The project name is lime-shadow.
New Fixed Point Results for F-Contractions in
ordered Dislocated b-Metric Spaces with
Applications
Djelloul Chennoufi1, Zahia Mostefaoui2, Said Beloul3*†
1Laboratory of Mathematical Analysis and Applications, Department of
Mathematics, Benyoucef Benkhedda University, Algiers 1, 02 Didouche
Mourad Street, Algiers, Algeria.
2Department of Mathematics, Higher Normal School of Kouba, Algeria.
3Department of Mathematics, El Oued University, Algeria.
*Corresponding author(s). E-mail(s): beloulsaid@gmail.com;
Contributing authors: chennoufi.dj@gmail.com;
zahia.moustefaoui@g.ens-kouba.dz;
†
Abstract
In this paper, we establish several existence and uniqueness results for com-
mon fixed points of two pairs of self-mappings in complete ordered dislocated
b-metric spaces by using the framework of F -contractions. Moreover, we apply
the obtained results to prove the existence of solutions for a boundary value
problem and provide illustrative example to demonstrate the effectiveness and
applicability of our theoretical findings.
MSC Classification: 47H10 , 54H25
1 Introduction
Fixed point theory has become one of the most active and influential areas of non-
linear analysis due to its wide range of applications in differential equations, integral
1
equations, optimization, approximation theory, game theory, and mathematical eco-
nomics. Over the years, many generalizations of metric spaces have been introduced in
order to accommodate situations where the classical metric axioms are too restrictive.
Among these generalizations, b-metric spaces which introduced by Bakhtin [5] and
developed by Czerwik [8] as a natural extension of a metric space, replaces the usual
triangle inequality by a relaxed form involving a constant s ≥1, later, some fixed
point results were obtained in such spaces, for single valued or set valued mappings,
for instance see [9, 10, 12, 19]. On other hand, the concept of a dislocated metric [3]
introduced to weaken the self-distance condition of a metric, allows the possibility
that the self-distance of a point may be nonzero. This feature has proved useful in
computer science, domain theory, and approximation processes, as well as in several
branches of nonlinear analysis. By combining the ideas of a dislocated metric and a
b-metric, one obtains the notion of a dislocated b-metric space [2], which extends
both structures simultaneously. Such spaces constitute an appropriate framework for
investigating fixed point problems under generalized contractive conditions, especially
when the geometry of the underlying space does not permit the use of standard
metric methods. Consequently, the study of fixed point theory in dislocated b-metric
spaces has become an important and growing topic. Another major development in
modern fixed point theory is the introduction of new contractive conditions that go
beyond the classical linear contraction. Among them, the notion of an F-contraction,
introduced by Wardowski [21], has generated substantial interest and has led to many
important extensions and applications and many authors investigate in this way
[4, 11, 13, 15, 18].
Motivated by these developments, the aim of the present paper is to investigate
fixed point results for F-contractions and related contractive mappings in the setting of
dislocated b-metric spaces. We establish new fixed point theorems for F-contractions
and related mappings in ordered dislocated b-metric spaces. We also discuss some
consequences and special cases that follow from our results. In the final section, we
provide an application to the existence of solution for a boundary value problem to
illustrating the effectiveness of the obtained theorems.
2 Preliminaries
In this section, we recall some basic definitions and auxiliary results that will be used
throughout the paper.
3 Main results
lime-shadow is a research project whose aim is to investigate fixed point results for F-contractions and related contractive mappings in the setting of dislocated b-metric spaces. The work establishes new fixed point theorems for F-contractions and related mappings in ordered dislocated b-metric spaces, and it also discusses some applications of these results. The project is motivated by prior work on extensions and applications of this kind, cited in the source as [4, 11, 13, 15, 18].
The evidence identifies no human roles, so no personas are derived here; this batch does not establish a persona catalog. The only named actor is the project itself, lime-shadow, which carries the stated research aim. Supporting actors are the mathematical objects the project works with: F-contractions, related contractive mappings, dislocated b-metric spaces, and ordered dislocated b-metric spaces. The authors cited as [4, 11, 13, 15, 18] are referenced as background, not as system actors.
