Existence of Positive Solution for a Third-Order BVP with Advanced Arguments and Stieltjes Integral Boundary Conditions

Abstract

A class of third-order boundary value problems with advanced arguments and Stieltjes integral boundary conditions is discussed. Some existence criteria of at least one positive solution are established. The main tool used is the Guo-Krasnoselskii fixed point theorem.

1. Introduction

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity-driven flows, and so on [1].

Recently, third-order boundary value problems (BVPs for short) with integral boundary conditions, which cover third-order multipoint BVPs as special cases, have attracted much attention from many authors; see [2–6] and the references therein. In particular, in 2012, by using a fixed point theorem due to Avery and Peterson [7], Jankowski [4] established the existence of at least three nonnegative solutions to the following BVP: 𝑢^′′′(𝑡)(𝑡)𝑓((𝛼(𝑡(),𝑢(0)+ℎ𝑡,𝑢)))=0,𝑡∈0,1=𝑢^′′(0)=0,𝑢(1)=𝛽𝑢𝜂+𝜆[𝑢], (1) where 𝜆 denoted a linear functional on 𝐶[0,1] given by 𝜆[𝑢]=¹₀𝑢(𝑡)𝑑Λ(𝑡) (2) involving a Stieltjes integral with a suitable function Λ of bounded variation. The measure 𝑑Λ could be a signed one. The situation with a signed measure 𝑑Λ was first discussed in [8, 9] for second-order differential equations; it was also discussed in [10, 11] for second-order impulsive differential equations. For some other related results, one can refer to [12–14].

Among the boundary conditions in (1), only 𝑢(1) is related to a Stieltjes integral. In this paper, we are concerned with the following third-order BVP with advanced arguments and Stieltjes integral boundary conditions: 𝑢^′′′(𝑡)((𝛼(𝑡(),𝑢(0)𝜂[𝑢]+𝑓𝑡,𝑢)))=0,𝑡∈0,1=𝛾𝑢+𝜆,𝑢^′′(0)𝑢(1)𝜂[𝑢].=0,=𝛽𝑢+𝜆 (3) Throughout this paper, we always assume that 𝛼∶[0,1]→[0,1] is continuous and 𝛼(𝑡)≥𝑡 for 𝑡∈[0,1], 0<𝜂<1, 0≤𝛾<𝛽<1, Λ is a suitable function of bounded variation, and 𝜆[𝑢] is defined as in (2). It is important to indicate that it is not assumed that 𝜆[𝑢] is positive to all positive 𝑢. Some existence criteria of at least one positive solution to the BVP (3) are obtained by using the following well-known Guo-Krasnoselskii fixed point theorem [15, 16].

Theorem 1. Let 𝐸 be a Banach space, and let 𝐾 be a cone in 𝐸. Assume that Ω₁ and Ω₂ are bounded open subsets of 𝐸 such that 0∈Ω₁, Ω₁⊂Ω₂, and let 𝑇∶𝐾∩(Ω₂⧵Ω₁)→𝐾 be a completely continuous operator such that either(1)‖𝑇𝑢‖≤‖𝑢‖ for 𝑢∈𝐾∩𝜕Ω₁ and ‖𝑇𝑢‖≥‖𝑢‖ for 𝑢∈𝐾∩𝜕Ω₂ or(2)‖𝑇𝑢‖≥‖𝑢‖ for 𝑢∈𝐾∩𝜕Ω₁ and ‖𝑇𝑢‖≤‖𝑢‖ for 𝑢∈𝐾∩𝜕Ω₂. Then, 𝑇 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

2. Preliminaries

Let Δ=1−𝛾−(𝛽−𝛾)𝜂. Then, Δ>0.

Lemma 2. For any 𝑦∈𝐶[0,1], the BVP 𝑢^′′′(𝑡)(𝑡)(),𝑢(0)𝜂[𝑢]=−𝑦,𝑡∈0,1=𝛾𝑢+𝜆,𝑢^′′(0)𝑢(1)𝜂[𝑢]=0,=𝛽𝑢+𝜆 (4) has the unique solution 𝑢(𝑡)=1−𝛽−𝛾𝜂+𝛽−𝛾𝑡Δ𝜆[𝑢]+𝛾+𝛽−𝛾𝑡Δ¹₀𝑘𝑦(𝑠)+𝜂,𝑠𝑑𝑠₁₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,𝑡∈[0,1], (5) where 𝑘()=1𝑡,𝑠2(1−𝑡)𝑡−𝑠²𝑡(),0≤𝑠≤𝑡≤1,1−𝑠²,0≤𝑡≤𝑠≤1. (6)

