Successive Iteration of Positive Solutions for Fourth-Order Two-Point Boundary Value Problems

Abstract

We are concerned with a fourth-order two-point boundary value problem. We prove the existence of positive solutions and establish iterative schemes for approximating the solutions. The interesting point of our method is that the nonlinear term is involved with all lower-order derivatives of unknown function, and the iterative scheme starts off with a known cubic function or the zero function. Finally we give two examples to verify the effectiveness of the main results.

1. Introduction

The bending of an elastic beam can be described with some fourth-order boundary value problems. Owing to their importance in engineering, physics, and material mechanics, boundary value problems for elastic beam equations have attracted much attention of many authors. Many methods, such as upper and lower solutions method, shooting method, coincidence degree theory, or fixed point theorem and so on, have been developed to derive existence of solutions for fourth-order differential equations with two-point boundary conditions; see [1–20] and many references cited therein. However, almost all of the papers we mentioned focused their attention on the existence of solutions or positive solutions. In the existing literature, there are few papers that concern with the computational methods of solutions or positive solutions; see [21–24]. In this paper, by the successively iterative technique, we study the existence and iteration of monotone positive solutions for the following fourth-order differential equation: 𝑢^′′′′(𝑡)=𝑞(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),𝑢^′′′(𝑡),a.e.𝑡∈[0,1], (1) together with two-point boundary conditions 𝑢(0)=𝑢^′(1)=𝑢^′′(0)=𝑢^′′′(1)=0. (2) Throughout this paper, the following conditions are satisfied: (𝐴1)  𝑓∈𝐶([0,1]×[0,∞)×[0,∞)×(−∞,0]×(−∞,0],[0,∞)); (𝐴2)  𝑞∈𝐿¹[0,1] is nonnegative and 0<∫¹₀𝑞(𝑠)𝑑𝑠<∞.

In material mechanics, (1) describes the deflection of a beam under a certain force. The boundary conditions of the forms (2) mean that the elastic beam is simply fixed at the end 𝑡=0 and fastened with a sliding clamp at the end 𝑡=1.

Fourth-order differential equations with boundary conditions (2) have been studied. For example, Graef and Yang in [7] studied the nonlinear fourth-order differential equation: 𝑢^′′′′(𝑡)=𝜆𝑎(𝑡)𝑓(𝑢(𝑡)),0<𝑡<1, (3) together with boundary conditions (2). The authors established some sufficient conditions for the existence and nonexistence of positive solutions. The main tool is the Kranosel’skii fixed point theorem. Using the Leggett-Williams fixed point theorem, Yang [18] established an existence criterion for triple positive solutions of the nonlinear fourth-order differential equation: 𝑢^′′′′(𝑡)=𝑔(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),0<𝑡<1, (4) with two-point boundary conditions (2). By applying iterative techniques, Sun and Wang [21] investigated the existence and iteration of monotone positive solution for the following elastic beam equation 𝑢^′′′′(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),0<𝑡<1, (5) with boundary conditions (2). By using Leray-Schauder nonlinear alternate, Leray-Schauder fixed point theorem and a fixed point theorem due to Avery and Peterson, Sun [15] presented some results on the existence and multiplicity of positive solutions to problem (1)-(2). Bai [3] proposed a constructive method which yields two monotone sequences that converge uniformly to extremal solutions to the following differential equations 𝑢^′′′′(𝑡)−𝑀𝑢^′′(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),𝑢^′′′(𝑡),0<𝑡<1, (6) or 𝑢^′′′′(𝑡)−𝑀𝑢^′′(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),0<𝑡<1, (7) with boundary conditions (2), if there exists an upper solution 𝛽 and a lower solution 𝛼 with 𝛽⩽𝛼. By placing some restrictions on the nonlinear term 𝑓, Bai [4] obtained the existence results for the fourth-order two-point boundary value problem (1)-(2) via the lower and upper solution method. In particular, a new truncating technique and an appropriate Nagumo-type condition are introduced and employed. In addition, Yao studied (1) with nonhomogeneous boundary condition 𝑢(0)=𝑎,𝑢^′(1)=𝑏,𝑢^′′(0)=𝑐,𝑢^′′′(1)=𝑑. (8) In [20] the author obtained local existence theorem via Leray-Schauder fixed point theorem. In [22] Yao constructed an iterative sequence by the monotonic technique and proved that the sequence approximates successively to the solution of the equation under suitable assumptions.

Motivated by the above-mentioned works, this paper investigates the existence and iteration of positive solutions for problem (1)-(2). It is noted that the successively iterative scheme starts off with the zero function or a simple cubic function, which is feasible for the computational purpose. At the end of the paper, two examples are included to illustrate the main results.

