Existence of Three Positive Solutions for 𝑚-Point Discrete Boundary Value Problems with 𝑝-Laplacian

Abstract

We consider the multi-point discrete boundary value problem with one-dimensional 𝑝-Laplacian operator Δ(𝜙_𝑝(Δ𝑢(𝑡−1))+𝑞(𝑡)𝑓(𝑡,𝑢(𝑡),Δ𝑢(𝑡))=0, 𝑡∈{1,…,𝑛−1} subject to the boundary conditions: 𝑢(0)=0, 𝑢(𝑛)=∑^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖), where 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠,𝑝>1,𝜉_𝑖∈{2,…,𝑛−2} with 1<𝜉₁<⋯<𝜉_𝑚−2<𝑛−1 and 𝑎_𝑖∈(0,1),0<∑^𝑚−2_𝑖=1𝑎_𝑖<1. Using a new fixed point theorem due to Avery and Peterson, we study the existence of at least three positive solutions to the above boundary value problem.

1. Introduction

The second-order differential and difference boundary value problems arise in many branches of both applied and basic mathematics and have been extensively studied in literature. We refer the reader to some recent results for second-order nonlinear two-point [1–6] and multipoint [7–9] boundary value problems. The main tools used in the above works are fixed point theorems.

Recently, Feng and Ge in [9] considered the following multipoint BVPs:

(𝜙_𝑝(𝑢(𝑡−1))^′)^′+𝑞(𝑡)𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡))=0,𝑡∈(0,1),𝑢(0)=0,𝑢(𝑛)=^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖). (1.1) The authors obtained sufficient conditions that guarantee the existence of at least three positive solutions by using fixed point theorems due to Avery-Peterson.

In this work, we study the existence of multiple positive solutions to the discrete boundary value problem for the one-dimensional 𝑝-Laplacian:

Δ(𝜙_𝑝(Δ𝑢(𝑡−1)))+𝑞(𝑡)𝑓(𝑡,𝑢(𝑡),Δ𝑢(𝑡))=0,𝑡∈{1,…,𝑛−1}𝑢(0)=0,𝑢(𝑛)=^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖), (1.2) where Δ𝑢(𝑡)=𝑢(𝑡+1)−𝑢(𝑡) for 𝑡∈{0,1,…,𝑛−1}, Δ²𝑢(𝑡)=𝑢(𝑡+2)−2𝑢(𝑡+1)+𝑢(𝑡), for 𝑡∈{0,1,…,𝑛−2}, and 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1, 𝜉_𝑖∈{2,…,𝑛−2} with 1<𝜉₁<⋯<𝜉_𝑚−2<𝑛−1.

In order to study the existence of at least three positive solutions to the above boundary value problem, we assume that 𝑎_𝑖, 𝑓,𝑞 satisfy the following.

(𝐻₁)𝑎_𝑖∈(0,1) satisfy 0<∑^𝑚−2_𝑖=1𝑎_𝑖<1.(𝐻₂)𝑓∶{1,…,𝑛−1}×[0,+∞)×𝑅→(0,+∞) is continuous.(𝐻₃)𝑞(𝑡)>0 for 𝑡∈{1,2,…,𝑛−1}.

We will depend on an application of a fixed point theorems due to Avery and Peterson, which deals with fixed points of a cone-preserving operator defined on an ordered Banach space to obtain our main results.

2. Preliminaries

For the convenience of readers, we provide some background material from the theory of cones in Banach spaces. In this section, we also state Avery-Peterson's fixed point theorem.

Definition 2.1. Let 𝐸 be a real Banach space over 𝑅. A nonempty convex closed set 𝑃⊂𝐸 is said to be a cone of 𝐸 if it satisfies the following conditions:(i)𝑎𝑢+𝑏𝑣∈𝑃 for all 𝑢,𝑣∈𝑝 and all 𝑎≥0, 𝑏≥0;(ii)𝑢,−𝑢∈𝑃 implies 𝑢=0.Every cone 𝑃⊂𝐸 induces an ordering in 𝐸 given by 𝑥≤𝑦 if and only if 𝑦−𝑥∈𝑃.

Definition 2.2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Definition 2.3. The map 𝛼 is said to be a nonnegative continuous concave functional on a cone 𝑃 of a real Banach space 𝐸 provided that 𝛼∶𝑃→[0,∞) is continuous and 𝛼(𝑡𝑥+(1−𝑡)𝑦)≥𝑡𝛼(𝑥)+(1−𝑡)𝛼(𝑦) (2.1) for all 𝑥,𝑦∈𝑃 and 0≤𝑡≤1. Similarly, we say that the map 𝛽 is a nonnegative continuous convex functional on a cone 𝑃 of a real Banach space 𝐸 provided that 𝛽∶𝑃→[0,∞) is continuous and 𝛽(𝑡𝑥+(1−𝑡)𝑦)≤𝑡𝛽(𝑥)+(1−𝑡)𝛽(𝑦) (2.2) for all 𝑥,𝑦∈𝑃 and 0≤𝑡≤1.

Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on a cone 𝑃, let 𝛼 be a nonnegative continuous concave functional on a cone 𝑃, and let 𝜓 be a nonnegative continuous functional on a cone 𝑃. Then for positive real numbers 𝑎,𝑏,𝑐, and 𝑑, we define the following convex sets:

𝑃(𝛾,𝑑)={𝑢∈𝑃∣𝛾(𝑢)<𝑑},𝑃(𝛾,𝛼,𝑏,𝑑)={𝑢∈𝑃∣𝑏≤𝛼(𝑢),𝛾(𝑢)≤𝑑},𝑃(𝛾,𝜃,𝛼,𝑏,𝑐,𝑑)={𝑢∈𝑃∣𝑏≤𝛼(𝑢),𝜃(𝑢)≤𝑐,𝛾(𝑢)≤𝑑} (2.3) and a closed set

𝑅(𝛾,𝜓,𝑎,𝑑)={𝑢∈𝑃∣𝑎≤𝜓(𝑢),𝛾(𝑢)≤𝑑}. (2.4)

To prove our results, we need the following fixed point theorem due to Avery and Peterson in [1].

Theorem 2.4. Let 𝑃 be a cone in a real Banach space 𝐸. Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on 𝑃, let 𝛼 be a nonnegative continuous concave functional on 𝑃, and let 𝜓 be a nonnegative continuous functional on 𝑃 satisfying 𝜓(𝜆𝑢)≤𝜆𝜓(𝑢) for 0≤𝜆≤1, such that for some positive numbers 𝑀 and 𝑑, 𝛼(𝑢)≤𝜓(𝑢),‖𝑢‖≤𝑀𝛾(𝑢) (2.5) for all 𝑢∈𝑃(𝛾,𝑑). Suppose that 𝑇∶𝑃(𝛾,𝑑)→𝑃(𝛾,𝑑) (2.6) is completely continuous and there exist positive numbers 𝑎,𝑏, and 𝑐 with 𝑎<𝑏 such that(𝑆₁){𝑢∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑐,𝑑)∣𝛼(𝑢)>𝑏}≠∅ and 𝛼(𝑇𝑢)>𝑏 for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑏,𝑐,𝑑);(𝑆₂)𝛼(𝑇𝑢)>𝑏 for 𝑢∈𝑃(𝛾,𝛼,𝑏,𝑑) with 𝜃(𝑇𝑢)>𝑐;(𝑆₃)0∈𝑅(𝛾,𝜓,𝑎,𝑑) and 𝜓(𝑇𝑢)<𝑎 for 𝑢∈𝑅(𝛾,𝜓,𝑎,𝑑) with 𝜓(𝑢)=𝑎.Then 𝑇 has at least three fixed points 𝑢₁,𝑢₂,𝑢₃∈𝑃(𝛾,𝑑), such that
𝛾(𝑢_𝑖)≤𝑑 for 𝑖=1,2,3,𝑏<𝛼(𝑢₁)𝑎<𝜓(𝑢₂), with 𝛼(𝑢₂)<𝑏,𝜓(𝑢₃)<𝑎.

3. Related Lemmas

Let the Banach space 𝐸={𝑢∶{0,1,…,𝑛}→𝑅} be endowed with the ordering 𝑥≤𝑦 if 𝑥(𝑡)≤𝑦(𝑡) for all 𝑡∈{0,1,…,𝑛}, and the maximum norm

‖𝑢‖=max{max_{𝑡∈{0,1,…,𝑛}}|𝑢(𝑡)|,max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|}. (3.1) Then, we define the cone 𝑃 in 𝐸 by

𝑃={𝑢∈𝐸∣𝑢(𝑡)≥0,𝑡∈{0,1,…,𝑛};𝑢(0)=0,Δ²𝑢(𝑡)≤0,𝑡∈{0,1,…,𝑛−2}}. (3.2) Let 𝑘 be a natural number, such that 𝑘<min{𝜉₁,𝑛−𝜉_𝑚−2}.

