Solutions for 𝑝-Laplacian Dynamic Delay Differential Equations on Time Scales

Abstract

Let T be a time scale. We study the existence of positive solutions for the nonlinear four-point singular boundary value problem with 𝑝-Laplacian dynamic delay differential equations on time scales, subject to some boundary conditions. By using the fixed-point index theory, the existence of positive solution and many positive solutions for nonlinear four-point singular boundary value problem with 𝑝-Laplacian operator is obtained.

1. Introduction

The study of dynamic equations on time scales goes back to its founder Hilger [1] and is a new area of still fairly theoretical exploration in mathematics. Boundary value problems for delay differential equations arise in a variety of areas of applied mathematics, physics, and variational problems of control theory (see [2, 3]). In recent years, many authors have begun to pay attention to the study of boundary-value problems or with 𝑝-Laplacian equations or with 𝑝-Laplacian dynamic equations on time scales (see [4–14] and the references therein).

In [7], Sun and Li considered the existence of positive solution of the following dynamic equations on time scales:𝑢^Δ∇(𝑡)+𝑎(𝑡)𝑓(𝑡,𝑢(𝑡))=0,𝑡∈(0,𝑇),(1.1)𝛽𝑢(0)−𝛾𝑢^Δ(0)=0,𝛼𝑢(𝜂)=𝑢(𝑇),(1.2) where 𝛽,𝛾≥0,𝛽+𝛾>0,𝜂∈(0,𝜌(𝑇)),0<𝛼<𝑇/𝜂. They obtained the existence of single and multiple positive solutions of the problem (1.1) and (1.2) by using fixed-point theorem and Leggett-Williams fixed-point theorem (see [15]), respectively.

In [4], Anderson discussed the following dynamic equation on time scales:𝑢^Δ∇(𝑡𝑢)+(0)𝑎=(𝑡)0𝑓,(𝑢(𝑡𝛼)𝑢)=(𝜂)0=,𝑢𝑡(∈𝑇)(.0,𝑇),(1.3) He obtained some results for the existence of one positive solution of the problem (1.3) based on the limits 𝑓₀=lim_𝑢→0⁺𝑓(𝑢)/𝑢 and 𝑓_∞=lim_𝑢→∞𝑓(𝑢)/𝑢.

In [5], Kaufmann studied the problem (1.3) and obtained the existence results of at least two positives solutions.

In [14], Wang et al. discussed the following dynamic equation on time scales by using Avery-Peterson fixed theorem (see [14]):𝜙_𝑝𝑢^^+𝑞(𝑡)𝑓𝑡,𝑢(𝑡),𝑢(𝑡−1),𝑢^(𝑡)=0,𝑡∈(0,1),(1.4)𝑢(𝑡)=𝜉(𝑡),−1≤𝑡≤0,𝑢(1)=0,(1.5)𝑢(𝑡)=𝜉(𝑡),−1≤𝑡≤0,𝑢^(1)=0.(1.5^) They obtained some results for the existence three positive solutions of the problem (1.4), (1.5) and (1.4), and (1.5^), respectively.

In [15], Lee and Sim discussed the following equation:𝜙_𝑝𝑢^^+𝜆ℎ(𝑢𝑡)(0𝑓)(=𝑢(𝑢𝑡)()1=)=0,0.a.e.𝑡∈(0,1),(1.6) By applying the global bifurcation theorem and figuring the shape of unbounded subcontinua of solutions, they obtain many different types of global existence results of positive solutions.

However, there are not many concerning the 𝑝-Laplacian problems on time scales. Especially, for the singular multipoint boundary value problems for 𝑝-Laplacian dynamic delay differential equations on time scales, with the author’s acknowledge, no one has studied the existence of positive solutions in this case.

Recently, in [16], we have studied the existence of positive solutions for the following nonlinear two-point singular boundary value problem with 𝑝-Laplacian operator:𝛼𝜙_𝑝𝜙_𝑝𝑢^^+𝑎(𝑡)𝑓(𝑢(𝑡))=0,0<𝑡<1,(𝑢(0))−𝛽𝜙_𝑝𝑢^(0)=0,𝛾𝜙_𝑝(𝑢(1))+𝛿𝜙_𝑝𝑢^(1)=0.(1.7) By using the fixed-point theorem of cone expansion and compression of norm type, the existence of positive solution and infinitely many positive solutions for nonlinear singular boundary value problem (1.7) with 𝑝-Laplacian operator is obtained.

Now, motivated by the results mentioned above, in this paper, we study the existence of positive solutions for the following nonlinear four-point singular boundary value problem with higher-order 𝑝-Laplacian dynamic delay differential equations operator on time scales (SBVP):𝜙_𝑝𝑢^Δ(𝑡)^∇+𝑔(𝑡)𝑓(𝑢(𝑡−𝜏),𝑢(𝑡))=0,0<𝑡<𝑇,𝜏>0,(1.8)𝛼𝜙𝑢_𝑝(𝑡)=𝜁(𝑡),−𝜏≤𝑡≤0,𝛾𝜙_𝑝(𝑢(0))−𝛽𝜙_𝑝𝑢^Δ(𝜉)=0,(𝑢(𝑇))+𝛿𝜙_𝑝𝑢^Δ(𝜂)=0,(1.9) or𝑢𝑢(𝑡()0=)−𝜁𝐵(𝑡)₀,−𝜏≤𝑡≤0,𝑢(𝑇)+𝐵₁𝑢^Δ(𝜉)=0,𝑢^Δ(𝜂)=0,(1.10) where 𝜙_𝑝(𝑠) is 𝑝-Laplacian operator, that is, 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠,𝑝>1, 𝜙_𝑞=𝜙^𝑝−1, 1/𝑝+1/𝑞=1. 𝜉,𝜂∈(0,𝑇), 𝜏∈[0,𝑇] is prescribed and 𝜉<𝜂, 𝑔∶(0,𝑇)→[0,∞), 𝛼>0,𝛽≥0,𝛾>0,𝛿≥0 and 𝐵₀,𝐵₁ are both nondecreasing continuous odd functions defined on (−∞,+∞).

In this paper, by constructing one integral equation which is equivalent to the problem (1.8), (1.9) and (1.8), and (1.10), we research the existence of positive solutions for nonlinear singular boundary value problem (1.8), (1.9) and (1.8), and (1.10) when 𝑔 and 𝑓 satisfy some suitable conditions.

Our main tool of this paper is the following fixed point index theory.

Theorem 1.1 (see [17, 18]). Suppose that 𝐸 is a real Banach space, 𝐾⊂𝐸 is a cone, let Ω_𝑟={𝑢∈𝐾∶‖𝑢‖≤𝑟}. Let operator 𝑇∶Ω_𝑟→𝐾 be completely continuous and satisfy 𝑇𝑥≠𝑥,∀𝑥∈𝜕Ω_𝑟. Then(i)if ‖𝑇𝑥‖≤‖𝑥‖,∀𝑥∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝐾)=1;(ii)if ‖𝑇𝑥‖≥‖𝑥‖,∀𝑥∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝐾)=0.

This paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we discuss the existence of single solution of the systems (1.8) and (1.9). In Section 4, we study the existence of at least two solutions of the systems (1.8) and (1.9). In Section 5, we discuss the existence of single and many solutions of the systems (1.8) and (1.10). In Section 6, we give two examples as the application.

2. Preliminaries and Lemmas

For convenience, we can found some basic definitions in [1, 19, 20].

In the rest of this paper, 𝐓 is closed subset of 𝑅 with 0∈𝐓_𝑘,𝑇∈𝐓^𝑘. And let 𝐵={𝑢∈𝐶_𝑙𝑑[−𝜏,𝑇]}, then 𝐵 is a Banach space with the norm ‖𝑢‖=max_{𝑡∈[−𝜏,𝑇]}|𝑢(𝑡)|. And let𝐾={𝑢∈𝐵∶𝑢(𝑡)≥0,𝑢(𝑡)isconcavefunction,𝑡∈[0,𝑇]}.(2.1) Obviously, 𝐾 is a cone in 𝐵. Set 𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖≤𝑟}.

Definition 2.1. 𝑢(𝑡) is called a solution of SBVP (1.8) and (1.9) if it satisfies the following:(1)𝑢∈𝐶[−𝜏,0]∩𝐶_𝑙𝑑(0,𝑇),(2)𝑢(𝑡)>0 for all 𝑡∈(0,𝑇) and satisfies conditions (1.9),(3)(𝜙_𝑝(𝑢^Δ(𝑡)))^∇=−𝑔(𝑡)𝑓(𝑢(𝑡−𝜏),𝑢(𝑡)) holds for 𝑡∈(0,𝑇). In the rest of the paper, we also make the following assumptions:(𝐻₁) 𝑓∈𝐶_𝑙𝑑([0,+∞)²,[0,+∞)),(𝐻₂)𝑔(𝑡)∈𝐶_𝑙𝑑((0,𝑇),[0,+∞)) and there exists 𝑡₀∈(0,𝑇), such that 𝑔𝑡₀>0,0<^𝑇0𝑔(𝑠)∇𝑠<+∞,(2.2)(𝐻₃)𝜁(𝑡)∈𝐶([−𝜏,0], 𝜁(𝑡)>0 on [−𝜏,0) and 𝜁(0)=0,(𝐻₄)𝐵₀,𝐵₁ are both increasing, continuous, odd functions defined on (−∞,+∞), and at least one of them satisfies the condition that there exists one 𝑏>0 such that 0<𝐵_𝑖(𝑣)≤𝑏𝑣,∀𝑣≥0,𝑖=0or1.(2.3)It is easy to check that condition (𝐻₂) implies that 0<^𝑇0𝜙_𝑞^𝑠0𝑔𝑠₁∇𝑠₁Δ𝑠<+∞.(2.4) We can easily get the following Lemmas.

Lemma 2.2. Suppose that condition (𝐻₂) holds. Then there exists a constant 𝜃∈(0,1/2) that satisfies 0<^{𝜃𝑇−𝜃}𝑔(𝑡)∇𝑡<∞.(2.5) Furthermore, the function 𝐴(𝑡)=^𝑡𝜃𝜙_𝑞^𝑡𝑠𝑔𝑠₁∇𝑠₁Δ𝑠+^{𝑡𝑇−𝜃}𝜙_𝑞^𝑠𝑡𝑔𝑠₁∇𝑠₁∇𝑠,𝑡∈[𝜃,𝑇−𝜃](2.6) is positive continuous functions on [𝜃,𝑇−𝜃]; therefore, 𝐴(𝑡) has minimum on [𝜃,𝑇−𝜃]. Hence, we suppose that there exists 𝐿>0 such that 𝐴(𝑡)≥𝐿,𝑡∈[𝜃,𝑇−𝜃].

Proof. At first, it is easily seen that 𝐴(𝑡) is continuous on [𝜃,𝑇−𝜃]. Nest, let 𝐴₁(𝑡)=^𝑡𝜃𝜙_𝑞^𝑡𝑠𝑔𝑠₁∇𝑠₁Δ𝑠,𝐴²(𝑡)=^{𝑡𝑇−𝜃}𝜙_𝑞^s_𝑡𝑔𝑠₁∇𝑠₁Δ𝑠.(2.7) Then, from condition (𝐻₂), we have the function 𝐴₁(𝑡) is strictly monotone nondecreasing on [𝜃,𝑇−𝜃] and 𝐴₁(𝜃)=0, the function 𝐴₂(𝑡) is strictly monotone nonincreasing on [𝜃,𝑇−𝜃] and 𝐴₂(𝑇−𝜃)=0, which implies 𝐿=min_{𝑡∈[𝜃,𝑇−𝜃]}𝐴(𝑡)>0. The proof is complete.

Lemma 2.3 (see [16]). Let 𝑢∈𝐾 and 𝜃 of Lemma 2.2, then 𝑢(𝑡)≥𝜃‖𝑢‖,𝑡∈[𝜃,𝑇−𝜃].(2.8)

Lemma 2.4. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃), and (𝐻₄) hold, 𝑢(𝑡)∈𝐵∩𝐶_𝑙𝑑(0,1) is a solution of the following boundary value problems: 𝜙_𝑝𝑢^Δ(𝑡)^∇+𝑔(𝑡)𝑓(𝑢(𝑡−𝜏)+ℎ(𝑡−𝜏),𝑢(𝑡))=0,0<𝑡<𝑇,(2.9)𝛼𝜙_𝑝𝑢(𝑡)=0,−𝜏≤𝑡≤0,𝛾𝜙_𝑝(𝑢(0))−𝛽𝜙_𝑝𝑢^Δ(𝜉)=0,(𝑢(𝑇))+𝛿𝜙_𝑝𝑢^Δ(𝜂)=0,(2.10) or 𝑢𝑢((0𝑡))=−0𝐵,₀−𝜏≤𝑡≤0,𝑢(𝑇)+𝐵₁𝑢^Δ(𝜉)=0,𝑢^Δ(𝜂)=0,(2.10^) where ℎ(𝑡)=𝜁0,(𝑡),−0𝜏≤≤𝑡≤𝑡≤𝑇.0,(2.11) Then, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡),−𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8) and (1.9) or (1.8) and (1.10).

Proof. It is easy to check that 𝑢(𝑡) satisfies (1.8) and (1.9) or (1.8) and (1.10).

So in the rest section of this paper, we focus on SBVP (2.9), (2.10), and (2.9), (2.10^).

