The Solutions of Sturm-Liouville Boundary-Value Problem for Fourth-Order Impulsive Differential Equation via Variational Methods

Abstract

The Sturm-Liouville boundary-value problem for fourth-order impulsive differential equations is studied. The existence results for one solution and multiple solutions are obtained. The main ideas involve variational methods and three critical points theory.

1. Introduction

The aim of the present paper is to study the following Sturm-Liouville boundary-value problem for the fourth-order impulsive differential equation: 𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)=𝜆𝑓(𝑡,𝑢(𝑡)),𝑡≠𝑡_𝑖,𝑡∈[0,𝑇],Δ𝑢^′′′𝑡_𝑖=𝜆𝐼_1𝑖𝑢𝑡_𝑖,𝑖=1,2,…,𝑙,−Δ𝑢^′′𝑡_𝑖=𝜆𝐼_2𝑖𝑢^′𝑡_𝑖(0),𝑖=1,2,…,𝑙,𝑎𝑢+𝑏𝑢^′′′(0)=0,𝑎𝑢(𝑇)−𝑏𝑢^′′′(𝑇)=0,𝑐𝑢^′(0)−𝑑𝑢^′′(0)=0,𝑐𝑢^′(𝑇)+𝑑𝑢^′′(𝑇)=0, (1) where 𝑎, 𝑏, 𝑐, and 𝑑 are real constants, 𝜆 is a positive parameter, 0=𝑡₀<𝑡₁<⋯<𝑡_𝑙<𝑡_𝑙+1=𝑇, 𝑎,𝑏,𝑐,𝑑>0, Δ𝑢^′′′(𝑡_𝑖)=𝑢^′′′(𝑡⁺_𝑖)−𝑢^′′′(𝑡⁻_𝑖), Δ𝑢^′′(𝑡_𝑖)=𝑢^′′(𝑡⁺_𝑖)−𝑢^′′(𝑡⁻_𝑖), 𝑢^′′′(𝑡⁺_𝑖), 𝑢^′′(𝑡⁺_𝑖)(𝑢^′′′(𝑡⁻_𝑖),𝑢^′′(𝑡⁻_𝑖)) denote the right (left) limits, respectively, of 𝑢^′′′(𝑡), 𝑢^′′(𝑡) at 𝑡=𝑡_𝑖, and 𝑓∈𝐶([0,𝑇]×𝑅,𝑅), 𝐼_1𝑖,𝐼_2𝑖∈𝐶(𝑅;𝑅), 𝑖=1,2,…,𝑙.

Recently, many authors have studied the existence of solutions for boundary-value problems with impulsive effects [1–16]. Variational methods are powerful tools for them. We refer the readers to [17–19] for related basic information.

In [10], the authors studied the following equation with impulsive effects: −𝑢^′′(𝑡)=𝜆𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡)),𝑡≠𝑡_𝑖,𝑡∈[0,𝑇],−Δ𝑢^′𝑡_𝑖=𝐼_𝑖𝑢𝑡_𝑖𝑢,𝑖=1,2,…,𝑙,_′(0)=0,𝑢(𝑇)=0. (2) By applying critical point theory to (2), several existence results are obtained when 𝑓 is imposed some assumptions and 𝜆 lies in suitable interval. In [8], the authors studied the existence of solutions for the following problem: 𝑢^(𝑖𝑣)(𝑡)+𝐴𝑢^′′(𝑡)+𝐵𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),a.e.𝑡∈[0,𝑇],−Δ𝑢^′′𝑡_𝑗=𝐼_1𝑗𝑢^′𝑡_𝑗,𝑗=1,2,…,𝑙,−Δ𝑢^′′′𝑡_𝑗=𝐼_2𝑗𝑢𝑡_𝑗𝑢(0)(𝑇),𝑗=1,2,…,𝑙,=𝑢=𝑢^′′0⁺=𝑢^′′(𝑇⁻)=0. (3) They essentially proved that when 𝑓, 𝐼_1𝑗, and 𝐼_2𝑗 satisfy some conditions, (3) has at least one solution or infinitely many classical solutions via variational methods.

To the best of our knowledge, besides [12, 13] for second-order differential equations, [8] for fourth-order differential equation, limited work has been done in the Sturm-Liouville boundary-value problem, let alone higher order. Motivated by the above facts, we study the existence of solutions for problem (1) by applying variational methods. With the impulse effects and the Sturm-Liouville boundary conditions taken into consideration, the corresponding variational functional 𝐽 will be more complicated than the ones of any fourth-order boundary-value problems before. In our study, some difficulties such as how to prove that the critical points of 𝐽 are just the solutions of problem (1) and how to prove the space 𝑋 and the functional 𝐽 to satisfy the conditions of the related theorems must be overcome. To verify that the weak solution of problem (1) is just the classical solution of (1), we construct a Fundamental Lemma 5, by which we can easily prove that the critical point of the functional is just the solution of problem (1).

This paper is organized as follows. In Section 2, we present some preliminaries and establish the variational structure. In Section 3, we discuss the existence results for one solution and multiple solutions. In Section 4, we discuss the existence results for positive solutions. In Section 5, we will give some examples.

2. Preliminaries and Variational Structure

First we present some theorems that will be needed in the proof of main results.

Theorem 1 (see Theorem 2.2 [19]). Let 𝐸 be a real Banach space and 𝐼∈𝐶¹(𝐸,𝑅) satisfying the Palais-Smale condition (PS). Suppose 𝐼(0)=0 and(C1)there are constants 𝜌,𝛼>0 such that 𝐼|_𝜕𝐵𝜌≥𝛼,(C2)there is an 𝑒∈𝐸⧵𝐵_𝜌 such that 𝐼(𝑒)≤0. Then 𝐼 possesses a critical value 𝑐≥𝛼. Moreover, 𝑐 can be characterized as 𝑐=inf_𝑔∈Γmax_{𝑢∈𝑔([0,1])}𝐼(𝑢), (4) where Γ=𝑔∈(𝐶[0,1],𝐸)∶𝑔(0)=0,𝑔(1)=𝑒. (5)

Theorem 2 (see Theorem 9.12 [19]). Let 𝐸 be an infinite dimensional Banach space and let 𝐼∈𝐶¹(𝐸,𝑅) be even, satisfy (PS), and 𝐼(0)=0. If 𝐸=𝑉⨁𝑋, where 𝑉 is finite dimensional, and 𝐼 satisfies that(C3)there exist constants 𝜌,𝛼>0 such that 𝐼|_{𝜕𝐵𝜌∩𝑋}≥𝛼,(C4)for each finite dimensional subspace 𝑊⊂𝐸, there is an 𝑅=𝑅(𝑊) such that 𝐼≤0 on 𝑊⧵𝐵_𝑅(𝑊).Then 𝐼 has unbounded sequence of critical values.

Let 𝑋 be a nonempty set and Φ,Ψ∶𝑋→𝑅 two functionals. For all 𝑟,𝑟₁,𝑟₂>inf_𝑋Φ, 𝑟₂>𝑟₁, 𝑟₃>0, we define 𝜑(𝑟)∶=inf_{𝑢∈Φ−1(]−∞,𝑟[)}sup_𝑢∈Φ⁻¹_{(]−∞,𝑟[)}Ψ(𝑢)−Ψ(𝑢)(𝑢),𝛽𝑟𝑟−Φ₁,𝑟₂∶=inf_{𝑢∈Φ−1(]−∞,𝑟1[)}sup_{𝑣∈Φ−1([𝑟1,𝑟2[)}Ψ(𝑣)−Ψ(𝑢)Φ(𝑣)(𝑢),𝛾𝑟−Φ₂,𝑟₃∶=sup_𝑢∈Φ⁻¹_{(]−∞,𝑟2+𝑟3[)}Ψ(𝑢)𝑟₃,𝛼𝑟₁,𝑟₂,𝑟₃∶=max𝜑𝑟₁,𝜑𝑟₂,𝛾𝑟₂,𝑟₃. (6)

Theorem 3 (see Theorem 2.1 [2]). Let 𝑋 be a reflexive real Banach space, Φ∶𝑋→𝑅 a convex, coercive, and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on 𝑋^∗, and Ψ∶𝑋→𝑅 a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact, such that(1)inf_𝑋Φ=Φ(0)=Ψ(0)=0;(2)for every 𝑢₁, 𝑢₂ such that Ψ(𝑢₁)≥0 and Ψ(𝑢₂)≥0 one has inf_[]𝑡∈0,1Ψ𝑡𝑢₁+(1−𝑡)𝑢₂≥0. (7)Assume that there are three positive constants 𝑟₁, 𝑟₂, 𝑟₃ with 𝑟₁<𝑟₂, such that(i)𝜑(𝑟₁)<𝛽(𝑟₁,𝑟₂);(ii)𝜑(𝑟₂)<𝛽(𝑟₁,𝑟₂);(iii)𝛾(𝑟₂,𝑟₃)<𝛽(𝑟₁,𝑟₂).Then, for each 𝜆∈]1/𝛽(𝑟₁,𝑟₂),1/𝛼(𝑟₁,𝑟₂,𝑟₃)[, the functional Φ−𝜆Ψ admits three distinct critical points 𝑢₁, 𝑢₂, 𝑢₃ such that 𝑢₁∈Φ⁻¹(]−∞,𝑟₁[), 𝑢₂∈Φ⁻¹([𝑟₁,𝑟₂[), and 𝑢₃∈Φ⁻¹(]−∞,𝑟₂+𝑟₃[).

