Existence of Solution to a Second-Order Boundary Value Problem via Noncompactness Measures

Abstract

The existence and uniqueness of the solutions to the Dirichlet boundary value problem in the Banach spaces is discussed by using the fixed point theory of condensing mapping, doing precise computation of measure of noncompactness, and calculating the spectral radius of linear operator.

1. Introduction

This paper is mainly concerned with the following second-order Dirichlet boundary value problem:−𝑢^′′(𝑡)((𝑡[],𝑢(0)(1)=𝑓𝑡,𝑢)),𝑡∈𝐼=0,1=𝑢=𝜃, (1.1) in a Banach space 𝐸, where 𝑓(𝑡,𝑥)∈𝐶(𝐼×𝐸,𝐸), 𝜃 is the zero element of 𝐸.

In the last several decades, there has been much attention focused on the boundary value problems for various nonlinear ordinary differential equations, difference equations, and functional differential equations, see [1–20] and the references therein. The existence of solutions for Neumann boundary value problems has been considerably investigated in many publications such as [2–5, 8–10]. Dirichlet boundary value problems have deserved the attention of many researchers, see [11–20] and the references therein.

In particular, the authors in [11] have studied the following two-point boundary value problem:𝑥^′′=𝐻𝑡,𝑥,𝑥^′,0<𝑡<1,𝑎𝑥(0)−𝑏𝑥^′(0)=𝑥₀,𝑐𝑥(1)+𝑑𝑥^′(1)=𝑥₁, (1.2) where 𝑎,𝑏,𝑐,𝑑≥0 and 𝑎𝑑+𝑏𝑐>0. They obtained the existence of solutions by means of the Darbo fixed point theorem and properties of the measure of noncompactness.

We would like to mention the results due to [11]. First, we point out that many authors applied the famous Sadovskii’s fixed point theorem to investigate similar problems and used the following hypothesis with respect to the Kuratowski measure of noncompactness 𝛼(⋅): there exists a constant 𝑘>0 such that for any bounded and equicontinuous set 𝐴,𝐵⊂𝐶(𝐼,𝐸) and 𝑡∈𝐼, 𝛼(𝐻(𝐼×𝐴×𝐵))≤𝑘max{𝛼(𝐴),𝛼(𝐵)}. What is more, they required a stronger condition, that is, ‖𝐻(𝑡,𝑥,𝑦)‖≤𝐿 for (𝑡,𝑥,𝑦)∈𝐼×𝐸×𝐸 and the constant 𝑘 satisfies 0<𝑘<1/2 (see Remarks 3.2–3.6).

The authors in [15, 18] have studied the following boundary value problem:𝑥^′′=𝑓𝑡,𝑥,𝑥^′,𝑐₁𝑥(𝑎)−𝑑₁𝑥^′(𝑎)=𝑥₁,𝑐₂𝑥(𝑏)+𝑑₂𝑥^′(𝑏)=𝑥₂, (1.3) where 𝑋 is a real Banach space, 𝐽=[𝑎,𝑏]⊂ℝ, 𝑓∶𝐽×𝑋²→𝑋 is continuous, 𝑐_𝑖,𝑑_𝑖∈ℝ, and 𝑥_𝑖∈𝑋 for 𝑖=1,2. They obtained the existence of solutions by means of Sadovskii’s fixed point theorem and properties of the measure of noncompactness.

Motivated by the above-mentioned work [11, 15, 18], the main aim of this paper is to study the existence and uniqueness of solutions for the problem (1.1) under the new conditions. The main new features presented in this paper are as follows First, the existence and uniqueness of solutions to Banach space’s Dirichlet boundary value problem is proved precisely calculating the spectral radius of linear operation. Second, the conditions imposed on the BVP (1.1) are weak. Third, the main tools used in the analysis are Sadovskii’s fixed point theorem and precise computation of measure of noncompactess. Our results can be seen as a supplement of the results in [11] (see Remarks 3.2–3.6).

This paper is organized as follows. In Section 2, we provide some basic definitions, preliminaries facts, and various lemmas which will be used throughout this paper. In Section 3, we give main results in this paper.

2. Preliminaries and Lemmas

Let 𝐸 be a real Banach space and 𝑃 be a cone in 𝐸 which defines a partial ordering in 𝐸 by 𝑥≤𝑦 if and only if 𝑦−𝑥∈𝑃· 𝑃 is said to be normal if there exists a positive constant 𝑁 such that 𝜃≤𝑥≤𝑦 implies ‖𝑥‖≤𝑁‖𝑦‖, where 𝜃 denotes the zero element of 𝐸, and the smallest 𝑁 is called the normal constant of 𝑃 (it is clear, 𝑁≥1). If 𝑥≤𝑦 and 𝑥≠𝑦, we write 𝑥<𝑦. For details on cone theory, see the monograph [7].

Let 𝐼=[0,1]. By 𝐶(𝐼,𝐸) we denote the Banach space of all continuous functions from 𝐼 into 𝐸 with the norm𝑦_𝑐∶=max𝑦(𝑡)∶𝑡∈𝐼. (2.1)

Definition 2.1 (see [7]). Assume that 𝑆 is a bounded set in 𝐸. Let 𝛼(𝑆)=inf{𝛿>0∶𝑆 be expressed as the union 𝑆=∪^𝑚_𝑖=1𝑆_𝑖 of a finite number of sets 𝑆_𝑖 with diameter diam(𝑆_𝑖)≤𝛿}.
𝛼(𝑆) is said to be the Kuratowski measure of noncompactness and is called the noncompactness measure for short. For details and properties of the noncompactness measure see [7].

