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R E S E A R C H

Open Access

Existence of a solution of integral equations

via fixed point theorem

Selma Gülyaz

1

, Erdal Karapınar

2*

, Vladimir Rakocevi´c

3

and Peyman Salimi

4

*Correspondence:

[email protected]; [email protected]

2Department of Mathematics,

Atilim University, ˙Incek, Ankara 06836, Turkey

Full list of author information is available at the end of the article

Abstract

In this paper, we establish a solution to the following integral equation:

u(t) =

T

0

G(t,s)f

(s,

u(s))ds for allt∈[0,T], ()

whereT> 0,f: [0,T]×R→RandG: [0,T]×[0,T]→[0,∞) are continuous functions. For this purpose, we also obtain some auxiliary fixed point results which generalize, improve and unify some fixed point theorems in the literature.

MSC: 47H10; 54H25

Keywords: cyclic representation; fixed point

1 Introduction and preliminaries

Fixed point theory is one of the most efficient tools in nonlinear functional analysis to solve the nonlinear differential and integral equations. The existence/uniqueness of a so-lution of differential/integral equations turns into the existence/uniqueness of a (common) fixed point of the operators which are obtained after suitable substitutions and elementary calculations; see,e.g., [–].

In this paper, we first obtain some fixed point theorems to solve the integral equa-tion menequa-tioned above. For the sake of completeness, we recollect some basic defi-nitions and elementary results. Let X be a nonempty set and T be a self-mapping on X. Then, the set of all fixed points of T on X is denoted by Fix(T)X. Let be

the set of all functions ψ : [,∞)×[,∞)→ [,∞) satisfying the following condi-tions:

() ψis continuous,

() ψ(t,t) = if and only ift=t= ,

() ψ(t,t)≤(t+t).

Cyclic mapping and cyclic contraction were introduced by Kirk-Srinavasan-Veeramani to improve the well-known Banach fixed point theorem. Later, various types of cyclic con-traction have been investigated by a number of authors; see,e.g., [, –].

Definition .[] Suppose that (X,d) is a metric space andTis a self-mapping onX. Let

mbe a natural number andXi,i= , . . . ,m, be nonempty sets. ThenY=

m

i=Xiis called a

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cyclic representation ofXwith respect toT if

T(X)⊂X, . . . , T(Xm–)⊂Xm, T(Xm)⊂Xm+,

whereXm+=X.

Definition .[] LetT:XX,r>  andη,ξ:X→[, +∞) be two functions. We say thatTisr-(η,ξ)-admissible if

(i) η(x)≥rfor somexXimpliesη(Tx)≥r, (ii) ξ(x)≤rfor somexXimpliesξ(Tx)≤r.

Definition . Let (X,d) be a metric space andT:YYbe a self-mapping, whereY=

m

i=Xiis a cyclic representation ofY with respect toT. Letη,ξ :Y →[, +∞) be two

functions. An operatorT:YYis called:

• a cyclic weakr-(η,ξ)-C-contractive mapping of the first kind if

η(x)η(y)d(Tx,Ty)≤ξ(x)ξ(y)

 

d(x,Ty) +d(y,Tx)–ψd(x,Ty),d(y,Tx) ()

holds for allxXiandyXi+, whereψ.

• a cyclic weakr-(η,ξ)-C-contractive mapping of the second kind if

η(x)η(y) +rd(Tx,Ty)≤ξ(x)ξ(y) +r[[d(x,Ty)+d(y,Tx)]–ψ(d(x,Ty),d(y,Tx))] ()

such thatr+r> holds for allxX

iandyXi+, whereψ. 2 Auxiliary fixed point results

We state the main result of this section as follows.

Theorem . Let(X,d)be a complete metric space,m∈N,X,X, . . . ,Xm be nonempty closed subsets of(X,d)and Y=mi=Xi.Suppose that T:YY is a cyclic weak r-(η,ξ) -C-contractive mapping of the first kind such that

(i) Tisr-(η,ξ)-admissible;

(ii) there existsx∈Ysuch thatη(x)≥randξ(x)≤r;

(iii) if{xn}is a sequence inYsuch thatη(xn)≥randξ(xn)≤rfor alln∈Nandxnx

asn→ ∞,thenη(x)≥randξ(x)≤r.

