Positive Solutions for Singular Complementary Lidstone Boundary Value Problems

Abstract

By using fixed-point theorems of a cone, we investigate the existence and multiplicity of positive solutions for complementary Lidstone boundary value problems: (−1)^𝑛𝑢^(2𝑛+1)(𝑡)=ℎ(𝑡)𝑓(𝑢(𝑡)), in 0<𝑡<1, 𝑢(0)=0, 𝑢^(2𝑖+1)(0)=𝑢^(2𝑖+1)(1)=0, 0≤𝑖≤𝑛−1, where 𝑛∈𝑁.

1. Introduction

In this paper, we are concerned with the existence of positive solutions for the following nonlinear differential equation:(−1)^𝑛𝑢^(2𝑛+1)(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)=ℎ)),in0<𝑡<1,=0,𝑢^(2𝑖+1)(0)=𝑢^(2𝑖+1)(1)=0,0≤𝑖≤𝑛−1, (1.1) where 𝑛∈𝑁, ℎ(𝑡)∈𝐶((0,1),[0,+∞)), and ℎ(𝑡) may be singular at 𝑡=0 or 𝑡=1; 𝑓∈𝐶([0,+∞),[0,+∞)).

Recently, on the boundary value problems of 2𝑛th-order ordinary differential equation (system) (−1)^𝑛𝑢^(2𝑛)(𝑡)=𝜆ℎ(𝑡)𝑓(𝑢(𝑡)), (1.2) many authors have established the existence and multiplicity of positive solutions of (1.2) by means of the method of upper and lower solutions and fixed point theorem, see [1–7] and references therein. More recently, the complementary Lidstone problem: (−1)^𝑛𝑢^(2𝑛+1)(𝑡)=ℎ𝑡,𝑢(𝑡),…,𝑢^(𝑞)(𝑡)𝑢(0),𝑛≥1,𝑞fixed,0≤𝑞≤2𝑛in0<𝑡<1,=𝑎₀,𝑢^(2𝑖+1)(0)=𝑎_𝑖,𝑢^(2𝑖+1)(1)=𝛽_𝑖,0≤𝑖≤𝑛−1, (1.3) was discussed in [8]. Here, ℎ∶[0,1]×𝑅^𝑞+1→𝑅 is continuous at least in the interior of the domain of interest. Existence and uniqueness criteria for the above problem are proved by the complementary Lidstone interpolating polynomial of degree 2𝑛. In [9], the authors have studied the existence of positive solutions of singular complementary Lidstone problems on the basis of a fixed-point theorem of cone compression type. As far as we know, no papers are concerned with the multiplicity of positive solutions for (1.1). Therefore, inspired by the above references, we will show the existence and multiplicity of positive solutions of (1.1). The proof of our results is based on the following fixed-point theorems in a cone

Let 𝐸 be a real Banach space with norm ‖⋅‖ and 𝑃⊂𝐸 a cone in 𝐸, 𝑃_𝑟={𝑥∈𝑃∶‖𝑥‖<𝑟}(𝑟>0). Then 𝑃_𝑟={𝑥∈𝑃∶‖𝑥‖≤𝑟}. A map 𝛼 is said to be a nonnegative continuous concave functional on 𝑃 if 𝛼∶𝑃→[0,+∞) is continuous and 𝛼𝑡𝑥+(1−𝑡)𝑦≥𝑡𝛼(𝑥)+(1−𝑡)𝛼𝑦 (1.4) for all 𝑥,𝑦∈𝑃 and 𝑡∈[0,1]. For numbers 𝑎,𝑏 such that 0<𝑎<𝑏 and 𝛼 is a nonnegative continuous concave functional on 𝑃, we define the convex set 𝑃(𝛼,𝑎,𝑏)={𝑥∈𝑃∶𝑎≤𝛼(𝑥),‖𝑥‖≤𝑏}. (1.5)

