Positive Solutions for Resonant and Nonresonant Nonlinear Third-Order Multipoint Boundary Value Problems

Abstract

Positive solutions for a kind of third-order multipoint boundary value problem under the nonresonant conditions and the resonant conditions are considered. In the nonresonant case, by using the Leggett-Williams fixed point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using the Leggett-Williams norm-type theorem due to O’Regan and Zima, the existence result of at least one positive solution is established. It is remarkable to point out that it is the first time that the positive solution is considered for the third-order boundary value problem at resonance. Some examples are given to demonstrate the main results of the paper.

1. Introduction

We consider the existence of positive solutions for third-order 𝑚-point boundary value problem: 𝑥^′′′(𝑡)((𝑡[],𝑥+𝑓𝑡,𝑥))=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=0,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖, (1) where 0<𝜉₁<𝜉₂<⋯<𝜉_𝑚−2<1, 0≤𝛽_𝑖≤1, 𝑖=1,2,…,𝑚−2, ∑^𝑚−2_𝑖=1𝛽_𝑖≤1, and 𝑓∈𝐶([0,1]×[0,∞),𝑅).

If condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1 holds, the problem is called resonant boundary value problem or boundary value problem at resonance. Otherwise, the associated problem is called nonresonant boundary value problem. Here, the condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1 is denoted by (1.2).

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, as the deflection of a curved beam having a constant or varying cross section, three-layer beam, and so on [1]. In recent years, the existence of positive solutions for nonresonant two-point or three-point boundary value problems (Bvp for short) for nonlinear third-order ordinary differential equations has been studied by several authors. For examples, Anderson [2] established the existence of at least three positive solutions to problem −𝑥^′′′(𝑡)(𝑥(𝑡(),𝑥(0)+𝑓))=0,𝑡∈0,1=𝑥^′𝑡₂=𝑥^′′(1)=0, (2) where 𝑓∶𝑅→[0,+∞) is continuous and 1/2≤𝑡₂<1.

By using the well-known Guo-Krasnoselski𝑖̆ fixed point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for third-order three-point problem: 𝑥^′′′(𝑡)(𝑡)𝑓((𝑡(),𝑥=𝑎𝑡,𝑥)),𝑡∈0,1_′′𝜂=0,𝑥(0)=𝑥(1)=0,𝜂∈(0,1). (3) For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [5–12] and references therein.

For resonant problem of second-order or higher-order differential equations, many existence results of solutions have been established, see [13–25]. In [25], the authors considered the problem 𝑥^′′′(𝑡)=𝑓𝑡,𝑥,𝑥^′(𝑡)(),𝑥+𝑒,𝑡∈0,1_′(0)=0,𝑥(1)=𝛽𝑥𝜂,𝑥(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥𝜉_𝑖. (4) By using the Mawhin continuation theorem, the existence results of solutions are established under the resonant condition 𝛽=1, ∑^𝑚−2_𝑖=1𝛼_𝑖=1, ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0 and 𝛽=1/𝜂, ∑^𝑚−2_𝑖=1𝛼_𝑖=1, ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0, respectively.

It is well known that the problem of existence for positive solution to nonlinear Bvp is very difficult when the resonant case is considered. Only few works gave the approach in this area for first- and second-order differential equations [26–31]. To our best knowledge, no paper deal with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [27–29], we study the existence of positive solution for problem (1) under the nonresonant condition ∑^𝑚−2_𝑖=1𝛽_𝑖<1 and resonant condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1, respectively. By using the Leggett-Williams fixed point theorem and its generalization [28, 30], we establish the existence results of positive solutions. The results obtained in this paper are interesting because(1)the results obtained in the nonresonant case are more general than those established before;(2)it is the first time that the positive solution is considered for third-order Bvp at resonance.

The rest of the paper is organized as follows. Some definitions and lemmas are given in Section 2. In Section 3, we consider the nonresonant case for problem (1). In Section 4, we discuss the existence of positive solution for problem (1) with resonant condition (1.2). Finally, in Section 5, we give some examples to illustrate the main results of the paper.

2. Background Definitions and Lemmas

For the convenience of the reader, we present here the necessary definitions and two-fixed point theorems.

Let 𝑋, 𝑌 be real Banach spaces. A nonempty convex closed set 𝐶⊂𝑋 is said to be a cone provided that(1)𝑎𝑥∈𝐶, for all 𝑥∈𝐶, 𝑎≥0,(2)𝑥,−𝑥∈𝐶 implies 𝑥=0.

Definition 1. The map 𝜓 is said to be a nonnegative continuous concave functional on 𝐶 provided that 𝜓∶𝐶→+∞ is continuous and 𝜓𝑡𝑥+(1−𝑡)𝑦≥𝑡𝜓(𝑥)+(1−𝑡)𝜓𝑦, (5) for all 𝑥,𝑦∈𝐶 and 𝑡∈[0,1].

