Positive Solutions for Third-Order Boundary-Value Problems with the Integral Boundary Conditions and Dependence on the First-Order Derivatives

Abstract

By using a fixed point theorem in a cone and the nonlocal third-order BVP's Green function, the existence of at least one positive solution for the third-order boundary-value problem with the integral boundary conditions 𝑥^′′′(𝑡)+𝑓(𝑡,𝑥(𝑡),𝑥^′(𝑡))=0, 𝑡∈𝐽, 𝑥(0)=0, 𝑥^′′(0)=0, and 𝑥(1)=∫¹₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡 is considered, where 𝑓 is a nonnegative continuous function, 𝐽=[0,1], and 𝑔∈𝐿[0,1]. The emphasis here is that 𝑓 depends on the first-order derivatives.

1. Introduction

Third-order boundary-value problems for differential equation play a very important role in a variety of different areas of applied mathematics and physics. Recently, third-order boundary-value problems have been many scholars' research object. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can produce boundary-value problems with integral boundary conditions [1–3]. For more information about the general theory of integral equations and their relation with boundary-value problems, we refer readers to the books of Corduneanu [4] and Agarwal and O'Regan [5].

Moreover, boundary-value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint, and nonlocal boundary-value problems as special cases. Such kind of BVPs in Banach space has been studied by some researchers [6–8].

By the fixed point index theory in cones [9], Zhang et al. [10] investigated the multiplicity of positive solutions for a class of nonlinear boundary-value problems of fourth-order differential equations with integral boundary conditions in ordered Banach spaces. Feng et al. [11] investigated the existence and multiplicity of positive solutions for a class of nonlinear boundary-value problems of second-order differential equations with integral boundary conditions in ordered Banach spaces. Guo et al. [12] investigated the existence of positive solutions for the third-order boundary-value problems with integral boundary conditions and dependence on the second derivatives. In [13], by using the fixed point theorem of cone expansion and compression of norm type, Zhang and Ge proved the existence and multiplicity of symmetric positive solutions for the fourth-order boundary-value problems with integral boundary conditions. By using Krasnoselskii's fixed point theorem, Wang et al. [14] investigated the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation with integral boundary conditions 𝑥⁽⁴⁾(𝑡)=𝜔(𝑡)𝑓𝑡,𝑥(𝑡),𝑥^′′(𝑡)𝑥(0),0<𝑡<1,=¹₀ℎ₁(𝑠)𝑥(𝑠)𝑥(1)=𝑑𝑠,₁₀𝑘₁(𝑠)𝑥(𝑠)𝑥𝑑𝑠,_′′(0)=¹₀ℎ₂(𝑠)𝑥^′′(𝑠)𝑥𝑑𝑠,_′′(1)=¹₀𝑘₂(𝑠)𝑥^′′(𝑠)𝑑𝑠, (1) where the arguments are based on Krasnoselskii's fixed point theorem for operators on a cone.

However, Zhao et al. [15] investigated the following third-order boundary-value problem with integral boundary conditions: 𝑥^′′′(𝑡)((𝑡𝑥(0)+𝑓𝑡,𝑥))=𝜃,𝑡∈𝐽,=𝜃,𝑥^′′(0)𝑥(1)==𝜃,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡, (2) under the assumptions(1)𝐽=[0,1], and 𝜃 is the zero element of 𝐸, (2)𝑓∶𝐶([0,1]×𝑃,𝑃), and 𝑔∈𝐿[0,1] is nonnegative,where 𝑃 is a cone in the real Banach 𝐸.

All the above works were done under the assumption that the first-order derivative 𝑥^′ is not involved explicitly in the nonlinear term 𝑓. In this paper, we are concerned with the existence of positive solutions for the third-order boundary-value problem with the integral boundary conditions 𝑥^′′′(𝑡)+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑥(0)=0,𝑡∈𝐽,=0,𝑥^′′(0)𝑥(1)==0,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡. (3)

Throughout, we assume(𝐻₁)  𝐽=[0,1], 𝑓∶[0,1]×𝑅²→𝑅⁺ is continuous, 𝑔∈𝐿[0,1], 𝑔(𝑡)≥0, and 𝜎∈[0,1), where 𝜎=∫¹₀𝑠𝑔(𝑠)𝑑𝑠.