Let the nonnegative continuous concave functional 𝛼, the nonnegative continuous convex functionals 𝜃, 𝛾, and the nonnegative continuous functional 𝜓 be defined on the cone 𝑃 by

𝛾(𝑢)=max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|,𝜓(𝑢)=𝜃(𝑢)=max_{𝑡∈{0,1,…,𝑛}}|𝑢(𝑡)|,𝛼(𝑢)=min_{𝑡∈{𝑘+1,…,𝑛−𝑘−1}}|𝑢(𝑡)| (3.3) for 𝑢∈𝑃.

In order to prove our main results, we need the following lemma.

Lemma 3.1. If 𝑢∈𝑃, then max_{𝑡∈{0,1,…,𝑛}}|𝑢(𝑡)|≤𝑛max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|,thatis,𝜃(𝑢)≤𝑛𝛾(𝑢). (3.4)

Proof. Suppose that the maximum of 𝑢 occurs at 𝑡₀∈{0,1,…,𝑛}; by the definition of the cone 𝑃, we know Δ𝑢(𝑡+1)≤Δ𝑢(𝑡), and then, 𝑢(𝑡₀)=𝑢(𝑡₀)−𝑢(0)=Δ𝑢(0)+Δ𝑢(1)+⋯+Δ𝑢(𝑡₀−1)≤𝑡₀Δ𝑢(0)≤𝑛Δ𝑢(0)≤𝑛max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|. (3.5) So, we have max_{𝑡∈{0,1,…,𝑛}}|𝑢(𝑡)|≤𝑛max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|. (3.6) The proof is complete.

By Lemma 3.1 and the definitions, the functionals defined above satisfy

1[𝑛/𝑘]+1𝜃(𝑢)≤𝛼(𝑢)≤𝜃(𝑢)=𝜓(𝑢),‖𝑢‖=max{𝜃(𝑢),𝛾(𝑢)}≤𝑛𝛾(𝑢) (3.7) for all 𝑢∈𝑃(𝛾,𝑑)⊂𝑃. Therefore, condition (2.5) is satisfied.

Now, we show that (1/([𝑛/𝑘]+1))𝜃(𝑢)≤𝛼(𝑢). Here, we also suppose 𝜃(𝑢)=𝑢(𝑡₀), and by the definitions of 𝛼 and the cone 𝑃, we can distinguish two cases.

(i) 𝛼(𝑢)=𝑢(𝑘+1), then we certainly have 𝑡₀≥𝑘+1, and 𝑢(𝑡₀𝑡)=Δ𝑢(0)+⋯+Δ𝑢(𝑘)+Δ𝑢(𝑘+1)+⋯+Δ𝑢(2𝑘)+Δ𝑢(2𝑘+1)+⋯+Δ𝑢(3𝑘)+⋯+Δ𝑢([⁰𝑘𝑡]𝑘)+Δ𝑢([⁰𝑘]𝑘+1)+⋯+Δ𝑢(𝑡₀𝑡−1)≤([⁰𝑘𝑛]+1)𝑢(𝑘+1)≤([𝑘]+1)𝑢(𝑘+1), (3.8) that is, 1/([𝑛/𝑘]+1)𝑢(𝑡₀)≤𝑢(𝑘+1).

(ii) 𝛼(𝑢)=𝑢(𝑛−𝑘−1), then 𝑡₀≤𝑛−𝑘−1, 𝑢(𝑛)≤𝑢(𝑛−𝑘−1) and 𝑢(𝑡₀)=𝑢(𝑛)+(−Δ𝑢(𝑛−1)−⋯−Δ𝑢(𝑛−𝑘−1)−Δ𝑢(𝑛−𝑘−2)−⋯−Δ𝑢(𝑛−2𝑘−1)−Δ𝑢(𝑛−2𝑘−2)−⋯−Δ𝑢(𝑛−[𝑛−𝑡₀𝑘]𝑘−1)−Δ𝑢(𝑛−[𝑛−𝑡₀𝑘]𝑘−2)−⋯−Δ𝑢(𝑡₀))≤𝑢(𝑛)+([𝑛−𝑡₀𝑘]+1)[𝑢(𝑛−𝑘−1)−𝑢(𝑛)]≤([𝑛−𝑡₀𝑘𝑛]+1)𝑢(𝑛−𝑘−1)≤([𝑘]+1)𝑢(𝑛−𝑘−1), (3.9) that is, (1/([𝑛/𝑘]+1))𝑢(𝑡₀)≤𝑢(𝑛−𝑘−1). So, we have (1/([𝑛/𝑘]+1))𝜃(𝑢)≤𝛼(𝑢).

