Existence of Solutions for a Periodic Boundary Value Problem via Generalized Weakly Contractions

Abstract

We discuss the existence of solutions for a periodic boundary value problem and for some polynomials. For this purpose, we present some fixed point theorems for weakly and generalized weakly contractive mappings in the setting of partially ordered complete metric spaces.

1. Introduction

Existence of solutions for a periodic boundary value problem by using upper and lower solution methods has attracted the attention of many authors (see, e.g., [1–5]).

We consider a special case of the following boundary value problem: 𝑢^′(𝑡)((𝑡[],𝑢(0)(𝑇)=𝑓𝑡,𝑢))if𝑡∈0,𝑇=𝑢+𝜁₀, (1) where 𝑇>0, 𝑓∶[0,𝑇]×ℝ→ℝ is a continuous map and 𝜁₀ is constant.

Obviously, if 𝜁₀=0, then the problem (1) becomes the following periodic boundary value problem: 𝑢^′(𝑡)((𝑡[],𝑢(0)(𝑇).=𝑓𝑡,𝑢))if𝑡∈0,𝑇=𝑢 (2)

Definition 1. A lower solution for (1) is a function 𝛼∈𝐶¹([0,𝑇]) such that 𝛼^′(𝑡)((𝑡[],𝛼(0)(𝑇)≤𝑓𝑡,𝛼))if𝑡∈0,𝑇≤𝛼+𝜁₀. (3)

Let 𝒜 stand for the class of functions 𝜙∶[0,+∞)→[0,+∞), which satisfy the following conditions: (i)𝜙is nondecreasing, (ii)𝜙(𝑥)<𝑥, for each 𝑥>0, (iii)𝛽(𝑥)=𝜙(𝑥)/𝑥∈𝒮.

Very recently, Amini-Harandi and Emami [1] proved the following existence theorem, which extended the main theorem of Harjani and Sadarangani [2].

Theorem 2. Consider problem (2), with 𝑓 being continuous. Suppose that there exists 𝜆>0 such that for 𝑥,𝑦∈ℝ with 𝑦≥𝑥, 0≤𝑓𝑡,𝑦+𝜆𝑦−𝑓(𝑡,𝑥)+𝜆𝑥≤𝜆𝜙𝑦−𝑥, (4) where 𝜙∈𝒜. Then, the existence of a lower solution for (2) provides the existence of a unique solution of (2).

In this paper, we solve (2) by extending a fixed point theorem in the context of partially ordered metric space. Our results improve/extend/generalize some results in the literature, in particular, the results of Amini-Harandi and Emami [1] and Harjani and Sadarangani [2]. Finally, in the last section, we prove the existence of a solution for some polynomials, as applications.

2. Preliminaries

In this section, we state a necessary background on the topic of fixed point theory, one of the core subjects of nonlinear analysis, for the sake of completeness of the paper. Fixed point theory has a wide potential application not only in the branches of mathematics, but also in several disciplines such as economics, computer science, and biology (see, e.g., [6, 7]). The most beautiful and elementary result in this direction is the Banach contraction mapping principle [8]. After this substantial result of Banach, several authors have extended this principle in many different ways (see, e.g., [1–7, 9–31]). In particular, the authors have introduced new type of contractions and researched the existence and uniqueness of the fixed point in various spaces. One of the important contraction types, a 𝜙-contraction, was introduced by Boyd and Wong [14]. In 1997, Alber and Duerre-Delabriere [10] defined the concept of a weak-𝜑-contraction which is a generalization of the 𝜙-contraction. A self-mapping 𝑓 on a metric space (𝑋,𝑑) is said to be weak-𝜑-contractive if there exists a map 𝜑∶[0,+∞)→[0,+∞) with 𝜑(0)=0 and 𝜑(𝑡)>0 for all 𝑡>0 such that 𝑑𝑓(𝑥),𝑓𝑦≤𝑑𝑥,𝑦−𝜑𝑑𝑥,𝑦, (5) for all 𝑥,𝑦∈𝑋.