To show the existence of positive solutions for (3), we define two positive continuous convex functionals. Then, by using the fixed point theorem [16] in a cone and the nonlocal third-order BVP's Green function, we give some new criteria for the existence of positive solutions for (3).

2. Preliminaries

Let 𝑌=𝐶[0,1] be the Banach space equipped with the norm ‖𝑥‖₀=max_𝑡∈[0,1]|𝑥(𝑡)|.

Lemma 1 (see [15]). Suppose (𝐻₁) holds. Then for any 𝑦(𝑡)∈𝐶[0,1], the problem 𝑥^′′′(𝑡)(𝑡)𝑥(0)+𝑦=0,𝑡∈𝐽,=0,𝑥^′′(0)𝑥(1)==0,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡 (4) has a unique solution 𝑥(𝑡)=¹₀𝐻(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (5) where 𝐻()()+𝑡𝑡,𝑠=𝐺𝑡,𝑠1−𝜎¹₀𝐺()𝑔(𝜏)𝐺()=1𝜏,𝑠𝑑𝜏,𝑡,𝑠2𝑡(1−𝑠)²−12(𝑡−𝑠)²1,0≤𝑠≤𝑡≤1,2𝑡(1−𝑠)²,0≤𝑡≤𝑠≤1. (6)

Lemma 2 (see [15]). For 𝑡,𝑠∈[0,1], one has 0≤𝐺(𝑡,𝑠)≤max_{0≤𝑡,𝑠≤1}𝐺(𝑡,𝑠)≤1/8.

Remark 3. When 𝑡, 𝑠∈(0,1), it is easy to check that 𝐺(𝑡,𝑠)>0.
In addition, for 0≤𝑠≤𝑡≤1, the maximum of 𝐺(𝑡,𝑠) occurs at 𝑡=(1+𝑠²)/2.

Lemma 4 (see [15]). Choose 𝛿∈(0,1/2) and 𝐽_𝛿=[𝛿,1−𝛿]; then for all 𝑡∈𝐽_𝛿, 𝑣, 𝑠∈[0,1], one has 𝐺(𝑡,𝑠)≥𝜌𝐺(𝑣,𝑠), (7) where 𝜌=4𝛿²(1−𝛿).

Remark 5. For 0≤𝑠≤𝑡≤1, denote 𝐺(𝑡,𝑠)=𝐺₁(𝑡,𝑠). Notice that 𝐺₁(𝑡,𝑠) is concave with respect to 𝑡; we have min_{𝑡∈𝐽𝛿,0≤𝑠≤𝑡}𝐺₁(𝑡,𝑠)=min𝐺₁(𝛿,𝑠),𝐺₁(=11−𝛿,𝑠)2𝛿²(1−𝛿). (8)

Lemma 6 (see [15]). Assume that (𝐻₁) holds; then (i)𝐻(𝑡,𝑠)≤(1/2)𝛾, 𝑡∈[0,1], (ii)𝐻(𝑡,𝑠)≥𝜌𝐻(𝑣,𝑠), 𝑡∈𝐽_𝛿, 𝑣,𝑠∈[0,1],where 𝛾=(1+∫¹₀(1−𝑠)𝑔(𝑠)𝑑𝑠)/(1−𝜎).

