Existence and Uniqueness of Positive Solution for a Boundary Value Problem of Fractional Order

Abstract

We are concerned with the existence and uniqueness of positive solutions for the following nonlinear fractional boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0, 0≤𝑡≤1, 3<𝛼≤4, 𝑢(0)=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0, where 𝐷^𝛼₀₊ denotes the standard Riemann-Liouville fractional derivative. Our analysis relies on a fixed point theorem in partially ordered sets. Some examples are also given to illustrate the results.

1. Introduction

Differential equations of fractional order occur more frequently in different research and engineering areas such as physics, chemistry, economics, and control of dynamical. Indeed, we can find numerous applications in viscoelasticity, electrochemistry control, porous media, and electromagnetism. (see, e.g., [1–7]).

For an extensive collection of results about this type of equations we refer the reader to the monograph by Kilbas and Trujillo [8], Samko et al. [9], Miller and Ross [10], and Podlubny [11].

On the other hand, some basic theory for the initial value problems of fractional differential equations involving the Riemann-Liouville differential operator has been discussed by Lakshmikantham and Vatsala [12], Lakshmikantham [13], El-Sayed and El-Maghrabi [14], Bai [15], Bai and Ge [16], Bai and Lü [17], Zhang [18], and Kempfle et al. [19, 20].

In [17] the authors studied the following two-point boundary value problem of fractional order:𝐷^𝛼₀⁺𝑢(𝑡)(𝑡)𝑓((𝑡𝑢(0)(1)+𝑎𝑡,𝑢))=0,0<𝑡<1,1<𝛼≤2,=𝑢=0,(1.1) and they proved the existence of positive solutions by means of the Krasnosel'skii fixed point theorem and Leggett-Williams fixed point theorem.

In [18] the author investigated the existence of solutions of ^𝑐𝐷^𝛼₀⁺𝑢(𝑡)((𝑡𝑢(0)(1)=𝑓𝑡,𝑢)),0<𝑡<1,1<𝛼≤2,=𝜈≠0,𝑢=𝜌≠0.(1.2) Since boundary values are nonzero, the Riemann-Liouville fractional derivative 𝐷^𝛼₀⁺ is not suitable and the author used the Caputo fractional derivative ^𝑐𝐷^𝛼₀⁺.

Motivated by these works, in this paper we discuss the existence and uniqueness of positive solutions for the following nonlinear boundary value problem of fractional order: 𝐷^𝛼₀⁺𝑢(𝑡)((𝑡𝑢(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,3<𝛼≤4,=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0.(1.3)

This problem was studied in [21], where the authors use lower and upper solution method and the Schauder fixed point theorem which cannot ensure the uniqueness of the solution. The practical relevance of 3<𝛼≤4 appears in problems related with other areas as physics and economics which can be modeled by these fractional boundary values problems. Particularly, these problems appear in the Hamiltonian formulation for the lagrangians depending on fractional derivatives of coordinates when the systems are nonconservative (see, e.g., [7]).

Our main interest in this paper is to give an alternative answer to the main results of the paper [21].

The main tool used in our study is a fixed point theorem in partially ordered sets which gives us uniqueness of the solution.

2. Preliminaries and Previous Results

For the convenience of the reader, we present here some definitions, lemmas and basic results that will be used in the proofs of our theorems.

Definition 2.1. The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑓∶(0,∞)→ℝ is given by 𝐼^𝛼₀⁺𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠(2.1) provided that the right-hand side is pointwise defined on (0,∞) and where Γ(𝛼) denotes the gamma function.

Definition 2.2. The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑓∶(0,∞)→ℝ is given by 𝐷^𝛼₀⁺𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀𝑓(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠,(2.2) where 𝑛=[𝛼]+1 and [𝛼] denotes the integer part of 𝛼.
The following two lemmas can be found in [17, 22].

Lemma 2.3. Let 𝛼>0 and 𝑢∈𝐶(0,1)∩𝐿¹(0,1). Then fractional differential equation 𝐷^𝛼₀⁺𝑢(𝑡)=0(2.3) has 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛(2.4) for some 𝑐_𝑖∈ℝ (𝑖=1,2,…𝑛) and 𝑛=[𝛼]+1 as unique solution.

Lemma 2.4. Assume that 𝑢∈𝐶(0,1)∩𝐿¹(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿¹(0,1). Then 𝐼^𝛼₀⁺𝐷^𝛼₀⁺𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛,(2.5) for some 𝑐_𝑖∈ℝ (𝑖=1,…,𝑛) and 𝑛=[𝛼]+1.

