Positive Solutions for a Mixed-Order Three-Point Boundary Value Problem for 𝑝-Laplacian

Abstract

The author investigates the existence and multiplicity of positive solutions for boundary value problem of fractional differential equation with 𝑝-Laplacian operator. The main tool is fixed point index theory and Leggett-Williams fixed point theorem.

1. Introduction

In this paper, we are interested in the existence of single and multiple positive solutions to nonlinear mixed-order three-point boundary value problem for 𝑝-Laplacian. Consider the following: 𝜑_𝑝𝐷^𝛼₀⁺𝑢(𝑡)^′+𝑎(𝑡)𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝐷^𝛼₀⁺𝑢(0)=𝑢(0)=𝑢^′′(0)=0,𝑢^′(1)=𝛾𝑢^′𝜂, (1)(2) where 𝜂,𝛾∈(0,1), 𝛼∈(2,3], 𝐷^𝛼₀⁺ is the Caputo's derivative, 𝜑_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, and 𝜑_𝑞=(𝜑_𝑝)⁻¹ with (1/𝑝)+(1/𝑞)=1. We assume the following conditions throughout.(H1)𝑓∈𝐶([0,1]×[0,∞),[0,∞)).(H2)𝑎∈𝐿¹(0,1) is nonnegative and 𝑎(𝑡)≢0 on any subinterval of (0,1).

The equation with a 𝑝-Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics, nonlinear elasticity and glaciology, combustion theory, population biology, nonlinear flow laws, and so on. Liang et al. in [1] used the fixed point theorem of Avery and Henderson to show the existence of at least two positive solutions. Zhao et al. [2] studied the existence of at least three positive solutions by using Leggett-Williams fixed point theorem. Chai [3] obtain results for the existence of at least one nonnegative solution and two positive solutions by using fixed point theorem on cone. Su et al. [4] studied the existence of one and two positive solutions by using the fixed point index theory. Su [5] studied the existence of one and two positive solution by using the method of defining operator by the reverse function of Green function and the fixed point index theory. Tang et al. [6] studied the existence of positive solutions of fractional differential equation with 𝑝-laplacian by using the coincidence degree theory.

Motivated by the above works, we obtain some sufficient conditions for the existence of at least one and three positive solutions for (1) and (2).

The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our results. In Section 3, we discuss the existence of at least one positive solution for (1) and (2). In Section 4, we discuss the existence of multiple positive solutions for (1) and (2). Finally, we give some examples to illustrate our results in Section 5.

2. Preliminaries

Definition 1. Let 𝐸 be a real Banach space. A nonempty closed convex set 𝐾⊂𝐸 is called cone if(1)𝑥∈𝐾, 𝜆≥0 then 𝜆𝑥∈𝐾,(2)𝑥∈𝐾, −𝑥∈𝐾 then 𝑥=0.

Definition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Remark 3. By the positive solution of (1) and (2) we understand a function 𝑢(𝑡) which is positive on [0,1] and satisfies the differential equation (1) and the boundary conditions (2).
We will consider the Banach space 𝐸=𝐶[0,1] equipped with standard norm: ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)| (3) The proof of existence of solution is based upon an application of the following theorem.

Theorem 4 (see [7, 8]). Let 𝐸 be a Banach space and let 𝐾 be a cone of 𝐸. For 𝑟>0, define 𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖≤𝑟} and assume that 𝑇∶𝐾_𝑟→𝐾 is a completely continuous operator such that 𝑇𝑢≠𝑢 for all 𝑢∈𝜕𝐾_𝑟. (1)If ‖𝑇𝑢‖≤‖𝑢‖ for all 𝑢∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=1.(2)If ‖𝑇𝑢‖≥‖𝑢‖ for all 𝑢∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=0.

Lemma 5 (see [9]). Let 𝑛∈ℕ with 𝑛≥2, 𝑛−1<𝛼≤𝑛. If 𝑢∈𝐶^𝑛−1[𝑎,𝑏] and 𝐷^𝛼_𝑎+𝑢∈𝐶(𝑎,𝑏), then 𝐼^𝛼_𝑎+𝐷^𝛼_𝑎+𝑢(𝑡)=𝑢(𝑡)−^𝑛−1_𝑘=0𝑢^(𝑘)(𝑎)()𝑘!𝑡−𝑎^𝑘 (4) holds on (𝑎,𝑏).

