Exact Multiplicity of Solutions for a Class of Singular Generalized One-Dimensional 𝑝-Laplacian Problem

Abstract

We describe the existence of positive solutions for a class of singular generalized one-dimensional 𝑝-Laplacian problem. By applying the related fixed point theory in cone, some new and general results on the existence of positive solutions to the singular generalized 𝑝-Laplacian problem are obtained. Note that the nonlinear term 𝑓 involves the first-order derivative explicitly.

1. Introduction

Recently, increasing attention is paid to question of positive solution for singular boundary value problems [1–11]. One notices that the singular boundary value problems for ordinary differential equation describe many phenomena in applied mathematics and physical science, which can be found in the theory of nonlinear diffusion generated by nonlinear sources and in the thermal ignition of gases [12–17]. Moreover, there are excellent results of nonsingular problem, see [18–21] and the references therein.

Moreover, the nonlocal 𝑝-Laplacian problems for ordinary differential equation have been studied extensively. There are many papers dealing with the existence of positive solutions for the nonlocal 𝑝-Laplacian boundary value problem, in which the nonlinear term 𝑓 is independent of the first-order derivative with different boundary conditions. For example, Ma et al. [22] established some existence results of positive solutions for the problem 𝜑_𝑝𝑢^′′(𝑡)𝑓()(),𝑢+𝑞𝑡,𝑢=0,𝑡∈0,1_′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑢^′𝜉_𝑖,𝑢(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑢𝜉_𝑖. (1) The main tool is the monotone iterative technique. By applications of fixed point index theory, Cheung and Ren [23] investigated two classes of quasilinear nonlocal problems: 𝜑_𝑝𝑢^′′(𝑡)𝑓()𝑢+ℎ𝑡,𝑢=0,0<𝑡<1,_′(0)=0,𝑢(1)=^𝑚−2_𝑖=1𝛼_𝑖𝑢𝜉_𝑖,𝑢(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑢𝜉_𝑖,𝑢^′(1)=0. (2) Sufficient conditions for the existence of twin positive solutions are established. In [16], Feng and Ge studied nonlocal problem for the one-dimensional 𝑝-Laplacian 𝜑_𝑝𝑢^′′+𝑞(𝑡)𝑓𝑡,𝑢,𝑢^′(),𝑢(0)(1)==0,𝑡∈0,1=0,𝑢^𝑚−2_𝑖=1𝑎_𝑖𝑢𝜉_𝑖. (3) They obtained sufficient conditions for the existence of at least three solutions to the above problem by using the Avery and Peterson fixed point theorem.

However, there are not many concerning the boundary value problems, in which a generalized 𝑝-Laplacian equation with a nonlocal boundary conditions. The motivation for the present work stems from many recent investigations in [16, 23–25] and references therein. Our purpose of this paper is to establish some sufficient conditions for the existence of triple positive solutions to a class of a singular generalized one-dimensional 𝑝-Laplacian problem 𝑟(𝑡)𝜑_𝑝𝑢^′′+ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)(),𝛼=0,𝑡∈𝑎,𝑏₁𝑢(𝑎)−𝛽₁𝑢^′(𝑎)=^𝑚−2_𝑖=1𝛾_𝑖𝑢𝜉_𝑖,𝑢^′(𝑏)=0, (𝒫) where 𝜑_𝑝(𝑢) is 𝑝-Laplacian operator; that is, 𝜑_𝑝(𝑢)=|𝑢|^𝑝−2𝑢, 𝑝>1, (𝜑_𝑝)⁻¹(𝑢)=𝜑_𝑞(𝑢), 1/𝑝+1/𝑞=1, and 𝛼₁, 𝛽₁, 𝛾_𝑖, 𝜉_𝑖, 𝑟, ℎ, 𝑓 satisfy(𝐴₁)𝛼₁,𝛽₁∈[0,∞), 𝛾_𝑖∈[0,∞) satisfy 𝛼₁>∑^𝑚−2_𝑖=1𝛾_𝑖, 𝑎<𝜉₁<⋯<𝜉_𝑚−2<𝑏(𝑚>3),(𝐴₂)𝑟(𝑡)∈𝐶¹(𝑎,𝑏), 𝑟(𝑡) is positive and nondecreasing on [𝑎,𝑏],(𝐴₃)ℎ∈𝐿¹((𝑎,𝑏),[0,∞)) may be singular at 𝑡=0 and/or 𝑡=1, and 0<∫_Γℎ(𝑡)𝑑𝑡<∞, where Γ⊆[𝑎,𝑏],(𝐴₄)𝑓∶[𝑎,𝑏]×[0,∞)×(−∞,∞)→[0,∞) is an 𝐿¹-Carathédory function; that is, for each (𝑡,𝑥,𝑦)∈[𝑎,𝑏]×[0,∞)×(−∞,∞), the mapping 𝑡→𝑓(𝑡,𝑥,𝑦) is Lebesgue measurable on [𝑎,𝑏]; for a.e. 𝑡∈[𝑎,𝑏], the mapping (𝑡,𝑥,𝑦)→𝑓(𝑡,𝑥,𝑦) is continuous on [0,∞)×(−∞,∞).

Under the above assumptions, some new and general results on the existence of multiple positive solutions to singular one-dimensional 𝑝-Laplacian are obtained. Our results develop some results of Cheung and Ren [23] and include and improve the main results of Feng and Ge [16] and Zhang [25].

