A Contraction Fixed Point Theorem in Partially Ordered Metric Spaces and Application to Fractional Differential Equations

Abstract

We generalize a fixed point theorem in partially ordered complete metric spaces in the study of A. Amini-Harandi and H. Emami (2010). We also give an application on the existence and uniqueness of the positive solution of a multipoint boundary value problem with fractional derivatives.

1. Introduction

Let 𝒮 denote the class of those functions 𝛽∶[0,+∞)→[0,1) which satisfies the condition𝛽𝑡_𝑛→1implies𝑡_𝑛→0. (1.1) The following generalization of Banach's contraction principle is due to Geraghty [1].

Theorem 1.1. Let (𝑀,𝑑) be a complete metric space and let 𝑓∶𝑀→𝑀 be a map. Suppose there exists 𝛽∈𝒮 such that for each 𝑥,𝑦∈𝑀,𝑑𝑓(𝑥),𝑓𝑦≤𝛽𝑑𝑥,𝑦𝑑𝑥,𝑦. (1.2) Then 𝑓 has a unique fixed point 𝑧∈𝑀, and {𝑓^𝑛(𝑥)} converges to 𝑧, for each 𝑥∈𝑀.

And then, Amini-Harandi and Emami [2] proved a version of Theorem 1.1 in the context of partially ordered complete metric spaces.

Theorem 1.2. Let (𝑀,⪯) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑀 such that (𝑀,𝑑) is a complete metric space. Let 𝑓∶𝑀→𝑀 be a nondecreasing mapping, and there exists an element 𝑥₀∈𝑀 with 𝑥₀⪯𝑓(𝑥₀). Suppose that there exists 𝛽∈𝒮 such that 𝑑𝑓(𝑥),𝑓𝑦≤𝛽𝑑𝑥,𝑦𝑑𝑥,𝑦,foreach𝑥,𝑦∈𝑀with𝑥≽𝑦. (1.3) Assume that either 𝑓 is continuous or 𝑀 is such that ifanincreasingsequence𝑥_𝑛→𝑥in𝑀,then𝑥_𝑛⪯𝑥,∀𝑛. (1.4) Besides, if forany𝑥,𝑦∈𝑀,thereexists𝑧∈𝑀whichiscomparableto𝑥and𝑦, (1.5) Then 𝑓 has a unique fixed point.

Theorem 2.2 was also applied to obtain the existence and uniqueness of the solution of a periodic boundary value problem by Amini-Harandi and Emami [2] and a singular fractional three-point boundary value problem by Cabrera et al. [3].

2. Main Results

In this section, we firstly define a class of functions 𝒜 by 𝛽∶[0,+∞)→[0,+∞), and there exists a constant 𝐾>0 such that sup_{𝑡∈[0,+∞)}𝛽(𝑡)≤𝐾. (2.1)

Remark 2.1. Function classes 𝒜 include all bounded functions on [0,+∞) with upper bound 𝐾 which are more extensive than those of 𝒮. For example, 𝛽(𝑡)=𝐾sin𝑡∈𝒜 but not in 𝒮.

Now, we give an extended version of Theorem 1.2 in the context of partially ordered complete metric spaces.

Theorem 2.2. Let (𝑀,⪯) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑀 such that (𝑀,𝑑) is a complete metric space. Let 𝑓∶𝑀→𝑀 be a nondecreasing mapping, and there exists s an element 𝑥₀∈𝑀 with 𝑥₀⪯𝑓(𝑥₀). Suppose that there exists a constant 𝜃∈(0,1/𝐾) and 𝛽∈𝒮 such that 𝑑𝑓(𝑥),𝑓𝑦≤𝜃𝛽𝜃𝑑𝑥,𝑦𝑑𝑥,𝑦,foreach𝑥,𝑦∈𝑀with𝑥≽𝑦. (2.2) Assume that either 𝑓 is continuous or ifanincreasingsequence𝑥_𝑛→𝑥in𝑀,then𝑥_𝑛⪯𝑥,∀𝑛. (2.3) Besides, if forany𝑥,𝑦∈𝑀,thereexists𝑧∈𝑀whichiscomparableto𝑥and𝑦, (2.4) then 𝑓 has a unique fixed point.