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Muse and headline. The muse is the fixed point itself: the single value a mapping returns unchanged. The headline is “Every contraction lands somewhere.” The product is lime-shadow, a reading and reference surface for the paper New Fixed Point Results for F-Contractions in ordered Dislocated b-Metric Spaces with Applications (Chennoufi, Mostefaoui, Beloul; MSC 47H10, 54H25).
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Domain and subject matter. The work concerns fixed point theory in ordered dislocated b-metric spaces using F-contractions (Wardowski). It establishes existence and uniqueness results for common fixed points of two pairs of self-mappings in complete ordered dislocated b-metric spaces, derives consequences and special cases, and applies the results to a boundary value problem, supported by an illustrative example. MSC classification: 47H10, 54H25.
Mathematical foundations referenced. b-metric spaces (Bakhtin; developed by Czerwik), replacing the triangle inequality with a relaxed form involving a constant s ≥ 1; dislocated metrics, which weaken the self-distance condition so a point's self-distance may be nonzero; dislocated b-metric spaces, combining both; and F-contractions.
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Theorem 1 Let (X, ≤, σb) be a partially ordered complete dislocated b-metric space, and let
S, T, A, B →X be self-mappings. Assume that the following conditions are satisfied:
• [H1] : SX ⊆BX, TX ⊆AX.
• [H2] : S, T are dominating, A, B are dominated.
• [H3] : (S, T) is an ordered F-contraction with respected to (A, B).
• [H4] : {xn} is nondecreasing , xn ≤yn for all n ∈N and yn →z, then xn ≤z.
• [H5] :(i) S, or A is continuous, (S, A) is compatible and (T, B) is weakly compatible.
(ii) T, or B is continuous, (T, B) is compatible and (S, A) is weakly compatible.
Then S, T, A and B has a common fixed point.
Proof Let x0 ∈X. Since SX ⊂BX there exists x1 ∈X such that y1 = Bx1 = Sx0. Let
y2 = Tx1, since TX ⊂AX, there exists x2 ∈X such that y2 = Ax2 = Tx1. Continuing this
process, we can construct sequence {yn} in X as follows :
y2n+1 = Sx2n = Bx2n+1,
y2n+2 = Tx2n+1 = Ax2n+2,
n ≥0.
(2)
4
If there exists n0 ∈N such that σb(y2n0, y2n0+1) = 0. Then y2n0 = y2n0+1, using (1) with
x = x2n0 and y = x2n0+1, we get
τ + F(sεσb(Sx2n0, Tx2n0+1)) ≤F(M(x2n0, x2n0+1)),
which implies that
τ + F(sεσb(y2n0+1, y2n0+2)) = τ + F(sεσb(Sx2n0, Tx2n0+1)) ≤F(M(x2n0, x2n0+1)),
so we have
M(x2n, x2n+1) = max{σb(Ax2n0, Bx2n0+1), σb(Ax2n0, Sx2n0), σb(Bx2n0+1, Tx2n0+1),
σb(Ax2n0, Tx2n0+1) + σb(Bx2n0+1, Sx2n0)
4s
}
= max{σb(y2n0, y2n0+1), σb(y2n0, y2n0+1), σb(y2n0+1, y2n0+2),
σb(y2n0, y2n0+2) + σb(y2n0+1, y2n0+1)
4s
}
= max{0, 0, σb(y2n0+1, y2n0+2), σb(y2n0, y2n0+2)
4s
}
≤max{σb(y2n0+1, y2n0+2), s[σb(y2n0, y2n0+1) + σb(y2n0+1, y2n0+2)]
4s
}
= σb(y2n0+1, y2n0+2),
since
σb(y2n0, y2n0+2) + σb(y2n0+1, y2n0+1)
4s
≤3sσb(y2n0, y2n0+1) + sσb(y2n0+1, y2n0+2)
4s
≤σb(y2n0+1, y2n0+2)
4s
.