Lemma 3.2. Assume that (𝐻₁)-(𝐻₃) hold. Then, for any 𝑥∈𝐸,𝑥(𝑡)≥0, Δ(𝜙_𝑝(Δ𝑢(𝑡−1))+𝑞(𝑡)𝑓(𝑡,𝑥(𝑡),Δ𝑥(𝑡))=0,𝑡∈{1,…,𝑛−1},𝑢(0)=0,𝑢(𝑛)=^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖) (3.10)(3.11) has a unique solution 𝑢(𝑡) given by 𝑢(𝑡)=^𝑡−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))) (3.12) or 𝑢(𝑡)=^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖)))−^𝑛−1_𝑗=𝑡𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.13) where 𝐴_𝑢 satisfies ^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖)))=^𝑛−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))). (3.14)

Proof. For any 𝑥∈𝐸 suppose that 𝑢 is a solution of the BVPs (3.7) and (3.11). According to the property of the difference operator, it follows that 𝜙_𝑝(Δ𝑢(1))−𝜙_𝑝𝜙(Δ𝑢(0))=−𝑞(1)𝑓(1,𝑥(1),Δ𝑥(1)),_𝑝(Δ𝑢(2))−𝜙_𝑝⋯𝜙(Δ𝑢(1))=−𝑞(2)𝑓(2,𝑥(2),Δ𝑥(2)),_𝑝(Δ𝑢(𝑡))−𝜙_𝑝(Δ𝑢(𝑡−1))=−𝑞(𝑡)𝑓(𝑡,𝑥(𝑡),Δ𝑥(𝑡)), (3.15) then 𝜙_𝑝(Δ𝑢(𝑡))−𝜙_𝑝(Δ𝑢(0))=−^𝑡_𝑖=1𝜙𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖)),_𝑝(Δ𝑢(𝑡))=𝜙_𝑝(Δ𝑢(0))−^𝑡_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖)). (3.16) Let 𝜙_𝑝(Δ𝑢(0))=𝐴_𝑢, Δ𝑢(𝑡)=𝜙_𝑝⁻¹(𝐴_𝑢−^𝑡_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.17) and by Δ𝑢(0)=𝑢(1)−𝑢(0)=𝜙_𝑝⁻¹(𝐴_𝑢),Δ𝑢(1)=𝑢(2)−𝑢(1)=𝜙_𝑝⁻¹(𝐴_𝑢−¹_𝑖=1⋯𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))),Δ𝑢(𝑡−1)=𝑢(𝑡)−𝑢(𝑡−1)=𝜙_𝑝⁻¹(𝐴_𝑢−^𝑡−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.18) then 𝑢(𝑡)−𝑢(0)=^𝑡−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))). (3.19) By Δ𝑢(𝑛−1)=𝑢(𝑛)−𝑢(𝑛−1)=𝜙_𝑝⁻¹(𝐴_𝑢−^𝑛−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))),Δ𝑢(𝑛−2)=𝑢(𝑛−1)−𝑢(𝑛−2)=𝜙_𝑝⁻¹(𝐴_𝑢−^𝑛−2_𝑖=1⋯𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))),Δ𝑢(𝑡)=𝑢(𝑡+1)−𝑢(𝑡)=𝜙_𝑝⁻¹(𝐴_𝑢−^𝑡_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.20) so that 𝑢(𝑛)−𝑢(𝑡)=^𝑛−1_𝑗=𝑡𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))),𝑢(𝑡)=𝑢(𝑛)−^𝑛−1_𝑗=𝑡𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.21) Using the boundary condition (3.11), we can easily obtain 𝑢(𝑡)=^𝑡−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))) (3.22) or 𝑢(𝑡)=^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖)))−^𝑛−1_𝑗=𝑡𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))), (3.23) where 𝐴_𝑢 satisfies (3.14).
On the other hand, it is easy to verify that if 𝑢 is the solution of (3.12) or (3.13), then 𝑢 is a solution of (3.10) and (3.11).

Lemma 3.3. For any 𝑢∈𝐸, 𝑢(𝑡)≥0, there exists a unique 𝐴_𝑢∈(−∞,+∞) satisfying (3.14). Moreover, there is a unique 𝑛₀∈{1,…,𝑛−1}, such that ^𝑛0−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))<𝐴_𝑢≤^𝑛0_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)). (3.24)

