/ / Article
Special Issue

## Variational Analysis, Optimization, and Fixed Point Theory 2014

View this Special Issue

Research Article | Open Access

Volume 2014 |Article ID 187031 | 10 pages | https://doi.org/10.1155/2014/187031

# Fixed Point Theorems for a Class of α-Admissible Contractions and Applications to Boundary Value Problem

Accepted12 Jun 2014
Published03 Jul 2014

#### Abstract

A class of α-admissible contractions defined via altering distance functions is introduced. The existence and uniqueness conditions for fixed points of such maps on complete metric spaces are investigated and related fixed point theorems are presented. The results are reconsidered in the context of partially ordered metric spaces and applied to boundary value problems for differential equations with periodic boundary conditions.

#### 1. Introduction and Preliminaries

Recent developments in fixed point theory have shown the significance of theoretical studies which are directly applicable in other areas. In particular, the problems related with existence and uniqueness of solutions of integral and differential equations are of particular importance. Differential and integral equations govern the behaviour of various real-life problems for which the question of existence and uniqueness of solutions is crucial. This fact motivates the intensive research activities in the area and the rapidly increasing number of publications .

The main goal of studies in fixed point theory is to improve the contractive conditions imposed on the mappings under consideration. Altering distance functions defined by Khan et al.  have been widely used for this reason both alone or combined with other auxiliary functions.

Definition 1. An altering distance function is a function which satisfies the following:(1)is continuous and nondecreasing;(2).

Admissible mappings have been defined recently by Samet et al.  and employed quite often in order to generalize the results on various contractions. We state next the definitions of -admissible mapping and triangular -admissible mappings.

Definition 2. A mapping is called -admissible if for all one has where is a given function.

Definition 3. A mapping is called triangular -admissible if it is -admissible and satisfies where and is a given function.

Inspired by this definition, we define the following weaker condition which proves to be sufficient in the forthcoming discussions.

Definition 4. A mapping is said to be weak triangular -admissible if it is -admissible and satisfies where and is a given function.

Weak triangular -admissible mappings satisfy a property stated in the following lemma and the proof of which easily follows from the definition and can be found in .

Lemma 5 (see ). Let be a weak triangular -admissible mapping. Assume that there exists such that . If , then for all with .

#### 2. Existence and Uniqueness Theorems on Complete Metric Spaces

In this section we present our main results which include theorems on existence and uniqueness of fixed points for a class of weak triangular -admissible mappings.

First we define the following two classes of contractions which we are going to investigate in this section and throughout the paper.

Definition 6. Let be a metric space, an altering distance function, and a continuous function satisfying for all .(I)A mapping belongs to class (I) if it satisfies where (II)A mapping belongs to class (II) if it satisfies where

Remark 7. Note that for all .

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

Theorem 8. Let be a complete metric space. Let be a continuous, weak triangular -admissible mapping such that where is an altering distance function, is a continuous function satisfying for all , and . If there exists such that , then has a fixed point.

Proof. Let satisfy and define the sequence as for .
If for some , then, obviously, is a fixed point of . Suppose that for all .
Note that, due to the fact that is -admissible and , we deduce
Substituting and in (8) and using (9) we get where Note that is smaller than both and . Then, can be either or . If for some , then the expression (10) implies that which contradicts the condition for . Hence for all and we have which results in since is nondecreasing. Thus, we conclude that the nonnegative sequence is decreasing. Therefore, there exists such that . Let in (10); we get By the hypothesis of the theorem, since , for all , this inequality is possible only if , and hence
Next, we will prove that is a Cauchy sequence. Suppose, on the contrary, that is not Cauchy. Then, for some , there exist subsequences and of such that for all , where, corresponding to each , we can choose as the smallest integer with for which (17) holds. Then Employing triangle inequality and making use of (17) and (18), we obtain Passing to limit as and using (16), we get From the triangular inequality, we also have Letting in the two inequalities above and using (16) and (20), we get In a similar way, by using the triangular inequality, we obtain that Taking limit as in the above two inequalities and regarding (16) and (20), we get Furthermore, the relations give by letting and taking into account (16) and (20). By the definition of and using limits found above, we get Indeed, since passing to the limit as in (28) and using (16), (20), (22), (24), and (26), we obtain Notice that since is weak triangular -admissible, we deduce from Lemma 5 that . Therefore, we can apply condition (8) to and to obtain Letting and taking into account (20) and (27), we have However, since , for , we deduce that , which contradicts the assumption that is not a Cauchy sequence. Thus, must be Cauchy. Due to the fact that is a complete metric space, there exists such that . Finally, the continuity of gives That is, is a fixed point of , which completes the proof.

One of the advantages of -admissible mappings is that the continuity of the map is no longer required for the existence of a fixed point provided that the space under consideration has the following property.(A)If is a sequence in such that then there exists a subsequence of for which

Bearing this fact in mind, we rewrite the statement of Theorem 8 in the light of this property.