Proof. By integrating the differential equation in (4) three times from 0 to 𝑡 and using the boundary condition 𝑢^′′(0)=0, we get 𝑢(𝑡)=𝑢(0)+𝑢^′(0)1𝑡−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝑑𝑠,𝑡∈[0,1]. (7) So, 𝑢^′(0)(1)(0)+1=𝑢−𝑢2¹₀(1−𝑠)²𝑦(𝑠)𝑑𝑠. (8) In view of (7), (8), and the boundary conditions 𝑢(0)=𝛾𝑢(𝜂)+𝜆[𝑢] and 𝑢(1)=𝛽𝑢(𝜂)+𝜆[𝑢], we have 𝑢(𝑡)=𝑡𝑢𝜂[𝑢]+𝛾+𝛽−𝛾+𝜆₁₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,𝑡∈[0,1]. (9) So, 𝑢𝜂=1Δ𝜆[𝑢]+1Δ¹₀𝑘𝜂,𝑠𝑦(𝑠)𝑑𝑠. (10) Substituting (10) into (9), we get 𝑢(𝑡)=1−𝛽−𝛾𝜂+𝛽−𝛾𝑡Δ𝜆[𝑢]+𝛾+𝛽−𝛾𝑡Δ¹₀𝑘𝑦(𝑠)+𝜂,𝑠𝑑𝑠₁₀𝑘(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,𝑡∈[0,1]. (11)

Lemma 3 (see [4]). Consider that 0≤𝑘(𝑡,𝑠)≤(1/2)(1+𝑠)(1−𝑠)², (𝑡,𝑠)∈[0,1]×[0,1].

Throughout, we assume that the following conditions are fulfilled: (C1) 𝑓∈𝐶([0,1]×[0,+∞),[0,+∞)),  (C2) 𝑑Λ(¹₀𝑡)≥0,¹₀(𝑡)𝜅(𝑠)=𝑡𝑑Λ≥0,₁₀𝑘(𝑡,𝑠)𝑑Λ(𝑡)≥0,𝑠∈[0,1]. (12)

For convenience, we denote 𝜌=1−𝛽−𝛾𝜂¹₀𝑑Λ(𝑡)+𝛽−𝛾¹₀(𝑡),𝜌𝑡𝑑Λ_′=𝛾¹₀𝑑Λ(𝑡)+𝛽−𝛾¹₀𝑡𝑑Λ(𝑡). (13) Obviously, 𝜌,𝜌^′≥0. In the remainder of this paper, we always assume that 𝜌<Δ.

Let 𝐶[0,1] be equipped with the maximum norm. Then, 𝐶[0,1] is a Banach space. Define 𝐾=𝑢∈𝐶[0,1]∶𝑢(𝑡)≥0,𝑡∈[0,1],min_{𝑡∈𝜂,1}𝑢(𝑡)≥Γ‖𝑢‖,𝜆[𝑢]≥0, (14) where 𝛽Γ=min1−𝜂,1−𝛽𝜂𝛽𝜂1−𝛾1−𝜂. (15) Then, 𝐾 is a cone in 𝐶[0,1].

Now, we define operators 𝑇 and 𝑆 on 𝐾 by (𝑡)=𝑡𝑇𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾Δ𝜆[𝑢]+(𝑡),[],(𝑡)=𝑡𝐹𝑢)(𝑡∈0,1𝑆𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾𝜆[]+(𝑡),[],Δ−𝜌𝐹𝑢𝐹𝑢)(𝑡∈0,1 (16) where (𝑡)=𝑡𝐹𝑢)(𝛾+𝛽−𝛾Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝑘(𝑡,𝑠)𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠,𝑡∈[0,1]. (17)

Lemma 4. Consider that 𝑇,𝑆∶𝐾→𝐾.