Using Lemma 2.4, in [21] the following result is proved.

Lemma 2.5. Given 𝑓∈𝐶[0,1] and 𝑓(𝑡)≥0, the unique nonnegative solution for 𝐷^𝛼₀⁺𝑢(𝑡)(𝑡)𝑢(0)+𝑓=0,0<𝑡<1,3<𝛼≤4,=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0(2.6) is 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠)𝑑𝑠,(2.7) where 𝐺()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑡,0≤𝑠≤𝑡≤1,_𝛼−1(1−𝑠)^𝛼−3Γ(𝛼),0≤𝑡≤𝑠≤1.(2.8)

In the sequel, we present the fixed-point theorems which we will use later. These results appear in [23].

Theorem 2.6. 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. Assume that 𝑋 satisfies the following condition 𝑖𝑓𝑥_𝑛isanondecreasingsequencein𝑋suchthat𝑥_𝑛→𝑥,then𝑥_𝑛≤𝑥∀𝑛∈ℕ.(2.9) Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping such that 𝑑𝑇𝑥,𝑇𝑦≤𝑑𝑥,𝑦−𝜓𝑑𝑥,𝑦,for𝑥≥𝑦,(2.10) where 𝜓∶[0,∞)→[0,∞) is a continuous and nondecreasing function such that 𝜓 is positive in (0,∞), 𝜓(0)=0 and lim_𝑡→∞𝜓(𝑡)=∞. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑇(𝑥₀) then 𝑇 has a fixed point.

Moreover, if (𝑋,≤) satisfies the following condition:for𝑥,𝑦∈𝑋thereexists𝑧∈𝑋whichiscomparableto𝑥and𝑦,(2.11) which appears in [24], the following result is proved [23].

Theorem 2.7. Adding condition (2.11) to the hypotheses of Theorem 2.6 one obtains the uniqueness of the fixed point.

Remark 2.8. In Theorems 2.6 and 2.7 the condition lim_𝑡→∞𝜓(𝑡)=∞ is redundant.

In our considerations we will work in the Banach space 𝐶[0,1]={𝑥∶[0,1]→ℝ,continuous} with the standard norm ‖𝑥‖=sup{|𝑥(𝑡)|∶𝑡∈[0,1]}.

Notice that this space can be equipped with a partial order given by 𝑥,𝑦∈𝐶[0,1],𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡),for𝑡∈[0,1].(2.12) In [24] it is proved that (𝐶[0,1],≤) with the classical metric given by 𝑑𝑥,𝑦=sup_0≤𝑡≤1𝑥(𝑡)−𝑦(𝑡)(2.13) satisfies condition (2.9) of Theorem 2.6. Moreover, for 𝑥,𝑦∈𝐶[0,1], as the function max(𝑥,𝑦)∈𝐶[0,1], (𝐶[0,1],≤) satisfies condition (2.11).

Finally, by ℱ we denote the class of functions 𝜓∶[0,∞)→[0,∞) continuous, nondecreasing, positive in (0,∞) and 𝜓(0)=0.

By 𝒥 we denote the class of functions 𝜑∶[0,∞)→[0,∞) continuous, nondecreasing, satisfying that 𝐼−𝜑∈ℱ, where 𝐼 denotes the identity mapping on [0,∞).

3. Main Result

The main result of the paper is the following.

Theorem 3.1. Problem (1.3) has a unique positive solution 𝑢(𝑡) if the following conditions are satisfied. (H1)𝑓∶[0,1]×[0,∞)→[0,∞) is continuous and nondecreasing with respect to the second argument.(H2)There exists 𝑡₀∈[0,1] such that 𝑓(𝑡₀,0)>0. (H3)There exists 0<𝜆≤(𝛼−2)Γ(𝛼+1)/2 such that, for 𝑥,𝑦∈[0,∞) with 𝑦≥𝑥 and 𝑡∈[0,1], 𝑓𝑡,𝑦−𝑓(𝑡,𝑥)≤𝜆⋅𝜓𝑦−𝑥,(3.1) where 𝜓∈𝒥.

Before the proof of Theorem 3.1, we will need some properties of Green's function appearing in Lemma 2.5.

Lemma 3.2. 𝐺(𝑡,𝑠)≥0, and 𝐺 is a continuous function on [0,1]×[0,1].