Lemma 7. If 𝑦∈𝐶[0,1], 𝑦(𝑡)≥0, then the unique solution 𝑥(𝑡) of problem (4) satisfies min_{𝑡∈𝐽𝛿}𝑥(𝑡)≥𝜌‖𝑥‖₀. (9)

Proof. By Lemmas 4 and 6 and (5), we get min_{𝑡∈𝐽𝛿}𝑥(𝑡)=min_{𝑡∈𝐽𝛿}¹₀𝐻(𝑡,𝑠)𝑦(𝑠)𝑑𝑠≥𝜌¹₀𝐻()𝑦(𝑠)(𝑣).𝑣,𝑠𝑑𝑠≥𝜌𝑥 (10)
For 𝑣∈[0,1], we have min_{𝑡∈𝐽𝛿}𝑥(𝑡)≥𝜌𝑥(𝑣). (11)
So, min_{𝑡∈𝐽𝛿}𝑥(𝑡)≥𝜌max_[]𝑣∈0,1𝑥(𝑣)=𝜌max_[]𝑣∈0,1|𝑥(𝑣)|=𝜌‖𝑥‖₀. (12)
The proof is completed.

Let 𝑋 be a Banach space and 𝐾⊂𝑋 a cone. Suppose 𝛼, 𝛽∶𝑋→𝑅⁺ are two continuous convex functionals satisfying 𝛼(𝜆𝑥)=|𝜆|𝛼(𝑥), 𝛽(𝜆𝑥)=|𝜆|𝛽(𝑥), for 𝑥∈𝑋, 𝜆∈𝑅, ‖𝑥‖≤𝑀max{𝛼(𝑥),𝛽(𝑥)}, for 𝑥∈𝑋, and 𝛼(𝑥)≤𝛼(𝑦) for 𝑥, 𝑦∈𝐾, 𝑥≤𝑦, where 𝑀>0 is a constant.

Theorem 8 (see [16]). Let 𝑟₂>𝑟₁>0, 𝐿>0 be constants and Ω_𝑖=𝑥∈𝑋∶𝛼(𝑥)<𝑟_𝑖,𝛽(𝑥)<𝐿,𝑖=1,2, (13)
two bounded open sets in 𝑋. Set 𝐷_𝑖=𝑥∈𝑋∶𝛼(𝑥)=𝑟_𝑖,𝑖=1,2. (14)
Assume 𝑇∶𝐾→𝐾 is a completely continuous operator satisfying (𝐴₁)𝛼(𝑇𝑥)<𝑟₁,  𝑥∈𝐷₁∩𝐾; 𝛼(𝑇𝑥)>𝑟₂, 𝑥∈𝐷₂∩𝐾; (𝐴₂)𝛽(𝑇𝑥)<𝐿, 𝑥∈𝐾; (𝐴₃) there is a 𝑝∈(Ω₂∩𝐾)⧵{0} such that 𝛼(𝑝)≠0 and 𝛼(𝑥+𝜆𝑝)≥𝛼(𝑥), for all 𝑥∈𝐾 and 𝜆≥0.
Then 𝑇 has at least one fixed point in (Ω₂⧵Ω₁)∩𝐾.

3. Main Results

Let 𝑋=𝐶¹[0,1] be the Banach space equipped with the norm ‖𝑥‖=max_𝑡∈[0,1]|𝑥(𝑡)|+max_𝑡∈[0,1]|𝑥^′(𝑡)|, and 𝐾={𝑥∈𝑋∶𝑥(𝑡)≥0, min_{𝑡∈𝐽𝛿}𝑥(𝑡)≥𝜌‖𝑥‖₀} is a cone in 𝑋.

Define two continuous convex functionals 𝛼(𝑥)=max_𝑡∈[0,1]|𝑥(𝑡)| and 𝛽(𝑥)=max_𝑡∈[0,1]|𝑥^′(𝑡)|, for each 𝑥∈𝑋; then ‖𝑥‖≤2max{𝛼(𝑥),𝛽(𝑥)} and 𝛼(𝜆𝑥)=|𝜆|𝛼(𝑥), 𝛽(𝜆𝑥)=|𝜆|𝛽(𝑥), for 𝑥∈𝑋, 𝜆∈𝑅; 𝛼(𝑥)≤𝛼(𝑦) for 𝑥, 𝑦∈𝐾, 𝑥≤𝑦.