Later, Zhang and Song [31] introduced the notion of a generalized weak-𝜑-contraction which is a natural extension of the weak-𝜑-contraction. A self-mapping 𝑓 on a metric space (𝑋,𝑑) is said to be generalized weak-𝜑-contractive if there exists a map 𝜑∶[0,+∞)→[0,+∞) with 𝜑(0)=0 and 𝜑(𝑡)>0 for all 𝑡>0 such that 𝑑𝑓(𝑥),𝑓𝑦≤𝑁𝑥,𝑦−𝜑𝑁𝑥,𝑦, (6) for all 𝑥,𝑦∈𝑋, where 𝑁𝑑(𝑥𝑦,𝑑𝑦(𝑥)𝑥,𝑦=max𝑥,𝑦,𝑑𝑥,𝑓),𝑑𝑦,𝑓𝑥,𝑓+𝑑𝑦,𝑓2. (7) For more details on weak 𝜑-contractions, we refer to, for example, [20, 21, 28].

On the other hand, the existence and uniqueness of a fixed point in the context of partially ordered metric spaces were first investigated in 1986 by Turinici [30]. After this pivotal paper, a number of results were reported in this direction with applications to matrix equations, ordinary differential equations, and integral equations (see, e.g., [1, 2, 4, 5, 7, 9, 11–13, 15–19, 22, 25–27]).

Recently, the main theorem of Geraghty [16, Theorem 2.1] is reproved by Amini-Harandi and Emami [1] in the context of partially ordered metric space. On the other hand, the main theorem of Amini-Harandi and Emami [1, Theorem 2.1] extends the theorem of Harjani and Sadarangani [2]. The authors in [1, 2] also proved the existence and uniqueness of a solution for a periodic boundary value problem.

Before stating the main theorem in [1], we recall the following class of functions introduced by Geraghty [16]. Let 𝒮 denote the set of all functions 𝜓∶[0,+∞)→[0,1) such that 𝜓𝑡_𝑛⟶1implies𝑡_𝑛⟶0. (8)

Theorem 3. Let (𝑋,⪯) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space. Let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that there exists an element 𝑥₀∈𝑋 with 𝑥₀⪯𝑓(𝑥₀). Suppose that there exists 𝛽∈𝒮 such that 𝑑𝑓(𝑥),𝑓𝑦≤𝛽𝑑𝑥,𝑦𝑑𝑥,𝑦𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑥,𝑦∈𝑋𝑤𝑖𝑡ℎ𝑥≽𝑦. (9) Assume that either (a)𝑓 is continuous or (b)for every nondecreasing sequence {𝑥_𝑛} if 𝑥_𝑛→𝑥, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ.
Moreover, if for each 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 which is comparable to 𝑥 and 𝑦, then 𝑓 has a unique fixed point.

Let 𝐹(𝑓) denote the set of fixed points of 𝑓.

We give the following classes of functions. Let Φ denote the set of all mappings 𝜑∶[0,+∞)→[0,+∞) verifying that 𝜑𝑡_𝑛⟶0implies𝑡_𝑛⟶0. (10) It is clear that if 𝜑∈Φ, we have that 𝜑(𝑡)=0implies𝑡=0. (11)

3. Some Auxiliary Fixed Point Theorems

In the following theorem, we prove the existence and uniqueness of a fixed point for generalized weak-𝜑-contractive mappings in partially ordered complete metric spaces.

Theorem 4. Let (𝑋,⪯) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space. Let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that there exists an element 𝑥₀∈𝑋 with 𝑥₀⪯𝑓(𝑥₀). Suppose that there exists 𝜑∈Φ such that 𝑑𝑓(𝑥),𝑓𝑦≤𝑁𝑥,𝑦−𝜑𝑁𝑥,𝑦, (12) for each 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦 (i.e., a generalized weak-𝜑-contraction).
Suppose also that either (a)𝑓 is continuous or (b)for every nondecreasing sequence {𝑥_𝑛} if 𝑥_𝑛→𝑥, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ.
Then 𝑓 has a fixed point. Moreover, if every 𝑥,𝑦∈𝐹(𝑓) is comparable, then the fixed point of 𝑓 is unique.

