Existence and Uniqueness of Positive Solutions for a Singular Fractional Three-Point Boundary Value Problem

Abstract

We investigate the existence and uniqueness of positive solutions for the following singular fractional three-point boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=𝑢^′(0)=𝑢^′′(0)=0,𝑢^′′(1)=𝛽𝑢^′′(𝜂), where 3<𝛼≤4, 𝐷^𝛼₀₊ is the standard Riemann-Liouville derivative and 𝑓∶(0,1]×[0,∞)→[0,∞) with lim_𝑡→0+𝑓(𝑡,⋅)=∞ (i.e., 𝑓 is singular at 𝑡=0). Our analysis relies on a fixed point theorem in partially ordered metric spaces.

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications (see, e.g., [1–5]).

Recently, many papers have appeared dealing with the existence of solutions of nonlinear fractional boundary value problems.

In [6], the authors studied the existence and multiplicity of positive solutions for the boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢=0, (1.1) where 1<𝛼≤2 and 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous, by using some fixed point theorem on cones.

In [7], the authors considered the following nonlinear fractional boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢′=𝑢^′′(0)=𝑢^′′(1)=0, (1.2) where 3<𝛼≤4 and 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous. They obtained their results by using lower and upper solution method and fixed point theorems.

In [8] the authors investigated the existence and uniqueness of positive and nondecreasing solutions for a class of singular fractional boundary value problems by using a fixed point theorem in partially ordered metric spaces.

Very recently, in [9] the authors studied the existence of solutions of the following three-point boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢^′(0)=𝑢^′′(0)=0,𝑢^′′(1)=𝛽𝑢^′′𝜂, (1.3) where 3<𝛼≤4, 0<𝜂<1, 0<𝛽𝜂^𝛼−3<1 and 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous.

Motivated by [8, 9], in this paper we discuss the existence and uniqueness of positive solutions for Problem (1.3) assuming that 𝑓∶(0,1]×[0,∞)→[0,∞) is such that lim_𝑡→0+𝑓(𝑡,⋅)=∞ (i.e., 𝑓 is singular at 𝑡=0). Our main tool is a fixed point theorem in partially ordered metric spaces which appears in [10].

2. Preliminaries and Basic Facts

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

Definition 2.1 (see [5]). 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 classical gamma function.

Definition 2.2 (see [2]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑓∶(0,∞)→ℝ is defined as 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 [2] and they are crucial in finding an integral representation of the boundary value problem (1.3).

Lemma 2.3 (see [2]). Assume that 𝑢∈𝐶(0,1)∩𝐿¹(0,1) and 𝛼>0.
Then the 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) where 𝑐_𝑖∈ℝ,𝑖=1,2,…,𝑛 and 𝑛=[𝛼]+1, as unique solution.

Lemma 2.4 (see [2]). Assume that 𝑢∈𝐶(0,1)∩𝐿¹(0,1) with 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=1,2,…,𝑛 and 𝑛=[𝛼]+1.

By using Lemma 2.4, in [9] the authors proved the following result.

Lemma 2.5 (see [9]). Let 0<𝜂<1 and 𝛽≠1/𝜂^𝛼−3 and ℎ∈𝐶[0,1].
Then the boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)+ℎ=0,0<𝑡<1,=𝑢^′(0)=𝑢^′′(0)=0,𝑢^′′(1)=𝛽𝑢^′′𝜂, (2.6) where 3<𝛼≤4, has as unique solution 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠+𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠ℎ(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)𝐻()=𝜕,0≤𝑡≤𝑠≤1,𝑡,𝑠²𝐺(𝑡,𝑠)𝜕𝑡²=(𝛼−1)(𝛼−2)Γ(𝛼)𝑡^𝛼−3(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−3(),0≤𝑠≤𝑡≤1,𝛼−1)(𝛼−2Γ(𝛼)𝑡^𝛼−3(1−𝑠)^𝛼−3,0≤𝑡≤𝑠≤1. (2.8)

Remark 2.6. In [9] it is proved that 𝐺 is a continuous function on [0,1]×[0,1], 𝐺(𝑡,𝑠)≥0, 𝐺(𝑡,1)=0 and sup_0≤𝑡≤1¹₀𝐺()2𝑡,𝑠𝑑𝑠=()Γ(),𝛼−2𝛼+1¹₀𝐻𝜂𝜂,𝑠𝑑𝑠=^𝛼−3(𝛼−1)1−𝜂Γ(𝛼). (2.9)

In the sequel, we present the fixed point theorem which we will use later. Previously, we present the following class of functions.

By 𝒮 we denote the class of functions 𝛽∶[0,∞)→[0,1) satisfying the following condition: 𝛽𝑡_𝑛⟶1implies𝑡_𝑛⟶0. (2.10) Examples of functions belonging to 𝒮 are 𝛽(𝑡)=𝑘𝑡 with 0≤𝑘<1 and 𝛽(𝑡)=1/(1+𝑡).

The fixed point theorem which we will use later appears in [10].

