On the Existence of Positive Solutions of Resonant and Nonresonant Multipoint Boundary Value Problems for Third-Order Nonlinear Differential Equations

Abstract

Positive solutions for a kind of third-order multipoint boundary value problem under the non-resonant conditions and the resonant conditions are considered. In the nonresonant case, by using Leggett-Williams fixed-point theorem, the existence of at least three positive solutions is obtained. In the resonant case, by using Leggett-Williams norm-type theorem due to O’Regan and Zima, existence result of at least one positive solution is established. The results obtained are valid and new for the problem discussed. Two examples are given to illustrate the main results.

1. Introduction

We consider the third-order 𝑚-point boundary value problem given by 𝑥^′′′(𝑡)((𝑡[]𝑥+𝑓𝑡,𝑥))=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖, (1) where 0<𝜉₁<𝜉₂<⋯<𝜉_𝑚−2<1, 0≤𝛼_𝑖, 𝛽_𝑖≤1, 𝑖=1,2,…,𝑚−2, ∑^𝑚−2_𝑖=1𝛼_𝑖<1, ∑^𝑚−2_𝑖=1𝛽_𝑖≤1, and 𝑓∈𝐶([0,1]×[0,∞),𝑅). For the convenience of writing later, we denote 𝜉₀=0,𝜉_𝑚−1=1, and 𝛼₀=𝛼_𝑚−1=𝛽₀=𝛽_𝑚−1=0.

If condition ∑^𝑚−2_𝑖=1𝛽_𝑖≠1 holds, the problem is nonresonant; that is, the associated linear problem 𝑥^′′′(𝑡)[]𝑥=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖 (1) has only zero solution, and the differential operator with boundary conditions is invertible. Otherwise, the problem is at resonance.

Third-order differential equations arise in many different areas of applied mathematics and physics, as the varying cross-section or deflection of a curved beam having a constant, three-layer beam and so on [1]. Recently, there have been extensive studies on positive solutions for nonresonant two-point or three-point boundary value problems for nonlinear third-order ordinary differential equations. 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, (3) where 𝑓∶𝑅→[0,+∞) is continuous and 1/2≤𝑡₂<1.

By using the well-known Guo-Krasnosel’skii fixed-point theorem [3], Palamides and Smyrlis [4] proved that there exists at least one positive solution for the 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). (4)

In another paper [5], Graef and Kong studied the existence of positive solutions for the third-order semipositone boundary value problem: 𝑢^′′′(𝑡)((𝑡(𝑡)(),𝑢(0)=𝜆𝑓𝑡,𝑢))+𝑒,𝑡∈0,1=𝑢^′𝑝=¹_𝑞𝑤(𝑠)𝑢^′′(𝑠)𝑑𝑠=0, (5) where 1/2<𝑝<𝑞<1 are constants and 𝜆>0 is a parameter. Also 𝑓∶(0,1)×[0,∞)→𝑅,𝑒∶(0,1)→𝑅, and 𝑒∈𝐿(0,1). Moreover 𝑤∶[𝑞,1]→[0,∞) are continuous functions. By using the Guo-Krasnosel’skii fixed-point theorem, the author established the existence of positive solutions. For more existence results of positive solutions for boundary value problems of third-order ordinary differential equations, one can see [6–12] and references therein.

For boundary value problems of second-order or higher-order differential equations at resonance, many existence results of solutions have been established; see [13–25]. In [25], the authors considered the following 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. (6) By using Mawhin continuation theorem [26], the existence results of solutions were obtained under the resonant conditions 𝛽=1, ∑^𝑚−2_𝑖=1𝛼_𝑖=1, and ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0 and  𝛽=1/𝜂, ∑^𝑚−2_𝑖=1𝛼_𝑖=1, and ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0, respectively.

It is well known that the problem of existence of positive solution for 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 [27–34]. To our best knowledge, few paper dealt with the existence result of positive solution for resonant third-order boundary value problems. Motivated by the approach in [28–30, 35], we study the positive solution for problem (1) under nonresonant condition ∑^𝑚−2_𝑖=1𝛽_𝑖<1 and resonant condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1, respectively. By using Leggett-Williams fixed-point theorem and its generalization [27, 29], we establish the existence results of positive solutions. The results obtained in this paper are interesting in the following aspects.(1)In the nonresonant case, Green’s function is established and the results obtained are more general than those of earlier work.(2)It is the first time that the positive solution is considered for third-order boundary value problem at resonance.

2. Background Definitions and Lemmas

Let 𝑋,𝑌 be real Banach spaces. A nonempty closed convex set 𝐶⊂𝑋 is said to be a cone provided that 𝑎𝑥∈𝐶, if 𝑥∈𝐶, 𝑎≥0 and 𝑥,−𝑥∈𝐶 implies 𝑥=0.

