≤ β(M (u, v, v))M (u, v, v) = β(S(u,
e v, v))S(u,
e v, v)
S(u,
e v, v),
< Ω−1 (1)S(u,
8
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
e v, v). So, we get S(u,
e v, v) < Ω−1 (1)S(u,
e v, v), a contradicbecause M (u, v, v) = S(u,
tion. Hence, u = v, and f has a unique fixed point. Conversely, if f has a unique fixed
point, then the set of fixed points of f is trivially well ordered.
2.2. Fixed point results via comparison functions. Let Ψ be the family of all
nondecreasing functions ψ : [0, ∞) → [0, ∞) such that
lim ψ n (t) = 0
n→∞
for all t > 0. It is easy to see that the following holds.
Lemma 2.4. If ψ ∈ Ψ, then the following are satisfied.
(a) ψ(t) < t for all t > 0;
(b) ψ(0) = 0.
e be an ordered S-metric
e
Definition 2.5. Let (X, , S)
space. A mapping f : X → X
e
is called an S-ψ-contraction
if there exists ψ ∈ Ψ such that
e x, f y, f z)) ≤ ψ(M (x, y, z))
Ω2 (2S(f
(2.7)
for all mutually comparable elements x, y, z ∈ X, where
n
o
e y, z), 1 Ω−1 [S(x,
e f x, f y)], S(y,
e f y, f z) .
M (x, y, z) = max S(x,
2
e be an ordered S-complete
e
e
Theorem 2.6. Let (X, , S)
S-metric
space. Let f : X → X
be an increasing mapping with respect to such that there exists an element x0 ∈ X
e
with x0 f x0 . Suppose that f is an S-ψ-contractive
mapping. If
(I) f is continuous, or
e has the s.l.c. property,
(II) (X, , S)
then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and
only if f has one and only one fixed point.
Proof. Put xn = f n x0 .
e n , xn+1 , xn+1 ) = 0. We assume that xn 6= xn+1 ,
Step I: We will show that lim S(x
n→∞
for all n ∈ N (otherwise there is nothing to prove). As in the proof of Theorem 2.3, we
have that the sequence {xn } is increasing. Hence, by (2.7) we have
e n , xn+1 , xn+1 ) = S(f
e xn−1 , f xn , f xn ) ≤ Ω(2S(f
e (xn−1 , f xn , f xn ))
S(x
e n−1 , xn , xn ))
≤ ψ(M (xn−1 , xn , xn )) ≤ ψ(S(x
e n−1 , xn , xn ),
< S(x
(2.8)
because
e n−1 , xn , xn ), S(x
e n , xn+1 , xn+1 )},
M (xn−1 , xn , xn+1 ) ≤ max{S(x
e n−1 , xn , xn ), S(x
e n , xn+1 , xn+1 )} = S(x
e n−1 , xn , xn ).
and it is easy to see that max{S(x
e n , xn+1 , xn+1 )} is decreasing. Then there exists
So, from (2.8) we conclude that {S(x
e
r ≥ 0 such that lim S(xn , xn+1 , xn+2 ) = r. It is easy to verify that
n→∞
e n−1 , xn , xn ) = 0.
r = limn→∞ S(x
e
Step 2. Similarly as in the proof of Theorem 2.3, if {xn } were not an S-Cauchy
sequence, then there would exist ε > 0 for which we can find two subsequences {xmi }
and {xni } of {xn } such that (2.5) and (2.6) hold. Then we would have
lim sup M (xmi , xni −1 , xni −1 ) ≤ Ω(ε).
i→∞
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
9
Now, from (2.7) and the mentioned inequalities, we have
1
Ω(ε) = Ω2 2 · Ω−1 ε
2
2
e m +1 , xn , xn ))]
≤ Ω [lim sup(2S(x
i
i
i
i→∞
≤ lim sup ψ(M (xmi , xni −1 , xni −1 )) < Ω(ε)
i→∞
e
e
which is a contradiction. Hence, {xn } is a S-Cauchy
sequence and S-completeness
of
e
X yields that {xn } S-converges
to a point u ∈ X.
