Positive Solution for the Nonlinear Hadamard Type Fractional Differential Equation with 𝑝-Laplacian

Abstract

We study the following nonlinear fractional differential equation involving the 𝑝-Laplacian operator 𝐷^𝛽(𝜑_𝑝(𝐷^𝛼𝑢(𝑡)))=𝑓(𝑡,𝑢(𝑡)), 1<𝑡<𝑒, 𝑢(1)=𝑢^′(1)=𝑢^′(𝑒)=0, 𝐷^𝛼𝑢(1)=𝐷^𝛼𝑢(𝑒)=0, where the continuous function 𝑓∶[1,𝑒]×[0,+∞)→[0,+∞), 2<𝛼≤3, 1<𝛽≤2. 𝐷^𝛼 denotes the standard Hadamard fractional derivative of the order 𝛼, the constant 𝑝>1, and the 𝑝-Laplacian operator 𝜑_𝑝(𝑠)=|𝑠|^𝑝−2𝑠. We show some results about the existence and the uniqueness of the positive solution by using fixed point theorems and the properties of Green's function and the 𝑝-Laplacian operator.

1. Introduction

Fractional differential equations have attracted more and more attention for their useful applications in various fields, such as economics, science, and engineering; see [1–4]. In the last few decades, much attention has been focused on the study of the existence and uniqueness of solutions for boundary value problems of Riemann-Liouville type or Caputo type fractional differential equations; see [5–19]. There are few papers devoted to the research of the 𝑝-Laplacian fractional differential equations; see [20–25].

By the use of the fixed point theorem on cones, Chai in [20] obtained the existence and multiplicity of positive solutions for a class of boundary value problem of fractional differential equation with 𝑝-Laplacian operator: 𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)=𝑓𝑡,𝑢)),0<𝑡<1,=0,𝑢+𝜎𝐷^𝛾₀₊𝑢(1)=0,𝐷^𝛼₀₊𝑢(0)=0, (1) where 1<𝛼≤2, 0<𝛽≤1, 0<𝛾≤1, 0≤𝛼−𝛾−1, 𝜎 is a positive constant number, and 𝐷^𝛼₀₊,𝐷^𝛽₀₊,𝐷^𝛾₀₊ are the standard Riemann-Liouville derivatives. 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1, 𝜙_𝑝⁻¹=𝜙_𝑞, 1/𝑝+1/𝑞=1.

Han et al. in [22] studied the following boundary value problem of nonlinear fractional differential equation with 𝑝-Laplacian operator: 𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓(𝑢)𝑢(0)(𝜉)𝜙+𝑎=0,0<𝑡<1,=𝛾𝑢+𝜆,_𝑝𝐷^𝛼₀₊𝑢(0)=𝜙_𝑝(𝐷^𝛼₀₊𝑢(1))^′=𝜙_𝑝𝐷^𝛼₀₊𝑢(1)^′′=0, (2) where 0<𝛼≤1, 2<𝛽≤3, 0≤𝛾<1, 0≤𝜉≤1, 𝜆>0 is a parameter, and 𝐷^𝛼₀₊,𝐷^𝛽₀₊ are the standard Caputo fractional derivatives. By the properties of Green function and Schauder fixed point theorem, several existence and nonexistence results and the uniqueness of positive solutions are acquired.

Liu et al. in [23] investigated the solvability of a fractional differential equation model involving the 𝑝-Laplacian operator with boundary value conditions as follows: 𝜑_𝑝^𝑐𝐷^𝛼𝑥(𝑡)^′((𝑡(),𝑥(0)=𝑓𝑡,𝑥)),𝑡∈0,1=𝑟₀𝑥(1),𝑥^′(0)=𝑟₁𝑥^′(1),𝑥^(𝑖)=0,𝑖=2,3,…,[𝛼]−1, (3) where 1<𝛼∈ℝ, 𝑟₀, 𝑟₁≠1, and ^𝑐𝐷^𝛼 is the standard Caputo derivative. By the means of the Banach contraction mapping principle, they obtained the existence and uniqueness of a solution for the model.

Lu et al. in [24] considered the following fractional boundary value problem with 𝑝-Laplacian operator: 𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)=𝑓𝑡,𝑢)),0<𝑡<1,=𝑢^′(0)=𝑢^′(1)=0,𝐷^𝛼₀₊𝑢(0)=𝐷^𝛽₀₊𝑢(1)=0, (4) where 2<𝛼≤3, 1<𝛽≤2, and 𝐷^𝛼₀₊,𝐷^𝛽₀₊ are the standard Riemann-Liouville fractional derivatives. By the properties of Green’s function, the Guo-Krasnosel’skii fixed point theorem, the Leggett-Williams fixed point theorem, and the upper and lower solutions method, some new results on the existence of positive solutions are gained.