The rest of the paper is organized as follows. In Section 2, we provide some background material from the theory of cones in the Banach spaces, which are useful later. Some lemmas and criteria for the existence of three positive solutions for one-dimensional 𝑝-Laplacian problems are established in Section 3. Finally, we give an example to illustrate our main results.

2. Preliminaries

Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on a cone 𝐾, 𝛼 a nonnegative continuous concave functional on 𝐾, 𝛽 a nonnegative continuous functional on 𝐾, and 𝑚₁,  𝑚₂,  𝑚₃,  and  𝑚₄ positive numbers. We define the following convex sets and a closed set 𝑃𝛾,𝑚₄=𝑢∈𝐾∶𝛾(𝑢)<𝑚₄,𝑃𝛾,𝛼,𝑚₂,𝑚₄=𝑢∈𝐾∶𝑚₂≤𝛼(𝑢),𝛾(𝑢)≤𝑚₄,𝑃𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄=𝑢∈𝐾∶𝑚₂≤𝛼(𝑢),𝜃(𝑢)≤𝑚₃,𝛾(𝑢)≤𝑚₄,𝑄𝛾,𝛽,𝑚₁,𝑚₄=𝑢∈𝐾∶𝑚₁≤𝛽(𝑢),𝛾(𝑢)≤𝑚₄. (4)

Now we state the fixed point theorem due to Avery and Peterson [26].

Lemma 1. Let 𝐾 be a cone in a real Banach space 𝐸. Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on 𝐾, 𝛼 a nonnegative continuous concave functional on 𝐾, and 𝛽 a nonnegative continuous functional on 𝐾 satisfying 𝛽(𝜌𝑢)≤𝜌𝛽(𝑢) for 0≤𝜌≤1, such that, for some positive numbers 𝜖 and 𝑚₄, 𝛼(𝑢)≤𝛽(𝑢),‖𝑢‖≤𝜖𝛾(𝑢)∀𝑢∈𝑃𝛾,𝑚₄. (5) Suppose that 𝑇∶𝑃(𝛾,𝑚₄)→𝑃(𝛾,𝑚₄) is completely continuous and there are positive numbers 𝑚₁, 𝑚₂, and 𝑚₃ with 𝑚₁<𝑚₂ such that(𝐵₁){𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄)∶𝛼(𝑢)>𝑚₂}≠∅, 𝛼(𝑇𝑢)>𝑚₂ for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₃,𝑚₄),(𝐵₂)𝛼(𝑇𝑢)>𝑚₂ for 𝑢∈𝑃(𝛾,𝛼,𝑚₂,𝑚₄) with 𝜃(𝑇𝑢)>𝑚₃,(𝐵₃)0∉𝑄(𝛾,𝛽,𝑚₁,𝑚₄) and 𝛽(𝑇𝑢)<𝑚₁ for 𝑢∈𝑄(𝛾,𝛽,𝑚₁,𝑚₄) with 𝛽(𝑢)=𝑚₁. Then 𝑇 has at least three fixed points 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝛾,𝑚₄) such that 𝛾𝑢_𝑖≤𝑚₄(),𝑚𝑖=1,2,3₂<𝛼𝑢₁,𝑚₁<𝛽𝑢₂𝑤𝑖𝑡ℎ𝛼𝑢₂<𝑚₂,𝛽𝑢₃<𝑚₁. (6)

Denote that 𝐿⁺[𝑎,𝑏]={𝑣∈𝐿[𝑎,𝑏]∶𝑣(𝑡)≥0,𝑡∈[𝑎,𝑏]}.

Lemma 2. Assume that (𝐴₁) and (𝐴₂) hold. If 𝑔∈𝐿⁺[𝑎,𝑏], then the boundary value problem 𝑟(𝑡)𝜑_𝑝𝑢^′′+𝑔(𝑡)=0,𝑡∈(𝑎,𝑏),𝛼₁𝑢(𝑎)−𝛽₁𝑢^′(𝑎)=^𝑚−2_𝑖=1𝛾_𝑖𝑢𝜉_𝑖,𝑢^′(𝑏)=0, (7)(8) has a unique solution 𝑢(𝑡)=^𝑡_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠𝑔(𝜏)1𝑑𝜏𝑑𝑠+𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎𝑔(𝑠)+𝑑𝑠_{𝑚−2𝑖=1}𝛾_𝑖^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠𝑔(𝜏)𝑑𝜏𝑑𝑠. (9)

Proof. For a.e. 𝑡∈[𝑎,𝑏], integrating (7) from 𝑡 to 𝑏, in view of (8), we have 𝑢^′(𝑡)=𝜑_𝑞1𝑟(𝑡)^𝑏_𝑡𝑔(𝑠)𝑑𝑠. (10) Integrating this from 𝑎 to 𝑡 yields 𝑢(𝑡)=𝑢(𝑎)+^𝑡_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠𝑔(𝜏)𝑑𝜏𝑑𝑠. (11) Again, it follows from (8) that 𝑢(𝑎)=1𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎𝑔(𝑠)+𝑑𝑠_{𝑚−2𝑖=1}𝛾_𝑖^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠𝑔(𝜏)𝑑𝜏𝑑𝑠. (12) The proof is complete.

Lemma 3. Assume that (𝐴₁) and (𝐴₂) hold. If 𝑔∈𝐿⁺[𝑎,𝑏], then the unique solution of the problems (7) and (8) satisfies 𝑢(𝑡)≥0,min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢(𝑡)≥𝜏₁max_{[]𝑡∈𝑎,𝑏}𝑢(𝑡), (13) where 𝜏₁=𝛽₁/(𝛽₁+(𝑏−𝑎)(𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖)),𝜎∈(𝑎,(𝑎+𝑏)/2).

