# Coupled Fixed-Point Theorems for Contractions in Partially Ordered Metric Spaces and Applications

Authors: M. Eshaghi Gordji; Y. J. Cho; S. Ghods; M. Ghods; M. Hadian Dehkordi
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

Bhaskar and Lakshmikantham (2006) showed the existence of coupled coincidence points of a mappingF from X×X into X and a mapping g from X into X with some applications. The aim of this paper is to extend the results of Bhaskar and Lakshmikantham and improve the recent fixed-point theorems due to Bessem Samet (2010). Indeed, we introduce the definition of generalized g-Meir-Keeler type contractions and prove some coupled fixed point theorems under a generalized g-Meir-Keeler-contractive condition. Also, some applications of the main results in this paper are given.


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## Body

## 1. Introduction

The Banach contraction principle [1] is a classical and powerful tool in nonlinear analysis and has been generalized by many authors (see [2–15] and others).Recently, Bhaskar and Lakshmikantham [16] introduced the notion of a coupled fixed-point of the given two variables mapping. More precisely, let X be a nonempty set and F:X×X→X be a given mapping. An element (x,y)∈X×X is called a coupled fixed-point of the mapping F if(1.1)F(x,y)=x,F(y,x)=y.They also showed the uniqueness of a coupled fixed-point of the mappingF and applied their theorems to the problems of the existence and uniqueness of a solution for a periodic boundary value problem.Theorem 1.1 (see Zeidler [15]).
Let(X,≤) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a continuous mapping having the mixed monotone property on X. Assume that there exists k∈[0,1) such that
(1.2)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)]
for all x≥u and y≤v. Moreover, if there exist x0,y0∈X such that
(1.3)x0≤F(x0,y0),y0≥F(y0,x0),
then there exist x,y∈X such that x=F(x,y) and y=F(y,x).Later, in [17], Lakshmikantham and Ćirić investigated some more coupled fixed-point theorems in partially ordered sets, and some others obtained many results on coupled fixed-point theorems in cone metric spaces, intuitionistic fuzzy normed spaces, ordered cone metric spaces and topological spaces (see, e.g., [18–25]).In [9], Meir and Keeler generalized the well-known Banach fixed-point theorem [1] as follows.Theorem 1.2 (Meir and Keeler [9]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that, for any ϵ>0, there exists δ(ϵ)>0 such that
(1.4)ϵ≤d(x,y)<ϵ+δ(ϵ)⟹d(T(x),T(y))<ϵ
for all x,y∈X. Then T  admits a unique fixed-point x0∈X and, for all x∈X, the sequence {Tn(x)} converges to x0.Proposition 1.3 (see [17]).
Let(X,d) be a partially ordered metric space and F:X×X→X be a given mapping. If the contraction (1.2) is satisfied, then F is a generalized Meir-Keeler type contraction.Motivated by the results of Bhaskar and Lakshmikantham [16], Lakshmikantham and Ćirić [17], and Samet [26], in this paper, we introduce the definition of g-Meir-Keeler-contractive mappings and prove some coupled fixed-point theorems under a generalized g-Meir-Keeler contractive condition.


## 2. Main Results

LetX be a nonempty set. We note that an element (x,y)∈X×X is called a coupled coincidence point of a mapping F:X×X→X and g:X→X if F(x,y)=g(x) and F(y,x)=g(y) for all x,y∈X. Also, we say that F and g are commutative (or commuting) if g(F(x,y))=F(g(x),g(y)) for all x,y∈X.We introduce the following two definitions.Definition 2.1.
Let(X,≤) be a partially ordered set and F:X×X→X and g:X→X. We say that F has the mixed strictg-monotone property if, for any x,y∈X,
(2.1)x1,x2∈X,g(x1)<g(x2)⟹F(x1,y)<F(x2,y),y1,y2∈X,g(y1)<g(y2)⟹F(x,y1)>F(x,y2).Definition 2.2.
Let(X,≤) be a partially ordered set and d be a metric on X. Let F:X×X→X and g:X→X be two given mappings. We say that F is a generalizedg-Meir-Keeler type contraction if, for all ϵ>0, there exists δ(ϵ)>0 such that, for all x,y,u,v∈X with g(x)≤g(u) and g(y)≥g(v),
(2.2)ϵ≤12[d(g(x),g(u))+d(g(y),g(v))]<ϵ+δ(ϵ)⟹d(F(x,y),F(u,v))<ϵ.Lemma 2.3.
