Existence of Positive Solutions for a Fourth-Order Periodic Boundary Value Problem

Abstract

The existence results of positive solutions are obtained for the fourth-order periodic boundary value problem 𝑢⁽⁴⁾−𝛽𝑢^′′+𝛼𝑢=𝑓(𝑡,𝑢,𝑢^′′), 0≤𝑡≤1, 𝑢^(𝑖)(0)=𝑢^(𝑖)(1),  𝑖=0,1,2,3, where 𝑓∶[0,1]×ℝ⁺×ℝ→ℝ⁺ is continuous, 𝛼,𝛽∈ℝ, and satisfy 0<𝛼<((𝛽/2)+2𝜋²)², 𝛽>−2𝜋²,(𝛼/𝜋⁴)+(𝛽/𝜋²)+1>0. The discussion is based on the fixed point index theory in cones.

1. Introduction

This paper concerns the existence of positive solutions for the fourth-order periodic boundary value problem (PBVP) 𝑢⁽⁴⁾(𝑡)−𝛽𝑢^′′(𝑡)+𝛼𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝑢,0≤𝑡≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3,(1.1) where 𝛼,𝛽∈ℝ and 𝑓∶[0,1]×ℝ⁺×ℝ→ℝ⁺ is continuous, ℝ⁺=[0,∞). PBVP (1.1) describes the deformations of an elastic beam in equilibrium state with periodic boundary condition. In the equation, the 𝑢^′′ denotes the bending moment term which represents bending effect. Owing to its importance in physics, the existence of solutions to this problem has been studied by some authors, see [1–6]. In practice, only its positive solutions are significant. In this paper, we discuss the existence of positive solutions of PBVP (1.1).

In [1, 2], Cabada and Lois obtained the maximum principles for fourth-order operator 𝐿_4,𝛼𝑢=𝑢⁽⁴⁾+𝛼𝑢 in periodic boundary condition and then they proved the existence of solutions and the validity of the monotone method in the presence of lower and upper solutions for the periodic boundary problem 𝑢⁽⁴⁾(𝑡)((𝑡𝑢=𝑔𝑡,𝑢)),0≤t≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3.(1.2) In [3], the present author established a strongly maximum principle for operator 𝐿₄𝑢=𝑢⁽⁴⁾−𝛽𝑢^′′+𝛼𝑢 in periodic boundary condition, and showed that if 𝛼,𝛽 satisfy the assumption 𝛽0<𝛼<2+2𝜋²²,𝛽>−2𝜋²,𝛼𝜋⁴+𝛽𝜋²+1>0,(1.3) then 𝐿₄ is strongly inverse positive in space𝐹₄=𝑢∈𝐶⁴[0,1]∣𝑢^(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2;𝑢⁽³⁾(0)≥𝑢⁽³⁾(1).(1.4) As an application of this strongly maximum principle, the author considered the existence of positive solutions for the special fourth-order periodic boundary problem 𝑢⁽⁴⁾(𝑡)−𝛽𝑢^′′(𝑡)(𝑡)((𝑡𝑢+𝛼𝑢=𝑔𝑡,𝑢)),0≤𝑡≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3,(1.5) and obtained the following result.

Theorem A. Let 𝑔∶[0,1]×ℝ⁺→ℝ⁺ be continuous and the assumption (1.3) hold. If 𝑔 satisfies one of the following conditions
(G1) 𝑔⁰<𝛼,𝑔_∞>𝛼;
(G2) 𝑔₀>𝛼,𝑔^∞<𝛼,
where 𝑔₀=liminf_𝑢→0+min_[]𝑡∈0,1𝑓(𝑡,𝑢)𝑢,𝑔⁰=limsup_𝑢→0+max_[]𝑡∈0,1𝑓(𝑡,𝑢)𝑢,𝑔_∞=liminf_𝑢→+∞min_[]𝑡∈0,1𝑓(𝑡,𝑢)𝑢,𝑔^∞=limsup_𝑢→+∞max_[]𝑡∈0,1𝑓(𝑡,𝑢)𝑢,(1.6) then PBVP (1.5) has at least one positive solution.

Based upon this strongly maximum principle, the authors of [4, 5] further consider the existence and multiplicity of positive solutions of PBVP (1.5). In [6], Bereanu obtained existence results for PBVP (1.5) by using the method of topological degree. However, all of these works are on the special equation (1.5), and few people consider the existence of positive solutions of PBVP (1.1) that nonlinearity 𝑓 contains the bending moment term 𝑢^′′. The purpose of this paper is to discuss the existence of positive solutions of PBVP (1.1).