Definition 2.2 (see [7]). The mapping 𝐴 is said to be a condensing operator if 𝐴 is continuous, bounded, and for any nonrelatively compact and bounded set 𝑆⊂𝐷, 𝛼(𝐴(𝑆))<𝛼(𝑆). (2.2)
The following lemmas are of great importance in the proof of our main results.

Lemma 2.3. Suppose that 𝑀∉{−𝑛²𝜋²∣𝑛=1,2,3,…}. Then for any ℎ∈𝐶(𝐼,𝐸), the linear boundary value problem −𝑢^′′(𝑡)(𝑡)(𝑡)[],𝑢(0)(1)+𝑀𝑢=ℎ,𝑡∈𝐼=0,1=𝑢=𝜃, (2.3) has a unique solution 𝑢∶=𝑇_𝑀ℎ∈𝐶²(𝐼,𝐸), and 𝑇_𝑀∶𝐶(𝐼,𝐸)↦𝐶(𝐼,𝐸) is bounded linear operator.

Proof. Note that 𝑀∉{−𝑛²𝜋²∣𝑛=1,2,3,…}, which assures second-order boundary value problem −𝛾^′′(𝑡)(𝑡)𝛾(0)(1)+𝑀𝛾=0,=𝛾=0, (2.4) has only a zero solution. To obtain a solution of the problem (2.3), we require a mapping whose kernel 𝐺_𝑀(𝑡,𝑠)∶𝐼×𝐼→ℝ is the Green’s function of the boundary value problem (2.4). Let 𝛽=|𝑀|, we consider three cases.
Case 1. if 𝑀>0, we have 𝐺_𝑀()=𝛽()𝑡,𝑠sinh𝛽𝑡sinh1−𝑠𝛽()𝛽sinh𝛽,if0≤𝑡≤𝑠≤1,sinh𝛽𝑠sinh1−𝑡𝛽sinh𝛽,if0≤𝑠≤𝑡≤1. (2.5)Case 2. if 𝑀=0, we have 𝐺₀()=𝑡()𝑠()𝑡,𝑠1−𝑠,if0≤𝑡≤𝑠≤1,1−𝑡,if0≤𝑠≤𝑡≤1. (2.6)Case 3. if 𝑀<0, 𝑀≠−𝑛²𝜋²,𝑛=1,2,3,…, we have 𝐺_𝑀()=𝛽()𝑡,𝑠sin𝛽𝑡sin1−𝑠𝛽()𝛽sin𝛽,if0≤𝑡≤𝑠≤1,sin𝛽𝑠sin1−𝑡𝛽sin𝛽,if0≤𝑠≤𝑡≤1. (2.7)
After direct computations, it is easy to see that 𝑢(𝑡)=¹₀𝐺_𝑀(𝑡,𝑠)ℎ(𝑠)𝑑𝑠∶=𝑇_𝑀ℎ(𝑡) (2.8) is continuously differentiable, and 𝑢(𝑡) is a solution of (2.3).
We now claim that solution of the boundary value problem (2.3) is unique. The proof is as follows. If possible, suppose that 𝑣(𝑡)∈𝐶²(𝐼,𝐸) is another solution of the problem (2.3). For any 𝜑∈𝐸^∗(𝐸^∗ denotes the dual space of 𝐸), let 𝑝(𝑡)=𝜑(𝑢(𝑡)−𝑣(𝑡)), thus we obtain 𝑝(𝑡)∈𝐶²(𝐼), 𝑝^′′(𝑡)=𝜑(𝑢^′′(𝑡)−𝑣^′′(𝑡)). By (2.3), we have −𝑝^′′(𝑡)(𝑡)𝑝(0)(1)+𝑀𝑝=0,=𝑝=0, (2.9) that is, 𝑝(𝑡) is a solution of the boundary value problem (2.4). However, on the other hand, problem (2.4) has only a zero solution, therefore we have 𝜑(𝑢(𝑡)−𝑣(𝑡))=0. Thus, we get 𝑢(𝑡)−𝑣(𝑡)=𝜃, hence 𝑢(𝑡)≡𝑣(𝑡) in 𝐼, which implies that solution of the problem (2.3) is unique, say, 𝑢∶=𝑇_𝑀ℎ, and 𝑇_𝑀∶𝐶(𝐼,𝐸)→𝐶(𝐼,𝐸).
It is easy to see that 𝑇_𝑀 is bounded linear operator. This completes the proof.

Remark 2.4. If 𝑀>−𝜋², it is easy to see that 𝐺_𝑀(𝑡,𝑠)≥0.

Lemma 2.5. Assume that 𝑀>−𝜋², and 𝑇_𝑀∶𝐶(𝐼,𝐸)↦𝐶(𝐼,𝐸) is given by (2.8). Then(1)the spectral radius 𝑟(𝑇_𝑀)=1/(𝑀+𝜋²);(2)If 𝐸 is an ordered Banach space, then 𝑇_𝑀 is a positive operator, that is, if ℎ≥0, then 𝑇_𝑀ℎ≥0.