Lemma 2.5. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃),or (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄), hold, 𝑢(𝑡)∈𝐵∩𝐶_𝑙𝑑(0,1) is a solution of boundary value problems (2.9), (2.10) or (2.9), (2.10^), respectively, if and only if 𝑢(𝑡)∈𝐵 is a solution of the following integral equation, respectively: 𝑢(𝑡)=⎧⎪⎨⎪⎩𝜁(𝑡),−𝜏≤𝑡≤0,^𝑡0𝑢(𝑡)=⎧⎪⎨⎪⎩𝜁(𝑡𝑤),(𝑠)Δ𝑠,0−𝜏≤≤𝑡≤𝑡≤𝑇,0,_t0𝑤(𝑠)Δ𝑠,0≤𝑡≤𝑇,(2.12) where 𝑤(𝑡)=⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩𝜙^𝑞𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑡0𝜙_𝑞^𝜎𝑠𝜙𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,0≤𝑡≤𝜎,_𝑞𝛾𝛿^𝜂𝜎+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝑡𝜙_𝑞^𝑠𝜎𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,𝜎≤𝑡≤𝑇,(2.13)𝑤(𝑡)=⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩𝐵⁰∘𝜙_𝑞^𝜚𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑡0𝜙_𝑞^𝜚𝑠𝐵𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,0≤𝑡≤𝜚,₁∘𝜙_𝑞^𝜂𝜚+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝑡𝜙_𝑞^𝑠𝜚𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,𝜚≤𝑡≤𝑇.(2.13^) Here 𝜎,𝜚 is unique solution of the equation, respectively, 𝑔₁(𝑡)=𝑔₂(𝑡),𝑔₁(𝑡)=𝑔₂(𝑡),(2.14) where 𝑔₁(𝑡)=𝜙_𝑞𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑡0𝜙_𝑞^𝜎𝑠𝑔𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,₂(𝑡)=𝜙_𝑞𝛾𝛿^𝜂𝜎+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝑡𝜙_𝑞^𝑠𝜎𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,𝑔₁(𝑡)=𝐵₀∘𝜙_𝑞^𝜚𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑡0𝜙_𝑞^𝜚𝑠𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠,𝑔₂(𝑡)=𝐵₁∘𝜙_𝑞^𝜂𝜚+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝑡𝜙_𝑞^𝑠𝜚𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠.(2.15) Equation 𝑔₁(𝑡)=𝑔₂(𝑡), 𝑔₁(𝑡)=𝑔₂(𝑡) has unique solution in (0,𝑇). Because 𝑔₁(𝑡), 𝑔₁(𝑡) is strictly monotone increasing on [0,𝑇), and 𝑔₁(0)=0, 𝑔₁(0)=0, 𝑔₂(𝑡),𝑔₂(𝑡) is strictly monotone decreasing on (0,𝑇], and 𝑔₂(𝑇)=0, 𝑔₂(𝑇)=0.

Proof. We only proof the first section of the results.
Necessity. Obviously, for 𝑡∈(−𝜏,0), we have 𝑢(𝑡)=𝜁(𝑡).
If 𝑡∈(0,𝑇), by the equation of the boundary condition and we have 𝑢^Δ(𝜉)≥0,𝑢^Δ(𝜂)≤0, then there exist is a constant 𝜎∈[𝜉,𝜂]⊂(0,𝑇) such that 𝑢^Δ(𝜎)=0.
Firstly, by integrating the equation of the problems (2.9) on (𝜎,𝑡), we have 𝜙_𝑝𝑢^Δ(𝑡)=𝜙_𝑝𝑢^Δ(𝜎)−^𝑡𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠,(2.16) then 𝑢^Δ(𝑡)=−𝜙_𝑞^𝑡𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠,(2.17) thus 𝑢(𝑡)=𝑢(𝜎)−^𝑡𝜎𝜙_𝑞^𝑠𝜎𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠.(2.18)
By 𝑢^Δ(𝜎)=0 and condition (2.16), 𝑡=𝜂 on (2.16), we have 𝜙_𝑝𝑢^Δ(𝜂)=−^𝜂𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠.(2.19) By the equation of the boundary condition (2.10), we have 𝜙_𝑝(𝑢(𝑇))=−𝛿𝛾𝜙^𝑝𝑢^Δ(𝜂),(2.20) then 𝑢(𝑇)=𝜙_𝑞𝛾𝛿^𝜂𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠.(2.21) Then, by (2.18) and let 𝑡=𝑇 on (2.18), we have 𝑢(𝜎)=𝜙_𝑞𝛾𝛿^𝜂𝜎+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝜎𝜙_𝑞^𝑠𝜎𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠.(2.22) Then 𝑢(𝑡)=𝜙_𝑞𝛾𝛿^𝜂𝜎+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝑡𝜙_𝑞^𝑠𝜎𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠.(2.23) Similarly, for 𝑡∈(0,𝜎), by integrating the equation of problems (2.9) on (0,𝜎), we have 𝑢(𝑡)=𝜙_𝑞𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠Δ𝑠_𝑡0𝜙_𝑞^𝜎𝑠𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠.(2.24) Therefore, for any 𝑡∈[0,𝑇], 𝑢(𝑡) can be expressed as equation 𝑢(𝑡)=⎧⎪⎨⎪⎩𝜁(𝑡),−𝜏≤𝑡≤0,^𝑡0𝑤(𝑠)Δ𝑠,0≤𝑡≤𝑇,(2.25) where 𝑤(𝑡) is expressed as (2.13).Sufficiency. Suppose that 𝑢(𝑡)=∫^𝑡0𝑤(𝑠)Δ𝑠_𝑛−2Δ𝑠,0≤𝑡≤𝑇. Then by (2.13), we have 𝑢^Δ(𝑡)=⎧⎪⎨⎪⎩𝜙^𝑞^𝜎𝑡−𝜙_𝑞𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠≥0,0≤𝑡≤𝜎,^𝑡𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠≤0,𝜎≤𝑡≤𝑇,(2.26) So, (𝜙_𝑝(𝑢^Δ)^∇+𝑔(𝑡)𝑓(𝑢(𝑡−𝜏)+ℎ(𝑡−𝜏),𝑢(𝑡))=0,0<𝑡<𝑇. These imply that (2.9) holds. Furthermore, by letting 𝑡=0 and 𝑡=𝑇 on (2.13) and (2.26), we can obtain the boundary value equations of (2.10). The proof is complete.

Now, we define an operetor equation 𝑇 given by(𝑇𝑢)(𝑡)=⎧⎪⎨⎪⎩𝜁(𝑡),−𝜏≤𝑡≤0,^𝑡0𝑤(𝑠)Δ𝑠,0≤𝑡≤𝑇,𝑇𝑢(𝑡)=⎧⎪⎨⎪⎩𝜁(𝑡),−𝜏≤𝑡≤0,^𝑡0𝑤(𝑠)Δ𝑠,0≤𝑡≤𝑇,(2.27) where 𝑤(𝑡),𝑤(𝑡) is given by (2.13) and (2.13^).

From the definition of 𝑇,𝑇 and above discussion, we deduce that for each 𝑢∈𝐾,𝑇𝑢,𝑇𝑢∈𝐾. Moreover, we have the following Lemma.

