Existence and Exact Asymptotic Behavior of Positive Solutions for a Fractional Boundary Value Problem

Abstract

We establish the existence and uniqueness of a positive solution 𝑢 for the fractional boundary value problem 𝐷^𝛼𝑢(𝑥)=−𝑎(𝑥)𝑢^𝜎(𝑥), 𝑥∈(0,1) with the condition lim_𝑥→0𝐷^𝛼−1𝑢(𝑥)=0,𝑢(1)=0, where 1<𝛼≤2, 𝜎∈(−1,1), and 𝑎 is a nonnegative continuous function on (0,1) that may be singular at 𝑥=0 or 𝑥=1.

1. Introduction

Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetism. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see[1–3]). In consequence, the subject of fractional differential equations is gaining much importance and attention. Motivated by the surge in the development of this subject, we consider the following problem: 𝐷^𝛼𝑢(𝑥)=−𝑎(𝑥)𝑢^𝜎(𝑥),𝑥∈(0,1),lim_𝑥→0+𝐷^𝛼−1𝑢(𝑥)=0,𝑢(1)=0,(1) where 1<𝛼≤2, −1<𝜎<1, 𝑎 is a nonnegative continuous function on (0,1) that may be singular at 𝑥=0 or 𝑥=1 and 𝐷^𝛼 is the Riemann-Liouville fractional derivative. Then we study the existence and exact asymptotic behavior of positive solutions for this problem.

We recall that for a measurable function 𝑣, the Riemann-Liouville fractional integral 𝐼_𝛽𝑣 and the Riemann-Liouville derivative 𝐷^𝛽𝑣 of order 𝛽>0 are, respectively, defined by 𝐼_𝛽𝑣(𝑥)=1Γ𝛽^𝑥₀(𝑥−𝑡)^𝛽−1𝑣(𝑡)𝐷𝑑𝑡,_𝛽𝑣(𝑥)=1Γ𝑑𝑛−𝛽𝑑𝑥^𝑛𝑥₀(𝑥−𝑡)^{𝑛−𝛽−1}𝑣(𝑡)=𝑑𝑑𝑡𝑑𝑥^𝑛𝐼_𝑛−𝛽𝑣(𝑥),(2) provided that the right-hand sides are pointwise defined on (0,1]. Here 𝑛=[𝛽]+1 and [𝛽] means the integral part of the number 𝛽 and Γ is the Euler Gamma function.

Moreover, we have the following well-known properties (see [2, 4]):(i)𝐼_𝛽𝐼_𝛾𝑣(𝑥)=𝐼_𝛽+𝛾𝑣(𝑥) for 𝑥∈[0,1], 𝑣∈𝐿¹((0,1]), 𝛽+𝛾≥1,(ii)𝐷^𝛽𝐼_𝛽𝑣(𝑥)=𝑣(𝑥) for a.e. 𝑥∈[0,1], where 𝑣∈𝐿¹((0,1]), 𝛽>0,(iii)𝐷^𝛽𝑣(𝑥)=0 if and only if 𝑣(𝑥)=∑^𝑛_𝑗=1𝑐_𝑗𝑡^𝛽−𝑗, where 𝑛=[𝛽]+1 and (𝑐₁,𝑐₂,…,𝑐_𝑛)∈ℝ^𝑛.

Several results are obtained for fractional differential equation with different boundary conditions (see [5–15] and the references therein), but none of them deals with the existence of a positive solution for problem (1).

Our aim in this paper is to establish the existence and uniqueness of a positive solution 𝑢∈𝐶_2−𝛼([0,1]) for problem (1) with a precise asymptotic behavior, where 𝐶_2−𝛼([0,1]) is the set of all functions 𝑓 such that 𝑡→𝑡^2−𝛼𝑓(𝑡) is continuous on [0,1]. Note that for 0<𝛼<2, the solution 𝑢 for problem (1) blows up at 𝑥=0.

