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

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 [1–7].

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. [8] have been widely used for this reason both alone or combined with other auxiliary functions.

Definition 1. An altering distance function is a function 𝜓∶[0,+∞)→[0,+∞) which satisfies the following:(1)𝜓is continuous and nondecreasing;(2)𝜓(𝑡)=0⇔𝑡=0.

Admissible mappings have been defined recently by Samet et al. [6] and employed quite often in order to generalize the results on various contractions. We state next the definitions of 𝛼-admissible mapping and triangular 𝛼-admissible mappings.

Definition 2. A mapping 𝑇∶𝑋→𝑋 is called 𝛼-admissible if for all 𝑥,𝑦∈𝑋 one has 𝛼𝑥,𝑦≥1⟹𝛼𝑇𝑥,𝑇𝑦≥1, (1) where 𝛼∶𝑋×𝑋→[0,∞) is a given function.

Definition 3. A mapping 𝑇∶𝑋→𝑋 is called triangular 𝛼-admissible if it is 𝛼-admissible and satisfies 𝛼𝛼()𝑥,𝑦≥1𝑦,𝑧≥1⟹𝛼𝑥,𝑧≥1, (2) where 𝑥,𝑦,𝑧∈𝑋 and 𝛼∶𝑋×𝑋→[0,∞) 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 𝑇∶𝑋→𝑋 is said to be weak triangular 𝛼-admissible if it is 𝛼-admissible and satisfies 𝛼(𝑥,𝑇𝑥)≥1⟹𝛼𝑥,𝑇²𝑥≥1, (3) where 𝑥∈𝑋 and 𝛼∶𝑋×𝑋→[0,∞) is a given function.

Weak triangular 𝛼-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 [9].

Lemma 5 (see [9]). Let 𝑇∶𝑋→𝑋 be a weak triangular 𝛼-admissible mapping. Assume that there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. If 𝑥_𝑛=𝑇^𝑛𝑥₀, then 𝛼(𝑥_𝑚,𝑥_𝑛)≥1 for all 𝑚,𝑛∈ℕ with 𝑚<𝑛.

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 𝛼-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 (𝑋,𝑑) be a metric space, 𝜓 an altering distance function, and 𝜙∶[0,+∞)→[0,+∞) a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0.(I)A mapping 𝑇∶𝑋→𝑋 belongs to class (I) if it satisfies 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋, (4)where 𝑀𝑑(),1𝑥,𝑦=max𝑥,𝑦,𝑑𝑥,𝑇𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥. (5)(II)A mapping 𝑇∶𝑋→𝑋 belongs to class (II) if it satisfies 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑁𝑥,𝑦,∀𝑥,𝑦∈𝑋, (6)where 𝑁𝑑,1𝑥,𝑦=max𝑥,𝑦2𝑑(),1𝑥,𝑇𝑥+𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥. (7)

Remark 7. Note that 𝑁(𝑥,𝑦)≤𝑀(𝑥,𝑦) for all 𝑥,𝑦∈𝑋.

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

Theorem 8. Let (𝑋,𝑑) be a complete metric space. Let 𝑇∶𝑋→𝑋 be a continuous, weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋, (8) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑀(𝑥,𝑦)=max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑥),𝑑(𝑦,𝑇𝑦),(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point.