Proof. The continuity of 𝐺 is easily checked. In order to prove the nonnegativness of 𝐺(𝑡,𝑠), for 0≤𝑡≤𝑠≤1, it is obvious that 𝐺()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)≥0.(3.2) In the case of 0≤𝑠≤𝑡≤1 with 𝑡≠0, we have 𝐺()=1𝑡,𝑠Γ(𝛼)𝑡^𝛼−1(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−1=1Γ(𝛼)𝑡^𝛼−1(1−𝑠)^𝛼−3−𝑠1−𝑡^𝛼−1.(3.3) As 𝑠≤𝑠/𝑡, we have 1−𝑠≥1−(𝑠/𝑡) and, consequently, (1−𝑠)^𝛼−3≥𝑠1−𝑡^𝛼−3.(3.4) Taking into account that the function 𝑔(𝛼)=𝑡^𝛼 with 𝛼>0 and 𝑡∈(0,1) is decreasing we have (1−𝑠)^𝛼−3≥𝑠1−𝑡^𝛼−3≥𝑠1−𝑡^𝛼−1.(3.5) The last inequality and (3.3) give us 𝐺(𝑡,𝑠)≥0 with 𝑡≠0. Finally, notice that 𝐺(0,𝑠)=0, and this finishes the proof.

Lemma 3.3. One has sup_[]𝑡∈0,1¹₀𝐺()2𝑡,𝑠𝑑𝑠=(𝛼−2)Γ(𝛼+1).(3.6)

Proof. Since ¹₀𝐺(𝑡,𝑠)𝑑𝑠=^𝑡₀𝐺(𝑡,𝑠)𝑑𝑠+¹_𝑡𝐺()=1𝑡,𝑠𝑑𝑠Γ(𝛼)^𝑡₀𝑡^𝛼−1(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−11𝑑𝑠+Γ(𝛼)¹_𝑡𝑡^𝛼−1(1−𝑠)^𝛼−3=1𝑑𝑠Γ(𝛼)𝑡^𝛼−1−1𝛼−2𝛼𝑡^𝛼(3.7) and if we put 𝜑(𝑡)=∫¹₀𝐺(𝑡,𝑠)𝑑𝑠=(1/Γ(𝛼))((𝑡^𝛼−1)/(𝛼−2)−(1/𝛼)𝑡^𝛼), then, as 𝜑^′(𝑡)=1Γ(𝛼)𝛼−1𝑡𝛼−2^𝛼−2−𝑡^𝛼−1>0,for𝑡>0,(3.8) we deduce that 𝜑(𝑡)=∫¹₀𝐺(𝑡,𝑠)𝑑𝑠 is strictly increasing and, consequently, sup_[]𝑡∈0,1¹₀𝐺(𝑡,𝑠)𝑑𝑠=¹₀𝐺()11,𝑠𝑑𝑠=Γ(𝛼)1−1𝛼−2𝛼=2𝛼()Γ(𝛼)=2𝛼−2()Γ().𝛼−2𝛼+1(3.9)

In the sequel, we give the proof of Theorem 3.1.