Definition 1. The map 𝜓 is a nonnegative continuous concave functional on 𝐶 if 𝜓∶𝐶→+∞ is continuous and 𝜓()𝑦(𝑥)+()𝜓𝑦,[].𝑡𝑥+1−𝑡≥𝑡𝜓1−𝑡𝑥,𝑦∈𝐶,𝑡∈0,1 (7)

Definition 2. Let constants 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) as follow: 𝐶_𝑟={‖𝑥‖},𝐶=(𝑥),‖𝑥‖.𝑥∈𝐶∣<𝑟𝜓,𝑎,𝑏𝑥∈𝐶∣𝑎≤𝜓≤𝑏 (8)

Lemma 3 (Leggett-Williams fixed-point theorem [35]). 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 operator 𝑇 has at least triple fixed-points 𝑥₁,𝑥₂, and 𝑥₃ with ‖𝑥₁‖<𝑎,𝑏<𝜓(𝑥₂),‖𝑥₃‖>𝑎, and 𝜓(𝑥₃)<𝑏.

Operator 𝐿∶dom𝐿⊂𝑋→𝑌 is called a Fredholm operator with index zero, which means that Im𝐿 is closed and dim Ker𝐿 = codim Im𝐿<∞, and there exist continuous projections 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌 such that Im𝑃=Ker𝐿 and Ker𝑄=Im𝐿. Furthermore, for dim Im 𝑄= codim Im𝐿, 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 the operator equation: 𝑥=(𝑃+𝐽𝑄𝑁)𝑥+𝐾_𝑃(𝐼−𝑄)𝑁𝑥. (9) Let 𝛾∶𝑋→𝐶 be a retraction, which means a continuous mapping such that 𝛾𝑥=𝑥 for all 𝑥∈𝐶 and Ψ∶=𝑃+𝐽𝑄𝑁+𝐾_𝑃(𝐼−𝑄)𝑁,Ψ_𝛾∶=Ψ∘𝛾. (10)

Lemma 4 (Leggett-Williams norm-type theorem [28]). Assume that 𝐶 is a cone in 𝑋 and that Ω₁ and Ω₂ are open bounded subsets of 𝑋 with Ω₁⊂Ω₂ and 𝐶∩(Ω₂⧵Ω₁)≠∅. Suppose that 𝐿∶dom𝐿⊂𝑋→𝑌 is a Fredholm operator of index zero and that(𝐶1)𝑄𝑁∶𝑋→𝑌 is bounded and continuous and 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) is 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. Main Results for Nonresonant Case

In this section we consider the positive solution for the nonresonant case with the condition 0<∑^𝑚−1_𝑖=0𝛽_𝑖<1 and we always suppose that 𝑓∈𝐶([0,1]×[0,∞),[0,∞)).

Firstly, we consider the third-order 𝑚-point boundary value problem given by 𝑥^′′′(𝑡)(𝑡)[]𝑥+𝑦=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖. (11)(12)

Lemma 5. Suppose 𝑦(𝑡)∈𝐶[0,1]. Then problem (11) and (12) is equivalent to 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (13) where for 𝑖=1,2,…,𝑚−1𝐺()=𝑎𝑡,𝑠¹𝑡+𝑎₀,𝑡≤𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖,−12𝑡²+𝑏₁𝑡+𝑏₀,𝑡≥𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖,𝑎₁=∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝑠−𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘,𝑏₁=𝑠−∑^𝑖−1_𝑘=0𝛼_𝑘𝑠−∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘,𝑎₀=^𝑖−1_𝑘=0𝛽_𝑘𝜉_𝑘∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝑠−𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘+^𝑚−1_𝑘=𝑖𝛽_𝑘𝜉_𝑘×−1𝑠−∑^𝑖−1_𝑘=0𝛼_𝑘𝑠−∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘+12𝑠²+12−12^𝑚−1_𝑘=𝑖𝛽_𝑘𝜉²_𝑘+𝑠²×1−^𝑚−1_𝑘=0𝛽_𝑘⁻¹,𝑏₀=^𝑖−1_𝑘=0𝛽_𝑘𝜉_𝑘∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝑠−𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘+12^𝑖−1_𝑘=0𝛽_𝑘𝑠²−12^𝑚−1_𝑘=𝑖𝛽_𝑘𝜉²_𝑘+12+^𝑚−1_𝑘=𝑖𝛽_𝑘𝜉_𝑘×∑−1𝑠−^𝑖−1_𝑘=0𝛼_𝑘𝑠−∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘×1−^𝑚−1_𝑘=0𝛽_𝑘⁻¹. (14)