Proof. Let 𝑢∈𝐾. Then, it is easy to verify that (𝑇𝑢)^′′(𝑡)=−^𝑡₀𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠≤0,𝑡∈[0,1], (18) which shows that 𝑇𝑢 is concave down on [0,1]. In view of (0)=𝛾𝐹𝑢)(Δ¹₀𝑘𝑓((𝛼(𝑠(1)=𝛽𝜂,𝑠𝑠,𝑢)))𝑑𝑠≥0,𝐹𝑢)(Δ¹₀𝑘𝜂,𝑠𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠≥0, (19) we have (0)=𝜂𝑇𝑢)(1−𝛽−𝛾Δ𝜆[𝑢]+(0)(1)=𝐹𝑢)(≥0,𝑇𝑢)(1+𝛽−𝛾1−𝜂Δ𝜆[𝑢]+(𝐹𝑢)(1)≥0. (20) So, (𝑇𝑢)(𝑡)≥0, 𝑡∈[0,1].
Now, we prove that min_{𝑡∈[𝜂,1]}(𝑇𝑢)(𝑡)≥Γ‖𝑇𝑢‖. To do it, we consider two cases.
Case  1. Let (𝑇𝑢)(𝜂)≤(𝑇𝑢)(1). Then min_{𝑡∈[𝜂,1]}(𝑇𝑢)(𝑡)=(𝑇𝑢)(𝜂), and there exists 𝑡∈[𝜂,1] such that ‖𝑇𝑢‖=(𝑇𝑢)(𝑡). Moreover, (𝑇𝑢)𝑡−(𝑇𝑢)(0)≤()𝜂−(0)𝑡−0𝑇𝑢𝑇𝑢)(.𝜂−0 (21) So, ‖‖≤1𝑇𝑢𝜂(𝑇𝑢)𝜂−1−𝜂𝜂(𝑇𝑢)(0), (22) which together with (𝑇𝑢)(0)=𝛾(𝑇𝑢)𝜂+𝜆[𝑢] (23) implies that ‖𝑇𝑢‖≤1−𝛾1−𝜂𝜂(𝑇𝑢)𝜂; (24) that is, min_{𝑡∈𝜂,1}(𝑡)≥𝜂𝑇𝑢)(‖‖.1−𝛾1−𝜂𝑇𝑢 (25)
Case  2. Let (𝑇𝑢)(𝜂)>(𝑇𝑢)(1) and ‖𝑇𝑢‖=(𝑇𝑢)(𝑡). Note that in this case min_{𝑡∈[𝜂,1]}(𝑇𝑢)(𝑡)=(𝑇𝑢)(1).
If 𝑡∈[0,𝜂], then (𝑇𝑢)(1)−(𝑇𝑢)𝑡1−𝑡≥(𝑇𝑢)(1)−(𝑇𝑢)𝜂.1−𝜂 (26) So, ‖‖≤1𝑇𝑢(𝜂−𝜂1−𝜂𝑇𝑢)(1),1−𝜂𝑇𝑢)( (27) which together with (𝜂=1𝑇𝑢)𝛽((𝑇𝑢)(1)−𝜆[𝑢]) (28) implies that ‖𝑇𝑢‖≤1−𝛽𝜂𝛽(1);1−𝜂𝑇𝑢)( (29) that is, min_{𝑡∈𝜂,1}(𝑡)≥𝛽𝑇𝑢)(1−𝜂‖‖.1−𝛽𝜂𝑇𝑢 (30)
If 𝑡∈(𝜂,1), then (𝑇𝑢)𝑡−(𝑇𝑢)𝜂≤()𝜂−(0)𝑡−𝜂𝑇𝑢𝑇𝑢)(.𝜂−0 (31) So, ‖‖≤1𝑇𝑢𝜂(𝑇𝑢)𝜂−1−𝜂𝜂(𝑇𝑢)(0), (32) which together with (23) and (28) implies that ‖𝑇𝑢‖≤1−𝛾1−𝜂(1);𝛽𝜂𝑇𝑢)( (33) that is, min_{𝑡∈[𝜂,1]}(𝑇𝑢)(𝑡)≥𝛽𝜂‖‖.1−𝛾1−𝜂𝑇𝑢 (34)
It follows from (25), (30), and (34) that min_{𝑡∈𝜂,1}(𝑇𝑢)(𝑡)≥Γ‖𝑇𝑢‖. (35)
Finally, we need to show that 𝜆[𝑇𝑢]≥0. In view of 𝜆[𝐹𝑢]=¹₀𝛾+𝛽−𝛾𝑡Δ¹₀𝑘𝑓((𝛼(𝑠(𝑡)+𝜂,𝑠𝑠,𝑢)))𝑑𝑠𝑑Λ₁₀₁₀𝑘()𝑓((𝛼(𝑠(𝑡)=𝜌𝑡,𝑠𝑠,𝑢)))𝑑𝑠𝑑Λ_′Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠≥0, (36) we have 𝜆[]=𝜌𝑇𝑢Δ𝜆[𝑢]+𝜆[𝐹𝑢]≥0. (37)
This shows that 𝑇∶𝐾→𝐾. Similarly, we can prove that 𝑆∶𝐾→𝐾.

