Research ArticleFixed Point Theorems for a Class of 120572-Admissible Contractionsand Applications to Boundary Value Problem

Hamed H Alsulami1 Selma Guumllyaz2 Erdal KarapJnar13 and EncJ M Erhan3

1 Nonlinear Analysis and Applied Mathematics (NAAM) Research Group King Abdulaziz University Jeddah Saudi Arabia2Department of Mathematics Cumhuriyet University Sivas Turkey3 Department of Mathematics Atilim University Incek 06836 Ankara Turkey

Correspondence should be addressed to Erdal Karapınar erdalkarapinaryahoocom

Received 15 May 2014 Accepted 12 June 2014 Published 3 July 2014

Academic Editor Jen-Chih Yao

Copyright copy 2014 Hamed H Alsulami et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

A class of 120572-admissible contractions defined via altering distance functions is introducedThe existence and uniqueness conditionsfor fixed points of such maps on complete metric spaces are investigated and related fixed point theorems are presentedThe resultsare reconsidered in the context of partially orderedmetric spaces and applied to boundary value problems for differential equationswith periodic boundary conditions

1 Introduction and Preliminaries

Recent developments in fixed point theory have shownthe significance of theoretical studies which are directlyapplicable in other areas In particular the problems relatedwith existence and uniqueness of solutions of integral anddifferential equations are of particular importance Differen-tial and integral equations govern the behaviour of variousreal-life problems for which the question of existence anduniqueness of solutions is crucial This fact motivates theintensive research activities in the area and the rapidlyincreasing number of publications [1ndash7]

The main goal of studies in fixed point theory is toimprove the contractive conditions imposed on themappingsunder consideration Altering distance functions defined byKhan et al [8] have been widely used for this reason bothalone or combined with other auxiliary functions

Definition 1 An altering distance function is a function 120595 [0 +infin) rarr [0 +infin) which satisfies the following

(1) 120595 is continuous and nondecreasing(2) 120595(119905) = 0 hArr 119905 = 0

Admissible mappings have been defined recently bySamet et al [6] and employed quite often in order to

generalize the results on various contractions We state nextthe definitions of 120572-admissible mapping and triangular 120572-admissible mappings

Definition 2 A mapping 119879 119883 rarr 119883 is called 120572-admissibleif for all 119909 119910 isin 119883 one has

120572 (119909 119910) ge 1 997904rArr 120572 (119879119909 119879119910) ge 1 (1)

where 120572 119883 times 119883 rarr [0infin) is a given function

Definition 3 A mapping 119879 119883 rarr 119883 is called triangular 120572-admissible if it is 120572-admissible and satisfies

120572 (119909 119910) ge 1

120572 (119910 119911) ge 1997904rArr 120572 (119909 119911) ge 1 (2)

where119909 119910 119911 isin 119883 and120572 119883times119883 rarr [0infin) is a given function

Inspired by this definition we define the followingweakercondition which proves to be sufficient in the forthcomingdiscussions

Definition 4 A mapping 119879 119883 rarr 119883 is said to be weaktriangular 120572-admissible if it is 120572-admissible and satisfies

120572 (119909 119879119909) ge 1 997904rArr 120572 (119909 1198792119909) ge 1 (3)

where 119909 isin 119883 and 120572 119883 times 119883 rarr [0infin) is a given function

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 187031 10 pageshttpdxdoiorg1011552014187031

2 Abstract and Applied Analysis

Weak triangular 120572-admissible mappings satisfy a prop-erty stated in the following lemma and the proof of whicheasily follows from the definition and can be found in [9]

Lemma 5 (see [9]) Let 119879 119883 rarr 119883 be a weak triangular120572-admissible mapping Assume that there exists 119909

0isin 119883 such

that 120572(1199090 1198791199090) ge 1 If 119909

119899= 1198791198991199090 then 120572(119909

119898 119909119899) ge 1 for all

119898 119899 isin N with119898 lt 119899

2 Existence and Uniqueness Theorems onComplete Metric Spaces

In this section we present our main results which includetheorems on existence and uniqueness of fixed points for aclass of weak triangular 120572-admissible mappings

First we define the following two classes of contractionswhich we are going to investigate in this section and through-out the paper

Definition 6 Let (119883 119889) be a metric space 120595 an alteringdistance function and 120601 [0 +infin) rarr [0 +infin) a continuousfunction satisfying 120595(119905) gt 120601(119905) for all 119905 gt 0

(I) A mapping 119879 119883 rarr 119883 belongs to class (I) if itsatisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (4)

where

119872(119909 119910) = max 119889 (119909 119910) 119889 (119909 119879119909) 119889 (119910 119879119910)

1

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(5)

(II) A mapping 119879 119883 rarr 119883 belongs to class (II) if itsatisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (6)

where

119873(119909 119910) = max 119889 (119909 119910) 12

[119889 (119909 119879119909) + 119889 (119910 119879119910)]

1

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(7)

Remark 7 Note that119873(119909 119910) le 119872(119909 119910) for all 119909 119910 isin 119883

Our first theorem gives conditions for the existence of afixed point for maps in class (I)

Theorem 8 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 120595 (119889 (119879x 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (8)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for

all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Let 1199090isin 119883 satisfy 120572(119909

0 1198791199090) ge 1 and define the

sequence 119909119899 as 119909119899= 119879119909119899minus1

for 119899 isin NIf1199091198990= 1199091198990+1

for some 1198990isin N then obviously119909

1198990= 1198791199091198990

is a fixed point of119879 Suppose that119889(119909119899 119909119899+1) gt 0 for all 119899 isin N

Note that due to the fact that 119879 is 120572-admissible and120572(1199090 1198791199090) = 120572(119909

0 1199091) ge 1 we deduce

120572 (1198791199090 1198791199091)

= 120572 (1199091 1199092) ge 1 997904rArr 120572 (119879119909

1 1198791199092)

= 120572 (1199092 1199093) ge 1 997904rArr sdot sdot sdot 997904rArr 120572 (119909

119899 119909119899+1) ge 1

(9)

Substituting 119909 = 119909119899and 119910 = 119909

119899minus1in (8) and using (9) we

get

120595 (119889 (119909119899+1 119909119899)) le 120572 (119909

119899 119909119899minus1) 120595 (119889 (119909

119899+1 119909119899))

= 120572 (119909119899 119909119899minus1) 120595 (119889 (119879119909

119899 119879119909119899minus1))

le 120601 (119872 (119909119899 119909119899minus1))

(10)

where

119872(119909119899 119909119899minus1)

= max 119889 (119909119899 119909119899minus1) 119889 (119909

119899 119879119909119899) 119889 (119909

119899minus1 119879119909119899minus1)

1

2

[119889 (119909119899 119879119909119899minus1) + 119889 (119909

119899minus1 119879119909119899)]

= max 119889 (119909119899 119909119899minus1) 119889 (119909

119899 119909119899+1) 119889 (119909

119899minus1 119909119899)

1

2

[119889 (119909119899 119909119899) + 119889 (119909

119899minus1 119909119899+1)]

= max119889 (119909119899 119909119899minus1) 119889 (119909

119899 119909119899+1)

119889 (119909119899minus1 119909119899+1)

2

(11)

Note that 119889(119909119899minus1 119909119899+1)2 le (12)[119889(119909

119899minus1 119909119899) + 119889(119909

119899 119909119899+1)]

is smaller than both 119889(119909119899minus1 119909119899) and 119889(119909

119899 119909119899+1) Then

119872(119909119899 119909119899minus1) can be either 119889(119909

119899minus1 119909119899) or 119889(119909

119899 119909119899+1) If

119872(119909119899 119909119899minus1) = 119889(119909

119899 119909119899+1) for some 119899 then the expression

(10) implies that

0 lt 120595 (119889 (119909119899+1 119909119899)) le 120601 (119889 (119909

119899 119909119899+1)) (12)

which contradicts the condition 120595(119905) gt 120601(119905) for 119905 gt 0 Hence119872(119909119899 119909119899minus1) = 119889(119909

119899 119909119899minus1) for all 119899 ge 1 and we have

0 lt 120595 (119889 (119909119899+1 119909119899)) le 120601 (119889 (119909

119899 119909119899minus1)) lt 120595 (119889 (119909

119899 119909119899minus1))

(13)

which results in

119889 (119909119899+1 119909119899) lt 119889 (119909

119899 119909119899minus1) (14)

Abstract and Applied Analysis 3

since 120595 is nondecreasing Thus we conclude that the non-negative sequence 119889(119909

119899+1 119909119899) is decreasing Therefore there

exists 119903 ge 0 such that lim119899rarrinfin

119889(119909119899+1 119909119899) = 119903 Let 119899 rarr infin in

(10) we get

120595 (119903) le 120601 (119903) (15)

By the hypothesis of the theorem since 120595(119905) gt 120601(119905) for all119905 gt 0 this inequality is possible only if 119903 = 0 and hence

lim119899rarrinfin

119889 (119909119899+1 119909119899) = 119903 = 0 (16)

Next we will prove that 119909119899 is a Cauchy sequence

Suppose on the contrary that 119909119899 is not Cauchy Then for

some 120576 gt 0 there exist subsequences 119909119898119896 and 119909

119899119896 of 119909

119899

such that

119899119896gt 119898119896gt 119896 119889 (119909

119899119896 119909119898119896) ge 120576 (17)

for all 119896 ge 1 where corresponding to each119898119896 we can choose

119899119896as the smallest integer with 119899

119896gt 119898119896for which (17) holds

Then

119889 (119909119899119896 119909119898119896minus1) lt 120576 (18)

Employing triangle inequality and making use of (17) and(18) we obtain

120576 le 119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

lt 119889 (119909119899119896 119909119899119896minus1) + 120576

(19)

Passing to limit as 119896 rarr infin and using (16) we get

lim119896rarrinfin

119889 (119909119899119896 119909119898119896) = 120576 (20)

From the triangular inequality we also have

119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896minus1)

le 119889 (119909119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896) + 119889 (119909

119898119896 119909119898119896minus1)

(21)

Letting 119896 rarr infin in the two inequalities above and using (16)and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896minus1) = 120576 (22)

In a similar way by using the triangular inequality we obtainthat

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896) le 119889 (119909

119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896)

(23)

Taking limit as 119896 rarr infin in the above two inequalities andregarding (16) and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896) = 120576 (24)