Theorem 2.7 (see [10]). 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 𝑇∶𝑋→𝑋 be a nondecreasing mapping such that there exists an element 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀. Suppose that there exists 𝛽∈𝒮 such that 𝑑𝑇𝑥,𝑇𝑦≤𝛽𝑑𝑥,𝑦𝑑𝑥,𝑦for𝑥,𝑦∈𝑋with𝑥≥𝑦. (2.11) Assume that either 𝑇 is continuous or 𝑋 is such that if𝑥_𝑛isanondecreasingsequencein𝑋suchthat𝑥_𝑛⟶𝑥then𝑥_𝑛≤𝑥∀𝑛∈ℕ. (2.12) Besides, if forall𝑥,𝑦∈𝑋thereexists𝑧∈𝑋whichiscomparableto𝑥,𝑦, (2.13) then 𝑇 has a unique fixed point.

In our considerations, we will work in the Banach space 𝐶[0,1]={𝑥∶[0,1]→ℝ,𝑥iscontinuous} 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]. (2.14) In [11] it is proved that (𝐶[0,1],≤) satisfies condition (2.12) of Theorem 2.7. Moreover, for 𝑥,𝑦∈𝐶[0,1], as the function max{𝑥,𝑦}∈𝐶[0,1], (𝐶[0,1],≤) satisfies condition (2.13).

3. Main Result

Our starting point of this section is the following lemma.

Lemma 3.1. Suppose that 0<𝜎<1, 3<𝛼≤4, and 𝐹∶(0,1]→ℝ  is a continuous function with lim_𝑡→0+𝐹(𝑡)=∞. If 𝑡^𝜎𝐹(𝑡) is a continuous function on [0,1] then the function defined by 𝐿(𝑡)=¹₀𝐺(𝑡,𝑠)𝐹(𝑠)𝑑s (3.1) is continuous on [0,1], where 𝐺(𝑡,𝑠) is the Green’s function appearing in Lemma 2.5.