Proof. Let 𝜑(𝑡)=^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝑡−^𝑗_𝑖=1−𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=0}𝜙_𝑝⁻¹(𝑡−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),𝑡∈(−∞,+∞), (3.25) so that 𝜑(0)=^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝑛−1_𝑗=0𝜙_𝑝⁻¹(−^𝑗_𝑖=1=𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=0}𝜙_𝑝⁻¹(^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑗_𝑖=1>𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑚−2𝑖=1}𝑎_𝑖[^𝑛−1_𝑗=0𝜙_𝑝⁻¹(^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))]≥0,𝜑(^𝑛−1_𝑖=1=𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝑛−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1<𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑚−2𝑖=1}𝑎_𝑖[^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝑛−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))]≤0. (3.26) By the continuity of 𝜑(𝑡), we know that there exists at least one 𝐴_𝑢∈(0,^𝑛−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))⊂(−∞,+∞) (3.27) satisfying (3.14).
On the other hand, 𝜑^′(𝑡)=^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0(𝜙_𝑝⁻¹)^′(𝑡−^𝑗_𝑖=1−𝑞(𝑡)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=0}(𝜙_𝑝⁻¹)^′(𝑡−^𝑗_𝑖=1<𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑚−2𝑖=1}𝑎_𝑖[^𝜉𝑖−1_𝑗=0(𝜙_𝑝⁻¹)^′(𝑡−^𝑗_𝑖=1−𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=0}(𝜙_𝑝⁻¹)^′(𝑡−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))]≤0. (3.28) Then 𝜑(𝑡) is strictly increasing on 𝑡∈(−∞,+∞).
So, there exists a unique 𝐴_𝑢∈(−∞,+∞) satisfying (3.14).
Moreover, there exists a 𝑛₀∈{1,…,𝑛−1}, such that ^𝑛0−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))<𝐴_𝑢≤^𝑛0_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)). (3.29)

Lemma 3.4. Assume that (𝐻₁)-(𝐻₃) hold. If 𝑥∈𝐸,𝑥(𝑡)≥0, then the unique solution 𝑢(𝑡) of the BVPs (3.10) and (3.11) has the following properties:(i)Δ²𝑢(𝑡)≤0, and 𝑢(𝑡)≥0;(ii)there exists a unique 𝑛₀∈{0,1,…,𝑛}, such that 𝑢(𝑛₀)=max_{𝑡∈{0,1,…,𝑛}}𝑢(𝑡), which 𝑛₀ is given in Lemma 3.3.

Proof. Suppose that 𝑢(𝑡) is the solution of (3.10) and (3.11). Then we have the following.(i)By Lemma 3.2, it is easy to see that Δ²𝑢(𝑡)≤0. Without loss of generality, we assume that 𝑢(𝑛)=min{𝑢(0),𝑢(𝑛)}. By Δ²𝑢(𝑡)≤0, we know that 𝑢(𝑡)≥𝑢(𝑛),𝑡∈{0,1,…,𝑛}. So we get 𝑢(𝑛)=∑^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖)≥∑^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝑛), that is, (1−∑^𝑚−2_𝑖=1𝑎_𝑖)𝑢(𝑛)≥0. Hence 𝑢(𝑛)≥0. So, from the concavity of 𝑢(𝑡), we know that 𝑢(𝑡)≥0,𝑡∈{0,1,…,𝑛}.(ii)From 𝑢(𝑛₀)=^𝑛0−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),𝑢(𝑛₀+1)=^𝑛0−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))+𝜙_𝑝⁻¹(𝐴_𝑢−^𝑛0_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))), (3.30) and by Lemma 3.3, we have 𝐴_𝑢−∑^𝑛0_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))<0, so that 𝑢(𝑛₀)>𝑢(𝑛₀+1).
Also 𝑢(𝑛₀−1)=^𝑛0−2_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))=𝑢(𝑛₀)−𝜙_𝑝⁻¹(𝐴_𝑢−^𝑛0−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))), (3.31) and one arrives at 𝑢(𝑛₀)≥𝑢(𝑛₀−1). So, 𝑢(𝑛₀)=max_{𝑡∈{0,1,…,𝑛}}𝑢(𝑡).
If there exist 𝑛₁,𝑛₂∈{0,1,…,𝑛},𝑛₁<𝑛₂ such that Δ𝑢(𝑛₁)=Δ𝑢(𝑛₂), then 0=𝜙_𝑝(Δ𝑢(𝑛₁))−𝜙_𝑝(Δ𝑢(𝑛₂))=^𝑛2_𝑛1+1𝑞(𝑖)𝑓(𝑖,𝑥(𝑖),Δ𝑥(𝑖))>0, (3.32) which is a contradiction.

Lemma 3.5. For any 𝑢∈𝑃, define the operator (𝑇𝑢)(𝑡)=^{𝑡−1𝑗=0}𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),0≤𝑡≤𝑛₀,_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1−𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=𝑡}𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),𝑛₀+1≤𝑡≤𝑛. (3.33) Then 𝑇∶𝑃→𝑃 is completely continuous.