Proof. We first show that 𝑓 has a fixed point. Since 𝑥₀⪯𝑓(𝑥₀) and 𝑓 is an increasing function, we obtain by induction that 𝑥₀⪯𝑓𝑥₀⪯𝑓²𝑥₀⪯𝑓³𝑥₀⪯⋯⪯𝑓^𝑛𝑥₀⪯⋯. (2.5) Put 𝑥_𝑛+1=𝑓^𝑛(𝑥₀), 𝑛=1,2,… For each integer 𝑛≥1, from (2.5), we have 𝑥_𝑛⪯𝑥_𝑛+1, then by (2.2) 𝑑𝑥_𝑛+1,𝑥_𝑛=𝑑𝑓^𝑛𝑥₀,𝑓^𝑛−1𝑥₀≤𝜃𝛽𝜃𝑑𝑥_𝑛,𝑥_𝑛−1𝑑𝑥_𝑛,𝑥_𝑛−1≤𝐾𝜃𝑑𝑥_𝑛,𝑥_𝑛−1. (2.6) If there exists 𝑛₀∈ℕ such that 𝑑(𝑥_𝑛0,𝑥_𝑛0−1)=0, then 𝑥_𝑛0=𝑓^𝑛0−1(𝑥₀)=𝑓(𝑥_𝑛0−1)=𝑥_𝑛0−1 and 𝑥_𝑛0−1 is a fixed point; in this case, the proof is finished. Otherwise, for any 𝑛∈ℕ, 𝑑(𝑥_𝑛,𝑥_𝑛−1)≠0. Then by (2.6), we have 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝐾𝜃𝑑𝑥_𝑛,𝑥_𝑛−1≤⋯≤(𝐾𝜃)^𝑛+1𝑑𝑥_𝑛0,𝑥_𝑛0−1→0,as𝑛→+∞, (2.7) that is, 𝑑𝑥_𝑛+1,𝑥_𝑛→0,as𝑛→+∞. (2.8)
Now, we show that 𝑥_𝑛 is a Cauchy sequence. By the triangle inequality and (2.2), we have 𝑑𝑥_𝑛,𝑥_𝑚≤𝑑𝑥_𝑛,𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑥_𝑚+1+𝑑𝑥_𝑚+1,𝑥_𝑚≤𝑑𝑥_𝑛,𝑥_𝑛+1+𝜃𝛽𝜃𝑑𝑥_𝑛,𝑥_𝑚𝑑𝑥_𝑛,𝑥_𝑚+𝑑𝑥_𝑚+1,𝑥_𝑚≤𝑑𝑥_𝑛,𝑥_𝑛+1+𝜃𝐾𝑑𝑥_𝑛,𝑥_𝑚+𝑑𝑥_𝑚+1,𝑥_𝑚, (2.9) and then 𝑑𝑥_𝑛,𝑥_𝑚≤(1−𝐾𝜃)⁻¹𝑑𝑥_𝑛,𝑥_𝑛+1+𝑑𝑥_𝑚+1,𝑥_𝑚→0,as𝑚,𝑛→+∞, (2.10) which implies that 𝑥_𝑛 is a Cauchy sequence in 𝑀. Since (𝑀,𝑑) is a complete metric space, then there exists a 𝑧∈𝑀 such that lim_𝑛→+∞𝑥_𝑛=𝑧. To prove that 𝑧 is a fixed point of 𝑓, if 𝑓 is continuous, then 𝑧=lim_𝑛→+∞𝑥_𝑛=lim_𝑛→+∞𝑓^𝑛𝑥₀=lim_𝑛→+∞𝑓^𝑛+1𝑥₀=𝑓lim_𝑛→+∞𝑓^𝑛𝑥₀=𝑓(𝑧), (2.11) hence 𝑧=𝑓(𝑧). If case (2.3) holds, then we claim that 𝑓(𝑧)=𝑧 still holds. In fact, 𝑑𝑓(𝑧),𝑧≤𝑑𝑓(𝑧),𝑓𝑥_𝑛+𝑑𝑓𝑥_𝑛,𝑧≤𝜃𝐾𝑑𝑧,𝑥_𝑛+𝑑𝑥_𝑛+1,𝑧, (2.12) since 𝑑(𝑧,𝑥_𝑛)→0, taking limit as 𝑛→+∞, then 𝑑(𝑓(𝑧),𝑧)≤0, this proves that 𝑑(𝑓(𝑧),𝑧)=0; consequently, 𝑓(𝑧)=𝑧.
Let 𝑦 be another fixed point of 𝑓. From (2.4) there exists 𝑥∈𝑀 which is comparable to 𝑦 and 𝑧. Monotonicity implies that 𝑓^𝑛(𝑥) is comparable to 𝑓^𝑛(𝑦)=𝑦 and 𝑓^𝑛(𝑧)=𝑧 for 𝑛=0,1,2… Moreover, 𝑑𝑧,𝑓^𝑛(𝑥)=𝑑𝑓^𝑛(𝑧),𝑓^𝑛(𝑥)≤𝐾𝜃𝑑𝑧,𝑓^𝑛−1(𝑥). (2.13) Taking limit, and then lim_𝑛→+∞𝑑(𝑧,𝑓^𝑛(𝑥))=0. Similar to [2], we have 𝑑(𝑧,𝑦)=0. The proof of the uniqueness of the fixed point is completed.