Motivated by the mentioned papers, we will consider the Hadamard fractional boundary value with 𝑝-Laplacian operator as below: 𝐷^𝛽𝜑_𝑝𝐷^𝛼𝑢(𝑡)((𝑡𝑢(1)=𝑓𝑡,𝑢)),1<𝑡<𝑒,=𝑢^′(1)=𝑢^′(𝑒)=0,𝐷^𝛼𝑢(1)=𝐷^𝛼𝑢(𝑒)=0, (5) where 2<𝛼≤3, 1<𝛽≤2, 𝜑_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, and 𝑓∶[1,𝑒]×[0,+∞)→[0,+∞) is a positive continuous function. Evidently, for any 𝑝>1, 𝜑_𝑝⁻¹=𝜑_𝑞, here 1/𝑝+1/𝑞=1. Here 𝐷^𝛼 is the standard Hadamard fractional derivative of order 𝛼 which is described as follows.

Definition 1 (see [1, Page 111]). The 𝛼th Hadamard fractional order derivative of a function 𝑢∶[1,+∞)↦ℝ is defined by 𝐷^𝛼𝑢(𝑡)=1Γ()𝑡𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₁𝑡log𝑠^{𝑛−𝛼−1}𝑢(𝑠)𝑠𝑑𝑠, (6) where 𝛼>0, 𝑛=[𝛼]+1, and [𝛼] denotes the largest integer which is less than or equal to 𝛼. Correspondingly, the 𝛼th Hadamard fractional order integral of 𝑢∶[1,+∞)↦ℝ is defined by 𝐼^𝛼𝑢(𝑡)=1Γ(𝛼)^𝑡₁𝑡log𝑠^𝛼−1𝑢(𝑠)𝑠𝑑𝑠, (7) where Γ is the gamma function.

To the best of our knowledge, there are few contributions to the Hadamard type with 𝑝-Laplacian operator; we fill the gap in this paper. In fact, we will discuss the existence and the uniqueness of the positive solutions of (5). The structure of this paper goes on as follows. In Section 2, we will introduce some basic lemmas that will be used. In Section 3, we first give some existence results including Theorems 10 and 11, Corollary 12, and Theorem 13. Then, we will prove Theorems 14 and 15 which reveal the uniqueness of the solution. In Section 4, we give two examples to illustrate our results.

2. Preliminary Results

In this section, we will first recall the following preliminary facts that will be used in our main results.

Lemma 2 (see [1, Theorem 2.3]). Let 𝛼>0, 𝑛=[𝛼]+1; then 𝐼^𝛼𝐷^𝛼𝑢(𝑡)=𝑢(𝑡)+^𝑛_𝑖=1𝑐_𝑖log𝑡^𝛼−𝑖, (8) where 𝑐_𝑖, 𝑖=1,2,…,𝑛, are some constants in ℝ.

The following lemma is the Schauder fixed point theorem which is well known; see Theorem 2.10 in [22].

Lemma 3. If  𝑈 is a nonempty closed, bounded, and convex subset of a Banach space 𝑋 and 𝑇∶𝑈→𝑈 is completely continuous, then 𝑇 has a fixed point in 𝑈.

Lemma 4 (see [24, Lemma 2.7]). Let 𝑋 be a Banach space, let 𝑃⊆𝑋 be a cone, and let Ω₁, Ω₂ be two bounded open balls of 𝐸 centered at the origin with Ω₁⊂Ω₂. Suppose that 𝑇∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is a completely continuous operator such that either(i)‖𝑇𝑥‖≤‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₁ and ‖𝑇𝑥‖≥‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₂, or(ii)‖𝑇𝑥‖≥‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₁ and ‖𝑇𝑥‖≤‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₂,holds. Then 𝑇 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

The following conclusion is the nonlinear alternative of Leray-Schauder type; see Lemma 2.6 in [10].

Lemma 5. Let 𝑋 be a Banach space with 𝐶⊆𝑋 being closed and convex. Assume that 𝑈 is a relatively open subset of 𝐶 with 0∈𝑈 and 𝐴∶𝑈→𝐶 is a continuous, compact map. Then either (1)𝐴 has a fixed point in 𝑈, or (2)there exists 𝑢∈𝜕𝑈 and 𝜆∈(0,1), with 𝑢=𝜆𝐴𝑢.

Next, we give several lemmas which will be applied in the proofs of our main results.

Lemma 6. Let 𝑢(𝑡) be the solution of the problem (5); then it can be described as below: 𝑢(𝑡)=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠=𝑌(𝑡), (9) where 𝐺(𝑡,𝑠)=log𝑡^𝛼−11−log𝑠^𝛼−2−log(𝑡/𝑠)^𝛼−1Γ(𝛼),𝑠≤𝑡,log𝑡^𝛼−11−log𝑠^𝛼−2Γ(𝛼)𝐻()=,𝑡≤𝑠,𝑡,𝑠log𝑡^𝛽−11−log𝑠^𝛽−1−log(𝑡/𝑠)^𝛽−1Γ𝛽,𝑠≤𝑡,log𝑡^𝛽−11−log𝑠^𝛽−1Γ𝛽,𝑡≤𝑠. (10)