In the following, we denote 𝜂₀=18+¹₀11−𝜎¹₀𝐺()𝑔(𝜏)𝜂𝜏,𝑠𝑑𝜏𝑑𝑠,₁=max_{[]𝑣∈0,1𝛿}^1−𝛿𝐻()𝜂𝑣,𝑠𝑑𝑠,₂=23+¹₀11−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏𝑑𝑠. (15)

We will suppose that there are 𝐿>𝑏>𝜌𝑏>𝑐>0 such that 𝑓(𝑡,𝑥,𝑦) satisfies the following growth conditions: (𝐻₂)𝑓(𝑡,𝑥,𝑦)<𝑐/𝜂₀, for (𝑡,𝑥,𝑦)∈[0,1]×[0,𝑐]×[−𝐿,𝐿], (𝐻₃)𝑓(𝑡,𝑥,𝑦)≥𝑏/𝜌𝜂₁, for (𝑡,𝑥,𝑦)∈[𝛿,1−𝛿]×[𝜌𝑏,𝑏]×[−𝐿,𝐿], (𝐻₄)𝑓(𝑡,𝑥,𝑦)<𝐿/𝜂₂, for (𝑡,𝑥,𝑦)∈[0,1]×[0,𝑏]×[−𝐿,𝐿].

Let 𝑓^∗=𝑓,∈[]×[]×(),𝑓,∈[]×()×(),𝑓𝑡,𝑥,𝑦𝑡,𝑥,𝑦𝑡,𝑥,𝑦0,10,𝑏−∞,∞𝑡,𝑏,𝑦𝑡,𝑥,𝑦0,1𝑏,∞−∞,∞₁=𝑓𝑡,𝑥,𝑦_∗,∈[]×[)×[],𝑓𝑡,𝑥,𝑦𝑡,𝑥,𝑦0,10,∞−𝐿,𝐿_∗(),∈[]×[)×(],𝑓𝑡,𝑥,−𝐿𝑡,𝑥,𝑦0,10,∞−∞,−𝐿_∗(𝑡,𝑥,𝐿),𝑡,𝑥,𝑦∈[0,1]×[0,∞)×[𝐿,∞). (16) We denote (𝑇𝑥)(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥,𝑥^′()𝑑𝑠,𝑇𝑥^′(𝑡)=¹₀𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥,𝑥^′𝑑𝑠. (17)

Lemma 9. Suppose (𝐻₁) holds. Then 𝑇∶𝐾→𝐾 is completely continuous.

