R E S E A R C H
Open Access
Fixed points of
(
ψ
,
φ
,
θ
)-contractive
mappings in partially ordered
b
-metric
spaces and application to quadratic integral
equations
Asadollah Aghajani
*and Reza Arab
*Correspondence: [email protected]
Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran
Abstract
We prove some coupled coincidence and coupled common fixed point theorems for mappings satisfying (ψ,φ,θ)-contractive conditions in partially ordered complete
b-metric spaces. The obtained results extend and improve many existing results from the literature. As an application, we prove the existence of a unique solution to a class of nonlinear quadratic integral equations.
MSC: Primary 47H10; secondary 54H25
Keywords: coupled common fixed point; coupled fixed point; coupled coincidence
point; mixedg-monotone property;b-metric space; partially ordered set
1 Introduction and preliminaries
In [, ], Czerwik introduced the notion of ab-metric space, which is a generalization of the usual metric space, and generalized the Banach contraction principle in the context of completeb-metric spaces. After that, many authors have carried out further studies onb-metric spaces and their topological properties (see,e.g., [–]). In this paper, some coupled coincidence and coupled common fixed point theorems for mappings satisfying (ψ,φ,θ)-contractive conditions in partially ordered completeb-metric spaces are proved. Also, we apply our results to study the existence of a unique solution to a large class of nonlinear quadratic integral equations. There are many papers in the literature concern-ing coupled fixed points introduced by Bhaskar and Lakshmikantham [] and their ap-plications in the existence and uniqueness of solutions for boundary value problems. A number of articles on this topic have been dedicated to the improvement and general-ization; see [–] and references therein. Also, to see some results on common fixed points for generalized contraction mappings, we refer the reader to [–]. For the sake of convenience, some definitions and notations are recalled from [, , ] and [].
Definition .[] LetXbe a (nonempty) set ands≥ be a given real number. A function
d:X×X−→R+is said to be ab-metric space iff for allx,y,z∈X, the following conditions
are satisfied:
(i) d(x,y) = iffx=y,
(ii) d(x,y) =d(y,x),
(iii) d(x,y)≤s[d(x,z) +d(z,y)].
The pair (X,d) is called ab-metric space with the parameters.
It should be noted that the class ofb-metric spaces is effectively larger than that of metric spaces since ab-metric is a metric whens= .
The following example shows that, in general, ab-metric need not necessarily be a met-ric (see also []).
Example .[] Let (X,d) be a metric space andρ(x,y) = (d(x,y))p, wherep> is a real number. Thenρis ab-metric withs= p–. However, if (X,d) is a metric space, then (X,ρ)
is not necessarily a metric space. For example, if X=Ris the set of real numbers and
d(x,y) =|x–y|is the usual Euclidean metric, thenρ(x,y) = (x–y)sis ab-metric onRwith
s= , but is not a metric onR.
Also, the following example of ab-metric space is given in [].
Example .[] LetXbe the set of Lebesgue measurable functions on [, ] such that
|f(x)|
dx<∞. DefineD:X×X−→[,∞) byD(f,g) =
|f(x) –g(x)|
dx. As ( |f(x) –
g(x)|dx) is a metric onX, then, from the previous example,Dis ab-metric onX, with
s= .
Khamsi [] also showed that each cone metric space over a normal cone has ab-metric structure.
Since, in general, a b-metric is not continuous, we need the following simple lemma about theb-convergent sequences in the proof of our main result.
Lemma .[] Let(X,d)be a b-metric space with s≥,and suppose that{xn}and{yn}
are b-convergent to x,y,respectively.Then we have
sd(x,y)≤lim infd(xn,yn)≤lim supd(xn,yn)≤s d(x,y).
In particular,if x=y,then we havelimd(xn,yn) = .Moreover,for each z∈X,we have
sd(x,z)≤lim infd(xn,z)≤lim supd(xn,z)≤sd(x,z).
In [], Lakshmikantham and Ćirić introduced the concept of mixedg-monotone prop-erty as follows.
Definition . [] Let (X,≤) be a partially ordered set and F : X× X −→X and
g :X−→X. We say F has the mixed g-monotone property if F is non-decreasing g -monotone in its first argument and is non-increasingg-monotone in its second argument, that is, for anyx,y∈X,
and
y,y∈X, gy≤gy ⇒ F(x,y)≥F(x,y).
Note that ifgis an identity mapping, thenFis said to have the mixed monotone property (see also []).
Definition .[] An element (x,y)∈X×Xis called a coupled coincidence point of a mappingF:X×X−→Xand a mappingg:X−→Xif
F(x,y) =gx, F(y,x) =gy.