Proof.  The proof is similar to Lemma  3.6 in [25], we omit it.

Let 𝐸 be the Banach space 𝐶¹[𝑎,𝑏] with the norm ‖𝑢‖=max{‖𝑢‖₀,‖𝑢^′‖₀}, where ‖𝑢‖₀=max_0≤𝑡≤1|𝑢(𝑡)|. Define a cone 𝐾⊂𝐸 by 𝐾=𝑢∈𝐸∶𝑢(𝑡)≥0for[],𝑢𝑡∈𝑎,𝑏_′′(𝑡)≤0,𝑢^′(𝑡)≥0for(),𝛼𝑡∈𝑎,𝑏₁𝑢(𝑎)−𝛽₁𝑢^′(𝑎)=^𝑚−2_𝑖=1𝛾_𝑖𝑢𝜉_𝑖,𝑢^′(𝑏)=0,min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢(𝑡)≥𝜏₁max_{[]𝑡∈𝑎,𝑏}𝑢(𝑡). (14) Let the assumptions (𝐴₁) and (𝐴₂) hold. Denote that 𝜆=^𝑏_𝑎𝑏−𝑠𝑟(𝑠)^1/(𝑝−1)1𝑑𝑠+𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝑏−𝑎𝑟(𝑎)^1/(𝑝−1)+^𝑚−2_𝑖=1𝛾_𝑖^𝜉𝑖_𝑎𝑏−𝑠𝑟(𝑠)^1/(𝑝−1)𝑑𝑠. (15) Evidently, 𝜆≥0. We take 𝑢₀(𝑡)=^𝑡_𝑎𝑚𝑏−𝑠𝑟(𝑠)^1/(𝑝−1)1𝑑𝑠+𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝑚𝑏−𝑎𝑟(𝑎)^1/(𝑝−1)+^𝑚−2_𝑖=1𝛾_𝑖^𝜉𝑖_𝑎𝑚𝑏−𝑠𝑟(𝑠)^1/(𝑝−1)𝑑𝑠, (16) where 𝑚 satisfies (𝑚)^1/(𝑝−1)∶=𝑚₂/𝜏²₁𝜆, 𝑚₂>0 is constant. It is not difficult to check that 𝑢₀(𝑡)≥0 for a.e. 𝑡∈[𝑎,𝑏], 𝑢^′′(𝑡)≤0, 𝑢^′(𝑡)≥0 for a.e. 𝑡∈(𝑎,𝑏), and 𝑢₀(𝑡) satisfies 𝛼₁𝑢₀(𝑎)−𝛽₁𝑢^′₀(𝑎)=∑^𝑚−2_𝑖=1𝛾_𝑖𝑢₀(𝜉_𝑖), 𝑢^′₀(𝑏)=0. That is to say 𝑢₀∈𝐾. Hence, 𝐾⧵{0}≠∅.

To obtain the existence of solutions for the problem (𝒫), the following priori estimate is needful.

Lemma 4. Assume that (𝐴₁) holds. If 𝑢∈𝐾, then max_{[]𝑡∈𝑎,𝑏}𝑢(𝑡)≤𝜏₂max_{[]𝑡∈𝑎,𝑏}𝑢^′(𝑡), (17) where 𝜏₂=(𝛽₁+𝛼₁(𝑏−𝑎))/(𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖).

Proof. The proof is similar to Lemma  3.1 in [21], we omit it.

Let the nonnegative continuous convex functionals 𝛾 and 𝜃, the nonnegative continuous concave functional 𝛼, and the nonnegative continuous functional 𝛽 be defined on cone 𝐾 by 𝛼(𝑢)=min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢(𝑡),𝛽(𝑢)=𝜃(𝑢)=max_{[]𝑡∈𝑎,𝑏}𝑢(𝑡),𝛾(𝑢)=max_{𝑡∈[𝑎,𝑏]}𝑢^′(𝑡). (18)

In the view of assumption (𝐴₃), that is, 0<∫^𝑏_𝑎ℎ(𝜏)𝑑𝜏<∞ and (𝐴₃). Thus, we can define an operator 𝑇 by (𝑇𝑢)(𝑡)∶=^𝑡_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝑑𝑠𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝑑𝜏_{𝑚−2𝑖=1}𝛾_𝑖^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)×𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠fora.e.𝑡∈[𝑎,𝑏]. (19)

Lemma 5. Assume that (𝐴₁)--(𝐴₄) hold. Then 𝑇∶𝐾→𝐾 is completely continuous.