Theorem 9. Let be a complete metric space. Assume that satisfies condition (A). Let be a weak triangular -admissible mapping such that where is an altering distance function, is a continuous function satisfying for all , and . If there exists such that , then has a fixed point.

Proof. Following the proof of Theorem 8, it is clear that the sequence defined by , for , converges to a limit . The only thing which remains to show is . Since , then condition (A) implies for all . Consequently, inequality (35) with and becomes where Passing to limit as and taking into account the continuity of and , we get From the condition , for , we conclude that and, hence, , which completes the proof.

Similar results can be stated for a map in the class (II). More precisely, conditions for existence of a fixed point of a map in class (II) are given in the next two theorems.

Theorem 10. Let be a complete metric space. Let be a continuous, weak triangular -admissible mapping such that where is an altering distance function, is a continuous function satisfying for all , and . If there exists such that , then has a fixed point.

Theorem 11. Let be a complete metric space. Assume that satisfies condition (A). Let be a weak triangular -admissible mapping such that where is an altering distance function, is a continuous function satisfying for all , and . If there exists such that , then has a fixed point.

Note that the proofs of Theorems 10 and 11 can be easily done by mimicking the proofs of Theorems 8 and 9, respectively.

We next discuss the conditions for the uniqueness of the fixed point. A sufficient condition for the uniqueness of the fixed point in Theorems 10 and 11 can be stated as follows.(B)Note, however, that this condition is not sufficient for the uniqueness of fixed point for maps of class (I).

Theorem 12. If condition (B) is added to the hypotheses of Theorem 10 (resp., Theorem 11), then the fixed point of is unique.

Proof. Since satisfies the hypothesis of Theorem 10 (resp., Theorem 11), then fixed point of exists. Suppose that we have two different fixed points; say . From condition (B), there exists , such that Then, since is -admissible, we have from (42) Thus, for the sequence defined as , we have where Observe that . Then we deduce .
Without loss of generality, we may assume that for all . If , then inequality (44) becomes That is, we have a contradiction. Then we should have for all , which results in due to the fact that for . On the other hand, since is nondecreasing, then for all . Thus, the sequence is a positive nonincreasing sequence and, hence, converges to a limit; say, . Taking limit as in (47) and regarding continuity of and , we deduce which is possible only if . Hence, we conclude that
In a similar way, we obtain From (49) and (50) it follows that , which completes the proof of uniqueness.

The theorems stated above have been inspired by the recent results of Yan et al. . They discussed contraction mappings defined on partially ordered complete metric spaces and their applications to boundary value problems. We state next a theorem which can be regarded as a generalization of the main result in  in complete metric spaces.

Theorem 13. Let be a complete metric space. Let be a weak triangular -admissible mapping such that where is an altering distance function and is a continuous function satisfying for all . Assume either that is continuous or that satisfies condition (A). If there exists such that , then has a fixed point. If, in addition, satisfies condition (B), then the fixed point is unique.

Proof of Theorem 13 can be done by following the lines of proofs of Theorems 8, 9, and 12. Hence, it is omitted.

Remark 14. Under the assumptions of Theorem 12, it can be proved that, for every , , where is the unique fixed point (i.e., the operator is Picard).

The contractions of classes (I) and (II) are quite general and many particular results can be concluded from Theorems 812. Below we state some of these conclusions.

Corollary 15. Let be a complete metric space. Let be a continuous, weak triangular -admissible mapping such that where and . If there exists such that , then has a fixed point.

Proof. Proof is obvious by choosing and in Theorem 8.

Corollary 16. Let be a complete metric space. Let be a continuous, weak triangular -admissible mapping such that for all , where . If there exists such that , then has a fixed point.

Proof. Due to the fact that the proof follows from Corollary 15.

#### 3. Fixed Points on Partially Ordered Metric Spaces

It has been pointed out in some studies that some results in metric spaces endowed with a partial order can be concluded from the fixed point results related with -admissible maps on metric spaces (see [9, 10]). In this section we give existence and uniqueness theorems on partially ordered metric spaces which can be regarded as consequences of the theorems presented in the previous section.

Recall that on a partially ordered set a map is nondecreasing if it satisfies for all such that .

Definition 17. Let be a metric space endowed with a partial order . If, for every nondecreasing sequence which converges to , there exists a subsequence of satisfying , then is said to be regular.

Our first theorem contains the conditions for existence of a fixed point for a map of class (I) defined on a partially ordered metric space.

Theorem 18. Let be a partially ordered complete metric space. Let be a nondecreasing mapping such that where is an altering distance function, is a continuous function satisfying for all , and , . Assume that there exists satisfying and that either is continuous or is regular. Then has a fixed point.

Proof. Define the map as It is clear that satisfies where is defined in (56). Let satisfy . Then, . On the other hand, since is nondecreasing, then is -admissible. Indeed, Note also that if then , and hence ; that is, if then and . Similar conclusion can be done if . Therefore, is weak triangular -admissible. If is continuous, then satisfies the conditions of Theorem 8 and, hence, has a fixed point.
Suppose now that is regular. Then, every nondecreasing sequence which converges to has a subsequence for which holds for all . Hence, implies for all . In other words, the set satisfies condition (A). By Theorem 9, the mapping has a fixed point.