Proof. We divide the proof into three cases.
Case 1 (𝑡₀=0). It is clear that 𝐿(0)=0.
Since 𝑡^𝜎𝐹(𝑡) is a continuous function on [0,1], we can find a constant 𝑀>0 such that |𝑡^𝜎𝐹(𝑡)|≤𝑀 for any 𝑡∈[0,1].
Then, we get |𝐿(𝑡)−𝐿(0)|=|𝐿(𝑡)|=¹₀𝐺(𝑡,𝑠)𝐹(𝑠)𝑑𝑠=¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)=𝑑𝑠_𝑡0𝑡^𝛼−1(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠+¹_𝑡𝑡^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)=𝑑𝑠₁₀𝑡^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)≤𝑑𝑠₁₀𝑡^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠+^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)≤𝑑𝑠𝑀𝑡^𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−3𝑠^−𝛼𝑀𝑑𝑠+Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝛼=𝑑𝑠𝑀𝑡^𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−3𝑠^−𝛼𝑑𝑠+𝑀𝑡^𝛼−1Γ(𝛼)^𝑡₀𝑠1−𝑡^𝛼−1𝑠^−𝛼=𝑑𝑠𝑀𝑡^𝛼−1Γ(𝛼)𝛽(1−𝜎,𝛼−2)+𝑀𝑡^𝛼−1Γ(𝛼)^𝑡₀𝑠1−𝑡^𝛼−1𝑠^−𝛼𝑑𝑠. (3.2) If in the integral ∫^𝑡₀(1−(𝑠/𝑡))^𝛼−1𝑠^−𝜎𝑑𝑠 we use the change of variables 𝑢=𝑠/𝑡 then we have ^𝑡₀𝑠1−𝑡^𝛼−1𝑠^−𝜎𝑑𝑠=𝑡^1−𝜎1₀(1−𝑢)^𝛼−1𝑢^−𝜎𝑑𝑢=𝑡^1−𝜎𝛽(1−𝜎,𝛼). (3.3) This and (3.2) give us |𝐿(𝑡)|≤𝑀𝑡^𝛼−1Γ(𝛼)𝛽(1−𝜎,𝛼−2)+𝑀𝑡^𝛼−𝜎Γ(𝛼)𝛽(1−𝜎,𝛼) (3.4) and letting 𝑡→0, we see that |𝐿(𝑡)|→0.
This proves the continuity of 𝐿 at 𝑡₀=0.
Case 2(𝑡₀∈(0,1)). We take 𝑡_𝑛→𝑡₀ and we have to prove that 𝐿(𝑡_𝑛)→𝐿(𝑡₀).
Without loss of generality, we can take 𝑡_𝑛>𝑡₀ (the same argument works for 𝑡_𝑛<𝑡₀).
In fact, 𝐿𝑡_𝑛−𝐿𝑡₀=^𝑡𝑛₀𝑡_𝑛^𝛼−1(1−𝑠)^𝛼−3−𝑡_𝑛−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)+𝑑𝑠_1𝑡𝑛𝑡_𝑛^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)−𝑑𝑠_𝑡00𝑡₀^𝛼−1(1−𝑠)^𝛼−3−𝑡₀−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)−𝑑𝑠_1𝑡0𝑡₀^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)=𝑑𝑠₁₀𝑡_𝑛^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠−^𝑡𝑛₀𝑡_𝑛−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠−¹₀𝑡₀^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)𝑑𝑠+^𝑡0₀𝑡₀−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)=𝑑𝑠₁₀𝑡_𝑛^𝛼−1−𝑡₀^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)−𝑑𝑠_𝑡00𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)−𝑑𝑠_{𝑡𝑛𝑡0}𝑡_𝑛−𝑠^𝛼−1Γ(𝛼)𝑠^−𝜎𝑠^𝜎𝐹(𝑠)≤𝑀𝑡𝑑𝑠_𝑛𝛼−1−𝑡₀^𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−3𝑠^−𝜎+𝑀𝑑𝑠Γ(𝛼)^𝑡0₀𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1𝑠^−𝜎+𝑀𝑑𝑠Γ(𝛼)^𝑡𝑛_𝑡0𝑡_𝑛−𝑠^𝛼−1𝑠^−𝜎≤𝑀𝑡𝑑𝑠_𝑛𝛼−1−𝑡₀^𝛼−1Γ(𝛼)𝛽()+𝑀1−𝜎,𝛼−2Γ(𝛼)𝐼¹_𝑛+𝑀Γ(𝛼)𝐼²_𝑛, (3.5) where 𝐼¹_𝑛=^𝑡0₀𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1𝑠^−𝜎𝐼𝑑𝑠,_2𝑛=^𝑡𝑛_𝑡0𝑡_𝑛−𝑠^𝛼−1𝑠^−𝜎𝑑𝑠. (3.6) In the sequel, we will prove that 𝐼¹_𝑛→0 when 𝑛→∞.
In fact, as 𝑡_𝑛→𝑡₀, then 𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1𝑠^−𝜎⟶0when𝑛⟶∞. (3.7) Moreover, 𝑡_𝑛−𝑠^𝛼−1−𝑡₀−𝑠^𝛼−1𝑠^−𝜎≤𝑡_𝑛−𝑠^𝛼−1+𝑡₀−𝑠^𝛼−1𝑠^−𝜎≤2𝑠^−𝜎 (3.8) and, as ¹₀2𝑠^−𝜎𝑠𝑑𝑠=2^−𝜎+1−𝜎+1¹₀=21−𝜎<∞, (3.9) we have that the sequence ((𝑡_𝑛−𝑠)^𝛼−1−(𝑡₀−𝑠)^𝛼−1)𝑠^−𝜎 converges pointwise to the zero function and |((𝑡_𝑛−𝑠)^𝛼−1−(𝑡₀−𝑠)^𝛼−1)𝑠^−𝜎| is bounded by a function belonging to 𝐿¹[0,1], then by Lebesgue’s dominated convergence theorem 𝐼¹_𝑛⟶0when𝑛⟶∞. (3.10) Now, we will prove that 𝐼²_𝑛→0 when 𝑛→∞.
In fact, as 𝐼²_𝑛=^𝑡𝑛_𝑡0𝑡_𝑛−𝑠^𝛼−1𝑠^−𝜎𝑑𝑠≤^𝑡𝑛_𝑡0𝑠^−𝜎1𝑑𝑠=𝑡1−𝜎^𝑛1−𝜎−𝑡₀^1−𝜎 (3.11) and letting 𝑛→∞ and, taking into account that 𝑡_𝑛→𝑡₀, from the last expression we get 𝐼²_𝑛⟶0when𝑛⟶∞. (3.12) Finally, from (3.5), (3.10), and (3.12) we get 𝐿𝑡_𝑛−𝐿𝑡₀⟶0when𝑛⟶∞. (3.13) This proves the continuity of 𝐿 at 𝑡₀.
Case3 (𝑡₀=1). It is easily checked that 𝐿(1)=0.
Now, following the same lines that in the proof of Case 1, we can demonstrate the continuity of 𝐿 at 𝑡₀=1.
This finishes the proof.

Lemma 3.2. Suppose that 0<𝜎<1, 3<𝛼≤4, 0<𝛽𝜂^𝛼−3<1, and 𝐹∶(0,1]→ℝ is a continuous function with lim_𝑡→0+𝐹(𝑡)=∞.
If 𝑡^𝜎𝐹(𝑡) is a continuous function on [0,1] then the function defined by 𝑁(𝑡)=𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝐹(𝑠)𝑑𝑠 (3.14) is continuous on [0,1], where 𝐻(𝑡,𝑠) is the function appearing in Lemma 2.5.

Proof. Since 𝑡^𝜎𝐹(𝑡) is continuous on [0,1], there exists a constant 𝑀>0 such that |𝑡^𝜎𝐹(𝑡)|≤𝑀 for any 𝑡∈[0,1].
Taking into account that |𝐻(≤2()𝑡,𝑠)|𝛼−1)(𝛼−2Γ(𝛼) (3.15) we have ¹₀𝐻𝜂,𝑠𝐹(𝑠)𝑑𝑠=¹₀𝐻𝜂,𝑠𝑠^−𝜎𝑠^𝜎𝐹(𝑠)≤()𝑑𝑠2𝑀𝛼−1)(𝛼−2Γ(𝛼)¹₀𝑠^−𝜎()𝑑𝑠=2𝑀𝛼−1)(𝛼−2Γ(𝛼)(1−𝛼)<∞, (3.16) and, consequently, the function 𝑁 is continuous at any point 𝑡∈[0,1].