Similarly, note that ifgis an identity mapping, then (x,y) is called a coupled fixed point of the mappingF(see also []).
Definition .[] An elementx∈Xis called a common fixed point of a mappingF:
X×X−→Xandg:X−→Xif
F(x,x) =gx=x. (.)
Definition .[] LetXbe a nonempty set andF:X×X−→Xandg:X−→X. One says thatFandgare commutative if for allx,y∈X,
F(gx,gy) =gF(x,y).
Definition .[] The mappingsFandg, whereF:X×X−→Xandg:X−→X, are said to be compatible if
lim
n−→∞d
gF(xn,yn)
,F(gxn,gyn)
=
and
lim
n−→∞d
gF(yn,xn)
,F(gyn,gxn)
= ,
whenever{xn}and{yn}are sequences inXsuch thatlimn−→∞F(xn,yn) =limn−→∞gxn=x andlimn−→∞F(yn,xn) =limn−→∞gyn=yfor allx,y∈X.
2 Main results
Throughout the paper, let be a family of all functionsψ: [,∞)−→[,∞) satisfying the following conditions:
(a) ψ is continuous, (b) ψ non-decreasing, (c) ψ(t) = if and only ift= .
(a) φis lower semi-continuous, (b) φ(t) = if and only ift= ,
andthe set of all continuous functionsθ: [,∞)−→[,∞) withθ(t) = if and only if
t= .
Let (X,d,≤) be a partially ordered b-metric space, and let T :X × X −→X and
g:X−→Xbe two mappings. Set
Ms,T,g(x,y,u,v) =max
d(gx,gu),d(gy,gv),dgx,T(x,y),
sd
gu,T(u,v),dgy,T(y,x), sd
gv,T(v,u),
d(gx,T(u,v)) +d(gu,T(x,y))
s ,
d(gy,T(v,u)) +d(gv,T(y,x)) s
and
NT,g(x,y,u,v) =min
dgx,T(x,y),dgu,T(u,v),dgu,T(x,y),dgx,T(u,v). Now, we introduce the following definition.
Definition . Let (X,d,≤) be a partially orderedb-metric space andψ∈,φ∈and
θ∈. We say thatT:X×X−→Xis an almost generalized (ψ,φ,θ)-contractive mapping with respect tog:X−→Xif there existsL≥ such that
ψsdT(x,y),T(u,v)≤ψMs,T,g(x,y,u,v)
–φMs,T,g(x,y,u,v)
+LθNT,g(x,y,u,v)
(.)
for allx,y,u,v∈Xwithgx≤guandgy≥gv.
Now, we establish some results for the existence of a coupled coincidence point and a coupled common fixed point of mappings satisfying almost generalized (ψ,φ,θ )-contractive condition in the setup of partially orderedb-metric spaces. The first result in this paper is the following coupled coincidence theorem.
Theorem . Suppose that(X,d,≤)is a partially ordered complete b-metric space.Let T :X×X−→X be an almost generalized(ψ,φ,θ)-contractive mapping with respect to g:X−→X,and T and g are continuous such that T has the mixed g-monotone property and commutes with g.Also,suppose T(X×X)⊆g(X).If there exists(x,y)∈X×X such
that gx≤T(x,y)and gy≥T(y,x),then T and g have coupled coincidence point in X.
Proof By the given assumptions, there exists (x,y)∈X×X such thatgx≤T(x,y)
andgy≥T(y,x). SinceT(X×X)⊆g(X), we can define (x,y)∈X×Xsuch thatgx=
T(x,y) andgy=T(y,x), thengx≤T(x,y) =gx andgy≥T(y,x) =gy. Also,
there exists (x,y)∈X×Xsuch thatgx=T(x,y) andgy=T(y,x). SinceT has the
mixedg-monotone property, we have
and
gy=T(y,x)≤T(y,x)≤T(y,x) =gy.
Continuing in this way, we construct two sequences{xn}and{yn}inXsuch that
gxn+=T(xn,yn) and gyn+=T(yn,xn) for alln= , , , . . . (.) for which
gx≤gx≤gx≤ · · · ≤gxn≤gxn+≤ · · ·,
gy≥gy≥gy≥ · · · ≥gyn≥gyn+≥ · · ·.
(.)