Proof. Let 𝐺(𝑡,𝑠) be Green’s function of problem −𝑥^′′′(𝑡)=0 with boundary condition (12). We can suppose 𝐺()=𝑎𝑡,𝑠²𝑡²+𝑎₁𝑡+𝑎₀𝑡≤𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖,𝑏𝑖=1,2,…,𝑚−1₂𝑡²+𝑏₁𝑡+𝑏₀𝑡≥𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖,𝑖=1,2,…,𝑚−1, (15) where 𝑎₀,𝑎₁,𝑎₂,𝑏₀,𝑏₁, and 𝑏₂ are undetermined coefficients.
Considering the properties of Green's function together with boundary condition (12), we have 𝑎₂𝑠²+𝑎₁𝑠+𝑎₀=𝑏₂𝑠²+𝑏₁𝑠+𝑏₀,2𝑎₂𝑠+𝑎₁=2𝑏₂𝑠+𝑏₁,2𝑎₂−2𝑏₂=1,𝑎₂𝑎=0,₁=^𝑖−1_𝑘=0𝛼_𝑘2𝑎₂𝜉_𝑘+𝑎₁+^𝑚−1_𝑘=𝑖𝛼_𝑘2𝑏₂𝜉_𝑘+𝑏₁,𝑏₂+𝑏₁+𝑏₀=^𝑖−1_𝑘=0𝛽_𝑘𝑎₂𝜉²_𝑘+𝑎₁𝜉_𝑘+𝑎₀+^𝑚−1_𝑘=𝑖𝛽_𝑘𝑏₂𝜉²_𝑘+𝑏₁𝜉_𝑘+𝑏₀. (16) The explicit expression of Green’s function is obtained by solving the linear function systems.

Lemma 6. Green’s function 𝐺(𝑡,𝑠) 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, 𝜕𝐺(𝑡,𝑠)𝜕𝑡=𝑎₁<0, (17) which means that 𝐺(𝑡,𝑠) is decreasing on variable 𝑡. Thus 𝐺(𝑡,𝑠)≥𝐺(𝑠,𝑠),0≤𝑡≤𝑠. (18) On the other hand, for 𝜉_𝑖−1≤𝑠≤𝜉_𝑖, 𝑖=1,2,…,𝑚−1, and 𝑡≥𝑠, 𝜕𝐺(𝑡,𝑠)𝜕𝑡=−𝑡+𝑏₁∑=−𝑡+𝑠−^𝑖−1_𝑘=0𝛼_𝑘𝑠−∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝜉_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘≤∑^𝑚−1_𝑘=𝑖𝛼_𝑘𝑠−𝜉_𝑘1−∑^𝑖−1_𝑘=0𝛼_𝑘<0. (19) Thus, 𝐺(𝑡,𝑠)≥𝐺(1,𝑠),𝑠≤𝑡≤1. (20) Furthermore, 𝜕𝐺(1,𝑠)𝜕𝑠≤−1−^𝑖−1_𝑘=0𝛼_𝑘^𝑚−1_𝑘=0𝛽_𝑘1−𝜉_𝑘<0. (21) Thus, for 𝑡,𝑠∈[0,1], 𝐺()()1𝑡,𝑠≥𝐺1,1=−2+𝑏₁+𝑏₀1=−2+1−∑^𝑚−1_𝑘=0𝛼_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘+12^𝑚−1_𝑘=0𝛽_𝑘+12−1−∑^𝑚−1_𝑘=0𝛼_𝑘1−∑^𝑚−1_𝑘=0𝛼_𝑘×1−^𝑚−1_𝑘=0𝛽_𝑘⁻¹=0. (22) This gives that 𝐺(𝑡,𝑠)≥0,𝑡,𝑠∈[0,1].

Lemma 7. If 𝑦(𝑡)≥0, 𝑡∈[0,1], and 𝑥(𝑡) is the solution of problem (11) and (12), then min_0≤𝑡≤1|𝑥(𝑡)|≥𝛿max_0≤𝑡≤1|𝑥(𝑡)|, (23) where 𝛿=(∑^𝑚−2_𝑖=1𝛽_𝑖(1−𝜉_𝑖))/(1−∑^𝑚−2_𝑖=1𝛽_𝑖𝜉_𝑖) is a positive constant.

Proof. From 𝑥^′′′(𝑡)=−𝑦(𝑡)≤0,𝑡∈[0,1], (24) we see that 𝑥^′′(𝑡) is decreasing on [0,1]. Considering 𝑥^′′(0)=0, we have 𝑥^′′(𝑡)≤0,𝑡∈(0,1). Next we claim that 𝑥^′(0)≤0. Suppose that, on the contrary, 𝑥^′(0)>0. We have 0=𝑥^′(0)−𝑥^′(0)=𝑥^′(0)−^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖>^𝑚−2_𝑖=1𝛼_𝑖𝑥^′(0)−𝑥^′𝜉_𝑖≥0,acontradiction. (25) Thus, max_0≤𝑡≤1𝑥(𝑡)=𝑥(0),min_0≤𝑡≤1𝑥(𝑡)=𝑥(1). (26) From the concavity of 𝑥(𝑡), we have 𝜉_𝑖(𝑥(1)−𝑥(0))≤𝑥𝜉_𝑖−𝑥(0). (27) Multiplying left and right sides by 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). (28) This completes the proof of Lemma 7.