Furthermore the relations

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896 119909119898119896minus1) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1)

(25)

give

lim119896rarrinfin

119889 (119909119899119896 119909119898119896minus1) = 120576 (26)

by letting 119896 rarr infin and taking into account (16) and (20) Bythe definition of119872(119909 119910) and using limits found above we get

lim119896rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = 120576 (27)

Indeed since

119872(119909119899119896minus1 119909119898119896minus1)

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119879119909119899119896minus1)

119889 (119909119898119896minus1 119879119909119898119896minus1)

1

2

[119889 (119909119899119896minus1 119879119909119898119896minus1) + 119889 (119909

119898119896minus1 119879119909119899119896minus1)]

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119909119899119896) 119889 (119909

119898119896minus1 119909119898119896)

1

2

[119889 (119909119899119896minus1 119909119898119896) + 119889 (119909

119898119896minus1 119909119899119896)]

(28)

passing to the limit as 119896 rarr infin in (28) and using (16) (20)(22) (24) and (26) we obtain

lim119899rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = max 120576 0 0 1

2

[120576 + 120576] = 120576 (29)

Notice that since 119879 is weak triangular 120572-admissible wededuce from Lemma 5 that 120572(119909

119899119896minus1 119909119898119896minus1) ge 1 Therefore we

can apply condition (8) to 119909119899119896minus1

and 119909119898119896minus1

to obtain

0 lt 120595 (119889 (119909119899119896 119909119898119896))

le 120572 (119909119899119896minus1 119909119898119896minus1

) 120595 (119889 (119909119899119896 119909119898119896))

le 120601 (119872(119909119899119896minus1 119909119898119896minus1))

(30)

Letting 119896 rarr infin and taking into account (20) and (27) wehave

0 lt 120595 (120576) le 120601 (120576) (31)

However since 120595(119905) gt 120601(119905) for 119905 gt 0 we deduce that 120576 = 0which contradicts the assumption that 119909

119899 is not a Cauchy

4 Abstract and Applied Analysis

sequence Thus 119909119899 must be Cauchy Due to the fact that

(119883 119889) is a complete metric space there exists 119906 isin 119883 suchthat lim

119899rarrinfin119909119899= 119906 Finally the continuity of 119879 gives

119906 = lim119899rarrinfin

119909119899= lim119899rarrinfin

119879119909119899minus1= 119879119906 (32)

That is 119906 is a fixed point of119879 which completes the proof

One of the advantages of 120572-admissible mappings is thatthe continuity of the map is no longer required for theexistence of a fixed point provided that the space underconsideration has the following property

(A) If 119909119899 is a sequence in119883 such that

119909119899997888rarr 119909 120572 (119909

119899 119909119899+1)) ge 1 forall119899 isin N (33)

then there exists a subsequence 119909119899119896 of 119909

119899 for which

120572 (119909119899119896 119909) ge 1 forall119896 isin N (34)

Bearing this fact in mind we rewrite the statement ofTheorem 8 in the light of this property

Theorem 9 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910))

forall119909 119910 isin 119883

(35)

where 120595 is an altering distance function 120601 [0 +infin) rarr [0

+infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Following the proof of Theorem 8 it is clear that thesequence 119909

119899 defined by 119909

119899= 119879119909119899minus1

for 119899 isin N convergesto a limit 119906 isin 119883 The only thing which remains to show is119879119906 = 119906 Since lim

119899rarrinfin119909119899= 119906 then condition (A) implies

120572(119909119899119896 119906) ge 1 for all 119896 isin N Consequently inequality (35)

with 119909 = 119909119899119896and 119910 = 119906 becomes

120595 (119889 (119909119899119896+1 119879119906))

le 120572 (119909119899119896 119906) 120595 (119889 (119909

119899119896+1 119879119906))

= 120572 (119909119899119896 119906) 120595 (119889 (119879119909

119899119896 119879119906))

le 120601 (119872(119909119899119896 119906))

(36)

where

119872(119909119899119896 119906)

= max 119889 (119909119899119896 119906) 119889 (119909

119899119896 119909119899119896+1)

119889 (119906 119879119906)

1

2

[119889 (119909119899119896 119879119906) + 119889 (119906 119909

119899119896+1)]

(37)

Passing to limit as 119896 rarr infin and taking into account the con-tinuity of 120595 and 120601 we get

120595 (119889 (119906 119879119906)) le 120601 (119889 (119906 119879119906)) (38)

From the condition 120595(119905) gt 120601(119905) for 119905 gt 0 we conclude that119889(119906 119879119906) = 0 and hence 119879119906 = 119906 which completes the proof

Similar results can be stated for a map 119879 119883 rarr 119883 in theclass (II) More precisely conditions for existence of a fixedpoint of amap in class (II) are given in the next two theorems

Theorem 10 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissible map-ping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (39)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Theorem 11 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (40)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Note that the proofs of Theorems 10 and 11 can beeasily done by mimicking the proofs of Theorems 8 and 9respectively

We next discuss the conditions for the uniqueness of thefixed point A sufficient condition for the uniqueness of thefixed point in Theorems 10 and 11 can be stated as follows

(B)

For 119909 119910 isin 119883 there exists 119911 isin 119883

such that 120572 (119909 119911) ge 1 120572 (119910 119911) ge 1(41)

Note however that this condition is not sufficient for theuniqueness of fixed point for maps of class (I)

Theorem 12 If condition (B) is added to the hypotheses ofTheorem 10 (resp Theorem 11) then the fixed point of 119879 isunique

Proof Since 119879 satisfies the hypothesis of Theorem 10 (respTheorem 11) then fixed point of 119879 exists Suppose that wehave two different fixed points say 119909 119910 isin 119883 From condition(B) there exists 119911 isin 119883 such that

120572 (119909 119911) ge 1 120572 (119910 119911) ge 1 (42)

Abstract and Applied Analysis 5

Then since 119879 is 120572-admissible we have from (42)

120572 (119909 119879119899119911) ge 1 120572 (119910 119879

119899119911) ge 1 forall119899 isin N (43)

Thus for the sequence 119911119899 isin 119883 defined as 119911

119899= 119879119899119911 we have

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119873 (119909 119911119899))

(44)

where

119873(119909 119911119899)

= max 119889 (119909 119911119899)

1

2

[119889 (119909 119879119909) + 119889 (119911119899 119879119911119899)]

1

2

[119889 (119909 119879119911119899) + 119889 (119911

119899 119879119909)]

= max119889 (119909 119911119899)

119889 (119911119899 119911119899+1)

2

1

2

[119889 (119909 119911119899+1) + 119889 (119911

119899 119909)]

(45)

Observe that 119889(119911119899 119911119899+1)2 le (119889(119911

119899 119909) + 119889(119909 119911

119899+1))2 Then

we deduce119873(119909 119910) = max119889(119909 119911119899) 119889(119909 119911

119899+1)

Without loss of generality we may assume that 119889(119909 119911119899) gt

0 for all 119899 isin N If 119873(119909 119910) = 119889(119909 119911119899+1) then inequality (44)

becomes

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899+1))

lt 120595 (119889 (119909 119911119899+1))

(46)

That is we have a contradiction Then we should have119873(119909 119910) = 119889(119909 119911

119899) for all 119899 isin N which results in

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899)) lt 120595 (119889 (119909 119911

119899))

(47)

due to the fact that 120595(119905) gt 120601(119905) for 119905 gt 0 On the other handsince120595 is nondecreasing then 119889(119909 119911

119899+1) le 119889(119909 119911

119899) for all 119899 isin

N Thus the sequence 119889(119909 119911119899) is a positive nonincreasing

sequence and hence converges to a limit say 119871 ge 0 Takinglimit as 119899 rarr infin in (47) and regarding continuity of 120595 and 120601we deduce

0 le 120595 (119871) le 120601 (119871) (48)

which is possible only if 119871 = 0 Hence we conclude that

lim119899rarrinfin

119889 (119909 119911119899) = 0 (49)

In a similar way we obtain

lim119899rarrinfin

119889 (119910 119911119899) = 0 (50)

From (49) and (50) it follows that 119909 = 119910 which completes theproof of uniqueness

The theorems stated above have been inspired by therecent results of Yan et al [7] They discussed contractionmappings defined on partially ordered complete metricspaces and their applications to boundary value problemsWestate next a theoremwhich can be regarded as a generalizationof the main result in [7] in complete metric spaces

Theorem 13 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a weak triangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910)) forall119909 119910 isin 119883 (51)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 satisfiescondition (A) If there exists 119909

0isin 119883 such that 120572(119909

0 1198791199090) ge 1

then 119879 has a fixed point If in addition 119883 satisfies condition(B) then the fixed point is unique

Proof ofTheorem 13 can be done by following the lines ofproofs of Theorems 8 9 and 12 Hence it is omitted

Remark 14 Under the assumptions of Theorem 12 it can beproved that for every 119909 isin 119883 lim

119899rarrinfin119879119899119909 = 119906 where 119906 is the

unique fixed point (ie the operator 119879 is Picard)

The contractions of classes (I) and (II) are quite generalandmany particular results can be concluded fromTheorems8ndash12 Below we state some of these conclusions

Corollary 15 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 (52)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 8

Corollary 16 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(53)

for all 119909 119910 isin 119883 where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 If there exists1199090isin 119883 such that 120572(119909

0 1198791199090) ge 1 then 119879 has a fixed point

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

Weak triangular 120572-admissible mappings satisfy a prop-erty stated in the following lemma and the proof of whicheasily follows from the definition and can be found in [9]

Lemma 5 (see [9]) Let 119879 119883 rarr 119883 be a weak triangular120572-admissible mapping Assume that there exists 119909

0isin 119883 such

that 120572(1199090 1198791199090) ge 1 If 119909

119899= 1198791198991199090 then 120572(119909

119898 119909119899) ge 1 for all

119898 119899 isin N with119898 lt 119899

2 Existence and Uniqueness Theorems onComplete Metric Spaces

In this section we present our main results which includetheorems on existence and uniqueness of fixed points for aclass of weak triangular 120572-admissible mappings

First we define the following two classes of contractionswhich we are going to investigate in this section and through-out the paper

Definition 6 Let (119883 119889) be a metric space 120595 an alteringdistance function and 120601 [0 +infin) rarr [0 +infin) a continuousfunction satisfying 120595(119905) gt 120601(119905) for all 119905 gt 0