Proof. Putting 𝑦(𝑡)=𝑓(𝑡,𝑢(𝑡)), we have 𝐷^𝛽(𝜑_𝑝(𝐷^𝛼𝑢(𝑡)))=𝑦(𝑡). By Lemma 2 and the fact that 1<𝛽≤2, 𝜑_𝑝𝐷^𝛼𝑢(𝑡)=𝐼^𝛽𝑦(𝑡)+𝑐₁log𝑡^𝛽−1+𝑐₂log𝑡^𝛽−2. (11) The boundary value hypotheses give 𝐷^𝛼𝑢(1)=𝐷^𝛼𝑢(𝑒)=0. So we can get that 𝑐₂=0,𝑐₁1=−Γ𝛽^𝑒₁1−log𝜏^𝛽−1𝑦(𝜏)𝑑𝜏𝜏. (12) Therefore, 𝜑_𝑝𝐷^𝛼𝑢(𝑡)=𝐼^𝛽𝑦(𝑡)−log𝑡^𝛽−1Γ𝛽^𝑒₁1−log𝜏^𝛽−1𝑦(𝜏)𝑑𝜏𝜏=1Γ𝛽^𝑡₁𝑡log𝜏^𝛽−1𝑦(𝜏)𝑑𝜏𝜏−log𝑡^𝛽−1Γ𝛽^𝑡₁1−log𝜏^𝛽−1𝑦(𝜏)𝑑𝜏𝜏−log𝑡^𝛽−1Γ𝛽^𝑒_𝑡1−log𝜏^𝛽−1𝑦(𝜏)𝑑𝜏𝜏=−^𝑡₁log𝑡^𝛽−11−log𝜏^𝛽−1−log𝑡−log𝜏^𝛽−1Γ𝛽×𝑦(𝜏)𝑑𝜏𝜏−^𝑒_𝑡log𝑡^𝛽−11−log𝜏^𝛽−1Γ𝛽𝑦(𝜏)𝑑𝜏𝜏=−^𝑒₁𝐻(𝑡,𝜏)𝑦(𝜏)𝑑𝜏𝜏. (13)
Notice the fact that 𝜑_𝑝⁻¹=𝜑_𝑞, 1/𝑝+1/𝑞=1; we have 𝐷^𝛼𝑢(𝑡)+𝜑_𝑞^𝑒₁𝐻(𝑡,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏=0. (14) Putting 𝑥(𝑡)=𝜑_𝑞(∫^𝑒₁𝐻(𝑡,𝜏)𝑓(𝜏,𝑢(𝜏))(𝑑𝜏/𝜏)), it follows from Lemma 2 that 𝑢(𝑡)=−𝐼^𝛼𝑥(𝑡)+𝐶₁log𝑡^𝛼−1+𝐶₂log𝑡^𝛼−2+𝐶₃log𝑡^𝛼−3. (15) This, combined with the fact that 𝑢(1)=𝑢^′(1)=𝑢^′(𝑒)=0, yields 𝐶₃=0,𝐶₂𝐶=0,₁=1Γ(𝛼)^𝑒₁1−log𝑠^𝛼−2𝑥(𝑠)𝑑𝑠𝑠. (16) Thus, 𝑢(𝑡)=^𝑒₁𝐺(𝑡,𝑠)𝑥(𝑠)𝑑𝑠𝑠=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠, (17) where 𝐺(𝑡,𝑠)=log𝑡^𝛼−11−log𝑠^𝛼−2−log(𝑡/𝑠)^𝛼−1Γ(𝛼),𝑠≤𝑡,log𝑡^𝛼−11−log𝑠^𝛼−2Γ(𝛼)𝐻()=,𝑡≤𝑠,𝑡,𝑠log𝑡^𝛽−11−log𝑠^𝛽−1−log(𝑡/𝑠)^𝛽−1Γ𝛽,𝑠≤𝑡,log𝑡^𝛽−11−log𝑠^𝛽−1Γ𝛽,𝑡≤𝑠. (18)
This completes the proof of Lemma 6.

Lemma 7. Suppose that 2<𝛼≤3, 1<𝛽≤2. Then the functions 𝐺(𝑡,𝑠) and 𝐻(𝑡,𝑠) defined in (10) have the following properties: (1)𝐺(𝑡,𝑠), 𝐻(𝑡,𝑠) are continuous on [1,𝑒]×[1,𝑒]; (2)for any 𝑡,𝑠∈[1,𝑒], 𝐺(𝑡,𝑠)≥0, 𝐻(𝑡,𝑠)≥0; (3)for any 𝑡,𝑠∈[1,𝑒], 𝐺(𝑡,𝑠)≤𝐺(𝑒,𝑠), 𝐻(𝑡,𝑠)≤𝐻(𝑠,𝑠); (4)there exist two positive functions 𝛾₁,𝛾₂∈𝐶(1,𝑒) such that min_{𝑒1/4≤𝑡≤𝑒3/4}𝐺(𝑡,𝑠)≥𝛾₁(𝑠)max_{1≤𝑡≤𝑒}𝐺(𝑡,𝑠)=𝛾₁(𝑠)𝐺(𝑒,𝑠),for𝑠∈(1,𝑒),min_{𝑒1/4≤𝑡≤𝑒3/4}𝐻(𝑡,𝑠)≥𝛾₂(𝑠)max_{1≤𝑡≤𝑒}𝐻(𝑡,𝑠)=𝛾₂(𝑠)𝐻(𝑠,𝑠),for𝑠∈(1,𝑒). (19)(20)