Lemma 5. The operators 𝑇 and 𝑆 have the same fixed points in 𝐾.

Proof. Suppose that 𝑢∈𝐾 is a fixed point of 𝑆. Then, 𝜆[𝑢]=¹₀1−𝛽−𝛾𝜂+𝛽−𝛾𝑡𝜆[]+(𝑡)(𝑡)=ΔΔ−𝜌𝐹𝑢𝐹𝑢)(𝑑Λ𝜆[],Δ−𝜌𝐹𝑢 (38) which shows that 𝜆[𝐹𝑢]=Δ−𝜌Δ𝜆[𝑢]. (39) So, 𝑢(𝑡)=(𝑡)=𝑡𝑆𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾𝜆[]+(𝑡)=𝑡Δ−𝜌𝐹𝑢𝐹𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾Δ𝜆[𝑢]+(𝑡)=(𝑡)[],𝐹𝑢)(𝑇𝑢)(,𝑡∈0,1 (40) which indicates that 𝑢 is a fixed point of 𝑇.
Suppose that 𝑢∈𝐾 is a fixed point of 𝑇. Then, 𝜆[𝑢]=¹₀1−𝛽−𝛾𝜂+𝛽−𝛾𝑡Δ𝜆[𝑢]+(𝑡)(𝑡)=𝜌𝐹𝑢)(𝑑ΛΔ𝜆[𝑢]+𝜆[𝐹𝑢], (41) which shows that 𝜆[𝑢]=Δ𝜆[].Δ−𝜌𝐹𝑢 (42) So, 𝑢(𝑡)=(𝑡)=𝑡𝑇𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾Δ𝜆[𝑢]+(𝑡)=𝑡𝐹𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾𝜆[]+(𝑡)=(𝑡)[],Δ−𝜌𝐹𝑢𝐹𝑢)(𝑆𝑢)(,𝑡∈0,1 (43) which indicates that 𝑢 is a fixed point of 𝑆.

Lemma 6. 𝑇,𝑆∶𝐾→𝐾 is completely continuous.

