Positive Solutions for a Class of Third-Order Three-Point Boundary Value Problem

Abstract

We investigate the problem of existence of positive solutions for the nonlinear third-order three-point boundary value problem 𝑢^′′′(𝑡)+𝜆𝑎(𝑡)𝑓(𝑢(𝑡))=0, 0<𝑡<1, 𝑢(0)=𝑢^′(0)=0, 𝑢^′′(1)=∝𝑢^′′(𝜂), where 𝜆 is a positive parameter, ∝∈(0,1), 𝜂∈(0,1), 𝑓∶(0,∞)→(0,∞), 𝑎∶(0,1)→(0,∞) are continuous. Using a specially constructed cone, the fixed point index theorems and Leray-Schauder degree, this work shows the existence and multiplicities of positive solutions for the nonlinear third-order boundary value problem. Some examples are given to demonstrate the main results.

1. Introduction

This paper deals with the following third-order nonlinear boundary value problem: 𝑢^′′′(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)+𝜆𝑎))=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′′(1)=𝛼𝑢^′′𝜂. (1.1) Third-order boundary value problems arise in a variety of different areas of applied mathematics and physics. In the few years, there has been increasing interest in studying certain third-order boundary value problems for nonlinear differential equation and have received much attention. To identify a few, we refer the reader to [1–6].

Recently, El-Shahed [1] discussed the following third-order two-point boundary value problem: 𝑢^′′′(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)+𝜆𝑎))=0,0<𝑡<1,=𝑢^′(0)=0,𝛼𝑢^′(1)+𝛽𝑢^′′(1)=0. (1.2) The methods employed in [1] are Kransnoselskii's fixed-point theorem of cone.

In later work, by placing restrictions on the nonlinear term 𝑓, Sun [2] studied the following boundary value problems and obtained the three solution via leggett-williams fixed point theorem: 𝑢^′′′(𝑡)=𝑎(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡)𝑢(0)𝜂,0<𝑡<1,=𝛿𝑢=0,𝑢^′𝜂=0,𝑢^′′(1)=0. (1.3)

The upper and lower solution is a powerful tool for proving existence for boundary value problems, Ma [7] studied the multiplicity of positive solutions of three-point boundary value problem of second-order ordinary differential equations. Du et al. [5] investigated a class of third-order nonlinear problem.

Motivated by the work of the above papers, the purpose of this article is to study the existence of solution for boundary value problem (1.1) using a new technique (different from the proof of [1, 2, 7]) and we get a new existence result. The tools are based on the fixed point index theorems and Leray-Schauder degree.

The paper is organized as follows: Section 2 states some definitions and some lemmas which are important to obtain our main result. Section 3 is devoted to the existence result of BVP (1.1). Section 4 gives some examples to illustrate our main results.

2. Preliminary

Definition 2.1. Let 𝐸 be a real Banach space. A nonempty closed convex set 𝐾⊂𝐸 is called a cone of 𝐸 if it satisfies the following two conditions:(1)𝑥∈𝐾, 𝜆≥0 implies 𝜆𝑥∈𝐾;(2)𝑥∈𝐾, −𝑥∈𝐾 implies 𝑥=0.

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

Lemma 2.3. Let 𝑦∈𝐶[0,1], then the following boundary value problem: 𝑢^′′′(𝑡)(𝑡)𝑢(0)+𝑦=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′′(1)=𝛼𝑢^′′𝜂, (2.1)(2.2) has the unique solution 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (2.3) where 𝐺()=−1𝑡,𝑠2(𝑡−𝑠)²+𝑡²2𝑡,𝑠≤𝜂,𝑠≤𝑡,₂2−1,𝑡≤𝑠≤𝜂,2(𝑡−𝑠)²+𝑡²2()𝑡1−𝛼,𝜂≤𝑠≤𝑡,₂2(1−𝛼),𝜂≤𝑠,𝑡≤𝑠. (2.4)