From (.) and (.) and inequality (.) with (x,y) = (xn,yn) and (u,v) = (xn+,yn+), we
obtain
ψd(gxn+,gxn+)
≤ψsd(gxn+,gxn+)
=ψsdT(xn,yn),T(xn+,yn+)
≤ψMs,T,g(xn,yn,xn+,yn+)
–φMs,T,g(xn,yn,xn+,yn+)
+LθNT,g(xn,yn,xn+,yn+)
, (.)
where
Ms,T,g(xn,yn,xn+,yn+) =max
d(gxn,gxn+),d(gyn,gyn+),d
gxn,T(xn,yn)
,
sd
gxn+,T(xn+,yn+)
,dgyn,T(yn,xn)
,
sd
gyn+,T(yn+,xn+)
,
d(gxn,T(xn+,yn+)) +d(gxn+,T(xn,yn))
s ,
d(gyn,T(yn+,xn+)) +d(gyn+,T(yn,xn)) s
=max
d(gxn,gxn+),d(gyn,gyn+),
sd(gxn+,gxn+),
sd(gyn+,gyn+), d(gxn,gxn+)
s ,
d(gyn,gyn+)
s
and
NT,g(xn,yn,xn+,yn+) =min
dgxn,T(xn,yn)
,dgxn+,T(xn+,yn+)
,
dgxn+,T(xn,yn)
,dgxn+,T(xn+,yn+)
= .
Since
d(gxn,gxn+)
s ≤
d(gxn,gxn+) +d(gxn+,gxn+)
≤max
d(gxn,gxn+),d(gxn+,gxn+)
and
d(gyn,gyn+)
s ≤
d(gyn,gyn+) +d(gyn+,gyn+)
≤max
d(gyn,gyn+),d(gyn+,gyn+)
,
then we get
Ms,T,g(xn,yn,xn+,yn+)≤max
d(gxn,gxn+),d(gxn+,gxn+),
d(gyn,gyn+),d(gyn+,gyn+)
,
NT,g(xn,yn,xn+,yn+) = .
(.)
By (.) and (.), we have
ψd(gxn+,gxn+)
≤ψmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
–φmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
. (.)
Similarly, we can show that
ψd(gyn+,gyn+)
≤ψmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
–φmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
. (.)
Now, denote
δn=max
d(gxn,gxn+),d(gyn,gyn+)
. (.)
Combining (.), (.) and the fact thatmax{ψ(a),ψ(b)}=ψ(max{a,b}) fora,b∈[, +∞), we have
ψ(δn+) =max
ψd(gxn+,gxn+)
,ψd(gyn+,gyn+)
. (.)
So, using (.), (.), (.) together with (.), we obtain
ψ(δn+)
≤ψmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
–φmaxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
. (.)
Now we prove that for alln∈N,
maxd(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)
=δn and δn+≤δn. (.)
Case . Ifmax{d(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)}=δn, then by
(.) we have
ψ(δn+)≤ψ(δn) –φ(δn) <ψ(δn), (.)
so (.) obviously holds.
Case . Ifmax{d(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)}=
d(gxn+,gxn+) > , then by (.) we have
ψd(gxn+,gxn+)
≤ψd(gxn+,gxn+)
–φd(gxn+,gxn+)
<ψd(gxn+,gxn+)
,
which is a contradiction.
Case . Ifmax{d(gxn,gxn+),d(gxn+,gxn+),d(gyn,gyn+),d(gyn+,gyn+)}=
d(gyn+,gyn+) > , then from (.) we have
ψd(gyn+,gyn+)
≤ψd(gyn+,gyn+)
–φd(gyn+,gyn+)
<ψd(gyn+,gyn+)
,
which is again a contradiction.
Thus, in all the cases, (.) holds for eachn∈N. It follows that the sequence{δn}is a monotone decreasing sequence of nonnegative real numbers and, consequently, there existsδ≥ such that
lim
n−→∞δn=δ. (.)
We show thatδ= . Suppose, on the contrary, thatδ> . Taking the limit asn−→ ∞in (.) and using the properties of the functionφ, we get
ψ(δ)≤ψ(δ) –φ(δ) <ψ(δ),
which is a contradiction. Thereforeδ= , that is,
lim
n−→∞δn=n−→∞lim max
d(gxn,gxn+),d(gyn,gyn+)
= ,
which implies that
lim
n−→∞d(gxn,gxn+) = and nlim−→∞d(gyn,gyn+) = . (.) Now, we claim that
lim
n,m−→∞max
d(gxn,gxm),d(gyn,gym)
= . (.)