3. Application to Fractional Differential Equations

In this section, we consider the unique positive solution for a general higher order fractional differential equation by using the generalized fixed point theorem −𝓓^𝛼𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝓓^𝜇1𝑥(𝑡),𝓓^𝜇2𝑥(𝑡),…,𝓓^𝜇𝑛−1𝑥(𝑡,𝓓_𝜇𝑖)𝑥(0)𝓓=0,1≤𝑖≤𝑛−1,_{𝜇𝑛−1+1}𝑥(0)=0,𝓓^𝜇𝑛−1𝑥(1)=^𝑚−2_𝑗=1𝑎_𝑗𝓓^𝜇𝑛−1𝑥𝜉_𝑗, (3.1) where 𝑛−1<𝛼≤𝑛,  𝑛∈ℕ and 𝑛≥3 with 0<𝜇₁<𝜇₂<⋯<𝜇_𝑛−2<𝜇_𝑛−1 and 𝑛−3<𝜇_𝑛−1<𝛼−2, 𝑎_𝑗∈ℝ, 0<𝜉₁<𝜉₂<⋯<𝜉_𝑚−2<1 satisfying 0<∑^𝑚−2_𝑗=1𝑎_𝑗𝜉^{𝛼−𝜇𝑛−1}_𝑗⁻¹<1 and 𝓓^𝛼 is the standard Riemann-Liouville derivative, 𝑓∈𝐶([0,1]×ℝ^𝑛,[0,+∞)). Recently, there has been a significant development in the study of fractional differential equations; for more details we refer the reader to [4–12] and the references cited therein.

For the convenience of the reader, we present some notations and lemmas which will be used in the proof of our results.

Definition 3.1 (see [13, 14]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝐼^𝛼𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠, (3.2) provided that the right-hand side is pointwise defined on (0,+∞).

Definition 3.2 (see [13, 14]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝓓^𝛼_𝐭𝑥(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑥(𝑠)𝑑𝑠, (3.3) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number 𝛼, provided that the right-hand side is pointwise defined on (0,+∞).