Proof. From (2.1), we have 𝑢(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝑑𝑠+𝐴𝑡²+𝐵𝑡+𝐶. (2.5) In particular, 𝑢(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝑑𝑠+𝐴𝑡²𝑢+𝐵𝑡+𝐶,_′(𝑡)=−𝑡^𝑡₀𝑦(𝑠)𝑑𝑠+^𝑡₀(𝑠)𝑢𝑠𝑦𝑑𝑠+2𝐴𝑡+𝐵,_′′(𝑡)=−^𝑡₀𝑦(𝑠)𝑑𝑠+2𝐴. (2.6) Combining this with boundary conditions (2.2), we conclude that ∫𝐴=¹⁰𝑦(𝑠)𝑑𝑠2()−1−𝛼𝛼∫^𝜂₀𝑦(𝑠)𝑑𝑠2(),1−𝛼𝐵=0,𝐶=0. (2.7) Therefore, BVP (2.1)-(2.2) has a unique solution: 𝑢(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝑑𝑠−𝛼𝑡²∫^𝜂₀𝑦(𝑠)𝑑𝑠2()+𝑡1−𝛼²∫¹₀𝑦(𝑠)𝑑𝑠2()=1−𝛼_𝑡0−12(𝑡−𝑠)²+𝑡²2𝑦(𝑠)𝑑𝑠+^𝜂_𝑡𝑡²2𝑦(𝑠)𝑑𝑠+¹_𝜂𝑡²2()𝑦(𝑠)1−𝛼𝑑𝑠,𝑡≤𝜂,_𝜂0−12(𝑡−𝑠)²+𝑡²2𝑦(𝑠)𝑑𝑠+^𝑡_𝜂−12(𝑡−𝑠)²+𝑡²2()𝑦(𝑠)+1−𝛼𝑑𝑠_1𝑡𝑡²2()𝑦(𝑠)=1−𝛼𝑑𝑠,𝑡≥𝜂,₁₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (2.8) The proof is completed.

Lemma 2.4. For all (𝑡,𝑠)∈[0,1]×[0,1], one has 𝐺(𝑡,𝑠)≥0.

Lemma 2.5. for all (𝑡,𝑠)∈[𝜏,1]×[0,1], one has 𝛾𝐺(1,𝑠)≤𝐺(𝑡,𝑠)≤𝐺(1,𝑠), (2.9) where 𝛾=𝛼𝜏²/2, and 𝜏 statisfies ∫¹_𝜏𝐺(𝑡,𝑠)𝑎(𝑠)𝑑𝑠>0.

Proof. For 𝑠≤𝑡,𝑠≤𝜂, 𝐺()1𝑡,𝑠=−2(𝑡−𝑠)²+𝑡²2=𝑠(2𝑡−𝑠)2(),𝐺()≤𝐺1,𝑠𝑡,𝑠𝐺()=1,𝑠2𝑡−𝑠=2−𝑠𝑡+𝑡−𝑠≥𝑡2−𝑠2. (2.10) For 𝑡≤𝑠≤𝜂, 𝐺()=𝑡𝑡,𝑠²2(),𝐺()≤𝐺1,𝑠𝑡,𝑠𝐺()=𝑡1,𝑠²/21/2=𝑡². (2.11) For 𝜂≤𝑠≤𝑡, 𝐺()1𝑡,𝑠=−2(𝑡−𝑠)²+𝑡²2()=1−𝛼𝛼𝑡²+2𝑡𝑠(1−𝛼)+𝑠²(1−𝛼)2()(),𝐺()1−𝛼≤𝐺1,𝑠𝑡,𝑠𝐺()=1,𝑠𝛼𝑡²+2𝑡𝑠(1−𝛼)+𝑠²(1−𝛼)𝛼+2𝑠(1−𝛼)+𝑠²(1−𝛼)≥𝛼𝑡². (2.12) For 𝜂≤𝑠,𝑡≤𝑠, 𝐺()=𝑡𝑡,𝑠²2()(),𝐺()1−𝛼≤𝐺1,𝑠𝑡,𝑠𝐺(1,𝑠)=𝑡². (2.13) Thus, 𝛼𝑡²2𝐺(1,𝑠)≤𝐺(𝑡,𝑠)≤𝐺(1,𝑠),for(𝑡,𝑠)∈[0,1]×[0,1]. (2.14) Therefore, 𝛾𝐺(1,𝑠)≤𝐺(𝑡,𝑠)≤𝐺(1,𝑠),∀(𝑡,𝑠)∈[𝜏,1]×[0,1]. (2.15) The proof is completed.