Proof. (1) and (2) are evident from the expression of 𝐺(𝑡,𝑠) and 𝐻(𝑡,𝑠). Since, for any fixed number 𝑠∈[1,𝑒], 𝐺(𝑡,𝑠) is an increasing function on [1,𝑒] and 𝐻(𝑡,𝑠) is a decreasing function on [𝑠,𝑒] and increasing on [1,𝑠], we get (3). To prove (4), suppose that 𝛾₁(𝑠)=(1/4)^𝛼−11−log𝑠^𝛼−2−1/4−log𝑠^𝛼−11−log𝑠^𝛼−2−1−log𝑠^𝛼−1,𝑠∈1,𝑒^1/4,(1/4)^𝛼−11−log𝑠^𝛼−21−log𝑠^𝛼−2−1−log𝑠^𝛼−1,𝑠∈𝑒^1/4,𝑒, (21) and put 𝑝₁(𝑡,𝑠)=log𝑡^𝛼−11−log𝑠^𝛼−2−log(𝑡/𝑠)^𝛼−1Γ(𝛼),𝑝₂(𝑡,𝑠)=log𝑡^𝛼−11−log𝑠^𝛼−2Γ(𝛼). (22) The monotonicity of 𝐺(𝑡,𝑠) gives min_{𝑒1/4≤𝑡≤𝑒3/4}𝐺()=𝑝𝑡,𝑠₁𝑒^1/4,𝑠,𝑠∈1,𝑒^1/4,𝑝₂𝑒^1/4,𝑠,𝑠∈𝑒^1/4,=1,𝑒Γ(𝛼)14^𝛼−11−log𝑠^𝛼−2−14−log𝑠^𝛼−1,𝑠∈1,𝑒^1/4,1Γ(𝛼)14^𝛼−11−log𝑠^𝛼−2,𝑠∈𝑒^1/4,𝑒,max_{1≤𝑡≤𝑒}𝐺()()=1𝑡,𝑠=𝐺𝑒,𝑠Γ(𝛼)1−log𝑠^𝛼−2−1−log𝑠^𝛼−1,𝑠∈(1,𝑒). (23) Which implies that (19) holds.
Similarly, by writing 𝑞₁(𝑡,𝑠)=log𝑡^𝛽−11−log𝑠^𝛽−1−log(𝑡/𝑠)^𝛽−1Γ𝛽,𝑞₂(𝑡,𝑠)=log𝑡^𝛽−11−log𝑠^𝛽−1Γ𝛽 (24) and applying the monotonicity of 𝐻(𝑡,𝑠), we have min_{𝑒1/4≤𝑡≤𝑒3/4}𝐻()=𝑞𝑡,𝑠₁𝑒^3/4,𝑠,𝑠∈1,𝑒^1/4,min𝑞₁𝑒^3/4,𝑞,𝑠₂𝑒^1/4,𝑠,𝑠∈𝑒^1/4,𝑒^3/4,𝑞₂𝑒^1/4,𝑠,𝑠∈𝑒^3/4,=𝑞,𝑒₁𝑒^3/4,𝑠,𝑠∈(1,𝑟],<?layoutcmd="vspace"value="15"?>𝑞₂𝑒^1/4[),=1,𝑠,𝑠∈𝑟,𝑒Γ𝛽34^𝛽−11−log𝑠^𝛽−1−34−log𝑠^𝛽−1(],1,𝑠∈1,𝑟Γ𝛽14^𝛽−11−log𝑠^𝛽−1,𝑠∈[𝑟,𝑒),max_{1≤𝑡≤𝑒}𝐻()()=1𝑡,𝑠=𝐻𝑠,𝑠Γ𝛽log𝑠1−log𝑠^𝛽−1,𝑠∈(1,𝑒), (25) where 𝑒^1/4<𝑟<𝑒^3/4 is the unique solution of the equation 341−log𝑠^𝛽−1−34−log𝑠^𝛽−1=141−log𝑠^𝛽−1. (26) Hence, setting 𝛾₂(𝑠)=(3/4)1−log𝑠^𝛽−1−3/4−log𝑠^𝛽−1log𝑠1−log𝑠^𝛽−1(],1,𝑠∈1,𝑟4log𝑠^𝛽−1,𝑠∈[𝑟,𝑒), (27) we obtain (20). This completes the proof of Lemma 7.

Let 𝑋=𝐶[1,𝑒], ‖𝑢‖=max_{1≤𝑡≤𝑒}|𝑢(𝑡)|. we define the cone 𝑃={𝑢∈𝑋∣𝑢(𝑡)≥0} and the bounded closed set 𝑈={𝑢∈𝑋∣0≤𝑢(𝑡)≤𝐾}.

The operator 𝑇∶𝑋→𝑋 is defined as the following form: (𝑡)(𝑡)=𝑇𝑢=𝑌_𝑒1𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠. (28) Evidently, the solutions of boundary value problem (5) are the corresponding fixed points of the operator 𝑇.

Lemma 8. Suppose that 𝑇∶𝑃→𝑋 is an operator as above; then 𝑇∶𝑃→𝑃 is completely continuous.