Proof of Theorem 3.1. Consider the cone 𝑃={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0}.(3.10) Obviously, 𝑃 is a closed set of 𝐶[0,1], and, thus, 𝑃 is a complete metric space with the distance given by 𝑑(𝑢,𝑣)=sup_𝑡∈[0,1]{|𝑢(𝑡)−𝑣(𝑡)|}. 𝑃 can be equipped with a partial order defined by 𝑥,𝑦∈𝑃,𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡),for𝑡∈[0,1].(3.11) Using a similar argument to that in [24], it can be proved that (𝑃,≤) satisfies condition (2.9) of Theorem 2.6. Moreover, as for 𝑥,𝑦∈𝑃 the function max(𝑥,𝑦)∈𝑃, (𝑃,≤) satisfies condition (2.11).
Now, we consider the operator 𝑇 defined on 𝑃 and given by (𝑇𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,for𝑢∈𝑃.(3.12) By (H1) and Lemma 3.2, 𝑇 applies 𝑃 into itself.
In the sequel we check that 𝑇 satisfies the assumptions of Theorem 2.6.
Firstly, we prove that 𝑇 is a nondecreasing operator.  In fact, by (H1), for 𝑢,𝑣∈𝑃 with 𝑢≥𝑣 and 𝑡∈[0,1], we have (𝑇𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠=(𝑇𝑣)(𝑡).(3.13) Now, we prove that 𝑇 satisfies the contractive condition appearing in Theorem 2.6.
In fact, for 𝑢,𝑣∈𝑃 and 𝑢≥𝑣, taking into account assumption (H3), we get 𝑑(𝑇𝑢,𝑇𝑣)=sup_[]𝑡∈0,1{|(𝑇𝑢)(𝑡)−(𝑇𝑣)(𝑡)|}=sup_[]𝑡∈0,1((𝑇𝑢)(𝑡)−(𝑇𝑣)(𝑡))=sup_[]𝑡∈0,1¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))−𝑓(𝑠,𝑣(𝑠))𝑑𝑠≤sup_[]𝑡∈0,1¹₀𝐺(𝑡,𝑠)𝜆⋅𝜓(𝑢(𝑠)−𝑣(𝑠))𝑑𝑠.(3.14) As 𝜓∈𝒥 and, thus, 𝜓 is nondecreasing and by Lemma 3.3, from the last inequality we obtain 𝑑(𝑇𝑢,𝑇𝑣)≤𝜆𝜓(𝑑(𝑢,𝑣))⋅sup_[]𝑡∈0,1¹₀𝐺()(𝑑(⋅2𝑡,𝑠𝑑𝑠=𝜆⋅𝜓𝑢,𝑣))()Γ().𝛼−2𝛼+1(3.15) Using the fact that 𝜆≤2/(𝛼−2)Γ(𝛼+1) (assumption (H3)), we have 𝑑(𝑇𝑢,𝑇𝑣)≤𝜓(𝑑(𝑢,𝑣))=𝑑(𝑢,𝑣)−𝑑(𝑢,𝑣)−𝜓(𝑑(𝑢,𝑣)).(3.16) Put 𝜑(𝑥)=𝑥−𝜓(𝑥), As 𝜓∈𝒥, this means that 𝜑∈ℱ. The last inequality gives us 𝑑(𝑇𝑢,𝑇𝑣)≤𝑑(𝑢,𝑣)−𝜑(𝑑(𝑢,𝑣)).(3.17) This proves that 𝑇 satisfies the contractive condition of Theorem 2.6.
Finally, as 𝐺(𝑡,𝑠)≥0 (Lemma 3.2) and 𝑓≥0 (assumption (H1)), we have (𝑇0)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,0)𝑑𝑠≥0,(3.18) where 0 denotes the zero function.
Now, Theorem 2.6 shows that problem (1.3) has at least one nonnegative solution. As (𝑃,≤) satisfies condition (2.11), we obtain the uniqueness of the solution.
In what follows, we will prove that this solution is positive (this means that 𝑥(𝑡)>0, for 𝑡∈(0,1)).
Finally, we will prove that the zero function is not the solution for problem (1.3). In fact, in contrary case, the zero function is a fixed point of 𝑇 and, thus, we have 0=¹₀𝐺(0,𝑠)𝑓(𝑠,0)𝑑𝑠,for𝑡∈[0,1].(3.19) The nonnegative character of the functions 𝐺 and 𝑓 and the last expression give us 𝐺(𝑡,𝑠)⋅𝑓(𝑠,0)=0a.e.(𝑠),for𝑡∈[0,1].(3.20) This and the fact that 𝐺(𝑡,𝑠)≠0 a.e. (𝑠) for any 𝑡∈[0,1] because 𝐺(𝑡,𝑠) is given by a polynomial implies 𝑓(𝑠,0)=0a.e.(𝑠).(3.21) Taking into account assumption (H2), 𝑓(𝑡₀,0)>0 for certain 𝑡₀∈[0,1]. By the continuity of 𝑓 we can find a set 𝐴⊂[0,1] with 𝑡₀∈𝐴 and 𝜇(𝐴)>0, where 𝜇 is the Lebesgue measure, such that 𝑓(𝑡,0)>0 for 𝑡∈𝐴. This contradicts (3.21).
This proves that the zero function is not the solution for problem (1.3). Now, we will prove that the solution 𝑥 is positive.
In the contrary case, we find 0<𝑡^∗<1 such that 𝑥(𝑡^∗)=0. As the solution 𝑥 is a fixed point of the operator 𝑇, this means that 𝑥𝑡^∗=¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠=0.(3.22) Since 𝑥∈𝑃 and, thus, 𝑥≥0 and by the fact that 𝑓 is nondecreasing in the second variable and 𝐺(𝑡,𝑠)≥0, we can get 0=𝑥𝑡^∗=¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,0)𝑑𝑠≥0,(3.23) and this inequality implies 𝑥𝑡^∗=¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,0)𝑑𝑠=0.(3.24)
Using a similar reasoning to the one above used we obtain a contradiction.
Therefore, 𝑥(𝑡)>0, for 𝑡∈[0,1].
This finishes the proof.