So we get
τ + F(σb(y2n0+1, y2n0+2) ≤τ + F(sεσb(y2n0+1, y2n0+2) ≤F(σb(y2n0+1, y2n0+2),
which implies that y2n0+1 = y2n0+2 and this yields that y2n0 = y2n0+1 = y2n0+2, so for all
n ∈N the sequence (yn) is constant then it is a Cauchy one.
Now, suppose that y2n ̸= y2n+1 for all n ∈N.By(1), we have
τ + F(sεσb(y2n+1, y2n+2)) ≤F(M(x2n, x2n+1)),
(3)
Therefore, since F is strictly increasing, we get
σb(y2n+1, y2n+2) ≤M(x2n, x2n+1),
where
M(x2n, x2n+1) = max
σb(Ax2n, Bx2n+1), σb(Ax2n, Sx2n), σb(Bx2n+1, Tx2n+1),
σb(Ax2n, Tx2n+1) + σb(Bx2n+1, Sx2n)
4s
= max
σb(y2n, y2n+1), σb(y2n, y2n+1), σb(y2n+1, y2n+2),
σb(y2n, y2n+2) + σb(y2n+1, y2n+1)
4s
= max {σb(y2n, y2n+1), σb(y2n+1, y2n+2)} ,
since
σb(y2n, y2n+2) + σb(y2n+1, y2n+1)
4s
≤3sσb(y2n, y2n+1) + sσb(y2n+1, y2n+2)
4s
.
5
Therefore
M(x2n, x2n+1) = max
σb(y2n, y2n+1), σb(y2n+1, y2n+2)
.
If σb(y2n, y2n+1) ≤σb(y2n+1, y2n+2), then (3) gives
τ + F
�sε σb(y2n+1, y2n+2)
≤F
�σb(y2n+1, y2n+2)
,
which is contradiction since τ > 0, then we have
τ + F
�sε σb(y2n+1, y2n+2)
≤F
�σb(y2n, y2n+1)
∀n ≥0.
(4)
Applying the same argument to the comparable pair (y2n+1, y2n+2),
τ + F
�sε σb(y2n+2, y2n+3)
≤F
�σb(y2n+1, y2n+2)
,
∀n ≥0.
(5)
By (4) and (5) , we have
F
�sε σb(yn+1, yn+2)
≤F
�σb(yn, yn+1)
−τ.
∀n ≥0.
(6)
Continuing in this manner and using (F4), for all n ∈N we get
F(snεσb(yn+1, yn+2)) ≤F(σb(y0, y1)) −nτ.
Passing to the limit in the last inequality we get
lim
n→∞F(snεσb(y2n, y2n+1)) = −∞,
from (F2) we get
lim
n→∞snεσb(yn, yn+1) = 0.
By the definition of the limit there exits n0 ∈N such that for all n ≥n0 we have
σb(yn, yn+1 < s−nε,
For n, m ∈N with m > n ≥1, we have
σb(xn, xm) ≤
m−1
X
i=n
si−n+1σb(yi, yi+1) ≤
m−1
X
i=n
siσb(yi, yi+1)
≤
m−1
X
i=n
si(1−ε) ≤
∞
X
i=1
si(1−ε) < ∞.
Then σb(yn, ym) →0 as n, m →∞, which implies that (yn) is a Cauchy sequence.
Then {xn} is a Cauchy sequence in X. Since (X, σb) is complete, there exists z ∈X such that
yn →z
as n →∞.
Since {xn} is nondecreasing and by hypothesis (H4), we have yn ≤z,
∀n ∈N.
Also we have
lim
n→∞Sx2n = lim
n→∞Bx2n+1 = lim
n→∞Tx2n+1 = lim
n→∞Ax2n+2 = z.