Step 3. u is a fixed point of f . This step is proved as in the proof of Theorem 2.3
with some elementary changes.
e y, z) = sinh(Sb (x, y, z)) then
If in the above theorem we take ψ(t) = sinh t and S(x,
we have the following corollary in the framework of Sb metric spaces.
Corollary 2.7. Let (X, , Sb ) be an ordered Sb -complete Sb -metric space with coefficient s > 1. Let f : X → X be an increasing mapping with respect to such that there
exists an element x0 ∈ X with x0 f x0 . Suppose that
sinh(s · sinh(s · 2 sinh(Sb (f x, f y, f z)))) ≤ sinh(M (x, y, z))
for all mutually comparable elements x, y, z ∈ X, where
Sb (x, f x, f y)
M (x, y, z) = max sinh(Sb (x, y, z)),
, sinh(Sb (y, f y, f z)) .
2s
If
(I) f is continuous, or
(II) (X, , Sb ) enjoys the s.l.c. property,
then f has a fixed point.
2.3. Fixed point results related to JS-contractions. Jleli et al. [8] introduced
the class Θ0 consisting of all functions θ : (0, ∞) → (1, ∞) satisfying the following
conditions:
(θ1 ) θ is non-decreasing;
(θ2 ) for each sequence {tn } ⊆ (0, ∞), lim θ(tn ) = 1 if and only if lim tn = 0;
n→∞
n→∞
(θ3 ) there exist r ∈ (0, 1) and ` ∈ (0, ∞] such that lim
t→0+
θ(t)−1
tr
= `;
(θ4 ) θ is continuous.
They proved the following result.
Theorem 2.8. [8, Corollary 2.1] Let (X, d) be a complete metric space and let T :
X → X be a given mapping. Suppose that there exist θ ∈ Θ0 and k ∈ (0, 1) such that
k
x, y ∈ X, d(T x, T y) 6= 0 =⇒ θ d(T x, T y) ≤ θ d(x, y) .
Then T has a unique fixed point.
From now on, we denote by Θ the set of all functions θ : [0, ∞) → [1, ∞) satisfying
the following conditions:
θ1 . θ is a continuous strictly increasing function;
θ2 . for each sequence {tn } ⊆ (0, ∞), lim θ(tn ) = 1 if and only if lim tn = 0;
n→∞
n→∞
10
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
Remark 2.9. [6] It is clear that f (t) = et does not belong to Θ0 , but it belongs to Θ.
2+2 ln(1+t)
2 cosh t
Other examples are f (t) = cosh t, f (t) = 1+cosh
t , f (t) = 1 + ln(1 + t), f (t) = 2+ln(1+t) ,
t
f (t) = ete and f (t) =
t
2ete
,
1+etet
for all t > 0.
e be an ordered S-metric
e
Definition 2.10. Let (X, , S)
space. A mapping f : X → X
e
is called an S rational-JS-contraction if
e x, f y, f z)]) ≤ θ(M (x, y, z))k
θ(Ω[2S(f
(2.9)
for all mutually comparable elements x, y, z ∈ X, where θ ∈ Θ, k ∈ [0, 1) and
e y, f y)S(z,
e z, f z)
e
e
S(y,
e y, z), S(x, x, f x)S(y, y, f y) ,
.
M (x, y, z) = max S(x,
e y, y) + S(x,
e z, z) 1 + S(y,
e f z, f z) + S(y,
e x, x)
1 + S(x,
e be an ordered S-complete
e
e
Theorem 2.11. Let (X, , S)
S-metric
space. Let f : X →
X be an increasing mapping with respect to such that there exists an element x0 ∈ X
e
with x0 f x0 . Suppose that f is an S-rational
JS-contractive mapping. If
(I) f is continuous, or
e enjoys the s.l.c. property,
(II) (X, , S)
then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and
only if f has one and only one fixed point.