Proof. For 𝑥∈𝐾, by Lemmas 2 and 4, it is obviously that 𝑇𝑥≥0.
By Lemma 7, we have min_{𝑡∈𝐽𝛿}𝑇𝑥(𝑡)≥𝜌‖𝑇𝑥‖₀. (18) So, we can get 𝑇(𝐾)⊂𝐾.
In the following, we will show that 𝑇∶𝐾→𝐾 is completely continuous.
At first we show that 𝑇∶𝐾→𝐾 is continuous.
Let 𝑥_𝑛, 𝑥^∗∈𝐾, it satisfies ‖𝑥_𝑛−𝑥^∗‖→0, (𝑛→∞), and then there is a constant 𝑀₀>0, such that max_𝑡∈[0,1]{|𝑥_𝑛(𝑡)|,|𝑥^∗(𝑡)|,|𝑥^′_𝑛(𝑡)|,|𝑥^∗′(𝑡)|}≤𝑀₀; then 𝑇𝑥_𝑛(𝑡)−𝑇𝑥^∗𝑡=()₁₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥_𝑛,𝑥^′_𝑛𝑑𝑠−¹₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥^∗,𝑥^∗′≤𝑑𝑠₁₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥_𝑛,𝑥^′_𝑛−𝑓₁𝑠,𝑥^∗,𝑥^∗′𝑑𝑠,𝑇𝑥_𝑛^′(𝑡)−𝑇𝑥^∗′(𝑡=₁₀)𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥,𝑥^′_𝑛−𝑑𝑠₁₀𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥^∗,𝑥^∗′≤𝑑𝑠₁₀𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥,𝑥^′_𝑛−𝑓₁𝑠,𝑥^∗,𝑥^∗′<𝑑𝑠₁₀12(1−𝑠)²+(×𝑓1−𝑠)₁𝑠,𝑥,𝑥^′_𝑛−𝑓₁𝑠,𝑥^∗,𝑥^∗′+𝑑𝑠₁₀11−𝜎¹₀𝐺()𝑔(𝜏)×𝑓𝜏,𝑠𝑑𝜏₁𝑠,𝑥,𝑥^′_𝑛−𝑓₁𝑠,𝑥^∗,𝑥^∗′𝑑𝑠. (19)
By 𝑓 which is uniformly continuous on [0,1]×[−𝑀₀,𝑀₀]×[−𝑀₀,𝑀₀], we get 𝑇𝑥_𝑛−𝑇𝑥^∗⟶0,(𝑛⟶∞). (20)
Next we show that 𝑇∶𝐾→𝐾 is compact.
Let 𝐵⊂𝐾 be bounded; then there is 𝑀>0, such that ‖𝑥‖≤𝑀. For 𝑥∈𝐵, we have |(𝑇𝑥)(𝑡)|=¹₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥,𝑥^′≤𝑑𝑠₁₀12𝛾𝑓₁𝑠,𝑥,𝑥^′=1𝑑𝑠2¹₀1+∫¹₀(1−𝑠)𝑔(𝑠)𝑑𝑠1−𝜎𝑑𝑠×𝐶^∗, (21) where 𝐶^∗=max{|𝑓₁(𝑡,𝑥,𝑥^′)|;𝑡∈[0,1],𝑥∈𝐵}.
Consider (𝑇𝑥)^′(𝑡=₁)₀𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥,𝑥^′=𝑑𝑠₁₀𝜕𝐺(𝑡,𝑠)+1𝜕𝑡1−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏×𝑓₁𝑠,𝑥,𝑥^′<𝑑𝑠₁₀12(1−𝑠)²+(+1−𝑠)𝑑𝑠₁₀11−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏𝑑𝑠×𝐶^∗=23+¹₀11−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏𝑑𝑠×𝐶^∗. (22)
It is clear that 𝑇(𝐵) is a bounded set in 𝐾, because 𝐻(𝑡,𝑠) is uniformly continuous on [0,1]×[0,1], for 𝜀>0, there exists 𝛿∈(0,𝜀), such that |𝐻(𝑡₁,𝑠)−𝐻(𝑡₂,𝑠)|<𝜀, and for 𝑡₁, 𝑡₂∈[0,1], |𝑡₁−𝑡₂|<𝛿.
For 𝑥∈𝐵, we have (𝑇𝑥)𝑡₁−(𝑇𝑥)𝑡₂=¹₀𝐻𝑡₁,𝑠𝑓₁𝑠,𝑥,𝑥^′−𝑑𝑠₁₀𝐻𝑡₂,𝑠𝑓₁𝑠,𝑥,𝑥^′≤𝑑𝑠₁₀𝐻𝑡₁,𝑠−𝐻𝑡₂,𝑠𝑑𝑠×𝐶^∗≤𝜀𝐶^∗,(𝑇𝑥)^′𝑡₁−(𝑇𝑥)^′𝑡₂=¹₀𝜕𝐻(𝑡,𝑠)𝜕𝑡_𝑡=𝑡1𝑓₁𝑠,𝑥,𝑥^′−𝑑𝑠₁₀𝜕𝐻(𝑡,𝑠)𝜕𝑡_𝑡=𝑡2𝑓₁𝑠,𝑥,𝑥^′=𝑑𝑠₁₀𝜕𝐺(𝑡,𝑠)𝜕𝑡_𝑡=𝑡1𝑓₁𝑠,𝑥,𝑥^′−𝑑𝑠₁₀𝜕𝐺(𝑡,𝑠)𝜕𝑡_𝑡=𝑡2𝑓₁𝑠,𝑥,𝑥^′=𝑑𝑠_𝑡10𝑠−𝑡₁𝑓₁𝑠,𝑥,𝑥^′+𝑑𝑠_𝑡20𝑡₂−𝑠𝑓₁𝑠,𝑥,𝑥^′≤1𝑑𝑠2𝑡₁−𝑡₂𝑡₁+𝑡₂×𝐶^∗≤𝜀𝐶^∗. (23)
Therefore 𝑇(𝐵) is equicontinuous. Using the Arzela-Ascoli theorem, a standard proof yields 𝑇∶𝐾→𝐾 which is completely continuous.