Let(X,≤) be a partially ordered set and d be a metric on X. Let F:X×X→X and g:X→X be two given mappings. If F is a generalized g-Meir-Keeler type contraction, then we have
(2.3)d(F(x,y),F(u,v))<12[d(g(x),g(u))+d(g(y),g(v))]
for all x,y,u,v with g(x)<g(u), g(y)≥g(v) or g(x)≤g(u), g(y)>g(v).Proof.
Letx,y,u,v∈X such that g(x)<g(u) and g(y)≥g(v) or g(x)≤g(u) and g(y)>g(v). Then d(g(x),g(u))+d(g(y),g(v))>0. Since F is a generalized g-Meir-Keeler type contraction, for ϵ=(1/2)[d(g(x),g(u))+d(g(y),g(v))], there exists δ(ϵ)>0 such that, for all x0,y0,u0,v0∈X with g(x0)≤g(u0) and g(y0)≥g(v0),
(2.4)ϵ≤12[d(g(x0),g(u0))+d(g(y0),g(v0))]<ϵ+δ(ϵ)⟹d(F(x0,y0),F(u0,v0))<ϵ.
Therefore, putting x0=x, y0=y, u0=u and v0=v, we have
(2.5)d(F(x,y),F(u,v))<12[d(g(x),g(u))+d(g(y),g(v))].
This completes the proof.From now on, we suppose that(X,≤) is a partially ordered set, and there exists a metric d on X such that (X,d) is a complete metric space.Theorem 2.4.
LetF:X×X→X and g:X→X be such that F(X×X)⊆g(X), g is continuous and commutative with F. Also, suppose that(a)
F has the mixed strict g-monotone property;(b)
F is a generalized g-Meir-keeler type contraction;(c)
there existx0,y0∈X such that g(x0)<F(x0,y0) and g(y0)>F(y0,x0).
Then there existx,y∈X such that g(x)=F(x,y) and g(y)=F(y,x); that is, F and g have a coupled coincidence in X×X.Proof.
Letx0,y0∈X be such that g(x0)<F(x0,y0) and g(y0)>F(y0,x0). Since F(X×X)⊆g(X), we can choose x1,y1∈X such that g(x1)=F(x0,y0) and g(y1)=F(y0,x0). Again, from F(X×X)⊆g(X), we can choose x2,y2∈X such that g(x2)=F(x1,y1) and g(y2)=F(y1,x1).
Continuing this process, we can construct the sequences{xn} and {yn} in X such that
(2.6)g(xn+1)=F(xn,yn),g(yn+1)=F(yn,xn)
for all n≥0.
Now, we show that(2.7)g(xn)<g(xn+1),g(yn)>g(yn+1)
for all n≥0. For n=0, we have
(2.8)g(x0)<F(x0,y0)=g(x1),g(y0)>F(y0,x0)=g(y1).
Since F has the mixed strict g-monotone property, then we have
(2.9)g(x0)<g(x1)⟹F(x0,y1)<F(x1,y1),g(y0)>g(y1)⟹F(x0,y0)<F(x0,y1).
It follows that F(x0,y0)<F(x1,y1), that is, g(x1)<g(x2).
Similarly, we have(2.10)g(y1)<g(y0)⟹F(y1,x0)<F(y0,x0),g(x1)>g(x0)⟹F(y1,x1)<F(y1,x0).
Thus it follows that F(y1,x1)<F(y0,x0), that is, g(y2)<g(y1).
Again, we have(2.11)g(x1)<g(x2)⟹F(x1,y2)<F(x2,y2),g(y1)>g(y2)⟹F(x1,y1)<F(x1,y2).
Thus it follows that F(x1,y1)<F(x2,y2), that is, g(x2)<g(x3).
Similarly, we have(2.12)g(y2)<g(y1)⟹F(y2,x1)<F(y1,x1),g(x2)>g(x1)⟹F(y2,x2)<F(y2,x1).
Thus it follows that F(y2,x2)<F(y1,x1), that is, g(y3)<g(y2).
Continuing this process for eachn≥1, we get the following:
(2.13)g(x0)<g(x1)<g(x2)<⋯<g(xn)<g(xn+1)<⋯,g(y0)>g(y1)>g(y2)>⋯g(yn)>g(yn+1)>⋯.
Denote that
(2.14)δn:=d(g(xn),g(xn+1))+d(g(yn),g(yn+1)).