Remark 3.3. Notice that the function 𝐻(𝑡,𝑠) appearing in Lemma 2.5 which is defined as 𝐻()=()𝑡,𝑠𝛼−1)(𝛼−2Γ(𝛼)𝑡^𝛼−3(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−3(),0≤𝑠≤𝑡≤1,𝛼−1)(𝛼−2Γ(𝛼)𝑡^𝛼−3(1−𝑠)^𝛼−3,0≤𝑡≤𝑠≤1, (3.17) is continuous function on [0,1]×[0,1] and, moreover, 𝐻(𝑡,𝑠)≥0.
In fact, for 0≤𝑡≤𝑠≤1 it is clear that 𝐻(𝑡,𝑠)≥0.
In the case, 0≤𝑠≤𝑡≤1, we have 𝐻()=()𝑡,𝑠𝛼−1)(𝛼−2Γ(𝛼)𝑡^𝛼−3(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−3=(𝛼−1)(𝛼−2)Γ(𝛼)(𝑡−𝑡𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−3≥0. (3.18) This proves the nonnegative character of the function 𝐻 on [0,1]×[0,1].

Lemma 3.4. Suppose that 0<𝜎<1. Then sup_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑠^−𝜎1𝑑𝑠=Γ(𝛼)𝛽(1−𝜎,𝛼−2)−𝛽(1−𝜎,𝛼), (3.19) where 𝐺(𝑡,𝑠) is the function appearing in Lemma 2.5.

Proof. By definition of 𝐺(𝑡,𝑠), we have ¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑑𝑠=^𝑡₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑑𝑠+¹_𝑡𝐺(𝑡,𝑠)𝑠^−𝜎=𝑑𝑠_𝑡0𝑡^𝛼−1(1−𝑠)^𝛼−3−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑠^−𝜎𝑑𝑠+¹_𝑡𝑡^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎=𝑑𝑠₁₀𝑡^𝛼−1(1−𝑠)^𝛼−3Γ(𝛼)𝑠^−𝜎𝑑𝑠−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑠^−𝛼=𝑡𝑑𝑠_𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−3𝑠^−𝜎1𝑑𝑠−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑑𝑠. (3.20) As we saw in Case 1 of Lemma 3.1. ^𝑡₀(𝑡−𝑠)^𝛼−1𝑠^−𝜎𝑡𝑑𝑠=^𝛼−𝜎Γ(𝛼)𝛽(1−𝜎,𝛼) (3.21) and, therefore, ^𝑡₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑡𝑑𝑠=^𝛼−1Γ(𝛼)𝛽()−𝑡1−𝜎,𝛼−2^𝛼−𝜎Γ(𝛼)𝛽(1−𝜎,𝛼). (3.22) Now, using elemental calculus it is easily seen that the function 𝜑(𝑡)=𝛽(1−𝜎,𝛼−2)Γ(𝛼)𝑡^𝛼−1−𝛽(1−𝜎,𝛼)Γ(𝛼)𝑡^𝛼−𝜎 (3.23) is increasing on the interval [0,1] and, therefore, sup_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑑𝑠=sup_0≤𝑡≤1𝜑(𝑡)(1)=1=𝜑Γ(𝛼)𝛽(1−𝜎,𝛼−2)−𝛽(1−𝜎,𝛼). (3.24)

Lemma 3.5. Suppose that 0<𝜎<1 then ¹₀𝐻𝜂,𝑠𝑠^−𝜎()𝑑𝑠=𝛼−1)(𝛼−2Γ(𝛼)𝜂^𝛼−3−𝜂^{𝛼−𝜎−2}𝛽(1−𝜎,𝛼−2), (3.25) where 𝐻(𝑡,𝑠) is the function appearing in Lemma 2.5.