Proof. First, by Lemma 4, we know that 𝑇(𝐾)⊂𝐾.
Next, we show that 𝑇 is compact. Let 𝐷⊂𝐾 be a bounded set. Then, there exists 𝑀₁>0 such that ‖𝑢‖≤𝑀₁ for any 𝑢∈𝐷. Since Λ is a function of bounded variation, there exists 𝑀₂>0 such that 𝑣_Δ′=∑^𝑛_𝑖=1|Λ(𝑡_𝑖)−Λ(𝑡_𝑖−1)|≤𝑀₂ for any partition Δ^′∶0=𝑡₀<𝑡₁<⋯<𝑡_𝑛−1<𝑡_𝑛=1. Let 𝑀₃=sup𝑓(𝑡,𝑢)∶(𝑡,𝑢)∈[0,1]×0,𝑀₁. (44) Then, for any 𝑢∈𝐷, ‖𝑇𝑢‖=max_[]𝑡∈0,1(𝑡)≤𝑇𝑢)(1+𝛽−𝛾1−𝜂Δ𝜆[𝑢]+𝛽Δ¹₀𝑘𝑓((𝛼(𝑠+1𝜂,𝑠𝑠,𝑢)))𝑑𝑠2¹₀(1+𝑠)(1−𝑠)²𝑓((𝛼(𝑠≤𝑠,𝑢)))𝑑𝑠1+𝛽−𝛾1−𝜂Δ𝑀₁𝑀₂+𝛽𝑀₃Δ¹₀𝑘5𝜂,𝑠𝑑𝑠+𝑀24³, (45) which shows that 𝑇(𝐷) is uniformly bounded.
On the other hand, for any 𝜀>0, since 𝑘(𝑡,𝑠) is uniformly continuous on [0,1]×[0,1], there exists 𝛿₁(𝜀)>0 such that for any 𝑡₁,𝑡₂∈[0,1] with |𝑡₁−𝑡₂|<𝛿₁(𝜀), 𝑘𝑡₁,𝑠−𝑘𝑡₂<𝜀,𝑠3𝑀₃,𝑠∈[0,1]. (46) Let 𝛿=min{𝛿₁(𝜀),𝜀Δ/3(𝛽−𝛾)𝑀₁𝑀₂,𝜀Δ/3(𝛽−𝛾)𝑀₃∫¹₀𝑘(𝜂,𝑠)𝑑𝑠}. Then, for any 𝑢∈𝐷, 𝑡₁,𝑡₂∈[0,1] with |𝑡₁−𝑡₂|<𝛿, we have (𝑇𝑢)𝑡₁−(𝑇𝑢)𝑡₂=𝛽−𝛾𝑡₁−𝑡₂Δ𝜆[𝑢]+𝛽−𝛾𝑡₁−𝑡₂Δ×¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝑘𝑡₁,𝑠−𝑘𝑡₂((𝛼(𝑠≤𝑡,𝑠×𝑓𝑠,𝑢)))𝑑𝑠𝛽−𝛾₁−𝑡₂Δ𝜆[𝑢]+𝛽−𝛾𝑡₁−𝑡₂Δ×¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝑘𝑡₁,𝑠−𝑘𝑡₂𝑓((𝛼(𝑠≤𝑡,𝑠𝑠,𝑢)))𝑑𝑠𝛽−𝛾₁−𝑡₂𝑀₁𝑀₂Δ+𝛽−𝛾𝑡₁−𝑡₂𝑀₃Δ¹₀𝑘𝜂,𝑠𝑑𝑠+𝑀₃¹₀𝑘𝑡₁,𝑠−𝑘𝑡₂,𝑠𝑑𝑠<𝜀, (47) which shows that 𝑇(𝐷) is equicontinuous. It follows from the Arzela-Ascoli theorem that 𝑇(𝐷) is relatively compact. Thus, we have shown that 𝑇 is a compact operator.
Finally, we prove that 𝑇 is continuous. Suppose that 𝑢_𝑛, 𝑢∈𝐾 and lim_𝑛→∞𝑢_𝑛=𝑢. Then, there exists 𝑀₄>0 such that ‖𝑢‖≤𝑀₄ and ‖𝑢_𝑛‖≤𝑀₄(𝑛=1,2,…). For any 𝜀>0, since 𝑓(𝑠,𝑥) is uniformly continuous on [0,1]×[0,𝑀₄], there exists 𝛿>0 such that for any 𝑥₁,𝑥₂∈[0,𝑀₄] with |𝑥₁−𝑥₂|<𝛿, 𝑓𝑠,𝑥₁−𝑓𝑠,𝑥₂<𝜀2𝛽/Δ∫¹₀𝑘(),[].𝜂,𝑠𝑑𝑠+5/12𝑠∈0,1 (48) At the same time, since lim_𝑛→∞𝑢_𝑛=𝑢, there exists positive integer 𝑁 such that for any 𝑛>𝑁, 𝑢_𝑛−𝑢<min𝛿,𝜀Δ21+𝛽−𝛾1−𝜂|Λ(1)−Λ(0)|. (49) It follows from (48) and (49) that for any 𝑛>𝑁, 𝑇𝑢_𝑛−𝑇𝑢=max_[]𝑡∈0,1𝑇𝑢_𝑛(𝑡)−(𝑡≤𝑇𝑢)()1+𝛽−𝛾1−𝜂Δ𝜆𝑢_𝑛[𝑢+𝛽−𝜆]Δ¹₀𝑘𝜂,𝑠𝑓𝑠,𝑢_𝑛(𝛼(𝑠((𝛼(𝑠+1))−𝑓𝑠,𝑢)))𝑑𝑠2¹₀(1+𝑠)(1−𝑠)²𝑓𝑠,𝑢_𝑛(𝛼(𝑠((𝛼(𝑠≤))−𝑓𝑠,𝑢)))𝑑𝑠1+𝛽−𝛾1−𝜂Δ𝑢_𝑛Λ(1)(0+−𝑢|−Λ)|₁₀𝛽Δ𝑘+1𝜂,𝑠2(1+𝑠)(1−𝑠)²×𝑓𝑠,𝑢_𝑛(𝛼(𝑠))−𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠<𝜀, (50) which indicates that 𝑇 is continuous. Therefore, 𝑇∶𝐾→𝐾 is completely continuous. Similarly, we can prove that 𝑆∶𝐾→𝐾 is also completely continuous.