Assume, on the contrary, that there exist> and subsequences{gxm(k)},{gxn(k)}of{gxn} and{gym(k)},{gyn(k)}of{gyn}withm(k) >n(k)≥ksuch that
maxd(gxn(k),gxm(k)),d(gyn(k),gym(k))
Additionally, corresponding ton(k), we may choosem(k) such that it is the smallest integer satisfying (.) andm(k) >n(k)≥k. Thus,
maxd(gxn(k),gxm(k)–),d(gyn(k),gym(k)–)
<. (.)
Using the triangle inequality in ab-metric space and (.) and (.), we obtain that
≤d(gxm(k),gxn(k))≤sd(gxm(k),gxm(k)–) +sd(gxm(k)–,gxn(k))
<sd(gxm(k),gxm(k)–) +s.
Taking the upper limit ask−→ ∞and using (.), we obtain
≤lim sup
k−→∞
d(gxn(k),gxm(k))≤s. (.)
Similarly, we obtain
≤lim sup
k−→∞
d(gyn(k),gym(k))≤s. (.)
Also,
≤d(gxn(k),gxm(k))≤sd(gxn(k),gxm(k)+) +sd(gxm(k)+,gxm(k))
≤sd(gxn(k),gxm(k)) +sd(gxm(k),gxm(k)+) +sd(gxm(k)+,gxm(k))
≤sd(gxn(k),gxm(k)) +
s+sd(gxm(k),gxm(k)+).
So, from (.) and (.), we have
s ≤lim supk−→∞ d(gxn(k),gxm(k)+)≤s
. (.)
Similarly, we obtain
s ≤lim supk−→∞
d(gyn(k),gym(k)+)≤s. (.)
Also,
≤d(gxm(k),gxn(k))≤sd(gxm(k),gxn(k)+) +sd(gxn(k)+,gxn(k))
≤sd(gxm(k),gxn(k)) +sd(gxn(k),gxn(k)+) +sd(gxn(k)+,gxn(k))
≤sd(gxm(k),gxn(k)) +
s+sd(gxn(k),gxn(k)+).
So, from (.) and (.), we have
s ≤lim supk−→∞ d(gxm(k),gxn(k)+)≤s
. (.)
In a similar way, we obtain
s ≤lim supk−→∞
Also,
d(gxn(k)+,gxm(k))≤sd(gxn(k)+,gxm(k)+) +sd(gxm(k)+,gxm(k)).
So, from (.) and (.), we have
s ≤lim sup
k−→∞ d(gxn(k)+,gxm(k)+). (.)
Similarly, we obtain
s ≤lim sup
k−→∞
d(gyn(k)+,gym(k)+). (.)
Linking (.), (.), (.), (.), (.), (.) together with (.), we get
s =min
,,
s+
s s ,
s+
s s
≤max
lim sup
k−→∞
d(gxn(k),gxm(k)),lim sup
k−→∞
d(gyn(k),gym(k)),
lim supk−→∞d(gxn(k),gxm(k)+) +lim supk−→∞d(gxm(k),gxn(k)+)
s ,
lim supk−→∞d(gyn(k),gym(k)+) +lim supk−→∞d(gym(k),gyn(k)+)
s
≤max
s,s,s
+s
s ,
s+s
s
=s.
So,
s ≤lim sup
k−→∞
Ms,T,g(xn(k),yn(k),xm(k),ym(k))≤s. (.)
Similarly, we have
s ≤lim infk−→∞Ms,T,g(xn(k),yn(k),xm(k),ym(k))≤s (.)
and
lim
k−→∞NT,g(xn(k),yn(k),xm(k),ym(k)) = . (.) Sincem(k) >n(k), from (.) we have
gxn(k)≤gxm(k), gyn(k)≥gym(k).
Thus,
ψsd(gxn(k)+,gxm(k)+)
=ψsdT(xn(k),yn(k)),T(xm(k),ym(k))
≤ψMs,T,g(xn(k),yn(k),xm(k),ym(k))
–φMs,T,g(xn(k),yn(k),xm(k),ym(k))
+LθNT,g(xn(k),yn(k),xm(k),ym(k))
,
ψsd(gyn(k)+,gym(k)+)
=ψsdT(yn(k),xn(k)),T(ym(k),xm(k))
≤ψMs,T,g(xn(k),yn(k),xm(k),ym(k))
–φMs,T,g(xn(k),yn(k),xm(k),ym(k))
+LθNT,g(xn(k),yn(k),xm(k),ym(k))
.
Sinceψis a non-decreasing function, we have
maxψsd(gx
n(k)+,gxm(k)+)
,ψsd(gy
n(k)+,gym(k)+)
=ψsmaxd(gxn(k)+,gxm(k)+),d(gyn(k)+,gym(k)+)
.