Proof. (1) Define operator ℒ∶𝒟(ℒ)↦𝐶(𝐼,𝐸) by ℒ𝑢∶=−𝑢^′′+𝑀𝑢, (2.10) where 𝒟(ℒ)={𝑢∈𝐶²(𝐼,𝐸)∣𝑢(0)=𝑢(1)=𝜃}. By Lemma 2.3, we have that 𝑇_𝑀 is bounded invertible operator of ℒ, and if 𝜆≠𝑀+𝑛²𝜋²,𝑛=1,2,3,…, then ℒ−𝜆𝐼 has a bounded invertible operator, thus 𝜆∈𝜌(ℒ).
Let 𝜆=𝜆_𝑛=𝑀+𝑛²𝜋², 𝜆_𝑛 is a eigenvalue of ℒ. For any 𝑥∈𝐸,𝑥≠𝜃, since sin√𝑀+𝑛²𝜋²𝑡 is eigenvector of 𝜆_𝑛, then the spectrum of operator ℒ is 𝜎(ℒ)={𝑀+𝑛²𝜋²∣𝑛=1,2,3,…}. By the spectral mapping theorem [21], we get 𝜎𝑇_𝑀={𝜃}∪1𝑀+𝑛²𝜋²∣𝑛∈𝒩, (2.11) so 𝑟(𝑇_𝑀)=1/(𝑀+𝜋²).
(2) If ℎ≥0, by definition of 𝐺_𝑀(𝑡,𝑠) and 𝑀>−𝜋², then we get 𝐺_𝑀(𝑡,𝑠)≥0, so by (2.8), we have 𝑇_𝑀ℎ≥0, that is, 𝑇_𝑀 is a positive operator. This achieves the proof.

Remark 2.6. In particular. If 𝐸=ℝ¹, 𝑀=0, then by (1) of Lemma 2.5, we get 𝑟(𝑇₀)=1/𝜋², where 𝑇₀ is an operator in 𝐶(𝐼): 𝑇₀𝜑(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝜑(𝑠)𝑑𝑠, (2.12) and ‖𝑇₀‖_𝑐=1/8. In fact: 𝑇₀𝜑(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝜑(𝑠)𝑑𝑠≤¹₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅𝜑_𝑐=12𝑡(𝑡−1)⋅𝜑_𝑐≤18⋅𝜑_𝑐. (2.13) This means that ‖𝑇₀𝜑‖_𝑐≤(1/8)‖𝜑‖_𝑐, therefore ‖𝑇₀‖_𝑐≤1/8. However, on the other hand, ‖𝑇₀(1)‖_𝑐=1/8. As a result, we obtain ‖𝑇₀‖_𝑐=1/8.

Lemma 2.7. Let 𝐽=[𝑎,𝑏], 𝑢∈𝐶(𝐽,𝐸), 𝜑∈𝐶(𝐽,ℝ⁺). Then ^𝑏_𝑎𝜑(𝑠)𝑢(𝑠)𝑑𝑠∈^𝑏_𝑎𝜑(𝑠)𝑑𝑠⋅co𝑢(𝐽), (2.14) where 𝑢(𝐽)={𝑢(𝑡)∣𝑡∈𝐽}, co𝑢(𝐽) is closed convex hull of 𝑢(𝐽).

Proof. If ∫^𝑏_𝑎𝜑(𝑠)𝑑𝑠=0, then (2.14) is true. We suppose that ∫^𝑏_𝑎𝜑(𝑠)𝑑𝑠>0, and take a partition of [𝑎,𝑏]: Δ_𝑛∶𝑎=𝑡₀^(𝑛)<𝑡₁^(𝑛)<𝑡₂^(𝑛)<⋯<𝑡_𝑚^(𝑛)=𝑏. (2.15) Let Δ𝑡_𝑖^(𝑛)=𝑡_𝑖^(𝑛)−𝑡^(𝑛)_𝑖−1, ‖Δ_𝑛‖=max{Δ𝑡_𝑖^(𝑛)∶1≤𝑖≤𝑚_𝑛}, by definition of Riemann integral, we get ^𝑏_𝑎𝜑(𝑠)𝑢(𝑠)𝑑𝑠=lim^𝑚_𝑛→∞^𝑛_𝑛=1𝜑𝑡_𝑖^(𝑛)𝑢𝑡_𝑖^(𝑛)Δ𝑡_𝑖^(𝑛),_𝑏𝑎𝜑(𝑡)𝑑𝑡=lim^𝑚_𝑛→∞^𝑛_𝑛=1𝜑𝑡_𝑖^(𝑛)Δ𝑡_𝑖^(𝑛). (2.16)
We take 𝑛 sufficiently large, such that ‖Δ_𝑛‖→0, then we get ∫^𝑏_𝑎𝜑(𝑠)𝑢(𝑠)𝑑𝑠∫^𝑏_𝑎𝜑(𝑠)𝑑𝑠=lim^𝑚_𝑛→∞^𝑛_𝑛=1𝜑𝑡_𝑖^(𝑛)𝑢𝑡_𝑖^(𝑛)Δ𝑡_𝑖^(𝑛)∑^𝑚𝑛_𝑛=1𝜑𝑡_𝑖^(𝑛)Δ𝑡_𝑖^(𝑛)∈co𝑢(𝐽). (2.17) This finishes the proof.