Proof. For any 𝑢∈𝐾, by the definition of 𝑇𝑢, we see that 𝑇𝑢∈𝐶¹[𝑎,𝑏], (𝜑_𝑝((𝑇𝑢)^′))^′∈𝐿¹[𝑎,𝑏]. From (𝐴₁)--(𝐴₄) and Lemma 3, we have (𝑇𝑢)(𝑡)≥0, (𝜑_𝑝((𝑇𝑢)^′))^′≤0 implies that (𝑇𝑢)(𝑡) is concave on [𝑎,𝑏], −(𝜑_𝑝((𝑇𝑢)^′))^′=ℎ(𝑡)𝑓(𝑡,(𝑇𝑢)(𝑡),(𝑇𝑢)^′(𝑡))≥0, 𝛼₁𝑇𝑢(𝑎)−𝛽₁𝑇𝑢^′(𝑎)=∑^𝑚−2_𝑖=1𝛾_𝑖𝑇𝑢(𝜉_𝑖), and 𝑇𝑢^′(𝑏)=0. So we can conclude that 𝑇𝑢∈𝐾. In what follows, we prove that 𝑇 is compact. Let 𝑅>0, there exists a set Ω such that Ω⊂{𝑢∈𝐾∶‖𝑢^′‖₀≤𝑅}. For any 𝑢∈Ω, the assumptions imply that 0≤^𝑏_𝑎ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑑𝑡≤max_{𝑡,𝑢,𝑢′∈[]𝑎,𝑏×[0,𝜏2𝑅]×[−𝑅,𝑅]}𝑓𝑡,𝑢,𝑢^′𝑏_𝑎ℎ(𝑡)𝑑𝑡<∞. (20) We take the arguments to show that the operator 𝑇 is completely continuous. In view of Lemma 4 and the continuity of 𝑓 in 𝑢 and 𝑢^′, assume that 𝑢_𝑛,𝑢₀∈𝐾 satisfy ‖𝑢_𝑛−𝑢₀‖→0(𝑛→∞). Moreover, 𝜑_𝑞(𝜐) is an increasing and continuously function, and we obtain that 𝑇𝑢_𝑛−𝑇𝑢₀₀=max_{𝑎≤𝑡≤𝑏}𝑇𝑢_𝑛(𝑡)−𝑇𝑢₀(𝑡≤_𝑏𝑎)𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢_𝑛(𝜏),𝑢^′_𝑛(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢₀(𝜏),𝑢^′₀(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢_𝑛(𝜏),𝑢^′_𝑛(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢₀(𝜏),𝑢^′₀(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢_𝑛(𝜏),𝑢^′_𝑛(𝜏)𝑑𝜏𝑑𝑠−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)×𝑓𝜏,𝑢₀(𝜏),𝑢^′₀(𝜏)(),𝑑𝜏𝑑𝑠⟶0𝑛⟶∞𝑇𝑢_𝑛^′−𝑇𝑢₀^′₀=max_{𝑎≤𝑡≤𝑏}𝑇𝑢_𝑛^′(𝑡)−𝑇𝑢₀^′(𝑡≤𝜑_𝑞)1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢_𝑛(𝜏),𝑢^′_𝑛(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢₀(𝜏),𝑢^′₀(𝜏)().𝑑𝜏⟶0𝑛⟶∞ (21) Therefore, 𝑇𝑢_𝑛−𝑇𝑢₀⟶0(𝑛⟶∞). (22) This means that the operator 𝑇∶𝐾→𝐾 is continuous.
Since ℎ(𝑡) may be singular at 𝑡=𝑎 and/or 𝑡=𝑏, choose two sequences {𝜇_𝑛}^∞_𝑛=1, {𝜈_𝑛}^∞_𝑛=1⊂(𝑎,𝑏) satisfying 𝜇_𝑛≤𝜈_𝑛 for any 𝑛≥1, such that 𝜇_𝑛→𝑎 and 𝛼_𝑛→𝑏 as 𝑛→∞, respectively. Define ℎ_𝑛(𝑡)=inf_{𝑎≤𝑡≤𝜇𝑛}ℎ(𝑡),𝑎≤𝑡≤𝜇_𝑛,ℎ(𝑡),𝜇_𝑛≤𝑡≤𝜈_𝑛,inf_{𝜈𝑛≤𝑡≤𝑏}ℎ(𝑡),𝜈_𝑛≤𝑡≤𝑏, (23) and an operator sequence 𝑇_𝑛 by 𝑇_𝑛𝑢(𝑡)=^𝑡_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝑑𝑠𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝛽₁𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝑑𝜏_{𝑚−2𝑖=1}𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)×𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠fora.e.𝑡∈[𝑎,𝑏]. (24) Clearly, ℎ_𝑛(𝑡)∶[𝑎,𝑏]→[0,∞) is a piecewise continuous function, and the operator 𝑇_𝑛∶𝐾→𝐾 is well defined. Further, we can see that 𝑇_𝑛∶𝐾→𝐾 is completely continuous.
Let 𝑟>0, 𝔹_𝑟∶={𝑢∈𝐾∶‖𝑢^′‖₀≤𝑟} and 𝑀_𝑟=max{𝑓(𝑡,𝑢,𝑢^′)∶𝑢∈𝐾,𝑡∈[𝑎,𝑏], 𝑢∈[0,𝜏₂𝑟],  𝑢^′∈[−𝑟,𝑟]}. We will prove that 𝑇_𝑛 approaches 𝑇 uniformly on 𝔹_𝑟. From the absolute continuity of integral, we obtain lim_{𝑛→∞Γ(𝑛)}𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠=0, (25) where Γ(𝑛)=[𝑎,𝜇_𝑛]∪[𝜈_𝑛,𝑏]. For any 𝑢∈𝔹_𝑟 and a.e. 𝑡∈[𝑎,𝜇_𝑛], one has that 𝑇_𝑛𝑢−𝑇𝑢₀=max_{𝑎≤𝑡≤𝜇𝑛}𝑇_𝑛𝑢(𝑡)(𝑡≤−𝑇𝑢)_𝜇𝑛𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)≤𝑑𝜏𝑑𝑠_Γ(𝑛)𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠−𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠. (26) For 𝑢∈𝔹_𝑟 and a.e. 𝑡∈[𝜈_𝑛,𝑏], we have the same result. It is easy to see that, for any 𝑢∈𝔹_𝑟 and a.e. 𝑡∈[𝜇_𝑛,𝜈_𝑛], there is ‖𝑇_𝑛𝑢−𝑇𝑢‖₀=0. Similarly, we can obtain that, for any 𝑢∈𝔹_𝑟 and a.e. 𝑡∈[𝑎,𝜇_𝑛], 𝑡∈[𝜇_𝑛,𝜈_𝑛], 𝑡∈[𝜈_𝑛,𝑏], respectively, 𝑇_𝑛𝑢^′−(𝑇𝑢)^′₀=max_𝑡𝑇_𝑛𝑢^′(𝑡)−(𝑇𝑢)^′(𝑡≤𝜑_𝑞)1𝑟(𝑎)^𝑏_𝑎ℎ_𝑛(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏−𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏. (27) In view of the continuity of function 𝜑_𝑞, from the above arguments, we have 𝑇_𝑛𝑢−𝑇𝑢⟶0(𝑛⟶∞). (28) That is to say, the sequence 𝑇_𝑛 is uniformly an approximate 𝑇 on any bounded subset of 𝐾. Applying the Arzela-Ascoli lemma to the operator 𝑇, we can conclude that 𝑇Ω is relatively compact; that is, 𝑇 is completely continuous.