Proof. Using the continuity of 𝑓 and the definition of 𝑇, it is easy to show that 𝑇∶𝑃→𝑃 is continuous. Next, we prove that 𝑇 is completely continuous.
Suppose that the sequence {𝑢_𝑖}⊆𝑃 is bounded, then there exists 𝑀>0, such that 𝑢_𝑖(𝑗)≤𝑀, for any 𝑖=1,2,…,𝑗=0,1,…,𝑛. By the continuity of 𝑓, 𝜙_𝑝⁻¹ and 𝐴_𝑢 are bounded, and we know that there exists 𝑀^′>0, such that |𝑇𝑢_𝑖(𝑡)|≤𝑀^′, for 𝑡∈{0,1,…,𝑛} and 𝑖=1,2,…,𝑛,…. In view of the bounded sequence {𝑇𝑢_𝑖(0)}, there exists {𝑢_𝑖0}⊆{𝑢_𝑖}, such that lim_𝑖→∞𝑇𝑢_𝑖0(0)=𝑎₀. For the bounded sequence {𝑇𝑢_𝑖0(1)}, there exists {𝑢_𝑖1}⊆{𝑢_𝑖0}, such that lim_𝑖→∞𝑇𝑢_𝑖1(1)=𝑎₁. By repetition in this way, we have that there exists {𝑢_𝑖𝑗}⊆{𝑢_𝑖𝑗−1} for 𝑗=2,3,…,𝑛, such that lim_𝑖→∞𝑇𝑢_𝑖𝑗(𝑗)=𝑎_𝑗. Let 𝑦={𝑎₀,𝑎₁,…,𝑎_𝑛}; by the definition of the norm on 𝐸, there exists {𝑢_𝑖𝑛}⊆{𝑢_𝑖}, such that lim_𝑖→∞𝑇𝑢_𝑖𝑛(𝑗)=𝑦.
Hence, 𝑇∶𝑃→𝑃 is completely continuous.

4. Existence of Triple Positive Solutions to (1.2)

We are now ready to apply Avery-Peterson's fixed point theorem to the operator 𝑇 to give sufficient conditions for the existence of at least three positive solutions to the BVPs (1.2).

Now for convenience we introduce the following notations. Let

𝐿=𝜙_𝑝⁻¹(^𝑛−1_𝑖=1𝑞(𝑖)),𝑀=min{^[𝑛/2]_𝑗=𝑘+1𝜙_𝑝⁻¹(^[𝑛/2]_𝑖=𝑗+1𝑞(𝑖)),^{𝑛−𝑘−1}_{𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖))},𝑁=max{^[𝑛/2]_𝑗=0𝜙_𝑝⁻¹(^[𝑛/2]+1_𝑖=𝑗+1𝑞(𝑖)),^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖))+^𝑛−1_{𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖))}. (4.1)

Theorem 4.1. Assume that conditions (𝐻₁)-(𝐻₃) hold. Let 0<𝑎<𝑏<([𝑛/𝑘]+1)𝑏<𝑑, and suppose that 𝑓 satisfies the following conditions:(𝐴₁)𝑓(𝑡,𝑢,𝑣)≤𝜙_𝑝(𝑑/𝐿), for (𝑡,𝑢,𝑣)∈{0,1,…,𝑛}×[0,𝑛𝑑]×[−𝑑,𝑑];(𝐴₂)𝑓(𝑡,𝑢,𝑣)>𝜙_𝑝(([𝑛/𝑘]+1)𝑏/𝑀), for (𝑡,𝑢,𝑣)∈{𝑘+1,…,𝑛−𝑘−1}×[𝑏,([𝑛/𝑘]+1)𝑏]×[−𝑑,𝑑];(𝐴₃)𝑓(𝑡,𝑢,𝑣)<𝜙_𝑝(𝑎/𝑁), for (𝑡,𝑢,𝑣)∈{0,1,…,𝑛}×[0,𝑎]×[−𝑑,𝑑].Then BVPs (1.2) have at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ such that max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢_𝑖(𝑡)|≤𝑑,𝑖=1,2,3,𝑏<min_{𝑡∈{𝑘+1,…,𝑛−𝑘−1}}|𝑢₁(𝑡)|,max_{𝑡∈{0,1,…,𝑛}}|𝑢₁(𝑡)|≤𝑏,𝑎<max_{𝑡∈{0,1,…,𝑛}}|𝑢₂𝑛(𝑡)|<([𝑘]+1)𝑏,withmin_{𝑡∈{𝑘+1,…,𝑛−𝑘−1}}|𝑢₂(𝑡)|<𝑏,max_{𝑡∈{0,1,…,𝑛}}|𝑢₃(𝑡)|<𝑎. (4.2)