Proof. Let 𝑥₀∈𝑋 satisfy 𝛼(𝑥₀,𝑇𝑥₀)≥1 and define the sequence {𝑥_𝑛} as 𝑥_𝑛=𝑇𝑥_𝑛−1 for 𝑛∈ℕ.
If 𝑥_𝑛0=𝑥_𝑛0+1 for some 𝑛₀∈ℕ, then, obviously, 𝑥_𝑛0=𝑇𝑥_𝑛0 is a fixed point of 𝑇. Suppose that 𝑑(𝑥_𝑛,𝑥_𝑛+1)>0 for all 𝑛∈ℕ.
Note that, due to the fact that 𝑇 is 𝛼-admissible and 𝛼(𝑥₀,𝑇𝑥₀)=𝛼(𝑥₀,𝑥₁)≥1, we deduce 𝛼𝑇𝑥₀,𝑇𝑥₁=𝛼𝑥₁,𝑥₂≥1⟹𝛼𝑇𝑥₁,𝑇𝑥₂=𝛼𝑥₂,𝑥₃≥1⟹⋯⟹𝛼𝑥_𝑛,𝑥_𝑛+1≥1. (9)
Substituting 𝑥=𝑥_𝑛 and 𝑦=𝑥_𝑛−1 in (8) and using (9) we get 𝜓𝑑𝑥_𝑛+1,𝑥_𝑛≤𝛼𝑥_𝑛,𝑥_𝑛−1𝜓𝑑𝑥_𝑛+1,𝑥_𝑛=𝛼𝑥_𝑛,𝑥_𝑛−1𝜓𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛−1≤𝜙𝑀𝑥_𝑛,𝑥_𝑛−1, (10) where 𝑀𝑥_𝑛,𝑥_𝑛−1=max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑇𝑥_𝑛,𝑑𝑥_𝑛−1,𝑇𝑥_𝑛−1,12𝑑𝑥_𝑛,𝑇𝑥_𝑛−1+𝑑𝑥_𝑛−1,𝑇𝑥_𝑛=max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛,12𝑑𝑥_𝑛,𝑥_𝑛+𝑑𝑥_𝑛−1,𝑥_𝑛+1=max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛+12. (11) Note that 𝑑(𝑥_𝑛−1,𝑥_𝑛+1)/2≤(1/2)[𝑑(𝑥_𝑛−1,𝑥_𝑛)+𝑑(𝑥_𝑛,𝑥_𝑛+1)] is smaller than both 𝑑(𝑥_𝑛−1,𝑥_𝑛) and 𝑑(𝑥_𝑛,𝑥_𝑛+1). Then, 𝑀(𝑥_𝑛,𝑥_𝑛−1) can be either 𝑑(𝑥_𝑛−1,𝑥_𝑛) or 𝑑(𝑥_𝑛,𝑥_𝑛+1). If 𝑀(𝑥_𝑛,𝑥_𝑛−1)=𝑑(𝑥_𝑛,𝑥_𝑛+1) for some 𝑛, then the expression (10) implies that 0<𝜓𝑑𝑥_𝑛+1,𝑥_𝑛≤𝜙𝑑𝑥_𝑛,𝑥_𝑛+1, (12) which contradicts the condition 𝜓(𝑡)>𝜙(𝑡) for 𝑡>0. Hence 𝑀(𝑥_𝑛,𝑥_𝑛−1)=𝑑(𝑥_𝑛,𝑥_𝑛−1) for all 𝑛≥1 and we have 0<𝜓𝑑𝑥_𝑛+1,𝑥_𝑛≤𝜙𝑑𝑥_𝑛,𝑥_𝑛−1<𝜓𝑑𝑥_𝑛,𝑥_𝑛−1, (13) which results in 𝑑𝑥_𝑛+1,𝑥_𝑛<𝑑𝑥_𝑛,𝑥_𝑛−1, (14) since 𝜓 is nondecreasing. Thus, we conclude that the nonnegative sequence 𝑑(𝑥_𝑛+1,𝑥_𝑛) is decreasing. Therefore, there exists 𝑟≥0 such that lim_𝑛→∞𝑑(𝑥_𝑛+1,𝑥_𝑛)=𝑟. Let 𝑛→∞ in (10); we get 𝜓(𝑟)≤𝜙(𝑟). (15) By the hypothesis of the theorem, since 𝜓(𝑡)>𝜙(𝑡), for all 𝑡>0, this inequality is possible only if 𝑟=0, and hence lim_𝑛→∞𝑑𝑥_𝑛+1,𝑥_𝑛=𝑟=0. (16)
Next, we will prove that {𝑥_𝑛} is a Cauchy sequence. Suppose, on the contrary, that {𝑥_𝑛} is not Cauchy. Then, for some 𝜀>0, there exist subsequences {𝑥_𝑚𝑘} and {𝑥_𝑛𝑘} of {𝑥_𝑛} such that 𝑛_𝑘>𝑚_𝑘>𝑘,𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≥𝜀 (17) for all 𝑘≥1, where, corresponding to each 𝑚_𝑘, we can choose 𝑛_𝑘 as the smallest integer with 𝑛_𝑘>𝑚_𝑘 for which (17) holds. Then 𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘−1<𝜀. (18) Employing triangle inequality and making use of (17) and (18), we obtain 𝜀≤𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘−1+𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘<𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘−1+𝜀. (19) Passing to limit as 𝑘→∞ and using (16), we get lim_𝑘→∞𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘=𝜀. (20) From the triangular inequality, we also have 𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘−1+𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1+𝑑𝑥_𝑚𝑘−1,𝑥_𝑚𝑘,𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1≤𝑑𝑥_𝑛𝑘−1,𝑥_𝑛𝑘+𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘+𝑑𝑥_𝑚𝑘,𝑥_𝑚𝑘−1. (21) Letting 𝑘→∞ in the two inequalities above and using (16) and (20), we get lim_𝑘→∞𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1=𝜀. (22) In a similar way, by using the triangular inequality, we obtain that 𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘−1+𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘,𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘≤𝑑𝑥_𝑛𝑘−1,𝑥_𝑛𝑘+𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘. (23) Taking limit as 𝑘→∞ in the above two inequalities and regarding (16) and (20), we get lim_𝑘→∞𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘=𝜀. (24) Furthermore, the relations 𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘−1+𝑑𝑥_𝑚𝑘−1,𝑥_𝑚𝑘,𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘−1≤𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘−1+𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1 (25) give lim_𝑘→∞𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘−1=𝜀 (26) by letting 𝑘→∞ and taking into account (16) and (20). By the definition of 𝑀(𝑥,𝑦) and using limits found above, we get lim_𝑘→∞𝑀𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1=𝜀. (27) Indeed, since 𝑀𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1=max𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1,𝑑𝑥_𝑛𝑘−1,𝑇𝑥_𝑛𝑘−1,𝑑𝑥_𝑚𝑘−1,𝑇𝑥_𝑚𝑘−1,12𝑑𝑥_𝑛𝑘−1,𝑇𝑥_𝑚𝑘−1+𝑑𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1=max𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1,𝑑𝑥_𝑛𝑘−1,𝑥_𝑛𝑘,𝑑𝑥_𝑚𝑘−1,𝑥_𝑚𝑘,12𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘+𝑑𝑥_𝑚𝑘−1,𝑥_𝑛𝑘, (28) passing to the limit as 𝑘→∞ in (28) and using (16), (20), (22), (24), and (26), we obtain lim_𝑛→∞𝑀𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−11=max𝜀,0,0,2[𝜀+𝜀]=𝜀. (29) Notice that since 𝑇 is weak triangular 𝛼-admissible, we deduce from Lemma 5 that 𝛼(𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1)≥1. Therefore, we can apply condition (8) to 𝑥_𝑛𝑘−1 and 𝑥_𝑚𝑘−1 to obtain 0<𝜓𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝛼𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1𝜓𝑑𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝜙𝑀𝑥_𝑛𝑘−1,𝑥_𝑚𝑘−1. (30) Letting 𝑘→∞ and taking into account (20) and (27), we have 0<𝜓(𝜀)≤𝜙(𝜀). (31) However, since 𝜓(𝑡)>𝜙(𝑡), for 𝑡>0, we deduce that 𝜀=0, which contradicts the assumption that {𝑥_𝑛} is not a Cauchy sequence. Thus, {𝑥_𝑛} must be Cauchy. Due to the fact that (𝑋,𝑑) is a complete metric space, there exists 𝑢∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑢. Finally, the continuity of 𝑇 gives 𝑢=lim_𝑛→∞𝑥_𝑛=lim_𝑛→∞𝑇𝑥_𝑛−1=𝑇𝑢. (32) That is, 𝑢 is a fixed point of 𝑇, which completes the proof.