Proof. First, we prove the existence of a fixed point of 𝑓. Since the self-mapping 𝑓 is nondecreasing and 𝑥₀⪯𝑓(𝑥₀), we get that 𝑥₀⪯𝑓𝑥₀⪯𝑓²𝑥₀⪯⋯⪯𝑓^𝑛𝑥₀⪯⋯. (13)
Define 𝑥_𝑛=𝑓^𝑛(𝑥₀), 𝑛=1,2,3,…. Then, expression (13) is equivalent to 𝑥_𝑛⪯𝑥_𝑛+1∀𝑛∈ℕ. (14) Assume that 𝑥_𝑛≠𝑥_𝑛+1 for each 𝑛∈ℕ. Otherwise, the proof is completed. From (12), we derive that 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝑁𝑥_𝑛,𝑥_𝑛−1−𝜑𝑁𝑥_𝑛,𝑥_𝑛−1, (15) where 𝑁𝑥_𝑛,𝑥_𝑛−1=max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛+1+𝑑𝑥_𝑛,𝑥_𝑛2=max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1. (16) If 𝑁(𝑥_𝑛,𝑥_𝑛−1)=𝑑(𝑥_𝑛,𝑥_𝑛+1) for some 𝑛, then from (15) and (16), we have 0<𝑑𝑥_𝑛+1,𝑥_𝑛≤𝑑𝑥_𝑛,𝑥_𝑛+1−𝜑𝑑𝑥_𝑛,𝑥_𝑛+1<𝑑𝑥_𝑛,𝑥_𝑛+1. (17) This is a contradiction. Hence, 𝑁(𝑥_𝑛,𝑥_𝑛−1)=𝑑(𝑥_𝑛,𝑥_𝑛−1) for all 𝑛≥1. So by (15) and (16), we have for all 𝑛≥1, 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝑑𝑥_𝑛,𝑥_𝑛−1−𝜑𝑑𝑥_𝑛,𝑥_𝑛−1<𝑑𝑥_𝑛,𝑥_𝑛−1. (18) 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. By using (18), we find that 0≤𝜑𝑑𝑥_𝑛,𝑥_𝑛−1≤𝑑𝑥_𝑛,𝑥_𝑛−1−𝑑𝑥_𝑛+1,𝑥_𝑛. (19) Taking 𝑛→∞ in (19), we get lim_𝑛→∞𝜑(𝑑(𝑥_𝑛,𝑥_𝑛−1))=0. Since 𝜑∈Φ, we obtain that lim_𝑛→∞𝑑(𝑥_𝑛,𝑥_𝑛−1)=0; that is, 𝑟=0.
We prove that the iterative sequence {𝑥_𝑛} is Cauchy. Take 𝑚>𝑛, then 𝑥_𝑛⪯𝑥_𝑚. From (12), we obtain that 𝑑𝑥_𝑚+1,𝑥_𝑛+1≤𝑁𝑥_𝑚,𝑥_𝑛−𝜑𝑁𝑥_𝑚,𝑥_𝑛, (20) and thus, 0≤𝜑𝑁𝑥_𝑚,𝑥_𝑛≤𝑁𝑥_𝑚,𝑥_𝑛−𝑑𝑥_𝑚+1,𝑥_𝑛+1, (21) where 𝑁𝑥_𝑚,𝑥_𝑛=max𝑑𝑥_𝑚,𝑥_𝑛,𝑑𝑥_𝑚,𝑥_𝑚+1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑚,𝑥_𝑛+1+𝑑𝑥_𝑛,𝑥_𝑚+12≤𝑑𝑥_𝑚,𝑥_𝑚+1+𝑑𝑥_𝑚+1,𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑥_𝑛. (22) Hence, by (21), 0≤𝜑𝑁𝑥_𝑚,𝑥_𝑛≤𝑑𝑥_𝑚,𝑥_𝑚+1+𝑑𝑥_𝑛+1,𝑥_𝑛. (23) This shows that lim_{𝑚,𝑛→∞}𝜑(𝑁(𝑥_𝑚,𝑥_𝑛))=0; that is, {𝑥_𝑛} is Cauchy. Since (𝑋,𝑑) is a complete metric space, then there exists 𝑥∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑥. Now, we prove that 𝑥 is a fixed point of 𝑓.
If (a) holds, that is, if 𝑓 is continuous, then 𝑥=lim_𝑛→∞𝑥_𝑛=lim_𝑛→∞𝑓𝑥_𝑛−1=𝑓(𝑥). (24) Suppose that (b) holds. By using (12), we derive that 0≤𝜑𝑁𝑥_𝑛,𝑥≤𝑁𝑥_𝑛,𝑥−𝑑𝑥_𝑛+1,𝑓(𝑥), (25) where 𝑁𝑥_𝑛,𝑥=max𝑑𝑥_𝑛,𝑥,𝑑𝑥_𝑛,𝑥_𝑛+1(𝑥,𝑑𝑥,𝑑𝑥,𝑓)_𝑛,𝑓(𝑥)+𝑑𝑥,𝑥_𝑛+12. (26) So lim_𝑛→∞𝑁(𝑥_𝑛,𝑥)=𝑑(𝑥,𝑓(𝑥)). Taking 𝑛→∞ in (25), we get lim_𝑛→∞𝜑(𝑁(𝑥_𝑛,𝑥))=0. Since 𝜑∈Φ, we conclude that lim_𝑛→∞𝑁(𝑥_𝑛,𝑥)=0. So 𝑑(𝑥,𝑓(𝑥))=0 and hence 𝑥=𝑓(𝑥).
Now, we show that this fixed point 𝑥 of the self-mapping 𝑓 is unique. If for each 𝑥,𝑦∈𝐹(𝑓), 𝑥 and 𝑦 are comparable, then the fixed point is unique. Let 𝑥,𝑦 be two fixed points of 𝑓. Then 𝑁(𝑥,𝑦)=𝑑(𝑥,𝑦) and from (12), we conclude that 𝜑(𝑑(𝑥,𝑦))=0. Thus, 𝑑(𝑥,𝑦)=0 and hence, 𝑥=𝑦. This completes the proof.

