Volume 2010, Article ID 916064,14pages doi:10.1155/2010/916064
Research Article
Contractive-Like Mapping Principles in Ordered Metric Spaces and Application to Ordinary
Differential Equations
J. Caballero, J. Harjani, and K. Sadarangani
Departamento de Matem´aticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain
Correspondence should be addressed to K. Sadarangani,[email protected] Received 25 November 2009; Revised 10 March 2010; Accepted 30 March 2010 Academic Editor: Tomonari Suzuki
Copyrightq2010 J. Caballero et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The purpose of this paper is to present a fixed point theorem for generalized contractions in partially ordered complete metric spaces. We also present an application to first-order ordinary differential equations.
1. Introduction
Existence of fixed point in partially ordered sets has been considered recently in 1–17.
Tarski’s theorem is used in 9 to show the existence of solutions for fuzzy equations and in11to prove existence theorems for fuzzy differential equations. In2,6,7,10,13some applications to ordinary differential equations and to matrix equations are presented. In3–
5,17some fixed point theorems are proved for a mixed monotone mapping in a metric space endowed with partial order and the authors apply their results to problems of existence and uniqueness of solutions for some boundary value problems.
In the context of ordered metric spaces, the usual contraction is weakened but at the expense that the operator is monotone. The main tool in the proof of the results in this context combines the ideas in the contraction principle with those in the monotone iterative technique 18.
LetSdenote the class of the class of the functionsβ :0,∞ → 0,1which satisfies the condition
βtn−→1⇒tn−→0. 1.1
In19the following generalization of Banach’s contraction principle appears.
Theorem 1.1. LetX, dbe a complete metric space and letT :X → Xbe a mapping satisfying d
Tx, Ty
≤β d
x, y
·d x, y
, forx, y∈X, 1.2
whereβ∈S. ThenThas a unique fixed pointz∈Xand{Tnx}converges tozfor eachx∈X.
Recently, in2the authors prove a version ofTheorem 1.1in the context of ordered complete metric spaces. More precisely, they prove the following result.
Theorem 1.2. LetX,≤be a partially ordered set and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetT :X → Xbe a nondecreasing mapping such that
d
Tx, Ty
≤β d
x, y
·d x, y
, forx, y∈X withx≤y, 1.3 whereβ∈S. Assume that eitherTis continuous orXsatisfies the following condition:
if{xn}is a nondecreasing sequence in X such that xn−→x, then xn≤x ∀n∈N. 1.4 Besides, suppose that for eachx, y ∈ Xthere existsz ∈ Xwhich is comparable toxandy. If there existsx0 ∈Xwithx0≤Tx0, thenThas a unique fixed point.
The purpose of this paper is to generalizeTheorem 1.2with the help of the altering functions.
We recall the definition of such functions.
Definition 1.3. An altering function is a function ψ : 0,∞ → 0,∞ which satisfies the following.
aψis continuous and nondecreasing.
bψt 0 if and only ift0.
Altering functions have been used in metric fixed point theory in recent papers20–
22.
In7the authors use these functions and they prove some fixed point theorems in ordered metric spaces.
2. Fixed Point Theorems
Definition 2.1. IfX,≤is a partially ordered set andT :X → X, we say thatT is monotone nondecreasing if forx, y∈X,
x≤y⇒Tx≤T y
. 2.1
This definition coincides with the notion of a nondecreasing function in the caseX R and≤represents the usual total order inR.
In the sequel, we prove the main result of the paper.
Theorem 2.2. LetX,≤be a partially ordered set and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. LetT :X → X be a continuous and nondecreasing mapping such that
ψ d
Tx, T
y
≤β d
x, y
·ψ d
x, y
, forx≥y, 2.2
whereψis an altering function andβ∈S.
If there existx0∈Xwithx0 ≤Tx0, thenThas a fixed point.
Proof. IfTx0 x0, then the proof is finished. Suppose thatx0< Tx0. Sincex0 < Tx0and Tis a nondecreasing mapping, we obtain by induction that
x0< Tx0≤T2x0≤T3x0≤ · · · ≤Tnx0≤Tn1x0≤ · · ·. 2.3
Putxn1 Txn. Taking into account thatβ∈Sand sincexn≤xn1for eachn∈N,then, by 2.2, we get
ψdxn1, xn ψdTxn, Txn−1
≤βdxn, xn−1·ψdxn, xn−1
≤ψdxn, xn−1.