One of the advantages of 𝛼-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 {𝑥_𝑛} is a sequence in 𝑋 such that 𝑥_𝑛⟶𝑥,𝛼𝑥_𝑛,𝑥_𝑛+1)≥1∀𝑛∈ℕ, (33)then there exists a subsequence {𝑥_𝑛𝑘} of {𝑥_𝑛} for which 𝛼𝑥_𝑛𝑘,𝑥≥1∀𝑘∈ℕ. (34)

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

Theorem 9. Let (𝑋,𝑑) be a complete metric space. Assume that 𝑋 satisfies condition (A). Let 𝑇∶𝑋→𝑋 be a weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋, (35) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑀(𝑥,𝑦)=max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑥),𝑑(𝑦,𝑇𝑦),(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point.

Proof. Following the proof of Theorem 8, it is clear that the sequence {𝑥_𝑛} defined by 𝑥_𝑛=𝑇𝑥_𝑛−1, for 𝑛∈ℕ, converges to a limit 𝑢∈𝑋. The only thing which remains to show is 𝑇𝑢=𝑢. Since lim_𝑛→∞𝑥_𝑛=𝑢, then condition (A) implies 𝛼(𝑥_𝑛𝑘,𝑢)≥1 for all 𝑘∈ℕ. Consequently, inequality (35) with 𝑥=𝑥_𝑛𝑘 and 𝑦=𝑢 becomes 𝜓𝑑𝑥_𝑛𝑘+1𝑥,𝑇𝑢≤𝛼_𝑛𝑘,𝑢𝜓𝑑𝑥_𝑛𝑘+1𝑥,𝑇𝑢=𝛼_𝑛𝑘,𝑢𝜓𝑑𝑇𝑥_𝑛𝑘𝑀𝑥,𝑇𝑢≤𝜙_𝑛𝑘,𝑢, (36) where 𝑀𝑥_𝑛𝑘𝑑𝑥,𝑢=max_𝑛𝑘,𝑢,𝑑𝑥_𝑛𝑘,𝑥_𝑛𝑘+1,𝑑(),1𝑢,𝑇𝑢2𝑑𝑥_𝑛𝑘,𝑇𝑢+𝑑𝑢,𝑥_𝑛𝑘+1. (37) Passing to limit as 𝑘→∞ and taking into account the continuity of 𝜓 and 𝜙, we get 𝜓(𝑑(𝑢,𝑇𝑢))≤𝜙(𝑑(𝑢,𝑇𝑢)). (38) From the condition 𝜓(𝑡)>𝜙(𝑡), for 𝑡>0, we conclude that 𝑑(𝑢,𝑇𝑢)=0 and, hence, 𝑇𝑢=𝑢, which completes the proof.

Similar results can be stated for a map 𝑇∶𝑋→𝑋 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 (𝑋,𝑑) be a complete metric space. Let 𝑇∶𝑋→𝑋 be a continuous, weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑁𝑥,𝑦,∀𝑥,𝑦∈𝑋, (39) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑁(𝑥,𝑦)=max{𝑑(𝑥,𝑦),(1/2)[𝑑(𝑥,𝑇𝑥)+𝑑(𝑦,𝑇𝑦)],(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point.

Theorem 11. Let (𝑋,𝑑) be a complete metric space. Assume that 𝑋 satisfies condition (A). Let 𝑇∶𝑋→𝑋 be a weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑁𝑥,𝑦,∀𝑥,𝑦∈𝑋, (40) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑁(𝑥,𝑦)=max{𝑑(𝑥,𝑦),(1/2)[𝑑(𝑥,𝑇𝑥)+𝑑(𝑦,𝑇𝑦)],(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 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)()For𝑥,𝑦∈𝑋,thereexists𝑧∈𝑋suchthat𝛼𝑥,𝑧≥1,𝛼𝑦,𝑧≥1. (41)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 𝑇 is unique.