Let us define the space 𝑋=𝑊^2,2([0,𝑇],𝑅) equipped with the norm ‖𝑢‖=^𝑇₀|𝑢(𝑡)|²+𝑢^′(𝑡)²+𝑢^′′(𝑡)²𝑑𝑡^1/2. (8) We set the functional 𝐽∶𝑋→𝑅 defined by 𝐽(𝑢)=12‖𝑢‖²_𝑋−𝜆^𝑇₀𝐹(𝑡,𝑢(𝑡))𝑑𝑡−𝜆^𝑙_𝑖=1^{𝑢(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠−𝜆^𝑙_𝑖=1^{𝑢′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠, (9) where ‖𝑢‖_𝑋=(∫^𝑇₀|𝑢^′′(𝑡)|²+|𝑢(𝑡)|²𝑑𝑡+(𝑎/𝑏)𝑢²(𝑇)+(𝑎/𝑏)𝑢²(0)+(𝑐/𝑑)|𝑢^′(𝑇)|²+(𝑐/𝑑)|𝑢^′(0)|²)^1/2, 𝐹(𝑡,𝑢)=∫^𝑢₀𝑓(𝑡,𝑠)𝑑𝑠. 𝐽 is differentiable for any 𝑢∈𝑋 and 𝐽^′(𝑢𝑣)=)(_𝑇0𝑢^′′𝑣^′′+𝑢𝑣𝑑𝑡−𝜆^𝑙_𝑖=1𝐼_1𝑖𝑢𝑡_𝑖𝑣𝑡_𝑖−𝜆^𝑙_𝑖=1𝐼_2𝑖𝑢^′𝑡_𝑖𝑣^′𝑡_𝑖−𝜆^𝑇₀𝑓((𝑡𝑣(𝑡)+𝑎𝑡,𝑢))𝑑𝑡𝑏𝑢(𝑇)𝑣(𝑇)+𝑎𝑏𝑢(0)𝑣(0)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)+𝑐𝑑𝑢^′(0)𝑣^′(0). (10) Set the usual norm of 𝐶¹([0,𝑇]), 𝐿²(0,𝑇), respectively, as follows: ‖𝑢‖_𝐶¹=maxmax_{[]𝑡∈0,𝑇}|𝑢(𝑡)|,max_{[]𝑡∈0,𝑇}𝑢^′(𝑡,‖𝑢‖_𝐿2)=^𝑇₀𝑢²(𝑡)𝑑𝑡^1/2. (11)

Lemma 4. The norm ‖𝑢‖_𝑋 is equivalent to the usual norm ‖𝑢‖.

Proof. First, we will show that there exists 𝑀₁>0 such that ‖𝑢‖²≤𝑀₁‖𝑢‖²_𝑋. Since 𝑢^′ is absolutely continuous in 𝑋, we have 𝑢^′(𝑡)=𝑢^′(0)+∫^𝑡₀𝑢^′′(𝑠)𝑑𝑠. So ^𝑇₀𝑢^′(𝑡)²𝑑𝑡=^𝑇₀𝑢^′(0)+^𝑡₀𝑢^′′(𝑠)𝑑𝑠²𝑢𝑑𝑡≤2𝑇^′(0)²+2𝑇^2𝑇₀𝑢^′′(𝑠)²𝑑𝑠, (12) which implies ‖𝑢‖²≤^𝑇₀|𝑢(𝑡)|²+𝑢^′′(𝑡)²𝑑𝑡+2𝑇𝑢^′(0)²+2𝑇^2𝑇₀𝑢^′′(𝑡)²≤𝑑𝑡1+2𝑇^2𝑇₀𝑢^′′(𝑡)²𝑑𝑠+^𝑇₀|𝑢(𝑡)|²𝑑𝑡+2𝑇𝑢^′(0)²≤max1+2𝑇²,2𝑇𝑑𝑐^𝑇₀𝑢^′′(𝑡)²+|𝑢(𝑡)|²𝑐𝑑𝑡+𝑑𝑢^′(0)²∶=𝑀₁‖𝑢‖²_𝑋, (13) where 𝑀₁=max{1+2𝑇²,2𝑇𝑑/𝑐}.
Next, we prove that there exists 𝑀₂>0 such that ‖𝑢‖²_𝑋≤𝑀₂‖𝑢‖².
Obviously, max_{𝑡∈[0,𝑇]}|𝑢(𝑡)|=𝑢(𝜉)=𝑢+∫^𝜉_𝜂𝑢^′(𝑠)𝑑𝑠≤𝑢+∫^𝑇₀|𝑢^′(𝑠)|𝑑𝑠, where 𝑢=∫^𝑇₀𝑢(𝑠)𝑑𝑠/𝑇=𝑢(𝜂). Thus, we have max_{𝑡∈[0,𝑇]}|𝑢(𝑡)|²≤2|𝑢|²+^𝑇₀𝑢^′(𝑠)𝑑𝑠²≤2𝑇^𝑇₀|𝑢(𝑠)|²𝑑𝑠+2𝑇^𝑇₀𝑢^′(𝑠)²𝑑𝑠. (14) Similar to the above proof, we have max_{𝑡∈[0,𝑇]}𝑢^′(𝑡)²≤2𝑇^𝑇₀𝑢^′(𝑠)²𝑑𝑠+2𝑇^𝑇₀𝑢^′′(𝑠)²𝑑𝑠. (15) By (14) and (15), we have ‖𝑢‖²_𝑋≤^𝑇₀|𝑢(𝑠)|²+𝑢^′′(𝑠)²+𝑑𝑠2𝑎𝑏2𝑇^𝑇₀|𝑢(𝑠)|²𝑑𝑠+2𝑇^𝑇₀𝑢^′(𝑠)²+𝑑𝑠2𝑐𝑑2𝑇^𝑇₀𝑢^′(𝑠)²𝑑𝑠+2𝑇^𝑇₀𝑢^′′(𝑠)²≤𝑑𝑠1+4𝑎𝑏𝑇^𝑇₀|𝑢(𝑠)|²𝑑𝑠+1+4𝑐𝑇𝑑^𝑇₀𝑢^′′(𝑠)²+𝑑𝑠4𝑎𝑇𝑏+4𝑐𝑑𝑇^𝑇₀𝑢^′(𝑠)²𝑑𝑠∶=𝑀₂‖𝑢‖², (16) where 𝑀₂=max{1+4𝑎/𝑏𝑇,1+4𝑐𝑇/𝑑,4𝑎𝑇/𝑏+4𝑐/𝑑𝑇}. By (13) and (16), the proof is complete.

Lemma 5 (Fundamental Lemma). Let 𝑢,𝑣∈𝐿¹([0,𝑇];𝑅). If for every 𝑓∈𝐶¹[0,𝑇] with 𝑓^′′∈𝐿¹[0,𝑇], 𝑓(0)=𝑓(𝑇)=𝑓^′(0)=𝑓^′(𝑇)=0 satisfying ∫^𝑇₀𝑢(𝑡)𝑓^′′(𝑡)𝑑𝑡=∫^𝑇₀𝑣(𝑡)𝑓(𝑡)𝑑𝑡, then there exist 𝐶₁,𝐶₂∈𝑅 such that 𝑢(𝑡)=∫^𝑡₀∫^𝑠₀𝑣(𝜃)𝑑𝜃𝑑𝑠+𝐶₁𝑡+𝐶₂ a.e. on [0,𝑇].