Proof. It is easy to see that 𝑇∶𝑃→𝑃 is continuous. Let Ω⊂𝑃 be a bounded set; then there is a positive constant 𝐴>0 such that ‖𝑢‖≤𝐴 for any 𝑢∈Ω. Write 𝐵=max_{1≤𝑡≤𝑒,0≤𝑢≤𝐴}𝑓(𝑡,𝑢(𝑡))+1. For any 𝑢∈Ω, we have |𝑇𝑢(𝑡)|=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤𝐵^𝑞−1𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏<+∞, (29) which shows that 𝑇Ω is uniformly bounded.
Next, the continuity of 𝐺(𝑡,𝑠) implies that, for any 𝜀>0, there exists a constant 𝛿 such that, for any 𝑡₁,𝑡₂∈[1,𝑒], if |𝑡₁−𝑡₂|<𝛿, then 𝐺𝑡₁,𝑠−𝐺𝑡₂<𝜀,𝑠𝐵^𝑞−1𝜑_𝑞∫^𝑒₁𝐻().𝜏,𝜏)(𝑑𝜏/𝜏 (30) Therefore, for any 𝑢∈Ω, 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂=^𝑒₁𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠×𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤𝐵^𝑞−1𝑒₁𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠×𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠<𝜀. (31) That is, 𝑇Ω is equicontinuity. By the means of Arzela-Ascoli theorem [26], we have that 𝑇∶𝑃→𝑃 is completely continuous. This completes the proof of Lemma 8.

In the final part of this section, we list the following basic properties of the 𝑝-Laplacian operator.

Lemma 9. (1) If 1<𝑝<2, 𝑥𝑦>0, and |𝑥|,|𝑦|≥𝑚>0, then 𝜑_𝑝(𝑥)−𝜑_𝑝𝑦≤𝑝−1𝑚^𝑝−2𝑥−𝑦. (32)
(2) If 𝑝>2,|𝑥|,|𝑦|≤𝑀, then 𝜑_𝑝(𝑥)−𝜑_𝑝𝑦≤𝑝−1𝑀^𝑝−2𝑥−𝑦. (33)

3. Proofs of the Main Results

In this section, first, we consider the existence of the solutions of problem (5).

Theorem 10. If  [max_{1≤𝑡≤𝑒,0≤𝑢≤𝐾}𝑓(𝑡,𝑢)]^𝑞−1∫^𝑒₁𝐺(𝑒,𝑠)(𝑑𝑠/𝑠)𝜑_𝑞(∫^𝑒₁𝐻(𝜏,𝜏)(𝑑𝜏/𝜏))≤𝐾, then the boundary value problem (5) has at least one positive solution.

Proof. For any 𝑢∈𝑈, by the assumption as above and the nonnegativeness of 𝐺(𝑡,𝑠), 𝐻(𝑡,𝑠), and 𝑓(𝑡,𝑢), we have 0≤𝑇𝑢(𝑡)=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝐾}𝑓(𝑡,𝑢)^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏≤𝐾. (34) Therefore, 𝑇 is a mapping from 𝑈 to 𝑈. This, combined with the continuity of 𝐺(𝑡,𝑠), 𝐻(𝑡,𝑠), and 𝑓(𝑡,𝑢), implies that 𝑇∶𝑈→𝑈 is continuous.
Let Ω⊂𝑈 be a bounded set; then there exists a positive constant 𝐴 such that ‖𝑢‖≤𝐴 for any 𝑢∈Ω. So we have, for any 𝑢∈Ω, |𝑇𝑢(𝑡)|=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝐴}𝑓(𝑡,𝑢)^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝐾}𝑓(𝑡,𝑢)^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏≤𝐾≤+∞. (35) Therefore, 𝑇Ω is uniformly bounded.
Since 𝐺(𝑡,𝑠) is continuous, for any 𝜀>0, there exists a constant 𝛿>0 satisfying that, for any 𝑡₁,𝑡₂∈[1,𝑒] and |𝑡₁−𝑡₂|<𝛿, 𝐺𝑡₁,𝑠−𝐺𝑡₂<𝜀,𝑠max_{1≤𝑡≤𝑒,0≤𝑢≤𝐴}𝑓(𝑡,𝑢)+1^𝑞−1𝜑_𝑞∫^𝑒₁𝐻().𝜏,𝜏)(𝑑𝜏/𝜏 (36) Then, for any 𝑢∈Ω, 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂≤^𝑒₁𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠×𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝐴}𝑓(𝑡,𝑢)+1^𝑞−1×𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏×^𝑒₁𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠𝑑𝑠𝑠<𝜀, (37) which shows that 𝑇Ω is equicontinuous. By Arzela-Ascoli theorem [26], 𝑇∶𝑈→𝑈 is a completely continuous operator. It follows from Lemma 3 that 𝑇 has a fixed point 𝑢 in 𝑈. That is, problem (5) has at least one positive solution. This completes the proof of Theorem 10.