Taking the upper limit ask−→ ∞and using (.) and (.), we get
ψ(s)≤ψ smax
lim sup
k−→∞
d(gxn(k)+,gxm(k)+),lim sup
k−→∞
d(gyn(k)+,gym(k)+)
≤ψ lim sup
k−→∞
Ms,T,g(xn(k),yn(k),xm(k),ym(k))
–φ lim inf
k−→∞ Ms,T,g(xn(k),yn(k),xm(k),ym(k))
+Lθ lim sup
k−→∞
NT,g(xn(k),yn(k),xm(k),ym(k))
≤ψ(s) –φ lim inf
k−→∞Ms,T,g(xn(k),yn(k),xm(k),ym(k))
,
which implies that
φ lim inf
k−→∞Ms,T,g(xn(k),yn(k),xm(k),ym(k))
= ,
so
lim inf
k−→∞Ms,T,g(xn(k),yn(k),xm(k),ym(k)) = ,
a contradiction to (.). Therefore, (.) holds and we have
lim
n,m−→∞d(gxn,gxm) = and n,mlim−→∞d(gyn,gym) = .
SinceXis a completeb-metric space, there existx,y∈Xsuch that
lim
n−→∞gxn+=x and nlim−→∞gyn+=y. (.)
From the commutativity ofTandg, we have
g(gxn+) =g
T(xn,yn)
=T(gxn,gyn), g(gyn+) =g
T(yn,xn)
Now, we shall show that
gx=T(x,y) and gy=T(y,x).
Lettingn−→ ∞in (.), from the continuity ofT andg, we get
gx= lim
n−→∞g(gxn+) =n−→∞lim T(gxn,gyn) =T nlim−→∞gxn,nlim−→∞gyn
=T(x,y),
gy= lim
n−→∞g(gyn+) =nlim−→∞T(gyn,gxn) =T nlim−→∞gyn,n−→∞lim gxn
=T(y,x).
This implies that (x,y) is a coupled coincidence point ofT and g. This completes the
proof.
Corollary . Let(X,d,≤)be a partially ordered complete b-metric space, and let T:
X×X−→X be a continuous mapping such that T has the mixed monotone property.
Suppose that there existψ∈,φ∈,θ∈and L≥such that
ψsdT(x,y),T(u,v)≤ψMs(x,y,u,v)
–φMs(x,y,u,v)
+LθN(x,y,u,v),
where
Ms(x,y,u,v) =max
d(x,u),d(y,v),dx,T(x,y),
sd
u,T(u,v),dy,T(y,x), sd
v,T(v,u),
d(x,T(u,v)) +d(u,T(x,y))
s ,
d(y,T(v,u)) +d(v,T(y,x)) s
and
N(x,y,u,v) =mindx,T(x,y),du,T(u,v),du,T(x,y),dx,T(u,v)
for all x,y,u,v∈X with x≤u and y≥v.If there exists(x,y)∈X×X such that x≤
T(x,y)and y≥T(y,x),then T has a coupled fixed point in X.
Proof Takeg=IXand apply Theorem ..
The following result is the immediate consequence of Corollary ..
Corollary . Let (X,d,≤) be a partially ordered complete b-metric space. Let T :
X×X−→X be a continuous mapping such that T has the mixed monotone property.
Suppose that there existsφ∈such that
dT(x,y),T(u,v)≤
sMs(x,y,u,v) –
sφ
Ms(x,y,u,v)
where
Ms(x,y,u,v) =max
d(x,u),d(y,v),dx,T(x,y),
sd
u,T(u,v),dy,T(y,x), sd
v,T(v,u),
d(x,T(u,v)) +d(u,T(x,y))
s ,
d(y,T(v,u)) +d(v,T(y,x)) s
for all x,y,u,v∈X with x≤u and y≥v.If there exists(x,y)∈X×X such that x≤
T(x,y)and y≥T(y,x),then T has a coupled fixed point in X.
3 Uniqueness of a common fixed point
In this section we shall provide some sufficient conditions under whichT andg have a unique common fixed point. Note that if (X,≤) is a partially ordered set, then we endow the productX×Xwith the following partial order relation, for all (x,y), (z,t)∈X×X,
(x,y)≤(z,t) ⇐⇒ x≤z, y≥t.
From Theorem ., it follows that the setC(T,g) of coupled coincidences is nonempty.
Theorem . By adding to the hypotheses of Theorem.,the condition:for every(x,y)
and(z,t)in X×X,there exists(u,v)∈X×X such that(T(u,v),T(v,u))is comparable to
(T(x,y),T(y,x))and to(T(z,t),T(t,z)),then T and g have a unique coupled common fixed point;that is,there exists a unique(x,y)∈X×X such that
x=gx=T(x,y), y=gy=T(y,x).