Lemma 2.6. If 𝑦∈𝐶[0,1] and 𝑦≥0, then the unique solution𝑢(𝑡)of the BVP (2.1)-(2.2) is nonnegative and satisfies min𝑢_{[]𝑡∈𝜏,1}(𝑡)≥𝛾‖𝑢‖. (2.16)

Proof. Let 𝑦∈𝐶⁺[0,1], it is obvious that it is nonnegative. For any 𝑡∈[0,1], by (2.3) and Lemma 2.5, it follows that 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠≤¹₀𝐺(1,𝑠)𝑦(𝑠)𝑑𝑠, (2.17) and thus, ‖𝑢‖≤¹₀𝐺(1,𝑠)𝑦(𝑠)𝑑𝑠. (2.18) On the other hand, (2.3) and Lemma 2.5 imply, for any 𝑡∈[𝜏,1], 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠≥𝛾¹₀𝐺(1,𝑠)𝑦(𝑠)𝑑𝑠. (2.19) Therefore, min_{[]𝑡∈𝜏,1}𝑢(𝑡)≥𝛾‖𝑢‖. (2.20) This completes the proof.

Let 𝐸=𝐶[0,1] with the usual normal ‖𝑢‖=max_𝑡∈[0,1]|𝑢(𝑡)|.

Define the cone 𝐾 by 𝐾=𝑢∈𝐶⁺[0,1]∶min_{[]𝑡∈𝜏,1}𝑢(𝑡)≥𝛾‖𝑢‖. (2.21)

Define an operator 𝑇 by 𝑇𝑢(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠. (2.22)

By Lemma 2.3, BVP (1.1) has a positive solution 𝑢=𝑢(𝑡) if and only if 𝑢 is a fixed point of 𝑇.

Lemma 2.7. Assume that 0<𝜆<∞. Then, 𝑇∶𝐾→𝐾 is completely continuous.

Proof. Firstly, it is easy to check that 𝑇∶𝐾→𝐾 is well defined. By Lemma 2.6, we know that 𝑇(𝐾)⊂𝐾.
Let Ω be any boundary subset of 𝐾, then there exists 𝑟>0, ‖𝑢‖≤𝑟, for all 𝑢∈Ω. Therefore, we have |𝑇𝑢|=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≤𝜆¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠. (2.23) So 𝑇Ω is boundary. Moreover, for any 𝑡₁,𝑡₂∈[0,1],|𝑡₁−𝑡₂|≤𝛿,𝛿>0, we have 𝑇𝑢𝑡₁−𝑇𝑢𝑡₂≤𝜆¹₀𝐺𝑡₁,𝑠−𝐺𝑡₂,𝑠𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠. (2.24) By the continuity of 𝑓 and 𝑎, we have 𝑎(𝑡) and 𝑓(𝑢(𝑡)) are boundary on 𝑢∈Ω,𝑡∈[0,1], which means that there exists a constant 𝑀^𝑓_𝑎>0, depending only on Ω such that 𝑎(𝑡)𝑓(𝑢(𝑡))<𝑀^𝑓_𝑎, (2.25) and thus for any 𝜀>0, 𝐺𝑡₁,𝑠−𝐺𝑡₂≤𝜀,𝑠𝜆𝑀^𝑓_𝑎,𝑇𝑢𝑡₁−𝑇𝑢𝑡₂<𝜀. (2.26) Therefore, we can get 𝑇Ω is equicontinuity. Thirdly, we prove that 𝑇 is continuous. Let 𝑢_𝑛→𝑢 as 𝑛→∞, 𝑢_𝑛∈𝐾. Then, the continuity of 𝑓, we can get 𝑇𝑢_𝑛(𝑡)−𝑇𝑢(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢_𝑛(𝑠)𝑑𝑠−𝜆¹₀𝐺()𝑎(𝑠)𝑓(𝑢(𝑠=𝜆𝑡,𝑠))𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢_𝑛(𝑠)(𝑢(𝑠≤𝜆−𝑓))𝑑𝑠₁₀𝐺(1,𝑠)𝑎(𝑠)𝑓𝑢_𝑛(𝑠)−𝑓(𝑢(𝑠))𝑑𝑠⟶0,𝑛⟶∞. (2.27) Then, 𝑇𝑢_𝑛(𝑡)→𝑇𝑢(𝑡). Therefore, 𝑇 is continuous.The operator 𝑇 is completely continuous by an application of the Ascoli-Arzela theorem. This completes the proof.