The following consequence of Theorem 4 plays a crucial role in the proof of our main result, Theorem 9.

Theorem 5. Let (𝑋,⪯) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space. Let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that there exists an element 𝑥₀∈𝑋 with 𝑥₀⪯𝑓(𝑥₀). Suppose that there exists 𝜑∈Φ such that 𝑑𝑓(𝑥),𝑓𝑦≤𝑑𝑥,𝑦−𝜑𝑑𝑥,𝑦, (27) for each 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦 (i.e., weak-𝜑-contraction). Suppose also that either (a)𝑓 is continuous or (b)for every nondecreasing sequence {𝑥_𝑛} if 𝑥_𝑛→𝑥, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ.
Then 𝑓 has a fixed point. Moreover, if for each 𝑥,𝑦∈𝐹(𝑓) there exists 𝑧∈𝑋 which is comparable to 𝑥 and 𝑦, then the fixed point of 𝑓 is unique.

Remark 6. In Theorem 4, if the condition “every 𝑥,𝑦∈𝐹(𝑓) is comparable” is replaced by the condition “for each 𝑥,𝑦∈𝐹(𝑓) there exists 𝑧∈𝑋 which is comparable to 𝑥 and 𝑦,” then we cannot conclude that the fixed point is unique. The following example illustrates our claim.