2.4
Using the fact thatψis nondecreasing, we have
dxn1, xn≤dxn, xn−1. 2.5
If there existsn0∈Nsuch thatdxn0, xn0−1 0, thenxn0 Txn0−1 xn0−1andxn0−1is a fixed point and the proof is finished. In another case, suppose that dxn1, xn/0 for alln ∈ N.
Then, taking into account2.5, the sequence{dxn1, xn}is decreasing and bounded below, so
nlim→ ∞dxn1, xn r ≥0 2.6
Assume thatr >0.
Then, from2.4, we have
ψdxn1, xn
ψdxn, xn−1 ≤βdxn, xn−1<1. 2.7
Lettingn → ∞in the last inequality and by the fact thatψis an altering function, we get 1≤ lim
n→ ∞βdxn, xn−1≤1 2.8
and, consequently, limn→ ∞βdxn, xn−1 1.Sinceβ ∈ Sthis implies that limn→ ∞dxn1, xn 0 and this contradicts our assumption thatr >0.Hence,
nlim→ ∞dxn1, xn 0. 2.9
In what follows, we will show that{xn}is a Cauchy sequence.
Suppose that{xn}is not a Cauchy sequence. Then, there exists >0 for which we can find subsequences{xmk}and{xnk}of{xn}withnk> mk> ksuch that
d
xnk, xmk
≥. 2.10
Further, corresponding to mk, we can choosenk in such a way that it is the smallest integer withnk> mkand satisfying2.10, then
d
xnk−1, xmk
< . 2.11
Using2.10,2.11, and the triangular inequality, we have ≤d
xnk, xmk
≤d
xnk, xnk−1 d
xnk−1, xmk
< d
xnk, xnk−1 .
2.12
Lettingk → ∞and using2.9, we get
k→ ∞limd
xnk, xmk
. 2.13
Again, the triangular inequality gives us d
xnk, xmk
≤d
xnk, xnk−1 d
xnk−1, xmk−1 d
xmk−1, xmk , d
xnk−1, xmk−1
≤d
xnk−1, xnk d
xnk, xmk d
xmk, xmk−1
. 2.14
Lettingk → ∞in the above two inequalities and using2.9and2.13, we have
klim→ ∞d
xnk−1, xmk−1
. 2.15
Asnk> mkandxnk−1 ≥xmk−1, by2.2, we obtain ψ
d
xnk, xmk ψ
d
Txnk−1, Txmk−1
≤β d
xnk−1, xmk−1
·ψ d
xnk−1, xmk−1
≤ψ d
xnk−1, xmk−1 .
2.16
Taking into account 2.13 and 2.15 and the fact thatψ is continuous and letting k → ∞in2.16, we get
ψ≤ lim
k→ ∞β d
xnk−1, xmk−1
·ψ≤ψ. 2.17
Asψis an altering function,ψ>0, the last inequality gives us
k→ ∞limβ d
xnk−1, xmk−1
1. 2.18
Sinceβ∈S, this means that
klim→ ∞d
xnk−1, xmk−1
0. 2.19
This fact and2.15give us0 which is a contradiction.
This shows that{xn}is a Cauchy sequence.
SinceX, dis a complete metric space, there exists z ∈ X such that limn→ ∞xn z.
Moreover, the continuity ofT implies that z lim
n→ ∞Txn lim
n→ ∞xn1 Tz, 2.20
and this proves thatzis a fixed point.
In what follows, we prove that Theorem 2.2 is still valid for T not necessarily continuous, assuming the following hypothesis inXwhich appears in10, Theorem 1:
if xnis a nondecreasing sequence inX such thatxn −→x,then xn≤x ∀n∈N. 2.21 Theorem 2.3. Let X,≤ be a partially ordered set and suppose that there exists a metricd in X such thatX, d is a complete metric space. Assume thatX satisfies2.21. LetT : X → X be a nondecreasing mapping such that
ψ d
Tx, T
y
≤β d
x, y
·ψ d
x, y
, forx≥y, 2.22
whereψis an altering function andβ∈S. If there existsx0∈Xwithx0≤Tx0, thenT has a fixed point.