Lemma 2.8 (see [7, 8]). Let 𝐸 be a real Banach space and let 𝐾 be a cone in 𝐸. For 𝑟≥0, define 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟}. Assume 𝑇∶𝐾_𝑟→𝐾 is a completely continuous operator such that 𝑇𝑥≠𝑥 for 𝑥∈𝜕𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}.(1)If ‖𝑇𝑥‖≥‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖𝑇,𝐾_𝑟,𝐾=0. (2.28)(2)If ‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖𝑇,𝐾_𝑟,𝐾=1. (2.29)

3. Main Results

Theorem 3.1. Assume that (A1)𝜆 is a positive parameter, 𝜂∈(0,1) and 𝛼∈(0,1);(A2)𝑎∶[0,1]→(0,∞) is continuous;(A3)𝑓∶[0,∞)→(0,∞) is continuous;(A4)𝑓_∞∶=lim_𝑢→∞(𝑓(𝑢)/𝑢)=∞.When 𝜆 is sufficiently small, (1.1) has at least one positive solution, whereas for 𝜆 is sufficiently large, (1.1) has no positive solution.

Proof. If 𝑞>0, then 𝛽𝑞=max_{‖𝑢‖𝑢∈𝐾,=𝑞}¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠>0. (3.1) For any number 0<𝑟₁, let 𝛿₁=𝑟₁/𝛽(𝑟₁), and set 𝐾_𝑟1=𝑢∈𝐾∶‖𝑢‖<𝑟₁. (3.2) Then, for 𝜆∈(0,𝛿₁) any 𝑢∈𝜕𝐾_𝑟1, we have 𝑇𝑢(𝑡)<𝛿₁¹₀𝐺(𝑡,𝑠)𝑓(𝑢(𝑠))𝑑𝑠≤𝛿₁𝛽𝑟₁=𝑟₁. (3.3) Thus, Lemma 2.8 implies 𝑖𝑇,𝐾_𝑟1,𝐾=1. (3.4) Since 𝑓_∞=∞, there is 𝑀>0, such that 𝑓(𝑢)≥𝜇𝑢, for 𝑢>𝑀, where 𝜇 is chosen so that 𝜆𝜇𝛾¹_𝜏𝐺(1,𝑠)𝑎(𝑠)𝑑𝑠>1. (3.5) Let 𝑟₂>𝑀/𝛾, and set 𝐾_𝑟2=𝑢∈𝐾∶‖𝑢‖<𝑟₂. (3.6) If 𝑢∈𝜕𝐾_𝑟2, then min_{[]𝑡∈𝜏,1}𝑢(𝑡)≥𝛾‖𝑢‖≥𝑀. (3.7) Therefore, 𝑇𝑢(1)=𝜆¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥𝜆¹_𝜏𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥𝜆¹_𝜏𝐺(1,𝑠)𝑎(𝑠)𝜇𝑢(𝑠)𝑑𝑠≥𝜆𝜇¹_𝜏𝐺(1,𝑠)𝑎(𝑠)𝑑𝑠𝛾‖𝑢‖≥𝜆𝜇𝛾¹_𝜏𝐺()𝑎(𝑠)‖𝑢‖>‖𝑢‖,1,𝑠𝑑𝑠 (3.8) which implies that ‖𝑇𝑢‖≥‖𝑢‖, (3.9) for 