Proof. Put xn = f n x0 .
e n , xn+1 , xn+1 ) = 0. Without loss of generality,
Step 1. We will show that lim S(x
n→∞
we may assume that xn 6= xn+1 , for all n ∈ N. Since xn−1 xn for each n ∈ N, then
by (2.9) we have
e n , xn+1 , xn+1 )) ≤ θ(Ω[2S(x
e n , xn+1 , xn+1 )]) = θ(S(f
e xn−1 , f xn , f xn ))
θ(S(x
e n−1 , xn , xn ))k ,
≤ θ(M (xn−1 , xn , xn ))k ≤ θ(S(x
(2.10)
because
n
e
e
e n−1 , xn , xn ), S(xn−1 , xn−1 , f xn−1 )S(xn , xn , f xn )
M (xn−1 , xn , xn ) = max S(x
e n−1 , xn , xn ) + S(x
e n−1 , xn , xn )
1 + S(x
o
e n , xn , f xn )S(x
e n , xn , f xn )
S(x
e n , f xn , f xn ) + S(x
e n , xn−1 , xn−1 )
1 + S(x
n
e
e
e n−1 , xn , xn ), S(xn−1 , xn−1 , xn )S(xn , xn , xn+1 )
= max S(x
e n−1 , xn , xn ) + S(x
e n−1 , xn , xn )
1 + S(x
o
e n , xn , xn+1 )S(x
e n , xn , xn+1 )
S(x
e n , xn+1 , xn+1 ) + S(x
e n , xn−1 , xn−1 )
1 + S(x
e n−1 , xn , xn ), S(x
e n , xn , xn+1 )}.
≤ max{S(x
From (2.10) we deduce that
e n , xn+1 , xn+1 ) ≤ θ S(x
e n−1 , xn , xn ) k .
θ S(x
Therefore,
e n , xn+1 , xn+1 ) ≤ θ S(x
e n−1 , xn , xn ) k ≤ · · · ≤ θ(S(x
e 0 , x1 , x1 ))kn .
1 ≤ θ S(x
Taking the limit as n → ∞ in (2.11) we have
e n , xn+1 , xn+1 ) = 1
lim θ S(x
n→∞
(2.11)
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
11
e n , xn+1 , xn+1 ) = 0. Therefore, we have
and since θ ∈ Θ we obtain limn→∞ S(x
e n+1 , xn , xn ) = 0.
lim S(x
n→∞
e
Step 2. Similarly as in the proof of Theorem 2.3, if {xn } were not an S-Cauchy
sequence, then there would exist ε > 0 for which we can find two subsequences {xmi }
and {xni } of {xn } such that (2.5) and (2.6) hold. Hence,
e m , xm , xn −1 ) < ε.
S(x
i
i
i
From the rectangular inequality, we get
e m +1 , xn , xn )].
e m , xn , xn ) ≤ Ω[S(x
e m , xm , xm +1 ) + 2S(x
ε ≤ S(x
i
i
i
i
i
i
i
i
i
By taking the upper limit as i → ∞, we get
1 −1
e m +1 , xn , xn ).
Ω (ε) ≤ lim sup S(x
i
i
i
2
i→∞
From the definition of M (x, y, z) and the above limits,
n
e m , xn −1 , xn −1 ),
lim sup M (xmi , xni −1 , xni −1 ) = lim sup max S(x
i
i
i
i→∞
i→∞
e m , xm , f xm )S(x
e n −1 , xn −1 , f xn −1 )
S(x
i
i
i
i
i
i
e m , xn −1 , xn −1 ) + S(x
e m , xn −1 , xn −1 )
1 + S(x
i
i
i
i
i
i
e n −1 , xn −1 , f xn −1 )S(x
e n −1 , xn −1 , f xn −1 ) o
S(x
i
i
i
i
i
i
e
e
1 + S(xni −1 , f xni −1 , f xni −1 ) + S(xni −1 , xmi , xmi )
≤ ε.