The strongly maximum principle implies that the fourth-order linear boundary value problem (LBVP) 𝐿₄𝑢∶=𝑢⁽⁴⁾−𝛽𝑢^′′𝑢+𝛼𝑢=0,0≤𝑡≤1,_(𝑖)(0)−𝑢^(𝑖)(1)𝑢=0,𝑖=0,1,2,₍₃₎(0)−𝑢⁽³⁾(1)=1(1.7) has a unique positive solution Φ∶[0,1]→(0,∞), see [3, Lemma 3]. This function has been introduced in [2, Lemma 2.1 and Remark 2.1]. Let 𝐼=[0,1], and set 𝜎=min_𝑡∈𝐼Φ(𝑡)max_𝑡∈𝐼Φ(𝑡),𝑀=max_𝑡∈𝐼Φ^′′(𝑡)min_𝑡∈𝐼Φ(𝑡).(1.8) Let 𝑓∶𝐼×ℝ⁺×ℝ→ℝ⁺ be continuous. To be convenient, we introduce the notations𝑓₀=liminf_𝑢→0+min_{|𝑣|≤𝑀𝑢,𝑡∈𝐼}𝑓(𝑡,𝑢,𝑣)𝑢,𝑓⁰=limsup_𝑢→0+max_{|𝑣|≤𝑀𝑢,𝑡∈𝐼}𝑓(𝑡,𝑢,𝑣)𝑢,𝑓_∞=liminf_𝑢→+∞min_{|𝑣|≤𝑀𝑢,𝑡∈𝐼}𝑓(𝑡,𝑢,𝑣)𝑢,𝑓^∞=limsup_𝑢→+∞max_{|𝑣|≤𝑀𝑢,𝑡∈𝐼}𝑓(𝑡,𝑢,𝑣)𝑢.(1.9)Our main result is as follows.

Theorem 1.1. Let 𝑓∶[0,1]×ℝ⁺×ℝ→ℝ⁺ be continuous, and let the assumption (1.3) hold. If 𝑓 satisfies one of the following conditions:
(F1) 𝑓⁰<𝛼,𝑓_∞>𝛼,
(F2) 𝑓₀>𝛼,𝑓^∞<𝛼,
then PBVP (1.1) has at least one positive solution.

Clearly, Theorem 1.1 is an extension of Theorem A. Since that 𝛼 is an eigenvalue of linear eigenvalue problem 𝑢⁽⁴⁾−𝛽𝑢^′′+𝛼𝑢=𝜆𝑢,(1.10) with periodic boundary condition, if one inequality in (F1) or (F2) of Theorem 1.1 is not true, the existence of solution to PBVP (1.1) cannot be guaranteed. Hence, (F1) and (F2) are the optimal conditions for the existence of the positive of PBVP (1.1).

In Theorem 1.1, the condition (F1) allows that 𝑓(𝑡,𝑢,𝑣) may be superlinear growth on 𝑢 and 𝑣, for example, 𝑓(𝑡,𝑢,𝑣)=𝑢²+𝑣², and the condition (F2) allows that 𝑓(𝑡,𝑢,𝑣) may be sublinear growth on 𝑢 and 𝑣, for example, 𝑓(𝑡,𝑢,𝑣)=√³𝑢²+𝑣².

The proof of Theorem 1.1 is based on the theory of the fixed point index in cones. Since the nonlinearity 𝑓 of PBVP (1.1) contains 𝑢^′′, the argument of Theorem A in [3] is not applicable to Theorem 1.1. We will prove Theorem 1.1 by choosing a proper cone of 𝐶²(𝐼) in Section 3. Some preliminaries to discuss PBVP (1.1) are presented in Section 2.