Proof. Since 𝑇 satisfies the hypothesis of Theorem 10 (resp., Theorem 11), then fixed point of 𝑇 exists. Suppose that we have two different fixed points; say 𝑥,𝑦∈𝑋. From condition (B), there exists 𝑧∈𝑋, such that 𝛼(𝑥,𝑧)≥1,𝛼𝑦,𝑧≥1. (42) Then, since 𝑇 is 𝛼-admissible, we have from (42) 𝛼𝑥,𝑇^𝑛𝑧≥1,𝛼𝑦,𝑇^𝑛𝑧≥1,∀𝑛∈ℕ. (43) Thus, for the sequence {𝑧_𝑛}∈𝑋 defined as 𝑧_𝑛=𝑇^𝑛𝑧, we have 0<𝜓𝑑𝑥,𝑧_𝑛+1≤𝛼𝑥,𝑧_𝑛𝜓𝑑𝑇𝑥,𝑇𝑧_𝑛≤𝜙𝑁𝑥,𝑧_𝑛, (44) where 𝑁𝑥,𝑧_𝑛=max𝑑𝑥,𝑧_𝑛,12𝑑(𝑥,𝑇𝑥)+𝑑𝑧_𝑛,𝑇𝑧_𝑛,12𝑑𝑥,𝑇𝑧_𝑛+𝑑𝑧_𝑛𝑑,𝑇𝑥=max𝑥,𝑧_𝑛,𝑑𝑧_𝑛,𝑧_𝑛+12,12𝑑𝑥,𝑧_𝑛+1+𝑑𝑧_𝑛,𝑥. (45) Observe that 𝑑(𝑧_𝑛,𝑧_𝑛+1)/2≤(𝑑(𝑧_𝑛,𝑥)+𝑑(𝑥,𝑧_𝑛+1))/2. Then we deduce 𝑁(𝑥,𝑦)=max{𝑑(𝑥,𝑧_𝑛),𝑑(𝑥,𝑧_𝑛+1)}.
Without loss of generality, we may assume that 𝑑(𝑥,𝑧_𝑛)>0 for all 𝑛∈ℕ. If 𝑁(𝑥,𝑦)=𝑑(𝑥,𝑧_𝑛+1), then inequality (44) becomes 0<𝜓𝑑𝑥,𝑧_𝑛+1≤𝛼𝑥,𝑧_𝑛𝜓𝑑𝑇𝑥,𝑇𝑧_𝑛≤𝜙𝑑𝑥,𝑧_𝑛+1<𝜓𝑑𝑥,𝑧_𝑛+1. (46) That is, we have a contradiction. Then we should have 𝑁(𝑥,𝑦)=𝑑(𝑥,𝑧_𝑛) for all 𝑛∈ℕ, which results in 0<𝜓𝑑𝑥,𝑧_𝑛+1≤𝛼𝑥,𝑧_𝑛𝜓𝑑𝑇𝑥,𝑇𝑧_𝑛≤𝜙𝑑𝑥,𝑧_𝑛<𝜓𝑑𝑥,𝑧_𝑛, (47) due to the fact that 𝜓(𝑡)>𝜙(𝑡) for 𝑡>0. On the other hand, since 𝜓 is nondecreasing, then 𝑑(𝑥,𝑧_𝑛+1)≤𝑑(𝑥,𝑧_𝑛) for all 𝑛∈ℕ. Thus, the sequence {𝑑(𝑥,𝑧_𝑛)} is a positive nonincreasing sequence and, hence, converges to a limit; say, 𝐿≥0. Taking limit as 𝑛→∞ in (47) and regarding continuity of 𝜓 and 𝜙, we deduce 0≤𝜓(𝐿)≤𝜙(𝐿), (48) which is possible only if 𝐿=0. Hence, we conclude that lim_𝑛→∞𝑑𝑥,𝑧_𝑛=0. (49)
In a similar way, we obtain lim_𝑛→∞𝑑𝑦,𝑧_𝑛=0. (50) From (49) and (50) it follows that 𝑥=𝑦, which completes the proof of uniqueness.

The theorems stated above have been inspired by the recent results of Yan et al. [7]. 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 [7] in complete metric spaces.

Theorem 13. Let (𝑋,𝑑) be a complete metric space. Let 𝑇∶𝑋→𝑋 be a weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑑𝑥,𝑦,∀𝑥,𝑦∈𝑋, (51) where 𝜓 is an altering distance function and 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0. Assume either that 𝑇 is continuous or that 𝑋 satisfies condition (A). If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point. If, in addition, 𝑋 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 𝑥∈𝑋, lim_𝑛→∞𝑇^𝑛𝑥=𝑢, where 𝑢 is the unique fixed point (i.e., the operator 𝑇 is Picard).

The contractions of classes (I) and (II) are quite general and many particular results can be concluded from Theorems 8–12. Below we state some of these conclusions.

Corollary 15. Let (𝑋,𝑑) be a complete metric space. Let 𝑇∶𝑋→𝑋 be a continuous, weak triangular 𝛼-admissible mapping such that 𝛼𝑥,𝑦𝑑𝑇𝑥,𝑇𝑦≤𝑘𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋, (52) where 0<𝑘<1 and 𝑀(𝑥,𝑦)=max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑥),𝑑(𝑦,𝑇𝑦),(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point.

Proof. Proof is obvious by choosing 𝜓(𝑡)=𝑡 and 𝜙(𝑡)=𝑘𝑡 in Theorem 8.

Corollary 16. Let (𝑋,𝑑) be a complete metric space. Let 𝑇∶𝑋→𝑋 be a continuous, weak triangular 𝛼-admissible mapping such that 𝛼𝑑()+𝑒𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝑎𝑑𝑥,𝑦+𝑏𝑑𝑥,𝑇𝑥+𝑐𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥, (53) for all 𝑥,𝑦∈𝑋, where 0<𝑎+𝑏+𝑐+𝑒<1. If there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, then 𝑇 has a fixed point.

Proof. Due to the fact that ()+𝑒𝑎𝑑𝑥,𝑦+𝑏𝑑𝑥,𝑇𝑥+𝑐𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥≤𝑘𝑀𝑥,𝑦, (54) 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 𝛼-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 (𝑋,⪯) a map 𝑇∶𝑋→𝑋 is nondecreasing if it satisfies 𝑇𝑥⪯𝑇𝑦 for all 𝑥,𝑦∈𝑋 such that 𝑥⪯𝑦.

Definition 17. Let (𝑋,𝑑,⪯) be a metric space endowed with a partial order ⪯. If, for every nondecreasing sequence {𝑥_𝑛}⊂𝑋 which converges to 𝑥∈𝑋, there exists a subsequence {𝑥_𝑛𝑘} of {𝑥_𝑛} satisfying 𝑥_𝑛𝑘⪯𝑥, then (𝑋,𝑑,⪯) 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 (𝑋,𝑑,⪯) be a partially ordered complete metric space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping such that 𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋,𝑤𝑖𝑡ℎ𝑥⪯𝑦, (55) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑀(𝑥,𝑦)=max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑥), 𝑑(𝑦,𝑇𝑦),(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. Assume that there exists 𝑥₀∈𝑋 satisfying 𝑥₀⪯𝑇𝑥₀ and that either 𝑇 is continuous or (𝑋,𝑑,⪯) is regular. Then 𝑇 has a fixed point.