Now, from (2.9) and the above inequalities, we have
h 1
i
e m +1 , xn , xn )])
θ Ω 2 · Ω−1 (ε) ≤ lim sup θ(Ω[2S(x
i
i
i
2
i→∞
≤ lim sup θ(M (xmi , xni −1 , xni −1 ))k
i→∞
k
≤ θ(ε)
e
which implies that ε = 0, a contradiction. So, we conclude that {xn } is an S-Cauchy
e
e
sequence. By S-completeness of X, it follows that {xn } S-converges to a point u ∈ X.
Step 3. u is a fixed point of f .
When f is continuous, the proof is straightforward.
Now, let (II) hold. Using the assumption on X we have xn u. By Lemma 1.12,
h Ω−1 [S(u,
h
i
e u, f u)] i
e n+1 , xn+1 , f u)
θ Ω 2·
≤ lim sup θ Ω 2S(x
2
n→∞
≤ lim sup θ(M (xn , xn , u))k ,
n→∞
where
n
e
e
e n , xn , u), S(xn , xn , f xn )S(xn , xn , f xn ) ,
lim M (xn , xn , u) = lim max S(x
n→∞
n→∞
e n , xn , xn ) + S(x
e n , u, u)
1 + S(x
o
e n , xn , f xn )S(u,
e u, f u)
S(x
= 0.
e n , f u, f u) + S(x
e n , xn , xn )
1 + S(x
e f u, f u) = 0, so, u = f u.
Therefore, we deduce that S(u,
12
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
Finally, suppose that the set of fixed points of f is well ordered. Assume, on contrary,
that u and v are two fixed points of f such that u 6= v . Then by (2.9), we have
e v, v)] = θ[S(f
e u, f v, f v)] ≤ θ(M (u, v, v))k = θ(S(u,
e v, v))k .
θ[S(u,
e v, v) = 0, a contradiction. Hence u = v, and f has a unique fixed
So, we get S(u,
point.
t
te
e y, z) = eSb (x,y,z) − 1 then we
If in the above theorem we take θ(t) = 2e tet and S(x,
1+e
have the following corollary in the framework of Sb metric spaces.
Corollary 2.12. Let (X, , Sb ) be an ordered Sb -complete Sb -metric space with coefficient s > 1. Let f : X → X be an increasing mapping with respect to such that there
exists an element x0 ∈ X with x0 f x0 . Suppose that
s
Sb (f x,f y,f z) −1]
s·[2eSb (f x,f y,f z) −1] −1]ees·[2e
−1
M (x,y,z)
2e[e
2eM (x,y,z)e
≤
S (f x,f y,f z) −1]
M (x,y,z)
s·[2eSb (f x,f y,f z) −1] −1]ees·[2e b
−1
1 + eM (x,y,z)e
1 + e[e
for all mutually comparable elements x, y, z ∈ X, where
n
[eSb (x,x,f x) − 1][eSb (y,y,f y) − 1]
,
M (x, y, z) = max eSb (x,y,z) − 1,
1 + eSb (x,y,y) − 1 + eSb (x,x,f y) − 1
[eSb (y,y,f y) − 1][eSb (z,z,f z) − 1] o
.
1 + eSb (y,f z,f z) − 1 + eSb (y,x,x) − 1
If
(I) f is continuous, or
(II) (X, , Sb ) enjoys the s.l.c. property,
then f has a fixed point.
3. Examples
Example 3.1. Let X = {0, 12 , 31 , 15 } be equipped with the partial order given as:
1 1 1 1 1 1 1 1 1 .
:= (0, 0), 0, , 0, , 0, , , , , , ,
2
3
5
2 2
3 3
5 5
Define a metric d on X by
(
0,
if x = y,
d(x, y) =
x + y, if x 6= y
e y, z) = sinh[d(x, z) + d(y, z)]. It is easy to see that (X, S)
e is an S-complete
e
and let S(x,
e
S-metric space.