Let the 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)𝑥=0,_′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖,𝑥(𝑡)isconcaveon[0,1]. (29)

Define the continuous nonnegative concave functional 𝜓∶𝐶→[0,∞) by 𝜓(𝑥)=min_0≤𝑡≤1𝑥(𝑡),𝑥∈𝐶. (30) Define the constants 𝑚^∗,𝑚_∗ by 𝑚^∗=max_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑑𝑠,𝑚_∗=min_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑑𝑠. (31)

Theorem 8. Suppose that there exist constants 0<𝑎<𝑏<𝑏/𝛿≤𝑐 such that (𝐴1)𝑓(𝑡,𝑥)<𝑎/𝑚^∗, (𝑡,𝑥)∈[0,1]×[0,𝑎],(𝐴2)𝑓(𝑡,𝑥)>𝑏/𝑚_∗, (𝑡,𝑥)∈[0,1]×[𝑏,𝑏/𝛿],(𝐴3)𝑓(𝑡,𝑥)<𝑐/𝑚^∗, (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>𝑎,min_0≤𝑡≤1𝑥₃<𝑏. (32)

Proof. The operator 𝑇∶𝐶→𝐸 is defined by 𝑇𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠. (33) It is clear that 𝑇∶𝐶→𝐶 and it is completely continuous.
Next, the conditions of Lemma 3 are checked. If 𝑥∈𝐶_𝑐, then ‖𝑥‖≤𝑐 and condition (𝐴3) implies that 𝑓()≤𝑐𝑡,𝑥𝑚^∗,0≤𝑡≤1. (34) Then ‖𝑇(𝑥)‖=max_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≤max_0≤𝑡≤1¹₀𝐺()𝑐𝑡,𝑠𝑑𝑠×𝑚^∗≤𝑐. (35) Thus, 𝑇∶𝐶_𝑐→𝐶_𝑐.

Similar to the proof above, we obtain that 𝑇∶𝐶_𝑎→𝐶_𝑎. Hence, condition (𝐻₂) of Lemma 3 is satisfied.

The fact that the constant function 𝑥(𝑡)=𝑏/𝛿∈{𝑃(𝜓,𝑏,𝑏/𝛿)∣𝜓(𝑥)>𝑏} implies that {𝑃(𝜓,𝑏,𝑏/𝛿)∣𝜓(𝑥)>𝑏}≠∅. If 𝑥∈𝑃(𝜓,𝑏,𝑏/𝛿), from assumption (𝐴2), 𝑓()≥𝑏𝑡,𝑥𝑚_∗. (36) Thus 𝜓(𝑇𝑥)=min_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠≥min_0≤𝑡≤1¹₀𝐺()𝑏𝑡,𝑠𝑑𝑠×𝑚_∗=𝑏, (37) which ensures that condition (𝐻₁) of Lemma 3 is satisfied. Finally we show that condition (𝐻₃) of Lemma 3 also holds. Suppose that 𝑥∈𝑃(𝜓,𝑏,𝑐) with ‖𝑇𝑥‖>𝑏/𝛿. Then 𝜓(𝑇𝑥)=min_0≤𝑡≤1(𝑡)‖‖𝑏𝑇𝑥≥𝛿×𝑇𝑥>𝛿×𝛿=𝑏. (38) 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 𝑥₁≤𝑎,𝑏<min_0≤𝑡≤1𝑥₂,𝑥₃>𝑎,min_0≤𝑡≤1𝑥₃<𝑏. (39)