Definition 2. Let 0<𝑎<𝑏 be given and let 𝜓 be a nonnegative continuous concave functional on the cone 𝐶. Define the convex sets 𝐶_𝑟 and 𝐶(𝜓,𝑎,𝑏) by 𝐶_𝑟={‖𝑥‖},𝐶=(𝑥),‖𝑥‖.𝑥∈𝐶∣<𝑟𝜓,𝑎,𝑏𝑥∈𝐶∣𝑎≤𝜓≤𝑏 (6)

Lemma 3 (the Leggett-Williams fixed point theorem [32]). Let 𝑇∶𝐶_𝑟→𝐶_𝑟 be a completely continuous operator and let 𝜓 be a nonnegative continuous concave functional on 𝐶 such that 𝜓(𝑥)≤‖𝑥‖ for all 𝑥∈𝐶_𝑟. Suppose that there exist 0<𝑎<𝑏<𝑑≤𝑐 such that(H₁){𝑥∈𝐶(𝜓,𝑏,𝑑)∣𝜓(𝑥)>𝑏}≠Ø and 𝜓(𝑇𝑥)>𝑏, for 𝑥∈𝐶(𝜓,𝑏,𝑑),(H₂)‖𝑇𝑥‖<𝑎 for ‖𝑥‖≤𝑎, (H₃)𝜓(𝑇𝑥)>𝑏 for 𝑥∈𝐶(𝜓,𝑏,𝑐) with ‖𝑇𝑥‖≥𝑑.Then, 𝑇 has at least three-fixed points 𝑥₁, 𝑥₂, and 𝑥₃ such that 𝑥₁<𝑎,𝑏<𝜓𝑥₂,𝑥₃>𝑎,𝜓𝑥₃<𝑏. (7)
Operator 𝐿∶dom𝐿⊂𝑋→𝑌 is called a Fredholm operator with index zero, that is, Im𝐿 is closed and dimKer𝐿=𝑐𝑜𝑑𝑖𝑚Im𝐿<∞, which implies that there exist continuous projections 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌 such that Im𝑃=Ker𝐿 and Ker𝑄=Im𝐿. Moreover, since dimIm𝑄=𝑐𝑜𝑑𝑖𝑚Im𝐿, there exists an isomorphism 𝐽∶Im𝑄→Ker𝐿. Denote by 𝐿_𝑃, the restriction of 𝐿 to Ker𝑃∩dom𝐿 to Im𝐿 and its inverse by 𝐾_𝑃, so 𝐾_𝑃∶Im𝐿→Ker𝑃∩dom𝐿 and the coincidence equation 𝐿𝑥=𝑁𝑥 is equivalent to 𝑥=(𝑃+𝐽𝑄𝑁)𝑥+𝐾_𝑃(𝐼−𝑄)𝑁𝑥. (8) Denote 𝛾∶𝑋→𝐶 to be a retraction, that is, a continuous mapping such that 𝛾𝑥=𝑥 for all 𝑥∈𝐶 and Ψ∶=𝑃+𝐽𝑄𝑁+𝐾_𝑃()Ψ𝐼−𝑄𝑁,_𝛾∶=Ψ∘𝛾. (9)

Lemma 4 (the Leggett-Williams norm-type theorem [27]). Let 𝐶 be a cone in 𝑋, and let Ω₁, Ω₂ be open bounded subsets of 𝑋 with Ω₁⊂Ω₂, 𝐶∩(Ω₂⧵Ω₁)≠∅. Assume that 𝐿∶dom𝐿⊂𝑋→𝑌 is a Fredholm operator of index zero and(C1)𝑄𝑁∶𝑋→𝑌 is continuous and bounded, 𝐾_𝑃(𝐼−𝑄)𝑁∶𝑋→𝑋 is compact on every bounded subset of 𝑋,(C2)𝐿𝑥≠𝜆𝑁𝑥 for all 𝑥∈𝐶∩𝜕Ω₂∩dom𝐿 and 𝜆∈(0,1),(C3)𝛾 maps subsets of Ω₂ into bounded subsets of C,(C4)𝑑_𝐵([𝐼−(𝑃+𝐽𝑄𝑁)𝛾]|_ker𝐿,Ker𝐿∩Ω₂,0)≠0, where 𝑑_𝐵 stands for the Brouwer degree,(C5) there exists 𝑢₀∈𝐶⧵{0} such that ‖𝑥‖≤𝜎(𝑢₀)‖Ψ𝑥‖ for 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, where 𝐶(𝑢₀)={𝑥∈𝐶∶𝜇𝑢₀≤𝑥} for some 𝜇>0 and 𝜎(𝑢₀) such that ‖𝑥+𝑢₀‖≥𝜎(𝑢₀)‖𝑥‖ for every 𝑥∈𝐶,(C6)(𝑃+𝐽𝑄𝑁)𝛾(𝜕Ω₂)⊂𝐶,(C7)Ψ_𝛾(Ω₂⧵Ω₁)⊂𝐶, then the equation 𝐿𝑥=𝑁𝑥 has a solution in the set 𝐶∩(Ω₂⧵Ω₁).

3. Positive Solution for the Nonresonant Problem

In this section, we suppose that𝑓∈𝐶([0,1]×[0,+∞),[0,+∞)) and ∑^𝑚−2_𝑖=1𝛽_𝑖<1. We begin with some preliminary results. Consider the problem 𝑥^′′′(𝑡)(𝑡)[],𝑥+𝑦=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=0,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖. (10)(11)

Lemma 5. Denote 𝜉₀=0, 𝜉_𝑚−1=1, 𝛽₀=𝛽_𝑚−1=0, then for 𝑦(𝑡)∈𝐶[0,1], problems (10) and (11) have the unique solution 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)^𝑠₀𝑦(𝜏)𝑑𝜏𝑑𝑠, (12) where 𝐺()=1𝑡,𝑠1−∑^𝑚−1_𝑘=0𝛽_𝑘×(1−𝑠)+∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝑠−𝜉_𝑘()+,𝑡≤𝑠,1−𝑡∑^𝑖−1_𝑘=0𝛽_𝑘(𝑡−𝑠)+∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝑡−𝜉_𝑘,𝑡≥𝑠, (13) for 𝜉_𝑖−1<𝑠<𝜉_𝑖, 𝑖=1,2,…,𝑚−1.