Define a self-map f by
0 12 31 51
f=
.
0 15 0 0
e enjoys the s.l.c. property.
We see that f is an increasing mapping w.r.t. and (X, , S)
Also, 0 f 0.
Define θ : [0, ∞) → [1, ∞) by θ(t) = cosh(t) and take k = 23 . One can easily check
that f is an Se rational JS-contractive mapping. Indeed, we have some cases as follows:
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
13
1. (x, y, z) = (0, 0, 21 ). Then,
i
h
i
h
e x, f y, f z))] = cosh sinh 2 sinh 2 f 0 + f 1 = cosh sinh 2 sinh 2 0 + 1
θ[Ω(2S(f
2
5
r
h
i
1 2
3
≈ 1.451 ≤ 1.465 ≈ cosh sinh 2 0 +
2
p
3
2
= θ(M (x, y)) .
2. (x, y, z) = (0, 12 , 21 ). Then,
i
h
i
h
e x, f y, f z))] = cosh sinh sinh 2 f 0 + f 1 = cosh sinh sinh 2 0 + 1
θ[Ω(2S(f
2
5
r
h
i2
1
3
≈ 1.087 ≤ 1.091 ≈ cosh sinh 0 +
2
p
3
2
= θ(M (x, y)) .
3. (x, y, z) = (0, 0, 31 ). Then,
i
h
e x, f y, f z))] = cosh sinh sinh 2 f 0 + f 1 = cosh[sinh sinh 2(0 + 0)]
θ[Ω(2S(f
3
r
h
1 i2
3
= 0 ≤ 1.172 ≈ cosh sinh 2 0 +
3
p
3
2
= θ(M (x, y)) .
4. (x, y, z) = (0, 51 , 51 ). Then,
h
i
e x, f y, f z))] = cosh sinh 2 sinh f 0 + f 1 = cosh[sinh sinh 2(0 + 0)]
θ[Ω(2S(f
5
r
h
1 i2
3
= 0 ≤ 1.0135 ≈ cosh sinh 0 +
5
p
3
2
= θ(M (x, y)) .
The rest of the cases can be checked similarly. Thus, all the conditions of Theorem
2.11 are satisfied and hence f has a fixed point. Indeed, 0 is the fixed point of f .
e
Example 3.2. Let X = [0, 1.5] be equipped with the S-metric
e y, z) = e|x−z|+|y−z| − 1
S(x,
for all x, y, z ∈ X, where Ω(t) = et − 1. Define a relation on X by x y iff y ≤ x, a
function f : X → X by
x x
f x = e− 2
8
and a function β by β(t) = 12 < 0.882 ≈ Ω−1 (1). For all mutually comparable elements
x, y, z ∈ X, we have
x
z
y
z
e x, f y, f z) = e|f x−f z|+|f y−f z| − 1 = e| x8 e− 2 − z8 e− 2 |+| y8 e− 2 − z8 e− 2 | − 1
S(f
1
1
1
≤ e 8 |x−z|+ 8 |y−z| − 1 ≤ (e|x−z|+|y−z| − 1)
8
1e
= S(x, y, z),
8
14
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
So,
1 S(x,y,z)
1 S(x,y,z)
e
e
1
1e
−1)
−1)
e
Ω2 [2Ssp(f
x, f y, f z)] ≤ Ω2 ( S(x,
y, z)) = e(e 4
− 1 ≤ e 2 (e 2
−1
4
1e
e y, z).