Since g(xn-1)<g(xn) and g(yn-1)>g(yn), it follows from (2.6) and Lemma 2.3 that
(2.15)d(g(xn),g(xn+1))=d(F(xn-1,yn-1),F(xn,yn))<12[d(g(xn-1),g(xn))+d(g(yn-1),g(yn))].
Since g(yn)<g(yn-1) and g(xn)>g(xn-1), it follows from (2.6) and Lemma 2.3 that
(2.16)d(g(yn+1),g(yn))=d(F(yn,xn),F(yn-1,xn-1))<12[d(g(yn),g(yn-1))+d(g(xn),g(xn-1))].
Thus it follows from (2.14)–(2.16) that δn<δn-1. This means that the sequence {δn/2} is monotone decreasing. Therefore, there exists δ*≥0 such that limn→∞δn/2=δ*, that is,
(2.17)lim⁡n→∞12[d(g(xn),g(xn+1))+d(g(yn),g(yn+1))]=δ*.
Now, we show thatδ*=0. Suppose that δ*>0 hold. Let δ*=ϵ. Then there exists a positive integer m such that
(2.18)ϵ≤12[d(g(xm),g(xm+1))+d(g(ym),g(ym+1))]<ϵ+δ(ϵ).
Then, by using (2.7) and the condition (b), we have
(2.19)d(F(xm,ym),F(xm+1,ym+1))<ϵ,
and so, by (2.6), we have
(2.20)d(g(xm+1),g(xm+2))<ϵ.
On the other hand, by (2.15), we have
(2.21)12[d(g(xm),g(xm+1))+d(g(ym),g(ym+1))]<ϵ,
which is a contradiction with (2.18). Thus we have ϵ=δ*=0, that is,
(2.22)lim⁡n→∞12[d(g(xn),g(xn+1))+d(g(yn),g(yn+1))]=0,
that is,
(2.23)lim⁡n→∞δn=0.
Now, we prove that{g(xn)} and {g(yn)} are Cauchy sequences in X. Suppose that at least one of {g(xn)} or {g(yn)} is not a Cauchy sequence. Then there exist ϵ>0 and two subsequences {lk}, {mk} of integers such that mk>lk≥k and
(2.24)d(g(xlk),g(xmk))≥ϵ2,d(g(ylk),g(ymk))≥ϵ2
for all k≥1. Thus we have
(2.25)rk=d(g(xlk),g(xmk))+d(g(ylk),g(ymk))≥ϵ
for all k≥1. Let mk be the smallest number exceeding lk such that (2.25) holds. Then we have
(2.26)d(g(xlk),g(xmk-1))+d(g(ylk),g(ymk-1))<ϵ.
Thus, from (2.14), (2.25), (2.26) and the triangle inequality, it follows that
(2.27)ϵ≤rk≤d(g(xlk),g(xmk-1))+d(g(xmk-1),g(xmk))+d(g(ylk),g(ymk-1))+d(g(ymk-1),g(ymk))<ϵ+δmk-1
and so
(2.28)ϵ≤lim⁡k→∞rk≤lim⁡k→∞(ϵ+δmk-1).
Hence, by (2.23), we have
(2.29)lim⁡k→∞rk=ϵ+.
It follows from (2.6), (2.14), and the triangle inequality that
(2.30)rk=d(g(xlk),g(xmk))+d(g(ylk),g(ymk))≤d(g(xlk),g(xlk+1))+d(g(xlk+1),g(xmk+1))+d(g(xmk+1),g(xmk))+d(g(ylk),g(ylk+1))+d(g(ylk+1),g(ymk+1))+d(g(ymk+1),g(ymk))=δlk+δmk+d(g(xlk+1),g(xmk+1))+d(g(ylk+1),g(ymk+1))=δlk+δmk+d(F(xlk,ylk),F(xmk,ymk))+d(F(ylk,xlk),F(ymk,xmk)).
Form (2.13) we have g(xlk)<g(xmk) and g(ylk)>g(ymk). Now, it follows from Lemma 2.3 and (2.30) that
(2.31)rk<δlk+δmk+d(g(xlk),g(xmk))+d(g(ylk),g(ymk)),
that is,
(2.32)rk<δlk+δmk+rk.
This is a contradiction. Therefore, {g(xn)} and {g(yn)} are Cauchy sequences. Since X is complete, there exist x,y∈X such that
(2.33)lim⁡n→∞g(xn)=x,lim⁡n→∞g(yn)=y.