Proof. Integrating both sides of (10) and considering the boundary condition 𝑥^′′(0)=0, we have −𝑥^′′(𝑡)=^𝑡₀𝑦(𝑠)𝑑𝑠. (14) Let 𝐺(𝑡,𝑠) be the Green function of problem −𝑥^′′(𝑡)𝑥=0,_′(0)=0,𝑥(1)=^𝑚−1_𝑖=0𝛽_𝑖𝑥𝜉_𝑖. (15)(16) From (15), we can suppose that 𝐺(𝑡,𝑠)=𝐴+𝐵𝑡𝑡≤𝑠,𝜉_𝑖−1<𝑠<𝜉_𝑖,𝐶+𝐷𝑡𝑡≥𝑠,𝜉_𝑖−1<𝑠<𝜉_𝑖. (17) For the definition and properties of the Green function together with (16), we have 𝐴+𝐵𝑠=𝐶+𝐷𝑠,𝐵−𝐷=1,𝐵=0,𝐶+𝐷=^𝑖−1_𝑘=0𝛽_𝑘𝐴+𝐵𝜉_𝑘+^𝑚−1_𝑘=𝑖𝛽_𝑘𝐶+𝐷𝜉_𝑘. (18) Hence, 𝐴=1−𝑠+∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝑠−𝜉_𝑘1−∑^𝑚−1_𝑖=0𝛽_𝑖,𝐶=1−∑^𝑖−1_𝑘=0𝛽_𝑘𝑠−∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝜉_𝑘1−∑^𝑚−1_𝑖=0𝛽_𝑖,𝐵=0,𝐷=−1. (19) Thus, 𝐺()=1𝑡,𝑠1−∑^𝑚−1_𝑘=0𝛽_𝑘×(1−𝑠)+∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝑠−𝜉_𝑘()+,𝑡≤𝑠,1−𝑡∑^𝑖−1_𝑘=0𝛽_𝑘(𝑡−𝑠)+∑^𝑚−1_𝑘=𝑖𝛽_𝑘𝑡−𝜉_𝑘,𝑡≥𝑠, (20) for 𝜉_𝑖−1<𝑠<𝜉_𝑖, 𝑖=1,2,…,𝑚−1.
Considering (14) together, we obtain that problems (10) and (11) have the unique solution 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)^𝑠₀𝑦(𝜏)𝑑𝜏𝑑𝑠. (21)

Lemma 6. The function 𝐺(𝑡,𝑠) established in Lemma 5 satisfies that 𝐺(𝑡,𝑠)≥0, 𝑡,𝑠∈[0,1].

Proof. For 𝜉_𝑖−1≤𝑠≤𝜉_𝑖, 𝑖=1,2,…,𝑚−1 and 𝑡≤𝑠, (1−𝑠)+^𝑚−1_𝑘=𝑖𝛽_𝑘𝑠−𝜉_𝑘≥^𝑚−1_𝑘=𝑖𝛽_𝑘1−𝜉_𝑘≥0. (22) For 𝜉_𝑖−1≤𝑠≤𝜉_𝑖, 𝑖=1,2,…,𝑚−1 and 𝑡≥𝑠, (1−𝑡)+^𝑖−1_𝑘=0𝛽_𝑘(𝑡−𝑠)+^𝑚−1_𝑘=𝑖𝛽_𝑘𝑡−𝜉_𝑘≥^𝑖−1_𝑘=0𝛽_𝑘(1−𝑠)+^𝑚−1_𝑘=𝑖𝛽_𝑘1−𝜉_𝑘≥0. (23) These ensures that 𝐺(𝑡,𝑠)≥0, 𝑡,𝑠∈[0,1].

Lemma 7. If 𝑦∈𝐶[0,1] and 𝑦≥0, then the unique solution 𝑥 of problems (10) and (11) satisfy min_0≤𝑡≤1𝑥(𝑡)≥𝛾max_0≤𝑡≤1𝑥(𝑡), (24) where 𝛾=(∑^𝑚−2_𝑖=1𝛽_𝑖(1−𝜉_𝑖))/(1−∑^𝑚−2_𝑖=1𝛽_𝑖𝜉_𝑖)>0 is a constant.

Proof. For 𝑥^′′′(𝑡)=−𝑦(𝑡)≤0, 𝑡∈[0,1], we get that 𝑥^′′(𝑡) is decreasing on [0,1]. Then, the condition 𝑥^′′(0)=0 ensures that have 𝑥^′′(𝑡)≤0,𝑡∈(0,1). This together with 𝑥^′(0)=0𝑥(𝑡) is concave and decreasing on [0,1]. Thus, max_0≤𝑡≤1𝑥(𝑡)=𝑥(0),min_0≤𝑡≤1𝑥(𝑡)=𝑥(1). (25) From the concavity of 𝑥(𝑡), we have 𝜉_𝑖(𝑥(1)−𝑥(0))≤𝑥𝜉_𝑖−𝑥(0). (26) Multiplying both sides with 𝛽_𝑖 and considering 𝑥(1)=∑^𝑚−2_𝑖=1𝛽_𝑖𝑥(𝜉_𝑖), we have 1−^𝑚−2_𝑖=1𝛽_𝑖𝜉_𝑖𝑥(1)≥^𝑚−2_𝑖=1𝛽_𝑖1−𝜉_𝑖𝑥(0). (27) This completes the proof of Lemma 7.