To state our result, we need some notations. We will use 𝒦 to denote the set of Karamata functions 𝐿 defined on (0,𝜂] by 𝐿(𝑡)∶=𝑐exp^𝜂_𝑡𝑧(𝑠)𝑠𝑑𝑠,(3) for some 𝜂>1, where 𝑐>0 and 𝑧∈𝐶([0,𝜂]) such that 𝑧(0)=0. It is clear that a function 𝐿 is in 𝒦 if and only if 𝐿 is a positive function in 𝐶¹((0,𝜂]) such that lim_𝑡→0+𝑡𝐿^′(𝑡)𝐿(𝑡)=0.(4) For two nonnegative functions 𝑓 and 𝑔 defined on a set 𝑆, the notation 𝑓(𝑥)≈𝑔(𝑥), 𝑥∈𝑆, means that there exists 𝑐>0 such that (1/𝑐)𝑓(𝑥)≤𝑔(𝑥)≤𝑐𝑓(𝑥), for all 𝑥∈𝑆. We denote also 𝑥⁺=max(𝑥,0) for 𝑥∈ℝ.

Throughout this paper we assume that 𝑎 is nonnegative on (0,1) and satisfies the following condition.

(𝐻₀)  𝑎∈𝐶((0,1)) such that 𝑎(𝑡)≈𝑡^−𝜆𝐿₁(𝑡)(1−𝑡)^−𝜇𝐿₂(1−𝑡),𝑡∈(0,1),(5) where 𝜆+(2−𝛼)𝜎≤1, 𝜇≤𝛼, 𝐿₁,𝐿₂∈𝒦 satisfying ^𝜂₀𝐿₁(𝑡)𝑡^{𝜆+(2−𝛼)𝜎}𝑑𝑡<∞,^𝜂₀𝐿₂(𝑡)𝑡^{𝜇−𝛼+1}𝑑𝑡<∞.(6) In the sequel, we introduce the function 𝜃 defined on (0,1) by 𝜃(𝑥)=()∫1−𝑥if𝜇<𝜎+𝛼−1,1−𝑥^𝜂_1−𝑥𝐿₂(𝑠)𝑠𝑑𝑠^1/(1−𝜎)()if𝜇=𝜎+𝛼−1,1−𝑥^{(𝛼−𝜇)/(1−𝜎)}𝐿₂(1−𝑥)^1/(1−𝜎)∫if𝜎+𝛼−1<𝜇<𝛼,_01−𝑥𝐿₂(𝑠)𝑠𝑑𝑠^1/(1−𝜎)if𝜇=𝛼.(7) Our main result is the following.

Theorem 1. Let 𝜎∈(−1,1) and assume that 𝑎 satisfies (H₀). Then problem (1) has a unique positive solution 𝑢∈𝐶_2−𝛼([0,1]) satisfying for 𝑥∈(0,1)𝑢(𝑥)≈𝑥^𝛼−2𝜃(𝑥).(8)

This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section 3, we give the proof of Theorem 1.

2. Technical Lemmas

To keep the paper self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].

Lemma 2. The following hold.(i)Let 𝐿∈𝒦 and 𝜀>0, then one has lim_𝑡→0+𝑡^𝜀𝐿(𝑡)=0.(9)(ii)Let 𝐿₁, 𝐿₂∈𝒦 and 𝑝∈ℝ. Then one has 𝐿₁+𝐿₂∈𝒦, 𝐿₁𝐿₂∈𝒦, and 𝐿^𝑝₁∈𝒦.

Example 3. Let 𝑚∈ℕ^∗. Let 𝑐>0, (𝜇₁,𝜇₂,…,𝜇_𝑚)∈ℝ^𝑚 and let 𝑑 be a sufficiently large positive real number such that the function 𝐿(𝑡)=𝑐^𝑚_𝑘=1log_𝑘𝑑𝑡^−𝜇𝑘(10) is defined and positive on (0,𝜂], for some 𝜂>1, where log_𝑘𝑥=log∘log∘⋯∘log𝑥 (𝑘 times). Then 𝐿∈𝒦.

Applying Karamata’s theorem (see [16, 17]), we get the following.