Lemma 2.8. Suppose that 𝐷 is a bounded set in 𝐸, then there exists a countable subset 𝐷₁ of 𝐷, such that 𝛼(𝐷)≤2𝛼𝐷₁. (2.18)

Proof. Let 𝛼(𝐷)>0, 𝐷≠∅. For 𝑟_𝑛=𝛼(𝐷)(1−1/𝑛), take 𝑥₁^(𝑛)∈𝐷, then 𝐷⧵𝐵(𝑥₁^(𝑛),𝑟_𝑛/2) is not a cover of 𝐷. Take 𝑥₂^(𝑛)∈𝐷⧵𝐵(𝑥₂^(𝑛),𝑟_𝑛/2), then 𝐵(𝑥₂^(𝑛),𝑟_𝑛/2) is not a cover of 𝐷. Continuing this process, take 𝑥^(𝑛)_𝑘+1∈𝐷⧵∪^𝑘_𝑖=1𝐵(𝑥_𝑖^(𝑛),𝑟_𝑛/2), then 𝐵(𝑥_𝑖^(𝑛),𝑟_𝑛/2)(𝑖=1,2,3,…,𝑘) is not a cover of 𝐷.
Set 𝐷_𝑛={𝑥_𝑘^(𝑛)∣𝑘=1,2,3,…}, then we get 𝑑(𝑠,𝑡)≥𝑟_𝑛/2, where 𝑑(𝑠,𝑡) denote the distance between two points 𝑠 and 𝑡 of 𝐷_𝑛. Thus it follows that 𝛼(𝐷_𝑛)≥𝑟_𝑛/2.
Setting 𝐷₁=∪^∞_𝑖=1𝐷_𝑛, choose 𝑛 sufficiently large such that 𝛼(𝐷₁)≥𝛼(𝐷_𝑛)≥𝑟_𝑛/2→𝛼(𝐷)/2, that is, 𝛼(𝐷)≤2𝛼(𝐷₁). The proof is completed.

Lemma 2.9. If 𝐵 is a bounded set in 𝐶(𝐼,𝐸), 𝐵(𝐼)={𝑢(𝑡)∣𝑢∈𝐵,𝑡∈𝐼}⊂𝐸. Then 𝛼(𝐵(𝐼))≤2𝛼(𝐵). (2.19)

Proof. For any 𝜀>0, there exists a partition 𝐵=∪^𝑛_𝑖=1𝐵_𝑖 such that diam𝐵_𝑖<𝛼(𝐵)+𝜀, (2.20) for 𝑖=1,2,3,…,𝑛. Choose 𝑢_𝑖∈𝐵_𝑖(𝑖=1,2,3,…,𝑛). Since 𝑢_𝑖 is uniformly continuous on 𝐼, there exists 𝛿>0, such that 𝑡′,𝑡^′′∈𝐼, and |𝑡′−𝑡^′′|<𝛿, we have 𝑢_𝑖𝑡^′−𝑢_𝑖𝑡^′′<𝜀. (2.21)
Let Δ∶0=𝑡₀<𝑡₁<𝑡₂<𝑡₃<⋯<𝑡_𝑚=1 be a partition of 𝐼, and ‖Δ‖<𝛿. Set 𝐼_𝑗=[𝑡_𝑗−1,𝑡_𝑗], 𝐷_𝑖𝑗=𝐵_𝑖(𝐼_𝑗)={𝑢(𝑡)∣𝑢∈𝐵_𝑖,𝑡∈𝐼_𝑗}. Clearly, we have 𝐷=𝐵(𝐼)=^𝑛_𝑖=1𝐵_𝑖(𝐼)=^𝑚𝑛_𝑗=1𝑖=1𝐵_𝑖𝐼_𝑗. (2.22)
For any 𝑢,𝑣∈𝐵_𝑖 and 𝑡,𝑠∈𝐼_𝑗, it follows from (2.20) and (2.21) that ‖𝑢(𝑡)−𝑣(𝑠)‖≤𝑢(𝑡)−𝑢_𝑖(𝑡)+𝑢_𝑖(𝑡)−𝑢_𝑖(𝑠)+𝑣(𝑠)−𝑢_𝑖(𝑠≤)𝑢−𝑢_𝑖𝑐+𝜀+𝑣−𝑢_𝑖𝑐≤2diam𝐵_𝑖(𝐵)+𝜀<2𝛼+3𝜀. (2.23) So, diam𝐷_𝑖𝑗=sup_{𝑢,𝑣∈𝐵𝑖;𝑡,𝑠∈𝐼𝑗}‖𝑢(𝑡)−𝑣(𝑠)‖≤2𝛼(𝐵)+3𝜀. (2.24) Thus it follows that 𝛼(𝐷)≤2𝛼(𝐵)+3𝜀. Therefore, by using the arbitrariness of 𝜀, we have 𝛼(𝐵(𝐼))≤2𝛼(𝐵). The lemma is proved.