Proof. Define 𝑤(𝑡)∈𝐶([0,𝑇];𝑅) by 𝑤(𝑡)=∫^𝑡₀∫^𝑠₀𝑣(𝜃)𝑑𝜃𝑑𝑠; we have ^𝑇₀𝑤(𝑡)𝑓^′′(𝑡)𝑑𝑡=^𝑇₀^𝑡₀^𝑠₀𝑣(𝜃)𝑑𝜃𝑑𝑠𝑓^′′(𝑡)𝑑𝑡. (17) By the Fubini theorem, we have ^𝑇₀𝑤(𝑡)𝑓^′′(𝑡)𝑑𝑡=^𝑇₀^𝑇_𝑠^𝑠₀𝑣(𝜃)𝑑𝜃𝑓^′′(𝑡)=𝑑𝑡𝑑𝑠_𝑇0𝑠0𝑣(𝜃)𝑑𝜃𝑓^′(𝑇)−𝑓^′(𝑠)𝑑𝑠=−^𝑇₀^𝑠₀𝑣(𝜃)𝑑𝜃𝑓^′(𝑠)𝑑𝑠=−^𝑇₀^𝑇_𝜃𝑣(𝜃)𝑓^′(𝑠)𝑑𝑠𝑑𝜃=−^𝑇₀𝑣(𝜃)𝑓(𝑇)(𝜃)=−𝑓𝑑𝜃_𝑇0𝑣(𝜃)𝑓(𝜃)𝑑𝜃. (18) So ^𝑇₀(𝑢(𝑡)−𝑤(𝑡))𝑓^′′(𝑡)𝑑𝑡=0. (19) In particular, we choose 𝑓(𝑡)=∫^𝑡₀∫^𝑠₀(𝑢(𝜃)−𝑤(𝜃)−𝐶₁𝜃−𝐶₂)𝑑𝜃𝑑𝑠, where 𝐶₁=12𝑇^3𝑇₀𝑡(𝑢(𝑡)(𝑡𝑇−𝑤))𝑑𝑡−2^𝑇₀𝑢(𝑠)(𝑠),𝐶−𝑤𝑑𝑠₂=4𝑇^𝑇₀𝑢(𝑠)(𝑠)6−𝑤𝑑𝑠−𝑇^2𝑇₀𝑡(𝑢−𝑤)𝑑𝑡. (20) By computation, 𝑓∈𝐶¹₀[0,𝑇], 𝑓^′′∈𝐿¹[0,𝑇], 𝑓(0)=𝑓(𝑇)=𝑓^′(0)=𝑓^′(𝑇)=0, and ^𝑇₀𝐶₁𝑡+𝐶₂𝑓^′′(𝑡)𝑑𝑡=0. (21) By (19) and (21), ∫^𝑇₀(𝑢(𝑡)−𝑤(𝑡)−𝐶₁𝑡−𝐶₂)𝑓^′′(𝑡)𝑑𝑡=0; that is, ∫^𝑇₀|𝑢(𝑡)−𝑤(𝑡)−𝐶₁𝑡−𝐶₂|²𝑑𝑡=0, which means 𝑢(𝑡)=𝑤(𝑡)+𝐶₁𝑡+𝐶₂. The proof is complete.

Definition 6. A function 𝑢∈𝑋 is said to be a weak solution of (1), if 𝑢 satisfies 𝐽^′(𝑢)(𝑣)=0 for all 𝑣∈𝑋.

Definition 7. A function 𝑢∈𝑋 is said to be a classical solution of problem (1) if 𝑢 satisfies the equation in (1) for a.e. 𝑡∈[0,𝑇]⧵{𝑡₁,𝑡₂,…,𝑡_𝑙} and the impulsive condition and boundary condition in (1). Moreover, 𝑢 is said to be a positive classical solution of problem (1) if 𝑢(𝑡)≥0, 𝑢(𝑡)≢0, 𝑡∈[0,𝑇].

Lemma 8. If 𝑢∈𝑋 is a weak solution of problem (1), then 𝑢 is a classical solution of problem (1).

Proof. By Definition 6, if 𝑢∈𝑋 is a weak solution of (1), then 𝐽^′(𝑢)(𝑣)=0 holds for all 𝑣∈𝑋 and hence for all 𝑣∈𝐶^∞₀(𝑡_𝑖,𝑡_𝑖+1),  𝑣^′∈𝐶^∞₀(𝑡_𝑖,𝑡_𝑖+1),  𝑣(𝑡)≡0,  𝑡∈[0,𝑡_𝑖]∪[𝑡_𝑖+1,𝑇]. So ∫^𝑡𝑖+1_𝑡𝑖𝑢^′′𝑣^′′+𝑢𝑣−𝜆𝑓(𝑡,𝑢)𝑣𝑑𝑡=0. By Lemma 5, we have 𝑢^′′(𝑡)=^𝑡_𝑡𝑖^𝑠₀−𝑢(𝜃)+𝜆𝑓(𝜃,𝑢(𝜃))𝑑𝜃𝑑𝑠+𝐶₁𝑡+𝐶₂ (22) for a.e. 𝑡∈[𝑡_𝑖,𝑡_𝑖+1] and some 𝐶₁,𝐶₂∈𝑅. So 𝑢⁽⁴⁾+𝑢−𝜆𝑓(𝑡,𝑢)=0, a.e. 𝑡∈[𝑡_𝑖,𝑡_𝑖+1], 𝑖=0,1,…,𝑙. Thus 𝑢 satisfies the equation in problem (1) and 𝑢∈𝐶⁴([0,𝑇]⧵{𝑡₁,𝑡₂,…,𝑡_𝑙}). By integration by parts for two times, we have that ^𝑇₀𝑢^′′𝑣^′′+𝑢𝑣𝑑𝑡−𝜆^𝑇₀𝑓(𝑡,𝑢(𝑡))𝑣(𝑡)𝑑𝑡−𝜆^𝑙_𝑖=1𝐼_1𝑖𝑢𝑡_𝑖𝑣𝑡_𝑖−𝜆^𝑙_𝑖=1𝐼_2𝑖𝑢^′𝑡_𝑖𝑣^′𝑡_𝑖+𝑎𝑏𝑢(𝑇)𝑣(𝑇)+𝑎𝑏𝑢(0)𝑣(0)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)+𝑐𝑑𝑢^′(0)𝑣^′(0)=^𝑙_𝑖=1𝑢^′′(𝑡)𝑣^′(𝑡)^𝑡𝑖+1_{𝑡=𝑡𝑖+}−^𝑙_𝑖=1𝑢^′′′(𝑡)𝑣(𝑡)^𝑡𝑖+1_{𝑡=𝑡𝑖+}+^𝑇₀𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)−𝜆𝑓(𝑡,𝑢(𝑡))𝑣(𝑡)𝑑𝑡−𝜆^𝑙_𝑖=1𝐼_1𝑖𝑢𝑡_𝑖𝑣𝑡_𝑖−𝜆^𝑙_𝑖=1𝐼_2𝑖𝑢^′𝑡_𝑖𝑣^′𝑡_𝑖+𝑎𝑏𝑢(𝑇)𝑣(𝑇)+𝑎𝑏𝑢(0)𝑣(0)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)+𝑐𝑑𝑢^′(0)𝑣^′(0)=^𝑙_𝑖=1Δ𝑢^′′′𝑡_𝑖−𝜆𝐼_1𝑖𝑢𝑡_𝑖𝑣𝑡_𝑖+^𝑇₀𝑢⁽⁴⁾(𝑡)(𝑡)((𝑡𝑣(𝑡)−+𝑢−𝜆𝑓𝑡,𝑢))𝑑𝑡_𝑙𝑖=1Δ𝑢^′′𝑡_𝑖+𝜆𝐼_2𝑖𝑢^′𝑡_𝑖𝑣^′𝑡_𝑖+𝑢^′′(𝑇)𝑣^′(𝑇)−𝑢^′′(0)𝑣^′(0)−𝑢^′′′(𝑇)𝑣(𝑇)+𝑢^′′′(0)𝑣(0)+𝑎𝑏𝑢(𝑇)𝑣(𝑇)+𝑎𝑏𝑢(0)𝑣(0)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)+𝑐𝑑𝑢^′(0)𝑣^′(0) (23) holds for all 𝑣∈𝑋. Since 𝑢 satisfies the equation of problem (1), (23) becomes ^𝑙_𝑖=1Δ𝑢^′′′𝑡_𝑖−𝜆𝐼_1𝑖𝑢𝑡_𝑖𝑣𝑡_𝑖−^𝑙_𝑖=1Δ𝑢^′′𝑡_𝑖+𝜆𝐼_2𝑖𝑢^′𝑡_𝑖𝑣^′𝑡_𝑖+𝑢^′′(𝑇)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)−𝑢^′′(0)−𝑐𝑑𝑢^′(0)𝑣^′(0)−𝑢^′′′(𝑇)−𝑎𝑏𝑢(𝑇𝑣(𝑇)+𝑢_′′′)(0)+𝑎𝑏𝑢(0)𝑣(0)=0 (24) for all 𝑣∈𝑋.
Next we will verify that 𝑢 satisfies impulsive condition in (1). If not, without loss of generality, we assume (24) holds for 𝑣(𝑡)≡0 for 𝑡∈[0,𝑡_𝑖]∪[𝑡_𝑖+1,𝑇], 𝑣^′(𝑡_𝑖)≠0. So (Δ𝑢^′′(𝑡_𝑖)+𝜆𝐼_2𝑖(𝑢^′(𝑡_𝑖)))𝑣^′(𝑡_𝑖)=0, and then Δ𝑢^′′(𝑡_𝑖)+𝜆𝐼_2𝑖(𝑢^′(𝑡_𝑖))=0. We assume 𝑣(𝑡)≡0 for 𝑡∈[0,𝑡_𝑖]∪[𝑡_𝑖+(𝑡_𝑖+1−𝑡_𝑖)/2,𝑇], 𝑣^′(𝑡_𝑖)≠0. So (24) becomes (Δ𝑢^′′(𝑡_𝑖)+𝜆𝐼_2𝑖(𝑢^′(𝑡_𝑖)))𝑣^′(𝑡_𝑖)=0, which means Δ𝑢^′′(𝑡_𝑖)+𝜆𝐼_2𝑖(𝑢^′(𝑡_𝑖)=0. Similarly, by choosing particular 𝑣∈𝑋, we can show that 𝑢 satisfies boundary conditions in problem (1).