𝑢∈𝜕𝐾_𝑟2. An application of Lemma 2.8 again shows that 𝑖𝑇,𝐾_𝑟2,𝐾=0. (3.10) Since we can adjust 𝑟₁,𝑟₂ so that 𝑟₁<𝑟₂, it follows the additivity of the fixed-point index that 𝑖𝑇,𝐾_𝑟2⧵𝐾_𝑟1,𝐾=−1. (3.11) Thus, 𝑇 has a fixed point in 𝐾_𝑟2⧵𝐾_𝑟1 which is the desired positive solution of (1.1).
We verify that BVP of (1.1) has no positive solution for 𝜆 large enough.
Otherwise, there exist 0<𝜆₁<𝜆₂<⋯<𝜆_𝑛<⋯, with lim_𝑛→∞𝜆_𝑛=+∞, such that for any positive integer 𝑛, the BVP, 𝑢^′′′(𝑡)+𝜆_𝑛𝑎(𝑡)𝑓(𝑢(𝑡𝑢(0)))=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′′(1)=𝛼𝑢^′′𝜂, (3.12) has a positive solution 𝑢_𝑛(𝑡). By (2.22), we have 𝑢_𝑛=𝜆_𝑛¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓𝑢_𝑛(𝑠)⟶+∞,(𝑛⟶∞). (3.13) Thus, 𝑢_𝑛⟶∞,(𝑛⟶∞). (3.14) Since 𝑓_∞, for 𝑐₀>0, there exists 𝑟₃>0, such that 𝑓(𝑢)/𝑢>𝑐₀, for 𝑢∈[𝑟₃,∞), which implies that 𝑓(𝑢)>𝑐₀𝑢,for𝑢∈𝑟₃,∞. (3.15) Let 𝑛 be large enough that ‖𝑢_𝑛‖≥𝑟₃, then 𝑢_𝑛≥𝑢_𝑛(1)=𝜆_𝑛¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓𝑢_𝑛(𝑠)𝑑𝑠≥𝜆_𝑛𝛾¹₀𝐺(1,𝑠)𝑎(𝑠)𝑑𝑠𝑐₀𝑢_𝑛>u_𝑛. (3.16) Choose 𝑛 so that 𝑐₀𝜆_𝑛𝛾∫¹₀𝐺(1,𝑠)𝑎(𝑠)𝑑𝑠>1 which is a contradiction. The proof is completed.

Theorem 3.2. Assume that (B1)𝜆 is a positive parameter; 𝜂∈(0,1) and 𝛼∈(0,1);(B2)𝑎∶[0,1]→(0,∞) is continuous and there exists 𝑚>0 such that 𝑎(𝑡)≥𝑚; (B3)𝑓∶[0,∞)→(0,∞) is continuous;(B4)𝑓_∞=lim_𝑢→∞(𝑓(𝑢)/𝑢)=0,𝑓₀=lim_𝑢→0(𝑓(𝑢)/𝑢)=0;(B5)there exists 𝜎>0, for 𝑢≥𝜎, such that 𝑓(𝑢)≥𝛽, where 𝛽>0, then there exists 𝛿₂>0, such that, for 𝜆>𝛿₂, BVP (1.1) has at least two positive solutions 𝑢¹_𝜆,𝑢²_𝜆 and max𝑢¹_𝜆>𝜎.