2. Preliminaries

Let 𝐶(𝐼) be the Banach space of all continuous functions on the unit interval 𝐼=[0,1] with the norm ‖𝑢‖_𝐶=max_{0≤𝑡≤𝜔}|𝑢(𝑡)|. Let 𝐶⁺(𝐼) denote the cone of all nonnegative functions in 𝐶(𝐼). Generaly, for 𝑛∈ℕ, we use 𝐶^𝑛(𝐼) to denote the Banach space of the 𝑛th-order continuous differentiable functions on 𝐼 with the norm ‖𝑢‖_𝐶^𝑛=∑^𝑛_𝑘=1‖𝑢^(𝑘)‖_𝐶. In 𝐶²(𝐼), we define a new norm by ‖𝑢‖_𝐶⁰²=‖𝑢‖_𝐶+𝑢^′′_𝐶.(2.1) Then ‖𝑢‖_𝐶⁰² is equivalent to ‖𝑢‖_𝐶². In fact, for every 𝑢∈𝐶²(𝐼), it is clear that ‖𝑢‖_𝐶⁰²≤‖𝑢‖_𝐶². On the other hand, by the Lagrange mean-value theorem, there exists 𝜉∈(0,1) such that 𝑢(1)−𝑢(0)=𝑢^′(𝜉). For 𝑡∈𝐼, we have𝑢^′(𝑡)≤𝑢^′(𝑡)−𝑢^′(𝜉)+𝑢^′(𝜉)=^𝑡_𝜉𝑢^′′(𝑠)+|𝑢(1)(0≤𝑑𝑠−𝑢)|₁₀𝑢^′′(𝑠|𝑢(1+|𝑢(0≤𝑢)𝑑𝑠+)|)|_′′𝐶+2‖𝑢‖_𝐶≤2‖𝑢‖_𝐶⁰².(2.2) Hence, ‖𝑢^′‖_𝐶≤2‖𝑢‖_𝐶⁰². By this, we have ‖𝑢‖_𝐶²=‖𝑢‖_𝐶+𝑢′_𝐶+𝑢^′′_𝐶=‖𝑢‖_𝐶⁰²+𝑢′_𝐶≤3‖𝑢‖_𝐶⁰².(2.3)

Therefore, the norms ‖𝑢‖_𝐶⁰² and ‖𝑢‖_𝐶² are equivalent.

Let 𝛼,𝛽∈ℝ satisfy the assumption (1.3). For ℎ∈𝐶(𝐼), we consider the fourth-order linear periodic boundary value problem (LPBVP) 𝑢⁽⁴⁾(𝑡)−𝛽𝑢^′′(𝑡)(𝑡)(𝑡)𝑢+𝛼𝑢=ℎ,0≤𝑡≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3.(2.4) Let Φ(𝑡) be the unique positive solution of LBVP(1.7), and set 𝐺()=Φ()Φ()𝑡,𝑠𝑡−𝑠,0≤𝑠≤𝑡≤1,1+𝑡−𝑠,0≤𝑡<𝑠≤1.(2.5) By [3, Lemma 1], we have the following result.

Lemma 2.1. Let 𝛼,𝛽∈ℝ satisfy the assumption (1.3). Then for every ℎ∈𝐶(𝐼), LPBVP (2.4) has a unique solution 𝑢(𝑡) which is given by 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠∶=𝑆ℎ(𝑡),𝑡∈ℝ.(2.6) Moreover, 𝑆∶𝐶(𝐼)→𝐶⁴(𝐼) is a linear bounded operator.

Let 𝜎 and 𝑀 be the positive constants given by (1.8). Choose a cone 𝐾 in 𝐶²(𝐼) by 𝐾=𝑢∈𝐶²(𝐼)∣𝑢(𝑡)≥𝜎‖𝑢‖_𝐶,𝑢^′′(𝑡)≤𝑀|𝑢(𝑡)|,𝑡∈𝐼.(2.7) We have the following.

Lemma 2.2. Let 𝛼,𝛽∈ℝ satisfy the assumption (1.3). Then for every ℎ∈𝐶⁺(𝐼), the solution of LPBVP (2.4) 𝑢=𝑆ℎ∈𝐾. Namely, 𝑆(𝐶⁺(𝐼))⊂𝐾.