Lemma 4. Let 𝜇∈ℝ and let 𝐿 be a function in 𝒦 defined on (0,𝜂]. One has the following.(i)If 𝜇<−1, then ∫^𝜂₀𝑠^𝜇𝐿(𝑠)𝑑𝑠 diverges and ∫^𝜂_𝑡𝑠^𝜇𝐿(𝑠)𝑑𝑠∼_𝑡→0⁺−(𝑡^1+𝜇𝐿(𝑡)/(𝜇+1)). (ii)If 𝜇>−1, then ∫^𝜂₀𝑠^𝜇𝐿(𝑠)𝑑𝑠 converges and ∫^𝑡₀𝑠^𝜇𝐿(𝑠)𝑑𝑠∼_𝑡→0⁺(𝑡^1+𝜇𝐿(𝑡)/(𝜇+1)).

Lemma 5. Let 𝐿∈𝒦 be defined on (0,𝜂]. Then one has lim_𝑡→0+𝐿(𝑡)∫^𝜂_𝑡(𝐿(𝑠)/𝑠)𝑑𝑠=0.(11) If further ∫^𝜂₀(𝐿(𝑠)/𝑠)𝑑𝑠 converges, then one has lim_𝑡→0+𝐿(𝑡)∫^𝑡₀(𝐿(𝑠)/𝑠)𝑑𝑠=0.(12)

Proof. We distinguish two cases.
Case  1. We suppose that ∫^𝜂₀(𝐿(𝑠)/𝑠)𝑑𝑠 converges. Since the function 𝑡→𝐿(𝑡)/𝑡 is nonincreasing in (0,𝜔], for some 𝜔<𝜂, it follows that, for each 𝑡≤𝜔, we have 𝐿(𝑡)≤^𝑡₀𝐿(𝑠)𝑠𝑑𝑠.(13) It follows that lim_𝑡→0⁺𝐿(𝑡)=0. So we deduce (11).
Now put 𝜑(𝑡)=𝐿(𝑡)𝑡,for𝑡∈0,𝜂.(14) Using that lim_𝑡→0⁺(𝑡𝜑^′(𝑡)/𝜑(𝑡))=−1, we obtain ^𝑡₀𝜑(𝑠)𝑑𝑠∼_𝑡→0⁺−^𝑡₀𝑠𝜑^′(𝑠)𝑑𝑠=−𝑡𝜑(𝑡)+^𝑡₀𝜑(𝑠)𝑑𝑠.(15) This implies that ^𝑡₀𝐿(𝑠)𝑠𝑑𝑠∼_𝑡→0⁺−𝐿(𝑡)+^𝑡₀𝐿(𝑠)𝑠𝑑𝑠.(16) So (12) holds.
Case  2. We suppose that ∫^𝜂₀(𝐿(𝑠)/𝑠)𝑑𝑠 diverges. We have, for some 𝜔<𝜂, ^𝜔_𝑡𝜑(𝑠)𝑑𝑠∼_𝑡→0⁺𝑡𝜑(𝑡)−𝜔𝜑(𝜔)+^𝜔_𝑡𝜑(𝑠)𝑑𝑠.(17) This implies that ^𝜔_𝑡𝐿(𝑠)𝑠𝑑𝑠∼_𝑡→0⁺𝐿(𝑡)−𝜔𝜑(𝜔)+^𝜔_𝑡𝐿(𝑠)𝑠𝑑𝑠.(18) This proves (11) and completes the proof.

Remark 6. Let 𝐿∈𝒦 defined on (0,𝜂]; then, using (4) and (11), we deduce that 𝑡→^𝜂_𝑡𝐿(𝑠)𝑠𝑑𝑠∈𝒦.(19) If further ∫^𝜂₀(𝐿(𝑠)/𝑠)𝑑𝑠 converges, we have by (11) that 𝑡→^𝑡₀𝐿(𝑠)𝑠𝑑𝑠∈𝒦.(20)

Lemma 7. Given 1<𝛼≤2 and 𝑓 is such that the function 𝑡→(1−𝑡)^𝛼−1𝑓(𝑡) is continuous and integrable on (0,1), then the boundary value problem 𝐷^𝛼𝑢(𝑡)=−𝑓(𝑡),𝑡∈(0,1),lim_𝑥→0𝐷^𝛼−1𝑢(𝑥)=0,𝑢(1)=0,(21) has a unique solution given by 𝑢(𝑥)=𝐺_𝛼𝑓(𝑥)∶=¹₀𝐺_𝛼(𝑥,𝑡)𝑓(𝑡)𝑑𝑡,(22) where 𝐺_𝛼()=1𝑥,𝑡Γ(𝛼)𝑥^𝛼−2(1−𝑡)^𝛼−1−(𝑥−𝑡)^+𝛼−1,(23) is the Green function for the boundary value problem (21).