(7)
We show that z is a common fixed point of S, T, A, and B. Since A is continuous, therefore,
lim
n→∞ASx2n = Az = lim
n→∞A2x2n+2.
(8)
Since the pair {S, A} is compatible, so,
lim
n→∞σb(SAx2n, ASx2n) = 0,
using Lemma 1, we get
1
s2 σb(Az, lim SAx2n) ≤lim
n→∞σb(SAx2n, ASx2n) = 0 ≤s2σb(Az, lim SAx2n),
6
which implies that
lim
n→∞SAx2n = Az.
(9)
Now put x = Ax2n and y = x2n+1 in (1) and suppose on the contrary that σb(Az, z) > 0;
we obtain
τ + F(sεσb(SAx2n, Tx2n+1)) ≤F(M(Ax2n, x2n+1))
(10)
where
M(Ax2n, x2n+1) = max
σb(A2x2n, Bx2n+1), σb(A2x2n, SAx2n), σb(Bx2n+1, Tx2n+1),
σb(A2x2n, Tx2n+1) + σb(Bx2n+1, SAx2n)
4s
.
Taking n →∞and using (7) and Lemma 1, we get
lim
n→∞M(Ax2n, x2n+1) = max
σb(Az, z), σb(Az, Az), σb(z, z), σb(Az, z) + σb(z, Az)
4s
= σb(Az, z),
Because
0 ≤σb(Az, Az) ≤s
σb(Az, Ax2n+1) + σb(Ax2n+1, Az)
= 2s σb(Ax2n+1, Az) n→∞
−−−−→0.
passing to the limit in (10) and using the continuity of F, we get
τ + F
�sεσb(Az, z)
≤F
�σb(Az, z)
.
(11)
This is a contradiction. Therefore, we get σb(Az, z) = 0, that is, Az = z. Again now put x = z
and y = x2n+1 in (1)and (2) and suppose on the contrary that (σb(Sz, z) > 0 , we obtain
τ + F(sε(σb(Sz, Tx2n+1)) ≤F(M(z, x2n+1))
(12)
where
M(z, x2n+1) = max
σb(Az, Bx2n+1), σb(Az, Sz), σb(Bx2n+1, Tx2n+1),
σb(Az, Tx2n+1) + σb(Bx2n+1, Sz)
4s
.
Letting n →∞on both sides in the above inequality, using Az = z, and (7) we obtain
M(Sz, z) = max{(σb(z, z), (σb(z, Sz), (σb(z, z),
σb(z, z) + σb(z, Sz)
4s
} = (σb(Sz, z),
passing to the limit in (10) and using the continuity of F and Lemma1, we get
τ + F(sε(σb(Sz, z)) ≤F((σb(Sz, z)).
(13)
This is a contradiction. Therefore, we get σb(Sz, z) = 0, that is, Sz = z.
On the other hand, since z = Sz ∈SX ⊂BX, there exists u ∈X such that z = Sz = Bu
. By It will need to prove Bu = Tu. Via z = Sz = Bu ≤u, it implies z ≤u. By utilizing (1),
we obtain
τ + F(sε(σb(z, Tu)) = τ + F(sε(σb(Sz, Tu)) ≤F(M(z, u))
where
M(z, u) = max{(σb(Az, Bu), (σb(Az, Sz), (σb(Bu, Tu), (σb(Az, Tu) + (σb(Bu, Sz)
4s
}.
= max{(σb(z, Bu), (σb(z, z), (σb(Bu, Tu), (σb(z, Tu) + (σb(Bu, z)
4s
} = σb(z, Tu),
Letting n →∞,we have
τ + F(sε(σb(z, Tu)) ≤(σb(z, Tu))
7
a contradiction, hence, (σb(z, Tu) = 0 implies Tu = z, and so Tu = z = Bu. Since the pair
(T, B) is weakly compatible, we have Tz = TBu = BTu = Bz. Now we prove that Tz = z.