Lemma 1.1 (see [10]). Let 𝐴∶𝑃_𝑐→𝑃_𝑐 be completely continuous and 𝛼 be a nonnegative continuous concave functional on 𝑃 such that 𝛼(𝑥)≤‖𝑥‖ for all 𝑥∈𝑃_𝑐. Suppose there exists 0<𝑑<𝑎<𝑏≤𝑐 such that(i){𝑥∈𝑃(𝛼,𝑎,𝑏)∶𝛼(𝑥)>𝑎}≠∅ and 𝛼(𝐴𝑥)>𝑎 for 𝑥∈𝑃(𝛼,𝑎,𝑏);(ii)‖𝐴𝑥‖<𝑑 for ‖𝑥‖≤𝑑;(iii)𝛼(𝐴𝑥)>𝑎 for 𝑥∈𝑃(𝛼,𝑎,𝑐) with ‖𝐴𝑥‖>𝑏.
Then 𝐴 has at least three fixed points 𝑥₁,𝑥₂,𝑥₃ satisfying 𝑥₁<𝑑,𝑎<𝛼𝑥₂,𝑥₃>𝑑,𝛼𝑥₃<𝑎. (1.6)

Lemma 1.2 (see [10]). Let 𝐸 be a Banach space, and let 𝑃⊂𝐸 be a closed, convex cone in 𝐸. Assume Ω₁, Ω₂ are bounded open subsets of 𝐸 with 0∈Ω₁,Ω₁⊂Ω₂, and let 𝐴∶𝑃∩(Ω₂⧵Ω₁)→𝑃 be a completely continuous operator such that either(i)‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝑃∩𝜕Ω₁ and ‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝑃∩𝜕Ω₂ or(ii)‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝑃∩𝜕Ω₁ and ‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝑃∩𝜕Ω₂.
Then 𝐴 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

This paper is organized as follows: in Section 2, some preliminaries are given; in Section 3, we give the existence results.

2. Preliminaries

First, it is clear to see that the boundary value problem (1.1),(−1)^𝑛𝑢^(2𝑛+1)(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)=ℎ)),in0<𝑡<1,=0,𝑢^(2𝑖+1)(0)=𝑢^(2𝑖+1)(1)=0,0≤𝑖≤𝑛−1, (2.1) is equivalent to the system𝑢^′(𝑡)()=𝑣,in0<𝑡<1,−1^𝑛𝑣^(2𝑛)(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)=ℎ)),in0<𝑡<1,=0,𝑣^(2𝑖)(0)=𝑣^(2𝑖)(1)=0,0≤𝑖≤𝑛−1. (2.2) Next, Problem (2.2) can be easily transformed into a nonlinear 2𝑛-order ordinary differential equation. Briefly, the initial value problem, 𝑢^′(𝑡)𝑢(0)=𝑣,in0<𝑡<1,=0, (2.3) can be solved as 𝑢(𝑡)=^𝑡₀𝑣(𝑠)𝑑𝑠. (2.4) Then, inserting (2.4) into the second equation of (1.1), we have(−1)^𝑛𝑣^(2𝑛)(𝑡)=ℎ(𝑡)𝑓^𝑡₀𝑣(𝑠)𝑣𝑑𝑠,in0<𝑡<1,_(2𝑖)(0)=𝑣^(2𝑖)(1)=0,0≤𝑖≤𝑛−1. (2.5) Finally, we only need to consider the existence of positive solutions of (2.5). The function 𝑣∈𝐶[0,1] is a positive solution of (2.5), if 𝑣 satisfies (2.5) and 𝑣≥0, 𝑡∈[0,1], 𝑣≠0.