Proof We know, from Theorem ., that there exists at least a coupled coincidence point. Suppose that (x,y) and (z,t) are coupled coincidence points ofT andg, that is,T(x,y) =
gx,T(y,x) =gy,T(z,t) =gzandT(t,z) =gt. We shall show thatgx=gzandgy=gt. By the assumptions, there exists (u,v)∈X×Xsuch that (T(u,v),T(v,u)) is comparable to (T(x,y),T(y,x)) and to (T(z,t),T(t,z)). Without any restriction of the generality, we can assume that
T(x,y),T(y,x)≤T(u,v),T(v,u) and T(z,t),T(t,z)≤T(u,v),T(v,u). Putu=u,v=vand choose (u,v)∈X×Xsuch that
gu=T(u,v), gv=T(v,u).
Forn≥, continuing this process, we can construct sequences{gun}and{gvn}such that
gun+=T(un,vn), gvn+=T(vn,un) for alln.
Further, setx=x,y=yandz=z,t=tand in the same way define sequences{gxn},
{gyn}and{gzn},{gtn}. Then it is easy to see that
for alln≥. Since (T(x,y),T(y,x)) = (gx,gy) = (gx,gy) is comparable to (T(u,v),T(v,u)) =
(gu,gv) = (gu,gv), then it is easy to show (gx,gy)≤(gu,gv). Recursively, we get that
(gxn,gyn)≤(gun,gvn) for alln. (.) Thus from (.) we have
ψd(gx,gun+)
≤ψsd(gx,gun+)
=ψsdT(x,y),T(un,vn)
≤ψMs,T,g(x,y,un,vn)
–φMs,T,g(x,y,un,vn)
+LθNT,g(x,y,un,vn)
,
where
Ms,T,g(x,y,un,vn) = max
d(gx,gun),d(gy,gvn),d
gx,T(x,y),
sd
gun,T(un,vn)
,dgy,T(y,x),
sd
gvn,T(vn,un)
,d(gx,T(un,vn)) +d(gun,T(x,y))
s ,
d(gy,T(vn,un)) +d(gvn,T(y,x)) s
≤maxd(gx,gun),d(gy,gvn),d(gy,gvn+),d(gx,gun+)
.
It is easy to show that
Ms,T,g(x,y,un,vn)≤max
d(gx,gun),d(gy,gvn)
and
NT,g(x,y,un,vn) = . Hence,
ψd(gx,gun+)
≤ψmaxd(gx,gun),d(gy,gvn)
–φmaxd(gx,gun),d(gy,gvn)
. (.)
Similarly, one can prove that
ψd(gy,gvn+)
≤ψmaxd(gx,gun),d(gy,gvn)
–φmaxd(gx,gun),d(gy,gvn)
. (.)
Combining (.), (.) and the fact thatmax{ψ(a),ψ(b)}=ψ(max{a,b}) fora,b∈[, +∞), we have
ψmaxd(gx,gun+),d(gy,gvn+)
=maxψd(gx,gun+)
,ψd(gy,gvn+)
≤ψmaxd(gx,gun),d(gy,gvn)
φmaxd(gx,gun),d(gy,gvn)
≤ψmaxd(gx,gun),d(gy,gvn)
. (.)
Using the non-decreasing property ofψ, we get that
maxd(gx,gun+),d(gy,gvn+)
≤maxd(gx,gun),d(gy,gvn)
implies thatmax{d(gx,gun),d(gy,gvn)}is a non-increasing sequence. Hence, there exists
r≥ such that
lim
n−→∞max
d(gx,gun),d(gy,gvn)
=r.
Passing the upper limit in (.) asn−→ ∞, we obtain
ψ(r)≤ψ(r) –φ(r),
which implies thatφ(r) = , and thenr= . We deduce that
lim
n−→∞max
d(gx,gun),d(gy,gvn)
= ,
which concludes
lim
n−→∞d(gx,gun) =n−→∞lim d(gy,gvn) = . (.)
Similarly, one can prove that
lim
n−→∞d(gz,gun) =nlim−→∞d(gt,gvn) = . (.) From (.) and (.), we havegx=gzandgy=gt. Sincegx=T(x,y) andgy=T(y,x), by the commutativity ofTandg, we have
g(gx) =gT(x,y)=T(gx,gy), g(gy) =gT(y,x)=T(gy,gx). (.)