Lemma 9. Let 𝑢∈𝑋; then ‖𝑢‖_𝐶¹≤𝑀‖𝑢‖_𝑋, where 𝑀=max{1/√𝑇,√𝑇}max{1+2𝑇²,2𝑇𝑑/𝑐}.

Proof. For any 𝑢∈𝑋, it follows from the mean-value theorem that 𝑢(𝜏)=1𝑇^𝑇₀𝑢(𝑠)𝑑𝑠 (25) for some 𝜏∈[0,𝑇]. Hence, for 𝑡∈[0,𝑇], using (25) and Hölder’s inequality, we have |𝑢(𝑡)|=𝑢(𝜏)+^𝑡_𝜏𝑢^′(𝑠)≤1𝑑𝑠√𝑇^𝑇0|𝑢(𝑠)|²𝑑𝑠^1/2+√𝑇^𝑇₀𝑢^′(𝑠)²𝑑𝑠^1/21≤max√𝑇,√𝑇‖𝑢‖. (26) Similarly, we have |𝑢^′(𝑡)|≤max{1/√𝑇,√𝑇}‖𝑢‖. So ‖𝑢‖_𝐶¹≤max{1/√𝑇,√𝑇}‖𝑢‖, which together with (13) yields the results.

Lemma 10. Suppose the following conditions hold.(H1)There exist constants 𝜇>2 and 𝑟≥0 such that, for |𝜉|≥𝑟, 0<𝜇𝐹(𝑡,𝜉)≤𝜉𝑓(𝑡,𝜉). (27)(H2)The impulsive functions 𝐼_1𝑖 satisfy sublinear growth; that is, there exist constants 𝛼_𝑖>0, 𝛽_𝑖>0, and 𝛾_𝑖∈[0,1), 𝑖=1,2,…,𝑙, such that 𝐼_1𝑖(𝑢)≤𝛼_𝑖+𝛽_𝑖|𝑢|^𝛾𝑖. (28)(H3)The impulsive functions 𝐼_2𝑖, 𝑖=1,2,…,𝑙, are bounded.(H4)𝑓(𝑡,𝑢)=𝑜(|𝑢|), 𝐼_1𝑖(𝑢)=𝑜(|𝑢|), 𝐼_2𝑖(𝑢)=𝑜(|𝑢|) as |𝑢|→0, 𝑖=1,2,…,𝑙.Then the functional 𝐽 defined by (9) is continuously differentiable. Moreover, it satisfies the Palais-Smale (PS) condition.

Proof. By the continuity of 𝑓, 𝐼_1𝑖 and 𝐼_2𝑖, 𝑖=1,2,…,𝑙, we know that 𝐽 is continuously differentiable. Next, we will prove that 𝐽 satisfies the Palais-Smale condition. Let {𝐽(𝑢_𝑘)} be a bounded sequence such that |𝐽^′(𝑢_𝑘)|→0 as 𝑘→∞. Then there exist two constants 𝑘,𝐶₁>0 such that for 𝑘 sufficiently large 𝐽𝑢_𝑘≤𝐶₁. (29) By (H3), there exists a constant 𝐶₂>0 such that 𝐼_2𝑖≤𝐶₂,𝑖=1,2,…,𝑙. (30) Then for 𝑘 sufficiently large, by (H1) (H2) and the definitions of 𝐽, 𝐽^′, we have 𝑢_𝑘²_𝑋=2𝐽𝑢_𝑘+2𝜆^𝑙_𝑖=1^{𝑢𝑘(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+2𝜆^𝑙_𝑖=1^{𝑢′𝑘(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠+2𝜆^𝑇₀𝐹𝑡,𝑢_𝑘(𝑡)𝑢𝑑𝑡≤2𝐽_𝑘+2𝜆^𝑙_𝑖=1^{𝑢𝑘(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+2𝜆^𝑙_𝑖=1^{𝑢′𝑘(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠+2𝜆𝜇^𝑇₀𝑓𝑡,𝑢_𝑘(𝑡)𝑢_𝑘(𝑡)𝑢𝑑𝑡=2𝐽_𝑘+2𝜆^𝑙_𝑖=1^{𝑢𝑘(𝑡𝑖)}₀𝐼_1𝑖(𝑡)𝑑𝑡+2𝜆^𝑙_𝑖=1^{𝑢′𝑘(𝑡𝑖)}₀𝐼_2𝑖(𝑠)2𝑑𝑠+𝜇𝑢_𝑘²_𝑋−2𝜆𝜇^𝑙_𝑖=1𝐼_1𝑖𝑢_𝑘𝑡_𝑖𝑢_𝑘𝑡_𝑖−2𝜆𝜇^𝑙_𝑖=1𝐼_2𝑖𝑢^′_𝑘𝑡_𝑖𝑢^′_𝑘𝑡_𝑖−2𝜇𝐽^′𝑢_𝑘𝑢_𝑘. (31) By Lemma 8, (29), (30), (31), (H2), and (H3), one has 12−1𝜇𝑢_𝑘²_𝑋=𝐽𝑢_𝑘−1𝜇𝐽^′𝑢_𝑘𝑢_𝑘+𝜆^𝑙_𝑖=1^{𝑢𝑘(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+𝜆^𝑙_𝑖=1^{𝑢′𝑘(𝑡𝑖)}₀𝐼_2𝑖(𝑠)−𝜆𝑑𝑠𝜇^𝑙_𝑖=1𝐼_1𝑖𝑢_𝑘𝑡_𝑖𝑢_𝑘𝑡_𝑖−𝜆𝜇^𝑙_𝑖=1𝐼_2𝑖𝑢^′_𝑘𝑡_𝑖𝑢^′_𝑘𝑡_𝑖≤𝐶₁+𝐶₁𝜇𝑢_𝑘+𝜆𝑢_𝑘𝐶1^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖𝑢_𝑘^𝛾𝑖_𝐶1+𝑙𝜆𝐶₂𝑢_𝑘𝐶1+𝜆𝜇𝑢_𝑘𝐶1^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖𝑢_𝑘^𝛾𝑖_𝐶1+𝑙𝜆𝐶₂𝜇𝑢_𝑘𝐶1=𝐶₁+𝐶₁𝜇𝑢_𝑘1+𝜆1+𝜇𝑀𝑢_𝑘^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖𝑀𝑢_𝑘^𝛾𝑖+𝑙𝜆𝐶₂11+𝜇𝑀𝑢_𝑘. (32) So {𝑢_𝑘} is bounded in 𝑋, which implies that the sequence {𝑢_𝑘} weakly converges to 𝑢.
Next we show that {𝑢_𝑘} strongly converges to 𝑢 in 𝑋: 𝐽^′𝑢_𝑘−𝐽^′(𝑢)𝑢_𝑘=𝑢−𝑢_𝑘−𝑢²_𝑋−𝜆^𝑙_𝑖=1𝐼_1𝑖𝑢_𝑘𝑡_𝑖−𝐼_1𝑖𝑢𝑡_𝑖𝑢_𝑘𝑡_𝑖−𝑢𝑡_𝑖−𝜆^𝑙_𝑖=1𝐼_2𝑖𝑢^′_𝑘𝑡_𝑖−𝐼_2𝑖𝑢^′𝑡_𝑖𝑢^′_𝑘𝑡_𝑖−𝑢^′𝑡_𝑖−𝜆^𝑇₀𝑓𝑡,𝑢_𝑘(𝑡)−𝑓(𝑡,𝑢(𝑡))𝑢_𝑘𝑡_𝑖−𝑢𝑡_𝑖𝑑𝑡. (33) Similar to the proof of Proposition 1.2 in [18], the weak convergence 𝑢_𝑘⇀𝑢 implies that {𝑢_𝑘} uniformly converges to 𝑢 in 𝐶([0,𝑇]). Since 𝑢_𝑘∈𝑋, 𝑢^′_𝑘 converges to 𝑢^′ in 𝐶[0,𝑇]. Thus 𝐽^′𝑢_𝑘−𝐽^′(𝑢)𝑢_𝑘𝜆−𝑢⟶0,_𝑙𝑖=1𝐼_1𝑖𝑢_𝑘𝑡_𝑖−𝐼_1𝑖𝑢𝑡_𝑖𝑢_𝑘𝑡_𝑖−𝑢𝑡_𝑖𝜆⟶0,_𝑙𝑖=1𝐼_2𝑖𝑢^′_𝑘𝑡_𝑖−𝐼_2𝑖𝑢^′𝑡_𝑖𝑢^′_𝑘𝑡_𝑖−𝑢^′𝑡_𝑖𝜆⟶0,_𝑇0𝑓𝑡,𝑢_𝑘(𝑡)−𝑓(𝑡,𝑢(𝑡))𝑢_𝑘𝑡_𝑖−𝑢𝑡_𝑖𝑑𝑡⟶0. (34) So ‖𝑢_𝑘−𝑢‖_𝑋→0 as 𝑘→∞. In other words, {𝑢_𝑘} converges strongly to 𝑢 in 𝑋.