Let Banach space 𝐸=𝐶[0,1] be endowed with the maximum norm. We define the cone 𝐶⊂𝐸 by 𝐶=𝑥∈𝐸∣𝑥(𝑡)≥0,𝑥^′′(0)=0,𝑥^′(0)𝑥(1)==0,_{𝑚−2𝑖=1}𝛽_𝑖𝑥𝜉_𝑖,𝑥(𝑡)isconcaveon[0,1]. (28)

Define the nonnegative continuous concave functional 𝜓∶𝐶→[0,∞) by 𝜓(𝑥)=min_0≤𝑡≤1𝑥(𝑡),𝑥∈𝐶. (29) Define the constants 𝜎, 𝛿 by 1𝜎=max_0≤𝑡≤1∫¹₀𝑠𝐺(𝑡,𝑠)𝑑𝑠,𝛿=¹₀𝑠𝐺(1,𝑠)𝑑𝑠. (30)

Theorem 8. Suppose that there exist constants 0<𝑎<𝑏<𝑏/𝛾≤𝑐 such that (A1)𝑓(𝑡,𝑥)<𝜎𝑎, (𝑡,𝑥)∈[0,1]×[0,𝑎],(A2)𝑓(𝑡,𝑥)>𝑏/𝛿, (𝑡,𝑥)∈[0,1]×[𝑏,(𝑏/𝛾)],(A3)𝑓(𝑡,𝑥)<𝜎𝑐, (𝑡,𝑥)∈[0,1]×[0,𝑐],then problem (1) has at least three positive solutions 𝑥₁, 𝑥₂, and 𝑥₃ satisfying 𝑥₁≤𝑎,𝑏<min_0≤𝑡≤1𝑥₂,𝑥₃>𝑎withmin_0≤𝑡≤1𝑥₃<𝑏. (31)

Proof. We define operator 𝑇∶𝐶→𝐸 by 𝑇𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)^𝑠₀𝑓(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠. (32) It is easy to check that 𝑇∶𝐶→𝐶 and is completely continuous.
Next, the conditions of Lemma 3 are checked. If 𝑥∈𝐶_𝑐, then ‖𝑥‖≤𝑐, and condition (A3) implies that 𝑓()‖𝑇(𝑥𝑡,𝑥≤𝜎𝑐,0≤𝑡≤1,)‖=max_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)^𝑠₀𝑓(𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠≤max_0≤𝑡≤1¹₀𝑠𝐺(𝑡,𝑠)𝑑𝑠×𝜎𝑐≤𝑐. (33) Thus, 𝑇∶𝐶_𝑐→𝐶_𝑐. In the same way, we see that 𝑇∶𝐶_𝑎→𝐶_𝑎. Hence, condition (H₂) of Lemma 3 is satisfied.
The fact that constant function 𝑥(𝑡)=(𝑏/𝛾)∈{𝐶(𝜓,𝑏,(𝑏/𝛾))∣𝜓(𝑥)>𝑏} implies that {𝐶(𝜓,𝑏,(𝑏/𝛾))∣𝜓(𝑥)>𝑏}≠∅. If 𝑥∈𝐶(𝜓,𝑏,(𝑏/𝛾)), from the assumption (A2), 𝑓(𝑡,𝑥)≥𝑏/𝛿. Thus, 𝜓(𝑇𝑥)=min_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)^𝑠₀𝑓((𝜏=𝜏,𝑥))𝑑𝜏𝑑𝑠₁₀𝐺(1,𝑠)^𝑠₀𝑓((𝜏≥𝜏,𝑥))𝑑𝜏𝑑𝑠₁₀()𝑏𝑠𝐺1,𝑠𝑑𝑠×𝛿≥𝑏, (34) which ensures that condition (H₁) of Lemma 3 is satisfied. Finally, we show that condition (H₃) of Lemma 3 also holds. Suppose that 𝑥∈𝐶(𝜓,𝑏,𝑐) with ‖𝑇𝑥‖>𝑏/𝛾, 𝜓(𝑇𝑥)=min_0≤𝑡≤1(𝑡)‖‖𝑏𝑇𝑥≥𝛾×𝑇𝑥>𝛾×𝛾=𝑏. (35) So, condition (H₃) of Lemma 3 is satisfied. Thus, an application of Lemma 3 implies that the nonresonant third-order boundary value problem (1) has at least three positive solutions 𝑥₁, 𝑥₂, and 𝑥₃ satisfying (31).

4. Positive Solution for Resonant Problem

In this section, the condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1 is considered. Obviously, problem (1) is at resonance under this condition. The norm-type Leggett-Williams fixed point theorem will be used to establish the existence results of positive solution. We define the Banach spaces 𝑋=𝑌=𝐶[0,1] endowed with the maximum norm.

Define linear operator 𝐿∶dom𝐿⊂𝑋→𝑌, 𝐿𝑥=−𝑥^′′′(𝑡), 𝑡∈[0,1], where dom𝐿=𝑥∈𝑋∣𝑥^′′′∈𝐶[0,1],𝑥^′′(0)𝑥=0,_′(0)=0,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖, (36) and 𝑁∶𝑋→𝑌 with (𝑁𝑥)(𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡∈[0,1]. (37) It is obvious that Ker𝐿={𝑥∈dom𝐿∶𝑥(𝑡)≡𝑐,𝑡∈[0,1]}. Denote the function 𝐺(𝑠),𝑠∈[0,1] as follows: 𝐺(𝑠)=(1−𝑠)²−^𝑚−1_𝑖=𝑘𝛽_𝑖𝜉_𝑖−𝑠²,𝜉_𝑘−1≤𝑠≤𝜉_𝑘,𝑘=1,2,…,𝑚−1. (38) Note that 𝐺(𝑠)≥0, 𝑠∈[0,1].