Let 𝐺_𝑛(𝑡,𝑠) be the Greens function of the following problem: (−1)^𝑛𝜔^(2𝑛)(𝑡)𝜔=0,in0<𝑡<1,_(2𝑖)(0)=𝜔^(2𝑖)(1)=0,0≤𝑖≤𝑛−1. (2.6) By induction, the Greens function 𝐺_𝑛(𝑡,𝑠) can be expressed as (see [2]) 𝐺_𝑖(𝑡,𝑠)=¹₀𝐺(𝑡,𝜉)𝐺_𝑖−1(𝜉,𝑠)𝑑𝜉,2≤𝑖≤𝑛, (2.7) where 𝐺₁()()=𝑡()𝑠()𝑡,𝑠=𝐺𝑡,𝑠1−𝑠,0≤𝑡≤𝑠≤1,1−𝑡,0≤𝑠≤𝑡≤1. (2.8) So it is easy to see that 𝐺_𝑛(𝑡,𝑠)>0,(𝑡,𝑠)∈(0,1)×(0,1). (2.9)

Lemma 2.1 (see [2]). (I) For any (𝑡,𝑠)∈[0,1]×[0,1], 𝐺_𝑛()≤1𝑡,𝑠6^𝑛−1𝑠(1−𝑠). (2.10)(II) Let 𝛿∈(0,1/2), then for any (𝑡,𝑠)∈[𝛿,1−𝛿]×[0,1], 𝐺_𝑛(𝑡,𝑠)≥𝜃_𝑛(𝛿)𝑠(1−𝑠)≥6^𝑛−1𝜃_𝑛(𝛿)max_0≤𝑡≤1𝐺_𝑛(𝑡,𝑠), (2.11) where 𝜃_𝑛(𝛿)=𝛿^𝑛((4𝛿³−6𝛿²+1)/6)^𝑛−1.

Therefore, the solution of (2.5) can be expressed as 𝑣(𝑡)=¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠. (2.12) We now define a mapping 𝑇∶𝐶[0,1]→𝐶[0,1] by 𝑇𝑣(𝑡)=¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠. (2.13)

Set 𝑃=𝑣∈𝐶[0,1]∶𝑣(𝑡)≥0,min_{𝑡∈[1/4,3/4]}𝑣(𝑡)≥𝜎‖𝑣‖, (2.14) where ‖𝑣‖=max_𝑡∈[0,1]|𝑣(𝑡)|, 𝜎=6^𝑛−1𝜃_𝑛(1/4)=11^𝑛−1/2^6𝑛−4. It is clear that 𝜎∈(0,1) by Lemma 2.1.

Lemma 2.2 (see [2]). 𝑇∶𝑃→𝑃 is completely continuous.

3. Main Result

Theorem 3.1. Assume that the following conditions hold(H1)ℎ(𝑡)∈𝐶((0,1),[0,+∞)) does not vanish identically on any subinterval of [0,1], 0<^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠≤¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠<+∞ (3.1)(H2)𝑓∶[0,+∞)→[0,+∞) is nondecreasing, and lim_𝑟→0+𝑓(𝑟)𝑟=0,lim_𝑟→+∞𝑓(𝑟)𝑟=+∞. (3.2) Then (2.5) or (1.1) has at least one positive solution.

Proof. Since lim_𝑟→0⁺(𝑓(𝑟))/𝑟=0, there exists 𝜂₁>0 such that 𝑓(𝑟)𝑟≤6^𝑛−1∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠,for0<𝑟≤𝜂₁. (3.3) Take 𝑅₁=𝜂₁, and set Ω₁={𝑣∈𝐸∶‖𝑣‖<𝑅₁}. Then, for forall𝑣∈𝑃∩𝜕Ω₁, we have 𝑇𝑣(𝑡)=¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓𝑅₁≤1𝑑𝑠6^𝑛−16^𝑛−1𝑅₁∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠¹⁰𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠≤𝑅₁=‖𝑣‖ (3.4) by (H2) and Lemma 2.1.
Consequently, ‖𝑇𝑣‖≤‖𝑣‖,∀𝑣∈𝜕𝑃∩𝜕Ω₁. (3.5)
On the other hand, since lim_𝑟→+∞(𝑓(𝑟))/𝑟=+∞, there exists 𝑅>0 such that 𝑓(𝑟)𝑟≥4𝜎𝜃_𝑛(1/4)∫^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠,for𝑟≥𝑅. (3.6) Choose 𝑅₂=max{𝑅₁,4𝑅/𝜎}+1, and set Ω₂={𝑣∈𝐸∶‖𝑣‖<𝑅₂}. Then, for ∀𝑣∈𝑃∩𝜕Ω₂, we have 1𝑇𝑣4=¹₀𝐺_𝑛14,𝑠ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_1/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓₀^2/4𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^2/4_1/4𝑣(𝜏)𝑑𝜏𝑑𝑠 (3.7) by (H2) and Lemma 2.1.
Since 𝑣∈𝑃∩𝜕Ω₂, then we have ^2/4_1/4𝑣(𝜏)𝑑𝜏≥^2/4_1/4min_{𝜏∈[1/4,3/4]}𝑣(𝜏)𝜎𝑑𝜏≥4‖𝑣‖≥𝜎4𝑅₂>𝑅. (3.8) So from (3.7), we get 1𝑇𝑣4≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^2/4_1/4𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠()ℎ(𝑠)𝑓𝜎1−𝑠4𝑅₂𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠()ℎ(𝑠)41−𝑠𝑑𝑠𝜎𝜃_𝑛(1/4)∫^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠𝜎𝑅₂4≥𝑅₂=‖𝑣‖. (3.9)
Consequently, ‖𝑇𝑣‖≥‖𝑣‖,∀𝑣∈𝜕𝑃∩𝜕Ω₂. (3.10) Therefore, by Lemma 1.2, (1.1) has at least one positive solution.