Lemma 6. The three-point boundary value problem (1)-(2) has a unique solution 𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏+𝜏,𝑢))𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠×𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝑑𝑠, (5) where 𝐺₁()=()𝑡()𝑡,𝑠𝛼−11−𝑠^𝛼−2−(𝑡−𝑠)^𝛼−1Γ(𝛼)()𝑡(),0≤𝑠≤𝑡≤1,𝛼−11−𝑠^𝛼−2Γ(𝛼)𝐺,0≤𝑡≤𝑠≤1,₂=()𝜂,𝑠𝛼−1)(1−𝑠^𝛼−2−(𝛼−1)𝜂−𝑠^𝛼−2Γ(𝛼)(),0≤𝑠≤𝜂≤1,𝛼−1)(1−𝑠^𝛼−2Γ(𝛼),0≤𝜂≤𝑠≤1. (6)

Proof. Integrating both sides of (1) on [0,1], we have 𝜑_𝑝𝐷^𝛼𝑢(𝑡)−𝜑_𝑝𝐷^𝛼𝑢(0)=−^𝑡₀𝑎(𝑠)𝑓(𝑠,𝑢)𝑑𝑠. (7) So 𝐷^𝛼𝑢(𝑡)=−𝜑_𝑞^𝑡₀𝑎(𝑠)𝑓(𝑠,𝑢)𝑑𝑠. (8) From Lemma 5, we have 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠+𝐴+𝐵𝑡+𝐶𝑡²,𝑢^′(𝑡)1=−Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()𝑢𝜏,𝑢𝑑𝜏𝑑𝑠+𝐵+2𝐶𝑡,_′′(𝑡)1=−Γ(𝛼−2)^𝑡₀(𝑡−𝑠)^𝛼−3𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠+2𝐶. (9) From (2), 𝐴=0 and 𝐶=0.
Now, consider the following: 𝑢^′(1)1=−Γ(𝛼−1)¹₀(1−𝑠)^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()𝑢𝜏,𝑢𝑑𝜏𝑑𝑠+𝐵,_′𝜂1=−Γ(𝛼−1)^𝜂₀𝜂−𝑠^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠+𝐵, (10) by the boundary value condition 𝑢^′(1)=𝛾𝑢^′(𝜂), we have 1𝐵=1−𝛾Γ(𝛼−1)¹₀(1−𝑠)^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()−𝛾𝜏,𝑢𝑑𝜏𝑑𝑠1−𝛾Γ(𝛼−1)^𝜂₀𝜂−𝑠^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠, (11) so 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()+𝑡𝜏,𝑢𝑑𝜏𝑑𝑠Γ()×1−𝛾𝛼−1₁₀(1−𝑠)^𝛼−2𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()−𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡Γ()×1−𝛾𝛼−1_𝜂0𝜂−𝑠^𝛼−2𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠. (12) Splitting the second integral in two parts of the form 𝑡Γ()+𝑘𝛼−1Γ()=𝑡1−𝛾𝛼−1Γ(),1−𝛾𝛼−1 (13) we have 𝑘=𝛾𝑡; thus, 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()+𝑡𝜏,𝑢𝑑𝜏𝑑𝑠Γ(𝛼−1)¹₀(1−𝑠)^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()+𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾Γ(𝛼−1)¹₀(1−𝑠)^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()−𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾Γ(𝛼−1)^𝜂₀𝜂−𝑠^𝛼−2𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠; (14) therefore, 𝑢(𝑡)=^𝑡₀𝑡(1−𝑠)^𝛼−2Γ()−()𝛼−1𝑡−𝑠^𝛼−1Γ(𝛼)𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()+𝜏,𝑢𝑑𝜏𝑑𝑠_1𝑡𝑡(1−𝑠)^𝛼−2Γ()𝜑𝛼−1^𝑞𝑠₀𝑎(𝜏)𝑓()+𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾^𝜂₀(1−𝑠)^𝛼−2Γ()−𝛼−1𝜂−𝑠^𝛼−2Γ()𝜑𝛼−1^𝑞×^𝑠₀𝑎(𝜏)𝑓()+𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹_𝜂(1−𝑠)^𝛼−2Γ()𝜑𝛼−1^𝑞𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠. (15) This completes the proof.

Lemma 7. Let 𝛽∈(0,1) be fixed. The kernel, 𝐺₁(𝑡,𝑠), satisfies the following properties.(1)0≤𝐺₁(𝑡,𝑠)≤𝐺₁(1,𝑠) for all 𝑠∈(0,1),(2)min_{𝛽≤𝑡≤1}𝐺₁(𝑡,𝑠)≥𝛽𝐺₁(1,𝑠) for all 𝑠∈[0,1].