Remark 11. By (H1), there exist 𝑎₁,𝑎₂>0 such that 𝐹(𝑡,𝜉)≥𝑎₁𝜉^𝜇−𝑎₂, (35) for all 𝑡∈[0,𝑇], 𝜉∈𝑅.

3. Existence Results for One Solution and Infinitely Many Solutions

Theorem 12. Suppose that (H1)–(H3) hold. Furthermore, we assume (H4) holds. Then problem (1) has at least one nontrivial solution.

Proof. We will use Theorem 1 to prove the theorem. By Lemma 10, we have known that 𝐽 satisfies the (PS) condition and it is obvious that 𝐽(0)=0. By (H4), for any 𝜀>0, there exists a 𝛿>0 such that |𝜉|≤𝛿, which implies 𝐹(≤1𝑡,𝜉)2𝜀𝜉²,^𝑙_𝑖=1^𝜉₀𝐼_1𝑖(𝑠)≤1𝑑𝑠2𝑙𝜀𝜉²,_{𝑙𝑖=1𝜉0}𝐼_2𝑖(𝑠)≤1𝑑𝑠2𝑙𝜀𝜉² (36) for all 𝑡∈[0,𝑇]. Consequently, by Lemma 9, one has, for ‖𝑢‖_𝑋≤𝛿/𝑀, ^𝑇₀𝐹((𝑡≤1𝑡,𝑢))𝑑𝑡2𝑀²𝑇𝜀‖𝑢‖²_𝑋,_{𝑙𝑖=1𝑢(𝑡𝑖)0}𝐼_1𝑖(𝑠)≤1𝑑𝑠2𝑀²𝑙𝜀‖𝑢‖²_𝑋,_{𝑙𝑖=1𝑢′(𝑡𝑖)0}𝐼_2𝑖(𝑠)≤1𝑑𝑠2𝑀²𝑙𝜀‖𝑢‖²_𝑋. (37) Thus ^𝑇₀𝐹(𝑡,𝑢(𝑡))𝑑𝑡+^𝑙_𝑖=1^{𝑢(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+^𝑙_𝑖=1^{𝑢′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)≤1𝑑𝑠2𝑀²(𝑇+2𝑙)𝜀‖𝑢‖²_𝑋=𝑜‖𝑢‖²_𝑋 (38) as ‖𝑢‖_𝑋→0. Therefore, 𝐽(𝑢)=12‖𝑢‖²_𝑋−𝜆^𝑙_𝑖=1^{𝑢(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠−𝜆^𝑙_𝑖=1^{𝑢′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠−𝜆^𝑇₀𝐹((𝑡≥1𝑡,𝑢))𝑑𝑡2‖𝑢‖²_𝑋+𝑜‖𝑢‖²_𝑋 (39) as 𝑢→0. Thus (C1) holds.
To verify (C2), we choose 𝑒(𝑡)∈𝑋, 𝜅∈𝑅 such that 𝐾₁=‖𝑒‖_𝑋>0, 𝐾₂=‖𝑒‖_𝐿²>0, 𝐾₁, 𝐾₂ are constants. Then by (H1), (H2), Remark 11, Lemma 8, and (30), one has 𝐽()=𝜅𝜅𝑒²2‖𝑒‖²_𝑋−𝜆^𝑙_𝑖=1^{𝜅𝑒(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠−𝜆^𝑙_𝑖=1^{𝜅𝑒′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠−𝜆^𝑇₀𝐹((𝑡≤𝜅𝑡,𝜅𝑒))𝑑𝑡₂2‖𝑒‖²_𝑋+𝜆^𝑙_𝑖=1𝛼_𝑖‖𝜅𝑒‖_𝐶¹+𝛽_𝑖‖𝜅𝑒‖^𝛾𝑖_𝐶⁺¹₁𝛾_𝑖+1+𝜆𝑙𝐶₂|𝜅|‖𝑒‖_𝐶¹−𝜆^𝑇₀𝑎₁|𝜅|^𝜇|𝑒(𝑡)|^𝜇−𝑎₂≤𝜅𝑑𝑡₂2𝐾²₁+𝜆^𝑙_𝑖=1𝛼_𝑖|𝜅|𝑀𝐾₁+𝛽_𝑖|𝜅|^𝛾𝑖+1𝑀^𝛾𝑖+1𝐾^𝛾𝑖₁⁺¹𝛾_𝑖+1+𝜆𝑙𝐶₂𝑀𝐾₁|𝜅|−𝜆^𝑇₀𝑎₁|𝜅|^𝜇|𝑒(𝑡)|^𝜇−𝑎₂𝑑𝑡. (40) By Hölder’s inequality, we have ^𝑇₀|𝑒(𝑡)|^𝜇𝑑𝑡≥^𝑇₀𝑒²(𝑡)𝑑𝑡𝑇^{(2−𝜇)/𝜇𝜇/2}=‖𝑒‖^𝜇_𝐿2𝑇^(2−𝜇)/2=𝐾^𝜇₂𝑇^(2−𝜇)/2. (41) Substituting (41) into (40), we have 𝐽()≤𝜅𝜅𝑒²2𝐾²₁+𝜆𝑀𝐾₁^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖|𝜅𝑒|^𝛾𝑖+1𝑀^𝛾𝑖+1𝐾^𝛾𝑖₁⁺¹𝛾_𝑖+1+𝑙𝐶₂𝐾₁𝑀|𝜅|−𝑎₁𝜆|𝜅|^𝜇𝐾^𝜇₂𝑇^(2−𝜇)/2+𝑎₂𝜆𝑇⟶−∞ (42) as |𝜅|→+∞. Hence (C2) holds. Therefore, applying Theorem 1, we deduce that 𝐽 admits a critical value 𝑐>0 characterized as in the statement of Theorem 1 to 𝐽, 𝐽 possesses critical value 𝑐>0 given by𝑐=inf_𝑔∈Γmax_{𝑢∈𝑔([0,1])}𝐽(𝑢), (43) where Γ=𝑔∈𝐶([0,1],𝐸)∶𝑔(0)=0,𝑔(1)=𝑒. (44) Let 𝑢^∗∈𝑋 be a critical point associated with the critical value 𝑐 of 𝐽 (i.e., 𝐽(𝑢^∗)=𝑐). Condition 𝑐>0 implies that 𝑢^∗≢0. Lemma 8 means that IBVP (1) has at least one nontrivial solution.

Theorem 13. Suppose that (H1)–(H4) hold. Moreover, assume that the nonlinearity 𝑓(𝑡,𝑢) and impulsive functions 𝐼_1𝑖, 𝐼_2𝑖 are all odd in 𝑢. Then IBVP (1) has infinitely many classical solutions.