Proof. Let 𝛿₂=(𝑀𝛾𝑚𝛽)⁻¹𝜎, then for 𝜆>𝛿₂, Lemma 2.7 implies that 𝑇∶𝐾→𝐾 is completely continuous. Considering (B4), there exists 0<𝑟<𝜎 such that 𝑓(𝑢)≤𝑢/2Λ𝜆, for 0≤𝑢≤𝑟, where Λ=∫¹₀𝐺(1,𝑠)𝑎(𝑠)𝑑𝑠.
So, for 𝑢∈𝜕𝐾_𝑟, we have from (2.4) (𝑇𝑢)(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠)𝑑𝑠≤𝜆¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≤𝜆¹₀𝐺()𝑎(s)‖𝑢‖1,𝑠𝑑𝑠=‖𝑢‖2Λ𝜆2<‖𝑢‖=𝑟. (3.17)
Consequently, for 𝑢∈𝜕𝐾_𝑟, we have ‖𝑇𝑢‖<‖𝑢‖, by Lemma 2.8, 𝑖𝑇,𝐾_𝑟,𝐾=1. (3.18) Now considering (B4), there exists ℎ>0, for 𝑢>ℎ, such that 𝑓(𝑢)≤𝑢/2Λ𝜆. Letting 𝜌=max_{0≤𝑢≤ℎ}𝑓(𝑢), then (𝑢)≤𝑢0≤𝑓2Λ𝜆+𝜌. (3.19) Choose 𝑅>max𝑟,2Λ𝜌𝜆. (3.20) So for 𝑢∈𝜕𝐾_𝑅, from (3.18) and (3.19), we have (𝑇𝑢)(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢)𝑑𝑠≤𝜆¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢)𝑑𝑠≤𝜆¹₀𝐺()𝑎(𝑠)11,𝑠𝑑𝑠‖𝑢‖<‖𝑢‖2Λ𝜆+𝜌2+𝑅2=‖𝑢‖, (3.21) That is, by Lemma 2.8, 𝑖𝑇,𝐾_𝑅,𝐾=1. (3.22) On the other hand, for 𝑢∈𝐾^𝑅_𝜎={𝑢∈𝐾∶‖𝑢‖≤𝑅,min_{𝑡∈𝐽𝜃}𝑢(𝑡)≥𝜎,𝜃∈(0,1/2),𝐽_𝜃=[𝜃,1−𝜃]}, (2.3) and (2.4) yield that ‖𝑇𝑢‖≤𝜆¹₀𝐺()𝑎(𝑠)1𝑡,𝑠𝑑𝑠‖𝑢‖2Λ𝜆+𝜌<𝑅. (3.23) Furthermore, for 𝑢∈𝐾^𝑅_𝜎, from (2.3) and (2.4), we obtain min_{𝑡∈𝐽𝜃}(𝑇𝑢)(𝑡)=min_{𝑡∈𝐽𝜃}𝜆¹₀𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥min_{𝑡∈𝐽𝜃}𝜆_𝜃^1−𝜃𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥𝜆𝛾_𝜃^1−𝜃𝐺(1,𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥𝜆𝑀𝛾𝑚𝛽>𝛿₂M𝛾𝑚𝛽=𝜎, (3.24) where 𝑀=∫_𝜃^1−𝜃𝐺(1,𝑠)𝑑𝑠. Let 𝑢₀≡(𝜎+𝑅)/2 and 𝐻(𝑡,𝑢)=(1−𝑡)𝑇𝑢+𝑡𝑢₀, then 𝐻∶[0,1]×𝐾^𝑅_𝜎→𝐾 is continuous, and from the analysis above, we obtain for (𝑡,𝑢)∈[0,1]×𝐾^𝑅_𝜎: 𝐻(𝑡,𝑢)∈𝐾^𝑅_𝜎. (3.25) Therefore, for 𝑢∈𝜕𝐾^𝑅_𝜎, we have 𝐻(𝑡,𝑢)≠𝑢. Hence, by the normality property and the homotopy invariance property of the fixed point index, we obtain 𝑖𝑇,𝐾^𝑅_𝜎,𝐾=𝑖𝑢₀,𝐾^𝑅_𝜎,𝐾=1. (3.26) Consequently, by the solution property of the fixed point index, 𝑇 has a fixed point 𝑢¹_𝜆and 𝑢¹_𝜆∈𝐾^𝑅_𝜎. By Lemma 2.4, it follows that 𝑢^𝜆₁ is a solution to BVP (1.1), and max_𝑡∈[0,1]𝑢¹_𝜆≥min_{𝑡∈𝐽𝜃}𝑢¹_𝜆>𝛾. (3.27) On the other hand, from (3.18) and (3.19) together with the additivity of the fixed point index, we get 𝑖𝑇,𝐾_𝑅⧵𝐾_𝑟∪𝐾^𝑅_𝜎=𝑖𝑇,𝐾_𝑅,𝐾−𝑖𝑇,𝐾^𝑅_𝜎,𝐾−𝑖𝑇,𝐾_𝑟,𝐾=1−1−1=−1. (3.28)
Hence, by the solution property of the fixed point index, 𝑇 has a fixed point 𝑢²_𝜆 and 𝑢²_𝜆∈𝐾_𝑅⧵(𝐾_𝑟∪𝐾^𝑅_𝜎). By Lemma 2.3, it follows that 𝑢²_𝜆 is also a solution to BVP (1.1), and 𝑢¹_𝜆≠𝑢²_𝜆. The proof is completed.