Denote the function 𝑈(𝑡,𝑠) and positive number 𝜅 as follows: 𝑈()=−1𝑡,𝑠6𝑠³+12𝑠²−121𝑠+6+4𝑡³+2324∫¹₀𝐺(𝑠)𝐺(𝑠),−1𝑑𝑠0≤𝑡≤𝑠≤1,6𝑠³−121𝑠+𝑠𝑡−2𝑡²+16+4𝑡³+2324∫¹₀𝐺(𝑠)𝐺(𝑠),𝑑𝑠0≤𝑠≤𝑡≤1,𝜅∶=min1,min_[]𝑠∈0,1∫¹₀𝐺(𝜏)𝑑𝜏𝐺(𝑠),min_{[]𝑡,𝑠∈0,1}1𝑈().𝑡,𝑠 (39)

Theorem 9. Assume that there exists positive constant 𝑅∈(0,∞) such that 𝑓∶[0,1]×[0,𝑅]→(−∞,+∞) is continuous and satisfies the following conditions:(S1)𝑓(𝑡,𝑥)≥−𝜅𝑥, for (𝑡,𝑥)∈[0,1]×[0,𝑅],(S2)𝑓(𝑡,𝑥)<0 for [𝑡,𝑥]∈[0,1]×[(1−(𝜅/2))𝑅,𝑅],(S3) there exists 𝑟∈(0,𝑅),𝑡₀∈[0,1], 𝑎∈(0,1],𝑀₁<𝑀∈(0,1) and continuous functions 𝑔∶[0,1]→[0,+∞), ℎ∶(0,𝑟]→[0,+∞) such that 𝑓(𝑡,𝑥)≥𝑔(𝑡)ℎ(𝑥), [𝑡,𝑥]∈[0,1]×(0,𝑟] and ℎ(𝑥)/𝑥^𝑎 is nonincreasing on (0,𝑟] with ℎ(𝑟)𝑟^𝑎1₀𝑈𝑡₀,𝑠𝑔(𝑠)𝑑𝑠≥1−𝑀𝑀^𝑎, (40)then problem (1) with resonant condition (1.2) has at least one positive solution.