Since {g(xn)} is monotone increasing and {g(yn)} is monotone decreasing, we have
(2.34)g(xn)<x,g(yn)>y
for all n≥1. Thus it follows from (2.33) and the continuity of g that
(2.35)lim⁡n→∞g(g(xn))=g(x),lim⁡n→∞g(g(yn))=g(y).
Thus, for all m≥1, there exists a positive integer n0 such that, for all n≥n0,
(2.36)d(g(g(xn)),g(x))<14m,d(g(g(yn)),g(y))<14m.
Hence, from (2.6), the commutativity of F and g and the triangle inequality, we have
(2.37)d(F(x,y),g(x))≤d(F(x,y),g(g(xn)))+d(g(g(xn)),g(x))=d(F(x,y),g(F(xn-1,yn-1)))+d(g(g(xn)),g(x))=d(F(x,y),F(g(xn-1),g(yn-1)))+d(g(g(xn)),g(x)).
Thus, it follows from (2.34), (2.36), and Lemma 2.3 that
(2.38)d(F(x,y),g(x))<12[d(g(g(xn-1)),g(x))+d(g(g(yn-1)),g(y))]+d(g(g(xn)),g(x))<18m+18m+14m=12m⟶0
as m→∞. Therefore, we have F(x,y)=g(x). Similarly, we can show that F(y,x)=g(y). This means that F and g have a coupled coincidence point in X×X. This completes the proof.Corollary 2.5.
LetF:X×X→X be a mapping satisfying the following conditions:(a)
F has the mixed strict monotone property;(b)
F is a generalized Meir-Keeler type contraction;(c)
there existsx0,y0∈X such that x0<F(x0,y0) and y0>F(y0,x0).
Then there existx,y∈X such that x=F(x,y) and y=F(y,x).Proof.
The conclusion follows from Theorem2.4 by putting g=I (: the identity mapping) on X.Now, we introduce the product spaceX×X with the following partial order: for all (x,y),(u,v)∈X×X,
(2.39)(u,v)≤(x,y)⟺u<x,v≥y.Theorem 2.6.
Suppose that all the hypotheses of Theorem2.4 hold and, further, for all (x,y),(x*,y*)∈X×X, there exists (u,v)∈X×X such that (F(u,v),F(v,u)) is comparable with (F(x,y),F(y,x)) and (F(x*,y*),F(y*,x*)). Then F and g have a unique coupled common fixed-point, that is, there exists a unique (x,y)∈X×X such that
(2.40)x=g(x)=F(x,y),y=g(y)=F(y,x).Proof.
By Theorem2.4, the set of coupled coincidences of the mapping F and g is nonempty.
First, we show that, if(x,y) and (x*,y*) are coupled coincidence points of F and g, that is, if
(2.41)g(x)=F(x,y),g(y)=F(y,x),g(x*)=F(x*,y*),g(y*)=F(y*,x*),
then we have
(2.42)g(x)=g(x*),g(y)=g(y*).
Put u0=u, v0=v and choose u1,v1∈X such that g(u1)=F(u0,v0) and g(v1)=F(v0,u0). Then, similarly as in the proof of Theorem 2.4, we can inductively define the sequences {g(un)} and {g(vn)} such that
(2.43)g(un+1)=F(un,vn),g(vn+1)=F(vn,un)
for all n≥0. Also, if we set x0=x, y0=y, x0*=x*, and y0*=y*, then we can define the sequences {g(xn)}, {g(yn)}, {g(xn*)}, and {g(yn*)} as follows:
(2.44)g(xn+1)=F(xn,yn),g(yn+1)=F(yn,xn),g(xn+1*)=F(xn*,yn*),g(yn+1*)=F(yn*,xn*)
for all n≥0. Since
(2.45)(F(x,y),F(y,x))=(g(x1),g(y1))=(g(x),g(y)),(F(u,v),F(v,u))=(g(u1),g(v1))
are comparable each other, then g(x)<g(u1) and g(y)≥g(v1). It is easy to show that (g(x),g(y)), and (g(un),g(vn)) are comparable each other, that is, g(x)<g(un) and g(y)≥g(vn) for all n≥1. Thus it follows from Lemma 2.3 that
(2.46)d(g(x),g(un+1))+d(g(y),g(vn+1))=d(F(x,y),F(un,vn))+d(F(y,x),F(vn,un))<12[d(g(x),g(un))+d(g(y),g(vn))]+12[d(g(y),g(vn))+d(g(x),g(un))]=d(g(x),g(un))+d(g(y),g(vn))
and so
(2.47)12[d(g(x),g(un+1))+d(g(y),g(vn+1))]<12n[d(g(x),g(u1))+d(g(y),g(v1))]⟶0
as n→∞. Therefore, we have
(2.48)lim⁡n→∞d(g(x),g(un+1))=0,lim⁡n→∞d(g(y),g(vn+1))=0.