Proof. Since 𝑢₀=−𝐼_𝛼𝑓 is a solution of the equation 𝐷^𝛼𝑢=−𝑓, then 𝐷^𝛼(𝑢+𝐼_𝛼𝑓)=0. Consequently there exist two constants 𝑐₁, 𝑐₂∈ℝ such that 𝑢(𝑥)+𝐼_𝛼𝑓(𝑥)=𝑐₁𝑥^𝛼−1+𝑐₂𝑥^𝛼−2. Using the fact that lim_𝑥→0𝐷^𝛼−1𝑢(𝑥)=0 and 𝑢(1)=0, we obtain 𝑐₁=0 and 𝑐₂=𝐼_𝛼𝑓(1). So 𝑢(𝑥)=1Γ(𝛼)𝑥^𝛼−21₀(1−𝑡)^𝛼−1𝑓(𝑡)−1𝑑𝑡Γ(𝛼)^𝑥₀(𝑥−𝑡)^𝛼−1𝑓(𝑡)=𝑑𝑡₁₀𝐺_𝛼(𝑥,𝑡)𝑓(𝑡)𝑑𝑡.(24) In the following, we give some estimates on the function 𝐺_𝛼. So, we need the following lemma.

Lemma 8. For 𝜆,𝜇∈(0,∞), and 𝑎,𝑡∈[0,1], one has 𝜇min1,𝜆1−𝑎𝑡^𝜆≤1−𝑎𝑡^𝜇𝜇≤max1,𝜆1−𝑎𝑡^𝜆.(25)

Proposition 9. On (0,1)×(0,1), one has 𝐺_𝛼(𝑥,𝑡)≈𝑥^𝛼−2(1−𝑡)^𝛼−2(1−max(𝑥,𝑡)).(26)

Proof. For 𝑥,𝑡∈(0,1)×(0,1), we have 𝐺_𝛼()=()𝑥,𝑡1−𝑡^𝛼−1𝑥^𝛼−2Γ(𝛼)()1−𝑥𝑥−𝑡⁺𝑥(1−𝑡)^𝛼−1.(27) Since (𝑥−𝑡)⁺/𝑥(1−𝑡)∈(0,1) for 𝑡∈(0,1), then, by applying Lemma 8 with 𝜇=𝛼−1 and 𝜆=1, we obtain 𝐺_𝛼(𝑥,𝑡)≈𝑥^𝛼−2(1−𝑡)^𝛼−1()1−𝑥−𝑡⁺1−𝑡=𝑥^𝛼−2(1−𝑡)^𝛼−2(1−max(𝑥,𝑡)),(28) which completes the proof.

In the sequel we put 𝑏(𝑡)=𝑡^−𝛽𝐿₃(𝑡)(1−𝑡)^−𝛾𝐿₄(1−𝑡),(29) where 𝐿₃,𝐿₄∈𝒦 and we aim to give some estimates on 𝑥^2−𝛼𝐺_𝛼𝑏(𝑥).