Lemma 2.10 (see [7]). Assume that 𝐻⊂𝐶[𝐽,𝐸] is bounded and equicontinuous. Then 𝛼(𝐻(𝑡)) is continuous on 𝐽 and 𝛼_𝐽𝑥(𝑡)𝑑𝑡∶𝑥∈𝐻≤_𝐽𝛼(𝐻(𝑡))𝑑𝑡. (2.25)

Lemma 2.11 (see [7]). Suppose that 𝐻 is a countable family of strongly measurable functions 𝑥∶𝐽↦𝐸. If there exists a function 𝑀∈𝐿[𝐽,ℝ⁺] such that ‖𝑥(𝑡)‖≤𝑀(𝑡) for a.e. 𝑡∈𝐽, then 𝛼(𝐻(𝑡))∈𝐿[𝐽,ℝ⁺] and 𝛼_𝐽𝑥(𝑡)𝑑𝑡∶𝑥∈𝐻≤2_𝐽𝛼(𝐻(𝑡))𝑑𝑡. (2.26)

Lemma 2.12. Assume that Ω₁ is equicontinuous in 𝐶(𝐼,𝐸). Then co(Ω₁) is equicontinuous.

Proof. For any 𝜀>0, it follows from the equicontinuity of Ω₁ that there exists 𝛿>0 such that |𝑡₁−𝑡₂|<𝛿 implies ‖𝑢(𝑡₁)−𝑢(𝑡₂)‖<𝜀/3 for all 𝑡₁,𝑡₂∈𝐼 and 𝑢∈Ω₁.
For any ℎ∈co(Ω₁), by virtue of definition of co(Ω₁), we have ∀𝑢₁,𝑢₂,…,𝑢_𝑛∈Ω₁,𝜆₁,𝜆₂,…,𝜆_𝑛>0,^𝑛_𝑖=1𝜆_𝑖=1,^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖−ℎ_𝑐<𝜀3. (2.27)
Thus, we get ℎ𝑡₁−ℎ𝑡₂≤ℎ𝑡₁−^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖𝑡₁+^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖𝑡₂−ℎ𝑡₂+^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖𝑡₁−𝑢_𝑖𝑡₂≤2^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖−ℎ_𝑐+^𝑛_𝑖=1𝜆_𝑖𝑢_𝑖𝑡₁−𝑢_𝑖𝑡₂<23𝜀+^𝑛_𝑖=1𝜆_𝑖⋅13𝜀=𝜀. (2.28) Hence, co(Ω₁) is equicontinuous. This finishes the proof.

Lemma 2.13 (see [7]). If 𝐻⊂𝐶[𝐼,𝐸] is bounded and equicontinuous, then 𝛼(𝐵(𝑡)) is continuous in 𝐼 and 𝛼(𝐵)=𝛼(𝐵(𝐼))=max_𝑥∈𝐼𝛼(𝐵(𝑡)).

Lemma 2.14 (see [7] (Sadovskii’s Theorem)). Assume that 𝐷 is a nonempty bounded, closed, and convex set. If a mapping 𝐴∶𝐷↦𝐷 is condensing, then 𝐴 has a fixed point in 𝐷.

3. Main Results

In this section, we present and prove our main results.