Lemma 2.6. 𝑇,𝑇∶𝐾→𝐾 is completely continuous.

Proof. We only proof the completely continuous of 𝑇.
Because (𝑇𝑢)^Δ(𝑡)=𝑤^Δ(𝑡)=⎧⎪⎨⎪⎩𝜙^𝑞^𝜎𝑡−𝜙_𝑞𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠≥0,0≤𝑡≤𝜎^𝑡𝜎𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠≤0,𝜎≤𝑡≤𝑇,(2.28) is continuous, decreasing on [0,𝑇], and satisfies that (𝑇𝑢)^Δ(𝜎)=0, then, 𝑇𝑢∈𝐾 for each 𝑢∈𝐾 and (𝑇𝑢)(𝜎)=max_{𝑡∈[0,𝑇]}(𝑇𝑢)(𝑡). This shows that 𝑇𝐾⊂𝐾. Furthermore, it is easy to check by Arzela-ascoli Theorem that 𝑇∶𝐾→𝐾 is completely continuous.

Lemma 2.7. Suppose that conditions (𝐻₁),(𝐻₂), and (𝐻₃) hold, the solution 𝑢(𝑡)∈𝐾 of problem (2.9) and (2.10) satisfy max_{0≤𝑡≤𝑇}||𝑢(𝑡−𝜏)+ℎ(𝑡−𝜏)||≤max_{−𝜏≤𝑡≤0}||𝜁(𝑡)||.(2.29)

Proof. Firstly, we can have max_{0≤𝑡≤𝑇}||𝑢(𝑡−𝜏)+ℎ(𝑡−𝜏)||≤=mmaxa_{0≤𝑡≤𝑇}x||𝑢_{−𝜏≤𝑡≤𝑇−𝜏}(𝑡−𝜏)||+max_{0≤𝑡≤𝑇}||ℎ(𝑡−𝜏)||=max_{−𝜏≤𝑡≤0}||𝑢(𝑡)||+max_{−𝜏≤𝑡≤𝑇−𝜏}||ℎ(𝑡)||||𝜁(𝑡)||.(2.30) The proof is complete.

For convenience, we set𝐻=max_{−𝜏≤𝑡≤0}||𝜁(𝑡)||,𝜃^∗=2𝜃_∗𝐿,=1𝑇+𝜙_𝑞(𝛽/𝛼)𝜙_𝑞∫^𝑇0𝑔(𝑟)∇𝑟,𝜃^∗∗=1(𝑏+1)𝜙_𝑞∫^T0𝑔(𝑟)∇𝑟,(2.31) where 𝐿 is the constant from Lemma 2.2. By Lemma 2.5, we can also set𝑓⁰=lim_𝑢2→0max_{0≤𝑢1≤𝐻}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1,𝑓^∞=lim_𝑢2→∞max_{0≤𝑢1≤𝐻}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1𝑓₀,=lim_𝑢2→0min_{0≤𝑢1≤𝐻}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1𝑓_∞=lim_𝑢2→∞min_{0≤𝑢1≤𝐻}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1.(2.32)

3. The Existence of Single Positive Solution to (1.8) and (1.9)

In this section, we present our main results.

Theorem 3.1. Suppose that condition (𝐻₁),(𝐻₂), and (𝐻₃) hold. Assume that 𝑓 also satisfies(𝐴₁):𝑓(𝑢₁,𝑢₂)≥(𝑚𝑟)^𝑝−1, for 𝜃𝑟≤𝑢₂≤𝑟, 0≤𝑢₁≤𝐻,(𝐴₂):𝑓(𝑢₁,𝑢₂)≤(𝑀𝑅)^𝑝−1, for 0≤𝑢₂≤𝑅,0≤𝑢₁≤𝐻,where 𝑚∈(𝜃^∗,∞), 𝑀∈(0,𝜃_∗).
Then, the SBVP (2.9), (2.10) has a solution 𝑢 such that ‖𝑢‖ lies between 𝑟 and 𝑅. Furthermore by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡),−𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8) and (1.9).

Proof of Theorem 3.1. Without loss of generality, we suppose that 𝑟<𝑅. For any 𝑢∈𝐾, by Lemma 2.3, we have 𝑢(𝑡)≥𝜃‖𝑢‖,𝑡∈[𝜃,𝑇−𝜃].(3.1) We define two open subset Ω₁ and Ω₂ of 𝐸: Ω₁={𝑢∈𝐾∶‖𝑢‖<𝑟},Ω₂={𝑢∈𝐾∶‖𝑢‖<𝑅}.(3.2) For any 𝑢∈𝜕Ω₁, by (3.1), we have 𝑟=‖𝑢‖≥𝑢(𝑡)≥𝜃‖𝑢‖=𝜃𝑟,𝑡∈[𝜃,𝑇−𝜃].(3.3) For 𝑡∈[𝜃,𝑇−𝜃] and 𝑢∈𝜕Ω₁, we shall discuss it from three perspectives.(i)If 𝜎∈[𝜃,𝑇−𝜃], thus for 𝑢∈𝜕Ω₁, by (𝐴₁) and Lemma 2.4, we have2‖𝑇𝑢‖=≥2(𝑇𝑢)(𝜎)_𝜎0𝜙_𝑞^𝜎𝑠+𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠_𝑇𝜎𝜙_𝑞^𝑠𝜎≥𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠_𝜎𝜃𝜙_𝑞^𝜎𝑠+𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠_{𝜎𝑇−𝜃}𝜙_𝑞^𝑠𝜎≥𝑚𝑟𝐴(𝜎)≥𝑚𝑟𝐿𝑔(≥𝑟)2𝑓𝑟(=𝑢(2𝑟‖−𝑢𝜏‖).+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠(3.4)(ii)If 𝜎∈(𝑇−𝜃,𝑇], thus for 𝑢∈𝜕Ω₁, by (𝐴₁) and Lemma 2.4, we have‖𝑇𝑢‖=≥(𝑇𝜙𝑢)_𝑞(𝜎)𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝜎0𝜙_𝑞^𝜎𝑠≥𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠_{𝜃𝑇−𝜃}𝜙_𝑞^{𝑠𝑇−𝜃}≥𝑚𝑟𝐴(𝑇−𝜃)≥𝑚𝑔(𝑟𝑟𝐿)𝑓≥(2𝑢𝑟(𝑟>−𝑟𝜏=)+‖𝑢ℎ‖(.𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠(3.5)(iii)If 𝜎∈(0,𝜃), thus for 𝑢∈𝜕Ω₁, by (𝐴₁) and Lemma 2.4, we have‖𝑇𝑢‖=≥(𝑇𝜙𝑢)_𝑞(𝜎)𝛾𝛿^𝜂𝜎+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇𝜎𝜙_𝑞^𝑠𝜎≥𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠_{𝜃𝑇−𝜃}𝜙_𝑞^𝑠𝜃≥𝑚𝑟𝐴(𝜃)≥𝑚𝑟𝑔𝐿(𝑟≥)𝑓2(𝑟𝑢>(𝑟𝑟−=𝜏‖)𝑢+‖ℎ.(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠(3.6)
Therefore, no matter under which condition, we all have ‖𝑇𝑢‖>‖𝑢‖,∀𝑢∈𝜕Ω₁.(3.7) Then by Theorem 1.1, we have 𝑖𝑇,Ω₁,𝐾=0.(3.8)
On the other hand, for 𝑢∈𝜕Ω₂, we have 𝑢(𝑡)≤‖𝑢‖=𝑅, and by (𝐴₂), we know that ‖𝑇𝑢‖=≤(𝑇𝜙𝑢)_𝑞(𝜎)𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇0𝜙_𝑞^𝜎𝑠≤𝑇+𝜙_𝑞𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠𝛼𝛽𝑀𝑅𝜙^𝑞^𝑇0𝑔(𝑟)∇𝑟≤𝑅=‖𝑢‖,(3.9) thus ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω₂.(3.10) Then, by Theorem 1.1, we have 𝑖𝑇,Ω₂,𝐾=1.(3.11)
Therefore, by (3.8), and (3.11), 𝑟<𝑅, we have 𝑖𝑇,Ω₂⧵Ω₁,𝐾=1.(3.12) Then operator 𝑇 has a fixed point 𝑢∈(Ω₁⧵Ω₂), and 𝑟≤‖𝑢‖≤𝑅. This completes the proof of Theorem 3.1.