2. Preliminaries

In this section, we introduce some necessary definitions and properties of the Green function.

Definition 1 (see [25, 26]). 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 imply 𝜆𝑥∈𝐾;(2)𝑥∈𝐾,−𝑥∈𝐾 imply 𝑥=0.

Definition 2 (see [25, 26]). An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

The Green’s function 𝐺∶[0,1]×[0,1]→[0,+∞) for problem (1)-(2) is given by 𝐺()=1𝑡,𝑠66𝑡−3𝑡²−𝑠²𝑠,0⩽𝑠⩽𝑡⩽1,6𝑠−3𝑠²−𝑡²𝑡,0⩽𝑡⩽𝑠⩽1, (9) so solving the boundary value problem (1)-(2) is equivalent to finding a solution to the integral equation 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠)𝑑𝑠. (10) Some properties of functions 𝐺(𝑡,𝑠) are summarized as follows.

Lemma 3. For all 𝑡,𝑠∈[0,1], the following inequalities hold true: 𝜕0⩽−³𝐺(𝑡,𝑠)𝜕𝑡³𝜕⩽1,0⩽−²𝐺(𝑡,𝑠)𝜕𝑡²()⩽𝑡,0⩽𝜕𝐺𝑡,𝑠⩽1𝜕𝑡21−𝑡²()⩽1,0⩽𝐺𝑡,𝑠63𝑡−𝑡³. (11)

Proof. For any fixed 𝑠∈[0,1], from (9) we obtain 𝜕³𝐺(𝑡,𝑠)𝜕𝑡³=𝜕0,0⩽𝑠<𝑡⩽1,−1,0⩽𝑡<𝑠⩽1,₂𝐺(𝑡,𝑠)𝜕𝑡²=−𝑠,0⩽𝑠⩽𝑡⩽1,−𝑡,0⩽𝑡⩽𝑠⩽1, (12) then, 𝜕0⩽−³𝐺(𝑡,𝑠)𝜕𝑡³𝜕⩽1,0⩽−²𝐺(𝑡,𝑠)𝜕𝑡²⩽𝑡,𝑡,𝑠∈[0,1]. (13)
For all 𝑡,𝑠∈[0,1], it follows from (9) that ()0⩽𝜕𝐺𝑡,𝑠=()1𝜕𝑡1−𝑡𝑠,0⩽𝑠⩽𝑡⩽1,𝑠−2𝑠²−12𝑡²⩽1,0⩽𝑡⩽𝑠⩽1,21−𝑡². (14)
For all 𝑡,𝑠∈[0,1], if 𝑠⩽𝑡, it yields 𝐺()=1𝑡,𝑠66𝑡−3𝑡²−𝑠²1𝑠=63−3(1−𝑡)²−𝑠²𝑠⩽163−𝑠²1𝑠⩽63𝑡−𝑡³. (15) While 𝑡⩽𝑠, we have 𝐺()=1𝑡,𝑠66𝑠−3𝑠²−𝑡²1𝑡=63−3(1−𝑠)²−𝑡²𝑡⩽163𝑡−𝑡³. (16) The proof is completed.

In what follows we will consider the Banach space 𝐸=𝐶³[0,1] equipped with norm ‖𝑢‖=maxmax_0⩽𝑡⩽1|𝑢(𝑡)|,max_0⩽𝑡⩽1𝑢^′(𝑡),max_0⩽𝑡⩽1𝑢^′′(𝑡),max_0⩽𝑡⩽1𝑢^′′′(𝑡). (17) Define the cone 𝐾⊂𝐶³[0,1] by 𝐾=𝑢∈𝐶³[0,1]∶𝑢(𝑡)⩾0,𝑢^′(𝑡)⩾0,𝑢^′′(𝑡)𝑢⩽0,_′′′(𝑡)⩽0,𝑡∈[0,1]. (18) For 𝑢∈𝐾, we give an operator 𝑇 by (𝑡)=𝑇𝑢)(₁₀𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠)𝑑𝑠. (19) Obviously, the fixed points of 𝑇 are monotone and nonnegative solutions of the problem (1)-(2).

Lemma 4. Consider that 𝑇∶𝐾→𝐾 is completely continuous.

Proof. Suppose that 𝑢∈𝐾. Using (𝐴1), (𝐴2), (19), and Lemma 3, we obtain (𝑇𝑢)(𝑡)⩾0, (𝑇𝑢)^′(𝑡)⩾0, (𝑇𝑢)^′′(𝑡)⩽0, (𝑇𝑢)^′′′(𝑡)⩽0, for all 𝑡∈[0,1]. Then 𝑇(𝐾)⊆𝐾. Using arguments similar to those of [21, 22], one can prove that 𝑇 is a completely continuous operator.

