Symmetric Positive Solutions for Nonlinear Singular Fourth-Order Eigenvalue Problems with Nonlocal Boundary Condition

Abstract

We investigate nonlinear singular fourth-order eigenvalue problems with nonlocal boundary condition 𝑢⁽⁴⁾(𝑡)−𝜆ℎ(𝑡)𝑓(𝑡,𝑢,𝑢^′′)=0, 0<𝑡<1, 𝑢(0)=𝑢(1)=∫¹₀𝑎(𝑠)𝑢(𝑠)𝑑𝑠, 𝑢^′′(0)=𝑢^′′(1)=∫¹₀𝑏(𝑠)𝑢^′′(𝑠)𝑑𝑠, where 𝑎,𝑏∈𝐿¹[0,1], 𝜆>0, ℎ may be singular at 𝑡=0 and/or 1. Moreover 𝑓(𝑡,𝑥,𝑦) may also have singularity at 𝑥=0 and/or 𝑦=0. By using fixed point theory in cones, an explicit interval for 𝜆 is derived such that for any 𝜆 in this interval, the existence of at least one symmetric positive solution to the boundary value problem is guaranteed. Our results extend and improve many known results including singular and nonsingular cases. The associated Green's function for the above problem is also given.

1. Introduction

Boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics and so on, and the existence of positive solutions for such problems has become an important area of investigation in recent years. To identify a few, we refer the reader to [1–7] and references therein.

At the same time, a class of boundary value problems with nonlocal boundary conditions appeared in heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics. Such problems include two-point, three-point, multipoint boundary value problems as special cases and have attracted the attention of Gallardo [1], Karakostas and Tsamatos [2], and Lomtatidze and Malaguti [3] (and see the references therein). For more information about the general theory of integral equations and their relation to boundary value problems we refer the reader to the book of Corduneanu [8] and Agarwal and O'Regan [9].

Motivated by the works mentioned above, in this paper, we study the existence of symmetric positive solutions of the following fourth-order nonlocal boundary value problem (BVP):

𝑢⁽⁴⁾(𝑡)−𝜆ℎ(𝑡)𝑓(𝑡,𝑢,𝑢^′′)=0,0<𝑡<1,𝑢(0)=𝑢(1)=¹₀𝑢𝑎(𝑠)𝑢(𝑠)𝑑𝑠,_′′(0)=𝑢^′′(1)=¹₀𝑏(𝑠)𝑢^′′(𝑠)𝑑𝑠, (1.1) where 𝑎,𝑏∈𝐿¹[0,1], 𝜆>0, ℎ may be singular at 𝑡=0 and/or 1. Moreover 𝑓(𝑡,𝑥,𝑦) may also have singularity at 𝑥=0 and/or 𝑦=0.

The main features of this paper are as follows. Firstly, comparing with [4–7], we discuss the boundary value problem with nonlocal boundary conditions, that is, BVP (1.1) including fourth-order two-point, three-point, multipoint boundary value problems as special cases. Secondly, comparing with [4–7], we discuss the boundary value problem when nonlinearity 𝑓 contains second-derivatives 𝑢^′′. Thirdly, here we not only allow ℎ have singularity at 𝑡=0 and/or 1 but also allow 𝑓(𝑡,𝑥,𝑦) have singularity at 𝑥=0 and/or 𝑦=0. Finally, in [4–7], authors only studied the existence of positive solutions. However, they did not further provide characters of positive solutions, such as symmetry. It is now natural to consider the existence of symmetric positive solutions. To our knowledge, no paper has considered the existence of symmetric positive solutions and nonlinearity 𝑓 with singularity at 𝑥=0 and (or) 𝑦=0 for fourth-order equation with nonlocal boundary condition. Hence we improve and generalize the results of [4–7] to some degree, and so it is interesting and important to study the existence of symmetric positive solutions for problem (1.1). The arguments are based upon a specially constructed cone and the fixed point theory for cones.

Let 𝐾 be a cone in a Banach space 𝐸 and let 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟},𝜕𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}, and 𝐾_𝑟,𝑅={𝑥∈𝐾∶𝑟≤‖𝑥‖≤𝑅}, where 0<𝑟<𝑅<+∞.

Our main tool of this paper is the following fixed point theorem.

Lemma 1.1 (see [10]). Let 𝐾 be a positive cone in real Banach space 𝐸, 0<𝑟<𝑅<+∞, and let 𝑇∶𝐾_𝑟,𝑅→𝐾 be a completely continuous operator and such that (i)‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾_𝑅;(ii)there exists 𝑒∈𝜕𝐾₁ such that 𝑥≠𝑇𝑥+𝑚𝑒 for any 𝑥∈𝜕𝐾_𝑟 and 𝑚>0.Then 𝑇 has a fixed point in 𝐾_𝑟,𝑅.

