Twin Positive Solutions of a Nonlinear 𝑚-Point Boundary Value Problem for Third-Order 𝑝-Laplacian Dynamic Equations on Time Scales

Abstract

Several existence theorems of twin positive solutions are established for a nonlinear 𝑚-point boundary value problem of third-order 𝑝-Laplacian dynamic equations on time scales by using a fixed point theorem. We present two theorems and four corollaries which generalize the results of related literature. As an application, an example to demonstrate our results is given. The obtained conditions are different from some known results.

1. Introduction

A time scale 𝐓 is a nonempty closed subset of ℝ. We make the blanket assumption that 0 and 𝑇 are points in 𝐓. By an interval (0,𝑇), we always mean the intersection of the real interval (0,𝑇) with the given time scale, that is, (0,𝑇)∩𝐓.

In this paper, we will be concerned with the existence of positive solutions of the 𝑝-Laplacian dynamic equations on time scales:(𝜙_𝑝(𝑢^Δ∇))^∇+𝑎(𝑡)𝑓(𝑡,𝑢(𝑡))=0,𝑡∈(0,𝑇),(1.1)𝜙_𝑝(𝑢^Δ∇(0))=^𝑚−2_𝑖=1𝑎_𝑖𝜙_𝑝(𝑢^Δ∇(𝜉_𝑖)),𝑢^Δ(0)=0,𝑢(𝑇)=^𝑚−2_𝑖=1𝑏_𝑖𝑢(𝜉_𝑖),(1.2)𝜙_𝑝(𝑠)where 𝑝 is 𝜙_𝑝(𝑠)=-Laplacian operator; that is, |𝑠, and

(H₁)|^𝑝−2 satisfy 𝑠,𝑝>1, and 𝜙^𝑝−1(H₂)=𝜙_𝑞,1/𝑝+1/𝑞= and there exists 1,0<𝜉₁ such that <⋯<𝜉_𝑚−2(H₃)<𝜌(𝑇).We point out that the 𝑎_𝑖,𝑏_𝑖∈[0,+∞),𝑖=1,2,…,-derivative and the 0<∑^{𝑚−2𝑖=1}𝑎_𝑖<1-derivative in (1.2) and the ∑^{𝑚−2𝑖=1}𝑏_𝑖<1; space in 𝑎(𝑡)∈𝐶_ld([0,𝑇],[0,+∞)) are defined in Section 2.

Recently, there has been much attention paid to the existence of positive solutions for third-order nonlinear boundary value problems of differential equations. For example, see [1–10] and the listed references. Anderson [2] considered the following third-order nonlinear problem:𝑡₀∈(𝜉_𝑚−2,𝑇)He used the Krasnoselskii and the Leggett and Williams fixed-point theorems to prove the existence of solutions to the nonlinear problem (1.3). Li [6] considered the existence of single and multiple positive solutions to the nonlinear singular third-order two-point boundary value problem:𝑎(𝑡₀)>0;Under various assumptions on 𝑓∈𝐶([0,𝑇]×[0,+∞),[0,+∞)) and Δ, they established intervals of the parameter ∇ which yield the existence of at least two and infinitely many positive solutions of the boundary value problem by using Krasnoselski's fixed-point theorem of cone expansion-compression type. Liu et al. [7] discussed the existence of at least one or two nondecreasing positive solutions for the following singular nonlinear third-order differential equations:𝐶_𝑙𝑑Green's function and the fixed-point theorem of cone expansion-compression type are utilized in their paper. In [8], Sun considered the following nonlinear singular third-order three-point boundary value problem:(H₂)He obtained various results on the existence of single and multiple positive solutions to the boundary value problem (1.6) by using a fixed-point theorem of cone expansion-compression type due to Krasnosel'skii. In [10], Zhou and Ma studied the existence and iteration of positive solutions for the following third-order generalized right-focal boundary value problem with 𝑥𝑥(^𝑡(₁𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡₁≤𝑡≤𝑡₃,)=𝑥^(𝑡₂)=0,𝛾𝑥(𝑡₃)+𝛿𝑥^(𝑡₃)=0.(1.3)-Laplacian operator:𝑢𝑢(0^()𝑡)=+𝑢𝜆_𝑎(𝑡)𝑓(𝑢(𝑡))=0,0<𝑡<1,(0)=𝑢^(1)=0.(1.4)They established a corresponding iterative scheme for (1.7) by using the monotone iterative technique.

On the other hand, the existence of positive solutions for third-order nonlinear boundary value problems of difference equations is also extensively studied by a number of authors (see [1, 3, 5, 9] and the listed references). The present work is motivated by a recent paper [4]. In [4], Henderson and Yin considered the existence of solutions for a third-order boundary value problem on a time-scale equation of the form𝑎which is uniform for the third-order difference equation and the third-order differential equation.

2. Preliminaries and Lemmas

For convenience, we list the following definitions which can be found in [4, 11–15].

Definition 2.1. Let 𝑓 be a time scale. For 𝜆 and 𝑥𝑥(^𝑎()𝑡)=+𝑥𝜆𝛼_(𝑡)𝑓(𝑡,𝑥(𝑡))=0,𝑎<𝑡<𝑏,(𝑎)=𝑥^(𝑏)=0.(1.5), define the forward jump operator 𝑢𝑢(^0()𝑡)=−𝑢𝜆_𝑎(𝑡)𝐹(𝑡,𝑢(𝑡))=0,0<𝑡<1,(𝜂)=𝑢^(1)=0.(1.6) and the backward jump operator 𝑝, respectively, by(𝑢𝜙(0_𝑝)(𝑢=^))^_𝑚(𝑡)=𝑞(𝑡)𝑓(𝑡,𝑢(𝑡)),0≤𝑡≤1,_𝑖=1𝛼_𝑖𝑢(𝜉_𝑖),𝑢^(𝜂)=0,𝑢^(1)=^𝑛_𝑖=1𝛽_𝑖𝑢^(𝜃_𝑖).(1.7)for all 𝑢^Δ3=𝑓𝑡,𝑢,𝑢^Δ,𝑢^ΔΔ,𝑡∈𝐓,(1.8). If 𝐓, 𝑡<sup𝐓 is said to be right-scattered, and if 𝑟>inf𝐓 is said to be left-scattered; if 𝜎, 𝜌 is said to be right-dense, and if 𝜌𝜎((𝑟𝑡))==isnufp𝜏𝜏∈∈𝐓𝐓∣∣𝜏𝜏><𝑡𝑟∈∈𝐓𝐓,(2.1) is said to be left-dense. If 𝑡,𝑟∈𝐓 has a right-scattered minimum 𝜎(𝑡)>𝑡, define 𝑡; otherwise set 𝜌(𝑟)<𝑟,𝑟. If 𝜎(𝑡)=𝑡 has a left-scattered maximum 𝑡, define 𝜌(𝑟)=𝑟,𝑟; otherwise set 𝐓.