3. Main Results

In this section, we give the main results of this paper.

Theorem 5. Assume that there exists 𝑅>0 such that(𝐻1)  𝑓(𝑡,𝑥₁,𝑦₁,𝑧₁,𝑤₁)⩽𝑓(𝑡,𝑥₂,𝑦₂,𝑧₂,𝑤₂) for any 0⩽𝑡⩽1, 0⩽𝑥₁⩽𝑥₂⩽𝑅, 0⩽𝑦₁⩽𝑦₂⩽𝑅, 0⩽−𝑧₁⩽−𝑧₂⩽𝑅, 0⩽−𝑤₁⩽−𝑤₂⩽𝑅;(𝐻2)  max_0⩽𝑡⩽1𝑓(𝑡,𝑅,𝑅,−𝑅,−𝑅)⩽Λ𝑅, where Λ=(∫¹₀𝑞(𝑠)𝑑𝑠)⁻¹;(𝐻3)  𝑓(𝑡,0,0,0,0)≢0.
Then problem (1)-(2) has two monotone positive solutions 𝑤^∗ and 𝑣^∗ satisfying ‖𝑤^∗‖⩽𝑅, ‖𝑣^∗‖⩽𝑅 and lim_𝑛→∞‖𝑤_𝑛−𝑤^∗‖=0, lim_𝑛→∞‖𝑣_𝑛−𝑣^∗‖=0, where 𝑤_𝑛=𝑇^𝑛𝑤₀, 𝑣_𝑛=𝑇^𝑛𝑣₀, 𝑛=1,2,…, 𝑤₀(𝑡)=(𝑅/6)(3𝑡−𝑡³),  and 𝑣₀(𝑡)=0, for 𝑡∈[0,1].