Proof. Let ℎ∈𝐶⁺(𝐼), 𝑢=𝑆ℎ. For every 𝑡∈𝐼, from (2.6) it follows that 0≤𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠≤max_𝑡∈𝐼Φ(𝑡)¹₀ℎ(𝑠)𝑑𝑠,(2.8) which implies that ‖𝑢‖_𝐶≤max_𝑡∈𝐼Φ(𝑡)¹₀ℎ(𝑠)𝑑𝑠.(2.9) By this and (2.6), we have 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠≥min_𝑡∈𝐼Φ(𝑡)¹₀ℎ(𝑠)𝑑𝑠=𝜎max_𝑡∈𝐼Φ(𝑡)¹₀ℎ(𝑠)𝑑𝑠≥𝜎‖𝑢‖_𝐶.(2.10) For 𝑡∈𝐼, by the definition of 𝐺 and Φ, we have 𝑢(𝑡)=^𝑡₀Φ(𝑡−𝑠)ℎ(𝑠)𝑑𝑠+¹_𝑡Φ(1+𝑡−𝑠)ℎ(𝑠)𝑑𝑠.(2.11) Making derivation to both sides of this equality, we have 𝑢^′′(𝑡)=^𝑡₀Φ^′′(𝑡−𝑠)ℎ(𝑠)𝑑𝑠+¹_𝑡Φ^′′(1+𝑡−𝑠)ℎ(𝑠)𝑑𝑠,(2.12) from which it follows that 𝑢^′′(𝑡)≤max_𝑡∈𝐼Φ^′′(𝑡)^𝑡₀ℎ(𝑠)𝑑𝑠+max_𝑡∈𝐼Φ^′′(𝑡)¹_𝑡ℎ(𝑠)𝑑𝑠=max_𝑡∈𝐼Φ^′′(𝑡)¹₀ℎ(𝑠)𝑑𝑠=𝑀min_𝑡∈𝐼Φ(𝑡)¹₀ℎ(𝑠)𝑑𝑠≤𝑀𝑢(𝑡),𝑡∈𝐼.(2.13) Therefore, 𝑢∈𝐾. This means that 𝑆(𝐶⁺(𝐼))⊂𝐾.

For every 𝑢∈𝐾, since 𝑓∶𝐼×ℝ⁺×ℝ→ℝ⁺ is continuous, we see that 𝐹(𝑢)∶=𝑓(⋅,𝑢(⋅),𝑢^′′(⋅))∈𝐶⁺(𝐼). By Lemma 2.2, 𝑤=𝑆(𝐹(𝑢))∈𝐾. Define an operator 𝐴∶𝐾→𝐾 by 𝐴𝑢(𝑡)=𝑆(𝐹(𝑢))=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′′(𝑠)𝑑𝑠.(2.14) We have the following.

Lemma 2.3. 𝐴∶𝐾→𝐾 is a completely continuous operator.

Proof. Let 𝐷⊂𝐾 be a bounded set in 𝐶²(𝐼). By the continuity of 𝑓∶𝐼×ℝ⁺×ℝ→ℝ⁺, 𝐹(𝐷) is a bounded set in 𝐶(𝐼). By the boundedness of the operator 𝑆∶𝐶(𝐼)→𝐶⁴(𝐼), 𝐴(𝐷)=𝑆(𝐹(𝐷)) is a bounded set in 𝐶⁴(𝐼). By the compactness of the embedding 𝐶⁴(𝐼)↪𝐶²(𝐼), 𝐴(𝐷) is a precompact set in 𝐶²(𝐼). So 𝐴∶𝐾→𝐾 is completely continuous.

By the definition of 𝑆 and 𝐾, the positive solution of PBVP (1.1) is equivalent to the nontrivial fixed point of 𝐴. We will find the nonzero fixed point of 𝐴 by using the fixed point index theory in cones.

We recall some concepts and conclusions on the fixed point index in [7, 8]. Let 𝐸 be a Banach space, and let 𝐾⊂𝐸 be a closed convex cone in 𝐸. Assume Ω is a bounded open subset of 𝐸 with boundary 𝜕Ω, and 𝐾∩Ω≠∅. Let 𝐴∶𝐾∩Ω→𝐾 be a completely continuous mapping. If 𝐴𝑢≠𝑢 for any 𝑢∈𝐾∩𝜕Ω, then the fixed point index 𝑖(𝐴,𝐾∩Ω,𝐾) has definition. One important fact is that if 𝑖(𝐴,𝐾∩Ω,𝐾)≠0, then 𝐴 has a fixed point in 𝐾∩Ω. The following two lemmas are needed in our argument.

Lemma 2.4 (see [8]). Let Ω be a bounded open subset of 𝐸 with 𝜃∈Ω, and let 𝐴∶𝐾∩Ω→𝐾 be a completely continuous mapping. If 𝜆𝐴𝑢≠𝑢 for every 𝑢∈𝐾∩𝜕Ω and 0<𝜆≤1, then 𝑖(𝐴,𝐾∩Ω,𝐾)=1.