Definition 2.2. For 𝑚 and 𝐓_𝑘=𝐓−{𝑚}, the delta derivative of 𝐓_𝑘=𝐓 at the point 𝐓 is defined to be the number 𝑀 (provided that it exists), with the property that for each 𝐓^𝑘=𝐓−{𝑀} there is a neighborhood 𝐓^𝑘=𝐓 of 𝑓∶𝐓→𝑅 such that𝑡∈𝐓^𝑘for all 𝑓.

For 𝑡 and 𝑓^Δ(𝑡), the nabla derivative of 𝜖>0 at 𝑈 is denoted by 𝑡 (provided that it exists), with the property that for each ||𝑓(𝜎(𝑡))−𝑓(𝑠)−𝑓^Δ(𝑡)(𝜎(𝑡)−𝑠)||≤𝜖||𝜎(𝑡)−𝑠||(2.2) there is a neighborhood 𝑠∈𝑈 of 𝑓∶𝐓→𝑅 such that𝑡∈𝐓_𝑘for all 𝑓.

Definition 2.3. A function 𝑡 is left-dense continuous (i.e., 𝑓^∇(𝑡)-continuous) if 𝜖>0 is continuous at each left-dense point in 𝑈, and its right-sided limit exists at each right-dense point in 𝑡.

Definition 2.4. If ||𝑓(𝜌(𝑡))−𝑓(𝑠)−𝑓^∇(𝑡)(𝜌(𝑡)−𝑠)||≤𝜖||𝜌(𝑡)−𝑠||(2.3), then one defines the delta integral by𝑠∈𝑈 If 𝑓, then one defines the nabla integral by𝑙𝑑

To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear BVP𝑓𝐓

Lemma 2.5. If 𝐓 and 𝜙^Δ(𝑡)=𝑓(𝑡), then for ^𝑏𝑎𝑓(𝑡)Δ𝑡=𝜙(𝑏)−𝜙(𝑎).(2.4) the BVP (2.6)-(2.7) has the unique solution𝐹^∇(𝑡)=𝑓(𝑡)where ^𝑏𝑎𝑓(𝑡)∇𝑡=𝐹(𝑏)−𝐹(𝑎).(2.5)

Proof. (i) Let (𝜙_𝑝(𝑢^Δ∇))^∇+ℎ(𝑡)=0,𝑡∈(0,𝑇),(2.6)𝜙_𝑝(𝑢^Δ∇(0))=^𝑚−2_𝑖=1𝑎_𝑖𝜙_𝑝(𝑢^Δ∇(𝜉_𝑖)),𝑢^Δ(0)=0,𝑢(𝑇)=^𝑚−2_𝑖=1𝑏_𝑖𝑢(𝜉_𝑖).(2.7) be a solution, then we will show that (2.8) holds. By taking the nabla integral of problem (2.6) on ∑^{𝑚−2𝑖=1}𝑎_𝑖≠1, we have∑^{𝑚−2𝑖=1}𝑏_𝑖≠1thenℎ∈𝐶_𝑙𝑑[0,𝑇]By taking the nabla integral of (2.11) on 𝑢(𝑡)=−^𝑡0(𝑡−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐶,(2.8), we can get𝐴=−∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀ℎ(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖𝐶=∫_𝑇0,(𝑇−𝑠)𝜙_𝑞∫^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞∫^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖.(2.9)By taking the delta integral of (2.12) on 𝑢, we can get(0,𝑡)Similarly, let 𝜙_𝑝(𝑢^Δ∇(𝑡))=−^𝑡0ℎ(𝜏)∇𝜏+𝐴(2.10) on (2.10), then we have 𝑢^Δ∇(𝑡)=𝜙_𝑞−^𝑡0ℎ(𝜏)∇𝜏+𝐴=−𝜙_𝑞^𝑡0ℎ(𝜏)∇𝜏−𝐴.(2.11); let (0,𝑡) on (2.10), then we have𝑢^Δ(𝑡)=−^𝑡0𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐵.(2.12) Let (0,𝑡) on (2.12), then we have𝑢(𝑡)=−^𝑡0(𝑡−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐵𝑡+𝐶.(2.13)Let 𝑡=0 on (2.13), then we have𝜙_𝑝(𝑢^Δ∇(0))=𝐴Similarly, let 𝑡=𝜉_𝑖 on (2.13), then we have𝜙_𝑝(𝑢^Δ∇(𝜉_𝑖))=−^𝜉𝑖₀ℎ(𝜏)∇𝜏+𝐴.(2.14)By the boundary condition (2.7), we can get𝑡=0𝑢^Δ(0)=𝐵.(2.15)Solving (2.19), we get𝑡=𝑇By the boundary condition (2.7), we can obtain𝑢(𝑇)=−^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐵𝑇+𝐶.(2.16)Substituting (2.20) in the above expression, one has𝑡=𝜉_𝑖
(ii) We show that the function 𝑢(𝜉_𝑖)=−^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐵𝜉_𝑖+𝐶.(2.17) given in (2.8) is a solution.
Let 𝐵=0,(2.18)𝐴=^𝑚−2_𝑖=1𝑎_𝑖−^𝜉𝑖₀ℎ(𝜏)∇𝜏+𝐴.(2.19) be as in (2.8). By [12, Theorem 2.10(iii)] and taking the delta derivative of (2.8), we have𝐴=−∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀ℎ(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖.(2.20)moreover, we get
−^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐶=^𝑚−2_𝑖=1𝑏_𝑖−^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠+𝐶.(2.21)Taking the nabla derivative of this expression yields 𝐶=∫^𝑇0(𝑇−𝑠)𝜙_𝑞∫^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞∫^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖.(2.22). Also, routine calculation verifies that 𝑢 satisfies the boundary value conditions in (2.7) so that 𝑢 given in (2.8) is a solution of (2.6) and (2.7). The proof is complete.