From (1) , we have
τ + F(sε(σb(z, Tz)) = τ + F(sε(σb(Sz, Tz)) ≤F(M(z, z))
where
M(z, z) = max{(σb(Az, Bz), (σb(Az, Sz), (σb(Bz, Tz),
(σb(Az, Tz) + (σb(Bz, Sz)
4s
}
= max{(σb(z, z), (σb(z, z), (σb(z, Tz),
(σb(z, Tz) + (σb(z, z)
4s
} = (σb(z, Tz).
This implies that (σb(z, Tz) = 0, so Tz = z. Therefore, we obtain Sz = Az = Tz = Bz = z,
so z is a common fixed point of S, T, A, and B.
We take a similar argument when S is continuous.
Similarly, condition (ii) holds ,we can also fin the common fixed point of of S, T, A and B.
Next, we prove that S, T, A, and B have a unique common fixed point. Suppose that ω ∈X
is also a common fixed point of S, T, A and B, then then from (1), we have
τ + F((σb(z, ω)) = τ + F((σb(Sz, Tω)) ≤F(M(z, ω)),
where
4
,
Tx = ln
1 + x
5
.
(14)
Clearly, S, T, A, and B are continuous self-mappings complying with SX = TX = AX =
BX = [0, ∞). We note that the pair {A, S} is compatible. Indeed, let {xn} be a sequence in
X satisfying limn→∞Axn = limn→∞Sxn = t for some t ∈X. Then
lim
n→∞|Sxn −t| = lim
n→∞|Axn −t| = 0,
(15)
and equivalently
lim
n→∞
ln
1 + xn
4
−t
= lim
n→∞|e5xn −1 −t| = 0,
(16)
implies
lim
n→∞||xn −(4et −4)| = lim
n→∞
xn −ln(t + 1)
5
= 0.
(17)
The uniqueness of limit gives 4et −4 = ln(t + 1)
5
; thus, t = 0 is the only possible solution.
Due to the continuity of A, S we have
lim
n→∞σb(ASxn, SAxn) = lim
n→∞|ASxn + SAxn|2 = |A(0) + S(0)|2 = |0 + 0|2 = 0.
(18)
Similarly, the pair {B, T} is compatible. Now for each x, y ∈X, consider
σb(Sx, Ty) =
ln
1 + x
4
1 + y
5
2
≤
x
4 + y
5
2
=
1
20(5x + 4y)
2
≤
1
202
(e5x −1) + (e4y −1)
2
=
1
202 σb(Ax, By)
≤
1
202 M(x, y).
Hence
σb(Sx, Ty) ≤
1
202 M(x, y),
Hence
202 σb(Sx, Ty) ≤M(x, y),
(ε = 2, s = 2)
Then
102(4 σb(Sx, Ty)) ≤M(x, y),
2 ln 10 + ln
�4σb(Sx, Ty)
≤ln
�M(x, y)
.
(19)
Define the function F : R+ →R by F(t) = ln(t), for all t ∈R+. Hence, for all x, y ∈X, such
that σb(Sx, Ty) > 0, with τ = 2 ln 10 we obtain
τ + F
�sεσb(Sx, Ty)
≤F
�M(x, y)
.
(20)
Thus, the contractive condition is satisfied for all x, y ∈X. Hence, all the hypotheses of the
theorem are satisfied. Note that S, T, A, and B have a unique common fixed point x = 0.
9
4 Application
In this section, we will give some applications in the existence and uniqueness of
solutions to several equations.
First of all, we consider the following equation:
d2x
dt2 = f(t, x(t)),
t ∈[a, b],
x(a) = x(b) = 0,
(21)
Theorem 3 For Equation (21), if the following condition holds:
|f(t, x(t)) + f(t, y(t))| ≤λ|x(t) + y(t)|,
(22)
where λ > 0 is a constant, f(t, 0) = 0, and λ(b −a)2 ≤21−ε
2 with ε > 1, then Equation (21)
has a unique solution in [a, b] provided that 1
8λ(b −a)2 ≤1.