3. Main Results

For convenience, we define 𝑓⁰=limsup_𝑥→0+max_[]𝑡∈0,1𝑓(𝑡,𝑥)𝑥,𝑓₀=liminf_𝑥→0+min_{𝑡∈𝜂,1}𝑓(𝑡,𝑥)𝑥,𝑓^∞=limsup_𝑥→+∞max_[]𝑡∈0,1𝑓(𝑡,𝑥)𝑥,𝑓_∞=liminf_𝑥→+∞min_{𝑡∈𝜂,1}𝑓(𝑡,𝑥)𝑥,𝐻₁=1+𝛽−𝛾1−𝜂𝜌Δ−𝜌Δ^′1₀𝑘𝜂,𝑠𝑑𝑠+¹₀𝜅(𝑠)+𝛽𝑑𝑠Δ¹₀𝑘5𝜂,𝑠𝑑𝑠+,𝐻24₂1=Γ𝜌Δ−𝜌Δ^′1_𝜂𝑘𝜂,𝑠𝑑𝑠+¹_𝜂𝜅(𝑠)+1𝑑𝑠Δ¹_𝜂𝑘𝜂,𝑠𝑑𝑠. (51)

Theorem 7. If 𝐻₁𝑓⁰<1<𝐻₂𝑓_∞, then the BVP (3) has at least one positive solution.

Proof. Since 𝐻₁𝑓⁰<1, there exists 𝜀₁>0 such that 𝐻₁𝑓⁰+𝜀₁≤1. (52) By the definition of 𝑓⁰, we may choose 𝜌₁>0 so that 𝑓(𝑡,𝑥)≤𝑓⁰+𝜀₁𝑥,𝑡∈[0,1],𝑥∈0,𝜌₁. (53) Let Ω₁={𝑢∈𝐶[0,1]∶‖𝑢‖<𝜌₁}. Then, for any 𝑢∈𝐾∩𝜕Ω₁, 0≤𝑢(𝑡)≤‖𝑢‖=𝜌₁,𝑡∈[0,1], (54) which together with (52) and (53) implies that (𝑡)=𝑡𝑆𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾×𝜌Δ−𝜌_′Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+𝑡𝑠,𝑢)))𝑑𝑠𝛾+𝛽−𝛾Δ¹₀𝑘𝜂,𝑠𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠+¹₀𝑘()𝑓((𝛼(𝑠≤𝑡,𝑠𝑠,𝑢)))𝑑𝑠1+𝛽−𝛾1−𝜂×𝜌Δ−𝜌_′Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+𝛽𝑠,𝑢)))𝑑𝑠Δ¹₀𝑘𝑓((𝛼(𝑠+1𝜂,𝑠𝑠,𝑢)))𝑑𝑠2¹₀(1+𝑠)(1−𝑠)²𝑓((𝛼(𝑠≤𝑓𝑠,𝑢)))𝑑𝑠₀+𝜀₁1+𝛽−𝛾1−𝜂×𝜌Δ−𝜌_′Δ¹₀𝑘𝑢(𝛼(𝑠+𝜂,𝑠))𝑑𝑠₁₀𝜅(𝑠)𝑢(𝛼(𝑠+𝛽))𝑑𝑠Δ¹₀𝑘𝑢(𝛼(𝑠+1𝜂,𝑠))𝑑𝑠2¹₀(1+𝑠)(1−𝑠)²𝑢(𝛼(𝑠≤𝑓))𝑑𝑠₀+𝜀₁‖𝑢‖1+𝛽−𝛾1−𝜂×𝜌Δ−𝜌_′Δ¹₀𝑘+𝜂,𝑠𝑑𝑠₁₀𝜅(𝑠)+𝛽𝑑𝑠Δ¹₀𝑘5𝜂,𝑠𝑑𝑠+24=𝐻₁𝑓⁰+𝜀₁‖𝑢‖≤‖𝑢‖,𝑡∈[0,1]. (55) This shows that ‖𝑆𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁. (56)
On the other hand, since 1<𝐻₂𝑓_∞, there exists 𝜀₂>0 such that 𝐻₂𝑓_∞−𝜀₂≥1. (57) By the definition of 𝑓_∞, we may choose 𝜌₂>0 so that 𝑓(𝑡,𝑥)≥𝑓_∞−𝜀₂𝑥,𝑡∈𝜂,1,𝑥∈𝜌₂,+∞. (58) Let 𝜌₂=max{2𝜌₁,𝜌₂/Γ} and Ω₂={𝑢∈𝐶[0,1]∶‖𝑢‖<𝜌₂}. Then, for any 𝑢∈𝐾∩𝜕Ω₂, 𝑢(𝑡)≥Γ‖𝑢‖=Γ𝜌₂≥𝜌₂,𝑡∈𝜂,1, (59) which together with (57) and (58) implies that ()𝜂=1𝑆𝑢𝜌Δ−𝜌Δ^′1₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+1𝑠,𝑢)))𝑑𝑠Δ¹₀𝑘𝑓((𝛼(𝑠≥1𝜂,𝑠𝑠,𝑢)))𝑑𝑠𝜌Δ−𝜌Δ^′1_𝜂𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠_1𝜂𝜅(𝑠)𝑓((𝛼(𝑠+1𝑠,𝑢)))𝑑𝑠Δ¹_𝜂𝑘𝑓((𝛼(𝑠≥𝑓𝜂,𝑠𝑠,𝑢)))𝑑𝑠_∞−𝜀₂1𝜌Δ−𝜌Δ^′1_𝜂𝑘𝑢(𝛼(𝑠+𝜂,𝑠))𝑑𝑠_1𝜂𝜅(𝑠)𝑢(𝛼(𝑠+1))𝑑𝑠Δ¹_𝜂𝑘𝑢(𝛼(𝑠≥𝑓𝜂,𝑠))𝑑𝑠_∞−𝜀₂Γ‖𝑢1‖𝜌Δ−𝜌Δ^′1_𝜂𝑘+𝜂,𝑠𝑑𝑠_1𝜂𝜅(𝑠)+1𝑑𝑠Δ¹_𝜂𝑘𝜂,𝑠𝑑𝑠=𝐻₂𝑓_∞−𝜀₂‖𝑢‖≥‖𝑢‖. (60) This indicates that ‖𝑆𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂. (61)
Therefore, it follows from (56), (61), and Theorem 1 that the operator 𝑆 has one fixed point 𝑢∈𝐾∩(Ω₂⧵Ω₁), which is a positive solution of the BVP (3).