Then T has a fixed point xni=Xi.Moreover,ifη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r for all x,yFix(T)Y,then T has a unique fixed point.

Proof Let there exist x∈ Y such that η(x)≥r and ξ(x)≤ r. Since T is r-(η,ξ

)-admissible, thenη(Tx)≥randξ(Tx)≤r. Again, since T isr-(η,ξ)-admissible, then

η(Tx)≥randξ(Tx)≤r. By continuing this process, we get

ηTnx

r and ξTnx

r for alln∈N. ()

On the other hand, sincex∈Y, there exists someisuch thatx∈Xi. NowT(Xi)⊆

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Txn=xn+, wherexnXin. Hence, forn≥, there existsin∈ {, , . . . ,m}such thatxnXin

andxn+Xin+. In casexn=xn+for somen= , , , . . . , then it is clear thatxnis a fixed point ofT. Now assume thatxn=xn+for alln. Hence, we haved(xn–,xn) >  for alln. Set dn:=d(xn,xn+). We shall show that the sequence{dn}is non-increasing. Due to () with x=xn–andy=xn, we get

rd(xn,xn+)

=rd(Txn–,Txn)≤η(xn–)η(xn)d(Txn–,Txn)

ξ(xn–)ξ(xn)

 

d(xn–,Txn) +d(xn,Txn–)

ψd(xn–,Txn),d(xn,Txn–)

=ξ(xn–)ξ(xn)

d(xn–,xn+) –ψ

d(xn–,xn+), 

r

d(xn–,xn+) –ψ

d(xn–,xn+),  ,

which implies

d(xn,xn+)≤

d(xn–,xn+) –ψ

d(xn–,xn+), 

≤

d(xn–,xn+)

≤ 

d(xn–,xn) +d(xn,xn+)

, ()

and sodndn–for alln∈N. Then there existd≥ such thatlimn→∞dn=d. Suppose,

on the contrary, thatd> . Also, taking limit asn→ ∞in (), we deduce

d≤

nlim→∞d(xn–,xn+)≤

(d+d) =d,

that is,

lim

n→∞d(xn–,xn+) = d. ()

Taking limit asn→ ∞in () and using (), we get

d≤

[d] –ψ(d, ).

Consequently, we haveψ(d, ) = , which yieldsd= . Hence

lim

n→∞d(xn,xn+) = . ()

We shall show that{xn}is a Cauchy sequence. To reach this goal, first we prove the

fol-lowing claim:

(K) For everyε> , there existsn∈Nsuch that ifr,qnwithrq≡(m), then

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Suppose, to the contrary, that there existsε>  such that for anyn∈N, we can find rn>qnnwithrnqn≡(m) satisfying

d(xqn,xrn)≥ε. ()

Now, we taken> m. Then, corresponding toqnn, one can choosernin such a way that

it is the smallest integer withrn>qnsatisfyingrnqn≡(m) andd(xqn,xrn)≥ε. Therefore, d(xqn,xrn–m) <ε. By using the triangular inequality,

εd(xqn,xrn)≤d(xqn,xrn–m) + m

i=

d(xrn–i,xrn–i–) <ε+ m

i=

d(xrn–i,xrn–i–).

Lettingn→ ∞in the last inequality, keeping () in mind, we derive that

lim

n→∞d(xqn,xrn) =ε. ()

Again,

εd(xqn,xrn)

d(xqn,xqn+) +d(xqn+,xrn+) +d(xrn+,xrn)

d(xqn,xqn+) +d(xqn+,xqn) +d(xqn,xrn) +d(xrn,xrn+) +d(xrn+,xrn).

Taking () and () into account, we get

lim

n→∞d(xqn+,xrn+) =ε ()

asn→ ∞in ().