Remark 1.2. If (i) and (ii) are satisfied for 𝑥∈𝜕𝐾_𝑟 and 𝑥∈𝜕𝐾_𝑅, respectively, then Lemma 1.1 is still true.

The following concept will also be utilized.

Definition 1.3. If 𝑢∶[0,1]→𝑅 is a continuous function and 𝑢(𝑡)=𝑢(1−𝑡) for 𝑡∈[0,1], then one says that 𝑢 is symmetric on [0,1].

2. Preliminaries and Lemmas

In this section, we present some lemmas that are important to prove our main results.

Lemma 2.1. Suppose that 𝑑∶=∫¹₀𝑚(𝑠)𝑑𝑠≠1, 𝑚∈𝐿¹[0,1], 𝑦∈𝐶[0,1], then BVP 𝑢^′′(𝑡)+𝑦(𝑡)=0,0<𝑡<1,𝑢(0)=𝑢(1)=¹₀𝑚(𝑠)𝑢(𝑠)𝑑𝑠 (2.1)(2.2) has a unique solution 𝑢(𝑡)=¹₀𝐻(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (2.3) where 1𝐻(𝑡,𝑠)=𝐺(𝑡,𝑠)+1−𝑑¹₀𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥,𝐺(𝑡,𝑠)=𝑡(1−𝑠),0≤𝑡≤𝑠≤1,𝑠(1−𝑡),0≤𝑠≤𝑡≤1. (2.4)

Proof. Integrating both sides of (2.1) on [0,𝑡], we have 𝑢^′(𝑡)=−^𝑡₀𝑦(𝑠)𝑑𝑠+𝐵. (2.5) Again integrating (2.5) from 0 to 𝑡, we get 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)𝑦(𝑠)𝑑𝑠+𝐵𝑡+𝐴. (2.6) In particular, 𝑢(1)=−¹₀(1−𝑠)𝑦(𝑠)𝑑𝑠+𝐵+𝐴,𝑢(0)=𝐴. (2.7) By (2.2) we get 𝐵=¹₀(1−𝑠)𝑦(𝑠)𝑑𝑠. (2.8) By 𝐺(𝑠,𝑥)=𝐺(𝑥,𝑠) and (2.6), we can get 𝐴=𝑢(0)=¹₀𝑚(𝑥)𝑢(𝑥)𝑑𝑥=¹₀𝑚(𝑥)(−^𝑥₀=(𝑥−𝑠)𝑦(𝑠)𝑑𝑠+𝐵𝑥+𝐴)𝑑𝑥₁₀𝑚(𝑥)(−^𝑥₀(𝑥−𝑠)𝑦(𝑠)𝑑𝑠+𝑥¹₀(1−𝑠)𝑦(𝑠)𝑑𝑠)𝑑𝑥+𝐴¹₀=𝑚(𝑥)𝑑𝑥₁₀𝑚(𝑥)(^𝑥₀𝑠(1−𝑥)𝑦(𝑠)𝑑𝑠+¹_𝑥=𝑥(1−𝑠)𝑦(𝑠)𝑑𝑠)𝑑𝑥+𝐴𝑑₁₀𝑚(𝑥)(¹₀=𝐺(𝑠,𝑥)𝑦(𝑠)𝑑𝑠)𝑑𝑥+𝐴𝑑₁₀(¹₀𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥)𝑦(𝑠)𝑑𝑠+𝐴𝑑. (2.9) So, 1𝐴=1−𝑑¹₀(¹₀𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥)𝑦(𝑠)𝑑𝑠. (2.10) By (2.6), (2.8), and (2.10), we have 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)𝑦(𝑠)𝑑𝑠+𝐵𝑡+𝐴=−^𝑡₀(𝑡−𝑠)𝑦(𝑠)𝑑𝑠+𝑡¹₀1(1−𝑠)𝑦(𝑠)𝑑𝑠+1−𝑑¹₀(¹₀=𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥)𝑦(𝑠)𝑑𝑠_𝑡0𝑠(1−𝑡)𝑦(𝑠)𝑑𝑠+¹_𝑡1𝑡(1−𝑠)𝑦(𝑠)𝑑𝑠+1−𝑑¹₀(¹₀=𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥)𝑦(𝑠)𝑑𝑠₁₀1𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+1−𝑑¹₀(¹₀=𝐺(𝑠,𝑥)𝑚(𝑥)𝑑𝑥)𝑦(𝑠)𝑑𝑠₁₀𝐻(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (2.11) This completes the proof of Lemma 2.1.