Here an example is given to illustrate the main results of this section. We consider the following nonresonant three-point boundary value problem: 𝑥^′′(𝑡)()()𝑥+𝑓𝑡,𝑥=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=12𝑥^′13(1)=1,𝑥2𝑥13, (40) where 𝑓()=1𝑡,𝑥𝑒10^𝑡+𝑥³+5𝜋1,0<𝑥<6𝑒10^𝑡+221𝜋,𝑥≥6. (41) Here 𝛼₁=1/2,  𝛽₁=1/2,  𝜉₁=1/3, and 𝐺()=1𝑡,𝑠2𝑠²+𝑠𝑡−103𝑠−133𝑡+21,0≤𝑠≤3−1,𝑡≤𝑠,2𝑡²1+2𝑠𝑡−3𝑡−1033𝑠+21,0≤𝑠≤3(),𝑡≥𝑠,𝑠−1²,13−1≤𝑠≤1,𝑡≤𝑠,2𝑡²+12𝑠²1+𝑠𝑡+1−2𝑠,3≤𝑠≤1,𝑡≥𝑠. (42) By a simple computation, we can get that 𝑚^∗=max_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑑𝑠=32,𝑚81_∗=min_0≤𝑡≤1¹₀𝐺(𝑡,𝑠)𝑑𝑠=16.81 (43) We choose 𝑎=1,𝑏=4, and 𝑐=32. It is easy to check that (1)𝑓(𝑡,𝑥)<81/32, [𝑡,𝑥]∈[0,1]×[0,1],(2)𝑓(𝑡,𝑥)>81/4, [𝑡,𝑥]∈[0,1]×[4,10],(3)𝑓(𝑡,𝑥)<81, [𝑡,𝑥]∈[0,1]×[0,32].

Thus all conditions of Theorem 8 hold. This ensures that problem (40) 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. (44)

4. Main Results for Resonant Case

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 spaces 𝑋=𝑌=𝐶[0,1] endowed with the maximum norm. It is well known that 𝑋 and 𝑌 are the Banach spaces.

Define the linear operator 𝐿∶dom𝐿⊂𝑋→𝑌 and (𝐿𝑥)(𝑡)=−𝑥^′′′(𝑡),𝑡∈[0,1], (45) where dom𝐿=𝑥∈𝑋∣𝑥^′′′∈𝐶[0,1],𝑥^′′(0)𝑥=0,_′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖 (46) and the nonlinear operator 𝑁∶𝑋→𝑌 with (𝑁𝑥)(𝑡)=𝑓(𝑡,𝑥(𝑡)),𝑡∈[0,1]. (47) It is obvious that Ker𝐿={𝑥∈dom𝐿∶𝑥(𝑡)≡𝑐,𝑡∈[0,1]}. Denote the function 𝐺(𝑠), 𝑠∈[0,1] as follow: 𝐺(𝑠)=(1−𝑠)²−^𝑚−2_𝑘=𝑖𝛽_𝑘𝜉_𝑘−𝑠²∑+21−^𝑚−2_𝑘=𝑖𝛽_𝑘𝜉_𝑘1−∑^𝑚−2_𝑘=1𝛼_𝑘^𝑚−2_𝑘=𝑖𝛼_𝑘𝜉_𝑘,𝜉−𝑠_𝑖−1≤𝑠≤𝜉_𝑖,𝑖=1,2,…,𝑚−2. (48) Define the function 𝑘(𝑡,𝑠) as follow:𝑘()=1𝑡,𝑠6(1−𝑠)³−12(𝑡−𝑠)²1−𝑠𝑡+2𝑠−𝑘₀−^{𝑖−1𝑘=0}𝛽_𝑘(1/2)𝑠²−𝜉_𝑘𝑠+(1/2)𝜉²_𝑘(1/2−𝑡)1−^𝑚−1_𝑖=0𝛽_𝑖𝜉_𝑖,𝑡≥𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖,16(1−𝑠)³1−𝑠𝑡+2𝑠−𝑘₀−^{𝑖−1𝑘=0}𝛽_𝑘(1/2)𝑠²−𝜉_𝑘𝑠+(1/2)𝜉²_𝑘(1/2−𝑡)1−^𝑚−1_𝑖=0𝛽_𝑖𝜉_𝑖,𝑡≤𝑠,𝜉_𝑖−1≤𝑠≤𝜉_𝑖, (49)for 𝑖=1,2,…,𝑚−1 and 𝑘₀=(1/2−𝑡)[1−(1/2)(1−∑^𝑚−1_𝑖=0𝛽_𝑖𝜉_𝑖)].

The function 𝑈(𝑡,𝑠) and positive number 𝜅 are given by 𝑈()()+𝐺(𝑠)𝑡,𝑠=𝑘𝑡,𝑠∫¹₀𝐺(𝑠)𝑑𝑠1−¹₀𝑘(),[],𝑡,𝑠𝑑𝑠𝑡,𝑠∈0,1𝜅∶=min1,min_[]𝑠∈0,1∫¹₀𝐺(𝑠)𝑑𝑠𝐺(𝑠),min_{[]𝑡,𝑠∈0,1}1𝑈(),∑𝑡,𝑠𝜎=1+^{𝑚−2𝑖=1}𝛼_𝑖𝜉²_𝑖1−∑^𝑚−2_𝑖=1𝛼_𝑖. (50)

Theorem 9. Suppose 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−(𝜅𝜎/2))𝑅,𝑅],(𝑆3)there exist 𝑟∈(0,𝑅),𝑀∈(0,1),𝑡₀∈[0,1],𝑎∈(0,1], and continuous functions 𝑔∶[0,1]⟶[0,+∞),ℎ∶(0,𝑟]⟶[0,+∞) (51)such that 𝑓(𝑡,𝑥)≥𝑔(𝑡)ℎ(𝑥),[𝑡,𝑥]∈[0,1]×(0,𝑟] (52) and ℎ(𝑥)/𝑥^𝑎 is nonincreasing on (0,𝑟] with ℎ(𝑟)𝑟^𝑎1₀𝑈𝑡₀,𝑠𝑔(𝑠)𝑑𝑠≥1−𝑀𝑀^𝑎. (53) Then resonant problem (1) has at least one positive solution.