Theorem 10. Suppose (𝐻₁)--(𝐻₄) hold. Then BVP (3) has at least one positive solution 𝑥(𝑡) satisfying 𝑐<𝛼(𝑥)<𝑏,𝛽(𝑥)<𝐿. (24)

Proof. Take Ω₁=𝑥∈𝑋∶|𝑥(𝑡)|<𝑐,𝑥(𝑡)^′,Ω<𝐿₂=𝑥∈𝑋∶|𝑥(𝑡)|<𝑏,𝑥(𝑡)^′<𝐿, (25) two bounded open sets in 𝑋, and 𝐷₁={𝑥∈𝑋∶𝛼(𝑥)=𝑐},𝐷₂={𝑥∈𝑋∶𝛼(𝑥)=𝑏}. (26)
By Lemma 9, 𝑇∶𝐾→𝐾 is completely continuous, and there is a 𝑝∈(Ω₂∩𝐾)⧵{0} such that 𝛼(𝑝)≠0 and 𝛼(𝑥+𝜆𝑝)≥𝛼(𝑥) for all 𝑢∈𝐾 and 𝜆≥0.
By (𝐻₂), for 𝑥∈𝐷₁∩𝐾 and 𝛼(𝑥)=𝑐, we get 𝛼(𝑇𝑥)=max_𝑡∈[0,1]¹₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥,𝑥^′𝑑𝑠=max_[]𝑡∈0,1¹₀𝐺()+𝑡𝑡,𝑠1−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏×𝑓₁𝑠,𝑥,𝑥^′𝑑𝑠<¹₀max_[]𝑡∈0,1𝐺()+𝑡𝑡,𝑠1−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏×𝑓₁𝑠,𝑥,𝑥^′<𝑑𝑠₁₀18𝑑𝑠+¹₀𝑡1−𝜎¹₀𝐺()𝑔(𝜏)×𝑐𝜏,𝑠𝑑𝜏𝑑𝑠𝜂₀=18+¹₀𝑡1−𝜎¹₀𝐺()𝑔(𝜏)×𝑐𝜏,𝑠𝑑𝜏𝑑𝑠𝜂₀=𝑐. (27)
By Lemma 7, for 𝑥∈𝐷₂∩𝐾 and 𝛼(𝑥)=𝑏, there is 𝑥(𝑡)≥𝜌𝛼(𝑥)=𝜌𝑏, 𝑡∈𝐽_𝛿.
So, by (𝐻₃), we get 𝛼(𝑇𝑥)=max_𝑡∈[0,1]¹₀𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥,𝑥^′𝑑𝑠>_𝛿^1−𝛿𝐻(𝑡,𝑠)𝑓₁𝑠,𝑥,𝑥^′𝑑𝑠>𝜌_𝛿^1−𝛿𝐻()𝑏𝑣,𝑠𝑑𝑠×𝜌𝜂₁. (28)
For 𝑣∈[0,1], we have 𝛼(𝑇𝑥)>𝜌_𝛿^1−𝛿𝐻()𝑏𝑣,𝑠𝑑𝑠×𝜌𝜂₁. (29)
So, 𝛼(𝑇𝑥)>𝜌max_{[]𝑣∈0,1𝛿}^1−𝛿𝐻()𝑏𝑣,𝑠𝑑𝑠×𝜌𝜂₁=𝑏. (30)
By (𝐻₄), for 𝑥∈𝐾, we have 𝛽(𝑇𝑥)=max_[]𝑡∈0,1¹₀𝜕𝐻(𝑡,𝑠)𝑓𝜕𝑡¹𝑠,𝑥,𝑥^′<𝑑𝑠₁₀𝜕𝐺(𝑡,𝑠)+1𝜕𝑡1−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏×𝑓₁𝑠,𝑥,𝑥^′=𝑑𝑠_𝑡012(1−𝑠)²−(𝑡−𝑠)𝑓₁𝑠,𝑥,𝑥^′+𝑑𝑠_1𝑡12(1−𝑠)²𝑓₁𝑠,𝑥,𝑥^′+𝑑𝑠₁₀11−𝜎¹₀𝐺(𝜏,𝑠)𝑔(𝜏)𝑑𝜏×𝑓₁𝑠,𝑥,𝑥^′<𝑑𝑠₁₀12(1−𝑠)²+(+1−𝑠)𝑑𝑠₁₀11−𝜎¹₀𝐺()𝑔(𝜏)×𝐿𝜏,𝑠𝑑𝜏𝑑𝑠𝜂₂=23+¹₀11−𝜎¹₀𝐺()𝑔(𝜏)×𝐿𝜏,𝑠𝑑𝜏𝑑𝑠𝜂₂=𝐿. (31)
Theorem 8 implies there is 𝑥∈(Ω₂⧵Ω₁)∩𝐾 such that 𝑥=𝑇𝑥. So, 𝑥(𝑡) is a positive solution for BVP (3) satisfying 𝑐<𝛼(𝑥)<𝑏,𝛽(𝑥)<𝐿. (32)
Thus, Theorem 10 is completed.