Theorem 8. If 𝐻₁𝑓^∞<1<𝐻₂𝑓₀, then the BVP (3) has at least one positive solution.

Proof. Since 𝐻₂𝑓₀>1, there exists 𝜀₃>0 such that 𝐻₂𝑓₀−𝜀₃≥1. (62) By the definition of 𝑓₀, we may choose 𝜌₃>0 so that 𝑓(𝑡,𝑥)≥𝑓₀−𝜀₃𝑥,𝑡∈𝜂,1,𝑥∈0,𝜌₃. (63) Let Ω₃={𝑢∈𝐶[0,1]∶‖𝑢‖<𝜌₃}. Then, for any 𝑢∈𝐾∩𝜕Ω₃, Γ‖𝑢‖≤𝑢(𝑡)≤‖𝑢‖=𝜌₃,𝑡∈𝜂,1, (64) which together with (62) and (63) implies that ()𝜂=1𝑆𝑢𝜌Δ−𝜌Δ^′1₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+1𝑠,𝑢)))𝑑𝑠Δ¹₀𝑘𝑓((𝛼(𝑠≥1𝜂,𝑠𝑠,𝑢)))𝑑𝑠𝜌Δ−𝜌Δ^′1_𝜂𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠_1𝜂𝜅(𝑠)𝑓((𝛼(𝑠+1𝑠,𝑢)))𝑑𝑠Δ¹_𝜂𝑘𝑓((𝛼(𝑠≥𝑓𝜂,𝑠𝑠,𝑢)))𝑑𝑠₀−𝜀₃1𝜌Δ−𝜌Δ^′1_𝜂𝑘𝑢(𝛼(𝑠+𝜂,𝑠))𝑑𝑠_1𝜂𝜅(𝑠)𝑢(𝛼(𝑠+1))𝑑𝑠Δ¹_𝜂𝑘𝑢(𝛼(𝑠≥𝑓𝜂,𝑠))𝑑𝑠₀−𝜀₃Γ‖𝑢1‖𝜌Δ−𝜌Δ^′1_𝜂𝑘𝜂,𝑠𝑑𝑠+¹_𝜂𝜅(𝑠)+1𝑑𝑠Δ¹_𝜂𝑘𝜂,𝑠𝑑𝑠=𝐻₂𝑓₀−𝜀₃‖𝑢‖≥‖𝑢‖. (65) This shows that ‖𝑆𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₃. (66)
On the other hand, since 𝐻₁𝑓^∞<1, there exists 𝜀₄>0 so that 𝐻₁𝑓^∞+𝜀₄<1. (67) By the definition of 𝑓^∞, we may choose 𝜌₄>0 such that 𝑓(𝑡,𝑥)≤𝑓^∞+𝜀₄𝑥,𝑡∈[0,1],𝑥∈𝜌₄,+∞, (68) which implies that 𝑓(𝑡,𝑥)≤𝑀+𝑓^∞+𝜀₄𝑥,𝑡∈[0,1],𝑥∈[0,+∞), (69) where 𝑀=max𝑓(𝑡,𝑥)∶𝑡∈[0,1],𝑥∈0,𝜌₄. (70) Now, we choose 𝜌₄=max{2𝜌₃,𝑀𝐻₁/(1−𝐻₁(𝑓^∞+𝜀₄))}, and let Ω₄={𝑢∈𝐶[0,1]∶‖𝑢‖<𝜌₄}. Then, for any 𝑢∈𝐾∩𝜕Ω₄, 0≤𝑢(𝑡)≤‖𝑢‖,𝑡∈[0,1], (71) which together with (69) implies that (𝑡)=𝑡𝑆𝑢)(1−𝛽−𝛾𝜂+𝛽−𝛾×𝜌Δ−𝜌_′Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+𝑡𝑠,𝑢)))𝑑𝑠𝛾+𝛽−𝛾Δ¹₀𝑘𝜂,𝑠𝑓(𝑠,𝑢(𝛼(𝑠)))𝑑𝑠+¹₀𝑘()𝑓((𝛼(𝑠≤𝑡,𝑠𝑠,𝑢)))𝑑𝑠1+𝛽−𝛾1−𝜂×𝜌Δ−𝜌_′Δ¹₀𝑘𝑓((𝛼(𝑠+𝜂,𝑠𝑠,𝑢)))𝑑𝑠₁₀𝜅(𝑠)𝑓((𝛼(𝑠+𝛽𝑠,𝑢)))𝑑𝑠Δ¹₀𝑘𝑓((𝛼(𝑠+1𝜂,𝑠𝑠,𝑢)))𝑑𝑠2¹₀(1+𝑠)(1−𝑠)²𝑓((𝛼(𝑠≤𝑓𝑠,𝑢)))𝑑𝑠𝑀+^∞+𝜀₄‖𝑢‖1+𝛽−𝛾1−𝜂×𝜌Δ−𝜌_′Δ¹₀𝑘+𝜂,𝑠𝑑𝑠₁₀𝜅(𝑠)+𝛽𝑑𝑠Δ¹₀𝑘5𝜂,𝑠𝑑𝑠+=𝑓24𝑀+^∞+𝜀₄‖𝑢‖𝐻₁≤‖𝑢‖,𝑡∈[0,1]. (72) This indicates that ‖𝑆𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₄. (73)
Therefore, it follows from (66), (73), and Theorem 1 that the operator 𝑆 has one fixed point 𝑢∈𝐾∩(Ω₄⧵Ω₃), which is a positive solution of the BVP (3).

