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
3and 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
cyclic representation ofXwith respect toT if
T(X)⊂X, . . . , T(Xm–)⊂Xm, T(Xm)⊂Xm+,
whereXm+=X.
Definition .[] LetT:X→X,r> andη,ξ:X→[, +∞) be two functions. We say thatTisr-(η,ξ)-admissible if
(i) η(x)≥rfor somex∈Ximpliesη(Tx)≥r, (ii) ξ(x)≤rfor somex∈Ximpliesξ(Tx)≤r.
Definition . Let (X,d) be a metric space andT:Y→Ybe a self-mapping, whereY=
m
i=Xiis a cyclic representation ofY with respect toT. Letη,ξ :Y →[, +∞) be two
functions. An operatorT:Y→Yis 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 allx∈Xiandy∈Xi+, 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 allx∈X
iandy∈Xi+, 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:Y→Y 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∈Nandxn→x
asn→ ∞,thenη(x)≥randξ(x)≤r.
Then T has a fixed point x∈ni=Xi.Moreover,ifη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r for all x,y∈Fix(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 someisuch thatx∈Xi. NowT(Xi)⊆
Txn=xn+, wherexn∈Xin. Hence, forn≥, there existsin∈ {, , . . . ,m}such thatxn∈Xin
andxn+∈Xin+. In casexn=xn+for somen= , , , . . . , then it is clear thatxnis 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 sodn≤dn–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,q≥nwithr–q≡(m), then
Suppose, to the contrary, that there existsε> such that for anyn∈N, we can find rn>qn≥nwithrn–qn≡(m) satisfying
d(xqn,xrn)≥ε. ()
Now, we taken> m. Then, corresponding toqn≥n, one can choosernin such a way that
it is the smallest integer withrn>qnsatisfyingrn–qn≡(m) andd(xqn,xrn)≥ε. Therefore, d(xqn,xrn–m) <ε. By using the triangular inequality,
ε≤d(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,
ε≤d(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
Lettingn→ ∞in () and (), we conclude that
lim
n→∞d(xrn,xqn+) =ε. ()
Sincexqnandxrnlie in different adjacently labeled setsXiandXi+for certain ≤i≤m,
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
ε≤ε–ψ(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,q≥nwithr–q≡(m), then
d(xr,xq)≤
ε
. ()
Sincelimn→∞d(xn,xn+) = , we also findn∈Nsuch that
d(xn,xn+)≤
ε
m ()
for anyn≥n. Suppose thatr,s≥max{n,n}ands>r. Then, there existsk∈ {, , . . . ,m}
such thats–r≡k(m). Therefore, s–r+ϕ≡(m) forϕ=m–k+ . So, we have, for
j∈ {, . . . ,m},s+j–r≡(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×
ε m≤
ε +m×
ε m=ε.
This proves that{xn}is a Cauchy sequence. SinceYis closed in (X,d), then (Y,d) is also
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 thatx∈Xi,Tx∈Xi+ and we take a subsequencexnk of
{xn}withxnk∈Xi–. 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,y∈Fix(T)Y such thatη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r, wherex=y. The
cyclic character ofTand the fact thatx,y∈Xare fixed points ofTimply thatx,y∈mi=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
Example . LetX=Rwith the metric d(x,y) =|x–y|for all x,y∈X. SupposeA=
(–∞, ] andA= [,∞) andY=
i=Ai. DefineT:Y→Yandη,ξ:Y→[,∞) by
Tx= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x+ √
x+ ifx∈(–∞, –],
sinx
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⊆Aand
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)≤ andxn→x
asn→ ∞. Thenxn∈[–, ]. So,x∈[–, ],i.e.,η(x)≥ andξ(x)≤.
Letx∈Aandy∈A. Now, ifx∈/[–, ] ory∈/[, ], thenη(x)η(y) = . Also, ifx∈[–, ]
andy∈[, ], thend(Tx,Ty) = . That is,η(x)η(y)d(Tx,Ty) = for allx∈Aand ally∈A.
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 allx∈Aandy∈A. 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:Y→Y 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∈Nandxn→x
asn→ ∞,thenη(x)≥randξ(x)≤r.