Lemma 2.6. Assume  𝑢^Δ(𝑡)=−^𝑡0𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴∇𝑠;(2.23)   holds. For 𝜙𝑢^Δ∇(𝑡)=−𝜙_𝑞^𝑡0ℎ(𝜏)∇𝜏−𝐴_𝑝,(𝑢^Δ∇)=−^𝑡0ℎ(𝜏)∇𝜏−𝐴.(2.24) and (𝜙_𝑝(𝑢^Δ∇))^∇=−ℎ(𝑡), the unique solution 𝑢 of (2.6) and (2.7) satisfies𝑢

Proof. Let(H₁)Sinceℎ∈𝐶_𝑙𝑑[0,𝑇]then ℎ≥0.
According to Lemma 2.5, we get𝑢If 𝑢(𝑡)≥0for𝑡∈[0,𝑇].(2.25), we have𝜑₀(𝑠)=𝜙_𝑞^𝑠0ℎ(𝜏)∇𝜏−𝐴.(2.26) So ∫^𝑠0ℎ(𝜏)∇𝜏−𝐴=∫^𝑠0ℎ(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀ℎ(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖≥0,(2.27).

Lemma 2.7. Assume  𝜑₀(𝑠)≥0  holds. If 𝑢(0)=𝐶=∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖≥∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝑇−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖=∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∫^𝑇𝜉_𝑖(𝑇−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖=^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠+∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝑇𝜉_𝑖(𝑇−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖𝑢(𝑇)=−_𝑇0≥0,=−_𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠+𝐶(𝑇−𝑠)𝜑₀(𝑠)∇𝑠+∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖≥−^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠+∫^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝑇−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖=∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝑇𝜉_𝑖(𝑇−𝑠)𝜑₀(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖≥0.(2.28) and 𝑡∈(0,𝑇), then the unique solution 𝑢(𝑡)=−^𝑡0(𝑡−𝑠)𝜑₀(𝑠)∇𝑠+11−∑^{𝑚−2𝑖=1}𝑏_𝑖[^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀≥−_𝑇0(𝜉_𝑖−𝑠)𝜑₀(𝑠)∇𝑠](𝑇−𝑠)𝜑₀(𝑠)∇𝑠+11−∑^{𝑚−2𝑖=1}𝑏_𝑖[^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀=1(𝑇−𝑠)𝜑₀(𝑠)∇𝑠]1−∑^{𝑚−2𝑖=1}𝑏_𝑖[−1−^𝑚−2_𝑖=1𝑏_𝑖)^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠+^𝑇0(𝑇−𝑠)𝜑₀(𝑠)∇𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀=1(𝑇−𝑠)𝜑₀(𝑠)∇𝑠]1−∑^{𝑚−2𝑖=1}𝑏_{𝑖𝑚−2𝑖=1}𝑏_𝑖^𝑇𝜉_𝑖(𝑇−𝑠)𝜑₀(𝑠)∇𝑠≥0.(2.29) of (2.6) and (2.7) satisfies𝑢(𝑡)≥0,𝑡∈[0,𝑇]where (𝐻₁)

Proof. It is easy to check that ℎ∈𝐶_𝑙𝑑[0,𝑇]; this implies thatℎ≥0It is easy to see that 𝑢 for any inf_{𝑡∈[0,𝑇]}𝑢(𝑡)≥𝛾‖‖𝑢‖‖,(2.30) with 𝛾=∑^{𝑚−2𝑖=1}𝑏_𝑖(𝑇−𝜉_𝑖)𝑇−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖,‖‖𝑢‖‖=max_{𝑡∈[0,𝑇]}||𝑢(𝑡)||.(2.31). Hence, 𝑢^Δ(𝑡)=−∫^𝑡0𝜑₀(𝑠)∇𝑠≤0 is a decreasing function on ‖‖𝑢‖‖=𝑢(0),min_{𝑡∈[0,𝑇]}𝑢(𝑡)=𝑢(𝑇).(2.32). This means that the graph of 𝑢^Δ(𝑡₂)≤𝑢^Δ(𝑡₁) is concave down on 𝑡₁,𝑡₂∈[0,𝑇].
For each 𝑡₁≤𝑡₂, we have𝑢^Δ(𝑡)that is,[0,𝑇]so that𝑢(𝑡)With the boundary condition (0,𝑇), we have𝑖∈{1,2,…,𝑚−2}This completes the proof.

Let the norm on 𝑢(𝑇)−𝑢(0)𝑇−0≥𝑢(𝑇)−𝑢(𝜉^𝑖)𝑇−𝜉_𝑖,(2.33) be the maximum norm. Then, the 𝑇𝑢(𝜉_𝑖)−𝜉_𝑖𝑢(𝑇)≥(𝑇−𝜉_𝑖)𝑢(0),(2.34) is a Banach space. It is easy to see that BVP (1.1)-(1.2) has a solution 𝑇^𝑚−2_𝑖=1𝑏_𝑖𝑢(𝜉_𝑖)−^𝑚−2_𝑖=1𝑏_𝑖𝜉_𝑖𝑢(𝑇)≥^𝑚−2_𝑖=1𝑏_𝑖(𝑇−𝜉_𝑖)𝑢(0).(2.35) if and only if 𝑢(𝑇)=∑^{𝑚−2𝑖=1}𝑏_𝑖𝑢(𝜉_𝑖) is a fixed point of the operator𝑢(𝑇)≥∑^{𝑚−2𝑖=1}𝑏_𝑖(𝑇−𝜉_𝑖)𝑇−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑢(0).(2.36)where𝐶_𝑙𝑑[0,𝑇]Denote𝐶_𝑙𝑑[0,𝑇]where 𝑢=𝑢(𝑡) is the same as in Lemma 2.7. It is obvious that 𝑢 is a cone in (𝐴𝑢)(𝑡)=−^𝑡0(𝑡−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠+𝐶,(2.37). By Lemma 2.7, 𝐴=−∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖𝐶=∫_𝑇0,(𝑇−𝑠)𝜙_𝑞∫^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞∫^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖.(2.38). So by applying Arzela-Ascoli theorem on time scales [16], we can obtain that 𝐾=𝑢∣𝑢∈𝐶_𝑙𝑑[0,𝑇],𝑢(𝑡)≥0,inf_{𝑡∈[0,𝑇]}𝑢(𝑡)≥𝛾‖‖𝑢‖‖,(2.39) is relatively compact. In view of Lebesgue's dominated convergence theorem on time scales [13], it is easy to prove that 𝛾 is continuous. Hence, 𝐾 is completely continuous.