3. Main Results

We are ready to apply the Avery and Peterson fixed point theorem to the operator 𝑇 to give sufficient conditions for the existence of at least three positive solutions to the problem (𝒫). For convenience, we introduce the following notation: 1𝐿=𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝛼𝑑𝜏,𝑀=¹−∑^𝑚−2_𝑖=1𝛾_𝑖𝛽₁,𝛽𝑁=𝑏−𝑎+¹+∑^𝑚−2_𝑖=1𝛾_𝑖𝜉_𝑖−𝑎𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖. (29)

Theorem 6. Suppose (𝐴₁)--(𝐴₄) hold and 𝑓(𝑡,0,0)≢0 for a.e. 𝑡∈[𝑎,𝑏]. If there exist positive numbers 𝑚₁, 𝑚₂, and 𝑚₄ with 𝑚₁<𝑚₂≤𝜏²₁𝜆((𝑏−𝑎)/𝑟(𝑎))^{−1/(𝑝−1)}𝑚₄ such that the following conditions are satisfied:(𝐶₁)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿)𝜑_𝑝(𝑚₄), (𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝜏₂𝑚₄]×[−𝑚₄,𝑚₄],(𝐶₂)𝑓(𝑡,𝜇,𝜈)>(1/𝐿)𝜑_𝑝(𝑚₂/𝜏₁𝑀), (𝑡,𝜇,𝜈)∈[𝜎,𝑏−𝜎]×[𝑚₂,𝑚₂𝜏₁⁻²]×[−𝑚₄,𝑚₄],(𝐶₃)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿)𝜑_𝑝(𝑚₁/𝑁), (𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝑚₁]×[−𝑚₄,𝑚₄], then the problem (𝒫) has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ satisfying max_{𝑡∈[𝑎,𝑏]}𝑢^′_𝑖(𝑡)≤𝑚₄𝑚𝑓𝑜𝑟𝑖=1,2,3,₂<min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢₁(𝑡),𝑚₁<max_{[]𝑡∈𝑎,𝑏}𝑢₂(𝑡)𝑤𝑖𝑡ℎmin_{[]𝑡∈𝜎,𝑏−𝜎}𝑢₂(𝑡)<𝑚₂,max_{[]𝑡∈𝑎,𝑏}𝑢₃(𝑡)<𝑚₁. (30)