Proposition 3.3 (see [13, 14]). Consider the following.(1) If 𝑥∈𝐿¹(0,1), 𝜈>𝜎>0, then 𝐼^𝜈𝐼^𝜎𝑥(𝑡)=𝐼^𝜈+𝜎𝑥(𝑡),𝓓^𝜎_𝐭𝐼^𝜈𝑥(𝑡)=𝐼^𝜈−𝜎𝑥(𝑡),𝓓^𝜎_𝐭𝐼^𝜎𝑥(𝑡)=𝑥(𝑡). (3.4)(2) If 𝜈>0,  𝜎>0, then 𝓓^𝜈_𝐭𝑡^𝜎−1=Γ(𝜎)Γ()𝑡𝜎−𝜈^{𝜎−𝜈−1}. (3.5)

Proposition 3.4 (see [13, 14]). Let 𝛼>0, and 𝑓(𝑥) is integrable, then 𝐼^𝛼𝓓^𝛼_𝐭𝑥(𝑡)=𝑓(𝑥)+𝑐₁𝑥^𝛼−1+𝑐₂𝑥^𝛼−2+⋯+𝑐_𝑛𝑥^𝛼−𝑛, (3.6) where 𝑐_𝑖∈ℝ  (𝑖=1,2,…,𝑛), 𝑛 is the smallest integer greater than or equal to 𝛼.

Lemma 3.5 (see [15]). Let 𝑥(𝑡)=𝐼^𝜇𝑛−1𝑢(𝑡), then BVP (3.1) is equivalent to the following BVP: −𝓓^{𝛼−𝜇𝑛−1}𝑢(𝑡)=𝑓𝑡,𝐼^𝜇𝑛−1𝑢(𝑡),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑡),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑡)(𝑡,𝑢(0),𝑢)=𝑢^′(0)=0𝑢(1)=^𝑚−2_𝑗=1𝑎_𝑗𝑢𝜉_𝑗. (3.7) Moreover, if 𝑣∈𝐶([0,1],[0,+∞)) is a solution of problem (3.7), then the function 𝑥(𝑡)=𝐼^𝜇𝑛−1𝑢(𝑡) is a positive solution of problem (3.1).

Let 𝑘()=(𝑡(𝑡,𝑠1−𝑠))^{𝛼−𝜇𝑛−1−1}−(𝑡−𝑠)^{𝛼−𝜇𝑛−1−1}Γ𝛼−𝜇_𝑛−1(𝑡(,0≤𝑠≤𝑡≤1,1−𝑠))^{𝛼−𝜇𝑛−1−1}Γ𝛼−𝜇_𝑛−1,0≤𝑡≤𝑠≤1, (3.8) from [15], the Green function of (3.7) is 𝐻()()+𝑡𝑡,𝑠=𝑘𝑡,𝑠^{𝛼−𝜇𝑛−1−1}1−∑^𝑚−2_𝑗=1𝑎_𝑗𝜉^{𝛼−𝜇𝑛−1}_𝑗^−1𝑚−2_𝑗=1𝑎_𝑗𝑘𝜉_𝑗,𝑠, (3.9) and with property ()≤10≤𝐻𝑡,𝑠Γ𝛼−𝜇_𝑛−1∑1+^{𝑚−2𝑗=1}𝑎_𝑗1−∑^𝑚−2_𝑗=1𝑎_𝑗𝜉^{𝛼−𝜇𝑛−1}_𝑗⁻¹=𝜂. (3.10)

In our considerations, we will work in the Banach space 𝐸=𝐶([0,1];ℝ) with the classical metric given by 𝑑(𝑥,𝑦)=sup_0≤𝑡≤1|𝑥(𝑡)−𝑦(𝑡)|. Notice that this space can be equipped with a partial order given by 𝑥,𝑦∈𝐶[0,1],𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡)forany𝑡∈[0,1]. (3.11) In [2], it is proved that (𝐶[0,1],≤) satisfies condition (2.3) of Theorem 2.2. Moreover, for 𝑥,𝑦∈𝐶[0,1], as the function max{𝑥,𝑦}∈𝐶[0,1],  (𝐶[0,1],≤) satisfies condition [15]. Consider the cone 𝑃={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0}. (3.12) Note that as 𝑃 is a closed set of 𝐶[0,1], 𝑃 is a complete metric space.