Lemma 2.8. 𝐶_𝑙𝑑[0,𝑇] is completely continuous.

Proof. First, we show that 𝐴(𝐾)⊂𝐾 maps bounded set into bounded set.
Assume 𝐴(𝐾) is a constant and 𝐴. Note that the continuity of 𝐴∶𝐾→𝐾 guarantees that there is 𝐴∶𝐾→𝐾 such that 𝐴 for 𝑐>0. So𝑢∈𝐾_𝑐={𝑢∈𝐾∶‖𝑢‖≤𝑐}That is, 𝑓 is uniformly bounded.
In addition, notice that for any 𝑐^>0, we have𝑓(𝑡,𝑢(𝑡))≤𝜙_𝑝(𝑐^)So, by applying Arzela-Ascoli theorem on time scales [16], we obtain that 𝑡∈[0,𝑇] is relatively compact.
Finally, we prove that ‖‖𝐴𝑢‖‖≤=m∫ax_{𝑡∈[0,𝑇]}||𝐴𝑢(𝑡)_𝑇0||≤𝐶(𝑇−𝑠)𝜙_𝑞∫^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖≤𝑐^∫^𝑇0(𝑇−𝑠)𝜙_𝑞∫^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏/1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖.(2.40) is continuous. Suppose that 𝐴𝐾_𝑐 and 𝑡₁,𝑡₂∈[0,𝑇] converges to ||𝐴𝑢(𝑡=₁)|||−_𝑡10𝐴𝑢(𝑡₂)||≤𝑐_(𝑡₂−𝑡₁)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠+^𝑡2_𝑡1(𝑡₂−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴∇𝑠|||||𝑡₁−𝑡₂||^𝑇0𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖+𝑇max_{𝑠∈[0,𝑇]}∇𝑠𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖.(2.41) uniformly on 𝐴𝐾_𝑐 Hence, 𝐴∶𝐾_𝑐→𝐾 is uniformly bounded and equicontinuous on {𝑢_𝑛}^∞𝑛=1⊂𝐾_𝑐 The Arzela-Ascoli theorem on time scales [16] tells us that there exists uniformly convergent subsequence in 𝑢_𝑛(𝑡) Let 𝑢^∗(𝑡) be a subsequence which converges to [0,𝑇]. uniformly on {𝐴𝑢_𝑛(𝑡)}^∞𝑛=1 In addition,[0,𝑇].Observe the expression of {𝐴𝑢_𝑛(𝑡)}^∞𝑛=1. and then letting {𝐴𝑢_𝑛(𝑚)(𝑡)}^∞𝑚=1 we obtain𝑣(𝑡)where[0,𝑇]. Here, we have used the Lebesgue dominated convergence theorem on time scales [13]. From the definition of 0≤𝐴𝑢_𝑛(𝑡)≤𝑐^∫^𝑇0(𝑇−𝑠)𝜙_𝑞∫^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏/1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖.(2.42), we know that {𝐴𝑢_𝑛(𝑚)(𝑡)}, on 𝑚→∞, This shows that each subsequence of 𝑣(𝑡)=−^𝑡0(𝑡−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢^∗(𝜏))∇𝜏−𝐴^∗∇𝑠+𝐶^∗,(2.43) uniformly converges to 𝐴^∗=−∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)𝑓(𝜏,𝑢^∗(𝜏))∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖𝐶_∗,=11−∑^{𝑚−2𝑖=1}𝑏_𝑖−^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢^∗(𝜏))∇𝜏−𝐴^∗∇𝑠_𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢^∗(𝜏))∇𝜏−𝐴^∗∇𝑠.(2.44) Therefore, the sequence 𝐴 uniformly converges to 𝑣(𝑡)=𝐴𝑢^∗(𝑡) This means that [0,𝑇]. is continuous at {𝐴𝑢_𝑛(𝑡)}^∞𝑛=1. So, 𝐴𝑢^∗(𝑡). is continuous on {𝐴𝑢_𝑛(𝑡)}^∞𝑛=1 since 𝐴𝑢^∗(𝑡). is arbitrary. Thus, 𝐴 is completely continuous. This proof is complete.

Lemma 2.9. Let𝑢^∗∈𝐾_𝑐For 𝐴, 𝐾_𝑐

Proof. Since𝑢^∗then 𝐴. For all 𝜑(𝑠)=𝜙_𝑞^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴.(2.45), we have𝜉_𝑖(𝑖=1,…,𝑚−2)In fact, let ^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜑(𝑠)∇𝑠≤𝜉^𝑖𝑇^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠.(2.46); taking the nabla derivative of this expression, we have^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−𝐴=^𝑠0𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖≥0,(2.47)Hence, 𝜑(𝑠)≥0 is a nondecreasing function on 𝑡∈(0,𝑇]. That is,∫^𝑡0(𝑡−𝑠)𝜑(𝑠)∇𝑠𝑡^∇=𝑡∫^𝑡0𝜑(𝑠)∇𝑠−∫^𝑡0(𝑡−𝑠)𝜑(𝑠)∇𝑠𝑡𝜌(𝑡)≥0.(2.48)For all 𝜓(𝑡)=𝑡∫^𝑡0𝜑(𝑠)∇𝑠−∫^𝑡0(𝑡−𝑠)𝜑(𝑠)∇𝑠,𝜓^∇(𝑡)=^𝑡0𝜑(𝑠)∇𝑠+𝑡𝜑(𝑡)−^𝑡0𝜑(𝑠)∇𝑠=𝑡𝜑(𝑡)≥0.(2.49)By (2.51), for 𝜓(𝑡), we have[0,𝑇]