Example 7. Let 𝑋={𝑥,𝑦,𝑧,𝑤} be endowed with the relation ⪯ given as follows: 𝑥⪯𝑧,𝑥⪯𝑤,𝑦⪯𝑧,𝑦⪯𝑤, (28) and 𝑎⪯𝑎 for each 𝑎∈𝑋. Obviously, (𝑋,⪯) is a partially ordered set. Also, we may endow 𝑋 with the following metric: 𝑑()()𝑑()𝑥,𝑧=𝑑𝑥,𝑤=𝑑𝑦,𝑧=𝑑𝑦,𝑤=𝑑𝑥,𝑦=1,𝑧,𝑤=2, (29) and 𝑑(𝑎,𝑎)=0 for each 𝑎∈𝑋. Define 𝑓∶𝑋→𝑋 by 𝑓(𝑥)=𝑥, 𝑓(𝑦)=𝑦, 𝑓(𝑧)=𝑤, and 𝑓(𝑤)=𝑧. Obviously, the mapping 𝑓 is nondecreasing and 𝑑𝑓(𝑎),𝑓(𝑏)≤𝑑(𝑎,𝑏)−𝜑(𝑑(𝑎,𝑏)), (30) for all 𝑎,𝑏∈𝑋 with 𝑎⪯𝑏, where 𝜑(𝑡)=(1/3)𝑡. Also 𝐹(𝑓)={𝑥,𝑦}, but 𝑥⪯𝑧 and 𝑦⪯𝑧.

Remark 8. If 𝛽∈𝒮, then 𝜑(𝑡)=𝑡−𝛽(𝑡)𝑡∈Φ. But if 𝜑∈Φ, then we can not conclude that the function 𝛽(𝑡)=𝜑(𝑡)1−𝑡,𝑡>00,𝑡=0 (31) belongs to 𝒮. Consider, for example, 𝜑(𝑡)=121𝑡,0≤𝑡<12,1≤𝑡. (32) which illustrates our claim. As a result, Theorem 5 is a proper extension of Theorem 3.

4. Applications

4.1. Solving a Boundary Value Problem

In this paragraph, we prove the existence of a solution of the problem (1).

Theorem 9. Consider problem (1) with 𝑓 being continuous. Suppose that there exists 𝜆>0 such that for 𝑥,𝑦∈ℝ with 𝑦≥𝑥𝑓(),0≤𝑓𝑡,𝑦+𝜆𝑦−𝑡,𝑥+𝜆𝑥≤𝜆𝑦−𝑥−𝜑𝑦−𝑥 (33) where 𝜑∈Φ and 𝑡↦𝑡−𝜑(𝑡) is nondecreasing. Then the existence of a lower solution for (1) provides the existence of a unique solution for (1).