Lemma 2.5 (see [8]). Let Ω be a bounded open subset of 𝐸, and let 𝐴∶𝐾∩Ω→𝐾 be a completely continuous mapping. If there exists an 𝑒∈𝐾⧵{𝜃} such that 𝑢−𝐴𝑢≠𝜏𝑒 for every 𝑢∈𝐾∩𝜕Ω and 𝜏≥0, then 𝑖(𝐴,𝐾∩Ω,𝐾)=0.

3. Proof of the Main Result

Proof of Theorem 1.1. Choose the working space 𝐸=𝐶²(𝐼) with the norm ‖𝑢‖_𝐶⁰². Let 𝐾 be the closed convex cone in 𝐶²(𝐼) defined by (2.7), and let 𝐴∶𝐾→𝐾 be the operator defined by (2.14). By Lemma 2.3 and the definition of 𝐾, the nonzero fixed of the operator 𝐴 is the positive solution of PBVP (1.1). Let 0<𝑟<𝑅<+∞, and set Ω₁=𝑢∈𝐶²(𝐼)∣‖𝑢‖_𝐶⁰²<𝑟,Ω₂=𝑢∈𝐶²(𝐼)∣‖𝑢‖_𝐶⁰²<𝑅.(3.1) We show that, if 𝑟 is small enough and 𝑅 large enough, the operator 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁) in either case that (F1) holds or (F2) holds.Case 1. Assume that (F1) holds.
Since 𝑓⁰<𝛼, by the definition of 𝑓⁰, we may choose 𝜀∈(0,𝛼) and 𝛿>0, such that𝑓(𝑡,𝑢,𝑣)≤(𝛼−𝜀)𝑢,𝑡∈𝐼,|𝑣|≤𝑀𝑢,0≤𝑢≤𝛿.(3.2) Let 𝑟∈(0,𝛿). We prove that 𝐴 satisfies the condition of Lemma 2.4 in 𝐾∩𝜕Ω₁; namely, 𝜆𝐴𝑢≠𝑢, for every 𝑢∈𝐾∩𝜕Ω₁ and 0<𝜆≤1. In fact, if there exist 𝑢₀∈𝐾∩𝜕Ω₁ and 0<𝜆₀≤1 such that 𝜆₀𝐴𝑢₀=𝑢₀, then by the definition of 𝐴 and Lemma 2.1, 𝑢₀∈𝐶⁴(𝐼) satisfies the differential equation 𝑢₀⁽⁴⁾(𝑡)−𝛽𝑢₀^′′(𝑡)+𝛼𝑢₀(𝑡)=𝜆₀𝑓𝑡,𝑢₀(𝑡),𝑢₀^′′(𝑡),𝑡∈𝐼,(3.3) and the periodic boundary condition 𝑢^(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3.(3.4) Since 𝑢₀∈𝐾∩𝜕Ω₁, by the definitions of 𝐾 and Ω₁, we have 𝑢₀^′′(𝑡)≤𝑀𝑢₀(𝑡),0<𝜎𝑢_0𝐶≤𝑢₀(𝑡)≤𝑢_0𝐶02=𝑟<𝛿,𝑡∈ℝ.(3.5) From this and (3.2), it follows that 𝑓𝑡,𝑢₀(𝑡),𝑢₀^′′(𝑡)≤(𝛼−𝜀)𝑢₀(𝑡),𝑡∈ℝ.(3.6) By this inequality and (3.3), we have 𝑢₀⁽⁴⁾(𝑡)−𝛽𝑢₀^′′(𝑡)+𝛼𝑢₀(𝑡)≤𝜆₀(𝛼−𝜀)𝑢₀(𝑡)≤(𝛼−𝜀)𝑢₀(𝑡),∈𝐼.(3.7) Integrating this inequality from 0 to 1 and using the periodic boundary condition (3.4), we obtain that 𝛼¹₀𝑢₀(𝑡)𝑑𝑡≤(𝛼−𝜀)¹₀𝑢₀(𝑡)𝑑𝑡.(3.8) Since ∫¹₀𝑢₀(𝑡)𝑑𝑡≥𝜎‖𝑢₀‖_𝐶>0, form this inequality it follows that 𝛼≤𝛼−𝜀, which is a contradiction. Hence, 𝐴 satisfies the condition of Lemma 2.4 in 𝐾∩𝜕Ω₁. By Lemma 2.4 we have 𝑖𝐴,𝐾∩Ω₁,𝐾=1.(3.9)