Let us denote Λ₁=^𝑒3/4_𝑒1/4𝛾₁(𝑠)𝐺(𝑒,𝑠)𝑑𝑠𝑠×𝜑_𝑞^𝑒3/4_𝑒1/4𝛾₂(𝜏)𝐻(𝜏,𝜏)𝑑𝜏𝜏⁻¹,Λ₂=^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏⁻¹. (38)

Theorem 11. Suppose that 𝑓(𝑡,𝑢)∶[1,𝑒]×[0,+∞) is a continuous function and there exist two constants 𝑟₂>𝑟₁>0 satisfying that (i)𝑓(𝑡,𝑢)≥𝜑_𝑝(Λ₁𝑟₁), for (𝑡,𝑢)∈[1,𝑒]×[0,𝑟₁]; (ii)𝑓(𝑡,𝑢)≤𝜑_𝑝(Λ₂𝑟₂), for (𝑡,𝑢)∈[1,𝑒]×[0,𝑟₂]. Then the boundary value problem (5) has at least one positive solution 𝑢 which satisfies that 𝑟₁≤‖𝑢‖≤𝑟₂.

Proof. Let Ω₁={𝑢∈𝑃∣‖𝑢‖<𝑟₁}. For any 𝑢∈𝜕Ω₁, we have 0≤𝑢(𝑡)≤𝑟₁ for 𝑡∈[1,𝑒]. By the assumption (i), for any 𝑡∈[𝑒^1/4,𝑒^3/4], |𝑇𝑢(𝑡)|=^𝑒₁𝐺(𝑡,𝑠)×𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≥Λ₁𝑟₁^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠≥Λ₁𝑟₁^𝑒3/4_𝑒1/4𝐺(𝑡,𝑠)𝜑_𝑞^𝑒3/4_𝑒1/4𝐻(𝑠,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠≥Λ₁𝑟₁^𝑒3/4_𝑒1/4𝛾₁(𝑠)𝐺(𝑒,𝑠)×𝜑_𝑞^𝑒3/4_𝑒1/4𝛾₂(𝜏)𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠=𝑟₁=‖𝑢‖. (39) Hence, ‖𝑇𝑢‖≥‖𝑢‖,for𝑢∈𝜕Ω₁. (40) Similarly, let Ω₂∶={𝑢∈𝑃∣‖𝑢‖<𝑟₂}. For any 𝑢∈𝜕Ω₂, we get 0≤𝑢(𝑡)≤𝑟₂, 𝑡∈[1,𝑒]. It follows from (ii) that for any 𝑡∈[1,𝑒], |𝑇𝑢(𝑡)|=^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤Λ₂𝑟₂^𝑒₁𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠≤Λ₂𝑟₂^𝑒₁𝐺(𝑒,𝑠)𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑑𝑠𝑠=𝑟₂=‖𝑢‖. (41) Therefore, ‖𝑇𝑢‖≤‖𝑢‖,for𝑢∈𝜕Ω₂. (42) By Lemmas 4 and 8, 𝑇 has a fixed point in Ω₂⧵Ω₁. Therefore, the boundary value problem (5) has one positive solution in Ω₂⧵Ω₁. This completes the proof of Theorem 11.

Corollary 12. Suppose that 𝑓(𝑡,𝑢)∶[1,𝑒]×[0,+∞) is a continuous function and there exist two constants 𝑟₂>𝑟₁>0 satisfying that (i)𝑓(𝑡,𝑢)≤𝜑_𝑝(Λ₂𝑟₁), for (𝑡,𝑢)∈[1,𝑒]×[0,𝑟₁]; (ii)𝑓(𝑡,𝑢)≥𝜑_𝑝(Λ₁𝑟₂), for (𝑡,𝑢)∈[1,𝑒]×[0,𝑟₂]. Then the boundary value problem (5) has at least one positive solution 𝑢 which satisfies that 𝑟₁≤‖𝑢‖≤𝑟₂.

Proof. The proof of Corollary 12 is similar to the one of Theorem 11. So we omit the detail.

Theorem 13. Suppose that 𝑓(𝑡,𝑢)∶[1,𝑒]×[0,+∞) is a positive continuous function and there exists a constant 𝑟>0 such that 𝑟>max_{1≤𝑡≤𝑒,0≤𝑢≤𝑟}𝑓(𝑡,𝑢)+1^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏. (43) Then the boundary value problem (5) has at least one positive solution.

Proof. Let 𝐸={𝑢∈𝑃∶‖𝑢‖<𝑟}. (44) From Lemma 8, we know 𝑇∶𝐸→𝑃 is completely continuous. Assume that there exist 𝑢∈𝐸, 𝜆∈(0,1) such that 𝑢=𝜆𝑇𝑢. Then we have |𝑢(𝑡=||≤)|𝜆𝑇𝑢_𝑒1𝐺(𝑡,𝑠)𝜑_𝑞^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝑟}𝑓(𝑡,𝑢)+1^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏. (45) Thus, ‖𝑢‖≤max_{1≤𝑡≤𝑒,0≤𝑢≤𝑟}𝑓(𝑡,𝑢)+1^𝑞−1×^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_𝑞^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏. (46) By (43), we can imply that ‖𝑢‖<𝑟, which means that 𝑢∉𝜕𝐸. That is to say, there is no 𝑢∈𝜕𝐸 such that 𝑢=𝜆𝑇𝑢 for some 𝜆∈(0,1). Therefore, by Lemma 5, we conclude that the problem (5) has at least one positive solution. This completes the proof of Theorem 13.