Proof. Firstly, we claim that Im𝐿=𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. (41) In fact, for each 𝑦∈{𝑦∈𝑌∣∫¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0}, we take 𝑥(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝑑𝑠. (42) It is easy to check that −𝑥^′′′(𝑡)=𝑦(𝑡),𝑥^′′(0)=0,𝑥^′(0)=0, and 𝑥(1)=∑^𝑚−2_𝑖=1𝛽_𝑖𝑥(𝜉_𝑖), which means 𝑥(𝑡)∈dom𝐿. Thus, 𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0⊂Im𝐿. (43) On the other hand, for each 𝑦(𝑡)∈Im𝐿, there exists 𝑥(𝑡)∈dom𝐿, −𝑥^′′′(𝑡)=𝑦(𝑡),𝑥^′′(0)=0,𝑥^′(0)𝑥(1)==0,_{𝑚−2𝑖=1}𝛽_𝑖𝑥𝜉_𝑖. (44) Integrating both sides on [0,𝑡], we have 𝑥(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)1𝑑𝑠+2𝑥^′′(0)𝑡²+𝑥^′(0)𝑡+𝑥(0). (45) Considering condition 𝑥^′′(0)=0,𝑥^′(0)=0,𝑥(1)=∑^𝑚−2_𝑖=1𝛽_𝑖𝑥(𝜉_𝑖), and ∑^𝑚−1_𝑖=0𝛽_𝑖=1, we conclude that ¹₀(1−𝑠)²𝑦(𝑠)𝑑𝑠−^𝑚−1_𝑖=0𝛽_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠²𝑦(𝑠)𝑑𝑠=0, (46) which equivalents to the conclusion that ∫¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. So, we have Im𝐿⊂𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. (47) Thus, Im𝐿=𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. (48) Clearly, dimKer𝐿=1 and Im𝐿 are closed. Next, we see that 𝑌=𝑌₁⊕Im𝐿, where 𝑌₁=𝑦₁∣𝑦₁=1∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑦(𝑠)𝑑𝑠,𝑦∈𝑌. (49) In fact, for each 𝑦(𝑡)∈𝑌, we have ¹₀𝐺(𝑠)𝑦(𝑠)−𝑦₁𝑑𝑠=0. (50) This shows that 𝑦−𝑦₁∈Im𝐿. Since 𝑌₁∩Im𝐿={0}, we have 𝑌=𝑌₁⊕Im𝐿. Thus, 𝐿 is a Fredholm operator with index zero.
Then, define the projections 𝑃∶𝑋→𝑋, 𝑄∶𝑌→𝑌 by 𝑃𝑥=¹₀𝑥(𝑠)1𝑑𝑠,𝑄𝑦=∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑦(𝑠)𝑑𝑠. (51) Clearly, Im𝑃=Ker𝐿, Ker𝑄=Im𝐿 and Ker𝑃={𝑥∈𝑋∶∫¹₀𝑥(𝑠)𝑑𝑠=0}. Note that for 𝑦∈Im𝐿, the inverse 𝐾_𝑃 of 𝐿_𝑃 is given by 𝐾_𝑃𝑦=¹₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (52) where 𝑘()=−1𝑡,𝑠6𝑠³+12𝑠²−121𝑠+6−1,0≤𝑡≤𝑠≤1,6𝑠³−121𝑠+𝑠𝑡−2𝑡²+16,0≤𝑠≤𝑡≤1. (53) In fact, it is easy to check that 𝐿𝐾_𝑃𝑦=−¹₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠^′′′(𝑡),𝐾=𝑦_𝑃(𝐿)(𝑥)=^𝑡₀−16𝑠³−121𝑠+𝑠𝑡−2𝑡²+16−𝑥^′′′(𝑠)+𝑑𝑠_1𝑡−16𝑠³+12𝑠²−121𝑠+6−𝑥^′′′(𝑠)(𝑡).𝑑𝑠=𝑥 (54)(55) Considering that 𝑓 can be extended continuously on [0,1]×(−∞,+∞); condition (C1) of Lemma 4 is fulfilled.
Define the cone of nonnegative functions 𝐶 and subsets of 𝑋Ω₁,Ω₂ by {(𝑡)[,Ω𝐶=𝑥∈𝑋∶𝑥≥0,𝑡∈0,1]}₁={|𝑥(𝑡‖𝑥‖[,Ω𝑥∈𝑋∶𝑟>)|>𝑀,𝑡∈0,1]}₂={𝑥∈𝑋∶|𝑥(𝑡)|<𝑅,𝑡∈[0,1]}. (56) Clearly, Ω₁ and Ω₂ are bounded and open sets, furthermore Ω₁={𝑥∈𝑋∶𝑟≥|𝑥(𝑡)|≥𝑀‖𝑥‖,𝑡∈[0,1]}⊂Ω₂,𝐶∩Ω₂⧵Ω₁≠∅. (57)
Let the isomorphism 𝐽=𝐼 and (𝛾𝑥)(𝑡)=|𝑥(𝑡)| for 𝑥∈𝑋. Then, it is easy to check that 𝛾 is a retraction and maps subsets of Ω₂ into bounded subsets of 𝐶, which means that condition (C3) of Lemma 4 is satisfied.
Next, we confirm that (C2) of Lemma 4 holds. For this purpose, suppose that there exists 𝑥₀∈𝐶∩𝜕Ω₂∩dom𝐿 and 𝜆₀∈(0,1) such that 𝐿𝑥₀=𝜆₀𝑁𝑥₀. Then, −𝑥₀^′′′(𝑡)=𝜆₀𝑓𝑡,𝑥₀, (58) for all 𝑡∈[0,1]. Thus, 𝑥₀^′′(𝑡)=−𝜆₀^𝑡₀𝑓𝑠,𝑥₀(𝑠)𝑥𝑑𝑠,_′0(𝑡)=−𝜆₀^𝑡₀(𝑡−𝑠)𝑓𝑠,𝑥₀(𝑠)𝑑𝑠. (59) Let 𝑥₀(𝑡₀)=‖𝑥₀‖=𝑅. We verify that 𝑡₀≠0 and 𝑡₀≠1. The step is divided into three cases:(1)we show that 𝑡₀≠1. Suppose, on the contrary, that 𝑥₀(𝑡) achieves maximum value 𝑅 only at 𝑡=1. Then, 𝑥(1)=∑^𝑚−2_𝑖=1𝛽_𝑖𝑥₀(𝜉_𝑖) in combination with ∑^𝑚−2_𝑖=1𝛽_𝑖=1 yields that max_{1≤𝑖≤𝑚−2}𝑥₀(𝜉_𝑖)≥𝑅, which is a contradiction,(2)we show that 𝑡₀≠0. Suppose, on the contrary, that 𝑥₀(𝑡) achieves maximum value 𝑅 at 𝑡=0. Then, 𝑥₀^′′′(0)=−𝜆₀𝑓(0,𝑅)>0 (60)which together with the condition 𝑥^′′(0)=𝑥^′(0)=0 yield that 𝑥(𝑡) is increasing near the point 𝑡=0. This contradicts to the fact that 𝑥₀(𝑡) achieves maximum value 𝑅 at 𝑡=0.
Thus, there exists 𝑡₀∈(0,1) such that 𝑥₀(𝑡₀)=𝑅=max_0≤𝑡≤1𝑥₀(𝑡). We may choose 𝜂<𝑡₀ nearest to 𝑡₀ with 𝑥₀^′′(𝜂)=0. From the mean value theory, we claim that there exists 𝜉∈(𝜂,𝑡₀) such that 𝑥₀𝜂=𝑥₀𝑡₀−𝑥^′₀(𝜉)𝑡₀−𝜂. (61) Here, 𝑥^′₀(𝜉)=−𝜆₀^𝜉₀(𝜉−𝑠)𝑓𝑠,𝑥₀𝑑𝑠≤𝜆₀𝜅^𝜉₀(𝜉−𝑠)𝑥₀(𝑠)𝑑𝑠≤𝜆₀𝜅𝑅^𝜉₀()1𝜉−𝑠𝑑𝑠=2𝜉²𝜆₀𝜅𝑅. (62) Thus, 𝑥₀𝜂=𝑥₀𝑡₀−𝑥^′₀(𝜉)𝑡₀1−𝜂≥𝑅−2𝜉²𝜆₀𝜅𝑡₀𝑅≥𝜅−𝜂1−2𝑅. (63) Then, 0≥𝑥₀^′′𝑡₀−𝑥₀^′′𝜂=−𝜆₀^𝑡0_𝜂𝑓𝑠,𝑥₀(𝑠)𝑑𝑠, (64) which contradict to condition (S2). Thus, (C2) holds.