Similarly, we can prove that(2.49)lim⁡n→∞d(g(x*),g(un+1))=0,lim⁡n→∞d(g(y*),g(vn+1))=0.
Thus, by the triangle inequality, (2.48) and (2.49), we have
(2.50)d(g(x),g(x*))≤d(g(x),g(un+1))+d(g(x*),g(un+1))⟶0,d(g(y),g(y*))≤d(g(y),g(vn+1))+d(g(y*),g(vn+1))⟶0
as n→∞, which imply that g(x)=g(x*) and g(y)=g(y*).
Now, we prove thatg(x)=x and g(y)=y. Denote that g(x)=z and g(x)=w. Since g(x)=F(x,y) and g(y)=F(y,x), by the commutativity of F and g, we have
(2.51)g(z)=g(g(x))=g(F(x,y))=F(g(x),g(y))=F(z,w),(2.52)g(w)=g(g(y))=g(F(y,x))=F(g(y),g(x))=F(w,z).
Thus, (z,w) is a coupled coincidence point of F and g.
Puttingx*=z and y*=w in (2.52), it follows from (2.42) that
(2.53)z=g(x)=g(x*)=g(z),w=g(y)=g(y*)=g(w)
and so, from (2.51) and (2.52),
(2.54)z=g(z)=F(z,w),w=g(w)=F(w,z).
Therefore, (z,w) is a coupled common fixed-point of F and g.
Finally, to prove the uniqueness of the coupled common fixed-point ofF and g, assume that (p,q) is another coupled common fixed-point of F and g. Then, by (2.42), we have p=g(p)=g(z)=z and q=g(q)=g(w)=w. This completes the proof.Corollary 2.7.
Suppose that all the hypotheses of Corollary2.5 hold and, further, for all (x,y) and (x*,y*)∈X×X, there exists (u,v)∈X×X that is comparable with (x,y) and (x*,y*). Then there exists a unique x∈X such that x=F(x,x).


## 3. Applications

Now, we give some applications of the main results in Section2.Theorem 3.1.
LetF:X×X→X and g:X→X be two given mappings. Assume that there exists a function φ:[0,+∞)→[0,+∞) satisfying the following conditions:(a)
φ(0)=0 and φ(t)>0 for any t>0;(b)
φ is nondecreasing and right continuous;(c)
for anyϵ>0, there exists δ(ϵ)>0 such that, for all x,y,u,v∈X with g(x)≤g(u) and g(y)≥g(v),(3.1)ϵ≤φ(12[d(g(x),g(u))+d(g(y),g(v))])<ϵ+δ(ϵ)⟹φ[d(F(x,y),F(u,v))]<ϵ.
Then F is a generalized g-Meir-Keeler type contraction.Proof.
For anyϵ>0, it follows from (a) that φ(ϵ)>0 and so there exists α>0 such that, for all u,v,u*,v⋆∈X with g(u)≤g(u*) and g(v)≥g(v*),
(3.2)φ(ϵ)≤φ(12[d(g(u),g(u*))+d(g(v),g(v*))])<φ(ϵ)+α⟹φ[d(F(u,v),F(u*,v*))]<φ(ϵ).
From the right continuity of φ, there exists δ>0 such that φ(ϵ+δ)<φ(ϵ)+α. For any x,y,u,v∈X such that g(x)≤g(u), g(y)≥g(v) and
(3.3)ϵ≤12[d(g(x),g(u))+d(g(y),g(v))]<ϵ+δ,
since φ is nondecreasing function, we get the following:
(3.4)φ(ϵ)≤φ(12[d(g(x),g(u))+d(g(y),g(v))])<φ(ϵ+α)<φ(ϵ)+α.
By (3.2), we have φ[d(F(x,y),F(u,v))]<φ(ϵ) and so d(F(x,y),F(u,v))<ϵ. Therefore, it follows that F is a generalized g-Meir-Keeler type contraction. This completes the proof.Corollary 3.2 (see [26, Theorem  3.1]).