Lemma 2.10 (See [17]). Let 𝜓(𝑡)≥0.(2.50) be a Banach space, and let 𝑡∈(0,𝑇] be a cone. Assume ∫^𝑡0(𝑡−𝑠)𝜑(𝑠)∇𝑠𝑡≤∫^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠𝑇.(2.51) are open bounded subsets of 𝜉_𝑖(𝑖=1,…,𝑚−2) with ^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜑(𝑠)∇𝑠≤𝜉^𝑖𝑇^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠.(2.52) and let
𝐸 be a completely continuous operator such that
(i)𝐾⊂𝐸 and Ω₁,Ω₂, or(ii)𝐸, and 0∈Ω₁,Ω₁⊂Ω₂,Then, 𝐹∶𝐾∩(Ω₂⧵Ω₁)→𝐾 has a fixed point in ‖𝐹𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁,

Now, we introduce the following notations. Let‖𝐹𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂For ‖𝐹𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁, and ‖𝐹𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂.,𝐹by Lemma 2.6, where 𝐾∩(Ω₂⧵Ω₁). and 𝐴₀=11−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖𝐵₀∇𝑠⁻¹,=𝑇∑^{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠⁻¹.(2.53) are well defined.

3. Main Results

Theorem 3.1. Assume  𝑙>0,Ω_𝑙={𝑢∈𝐾∶‖𝑢‖<𝑙}, 𝜕Ω_𝑙={𝑢∈𝐾∶‖𝑢‖=𝑙}, and 𝛼(𝑙)=sup‖‖𝐴𝑢‖‖∶𝑢∈𝜕Ω_𝑙,𝛽(𝑙)=inf‖‖𝐴𝑢‖‖∶𝑢∈𝜕Ω_𝑙,(2.54)  hold, and assume that the following conditions hold:
(A₁)𝛼, and 𝛽(A₂)(𝐻₁);(A₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝_𝑖∈𝐶([0,+∞),[0,+∞)),𝑖=1,2(A₄)there exists lim_𝑙→0+𝑝₁(𝑙)𝑙^𝑝−1<𝐴^0𝑝−1,lim_𝑙→∞𝑝₂(𝑙)𝑙^𝑝−1<𝐴^0𝑝−1;(3.1) such that 𝑘_𝑖∈𝐿¹([0,𝑇],[0,+∞)),𝑖=1,2Then, problem (1.1)-(1.2) has at least two positive solutions 0<𝑐₁≤𝑐₂ satisfying 0≤𝜆₂<𝑝−1<𝜆₁.

Theorem 3.2. Assume  𝑓𝑓((𝑡𝑡,,𝑙𝑙))≤≤𝑝𝑝₁₂(𝑙)+𝑘₁(𝑡)𝑙^𝜆1,(𝑡,𝑙)∈[0,𝑇]×[0,𝑐₁],(𝑙)+𝑘₂(𝑡)𝑙^𝜆2,(𝑡,𝑙)∈[0,𝑇]×[𝑐₂,+∞);(3.2), 𝑏>0, and min𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[𝛾𝑏,𝑏]≥(𝑏𝐵₀)^𝑝−1.(3.3)  hold, and assume that the following conditions hold:
(B₁)𝑢^∗1,𝑢^∗2, and 0<‖𝑢^∗1‖<𝑏<‖𝑢^∗2‖(B₂)(𝐻₁); (B₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝_𝑖∈𝐶([0,+∞),[0,+∞)),𝑖=3,4(B₄)there exists lim_𝑙→0+𝑝₃(𝑙)𝑙^𝑝−1>𝐵⁰𝛾^𝑝−1,lim_𝑙→∞𝑝₄(𝑙)𝑙^𝑝−1>𝐵⁰𝛾^𝑝−1;(3.4) such that 𝑘_𝑖∈𝐿¹([0,𝑇],[0,+∞)),𝑖=3,4Then, problem (1.1)-(1.2) has at least two positive solutions 0<𝑐₃≤𝑐₄ satisfying 0≤𝜆₄<𝑝−1<𝜆₃.