(I) A mapping 119879 119883 rarr 119883 belongs to class (I) if itsatisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (4)

where

119872(119909 119910) = max 119889 (119909 119910) 119889 (119909 119879119909) 119889 (119910 119879119910)

1

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(5)

(II) A mapping 119879 119883 rarr 119883 belongs to class (II) if itsatisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (6)

where

119873(119909 119910) = max 119889 (119909 119910) 12

[119889 (119909 119879119909) + 119889 (119910 119879119910)]

1

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(7)

Remark 7 Note that119873(119909 119910) le 119872(119909 119910) for all 119909 119910 isin 119883

Our first theorem gives conditions for the existence of afixed point for maps in class (I)

Theorem 8 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 120595 (119889 (119879x 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (8)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for

all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Let 1199090isin 119883 satisfy 120572(119909

0 1198791199090) ge 1 and define the

sequence 119909119899 as 119909119899= 119879119909119899minus1

for 119899 isin NIf1199091198990= 1199091198990+1

for some 1198990isin N then obviously119909

1198990= 1198791199091198990

is a fixed point of119879 Suppose that119889(119909119899 119909119899+1) gt 0 for all 119899 isin N

Note that due to the fact that 119879 is 120572-admissible and120572(1199090 1198791199090) = 120572(119909

0 1199091) ge 1 we deduce

120572 (1198791199090 1198791199091)

= 120572 (1199091 1199092) ge 1 997904rArr 120572 (119879119909

1 1198791199092)

= 120572 (1199092 1199093) ge 1 997904rArr sdot sdot sdot 997904rArr 120572 (119909

119899 119909119899+1) ge 1

(9)

Substituting 119909 = 119909119899and 119910 = 119909

119899minus1in (8) and using (9) we

get

120595 (119889 (119909119899+1 119909119899)) le 120572 (119909

119899 119909119899minus1) 120595 (119889 (119909

119899+1 119909119899))

= 120572 (119909119899 119909119899minus1) 120595 (119889 (119879119909

119899 119879119909119899minus1))

le 120601 (119872 (119909119899 119909119899minus1))

(10)

where

119872(119909119899 119909119899minus1)

= max 119889 (119909119899 119909119899minus1) 119889 (119909

119899 119879119909119899) 119889 (119909

119899minus1 119879119909119899minus1)

1

2

[119889 (119909119899 119879119909119899minus1) + 119889 (119909

119899minus1 119879119909119899)]

= max 119889 (119909119899 119909119899minus1) 119889 (119909

119899 119909119899+1) 119889 (119909

119899minus1 119909119899)

1

2

[119889 (119909119899 119909119899) + 119889 (119909

119899minus1 119909119899+1)]

= max119889 (119909119899 119909119899minus1) 119889 (119909

119899 119909119899+1)

119889 (119909119899minus1 119909119899+1)

2

(11)

Note that 119889(119909119899minus1 119909119899+1)2 le (12)[119889(119909

119899minus1 119909119899) + 119889(119909

119899 119909119899+1)]

is smaller than both 119889(119909119899minus1 119909119899) and 119889(119909

119899 119909119899+1) Then

119872(119909119899 119909119899minus1) can be either 119889(119909

119899minus1 119909119899) or 119889(119909

119899 119909119899+1) If

119872(119909119899 119909119899minus1) = 119889(119909

119899 119909119899+1) for some 119899 then the expression

(10) implies that

0 lt 120595 (119889 (119909119899+1 119909119899)) le 120601 (119889 (119909

119899 119909119899+1)) (12)

which contradicts the condition 120595(119905) gt 120601(119905) for 119905 gt 0 Hence119872(119909119899 119909119899minus1) = 119889(119909

119899 119909119899minus1) for all 119899 ge 1 and we have

0 lt 120595 (119889 (119909119899+1 119909119899)) le 120601 (119889 (119909

119899 119909119899minus1)) lt 120595 (119889 (119909

119899 119909119899minus1))

(13)

which results in

119889 (119909119899+1 119909119899) lt 119889 (119909

119899 119909119899minus1) (14)

Abstract and Applied Analysis 3

since 120595 is nondecreasing Thus we conclude that the non-negative sequence 119889(119909

119899+1 119909119899) is decreasing Therefore there

exists 119903 ge 0 such that lim119899rarrinfin

119889(119909119899+1 119909119899) = 119903 Let 119899 rarr infin in

(10) we get

120595 (119903) le 120601 (119903) (15)

By the hypothesis of the theorem since 120595(119905) gt 120601(119905) for all119905 gt 0 this inequality is possible only if 119903 = 0 and hence

lim119899rarrinfin

119889 (119909119899+1 119909119899) = 119903 = 0 (16)

Next we will prove that 119909119899 is a Cauchy sequence

Suppose on the contrary that 119909119899 is not Cauchy Then for

some 120576 gt 0 there exist subsequences 119909119898119896 and 119909

119899119896 of 119909

119899

such that

119899119896gt 119898119896gt 119896 119889 (119909

119899119896 119909119898119896) ge 120576 (17)

for all 119896 ge 1 where corresponding to each119898119896 we can choose

119899119896as the smallest integer with 119899

119896gt 119898119896for which (17) holds

Then

119889 (119909119899119896 119909119898119896minus1) lt 120576 (18)

Employing triangle inequality and making use of (17) and(18) we obtain

120576 le 119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

lt 119889 (119909119899119896 119909119899119896minus1) + 120576

(19)

Passing to limit as 119896 rarr infin and using (16) we get

lim119896rarrinfin

119889 (119909119899119896 119909119898119896) = 120576 (20)

From the triangular inequality we also have

119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896minus1)

le 119889 (119909119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896) + 119889 (119909

119898119896 119909119898119896minus1)

(21)

Letting 119896 rarr infin in the two inequalities above and using (16)and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896minus1) = 120576 (22)

In a similar way by using the triangular inequality we obtainthat

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896) le 119889 (119909

119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896)

(23)

Taking limit as 119896 rarr infin in the above two inequalities andregarding (16) and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896) = 120576 (24)

Furthermore the relations

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896 119909119898119896minus1) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1)

(25)

give

lim119896rarrinfin

119889 (119909119899119896 119909119898119896minus1) = 120576 (26)

by letting 119896 rarr infin and taking into account (16) and (20) Bythe definition of119872(119909 119910) and using limits found above we get

lim119896rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = 120576 (27)

Indeed since

119872(119909119899119896minus1 119909119898119896minus1)

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119879119909119899119896minus1)

119889 (119909119898119896minus1 119879119909119898119896minus1)

1

2

[119889 (119909119899119896minus1 119879119909119898119896minus1) + 119889 (119909

119898119896minus1 119879119909119899119896minus1)]

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119909119899119896) 119889 (119909

119898119896minus1 119909119898119896)

1

2

[119889 (119909119899119896minus1 119909119898119896) + 119889 (119909

119898119896minus1 119909119899119896)]

(28)

passing to the limit as 119896 rarr infin in (28) and using (16) (20)(22) (24) and (26) we obtain

lim119899rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = max 120576 0 0 1

2

[120576 + 120576] = 120576 (29)

Notice that since 119879 is weak triangular 120572-admissible wededuce from Lemma 5 that 120572(119909

119899119896minus1 119909119898119896minus1) ge 1 Therefore we

can apply condition (8) to 119909119899119896minus1

and 119909119898119896minus1

to obtain

0 lt 120595 (119889 (119909119899119896 119909119898119896))

le 120572 (119909119899119896minus1 119909119898119896minus1

) 120595 (119889 (119909119899119896 119909119898119896))

le 120601 (119872(119909119899119896minus1 119909119898119896minus1))

(30)

Letting 119896 rarr infin and taking into account (20) and (27) wehave

0 lt 120595 (120576) le 120601 (120576) (31)

However since 120595(119905) gt 120601(119905) for 119905 gt 0 we deduce that 120576 = 0which contradicts the assumption that 119909

119899 is not a Cauchy

4 Abstract and Applied Analysis

sequence Thus 119909119899 must be Cauchy Due to the fact that

(119883 119889) is a complete metric space there exists 119906 isin 119883 suchthat lim

119899rarrinfin119909119899= 119906 Finally the continuity of 119879 gives

119906 = lim119899rarrinfin

119909119899= lim119899rarrinfin

119879119909119899minus1= 119879119906 (32)

That is 119906 is a fixed point of119879 which completes the proof

One of the advantages of 120572-admissible mappings is thatthe continuity of the map is no longer required for theexistence of a fixed point provided that the space underconsideration has the following property

(A) If 119909119899 is a sequence in119883 such that

119909119899997888rarr 119909 120572 (119909

119899 119909119899+1)) ge 1 forall119899 isin N (33)

then there exists a subsequence 119909119899119896 of 119909

119899 for which

120572 (119909119899119896 119909) ge 1 forall119896 isin N (34)

Bearing this fact in mind we rewrite the statement ofTheorem 8 in the light of this property

Theorem 9 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910))

forall119909 119910 isin 119883

(35)

where 120595 is an altering distance function 120601 [0 +infin) rarr [0

+infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Following the proof of Theorem 8 it is clear that thesequence 119909

119899 defined by 119909

119899= 119879119909119899minus1

for 119899 isin N convergesto a limit 119906 isin 119883 The only thing which remains to show is119879119906 = 119906 Since lim

119899rarrinfin119909119899= 119906 then condition (A) implies

120572(119909119899119896 119906) ge 1 for all 119896 isin N Consequently inequality (35)

with 119909 = 119909119899119896and 119910 = 119906 becomes

120595 (119889 (119909119899119896+1 119879119906))

le 120572 (119909119899119896 119906) 120595 (119889 (119909

119899119896+1 119879119906))

= 120572 (119909119899119896 119906) 120595 (119889 (119879119909

119899119896 119879119906))

le 120601 (119872(119909119899119896 119906))

(36)

where

119872(119909119899119896 119906)

= max 119889 (119909119899119896 119906) 119889 (119909

119899119896 119909119899119896+1)

119889 (119906 119879119906)

1

2

[119889 (119909119899119896 119879119906) + 119889 (119906 119909

119899119896+1)]

(37)

Passing to limit as 119896 rarr infin and taking into account the con-tinuity of 120595 and 120601 we get