It is well known that the BVP (3.7) is equivalent to the integral equation 𝑢(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠),𝑢(𝑠)𝑑𝑠. (3.13) Now, for 𝑢∈𝑃 we define the operator 𝑇 by (𝑇𝑢)(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠),𝑢(𝑠)𝑑𝑠. (3.14) Then from the assumption on 𝑓 and (3.10), we have 𝑇(𝑃)⊂𝑃. (3.15)

We also introduce the following class of nondecreasing functions ℬ by 𝜙∶[0,+∞)→[0,+∞) satisfying the following: 𝜙(𝑥)≤𝐾𝑥,forany𝑥>0. (3.16) Clearly, if 𝜙∈ℬ, then 𝜙(𝑥)/𝑥∈𝒜. The standard functions 𝜙∈ℬ, for example, 𝜙(𝑥)=𝐾𝑥, 𝜙(𝑥)=(2𝐾/𝜋)𝑥arctan𝑥, and 𝜙(𝑥)=𝐾𝑥²/(1+𝑥).

Theorem 3.6. Suppose 𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛) is nondecreasing in 𝑥_𝑖 on [0,+∞); moreover, there exist 𝑛 positive constants 𝜌_𝑖, 𝑖=1,2,…,𝑛 that satisfy max𝜌₁,𝜌₂,…,𝜌_𝑛≤𝑛𝜂⁻¹, (3.17) and there exist a function 𝜙∈ℬ and constants 0<𝜃₁<Γ(𝜇_𝑛−1)/𝐾, 0<𝜃_𝑛<1/𝐾, 0<𝜃_𝑖<Γ(𝜇_𝑛−1−𝜇_𝑖)/𝐾, 𝑖=1,2,…,𝑛−2 such that 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛≤^𝑛_𝑖=1𝜌_𝑖𝜙𝜃_𝑖𝑥_𝑖−𝑦_𝑖, (3.18) for 𝑥_𝑖,𝑦_𝑖∈[0,+∞), 𝑖=1,2,…,𝑛 with 𝑥_𝑖≥𝑦_𝑖 and 𝑡∈[0,1]. Then problem (3.1) has a unique nonnegative solution.

Proof. We check that the hypotheses in Theorem 2.2 are satisfied.
Firstly, the operator 𝑇 is nondecreasing by the hypothesis. Then for any 𝑢≥𝑣, we have (𝑇𝑢)(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠)(𝑠)≥,𝑢𝑑𝑠₁₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝜇𝑛−1𝑣(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑣(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑣(𝑠)(𝑠=(𝑡).,𝑣)𝑑𝑠𝑇𝑣)( (3.19) Noticing that 𝑢(𝑠)(𝑠)(),𝐼−𝑣≤𝑑𝑢,𝑣_𝜇𝑛−1𝑢(𝑠)−𝐼^𝜇𝑛−1𝑣(𝑠)≤^𝑡₀(𝑡−𝑠)^{𝜇𝑛−1−1}|𝑢(𝑠)−𝑣(𝑠)|Γ𝜇_𝑛−1𝑑()𝑑𝑠≤𝑢,𝑣Γ𝜇_𝑛−1,𝐼^{𝜇𝑛−1−𝜇𝑖}𝑢(𝑠)−𝐼^{𝜇𝑛−1−𝜇𝑖}𝑣(𝑠)≤^𝑡₀(𝑡−𝑠)^{𝜇𝑛−1−𝜇𝑖−1}|𝑢(𝑠)−𝑣(𝑠)|Γ𝜇_𝑛−1−𝜇_𝑖≤𝑑()𝑑𝑠𝑢,𝑣Γ𝜇_𝑛−1−𝜇_𝑖,𝑖=1,2,…,𝑛−2, (3.20) and take 𝜃𝜃=max¹Γ𝜇_𝑛−1,𝜃₂Γ𝜇_𝑛−1−𝜇₁𝜃,…,^𝑛−1Γ𝜇_𝑛−1−𝜇_𝑛−2,𝜃_𝑛. (3.21)
Thus, for any 𝑢≥𝑣, we have 𝑑(𝑇𝑢,𝑇𝑣)=max_[]𝑡∈0,1|𝑇𝑢(𝑡)−𝑇𝑣(𝑡)|≤𝜂¹₀𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠),𝑢(𝑠)𝑑𝑠−𝑓𝑠,𝐼^𝜇𝑛−1𝑣(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑣(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑣(𝑠),𝑣(𝑠)≤𝜂¹₀𝜌₁𝜙𝜃₁𝐼^𝜇𝑛−1𝑢(𝑠)−𝐼^𝜇𝑛−1𝑣(𝑠)+𝜌₂𝜙𝜃₂𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠)−𝐼^{𝜇𝑛−1−𝜇1}𝑣(𝑠)+⋯+𝜌_𝑛−1𝜙𝜃_𝑛−1𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠)−𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑣(𝑠)+𝜌_𝑛𝜙𝜃_𝑛(𝑢(𝑠)(𝑠𝜌−𝑣))𝑑𝑠≤𝑛𝜂max₁,𝜌₂,…,𝜌_𝑛𝜙(𝜃𝑑(𝑢,𝑣))≤𝜙(𝜃𝑑(𝑢,𝑣)). (3.22) For 𝑢≠𝑣, we have 𝑑()≤𝜙((𝑇𝑢,𝑇𝑣𝜃𝑑𝑢,𝑣))()()(((),𝜃𝑑𝑢,𝑣𝜃𝑑𝑢,𝑣=𝛽𝜃𝑑𝑢,𝑣))𝜃𝑑𝑢,𝑣 (3.23) and this inequality is obviously satisfied for 𝑢=𝑣. Thus, we have 𝑑()≤𝜙((𝑇𝑢,𝑇𝑣𝜃𝑑𝑢,𝑣))()()((()𝜃𝑑𝑢,𝑣𝜃𝑑𝑢,𝑣=𝛽𝜃𝑑𝑢,𝑣))𝜃𝑑𝑢,𝑣,forany𝑢,𝑣∈𝑃with𝑢≥𝑣. (3.24)
Finally, since the zero function satisfies 0≤𝑇0, Theorem 2.2 tells us that the operator 𝑇 has a unique fixed point in 𝑃, or, equivalently, the BVP (3.1) has a unique nonnegative solution 𝑥 in 𝐶[0,1].