Proof. Define 𝜁=𝜁₀/𝑇. Then, problem (1) becomes as follows 𝑢^′(𝑡)((𝑡[],𝑢(0)(𝑇)=𝑓𝑡,𝑢))if𝑡∈0,𝑇=𝑢+𝜁𝑇. (34) Suppose 𝑦(𝑡)=𝑢(𝑡)+𝜁𝑡. So 𝑦^′(𝑡)=𝑢^′(𝑡)+𝜁 and hence problem (34) can be rewritten as 𝑦^′(𝑡)(𝑡)[],𝑦(0)(𝑇),=ℎ𝑡,𝑦if𝑡∈0,𝑇=𝑦 (35) where ℎ∶𝐼×ℝ→ℝ is defined by ℎ(𝑡,𝑧)=𝑓(𝑡,𝑧−𝜁𝑡)+𝜁 and 𝐼=[0,𝑇]. Obviously, ℎ is continuous. Also the lower solution of (34) is replaced by the lower solution of (35). Now we prove that the problem (35) has a unique solution. Obviously, if 𝑥,𝑦∈ℝ and 𝑦≥𝑥, then for every 𝑡∈𝐼, 𝑦−𝜁𝑡≥𝑥−𝜁𝑡 and hence from (33), −𝑓()(−()−().0≤𝑓𝑡,𝑦−𝜁𝑡+𝜆𝑦−𝜁𝑡𝑡,𝑥−𝜁𝑡+𝜆𝑥−𝜁𝑡)≤𝜆𝑦−𝜁𝑡𝑥−𝜁𝑡−𝜑𝑦−𝜁𝑡𝑥−𝜁𝑡 (36) Inequality (36) implies that if 𝑥,𝑦∈ℝ, 0≤ℎ𝑡,𝑦+𝜆𝑦−[ℎ(𝑡,𝑥)+𝜆𝑥]≤𝜆𝑦−𝑥−𝜑𝑦−𝑥. (37) Problem (35) is equivalent to the following integral equation: 𝑦(𝑡)=^𝑇₀𝐺(𝑡,𝑠)ℎ𝑠,𝑦(𝑠)+𝜆𝑦(𝑠)𝑑𝑠, (38) where 𝐺()=𝑒𝑡,𝑠𝑒^{𝜆(𝑇+𝑠−𝑡)𝜆𝑡}𝑒−10≤𝑠<𝑡≤𝑇,_{𝜆(𝑠−𝑡)}𝑒^𝜆𝑡−10≤𝑡<𝑠≤𝑇. (39) Let 𝐶(𝐼,ℝ) be the set of continuous functions defined on 𝐼=[0,𝑇]. Consider 𝐹∶𝐶(𝐼,ℝ)→𝐶(𝐼,ℝ) given by 𝐹𝑦(𝑡)=^𝑇₀𝐺(𝑡,𝑠)ℎ𝑠,𝑦(𝑠)+𝜆𝑦(𝑠)𝑑𝑠. (40) Note that if 𝑦∈𝐶(𝐼,ℝ) is a fixed point of 𝐹, then 𝑦∈𝐶¹(𝐼,ℝ) is a solution of (35). Now, we check that hypotheses of Theorem 5 are satisfied.
Take 𝑋=𝐶(𝐼,ℝ). The space 𝑋 can be equipped with a partial order ≤ given by 𝑥,𝑦∈𝐶(𝐼,ℝ),𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡),∀𝑡∈𝐼. (41) Also, 𝑋 can be equipped with the following metric: 𝑥,𝑦∈𝐶(𝐼,ℝ),𝑑𝑥,𝑦=sup_𝑡∈𝐼𝑥(𝑡)−𝑦(𝑡). (42) We have that (𝑋,𝑑) is complete. For every 𝑦≥𝑥 and for every 𝑡∈𝐼, we have 𝑦−𝑡𝜁≥𝑥−𝑡𝜁 and by hypothesis, 𝑓𝑡,𝑦−𝑡𝜁+𝜆𝑦−𝑡𝜁≥𝑓(𝑡,𝑥−𝑡𝜁)+𝜆(𝑥−𝑡𝜁). (43) Therefore, ℎ𝑡,𝑦+𝜆𝑦≥ℎ(𝑡,𝑥)+𝜆𝑥, (44) and since 𝐺(𝑡,𝑠)>0 for (𝑡,𝑠)∈𝐼×𝐼, hence 𝐹𝑦(𝑡)≥(𝐹𝑥)(𝑡), (45) for all 𝑥,𝑦∈𝐶(𝐼,ℝ) with 𝑦≥𝑥.
Also, for all 𝑥,𝑦∈𝐶(𝐼,ℝ) with 𝑦≥𝑥, we find (using the fact that 𝑡↦𝑡−𝜑(𝑡) is nondecreasing) 𝑑𝐹𝑦,𝐹𝑥=sup_𝑡∈𝐼𝐹𝑦(𝑡)−(𝐹𝑥)(𝑡)≤sup_𝑡∈𝐼^𝑇₀𝐺()×ℎ(𝑠(𝑠)((𝑠(𝑠𝑡,𝑠𝑠,𝑦)+𝜆𝑦−ℎ𝑠,𝑥))−𝜆𝑥)𝑑𝑠≤sup_𝑡∈𝐼^𝑇₀𝐺()𝜆𝑦(𝑠)(𝑠)𝑦(𝑠)(𝑠)𝑑𝑑𝑡,𝑠−𝑥−𝜑−𝑥𝑑𝑠≤𝜆𝑦,𝑥−𝜑𝑦,𝑥sup_𝑡∈𝐼^𝑇₀𝐺()𝑑𝑑𝑡,𝑠𝑑𝑠=𝜆𝑦,𝑥−𝜑𝑦,𝑥×sup_𝑡∈𝐼1𝑒^𝜆𝑇1−1𝜆𝑒^{𝜆(𝑇+𝑠−𝑡)𝑡}₀+1𝜆𝑒^{𝜆(𝑠−𝑡)𝑇}_𝑡𝑑𝑑1=𝜆𝑦,𝑥−𝜑𝑦,𝑥𝜆𝑒^𝜆𝑇𝑒−1^𝜆𝑇𝑑.−1=𝑑𝑦,𝑥−𝜑𝑦,𝑥 (46) Finally, let 𝛼(𝑡) be a lower solution for (35). We can show that 𝛼≤𝐹𝛼 by a method similar to that in [1, 2]. Also, 𝑋 is totally ordered. Hence, due to Theorem 5, 𝐹 has a unique fixed point. Therefore, problem (35) has a unique solution 𝑦∈𝐶¹(𝐼,ℝ). Thus, 𝑥(𝑡)=𝑦(𝑡)−𝜁𝑡 is the unique solution of (34) and this completes the proof.