Theorem 3.2. Suppose that condition (𝐻₁),(𝐻₂), and (𝐻₃) hold. Assume that 𝑓 also satisfies(𝐴₃)∶𝑓⁰=𝜑∈[0,(𝜃_∗/4)^𝑝−1), (𝐴₄)∶𝑓_∞=𝜆∈((2𝜃^∗/𝜃)^𝑝−1,∞). Then, the SBVP (2.9), (2.10) has a solution 𝑢 which is bounded in ‖⋅‖. Furthermore, by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡),−𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.9).

Proof of Theorem 3.2. First, by 𝑓⁰=𝜑∈[0,(𝜃_∗/4)^𝑝−1), for 𝜖=(𝜃_∗/4)^𝑝−1−𝜑, there exists an adequately small positive number 𝜌, as 0≤𝑢₂≤𝜌,𝑢₂≠0,𝑢₁≤𝐻, we have 𝑓𝑢₁,𝑢₂≤(𝜑+𝜖)𝑢₂^𝑝−1≤𝜃^∗4^𝑝−1𝜌^𝑝−1=𝜃^∗4𝜌^𝑝−1.(3.13) Then let 𝑅=𝜌,𝑀=𝜃_∗/4∈(0,𝜃_∗), thus by (3.13), 𝑓𝑢₁,𝑢₂≤(𝑀𝑅)^𝑝−1,0≤𝑢₂≤𝑅.(3.14) So condition (𝐴₂) holds.
Next, by condition (𝐴₄), 𝑓_∞=𝜆∈((2𝜃^∗/𝜃)^𝑝−1,∞), then for 𝜖=𝜆−(2𝜃^∗/𝜃)^𝑝−1, there exists an appropriately big positive number 𝑟≠𝑅, as 𝑢₂≥𝜃𝑟,𝑢₁≤𝐻, we have 𝑓𝑢₁,𝑢₂≥(𝜆−𝜖)𝑢₂^𝑝−1≥2𝜃^∗𝜃^𝑝−1(𝜃𝑟)^𝑝−1=2𝜃^∗𝑟^𝑝−1.(3.15) Let 𝑚=2𝜃^∗>𝜃^∗, thus by (3.15), condition (𝐴₁) holds. Therefore, by Theorem 3.1, we know that the results of Theorem 3.2 hold. The proof of Theorem 3.2 is complete.

Theorem 3.3. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃) hold. Assume that 𝑓 also satisfies(𝐴₅):𝑓^∞=𝜆∈[0,(𝜃_∗/4)^𝑝−1), (𝐴₆):𝑓₀=𝜑∈((2𝜃^∗/𝜃)^𝑝−1,∞). Then, the SBVP (2.9), (2.10) has a solution 𝑢 which is bounded in ‖⋅‖. Furthermore by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡),−𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.9).

Proof of Theorem 3.3. First, by condition (𝐴₆), 𝑓₀=𝜑∈((2𝜃^∗/𝜃)^𝑝−1,∞), then for 𝜖=𝜑−(2𝜃^∗/𝜃)^𝑝−1, there exists an adequately small positive number 𝑟, as 0≤𝑢₂≤𝑟,𝑢₂≠0,𝑢₁≤𝐻, we have 𝑓𝑢₁,𝑢₂≥(𝜑−𝜖)𝑢₂^𝑝−1=2𝜃^∗𝜃^𝑝−1𝑢₂^𝑝−1,(3.16) thus when 𝜃𝑟≤𝑢₂≤𝑟,𝑢₁≤𝐻, we have 𝑓𝑢₁,𝑢₂≥2𝜃^∗𝜃^𝑝−1(𝜃𝑟)^𝑝−1=2𝜃^∗𝑟^𝑝−1.(3.17) Let 𝑚=2𝜃^∗>𝜃^∗, so by (3.17), condition (𝐴₁) holds.
Next, by condition (𝐴₅): 𝑓^∞=𝜆∈[0,(𝜃_∗/4)^𝑝−1), then for 𝜖=(𝜃_∗/4)^𝑝−1−𝜆, there exists an suitably big positive number 𝜌≠𝑟, as 𝑢₂≥𝜌,𝑢₁≤𝐻, we have 𝑓𝑢₁,𝑢₂≤(𝜆+𝜖)𝑢₂^𝑝−1≤𝜃^∗4^𝑝−1𝑢₂^𝑝−1.(3.18) If 𝑓 is unbounded, by the continuity of 𝑓 on [0,∞)², then exists constant 𝑅(≠𝑟)≥𝜌, and a point (𝑢₀₁,𝑢₀₂)∈[0,∞)² such that 𝑓𝑢𝜌≤𝑢₀₂≤𝑅,₁,𝑢₂≤𝑓𝑢₀₁,𝑢₀₂,0≤𝑢₂≤𝑅,𝑢₁≤𝐻.(3.19) Thus, by 𝜌≤𝑢₀₂≤𝑅,𝑢₁≤𝐻, we know 𝑓𝑢₁,𝑢₂≤𝑓𝑢₀₁,𝑢₀₂≤𝜃^∗4^𝑝−1𝑢₀₂^𝑝−1≤𝜃^∗4𝑅^𝑝−1.(3.20) Choose 𝑀=𝜃_∗/4∈(0,𝜃_∗). Then, we have 𝑓𝑢₁,𝑢₂≤(𝑀𝑅)^𝑝−1,0≤𝑢₂≤𝑅,𝑢₁≤𝐻.(3.21) If 𝑓 is bounded, we suppose 𝑓(𝑢₁,𝑢₂)≤𝑀^𝑝−1,𝑢₂∈[0,∞),𝑀∈𝑅₊, there exists an appropriately big positive number 𝑅>(4/𝜃_∗)𝑀, then choose 𝑀=𝜃_∗/4∈(0,𝜃_∗), we have 𝑓𝑢₁,𝑢₂≤𝑀^𝑝−1≤𝜃^∗4𝑅^𝑝−1=(𝑀𝑅)^𝑝−1,0≤𝑢₂≤𝑅,𝑢₁≤𝐻.(3.22) Therefore, condition (𝐴₂) holds. Therefore, by Theorem 3.1, we know that the results of Theorem 3.3 holds. The proof of Theorem 3.3 is complete.