Also we have the following inequalities:

d(xqn,xrn+)≤d(xqn,xrn) +d(xrn,xrn+) ()

and

d(xqn,xrn)≤d(xqn,xrn+) +d(xrn,xrn+). ()

Lettingn→ ∞in () and (), we derive that

lim

n→∞d(xqn,xrn+) =ε. ()

Again, we have

d(xrn,xqn+)≤d(xrn,xrn+) +d(xrn+,xqn+) ()

and

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Lettingn→ ∞in () and (), we conclude that

lim

n→∞d(xrn,xqn+) =ε. ()

Sincexqnandxrnlie in different adjacently labeled setsXiandXi+for certain ≤im,

using the fact thatT is a cyclic weakr-(η,ξ)-C-contractive mapping of the first kind, we have

rd(xrn+,xqn+)

=rd(Txrn,Txqn)≤η(xrn)η(xqn)d(Txrn,Txqn)

ξ(xrn)ξ(xqn)

 

d(xrn,Txqn) +d(xqn,Txrn)

ψd(xrn,Txqn),d(xqn,Txrn)

r

 

d(xrn,xqn+) +d(xqn,xrn+)

ψd(xrn,xqn+),d(xqn,xrn+) , ()

which implies

d(xrn+,xqn+)≤

 

d(xrn,xqn+) +d(xqn,xrn+)

ψd(xrn,xqn+),d(xqn,xrn+)

.

Lettingn→ ∞in the inequality above and keeping the expressions (), (), (), (), () in mind, we conclude that

εεψ(ε,ε).

Thus, we haveψ(ε,ε) = , which yields thatε= . Hence, (K) is satisfied.

We shall show that the sequence{xn}is Cauchy. Fixε> . By the claim, we findn∈N

such that ifr,qnwithrq≡(m), then

d(xr,xq)≤

ε

. ()

Sincelimn→∞d(xn,xn+) = , we also findn∈Nsuch that

d(xn,xn+)≤

ε

m ()

for anynn. Suppose thatr,s≥max{n,n}ands>r. Then, there existsk∈ {, , . . . ,m}

such thatsrk(m). Therefore, sr+ϕ≡(m) forϕ=mk+ . So, we have, for

j∈ {, . . . ,m},s+jr≡(m)

d(xr,xs)≤d(xr,xs+j) +d(xs+j,xs+j–) +· · ·+d(xs+,xs).

By () and () and from the last inequality, we get

d(xr,xs)≤

ε +j×

εm

ε +m×

εm=ε.

This proves that{xn}is a Cauchy sequence. SinceYis closed in (X,d), then (Y,d) is also

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we prove thatxis a fixed point ofT. In fact, sincelimn→∞xn=xand, asY=

m

i=Xiis a

cyclic representation ofYwith respect toT, the sequence{xn}has infinite terms in each Xifori∈ {, , . . . ,m}. Suppose thatxXi,TxXi+ and we take a subsequencexnk of

{xn}withxnkXi–. Now from (iii) we haveη(x)≥randξ(x)≤r. By using the contractive

condition, we can obtain

rd(Tx,Txnk)≤η(x)η(xnk)d(Tx,Txnk)

ξ(x)ξ(xnk)

 

d(x,Txnk) +d(xnk,Tx)

ψd(x,Txnk),d(xnk,Tx)

r

 

d(x,Txnk) +d(xnk,Tx)

ψd(x,Txnk),d(xnk,Tx) , ()

which implies

d(Tx,xnk+)≤

 

d(x,xnk+) +d(xnk,Tx)

ψd(x,xnk+),d(xnk,Tx)

.

Passing to the limit ask→ ∞in the last inequality, we get

d(x,Tx)≤ 

d(x,Tx) –ψ

,d(x,Tx)

≤ 

d(x,Tx),

which impliesd(x,Tx) = ,i.e.,x=Tx. Finally, to prove the uniqueness of the fixed point, suppose thatx,yFix(T)Y such thatη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r, wherex=y. The

cyclic character ofTand the fact thatx,yXare fixed points ofTimply thatx,ymi=Xi.

Suppose thatx=y. That is,d(x,y) > . Using the contractive condition, we obtain

rd(Tx,Ty)≤η(x)η(y)d(Tx,Ty)

ξ(x)ξ(y)

 

d(x,Ty) +d(y,Tx)–ψd(x,Ty),d(y,Tx)

r

 

d(x,Ty) +d(y,Tx)–ψd(x,Ty),d(y,Tx) ,

which implies

d(x,y)≤d(x,y) –ψd(x,y),d(x,y).