Proof. (1) As 2<𝛼≤3 and 0≤𝑠≤𝑡≤1, we have (𝛼−1)𝑡(1−𝑠)^𝛼−2>𝑡(1−𝑠)^𝛼−2≥(𝑡−𝑠)(𝑡−𝑠)^𝛼−2=(𝑡−𝑠)^𝛼−1; (16) thus, 𝐺₁(𝑡,𝑠)>0. Note 𝜕𝐺₁(𝑡,𝑠)/𝜕𝑡≥0 then, 𝐺₁(𝑡,𝑠) is increasing as a function of 𝑡; therefore, 𝐺₁(𝑡,𝑠)≤𝐺₁(1,𝑠)∀𝑠∈[0,1]. (17)
(2) For 𝛽≤𝑡≤1, we have min_{𝛽≤𝑡≤1}𝐺₁(𝑡,𝑠)=𝐺₁𝛽,𝑠, (18) where 𝐺₁=()𝛽()𝛽,𝑠𝛼−11−𝑠^𝛼−2−𝛽−𝑠^𝛼−1Γ(𝛼)()𝛽()0≤𝑠≤𝛽,𝛼−11−𝑠^𝛼−2Γ(𝛼)𝛽≤𝑠≤1. (19) (a) If 0<𝑠≤𝛽, min_{𝛽≤𝑡≤1}𝐺₁()=𝛽()𝑡,𝑠𝛼−1)(1−𝑠^𝛼−2Γ(𝛼)−𝛽−𝑠^𝛼−1Γ(𝛼); (20) on the other hand, 𝛽𝐺₁()=𝛽()1,𝑠𝛼−1)(1−𝑠^𝛼−2Γ(𝛼)−𝛽(1−𝑠)^𝛼−1Γ(𝛼). (21) Since 2<𝛼≤3 and(i)𝛼−1>1, 𝛽∈(0,1)⇒𝛽^𝛼−1<𝛽,(ii)𝑠≤𝛽⇒𝑠/𝛽≤1⇒1−(𝑠/𝛽)≥0, (iii)𝛽<1⇒1<1/𝛽⇒−𝑠(1/𝛽)<−𝑠⇒1−𝑠(1/𝛽)<1−𝑠;
thus, we have 𝑠1−𝛽^𝛼−1<(1−𝑠)^𝛼−1. (22) From (20), we obtain 𝛽−𝑠^𝛼−1=𝛽𝑠1−𝛽^𝛼−1=𝛽^𝛼−1𝑠1−𝛽^𝛼−1𝑠≤𝛽1−𝛽^𝛼−1<𝛽(1−𝑠)^𝛼−1. (23) It follows from (20), (21), and (23), that item 2 in the proof hold.
(b) If 𝛽≤𝑠<1, min_{𝛽≤𝑡≤1}𝐺₁()=𝛽()𝑡,𝑠𝛼−1)(1−𝑠^𝛼−2Γ(𝛼),𝛽𝐺₁()=𝛽()1,𝑠𝛼−1)(1−𝑠^𝛼−2Γ(𝛼)−𝛽(1−𝑠)^𝛼−1Γ(𝛼). (24) It follows from (24) that item 2 in the proof holds.

Lemma 8 (see [10]). The unique solution 𝑢(𝑡) of (1), (2) is nonnegative and satisfies min_{𝛽≤𝑡≤1}𝑢(𝑡)≥𝛽‖𝑢‖. (25)

Define the cone 𝐾 by 𝐾=𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0,min_{𝛽≤𝑡≤1}𝑢(𝑡)≥𝛽‖𝑢‖ (26) and the operator 𝑇∶𝐾→𝐸 by 𝑇𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏+𝜏,𝑢))𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝑑𝑠. (27)

Remark 9. By Lemma 6, the problem (1)-(2) has a positive solution 𝑢(𝑡) if and only if 𝑢 is a fixed point of 𝑇.

Lemma 10. 𝑇 is completely continuous and 𝑇(𝐾)⊆𝐾.