Proof. By definition of 𝐻(𝑡,𝑠), we have ¹₀𝐻𝜂,𝑠𝑠^−𝜎𝑑𝑠=^𝜂₀𝐻𝜂,𝑠𝑠^−𝜎𝑑𝑠+¹_𝜂𝐻𝜂,𝑠𝑠^−𝜎=𝑑𝑠_𝜂0(𝛼−1)(𝛼−2)Γ(𝛼)𝜂^𝛼−3(1−𝑠)^𝛼−3−𝜂−𝑠^𝛼−3𝑠^−𝜎+𝑑𝑠_1𝜂(𝛼−1)(𝛼−2)Γ(𝛼)𝜂^𝛼−3(1−𝑠)^𝛼−3𝑠^−𝜎=𝑑𝑠₁₀(𝛼−1)(𝛼−2)Γ(𝛼)𝜂^𝛼−3(1−𝑠)^𝛼−3𝑠^−𝜎𝑑𝑠−^𝜂₀(𝛼−1)(𝛼−2)Γ(𝛼)𝜂−𝑠^𝛼−3𝑠^−𝜎=()𝑑𝑠𝛼−1)(𝛼−2Γ(𝛼)𝜂^𝛼−31₀(1−𝑠)^𝛼−3𝑠^−𝜎()𝑑𝑠−𝛼−1)(𝛼−2Γ(𝛼)^𝜂₀𝜂−𝑠^𝛼−3𝑠^−𝜎=()𝑑𝑠𝛼−1)(𝛼−2Γ(𝛼)𝜂^𝛼−3𝛽()−()1−𝜎,𝛼−2𝛼−1)(𝛼−2Γ(𝛼)^𝜂₀𝜂−𝑠^𝛼−3𝑠^−𝜎𝑑𝑠. (3.26) By a similar argument that the one used in the Case 1 of Lemma 3.1, we have ¹₀𝐻𝜂,𝑠𝑠^−𝜎()𝑑𝑠=𝛼−1)(𝛼−2Γ(𝛼)𝜂^𝛼−3𝛽()−()1−𝜎,𝛼−2𝛼−1)(𝛼−2Γ(𝛼)𝜂^{𝛼−𝜎−2}𝛽()=()1−𝜎,𝛼−2𝛼−1)(𝛼−2Γ(𝛼)𝜂^𝛼−3−𝜂^{𝛼−𝜎−2}𝛽(1−𝜎,𝛼−2). (3.27) This finishes the proof.

By commodity, we denote by 𝐾 the constant given by 1𝐾=Γ(𝛼)𝛽𝜂1+^𝛼−3−𝜂^{𝛼−𝜎−2}1−𝛽𝜂^𝛼−3𝛽(1−𝜎,𝛼−2)−𝛽(1−𝜎,𝛼). (3.28) Moreover, we introduce the following class of functions which will be used in the main result of the paper. By 𝒜 we denote the class of functions 𝜙∶[0,∞)→[0,∞) satisfying the following: (i)𝜙is nondecreasing. (ii)𝜙(𝑥)<𝑥 for any 𝑥>0. (iii)𝛽(𝑥)=𝜙(𝑥)/𝑥∈𝒮, where 𝒮 is the class of functions introduced in Remark 2.6.

Theorem 3.6. Let 0<𝜎<1, 3<𝛼≤4, 0<𝜂<1, 0<𝛽𝜂^𝛼−3<1, and 𝑓∶(0,1]×[0,∞)→[0,∞) is a continuous function with lim_𝑡→0+𝑓(𝑡,⋅)=∞ and such that 𝑡^𝜎𝑓(𝑡,𝑦) is a continuous function on [0,1]×[0,∞). Assume that there exists 0<𝜆≤1/𝐾 such that for 𝑥,𝑦∈[0,∞) with 𝑦≥𝑥 and 𝑡∈[0,1]0≤𝑡^𝜎𝑓𝑡,𝑦−𝑓(𝑡,𝑥)≤𝜆𝜙𝑦−𝑥, (3.29) where 𝜙∈𝒜.
Then Problem (1.3) has a unique positive solution (this means that 𝑥(𝑡)>0 for 𝑡∈(0,1)).