Proof. Let 𝐾_𝑅={𝑢∈𝐾∶‖𝑢‖⩽𝑅}; it follows that 𝑇∶𝐾_𝑅→𝐾_𝑅. In fact, if 𝑢∈𝐾_𝑅, then ‖𝑢‖⩽𝑅; thus, for any 𝑡∈[0,1], 0⩽𝑢(𝑡)⩽max_0⩽𝑡⩽1𝑢(𝑡)⩽‖𝑢‖⩽𝑅,0⩽𝑢^′(𝑡)⩽max_0⩽𝑡⩽1𝑢^′(𝑡)⩽‖𝑢‖⩽𝑅,0⩽−𝑢^′′(𝑡)⩽max_0⩽𝑡⩽1−𝑢^′′(𝑡)⩽‖𝑢‖⩽𝑅,0⩽−𝑢^′′′(𝑡)⩽max_0⩽𝑡⩽1−𝑢^′′′(𝑡)⩽‖𝑢‖⩽𝑅, (20) which together with the conditions (𝐻1) and (𝐻2) implies that (𝑡)=0⩽𝑇𝑢)(₁₀𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠⩽1)𝑑𝑠63𝑡−𝑡³¹₀𝑞(𝑠)𝑓()⩽1𝑠,𝑅,𝑅,−𝑅,−𝑅𝑑𝑠2Λ𝑅¹₀𝑞(𝑠)1𝑑𝑠=2[],()𝑅,𝑡∈0,10⩽𝑇𝑢^′(𝑡)=¹₀𝜕𝐺(𝑡,𝑠)𝑞(𝑠)𝑓(𝑠)𝜕𝑡𝑠,𝑢,𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠)⩽1𝑑𝑠21−𝑡²¹₀𝑞(𝑠)𝑓()⩽1𝑠,𝑅,𝑅,−𝑅,−𝑅𝑑𝑠2Λ𝑅¹₀𝑞(𝑠)1𝑑𝑠=2[],()𝑅,𝑡∈0,10⩽−𝑇𝑢^′′(𝑡)=−¹₀𝜕²𝐺(𝑡,𝑠)𝜕𝑡²𝑞(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠)𝑑𝑠⩽𝑡¹₀𝑞(𝑠)𝑓(𝑠,𝑅,𝑅,−𝑅,−𝑅)𝑑𝑠⩽Λ𝑅¹₀𝑞(𝑠)[],()𝑑𝑠=𝑅,𝑡∈0,10⩽−𝑇𝑢^′′′(𝑡)=−¹₀𝜕³𝐺(𝑡,𝑠)𝜕𝑡³𝑞(𝑠)𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠),𝑢^′′(𝑠),𝑢^′′′(𝑠)⩽𝑑𝑠₁₀𝑞(𝑠)𝑓(𝑠,𝑅,𝑅,−𝑅,−𝑅)𝑑𝑠⩽Λ𝑅¹₀𝑞(𝑠)𝑑𝑠=𝑅,𝑡∈[0,1]. (21) Hence, we have shown that ‖𝑇𝑢‖⩽𝑅, that is, 𝑇∶𝐾_𝑅→𝐾_𝑅.
(1) Let 𝑤₀(𝑡)=(𝑅/6)(3𝑡−𝑡³),𝑡∈[0,1]; then ‖𝑤₀‖=𝑅 and 𝑤₀∈𝐾_𝑅. Denote 𝑤_𝑛=𝑇^𝑛𝑤₀,𝑛=1,2,3,…. Since 𝑇∶𝐾_𝑅→𝐾_𝑅, we have 𝑤_𝑛∈𝑇(𝐾_𝑅)⊆𝐾_𝑅,𝑛=1,2,3,…. Since 𝑇 is completely continuous, we assert that {𝑤_𝑛}^∞_𝑛=1 is a sequentially compact set.
Since 𝑤₁=𝑇𝑤₀∈𝐾_𝑅, by Lemma 3, (𝐻1), and (𝐻2), we obtain 𝑤₁(𝑡)=𝑇𝑤₀(𝑡)=¹₀𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑤₀(𝑠),𝑤^′₀(𝑠),𝑤₀^′′(𝑠),𝑤₀^′′′(𝑠)=𝑑𝑠₁₀𝐺()𝑞(𝑠)𝑓𝑅𝑡,𝑠𝑠,63𝑠−𝑠³,𝑅21−𝑠²⩽1,−𝑅𝑠,−𝑅𝑑𝑠63𝑡−𝑡³¹₀𝑞(𝑠)𝑓()⩽𝑅𝑠,𝑅,𝑅,−𝑅,−𝑅𝑑𝑠63𝑡−𝑡³Λ¹₀𝑞(𝑠)𝑅𝑑𝑠=63𝑡−𝑡³=𝑤₀(𝑡),[],𝑤𝑡∈0,1_′1(𝑡)=𝑇𝑤₀^′(𝑡)=¹₀𝜕𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝜕𝑡𝑠,𝑤₀(𝑠),𝑤^′₀(𝑠),𝑤₀^′′(𝑠),𝑤₀^′′′(𝑠)𝑑𝑠=¹₀𝜕𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑅𝜕𝑡𝑠,63𝑠−𝑠³,𝑅21−𝑠²⩽1,−𝑅𝑠,−𝑅𝑑𝑠21−𝑡²¹₀𝑞(𝑠)𝑓()⩽𝑅𝑠,𝑅,𝑅,−𝑅,−𝑅𝑑𝑠21−𝑡²Λ¹₀𝑞(𝑠)𝑅𝑑𝑠=21−𝑡²=𝑤^′₀(𝑡),𝑡∈[0,1],−𝑤₁^′′(𝑡)=−𝑇𝑤₀^′′(𝑡)=−¹₀𝜕²𝐺(𝑡,𝑠)𝜕𝑡²𝑞(𝑠)𝑓𝑠,𝑤₀(𝑠),𝑤^′₀(𝑠),𝑤₀^′′(𝑠),𝑤₀^′′′(𝑠)𝑑𝑠=−¹₀𝜕²𝐺(𝑡,𝑠)𝜕𝑡²𝑞(𝑠)𝑓𝑅𝑠,63𝑠−𝑠³,𝑅21−𝑠²,−𝑅𝑠,−𝑅𝑑𝑠⩽𝑡¹₀𝑞(𝑠)𝑓(𝑠,𝑅,𝑅,−𝑅,−𝑅)𝑑𝑠⩽𝑡Λ𝑅¹₀𝑞(𝑠)𝑑𝑠=𝑅𝑡=−𝑤₀^′′(𝑡),𝑡∈[0,1],−𝑤₁^′′′(𝑡)=−𝑇𝑤₀^′′′(𝑡)=−¹₀𝜕³𝐺(𝑡,𝑠)𝜕𝑡³𝑞(𝑠)𝑓𝑠,𝑤₀(𝑠),𝑤^′₀(𝑠),𝑤₀^′′(𝑠),𝑤₀^′′′(𝑠)𝑑𝑠=−¹₀𝜕³𝐺(𝑡,𝑠)𝜕𝑡³𝑞(𝑠)𝑓𝑅𝑠,63𝑠−𝑠³,𝑅21−𝑠²⩽,−𝑅𝑠,−𝑅𝑑𝑠₁₀𝑞(𝑠)𝑓(𝑠,𝑅,𝑅,−𝑅,−𝑅)𝑑𝑠⩽Λ𝑅¹₀𝑞(𝑠)𝑑𝑠=𝑅=−𝑤₀^′′′(𝑡),𝑡∈[0,1]. (22) Thus, 