Proof. We apply Theorem 2 to complete the proof. Clearly 𝐽∈𝐶¹(𝑋,𝑅) is even and 𝐽(0)=0. Lemma 10 shows that 𝐽 satisfies (PS) condition. The arguments of Theorem 12 show that 𝐽 satisfies (C3) in Theorem 2. To verify (C4), let 𝑊 be any finite dimensional space in 𝑋. For any 𝑢∈𝑊, by (H1), (H2), Remark 11, Lemma 4, and (30), one has 𝐽(𝑢)=12‖𝑢‖²_𝑋−𝜆^𝑙_𝑖=1^{𝑢(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠−𝜆^𝑙_𝑖=1^{𝑢′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠−𝜆^𝑇₀𝐹((𝑡≤1𝑡,𝑢))𝑑𝑡2‖𝑢‖²_𝑋+𝜆^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖|𝑢|^𝛾𝑖+1𝛾_𝑖+1+𝑙𝜆𝐶₂‖𝑢‖_𝐶¹−𝜆^𝑇₀𝑎₁|𝑢(𝑡)|^𝜇−𝑎₂≤1𝑑𝑡2‖𝑢‖²_𝑋+𝜆𝑀‖𝑢‖_𝑋^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖𝑀^𝛾𝑖+1‖𝑢‖^𝛾𝑖_𝑋⁺¹𝛾_𝑖+1+𝑙𝜆𝐶₂𝑀‖𝑢‖_𝑋−𝜆^𝑇₀𝑎₁|𝑢(𝑡)|^𝜇−𝑎₂𝑑𝑡. (45) For finite dimensional space 𝑊, the norm ‖⋅‖_𝑋 is equivalent to ‖⋅‖_𝑊.
So there exists 𝑐₀>0 satisfying ‖𝑢‖_𝑋≤𝑐₀‖𝑢‖_𝐿^𝜇 (46) for 𝑢∈𝑊. Thus, ^𝑇₀𝑎₁|𝑢(𝑡)|^𝜇𝑑𝑡≥𝑎₁𝑐₀^−𝜇‖𝑢‖^𝜇_𝑋. (47) By (45) (46) (47) we have 𝐽(𝑢)≤12‖𝑢‖²_𝑋+𝜆^𝑙_𝑖=1𝛼_𝑖+𝛽_𝑖𝑀^𝛾𝑖+1‖𝑢‖^𝛾𝑖_𝑋⁺¹𝛾_𝑖+1+𝑙𝜆𝐶₂𝑀‖𝑢‖_𝑋−𝜆𝑎₁𝑐₀^−𝜇‖𝑢‖^𝜇_𝑋+𝜆𝑎₂𝑇⟶−∞ (48) as ‖𝑢‖_𝑋→+∞. That is, there exists 𝑅>0 such that 𝐽(𝑢)<0 for 𝑢∈𝑊⧵𝐵_𝑅(𝑊). The proof is complete.

4. Existence Result for Three Nonnegative Solutions

In this part, we need the following conditions.(H5)For 𝐼_1𝑖, 𝐼_2𝑖, and 𝑓,(C5)𝑓∈𝐶([0,𝑇]×[0,+∞);[0,+∞)), 𝐼_1𝑖∈𝐶(𝑅;𝑅), 𝐼_2𝑖∈𝐶(𝑅;𝑅), 𝑖=1,2,…,𝑙;(C6)𝑓(𝑡,0)=𝐼_1𝑖(0)=𝐼_2𝑖(0)=0 for almost every 𝑡∈[0,𝑇] and 𝐼_2𝑖(𝑥)𝑥≥0 for all 𝑥∈𝑅.

Lemma 14 (see Lemma 2.2 [12]). For 𝑢∈𝑋, let 𝑢^±=max{±𝑢,0}. Then the following five properties hold:(i)𝑢∈𝑋⇒𝑢⁺, 𝑢⁻∈𝑋;(ii)𝑢=𝑢⁺−𝑢⁻;(iii)‖𝑢⁺‖_𝑋≤‖𝑢‖_𝑋;(iv)if (𝑢_𝑛) uniformly converges to u in 𝐶([0,𝑇]), then (𝑢⁺_𝑛) uniformly converges to 𝑢⁺ in 𝐶([0,𝑇]);(v)𝑢⁺(𝑡)𝑢⁻(𝑡)=0, (𝑢⁺)^′(𝑡)(𝑢⁻)^′(𝑡)=0 for a.e. 𝑡∈[0,𝑇].

Lemma 15. If 𝑢∈𝐶([0,𝑇]) is a classical solution of problem 𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)=𝜆𝑓𝑡,𝑢⁺(𝑡),𝑡≠𝑡_𝑖,𝑡∈[0,𝑇],Δ𝑢^′′′𝑡_𝑖=𝜆𝐼_1𝑖𝑢⁺𝑡_𝑖,𝑖=1,2,…,𝑙,−Δ𝑢^′′𝑡_𝑖=𝜆𝐼_2𝑖𝑢^+′𝑡_𝑖(0),𝑖=1,2,…,𝑙,𝑎𝑢+𝑏𝑢^′′′(0)=0,𝑎𝑢(𝑇)−𝑏𝑢^′′′(𝑇)=0,𝑐𝑢^′(0)−𝑑𝑢^′′(0)=0,𝑐𝑢^′(𝑇)+𝑑𝑢^′′(𝑇)=0, (49) then 𝑢(𝑡)≥0 for 𝑡∈[0,𝑇], and hence it is a nonnegative classical solution of (1).

Proof. Since 𝑢∈𝐶[0,𝑇] and 𝑓∈𝐶([0,𝑇]×[0,+∞);[0,+∞)), we have 𝑢⁽⁴⁾∈𝐶[0,𝑇]⧵{𝑡₁,𝑡₂,…,𝑡_𝑖}. If 𝑢∈𝐶([0,𝑇]) is a classical solution of problem (49), by Lemma 14, (H5) and boundary conditions, we have 0=^𝑇₀𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)−𝜆𝑓𝑡,𝑢⁺(𝑡)𝑢⁻(𝑡)𝑑𝑡=−𝜆^𝑙_𝑖=1𝐼_1𝑖𝑢⁺𝑡_𝑖𝑢⁻𝑡_𝑖−𝜆^𝑙_𝑖=1𝐼_2𝑖𝑢^+′𝑡_𝑖(𝑢⁻)^′𝑡_𝑖+𝑢^′′′(𝑇)𝑢⁻(𝑇)−𝑢^′′′(0)𝑢⁻(0)−𝑢^′′(𝑇)(𝑢⁻)^′(𝑇)+𝑢^′′(0)(𝑢⁻)^′(0)+^𝑇₀𝑢^′′(𝑡)(𝑢⁻)^′′(𝑡)+𝑑𝑡_𝑇0𝑢(𝑡)(𝑢⁻)(𝑡)𝑑𝑡−𝜆^𝑇₀𝑓𝑡,𝑢⁺(𝑡)𝑢⁻(𝑡)=𝑎𝑑𝑡𝑏𝑢(𝑇)𝑢⁻(𝑇)+𝑎𝑏𝑢(0)𝑢⁻(0)+𝑐𝑑𝑢^′(𝑇)(𝑢⁻)^′(𝑇)+𝑐𝑑𝑢^′(0)(𝑢⁻)^′(0)+^𝑇₀𝑢^′′(𝑡)(𝑢⁻)^′′(𝑡)+𝑢(𝑡)(𝑢⁻)(𝑡)𝑑𝑡≤−‖𝑢⁻‖²_𝑋. (50) So 𝑢⁻(𝑡)=0 for 𝑡∈[0,𝑇]; that is, 𝑢(𝑡)≥0. The proof is complete.

Remark 16. By Lemmas 14 and 15, in order to obtain the nonnegative solutions of (1), it is sufficient to show the existence of solutions of (49).
For each 𝑢∈𝑋, set Φ(𝑢)=12‖𝑢‖²_𝑋,Ψ(𝑢)=^𝑙_𝑖=1^{𝑢+(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+^𝑙_𝑖=1^{(𝑢+)′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)+𝑑𝑠_𝑇0𝐹𝑡,𝑢⁺(𝑡)𝐽(𝑢)(𝑢)(𝑢).𝑑𝑡,=Φ−𝜆Ψ (51)(52) It is obvious that Φ, Ψ, and 𝐽 are differentiable for any 𝑢∈𝑋. Then we have Φ^′(𝑢)(𝑣)=^𝑇₀𝑢^′′𝑣^′′𝑎+𝑢𝑣𝑑𝑡+𝑏𝑢(𝑇)𝑣(𝑇)+𝑎𝑏𝑢(0)𝑣(0)+𝑐𝑑𝑢^′(𝑇)𝑣^′(𝑇)+𝑐𝑑𝑢^′(0)𝑣^′(0),Ψ^′(𝑢)(𝑣)=^𝑙_𝑖=1𝐼_1𝑖𝑢⁺𝑡_𝑖𝑣𝑡_𝑖+^𝑙_𝑖=1𝐼_2𝑖𝑢^+′𝑡_𝑖𝑣^′𝑡_𝑖+^𝑇₀𝑓𝑡,𝑢⁺(𝑡)𝑣(𝑡)𝑑𝑡. (53)(54)

Definition 17. A function 𝑢∈𝑋 is said to be a weak solution of (49), if 𝑢 satisfies 𝐽^′(𝑢)(𝑣)=0 for all 𝑣∈𝑋.