4. The Existence of Many Positive Solutions to (1.8) and (1.9)

Next, we will discuss the existence of many positive solutions.

Theorem 4.1. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃), and (𝐴₂) in Theorem 3.1 hold. Assume that 𝑓 also satisfies(𝐴₇):𝑓₀=+∞, (𝐴₈):𝑓_∞=+∞. Then, the SBVP (2.9), (2.10) has at last two solutions 𝑢₁,𝑢₂ such that 0<‖‖𝑢₁‖‖<𝑅<‖‖𝑢₂‖‖.(4.1) Furthermore, by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.9).

Proof of Theorem 4.1. First, by condition (𝐴₇), for any 𝑁>2/𝜃𝐿, there exists a constant 𝜌_∗∈(0,𝑅) such that 𝑓𝑢₁,𝑢₂≥𝑁𝑢₂^𝑝−1,0<𝑢₂≤𝜌_∗,𝑢₁≤𝐻.(4.2) Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌_∗}, for any 𝑢∈𝜕Ω_𝜌∗, by (4.2) and Lemma 2.3, similar to the previous proof of Theorem 3.1, we can have from three perspectives ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗.(4.3) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝜌∗,𝐾=0.(4.4) Next, by condition (𝐴₈), for any 𝑁>2/𝜃𝐿, there exists a constant 𝜌₀>0 such that 𝑓𝑢₁,𝑢₂≥𝑁𝑢₂^𝑝−1,𝑢₂>𝜌₀,𝑢₁≤𝐻.(4.5) We choose a constant 𝜌^∗>max{𝑅,𝜌₀/𝜃}, obviously 𝜌_∗<𝑅<𝜌^∗. Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌^∗}. For any 𝑢∈𝜕Ω_𝜌∗, by Lemma 2.3, we have 𝑢(𝑡)≥𝜃‖𝑢‖=𝜃𝜌^∗>𝜌₀,𝑡∈[𝜃,𝑇−𝜃].(4.6) Then by (4.5) and also similar to the previous proof of Theorem 3.1, we can also have from three perspectives ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗.(4.7) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝜌∗,𝐾=0.(4.8)
Finally, set Ω_𝑅={𝑢∈𝐾∶‖𝑢‖<𝑅}, For any 𝑢∈𝜕Ω_𝑅, by (𝐴₂), Lemma 2.3 and also similar to the latter proof of Theorem 3.1, we can also have ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝑅.(4.9) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝑅,𝐾=1.(4.10) Therefore, by (4.4), (4.8), (4.10), 𝜌_∗<𝑅<𝜌^∗, we have 𝑖𝑇,Ω_𝑅⧵Ω_𝜌∗,𝐾=1,𝑖𝑇,Ω_𝜌∗⧵Ω_𝑅,𝐾=−1.(4.11) Then 𝑇 has fixed-point 𝑢₁∈Ω_𝑅⧵Ω_𝜌∗, and fixed-point 𝑢₂∈Ω_𝜌∗⧵Ω_𝑅. Obviously, 𝑢₁,𝑢₂ are all positive solutions of problem (2.9), (2.10) and 𝜌_∗<‖𝑢₁‖<𝑅<‖𝑢₂‖<𝜌^∗. The proof of Theorem 4.1 is complete.

Theorem 4.2. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃), and (𝐴₁) in Theorem 3.1 hold. Assume that 𝑓 also satisfies(𝐴₉):𝑓⁰=0, (𝐴₁₀):𝑓^∞=0. Then, the SBVP (2.9), (2.10) has at last two solutions 𝑢₁,𝑢₂ such that 0<‖𝑢₁‖<𝑟<‖𝑢₂‖. Furthermore, by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Proof of Theorem 4.2. First, by 𝑓⁰=0, for 𝜖₁∈(0,𝜃_∗), there exists a constant 𝜌_∗∈(0,𝑟) such that 𝑓𝑢₁,𝑢₂≤𝜖₁𝑢₂^𝑝−1,0<𝑢₂≤𝜌_∗,𝑢₁≤𝐻.(4.12) Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌_∗}, for any 𝑢∈𝜕Ω_𝜌∗, by (4.12), we have ‖𝑇𝑢‖=≤(𝑇𝜙𝑢)_𝑞(𝜎)𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇0𝜙_𝑞^𝜎𝑠≤𝜙_𝑞𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠𝛼𝛽^𝜎𝜉+𝑇𝜙_𝑞𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠^𝑇0≤𝑇+𝜙_𝑞𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟𝛼𝛽𝜖¹𝜌_∗𝜙_𝑞^𝑇0𝑔(𝑟)∇𝑟≤𝜌_∗=‖𝑢‖,(4.13) that is ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗.(4.14) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝜌∗,𝐾=1.(4.15)
Next, let 𝑓^∗(𝑥)=max_{0≤𝑢2≤𝑥,𝑢1≤𝐻}𝑓(𝑢₁,𝑢₂), and note that 𝑓^∗(𝑥) is monotone increasing with respect to 𝑥≥0. Then from 𝑓^∞=0, it is easy to see that lim_𝑥→∞𝑓^∗(𝑥)𝑥^𝑝−1=0.(4.16) Therefore, for any 𝜖₂∈(0,𝜃_∗), there exists a constant 𝜌^∗>𝑟 such that 𝑓^∗(𝑥)≤𝜖₂𝑥^𝑝−1,𝑥≥𝜌^∗.(4.17) Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌^∗}, for any 𝑢∈𝜕Ω_𝜌∗, by (4.17), we have ‖𝑇𝑢‖=≤(𝑇𝜙𝑢)_𝑞(𝜎)𝛼𝛽^𝜎𝜉+𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠_𝑇0𝜙_𝑞^𝜎𝑠≤𝜙_𝑞𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟Δ𝑠𝛼𝛽^𝜎𝜉+𝑇𝜙_𝑞𝑔(𝑠)𝑓(𝑢(𝑠−𝜏)+ℎ(𝑠−𝜏),𝑢(𝑠))∇𝑠^𝑇0≤𝑇+𝜙_𝑞𝑔(𝑟)𝑓(𝑢(𝑟−𝜏)+ℎ(𝑟−𝜏),𝑢(𝑟))∇𝑟𝛼𝛽𝜙^𝑞^𝑇0≤𝑇+𝜙_𝑞𝑔(𝑟)𝑓^∗𝜌^∗∇𝑟𝛼𝛽𝜖²𝜌^∗𝜙_𝑞^𝑇0𝑔(𝑟)∇𝑟≤𝑟^∗=‖𝑢‖,(4.18) that is ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗.(4.19) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝜌∗,𝐾=1.(4.20)
Finally, set Ω_𝑟={𝑢∈𝐾∶‖𝑢‖<𝑟}. For any 𝑢∈𝜕Ω_𝑟, by (𝐴₁), Lemma 2.3 and also similar to the previous proof of Theorem 3.1, we can also have ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝜕Ω_𝑟.(4.21) Then by Theorem 1.1, we have 𝑖𝑇,Ω_𝑟,𝐾=0.(4.22) Therefore, by (4.15), (4.20), (4.22), 𝜌_∗<𝑟<𝜌^∗, we have 𝑖𝑇,Ω_𝑟⧵Ω_𝜌∗,𝐾=−1,𝑖𝑇,Ω_𝜌∗⧵Ω_𝑟,𝐾=1.(4.23) Then 𝑇 have fixed point 𝑢₁∈Ω_𝑟⧵Ω_𝜌∗, and fixed point 𝑢₂∈Ω_𝜌∗⧵Ω_𝑟. Obviously, 𝑢₁,𝑢₂ are all positive solutions of problem (1.8), (1.9) and 𝜌_∗<‖𝑢₁‖<𝑟<‖𝑢₂‖<𝜌^∗. The proof of Theorem 4.2 is complete.