Now we turn to the uniqueness of solution for boundary value problem (5).

Theorem 14. Suppose that 𝑝>2. If there exists a nonnegative function 𝑔 satisfying that (1)for any (𝑡,𝑢)∈[1,𝑒]×[0,+∞), 𝑓(𝑡,𝑢)≥𝑔(𝑡); (2)𝑁=∫^𝑒3/4_𝑒1/4𝛾₂(𝜏)𝐻(𝜏,𝜏)𝑔(𝜏)(𝑑𝜏/𝜏)>0; (3)for any 𝑡∈[1,𝑒], 𝑢,𝑣∈[0,+∞), |𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)|≤𝐿|𝑢−𝑣|, where 0<𝐿<^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑞−1𝑁^𝑞−2−1, (47)then the boundary value problem (5) has a unique solution.

Proof. Assume that 𝑢,𝑣 are two positive solutions of problem (5). It is easy to see that ^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏≥^𝑒3/4_𝑒1/4𝛾₂(𝜏)𝐻(𝜏,𝜏)𝑔(𝜏)𝑑𝜏𝜏=𝑁>0; (48) then by the fact 𝑝>2 (i.e., its dual exponent 1<𝑞<2) and Lemma 9, we have |𝑢(𝑡)−𝑣(𝑡)|≤^𝑒₁𝐺(𝑡,𝑠)𝑞−1𝑁^𝑞−2×^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝜏−^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑣(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤^𝑒₁𝐺(𝑡,𝑠)𝑞−1𝑁^𝑞−2×^𝑒₁𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))−𝑓(𝜏,𝑣(𝜏))𝑑𝜏𝜏𝑑𝑠𝑠≤^𝑒₁𝐺(𝑒,𝑠)𝑞−1𝑁^𝑞−2×^𝑒₁𝐻(𝜏,𝜏)𝐿‖𝑢−𝑣‖𝑑𝜏𝜏𝑑𝑠𝑠=^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠×^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑞−1𝑁^𝑞−2𝐿‖𝑢−𝑣‖=𝛿₁‖𝑢−𝑣‖, (49) where 𝛿₁=∫^𝑒₁𝐺(𝑒,𝑠)(𝑑𝑠/𝑠)∫^𝑒₁𝐻(𝜏,𝜏)(𝑑𝜏/𝜏)(𝑞−1)𝑁^𝑞−2𝐿. So we can get ‖𝑢−𝑣‖≤𝛿₁‖𝑢−𝑣‖. (50) By the third hypothesis, 0<𝛿₁<1, which implies that 𝑢(𝑡)=𝑣(𝑡). And this completes the proof of Theorem 14.

By using the same way, we can prove the last one of our main uniqueness results.

Theorem 15. Suppose that 1<𝑝<2. If there exists a nonnegative function ℎ satisfying that (1)for any (𝑡,𝑢)∈[1,𝑒]×[0,+∞), 𝑓(𝑡,𝑢)≤ℎ(𝑡); (2)𝑀=∫^𝑒₁𝐻(𝜏,𝜏)ℎ(𝜏)(𝑑𝜏/𝜏)>0; (3)for any 𝑡∈[1,𝑒], 𝑢,𝑣∈[0,+∞), |𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)|≤𝐿|𝑢−𝑣|, where 0<𝐿<^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑞−1𝑀^𝑞−2−1, (51)then the boundary value problem (5) has a unique solution.

4. Examples

In this section we give several examples to illustrate our main results.

Example 16. Consider the boundary value problem: 𝐷^3/2𝜑_5/2𝐷^5/2𝑢(𝑡)=1−log𝑡^1/21+sin²𝑢𝑢(1),1<𝑡<𝑒,=𝑢^′(1)=𝑢^′(𝑒)𝐷=0,_5/2𝑢(1)=𝐷^5/2𝑢(𝑒)=0. (52) Then the boundary value problem has a unique positive solution.