𝑤₁(𝑡)⩽𝑤₀(𝑡),𝑤^′₁(𝑡)⩽𝑤^′₀(𝑡),−𝑤₁^′′(𝑡)⩽−𝑤₀^′′(𝑡),−𝑤₁^′′′(𝑡)⩽−𝑤₀^′′′(𝑡),𝑡∈[0,1]. (23) Furthermore, it follows from (𝐻1) that 𝑤₂(𝑡)=𝑇𝑤₁(𝑡)⩽𝑇𝑤₀(𝑡)=𝑤₁(𝑡)[],𝑤,𝑡∈0,1_′2(𝑡)=𝑇𝑤₁^′(𝑡)⩽𝑇𝑤₀^′(𝑡)=𝑤^′₁(𝑡),𝑡∈[0,1],−𝑤₂^′′(𝑡)=−𝑇𝑤₁^′′(𝑡)⩽−𝑇𝑤₀^′′(𝑡)=−𝑤₁^′′(𝑡),𝑡∈[0,1],−𝑤₂^′′′(𝑡)=−𝑇𝑤₁^′′′(𝑡)⩽−𝑇𝑤₀^′′′(𝑡)=−𝑤₁^′′′(𝑡),𝑡∈[0,1]. (24) By the induction, we obtain 𝑤_𝑛+1(𝑡)⩽𝑤_𝑛(𝑡),𝑤^′_𝑛+1(𝑡)⩽𝑤^′_𝑛(𝑡),−𝑤^′′_𝑛+1(𝑡)⩽−𝑤_𝑛^′′(𝑡),−𝑤^′′′_𝑛+1(𝑡)⩽−𝑤_𝑛^′′′(𝑡),𝑡∈[0,1],𝑛=0,1,2,…. (25) Hence, there exists 𝑤^∗∈𝐾_𝑅 such that lim_𝑛→∞‖𝑤_𝑛−𝑤^∗‖=0. Applying the continuity of 𝑇 and using 𝑤_𝑛+1=𝑇𝑤_𝑛, we obtain 𝑇𝑤^∗=𝑤^∗, which implies that 𝑤^∗ is a nonnegative solution of problem (1)-(2). Furthermore, by (𝐻3) we know that the zero function is not a solution of problem (1)-(2). Thus, max_0⩽𝑡⩽1𝑤^∗(𝑡)>0. From the definition of the cone 𝐾, it yields 𝑤^∗(𝑡)⩾𝑡‖𝑤^∗‖>0,(𝑤^∗)^′(𝑡)⩾0,𝑡∈(0,1); that is, 𝑤^∗(𝑡) is a monotone positive solution of problem (1)-(2).
(2) Let 𝑣₀(𝑡)=0,𝑡∈[0,1]; then 𝑣₀∈𝐾_𝑅. Let 𝑣_𝑛=𝑇^𝑛𝑣₀,𝑛=1,2,3,…. With respect to 𝑇∶𝐾_𝑅→𝐾_𝑅, we obtain 𝑣_𝑛∈𝑇(𝐾_𝑅)⊆𝐾_𝑅,𝑛=1,2,3,…. Since 𝑇 is completely continuous, we assert that {𝑣_𝑛}^∞_𝑛=1 is a sequentially compact set.
Since 𝑣₁=𝑇𝑣₀=𝑇0∈𝐾_𝑅, we obtain 𝑅⩾𝑣₁(𝑡)=𝑇𝑣₀(𝑡)=(𝑇0)(𝑡)⩾0=𝑣₀(𝑡),𝑡∈[0,1],𝑅⩾𝑣^′₁(𝑡)=𝑇𝑣₀^′(𝑡)=(𝑇0)^′(𝑡)⩾0=𝑣^′₀(𝑡),𝑡∈[0,1],𝑅⩾−𝑣₁^′′(𝑡)=−𝑇𝑣₀^′′(𝑡)=−(𝑇0)^′′(𝑡)⩾0=−𝑣₀^′′(𝑡),𝑡∈[0,1],𝑅⩾−𝑣₁^′′′(𝑡)=−𝑇𝑣₀^′′′(𝑡)=−(𝑇0)^′′′(𝑡)⩾0=−𝑣₀^′′′(𝑡),𝑡∈[0,1]. (26) So by (𝐻1), we have 𝑣₂(𝑡)=𝑇𝑣₁(𝑡)⩾𝑇𝑣₀(𝑡)=𝑣₁(𝑡)[],𝑣,𝑡∈0,1_′2(𝑡)=𝑇𝑣₁^′(𝑡)⩾𝑇𝑣₀^′(𝑡)=𝑣^′₁(𝑡),𝑡∈[0,1],−𝑣₂^′′(𝑡)=−𝑇𝑣₁^′′(𝑡)⩾−𝑇𝑣₀^′′(𝑡)=−𝑣₁^′′(𝑡),𝑡∈[0,1],−𝑣₂^′′′(𝑡)=−𝑇𝑣₁^′′′(𝑡)⩾−𝑇𝑣₀^′′′(𝑡)=−𝑣₁^′′′(𝑡),𝑡∈[0,1]. (27) By the induction, we further obtain 𝑣_𝑛+1(𝑡)⩾𝑣_𝑛(𝑡),𝑣^′_𝑛+1(𝑡)⩾𝑣^′_𝑛(𝑡),−𝑣^′′_𝑛+1(𝑡)⩾−𝑣_𝑛^′′(𝑡),−𝑣^′′′_𝑛+1(𝑡)⩾−𝑣_𝑛^′′′(𝑡),𝑡∈[0,1],𝑛=0,1,2,…. (28) Hence, there exists 𝑣^∗∈𝐾_𝑅 such that lim_𝑛→∞‖𝑣_𝑛−𝑣^∗‖=0, and 𝑣^∗ is a monotone positive solution of problem (1)-(2). The proof is completed.