Proof. By Lemma 8, 𝑇(𝐾)⊆𝐾. In view of the assumption of nonnegativeness and continuity of functions 𝐺_𝑖(𝑥,𝑦) with 𝑖=1,2 and 𝑎(𝑡)𝑓(𝑡,𝑢(𝑡)), we conclude that 𝑇∶𝐾→𝐾 is continuous.
Let Ω⊆𝐾 be bounded; that is, there exists 𝑀>0 such that ‖𝑢‖≤𝑀 for all 𝑢∈Ω.
Let 𝐿=max_{0≤𝑡≤1,0≤𝑢≤𝑀}𝑓(𝑡,𝑢). (28)
Then from 𝑢∈Ω and from Lemmas 6 and 7, we have |𝑇𝑢(𝑡)|=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()+𝜏,𝑢𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠×𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()≤𝜏,𝑢𝑑𝜏𝑑𝑠₁₀𝐺₁(1,𝑠)𝜑_𝑞¹₀𝑎(𝜏)+𝛾𝐿𝑑𝜏𝑑𝑠1−𝛾¹₀𝐺₂𝜂,𝑠×𝜑_𝑞¹₀𝑎(𝜏)𝐿𝑑𝜏𝑑𝑠≤𝐿^𝑞−1𝜑_𝑞¹₀𝑎(𝜏)×𝑑𝜏₁₀(𝛼−1)Γ(𝛼)(1−𝑠)^𝛼−2𝛾𝑑𝑠+×1−𝛾₁₀(𝛼−1)Γ(𝛼)(1−𝑠)^𝛼−2=()𝐿𝑑𝑠𝛼−1^𝑞−1Γ(𝛼)𝜑1−𝛾^𝑞1₀𝑎(𝜏)𝑑𝜏¹₀(1−𝑠)^𝛼−2=𝐿𝑑𝑠_𝑞−1Γ(𝛼)𝜑1−𝛾^𝑞1₀𝑎(𝜏)𝑑𝜏≐𝑙. (29) Hence, 𝑇(Ω) is bounded.
On the other hand, let 𝑢∈Ω, 𝑡₁,𝑡₂∈[0,1] with 𝑡₁<𝑡₂; Then 𝑇𝑢𝑡₂−𝑇𝑢𝑡₁≤𝐿^𝑞−11₀𝐺₁𝑡₂,𝑠−𝐺₁𝑡₂,𝑠𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓()+𝐿𝜏,𝑢𝑑𝜏𝑑𝑠_𝑞−1𝛾𝑡₂−𝑡₁×1−𝛾₁₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠. (30) The continuity of 𝐺₁ implies that the right-side of the above inequality tends to zero if 𝑡₂→𝑡₁. Therefore, 𝑇 is completely continuous by Arzela-Ascoli theorem.
We introduce the notation 𝑓_𝑎∶=liminf_𝑢→𝑎min_{𝛽≤𝑡≤1}𝑓(𝑡,𝑢)𝑢^𝑝−1,𝑓^𝑏∶=limsup_𝑢→𝑏max_0≤𝑡≤1𝑓(𝑡,𝑢)𝑢^𝑝−1, (31) where 𝑎,𝑏=0⁺,∞. Consider the following: Λ₁=𝛽¹_𝛽𝐺₁(1,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝛾𝑑𝜏𝑑𝑠+×1−𝛾_1𝛽𝐺₂(𝜂,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑑𝜏𝑑𝑠⁻¹,Λ₂=1Γ(𝛼)𝜑1−𝛾^𝑞1₀𝑎(𝜏)𝑑𝜏⁻¹. (32)

3. Single Solutions

In what follows, the number 𝛽∈(0,1).

Theorem 11. Suppose that conditions (H1) and (H2) hold. Assume that 𝑓 also satisfy  (𝐴₁)𝑓(𝑡,𝑢)≥(𝑚𝑟)^𝑝−1, 𝛽𝑟≤𝑢≤𝑟, 𝛽≤𝑡≤1, (𝐴₂)𝑓(𝑡,𝑢)≤(𝑀𝑅)^𝑝−1, 0≤𝑢≤𝑅, 0≤𝑡≤1,
where 𝑚∈(Λ₁,∞) and 𝑀∈(0,Λ₂). Then (1)-(2) has at least one positive solution 𝑢 such that 𝑟≤‖𝑢‖≤𝑅.