120595 (119889 (119906 119879119906)) le 120601 (119889 (119906 119879119906)) (38)

From the condition 120595(119905) gt 120601(119905) for 119905 gt 0 we conclude that119889(119906 119879119906) = 0 and hence 119879119906 = 119906 which completes the proof

Similar results can be stated for a map 119879 119883 rarr 119883 in theclass (II) More precisely conditions for existence of a fixedpoint of amap in class (II) are given in the next two theorems

Theorem 10 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissible map-ping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (39)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Theorem 11 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (40)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Note that the proofs of Theorems 10 and 11 can beeasily done by mimicking the proofs of Theorems 8 and 9respectively

We next discuss the conditions for the uniqueness of thefixed point A sufficient condition for the uniqueness of thefixed point in Theorems 10 and 11 can be stated as follows

(B)

For 119909 119910 isin 119883 there exists 119911 isin 119883

such that 120572 (119909 119911) ge 1 120572 (119910 119911) ge 1(41)

Note however that this condition is not sufficient for theuniqueness of fixed point for maps of class (I)

Theorem 12 If condition (B) is added to the hypotheses ofTheorem 10 (resp Theorem 11) then the fixed point of 119879 isunique

Proof Since 119879 satisfies the hypothesis of Theorem 10 (respTheorem 11) then fixed point of 119879 exists Suppose that wehave two different fixed points say 119909 119910 isin 119883 From condition(B) there exists 119911 isin 119883 such that

120572 (119909 119911) ge 1 120572 (119910 119911) ge 1 (42)

Abstract and Applied Analysis 5

Then since 119879 is 120572-admissible we have from (42)

120572 (119909 119879119899119911) ge 1 120572 (119910 119879

119899119911) ge 1 forall119899 isin N (43)

Thus for the sequence 119911119899 isin 119883 defined as 119911

119899= 119879119899119911 we have

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119873 (119909 119911119899))

(44)

where

119873(119909 119911119899)

= max 119889 (119909 119911119899)

1

2

[119889 (119909 119879119909) + 119889 (119911119899 119879119911119899)]

1

2

[119889 (119909 119879119911119899) + 119889 (119911

119899 119879119909)]

= max119889 (119909 119911119899)

119889 (119911119899 119911119899+1)

2

1

2

[119889 (119909 119911119899+1) + 119889 (119911

119899 119909)]

(45)

Observe that 119889(119911119899 119911119899+1)2 le (119889(119911

119899 119909) + 119889(119909 119911

119899+1))2 Then

we deduce119873(119909 119910) = max119889(119909 119911119899) 119889(119909 119911

119899+1)

Without loss of generality we may assume that 119889(119909 119911119899) gt

0 for all 119899 isin N If 119873(119909 119910) = 119889(119909 119911119899+1) then inequality (44)

becomes

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899+1))

lt 120595 (119889 (119909 119911119899+1))

(46)

That is we have a contradiction Then we should have119873(119909 119910) = 119889(119909 119911

119899) for all 119899 isin N which results in

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899)) lt 120595 (119889 (119909 119911

119899))

(47)

due to the fact that 120595(119905) gt 120601(119905) for 119905 gt 0 On the other handsince120595 is nondecreasing then 119889(119909 119911

119899+1) le 119889(119909 119911

119899) for all 119899 isin

N Thus the sequence 119889(119909 119911119899) is a positive nonincreasing

sequence and hence converges to a limit say 119871 ge 0 Takinglimit as 119899 rarr infin in (47) and regarding continuity of 120595 and 120601we deduce

0 le 120595 (119871) le 120601 (119871) (48)

which is possible only if 119871 = 0 Hence we conclude that

lim119899rarrinfin

119889 (119909 119911119899) = 0 (49)

In a similar way we obtain

lim119899rarrinfin

119889 (119910 119911119899) = 0 (50)

From (49) and (50) it follows that 119909 = 119910 which completes theproof of uniqueness

The theorems stated above have been inspired by therecent results of Yan et al [7] They discussed contractionmappings defined on partially ordered complete metricspaces and their applications to boundary value problemsWestate next a theoremwhich can be regarded as a generalizationof the main result in [7] in complete metric spaces

Theorem 13 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a weak triangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910)) forall119909 119910 isin 119883 (51)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 satisfiescondition (A) If there exists 119909

0isin 119883 such that 120572(119909

0 1198791199090) ge 1

then 119879 has a fixed point If in addition 119883 satisfies condition(B) then the fixed point is unique

Proof ofTheorem 13 can be done by following the lines ofproofs of Theorems 8 9 and 12 Hence it is omitted

Remark 14 Under the assumptions of Theorem 12 it can beproved that for every 119909 isin 119883 lim

119899rarrinfin119879119899119909 = 119906 where 119906 is the

unique fixed point (ie the operator 119879 is Picard)

The contractions of classes (I) and (II) are quite generalandmany particular results can be concluded fromTheorems8ndash12 Below we state some of these conclusions

Corollary 15 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 (52)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 8

Corollary 16 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(53)

for all 119909 119910 isin 119883 where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 If there exists1199090isin 119883 such that 120572(119909

0 1198791199090) ge 1 then 119879 has a fixed point

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

since 120595 is nondecreasing Thus we conclude that the non-negative sequence 119889(119909

119899+1 119909119899) is decreasing Therefore there

exists 119903 ge 0 such that lim119899rarrinfin

119889(119909119899+1 119909119899) = 119903 Let 119899 rarr infin in

(10) we get

120595 (119903) le 120601 (119903) (15)

By the hypothesis of the theorem since 120595(119905) gt 120601(119905) for all119905 gt 0 this inequality is possible only if 119903 = 0 and hence

lim119899rarrinfin

119889 (119909119899+1 119909119899) = 119903 = 0 (16)

Next we will prove that 119909119899 is a Cauchy sequence

Suppose on the contrary that 119909119899 is not Cauchy Then for

some 120576 gt 0 there exist subsequences 119909119898119896 and 119909

119899119896 of 119909

119899

such that

119899119896gt 119898119896gt 119896 119889 (119909

119899119896 119909119898119896) ge 120576 (17)

for all 119896 ge 1 where corresponding to each119898119896 we can choose

119899119896as the smallest integer with 119899

119896gt 119898119896for which (17) holds

Then

119889 (119909119899119896 119909119898119896minus1) lt 120576 (18)

Employing triangle inequality and making use of (17) and(18) we obtain

120576 le 119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

lt 119889 (119909119899119896 119909119899119896minus1) + 120576

(19)

Passing to limit as 119896 rarr infin and using (16) we get

lim119896rarrinfin

119889 (119909119899119896 119909119898119896) = 120576 (20)

From the triangular inequality we also have

119889 (119909119899119896 119909119898119896)

le 119889 (119909119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896minus1)

le 119889 (119909119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896) + 119889 (119909

119898119896 119909119898119896minus1)

(21)

Letting 119896 rarr infin in the two inequalities above and using (16)and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896minus1) = 120576 (22)

In a similar way by using the triangular inequality we obtainthat

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896)

119889 (119909119899119896minus1 119909119898119896) le 119889 (119909

119899119896minus1 119909119899119896) + 119889 (119909

119899119896 119909119898119896)

(23)

Taking limit as 119896 rarr infin in the above two inequalities andregarding (16) and (20) we get

lim119896rarrinfin

119889 (119909119899119896minus1 119909119898119896) = 120576 (24)

Furthermore the relations

119889 (119909119899119896 119909119898119896) le 119889 (119909

119899119896 119909119898119896minus1) + 119889 (119909

119898119896minus1 119909119898119896)

119889 (119909119899119896 119909119898119896minus1) le 119889 (119909

119899119896 119909119899119896minus1) + 119889 (119909

119899119896minus1 119909119898119896minus1)

(25)

give

lim119896rarrinfin

119889 (119909119899119896 119909119898119896minus1) = 120576 (26)

by letting 119896 rarr infin and taking into account (16) and (20) Bythe definition of119872(119909 119910) and using limits found above we get

lim119896rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = 120576 (27)

Indeed since

119872(119909119899119896minus1 119909119898119896minus1)

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119879119909119899119896minus1)

119889 (119909119898119896minus1 119879119909119898119896minus1)

1

2

[119889 (119909119899119896minus1 119879119909119898119896minus1) + 119889 (119909

119898119896minus1 119879119909119899119896minus1)]

= max 119889 (119909119899119896minus1 119909119898119896minus1) 119889 (119909

119899119896minus1 119909119899119896) 119889 (119909

119898119896minus1 119909119898119896)

1

2

[119889 (119909119899119896minus1 119909119898119896) + 119889 (119909

119898119896minus1 119909119899119896)]

(28)

passing to the limit as 119896 rarr infin in (28) and using (16) (20)(22) (24) and (26) we obtain

lim119899rarrinfin

119872(119909119899119896minus1 119909119898119896minus1) = max 120576 0 0 1

2

[120576 + 120576] = 120576 (29)

Notice that since 119879 is weak triangular 120572-admissible wededuce from Lemma 5 that 120572(119909

119899119896minus1 119909119898119896minus1) ge 1 Therefore we

can apply condition (8) to 119909119899119896minus1

and 119909119898119896minus1

to obtain

0 lt 120595 (119889 (119909119899119896 119909119898119896))

le 120572 (119909119899119896minus1 119909119898119896minus1

) 120595 (119889 (119909119899119896 119909119898119896))

le 120601 (119872(119909119899119896minus1 119909119898119896minus1))

(30)

Letting 119896 rarr infin and taking into account (20) and (27) wehave

0 lt 120595 (120576) le 120601 (120576) (31)

However since 120595(119905) gt 120601(119905) for 119905 gt 0 we deduce that 120576 = 0which contradicts the assumption that 119909

119899 is not a Cauchy

4 Abstract and Applied Analysis

sequence Thus 119909119899 must be Cauchy Due to the fact that

(119883 119889) is a complete metric space there exists 119906 isin 119883 suchthat lim

119899rarrinfin119909119899= 119906 Finally the continuity of 119879 gives

119906 = lim119899rarrinfin

119909119899= lim119899rarrinfin

119879119909119899minus1= 119879119906 (32)

That is 119906 is a fixed point of119879 which completes the proof

One of the advantages of 120572-admissible mappings is thatthe continuity of the map is no longer required for theexistence of a fixed point provided that the space underconsideration has the following property