On the other hand, since 𝑓_∞>𝛼, by the definition of 𝑓_∞, there exist 𝜀₁>0 and 𝐻>0 such that𝑓(𝑡,𝑢,𝑣)≥𝛼+𝜀₁𝑥,𝑡∈𝐼,|𝑣|≤𝑀𝑢,𝑢≥𝐻.(3.10) Choose 𝑅>max{(1+𝑀/𝜎)𝐻,𝛿}, and let 𝑒(𝑡)≡1. Clearly, 𝑒∈𝐾⧵{𝜃}. We show that 𝐴 satisfies the condition of Lemma 2.5 in 𝐾∩𝜕Ω₂; namely, 𝑢−𝐴𝑢≠𝜏𝑒, for every 𝑢∈𝐾∩𝜕Ω₂ and 𝜏≥0. In fact, if there exist 𝑢₁∈𝐾∩𝜕Ω₂ and 𝜏₁≥0 such that 𝑢₁−𝐴𝑢₁=𝜏₁𝑒, since 𝑢₁−𝜏₁𝑒=𝐴𝑢₁, by definition of 𝐴 and Lemma 2.1, 𝑢₁(𝑡)∈𝐶⁴(𝐼) satisfies the differential equation 𝑢₁⁽⁴⁾(𝑡)−𝛽𝑢₁^′′(𝑡)+𝛼𝑢₁(𝑡)−𝜏₁=𝑓𝑡,𝑢₁(𝑡),𝑢₁^′′(𝑡),𝑡∈𝐼,(3.11) and the periodic boundary condition (3.4). Since 𝑢₁∈𝐾∩𝜕Ω₂, by the definition of 𝐾, we have 𝑢₁(𝑡)≥𝜎𝑢_1𝐶,𝑢₁^′′(𝑡)≤𝑀𝑢₁(𝑡),𝑡∈𝐼.(3.12) By the second inequality of (3.12), we have 𝑢_1𝐶02=𝑢_1𝐶+𝑢₁^′′_𝐶≤𝑢_1𝐶+𝑀𝑢_1𝐶=(1+𝑀)𝑢_1𝐶.(3.13) Consequently, 𝑢_1𝐶≥1𝑢1+𝑀^1𝐶02.(3.14) By (3.14) and the first inequality of (3.12), we have 𝑢₁(𝑡)≥𝜎𝑢_1𝐶≥𝜎𝑢1+𝑀^1𝐶02=𝜎1+𝑀𝑅>𝐻,𝑡∈𝐼.(3.15) From this, the second inequality of (3.12), and (3.10), it follows that 𝑓𝑡,𝑢₁(𝑡),𝑢₁^′′(𝑡)≥𝛼+𝜀₁𝑢₁(𝑡),𝑡∈𝐼.(3.16) By this and (3.11), we have 𝑢₁⁽⁴⁾(𝑡)−𝛽𝑢₁^′′(𝑡)+𝛼𝑢₁(𝑡)−𝜏₁≥𝛼+𝜀₁𝑢₁(𝑡),𝑡∈𝐼.(3.17) Integrating this inequality on 𝐼 and using the periodic boundary condition (3.4), we get that 𝛼¹₀𝑢₁(𝑡)𝑑𝑡−𝛼𝜏₁≥𝛼+𝜀₁¹₀𝑢₁(𝑡)𝑑𝑡.(3.18) Since ∫¹₀𝑢₁(𝑡)𝑑𝑡≥𝜎‖𝑢₁‖_𝐶>0, from this inequality it follows that 𝛼≥𝛼+𝜀₁, which is a contradiction. This means that 𝐴 satisfies the condition of Lemma 2.5 in 𝐾∩𝜕Ω₂. By Lemma 2.4, 𝑖𝐴,𝐾∩Ω₂,𝐾=0.(3.19)
Now, by the additivity of fixed point index, (3.9), and (3.19), we have𝑖𝐴,𝐾∩Ω₂⧵Ω₁,𝐾=𝑖𝐴,𝐾∩Ω₂,𝐾−𝑖𝐴,𝐾∩Ω₁,𝐾=−1.(3.20) Hence, 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁), which is the positive solution of PBVP (1.1).Case 2. Assume that (F2) holds.
By the assumption of 𝑓₀>𝛼 and the definition of 𝑓₀, there exist 𝜀>0 and 𝛿>0, such that𝑓(𝑡,𝑢,𝑣)≥(𝛼+𝜀)𝑢,𝑡∈𝐼,|𝑣|≤𝑀𝑢,0≤𝑢≤𝛿.(3.21) Let 𝑟∈(0,𝛿), and let 𝑒(𝑡)≡1. We prove that 𝐴 satisfies the hypothesis of Lemma 2.5 in 𝐾∩𝜕Ω₁; namely, 𝑢−𝐴𝑢≠𝜏𝑒 for every 𝑢∈𝐾∩𝜕Ω₁ and 𝜏≥0. In fact, if there exist 𝑢₀∈𝐾∩𝜕Ω₁ and 𝜏₀≥0 such that 𝑢₀−𝐴𝑢₀=𝜏₀𝑒, since 𝑢₀−𝜏₀𝑒=𝐴𝑢₀, by the definition of 𝐴 and Lemma 2.1, 𝑢₀(𝑡)∈𝐶⁴(𝐼) satisfies the differential equation 𝑢₀⁽⁴⁾(𝑡)−𝛽𝑢₀^′′(𝑡)+𝛼𝑢₀(𝑡)−𝜏₀=𝑓𝑡,𝑢₀(𝑡),𝑢₀^′′(𝑡),𝑡∈𝐼,(3.22) and the periodic boundary condition (3.4). Since 𝑢₀∈𝐾∩𝜕Ω₁, by the definitions of 𝐾 and Ω₁, 𝑢₀ satisfies (3.5). From (3.5) and (3.22), it follows that 𝑓𝑡,𝑢₀(𝑡),𝑢₀^′(𝑡)≥(𝛼+𝜀)𝑢₀(𝑡),𝑡∈𝐼.