Proof of Theorem 3.1. Let𝑓𝑓((𝑡𝑡,,𝑙𝑙))≥≥𝑝𝑝₃₄(𝑙)−𝑘₃(𝑡)𝑙^𝜆3,(𝑡,𝑙)∈[0,𝑇]×[0,𝑐₃],(𝑙)−𝑘₄(𝑡)𝑙^𝜆4,(𝑡,𝑙)∈[0,𝑇]×[𝑐₄,+∞);(3.5)then there exist 𝑎>0 and max𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[0,𝑎]≤(𝑎𝐴₀)^𝑝−1.(3.6) such that𝑢^∗3,𝑢^∗4If 0<‖𝑢^∗3‖<𝑎<‖𝑢^∗4‖, then 𝜖=12min𝐴^0𝑝−1−lim_𝑙→0+𝑝₁(𝑙)𝑙^𝑝−1,𝐴^0𝑝−1−lim_𝑙→∞𝑝₂(𝑙)𝑙^𝑝−1,(3.7). By condition 0<𝑎₁≤𝑐₁, we have𝑐₂≤𝑎₂<+∞so that𝑝𝑝₁₂(𝑙)≤(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1,0≤𝑙≤𝑎₁,(𝑙)≤(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1,𝑎₂≤𝑙≤+∞.(3.8)Therefore,0≤𝑙≤𝑎₁,𝑢∈𝜕Ω_𝑙It follows that0≤𝑢(𝑡)≤𝑙,0≤𝑡≤𝑇Noticing (A₃), we have𝑓(𝑡,𝑢(𝑡))≤≤(𝑝𝐴₁(𝑢(_0𝑝−1𝑡))+𝑘₁(𝑡)𝑢^𝜆1(𝑡)≤(𝐴_0𝑝−1−𝜖)𝑢^𝑝−1(𝑡)+𝑘₁(𝑡)𝑢^𝜆1(𝑡)−𝜖)‖‖𝑢‖‖^𝑝−1+𝑘₁(𝑡)‖‖𝑢‖‖^𝜆1=(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1+𝑘₁(𝑡)𝑙^𝜆1(3.9)Therefore, there exist ^𝑠0𝑎(=𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏−_𝑠0𝐴𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)𝑓(𝜏,𝑢(𝜏))∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖≤^𝑠0𝑎(𝜏)(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1+𝑘₁(𝜏)𝑙^𝜆1∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1+𝑘₁(𝜏)𝑙^𝜆1∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖.(3.10) such that ‖‖𝐴𝑢‖‖≤𝐶=11−∑^{𝑚−2𝑖=1}𝑏_𝑖(^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀≤1(𝜉_𝑖−𝑠)𝜑(𝑠)∇𝑠)1−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0≤1(𝑇−𝑠)𝜑(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖×^𝑇0(𝑇−𝑠)𝜙_𝑞+∑^𝑠0𝑎(𝜏)[(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1+𝑘₁(𝜏)𝑙^𝜆1]∇𝜏_{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)[(𝐴^0𝑝−1−𝜖)𝑙^𝑝−1+𝑘₁(𝜏)𝑙^𝜆1]∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠.(3.11). It implies that 𝛼(𝑙)𝑙≤1−∑1^{𝑚−2𝑖=1}𝑏_𝑖×^𝑇0(𝑇−𝑠)𝜙_𝑞+∑^𝑠0𝑎(𝜏)[𝐴^0𝑝−1−𝜖+𝑘₁(𝜏)𝑙^{𝜆1−𝑝+1}]∇𝜏_{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)[𝐴^0𝑝−1−𝜖+𝑘₁(𝜏)𝑙^{𝜆1−𝑝+1}]∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠.(3.12).
If 𝜆₁−𝑝+1>0 and lim_𝑙→0+𝛼(𝑙)𝑙≤1−∑1^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)(𝐴^0𝑝−1−𝜖)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)(𝐴^0𝑝−1−𝜖)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=(𝐴_0𝑝−1∇𝑠−𝜖)^1/^𝑝−11−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=(𝐴_0𝑝−1∇𝑠−𝜖)^1/^𝑝−1𝐴⁰⁻¹=(1−𝐴^{0−(𝑝−1)}𝜖)^1/^𝑝−1<1.(3.13), then 0<𝑎₁<𝑎₁. Similar to the above argument, noticing that 𝛼(𝑎₁)<𝑎₁, we can get ‖𝐴𝑢‖<‖𝑢‖,𝑢∈𝜕Ω_𝑎1. Therefore, there exist 𝑎₂≤𝑙<+∞ such that 𝑢∈𝜕Ω_𝑙. It implies that 0≤𝑢(𝑡)≤𝑙.
On the other hand, since 𝜆₂−𝑝+1<0 is continuous, by condition lim_𝑙→∞(𝛼(𝑙)/𝑙)<1, there exist 0<𝑎₂<𝑎₂ such that𝛼(𝑎₂)<𝑎₂If ‖𝐴𝑢‖<‖𝑢‖,𝑢∈𝜕Ω_𝑎2, then 𝑓∶[0,𝑇]×[0,+∞)→[0,+∞). Applying Lemma 2.9, it follows that(A₄) In the same way, we can prove that if 𝑎₁<𝑏₁<𝑏<𝑏₂<𝑎₂, then min𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[𝛾𝑏_𝑖,𝑏_𝑖]≥(𝑏_𝑖𝐵₀)^𝑝−1,𝑖=1,2.(3.14).

Now, we consider the operator 𝑢∈𝜕Ω_𝑏1 on 𝛾𝑏₁≤𝑢(𝑡)≤𝑏₁,0≤𝑡≤𝑇 and ‖‖𝐴𝑢‖‖≥=m−ax_{0≤𝑡≤𝑇𝑇0}||(𝐴𝑢)(𝑡)||(𝑇−𝑠)𝜑(𝑠)∇𝑠+11−∑^{𝑚−2𝑖=1}𝑏_𝑖(^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠−^𝑚−2_𝑖=1𝑏_𝑖^𝜉𝑖₀=∑(𝜉_𝑖−𝑠)𝜑(𝑠)∇𝑠)_{𝑚−2𝑖=1}𝑏_𝑖1−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖∫^𝜉𝑖₀(𝜉_𝑖−𝑠)𝜑(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖≥∑^{𝑚−2𝑖=1}𝑏_𝑖1−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0(𝑇−𝑠)𝜑(𝑠)∇𝑠−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)^𝑇0=𝑇∑(𝑇−𝑠)𝜑(𝑠)∇𝑠_{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)^𝑇0≥𝑇∑(𝑇−𝑠)𝜑(𝑠)∇𝑠_{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)(𝑏₁𝐵₀)^𝑝−1∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)(𝑏₁𝐵₀)^𝑝−1∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=𝑇∑∇𝑠_{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)𝑏¹𝐵₀^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=𝑏₁∇𝑠𝐵₀𝐵⁰⁻¹=𝑏₁=‖‖𝑢‖‖.(3.15), respectively. By Lemma 2.10, we assert that the operator 𝑢∈𝜕Ω_𝑏2 has two fixed points ‖𝐴𝑢‖≥‖𝑢‖ such that 𝐴 and Ω_𝑏1⧵Ω_𝑎1. Therefore, Ω_𝑎2⧵Ω_𝑏2, are positive solutions of problem (1.1)-(1.2).