(A) If 119909119899 is a sequence in119883 such that

119909119899997888rarr 119909 120572 (119909

119899 119909119899+1)) ge 1 forall119899 isin N (33)

then there exists a subsequence 119909119899119896 of 119909

119899 for which

120572 (119909119899119896 119909) ge 1 forall119896 isin N (34)

Bearing this fact in mind we rewrite the statement ofTheorem 8 in the light of this property

Theorem 9 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910))

forall119909 119910 isin 119883

(35)

where 120595 is an altering distance function 120601 [0 +infin) rarr [0

+infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Following the proof of Theorem 8 it is clear that thesequence 119909

119899 defined by 119909

119899= 119879119909119899minus1

for 119899 isin N convergesto a limit 119906 isin 119883 The only thing which remains to show is119879119906 = 119906 Since lim

119899rarrinfin119909119899= 119906 then condition (A) implies

120572(119909119899119896 119906) ge 1 for all 119896 isin N Consequently inequality (35)

with 119909 = 119909119899119896and 119910 = 119906 becomes

120595 (119889 (119909119899119896+1 119879119906))

le 120572 (119909119899119896 119906) 120595 (119889 (119909

119899119896+1 119879119906))

= 120572 (119909119899119896 119906) 120595 (119889 (119879119909

119899119896 119879119906))

le 120601 (119872(119909119899119896 119906))

(36)

where

119872(119909119899119896 119906)

= max 119889 (119909119899119896 119906) 119889 (119909

119899119896 119909119899119896+1)

119889 (119906 119879119906)

1

2

[119889 (119909119899119896 119879119906) + 119889 (119906 119909

119899119896+1)]

(37)

Passing to limit as 119896 rarr infin and taking into account the con-tinuity of 120595 and 120601 we get

120595 (119889 (119906 119879119906)) le 120601 (119889 (119906 119879119906)) (38)

From the condition 120595(119905) gt 120601(119905) for 119905 gt 0 we conclude that119889(119906 119879119906) = 0 and hence 119879119906 = 119906 which completes the proof

Similar results can be stated for a map 119879 119883 rarr 119883 in theclass (II) More precisely conditions for existence of a fixedpoint of amap in class (II) are given in the next two theorems

Theorem 10 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissible map-ping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (39)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Theorem 11 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (40)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Note that the proofs of Theorems 10 and 11 can beeasily done by mimicking the proofs of Theorems 8 and 9respectively

We next discuss the conditions for the uniqueness of thefixed point A sufficient condition for the uniqueness of thefixed point in Theorems 10 and 11 can be stated as follows

(B)

For 119909 119910 isin 119883 there exists 119911 isin 119883

such that 120572 (119909 119911) ge 1 120572 (119910 119911) ge 1(41)

Note however that this condition is not sufficient for theuniqueness of fixed point for maps of class (I)

Theorem 12 If condition (B) is added to the hypotheses ofTheorem 10 (resp Theorem 11) then the fixed point of 119879 isunique

Proof Since 119879 satisfies the hypothesis of Theorem 10 (respTheorem 11) then fixed point of 119879 exists Suppose that wehave two different fixed points say 119909 119910 isin 119883 From condition(B) there exists 119911 isin 119883 such that

120572 (119909 119911) ge 1 120572 (119910 119911) ge 1 (42)

Abstract and Applied Analysis 5

Then since 119879 is 120572-admissible we have from (42)

120572 (119909 119879119899119911) ge 1 120572 (119910 119879

119899119911) ge 1 forall119899 isin N (43)

Thus for the sequence 119911119899 isin 119883 defined as 119911

119899= 119879119899119911 we have

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119873 (119909 119911119899))

(44)

where

119873(119909 119911119899)

= max 119889 (119909 119911119899)

1

2

[119889 (119909 119879119909) + 119889 (119911119899 119879119911119899)]

1

2

[119889 (119909 119879119911119899) + 119889 (119911

119899 119879119909)]

= max119889 (119909 119911119899)

119889 (119911119899 119911119899+1)

2

1

2

[119889 (119909 119911119899+1) + 119889 (119911

119899 119909)]

(45)

Observe that 119889(119911119899 119911119899+1)2 le (119889(119911

119899 119909) + 119889(119909 119911

119899+1))2 Then

we deduce119873(119909 119910) = max119889(119909 119911119899) 119889(119909 119911

119899+1)

Without loss of generality we may assume that 119889(119909 119911119899) gt

0 for all 119899 isin N If 119873(119909 119910) = 119889(119909 119911119899+1) then inequality (44)

becomes

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899+1))

lt 120595 (119889 (119909 119911119899+1))

(46)

That is we have a contradiction Then we should have119873(119909 119910) = 119889(119909 119911

119899) for all 119899 isin N which results in

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899)) lt 120595 (119889 (119909 119911

119899))

(47)

due to the fact that 120595(119905) gt 120601(119905) for 119905 gt 0 On the other handsince120595 is nondecreasing then 119889(119909 119911

119899+1) le 119889(119909 119911

119899) for all 119899 isin

N Thus the sequence 119889(119909 119911119899) is a positive nonincreasing

sequence and hence converges to a limit say 119871 ge 0 Takinglimit as 119899 rarr infin in (47) and regarding continuity of 120595 and 120601we deduce

0 le 120595 (119871) le 120601 (119871) (48)

which is possible only if 119871 = 0 Hence we conclude that

lim119899rarrinfin

119889 (119909 119911119899) = 0 (49)

In a similar way we obtain

lim119899rarrinfin

119889 (119910 119911119899) = 0 (50)

From (49) and (50) it follows that 119909 = 119910 which completes theproof of uniqueness

The theorems stated above have been inspired by therecent results of Yan et al [7] They discussed contractionmappings defined on partially ordered complete metricspaces and their applications to boundary value problemsWestate next a theoremwhich can be regarded as a generalizationof the main result in [7] in complete metric spaces

Theorem 13 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a weak triangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910)) forall119909 119910 isin 119883 (51)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 satisfiescondition (A) If there exists 119909

0isin 119883 such that 120572(119909

0 1198791199090) ge 1

then 119879 has a fixed point If in addition 119883 satisfies condition(B) then the fixed point is unique

Proof ofTheorem 13 can be done by following the lines ofproofs of Theorems 8 9 and 12 Hence it is omitted

Remark 14 Under the assumptions of Theorem 12 it can beproved that for every 119909 isin 119883 lim

119899rarrinfin119879119899119909 = 119906 where 119906 is the

unique fixed point (ie the operator 119879 is Picard)

The contractions of classes (I) and (II) are quite generalandmany particular results can be concluded fromTheorems8ndash12 Below we state some of these conclusions

Corollary 15 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 (52)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 8

Corollary 16 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(53)

for all 119909 119910 isin 119883 where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 If there exists1199090isin 119883 such that 120572(119909

0 1198791199090) ge 1 then 119879 has a fixed point

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

sequence Thus 119909119899 must be Cauchy Due to the fact that

(119883 119889) is a complete metric space there exists 119906 isin 119883 suchthat lim

119899rarrinfin119909119899= 119906 Finally the continuity of 119879 gives

119906 = lim119899rarrinfin

119909119899= lim119899rarrinfin

119879119909119899minus1= 119879119906 (32)

That is 119906 is a fixed point of119879 which completes the proof

One of the advantages of 120572-admissible mappings is thatthe continuity of the map is no longer required for theexistence of a fixed point provided that the space underconsideration has the following property

(A) If 119909119899 is a sequence in119883 such that

119909119899997888rarr 119909 120572 (119909

119899 119909119899+1)) ge 1 forall119899 isin N (33)

then there exists a subsequence 119909119899119896 of 119909

119899 for which

120572 (119909119899119896 119909) ge 1 forall119896 isin N (34)

Bearing this fact in mind we rewrite the statement ofTheorem 8 in the light of this property

Theorem 9 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910))

forall119909 119910 isin 119883

(35)

where 120595 is an altering distance function 120601 [0 +infin) rarr [0

+infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909) 119889(119910 119879119910)(12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883 such that

120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Following the proof of Theorem 8 it is clear that thesequence 119909

119899 defined by 119909

119899= 119879119909119899minus1

for 119899 isin N convergesto a limit 119906 isin 119883 The only thing which remains to show is119879119906 = 119906 Since lim

119899rarrinfin119909119899= 119906 then condition (A) implies

120572(119909119899119896 119906) ge 1 for all 119896 isin N Consequently inequality (35)

with 119909 = 119909119899119896and 119910 = 119906 becomes

120595 (119889 (119909119899119896+1 119879119906))

le 120572 (119909119899119896 119906) 120595 (119889 (119909

119899119896+1 119879119906))

= 120572 (119909119899119896 119906) 120595 (119889 (119879119909

119899119896 119879119906))

le 120601 (119872(119909119899119896 119906))

(36)

where

119872(119909119899119896 119906)

= max 119889 (119909119899119896 119906) 119889 (119909

119899119896 119909119899119896+1)

119889 (119906 119879119906)

1

2

[119889 (119909119899119896 119879119906) + 119889 (119906 119909

119899119896+1)]

(37)

Passing to limit as 119896 rarr infin and taking into account the con-tinuity of 120595 and 120601 we get

120595 (119889 (119906 119879119906)) le 120601 (119889 (119906 119879119906)) (38)

From the condition 120595(119905) gt 120601(119905) for 119905 gt 0 we conclude that119889(119906 119879119906) = 0 and hence 119879119906 = 119906 which completes the proof

Similar results can be stated for a map 119879 119883 rarr 119883 in theclass (II) More precisely conditions for existence of a fixedpoint of amap in class (II) are given in the next two theorems

Theorem 10 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissible map-ping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (39)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Theorem 11 Let (119883 119889) be a complete metric space Assumethat 119883 satisfies condition (A) Let 119879 119883 rarr 119883 be a weaktriangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 (40)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Note that the proofs of Theorems 10 and 11 can beeasily done by mimicking the proofs of Theorems 8 and 9respectively

We next discuss the conditions for the uniqueness of thefixed point A sufficient condition for the uniqueness of thefixed point in Theorems 10 and 11 can be stated as follows

(B)

For 119909 119910 isin 119883 there exists 119911 isin 119883

such that 120572 (119909 119911) ge 1 120572 (119910 119911) ge 1(41)