Remark 10. If the mapping 𝑓∶[0,𝑇]×ℝ→ℝ satisfies the condition (33), then for 𝑥,𝑦∈ℝ with 𝑦≥𝑥 and for 𝑡∈[0,𝑇], −𝜆𝑦−𝑥≤𝑓𝑡,𝑦−𝑓(𝑡,𝑥)≤−𝜆𝜑𝑦−𝑥≤0. (47) Hence, for all 𝑥,𝑦∈ℝ and all 𝑡∈[0,𝑇], we have 𝑓𝑡,𝑦−𝑓(𝑡,𝑥)≤𝜆𝑦−𝑥. (48) Therefore, by using Banach contraction principle, for every 𝜂∈ℝ, the problem 𝑢^′(𝑡)((𝑡[]𝑢(0)=𝑓𝑡,𝑢))if𝑡∈0,𝑇=𝜂 (49) has a unique solution 𝑢_𝜂∈𝐶¹([0,𝑇]). So there exists a unique 𝜂∈ℝ such that 𝑢_𝜂 is a solution of (1).
Now let 𝑓∶[0,𝑇]×ℝ→ℝ be a mapping such that for all 𝑥,𝑦∈ℝ and all 𝑡∈[0,𝑇], 𝑓𝑡,𝑦−𝑓(𝑡,𝑥)≤𝑅𝑦−𝑥, (50) for some 𝑅>0. We know that for every 𝜂∈ℝ, problem (49) has a unique solution 𝑢_𝜂∈𝐶¹([0,𝑇]).

Question 1. It is natural to ask whether there is an 𝜂∈ℝ where 𝑢_𝜂 is a solution of problem ((2), i.e., (𝑢_𝜂(0)=𝑢_𝜂(𝑇))?

The following example shows that the above question is not true.

Example 11. Let 𝑓∶[0,𝑇]×ℝ→ℝ be defined by 𝑓(𝑡,𝑥)=𝑡+|𝑥|. Obviously, (50) holds for 𝑅=1. Let 𝑢∈𝐶¹([0,𝑇]) be a solution for problem (2). From 𝑢^′(𝑡)=𝑓(𝑡,𝑢(𝑡))=𝑡+|𝑢(𝑡)|, we conclude that 𝑢^′(𝑡)>0 for all 𝑡>0. Hence, 𝑢 is monotone nondecreasing. Using 𝑢(0)=𝑢(𝑇), we conclude that 𝑢≡0. Since 𝑢^′(𝑡)=𝑓(𝑡,𝑢(𝑡))=𝑡+|𝑢(𝑡)| and 𝑢≡0, then 𝑡=0 for all 𝑡∈[0,𝑇] and this is a contradiction. So, problem (2) has no solution.