Then T has a fixed point x∈ni=Xi.Moreover,ifη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r for all x,y∈Fix(T)Y,then T has a unique fixed point.
Proof By a similar method as in the proof of Theorem ., we have
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 sodn≤dn–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,q≥nwithr–q≡(m), then
d(xr,xq) <ε.
Suppose, to the contrary, that there exists ε> such that for anyn∈Nwe can find
rn>qn≥nwithrn–qn≡(m) satisfying
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 ≤i≤m,
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
ε≤ε–ψ(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,q≥nwithr–q≡(m), then
d(xr,xq)≤
ε
. ()
Sincelimn→∞d(xn,xn+) = , we also findn∈Nsuch that
d(xn,xn+)≤
ε
m ()
for anyn≥n. Suppose thatr,s≥max{n,n}ands>r. Then there existsk∈ {, , . . . ,m}
such thats–r≡k(m). Therefore, s–r+ϕ≡(m) forϕ=m–k+ . So, we have, for
j∈ {, . . . ,m},s+j–r≡(m),
By () and () and from the last inequality, we get
d(xr,xs)≤
ε +j×
ε m
≤ε +m×
ε m=ε.
This proves that{xn}is a Cauchy sequence. SinceYis closed in (X,d), then (Y,d) is also
complete, there existsx∈Y=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 thatx∈Xi,Tx∈Xi+ and we take a subsequencexnk of
{xn}withxnk∈Xi–. 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,y∈Fix(T)Y such thatη(x)≥r,η(y)≥r,ξ(x)≤r,ξ(y)≤r, wherex=y. The
cyclic character ofTand the fact thatx,y∈Xare 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
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) =u–v∞ for allu,v∈X.
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)∈Rsuch 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, x≥y ⇒ f(s,x)≤f(s,y). ()
LetZ:={u∈X:u≤β} ∪ {u∈X:u≥α}. There exist≤r< andθ,π:Z→R
such that ifθ(x)≥andθ(y)≥with (x≤βandy≥α) or (x≥αandy≤β),
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
Tπ(y)G(t,s)ds ∞≤
()
for allx∈Z, whereθ(x)≥. Suppose that
θ(x)≥ ⇒ θ(Tx)≥ forx∈ {u∈X:u≤β} ∪ {u∈X:u≥α}. ()
() If{xn}is a sequence in{u∈X:u≤β} ∪ {u∈X:u≥α}such thatθ(xn)≥for all n∈Nandxn→xasn→ ∞, thenθ(x)≥.
Theorem . Under assumptions()-(), integral equation()has a solution in {u∈ C([,T]) :α(t)≤u(t)≤β(t)for all t∈[,T]}.
Proof Define the closed subsets ofX,AandAby
A={u∈X:u≤β}
and
A={u∈X:u≥α}.
Also define the mappingT:X→Xby
Tu(t) =
T
G(t,s)fs,u(s)ds for allt∈[,T].
Let us prove that
T(A)⊆A and T(A)⊆A. ()
Supposeu∈A, 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)ds≥α(t)
for allt∈[,T]. Then we haveTu∈A.
Similarly, letu∈A, 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)ds≤β(t)
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
u–Tv∞+v–Tu∞
T
π(v)G(t,s)ds ∞
≤ r
u–Tv∞+v–Tu∞,
which implies
Tu–Tv∞≤ r
u–Tv∞+v–Tu∞.
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=
u∈C[,T]:α≤u(t)≤βfor allt∈[,T].
That is,z*∈A
∩Ais the solution to ().
Example . In this example, we denote byX=C([, ]) the set of real continuous func-tions on [, ]. We endowXwith the metric
Consider the following continuous functions:
f(t,x) =
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x ifx∈(–∞, ),
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:={u∈X:u≤β} ∪ {u∈X:u≥α}=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 (x≤βandy≥α) or (x≥αandy≤β).
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
Tπ(y)G(t,s)ds ∞≤
Assume that{xn}is a sequence inXsuch thatθ(xn)≥ for alln∈Nandxn→xasn→ ∞.
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{u∈C([, ]) : ≤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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