Proof. The BVPs (1.2) have a solution 𝑢=𝑢(𝑡) if and only if 𝑢 solves the operator equation 𝑢=𝑇𝑢. Thus we set out to verify that the operator 𝑇 satisfies Avery-Peterson's fixed point theorem which will prove the existence of three fixed points of 𝑇 which satisfy the conclusion of the theorem. Now the proof is divided into some steps.
(1) We will show that (𝐴₁) implies that 𝑇∶𝑃(𝑟,𝑑)→𝑃(𝑟,𝑑).
In fact, for 𝑢∈𝑃(𝑟,𝑑), there is 𝛾(𝑢)=max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢(𝑡)|≤𝑑. With Lemma 3.1, there is max_{𝑡∈{0,1,…,𝑛}}|𝑢(𝑡)|≤𝑛𝑑, and then condition (𝐴₁) implies 𝑓(𝑡,𝑢,𝑣)≤𝜙_𝑝(𝑑/𝐿). On the other hand, for 𝑢∈𝑃, there is 𝑇𝑢∈𝑃, then 𝑇𝑢 is concave on 𝑡∈{0,1,…,𝑛}, and max|Δ𝑇𝑢(𝑡)|=max{|Δ𝑇𝑢(0)|,|Δ𝑇𝑢(𝑛−1)|}, and so 𝛾(𝑇𝑢)=max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑇𝑢(𝑡)|=max{|Δ𝑇𝑢(0)|,|Δ𝑇𝑢(𝑛−1)|}=max{𝜙_𝑝⁻¹(𝐴_𝑢),|𝜙_𝑝⁻¹(𝐴_𝑢−^𝑛−1_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))|}≤𝜙_𝑝⁻¹(^𝑛−1_𝑖=1𝑞(𝑖)𝜙_𝑝(𝑑𝐿𝑑))=𝐿𝐿=𝑑. (4.3) Thus, 𝑇∶𝑃(𝑟,𝑑)→𝑃(𝑟,𝑑) holds.
(2)We show that condition (𝑆₁) in Theorem 2.4 holds.
We take 𝑢(𝑡)=([𝑛/𝑘]+1)𝑏, for 𝑡∈{1,2,…,𝑛}, and 𝑢(0)=0. It is easy to see that 𝑢(𝑡)∈𝑃(𝛾,𝜃,𝛼,𝑏,([𝑛/𝑘]+1)𝑏,𝑑) and 𝛼(𝑢)=([𝑛/𝑘]+1)𝑏>𝑏. Hence {𝑢∈𝑃(𝛾,𝜃,𝛼,𝑏,([𝑛/𝑘]+1)𝑏,𝑑)∣𝛼(𝑢)>𝑏}≠∅. Thus for 𝑢∈𝑃(𝛾,𝜃,𝛼,𝑏,([𝑛/𝑘]+1)𝑏,𝑑), there is 𝑏≤𝑢(𝑡)≤([𝑛/𝑘]+1)𝑏, |Δ𝑢(𝑡)|≤𝑑. Hence by condition (𝐴₂) of this theorem, one has 𝑓(𝑡,𝑢,𝑣)>𝜙_𝑝(([𝑛/𝑘]+1)𝑏/𝑀) for 𝑡∈[𝑘+1,𝑛−𝑘−1]. By Lemma 3.4 and combining the conditions on 𝛼 and 𝑃, we have 𝛼(𝑇𝑢)=min_{𝑘+1≤𝑡≤𝑛−𝑘−1}|(𝑇𝑢)(𝑡)|≥1([𝑛/𝑘]+1)max_{𝑡∈{0,…,𝑛}}1|𝑇𝑢(𝑡)|=([𝑛/𝑘]+1)(𝑇𝑢)(𝑛₀)=1([𝑛/𝑘]+1)min{^𝑛0−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))−^𝑛−1_𝑗=𝑛0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))}≥1([𝑛/𝑘]+1)min{^𝑛0−1_𝑗=𝑘+1𝜙_𝑝⁻¹(^𝑛0−1_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),_{𝑛−𝑘−1𝑗=𝑛0}𝜙_𝑝⁻¹(^𝑗_𝑖=𝑛0+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))}≥1([𝑛/𝑘]+1)min{^[𝑛/2]_𝑗=𝑘+1𝜙_𝑝⁻¹(^[𝑛/2]_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),_{𝑛−𝑘−1𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))}>1([𝑛/𝑘]+1)min{^[𝑛/2]_𝑗=𝑘+1𝜙_𝑝⁻¹(^[𝑛/2]_𝑖=𝑗+1𝑞(𝑖)𝜙_𝑝(([𝑛/𝑘]+1)𝑏𝑀)),_{𝑛−𝑘−1𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖)𝜙_𝑝(([𝑛/𝑘]+1)𝑏𝑀))}=1([𝑛/𝑘]+1)([𝑛/𝑘]+1)𝑏𝑀min{^[𝑛/2]_𝑗=𝑘+1𝜙_𝑝⁻¹(^[𝑛/2]_𝑖=𝑗+1𝑞(𝑖)),^{𝑛−𝑘−1}_{𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖))}=𝑏𝑀𝑀=𝑏. (4.4)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) Therefore