It is easy to verify the following properties of 𝐻(𝑡,𝑠) and 𝐺(𝑡,𝑠).

Lemma 2.2. If 𝑚(𝑡)>0, and 𝑑∶=∫¹₀𝑚(𝑠)𝑑𝑠∈(0,1), then (1)𝐻(𝑡,𝑠)≥0, 𝑡,𝑠∈[0,1], 𝐻(𝑡,𝑠)>0, 𝑡,𝑠∈(0,1);(2)𝐺(1−𝑡,1−𝑠)=𝐺(𝑡,𝑠), 𝐺(𝑡,𝑡)≤𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠), 𝑡,𝑠∈[0,1];(3)𝛾₁𝐻(𝑠,𝑠)≤𝐻(𝑡,𝑠)≤𝐻(𝑠,𝑠), where 𝛾₁ = 𝜂₁/(1−𝑑+𝜂₁)∈(0,1), 𝜂₁ = ∫¹₀𝐺(𝑥,𝑥)𝑚(𝑥)𝑑𝑥.

So we may denote Green's function of boundary value problem

−𝑢^′′(𝑡)=0,0<𝑡<1,𝑢(0)=𝑢(1)=¹₀𝑎(𝑠)𝑢(𝑠)𝑑𝑠,−𝑢^′′(𝑡)=0,0<𝑡<1,𝑢(0)=𝑢(1)=¹₀𝑏(𝑠)𝑢(𝑠)𝑑𝑠 (2.12) by 𝐻₁(𝑡,𝑠) and 𝐻₂(𝑡,𝑠), respectively. By Lemma 2.1, we know that 𝐻₁(𝑡,𝑠) and 𝐻₂(𝑡,𝑠) can be written by

𝐻₁1(𝑡,𝑠)=𝐺(𝑡,𝑠)+1−∫¹₀𝑎(𝑠)𝑑𝑠¹⁰𝐻𝐺(𝑠,𝑥)𝑎(𝑥)𝑑,₂1(𝑡,𝑠)=𝐺(𝑡,𝑠)+1−∫¹₀𝑏(𝑠)𝑑𝑠¹⁰𝐺(𝑠,𝑥)𝑏(𝑥)𝑑𝑥. (2.13) Obviously, 𝐻₁(𝑡,𝑠) and 𝐻₂(𝑡,𝑠) have the same properties with 𝐻(𝑡,𝑠) in Lemma 2.2.

Remark 2.3. For notational convenience, we introduce the following constants: 𝛼=¹₀𝑎(𝑠)𝑑𝑠,𝛽=¹₀𝑏(𝑠)𝑑𝑠,𝜇=max_0≤𝑡≤1¹₀𝐻₁𝜂(𝑡,𝑠)𝑑𝑠,𝛾=1−𝛽+𝜂∈(0,1),𝜂=¹₀𝐺(𝑥,𝑥)𝑏(𝑥)𝑑𝑥,𝐿=(1+𝜇)¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑑𝑠,𝑙=min_𝑡∈[0,1]¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑑𝑠. (2.14) Obviously, 𝜇>0, 0<𝑙<𝐿<∞.

Now we define an integral operator 𝑆∶𝐶[0,1]→𝐶[0,1] by

(𝑆𝑣)(𝑡)=¹₀𝐻₁(𝑡,𝜏)𝑣(𝜏)𝑑𝜏. (2.15) Then, we have

(𝑆𝑣)^′′(𝑡)=−𝑣(𝑡),0<𝑡<1,𝑆𝑣(0)=𝑆𝑣(1)=¹₀𝑎(𝑠)𝑆𝑣(𝑠)𝑑𝑠. (2.16)

Lemma 2.4. The fourth-order nonlocal boundary value problem (1.1) has a positive solution if and only if the following integral-differential boundary value problem 𝑣^′′(𝑡)+𝜆ℎ(𝑡)𝑓(𝑡,𝑆𝑣(𝑡),−𝑣(𝑡))=0,0<𝑡<1,𝑣(0)=𝑣(1)=¹₀𝑏(𝑠)𝑣(𝑠)𝑑𝑠 (2.17) has a positive solution.

Proof. In fact, if 𝑢 is a positive solution of (BVP) (1.1), let 𝑢(𝑡)=𝑆𝑣(𝑡), then 𝑣=−𝑢^′′. This implies that 𝑢^′′=−𝑣 is a solution of (2.17). Conversely, if 𝑣 is a positive solution of (2.17). Let 𝑢(𝑡)=𝑆𝑣(𝑡), by (2.16), 𝑢^′′(𝑡)=(𝑆𝑣)^′′(𝑡)=−𝑣. Thus 𝑢(𝑡)=𝑆𝑣(𝑡) is a positive solution of (BVP) (1.1). This completes the proof of Lemma 2.4.