Proof of Theorem 3.2. Let𝐴then there exist 𝑢^∗1,𝑢^∗2∈𝐾 and 𝑎₁≤‖𝑢^∗1‖≤𝑏₁ such that𝑏₂≤‖𝑢^∗2‖≤𝑎₂ If 𝑢^∗𝑖,𝑖=1,2, then 𝜖=12minlim_𝑙→0⁺𝑝₃(𝑙)𝑙^𝑝−1−𝐵⁰𝛾^𝑝−1,lim_𝑙→∞𝑝₄(𝑙)𝑙^𝑝−1−𝐵⁰𝛾^𝑝−1,(3.16). By Lemma 2.9 and condition 0<𝑏₃≤𝑐₃, we have𝑐₄≤𝑏₄<+∞It follows that𝑝₃(𝑙)≥𝐵⁰𝛾^𝑝−1𝑝₄+𝜖𝑙^𝑝−1,0≤𝑙≤𝑏₃,(𝑙)≥𝐵⁰𝛾^𝑝−1+𝜖𝑙^𝑝−1,𝑏₄≤𝑙≤+∞.(3.17)Noticing 0≤𝑙≤𝑏₃,𝑢∈𝜕Ω_𝑙, we get𝛾𝑙≤𝑢(𝑡)≤𝑙,0≤𝑡≤𝑇Therefore, there exists (B₃) with ‖‖𝐴𝑢‖‖≥=m𝑇∑ax_{0≤𝑡≤𝑇}||(𝐴𝑢)(𝑡)_{𝑚−2𝑖=1}||𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)^𝑇0≥𝑇∑(𝑇−𝑠)𝜑(𝑠)∇𝑠_{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖×_𝑇0𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)(𝑇−𝑠)𝜙_𝑞+∑^𝑠0𝑎(𝜏)𝑝₃(𝑢(𝜏))−𝑘₃(𝜏)𝑢^𝜆3∇𝜏_{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)[𝑝₃(𝑢(𝜏))−𝑘₃(𝜏)𝑢^𝜆3]∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖≥𝑇∑∇𝑠_{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖×_𝑇0𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)𝐵⁰𝛾^𝑝−1+∑+𝜖(𝛾𝑙)^𝑝−1−𝑘₃(𝜏)𝑙^𝜆3∇𝜏_{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)(𝐵₀/𝛾)^𝑝−1+𝜖(𝛾𝑙)^𝑝−1−𝑘₃(𝜏)𝑙^𝜆3∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠.(3.18) such that 𝛽(𝑙)𝑙≥𝑇∑^{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖×_𝑇0𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)(𝑇−𝑠)𝜙_𝑞+∑^𝑠0𝑎(𝜏)[𝐵^0𝑝−1+𝛾^𝑝−1𝜖−𝑘₃(𝜏)𝑙^{𝜆3−𝑝+1}]∇𝜏_{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)[𝐵^0𝑝−1+𝛾^𝑝−1𝜖−𝑘₃(𝜏)𝑙^{𝜆3−𝑝+1}]∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖∇𝑠.(3.19). It implies that 𝜆₃−𝑝+1>0 for lim_𝑙→0+𝛽(𝑙)𝑙≥𝑇∑^{𝑚−2𝑖=1}𝑏_𝑖−∑^{𝑚−2𝑖=1}𝑏_𝑖𝜉_𝑖×_𝑇0𝑇(1−∑^{𝑚−2𝑖=1}𝑏_𝑖)(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)(𝐵^0𝑝−1+𝛾^𝑝−1𝜖)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)(𝐵^0𝑝−1+𝛾^𝑝−1𝜖)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=(𝐵_0𝑝−1∇𝑠+𝛾^𝑝−1𝜖)^1/^𝑝−1𝐵⁰⁻¹=(1+𝛾^𝑝−1𝐵^{0−(𝑝−1)}𝜖)^1/^𝑝−1>1.(3.20).
If 𝑏₃ and 0<𝑏₃<𝑎, then 𝛽(𝑏₃)>𝑏₃. Similar to the above argument, noticing that ‖𝐴𝑢‖>‖𝑢‖, we can get 𝑢∈𝜕Ω_𝑏3.
Therefore, there exist 𝑏₄≤𝛾𝑙<+∞ with 𝑢∈𝜕Ω_𝑙 such that 𝑏₄≤𝛾𝑙≤𝑢(𝑡)≤𝑙,0≤𝑡≤𝑇. It implies that 𝜆₄−𝑝+1<0 for lim_𝑙→+∞(𝛽(𝑙)/𝑙)>1.
By condition 𝑏₄, we can see that there exist 0<𝑏₄<+∞ such that𝛽(𝑏₄)>𝑏₄If ‖𝐴𝑢‖>‖𝑢‖, then 𝑢∈𝜕Ω_𝑏4, (B₄), and 𝑏₃<𝑎₃<𝑎<𝑎₄<𝑏₄. It follows thatmax𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[0,𝑎_𝑖]≤(𝑎_𝑖𝐴₀)^𝑝−1,𝑖=3,4.(3.21)Similarly, if 𝑢∈𝜕Ω_𝑎3, then 0≤𝑢(𝑡)≤𝑎₃.

Now, we study the operator 0≤𝑡≤𝑇 on 𝑓(𝑡,𝑢(𝑡))≤(𝑎₃𝐴₀)^𝑝−1 and ‖‖𝐴𝑢‖‖≤11−∑^{𝑚−2𝑖=1}𝑏_𝑖^𝑇0≤1(𝑇−𝑠)𝜑(𝑠)∇𝑠1−∑^{𝑚−2𝑖=1}𝑏_𝑖𝑎₃𝐴₀^𝑇0(𝑇−𝑠)𝜙_𝑞^𝑠0𝑎(𝜏)∇𝜏+∑^{𝑚−2𝑖=1}𝑎_𝑖∫^𝜉𝑖₀𝑎(𝜏)∇𝜏1−∑^{𝑚−2𝑖=1}𝑎_𝑖=𝑎₃∇𝑠=‖‖𝑢‖‖.(3.22), respectively. By Lemma 2.10, we assert that the operator 𝑢∈𝜕Ω_𝑎4 has two fixed points ‖𝐴𝑢‖≤‖𝑢‖ such that 𝐴 and Ω_𝑎3⧵Ω_𝑏3. Therefore, Ω_𝑏4⧵Ω_𝑎4, are positive solutions of problem (1.1)-(1.2).

4. Further Discussion

If the conditions of Theorems 3.1 and 3.2 are weakened, we will get the existence of single positive solution of problem (1.1)-(1.2).