Proof. Without loss of generality, we suppose that 𝑟<𝑅. For any 𝑢∈𝐾, we have 𝑢(𝑡)≥𝛽‖𝑢‖,𝛽≤𝑡≤1. (33) We define two open subsets of 𝐸, Ω₁={𝑢∈𝐾∶‖𝑢‖<𝑟},Ω₂={𝑢∈𝐾∶‖𝑢‖<𝑅}. (34) For 𝑢∈𝜕Ω₁, by (33), we have 𝑟=‖𝑢‖≥𝑢(𝑡)≥𝛽‖𝑢‖=𝛽𝑟,𝑡∈𝛽,1. (35) For 𝑡∈[𝛽,1], by (𝐴₁), (27), and Lemma 7, we have 𝑇𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏+𝜏,𝑢))𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠𝜑_𝑞×^𝑠₀𝑎(𝜏)𝑓((𝜏≥𝜏,𝑢))𝑑𝜏𝑑𝑠_1𝛽𝛽𝐺₁(1,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)(𝑚𝑟)^𝑝−1+𝑑𝜏𝑑𝑠𝛾𝛽1−𝛾¹_𝛽𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)(𝑚𝑟)^𝑝−1𝑑𝜏𝑑𝑠=𝑟𝑚𝛽¹_𝛽𝐺₁(1,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)+𝛾𝑑𝜏𝑑𝑠1−𝛾¹_𝛽𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)‖𝑢‖.𝑑𝜏𝑑𝑠>𝑟= (36) Therefore, ‖𝑇𝑢‖>‖𝑢‖,∀𝑢∈𝜕Ω₁∩𝐾. (37) Then by Theorem 4, 𝑖𝑇,Ω₁,𝐾=0. (38)
On the other hand, as 𝑢∈𝜕Ω₂, we have 0≤𝑢(𝑡)≤‖𝑢‖=𝑅, (39) thus, by (𝐴₂), (27), Lemma 7, and (29), we have 𝑇𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏+𝜏,𝑢))𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝑑𝑠≤𝑀𝑅¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)+𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑑𝜏𝑑𝑠≤𝑀𝑅¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞¹₀𝑎(𝜏)+𝑑𝜏𝑑𝑠𝛾𝑡1−𝛾¹₀𝐺₂𝜂,𝑠𝜑_𝑞¹₀𝑎(𝜏)≤𝑑𝜏𝑑𝑠𝑀𝑅Γ(𝛼)𝜑1−𝛾^𝑞1₀𝑎(𝜏)‖𝑢‖.𝑑𝜏𝑑𝑠<𝑅= (40) Therefore, ‖𝑇𝑢‖<‖𝑢‖,∀𝑢∈𝜕Ω₂∩𝐾. (41) Then by Theorem 4, 𝑖𝑇,Ω₁,𝐾=1. (42) By (38) and (42): 𝑖𝑇,Ω₂⧵Ω₁,𝐾=1. (43) Then, 𝑇 has a fixed point 𝑢∈Ω₂⧵Ω₁, 𝑢 is positive solution of problem (1)-(2), and 𝑟<‖𝑢‖<𝑅.

Corollary 12. Suppose that conditions (H1) and (H2) hold. Assume that 𝑓 also satisfy  (𝐴₃)  𝑓_∞=𝜆∈((2Λ₁/𝛽)^𝑝−1,∞), (𝐴₄)  𝑓⁰=𝜓∈[0,(Λ₂/2)^𝑝−1).
Then (1)-(2) has at least one positive solution 𝑢 such that 𝑟≤‖𝑢‖≤𝑅.

Proof. By (𝐴₄), for 𝜀=(Λ₂/2)^𝑝−1−𝜓, there exists a suitably small positive number 𝐻₁, as 0<𝑢≤𝐻₁ and 0≤𝑡≤1, such that 𝑓(𝑡,𝑢)≤𝜓+𝜀𝑢^𝑝−1≤Λ₂2^𝑝−1𝑢^𝑝−1≤Λ₂2𝐻₁^𝑝−1. (44) Let 𝑅=𝐻₁ and 𝑀=(Λ₂/2)∈(0,Λ₂), then, by (44), condition (𝐴₂) holds.
By (𝐴₃), for 𝜀=𝜆−(2Λ₁/𝛽)^𝑝−1, there exists a sufficiently large 𝑟≠𝑅 such that 𝑓(𝑡,𝑢)𝑢^𝑝−1≥𝜆−𝜀=2Λ₁𝛽^𝑝−1,𝛽𝑟≤𝑢<∞,𝛽≤𝑡≤1. (45) Thus, when 𝛽𝑟≤𝑢≤𝑟, one has 𝑓(𝑡,𝑢)≥2Λ₁𝛽^𝑝−1𝑢^𝑝−1≥2Λ₁𝑟^𝑝−1. (46) Let 𝑚=2Λ₁∈(Λ₁,∞). Then, by (46), condition (𝐴₁) holds.
Hence, from Theorem 11 the desired result hold.

Corollary 13. Suppose that conditions (H1) and (H2) hold. Assume that 𝑓 also satisfy  (𝐴₅)  𝑓₀=𝜆∈((2Λ₁/𝛽)^𝑝−1,∞), (𝐴₆)  𝑓^∞=𝜓∈[0,(Λ₂/2)^𝑝−1).
Then (1)-(2) has at least one positive solution 𝑢 such that 𝑟≤‖𝑢‖≤𝑅.