Note however that this condition is not sufficient for theuniqueness of fixed point for maps of class (I)

Theorem 12 If condition (B) is added to the hypotheses ofTheorem 10 (resp Theorem 11) then the fixed point of 119879 isunique

Proof Since 119879 satisfies the hypothesis of Theorem 10 (respTheorem 11) then fixed point of 119879 exists Suppose that wehave two different fixed points say 119909 119910 isin 119883 From condition(B) there exists 119911 isin 119883 such that

120572 (119909 119911) ge 1 120572 (119910 119911) ge 1 (42)

Abstract and Applied Analysis 5

Then since 119879 is 120572-admissible we have from (42)

120572 (119909 119879119899119911) ge 1 120572 (119910 119879

119899119911) ge 1 forall119899 isin N (43)

Thus for the sequence 119911119899 isin 119883 defined as 119911

119899= 119879119899119911 we have

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119873 (119909 119911119899))

(44)

where

119873(119909 119911119899)

= max 119889 (119909 119911119899)

1

2

[119889 (119909 119879119909) + 119889 (119911119899 119879119911119899)]

1

2

[119889 (119909 119879119911119899) + 119889 (119911

119899 119879119909)]

= max119889 (119909 119911119899)

119889 (119911119899 119911119899+1)

2

1

2

[119889 (119909 119911119899+1) + 119889 (119911

119899 119909)]

(45)

Observe that 119889(119911119899 119911119899+1)2 le (119889(119911

119899 119909) + 119889(119909 119911

119899+1))2 Then

we deduce119873(119909 119910) = max119889(119909 119911119899) 119889(119909 119911

119899+1)

Without loss of generality we may assume that 119889(119909 119911119899) gt

0 for all 119899 isin N If 119873(119909 119910) = 119889(119909 119911119899+1) then inequality (44)

becomes

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899+1))

lt 120595 (119889 (119909 119911119899+1))

(46)

That is we have a contradiction Then we should have119873(119909 119910) = 119889(119909 119911

119899) for all 119899 isin N which results in

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899)) lt 120595 (119889 (119909 119911

119899))

(47)

due to the fact that 120595(119905) gt 120601(119905) for 119905 gt 0 On the other handsince120595 is nondecreasing then 119889(119909 119911

119899+1) le 119889(119909 119911

119899) for all 119899 isin

N Thus the sequence 119889(119909 119911119899) is a positive nonincreasing

sequence and hence converges to a limit say 119871 ge 0 Takinglimit as 119899 rarr infin in (47) and regarding continuity of 120595 and 120601we deduce

0 le 120595 (119871) le 120601 (119871) (48)

which is possible only if 119871 = 0 Hence we conclude that

lim119899rarrinfin

119889 (119909 119911119899) = 0 (49)

In a similar way we obtain

lim119899rarrinfin

119889 (119910 119911119899) = 0 (50)

From (49) and (50) it follows that 119909 = 119910 which completes theproof of uniqueness

The theorems stated above have been inspired by therecent results of Yan et al [7] They discussed contractionmappings defined on partially ordered complete metricspaces and their applications to boundary value problemsWestate next a theoremwhich can be regarded as a generalizationof the main result in [7] in complete metric spaces

Theorem 13 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a weak triangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910)) forall119909 119910 isin 119883 (51)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 satisfiescondition (A) If there exists 119909

0isin 119883 such that 120572(119909

0 1198791199090) ge 1

then 119879 has a fixed point If in addition 119883 satisfies condition(B) then the fixed point is unique

Proof ofTheorem 13 can be done by following the lines ofproofs of Theorems 8 9 and 12 Hence it is omitted

Remark 14 Under the assumptions of Theorem 12 it can beproved that for every 119909 isin 119883 lim

119899rarrinfin119879119899119909 = 119906 where 119906 is the

unique fixed point (ie the operator 119879 is Picard)

The contractions of classes (I) and (II) are quite generalandmany particular results can be concluded fromTheorems8ndash12 Below we state some of these conclusions

Corollary 15 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 (52)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 8

Corollary 16 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(53)

for all 119909 119910 isin 119883 where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 If there exists1199090isin 119883 such that 120572(119909

0 1198791199090) ge 1 then 119879 has a fixed point

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Then since 119879 is 120572-admissible we have from (42)

120572 (119909 119879119899119911) ge 1 120572 (119910 119879

119899119911) ge 1 forall119899 isin N (43)

Thus for the sequence 119911119899 isin 119883 defined as 119911

119899= 119879119899119911 we have

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119873 (119909 119911119899))

(44)

where

119873(119909 119911119899)

= max 119889 (119909 119911119899)

1

2

[119889 (119909 119879119909) + 119889 (119911119899 119879119911119899)]

1

2

[119889 (119909 119879119911119899) + 119889 (119911

119899 119879119909)]

= max119889 (119909 119911119899)

119889 (119911119899 119911119899+1)

2

1

2

[119889 (119909 119911119899+1) + 119889 (119911

119899 119909)]

(45)

Observe that 119889(119911119899 119911119899+1)2 le (119889(119911

119899 119909) + 119889(119909 119911

119899+1))2 Then

we deduce119873(119909 119910) = max119889(119909 119911119899) 119889(119909 119911

119899+1)

Without loss of generality we may assume that 119889(119909 119911119899) gt

0 for all 119899 isin N If 119873(119909 119910) = 119889(119909 119911119899+1) then inequality (44)

becomes

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899+1))

lt 120595 (119889 (119909 119911119899+1))

(46)

That is we have a contradiction Then we should have119873(119909 119910) = 119889(119909 119911

119899) for all 119899 isin N which results in

0 lt 120595 (119889 (119909 119911119899+1))

le 120572 (119909 119911119899) 120595 (119889 (119879119909 119879119911

119899))

le 120601 (119889 (119909 119911119899)) lt 120595 (119889 (119909 119911

119899))

(47)

due to the fact that 120595(119905) gt 120601(119905) for 119905 gt 0 On the other handsince120595 is nondecreasing then 119889(119909 119911

119899+1) le 119889(119909 119911

119899) for all 119899 isin

N Thus the sequence 119889(119909 119911119899) is a positive nonincreasing

sequence and hence converges to a limit say 119871 ge 0 Takinglimit as 119899 rarr infin in (47) and regarding continuity of 120595 and 120601we deduce

0 le 120595 (119871) le 120601 (119871) (48)

which is possible only if 119871 = 0 Hence we conclude that

lim119899rarrinfin

119889 (119909 119911119899) = 0 (49)

In a similar way we obtain

lim119899rarrinfin

119889 (119910 119911119899) = 0 (50)

From (49) and (50) it follows that 119909 = 119910 which completes theproof of uniqueness

The theorems stated above have been inspired by therecent results of Yan et al [7] They discussed contractionmappings defined on partially ordered complete metricspaces and their applications to boundary value problemsWestate next a theoremwhich can be regarded as a generalizationof the main result in [7] in complete metric spaces

Theorem 13 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a weak triangular 120572-admissible mapping such that

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910)) forall119909 119910 isin 119883 (51)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 satisfiescondition (A) If there exists 119909

0isin 119883 such that 120572(119909

0 1198791199090) ge 1

then 119879 has a fixed point If in addition 119883 satisfies condition(B) then the fixed point is unique

Proof ofTheorem 13 can be done by following the lines ofproofs of Theorems 8 9 and 12 Hence it is omitted

Remark 14 Under the assumptions of Theorem 12 it can beproved that for every 119909 isin 119883 lim

119899rarrinfin119879119899119909 = 119906 where 119906 is the

unique fixed point (ie the operator 119879 is Picard)

The contractions of classes (I) and (II) are quite generalandmany particular results can be concluded fromTheorems8ndash12 Below we state some of these conclusions

Corollary 15 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 (52)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910) + 119889(119910 119879119909)] If there exists 119909

0isin 119883

such that 120572(1199090 1198791199090) ge 1 then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 8

Corollary 16 Let (119883 119889) be a complete metric space Let 119879 119883 rarr 119883 be a continuous weak triangular 120572-admissiblemapping such that

120572 (119909 119910) 119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

(53)

for all 119909 119910 isin 119883 where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 If there exists1199090isin 119883 such that 120572(119909

0 1198791199090) ge 1 then 119879 has a fixed point

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(54)

the proof follows from Corollary 15

3 Fixed Points on Partially OrderedMetric Spaces

It has been pointed out in some studies that some results inmetric spaces endowed with a partial order can be concludedfrom the fixed point results related with 120572-admissible mapsonmetric spaces (see [9 10]) In this section we give existenceand uniqueness theorems on partially ordered metric spaceswhich can be regarded as consequences of the theoremspresented in the previous section

Recall that on a partially ordered set (119883 ⪯) a map 119879 119883 rarr 119883 is nondecreasing if it satisfies 119879119909 ⪯ 119879119910 for all119909 119910 isin 119883 such that 119909 ⪯ 119910

Definition 17 Let (119883 119889 ⪯) be a metric space endowed with apartial order⪯ If for every nondecreasing sequence 119909

119899 sub 119883

which converges to 119909 isin 119883 there exists a subsequence 119909119899119896 of

119909119899 satisfying 119909

119899119896⪯ 119909 then (119883 119889 ⪯) is said to be regular

Our first theorem contains the conditions for existenceof a fixed point for a map of class (I) defined on a partiallyordered metric space

Theorem 18 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(55)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905)

for all 119905 gt 0 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

Proof Define the map 120572 119883 times 119883 rarr [0infin) as

120572 (119909 119910) =

1 if 119909 ⪯ 119910 or 119910 ⪯ 1199090 otherwise

(56)

It is clear that 119879 satisfies

120572 (119909 119910) 120595 (119889 (119879119909 119879119910)) le 120601 (119872 (119909 119910)) forall119909 119910 isin 119883 (57)

where 120572 is defined in (56) Let 1199090isin 119883 satisfy 119909

0⪯ 1198791199090 Then

120572(1199090 1198791199090) ge 1 On the other hand since 119879 is nondecreasing

then 119879 is 120572-admissible Indeed120572 (119909 119910) ge 1 997904rArr 119909 ⪯ 119910 or

119910 ⪯ 119909 997904rArr 119879119909 ⪯ 119879119910 or

119879119910 ⪯ 119879119909 997904rArr 120572 (119879119909 119879119910) ge 1

(58)