4. Examples

Example 4.1. We consider the following third-order boundary value problems: 𝑢^′′′(𝑡)+𝜆(2𝑡+1)𝑒^𝑢𝑢(0)=0,=𝑢^′(0)=0,𝑢^′′(1)=34𝑢^′′14, (4.1) here 𝜂=1/4,𝛼=3/4,𝑓(𝑢(𝑡))=𝑒^𝑢,𝑎(𝑡)=2𝑡+1,𝑓_∞=lim_𝑢→∞(𝑓(𝑢)/𝑢)=∞,𝑓 is continuous, 𝑎(𝑡) is continuous. By direct calculations, we obtain that 𝜆<𝑟₁(1−𝛼),for𝑟₁>0. Therefore, by Theorem 3.1, there exists at least one solution 𝑢(𝑡) for BVP (4.1), whereas for 𝜆 large enough, (4.1) has no solution.

Example 4.2. Consider the following third-order ordinary differential equation: 𝑢^′′′()𝑓(𝑢(𝑡𝑢(0)+𝜆2𝑡+1))=0,=𝑢^′(0)=0,𝑢^′′(1)=14𝑢^′′12, (4.2) where 𝑓(𝑢(𝑡=𝑢))²𝑒^−𝑢𝑎,if𝑢≤𝑎,_3/2√𝑢𝑒^−𝑎,if𝑢>𝑎, (4.3)𝑓 is continuous, 𝑎(𝑡) is continuous. Here, 𝑚=1, 𝛼=1/4,𝛽=𝑎²𝑒^−𝑎,𝜎=𝑎,𝑎>0. Choose 𝛿₂=6𝑎/(2𝜃³−3𝜃²+3𝜃−1),𝜃∈(0,1/2),𝜏∈(0,1), when 𝜆>𝛿₂, by Theorem 3.2, there exist at least two solutions 𝑢¹_𝜆(𝑡),𝑢²_𝜆(𝑡) for BVP (4.1).

Acknowledgment

This project is sponsored by the Natural Science Foundation of China (11101349, 11071205), the NFS of Jiangsu Province (BK2011042), the NSF of Education Department of Jiangsu Province (11KJB11003), and Jiangsu Government Scholarship Program.