4. Example

Example 1. Consider the following boundary-value problem 𝑥^′′′(𝑡)+𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)𝑥(0)=0,0<𝑡<1,=0,𝑥^′′(0)𝑥(1)==0,₁₀𝑥(𝑡)𝑑𝑡, (33) where 𝑓=𝑡𝑡,𝑥,𝑦3,∈[]×[]×[],𝑥+2𝑥+cos𝑦𝑡,𝑥,𝑦0,10,0.5−3667,3667109𝑡3()()+𝑡𝑥−0.5+25742𝑥−0.56,∈[]×[]×[],𝑡+1+cos𝑦𝑡,𝑥,𝑦0,10.5,0.6−3667,36673()()+,∈[]×[]×[].11−𝑥+222𝑥+11cos𝑦𝑡,𝑥,𝑦0,10.6,11−3667,3667 (34)
In this problem, we know that 𝑔(𝑡)=1; then we can get 𝜎=∫¹₀𝑠𝑔(𝑠)𝑑𝑠=1/2. Choose 𝛿=1/8∈(0,1/2); then 𝜌=4𝛿²(1−𝛿)=7/128.
Furthermore, we obtain 𝜂₀=524,𝜌𝜂₁=358192,𝜂₂=34. (35)
If we take 𝑐=0.5, 𝑏=11, and 𝐿=3667, then we get 𝜌𝑏≈0.601>0.6: 𝑓=𝑡𝑡,𝑥,𝑦3𝑐𝑥+2𝑥+cos𝑦≤2.17<𝜂₀≈2.4, (36) for (𝑡,𝑥,𝑦)∈[0,1]×[0,0.5]×[−3667,3667], 𝑓=𝑡𝑡,𝑥,𝑦3()()+𝑏11−𝑥+222𝑥+11cos𝑦≥2575.2>𝜌𝜂₁≈2574.1, (37) for (𝑡,𝑥,𝑦)∈[𝛿,1−𝛿]×[𝛿𝑏,11]×[−3667,3667], 𝑓𝐿𝑡,𝑥,𝑦≤4888.8<𝜂₂≈4889.3, (38) for (𝑡,𝑥,𝑦)∈[0,1]×[0,11]×[−3667,3667].
Then all the conditions of Theorem 10 are satisfied. Therefore, by Theorem 10 we know that boundary-value problem (33) has at least one positive solution 𝑥(𝑡) satisfying 0.5<𝛼(𝑥)<11,𝛽(𝑥)<3667. (39)

Acknowledgments

The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2013208147).