Proof. Firstly we prove that Im𝐿=𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. (54) Indeed, for each 𝑦∈{𝑦∈𝑌∣∫¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0}, we choose 𝑥(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)−𝑡𝑑𝑠1−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑦(𝑠)𝑑𝑠. (55) We can check that −𝑥^′′′(𝑡)=𝑦(𝑡),𝑥^′′(0)=0,𝑥^′′(0)𝑥=0,_′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖, (56) which means 𝑥(𝑡)∈dom𝐿. Thus 𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0⊂Im𝐿. (57) On the other hand, for each 𝑦(𝑡)∈Im𝐿, there exists 𝑥(𝑡)∈dom𝐿 such that −𝑥^′′′(𝑡)=𝑦(𝑡),𝑥^′′(0)=0,𝑥^′′(0)𝑥=0,_′(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥^′𝜉_𝑖,𝑥(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑥𝜉_𝑖. (58) 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). (59) Considering the boundary condition together with the resonant condition ∑^𝑚−1_𝑖=0𝛽_𝑖=1, we have ¹₀(1−𝑠)²𝑦(𝑠)𝑑𝑠−^𝑚−2_𝑖=1𝛽_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠²𝑦(𝑠)𝑑𝑠+21−∑^𝑚−2_𝑖=1𝛽_𝑖𝜉_𝑖1−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑦(𝑠)𝑑𝑠=0. (60) Thus, Im𝐿=𝑦∈𝑌∣¹₀𝐺(𝑠)𝑦(𝑠)𝑑𝑠=0. (61) It is obvious that dim Ker𝐿=1 and Im𝐿 is closed.
Secondly we see 𝑌=𝑌₁⊕Im𝐿, where 𝑌₁=𝑦₁∣𝑦₁=1∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑦(𝑠)𝑑𝑠,𝑦∈𝑌. (62) In fact, for each 𝑦(𝑡)∈𝑌, we have ¹₀𝐺(𝑠)𝑦(𝑠)−𝑦₁𝑑𝑠=0. (63) This induces that 𝑦−𝑦₁∈Im𝐿. Since 𝑌₁∩Im𝐿={0}, we have 𝑌=𝑌₁⨁Im𝐿. Thus 𝐿 is a Fredholm operator with index zero.
Define two projections 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌 by 𝑃𝑥=¹₀𝑥(𝑠)1𝑑𝑠,𝑄𝑦=∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑦(𝑠)𝑑𝑠. (64) Clearly, Im𝑃=Ker𝐿 and Ker𝑄=Im𝐿. Note that, for 𝑦∈Im𝐿, the inverse 𝐾_𝑃 of 𝐿_𝑃 is given by 𝐾_𝑃𝑦=^𝑡₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (65) In fact, It is easy to check that 𝐿𝐾_𝑃𝑦=−¹₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠^′′′(𝑡)𝐾=𝑦_𝑃(𝐿)(𝑥)=¹₀𝑘(𝑡,𝑠)−𝑥^′′′(𝑠)𝑑𝑠=𝑥(𝑡). (66)
Next we will check that every condition of Lemma 4 is fulfilled. Remark that 𝑓 can be extended continuously on [0,1]×(−∞,+∞) and condition (𝐶1) of Lemma 4 is fulfilled.