Theorem 3.2. Assume (H1) holds. In addition, suppose that the following conditions hold:(H3)𝑓∶[0,+∞)→[0,+∞) is nondecreasing,lim_𝑟→0+𝑓(𝑟)𝑟=+∞,lim_𝑟→+∞𝑓(𝑟)𝑟=0. (3.11) Then (2.5) or (1.1) has at least one positive solution.

Proof. Since lim_𝑟→0⁺(𝑓(𝑟))/𝑟=+∞, there exists 𝜂>0 such that 𝑓(𝑟)𝑟≥4𝜎𝜃_𝑛(1/4)∫^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠,for0<𝑟≤𝜂. (3.12) Take 𝑅₁∈(0,𝜂), and set Ω₁={𝑣∈𝐸∶‖𝑣‖<𝑅₁}. Then, for forall𝑣∈𝑃∩𝜕Ω₁, we have 1𝑇𝑣4=¹₀𝐺_𝑛14,𝑠ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_1/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓₀^2/4𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^2/4_1/4𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^2/4_1/4min_{𝜏∈[1/4,3/4]}𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠()ℎ(𝑠)𝑓𝜎1−𝑠4𝑅₁𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠()ℎ(𝑠)41−𝑠𝑑𝑠𝜎𝜃_𝑛(1/4)∫^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠𝜎𝑅₁4≥𝑅₁=‖𝑣‖ (3.13) by (H3) and Lemma 2.1.
Consequently, ‖𝑇𝑣‖≥‖𝑣‖,∀𝑣∈𝜕𝑃∩𝜕Ω₁. (3.14)
On the other hand, since lim_𝑟→+∞(𝑓(𝑟))/𝑟=0, there exists 𝑅>0 such that 𝑓(𝑟)𝑟≤6^𝑛−1∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠,for𝑟≥𝑅. (3.15) Choose 𝑅₂=max{𝑅₁,𝑅}+1, and set Ω₂={𝑣∈𝐸∶‖𝑣‖<𝑅₂}. Then, for forall𝑣∈𝑃∩𝜕Ω₂, we have 𝑇𝑣(𝑡)=¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓𝑅₂≤1𝑑𝑠6^𝑛−16^𝑛−1𝑅₂∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠¹⁰𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠≤𝑅₂=‖𝑣‖. (3.16) Consequently, ‖𝑇𝑣‖≤‖𝑣‖,∀𝑣∈𝜕𝑃∩𝜕Ω₂. (3.17) Therefore, by Lemma 1.2, (1.1) has at least one positive solution.