Theorem 3.1. Let 𝐸 be a Banach space. Suppose that 𝑓(𝑡,𝑥)∈𝐶(𝐼×𝐸,𝐸) and the following conditions hold:(H₁) there exist two positive numbers 𝑐₀ and 𝑐₁, such that𝑓(𝑡,𝑥)≤𝑐₀+𝑐₁‖𝑥‖,∀𝑡∈𝐼,𝑥∈𝐸, (3.1)(H₂) for any a bounded set 𝐷 in 𝐸, there exists a constant 𝐿>0 such that𝛼𝑓(𝐼×𝐷)≤𝐿𝛼(𝐷), (3.2)(H₃) there exist two positive numbers 𝑐₁ and 𝐿 with 𝑐₁<𝜋², 𝐿<4. Then problem (1.1) has at least one solution.
Proof. Define the integral operator 𝐴∶𝐶(𝐼,𝐸)↦𝐶(𝐼,𝐸)(𝐴𝑢)(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝑇₀𝑓(⋅,𝑢). (3.3) Then 𝐴∶𝐶(𝐼,𝐸)↦𝐶(𝐼,𝐸) is continuous, and it is clear that 𝑢 is a solution of the problem (1.1) if and only if 𝑢 is a fixed point of 𝐴.
We now show that 𝐴 is a condensing operator. Let 𝐵 be bounded in 𝐶(𝐼,𝐸), by (H₁), we claim that {−(𝐴𝑢)^′′∣𝑢∈𝐵} is bounded. Since (𝐴𝑢)(0)=0, we know that {(𝐴𝑢)′∣𝑢∈𝐵} is bounded, this means that 𝐴(𝐵) is equicontinuous. Therefore, it follows from Lemma 2.13 that 𝛼(𝐴(𝐵))=max_𝑥∈𝐼𝛼(𝐴(𝐵)(𝑡)).
For any 𝑢∈𝐵, 𝑡∈𝐼, from Lemma 2.7, we have (𝐴𝑢)(𝑡)=¹₀𝐺₀()𝑓((𝑠∈𝑡,𝑠𝑠,𝑢))𝑑𝑠₁₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅𝑓((𝑠⊂co𝑠,𝑢))∣𝑠∈𝐼₁₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅co𝑓(𝐼×𝐵(𝐼)). (3.4) Hence, (𝐴(𝐵))(𝑡)⊂¹₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅co𝑓(𝐼×𝐵(𝐼)). (3.5)
Using the properties of the noncompactness measure together with (H₂), we obtain 𝛼(𝐴(𝐵)(𝑡))≤¹₀𝐺₀()𝑓((𝐼(𝐵(𝐼⋅𝑡,𝑠𝑑𝑠⋅𝛼𝐼×𝐵))≤𝐿𝛼))¹₀𝐺₀()=1𝑡,𝑠𝑑𝑠2𝑡()(𝐵(𝐼≤11−𝑡⋅𝐿𝛼))8𝐿𝛼(𝐵(𝐼)). (3.6)
By Lemma 2.9, we have 𝛼(𝐴(𝐵𝑡≤1)())8(𝐵(𝐼≤1𝐿⋅𝛼))8(𝐵)=1𝐿⋅2𝛼4𝐿⋅𝛼(𝐵). (3.7)
Hence, 𝛼(𝐴(𝐵≤1))4𝐿⋅𝛼(𝐵). (3.8) By (H₃), we get 0<𝐿/4<1, therefore 𝐴 is condensing.
Let Ω∶=𝐵_𝑐(𝜃,𝑅)={𝑢∈𝐶(𝐼,𝐸)∶‖𝑢‖<𝑅}, we will prove that 𝑢−𝜆𝐴𝑢≠𝜃, 0<𝜆≤1,𝑢∈𝜕Ω for 𝑅 sufficiently large. By means of the homotopy invariance theorem, we have deg(𝐼−𝐴,Ω,𝜃)=1. By virtue of the solvability of Kronecker [6], we know that 𝐴 has a fixed point in Ω, and the fixed point of 𝐴 is a solution of the problem (1.1).
Indeed. If there exists a constant 𝜆₀∈(0,1], 𝜆₀∈𝜕Ω such that 𝑢₀−𝜆₀𝐴𝑢₀=𝜃, then 𝑢₀ satisfies 𝑢₀=𝜆₀¹₀𝐺₀(𝑡,𝑠)𝑓𝑠,𝑢₀(𝑠)𝑑𝑠. (3.9)
Let 𝜑₀(𝑡)=‖𝑢₀(𝑡)‖, and 𝑇₀ is an operator in 𝐶(𝐼) defined by 𝑇₀𝜑(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝜑(𝑠)𝑑𝑠. (3.10) Then by Lemma 2.5 we have 𝑟(𝑇₀)=1/𝜋².
By (3.9) and (H₁), we have 𝜑₀(𝑡)≤¹₀𝐺₀(𝑡,𝑠)𝑐₀+𝑐₁𝜑₀(𝑠)𝑑𝑠=𝑐₀𝑇₀(1)+𝑐₁𝑇₀𝜑₀(𝑡)≤𝑐₀𝑇₀+𝑐₁𝑇₀𝜑₀(𝑡)≤𝑐₀+𝑐₁𝑇₀𝜑₀(𝑡). (3.11)
Thus, 𝜑₀≤𝑐₀𝑇₀+𝑐₁𝑇₀𝜑₀≤𝑐₀+𝑐₁𝑇₀𝜑₀, (3.12) continuing this process, by induction, we obtain 𝜑₀≤𝑐₀+𝑐₁𝑇₀𝑐₀+𝑐₁𝑇₀𝜑₀=𝑐₀+𝑐₀𝑐₁𝑇₀+𝑐²₁𝑇²₀𝜑₀≤⋯≤𝑐₀1+𝑐₁𝑇₀+𝑐²₁𝑇²₀+⋯+𝑐₁^𝑛−1𝑇₀^𝑛−1+𝑐^𝑛₁𝑇^𝑛₀𝜑₀, (3.13) which implies that 𝜑₀≤𝑐₀^𝑛−1_𝑛=0𝑐^𝑛₁𝑇^𝑛₀+𝑐^𝑛₁𝑇^𝑛₀𝜑₀. (3.14)
By the Gelfand theorem [22], we have 𝑟𝑇₀=lim_𝑛→∞^𝑛𝑇^𝑛₀=1𝜋². (3.15)
Set 𝑐₂=(𝑐₁+𝜋²)/2∈(𝑐₁,𝜋²), then we have 1/𝑐₂>1/𝜋². By (3.15), there exists an integer 𝑁₀, such that ^𝑛‖𝑇^𝑛₀‖<1/𝑐₂ as 𝑛≥𝑁₀, that is, ‖𝑇^𝑛₀‖<1/𝑐^𝑛₂, this means that 𝑐^𝑛₁‖𝑇^𝑛₀‖<(𝑐₁/𝑐₂)^𝑛. In view of series ∑^∞_𝑛=0(𝑐₁/𝑐₂)^𝑛 converges, we know that ∑^∞_𝑛=0𝑐^𝑛₁‖𝑇^𝑛₀‖ also converges.
Denote 𝑅₀=𝑐₀∑^∞_𝑛=0𝑐^𝑛₁‖𝑇^𝑛₀‖. By (3.14), we get 𝜑₀(𝑡)≤𝑅₀, which implies that ‖𝑢₀(𝑡)‖≤𝑅₀, hence ‖𝑢₀‖_𝑐≤𝑅₀.
Take 𝑅>𝑅₀, then we have 𝑢−𝜆𝐴𝑢≠𝜃, forall𝜆∈(0,1],𝑢∈𝜕Ω. Thus problem (1.1) has at least one solution. This proves the theorem.