Thenψ(d(x,y),d(x,y)) =  and sod(x,y) = ,i.e.,x=y, which is a contradiction. This

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Example . LetX=Rwith the metric d(x,y) =|xy|for all x,yX. SupposeA=

(–∞, ] andA= [,∞) andY=

i=Ai. DefineT:YYandη,ξ:Y→[,∞) by

Tx= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x+

x+ ifx∈(–∞, –],

sinx

x ifx∈[–, –), –x ifx∈[–, –),  ifx∈[–, ], –lnx ifx∈(, ),

–x

(–x)(–x) ifx∈[, ),

 √

 –x ifx∈[,∞),

η(x) =

 ifx∈[–, ],  otherwise,

ξ(x) =

 ifx∈[–, ],  otherwise.

Also, defineψ : [,∞)→[,∞) byψ(t,s) =(t+s). Clearly,TA⊆A,TA⊆Aand

η()≥ andξ()≤. Letη(x)≥, thenx∈[–, ]. On the other hand,Tw∈[–, ] for all

w∈[–, ],i.e.,η(Tx)≥. Similarly,ξ(x)≤ impliesξ(Tx)≤. Therefore,Tis anr-(η,ξ )-admissible mapping. Let{xn}be a sequence inXsuch thatη(xn)≥,ξ(xn)≤ andxnx

asn→ ∞. Thenxn∈[–, ]. So,x∈[–, ],i.e.,η(x)≥ andξ(x)≤.

LetxAandyA. Now, ifx∈/[–, ] ory∈/[, ], thenη(x)η(y) = . Also, ifx∈[–, ]

andy∈[, ], thend(Tx,Ty) = . That is,η(x)η(y)d(Tx,Ty) =  for allxAand allyA.

Hence,

η(x)η(y)d(Tx,Ty) = ≤ξ(x)ξ(y)

 

d(x,Ty) +d(y,Tx)–ψd(x,Ty),d(y,Tx)

for allxAandyA. ThenTis a cyclic weakr-(η,ξ)-C-contractive mapping of the first

kind. Therefore all the conditions of Theorem . hold andThas a fixed point inA∩A.

Here,x=  is a fixed point ofT.

Theorem . Let(X,d)be a complete metric space,m∈N,X,X, . . . ,Xm be nonempty closed subsets of(X,p)and Y =mi=Xi.Suppose that T:YY is a cyclic weak r-(η,ξ) -C-contractive mapping of the second kind such that

(i) Tisr-(η,ξ)-admissible;

(ii) there existsx∈Ysuch thatη(x)≥randξ(x)≤r;

(iii) if{xn}is a sequence inYsuch thatη(xn)≥randξ(xn)≤rfor alln∈Nandxnx

asn→ ∞,thenη(x)≥randξ(x)≤r.

Then T has a fixed point xni=Xi.Moreover,ifη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r for all x,yFix(T)Y,then T has a unique fixed point.

Proof By a similar method as in the proof of Theorem ., we have

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We shall show that the sequence{dn:=d(xn,xn+)}is non-increasing. Due to () withx= xn–andy=xn, we get

r+rd(xn,xn+)=r+rd(Txn–,Txn)≤η(xn–)η(xn) +r

d(Txn–,Txn)

ξ(xn–)ξ(xn) +r [

[d(xn–,Txn)+d(xn,Txn–)]–ψ(d(xn–,Txn),d(xn,Txn–))]

=ξ(xn–)ξ(xn) +r [

d(xn–,xn+)–ψ(d(xn–,xn+),)]

r+r[d(xn–,xn+)–ψ(d(xn–,xn+),)] ,

which implies

d(xn,xn+)≤

d(xn–,xn+) –ψ

d(xn–,xn+), 

≤

d(xn–,xn+)

≤ 

d(xn–,xn) +d(xn,xn+)

, ()

and sodndn–for alln∈N. Then there existsd≥ such thatlimn→∞dn=d. We shall

show thatd=  by the method ofreductio ad absurdum. Suppose thatd> . By letting

n→ ∞in (), we deduce

d≤

nlim→∞d(xn–,xn+)≤

(d+d) =d,

that is,

lim

n→∞d(xn–,xn+) = d. ()

Taking limit asn→ ∞in () and using (), we get

d≤

[d] –ψ(d, ).