Denotegx=aandgy=b. Then from (.) we have
g(a) =T(a,b), g(b) =T(b,a). (.)
Thus, (a,b) is a coupled coincidence point. It follows thatga=gzandgb=gy, that is,
g(a) =a, g(b) =b. (.)
From (.) and (.), we obtain
Therefore, (a,b) is a coupled common fixed point ofT andg. To prove the uniqueness of the point (a,b), assume that (c,d) is another coupled common fixed point ofT andg. Then we have
c=gc=T(c,d), d=gd=T(d,c).
Since (c,d) is a coupled coincidence point ofTandg, we havegc=gx=aandgd=gy=b. Thusc=gc=ga=aandd=gd=gb=b, which is the desired result.
Theorem . In addition to the hypotheses of Theorem.,if gxand gyare comparable,
then T and g have a unique common fixed point,that is,there exists x∈X such that x=
gx=T(x,x).
Proof Following the proof of Theorem .,Tandghave a unique coupled common fixed point (x,y). We only have to show thatx=y. Sincegxandgyare comparable, we may
assume thatgx≤gy. By using the mathematical induction, one can show that
gxn≤gyn for alln≥, (.)
where{gxn}and{gyn}are defined by (.). From (.) and Lemma ., we have
ψsd(x,y)=ψ
s sd(x,y)
≤lim sup
n−→∞
ψsd(gxn+,gyn+)
=lim sup
n−→∞
ψsdT(xn,yn),T(yn,xn)
≤ lim sup
n−→∞ ψ
Ms,T,g(xn,yn,yn,xn)
–lim inf
n−→∞ φ
Ms,T,g(xn,yn,yn,xn)
+lim sup
n−→∞ Lθ
NT,g(xn,yn,yn,xn)
≤ψd(x,y)–lim inf
n−→∞φ
Ms(xn,yn,yn,xn)
<ψd(x,y),
a contradiction. Therefore,x=y, that is,T andghave a common fixed point.
Remark . Since ab-metric is a metric whens= , from the results of Jachymski [], the condition
ψdF(x,y),F(u,v)≤ψmaxd(gx,gu),d(gy,gv)–φmaxd(gx,gu),d(gy,gv) is equivalent to
dF(x,y),F(u,v)≤ϕmaxd(gx,gu),d(gy,gv),
where ψ ∈, φ∈andϕ : [,∞)−→[,∞) is continuous,ϕ(t) <t for allt> and
4 Application to integral equations
Here, in this section, we wish to study the existence of a unique solution to a nonlinear quadratic integral equation, as an application to our coupled fixed point theorem. Con-sider the nonlinear quadratic integral equation
x(t) =h(t) +λ
k(t,s)f
s,x(s)ds
k(t,s)f
s,x(s)ds, t∈I= [, ],λ≥. (.)
Letdenote the class of those functionsγ : [, +∞)−→[, +∞) which satisfy the follow-ing conditions:
(i) γ is non-decreasing and(γ(t))p≤γ(tp)for allp≥.
(ii) There existsφ∈such thatγ(t) =t–φ(t)for allt∈[, +∞). For example,γ(t) =kt, where ≤k< andγ(t) =t+t are in.
We will analyze Eq. (.) under the following assumptions:
(a) fi:I×R−→R(i= , ) are continuous functions,fi(t,x)≥and there exist two func-tionsmi∈L(I)such thatfi(t,x)≤mi(t)(i= , ).
(a) f(t,x)is monotone non-decreasing inxandf(t,y)is monotone non-increasing iny
for allx,y∈Randt∈I.
(a) h:I−→Ris a continuous function.
(a) ki:I×I−→R(i= , ) are continuous int∈Ifor everys∈Iand measurable ins∈I for allt∈Isuch that
ki(t,s)mi(s)ds≤K, i= , ,
andki(t,x)≥.
(a) There exist constants≤Li< (i= , ) andγ ∈such that for allx,y∈Randx≥y, fi(t,x) –fi(t,y)≤Liγ(x–y) (i= , ).
(a) There existα,β∈C(I)such that
α(t)≤h(t) +λ
k(t,s)f
s,α(s)ds
k(t,s)f
s,β(s)ds
≤h(t) +λ
k(t,s)f
s,β(s)ds
k(t,s)f
s,α(s)ds≤β(t).
(a) max{Lp,L
p
}λpKp≤p–.
Consider the spaceX=C(I) of continuous functions defined onI= [, ] with the standard metric given by
ρ(x,y) =sup
t∈I
x(t) –y(t) forx,y∈C(I).