Proof. By (𝐴₅), for 𝜀=𝜆−(2Λ₁/𝛽)^𝑝−1, there exists a sufficiently small 𝑟>0, such that 𝑓(𝑡,𝑢)𝑢^𝑝−1≥𝜆−𝜀=2Λ₁𝛽^𝑝−1,0<𝑢<𝑟,𝛽≤𝑡≤1. (47) Thus, when 𝛽𝑟≤𝑢≤𝑟, we have 𝑓(𝑡,𝑢)≥2Λ₁𝛽^𝑝−1𝑢^𝑝−1≥2Λ₁𝑟^𝑝−1. (48) Let 𝑚=2Λ₁∈(Λ₁,∞). Then by (48), condition (𝐴₁) holds.
By (𝐴₆), for 𝜀=(Λ₂/2)^𝑝−1−𝜓, there exist a sufficiently large 𝐻₂≠𝑟 such that 𝑓(𝑡,𝑢)𝑢^𝑝−1Λ≤𝜓+𝜀=²2^𝑝−1,𝐻₂≤𝑢<∞,0≤𝑡≤1. (49) We consider the following two cases.
(a) Suppose that 𝑓(𝑡,𝑢) is unbounded, then we know from (𝐻1) that there is a 𝑅≠𝑟(>𝐻₂) such that 𝑓(𝑡,𝑢)≤𝑓(𝑡,𝑅),0≤𝑢≤𝑅,0≤𝑡≤1. (50) Since 𝑅>𝐻₂, then from (49) and (50) we have 𝑓()()≤Λ𝑡,𝑢≤𝑓𝑡,𝑅²2𝑅^𝑝−1,𝐻₂≤𝑢≤𝑅,0≤𝑡≤1. (51) Letting 𝑀=(Λ₂/2)∈(0,Λ₂), we have 𝑓(𝑡,𝑢)≤(𝑀𝑅)^𝑝−1,0≤𝑢≤𝑅,0≤𝑡≤1. (52) Thus, (𝐴₂) holds.
(b) Suppose that 𝑓(𝑡,𝑢) is bounded, say 𝑓(𝑡,𝑢)≤𝐿^𝑝−1,0≤𝑢<∞,0≤𝑡≤1. (53) In this case, taking sufficiently large 𝑅≥(2/Λ₂)𝐿, then letting 𝑀=Λ₂/2∈(0,Λ₂), we have 𝑓(𝑡,𝑢)≤𝐿^𝑝−1≤Λ₂2𝑅^𝑝−1≤(𝑀𝑅)^𝑝−1,0≤𝑢≤𝑅,0≤𝑡≤1. (54) Thus, (𝐴₂) holds.
Hence, from Theorem 11 the desired result holds.

4. Multiple Solutions

To show the existence of multiple solutions we will use the Leggett-Williams fixed point theorem [11]. To this end define the following subsets of a cone 𝐾 as 𝐾_𝑐={‖𝑢‖},𝐾=(𝑢),‖𝑢‖.𝑢∈𝐾∶<𝑐𝜓,𝑏,𝑑𝑢∈𝐾∶𝑏≤𝜓≤𝑑 (55)

Definition 14. A map 𝛼∶𝐾→[0,+∞) is said to be a nonnegative continuous concave functional on a cone 𝐾 of a real Banach space 𝐸 if 𝛼 is continuous and 𝛼𝑡𝑥+(1−𝑡)𝑦≥𝑡𝛼(𝑥)+(1−𝑡)𝛼𝑦 (56)
for all 𝑥,𝑦∈𝐾 and 𝑡∈[0,1].

Theorem 15 (see [11]). Suppose 𝑇∶𝐾_𝑐→𝐾_𝑐 is completely continuous and suppose that there exists a concave positive functional 𝜓 on 𝐾 such that 𝜓(𝑢)≤‖𝑢‖ for 𝑢∈𝐾_𝑐. Suppose that there exist constants 0<𝑎<𝑏<𝑑≤𝑐 such that(B1){𝑢∈𝐾(𝜓,𝑏,𝑑)∶𝜓(𝑢)>𝑏}≠∅ and 𝜓(𝑇𝑢)>𝑏 if 𝑢∈𝐾(𝜓,𝑏,𝑑),(B2)‖𝑇𝑢‖<𝑎 if 𝑢∈𝐾_𝑎,(B3)𝜓(𝑇𝑢)>𝑏 for 𝑢∈𝐾(𝜓,𝑏,𝑐) with ‖𝑇𝑢‖>𝑑.
Then, 𝑇 has at least three fixed points 𝑢₁, 𝑢₂, and 𝑢₃ such that ‖𝑢₁‖<𝑎, 𝑏<𝜓(𝑢₂) and ‖𝑢₃‖>𝑎 with 𝜓(𝑢₃)<𝑏.