Theorem 3.3. Assume that (H1) holds. In addition, the function 𝑓 is nondecreasing and satisfies the following growth conditions:(H4)lim_𝑟→∞𝑓(𝑟)sup𝑟<126^𝑛−1∫¹₀𝑠()ℎ(𝑠);1−𝑠𝑑𝑠 (3.18)(H5)lim_𝑟→0𝑓(𝑟)sup𝑟<6^𝑛−1∫¹₀𝑠()ℎ(𝑠);1−𝑠𝑑𝑠 (3.19)(H6) there exists a constant 𝑎>0 such that 𝑓(𝑟)>𝑎𝜃_𝑛(1/4)∫^3/4_2/4𝑠()ℎ(𝑠)𝑎1−𝑠𝑑𝑠,for𝑟∈4,𝑎.4𝜎 (3.20)Then (1.1) has at least three positive solutions.

Proof. For the sake of applying the Leggett-Williams fixed-point theorem, define a functional 𝛼(𝑢) on cone 𝑃 by 𝛼(𝑣)=min_{1/4≤𝑡≤3/4}𝑣(𝑡),∀𝑣∈𝑃. (3.21) Evidently, 𝛼∶𝑃→𝑅⁺ is a nonnegative continuous and concave. Moreover, 𝛼(𝑣)≤‖𝑣‖ for each 𝑣∈𝑃.
Now we verify that the assumption of Lemma 1.1 is satisfied.
Firstly, it can verify that there exists a positive number 𝑐 with 𝑐≥𝑏=𝑎/𝜎 such that 𝑇∶𝑃_𝑐→𝑃_𝑐.
By (H4), it is easy to see that there exists 𝜏>0 such that 𝑓(𝑟)𝑟<126^𝑛−1∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠,∀𝑟≥𝜏. (3.22)
Set 𝑀₁=𝑓(𝜏), and take 𝑐>max𝑏,2𝑀₁6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠. (3.23) If 𝑣∈𝑃_𝑐, then ‖𝑇𝑣‖=max_𝑡∈[0,1]¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠()ℎ(𝑠)𝑓𝑣<11−𝑠(‖‖)𝑑𝑠6^𝑛−11₀𝑠()ℎ(𝑠)61−𝑠𝑑𝑠^𝑛−1‖𝑣‖2∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠+𝑀₁<𝑐 (3.24) by (H1) and (H3).
Next, from (H5), there exists 𝑑∈(0,𝑎) such that 𝑓(𝑟)𝑟<6^𝑛−1∫¹₀𝑠()ℎ(𝑠)[].1−𝑠𝑑𝑠,∀𝑟∈0,𝑑 (3.25) Then for each 𝑣∈𝑃_𝑑, we have ‖𝑇𝑣‖=max_𝑡∈[0,1]¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)≤1𝑑𝜏𝑑𝑠6^𝑛−11₀𝑠()ℎ(𝑠)𝑓𝑣<11−𝑠(‖‖)𝑑𝑠6^𝑛−11₀𝑠()ℎ(𝑠)61−𝑠𝑑𝑠^𝑛−1‖𝑣‖∫¹₀𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠≤𝑑. (3.26)
Finally, we will show that {𝑣∈𝑃(𝛼,𝑎,𝑏)∶𝛼(𝑣)>𝑎}≠∅ and 𝛼(𝑇𝑣)>𝑎 for all 𝑣∈𝑃(𝛼,𝑎,𝑏).
In fact, 𝑣(𝑡)=𝑎+𝑏2∈{𝑣∈𝑃(𝛼,𝑎,𝑏)∶𝛼(𝑣)>𝑎}. (3.27) For 𝑣∈𝑃(𝛼,𝑎,𝑏), we have 𝑏≥‖𝑣‖≥𝑣≥min_{𝑡∈[1/4,3/4]}𝑣(𝑡)≥𝑎, (3.28) for all 𝑡∈[1/4,3/4]. Then we have min_{𝑡∈[1/4,3/4]}𝑇𝑣(𝑡)=min_{𝑡∈[1/4,3/4]}¹₀𝐺_𝑛(𝑡,𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_1/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^𝑠₀𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓₀^2/4𝑣(𝜏)𝑑𝜏𝑑𝑠≥𝜃_𝑛14^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑓^2/4_1/4𝑣(𝜏)𝑑𝜏𝑑𝑠>𝜃_𝑛14^3/4_2/4𝑠()ℎ(𝑠)𝑎1−𝑠𝑑𝑠𝜃_𝑛(1/4)∫^3/4_2/4𝑠(1−𝑠)ℎ(𝑠)𝑑𝑠=𝑎 (3.29) by (H6). In addition, for each 𝑣∈𝑃(𝛼,𝑎,𝑐) with ‖𝑇𝑣‖>𝑏, we have min_{𝑡∈[1/4,3/4]}(𝑇𝑣)(𝑡)≥𝜎‖𝑇𝑣‖>𝜎𝑏≥𝑎. (3.30)
Above all, we know that the conditions of Lemma 1.1 are satisfied. By Lemma 1.1, the operator 𝑇 has at least three fixed points 𝑣_𝑖(𝑖=1,2,3) such that 𝑣₁<𝑑,𝑎<min_{𝑡∈[1/4,3/4]}𝑣₂(𝑡)𝑣₃>𝑑withmin_{𝑡∈[1/4,3/4]}𝑣₃(𝑡)<𝑎. (3.31) The proof is complete.