Proof. Consider the cone: 𝑃={𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0}. (3.30) Since 𝑃 is a closed set of 𝐶[0,1], 𝑃 is a complete metric space with the distance given by 𝑑(𝑢,𝑣)=sup_0≤𝑡≤1|𝑢(𝑡)−𝑣(𝑡)|, for 𝑢,𝑣∈𝑃.
It is easily checked that 𝑃 satisfies conditions (2.12) and (2.13) of Theorem 2.7.
Now, for 𝑢∈𝑃 we define the operator 𝑇 by (𝑇𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠. (3.31) By Lemmas 3.1 and 3.2, for 𝑢∈𝑃 we have 𝑇𝑢∈𝐶[0,1].
Moreover, in view of the nonnegative character of 𝐺(𝑡,𝑠), 𝐻(𝜂,𝑠), and 𝑓(𝑠,𝑥), we have that 𝑇𝑢∈𝑃 for 𝑢∈𝑃.
In what follows, we check that assumptions in Theorem 2.7 are satisfied.
Firstly, we will prove that 𝑇 is nondecreasing.
In fact, by (3.29), for 𝑢≥𝑣 we have (𝑇𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝑓((𝑠=𝜂,𝑠𝑠,𝑢))𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑠^𝜎𝑓((𝑠+𝑠,𝑢))𝑑𝑠𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑠^−𝜎𝑠^𝜎𝑓((𝑠≥𝑠,𝑢))𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑠^𝜎𝑓((𝑠+𝑠,𝑣))𝑑𝑠𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑠^−𝜎𝑠^𝜎𝑓((𝑠=𝑠,𝑣))𝑑𝑠₁₀𝐺()𝑓((𝑠+𝑡,𝑠𝑠,𝑣))𝑑𝑠𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑓(𝑠,𝑣(𝑠))𝑑𝑠=(𝑇𝑣)(𝑡). (3.32) This proves that 𝑇 is a nondecreasing operator.
On the other hand, for 𝑢≥𝑣 and 𝑢≠𝑣, we have 𝑑(𝑇𝑢,𝑇𝑣)=sup_0≤𝑡≤1|(𝑇𝑢)(𝑡)−(𝑇𝑣)(𝑡)|=sup_0≤𝑡≤1((𝑇𝑢)(𝑡)−(𝑇𝑣)(𝑡))=sup_0≤𝑡≤1¹₀𝐺()𝑓((𝑠((𝑠+𝑡,𝑠𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑠𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑓(𝑠,𝑢(𝑠))−𝑓(𝑠,𝑣(𝑠))𝑑𝑠≤sup_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝑠^𝜎𝑓((𝑠((𝑠+𝛽𝑠,𝑢))−𝑓𝑠,𝑣))𝑑𝑠(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑠^−𝜎𝑠^𝜎𝑓(𝑠,𝑢(𝑠))−𝑓(𝑠,𝑣(𝑠))𝑑𝑠≤sup_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝜆𝜙(𝑢(𝑠)(𝑠+𝛽−𝑣))𝑑𝑠(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑠^−𝜎𝜆𝜙(𝑢(𝑠)−𝑣(𝑠))𝑑𝑠. (3.33) Since 𝜙 is nondecreasing, the last inequality implies 𝑑(𝑇𝑢,𝑇𝑣)≤𝜆𝜙(𝑑(𝑢,𝑣))sup_0≤𝑡≤0¹₀𝐺(𝑡,𝑠)𝑠^−𝜎+𝛽𝑑𝑠(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−3𝜆𝜙(𝑑(𝑢,𝑣))¹₀𝐻𝜂,𝑠𝑠^−𝜎(𝑑(𝑑𝑠=𝜆𝜙𝑢,𝑣))sup_0≤𝑡≤0¹₀𝐺(𝑡,𝑠)𝑠^−𝜎𝛽𝑑𝑠+(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑠^−𝜎𝑑𝑠. (3.34) Now, from Lemmas 3.4 and 3.5 it follows: 𝑑()(𝑑(1𝑇𝑢,𝑇𝑣≤𝜆𝜙𝑢,𝑣))Γ(𝛼)𝛽()()+𝛽1−𝜎,𝛼−2−𝛽1−𝜎,𝛼(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−3×(𝛼−1)(𝛼−2)Γ(𝛼)𝜂^𝛼−3−𝜂^{𝛼−𝜎−2}𝛽((𝑑(11−𝜎,𝛼−2)=𝜆𝜙𝑢,𝑣))Γ(𝛼)𝛽()()+𝛽𝜂1−𝜎,𝛼−2−𝛽1−𝜎,𝛼_𝛼−3−𝜂^{𝛼−𝜎−2}1−𝛽𝜂^𝛼−3Γ(𝛼)𝛽((𝑑(11−𝜎,𝛼−2)=𝜆𝜙𝑢,𝑣))Γ(𝛼)𝛽𝜂1+^𝛼−3−𝜂^{𝛼−𝜎−2}1−𝛽𝜂^𝛼−3𝛽()((𝑑(1−𝜎,𝛼−2−𝛽1−𝜎,𝛼)=𝜆𝜙𝑢,𝑣))𝐾. (3.35) Since 0<𝜆≤1/𝐾, from the last inequality we obtain 𝑑(𝑇𝑢,𝑇𝑣)≤𝜆𝜙(𝑑(𝑢,𝑣))𝐾≤𝜙(𝑑(𝑢,𝑣)), (3.36) and, since 𝑢≠𝑣, 𝑑()≤𝜙(𝑑(𝑇𝑢,𝑇𝑣𝑢,𝑣))𝑑()𝑑()(𝑑(𝑑().𝑢,𝑣𝑢,𝑣=𝛽𝑢,𝑣))𝑢,𝑣 (3.37) Since this inequality is obviously satisfied for 𝑢=𝑣, we have 𝑑(𝑇𝑢,𝑇𝑣)≤𝛽(𝑑(𝑢,𝑣))𝑑(𝑢,𝑣)forany𝑢,𝑣∈𝑃with𝑢≥𝑣. (3.38) Finally, since the zero function satisfies 0≤𝑇0, Theorem 2.7 says us that the operator 𝑇 has a unique fixed point in 𝑃, or, equivalently, Problem (1.3) has a unique nonnegative solution 𝑥 in 𝐶[0,1].
Now, we will prove that 𝑥 is a positive solution.
In contrary case, we can find 0<𝑡^∗<1 such that 𝑥(𝑡^∗)=0.
Taking into account that the nonnegative solution 𝑥 of Problem (1.3) is a fixed point of the operator, we have 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠+𝛽𝑡^𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠 (3.39) and, particularly, 𝑥𝑡^∗=¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠+𝛽𝑡^∗𝛼−1(𝛼−1)(𝛼−2)1−𝛽𝜂^𝛼−31₀𝐻𝜂,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠=0. (3.40) Since both summands in the right hand are nonnegative (see Remarks 2.6 and 3.3) we have ¹₀𝐺𝑡^∗,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠=0,₁₀𝐻𝜂,𝑠𝑓(𝑠,𝑥(𝑠))𝑑𝑠=0. (3.41) Given the nonnegative character of 𝐺(𝑡,𝑠), 𝐻(𝜂,𝑠), and 𝑓(𝑠,𝑢), we have 𝐺𝑡^∗𝑓((𝑠(𝑠),𝐻𝑓((𝑠(𝑠).,𝑠𝑠,𝑥))=0a.e.𝜂,𝑠𝑠,𝑥))=0a.e. (3.42) Taking into account that lim_𝑡→0+𝑓(𝑡,0)=∞, this means that for 𝑀>0 we can find 𝛿>0 such that for 𝑠∈[0,1]∩(0,𝛿) we have 𝑓(𝑠,0)>𝑀. Notice that [0,1]∩(0,𝛿)⊂{𝑠∈[0,1]∶𝑓(𝑠,𝑥(𝑠))>𝑀} and 𝜇([0,1]∩(0,𝛿))>0, where 𝜇 is the Lebesgue measure on [0,1].
This and (3.42) give us that 𝐺𝑡^∗(𝑠),𝐻(𝑠),,𝑠=0a.e.𝜂,𝑠=0a.e. (3.43) and this is a contradiction since 𝐺(𝑡^∗,𝑠) and 𝐻(𝜂,𝑠) are rational functions in the variable 𝑠.
Therefore, 𝑥(𝑡)>0 for 𝑡∈(0,1).
This finishes the proof.