Proof. Define the map 𝛼∶𝑋×𝑋→[0,∞) as 𝛼𝑥,𝑦=1if𝑥⪯𝑦or𝑦⪯𝑥0otherwise. (56) It is clear that 𝑇 satisfies 𝛼𝑥,𝑦𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋, (57) where 𝛼 is defined in (56). Let 𝑥₀∈𝑋 satisfy 𝑥₀⪯𝑇𝑥₀. Then, 𝛼(𝑥₀,𝑇𝑥₀)≥1. On the other hand, since 𝑇 is nondecreasing, then 𝑇 is 𝛼-admissible. Indeed, 𝛼𝑥,𝑦≥1⟹𝑥⪯𝑦or𝑦⪯𝑥⟹𝑇𝑥⪯𝑇𝑦or𝑇𝑦⪯𝑇𝑥⟹𝛼𝑇𝑥,𝑇𝑦≥1. (58) Note also that if 𝑥⪯𝑇𝑥 then 𝑇𝑥⪯𝑇²𝑥, and hence 𝑥⪯𝑇²𝑥; that is, if 𝛼(𝑥,𝑇𝑥)≥1 then 𝛼(𝑇𝑥,𝑇²𝑥)≥1 and 𝛼(𝑥,𝑇²𝑥)≥1. Similar conclusion can be done if 𝑥⪰𝑇𝑥. Therefore, 𝑇 is weak triangular 𝛼-admissible. If 𝑇 is continuous, then 𝑇 satisfies the conditions of Theorem 8 and, hence, has a fixed point.
Suppose now that 𝑋 is regular. Then, every nondecreasing sequence {𝑥_𝑛} which converges to 𝑥 has a subsequence {𝑥_𝑛𝑘} for which 𝑥_𝑛𝑘⪯𝑥 holds for all 𝑘∈ℕ. Hence, 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 implies 𝛼(𝑥_𝑛𝑘,𝑥)≥1 for all 𝑘∈ℕ. In other words, the set 𝑋 satisfies condition (A). By Theorem 9, the mapping 𝑇 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 (𝑋,𝑑,⪯) be a partially ordered complete metric space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping such that 𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑁𝑥,𝑦,∀𝑥,𝑦∈𝑋,𝑤𝑖𝑡ℎ𝑥⪯𝑦, (59) where 𝜓 is an altering distance function, 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0, and 𝑁(𝑥,𝑦)=max{𝑑(𝑥,𝑦),(1/2)[𝑑(𝑥,𝑇𝑥)+𝑑(𝑦,𝑇𝑦)],(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. Assume that there exists 𝑥₀∈𝑋 satisfying 𝑥₀⪯𝑇𝑥₀ and that either 𝑇 is continuous or (𝑋,𝑑,⪯) is regular. Then 𝑇 has a fixed point.

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

(C) For all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 which is comparable to both 𝑥 and 𝑦.

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

Proof. Note that if 𝛼 is the map defined in (56) condition (C) is equivalent to condition (B). Indeed, assume that condition (B) holds. Then for all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1, where 𝛼 is defined in (56). This means that both 𝑥 and 𝑦 are comparable to 𝑧; that is, condition (C) also holds. If, on the other hand, condition (C) is satisfied then for all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 which is comparable to both 𝑥 and 𝑦. Then 𝛼(𝑥,𝑧)=1≥1 and 𝛼(𝑦,𝑧)=1≥1; that is, condition (B) is also satisfied. Hence, by Theorem 12, the fixed point of the map 𝑇 is unique.

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

Corollary 21. Let (𝑋,𝑑,⪯) be a partially ordered and complete metric space. Let 𝑇∶𝑋→𝑋 be nondecreasing mapping such that 𝑑𝑇𝑥,𝑇𝑦≤𝑘𝑀𝑥,𝑦,∀𝑥,𝑦∈𝑋𝑤𝑖𝑡ℎ𝑥⪯𝑦, (60) where 0<𝑘<1 and 𝑀(𝑥,𝑦)=max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑥), 𝑑(𝑦,𝑇𝑦),(1/2)[𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥)]}. Assume that there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀. If 𝑇 is continuous or 𝑋 is regular, then 𝑇 has a fixed point.

Proof. Proof is obvious by choosing 𝜓(𝑡)=𝑡 and 𝜙(𝑡)=𝑘𝑡 in Theorem 18.

Corollary 22. Let (𝑋,𝑑,⪯) be a partially ordered and complete metric space. Let 𝑇∶𝑋→𝑋 be nondecreasing mapping such that 𝑑()+𝑒𝑇𝑥,𝑇𝑦≤𝑎𝑑𝑥,𝑦+𝑏𝑑𝑥,𝑇𝑥+𝑐𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥,∀𝑥,𝑦∈𝑋𝑤𝑖𝑡ℎ𝑥⪯𝑦, (61) where 0<𝑎+𝑏+𝑐+𝑒<1. Assume that there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀. If 𝑇 is continuous or 𝑋 is regular, then 𝑇 has a fixed point.

Proof. Due to the fact that ()+𝑒𝑎𝑑𝑥,𝑦+𝑏𝑑𝑥,𝑇𝑥+𝑐𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥≤𝑘𝑀𝑥,𝑦, (62) 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 𝛼 defined in (56). It is actually the main result of Yan et al. [7].