Proof. Following the proof ofTheorem 2.2, we only have to check thatTz z. Asxnis a nondecreasing sequence inXand limn→ ∞xnzthen, by2.21, we havexn≤zfor alln∈N, and, consequently,
ψ d
xn1, fz
ψdTxn, Tz≤βdxn, z·ψdxn, z≤ψdxn, z. 2.23
Lettingn → ∞and using the continuity ofψ, we have
0≤ψdz, Tz≤ψ0 0, 2.24
or, equivalently,
ψdz, Tz 0. 2.25
Asψis an altering function, this gives usdz, Tz 0 and, thus,Tz z.
Now, we present an example where it can be appreciated that the hypotheses in Theorems2.2and2.3do not guarantee uniqueness of the fixed point. This example appears in10.
LetX{1,0,0,1} ⊂R2and consider the usual order x, y
≤z, t⇐⇒x≤z, y≤t. 2.26 X,≤is a partially ordered set whose different elements are not comparable. Besides,X, d2 is a complete metric space considering d2 as the Euclidean distance. The identity map Tx, y x, yis trivially continuous and nondecreasing and condition2.2ofTheorem 2.2 is satisfied since elements inXare only comparable to themselves. Moreover,1,0≤T1,0 1,0andThas two fixed points inX.
In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems2.2and2.3. This condition appears in16and says that
forx, y∈X, there exists a lower bound or an upper bound. 2.27 In10it is proved that condition2.27is equivalent to
forx, y∈X, there existsz∈X which is comparable to xand y. 2.28 Theorem 2.4. Adding condition 2.28 to the hypotheses of Theorem 2.2 (resp.,Theorem 2.3), we obtain uniqueness of the fixed point off.
Proof. Suppose that there existy, z∈Xwhich are fixed points ofTandy /z. We distinguish two cases.
Case 1. If yand zare comparable, thenTny y and Tnz z are comparable for n 0,1,2, . . . .Using the contractive condition appearing in Theorem 2.2orTheorem 2.3and the fact thatβ∈S, we get
ψ d
y, z ψ
d Tn
y
, Tnz
≤β d
Tn−1 y
, Tn−1z
·ψ d
Tn−1 y
, Tn−1z
≤β d
y, z
·ψ d
y, z
< ψ d
y, z ,
2.29
which is a contradiction.
Case 2. Using condition2.28, there existsx∈X comparable toyandz. Monotonicity ofT implies thatTnxis comparable toTny yand toTnz z, forn0,1,2, . . . .Moreover, asβ∈S, we get
ψdz, Tnx ψdTnz, Tnx
≤β d
Tn−1z, Tn−1x
·ψ d
Tn−1z, Tn−1x β
d
z, Tn−1x
·ψ d
z, Tn−1x
≤ψ d
z, Tn−1x .
2.30
Sinceψis nondecreasing the above inequality gives us dz, Tnx≤d
z, Tn−1x
. 2.31
Thus, limn→ ∞dz, Tnx γ≥0.
Assume thatγ >0.
Taking into account thatψ is an altering function and lettingn → ∞in2.30, we obtain
ψ γ
≤ lim
n→ ∞β d
z, Tn−1x
·ψ γ
≤ψ γ
, 2.32
and this implies that limn→ ∞βdz, Tn−1x 1.
Sinceβ∈Sthen we get
nlim→ ∞d
z, Tn−1x
0, 2.33
and, consequently,γ 0, which is a contradiction.
Hence, limn→ ∞dz, Tnx 0.
Analogously, it can be proved that
nlim→ ∞d
y, Tnx
0. 2.34
Finally, as
d z, y
≤dz, Tnx d
Tnx, y
2.35 and taking limit, we obtaindz, y 0.
This finishes the proof.
Remark 2.5. Under the assumptions ofTheorem 2.4, it can be proved that for everyx ∈ X, limn→ ∞Tnx z, wherezis the fixed pointi.e., the operatorTis Picard.