LetF:X×X→X be a given mapping. Assume that there exists a function φ:[0,+∞)→[0,+∞) satisfying the following conditions:(a)
φ(0)=0 and φ(t)>0 for any t>0;(b)
φ is nondecreasing and right continuous;(c)
for anyϵ>0, there exists δ(ϵ)>0 such that x≤u, y≥v and
(3.5)ϵ≤φ(12[d(x,u)+d(y,v)])<ϵ+δ(ϵ)⟹φ[d(F(x,y),F(u,v))]<ϵ.Then F is a generalized Meir-Keeler type contraction.The following result is an immediate consequence of Theorems2.4 and 3.1.Corollary 3.3.
LetF:X×X→X and g:X→X be two given mappings such that F(X×X)⊆g(X), g is continuous and commutative with F. Also, suppose that(a)
F has the mixed strict g-monotone property;(b)
for anyϵ>0, there exists δ(ϵ)>0 such that, for all x,y,u,v∈X with g(x)≤g(u) and g(y)≥g(v),
(3.6)ϵ≤∫0(1/2)[d(g(x),g(u))+d(g(y),g(v))]φ(t)dt<ϵ+δ(ϵ)⟹∫0d(F(x,y),F(u,v))φ(t)dt<ϵ,
where φ is a locally integrable function from [0,+∞) into itself satisfying the following condition:
(3.7)∫0sφ(t)dt>0
for all s>0;(c)
there existx0,y0∈X such that g(x0)<F(x0,y0) and g(y0)>F(y0,x0).Then there exists (x,y)∈X×X such that g(x)=F(x,y) and g(y)=F(y,x). Moreover, if g(x0) and g(y0) are comparable to each other, then F and g have a unique coupled common fixed-point in X×X.Corollary 3.4.
LetF:X×X→X be a mapping satisfying the following conditions:(a)
F has the mixed strict monotone property;(b)
for anyϵ>0, there exists δ(ϵ)>0 such that x≤u, y≥v and
(3.8)ϵ≤∫0(1/2)[d(x,u)+d(y,v)]φ(t)dt<ϵ+δ(ϵ)⟹∫0[d(F(x,y),F(u,v))]φ(t)dt<ϵ,
where φ is a locally integrable function from [0,+∞) into itself satisfying
(3.9)∫0sφ(t)dt>0
for all s>0;(c)
there existx0,y0∈X such that x0<F(x0,y0) and y0>F(y0,x0).Then there exists (x,y)∈X×X such that x=F(x,y) and y=F(y,x). Moreover, if x0 and y0 are comparable to each other, then F has a unique coupled common fixed-point in X×X.Corollary 3.5.
LetF:X×X→X and g:X→X be two given mappings such that F(X×X)⊆g(X), g is continuous and commutes with F. Also, suppose that(a)
F has the mixed strict g-monotone property;(b)
for anyx,y,u,v∈X with g(x)≤g(u) and g(y)≥g(v),
(3.10)∫0[d(F(x,y),F(u,v))]φ(t)dt≤k∫0(1/2)[d(g(x),g(u))+d(g(y),g(v))]φ(t)dt,
where k∈(0,1) and φ is a locally integrable function from [0,+∞) into itself satisfying
(3.11)∫0sφ(t)dt>0
for all s>0;(c)
there existx0,y0∈X such that g(x0)<F(x0,y0) and g(y0)>F(y0,x0).Then there exists (x,y)∈X×X such that g(x)=F(x,y) and g(y)=F(y,x). Moreover, if g(x0) and g(y0) are comparable to each other, then F and g have a unique coupled common fixed-point in X×X.Proof.
For anyϵ>0, if we take δ(ϵ)=(1/k-1)ϵ and apply Corollary 3.3, then we can get the conclusion.Corollary 3.6.
LetF:X×X→X be a mapping satisfying the following conditions:(a)
F has the mixed strict monotone property,(b)
for anyx,y,u,v∈X with x≤u and y≥v,
(3.12)∫0d(F(x,y),F(u,v))φ(t)dt≤k∫0(1/2)[d(x,u)+d(y,v)]φ(t)dt,
where k∈(0,1) and φ is a locally integrable function from [0,+∞) into itself satisfying
(3.13)∫0sφ(t)dt>0
for all s>0;(c)
there existx0,y0∈X such that x0<F(x0,y0) and y0>F(y0,x0).Then there exist x,y∈X such that x=F(x,y) and y=F(y,x). Moreover, if x0 and y0 are comparable to each other, then F has a unique coupled common fixed-point in X×X.Finally, by using the above results, we show the existence of solutions for the following integral equation:(3.14)(x(t),y(t))=(∫0TG(t,s)[(f(s,x(s))+λx(s))-(f(s,y(s))+λy(s))]ds,∫0TG(t,s)[(f(s,y(s))+λy(s))-(f(s,x(s))+λx(s))]ds),
where x,y∈C(I,ℝ) (: the set of continuous functions from I into ℝ), T>0, f:I×ℝ→ℝ is a continuous function and
(3.15)G(t,s)={eλ(T+s-t)eλT-1,if0≤s<t≤T;eλ(s-t)eλT-1,if0≤t<s≤T.Definition 3.7.