Proposition 10. Assume that 𝐿₃,𝐿₄∈𝒦,𝛽≤1,𝛾≤𝛼 with ^𝜂₀𝑡^−𝛽𝐿₃(𝑡)𝑑𝑡<∞,^𝜂₀𝑡^{𝛼−1−𝛾}𝐿₄(𝑡)𝑑𝑡<∞.(30) Then for 𝑥∈(0,1), 𝑥^2−𝛼𝐺_𝛼𝑏(𝑥)≈()∫1−𝑥if𝛾<𝛼−1,1−𝑥^𝜂_1−𝑥𝐿₄(𝑡)𝑡()𝑑𝑡if𝛾=𝛼−1,1−𝑥^𝛼−𝛾𝐿₄()∫1−𝑥if𝛼−1<𝛾<𝛼,_01−𝑥𝐿₄(𝑡)𝑡𝑑𝑡if𝛾=𝛼.(31)

Proof. Using Proposition 9, we have 𝑥^2−𝛼𝐺_𝛼𝑏(𝑥)≈(1−𝑥)^𝑥₀(1−𝑡)^{𝛼−2−𝛾}𝑡^−𝛽𝐿₃(𝑡)𝐿₄()+1−𝑡𝑑𝑡_1𝑥(1−𝑡)^{𝛼−1−𝛾}𝑡^−𝛽𝐿₃(𝑡)𝐿₄()=()𝐼(𝑥)(𝑥).1−𝑡𝑑𝑡1−𝑥+𝐽(32) For 0<𝑥≤1/2, we use Lemma 4 and hypotheses (30) to deduce that 𝐼(𝑥)≈𝑥^1−𝛽𝐿₃(𝑥)∫if𝛽<1,_𝑥0𝐿₃(𝑡)𝑡𝐽(𝑥)≈𝑑𝑡if𝛽=1,_11/2(1−𝑡)^{𝛼−1−𝛾}𝐿₄(1−𝑡)𝑑𝑡+_𝑥^1/2𝑡^−𝛽𝐿₃(𝑡)𝑑𝑡≈1+_𝑥^1/2𝑡^−𝛽𝐿₃(𝑡)𝑑𝑡≈1.(33) Hence, it follows from Lemma 2 and hypothesis (30) that, for 0<𝑥≤1/2, we have 𝑥^2−𝛼𝐺_𝛼𝑏(𝑥)≈1.(34) Now, for 1/2≤𝑥<1, we use again Lemma 4 and hypothesis (30) to deduce that 𝐼(𝑥)≈₀^1/2𝑡^−𝛽𝐿₃(𝑡)𝑑𝑡+^𝑥_1/2(1−𝑡)^{𝛼−2−𝛾}𝐿₄(1−𝑡)𝑑𝑡≈1+^1/2_1−𝑥𝑡^{𝛼−2−𝛾}𝐿₄(𝑡)≈∫𝑑𝑡1if𝛾<𝛼−1,_𝜂1−𝑥𝐿₄(𝑡)𝑡()𝑑𝑡if𝛾=𝛼−1,1−𝑥^{𝛼−1−𝛾}𝐿₄()𝐽(𝑥)≈1−𝑥if𝛾>𝛼−1,_01−𝑥𝑡^{𝛼−1−𝛾}𝐿₄(𝑡)≈()𝑑𝑡1−𝑥^𝛼−𝛾𝐿₄()∫1−𝑥if𝛾<𝛼,_01−𝑥𝐿₄(𝑡)𝑡𝑑𝑡if𝛾=𝛼.(35) Hence, it follows from Lemmas 2 and 5 that, for 𝑥∈[1/2,1), we have 𝑥^2−𝛼𝐺_𝛼𝑏(𝑥)≈()∫1−𝑥if𝛾<𝛼−1,1−𝑥^𝜂_1−𝑥𝐿₄(𝑡)𝑡()𝑑𝑡if𝛾=𝛼−1,1−𝑥^𝛼−𝛾𝐿₄()∫1−𝑥if𝛼−1<𝛾<𝛼,_01−𝑥𝐿₄(𝑡)𝑡𝑑𝑡if𝛾=𝛼.(36)
This together with (34) implies that (36) holds on (0,1).

3. Proof of Theorem 1

We begin this section by giving a preliminary result that will play a crucial role in the proof of Theorem 1.