Proof Equation (21) is equivalent to the equation below:
x(t) =
Z b
a
Ω(t, r)f(r, x(r))ds,
(23)
where
Ω(t, r) =
( r−a
a−b (b −t), a ≤r ≤t,
t−a
a−b(b −r), t ≤r ≤b.
(24)
We use reverse thinking to prove (23). That is to say, we utilize (23) to show (21). Indeed,
by using (23) and (24), we have
x(t) =
Z b
a
Ω(t, r)f(r, x(r)) dr
=
Z r
a
Ω(t, r)f(r, x(r)) dr +
Z b
t
Ω(t, r)f(r, x(r)) dr
=
Z r
In order to find the solution of (23), we need to look for the fixed point of the following
mapping:
Sx(t) =
Z b
a
Ω(t, r)f(r, x(r)) dr.
∀t ∈[a, b].
(26)
Since f is continuous, then S is a self-mapping on X. Via (22) it is not hard to verify that
|Sx(t) + Ty(t)|2 =
Z b
a
Ω(t, s)f(s, x(s)) ds +
Z b
a
Ω(t, s)f(s, y(s)) ds
2
≤
Z b
a
|Ω(t, r)||f(r, x(r)) + f(r, y(r))| dr
!2
≤
Z b
a
|Ω(t, s)λ|x(r) + y(r)| dr
!2
≤λ2
Z b
a
|Ω(t, r)| dr
!2
max
r∈[a,b] |x(r) + y(r)|2
= λ2
1
2(t −a)(b −t)
2
σb(Ax, By)
≤1
64λ2(b −a)4σb(Ax, By)
≤1
26
22−ε
σb(Ax, By)
≤
1
24+ε σb(Ax, By)
≤
1
24+ε M(x, y)
Then
42 �2εσb(Sx, Ty)
≤M(x, y)
2 ln 4 + ln
�2εσb(Sx, Ty)
≤ln
�M(x, y)
.
(27)
Define the function F : R+ →R by F(t) = ln(t), for all t ∈R+. Hence, for all x, y ∈X, such
that σb(Sx, Ty) > 0, with τ = 2 ln 4 we obtain
τ + F
�sεσb(Sx, Ty)
≤F
�M(x, y)
.
(28)
11
Where S = T. Since f(t, 0) = 0, then by ((22)), it is easy to see that
|Sx| =
Z b
a
Ω(t, r)f(r, x(r)) dr
≤
Z b
a
|Ω(t, r)||f(r, x(r)) + 0| dr
=
Z b
a
|Ω(t, r)||λ(r, x(r)) + f(r, 0)| dr
≤
Z b
a
|Ω(t, r)|k|x(r) + 0| dr
≤λ
Z b
a
|Ω(t, r)| max
r∈[a,b] |x(r)| dr
= λ
Z b
a
|Ω(t, r)| dr max
r∈[a,b] |x(r)|
= 1
2λ(t −a)(b −t) max
r∈[a,b] |x(r)|
≤1
8λ(b −a)2 max
r∈[a,b] |x(r)|
≤max
r∈[a,b] |x(r)|,
so x ≤Sx, that is, S, T are dominating. Let A = B = I be an identity mapping on X.
Accordingly, A, B are dominated. Consequently, all the conditions of Theorem are satisfied.
Hence, by Theorem 1, f has a unique fixed point in X. Thus, (23) has a unique solution in
[a, b]. In other words, (21) has a unique solution in [a, b].
□
Conclusion
In this paper, we discussed the existence and uniqueness of common fixed point for
two pairs of self mappings under Wordowski contractive condition and suitable oth-
ers in dislocated ordered b-metric spaces. The obtained results are formulated in a
general framework and are designed to cover different types of contractive conditions
associated with F-contractions. In particular, our results improve and generalize a
number of earlier fixed point theorems proved in metric, b-metric, dislocated metric,
and dislocated b-metric spaces.