Lemma 18. If 𝑢∈𝑋 is a weak solution of (49), then 𝑢 is a classical solution of (49).

Proof. It is similar to the proof of Lemma 8, so we omit it here.

Lemma 19. Φ^′∶𝑋→𝑋^∗ admits a continuous inverse on 𝑋^∗.

Proof. First, for every 𝑢∈𝑋⧵{0}, by (53), we have lim_{‖𝑢‖𝑋→+∞}Φ^′(𝑢)(𝑢)‖𝑢‖_𝑋=lim_{‖𝑢‖𝑋→+∞}‖𝑢‖²_𝑋‖𝑢‖_𝑋=+∞, (55) which means that Φ^′ is coercive. Furthermore, given 𝑢,𝑣∈𝑋, one has Φ^′(𝑢)−Φ^′(𝑣)(𝑢−𝑣)≥‖𝑢−𝑣‖²_𝑋, (56) so Φ^′ is uniformly monotone. By Theorem 26.A(d) of [20], we have that (Φ^′)⁻¹ exists and (Φ^′)⁻¹ is continuous on 𝑋^∗. Thus, Φ^′∶𝑋→𝑋^∗ admits a continuous inverse on 𝑋^∗. The proof is complete.

Lemma 20. Ψ^′∶𝑋→𝑋^∗ is a continuous and compact operator.

Proof. First we will show that Ψ^′ is strongly continuous on 𝑋. Let 𝑢_𝑛⇀𝑢 as 𝑛→∞ on 𝑋. By [20], we have (𝑢_𝑛) converges uniformly to 𝑢 on [0,𝑇] as 𝑛→∞. Since 𝑓 is continuous, one has 𝑓(𝑡,𝑢_𝑛)→𝑓(𝑡,𝑢) as 𝑛→∞. Furthermore, 𝐼_1𝑖, 𝐼_2𝑖 are all continuous. So Ψ^′(𝑢_𝑛)→Ψ^′(𝑢), which implies that Ψ^′ is continuous and that Ψ^′ is a compact operator by Proposition 26.2 of [20]. The proof is complete.

Theorem 21. Suppose that the condition (H5) holds. Let 𝑘∶={2𝑀²(1024/𝑇³+(83/240)𝑇)}⁻¹, 𝐼(𝑠)∶=∑^𝑙_𝑖=1|𝐼_1𝑖(𝑠)|+∑^𝑙_𝑖=1|𝐼_2𝑖(𝑠)|. There exist four positive constants 𝑚, 𝑛, 𝑝, 𝑞, with √𝑘𝑚<𝑛<√𝑘𝑝<√𝑘𝑞 such that(H6)(C7)(∫^𝑚₀𝐼(𝑠)𝑑𝑠+∫^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡)/𝑚²<𝑘((−∫^𝑚₀𝐼(𝑠)𝑑𝑠+∫^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡−∫^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡)/𝑛²), (C8)(∫^𝑝₀𝐼(𝑠)𝑑𝑠+∫^𝑇₀𝐹(𝑡,𝑝)𝑑𝑡)/𝑝²<𝑘((−∫^𝑚₀𝐼(𝑠)𝑑𝑠+∫^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡−∫^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡)/𝑛²), (C9)(∫^𝑞₀𝐼(𝑠)𝑑𝑠+∫^𝑇₀𝐹(𝑡,𝑞)𝑑𝑡)/(𝑞²−𝑝²)<𝑘((−∫^𝑚₀𝐼(𝑠)𝑑𝑠+∫^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡−∫^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡)/𝑛²). Then, for every 𝑛𝜆∈²2𝑀²𝑘−^𝑚₀𝐼(𝑠)𝑑𝑠+^3𝑇/4_𝑇/4𝐹()−𝑡,𝑛𝑑𝑡_𝑇0𝐹(𝑡,𝑚)𝑑𝑡⁻¹,𝑚min²2𝑀^2𝑚₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡⁻¹,𝑝²2𝑀^2𝑝₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑝𝑑𝑡⁻¹,𝑞²−𝑝²2𝑀^2𝑞₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑞𝑑𝑡⁻¹, (57) the problem (1) has at least three distinct nonnegative classical solutions 𝑢_𝑖(𝑖=1,2,3), such that ‖𝑢_𝑖‖_𝐶¹<𝑞, which means that the problem (1) has at least two distinct positive classical solutions.

Proof. The proof is based on Theorem 3. First, we will prove that Φ and Ψ satisfy the hypotheses in Theorem 3. On the one hand, Φ is coercive and its Gâteaux derivative admits a continuous inverse by Lemma 19. On the other hand, Φ is obviously convex. Ψ’s Gâteaux derivative is continuous and compact by Lemma 20. In addition, inf_𝑋Φ=Φ(0)=Ψ(0)=0. By (H5), we have ^𝑠₀𝐼_1𝑖(𝑠)𝑑𝑠≥0,^𝑠₀𝐼_2𝑖(𝑠)𝐹()=𝑑𝑠≥0,𝑡,𝑠^𝑠₀𝑓(𝑡,𝑠)𝑑𝑠≥0, (58) which deduces that Ψ(𝑢)≥0 for all 𝑢∈𝑋.
Next, we will verify the conditions (i), (ii), (iii) in Theorem 3. First, we define 𝑣(𝑡)=32𝑛𝑇²𝑡²𝑇,𝑡∈0,8,−32𝑛𝑇²𝑇𝑡−4²𝑇+𝑛,𝑡∈8,𝑇4,𝑇𝑛,𝑡∈4,3𝑇4,−32𝑛𝑇²𝑡−3𝑇4²+𝑛,𝑡∈3𝑇4,7𝑇8,32𝑛𝑇²(𝑡−𝑇)²,𝑡∈7𝑇8,𝑇. (59) It is easy to verify that 𝑣⁺=𝑣,𝑣⁻=0. (60) By computing, Φ(𝑣)=1024𝑇³+83𝑇𝑛240²=𝑛²2𝑀²𝑘. (61) Let 𝑟₁=𝑚²/2𝑀², 𝑟₂=𝑝²/2𝑀², and 𝑟₃=(𝑞²−𝑝²)/2𝑀². By √𝑘𝑚<𝑛<√𝑘𝑝<√𝑘𝑞, one has 𝑟₁<Φ(𝑣)<𝑟₂, which means that 𝑣∈Φ⁻¹([𝑟₁,𝑟₂[) and 𝑟₃>0. When Φ(𝑢)<𝑟₁, by Lemma 9 and (51), we have maxmax_{[]𝑡∈0,𝑇}𝑢⁺(𝑡),max_{[]𝑡∈0,𝑇}𝑢^+′(𝑡𝑢)≤𝑀⁺_𝑋≤𝑀‖𝑢‖_𝑋≤2𝑀²Φ(𝑢)<𝑚. (62) In view of (51) and 𝐼(𝑠), we have sup_{𝑢∈Φ−1(]−∞,𝑟1[)}Ψ(𝑢)≤max_𝜉^𝑙_{≤𝑚𝑖=1}^𝜉₀𝐼_1𝑖(𝑠)𝑑𝑠+max_𝜉^𝑙_{≤𝑚𝑖=1}^𝜉₀𝐼_2𝑖(𝑠+)𝑑𝑠_𝑇0max_𝜉≤𝑚𝐹(𝑡,𝜉)𝑑𝑡=^𝑚₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡. (63) Similarly, we have sup_{𝑢∈Φ−1(]−∞,𝑟2[)}Ψ(𝑢)≤^𝑝₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑝𝑑𝑡,sup_{𝑢∈Φ−1(]−∞,𝑟2+𝑟3[)}Ψ(𝑢)≤^𝑞₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑞𝑑𝑡. (64) Therefore, taking into consideration that 0∈Φ⁻¹(]−∞,𝑟₁[) and 0∈Φ⁻¹(]−∞,𝑟₂[), by (63) and (64), we have 𝜑𝑟₁≤sup_𝑢∈Φ⁻¹_{(]−∞,𝑟1[)}Ψ(𝑢)𝑟₁≤2𝑀²𝑚^2𝑚₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹(),𝜑𝑟𝑡,𝑚𝑑𝑡₂≤sup_𝑢∈Φ⁻¹_{(]−∞,𝑟2[)}Ψ(𝑢)𝑟₂≤2𝑀²𝑝^2𝑝₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹,𝛾𝑟𝑡,𝑝𝑑𝑡₂,𝑟₃=sup_𝑢∈Φ⁻¹_{(]−∞,𝑟2+𝑟3[)}Ψ(𝑢)𝑟₃≤2𝑀²𝑞²−𝑝^2𝑞₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑞𝑑𝑡. (65) Furthermore, by (60) and the definition of 𝑣, Ψ(𝑣)=^𝑙_𝑖=1^{𝑣(𝑡𝑖)}₀𝐼_1𝑖(𝑠)𝑑𝑠+^𝑙_𝑖=1^{𝑣′(𝑡𝑖)}₀𝐼_2𝑖(𝑠)𝑑𝑠+^𝑇₀𝐹(𝑡,𝑣(𝑡≥))𝑑𝑡_{𝑙𝑖=1min𝑡∈[0,𝑇]0𝑣(𝑡)}𝐼_1𝑖(𝑠)𝑑𝑠+^𝑙_𝑖=1^{min𝑡∈[0,𝑇]|𝑣′}₀^(𝑡)|𝐼_2𝑖(𝑠+)𝑑𝑠_{3𝑇/4𝑇/4}𝐹(𝑡,𝑣(𝑡))𝑑𝑡=^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡. (66) Taking 𝑣∈Φ⁻¹([𝑟₁,𝑟₂[) into consideration, by (61), (63), (66), (C9), and Φ(𝑢)≥0, one has 𝛽𝑟₁,𝑟₂≥inf_{𝑢∈Φ−1(]−∞,𝑟1[)}Ψ(𝑣)−Ψ(𝑢)Φ(𝑣)(𝑢)≥∫−Φ_{3𝑇/4𝑇/4}𝐹(𝑡,𝑛)𝑑𝑡−∫^𝑚₀𝐼(𝑠)𝑑𝑠+∫^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡𝑛²/2𝑀²𝑘=2𝑀²𝑘𝑛²−^𝑚₀𝐼(𝑠)𝑑𝑠+^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡−^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡. (67) By (65), (67), and (C7)–(C9) of (H6), we have 𝛼𝑟₁,𝑟₂,𝑟₃<𝛽𝑟₁,𝑟₂, (68) which yields the conditions in Theorem 3. By Theorem 3, it follows that, for each 𝑛𝜆∈²2𝑀²𝑘−^𝑚₀𝐼(𝑠)𝑑𝑠+^3𝑇/4_𝑇/4𝐹()−𝑡,𝑛𝑑𝑡_𝑇0𝐹(𝑡,𝑚)𝑑𝑡⁻¹,𝑚min²2𝑀^2𝑚₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹(𝑡,𝑚)𝑑⁻¹,𝑝²2𝑀^2𝑝₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑝𝑑𝑡⁻¹,𝑞²−𝑝²2𝑀^2𝑞₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑞𝑑𝑡⁻¹, (69) the functional 𝐽=Φ−𝜆Ψ has three distinct critical points 𝑢_𝑖(𝑖=1,2,3) in 𝑋 with Φ(𝑢_𝑖)<𝑟₂+𝑟₃. By Lemma 9 and (51) 𝑢⁺_𝑖𝐶1=maxmax_{[]𝑡∈0,𝑇}𝑢⁺_𝑖(𝑡),max_{[]𝑡∈0.𝑇}𝑢⁺_𝑖^′(𝑡𝑢)≤𝑀⁺_𝑖𝑋≤𝑀𝑢_𝑖𝑋≤2𝑀²Φ𝑢_𝑖<𝑞. (70) By Remark 16, 𝑢_𝑖(𝑖=1,2,3) are three positive solutions of (1). The proof is complete.