Proof. It is known that boundary value problem (𝒫) has a solution 𝑢 if and only if 𝑢 solves the operator equation 𝑢=𝑇𝑢. By the definition of operator 𝑇, from Lemma 4 and the concavity of 𝑢, the functionals defined above satisfy 𝜎(𝑏−𝜎)𝜃(𝑢)≤𝛼(𝑢)≤𝜃(𝑢)=𝛽(𝑢), ‖𝑢‖=max{𝜃(𝑢),𝛾(𝑢)}≤max{1,𝜏₂}𝛾(𝑢) for any 𝑢∈𝑃(𝛾,𝑚₄)⊂𝐾. We also note that 𝛽(𝜌𝑢)≤𝜌𝛽(𝑢) for 0≤𝜌≤1, which suffices to show that the conditions of Lemma 1 hold with respect to 𝑇. First of all, we show that if the assumption (𝐶₁) is satisfied, then 𝑇∶𝑃(𝛾,𝑚₄)⟶𝑃(𝛾,𝑚₄). (31) In fact, for 𝑢∈𝑃(𝛾,𝑚₄), there is 𝛾(𝑢)=max_{𝑡∈[𝑎,𝑏]}|𝑢^′(𝑡)|≤𝑚₄. With Lemma 4, there is max_{𝑡∈[𝑎,𝑏]}𝑢(𝑡)≤𝜏₂max_{𝑡∈[𝑎,𝑏]}|𝑢^′(𝑡)|≤𝜏₂𝑚₄, and the assumption (𝐶₁) implies that 𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡))≤(1/𝐿)𝜑_𝑝(𝑚₄) for a.e. 𝑡∈[𝑎,𝑏]. On the other hand, for 𝑢∈𝐾, we obtain that 𝑇𝑢∈𝐾, and so (𝑇𝑢)(𝑡) is concave on [𝑎,𝑏], and there is max_{𝑡∈[𝑎,𝑏]}|(𝑇𝑢)^′(𝑡)|=|(𝑇𝑢)^′(𝑎)|. By using of 𝑟(𝑡) is positive and monotone increasing on a.e. [𝑎,𝑏], and so 𝛾(𝑇𝑢)=max_{[]𝑡∈𝑎,𝑏}(𝑇𝑢)^′(𝑡)=𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏≤𝜑_𝑞1𝐿𝜑_𝑝𝑚₄1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑑𝜏=𝑚₄. (32) Therefore, (31) is satisfied.
To check if the condition (𝐵₁) in Lemma 1 is fulfilled, we choose 𝑢₀(𝑡), where 𝑢₀(𝑡) is defined by (16). From the hypothesis conditions of the theorem, we have 𝛼𝑢₀=min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢₀(𝑡)=𝑢₀(𝜎)≥𝜏₁(𝑚)^1/(𝑝−1)𝜆≥𝑚₂𝜏₁⁻¹>𝑚₂,𝜃𝑢₀=max_{[]𝑡∈𝑎,𝑏}𝑢₀(𝑡)=𝑢₀(𝑏)=(𝑚)^1/(𝑝−1)𝜆>𝑚₂𝜏₁⁻²>𝑚₂,𝛾𝑢₀=max_{𝑡∈[𝑎,𝑏]}𝑢₀^′(𝑡)=𝑢₀^′(𝑎=)𝑚𝑏−𝑎𝑟(𝑎)^1/(𝑝−1)=𝑚₂𝜏₁⁻²𝜆𝑏−𝑎𝑟(𝑎)^1/(𝑝−1)≤𝑚₄. (33)(34) Equation (33) implies that 𝛼(𝑢₀)≥𝜏₁𝜃(𝑢₀). That is, min_{𝑡∈[𝜎,𝑏−𝜎]}𝑢₀(𝑡)≥𝜏₁max_{𝑡∈[𝑎,𝑏]}𝑢₀(𝑡). So for 𝑢₀∈𝐾, 𝑢₀∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄) and {𝑢₀∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄)∶𝛼(𝑢₀)>𝑚₂}≠∅ hold. Hence, for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄), we see that 𝑚₂≤𝑢(𝑡)≤𝑚₂𝜏₁⁻² and |𝑢^′(𝑡)|≤𝑚₄ for a.e. 𝑡∈[𝜎,𝑏−𝜎]. Hence by the assumption (𝐶₂), one has that 𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡))>(1/𝐿)𝜑_𝑝(𝑚₂𝜏₁⁻¹/𝑀) for a.e. 𝑡∈[𝜎,𝑏−𝜎]. By the definition of the functional 𝛼, we obtain 𝛼(𝑇𝑢)=min_{[]𝑡∈𝜎,𝑏−𝜎}(𝑇𝑢)(𝑡)≥𝜏₁max_{[]𝑡∈𝑎,𝑏}(𝑇𝑢)(𝑡)=𝜏₁(𝑇𝑢)(𝑏)=𝜏₁^𝑏_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖𝜑_𝑞×1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)×𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)≥𝜏𝑑𝜏𝑑𝑠₁𝛽₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖𝜑_𝑞1𝐿𝜑_𝑝𝑚₂𝜏₁⁻¹𝑀1𝑟(𝑎)×^𝑏_𝑎ℎ(𝜏)𝑑𝜏=𝑚₂. (35) Therefore, we get 𝛼(𝑇𝑢)>𝑚₂ for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑚₂,𝑚₂𝜏₁⁻²,𝑚₄).
Next, we show that the condition (𝐵₂) in Lemma 1 holds. In fact, if 𝑢∈𝑃(𝛾,𝛼,𝑚₂,𝑚₄) with 𝜃(𝑇𝑢)>𝑚₂𝜏₁⁻², then 𝛼(𝑇𝑢)=min_{[]𝑡∈𝜎,𝑏−𝜎}(𝑇𝑢)(𝑡)≥𝜏₁max_{[]𝑡∈𝑎,𝑏}(𝑇𝑢)(𝑡)=𝜏₁𝜃(𝑇𝑢)>𝑚₂𝜏₁⁻¹>𝑚₂. (36)
Finally, we assert that the condition (𝐵₃) in Lemma 1 also holds. Since 𝛽(0)=0<𝑚₁, there holds that 0∉𝑄(𝛾,𝛽,𝑚₁,𝑚₄). Assume that 𝑢∈𝑄(𝛾,𝛽,𝑚₁,𝑚₄) with 𝛽(𝑢)=𝑚₁. Then, by the assumption (𝐶₃) we get 𝛽(𝑇𝑢)=max_{[]𝑡∈𝑎,𝑏}(𝑡)=(𝑏)=𝑇𝑢)(𝑇𝑢)(_𝑏𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖𝜑_𝑞×1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑠)^𝑏_𝑠ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠≤^𝑏_𝑎𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖×𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)≤𝑑𝜏𝑑𝑠_𝑏𝑎𝜑_𝑞1𝐿𝜑_𝑝𝑚₁𝑁1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)+𝛽𝑑𝜏𝑑𝑠₁𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖𝜑_𝑞1𝐿𝜑_𝑝𝑚₁𝑁1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)+1𝑑𝜏𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖^𝑚−2_𝑖=1𝛾_𝑖×^𝜉𝑖_𝑎𝜑_𝑞1𝐿𝜑_𝑝𝑚₁𝑁1𝑟(𝑎)^𝑏_𝑎ℎ(𝜏)≤𝑚𝑑𝜏𝑑𝑠₁𝑁𝛽𝑏−𝑎+¹+∑^𝑚−2_𝑖=1𝛾_𝑖𝜉_𝑖−𝑎𝛼₁−∑^𝑚−2_𝑖=1𝛾_𝑖=𝑚₁. (37) Thus, the conditions (𝐵₁)--(𝐵₃) in Lemma 1 hold, and so the problem (𝒫) has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ such that (30) holds.