Proposition 11. Assume that the function 𝑎 satisfies (𝐻₀) and put 𝜔(𝑡)=𝑎(𝑡)𝑡^{(𝛼−2)𝜎}𝜃^𝜎(𝑡) for 𝑡∈(0,1). Then one has, for 𝑥∈(0,1), 𝑥^2−𝛼𝐺_𝛼𝜔(𝑥)≈𝜃(𝑥).(37)

Proof. For 𝑡∈(0,1), we have 𝜔(𝑡)=𝑎(𝑡)𝑡^{(𝛼−2)𝜎}𝜃^𝜎(𝑡)=𝑡^{−𝜆−(2−𝛼)𝜎}(1−𝑡)^−𝜇+𝜎𝐿₁(𝑡)𝐿₂()𝑡1−𝑡if𝜇<𝜎+𝛼−1,_{−𝜆−(2−𝛼)𝜎}(1−𝑡)^−𝜇+𝜎𝐿₁(𝑡)𝐿₂()×∫1−𝑡_𝜂1−𝑡𝐿₂(𝑠)𝑠𝑑𝑠^{𝜎/(1−𝜎)}𝑡if𝜇=𝜎+𝛼−1,_{−𝜆−(2−𝛼)𝜎}(1−𝑡)^{−(𝜇−𝜎𝛼)/(1−𝜎)}𝐿₁(𝑡)×𝐿₂(1−𝑡)^{𝜎/(1−𝜎)}𝑡if𝜎+𝛼−1<𝜇<𝛼,_{−𝜆−(2−𝛼)𝜎}(1−𝑡)^−𝜇𝐿₁(𝑡)𝐿₂()×∫1−𝑡_01−𝑡𝐿₂(𝑠)𝑠𝑑𝑠^{𝜎/(1−𝜎)}if𝜇=𝛼.(38) So, we can see that 𝜔(𝑡)=𝑡^−𝛽(1−𝑡)^−𝛾𝐿₁(𝑡)𝐿₂(1−𝑡),(39) where 𝛽≤1, 𝛾≤𝛼 and, according to Lemma 2, the functions 𝑡→𝐿₁(𝑡) and 𝑡→𝐿₂(𝑡) are in 𝒦. Moreover, using Lemma 4, we have ∫^𝜂₀𝑡^−𝛽𝐿₁(𝑡)𝑑𝑡<∞ and ∫^𝜂₀𝑡^−𝛾𝐿₂(𝑡)𝑑𝑡<∞. So the result follows from Proposition 10.