Corollary 23. Let (𝑋,𝑑,⪯) be a partially ordered and complete metric space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping such that 𝜓𝑑𝑇𝑥,𝑇𝑦≤𝜙𝑑𝑥,𝑦,∀𝑥,𝑦∈𝑋𝑓𝑜𝑟𝑤ℎ𝑖𝑐ℎ𝑥⪯𝑦, (63) where 𝜓 is an altering distance function and 𝜙∶[0,+∞)→[0,+∞) is a continuous function satisfying 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0. Assume either that 𝑇 is continuous or that 𝑋 is regular. If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point. If, in addition, 𝑋 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 [1]. Specifically, we study the existence and uniqueness of a solution for the following first-order periodic boundary value problem: 𝑢^′(𝑡)((𝑡[],𝑢(0)(𝑇),=𝑓𝑡,𝑢)),if𝑡∈𝐼=0,𝑇=𝑢 (64) where 𝑇>0 and 𝑓∶𝐼×ℝ→ℝ is a continuous function. Let 𝐶(𝐼) denote the space of continuous functions defined on 𝐼. Clearly, the function 𝑑𝑥,𝑦=sup_𝑡∈𝐼𝑥(𝑡)−𝑦(𝑡),𝑥,𝑦∈𝐶(𝐼) (65) defines a metric on 𝐶(𝐼). Moreover, (𝐶(𝐼),𝑑) is a complete metric space. Define also a partial order on 𝐶(𝐼) by 𝑥,𝑦∈𝐶(𝐼),𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡)∀𝑡∈𝐼. (66)

It is easy to see that 𝐶(𝐼) satisfies condition (C). Indeed, for 𝑥,𝑦∈𝐶(𝐼), we have 𝑥≤𝑧 and 𝑦≤𝑧, where 𝑧=max_𝑡∈𝐼𝑥(𝑡)(𝑡=𝑥(𝑡)(𝑡)(𝑡)𝑦(𝑡)(𝑡)(𝑡),𝑦)if𝑥≥𝑦if𝑦≥𝑥 (67) is also in 𝐶(𝐼) whenever 𝑥,𝑦∈𝐶(𝐼). Nieto and Rodríguez-López [1] proved that (𝐶(𝐼),𝑑,≤), where the metric 𝑑 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 𝛽∈𝐶¹(𝐼) satisfying 𝛽^′(𝑡)(𝑡)𝛽(0)(𝑇).≤𝑓𝑡,𝛽,for𝑡∈𝐼,≤𝛽 (68)

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

Theorem 25. Let the function 𝑓 in the problem (64) be continuous and let 0≤𝑓𝑡,𝑦+𝜆𝑦−𝑓(𝑡,𝑥)+𝜆𝑥≤𝛾(𝑥−𝑦)²(𝑥−𝑦)²+1^1/2 (69) hold for 𝑥,𝑦∈ℝ with 𝑦≥𝑥 and for some 𝜆,𝛾>0 satisfying 𝛾≤2𝜆(𝑒^𝜆𝑇−1)𝑇(𝑒^𝜆𝑇+1)^1/2. (70) If the problem (64) has a lower solution, then it has a unique solution.