Remark 10. The sign of third-order derivative of a function 𝑦(𝑡) at point 𝑡₀ cannot be confirmed even if 𝑡₀ is a maximal value of 𝑦(𝑡). Thus, the method in [29] is not applicable directly to problem (1). In our opinion, it is the key that the conditions (S2) in this paper are stronger than that in [29].
For 𝑥∈Ker𝐿∩Ω₂, define 𝐻()|𝑥|−𝜆𝑥,𝜆=𝑥−𝜆∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑓(𝑠,|𝑥|)𝑑𝑠, (65) where 𝑥=𝑐∈Ker𝐿∩Ω₂ and 𝜆∈[0,1]. Suppose that 𝐻(𝑥,𝜆)=0. In view of (S1), we obtain |𝑐|+𝜆𝑐=𝜆∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑓(|𝑐|𝑐|−𝜆𝑠,|)𝑑𝑠≥𝜆∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝜅|𝑐||𝑐)𝑑𝑠=𝜆|(1−𝜅≥0. (66) Hence, 𝐻(𝑥,𝜆)=0 implies 𝑐≥0. Furthermore, if 𝐻(𝑅,𝜆)=0, we get 0≤𝑅(1−𝜆)¹₀𝐺(𝑠)𝑑𝑠=𝜆¹₀𝐺(𝑠)𝑓(𝑠,𝑅)𝑑𝑠, (67) contradicting to (S2). Thus, 𝐻(𝑥,𝜆)≠0 for 𝑥∈𝜕Ω₂ and 𝜆∈[0,1]. Therefore, 𝑑_𝐵𝐻(𝑥,0),Ker𝐿∩Ω₂,0=𝑑_𝐵𝐻(𝑥,1),Ker𝐿∩Ω₂,0=𝑑_𝐵𝐼,Ker𝐿∩Ω₂,0=1. (68) This ensures 𝑑_𝐵𝐼−(𝑃+𝐽𝑄𝑁)𝛾_Ker𝐿,Ker𝐿∩Ω₂,0=𝑑_𝐵𝐻(𝑥,1),Ker𝐿∩Ω₂,0≠0. (69) Let 𝑥∈Ω₂⧵Ω₁ and 𝑡∈[0,1]. From condition (S1), we see Ψ_𝛾𝑥(𝑡)=¹₀|𝑥(𝑡+1)|𝑑𝑡∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑓(|𝑥(𝑠+𝑠,)|)𝑑𝑠₁₀𝑘()𝑓(|𝑥(𝑠−1𝑡,𝑠𝑠,)|)∫¹₀𝐺(𝑠)𝑑𝑠×¹₀𝐺(𝜏)𝑓(|𝑥(𝜏=𝜏,)|)𝑑𝜏𝑑𝑠₁₀|𝑥(𝑡)|𝑑𝑡+¹₀𝑈()𝑓(|𝑥(𝑠≥𝑡,𝑠𝑠,)|)𝑑𝑠₁₀|𝑥(𝑠)|𝑑𝑠−𝜅¹₀𝑈(𝑥(𝑠=𝑡,𝑠)|)|𝑑𝑠₁₀(1−𝜅𝑈(𝑡,𝑠))|𝑥(𝑠)|𝑑𝑠≥0. (70) Hence, Ψ_𝛾(Ω₂⧵Ω₁)⊂𝐶. Moreover, for 𝑥∈𝜕Ω₂, we have (𝑃+𝐽𝑄𝑁)𝛾𝑥=¹₀|𝑥(𝑠+1)|𝑑𝑠∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑓(|𝑥(𝑠≥𝑠,)|)𝑑𝑠₁₀𝜅1−∫¹₀𝐺(𝑠)𝐺(𝑠)|𝑥(𝑠𝑑𝑠)|𝑑𝑠≥0, (71) which means (𝑃+𝐽𝑄𝑁)𝛾(𝜕Ω₂)⊂𝐶. These ensure that (C6) and (C7) of Lemma 4 hold.
At last, we confirm that (C5) is satisfied. Taking 𝑢₀(𝑡)≡1 on [0,1], we see 𝑢₀∈𝐶⧵{0},𝐶𝑢₀={𝑥∈𝐶𝑥(𝑡)>0on[0,1]}, (72) and we can take 𝜎(𝑢₀)=1. Let 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, we have 𝑥(𝑡)[]‖𝑥‖𝑥(𝑡)‖𝑥‖[].>0,𝑡∈0,1,0<≤𝑟,≥𝑀on0,1 (73) Therefore, in view of (S3), we obtain, for all 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, (Ψ𝑥)𝑡₀=¹₀𝑥(𝑠)𝑑𝑠+¹₀𝑈𝑡₀𝑓((𝑠‖𝑥‖+,𝑠𝑠,𝑥))𝑑𝑠≥𝑀¹₀𝑈𝑡₀𝑔(𝑠)ℎ(𝑥(𝑠‖𝑥‖+,𝑠))𝑑𝑠=𝑀¹₀𝑈𝑡₀𝑔(𝑠)ℎ(𝑥(𝑠,𝑠))𝑥^𝑎(𝑠)𝑥^𝑎(𝑠)‖𝑥‖+ℎ(𝑟)𝑑𝑠≥𝑀𝑟^𝑎1₀𝑈𝑡₀,𝑠𝑔(𝑠)𝑀^𝑎‖𝑥‖^𝑎‖𝑥‖+(𝑥‖=‖𝑥‖.𝑑𝑠≥𝑀1−𝑀)‖ (74) So, ‖𝑥‖≤𝜎(𝑢₀)‖Ψ𝑥‖, for all 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, which means (C5) of Lemma 4 holds.
Thus, by Lemma 4, we confirm that the equation 𝐿𝑥=𝑁𝑥 has a solution 𝑥∈𝐶∩(Ω₂⧵Ω₁), which implies that nonlinear third-order multipoint boundary value problem (1) with resonance condition (1.2) has at least one positive solution.