Note also that if 119909 ⪯ 119879119909 then 119879119909 ⪯ 1198792119909 and hence 119909 ⪯ 1198792119909that is if120572(119909 119879119909) ge 1 then120572(119879119909 1198792119909) ge 1 and120572(119909 1198792119909) ge 1Similar conclusion can be done if119909 ⪰ 119879119909Therefore119879 is weaktriangular120572-admissible If119879 is continuous then119879 satisfies theconditions of Theorem 8 and hence has a fixed point

Suppose now that119883 is regularThen every nondecreasingsequence 119909

119899 which converges to 119909 has a subsequence 119909

119899119896

for which 119909119899119896⪯ 119909 holds for all 119896 isin N Hence 120572(119909

119899 119909119899+1) ge 1

implies 120572(119909119899119896 119909) ge 1 for all 119896 isin N In other words the set

119883 satisfies condition (A) ByTheorem 9 the mapping 119879 has afixed point

Analogously we state conditions for existence of fixedpoints for maps from class (II) on partially ordered metricspaces

Theorem 19 Let (119883 119889 ⪯) be a partially ordered completemetric space Let 119879 119883 rarr 119883 be a nondecreasing mappingsuch that

120595 (119889 (119879119909 119879119910)) le 120601 (119873 (119909 119910)) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(59)

where 120595 is an altering distance function 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) forall 119905 gt 0 and 119873(119909 119910) = max119889(119909 119910) (12)[119889(119909 119879119909) +119889(119910 119879119910)] (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists1199090isin 119883 satisfying 119909

0⪯ 1198791199090and that either 119879 is continuous or

(119883 119889 ⪯) is regular Then 119879 has a fixed point

The uniqueness of a fixed point on partially orderedmetric spaces requires an additional assumption on the set119883 This assumption reads as follows

(C) For all 119909 119910 isin 119883 there exists 119911 isin 119883 which iscomparable to both 119909 and 119910

Theorem 20 Adding condition (C) to the hypothesis ofTheorem 19 one obtains the uniqueness of the fixed point

Proof Note that if 120572 is the map defined in (56) condition (C)is equivalent to condition (B) Indeed assume that condition(B) holds Then for all 119909 119910 isin 119883 there exists 119911 isin 119883 suchthat 120572(119909 119911) ge 1 and 120572(119910 119911) ge 1 where 120572 is defined in (56)This means that both 119909 and 119910 are comparable to 119911 that iscondition (C) also holds If on the other hand condition (C)is satisfied then for all 119909 119910 isin 119883 there exists 119911 isin 119883 whichis comparable to both 119909 and 119910 Then 120572(119909 119911) = 1 ge 1 and120572(119910 119911) = 1 ge 1 that is condition (B) is also satisfied Henceby Theorem 12 the fixed point of the map 119879 is unique

Some consequences of the results stated above can beeasily concluded We next present two of them

Corollary 21 Let (119883 119889 ⪯) be a partially ordered and completemetric space Let 119879 119883 rarr 119883 be nondecreasing mapping suchthat

119889 (119879119909 119879119910) le 119896119872(119909 119910) forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910 (60)

where 0 lt 119896 lt 1 and 119872(119909 119910) = max119889(119909 119910) 119889(119909 119879119909)119889(119910 119879119910) (12)[119889(119909 119879119910)+119889(119910 119879119909)] Assume that there exists

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

1199090isin 119883 such that 119909

0⪯ 1198791199090 If 119879 is continuous or 119883 is regular

then 119879 has a fixed point

Proof Proof is obvious by choosing 120595(119905) = 119905 and 120601(119905) = 119896119905 inTheorem 18

Corollary 22 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be nondecreasing mappingsuch that

119889 (119879119909 119879119910)

le 119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)]

forall119909 119910 isin 119883 119908119894119905ℎ 119909 ⪯ 119910

(61)

where 0 lt 119886 + 119887 + 119888 + 119890 lt 1 Assume that there exists 1199090isin 119883

such that 1199090⪯ 1198791199090 If 119879 is continuous or 119883 is regular then 119879

has a fixed point

Proof Due to the fact that

119886119889 (119909 119910) + 119887119889 (119909 119879119909) + 119888119889 (119910 119879119910)

+

119890

2

[119889 (119909 119879119910) + 119889 (119910 119879119909)] le 119896119872(119909 119910)

(62)

the proof follows from Corollary 21

Our last result is a consequence ofTheorem 13 onpartiallyorderedmetric spaces and follows easily by using the function120572 defined in (56) It is actually themain result of Yan et al [7]

Corollary 23 Let (119883 119889 ⪯) be a partially ordered and com-plete metric space Let 119879 119883 rarr 119883 be a nondecreasingmapping such that

120595 (119889 (119879119909 119879119910)) le 120601 (119889 (119909 119910))

forall119909 119910 isin 119883 119891119900119903 119908ℎ119894119888ℎ 119909 ⪯ 119910

(63)

where 120595 is an altering distance function and 120601 [0 +infin) rarr[0 +infin) is a continuous function satisfying 120595(119905) gt 120601(119905) for all119905 gt 0 Assume either that 119879 is continuous or that 119883 is regularIf there exists 119909

0isin 119883 such that 119909

0⪯ 1198791199090 then 119879 has a fixed

point If in addition 119883 satisfies condition (C) then the fixedpoint is unique

4 Application to OrdinaryDifferential Equations

In this section we discuss application of our results toexistence and uniqueness of solutions of boundary valueproblems We present two examples first of which has beeninspired by [1] Specifically we study the existence anduniqueness of a solution for the following first-order periodicboundary value problem

1199061015840(119905) = 119891 (119905 119906 (119905)) if 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(64)

where 119879 gt 0 and 119891 119868 times R rarr R is a continuous functionLet119862(119868) denote the space of continuous functions defined on119868 Clearly the function

119889 (119909 119910) = sup119905isin119868

1003816100381610038161003816119909 (119905) minus 119910 (119905)

1003816100381610038161003816 119909 119910 isin 119862 (119868) (65)

defines a metric on 119862(119868) Moreover (119862(119868) 119889) is a completemetric space Define also a partial order on 119862(119868) by

119909 119910 isin 119862 (119868) 119909 le 119910 lArrrArr 119909 (119905) le 119910 (119905) forall119905 isin 119868 (66)

It is easy to see that 119862(119868) satisfies condition (C) Indeedfor 119909 119910 isin 119862(119868) we have 119909 le 119911 and 119910 le 119911 where

119911 = max119905isin119868

119909 (119905) 119910 (119905) =

119909 (119905) if 119909 (119905) ge 119910 (119905)119910 (119905) if 119910 (119905) ge 119909 (119905)

(67)

is also in 119862(119868) whenever 119909 119910 isin 119862(119868) Nieto and Rodrıguez-Lopez [1] proved that (119862(119868) 119889 le) where the metric 119889 andthe partial order le are defined above satisfies regularitycondition given in Definition 17 We now define a lowersolution for the problem (64)

Definition 24 A lower solution for (64) is a function 120573 isin1198621(119868) satisfying

1205731015840(119905) le 119891 (119905 120573 (119905)) for 119905 isin 119868

120573 (0) le 120573 (119879)

(68)

Our next theorem gives conditions for existence anduniqueness of solution to the problem (64)

Theorem 25 Let the function 119891 in the problem (64) becontinuous and let

0 le 119891 (119905 119910) + 120582119910 minus [119891 (119905 119909) + 120582119909] le 120574(

(119909 minus 119910)2

(119909 minus 119910)2+ 1

)

12

(69)

hold for 119909 119910 isin R with 119910 ge 119909 and for some 120582 120574 gt 0 satisfying

120574 le (

2120582(119890120582119879minus 1)

119879(119890120582119879+ 1)

)

12

(70)

If the problem (64) has a lower solution then it has a uniquesolution

Proof Rewrite the problem (64) as

1199061015840(119905) + 120582119906 (119905) = 119891 (119905 119906 (119905)) + 120582119906 (119905) for 119905 isin 119868 = [0 119879]

119906 (0) = 119906 (119879)

(71)

Clearly the problem (64) is equivalent to the integralequation

119906 (119905) = int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 (72)

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119890120582(119879+119904minus119905)

119890120582119879minus 1

0 le 119904 lt 119905 le 119879

119890120582(119904minus119905)

119890120582119879minus 1

0 le 119905 lt 119904 le 119879

(73)

Define the mapA 119862(119868) rarr 119862(119868) by

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904 119905 isin 119868 (74)

It is obvious that a fixed point 119906 isin 119862(119868) of A is a solution of(64)Wewill show that themapping119860 satisfies the conditionsof Corollary 23

First note thatA is nondecreasing by the hypothesis thatis for 119906 ge V

119891 (119905 119906) + 120582119906 ge 119891 (119905 V) + 120582V (75)

And hence

(A119906) (119905) = int119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)] 119889119904

ge int

119879

0

119866 (119905 119904) [119891 (119904 V (119904)) + 120582V (119904)] 119889119904 = (AV) (119905)

(76)

since 119866(119905 119904) gt 0 for (119905 119904) isin 119868 times 119868Note also that for 119906 ge V we have

119889 (A119906AV) = sup119905isin119868

|(A119906) (119905) minus (AV) (119905)|

= sup119905isin119868

((A119906) (119905) minus (AV) (119905))

= sup119905isin119868

int

119879

0

119866 (119905 119904) [119891 (119904 119906 (119904)) + 120582119906 (119904)

minus119891 (119904 V (119904)) + 120582V (119904)] 119889119904

le sup119905isin119868

int

119879

0

119866 (119905 119904) 120574radic[119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904

(77)

Employing the Cauchy-Schwarz inequality for the lastintegral we obtain

int

119879

0

119866 (119905 119904) 120574radic(119906(119904) minus V(119904))2

(119906(119904) minus V(119904))2 + 1119889119904

le (int

119879

0

119866 (119905 119904)2119889119904)

12

(int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904)

12

(78)

The first integral on the right-hand side can be calculated eas-ily and gives

int

119879

0

119866 (119905 119904)2119889119904 = int

119905

0

119866 (119905 119904)2119889119904 + int

119879

119905

119866 (119905 119904)2119889119904

= int

119905

0

1198902120582(119879+119904minus119905)