Theorem 16. Suppose that there exist 𝑎, 𝑏, 𝑐 with 0<𝑎<𝛽𝑏<𝑏≤𝑐 such that(C1)𝑓(𝑡,𝑢)<(𝑎Λ₂)^𝑝−1, (𝑡,𝑢)∈[0,1]×[0,𝑎],(C2)𝑓(𝑡,𝑢)>(𝑏Λ₁)^𝑝−1, (𝑡,𝑢)∈[𝛽,1]×[𝛽𝑏,𝑏],(C3)𝑓(𝑡,𝑢)<(𝑐Λ₂)^𝑝−1, (𝑡,𝑢)∈[0,1]×[0,𝑐].
Then (1)-(2) has at least three positive solutions.

Proof. By Lemma 10, 𝑇∶𝐾→𝐾 is completely continuous.
Let 𝜓(𝑢)=min_{𝛽≤𝑡≤1}𝑢(𝑡). (57) It is obvious that 𝜓 is a nonnegative continuous concave functional on 𝐾 with 𝜓(𝑢)≤‖𝑢‖, for 𝑢∈𝐾_𝑐. Now we will show that the conditions of Theorem 15 are satisfied. For 𝑢∈𝐾_𝑐, then ‖𝑢‖≤𝑐. For 𝑡∈[0,1] by (27), (29), and (𝐶3), one has 𝑇𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢)𝑑𝜏𝑑𝑠+𝛾𝑡×1−𝛾₁₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓()<𝜏,𝑢𝑑𝜏𝑑𝑠₁₀𝐺₁(𝑡,𝑠)𝜑_𝑞¹₀𝑎(𝜏)𝑐Λ₂^𝑝−1𝛾𝑑𝜏𝑑𝑠+×1−𝛾₁₀𝐺₂𝜂,𝑠𝜑_𝑞¹₀𝑎(𝜏)𝑐Λ₂^𝑝−1<𝑑𝜏𝑑𝑠𝑐Λ₂Γ(𝛼)𝜑1−𝛾^𝑞1₀𝑎(𝜏)𝑑𝜏=𝑐. (58) This implies 𝑇∶𝐾_𝑐→𝐾_𝑐. By the same method, if 𝑢∈𝐾_𝑎, then we can get ‖𝑇𝑢‖<𝑎 and therefore (𝐵2) is satisfied. Next, we assert that {𝑢∈𝐾(𝜓,𝛽𝑏,𝑏)∶𝜓(𝑢)>𝛽𝑏}≠𝜙 and 𝜓(𝑇𝑢)>𝛽𝑏 for all 𝑢∈𝐾(𝜓,𝛽𝑏,𝑏). In fact, the constant function 𝛽𝑏+𝑏2∈𝑢∈𝐾𝜓,𝛽𝑏,𝑏∶𝜓(𝑢)>𝛽𝑏. (59) On the other hand, for 𝑢∈𝐾(𝜓,𝛽𝑏,𝑏), we have 𝛽𝑏≤𝜓(𝑢)=min_{𝛽≤𝑡≤1}𝑢(𝑡)≤𝑢(𝑡)≤‖𝑢‖≤𝑏,∀𝑡∈𝛽,1. (60) Thus, in view of (27), Lemma 7, and (𝐶2), one has 𝜓(𝑇𝑢)=min_{𝛽≤𝑡≤1}¹₀𝐺₁(𝑡,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝑑𝑠+𝛾𝑡×1−𝛾₁₀𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏≥𝜏,𝑢))𝑑𝜏𝑑𝑠_1𝛽𝛽𝐺₁(1,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓((𝜏+𝜏,𝑢))𝑑𝜏𝑑𝑠𝛽𝛾1−𝛾¹_𝛽𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏𝑑𝑠>𝑏Λ₁𝛽¹_𝛽𝐺₁(1,𝑠)𝜑_𝑞^𝑠₀𝑎(𝜏)+𝛾𝑑𝜏𝑑𝑠1−𝛾¹_𝛽𝐺₂𝜂,𝑠𝜑_𝑞^𝑠₀𝑎(𝜏)𝑑𝜏𝑑𝑠=𝑏>𝛽𝑏 (61) as required and therefore (𝐵1) is satisfied.
Finally, we assert that if 𝑢∈𝐾(𝜓,𝛽𝑏,𝑐) with ‖𝑇𝑢‖>𝑏 then 𝜓(𝑢)>𝛽𝑏. To see this, suppose that 𝑢∈𝐾(𝜓,𝛽𝑏,𝑐) and ‖𝑇𝑢‖>𝑏, then it follows from Lemma 10 that 𝜓(𝑇𝑢)=min_{𝛽≤𝑡≤1}(𝑇𝑢)(𝑡)≥𝛽‖𝑇𝑢‖>𝛽𝑏. (62) Thus, (𝐵3) is satisfied.
Therefore, by the conclusion of Theorem 15, the operator 𝑇 has at least three fixed points. This implies that (1)-(2) has at least three solutions.