Acknowledgements.
The authors thank the referees for their careful reading of
the manuscript and insightful comments.
Author contributions.
All authors contributed equally in the writing of this paper.
All authors read and approved the final version of this manuscript.
Declarations
Conflict of interest.
All the authors declare that there is no conflict of interests.
Funding.
All the authors declare that there is no funding.
Keywords: common fixed point, F-contraction, weakly compatible, ordered
dislocated b-metric spaces
Definition 1 [8] Let X be a nonempty set and let s ≥1 be a real number. A function
bd : X × X →[0, ∞) is called a b-metric on X if, for all x, y, z ∈X, the following conditions
are satisfied:
bd(x, y) + bd(y, z)
.
Then the triple (X, bd, s) is called a b-metric space.
2
It is well known that every metric space is a b-metric space, but the converse is
not necessarily true.
Lemma 1 [1] Let (X, bd) be a b-metric space with s ≥1, and suppose that {xn} and {yn}
are b-convergent to x, y respectively. Then the following inequality holds.
1
s2 bd(x, y) ≤lim
n→∞inf bd(xn, yn) ≤lim
n→∞sup bd(xn, yn) ≤s2bd(x, y).
If x = y, then we have limnto∞bd(xn, yn) = 0 and for each z ∈X we have
1
sbd(x, z) ≤lim
n→∞inf bd(xn, z) ≤lim
n→∞sup bd(xn, z) ≤sd(x, z).
Definition 2 [2, 10] Let X be a nonempty set and let (s ≥1 be a constant. A function
σb : X × X →R+ is called a dislocated b-metric on X if, for all x, y, z ∈X, the following
conditions hold:
σb(x, y) + σb(y, z)
.
In this case, the triple (X, σb, s) is called a dislocated b-metric space with coefficient s.
In a dislocated b-metric space (X, σb, s)), if σb(x, y) = 0, then x = y. However, the
converse need not be true, since it may happen that σb(x, x) > 0 for some x ∈X.
Example 1 [2] Let X = R+ and let p ≥1 be a constant. Define σb : X × X →R+ by either
σb(x, y) = (x + y)p, or [ σb(x, y) =
�max x, y
p. Then X, σb, s) is a b-metric-like space with
coefficient s = 2p−1.
Clearly, (X, σb, s) is neither a b-metric space nor a metric-like space.
Definition 3 [1, 17] Let (X, ≤, σb) be a partially ordered dislocated b-metric space.
Definition 4 [10] Let (X, ⪯, σb) be a partially ordered dislocated b-metric space. Two self-
mappings S, T are said to be compatible if
lim
n→∞σb
�TSyn, STyn
= 0, whenever {yn} is a
sequence in X) such that lim
n→∞Tyn = lim
n→∞Syn = t for some t ∈X.
Definition 5 Let (X, ⪯, σb) be a partially ordered dislocated b-metric space. Two self-
mappings S, T are said to be weakly compatible if they commute at their coincidence points,
that is, if for any z ∈X satisfying Tz = Sz, one has STz = TSz.
3
Definition 6 [21] Let F : (0, +∞) →R be a function satisfying the following conditions:
(F1) F is strictly increasing;
(F2) for every sequence {un} ⊂(0, +∞), limn→∞un = 0
⇐⇒
limn→∞F(un) =
−∞;
(F3) There exists k ∈(0, 1) such that limα→0+ αkF(α) = 0.
In [16], the authors have replaced the property (F3) by the following property:
(F ′
Cosentino et al. [7] introduced the following condition:
(F4) : if (un) is a sequence such that τ + F(sun) ≤F(un−1), for all n ∈N, s ≥1 and
for some τ > 0, then τ + F(snun) ≤F(sn−1un−1), for all n ∈N.
We denote by F the class of all functions F : (0, +∞) →R) satisfying conditions
(F1)), (F2), (F ′
Example 2 Let Fi : (0, +∞) →R), (i ∈{1, 2, 3}), be defined by
F1(t) = ln t, F2(t) = t + ln t, F3(t) = −1
√
t.