Alower solution for the integral equation (3.14) is an element (α,β)∈C1(I,ℝ)×C1(I,ℝ) such that
(3.16)α′(t)+λβ(t)≤f(t,α(t))-f(t,β(t)),α(0)<α(T),β′(t)+λα(t)≥f(t,β(t))-f(t,α(t)),β(0)≥β(T),
where C1(I,ℝ) denotes the set of differentiable functions from I into ℝ.Now, we prove the existence of solutions for the integral equation (3.14) by using the existence of a lower solution for the integral equation (3.14).Theorem 3.8.
Let𝒜 be the class of the functions φ:[0,∞)→[0,∞) satisfying the following conditions:(a)
φ is increasing;(b)
for anyx≥0, there exists k∈[0,1) such that φ(x)<(k/2)x.In the integral equation (3.14), suppose that there exists λ>0 such that, for all x,y∈ℝ with y>x,
(3.17)0<f(t,y)+λy-[f(t,x)+λx]≤λφ(y-x),
where φ∈𝒜. If a lower solution of the integral equation (3.14) exists, then a solution of the integral equation (3.14) exists.Proof.
Define a mappingF:C(I,ℝ)×C(I,ℝ)→C(I,ℝ) by
(3.18)F(x(t),y(t))=∫0TG(t,s)[(f(s,x(s))+λx(s))-(f(s,y(s))+λy(s))]ds.
Note that, if(x(t),y(t))∈C(I,ℝ)×C(I,ℝ) is a coupled fixed-point of F, then (x(t),y(t)) is a solution of the integral equation (3.14).
Now, we check the hypotheses in Corollary2.5 as follows:(1)
X×X=C(I,ℝ)×C(I,ℝ) is a partially ordered set if we define the order relation in X×X as follows:
(3.19)(u(t),v(t))≤(x(t),y(t))iffu(t)<x(t),v(t)≥y(t)
for all (x(t),y(t)),(u(t),v(t))∈X×X and t∈I.(2)
(X,d) is a complete metric space if we define a metric d as follows:
(3.20)d(x(t),y(t))=sup⁡t∈I{|x(t)-y(t)|:x(t),y(t)∈X}.(3)
The mappingF has the mixed strict monotone property. In fact, by hypothesis, if x2>x1, then we have
(3.21)f(t,x2)+λx2>f(t,x1)+λx1,which implies that, for any t∈I,
(3.22)∫0T[f(s,x2(s))+λx2(s)-f(s,y(s))-λy(s)]G(t,s)ds>∫0T[f(s,x1(s))+λx1(s)-f(s,y(s))-λy(s)]G(t,s)ds,that is,
(3.23)F(x2(t),y(t))>F(x1(t),y(t)).
Similarly, ify1<y2, then we have
(3.24)f(t,y2)+λy2>f(t,y1)+λy1,
which implies that, for any t∈I,
(3.25)∫0T[f(s,x(s))+λx(s)-f(s,y2(s))-λy2(s)]G(t,s)ds<∫0T[f(s,x(s))+λx(s)-f(s,y1(s))-λy1(s)]G(t,s)ds,
that is,
(3.26)F(x(t),y2(t))<F(x(t),y1(t)).
Now, we show thatF satisfies (1.2). In fact, let (x,y)≤(u,v) and t∈I. Then we have
(3.27)d(F(x(t),y(t)),F(u(t),v(t)))=sup⁡{|F(x(t),y(t))-F(u(t),v(t))|:t∈I}=sup⁡t∈I{|∫0TG(t,s)[f(s,x(s))+λx(s)-f(s,y(s))-λy(s)]ds-∫0TG(t,s)[f(s,u(s))+λu(s)-f(s,v(s))-λv(s)]ds|∫0T}≤sup⁡t∈I∫0TG(t,s)|f(s,x(s))+λx(s)-f(s,u(s))-λu(s)+f(s,v(s))+λv(s)-f(s,y(s))-λy(s)|ds.