Corollary 6. Suppose that (𝐻₁) and (𝐻₃) hold. Assume (𝐻4)lim_𝑥→+∞max_0⩽𝑡⩽1𝑓(𝑡,𝑥,𝑥,−𝑥,−𝑥)/𝑥<Λ.
Then problem (1)-(2) has at least two monotone positive solutions 𝑤^∗ and 𝑣^∗ satisfying the conclusion of Theorem 5.

Remark 7. 𝑤^∗ and 𝑣^∗ may coincide with each other and then the problem (1)-(2) has only one monotone positive solution in 𝐾_𝑅.

Remark 8. By the proof of Theorem 5, {𝑤_𝑛}^∞_𝑛=1 and {𝑣_𝑛}^∞_𝑛=1 converge in 𝐶³[0,1] to the monotone positive solution 𝑤^∗ and 𝑣^∗ of the problem (1)-(2), respectively. In addition, the sequences {𝑤_𝑛}^∞_𝑛=1,{𝑣_𝑛}^∞_𝑛=1,{𝑤^′_𝑛}^∞_𝑛=1, and {𝑣^′_𝑛}^∞_𝑛=1 are increasing, while the sequences {𝑤_𝑛^′′}^∞_𝑛=1,{𝑣_𝑛^′′}^∞_𝑛=1,{𝑤_𝑛^′′′}^∞_𝑛=1, and {𝑣_𝑛^′′′}^∞_𝑛=1 are decreasing.

Remark 9. The iterative schemes in Theorem 5 start off with a known cubic function or the zero function, which is feasible for the computational purpose.

4. Examples

In this section we provide two examples to illustrate the application of our results.