Corollary 4.1. Assume  𝐴, 𝑢^∗3,𝑢^∗4∈𝐾, and 𝑏₃≤‖𝑢^∗3‖≤𝑎₃  hold, and assume that the following conditions hold:
(C₁)𝑎₄≤‖𝑢^∗4‖≤𝑏₄, and 𝑢^∗𝑖,𝑖=3,4(C₂)(𝐻₁);(C₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝₁∈𝐶([0,+∞),[0,+∞))(C₄)there exists lim_𝑙→0⁺(𝑝₁(𝑙)/𝑙^𝑝−1)<𝐴^0𝑝−1; such that 𝑘₁∈𝐿¹([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.2. Assume  𝑐₁>0, 𝜆₁>𝑝−1, and 𝑓(𝑡,𝑙)≤𝑝₁(𝑙)+𝑘₁(𝑡)𝑙^𝜆1,(𝑡,𝑙)∈[0,𝑇]×[0,𝑐₁];(4.1)  hold, and assume that the following conditions hold:
(D₁)𝑏>0, and min𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[𝛾𝑏,𝑏]≥(𝑏𝐵₀)^𝑝−1.(4.2)(D₂)(𝐻₁);(C₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝₂∈𝐶([0,+∞),[0,+∞))(D₄)there exists lim_𝑙→∞(𝑝₂(𝑙)/𝑙^𝑝−1)<𝐴^0𝑝−1; such that 𝑘₂∈𝐿¹([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.3. Assume  𝑐₂>0, 0≤𝜆₂<𝑝−1, and 𝑓(𝑡,𝑙)≤𝑝₂(𝑙)+𝑘₂(𝑡)𝑙^𝜆2,(𝑡,𝑙)∈[0,𝑇]×[𝑐₂,+∞);(4.3)   hold, and assume that the following conditions hold:
(E₁)𝑏>0, and min𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[𝛾𝑏,𝑏]≥(𝑏𝐵₀)^𝑝−1.(4.4)(E₂)(𝐻₁);(E₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝₃∈𝐶([0,+∞),[0,+∞))(E₄)there exists lim_𝑙→0+(𝑝₃(𝑙)/𝑙^𝑝−1)>(𝐵₀/𝛾)^𝑝−1; such that 𝑘₃∈𝐿¹([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

Corollary 4.4. Assume  𝑐₃>0, 𝜆₃>𝑝−1, and 𝑓(𝑡,𝑙)≥𝑝₃(𝑙)−𝑘₃(𝑡)𝑙^𝜆3,(𝑡,𝑙)∈[0,𝑇]×[0,𝑐₃];(4.5)  hold, and assume that the following conditions hold:
(F₁)𝑎>0, and max𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[0,𝑎]≤(𝑎𝐴₀)^𝑝−1.(4.6)(F₂)(𝐻₁);(F₃)there exist (𝐻₂) and (𝐻₃) such that 𝑝₄∈𝐶([0,+∞),[0,+∞))(F₄)there exists lim_𝑙→∞(𝑝₄(𝑙)/𝑙^𝑝−1)>(𝐵₀/𝛾)^𝑝−1; such that 𝑘₄∈𝐿¹([0,𝑇],[0,+∞))Then, problem (1.1)-(1.2) has at least one positive solution.

The proof of the above results is similar to those of Theorems 3.1 and 3.2; thus we omit it.

5. Some Examples

In this section, we present a simple example to explain our results. We only study the case 𝑐₄>0.

Let 0≤𝜆₄<𝑝−1. Consider the following BVP:𝑓(𝑡,𝑙)≥𝑝₄(𝑙)−𝑘₄(𝑡)𝑙^𝜆4,(𝑡,𝑙)∈[0,𝑇]×[𝑐₄,+∞);(4.7) where𝑎>0It is easy to check that max𝑓(𝑡,𝑙)∶(𝑡,𝑙)∈[0,𝑇]×[0,𝑎]≤(𝑎𝐴₀)^𝑝−1.(4.8) is continuous. In this case, 𝐓=𝑅,(0,𝑇)=(0,1), and 𝑓(𝑡,0)≡0, and it follows from a direct calculation that𝜙(𝜙₃(𝑢^))^+𝑓(𝑡,𝑢(𝑡))=0,𝑡∈(0,1),₃(𝑢^0)=12𝜙³𝑢^21,𝑢^(0)=0,𝑢(1)=12𝑢21,(5.1)We have𝑓(𝑡,𝑢)=⎧⎪⎪⎪⎨⎪⎪⎪⎩223𝑢³+min1√𝑡(1−𝑡),2𝑢√27245𝑢,+√𝑢⁵,((𝑡𝑡,,𝑢𝑢))∈∈[[00,,11]]××[[01,,13]],,36min√𝑡(11−𝑡),23𝑢√𝑢³,(𝑡,𝑢)∈[0,1]×[3,+∞).(5.2)Choosing 𝑓∶[0,1]×[0,+∞)→[0,+∞), and 𝑝=3,𝑎(𝑡)≡1,𝑚=3,𝑎₁=𝑏₁=1/2, it is easy to check that𝜉₁=1/2It follows that 𝐴₀=11−1/2∫¹⁰(1−𝑠)𝜙_𝑞𝑠+1/2⋅1/21−1/2𝑑𝑠⁻¹𝛾=𝑏₁=1.1062,(1−𝜉₁)1−𝑏₁𝜉₁=1/2(1−1/2)1−1/2⋅1/2=13.(5.3) satisfies the conditions 𝐵₀=𝑏¹−𝑏₁𝜉₁1−𝑏₁^𝑇0(1−𝑠)𝜙_𝑞𝑠+𝑎¹𝜉₁1−𝑎₁𝑑𝑠⁻¹=1/2−1/2⋅1/21−1/2¹⁰(1−𝑠)𝑠+1/2⋅1/21−1/2^1/2𝑑𝑠⁻¹=4.4248<5.(5.4) of Theorem 3.1; then problem (5.1) has at least two positive solutions.

Acknowledgment

Project supported by the National Natural Science Foundation of China (Grant NO. 10471040).