(3.23) By this inequality and (3.22), we have 𝑢₀⁽⁴⁾(𝑡)−𝛽𝑢₀^′′(𝑡)+𝛼𝑢₀(𝑡)−𝜏₀≥(𝛼+𝜀)𝑢₀(𝑡),𝑡∈𝐼.(3.24) Integrating this inequality on 𝐼 and using the periodic boundary condition (3.4), we have 𝛼¹₀𝑢₀(𝑡)𝑑𝑡−𝛼𝜏₀≥(𝛼+𝜀)¹₀𝑢₀(𝑡)𝑑𝑡.(3.25) Since ∫¹₀𝑢₀(𝑡)𝑑𝑡≥𝜎‖𝑢₀‖_𝐶>0, from this inequality, it follows that 𝛼≥𝛼+𝜀, which is a contradiction. Hence 𝐴 satisfies the hypothesis of Lemma 2.5 in 𝐾∩𝜕Ω₁. By Lemma 2.5, 𝑖𝐴,𝐾∩Ω₁,𝐾=0.(3.26)
Since 𝑓^∞<𝛼, by the definition of 𝑓^∞, there exist 𝜀₁∈(0,𝛼) and 𝐻>0 such that𝑓(𝑡,𝑢,𝑣)≤𝛼−𝜀₁𝑢,𝑡∈𝐼,|𝑣|≤𝑀𝑢,𝑢≥𝐻.(3.27) Choosing 𝑅>max{(1+𝑀/𝜎)𝐻,𝛿}, we show that 𝐴 satisfies the condition of Lemma 2.4 in 𝐾∩𝜕Ω₂; namely, 𝜆𝐴𝑢≠𝑢, for every 𝑢∈𝐾∩𝜕Ω₂ and 0<𝜆≤1. In fact, if there exist 𝑢₁∈𝐾∩𝜕Ω₂ and 0<𝜆₁≤1 such that 𝜆₁𝐴𝑢₁=𝑢₁, then by the definition of 𝐴 and Lemma 2.1, 𝑢₁∈𝐶⁴(𝐼) satisfies the differential equation 𝑢₁⁽⁴⁾(𝑡)−𝛽𝑢₁^′′(𝑡)+𝛼𝑢₁(𝑡)=𝜆₁𝑓𝑡,𝑢₁(𝑡),𝑢₁^′′(𝑡),𝑡∈𝐼,(3.28) and the periodic boundary condition (3.4). Since 𝑢₁∈𝐾∩𝜕Ω₂, by the definition of 𝐾, 𝑢₁ satisfies (3.12). From (3.12), we can show that 𝑢₁ satisfies (3.14). By (3.14) and the first inequality of (3.12), we have 𝑢₁(𝑡)≥𝜎𝑢_1𝐶≥𝜎𝑢1+𝑀^1𝐶02=𝜎1+𝑀𝑅>𝐻,𝑡∈ℝ.(3.29)From this, the second inequality of (3.12), and (3.27), it follows that 𝑓𝑡,𝑢₁(𝑡),𝑢₁^′′(𝑡)≤𝛼−𝜀₁𝑢₁(𝑡),𝑡∈𝐼.(3.30) By this inequality and (3.28), we have 𝑢₁⁽⁴⁾(𝑡)−𝛽𝑢₁^′′(𝑡)+𝛼𝑢₁(𝑡)≤𝜆₁𝛼−𝜀₁𝑢₁(𝑡)≤𝛼−𝜀₁𝑢₁(𝑡),𝑡∈𝐼.(3.31) Integrating this inequality on 𝐼 and using the periodic boundary condition (3.4), we obtain that 𝛼¹₀𝑢₁(𝑡)𝑑𝑡≤𝛼−𝜀₁¹₀𝑢₁(𝑡)𝑑𝑡.(3.32) Since ∫¹₀𝑢₁(𝑡)𝑑𝑡≥𝜎‖𝑢₁‖_𝐶>0, form this inequality it follows that 𝛼≤𝛼−𝜀₁, which is a contradiction. This means that 𝐴 satisfies the condition of Lemma 2.4 in 𝐾∩𝜕Ω₂. By Lemma 2.4, 𝑖𝐴,𝐾∩Ω₂,𝐾=1.(3.33)
From (3.26) and (3.33), it follows that𝑖𝐴,𝐾∩Ω₂⧵Ω₁,𝐾=𝑖𝐴,𝐾∩Ω₂,𝐾−𝑖𝐴,𝐾∩Ω₁,𝐾=1.(3.34) Hence, 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁), which is the positive solution of PBVP (1.1).
The proof of Theorem 1.1 is completed.