Example 1. Consider the problem 𝑢^′′′′(𝑡)=121𝑡+𝑢18²(𝑡)+16𝑢^′(𝑡)−16𝑢^′′(𝑡)−16𝑢^′′′(𝑡)[],𝑢(0),𝑡∈0,1=𝑢^′(1)=𝑢^′′(0)=𝑢^′′′(1)=0. (29) For this problem, 𝑓(𝑡,𝑥,𝑦,𝑧,𝑤)=𝑡+(1/9)𝑥²+(1/3)𝑦−(1/3)𝑧−(1/3)𝑤 and 𝑞(𝑡)=1/2. By the calculations, the conditions (𝐴1),(𝐴2), and (𝐻1)--(𝐻3) of Theorem 5 hold for 𝑅=6. From Theorem 5, we have that problem (29) has two monotone positive solutions 𝑤^∗ and 𝑣^∗ satisfying ‖𝑤^∗‖⩽6,‖𝑣^∗‖⩽6 and lim_𝑛→∞‖𝑤_𝑛−𝑤^∗‖=0,lim_𝑛→∞‖𝑣_𝑛−𝑣^∗‖=0.
Moreover, the two iterative schemes are 𝑤₀(𝑡)=3𝑡−𝑡³[],𝑤,𝑡∈0,1_𝑛+1(𝑡)=112^𝑡₀(𝑡−𝑠)³×1𝑠+9𝑤²_𝑛(𝑠)+13𝑤^′_𝑛(𝑠)−13𝑤_𝑛^′′(𝑠)−13𝑤_𝑛^′′′(𝑠+𝑡)𝑑𝑠12¹₀6𝑠−3𝑠²−𝑡²×1𝑠+9𝑤²_𝑛(𝑠)+13𝑤^′_𝑛(𝑠)−13𝑤_𝑛^′′(𝑠)−13𝑤_𝑛^′′′(𝑠[]𝑣)𝑑𝑠,𝑡∈0,1,𝑛=0,1,2,…,₀(𝑡)[],𝑣=0,𝑡∈0,1_𝑛+1(𝑡)=112^𝑡₀(𝑡−𝑠)³×1𝑠+9𝑣²_𝑛(𝑠)+13𝑣^′_𝑛(𝑠)−13𝑣_𝑛^′′(𝑠)−13𝑣_𝑛^′′′(𝑠)+𝑡𝑑𝑠12¹₀6𝑠−3𝑠²−𝑡²×1𝑠+9𝑣²_𝑛(𝑠)+13𝑣^′_𝑛(𝑠)−13𝑣_𝑛^′′(𝑠)−13𝑣_𝑛^′′′(𝑠)[]𝑑𝑠,𝑡∈0,1,𝑛=0,1,2,…. (30)
Using Matlab 7.5, the second and third terms of the two schemes are as follows: 𝑤₁(𝑡)=35594536𝑡−2761𝑡7560³+1𝑡16⁴+1𝑡80⁵−1𝑡5040⁸+1𝑡90720¹⁰[],𝑤,𝑡∈0,1₂(𝑡)=548𝑡−33627935𝑡1111065984³−134954777𝑡11110659840⁵[phone]𝑡8888527872⁶[phone]𝑡9258883200⁷−22814857𝑡7407106560⁸−31759981𝑡24690355200⁹+3292141𝑡4115059200¹⁰−1777𝑡32659200¹¹−735019𝑡35554111488¹²+48893𝑡49380710400¹³−1227799𝑡888852787200¹⁴+280289𝑡444426393600¹⁵+10099𝑡148142131200¹⁶−223𝑡5290790400¹⁷+1𝑡1567641600¹⁸+19𝑡98761420800¹⁹+1𝑡10973491200²⁰+19𝑡444426393600²¹−1𝑡98761420800²²−1𝑡888852787200²³+1𝑡3555411148800²⁴[],𝑣,𝑡∈0,1₁(𝑡)=5148𝑡−𝑡24³+1𝑡240⁵[],𝑣,𝑡∈0,1₂(𝑡)=548𝑡−5159𝑡124416³+4859𝑡1244160⁵+25𝑡995328⁶+1𝑡7776⁷−5𝑡248832⁸−19𝑡622080⁹+1𝑡165888¹⁰+11𝑡3110400¹¹−1𝑡1244160¹²−1𝑡6220800¹³+1𝑡24883200¹⁴,𝑡∈[0,1]. (31)

Example 2 (see [22]). Consider the problem 𝑢^′′′′(𝑡)=min{𝑡,1−𝑡}48||×𝑢(𝑡)1−2𝑡2++𝑢^′(𝑡)³−𝑢^′′(𝑡)+𝑢^′′′(𝑡)³,1𝑡∈0,212,𝑢(0),1=𝑢^′(1)=𝑢^′′(0)=𝑢^′′′(1)=0. (32) In this problem, 𝑓(𝑡,𝑥,𝑦,𝑧,𝑤)=(1/48)[2+(𝑥+𝑦)³−(𝑧+𝑤)³] and 𝑞(𝑡)=(min{𝑡,1−𝑡})/|1−2𝑡|. Obviously, 𝑞(𝑡) and 𝑓(𝑡,𝑥,𝑦,𝑧,𝑤) satisfy conditions (𝐴1) and (𝐴2). In addition, 𝑓(𝑡,𝑥,𝑦,𝑧,𝑤) is increasing with regard to 𝑥 and 𝑦 and is decreasing with 𝑧 and 𝑤, and 𝑓(𝑡,0,0,0,0)≢0. Moreover, max_0⩽𝑡⩽1𝑓1𝑡,,130130,−130,−=1302482+<333752×1,30₁₀𝑞(𝑡)𝑑𝑡=¹₀min{𝑡,1−𝑡}||21−2𝑡𝑑𝑡=3. (33) Therefore, assumptions (𝐻1)--(𝐻3) of Theorem 5 are satisfied with 𝑅=1/30. From Theorem 5, problem (32) has two monotone positive solutions 𝑤^∗ and 𝑣^∗ satisfying ‖𝑤^∗‖⩽1/30,‖𝑣^∗‖⩽1/30 and lim_𝑛→∞‖𝑤_𝑛−𝑤^∗‖=0, lim_𝑛→∞‖𝑣_𝑛−𝑣^∗‖=0.
Moreover, the two iterative schemes are 𝑤₀(𝑡)=11803𝑡−𝑡³[],𝑤,𝑡∈0,1_𝑛+1(𝑡)=1144^𝑡₀(𝑡−𝑠)³min{𝑠,1−𝑠}||×𝑤1−2𝑠2+_𝑛(𝑠)+𝑤^′_𝑛(𝑠)³−𝑤_𝑛^′′(𝑠)+𝑤_𝑛^′′′(𝑠)³+𝑡𝑑𝑠144¹₀6𝑠−3𝑠²−𝑡²min{𝑠,1−𝑠}||×1−2𝑠2+𝑤_𝑛(𝑠)+𝑤^′_𝑛(𝑠)³−𝑤_𝑛^′′(𝑠)+𝑤_𝑛^′′′(𝑠)³[]𝑣𝑑𝑠,𝑡∈0,1,𝑛=0,1,2,…,₀(𝑡)[],𝑣=0,𝑡∈0,1_𝑛+1(𝑡)=1144^𝑡₀(𝑡−𝑠)³min{𝑠,1−𝑠}||×𝑣1−2𝑠2+_𝑛(𝑠)+𝑣^′_𝑛(𝑠)³−𝑣_𝑛^′′(𝑠)+𝑣_𝑛^′′′(𝑠)³+𝑡𝑑𝑠144¹₀6𝑠−3𝑠²−𝑡²min{𝑠,1−𝑠}||×𝑣1−2𝑠2+_𝑛(𝑠)+𝑣^′_𝑛(𝑠)³−𝑣_𝑛^′′(𝑠)+𝑣_𝑛^′′′(𝑠)³[]𝑑𝑠,𝑡∈0,1,𝑛=0,1,2,…. (34)
Using Mathlab 7.5, the second terms of the two schemes are given as follows:

𝑤₁(𝑡)=−8449074379661+23951656146600000441468132942481𝑡−1951616426760000010842948935033𝑡1951616426760000²−785738110303𝑡169705776240000³[phone]+23951656146600000122149319510627𝑡+602213068828800002602305979889593𝑡702581913633600000²−265540119476783𝑡150553267207200000³−13518333667229𝑡19161324917280000⁴−1855160149𝑡117096985605600000⁵−74353511𝑡17564547840840000⁶−26984453𝑡175645478408400000⁷+1291391𝑡117096985605600000⁸−4131727𝑡38322649834560000⁹−31823𝑡2105640100800000¹⁰+3953𝑡491316023520000¹¹+509𝑡216757069200000¹²+1𝑡5704133400000¹³×√11−2𝑡,for𝑡∈0,2,𝑤₁(𝑡)=−8449074379661+23951656146600000441468132942481𝑡−1951616426760000010842948935033𝑡1951616426760000²−785738110303𝑡169705776240000³+−929396364017203+210774574090080000062633084321047𝑡−22910279792400000817863130365013𝑡140516382726720000²+2416385432114987𝑡526936435225200000³−10620690488597𝑡15055326720720000⁴−568554781𝑡117096985605600000⁵−686043443𝑡175645478408400000⁶−546781𝑡3054703972320000⁷[phone]𝑡234193971211200000⁸−1594511𝑡19161324917280000⁹−23767𝑡867028276800000¹⁰+11287𝑡2456580117600000¹¹+13𝑡6193059120000¹²+1𝑡5704133400000¹³×√12𝑡−1,for𝑡∈2,𝑣,1₁(𝑡)1=−[phone]𝑡−𝑡180²−1𝑡216³+1−283523111340𝑡+𝑡270²−1𝑡576³−2𝑡2835⁴√1−2𝑡,for𝑡∈0,12,𝑣₁(𝑡)1=−[phone]𝑡−𝑡180²−1𝑡216³+−1+22683111340𝑡−11𝑡1890²+13𝑡2835³−2𝑡2835⁴√2𝑡−1,for𝑡∈12,1. (35)

Remark 10. This example was carried out firstly in [22], in which the author showed that the solution 𝑢^∗∈𝐶³[0,1] is nontrivial, ‖𝑢^∗‖⩽1, and the iterative scheme starts off with 𝑢₀(𝑡)7=−24¹₀𝐺(){}𝑡,𝑠min𝑠,1−𝑠|1−2𝑠|𝑑𝑠. (36)

In the present paper, we proved that the solution 𝑢^∗∈𝐶³[0,1] is monotone positive, ‖𝑢^∗‖⩽1/30, and the iterative scheme starts off with 𝑢₀(𝑡)=(1/180)(3𝑡−𝑡³) or 𝑢₀(𝑡)=0, which is much more simple than that of [22].

Acknowledgments

The authors would like to thank the reviewer for his/her careful reading, suggestions, and comments which improved the presentation of this paper. This work is supported by the Natural Science Foundation of Zhejiang Province of China (LY12A01012).