≤ S(x,
y, z) = β(M (x, y, z))S(x,
2
Therefore, from Theorem 2.3, f has a fixed point.
e
Example 3.3. Let X = [0, 3] be equipped with the S-metric
e y, z) = e|x−z|+|y−z| − 1
S(x,
for all x, y, z ∈ X, where Ω(t) = et − 1. Define a relation on X by x y iff y ≤ x, a
function f : X → X by
f x = ln(x + 12)
1 t
and a function ψ by ψ(t) = 4 (e −1). For all mutually comparable elements x, y, z ∈ X,
by Mean Value Theorem, we have
e x, f y, f z) = e| ln(x+12)−ln(z+12)|+| ln(y+12)−ln(z+12)| − 1
S(f
x + 12 y − z
·
−1
z + 12 z + 12
1
1
≤ (|x − z| · |y − z| − 1) ≤ (e|x−z|+|y−z| − 1)
12
12
1 e
= S(x, y, z).
12
y+12
x+12
= e| ln z+12 |+| ln z+12 | − 1 =
So,
1 S(x,y,z)
e
e y, z)) = e(e 6
e x, f y, f z)] ≤ Ω2 ( 1 S(x,
Ω2 [2S(f
6
≤e
1e
S(x,y,z)
4
−1)
−1
e y, z)).
− 1 = ψ(S(x,
Therefore, from Theorem 2.6, f has a fixed point.
Example 3.4. Let Se : X × X × X → R+ be defined on X = [0, 20] by
e y, z) = e 13 (|x−y|+|y−z|+|z−x|) − 1
S(x,
e is an S-complete
e
e
for all x, y, z ∈ X. Then (X, S)
S-metric
space with Ω(t) = et − 1.
t
te
Define θ ∈ Θ by θ(t) = e . Let X be endowed with the standard order ≤ and let
x
). It is easy to see that f is an ordered
f : X → X be defined by f x = arctan( 32
increasing and continuous self map on X and 0 ≤ f 0. For any mutually comparable
x, y, z ∈ X we have
e x, f y, f z) = e 13 (|f x−f y|+|f y−f z|+|f z−f x|) − 1
S(f
y
y
= e 3 (|arctan 32 −arctan 32 |+|arctan 32 −arctan 32 |+|arctan 32 −arctan 32 |) − 1
y
y
1
x
z
z
x
1 1 (|x−y|+|y−z|+|z−x|)
≤ e 3 (| 32 − 32 |+| 32 − 32 |+| 32 − 32 |) − 1 ≤
e3
−1
32
1 e
= S(x, y, z).
32
1
x
z
z
So,
1
e x, f y, f z)] = e2S(f x,f y,f z) − 1 ≤ e 16 S(x,y,z) − 1
Ω[2S(f
1e
≤ S(x,
y, z).
8
e
e
x
MODIFIED S-METRIC SPACES AND SOME FIXED POINT THEOREMS · · ·
15
Therefore,
e x,f y,f z)]
Ω[2S(f
1
1 S(x,y,z)
e
e x, f y, f z)]) = eΩ[2S(f x,f y,f z)]e
θ(Ω[2S(f
≤ e 8 S(x,y,z)e 8
1
e
√1
e
S(x,y,z)
e y, z))] √2 .
2 = [θ(S(x,
≤ eS(x,y,z)e
e
e
Thus, (2.9) is satisfied with k = √12 . Hence, all the conditions of Theorem 2.11 are
satisfied. We see that 0 is the unique fixed point of f .
4. Application
In this section we present an application of Theorem 2.6. This application is inspired
by [13].
Let X = C(I) be the set of all real continuous functions on I = [0, T ]. Obviously,
this set with the p-metric given by
d(x, y) = sinh(max | x(t) − y(t) |)
t∈I
for all x, y ∈ X, is a p-complete p-metric space with Ω(t) = sinh t (in the sense of [14]).
e y, z) = d(x, z)+d(y, z) is an S-metric
e
e
Then S(x,
on X as sinh t is super-additive. (X, S)
can also be equipped with a tripled order given by
x y iff x(t) ≤ y(t) for all t ∈ I.
e has the s.l.c. property.