Proof of Theorem 1. From Proposition 11, there exists 𝑀>1 such that for each 𝑥∈(0,1)1𝑀𝜃(𝑥)≤𝑥^2−𝛼𝐺_𝛼𝜔(𝑥)≤𝑀𝜃(𝑥),(40) where 𝜔(𝑡)=𝑎(𝑡)𝑡^{(𝛼−2)𝜎}𝜃^𝜎(𝑡).
Put 𝑐₀=𝑀^{1/(1−|𝜎|)} and let :1Λ=𝑣∈𝐶([0,1])𝑐₀𝜃≤𝑣≤𝑐₀𝜃.(41) In order to use a fixed point theorem, we denote 𝑎(𝑡)=𝑎(𝑡)𝑡^{(𝛼−2)𝜎} and we define the operator 𝑇 on Λ by 𝑇𝑣(𝑥)=𝑥^2−𝛼𝐺_𝛼𝑎𝑣^𝜎(𝑥).(42)For this choice of 𝑐₀, we can easily prove that for 𝑣∈Λ, we have 𝑇𝑣≤𝑐₀𝜃 and 𝑇𝑣≥(1/𝑐₀)𝜃.
Now, we have (𝑥)=𝑥𝑇𝑣^2−𝛼Γ(𝛼)¹₀𝐺_𝛼(𝑥,𝑡)𝑎(𝑡)𝑣^𝜎(𝑡)=1𝑑𝑡Γ(𝛼)¹₀(1−𝑡)^𝛼−1−𝑥^2−𝛼(𝑥−𝑡)^+𝛼−1×𝑎(𝑡)𝑣^𝜎(𝑡)𝑑𝑡.(43) Since the function (𝑥,𝑡)→(1−𝑡)^𝛼−1−𝑥^2−𝛼((𝑥−𝑡)⁺)^𝛼−1 is continuous on [0,1]×[0,1] and the function 𝑡→(1−𝑡)^𝛼−1𝑎(𝑡)𝜃^𝜎(𝑡) is integrable on (0,1), we deduce that the operator 𝑇 is compact from Λ to itself. It follows by the Schauder fixed point theorem that there exists 𝑣∈Λ such that 𝑇𝑣=𝑣. Put 𝑢(𝑥)=𝑥^𝛼−2𝑣(𝑥). Then 𝑢∈𝐶_2−𝛼([0,1]) and 𝑢 satisfies the equation 𝑢(𝑥)=𝐺_𝛼𝑎𝑢^𝜎(𝑥).(44) Since the function 𝑡→(1−𝑡)^𝛼−1𝑎(𝑡)𝑢^𝜎(𝑡) is continuous and integrable on (0,1), then by Lemma 7 the function 𝑢 is a positive continuous solution of problem (1).
Finally, let us prove that 𝑢 is the unique positive continuous solution satisfying (8). To this aim, we assume that (1) has two positive solutions 𝑢,𝑣∈𝐶_2−𝛼([0,1]) satisfying (8) and consider the nonempty set 𝐽={𝑚≥1∶1/𝑚≤𝑢/𝑣≤𝑚} and put 𝑐=inf𝐽. Then 𝑐≥1 and we have (1/𝑐)𝑣≤𝑢≤𝑐𝑣. It follows that 𝑢^𝜎≤𝑐^|𝜎|𝑣^𝜎 and consequently −𝐷^𝛼𝑐^|𝜎|𝑣−𝑢=𝑎𝑐^|𝜎|𝑣^𝜎−𝑢^𝜎≥0,lim_𝑡→0+𝐷^𝛼−1𝑐^|𝜎|𝑣−𝑢(𝑡)=0,𝑐^|𝜎|𝑣−𝑢(1)=0.(45) Which implies by Lemma 7 that 𝑐^|𝜎|𝑣−𝑢=𝐺_𝛼(𝑎(𝑐^|𝜎|𝑣^𝜎−𝑢^𝜎))≥0. By symmetry, we obtain also that 𝑣≤𝑐^|𝜎|𝑢. Hence 𝑐^|𝜎|∈𝐽 and 𝑐≤𝑐^|𝜎|. Since |𝜎|<1, then 𝑐=1 and consequently 𝑢=𝑣.

Example 12. Let 𝜎∈(−1,1) and 𝑎 be a positive continuous function on (0,1) such that 𝑎(𝑡)≈(1−𝑡)^−𝜇3log1−𝑡^−𝛽,(46) where 𝜇<𝛼 and 𝛽∈ℝ or 𝜇=𝛼 and 𝛽>1. Then, using Theorem 1, problem (1) has a unique positive continuous solution 𝑢 satisfying the following estimates.(i)If 𝜇<𝜎+𝛼−1 or 𝜇=𝜎+𝛼−1 and 𝛽>1, then for 𝑥∈(0,1), 𝑢(𝑥)≈𝑥^𝛼−2(1−𝑥).(47)(ii)If 𝜇=𝜎+𝛼−1and𝛽=1, then for 𝑥∈(0,1), 𝑢(𝑥)≈𝑥^𝛼−2(31−𝑥)loglog1−𝑥^1/(1−𝜎).(48)(iii)If 𝜇=𝜎+𝛼−1and𝛽<1, then for 𝑥∈(0,1), 𝑢(𝑥)≈𝑥^𝛼−2(31−𝑥)log1−𝑥^{(1−𝛽)/(1−𝜎)}.(49)(iv)If 𝜎+𝛼−1<𝜇<𝛼, then for 𝑥∈(0,1), 𝑢(𝑥)≈𝑥^𝛼−2(1−𝑥)^{(𝛼−𝜇)/(1−𝜎)}3log1−𝑥^{−𝛽/(1−𝜎)}.(50)(v)If 𝜇=𝛼and𝛽>1, then for 𝑥∈(0,1), 𝑢(𝑥)≈𝑥^𝛼−23log1−𝑥^{(1−𝛽)/(1−𝜎)}.(51)

Acknowledgment

The authors thank the anonymous referees for a careful reading of the paper and for their helpful suggestions.