Example 12. Let 𝑓∶[0,𝑇]×ℝ→ℝ be defined by 𝑓(𝑡,𝑥)=exp(𝑡)−(1/2)𝑥 and let 𝜑∶[0,+∞)→[0,+∞) be defined by 𝜑(𝑡)=(1/3)𝑡. Take 𝜆=1. One can show that inequality (33) holds. Suppose that 𝛼∶ℝ→ℝ is defined by 𝛼(𝑡)=0. Obviously, 𝛼 is a lower solution of problem (2). Hence, problem (2) has a unique solution, which is 𝑢(𝑡)=23exp(𝑡)+𝐶exp−12𝑡, (51) where 𝐶=2(exp(𝑇)−1)/3(1−exp((−1/2)𝑡)).

4.2. Solving Some Polynomials

In this paragraph, we prove the existence and uniqueness of a solution of some polynomials.

Theorem 13. Let 𝑎₀,𝑎₁,…,𝑎_𝑘−1∈[0,+∞) be such that 𝑎₁+𝑎₂+⋯+𝑎_𝑘−1<1 and 𝑎₀≥1. Then, 𝑦^𝑘=𝑎_𝑘−1𝑦^𝑘−1+𝑎_𝑘−2𝑦^𝑘−2+⋯+𝑎₁𝑦+𝑎₀ (52) has a unique solution on [√^𝑘𝑎₀,+∞).

Proof. Suppose that 𝑓∶[𝑎₀,+∞)→[𝑎₀,+∞) is defined by 𝑓(𝑥)=𝑎_𝑘−1^𝑘𝑥^𝑘−1+𝑎_𝑘−2^𝑘𝑥^𝑘−2+⋯+𝑎√₁^𝑘𝑥+𝑎₀. (53) If 𝑥≤𝑦, then 𝑓(𝑥)≤𝑓(𝑦). So 𝑓 is nondecreasing. Also for 𝑥,𝑦∈[𝑎₀,+∞) with 𝑥≤𝑦, we derive that 0≤𝑓𝑦−𝑓(𝑥)=𝑎_𝑘−1^𝑘𝑦^𝑘−1−^𝑘𝑥^𝑘−1+𝑎_𝑘−2^𝑘𝑦^𝑘−2−^𝑘𝑥^𝑘−2+⋯+𝑎√₁^𝑘𝑦−√^𝑘𝑥. (54) Suppose that 𝑔_𝑖∶[1,+∞)→ℝ is defined by 𝑔_𝑖(𝑡)=𝑡−𝑡^{1−𝑖/𝑘}, for 𝑖=1,2,…,𝑘−1. Since 𝑔^′_𝑖(𝑡)=1−(1−𝑖/𝑘)1/𝑡^𝑖/𝑘≥0, then 𝑔_𝑖 is monotone nondecreasing. Hence, if 1≤𝑥≤𝑦, then 𝑔_𝑖(𝑥)≤𝑔_𝑖(𝑦). So, 𝑦^{1−𝑖/𝑘}−𝑥^{1−𝑖/𝑘}≤𝑦−𝑥. Therefore, from (54), we get 0≤𝑓𝑦−𝑓(𝑥)≤𝑦−𝑥−𝜑𝑦−𝑥, (55) where 𝜑(𝑡)=[1−(𝑎_𝑘−1+𝑎_𝑘−2+⋯+𝑎₁)]𝑡. Also 𝑎₀≤𝑓(𝑎₀). Thus, using Theorem 5, the mapping 𝑓 has a unique fixed point 𝑥∈[𝑎₀,+∞). Moreover, the sequence {𝑓^𝑛(𝑎₀)} converges to this fixed point. Note that here the space 𝑋 is taken to be [𝑎₀,+∞), which is equipped with the usual Euclidian metric and the usual partial order.
On the other hand, there exists a unique 𝑦∈[√^𝑘𝑎₀,+∞) such that 𝑦^𝑘=𝑥. So, from 𝑥=𝑓(𝑥), we have 𝑦^𝑘=𝑓(𝑦^𝑘) and therefore we find 𝑦^𝑘=𝑎_𝑘−1𝑦^𝑘−1+𝑎_𝑘−2𝑦^𝑘−2+⋯+𝑎₁𝑦+𝑎₀. (56) Also the sequence {^𝑘𝑓^𝑛(𝑎₀)} converges to 𝑦 and this completes the proof.