Analogously, we state conditions for existence of fixed points for maps from class (II) on partially ordered metric spaces.

Theorem 19. Let be a partially ordered complete metric space. Let be a nondecreasing mapping such that where is an altering distance function, is a continuous function satisfying for all , and . Assume that there exists satisfying and that either is continuous or is regular. Then has a fixed point.

The uniqueness of a fixed point on partially ordered metric spaces requires an additional assumption on the set . This assumption reads as follows.

(C) For all there exists which is comparable to both and .

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

Proof. Note that if is the map defined in (56) condition (C) is equivalent to condition (B). Indeed, assume that condition (B) holds. Then for all there exists such that and , where is defined in (56). This means that both and are comparable to ; that is, condition (C) also holds. If, on the other hand, condition (C) is satisfied then for all there exists which is comparable to both and . Then and ; that is, condition (B) is also satisfied. Hence, by Theorem 12, the fixed point of the map is unique.

Some consequences of the results stated above can be easily concluded. We next present two of them.

Corollary 21. Let be a partially ordered and complete metric space. Let be nondecreasing mapping such that where and , . Assume that there exists such that . If is continuous or is regular, then has a fixed point.

Proof. Proof is obvious by choosing and in Theorem 18.

Corollary 22. Let be a partially ordered and complete metric space. Let be nondecreasing mapping such that where . Assume that there exists such that . If is continuous or is regular, then has a fixed point.

Proof. Due to the fact that the proof follows from Corollary 21.

Our last result is a consequence of Theorem 13 on partially ordered metric spaces and follows easily by using the function defined in (56). It is actually the main result of Yan et al. .

Corollary 23. Let be a partially ordered and complete metric space. Let be a nondecreasing mapping such that where is an altering distance function and is a continuous function satisfying for all . Assume either that is continuous or that is regular. If there exists such that , then has a fixed point. If, in addition, satisfies condition (C), then the fixed point is unique.

#### 4. Application to Ordinary Differential Equations

In this section, we discuss application of our results to existence and uniqueness of solutions of boundary value problems. We present two examples, first of which has been inspired by . Specifically, we study the existence and uniqueness of a solution for the following first-order periodic boundary value problem: where and is a continuous function. Let denote the space of continuous functions defined on . Clearly, the function defines a metric on . Moreover, is a complete metric space. Define also a partial order on by

It is easy to see that satisfies condition (C). Indeed, for , we have and , where is also in whenever . Nieto and Rodríguez-López  proved that , where the metric and the partial order are defined above, satisfies regularity condition given in Definition 17. We now define a lower solution for the problem (64).

Definition 24. A lower solution for (64) is a function satisfying

Our next theorem gives conditions for existence and uniqueness of solution to the problem (64).

Theorem 25. Let the function in the problem (64) be continuous and let hold for with and for some satisfying If the problem (64) has a lower solution, then it has a unique solution.

Proof. Rewrite the problem (64) as
Clearly, the problem (64) is equivalent to the integral equation where is the Green function given by Define the map by It is obvious that a fixed point of is a solution of (64). We will show that the mapping satisfies the conditions of Corollary 23.
First, note that is nondecreasing by the hypothesis; that is, for , And, hence, since for .
Note also that, for , we have
Employing the Cauchy-Schwarz inequality for the last integral, we obtain The first integral on the right-hand side can be calculated easily and gives For the second integral in (78) we obtain the following estimate:
Making use of the inequalities in (77), (78), and (80) and the integral in (79) we deduce And, consequently, This last inequality can be also written as since is positive. The constant , on the other hand, satisfies and, therefore, inequality (83) implies and, thus, Define the functions and as and . It is clear that is an altering distance function and also that for all . Then (86) becomes for all ; that is, satisfies the contractive condition of Corollary 23. Finally, let be a lower solution for (86). We will show that . Multiplying the inequality by , we obtain which upon integration gives Since , the last inequality implies and, hence, From this inequality and inequality (90) we obtain and, consequently, We have shown that all conditions of Corollary 23 hold. Therefore has a unique fixed point, or, equivalently, the problem (64) has a unique solution.

As a second example we discuss the existence and uniqueness of solution for a second-order boundary value problem. More precisely, we consider

The problem (95) is equivalent to the integral equation where is the Green function given by In what follows, our last theorem gives conditions for the existence and uniqueness of solution to the problem (95) or, equivalently, the problem (96).

Theorem 26. Let be a continuous function which is nondecreasing in its second variable. Suppose that for all satisfying , there exists a constant such that Then the problem (95) has a unique nonnegative solution.

Proof. Consider the space and define a metric as usual. Obviously, is a complete metric space. Define also the usual partial order on ; that is, As in the previous example, a solution of the problem (95) is a fixed point of the operator defined by where is the Green function in (97). Since is nondecreasing with respect to its second variable, then for with we have for all ; that is, is nondecreasing. On the other hand, using the condition in (98), we estimate