In fact, forx∈Xandxcomparable tozthen using the same argument that is in Case 1ofTheorem 2.4can prove that limn→ ∞dz, Tnx 0 and, hence, limn→ ∞Tnx z.
Ifxis not comparable toz, we take thaty∈Xis comparable toxandz. Using a similar argument that is in Case2ofTheorem 2.4, we obtain
nlim→ ∞d z, Tn
y
0, lim
n→ ∞d
Tnx, Tn y
0. 2.36
Finally,
dz, Tnx≤d z, Tn
y d
Tn y
, Tnx
, 2.37
and taking limit asn → ∞, we obtain limn→ ∞dz, Tnx 0 or, equivalently, limn→ ∞Tnx z.
Remark 2.6. Notice that ifX,≤is totally ordered, condition2.28is obviously satisfied.
Remark 2.7. Considering ψ the identity mapping in Theorem 2.4, we obtain Theorem 1.2, being the main result of2.
3. Application to Ordinary Differential Equations
In this section we present an example where our results can be applied.
This example is inspired by10.
We study the existence of solution for the following first-order periodic problem ut ft, ut, t∈0, T,
u0 uT, 3.1
whereT >0 andf:I×R → Ris a continuous function.
Previously, we considered the spaceCI I 0, Tof continuous functions defined onI. Obviously, this space with the metric given by
d x, y
supxt−yt:t∈I
, forx, y∈ CI, 3.2
is a complete metric space.CIcan also be equipped with a partial order given by
x, y∈ CI, x≤y⇐⇒xt≤yt, for t∈I. 3.3
Clearly,CI,≤satisfies condition2.28, since forx, y∈ CIthe function max{x, y} ∈ CI.
Moreover, in10it is proved thatCI,≤with the above-mentioned metric satisfies condition2.21.
Now, letAdenote the class of functionsφ:0,∞ → 0,∞satisfying the following.
iφis nondecreasing.
iiφx< x, forx >0.
iiiβx φx/x∈S,
whereSis the class of functions defined inSection 1.
Examples of such functions are φt μ·t, with 0 ≤ μ < 1,φt t/1t, and φt ln1t.
Recall now the following definition
Definition 3.1. A lower solution for3.1is a functionα∈ C1Isuch that αt≤ft, αt fort∈I,
α0≤αT. 3.4
Now, we present the following theorem about the existence of solution for problem 3.1in presence of a lower solution.
Theorem 3.2. Consider problem3.1withf :I×R → Rcontinuous and suppose that there exist λ, α >0 with
α≤ 2λ
eλT−1 T
eλT1 1/2
, 3.5
such that forx, y∈Rwithx≤y
0≤f t, y
λy−
ft, x λx
≤α y−x
φ y−x
, 3.6
where φ ∈ A. Then the existence of a lower solution for3.1provides the existence of a unique solution of 3.1.
Proof. Problem3.1can be written as
ut λut ft, ut λut fort∈I 0, T,
u0 uT. 3.7
This problem is equivalent to the integral equation
ut T
0
Gt, s
fs, us λus
ds, 3.8
whereGt, sis the Green function given by
Gt, s
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
eλTs−t
eλT−1 0≤s < t≤T, eλs−t
eλT−1 0≤t < s≤T.
3.9
DefineF :CI → CIby
Fut T
0
Gt, s
fs, us λus
ds. 3.10
Notice that ifu∈ CIis a fixed point ofF, thenu∈ C1Iis a solution of3.1.
In the sequel, we check that hypotheses inTheorem 2.4are satisfied.
The mappingFis nondecreasing since, by hypothesis, foru≥v
ft, u λu≥ft, v λv, 3.11
and this implies, taking into account thatGt, s>0 fort, s∈I×I, that Fut
T
0
Gt, s
fs, us λus ds≥
T
0
Gt, s
fs, vs λvs
ds Fvt.
3.12
Besides, foru≥v, we have dFu, Fv sup
t∈I |Fut−Fvt|
sup
t∈I Fut−Fvt
sup
t∈I
T
0
Gt, s
fs, us λus−fs, vs−λvs ds
≤sup
t∈I
T
0
Gt, sα
us−vsφus−vsds
αsup
t∈I
T
0
Gt, s
us−vsφus−vsds.