Thus, we haveψ(d, ) =  and henced= , which is a contradiction. Consequently, we have

lim

n→∞dn=nlim→∞d(xn,xn+) = . ()

We shall show that{xn}is a Cauchy sequence. To reach this goal, first we prove the

fol-lowing claim:

(K) For everyε> , there existsn∈Nsuch that ifr,qnwithrq≡(m), then

d(xr,xq) <ε.

Suppose, to the contrary, that there exists ε>  such that for anyn∈Nwe can find

rn>qnnwithrnqn≡(m) satisfying

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Following the related lines in Theorem ., we deduce

lim

n→∞d(xqn,xrn) =ε, ()

lim

n→∞d(xqn+,xrn+) =ε, ()

lim

n→∞d(xqn,xrn+) =ε ()

and

lim

n→∞d(xrn,xqn+) =ε. ()

Sincexqn andxrnlie in different adjacently labeled setsXiandXi+for certain ≤im,

using the fact that a cyclic weak r-(η,ξ)-C-contractive mapping of the second kind, we have

r+rd(xrn+,xqn+)

=r+rd(Txrn,Txqn)η(x

rn)η(xqn) +r

d(Txrn,Txqn)

ξ(xrn)ξ(xqn) +r [

[d(xrn,Txqn)+d(xqn,Txrn)]–ψ(d(xrn,Txqn),d(xqn,Txrn))]

r+r[

[d(xrn,xqn+)+d(xqn,xrn+)]–ψ(d(xrn,xqn+),d(xqn,Txrn+))],

which implies

d(xrn+,xqn+)≤

 

d(xrn,xqn+) +d(xqn,xrn+)

ψd(xrn,xqn+),d(xqn,Txrn+)

.

Lettingn→ ∞in the inequality above and by applying () (), (), (), (), we de-duce that

εεψ(ε,ε).

Consequently, we haveψ(ε,ε) = , and henceε= . As a result, we conclude that (K) is satisfied. We assert that the sequence{xn}is Cauchy. Fixε> . By the claim, we findn∈N

such that ifr,qnwithrq≡(m), then

d(xr,xq)≤

ε

. ()

Sincelimn→∞d(xn,xn+) = , we also findn∈Nsuch that

d(xn,xn+)≤

ε

m ()

for anynn. Suppose thatr,s≥max{n,n}ands>r. Then there existsk∈ {, , . . . ,m}

such thatsrk(m). Therefore, sr+ϕ≡(m) forϕ=mk+ . So, we have, for

j∈ {, . . . ,m},s+jr≡(m),

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By () and () and from the last inequality, we get

d(xr,xs)≤

ε +j×

εm

ε +m×

εm=ε.

This proves that{xn}is a Cauchy sequence. SinceYis closed in (X,d), then (Y,d) is also

complete, there existsxY=mi=Xisuch thatlimn→∞xn=xin (Y,d). In what follows,

we prove thatxis a fixed point ofT. In fact, sincelimn→∞xn=xand, asY= m

i=Xiis a

cyclic representation ofYwith respect toT, the sequence{xn}has infinite terms in each Xifori∈ {, , . . . ,m}. Suppose thatxXi,TxXi+ and we take a subsequencexnk of

{xn}withxnkXi–. Now from (iii) we haveη(x)≥randξ(x)≤r. By using the contractive

condition, we can obtain

r+rd(Tx,Txnk)η(x)η(x nk) +r

d(Tx,Txnk)

ξ(x)ξ(xnk) +r [

[d(x,Txnk)+d(xnk,Tx)]–ψ(d(x,Txnk),d(xnk,Tx))]

r+r[

[d(x,Txnk)+d(xnk,Tx)]–ψ(d(x,Txnk),d(xnk,Tx))], ()

which implies

d(Tx,xnk+)≤

 

d(x,xnk+) +d(xnk,Tx)

ψd(x,xnk+),d(xnk,Tx)

.

Passing to the limit ask→ ∞in the last inequality, we get

d(x,Tx)≤ 

d(x,Tx) –ψ

,d(x,Tx)≤

d(x,Tx),

which impliesd(x,Tx) = ,i.e.,x=Tx. Finally, to prove the uniqueness of the fixed point, suppose thatx,yFix(T)Y such thatη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r, wherex=y. The

cyclic character ofTand the fact thatx,yXare fixed points ofTimply thatx,y∈mi=Xi.