In order to present an example which illustrates our results, we need to prove some properties about the hyperbolic tangent function.

Previously, we recalled some definitions.

Definition 3.7. A function 𝑓∶[0,∞)→[0,∞) is said to be subadditive if it satisfies 𝑓𝑥+𝑦≤𝑓(𝑥)+𝑓𝑦forany𝑥,𝑦∈[0,∞). (3.44) An example of subadditive function is the square root function, that is, 𝑓(𝑥)=√𝑥.

Remark 3.8. Suppose that 𝑓∶[0,∞)→[0,∞) is subadditive and 𝑦≤𝑥 then 𝑓(𝑥)−𝑓𝑦≤𝑓𝑥−𝑦. (3.45) In fact, since𝑓(𝑥)=𝑓𝑥−𝑦+𝑦≤𝑓𝑥−𝑦+𝑓𝑦 (3.46)this inequality implies that𝑓(𝑥)−𝑓𝑦≤𝑓𝑥−𝑦. (3.47)
Recall that a function 𝑓∶[0,∞)→[0,∞) is concave if for any 𝑥,𝑦∈[0,∞) and 𝜆∈[0,1]. 𝑓𝜆𝑥+(1−𝜆)𝑦≥𝜆𝑓(𝑥)+(1−𝜆)𝑓𝑦. (3.48)

Lemma 3.9. Let 𝑓∶[0,∞)→[0,∞) be a concave function with 𝑓(0)=0. Then 𝑓 is subadditive.

Proof. We take 𝑥,𝑦∈[0,∞).
Since 𝑓 is concave and 𝑓(0)=0, we get 𝑓(𝑥)𝑦=𝑓𝑥𝑥+𝑦0+≥𝑦𝑥+𝑦𝑥+𝑦𝑓(0)+𝑥𝑥+𝑦𝑓=𝑥𝑥+𝑦𝑥+𝑦𝑓,𝑓𝑦𝑥𝑥+𝑦𝑥+𝑦=𝑓𝑦𝑥+𝑦0+≥𝑥𝑥+𝑦𝑥+𝑦𝑓(0)+𝑦𝑥+𝑦𝑓=𝑦𝑥+𝑦𝑥+𝑦𝑓.𝑥+𝑦𝑥+𝑦 (3.49) Adding these inequalities, we have 𝑓(𝑥)𝑦≥𝑥+𝑓𝑓+𝑦𝑥+𝑦𝑥+𝑦𝑓.𝑥+𝑦𝑥+𝑦=𝑓𝑥+𝑦 (3.50) This proves the lemma.

In what follows, we will prove that the function 𝑓(𝑥)𝑒=tanh𝑥=^2𝑥−1𝑒^2𝑥+1 (3.51) belongs to the class 𝒜 previously defined.

Lemma 3.10. The function 𝑓∶[0,∞)→[0,∞) defined as 𝑓(𝑥)𝑒=tanh𝑥=^2𝑥−1𝑒^2𝑥+1 (3.52) satisfies: (a)𝑓∈𝒜. (b)𝑓 is subadditive.