Then Fi ∈F for each (i ∈1, 2, 3).
Definition 7 Let (X, ⪯, σb) be a partially ordered complete dislocated b-metric space, a pair
(S, T) of self mappings is said to be an ordered F-contractive pair with respected to (A, B)
if and only if for any comparable elements x, y ∈X, there exists τ > 0 and F ∈F such that
τ + F(sεσ(Sx, Ty)) ≤F(M(x, y)),
(1)
where s, ε > 1 and
M(x, y) = max{σb(Ax, By), σb(Ax, Sx), σb(By, Ty), 1
4s(σb(Ax, Ty) + σb(By, Sx))}.
M(z, ω) = max{(σb(Az, Bω), (σb(Az, Sz), (σb(Bω, Tω),
(σb(Az, Tω) + (σb(Bω, Sz)
4s
}
= max{(σb(z, ω), (σb(z, z), (σb(ω, ω),
(σb(z, ω) + (σb(ω, z)
4s
}.
Thus, we have
σb(z, ω) < (σb(z, ω),
which is a contradiction. Hence, z = ω and z is a unique common fixed point of the four
maps S, T, A, and B.
□
If S = T and A = B, we get the following corollary.
Corollary 2 Let (X, ≤, σb) be a partially ordered complete dislocated b-metric space, and
let S and A be self-mappings. Suppose that S is an ordered F-contraction with respect to A.
Assume, moreover, that the following conditions are satisfied:
• [H1] : SX ⊆AX.
• [H2] : S is dominating and A is dominated.
• [H3] : {xn} is nondecreasing, xn ≤yn for all n ∈N and yn →z, then xn ≤z.
• [H4] : S, or A is continuous, (S, A) is compatible.
Then S, T, A and B has a common fixed point.
Example 3 Let X = [0, ∞) with the partial order as
x ⪯y ⇐⇒x ≥y.
8
and define σb : X × X →R+ by σb(x, y) = |x + y|2, so (X, σb, 2) is a complete b-metric like
space. Define the mappings S, T, A, B : X →X, for all x ∈X, by
Ax = e5x −1,
Bx = e4x −1,
Sx = ln
1 + x
a
r −a
a −b (b −t)f(r, x(r)) dr +
Z b
t
t −a
a −b(b −r)f(r, x(r)) dr
= b −t
a −b
Z r
a
(r −a)f(r, x(r)) dr + t −a
a −b
Z b
t
(b −r)f(r, x(r)) dr,
(25)
then
dx
dt = −
1
a −b
Z r
a
(r −a)f(r, x(r)) dr + b −t
a −b(t −a)f(t, x(t))
1
a −b
Z b
t
(b −r)f(r, x(r)) dr −t −a
a −b(b −t)f(t, x(t))
= −
1
a −b
Z t
a
(r −a)f(r, x(r)) dr +
1
a −b
Z b
t
(b −r)f(r, x(r)) dr,
which infers that
d2x
dt2 = −
1
a −b(t −a)f(t, x(t)) −
1
a −b(b −t)f(t, x(t)) = f(t, x(t)).
By (3.5), x(a) = x(b) = 0. As a consequence, (3.1) holds.
10
Let X = C[a, b] with partial order as :
x ⪯y ⇐⇒
max
t∈[a,b] |x(t)| ≥max
t∈[a,b] |y(t)|.
Define a mapping σb : X × X →[0, +∞) by
σb(x, y) = max
t∈[a,b] |x(t) + y(t)|2.
It is evident that (X, σb) is a complete dislocated b-metric space with s = 2.
We introduce the self-mappings S, T : X →X defined by:
Sx(t) = Tx(t) =
Z b
a
x(t) =
Z b
a
Ω(t, r)f(r, x(r)) dr
and let A, B : X →X be the identity mappings Ax = Bx = x.
12
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