Remark 22. If we choose different 𝑣, then the constrictions on 𝐹, 𝐼_𝑖 are different.

5. Examples

Example 1. Let 𝑇>0, 𝑡_𝑖∈(0,𝑇), 𝑎,𝑏,𝑐,𝑑>0, 𝑥₁,𝑥₂,𝑥₃,𝑦_𝑖,𝑧_𝑖∈𝐶([0,𝑇],𝑅⁺), 𝑖=1,2,…,𝑙. Consider the following Sturm-Liouville boundary-value problem with impulse: 𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)=𝜆𝑓₁(𝑡,𝑢),𝑡∈[0,𝑇]⧵𝑡₁,𝑡₂,…,𝑡_𝑙,Δ𝑢^′′′𝑡_𝑖=𝜆𝐼_𝑖𝑢𝑡_𝑖,𝑖=1,2,…,𝑙,−Δ𝑢^′′𝑡_𝑖=−𝜆𝑧_𝑖(𝑡)sin𝑢^′3𝑡_𝑖(0),𝑖=1,2,…,𝑙,𝑎𝑢+𝑏𝑢^′′′(0)=0,𝑎𝑢(𝑇)−𝑏𝑢^′′′(𝑇)=0,𝑐𝑢^′(0)−𝑑𝑢^′′(0)=0,𝑐𝑢^′(𝑇)+𝑑𝑢^′′(𝑇)=0, (71) where 𝑓₁(𝑡,𝑢)=𝑥₁(𝑡)𝑢³+𝑥₂(𝑡)sin𝑢+𝑥₃(𝑡)|𝑢|,𝐼exp_𝑖(𝑢)=𝑦_𝑖(𝑡)𝑢³,|𝑢|𝑦≤1,_𝑖(𝑡)𝑢^1/3,|𝑢|>1. (72)

By computing, 𝐹(𝑡,𝑢)=(1/4)𝑥₁(𝑡)𝑢⁴−𝑥₂(𝑡)cos𝑢+𝑥₃(𝑡)sgn𝑢exp|𝑢|. Let 𝜇=6; there exists 𝑟≥0 such that, for |𝜉|≥𝑟, 0<𝜇𝐹(𝑡,𝜉)≤𝜉𝑓(𝑡,𝜉). The conditions (H1)–(H4) are satisfied. Applying Theorem 1, problem (71) has at least one nontrivial solution.

Furthermore, the functions 𝑓(𝑡,𝑢), 𝐼_1𝑖, 𝐼_2𝑖 are all odd in 𝑢. Applying Theorem 2, problem (71) has infinitely many classical solutions.

Example 2. Let 𝑇=1, 𝑡_𝑖∈(0,1), 𝑖=1,2,…,𝑙, 𝑎,𝑏,𝑐,𝑑>0. Consider the following Sturm-Liouville boundary-value problem: 𝑢⁽⁴⁾(𝑡)+𝑢(𝑡)=𝜆𝑓₂(𝑡,𝑢(𝑡)),𝑡∈[0,1]⧵𝑡₁,𝑡₂,…,𝑡_𝑙,Δ𝑢^′′′𝑡_𝑖=𝜆𝐼_1𝑖𝑢𝑡_𝑖,𝑖=1,2,…,𝑙,−Δ𝑢^′′𝑡_𝑖=𝜆𝐼_2𝑖𝑢^′𝑡_𝑖(0),𝑖=1,2,…,𝑙,𝑎𝑢+𝑏𝑢^′′′(0)=0,𝑎𝑢(1)−𝑏𝑢^′′′(1)=0,𝑐𝑢^′(0)−𝑑𝑢^′′(0)=0,𝑐𝑢^′(1)+𝑑𝑢^′′(1)=0, (73) where 𝑓₂(𝑡,𝑠)=𝐼_1𝑖(𝑠)=𝐼_2𝑖(𝑠)=𝑡()()0,0<𝑠≤1,𝑠−1,1<𝑠≤2,−𝑡𝑠−3,2<𝑠≤3,0,𝑠>3,forevery𝜆∈7680,245843,+∞ (74) problem (73) has at least two distinct positive classical solutions.

In fact, (H5) is fulfilled. By computing, 𝑀=max{3,2𝑑/𝑐}, so 𝑘=min{40/737529,30𝑐²/245843𝑑²}. Considering √𝑘𝑚<𝑛<√𝑘𝑝<√𝑘𝑞, without loss of generality, we choose 𝑚=1/2, 𝑛=2 and sufficiently large 𝑝, 𝑞. Then we have 𝑛²2𝑀²𝑘−^𝑚₀𝐼(𝑠)𝑑𝑠+^3𝑇/4_𝑇/4𝐹(𝑡,𝑛)𝑑𝑡−^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡⁻¹=161280,𝑚245843₂2𝑀^2𝑚₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹(𝑡,𝑚)𝑑𝑡⁻¹𝑝=∞,₂2𝑀^2𝑝₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑝𝑑𝑡⁻¹𝑞issufficientlylarge,₂−𝑝²2𝑀^2𝑞₀𝐼(𝑠)𝑑𝑠+^𝑇₀𝐹𝑡,𝑞𝑑𝑡⁻¹issufficientlylarge, (75) and that (H6) is satisfied. Applying Theorem 21, problem (73) has at least two distinct positive classical solutions for every 𝜆∈[161280/245843,+∞).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by National Science Foundation for Young Scholars (11001028) and Beijing Higher Education Young Elite Teacher Project YETP0458.