Proof. Rewrite the problem (64) as 𝑢^′(𝑡)(𝑡)((𝑡(𝑡)[],𝑢(0)(𝑇).+𝜆𝑢=𝑓𝑡,𝑢))+𝜆𝑢,for𝑡∈𝐼=0,𝑇=𝑢 (71)
Clearly, the problem (64) is equivalent to the integral equation 𝑢(𝑡)=^𝑇₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))+𝜆𝑢(𝑠)𝑑𝑠, (72) where 𝐺(𝑡,𝑠) is the Green function given by 𝐺()=𝑒𝑡,𝑠𝑒^{𝜆(𝑇+𝑠−𝑡)𝜆𝑇}𝑒−1,0≤𝑠<𝑡≤𝑇,_{𝜆(𝑠−𝑡)}𝑒^𝜆𝑇−1,0≤𝑡<𝑠≤𝑇. (73) Define the map 𝒜∶𝐶(𝐼)→𝐶(𝐼) by (𝒜𝑢)(𝑡)=^𝑇₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))+𝜆𝑢(𝑠)𝑑𝑠,𝑡∈𝐼. (74) It is obvious that a fixed point 𝑢∈𝐶(𝐼) of 𝒜 is a solution of (64). We will show that the mapping 𝐴 satisfies the conditions of Corollary 23.
First, note that 𝒜 is nondecreasing by the hypothesis; that is, for 𝑢≥𝑣, 𝑓(𝑡,𝑢)+𝜆𝑢≥𝑓(𝑡,𝑣)+𝜆𝑣. (75) And, hence, (𝒜𝑢)(𝑡)=^𝑇₀𝐺(𝑓((𝑠(𝑠≥𝑡,𝑠)𝑠,𝑢))+𝜆𝑢)𝑑𝑠_𝑇0𝐺(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))+𝜆𝑣(𝑠)𝑑𝑠=(𝒜𝑣)(𝑡), (76) since 𝐺(𝑡,𝑠)>0 for (𝑡,𝑠)∈𝐼×𝐼.
Note also that, for 𝑢≥𝑣, we have 𝑑(𝒜𝑢,𝒜𝑣)=sup_𝑡∈𝐼|(𝒜𝑢)(𝑡)−(𝒜𝑣)(𝑡)|=sup_𝑡∈𝐼((𝒜𝑢)(𝑡)−(𝒜𝑣)(𝑡))=sup_𝑡∈𝐼^𝑇₀𝐺(𝑓((𝑠(𝑠)((𝑠(𝑠𝑡,𝑠)𝑠,𝑢))+𝜆𝑢−𝑓𝑠,𝑣))+𝜆𝑣)𝑑𝑠≤sup_𝑡∈𝐼^𝑇₀𝐺(𝑡,𝑠)𝛾[𝑢(𝑠)−𝑣(𝑠)]²[𝑢(𝑠)−𝑣(𝑠)]²+1𝑑𝑠. (77)
Employing the Cauchy-Schwarz inequality for the last integral, we obtain ^𝑇₀𝐺(𝑡,𝑠)𝛾(𝑢(𝑠)−𝑣(𝑠))²(𝑢(𝑠)−𝑣(𝑠))²≤+1𝑑𝑠_𝑇0𝐺(𝑡,𝑠)²𝑑𝑠^1/2𝑇₀𝛾²[𝑢(𝑠)−𝑣(𝑠)]²[𝑢(𝑠)−𝑣(𝑠)]²+1𝑑𝑠^1/2. (78) The first integral on the right-hand side can be calculated easily and gives ^𝑇₀𝐺(𝑡,𝑠)²𝑑𝑠=^𝑡₀𝐺(𝑡,𝑠)²𝑑𝑠+^𝑇_𝑡𝐺(𝑡,𝑠)²=𝑑𝑠_𝑡0𝑒^{2𝜆(𝑇+𝑠−𝑡)}𝑒^𝜆𝑇−1²𝑑𝑠+^𝑇_𝑡𝑒^{2𝜆(𝑠−𝑡)}𝑒^𝜆𝑇−1²=𝑒𝑑𝑠_2𝜆𝑇−12𝜆𝑒^𝜆𝑇−1²=𝑒^𝜆𝑇+12𝜆𝑒^𝜆𝑇.−1 (79) For the second integral in (78) we obtain the following estimate: ^𝑇₀𝛾²[𝑢(𝑠)−𝑣(𝑠)]²[𝑢(𝑠)−𝑣(𝑠)]²+1𝑑𝑠≤𝛾²sup_𝑡∈𝐼|𝑢(𝑡)−𝑣(𝑡)|²|𝑢(𝑡)−𝑣(𝑡)|²+1⋅𝑇=𝛾²𝑑(𝑢,𝑣)²𝑑(𝑢,𝑣)²+1𝑇. (80)
Making use of the inequalities in (77), (78), and (80) and the integral in (79) we deduce 𝑑()≤𝑒𝒜𝑢,𝒜𝑣_𝜆𝑇+12𝜆𝑒^𝜆𝑇−1^1/2⋅𝛾²𝑑(𝑢,𝑣)²𝑑(𝑢,𝑣)²𝑇+1^1/2=𝑒^𝜆𝑇+12𝜆𝑒^𝜆𝑇−1^1/2𝛾𝑑(𝑢,𝑣)²𝑑(𝑢,𝑣)²+1^1/2√𝑇. (81) And, consequently, 𝑑(𝒜𝑢,𝒜𝑣)²≤𝛾²𝑒^𝜆𝑇+12𝜆𝑒^𝜆𝑇−1𝑑(𝑢,𝑣)²𝑑(𝑢,𝑣)²+1𝑇. (82) This last inequality can be also written as 2𝜆𝑒^𝜆𝑇−1𝑑(𝑢,𝑣)²+1𝑑(𝒜𝑢,𝒜𝑣)²≤𝛾²𝑒^𝜆𝑇+1𝑑(𝑢,𝑣)²𝑇, (83) since 𝜆 is positive. The constant 𝛾, on the other hand, satisfies 𝑒𝛾≤2𝜆^𝜆𝑇−1𝑇𝑒^𝜆𝑇+1^1/2, (84) and, therefore, inequality (83) implies 2𝜆𝑒^𝜆𝑇−1𝑑(𝑢,𝑣)²+1𝑑(𝒜𝑢,𝒜𝑣)²≤2𝜆𝑒^𝜆𝑇−1𝑇𝑒^𝜆𝑇𝑒+1^𝜆𝑇+1𝑑(𝑢,𝑣)²𝑇, (85) and, thus, 𝑑(𝒜𝑢,𝒜𝑣)²≤𝑑(𝑢,𝑣)²𝑑(𝑢,𝑣)².+1 (86) Define the functions 𝜓 and 𝜙 as 𝜓(𝑡)=𝑡² and 𝜙(𝑡)=𝑡²/(𝑡²+1). It is clear that 𝜓 is an altering distance function and also that 𝜓(𝑡)>𝜙(𝑡) for all 𝑡>0. Then (86) becomes 𝜓(𝑑(𝒜𝑢,𝒜𝑣))≤𝜙(𝑑(𝑢,𝑣)), (87) for all 𝑢≥𝑣; that is, 𝒜 satisfies the contractive condition of Corollary 23. Finally, let 𝛽(𝑡) be a lower solution for (86). We will show that 𝛽≤𝒜𝛽. Multiplying the inequality 𝛽^′(𝑡)+𝜆𝛽(𝑡)≤𝑓𝑡,𝛽(𝑡)+𝜆𝛽(𝑡),for𝑡∈𝐼, (88) by 𝑒^𝜆𝑡, we obtain 𝛽(𝑡)𝑒^𝜆𝑡′≤𝑓𝑡,𝛽(𝑡)+𝜆𝛽(𝑡)𝑒^𝜆𝑡,for𝑡∈𝐼, (89) which upon integration gives 𝛽(𝑡)𝑒^𝜆𝑡≤𝛽(0)+^𝑡₀𝑓𝑠,𝛽(𝑠)+𝜆𝛽(𝑠)𝑒^𝜆𝑠𝑑𝑠,for𝑡∈𝐼. (90) Since 𝛽(0)≤𝛽(𝑇), the last inequality implies 𝛽(0)𝑒^𝜆𝑡≤𝛽(𝑇)𝑒^𝜆𝑇≤𝛽(0)+^𝑇₀𝑓𝑠,𝛽(𝑠)+𝜆𝛽(𝑠)𝑒^𝜆𝑠𝑑𝑠, (91) and, hence, 𝛽(0)≤^𝑇₀𝑒^𝜆𝑠𝑒^𝜆𝑇𝑓(𝑠(𝑠−1𝑠,𝛽)+𝜆𝛽)𝑑𝑠. (92) From this inequality and inequality (90) we obtain 𝛽(𝑡)𝑒^𝜆𝑡≤^𝑡₀𝑒^{𝜆(𝑇+𝑠)}𝑒^𝜆𝑇𝑓(𝑠(𝑠+−1𝑠,𝛽)+𝜆𝛽)𝑑𝑠_𝑇𝑡𝑒^𝜆𝑠𝑒^𝜆𝑇𝑓(𝑠)(𝑠−1𝑠,𝛽+𝜆𝛽)𝑑𝑠, (93) and, consequently, 𝛽(𝑡)≤^𝑡₀𝑒^{𝜆(𝑇+𝑠−𝑡)}𝑒^𝜆𝑇𝑓(𝑠(𝑠+−1𝑠,𝛽)+𝜆𝛽)𝑑𝑠_𝑇𝑡𝑒^{𝜆(𝑠−𝑡)}𝑒^𝜆𝑇𝑓(𝑠)(𝑠=−1𝑠,𝛽+𝜆𝛽)𝑑𝑠_𝑇0𝐺(𝑓(𝑠)(𝑠=(𝑡)𝑡,𝑠)𝑠,𝛽+𝜆𝛽)𝑑𝑠𝒜𝛽,for𝑡∈𝐼. (94) We have shown that all conditions of Corollary 23 hold. Therefore 𝒜 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 −𝑑²𝑢𝑑𝑡²()[)[],𝑢(0)(1)=𝑓𝑡,𝑢,𝑢∈𝐼=0,∞,𝑡∈0,1=𝑢=0. (95)