5. Examples

In this section, we give two examples to illustrate the main results of the paper. First, we consider the nonresonant four-point boundary value problem 𝑥^′′′(𝑡)()(),𝑥+𝑓𝑡,𝑥=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)𝑥(1)=1=0,2𝑥13+13𝑥23, (75) where 𝑓()=1𝑡,𝑥𝑒100^𝑡+𝑥³𝜋1,0<𝑥<6,𝑒100^𝑡+216𝜋,𝑥≥6. (76) Here, 𝛽₁=1/2, 𝛽₂=1/3, 𝜉₁=1/3, 𝜉₂=2/3, and 𝐺(𝑡,𝑠)=−𝑠+1131,𝑡≤𝑠,0≤𝑠≤3,−𝑡+1131,𝑡≥𝑠,0≤𝑠≤3,−4𝑠+1431,𝑡≤𝑠,32≤𝑠≤3,−3𝑠−2𝑡+1431,𝑡≥𝑠,32≤𝑠≤3,26−6𝑠,𝑡≤𝑠,32≤𝑠≤1,6−𝑡−5𝑠,𝑡≥𝑠,3≤𝑠≤1. (77) By a simply computation, we can get that 𝑚=143162,𝛿=898162,𝛾=.11 (78)

We choose 𝑎=1, 𝑏=4, and 𝑐=162. It is easy to check that (1)𝑓(𝑡,𝑥)<143/162, [𝑡,𝑥]∈[0,1]×[0,1],(2)𝑓(𝑡,𝑥)>648/89, [𝑡,𝑥]∈[0,1]×[4,11/2],(3)𝑓(𝑡,𝑥)<143,[𝑡,𝑥]∈[0,1]×[0,162]. Thus, all the conditions of Theorem 8 are satisfied. This ensures that problem (75) has at least three positive solutions 𝑥₁, 𝑥₂, and 𝑥₃ satisfying max_0≤𝑡≤1𝑥₁≤1,min_0≤𝑡≤1𝑥₂>4,max_0≤𝑡≤1𝑥₃>1,min_0≤𝑡≤1𝑥₃<4. (79) Next, we consider the resonant third-order four-point boundary value problem 𝑥^′′′(𝑡)+−12𝑡²+125𝑡+𝑥16²−4𝑥+115√𝑥²[],𝑥−6𝑥+10=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)(0)2=0,𝑥=𝑥3, (80) where 𝛽=1, 𝜉=2/3 and 𝑓()=−1𝑡,𝑥2𝑡²+125𝑡+×𝑥16₂−4𝑥+115√𝑥²−6𝑥+10. (81) Here, 𝐺(𝑠)=59−232𝑠,0≤𝑠≤3,(1−𝑠)²,23≤𝑠≤1. (82) By a simple computation, we have ¹₀𝐺(𝑠)𝑑𝑠=1981,𝜅=1945,¹₀𝑈(0,𝑠)𝑑𝑠=1. (83) Choose 𝑅=1, 𝑟=1/4, 𝑡₀=0, 𝑎=1, and 𝑀=1/2.

We take 𝑔(𝑡)1=−2𝑡²+135𝑡+[],ℎ(𝑥)=√16,𝑡∈0,1𝑥²1−6𝑥+10,𝑥∈0,4. (84) Then, 7(𝑡)≤48≤𝑔53<14419[],𝑥45,𝑡∈0,1₂−4𝑥+115≥−𝑥,𝑥∈[0,1]. (85) It is easy to check that (1)𝑓(𝑡,𝑥)>−(19/45)𝑥, for all (𝑡,𝑥)∈[0,1]×[0,1],(2)𝑓(𝑡,𝑥)<0, for all (𝑡,𝑥)∈[0,1]×[71/90,1],(3)𝑓(𝑡,𝑥)≥(101/80)(−(1/2)𝑡²+(1/3)𝑡+(5/16))√𝑥²−6𝑥+10≥𝑔(𝑡)ℎ(𝑥),[𝑡,𝑥]∈[0,1]×(0,1/4] and ℎ(𝑥)/𝑥=√𝑥²−6𝑥+10/𝑥is nonincreasing on (0,1/4] with ℎ(𝑟)𝑟^𝑎1₀𝑈()𝑔(𝑠)7√0,𝑠𝑑𝑠>13748>1=1−𝑀𝑀^𝑎. (86) Then, all conditions of Theorem 9 are satisfied. This ensures that the resonant problem has at least one solution, positive on [0,1].

Remark 11. The established existence results of positive solutions for third-order boundary value problems in [2, 3, 5–12], for examples, are not applicable to the problem (75) or (5.2).

Acknowledgments

The paper is supported by the Natural Science Foundation of China (no. 11201109), the Natural Science Foundation of Anhui Educational Department (KJ2012Z335 and KJ2012B144), the Excellent Talents Foundation of University of Anhui Province (2012SQRL165), and the NSF of Hefei Normal University (2012Kj09).