Similar to the above arguments, we will discuss the problem 𝑟(𝑡)𝜑_𝑝𝑢^′′+ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)(),𝑢=0,𝑡∈𝑎,𝑏_′(𝑎)=0,𝛼₂𝑢(𝑏)+𝛽₂𝑢^′(𝑏)=^𝑚−2_𝑖=1𝛿_𝑖𝑢𝜉_𝑖. 𝒫^∗ Now, we replace the assumption (𝐴₁) with the following:

(𝐴^∗₁) 𝛼₂,𝛽₂∈[0,∞), 𝛿_𝑖∈[0,∞) satisfy 𝛼₂>∑^𝑚−2_𝑖=1𝛿_𝑖, 𝑎<𝜉₁<⋯<𝜉_𝑚−2<𝑏(𝑚>3).

We only give the main results of the problem 𝒫^∗, and the proofs are omitted.

Lemma 7. Assume that (𝐴^∗₁), (𝐴₂), (𝐴₃), and (𝐴₄) hold. Then the problem 𝒫^∗ has a unique solution 𝑢(𝑡)=^𝑏_𝑡𝜑_𝑞1𝑟(𝑠)^𝑠_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝑑𝑠𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖×𝛽₂𝜑_𝑞1𝑟(𝑏)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+𝑑𝜏_{𝑚−2𝑖=1}𝛿_𝑖^𝑏_𝜉𝑖𝜑_𝑞×1𝑟(𝑠)^𝑠_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠. (38)

The cone is defined by 𝐾^∗=𝑢∈𝐸∶𝑢(𝑡)≥0for𝑡∈[𝑎,𝑏],𝑢^′′(𝑡)𝑢≤0,_′(𝑡)≥0for𝑡∈(𝑎,𝑏),𝑢^′(𝑎)𝛼=0,₂𝑢(𝑏)+𝛽₂𝑢^′(𝑏)=^𝑚−2_𝑖=1𝛿_𝑖𝑢𝜉_𝑖andmin_{[]𝑡∈𝜎,𝑏−𝜎}𝑢(𝑡)≥𝜏₁max_{[]𝑡∈𝑎,𝑏}𝑢(𝑡). (39)

Lemma 8. Assume that (𝐴₁) holds. If 𝑢∈𝐾^∗, then min_{[]𝑡∈𝜎,𝑏−𝜎}𝑢(𝑡)≥𝜏₃‖𝑢‖₀,‖𝑢‖₀≤𝜏₄𝑢^′₀, (40) where 𝜏₃=𝛽₂/(𝛽₂+(𝑏−𝑎)(𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖)) and 𝜏₄=(𝛽₂+𝛼₂(𝑏−𝑎))/(𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖).

In what follows, we define integral operators 𝑇 and 𝐿^∗, 𝑀^∗, and 𝑁^∗: (𝑇𝑢)(𝑡)∶=^𝑏_𝑡𝜑_𝑞1𝑟(𝑠)^𝑠_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)+1𝑑𝜏𝑑𝑠𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖×𝛽₂𝜑_𝑞1𝑟(𝑏)^𝑏_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏+^𝑚−2_𝑖=1𝛿_𝑖×^𝑏_𝜉𝑖𝜑_𝑞1𝑟(𝑠)^𝑠_𝑎ℎ(𝜏)𝑓𝜏,𝑢(𝜏),𝑢^′(𝜏)𝑑𝜏𝑑𝑠fora.e[],𝐿.𝑡∈𝑎,𝑏_∗=1𝑟(𝑏)^𝑏_𝑎ℎ(𝜏)𝑀𝑑𝜏,_∗=𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖𝛽₂,𝑁^∗𝛽=𝑏−𝑎+²+∑^𝑚−2_𝑖=1𝛿_𝑖𝑏−𝜉_𝑖𝛼₂−∑^𝑚−2_𝑖=1𝛿_𝑖. (41)

Theorem 9. Suppose that (𝐴^∗₁), (𝐴₂), (𝐴₃), and (𝐴₄) hold and 𝑓(𝑡,0,0)≢0 for a.e. 𝑡∈[𝑎,𝑏]. If there exist positive numbers 𝜆^′, 𝑚₁, 𝑚₂, and 𝑚₄ with 𝑚₁<𝑚₂≤𝜏²₃𝜆^′((𝑏−𝑎)/𝑟(𝑏))^{−1/(𝑝−1)}𝑚₄ such that the following conditions are satisfied:(𝐷₁)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿^∗)𝜑_𝑝(𝑚₄/𝐿^∗),(𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝜏₄𝑚₄]×[−𝑚₄,𝑚₄],(𝐷₂)𝑓(𝑡,𝜇,𝜈)>(1/𝐿^∗)𝜑_𝑝(𝑚₂/𝑀^∗),(𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[𝑚₂,𝑚₂𝜏₃⁻¹]×[−𝑚₄,𝑚₄],(𝐷₃)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿^∗)𝜑_𝑝(𝑚₁/𝑁^∗),(𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝑚₁]×[−𝑚₄,𝑚₄],then the problem 𝒫^∗ has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ satisfying max_{𝑡∈[𝑎,𝑏]}𝑢^′_𝑖(𝑡)≤𝑚₄𝑚𝑓𝑜𝑟𝑖=1,2,3,₂<min_{[]𝑡∈𝑎,𝑏}𝑢₁(𝑡),𝑚₁<max_{[]𝑡∈𝑎,𝑏}𝑢₂(𝑡)𝑤𝑖𝑡ℎmin_{[]𝑡∈𝑎,𝑏}𝑢₂(𝑡)<𝑚₂,max_{[]𝑡∈𝑎,𝑏}𝑢₃(𝑡)<𝑚₁. (42)