The problem (95) is equivalent to the integral equation 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑡∈[0,1], (96) where 𝐺(𝑡,𝑠) is the Green function given by 𝐺()=𝑡()𝑠()𝑡,𝑠1−𝑠,0≤𝑡<𝑠≤1,1−𝑡,0≤𝑠<𝑡≤1. (97) 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 𝑓∶[0,1]×ℝ→[0,∞) be a continuous function which is nondecreasing in its second variable. Suppose that for all 𝑢,𝑣∈ℝ satisfying 𝑢≤𝑣, there exists a constant 0<𝛾<8 such that 𝑓()()()𝑡,𝑣−𝑓𝑡,𝑢≤𝛾𝑢−𝑣²(𝑢−𝑣)²+1^1/2. (98) Then the problem (95) has a unique nonnegative solution.

Proof. Consider the space 𝑌={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0} and define a metric 𝑑(𝑢,𝑣)=sup{|𝑢(𝑡)−𝑣(𝑡)|∶𝑡∈[0,1]} as usual. Obviously, (𝑌,𝑑) is a complete metric space. Define also the usual partial order on 𝑌; that is, 𝑢≤𝑣⟹𝑢(𝑡)≤𝑣(𝑡)∀𝑡∈[0,1]. (99) As in the previous example, a solution of the problem (95) is a fixed point of the operator 𝒜∶𝑌→𝑌 defined by (𝒜𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑡∈[0,1], (100) where 𝐺(𝑡,𝑠) is the Green function in (97). Since 𝑓 is nondecreasing with respect to its second variable, then for 𝑢,𝑣∈𝑌 with 𝑣≥𝑢 we have (𝒜𝑣)(𝑡)=¹₀𝐺()𝑓((𝑠≥𝑡,𝑠𝑠,𝑣))𝑑𝑠₁₀𝐺()𝑓((𝑠=(𝑡),𝑡,𝑠𝑠,𝑢))𝑑𝑠𝒜𝑢)( (101) for all 𝑡∈[0,1]; that is, 𝒜 is nondecreasing. On the other hand, using the condition in (98), we estimate 𝑑(𝒜𝑣,𝒜𝑢)=sup_𝑡∈[0,1]|(𝒜𝑣)(𝑡)−(𝒜𝑢)(𝑡)|=sup_𝑡∈[0,1]((𝒜𝑣)(𝑡)−(𝒜𝑢)(𝑡))=sup_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))−𝑓(𝑠,𝑢(𝑠))𝑑𝑠≤sup_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝛾(𝑣(𝑠)−𝑢(𝑠))²(𝑣(𝑠)−𝑢(𝑠))²+1𝑑𝑠≤𝛾sup_𝑡∈[0,1]|𝑣(𝑡)−𝑢(𝑡)|²|𝑣(𝑡)−𝑢(𝑡)|²+1sup_𝑡∈[0,1]¹₀𝐺(𝑡,𝑠)𝑑𝑠. (102) Moreover, calculating the integral of Green's function, which is ¹₀𝐺()𝑡𝑡,𝑠𝑑𝑠=−²2+𝑡2, (103) we obtain sup_𝑡∈[0,1]¹₀𝐺()1𝑡,𝑠𝑑𝑠=8. (104) Therefore, inequality (102) and the assumption 0<𝛾<8 result in 𝑑()≤𝛾𝒜𝑣,𝒜𝑢8sup_𝑡∈[0,1]|𝑣(𝑡)−𝑢(𝑡)|²|𝑣(𝑡)−𝑢(𝑡)|²+1≤sup_𝑡∈[0,1]|𝑣(𝑡)−𝑢(𝑡)|²|𝑣(𝑡)−𝑢(𝑡)|²=+1𝑑(𝑣,𝑢)²𝑑(𝑣,𝑢)²,+1 (105) or, shortly, in 𝑑(𝒜𝑣,𝒜𝑢)²≤𝑑(𝑣,𝑢)²𝑑(𝑣,𝑢)².+1 (106) Define the functions 𝜓(𝑡)=𝑡², 𝜙(𝑡)=𝑡²/(𝑡²+1) as in the previous example, and note that these functions satisfy the conditions of Corollary 23. Inequality (106) becomes 𝜓(𝑑(𝒜𝑣,𝒜𝑢))≤𝜙(𝑑(𝑣,𝑢)). (107) Finally, since both functions 𝑓 and 𝐺 are nonnegative, we get (𝒜0)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,0)𝑑𝑠≥0. (108) Therefore, all the conditions of Corollary 23 hold and, hence, the operator 𝒜 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 interests regarding the publication of this paper.

Authors’ Contribution

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