Remark 3.2. In [11], the nonlinear term 𝑓(𝑡,𝑢,𝑢^′) is bounded, if 𝑓(𝑡,𝑢,𝑢^′)=𝑓(𝑡,𝑢), in our result, the nonlinear term 𝑓(𝑡,𝑢) may no more than a linear growth.

Remark 3.3. In [11], if 𝑓(𝑡,𝑢,𝑢^′)=𝑓(𝑡,𝑢), the growth restriction of 𝐿 for (H₂) is 0<𝐿<1/2. However, in our result, 𝐿 satisfies 0<𝐿<4.

Theorem 3.4. Let 𝐸 be a Banach space, and 𝑓(𝑡,𝑥)∈𝐶(𝐼×𝐸,𝐸). Assume that condition (H₁) and the following conditions hold:(H₂)^′ for all 𝑡∈𝐼, for any a bounded set 𝐷 in 𝐸, there exists a constant 𝐿>0 such that𝛼𝑓(𝑡,𝐷)≤𝐿𝛼(𝐷), (3.16)(H₃)^′ there exist two positive numbers 𝑐₁ and 𝐿 with 0<𝑐₁<𝜋², 𝐿<2. Then problem (1.1) has at least one solution.

Proof. Assume that the operator 𝐴 is defined the same as in Theorem 3.1. We show that the operator 𝐴 is condensing. In fact, for a bounded set 𝐵∈𝐶(𝐼,𝐸), there exists a countable subset 𝐵₁={𝑢_𝑛}, such that 𝛼(𝐴(𝐵))≤2𝛼(𝐴(𝐵₁)). However, on the other hand, we have 𝛼𝐴𝐵₁=max_𝑡∈𝐼𝛼𝐴𝐵₁(𝑡). (3.17)
By Lemma 2.11 and (H₂)^′, we obtain 𝛼𝐴𝐵₁=𝛼¹₀𝐺₀(𝑡,𝑠)𝑓𝑠,𝑢_𝑛(𝑠)𝑑𝑠∣𝑛∈ℕ≤2¹₀𝐺₀(𝑡,𝑠)𝛼𝑓𝑠,𝑢_𝑛(𝑠)∣𝑛∈ℕ𝑑𝑠≤2¹₀𝐺₀(𝑡,𝑠)𝐿𝛼𝐵₁(𝑠)𝑑𝑠≤2𝐿¹₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅𝛼𝐵₁≤14𝐿𝛼𝐵₁. (3.18) Thus, 𝛼(𝐴(𝐵≤1))2𝐿𝛼(𝐵). (3.19) By (H₃)^′, we get 0<𝐿/2<1, therefore 𝐴 is condensing. By using the same arguments of Theorem 3.1, we can obtain the conclusion of Theorem 3.4. The detailed proof is omitted here. The proof is achieved.

Next, we establish a uniqueness of solution for the problem (1.1).

Theorem 3.5. Let 𝐸 be a Banach space. Suppose that 𝑓(𝑡,𝑥)∈𝐶(𝐼×𝐸,𝐸) and that there exists a constant 𝐿 with 0<𝐿<𝜋² such that 𝑓𝑡,𝑢₂−𝑓𝑡,𝑢₁≤𝐿𝑢₂−𝑢₁,∀𝑢₁,𝑢₂∈𝐸. (3.20) Then problem (1.1) has a unique solution.

Proof. Assume that operator 𝐴 is defined the same as in Theorem 3.1, and the fixed point of 𝐴 is a solution of the problem (1.1).
We will prove that for sufficiently large 𝑛 the operator 𝐴^𝑛 is a contraction operator. Indeed, by the definition of 𝐴 and (3.20), we have the estimate 𝐴^𝑛𝑢₂(𝑡)−𝐴^𝑛𝑢₁(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝑓𝑠,𝐴^𝑛−1𝑢₂(𝑠)−𝑓𝑠,𝐴^𝑛−1𝑢₁(𝑠)𝑑𝑠≤𝐿¹₀𝐺₀(𝑡,𝑠)𝐴^𝑛−1𝑢₂(𝑠)−𝐴^𝑛−1𝑢₁(𝑠)𝑑𝑠=𝐿𝑇₀𝐴^𝑛−1𝑢₂(𝑠)−𝐴^𝑛−1𝑢₁(𝑠). (3.21)
By induction, we have 𝐴^𝑛𝑢₂(𝑡)−𝐴^𝑛𝑢₁(𝑡)≤𝐿^𝑛⋅𝑇^𝑛₀⋅𝑢₂(𝑠)−𝑢₁(𝑠)≤𝐿^𝑛⋅𝑇^𝑛₀⋅𝑢₂−𝑢_1𝑐. (3.22) Thus, 𝐴^𝑛𝑢₂−𝐴^𝑛𝑢_1𝑐≤𝐿^𝑛⋅𝑇^𝑛₀⋅𝑢₂−𝑢_1𝑐. (3.23)
Moreover, we can choose 𝑛 to be sufficiently large such that ^𝑛‖𝑇^𝑛₀‖ tends to 𝑟(𝑇₀)=1/𝜋².
Further, take 𝐿₁∈(𝐿,𝜋²), there exists an integer 𝑁₀, such that ‖𝑇₀‖≤1/𝐿^𝑛₁ as 𝑛≥𝑁₀. By (3.23), we obtain 𝐴^𝑛𝑢₂−𝐴^𝑛𝑢_1𝑐≤𝐿^𝑛⋅𝑇^𝑛₀⋅𝑢₂−𝑢_1𝑐≤𝐿𝐿₁^𝑛⋅𝑢₂−𝑢_1𝑐, (3.24) which implies that 𝐴^𝑛 is a contraction mapping by 𝐿/𝐿₁<1. By the contraction mapping principle, we conclude that there exists a unique fixed point for 𝐴, this proves that problem (1.1) has a unique solution. This completes the proof.