Similar to Theorem 3.1, we also obtain the following Theorems.

Theorem 4.3. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃) and (𝐴₂) in Theorem 3.1, (𝐴₄) in Theorem 3.2 and (𝐴₆) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.10) has at last two solutions 𝑢₁,𝑢₂ such that 0<‖𝑢₁‖<𝑅<‖𝑢₂‖. Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 4.4. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃) and (𝐴₁) in Theorem 3.1, (𝐴₃) in Theorem 3.2 and (𝐴₅) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.10) have at last two solutions 𝑢₁,𝑢₂ such that 0<‖𝑢₁‖<𝑟<‖𝑢₂‖. Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡), 𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

5. The Existence of Many Positive Solutions to (1.8) and (1.10)

In the following, we will deal with problem (1.8), (1.10), the method is similar to that in Sections 3 and 4, so we omit many proof in this section.

Theorem 5.1. Suppose that condition (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) hold. Assume that 𝑓 also satisfies(𝐴^1):𝑓(𝑢₁,𝑢₂)≥(𝑚𝑟)^𝑝−1, for 𝜃𝑟≤𝑢₂≤𝑟,0≤𝑢₁≤𝐻,(𝐴^2):𝑓(𝑢₁,𝑢₂)≤(𝑀𝑅)^𝑝−1, for 0≤𝑢₂≤𝑅,0≤𝑢₁≤𝐻,where 𝑚∈(𝜃^∗,∞),𝑀∈(0,𝜃_∗∗). Then, the SBVP (2.9), (2.13) has a solution 𝑢 such that ‖𝑢‖ lies between 𝑟 and 𝑅. Furthermore by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.2. Suppose that condition (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) hold. Assume that 𝑓 also satisfies(𝐴^3)∶𝑓⁰=𝜑∈[0,(𝜃_∗∗/4)^𝑝−1), (𝐴^4):𝑓_∞=𝜆∈((2𝜃^∗/𝜃)^𝑝−1,∞). Then, the SBVP (2.9), (2.13) has a solution 𝑢 which is bounded in ‖⋅‖. Furthermore by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.3. Suppose that condition (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) hold. Assume that 𝑓 also satisfies(𝐴^5):𝑓^∞=𝜆∈[0,(𝜃_∗∗/4)^𝑝−1), (𝐴^6):𝑓₀=𝜑∈((2𝜃^∗/𝜃)^𝑝−1,∞). Then, the SBVP (2.9), (2.13) has a solution 𝑢 which is bounded in ‖⋅‖. Furthermore by Lemma 2.4, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.4. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) and (𝐴^2) in Theorem 5.1 hold. Assume that 𝑓 also satisfies(𝐴^7):𝑓₀=+∞, (𝐴^8):𝑓_∞=+∞. Then, the SBVP (2.9), (2.13) has at least two solutions 𝑢₁, 𝑢₂ such that 0<‖‖𝑢₁‖‖<𝑅<‖‖𝑢₂‖‖.(5.1) Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.5. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) and (𝐴^1) in Theorem 5.1 hold. Assume that 𝑓 also satisfies(𝐴^9):𝑓⁰=0, (𝐴^10):𝑓^∞=0. Then, the SBVP (2.9), (2.13) has at least two solutions 𝑢₁, 𝑢₂ such that 0<‖𝑢₁‖<𝑟<‖𝑢₂‖. Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.6. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) and (𝐴₂) in Theorem 5.1, (𝐴₄) in Theorem 5.2 and (𝐴^6) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.13) has at least two solutions 𝑢₁,𝑢₂ such that 0<‖𝑢₁‖<𝑅<‖𝑢₂‖. Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

Theorem 5.7. Suppose that conditions (𝐻₁),(𝐻₂),(𝐻₃),(𝐻₄) and (𝐴^1) in Theorem 5.1, (𝐴^3) in Theorem 5.2 and (𝐴^5) in Theorem 3.3 hold. Then, the SBVP (2.9), (2.13) has at least two solutions 𝑢₁,𝑢₂ such that 0<‖𝑢₁‖<𝑟<‖𝑢₂‖. Furthermore by Lemma 2.4, 𝑢₁(𝑡)=𝑢₁(𝑡)+ℎ(𝑡),𝑢₂(𝑡)=𝑢₂(𝑡)+ℎ(𝑡), −𝜏≤𝑡≤𝑇 is a positive solution to the SBVP (1.8), (1.10).

6. Application

In the section, we present two simple examples to explain our result.