Example 9. Consider the following BVP: 𝑢^′′′(𝑡)√+𝑡3359𝑢𝑡913𝑒^{√𝑢(𝑡)}+5881𝑢√𝑡²17𝑢√𝑡(),𝑢(0)=1+1=0,𝑡∈0,14𝑢12+¹₀𝑢(𝑡)(2𝑡−1)𝑑𝑡,𝑢^′′(0)𝑢(1)=1=0,2𝑢12+¹₀𝑢(𝑡)(2𝑡−1)𝑑𝑡. (74)
In view of 𝑑Λ(𝑡)=(2𝑡−1)𝑑𝑡, we have 𝑑Λ(¹₀𝑡)=0,¹₀(𝑡)=1𝑡𝑑Λ6(𝑠)=1,𝜅𝑠12²(1−𝑠)²,𝑠∈[0,1]. (75) At the same time, since 𝛾=1/4 and 𝛽=𝜂=1/2, a simple calculation shows that 5Δ=8,𝜌=𝜌^′=1224,Γ=7₁₀,𝑘1𝜂,𝑠𝑑𝑠=,16¹_𝜂𝑘1𝜂,𝑠𝑑𝑠=,96₁₀𝜅(𝑠)1𝑑𝑠=,360¹_𝜂𝜅(𝑠)1𝑑𝑠=.720 (76) So, 𝐻₁=9133360,𝐻₂=17.2940 (77) If we let 𝑓(𝑡,𝑥)=𝑡[(3359𝑥/913𝑒^𝑥)+(5881𝑥²/17(𝑥+1))], (𝑡,𝑥)∈[0,1]×[0,+∞), then it is easy to compute that 𝑓⁰=3359913,𝑓_∞=5881,34 (78) which together with (77) implies that 𝐻₁𝑓⁰=33593360<1<𝐻₂𝑓_∞=5881.5880 (79) Therefore, it follows from Theorem 7 that the BVP (74) has at least one positive solution.

Acknowledgment

This paper is supported by the Natural Science Foundation of Gansu Province of China (1208RJZA240).