From Theorem 6 or Theorem 9, we can see that, when assumptions like (𝐶₁), (𝐶₂), and (𝐶₃) are imposed appropriately on the nonlinear term 𝑓, we also establish the existence results of an arbitrary odd number of positive solutions of the problem (𝒫) or 𝒫^∗. Here, we only present the form of Theorem 6.

Theorem 10. Suppose that (𝐴₁)--(𝐴₄) hold and 𝑓(𝑡,0,0)≢0 for a.e. 𝑡∈[𝑎,𝑏]. If there exist positive numbers 𝜆, 𝑚_1𝑖, 𝑚_2𝑖, and 𝑚_4𝑖𝑖=1,2,…, with 𝑚₁₁<𝑚₂₁≤𝜏²₁𝜆((𝑏−𝑎)/𝑟(𝑎))^1/(𝑝−1)𝑚₄₁<𝑚₁₂<𝑚₂₂≤𝜏²₁𝜆((𝑏−𝑎)/𝑟(𝑎))^1/(𝑝−1)𝑚₄₂<⋯<𝑚_1𝑛<𝑚_2𝑛≤𝜏²₁𝜆((𝑏−𝑎)/𝑟(𝑎))^1/(𝑝−1)𝑚_4𝑛, (43) such that the following conditions are satisfied:(𝐶₁)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿)𝜑_𝑝(𝑚_4𝑖), (𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝜏₂𝑚_4𝑖]×[−𝑚_4𝑖,𝑚_4𝑖],(𝐶₂)𝑓(𝑡,𝜇,𝜈)>(1/𝐿)𝜑_𝑝(𝑚_2𝑖/𝜏₁𝑀), (𝑡,𝜇,𝜈)∈[𝜎,𝑏−𝜎]×[𝑚_2𝑖,𝑚_2𝑖𝜏₁⁻²]×[−𝑚_4𝑖,𝑚_4𝑖],(𝐶₃)𝑓(𝑡,𝜇,𝜈)≤(1/𝐿)𝜑_𝑝(𝑚₁/𝑁), (𝑡,𝜇,𝜈)∈[𝑎,𝑏]×[0,𝑚_1𝑖]×[−𝑚_4𝑖,𝑚_4𝑖], then the problem (𝒫) has at least 2𝑛−1 positive solutions.

4. Example

Let 𝑝=3, 𝑚=4, and 𝜎=1/4. Consider the boundary value problem 𝑟(𝑡)𝜑_𝑝𝑢^′′+ℎ(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)(0)=0,0<𝑡<1,2𝑢−𝑢^′(0)=12𝑢14+12𝑢12,𝑢^′(1)=0, (44) where 𝑟(𝑡)(𝑡)=1=𝑡+1,ℎ𝑡(),𝑓=1−𝑡𝑡,𝜇,𝜈10⁻⁴𝑡+2𝜇⁸+𝜈3×10⁴³×10⁻⁵,𝜇≤4,10⁻⁴𝜈𝑡+131072+3×10⁴³×10⁻⁵,𝜇>4. (45)

It is easy to check that (𝐴₁)--(𝐴₄) hold. By some calculations, we have 𝜏₁=1/2, 𝜏₂=3, 𝜆≈1.615, 𝐿≈3.142, 𝑀=1, and 𝑁=2.375. If we choose 𝑚₁=1/2, 𝑚₂=1, and 𝑚₄=3×10⁴, then 𝑓(𝑡,𝜇,𝜈) satisfies 𝑓≤1𝑡,𝜇,𝜈𝐿𝜑₃𝑚₄≈2.866×10⁸,𝑡,𝜇,𝜈∈[0,1]×0,9×10⁴×−3×10⁴,3×10⁴,𝑓>1𝑡,𝜇,𝜈𝐿𝜑₃𝑚₂𝜏₁𝑀∈1≈1.274,𝑡,𝜇,𝜈4,34×[1,4]×−3×10⁴,3×10⁴,𝑓≤1𝑡,𝜇,𝜈𝐿𝜑₃𝑚₁𝑁∈[]×1≈0.014,𝑡,𝜇,𝜈0,10,2×−3×10⁴,3×10⁴. (46) Thus all assumptions of Theorem 6 hold, and so the problem (44) has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ such that max_𝑡∈[0,1]|𝑢^′_𝑖(𝑡)|≤3×10⁴ for 𝑖=1,2,3, 1<min_{(1/4)≤𝑡≤(3/4)}𝑢₁(𝑡), 1/2<max_𝑡∈[0,1]𝑢₂(𝑡)  with min_{(1/4)≤𝑡≤(3/4)}𝑢₂(𝑡)<1, and max_𝑡∈[0,1]𝑢₃(𝑡)<1/2.