Remark 3.6. By the direct application of the Banach contraction mapping principle, the conclusion of Theorem 3.5 holds true under the condition 0<𝐿<8. However, we require the condition 0<𝐿<𝜋², here 𝜋² is optimum.
The following theorem is concerned with the existence of positive solutions for problem (1.1).

Theorem 3.7. Let 𝐸 be an ordered Banach space, 𝐾 be a normal cone with positive elements. Suppose that 𝑓(𝑡,𝑥)∈𝐶(𝐼×𝐸,𝐸) satisfy the following conditions:(P₁) there exists a constant 𝑐 with 0<𝑐<𝜋², and ℎ₀∈𝐶(𝐼,𝐾), such that𝜃≤𝑓(𝑡,𝑥)≤𝑐𝑥+ℎ₀(𝑡),∀𝑥≥𝜃, (3.25)(P₂) for any a bounded set 𝐷 in 𝐸, there exists a constant 𝐿 with 0<𝐿<8 such that𝛼𝑓(𝐼×𝐷)≤𝐿𝛼(𝐷). (3.26) Then problem (1.1) has at least one positive solution.

Proof. Consider the linear boundary value problem −𝑢^′′−𝑐𝑢=ℎ₀,𝑢(0)=𝑢(1)=𝜃. (3.27) By Lemma 2.3, the linear boundary value problem (3.27) has unique a positive solution 𝑢₀∈𝐶(𝐼,𝐾).
Set 𝐷=[𝜃,𝑢₀]⊂𝐶(𝐼,𝐸), then 𝐷 is bounded and convex closed set in 𝐶(𝐼,𝐸). For any 𝑢∈𝐷, by (P₁), we have 𝜃≤𝑓(𝑡,𝑢(𝑡))≤𝑐𝑢(𝑡)+ℎ₀(𝑡)≤𝑐𝑢₀(𝑡)+ℎ₀(𝑡). (3.28)
Multiply by 𝐺₀(𝑡,𝑠) and integrate from 0 to 1, we obtain 𝜃≤𝐴𝑢(𝑡)≤¹₀𝐺₀(𝑡,𝑠)𝑐𝑢₀(𝑠)+ℎ₀(𝑠)𝑑𝑠=𝑢₀(𝑡), (3.29) that is, 𝐴𝑢∈𝐷, so 𝐴(𝐷)⊂𝐷. By the proof of Theorem 3.1, it follows that 𝐴 is equicontinuous. Thus, by Lemma 2.12, we know that  Ω₀=co(𝐴(𝐷)) is equicontinuous.
Next we show that 𝐴∶Ω₀→Ω₀ is condensing. For any 𝐵⊂Ω₀, then 𝐵 is bounded and equicontinuous, therefore 𝐴(𝐵)⊂Ω₀ is bounded and equicontinuous. By Lemma 2.13, we have 𝛼(𝐴(𝐵))=max_𝑡∈𝐼𝛼(𝐴(𝐵(𝑡))). (3.30)
However, on the other hand, we have 𝛼(𝐴(𝐵(𝑡)))=𝛼¹₀𝐺₀(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠∣𝑢∈𝐵. (3.31)
Thus, For any 𝑢∈𝐵, 𝑡∈𝐼, we acquire 𝐴𝑢(𝑡)=¹₀𝐺₀()𝑓((𝑠∈𝑡,𝑠𝑠,𝑢))𝑑𝑠₁₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅𝑓((𝑠⊂co𝑠,𝑢))∣𝑠∈𝐼₁₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅co𝑓(𝐼×𝐵(𝐼)). (3.32)
Hence, 𝐴(𝐵)(𝑡)⊂¹₀𝐺₀(𝑡,𝑠)𝑑𝑠⋅co𝑓(𝐼×𝐵(𝐼)). (3.33)
Further, we obtain 𝛼(A(𝐵)(𝑡))≤¹₀𝐺₀()𝑓((𝐼≤𝐿𝑡,𝑠𝑑𝑠⋅𝛼𝐼×𝐵))2𝑡()(𝐵(𝐼≤𝐿1−𝑡⋅𝛼))8⋅𝛼(𝐵). (3.34)
This means that 𝛼(𝐴(𝐵≤𝐿))8⋅𝛼(𝐵). (3.35)
By (P₂), we have 0<𝐿/8<1, so 𝐴 is condensing. Applying Lemma 2.14, we conclude that 𝐴 has a fixed point which is a solution of problem (1.1). The proof of the theorem is completed.

Acknowledgments

Wen-Xue Zhou's work was supported by NNSF of China (10901075), Program for New Century Excellent Talents in University (NCET-10-0022), the Key Project of Chinese Ministry of Education (210226), and NSF of Gansu Province of China (1107RJZA091).