5. Examples

Example 1. Consider the boundary value problem with 𝑝-Laplacian: 𝜑_𝑝𝐷₀^5/2₊𝑢(𝑡)^′+14𝑡^−1/2𝑎𝑢^1/2𝑒^2𝑢𝑏+𝑒^𝑢+𝑒^2𝑢𝐷=0,0<𝑡<1,_05/2+𝑢(0)=𝑢(0)=𝑢^′′(0)=0,𝑢^′(1)=89𝑢^′12, (63) where 𝛼=5/2, 𝜂=1/2, 𝛾=8/9, 𝛽=1/2, 𝑝=3/2, 𝑞=3, 𝑓(𝑡,𝑢)=𝑎𝑢^1/2𝑒^2𝑢/(𝑏+𝑒^𝑢+𝑒^2𝑢), 𝑎(𝑡)=(1/4)𝑡^−1/2, and 𝜑₃(∫¹₀𝑎(𝑡)𝑑𝑡)=1/4, 𝜑₃(∫^𝑠₀𝑎(𝑡)𝑑𝑡)=(1/4)𝑠.
Then 𝑓⁰𝑎=𝜓=𝑏+2,𝑓_∞=𝜆=𝑎. (64) Next, Λ₁=12¹_1/2𝐺₁()11,𝑠4𝑠𝑑𝑠+8¹_1/2𝐺₂121,𝑠4𝑠𝑑𝑠⁻¹=9352𝜋⁻¹≈4.873999426. (65) Then, for 𝑎=5 we have 𝑓_∞∈((2Λ₁/𝛽)^𝑝−1,∞), so condition (𝐴₃) holds.
Now, Consider Λ₂=√𝜋3≈0.5908179503. (66) Then, for 𝑎=5 and 𝑏=8 we have 𝑓⁰∈[0,(Λ₂/2)^𝑝−1), so condition (𝐴₄) holds. Therefore, by Corollary 12, (63) has at least one positive solution.

Example 2. Consider the boundary value problem with 𝑝-Laplacian: 𝜑_𝑝𝐷₀^5/2₊𝑢(𝑡)^′+14𝑡^−1/2𝑓()𝐷𝑡,𝑢=0,0<𝑡<1,_05/2+𝑢(0)=𝑢(0)=𝑢^′′(0)=0,𝑢^′(1)=89𝑢^′12, (67) where 𝛼=5/2, 𝜂=1/2, 𝛾=8/9, 𝛽=1/2, 𝑝=3/2, 𝑞=3, 𝑎(𝑡)=(1/4)𝑡^−1/2, 𝜑₃(∫¹₀𝑎(𝑡)𝑑𝑡)=1/4, and 𝜑₃(∫^𝑠₀𝑎(𝑡)𝑑𝑡)=(1/4)𝑠.
Let 𝑓()=𝑡𝑡,𝑢40+14𝑢²𝑡,𝑢≤113+40+𝑢^1/4,𝑢>1. (68) By Example 1, we have Λ₁≈4.873999426 and Λ₂≈0.5908179503.
Choosing 𝑎=1/14, 𝑏=18, and 𝑐=1296, then 𝑓()=𝑡𝑡,𝑢40+14𝑢²≈0.096…<𝑎Λ₂^1/2()∈[]×1≈0.20…,𝑡,𝑢0,10,,𝑓()=𝑡14𝑡,𝑢40+13+𝑢^1/4≈14.7…>9Λ₁^1/2()∈1≈9.3…,𝑡,𝑢2×[],𝑓()=𝑡,19,18𝑡,𝑢40+13+𝑢^1/4≈19…<1296Λ₂^1/2()∈[]×[].≈27…,𝑡,𝑢0,10,1296 (69) Then the conditions (𝐶1)--(𝐶3) are satisfied. Therefore, it follows from Theorem 16 that (67) has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ such that 𝑢₁<1𝑢14,18<𝜓₂,𝑢₃>114with𝜓𝑢₃<18. (70)

Acknowledgment

Francisco J. Torres was partially supported by DIUDA 221231, Universidad de Atacama.