Moreover, as in [13] it can be proved that (C(I), 3 , S)
Consider the first-order periodic boundary value problem
(
x0 (t) = f (t, x(t)),
x(0) = x(T ),
(4.1)
where t ∈ I and f : I × R → R is a continuous function. A lower solution for (4.1) is a
function α ∈ C 1 (I) such that
(
α0 (t) ≤ f (t, α(t))
α(0) ≤ α(T ),
where t ∈ I. Assume that there exists λ > 0 such that for all x, y ∈ X and t ∈ I, we
have
λ
(4.2)
|f (t, x(t)) + λx(t) − f (t, y(t)) − λy(t)| ≤ |x(t) − y(t)|.
8
We will show that the existence of a lower solution for (4.1) provides the existence of
a unique solution of (4.1). Problem (4.1) can be rewritten as
(
x0 (t) + λx(t) = f (t, x(t)) + λx(t) ≡ F (t, x(t))
x(0) = x(T ),
It is well known that this problem is equivalent to the integral equation
Z T
x(t) =
G(t, s)F (s, x(s)) ds,
0
where G is the Green’s function given as
( λ(T +s−t)
e
G(t, s) =
eλT −1
eλ(s−t)
eλT −1
,
, 0 ≤ s ≤ t ≤ T,
0 ≤ t ≤ s ≤ T.
16
MUSTAFA, SHAHKOOHI, PARVANEH, KADELBURG, JARADAT
Easy calculation gives
Z
T
G(t, s) ds =
0
1
.
λ
Now define an operator H : C(I) → C(I) as
Z T
G(t, s)F (s, x(s)) ds.
Hx(t) =
(4.3)
0
It is easy to see that the mapping H is increasing w.r.t. . Note that if u ∈ C 1 (I) is a
fixed point of H then u is a solution of (4.1).
Let x, z ∈ X be comaprable. Then we have
d(Hx, Hz) = sinh max Hx(t) − Hz(t)
t∈I
Z T
Z T
= sinh max
G(t, s)F (s, x(s)) ds −
G(t, s)F (s, z(s)) ds
t∈I
Z
≤ sinh max
t∈I
≤ sinh
0
T
0
G(t, s) · F (s, x(s)) − F (s, z(s)) ds
0
1
λ
max |x(s) − z(s)| = sinh
sinh−1 d(x, z)
λ s∈I 8
8
1
1
≤ d(x, z).
8
Therefore,
1
1e
e
S(Hx,
Hy, Hz) = d(Hx, Hz) + d(Hy, Hz) ≤ (d(x, z) + d(y, z)) = S(x,
y, z).
8
8
So,
1e
1e
e
Ω2 (2S(Hx,
Hy, Hz)) ≤ Ω2 ( S(x,
y, z)) = sinh(sinh( S(x,
y, z)))
4
4
e y, z)) ≤ ψ(M (x, y, z)),
= ψ(S(x,
where
n
o
−1 e
e y, z), sinh (S(x, f x, f y)) , S(y,
e f y, f z) ,
M (x, y) = max S(x,
2
and ψ(t) = sinh(sinh(t/4)). Finally, if there exists a lower solution α of 4.1, the
hypotheses of Theorem 2.6 are satisfied. Therefore, there exists a fixed point x̂ ∈ C(I)
of H, which is a solution of the given boundary value problem.
Conflicts of interest: The authors declare no conflict of interest.
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1
Department of Mathematics, Statistics and Physics, Qatar University, Doha, Qatar.
E-mail address: zead@qu.edu.qa
2
Department of Mathematics, Aliabad Katool Branch, Islamic Azad University, Aliabad katool, Iran.
E-mail address: rog.jalal@gmail.com
3
Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, GilanE-Gharb, Iran
E-mail address: vahid.parvaneh@kiau.ac.ir
4
Faculty of Mathematics, University of Belgrade, Studentski trg 16, 11000 Beograd,
Serbia
E-mail address: kadelbur@matf.bg.ac.rs
5
Department of Mathematics, Statistics and Physics, Qatar University, Doha, Qatar.
E-mail address: mmjst4@qu.edu.qa
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