Suppose thatx=y. That is,d(x,y) > . Using the contractive condition, we obtain

r+rd(Tx,Ty)≤η(x)η(y) +rd(Tx,Ty)

ξ(x)ξ(y) +r[[d(x,Ty)+d(y,Tx)]–ψ(d(x,Ty),d(y,Tx))]

r+r[[d(x,Ty)+d(y,Tx)]–ψ(d(x,Ty),d(y,Tx))],

which implies

d(x,y)≤d(x,y) –ψd(x,y),d(x,y).

Hence, we obtainψ(d(x,y),d(x,y)) = , which impliesd(x,y) = , that is,x=ya

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3 Existence of solutions of an integral equation

ForT> , we denote byX=C([,T]) the set of real continuous functions on [,T]. We endowXwith the metric

d∞(u,v) =uv∞ for allu,vX.

It is evident that (X,d∞) is a complete metric space. Consider the integral equation

u(t) =

T

G(t,s)fs,u(s)ds for allt∈[,T], ()

() f: [,T]×R→RandG: [,T]×[,T]→[,∞)are continuous functions. () Let(α,β)∈X,(α

,β)∈Rsuch that

α≤α(t)≤β(t)≤β for allt∈[,T]. ()

Assume that for allt∈[,T], we have

α(t)≤

T

G(t,s)fs,β(s)ds ()

and

β(t)≥

T

G(t,s)fs,α(s)ds. ()

Let for alls∈[,T],f(s,·)be a decreasing function, that is,

x,y∈R, xyf(s,x)≤f(s,y). ()

LetZ:={uX:uβ} ∪ {uX:uα}. There exist≤r< andθ,π:Z→R

such that ifθ(x)≥andθ(y)≥with (xβandyα) or (xαandyβ),

then for everys∈[,T], we have

fs,x(s)–fs,y(s)≤r|π(y)|

x(s) –Ty(s)+y(s) –Tx(s). ()

() Assume that

(y)G(t,s)ds ∞≤

 ()

for allxZ, whereθ(x)≥. Suppose that

θ(x)≥ ⇒ θ(Tx)≥ forx∈ {uX:uβ} ∪ {uX:uα}. ()

() If{xn}is a sequence in{uX:uβ} ∪ {uX:uα}such thatθ(xn)≥for all n∈Nandxnxasn→ ∞, thenθ(x)≥.

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Theorem . Under assumptions()-(), integral equation()has a solution in {uC([,T]) :α(t)≤u(t)≤β(t)for all t∈[,T]}.

Proof Define the closed subsets ofX,AandAby

A={uX:uβ}

and

A={uX:uα}.

Also define the mappingT:XXby

Tu(t) =

T

G(t,s)fs,u(s)ds for allt∈[,T].

Let us prove that

T(A)⊆A and T(A)⊆A. ()

SupposeuA, that is,

u(s)≤β(s) for alls∈[,T].

Applying condition (), sinceG(t,s)≥ for allt,s∈[,T], we obtain that

G(t,s)fs,u(s)≥G(t,s)fs,β(s) for allt,s∈[,T].

The above inequality with condition () imply that

T

G(t,s)fs,u(s)ds

T

G(t,s)fs,β(s)dsα(t)

for allt∈[,T]. Then we haveTuA.

Similarly, letuA, that is,

u(s)≥α(s) for alls∈[,T].

Using condition (), sinceG(t,s)≥ for allt,s∈[,T], we obtain that

G(t,s)fs,u(s)≤G(t,s)fs,α(s) for allt,s∈[,T].

The above inequality with condition () imply that

T

G(t,s)fs,u(s)ds

T

G(t,s)fs,α(s)dsβ(t)

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Now, let (u,v)∈A×A, that is, for allt∈[,T],

u(t)≤β(t), v(t)≥α(t).

This implies from condition () that for allt∈[,T],

u(t)≤β, v(t)≥α.