Since the function φ(x) is increasing and (x,y)≤(u,v), we have
(3.28)φ(x(s)-u(s))≤φ(d(x(s),u(s))),φ(v(s)-y(s))≤φ(d(v(s),y(s))),
we obtain the following:
(3.29)d(F(x(t),y(t)),F(u(t),v(t)))≤sup⁡t∈I∫0TG(t,s)|λφ(x(s)-u(s))+λφ(v(s)-y(s))|ds≤λsup⁡t∈I∫0TG(t,s)|φ(d(x(s),u(s)))+φ(d(v(s),y(s)))|ds=λ(φ(d(x(s),u(s)))+φ(d(v(s),y(s))))⋅sup⁡t∈I∫0TG(t,s)ds=λ(φ(d(x(s),u(s)))+φ(d(v(s),y(s))))⋅sup⁡t∈I1eλT-1([1λeλ(T+s-t)]0t+[1λeλ(s-t)]tT)=λ(φ(d(x(s),u(s)))+φ(d(v(s),y(s))))⋅1λeλT-1(eλT-1)=φ(d(x(s),u(s)))+φ(d(v(s),y(s)))<k2[d(x(s),u(s))+d(v(s),y(s))]≤k2sup⁡{|x(t)-u(t)|:t∈I}+k2sup⁡{|v(t)-y(t)|:t∈I}=k2[d(x(t),u(t))+d(y(t),v(t))].
Then, by Proposition 1.3, F is a generalized Meir-Keeler type contraction.
Finally, let(α(t),β(t))∈C1(I,ℝ)×C1(I,ℝ) be a lower solution for the integral equation (3.14). Then we show that
(3.30)α<F(α,β),β≥F(β,α).
Indeed, we have α′(t)+λβ(t)≤f(t,α(t))-f(t,β(t)) for any t∈I and so
(3.31)α′(t)+λα(t)≤f(t,α(t))-f(t,β(t))+λα(t)-λβ(t)
for any t∈I. Multiplying by eλt in (3.31), we get the following:
(3.32)(α(t)eλt)′≤[(f(t,α(t))+λα(t))-(f(t,β(t))+λβ(t))]eλt
for any t∈I, which implies that
(3.33)α(t)eλt≤α(0)+∫0t[(f(s,α(s))+λα(s))-f(s,β(s))-λβ(s)]eλsds
for any t∈I. This implies that
(3.34)α(0)eλt<α(T)eλT≤α(0)+∫0T[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]eλsds
and so
(3.35)α(0)<∫0TeλseλT-1[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds.
Thus it follows from (3.35) and (3.33) that
(3.36)α(t)eλt<∫tTeλseλT-1[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds+∫0teλ(T+s)eλT-1[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds,
and so
(3.37)α(t)<∫0teλ(T+s-t)eλT-1[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds+∫tTeλ(s-t)eλT-1[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds.
Hence we have
(3.38)α(t)<∫0TG(t,s)[f(s,α(s))+λα(s)-f(s,β(s))-λβ(s)]ds=F(α(t),β(t))
for any t∈I.
Similarly, we haveβ(t)≥F(β(t),α(t)). Therefore, by Corollary 2.5, F has a coupled fixed-point.Example 3.9.
In the integral equation (3.14), we put λ=1.5, f(u,v)=u-v for all (u,v)∈I×ℝ and T=0.5. Then f is a continuous function, and we have
(3.39)(x(t),y(t))=(∫00.5G(t,s)[0.5x(s)-0.5y(s)]ds,∫00.5G(t,s)[0.5y(s)-0.5x(s)]ds),
where x,y∈C(I,ℝ), and
(3.40)G(t,s)={e1.5(0.5+s-t)e0.75-1,if0≤s<t≤0.5,e1.5(s-t)e0.75-1,if0≤t<s≤0.5.
Also, (α(t),β(t))=(-2e-0.5t,3e-0.5t) is a lower solution of (3.39). Moreover, if we define φ(x)=x/3 for all x∈[0,∞), then φ is increasing and, for any x>0, there exists k=1/1.1∈[0,1) such that φ(x)=x/3<(k/2)x=x/2.2. For all x,y∈ℝ with y>x, we have
(3.41)0<f(t,y)+λy-[f(t,x)+λx]=0.5(y-x)≤λφ(y-x)=1.5y-x3=0.5(y-x).
Therefore, all the conditions of Theorem 3.8 hold, and a solution of (3.39) exists.



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*Source: 150363-2012-03-15.xml*