Theorem 3.7. If the assumptions of Theorem 3.6 are satisfied, and there exists 𝑡₀∈[0,1] such that 𝑓(𝑡₀,0,…,0)≠0, then the unique solution of (3.1) is positive (a positive solution means a solution satisfying 𝑥(𝑡)>0 for 𝑡∈(0,1)).

Proof. By Theorem 3.6, the problem (3.1) has a unique nonnegative solution. We prove the nonnegative solution is also positive.
Otherwise, there exists 0<𝑡^∗<1 such that 𝑥(𝑡^∗)=0, and 𝑥𝑡^∗=¹₀𝐻𝑡^∗,𝑠𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠),𝑢(𝑠)𝑑𝑠=0. (3.25) Then 0=𝑥𝑡^∗=¹₀𝐻𝑡^∗,𝑠𝑓𝑠,𝐼^𝜇𝑛−1𝑢(𝑠),𝐼^{𝜇𝑛−1−𝜇1}𝑢(𝑠),…,𝐼^{𝜇𝑛−1−𝜇𝑛−2}𝑢(𝑠)(𝑠)≥,𝑢𝑑𝑠₁₀𝐻𝑡^∗,𝑠𝑓(𝑠,0,…,0)𝑑𝑠≥0. (3.26) Consequently, ¹₀𝐻𝑡^∗,𝑠𝑓(𝑠,0,…,0)𝑑𝑠=0, (3.27) this yields 𝐻𝑡^∗,𝑠𝑓(𝑠,0,…,0)=0,a.e.𝑠∈[0,1]. (3.28) Note that 𝐻(𝑡^∗,𝑠)>0,𝑠∈(0,1), then we have 𝑓(𝑠,0,…,0)=0,a.e.𝑠∈[0,1]. (3.29)
But on the other hand, since 𝑓(𝑡₀,0,…,0)≠0, 𝑡₀∈[0,1], we have 𝑓(𝑡₀,0,…,0)>0, by the continuity of 𝑓, we can find a set Ω⊂[0,1] satisfying 𝑡₀∈Ω and the Lebesgue measure 𝜇(Ω)>0 such that 𝑓(𝑡,0,…,0)>0 for any 𝑡∈Ω. This contradicts to (3.29). Therefore, 𝑥(𝑡)>0, that is, 𝑥(𝑡) is positive solution of (3.1).