Example 3.1. Consider the superlinear fourth-order periodic boundary problem 𝑢⁽⁴⁾−𝑢^′′+𝑢=𝑎₁(𝑡)𝑢²+𝑎₂(𝑡)𝑢^′′2𝑢,0≤𝑡≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3,(3.35) where 𝑎₁,𝑎₂∈𝐶(𝐼) and 𝑎₁(𝑡),𝑎₂(𝑡)>0 for 𝑡∈𝐼. It is easy to verify that 𝛼=1 and 𝛽=1 satisfy the assumption (P). 𝑓(𝑡,𝑢,𝑣)=𝑎₁(𝑡)𝑢²+𝑎₂(𝑡)𝑣² satisfies the condition (F1), in which 𝑓⁰=0 and 𝑓_∞=+∞. Hence, by Theorem 1.1, (3.35) has at least one positive solution.

Example 3.2. Consider the sublinear fourth-order periodic boundary problem 𝑢⁽⁴⁾+𝑢^′′+𝑢=𝑏₁(𝑡)√𝑢+𝑏₂𝑢^′′𝑢,0≤𝑡≤1,_(𝑖)(0)=𝑢^(𝑖)(1),𝑖=0,1,2,3,(3.36) where 𝑏₁,𝑏₂∈𝐶(𝐼) and 𝑏₁(𝑡),𝑏₂(𝑡)>0 for 𝑡∈𝐼. For PBVP (3.36), it is easy to verify that 𝛼=1 and 𝛽=−1 satisfy the assumption (1.3), and 𝑓(𝑡,𝑢,𝑣)=𝑏₁(𝑡)√𝑢+𝑏₂(𝑡)|𝑣| satisfies the condition (F2) with 𝑓₀=+∞ and 𝑓^∞=0. By Theorem 1.1, (3.36) has a positive solution.
Since (3.35) and (3.36) have nonlinear terms of 𝑢^′′, which are not in the range considered by [1–6], the existence results in Example 3.1, and Example 3.2 cannot be obtained from [1–6].

Acknowledgment

The author thanks to the referee for the helpful comments and suggestions. Research supported by NNSFs of China (10871160, 11061031) and Project of NWNU-KJCXGC-3-47.