Define the set of nonnegative functions 𝐶 and subsets of X Ω₁,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]}. (67) Remark that Ω₁ and Ω₂ are open and bounded sets. Furthermore Ω₁={𝑥∈𝑋∶𝑟≥|𝑥(𝑡)|≥𝑀‖𝑥‖,𝑡∈[0,1]}⊂Ω₂,𝐶∩Ω₂⧵Ω₁≠∅. (68)
Let the isomorphism 𝐽=𝐼 and (𝛾𝑥)(𝑡)=|𝑥(𝑡)| for 𝑥∈𝑋. Then 𝛾 is a retraction and maps subsets of Ω₂ into bounded subsets of 𝐶, which ensures that condition (𝐶3) of Lemma 4 is fulfilled.
Then we prove that (𝐶2) of Lemma 4 is fulfilled. For this purpose, suppose that there exist 𝑥₀∈𝐶∩𝜕Ω₂∩dom𝐿 and 𝜆₀∈(0,1) such that 𝐿𝑥₀=𝜆₀𝑁𝑥₀. Then −𝑥^′′′₀(𝑡)=𝜆₀𝑓𝑡,𝑥₀ (69) for all 𝑡∈[0,1]. Thus 𝑥^′′₀(𝑡)=−𝜆₀^𝑡₀𝑓𝑠,𝑥₀(𝑠)𝑥𝑑𝑠,_′0(𝑡)=−𝜆₀^𝑡₀(𝑡−𝑠)𝑓𝑠,𝑥₀(𝑠)𝑑𝑠−𝜆₀11−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑓𝑠,𝑥₀(𝑠)𝑑𝑠. (70)(71) Let 𝑥₀(𝑡₀)=‖𝑥₀‖=𝑅. The proof is divided into three cases.(1)We show that 𝑡₀≠1. Suppose, on the contrary, that 𝑥₀(𝑡) achieves maximum value 𝑅 only at 𝑡₀=1. Then the boundary condition 𝑥(1)=∑^𝑚−2_𝑖=1𝛽_𝑖𝑥₀(𝜉_𝑖) in combination with the resonant condition ∑^𝑚−2_𝑖=1𝛽_𝑖=1 yields that max_{1≤𝑖≤𝑚−2}𝑥₀(𝜉_𝑖)≥𝑅, which is a contradiction.(2)We claim that 𝑡₀≠0. Suppose, on the contrary, that 𝑥₀(𝑡) achieves maximum value 𝑅 at 𝑡₀=0. From condition (𝑆2), there exists 𝑡₁>0 near to zero such that 𝑓𝑡,𝑥₀(𝑡)<0,𝑡∈0,𝑡₁. (72) Then 𝑥^′₀(𝑡)=−𝜆₀^𝑡₀(𝑡−𝑠)𝑓𝑠,𝑥₀(𝑠)𝑑𝑠−𝜆₀11−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑓(𝑠,𝑥)𝑑𝑠>0,𝑡∈0,𝑡₁, (73) which contradicts the fact that 𝑥₀(𝑡) achieves maximum value at 𝑡₀=0.(3)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 𝑥^′′₀(𝜂)=0. From the mean value theory, we claim that there exists 𝜉∈(𝜂,𝑡₀) such that 𝑥₀𝜂=𝑥₀𝑡₀−𝑥^′₀(𝜉)𝑡₀−𝜂. (74)However, 𝑥^′₀(𝜉)=−𝜆₀^𝜉₀(𝜉−𝑠)𝑓𝑠,𝑥₀𝑑𝑠−𝜆₀11−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑓(𝑠,𝑥)𝑑𝑠≤𝜆₀𝜅^𝜉₀(𝜉−𝑠)𝑥₀(𝑠)𝑑𝑠+𝜆₀𝜅11−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖−𝑠𝑥₀𝑑𝑠≤𝜆₀𝜅𝑅^𝜉₀(𝜉−𝑠)𝑑𝑠+𝜆₀1𝜅𝑅1−∑^𝑚−2_𝑖=1𝛼_𝑖^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀𝜉_𝑖=1−𝑠𝑑𝑠2𝜆₀𝜅𝑅𝜉²+∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖1−∑^𝑚−2_𝑖=1𝛼_𝑖. (75) Thus 𝑥₀𝜂=𝑥₀𝑡₀−𝑥^′₀(𝜉)𝑡₀1−𝜂≥𝑅−2𝜆₀𝜅𝜉²+∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖1−∑^𝑚−2_𝑖=1𝛼_𝑖𝑡₀𝑅≥𝜅−𝜂1−2∑1+^{𝑚−2𝑖=1}𝛼_𝑖𝜉²_𝑖1−∑^𝑚−2_𝑖=1𝛼_𝑖𝑅=1−𝜅𝜎2𝑅. (76) Then, considering assumption (𝑆2), we have 0≥𝑥^′′₀𝑡₀−𝑥^′′₀𝜂=−𝜆₀^𝑡0_𝜂𝑓𝑠,𝑥₀(𝑠)𝑑𝑠>0, (77) which is a contradiction. Thus (𝐶2) of Lemma 4 is fulfilled.