Example 3.8. Consider the following fractional boundary value problem: −𝓓^5/2𝑥(𝑡)=𝑒^𝑡+1𝑥(𝑡)+1102sin²𝓓^1/8𝑥(𝑡+1)3cos²𝓓^1/4𝑥(𝑡𝓓),0<𝑡<1,_1/4𝑥(0)=𝓓^5/4𝑥(0)=0,𝓓^1/4𝑥(1)=14𝓓^1/4𝑥14+12𝓓^1/4𝑥34. (3.30) Then the BVP (3.30) has a unique positive solution.

Proof. Since ^𝑚−2_𝑗=1𝑎_𝑗𝜉^{𝛼−𝜇𝑛−1}_𝑗⁻¹=1414^5/4+1234^5/41=0.39316<1,𝜂=Γ𝛼−𝜇_𝑛−1∑1+^{𝑚−2𝑗=1}𝑎_𝑗1−∑^𝑚−2_𝑗=1𝑎_𝑗𝜉^{𝛼−𝜇𝑛−1}_𝑗⁻¹=0.10964. (3.31) Thus, 𝑛𝜂⁻¹=3.0405. (3.32)
Let 𝐾=15 and 𝜙(𝑥)=15𝑥, and take 𝑓𝑡,𝑥₁,𝑥₂,𝑥₃=𝑒^𝑡+1𝑥10¹+12sin²𝑥₂+13cos²𝑥₃,𝑡,𝑥₁,𝑥₂,𝑥₃∈[0,1]×[0,+∞)³. (3.33) Then, for any 𝑥₁≥𝑦₁, 𝑥₂≥𝑦₂, 𝑥₃≥𝑦₃, 𝑓𝑡,𝑥₁,𝑥₂,𝑥₃−𝑓𝑡,𝑦₁,𝑦₂,𝑦₃=𝑥₁−𝑦₁+10sin²𝑥₂−sin²𝑦₂2+cos²𝑥₃−cos²𝑦₃3≤𝑥₁−𝑦₁+𝑥10²−𝑦₂2+𝑥₃−𝑦₃3=2175×15×4𝑥₁−𝑦₁+1110×15×3𝑥₂−𝑦₃+491×15×𝑥20³−𝑦₃=2𝜙1754𝑥₁−𝑦₁+1𝜙1103𝑥₂−𝑦₃+49𝜙1𝑥20³−𝑦₃=³_𝑖=1𝜌_𝑖𝜙𝜃_𝑖𝑥_𝑖−𝑦_𝑖, (3.34) where 𝜌₁=275,𝜌₂=110,𝜌₃=49,𝜃₁=14,𝜃₂=13,𝜃₃=1.20 (3.35) Thus, 𝜙∈ℬ and all of the conditions of Theorem 3.6 are satisfied.
On the other hand, 𝑓(0,0,…,0)=4/3≠0, by Theorems 3.6 and 3.7, the BVP (19) has a unique positive solution.

Remark 3.9. In Example 3.8, 𝛽(𝑥)=𝜙(𝑥)/𝑥=𝐾=15 which does not possess property 𝛽∶[0,+∞)→[0,1) and 𝛽𝑡_𝑛→1implies𝑡_𝑛→0. (3.36) Thus, the unique positive solution of BVP (3.30) cannot be obtained via Theorem 1.2, but we obtain the unique positive solution of BVP (3.30) by using the generalized fixed point Theorem 2.2, which implies that Theorem 2.2 is an essential promotion of Theorem 1.2.

Acknowledgment

This work is supported by the Natural Sciences of Education and Science Office Bureau of Sichuan Province of China under Grant no. 2010JY0J41,11ZB152.