Remark 3.4. In Theorem 3.1, condition (H2) seems to be a strong condition in order to obtain a positive solution for problem (1.3), but when there is uniqueness of solution one will see that this condition is a very adjusted one. More precisely, under the assumption that problem (1.3) has a unique nonnegative solution 𝑥(𝑡) one has 𝑓𝑡₀,0>0forcertain𝑡₀∈[0,1]ifandonlyif𝑥(𝑡)isapositivesolution.(3.25) In fact, if 𝑓(𝑡₀,0)>0 for certain 𝑡₀∈[0,1] the argument used in the proof of Theorem 3.1 give us that 𝑥(𝑡) is a positive solution.
For the other implication, suppose that 𝑓(𝑡,0)=0 for any 𝑡∈[0,1]. Under this assumption, our problem (1.3) admits as solutions the function 𝑥(𝑡) and the zero function and this contradicts the hypothesis about uniqueness of solution to problem (1.3). Therefore, 𝑓(𝑡₀,0)>0 for certain 𝑡₀∈[0,1].

Remark 3.5. Notice that the assumptions in Theorem 3.1 are invariant by additive perturbations. More precisely, if 𝑓(𝑡,0)=0 for any 𝑡∈[0,1] and 𝑓 satisfies (H1) and (H3) of Theorem 3.1, then 𝑔(𝑡,𝑢)=𝑎(𝑡)+𝑓(𝑡,𝑢), with 𝑎∶[0,1]→[0,∞) a nondecreasing continuous function with 𝑎(𝑡₀)≠0 for certain 𝑡₀∈[0,1], satisfies (H1), (H2), and (H3) of Theorem 3.1 and the following nonlinear boundary value problem of fractional order: 𝐷^𝛼₀⁺𝑢(𝑡)((𝑡𝑢(0)+𝑔𝑡,𝑢))=0,0<𝑡<1,3<𝛼≤4,=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0(3.26) has a unique positive solution (by Theorem 3.1).

In the sequel we present an example where the results can be applied.

Example 3.6. Consider the fractional boundary value problem 𝐷₀^7/2₊𝑢(𝑡)+𝑡²((𝑡𝑢(0)+1ln2+𝑢))=0,0<𝑡<1,=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0.(3.27) In this case, 𝑓(𝑡,𝑢)=(𝑡²+1)ln(2+𝑢) for (𝑡,𝑢)∈[0,1]×[0,∞). Obviously, 𝑓 is a continuous function and 𝑓(𝑡,0)=(𝑡²+1)ln2≠0 for 𝑡∈[0,1]. As 𝜕𝑓/𝜕𝑢=(𝑡²+1)(1/(2+𝑢))>0 for 𝑢∈[0,∞), 𝑓 is nondecreasing with respect to the second variable.
Besides, for 𝑢≥𝑣 and 𝑡∈[0,1], we have 𝑡²+1[ln(2+𝑢)−ln(2+𝑣)]=𝑡²+1⋅ln2+𝑢=𝑡2+𝑣₂+1ln2+𝑣+𝑢−𝑣=𝑡2+𝑣²+1ln1+𝑢−𝑣≤𝑡2+𝑣₂+1ln(1+(𝑢−𝑣))≤2ln(1+(𝑢−𝑣)).(3.28) A straightforward calculation gives us that 𝜓(𝑥)=ln(1+𝑥) satisfies that 𝜓∈𝒥.
Moreover, in this case 𝜆=2, 𝛼=7/2 and we have (𝛼−2)Γ(𝛼+1)2=(7/2−2)Γ(7/2+1)2=34Γ72=3+14⋅727⋅Γ2≈8.7228>2=𝜆.(3.29) Finally, Theorem 3.1 proves the existence and uniqueness of a positive solution for problem (3.27).

4. A Final Remark

In connection with problem (1.3), the main result in [21] is the following.