This space can also be equipped with a partial order given by
Now, forp≥, we define
d(x,y) =ρ(x,y)p= sup
t∈I
x(t) –y(t)p=sup
t∈I
x(t) –y(t)p forx,y∈C(I).
It is easy to see that (X,d) is a completeb-metric space withs= p–[].
Also,X×X=C(I)×C(I) is a partially ordered set if we define the following order rela-tion:
(x,y), (u,v)∈X×X, (x,y)≤(u,v) ⇐⇒ x≤u and y≥v.
For anyx,y∈Xand eacht∈I,max{x(t),y(t)}andmin{x(t),y(t)}belong toXand are up-per and lower bounds ofx,y, respectively. Therefore, for every (x,y), (u,v)∈X×X, one can take (max{x,u},min{y,v})∈X×Xwhich is comparable to (x,y) and (u,v). Now, we formulate the main result of this section.
Theorem . Under assumptions(a)-(a),Eq. (.)has a unique solution in C(I).
Proof We consider the operatorT:X×X−→Xdefined by
T(x,y)(t) =h(t) +λ
k(t,s)f
s,x(s)ds
k(t,s)f
s,y(s)ds fort∈I.
By virtue of our assumptions,Tis well defined (this means that ifx,y∈X, thenT(x,y)∈X). Firstly, we prove thatT has the mixed monotone property. In fact, forx≤xandt∈I,
we have
T(x,y)(t) –T(x,y)(t) =h(t) +λ
k(t,s)f
s,x(s)
ds
k(t,s)f
s,y(s)ds
–h(t) –λ
k(t,s)f
s,x(s)
ds
k(t,s)f
s,y(s)ds
=λ
k(t,s)
f
s,x(s)
–f
s,x(s)
ds
k(t,s)f
s,y(s)ds
≤.
Similarly, ify≥y andt∈I, thenT(x,y)(t)≤T(x,y)(t). Therefore,T has the mixed
monotone property. Also, for (x,y)≤(u,v), that is,x≤uandy≥v, we have T(x,y)(t) –T(u,v)(t)
≤λ
k(t,s)f
s,x(s)ds
k(t,s)
f
s,y(s)–f
s,v(s)ds
+λ
k(t,s)f
s,v(s)ds
k(t,s)
f
s,x(s)–f
s,u(s)ds
≤λ
k(t,s)f
s,x(s)ds
k(t,s)f
s,y(s)–f
s,v(s)ds
+λ
k(t,s)f
s,v(s)ds
k(t,s)f
s,x(s)–f
≤λ
k(t,s)m(s)ds
k(t,s)Lγ
y(s) –v(s)ds
+λ
k(t,s)m(s)ds
k(t,s)Lγ
u(s) –x(s)ds.
Since the functionγ is non-decreasing andx≤uandy≥v, we have
γu(s) –x(s)≤γ sup
t∈I
x(s) –u(s)=γρ(x,u)
and
γy(s) –v(s)≤γ sup
t∈I
y(s) –v(s)=γρ(y,v),
hence
T(x,y)(t) –T(u,v)(t)≤λK
k(t,s)Lγ
ρ(y,v)ds+λK
k(t,s)Lγ
ρ(u,x)ds
≤λKmax{L,L}
γρ(u,x)+γρ(y,v). Then we can obtain
dT(x,y),T(u,v)=sup
t∈I
T(x,y)(t) –T(u,v)(t)p
≤λKmax{L,L}
γρ(u,x)+γρ(y,v)p =λpKpmaxLp
,L
p
γρ(u,x)+γρ(y,v)p,
and using the fact that (a+b)p≤p–(ap+bp) fora,b∈(, +∞) andp> , we have
dT(x,y),T(u,v)≤p–λpKpmaxLp,Lpγρ(u,x)p+γρ(y,v)p
≤p–λpKpmaxLp,Lpγd(u,x)+γd(y,v)
≤pλpKpmaxLp,LpγMs(x,y,u,v)
≤pλpKpmaxLp,LpMs(x,y,u,v) –φ
Ms(x,y,u,v)
≤
p–Ms(x,y,u,v) –
p–φ
Ms(x,y,u,v)
.
This proves that the operator T satisfies the contractive condition (.) appearing in Corollary ..
Finally, letα,βbe the functions appearing in assumption (a); then, by (a), we get
α≤T(α,β)≤T(β,α)≤β.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Each author contributed equally in the development of this manuscript. Both authors read and approved the final version of this manuscript.
Received: 16 April 2013 Accepted: 20 August 2013 Published:07 Nov 2013
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