Example 6.1. Let 𝐓={1−(1/2)^𝐍}∪{1}, where 𝐍 denotes the set of all nonnegative integers. Consider the following 3-order singular boundary value problem (SBVP) with 𝑝-Laplacian 𝜙_𝑝𝑢^Δ^∇(𝑡)+210𝑡^−1/2𝑢^1/2(𝑡)⋅15+(94/5)𝑒^2𝑢(𝑡)𝑢(𝑡)=𝑡₂120𝑢(𝑡−1)+7𝑒^𝑢(𝑡)+𝑒^2𝑢(𝑡)=0,0<𝑡<1,𝜙−1,−1≤𝑡≤0,_𝑝(𝑢(0))−𝜙_𝑝𝑢^Δ14=0,𝜙^𝑝(𝑢(1))+𝛿𝜙_𝑝𝑢^Δ12=0,(6.1) where 𝛼=𝛾=1,𝛽=1,𝑝=32,𝜏=𝛿𝑇≥=01,.𝜉=14,𝜂=21,𝜃=41,(6.2) So, by Lemma 2.4, we discuss the following SBVP: 𝜙_𝑝𝑢^Δ^∇(𝑡)+210𝑡^−1/2𝑢^1/2(𝑡)⋅15+(94/5)𝑒^2𝑢(𝑡)𝜙𝑢1(𝑡2)0=[𝑢0(,𝑡−−11)+≤ℎ𝑡≤(𝑡−0,_𝑝1)]+7𝑒^𝑢(𝑡)+𝑒^2𝑢(𝑡)=0,0<𝑡<1,(𝑢(0))−𝜙_𝑝𝑢^Δ14=0,𝜙^𝑝(𝑢(1))+𝛿𝜙_𝑝𝑢^Δ12=0,(6.3) where ℎ(𝑡)=𝑡²0−,1,−01≤≤𝑡𝑡≤≤10,,𝑔(𝑡)=120𝑡^−1/2𝑓𝑢,𝜁(𝑡)=𝑡²−1,₁,𝑢₂=𝑢₂^1/215+120𝑢(₁9+4/57)𝑒𝑒^𝑢2+^2𝑢2𝑒^2𝑢2.(6.4) Then obviously, 𝑞=3,𝐻=max_{−1≤𝑡≤0}||𝜁(𝑡)||=1,𝑓⁰=𝜑=lim_𝑢2→0+max_{0≤𝑢1≤1}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1=51𝑓20_∞,=𝜆=lim_𝑢2→∞min_{0≤𝑢1≤1}𝑓𝑢₁,𝑢₂𝑢^2𝑝−1=955,^𝑇0𝑔(𝑡)∇𝑡=110,(6.5) so conditions (𝐻₁),(𝐻₂),(𝐻₃) hold.
Next, 𝜃_∗=1𝑇+𝜙_𝑞(𝛽/𝛼)𝜙_𝑞∫^𝑇0𝑔(𝑟)∇𝑟=50,(6.6) then (𝜃_∗/4)^𝑝−1=5√2/2>51/20, that is, 𝜑∈[0,(𝜃_∗/4)^𝑝−1), so condition (𝐴₃) holds.
For 𝜃=1/4, it is easy to see by calculating that 𝐿=min_{𝑡∈[𝜃,𝑇−𝜃]}𝐴(𝑡)=116376+√33.(6.7) Because of 2𝜃𝜃^∗^𝑝−1=96×17+12√3^1/2<955,(6.8) then 𝜆∈2𝜃^∗𝜃^𝑝−1,∞,(6.9) so condition (𝐴₄) holds. Then by Theorem 3.2, SBVP (6.3) has at least a positive solution 𝑢(𝑡). So, 𝑢(𝑡)=𝑢(𝑡)+ℎ(𝑡),−1<𝑡<1 is the positive solution of SBVP (6.1).

Example 6.2. Consider the following 3-order singular boundary value problem (SBVP) with 𝑝-Laplacian: 𝜙_𝑝𝑢^Δ^∇(𝑡)+164𝜋⁴𝑡𝑢^−1/2((𝑡1)−=𝑡−)𝑡𝑒𝑢_𝑡(𝑡−1)+𝑢²(𝑡)+𝑢⁴(𝑡)=0,0<𝑡<1,2𝜙_𝑝,−1≤𝑡≤0,(𝑢(0))−𝜙_𝑝𝑢^Δ14=0,𝜙^𝑝(𝑢(1))+𝛿𝜙_𝑝𝑢^Δ12=0,(6.10) where 𝛽=𝛾=1,𝛼=2,𝑝=4,𝛿≥0,𝑝=4,𝜉=1𝜃=14,𝜂=31,4,𝜏=𝑇=1.(6.11) So, by Lemma 2.4, we discuss the following SBVP: 𝜙_𝑝𝑢^Δ^∇(𝑡)+164𝜋⁴2𝜙_𝑝𝑡^−1/2(1−𝑡)𝑢([𝑡𝑢)(=𝑡−0,1)+−ℎ1(≤𝑡−𝑡≤1)0],+𝑢²(𝑡)+𝑢⁴(𝑡)=0,0<𝑡<1,(𝑢(0))−𝜙_𝑝𝑢^Δ14=0,𝜙^𝑝(𝑢(1))+𝛿𝜙_𝑝𝑢^Δ12=0,(6.12) where 𝑔(𝑡)=ℎ(𝑡)1=𝜁0,(𝑡),−01≤≤𝑡≤𝑡≤1,0,𝜁(𝑡)=−𝑡𝑒^𝑡,64𝜋⁴𝑡^−1/2(1−𝑡),𝑓𝑢₁,𝑢₂=𝑢₁+𝑢²²+𝑢⁴².(6.13) Then obviously, 𝑞=43,¹⁰𝑔(𝑡)∇𝑡=164𝜋³,𝐻=max_{−1≤𝑡≤0}||𝜁(𝑡)||=𝑒,𝑓_∞=+∞,𝑓₀=+∞,(6.14) so conditions (𝐻₁),(𝐻₂),(𝐻₃),(𝐴₇),(𝐴₈) hold.
Next, 𝜙_𝑞¹⁰𝑎(𝑡)∇𝑡=14𝜋,𝜃^∗=4𝜋1+√³4,(6.15) we choose 𝑅=3,𝑀=2, and for 𝜃=1/4, because of the monotone increasing of 𝑓(𝑢₁, 𝑢₂, 𝑢₃) on [0,∞)³, then 𝑓𝑢₁,𝑢₂≤𝑓(𝑒,3)=𝑒+90,0≤𝑢₂≤3,0≤𝑢₁≤𝑒.(6.16) Therefore, by 𝑀∈0,𝜃_∗,(𝑀𝑅)^𝑝−1=(6)³=216,(6.17) we know that 𝑓𝑢₁,𝑢₂,𝑢₃≤(𝑀𝑅)^𝑝−1,0≤𝑢₂≤3,0≤𝑢₁≤𝑒,(6.18) so condition (𝐴₂) holds. Then by Theorem 4.1, SBVP (6.12) has at least two positive solutions 𝑣₁,𝑣₂ and 0<‖𝑣₁‖<3<‖𝑣₂‖. Then, by Lemma 2.4, 𝑣₁(𝑡)=𝑣₁(𝑡)+ℎ(𝑡),𝑣₂(𝑡)=𝑣₂(𝑡)+ℎ(𝑡),𝑡∈(−1,1) are the positive solutions of the SBVP (6.10).

Acknowledgments

The first the second authors were supported financially by Shandong Province Natural Science Foundation (ZR2009AQ004), NSFC (11026108, 11071141), and the third author was supported by Shandong Province planning Foundation of Social Science (09BJGJ14), and Shandong Province Natural Science Foundation (Z2007A04).