Proof. (a) Since 𝑓′(𝑥)=4𝑒^2𝑥/(𝑒^2𝑥+1)²>0 for 𝑥>0, 𝑓 is nondecreasing.
Moreover, the function 𝑔(𝑥)𝑒=𝑥−tanh𝑥=𝑥−^2𝑥−1𝑒^2𝑥+1 (3.53) has as derivative 𝑔^′(𝑥)=𝑒^2𝑥−1²𝑒^2𝑥+1²>0for𝑥>0 (3.54) and, consequently, 𝑔 is strictly nondecreasing on (0,∞).
Since 𝑔(0)=0, we have 0=𝑔(0)<𝑔(𝑥) for 𝑥>0 or, equivalently, 𝑓(𝑥)=tanh𝑥<𝑥 for 𝑥>0.
In order to prove that 𝛽(𝑥)=tanh𝑥/𝑥∈𝒮, notice that if 𝛽(𝑡_𝑛)→1 then the sequence (𝑡_𝑛) is a bounded sequence.
In fact, in contrary case 𝑡_𝑛→∞ and we have 𝛽𝑡_𝑛=tanh𝑡_𝑛𝑡_𝑛⟶0 (3.55) which contradicts the fact that 𝛽(𝑡_𝑛)→1.
Now, we suppose that 𝛽(𝑡_𝑛)→1 and 𝑡_𝑛↛0.
Then, we can find 𝜀>0 such that for each 𝑛∈ℕ there exists 𝜚_𝑛≥𝑛 with 𝑡_𝜚𝑛≥𝜀.
Since (𝑡_𝑛) is a bounded sequence (because 𝛽(𝑡_𝑛)→1) we can find a subsequence of (𝑡_𝜚𝑛), which we will denote of the same way, such that 𝑡_𝜚𝑛→𝑎.
As 𝛽(𝑡_𝑛)→1, it follows that 𝛽𝑡_𝜚𝑛=tanh𝑡_𝜚𝑛𝑡_𝜚𝑛⟶tanh𝑎𝑎=1 (3.56) and, as the unique solution of the equation tanh𝑥=𝑥 on [0,∞) is 𝑥₀=0, we deduce that 𝑎=0.
Therefore, 𝑡_𝜚𝑛→0 and this implies that there exists 𝑛₀∈ℕ such that 𝑡_𝜚𝑛<𝜀 for 𝑛≥𝑛₀.
This contradicts the fact that 𝑡_𝜚𝑛≥𝜀 for any 𝑛∈ℕ.
Therefore, 𝑡_𝑛→0.
This proves that 𝛽(𝑥)=tanh𝑥/𝑥∈𝒮.
Therefore, 𝑓∈𝒜.
(b) Since tanh0=0 and (tanh𝑥)^′′=8𝑒^2𝑥1−𝑒^2𝑥𝑒^2𝑥+1³<0for𝑥>0, (3.57) this means that 𝑓(𝑥)=tanh𝑥 is a concave function with tanh0=0 and, by Lemma 3.9, 𝑓(𝑥)=tanh𝑥 is subadditive.

Remark 3.11. By Remark 3.8 and by (b) of Lemma 3.9, for 𝑥,𝑦∈[0,∞) with 𝑦≤𝑥tanh𝑥−tanh𝑦≤tanh𝑥−𝑦. (3.58)

Now, we present an example which illustrates our result.

Example 3.12. Consider the following singular fractional boundary value problem 𝐷₀^7/2₊𝑢(𝑡)+𝜆𝑡²+1tanh𝑢(𝑡)𝑡^1/2𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=𝑢^′′(0)=0,𝑢^′′(1)=𝑢^′′14. (3.59) In this case, 𝜎=1/2, 𝜂=1/4, 𝛽=1 and 𝛼=7/2.
Moreover, in this case 𝑓(𝑡,𝑢)=𝜆(𝑡²+1)tanh𝑢/𝑡^1/2 for (𝑡,𝑢)∈(0,1]×[0,∞).
Notice that 𝑓 is continuous in (0,1]×[0,∞) and lim_𝑡→0+𝑓(𝑡,⋅)=∞.
Now, we check that 𝑓(𝑡,𝑢) satisfies assumptions appearing in Theorem 3.6.
It is clear that 𝑡^1/2𝑓(𝑡,𝑢)=𝜆(𝑡²+1)tanh𝑢 is a continuous function on [0,1]×[0,∞).
Moreover, by Lemma 3.10 and Remark 3.11, for 𝑢≥𝑣 and 𝑡∈[0,1] we have 0≤𝑡^1/2𝑓()()𝑡𝑡,𝑢−𝑓𝑡,𝑣=𝜆²()𝑡+1tanh𝑢−tanh𝑣≤𝜆²+1tanh(𝑢−𝑣)≤2𝜆tanh(𝑢−𝑣), (3.60) where 𝑓(𝑥)=tanh𝑥 is a function belonging to 𝒜 (see, Lemma 3.10).
Finally, Theorem 3.6 says us that Problem (3.59) has a unique positive solution for 12𝜆≤𝐾=Γ(7/2)1+(1/4)^1/2−1/4/1−(1/4)^1/2𝛽()()=1/2,3/2−𝛽1/2,7/2307√𝜋. (3.61) Or, equivalently, for 𝜆≤15/7√𝜋.

Acknowledgment

This research was partially supported by “Universidad de Las Palmas de Gran Canaria”, Project ULPGC 2010–006.