3.13
Using the Cauchy-Schwarz inequality in the last integral, we get T
0
Gt, sα
us−vsφus−vsds
≤ T
0
Gt, s2ds
1/2 T 0
us−vsφus−vsds
1/2
.
3.14
The first integral gives us T
0
Gt, s2ds t
0
Gt, s2ds T
t
Gt, s2ds
t
0
e2λTs−t eλT−12ds
T
t
e2λs−t eλT−12ds 1
2λ
eλT−12
e2λT−1
eλT1 2λ
eλT−1.
3.15
Asφis nondecreasing, the second integral in3.14can be estimated by T
0
us−vsφus−vsds≤T· u−v ·φu−v du, v·φdu, v·T. 3.16
Taking into account3.14,3.15, and3.16, from3.13we get
dFu, Fv≤α· eλT1 2λ
eλT−1 1/2
·du, v1/2·φdu, v1/2·T1/2
α· T·
eλT1 2λ
eλT−1 1/2
·du, v1/2·φdu, v1/2.
3.17
Sinceα≤2λeλT−1/T·eλT11/2, the last inequality gives us
dFu, Fv≤du, v1/2·φdu, v1/2 3.18
or, equivalently,
dFu, Fv≤du, v·
φdu, v
du, v 1/2
. 3.19
This implies that
dFu, Fv2≤ φdu, v
du, v ·du, v2. 3.20
Puttingψx x2, which is an altering function, andβφx/x∈Sbecauseφ∈ A, we have ψdFu, Fv≤βdu, v·ψdu, v foru≥v. 3.21 This proves that the operatorFsatisfies condition2.2ofTheorem 2.2.
Finally, lettingαtbe a lower solution for3.1, we claim thatα≤Fα.
In fact,
αt λαt≤ft, αt λαt, fort∈I. 3.22
Multiplying byeλt,
αteλt
≤
ft, αt λαt
eλt, fort∈I, 3.23
and this gives us
αteλt ≤α0 t
0
fs, αs λαs
eλsds, fort∈I. 3.24
Asα0≤αT, the last inequality implies that
α0eλT ≤αTeλT≤α0 T
0
fs, αs λαs
eλsds 3.25
and so
α0≤ T
0
eλs eλT−1
fs, αs λαs
ds. 3.26
This and3.24give us
αteλt ≤ t
0
eλTs eλT−1
fs, αs λαs ds
T
t
eλs eλT−1
fs, αs λαs
ds, 3.27
and, consequently,
αt≤ t
0
eλTs−t eλT−1
fs, αs λαs ds
T
t
eλs−t eλT−1
fs, αs λαs ds
T
0
Gt, s
fs, αs λαs ds Fαt, fort∈I.
3.28
Finally,Theorem 2.4gives thatFhas a unique fixed point.
Remark 3.3. Notice that ifφ ∈ A, thenϕx
xφx ∈ A. In fact, as φ ∈ A, thenφ is nondecreasing and, consequently,ϕis also nondecreasing.
Moreover, asφx< x, thenxφx< x2, and, thus,
xφx< x.
Finally, asϕx/x
xφx/x
φx/x, and asβx φx/x∈S, then it is easily seen thatϕx/x∈S.
Example 3.4. Considerφ0:0,∞ → 0,∞given by
φ0t
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
0, 0≤t≤2, 1
2t−1, 2< t≤4, 1
4t, 4< t.
3.29
It is easily seen thatφ0∈ A. Taking into accountRemark 3.3,φx
xφ0x∈ A.
Now, we consider problem3.1with f : I ×R → Rcontinuous and suppose that there existλ, α >0 with
α≤ 2λ
eλT−1 T
eλT1 1/2
3.30
such that forx, y∈Rwithy≥x 0≤f
t, y
λy−
ft, x λx
≤α y−x
φ0
y−x αφ
y−x
, 3.31
whereφ0is the function above mentioned.
This example can be treated by ourTheorem 3.2but it cannot be covered by the results of6becauseψx x−φxis not increasing.
Acknowledgments
This research was partially supported by “Ministerio de Educaci ´on y Ciencia”, Project MTM 2007/65706. This work is dedicated to Professor W. Takahashi on the occasion of his retirement.
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