Example 3.4. If 𝑛=1, then consider the boundary value problem: −𝑢^′′′(𝑡)=𝑢²𝑡()𝑢(0)1−𝑡,in0<𝑡<1,=0,𝑢^′(0)=𝑢^′(1)=0. (3.32)

Example 3.5. If 𝑛=2, then consider the boundary value problem: 𝑢⁽⁵⁾(𝑡)=𝑢^1/3𝑡()𝑢(0)(0)1−𝑡,in0<𝑡<1,=0,𝑢′=𝑢^′(1)=𝑢^′′′(0)=𝑢^′′′(1)=0. (3.33) It is obvious to see that Examples 3.4 and 3.5 satisfy the assumptions of Theorems 3.1 and 3.2.

Example 3.6. Let 𝑛=1, ℎ(𝑡)=1/(𝑡(1−𝑡)). Then let us consider the following problem: −𝑢⁽³⁾(𝑡)=1𝑡()𝑓(𝑢)𝑢(0)(0)(1)1−𝑡,in0<𝑡<1,=0,𝑢′=𝑢′=0, (3.34) where the function 𝑓 is defined as follows: 𝑓(𝑢)=(𝑢−1)^1/2+65,𝑢≥1,195𝑢²2+𝑢,0≤𝑢<1. (3.35) It is obvious that 𝑓 is continuous and (H1) holds. On the other hand, since [𝑢²/(2+𝑢)]^′=(𝑢²+4𝑢)/(2+𝑢)²≥0, for 0≤𝑢, it is clear to see that 195𝑢²/(2+𝑢) is nondecreasing for 0≤𝑢<1, and (𝑢−1)^1/2+65 is also nondecreasing for 𝑢≥1. In addition, lim_𝑟→∞𝑓(𝑟)sup𝑟=lim_𝑟→∞()sup𝑟−1^1/2+65𝑟1=0<26^𝑛−1∫¹₀𝑠((=11−𝑠)(1/𝑠1−𝑠))𝑑𝑠2,lim_𝑟→0𝑓(𝑟)sup𝑟=lim_𝑟→0sup195𝑟²𝑟()62+𝑟=0<^𝑛−1∫¹₀𝑠(1−𝑠)(1/𝑠(1−𝑠))𝑑𝑠=1. (3.36) So (H4) and (H5) hold.
Finally, choosing 𝑎=4, then for 𝑟∈[1,1/𝜎], we have 𝑓(𝑟)𝑎≥65>64=𝜃_𝑛(1/4)∫^3/4_2/4𝑠()ℎ(𝑠).1−𝑠𝑑𝑠 (3.37) Therefore (H6) hold.

Acknowledgments

The authors are grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. The authors also would like to thank Professor Ravi P. Agarwal for bringing several references to his attention.