(119890120582119879minus 1)2119889119904 + int

119879

119905

1198902120582(119904minus119905)

(119890120582119879minus 1)2119889119904

=

1198902120582119879minus 1

2120582 (119890120582119879minus 1)2

=

119890120582119879+ 1

2120582 (119890120582119879minus 1)

(79)

For the second integral in (78) we obtain the following esti-mate

int

119879

0

1205742 [119906 (119904) minus V (119904)]2

[119906 (119904) minus V (119904)]2 + 1119889119904 le 120574

2sup119905isin119868

|119906 (119905) minus V (119905)|2

|119906 (119905) minus V (119905)|2 + 1sdot 119879

= 1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879

(80)

Making use of the inequalities in (77) (78) and (80) andthe integral in (79) we deduce

119889 (A119906AV)

le (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

sdot (1205742 119889 (119906 V)2

119889 (119906 V)2 + 1119879)

12

= (

119890120582119879+ 1

2120582 (119890120582119879minus 1)

)

12

120574(

119889 (119906 V)2

119889 (119906 V)2 + 1)

12

radic119879

(81)

And consequently

119889(A119906AV)2 le 1205742119890120582119879+ 1

2120582 (119890120582119879minus 1)

119889(119906 V)2

119889(119906 V)2 + 1119879 (82)

This last inequality can be also written as

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le 1205742(119890120582119879+ 1) 119889(119906 V)2119879

(83)

since 120582 is positiveThe constant 120574 on the other hand satisfies

120574 le (

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

)

12

(84)

and therefore inequality (83) implies

(2120582 (119890120582119879minus 1)) (119889(119906 V)2 + 1) 119889(A119906AV)2

le

2120582 (119890120582119879minus 1)

119879 (119890120582119879+ 1)

(119890120582119879+ 1) 119889(119906 V)2119879

(85)

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

and thus

119889(A119906AV)2 le119889 (119906 V)2

119889 (119906 V)2 + 1 (86)

Define the functions 120595 and 120601 as 120595(119905) = 1199052 and 120601(119905) = 1199052(1199052 +1) It is clear that 120595 is an altering distance function and alsothat 120595(119905) gt 120601(119905) for all 119905 gt 0 Then (86) becomes

120595 (119889 (A119906AV)) le 120601 (119889 (119906 V)) (87)

for all 119906 ge V that is A satisfies the contractive condition ofCorollary 23 Finally let 120573(119905) be a lower solution for (86) Wewill show that 120573 le A120573 Multiplying the inequality

1205731015840(119905) + 120582120573 (119905) le 119891 (119905 120573 (119905)) + 120582120573 (119905) for 119905 isin 119868 (88)

by 119890120582119905 we obtain

(120573(119905)119890120582119905)

1015840

le [119891 (119905 120573 (119905)) + 120582120573 (119905)] 119890120582119905 for 119905 isin 119868 (89)

which upon integration gives

120573 (119905) 119890120582119905le 120573 (0) + int

119905

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904 for 119905 isin 119868

(90)

Since 120573(0) le 120573(119879) the last inequality implies

120573 (0) 119890120582119905le 120573 (119879) 119890

120582119879

le 120573 (0) + int

119879

0

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119890120582119904119889119904

(91)

and hence

120573 (0) le int

119879

0

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904 (92)

From this inequality and inequality (90) we obtain

120573 (119905) 119890120582119905le int

119905

0

119890120582(119879+119904)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582119904

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

(93)

and consequently

120573 (119905) le int

119905

0

119890120582(119879+119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

+ int

119879

119905

119890120582(119904minus119905)

119890120582119879minus 1

[119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= int

119879

0

119866 (119905 119904) [119891 (119904 120573 (119904)) + 120582120573 (119904)] 119889119904

= (A120573) (119905) for 119905 isin 119868

(94)

We have shown that all conditions of Corollary 23 holdTherefore A has a unique fixed point or equivalently theproblem (64) has a unique solution

As a second examplewe discuss the existence and unique-ness of solution for a second-order boundary value problemMore precisely we consider

minus

1198892119906

1198891199052= 119891 (119905 119906) 119906 isin 119868 = [0infin) 119905 isin [0 1]

119906 (0) = 119906 (1) = 0

(95)

The problem (95) is equivalent to the integral equation

119906 (119905) = int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (96)

where 119866(119905 119904) is the Green function given by

119866 (119905 119904) =

119905 (1 minus 119904) 0 le 119905 lt 119904 le 1

119904 (1 minus 119905) 0 le 119904 lt 119905 le 1

(97)

In what follows our last theorem gives conditions for theexistence and uniqueness of solution to the problem (95) orequivalently the problem (96)

Theorem 26 Let 119891 [0 1] times R rarr [0infin) be a continuousfunction which is nondecreasing in its second variable Supposethat for all 119906 V isin R satisfying 119906 le V there exists a constant0 lt 120574 lt 8 such that

119891 (119905 V) minus 119891 (119905 119906) le 120574((119906 minus V)2

(119906 minus V)2 + 1)

12

(98)

Then the problem (95) has a unique nonnegative solution

Proof Consider the space 119884 = 119906 isin 119862[0 1] 119906(119905) ge 0 anddefine ametric 119889(119906 V) = sup|119906(119905)minusV(119905)| 119905 isin [0 1] as usualObviously (119884 119889) is a complete metric space Define also theusual partial order on 119884 that is

119906 le V 997904rArr 119906 (119905) le V (119905) forall119905 isin [0 1] (99)

As in the previous example a solution of the problem (95) isa fixed point of the operatorA 119884 rarr 119884 defined by

(A119906) (119905) = int1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 119905 isin [0 1] (100)

where 119866(119905 119904) is the Green function in (97) Since 119891 isnondecreasing with respect to its second variable then for119906 V isin 119884 with V ge 119906 we have

(AV) (119905) = int1

0

119866 (119905 119904) 119891 (119904 V (119904)) 119889119904

ge int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

= (A119906) (119905)

(101)

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

for all 119905 isin [0 1] that is A is nondecreasing On the otherhand using the condition in (98) we estimate

119889 (AVA119906)

= sup119905isin[01]

|(AV) (119905) minus (A119906) (119905)|

= sup119905isin[01]

((AV) (119905) minus (A119906) (119905))

= sup119905isin[01]

int

1

0

119866 (119905 119904) [119891 (119904 V (119904)) minus 119891 (119904 119906 (119904))] 119889119904

le sup119905isin[01]

int

1

0

119866 (119905 119904) 120574radic(V (119904) minus 119906 (119904))2

(V (119904) minus 119906 (119904))2 + 1119889119904

le 120574 sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904

(102)

Moreover calculating the integral of Greenrsquos function whichis

int

1

0

119866 (119905 119904) 119889119904 = minus

1199052

2

+

119905

2

(103)

we obtain

sup119905isin[01]

int

1

0

119866 (119905 119904) 119889119904 =

1

8

(104)

Therefore inequality (102) and the assumption 0 lt 120574 lt 8result in

119889 (AVA119906) le120574

8

sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V (119905) minus 119906 (119905)|2 + 1

le sup119905isin[01]

radic|V (119905) minus 119906 (119905)|2

|V(119905) minus 119906(119905)|2 + 1

= radic119889 (V 119906)2

119889 (V 119906)2 + 1

(105)

or shortly in

119889(AVA119906)2 le119889(V 119906)2

119889(V 119906)2 + 1 (106)

Define the functions 120595(119905) = 1199052 120601(119905) = 1199052(1199052 + 1) as in theprevious example and note that these functions satisfy theconditions of Corollary 23 Inequality (106) becomes

120595 (119889 (AVA119906)) le 120601 (119889 (V 119906)) (107)

Finally since both functions 119891 and 119866 are nonnegative we get

(A0) (119905) = int1

0

119866 (119905 119904) 119891 (119904 0) 119889119904 ge 0 (108)

Therefore all the conditions of Corollary 23 hold and hencethe operatorA has a unique fixed point that is the problem(95) has a unique nonnegative solution

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Authorsrsquo Contribution

All authors contributed equally and significantly in writingthis paper All authors read and approved the final paper

References

[1] J J Nieto and R Rodrıguez-Lopez ldquoContractive mappingtheorems in partially ordered sets and applications to ordinarydifferential equationsrdquo Order vol 22 no 3 pp 223ndash239 2005

[2] J Harjani andK Sadarangani ldquoFixed point theorems for weaklycontractive mappings in partially ordered setsrdquoNonlinear Anal-ysis Theory Methods amp Applications vol 71 no 7-8 pp 3403ndash3410 2009

[3] J Harjani and K Sadarangani ldquoGeneralized contractions inpartially ordered metric spaces and applications to ordinarydifferential equationsrdquo Nonlinear Analysis Theory Methods ampApplications vol 72 no 3-4 pp 1188ndash1197 2010

[4] J Caballero J Harjani and K Sadarangani ldquoContractive-likemapping principles in ordered metric spaces and applicationto ordinary differential equationsrdquo Fixed Point Theory andApplications vol 2010 Article ID 916064 14 pages 2010

[5] A Amini-Harandi and H Emami ldquoA fixed point theoremfor contraction type maps in partially ordered metric spacesand application to ordinary differential equationsrdquo NonlinearAnalysis Theory Methods amp Applications vol 72 no 5 pp2238ndash2242 2010

[6] B Samet C Vetro and P Vetro ldquoFixed point theorems for120572-120595-contractive type mappingsrdquo Nonlinear Analysis TheoryMethods and Applications vol 75 no 4 pp 2154ndash2165 2012

[7] F Yan Y Su andQ Feng ldquoAnew contractionmapping principlein partially ordered metric spaces and applications to ordinarydifferential equationsrdquo Fixed PointTheory and Applications vol2012 article 152 2012

[8] M S Khan M Swaleh and S Sessa ldquoFixed point theorems byaltering distances between the pointsrdquo Bulletin of the AustralianMathematical Society vol 30 no 1 pp 1ndash9 1984

[9] E Karapınar P Kumam and P Salimi ldquoOn 120572120595-Meir-Keelercontractivemappingsrdquo Fixed PointTheory andApplications vol2013 article 94 2013

[10] E Karapınar and B Samet ldquoGeneralized 120572-120595 contractive typemappings and related fixed point theorems with applicationsrdquoAbstract and Applied Analysis vol 2012 Article ID 793486 17pages 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of