Now, by conditions () and (), we have, for alls∈[,T],

Tu(t) –Tv(t)=

T

G(t,s)fs,u(s)–fs,v(s)ds

T

G(t,s)fs,u(s)–fs,v(s)ds

T

G(t,s)r|π(y)|

u(s) –Tv(s)+v(s) –Tu(s)ds

r

uTv∞+vTu

T

π(v)G(t,s)ds

r

uTv∞+vTu∞,

which implies

TuTv∞≤ r

uTv∞+vTu∞.

Defineη,ξ:Z→[,∞) byη(u) =,, otherwiseθ(u)≥, andξ(u) = . Further,ψ(t,t) =(–r) (t+t).

Hence,

η(u)η(v)d∞(Tu,Tv)≤ r

d∞(u,Tv) +d∞(v,Tu)

for all (u,v)∈A×A. By a similar method, we can show that the above inequality holds

if (u,v)∈A×A. Now, all the conditions of Theorem . hold andThas a fixed pointz*

in

A∩A=

uC[,T]:αu(t)≤βfor allt∈[,T].

That is,z*A

∩Ais the solution to ().

Example . In this example, we denote byX=C([, ]) the set of real continuous func-tions on [, ]. We endowXwith the metric

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Consider the following continuous functions:

f(t,x) =

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

xifx(–, ),

 ifx∈[, ],

x–  ifx(, ),

 ifx∈[,√e– ], x+  –e ifx(e– ,)

for allt∈[, ]

and

G(t,s) = t  +te

s for alls,t[, ]×[, ].

Letα(t) =  andβ(t) = . Then, for (α,β) = (, )∈R, we have

α≤α(t)≤β(t)≤β;

α(t) = ≤

G(t,s)fs,β(s)ds= 

and

β(t) = ≥

G(t,s)fs,α(s)ds= .

Also,Z:={uX:uβ} ∪ {uX:uα}=X. Defineθ,π:Z→Rby

θx(t)=

 if ≤x(t)≤ for allt∈[, ]

–, otherwise and π(x) =

e– .

Clearly,θ()≥. Also, ifθ(x(t))≥, then ≤x(t)≤. On the other hand,

Tu(t) =

G(t,s)fs,u(s)ds= 

for all ≤u(t)≤. That is,θ(Tx(t))≥. Hence,θ(x)≥ impliesθ(Tx)≥.

Assumeθ(x(s))≥ andθ(y(s))≥ with (xβandyα) or (xαandyβ).

Thus, ≤x(s)≤ and ≤y(s)≤, which impliesf(s,x(s)) =f(s,y(s)) = . That is,

fs,x(s)–fs,y(s)= ≤r|π(y)|

x(s) –Ty(s)+y(s) –Tx(s)

for alls∈[, ], where ≤r< . Further,

π(y)G(t,s)ds=

e– 

t

 +te

sds= t

 +t≤,

and so

(y)G(t,s)ds ∞≤

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Assume that{xn}is a sequence inXsuch thatθ(xn)≥ for alln∈Nandxnxasn→ ∞.

Then ≤xn≤. So, ≤x≤. That is,θ(x)≥.

Therefore, all of the conditions of Theorem . are satisfied. Then the integral equation

u(t) = t  +t

esfs,u(s)ds

has a solution in{uC([, ]) : ≤u(t)≤ for allt∈[, ]}. Here,u(t) =  is a solution. But if we chosex(t) =  andy(t) =

e– , thenf(s,x

(s)) =  andf(s,y(s)) = . That

is,

fs,x(s)

fs,y(s)= .

Also,

lnx(s) –y(s) 

+ =

ln –√e– + =lne= ,

and so

fs,x(s)

fs,y(s)=  >  =

lnx(s) –y(s) 

+ .

That is, Theorem . of [] cannot be applied to this example.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

Author details

1Department of Mathematics, Cumhuriyet University, Sivas, Turkey.2Department of Mathematics, Atilim University, ˙Incek,

Ankara 06836, Turkey. 3Faculty of Sciences and Mathematics, University of Nis, Visegradska 33, Nis, 18000, Serbia.4Young

Researchers and Elite Club, Rasht Branch, Islamic Azad University, Rasht, Iran.

Acknowledgements

The authors thank the anonymous referees for their remarkable comments, suggestions and ideas that helped to improve this paper. The third author (V Rakocevi´c) is supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.

Received: 14 May 2013 Accepted: 18 October 2013 Published:11 Nov 2013 References

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