Proof. Since 𝛼=5/2, 𝛽=3/2, a straightforward calculation gives ^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠=^𝑒₁1−log𝑠^𝛼−2−1−log𝑠^𝛼−1Γ(𝛼)𝑑𝑠𝑠=1Γ(𝛼)^𝑒₁1−log𝑠^𝛼−2𝑑𝑠𝑠−1Γ(𝛼)^𝑒₁1−log𝑠^𝛼−1𝑑𝑠𝑠=1Γ(𝛼)¹₀(1−𝑡)^𝛼−2−1𝑑𝑡Γ(𝛼)¹₀(1−𝑡)^𝛼−1=1𝑑𝑡()Γ()=1𝛼−1𝛼+1()Γ()=3/27/21645√𝜋_𝑒1,𝐻(𝜏,𝜏)𝑑𝜏𝜏=^𝑒₁log𝜏^𝛽−11−log𝜏^𝛽−1Γ𝛽𝑑𝜏𝜏=1Γ𝛽¹₀𝑡^𝛽−1(1−𝑡)^𝛽−1=Γ𝛽𝑑𝑡Γ=Γ()2𝛽3/2Γ(3)=√𝜋4. (53)
Taking 𝐾=1,𝑓(𝑡,𝑢)=(1−log𝑡)^1/2(1+sin²𝑢), and 𝑝=5/2>2 (its dual exponent 𝑞=5/3), we have max_{1≤𝑡≤𝑒,0≤𝑢≤1}𝑓(𝑡,𝑢)=1+sin²1<2,max_{1≤𝑡≤𝑒,0≤𝑢≤1}𝑓(𝑡,𝑢)^{(5/3)−1𝑒}₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑_5/3^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏<2^2/3⋅1645√𝜋⋅√𝜋4^2/3≈0.185<1. (54) By Theorem 10, the boundary value problem (52) has at least one positive solution.
Choosing the nonnegative function 𝑔(𝑡)=(1−log𝑡)^1/2, for any (𝑡,𝑢)∈[1,𝑒]×[0,+∞), we gain that 𝑓(𝑡,𝑢)=(1−log𝑡)^1/2(1+sin²𝑢)≥𝑔(𝑡). Then 𝑁=^𝑒3/4_𝑒1/4𝛾₂(𝜏)𝐻(𝜏,𝜏)𝑔(𝜏)𝑑𝜏𝜏=^𝑟_𝑒1/41Γ()33/241−log𝜏^3/2−1−34−log𝜏^3/2−1⋅1−log𝜏^1/2𝑑𝜏𝜏+^𝑒3/4_𝑟1Γ()13/24^3/2−11−log𝜏^3/2−1⋅1−log𝜏^1/2𝑑𝜏𝜏≥^𝑒3/4_𝑟1√𝜋1−log𝜏𝑑𝜏𝜏=12√𝜋1−log𝑟²−132√𝜋=12√𝜋√1−1+36²−132√𝜋=196√𝜋, (55) where 𝑟=𝑒^√1−(3/6) is the solution of (26) when 𝛽=3/2. For any 𝑡∈[1,𝑒], 𝑢,𝑣∈[0,+∞), taking 𝐿=2, we obtain 𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)=1−log𝑡^1/2sin²𝑢−sin²𝑣≤2|𝑢−𝑣|=𝐿|𝑢−𝑣|. (56) Thus, ^𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑞−1𝑁^𝑞−2−1=1645√𝜋⋅√𝜋4⋅23𝑁^−1/3−1≥1358196√𝜋^1/3≈3.045>2=𝐿. (57) By Theorem 14, the boundary value problem (52) has a unique solution.

Example 17. Consider the following nonlinear boundary value problem: 𝐷^3/2𝜑_3/2𝐷^5/2𝑢(𝑡)=log^1/2𝑡sin²𝑢(1)𝑢,1<𝑡<𝑒,=𝑢^′(1)=𝑢^′(𝑒)=0,𝐷^5/2𝑢(1)=𝐷^5/2𝑢(𝑒)=0. (58) Then the boundary value problem has a unique positive solution.

Proof. Taking 𝑟=1, since 𝛼=5/2, 𝛽=3/2, 𝑝=3/2<2 (its dual exponent 𝑞=3) and 𝑓(𝑡,𝑢)=log^1/2𝑡sin²𝑢, we obtain max_{1≤𝑡≤𝑒,0≤𝑢≤1}𝑓(𝑡,𝑢)=sin²1<1,max_{1≤𝑡≤𝑒,0≤𝑢≤1}𝑓(𝑡,𝑢)+1^3−1𝑒₁𝐺(𝑒,𝑠)𝑑𝑠𝑠𝜑₃^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏<4⋅1645√𝜋⋅𝜋16≈0.158<1. (59)(60) By means of Theorem 13, the boundary value problem (58) has at least one positive solution.
Taking the nonnegative function ℎ(𝑡)=log^1/2𝑡, for (𝑡,𝑢)∈[1,𝑒]×[0,+∞), it is easy to obtain 𝑓(𝑡,𝑢)=log^1/2𝑡sin²𝑢≤ℎ(𝑡) and 𝑀=^𝑒₁𝐻(𝜏,𝜏)ℎ(𝜏)𝑑𝜏𝜏=1Γ(3/2)^𝑒₁log𝜏1−log𝜏^1/2𝑑𝜏𝜏=815√𝜋. (61) Choosing 𝐿=2, for any 𝑡∈[1,𝑒], 𝑢,𝑣∈[0,+∞), we have 𝑓(𝑡,𝑢)−𝑓(𝑡,𝑣)=log^1/2𝑡sin²𝑢−sin²𝑣≤2|𝑢−𝑣|=𝐿|𝑢−𝑣_𝑒1|,𝐺(𝑒,𝑠)𝑑𝑠𝑠^𝑒₁𝐻(𝜏,𝜏)𝑑𝜏𝜏𝑞−1𝑀^𝑞−2−1=1645√𝜋⋅√𝜋48⋅2⋅15√𝜋⁻¹≈18.694>2=𝐿. (62) From Theorem 15, the boundary value problem (58) has a unique solution.

Acknowledgments

The authors would like to thank the referees for their detailed and helpful suggestions for revising this paper. This work was supported by the National Natural Science Foundation of China (Grant nos. 11226167 and 11361020), the Natural Science Foundation of Hainan Province (no. 111005), and the Ph.D. Scientific Research Starting Foundation of Hainan Normal University (no. HSBS1016).