we have 𝑛𝛼(𝑇𝑢)>𝑏∀𝑢∈𝑃(𝛾,𝜃,𝛼,𝑏,([𝑘]+1)𝑏,𝑑). (4.5) Consequently, condition (𝑆₁) in Theorem 2.4 is satisfied.
(3)We now prove that (𝑆₂) in Theorem 2.4 holds.
With (3.7), we have 1𝛼(𝑇𝑢)≥1([𝑛/𝑘]+1)𝜃(𝑇𝑢)>𝑛([𝑛/𝑘]+1)([𝑘]+1)𝑏=𝑏 (4.6) for 𝑢∈𝑃(𝛾,𝛼,𝑏,𝑑) with 𝜃(𝑇𝑢)>([𝑛/𝑘]+1)𝑏. Hence, condition (𝑆₂) in Theorem 2.4 is satisfied.
(4)Finally, we prove that (𝑆₃) in Theorem 2.4 is also satisfied.
Since 𝜓(0)=0<𝑎, so 0∈𝑅(𝛾,𝜓,𝑎,𝑑). Suppose that 𝑢∈𝑅(𝛾,𝜓,𝑎,𝑑) with 𝜓(𝑢)=𝑎. Then, by the condition (𝑆₃) of this theorem, 𝜓(𝑇𝑢)=max_{𝑡∈{0,1,…,𝑛}}|𝑇𝑢(𝑡)|=𝑇𝑢(𝑛₀)=max{^𝑛0−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1−𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=𝑛0}𝜙_𝑝⁻¹(𝐴_𝑢−^𝑗_𝑖=1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))}≤max{^[𝑛/2]_𝑗=0𝜙_𝑝⁻¹(^[𝑛/2]+1_𝑖=𝑗+1𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖))),_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1+𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))_{𝑛−1𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖)𝑓(𝑖,𝑢(𝑖),Δ𝑢(𝑖)))}≤max{^[𝑛/2]_𝑗=0𝜙_𝑝⁻¹(^[𝑛/2]+1_𝑖=𝑗+1𝑞(𝑖)𝜙_𝑝(𝑎𝑁)),_{𝑚−2𝑖=1}𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖)𝜙_𝑝(𝑎𝑁))+^𝑛−1_{𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}𝑞(𝑖)𝜙_𝑝(𝑎𝑁=𝑎))}𝑁max{^[𝑛/2]_𝑗=0𝜙_𝑝⁻¹(^[𝑛/2]+1_𝑖=𝑗+1𝑞(𝑖)),^𝑚−2_𝑖=1𝑎_𝑖^𝜉𝑖−1_𝑗=0𝜙_𝑝⁻¹(^𝑛−1_𝑖=𝑗+1𝑞(𝑖))+^𝑛−1_{𝑗=[𝑛/2]}𝜙_𝑝⁻¹(^𝑗_{𝑖=[𝑛/2]+1}=𝑎𝑞(𝑖))}𝑁𝑁=𝑎. (4.7) Thus condition (𝑆₃) in Theorem 2.4 holds.

Therefore an application of Theorem 2.4 implies that BVPs (1.2) have at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ such that

max_{𝑡∈{0,1,…,𝑛−1}}|Δ𝑢_𝑖(𝑡)|≤𝑑,𝑖=1,2,3,𝑏<min_{𝑡∈{𝑘+1,…,𝑛−𝑘−1}}|𝑢₁(𝑡)|,max_{𝑡∈{0,1,…,𝑛}}|𝑢₁(𝑡)|≤𝑏,𝑎<max_{𝑡∈{0,1,…,𝑛}}|𝑢₂𝑛(𝑡)|<([𝑘]+1)𝑏,withmin_{𝑡∈{𝑘+1,…,𝑛−𝑘−1}}|u₂(𝑡)|<𝑏,max_{𝑡∈{0,1,…,𝑛}}|𝑢₃(𝑡)|<𝑎. (4.8)

Acknowledgments

The project is supported by the Natural Science Foundation of China (10971045), and the Natural Science Foundation of Hebei Province (A2009000664).