Remark 10. The sign of third-order derivative of a function ℎ(𝑡) at point 𝑡₀ cannot be confirmed when 𝑡₀ is a maximal value of ℎ(𝑡). Thus the methods in [28] are not applicable directly to this problem.

For 𝑥∈Ker𝐿∩Ω₂, define the projection 𝐻(𝑥,𝜆) as follows: 𝐻()|𝑥|−𝜆𝑥,𝜆=𝑥−𝜆∫¹₀𝐺(𝑠)𝑑𝑠¹⁰𝐺(𝑠)𝑓(𝑠,|𝑥|)𝑑𝑠, (78) where 𝜆∈[0,1] and 𝑥∈Ker𝐿∩Ω₂. Suppose 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. (79) Hence 𝐻(𝑥,𝜆)=0 implies 𝑐≥0. Hence, if 𝐻(𝑅,𝜆)=0, we get 0≤𝑅(1−𝜆)¹₀𝐺(𝑠)𝑑𝑠=𝜆¹₀𝐺(𝑠)𝑓(𝑠,𝑅)𝑑𝑠, (80) contradicting (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. (81) This ensures 𝑑_𝐵𝐼−(𝑃+𝐽𝑄𝑁)𝛾|_Ker𝐿,Ker𝐿∩Ω₂,0=𝑑_𝐵𝐻(𝑥,1),Ker𝐿∩Ω₂,0≠0. (82) Let 𝑥∈Ω₂⧵Ω₁ and 𝑡∈[0,1]. From condition (𝑆1), we see Ψ_𝛾𝑥(𝑡)=¹₀|𝑥(𝑡1)|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. (83) 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, (84) 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∈𝐶∣𝑥(𝑡)>0on[0,1]} (85) and we can choose 𝜎(𝑢₀)=1. For 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, we have 𝑥(𝑡)[]‖𝑥‖𝑥(𝑡)‖𝑥‖[].>0,𝑡∈0,1,0<≤𝑟,≥𝑀on0,1 (86) 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)‖ (87) So ‖𝑥‖≤𝜎(𝑢₀)‖Ψ𝑥‖ for all 𝑥∈𝐶(𝑢₀)∩𝜕Ω₁, which means (𝐶5) of Lemma 4 holds.

Thus with the application of Lemma 4, we confirm that the equation 𝐿𝑥=𝑁𝑥 has a solution 𝑥∈𝐶∩(Ω₂⧵Ω₁), which implies that the resonant problem (1) has at least one positive solution.

Finally an example is given to illustrate the main results of the resonance case. We investigate the resonant third-order three-point boundary value problem: 𝑥^′′′(𝑡)+−12𝑡²+125𝑡+𝑥16²−4𝑥+115×√𝑥²[]𝑥−6𝑥+10=0,𝑡∈0,1_′′(0)=0,𝑥^′(0)=12𝑥^′23(0)2,𝑥=𝑥3, (88) where 𝛼₁=1/2,  𝛽=1,  𝜉=2/3, and 𝑓()=−1𝑡,𝑥2𝑡²+125𝑡+𝑥16²−4𝑥+115×√𝑥²−6𝑥+10. (89) Here 𝐺(𝑠)=41−32𝑠,0≤𝑠≤3(1−𝑠)²,23𝑘()=1≤𝑠≤1,𝑡,𝑠6(1−𝑠)³−12(𝑡−𝑠)²1−𝑠𝑡+25𝑠−6122−𝑡,𝑡≥𝑠,0≤𝑠≤316(1−𝑠)³+1−𝑠𝑡25𝑠−6122−𝑡,𝑡≤𝑠,0≤𝑠≤3.16(1−𝑠)³−12(𝑡−𝑠)²1−𝑠𝑡+21𝑠−2×3−𝑡2𝑠²3−2𝑠+22,𝑡≥𝑠,31≤𝑠≤16(1−𝑠)³1−𝑠𝑡+2𝑠−12×3−𝑡2𝑠²3−2𝑠+22,𝑡≤𝑠,3≤𝑠≤1. (4) By a simple computation, we have ¹₀𝐺(𝑠)𝑑𝑠=3181,𝜎=139,𝜅=31,81¹₀𝑈(0,𝑠)𝑑𝑠=1. (91) 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. (92) Then, 7(𝑡)≤48≤𝑔53<14431[],𝑥81,𝑡∈0,1₂−4𝑥+115≥−𝑥,𝑥∈[0,1]. (93) It is easy to check that (1)𝑓(𝑡,𝑥)>−(31/81)𝑥, for all (𝑡,𝑥)∈[0,1]×[0,1],(2)𝑓(𝑡,𝑥)<0, for all (𝑡,𝑥)∈[0,1]×[1055/1458,1],(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/𝑥 is nonincreasing on (0,1/4] with ℎ(𝑟)𝑟^𝑎1₀𝑈()𝑔(𝑠)7√0,𝑠𝑑𝑠>13748>1=1−𝑀𝑀^𝑎. (94)Then all conditions of Theorem 9 are satisfied. This ensures that the resonant problem has at least one solution, positive on [0,1].

Conflict of Interests

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

Acknowledgments

The project is supported by the National Natural Science Foundation of China (11201109) and the Natural Science Foundation of Anhui Educational Department (KJ2012B144 and KJ2012Z335).