Theorem 4.1 (see [21, Theorem  3.1]). Problem (1.3) has a positive solution 𝑢(𝑡) if the following conditions are satisfied: (𝐻_𝑓)𝑓(𝑡,𝑢)∈𝒞([0,1]×[0,∞),ℝ⁺) is nondecreasing relative to 𝑢, 𝑓(𝑡,𝑝(𝑡))≠0 for 𝑡∈(0,1), where 𝑝(𝑡)=∫¹₀𝐺(𝑡,𝑠)𝑑𝑠=(1/Γ(𝛼))((𝑡^𝛼−1)/(𝛼−2)−(1/𝛼)𝑡^𝛼), and there exists a positive constant 𝜇<1 such that 𝑘^𝜇𝑓(𝑡,𝑢)≤𝑓(𝑡,𝑘𝑢),∀0≤𝑘≤1.(4.1)

In the sequel, we present an example which can be treated by Theorem 3.1 and it cannot be covered by Theorem 4.1.

Example 4.2. Consider the fractional boundary value problem 𝐷_𝑂^7/2₊𝑢(𝑡)+𝑡²(𝑡)𝑢(0)+1𝜌𝑢+𝑐=0,0≤𝑡≤1,=𝑢^′(0)=𝑢^′′(0)=𝑢^′′(1)=0,(4.2) with 𝑐>0 and 0<𝜌<1.
In this case, 𝑓(𝑡,𝑢)=(𝑡²+1)(𝜌𝑢+𝑐), for (𝑡,𝑢)∈[0,1]×[0,∞). Obviously, 𝑓 is continuous and nondecreasing with respect to the second variable since 𝜕𝑓/𝜕𝑢=𝜌(𝑡²+1)>0.
Besides, if 𝑢≥𝑣 and 𝑡∈[0,1], we have 𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)=𝑡²=𝑡+1𝜌𝑢+𝑐−𝜌𝑣+𝑐₂+1𝜌(𝑢−𝑣)≤2𝜌(𝑢−𝑣).(4.3) In this case, 𝜓(𝑥)=𝜌𝑥 and it is easily seen that 𝜑(𝑥)=𝑥−𝜓(𝑥)=(1−𝜌)𝑥 belongs to ℱ.
Moreover, in this case, 𝜆=2 and, as 𝛼=7/2, we have (𝛼−2)Γ(𝛼+1)2=(7/2−2)Γ(7/2+1)2=34Γ72=3+14⋅727⋅Γ2≈8.7228>2=𝜆.(4.4) As 𝑓(𝑡,0)=𝑐(𝑡²+1)>0 for any 𝑡∈[0,1], Theorem 3.1 gives us the existence and uniqueness of positive solution for problem (4.2).
On the other hand, we will show that 𝑓(𝑡,𝑢)=(𝑡²+1)(𝜌𝑢+𝑐) with 0<𝜌<1 and 𝑐>0 does not satisfy (𝐻_𝑓) of Theorem 4.1. In fact, suppose that there exists 0<𝜇<1 such that 𝑘^𝜇𝑓(𝑡,𝑢)≤𝑓(𝑡,𝑘𝑢),forany0≤𝑘≤1.(4.5) This implies that 𝑘^𝜇≤𝑓(𝑡,𝑘𝑢)𝑓()=𝑡𝑡,𝑢²+1𝜌𝑘𝑢+𝑐𝑡²=+1𝜌𝑢+𝑐𝜌𝑘𝑢+𝑐𝜌𝑢+𝑐.(4.6) Notice that lim_𝑢→∞(𝜌𝑘𝑢+𝑐/𝜌𝑢+𝑐)=𝑘, and, consequently, taking limit as 𝑢→∞ in the last inequality, we get 𝑘^𝜇≤𝑘,(4.7) this is false because 0<𝜇<1 and the function ℎ(𝛼)=𝑘^𝛼 is decreasing when 0<𝑘<1.  Therefore, problem (4.2) can be covered by Theorem 3.1 and it cannot be studied by Theorem 4.1.

5. Conclusions

Our main contribution in this paper is to prove under certain assumptions the existence and the uniqueness of positive solution for problem (1.3) which was treated in [21]. In [21] the question of uniqueness of solution was not considered. Moreover, we present an example which can be covered by the results of this paper and cannot be treated by the ones obtained in [21].

Acknowledgment

This paper was partially supported by Ministerio de Educación y Ciencia Project MTM 2007/65706.