So we will concentrate our study on (2.17). Let 𝐶⁺[0,1]={𝑥∈𝐶[0,1]∶𝑥≥0} and

𝐾={𝑥∈𝐶⁺[0,1]∶𝑥(𝑡)issymmetricandconcavefunctionon[0,1],min_𝑡∈[0,1]𝑥(𝑡)≥𝛾‖𝑥‖}; (2.18)‖⋅‖ is the supremum norm on 𝐶⁺[0,1]. It is easy to see that 𝐾 is a cone in 𝐶[0,1] and 𝐾_𝑟,𝑅⊂𝐾⊂𝐶⁺[0,1]. Now we define an operator 𝑇∶𝐾⧵{0}→𝐶⁺[0,1] by

(𝑇𝑣)(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠,𝑡∈[0,1]. (2.19) Clearly 𝑣 is a solution of the BVP (2.17) if and only if 𝑣 is a fixed point of the operator 𝑇.

In the rest of the paper, we make the following assumptions:

(𝐻₁)𝑎,𝑏∈𝐿¹[0,1], 𝑎(𝑡)≥0, 𝑏(𝑡)≥0, 𝑎(1−𝑡)=𝑎(𝑡),𝑏(1−𝑡)=𝑏(𝑡), 𝛼,𝛽∈(0,1);(𝐻₂)ℎ∈𝐶((0,1),[0,+∞)), ℎ(1−𝑡)=ℎ(𝑡), 0<∫¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑑𝑠<+∞;(𝐻₃)𝑓(𝑡,𝑢,𝑣)∈𝐶((0,1)×(0,+∞)×(−∞,0),[0,+∞)), 𝑓((1−𝑡),𝑢,𝑣)=𝑓(𝑡,𝑢,𝑣), and for any 0<𝑟₁<𝑅₁<+∞, 0<𝑟₂<𝑅₂<+∞lim_𝑛→+∞sup_{𝑢∈𝐾𝑟11,𝑅,−𝑣∈𝐾𝑟22,𝑅𝐷(𝑛)}𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑢(𝑠),𝑣(𝑠))𝑑𝑠=0, (2.20) where 𝐷(𝑛)=[0,1/𝑛]∪[(𝑛−1)/𝑛,1].

Remark 2.5. If (𝐻₁) holds, then for all 𝑡,𝑠∈[0,1], we have 𝐻₁(1−𝑡,1−𝑠)=𝐻₁(𝑡,𝑠),𝐻₂(1−𝑡,1−𝑠)=𝐻₂(𝑡,𝑠). (2.21)

Lemma 2.6. Assume that conditions (𝐻₁), (𝐻₂), and (𝐻₃) hold. Then 𝑇∶𝐾_𝑟,𝑅→𝐶⁺[0,1] is a completely continuous operator.

Proof. Firstly, for any 𝑟>0, we will show sup_{𝑣∈𝜕𝐾𝑟}𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠<+∞. (2.22) At the same time, this implies that 𝑇∶𝐾⧵{0}→𝐶⁺[0,1] is well defined.
In fact, by (𝐻₃), for any 𝑟>0, there exists a natural number 𝑚 such that
sup_{𝑣∈𝜕𝐾𝑟}𝜆_𝐷(𝑚)𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠<1. (2.23) For any 𝑣∈𝜕𝐾_𝑟, let 𝑣(𝑡₀)=max_𝑡∈[0,1]|𝑣(𝑡)| = 𝑟. It follows from the concavity of 𝑣(𝑡) on [0,1] that 𝑣(𝑡)≥𝑟𝑡𝑡₀,0≤𝑡≤𝑡₀,𝑟1−𝑡₀(1−𝑡),𝑡₀≤𝑡≤1. (2.24) So we obtain 𝑣(𝑡)≥𝑟𝑡,0≤𝑡≤𝑡₀,𝑟(1−𝑡),𝑡₀≤𝑡≤1. (2.25) Consequently, from (2.25) for any 𝑡∈[1/𝑚,(𝑚−1)/𝑚], we have 𝑟/𝑚≤𝑣(𝑡)≤𝑟 and 𝑙_𝑚𝑟𝑚=𝑟𝑚min_{𝑡∈[1/𝑚,(𝑚−1)/𝑚]}¹₀𝐻₁(𝑡,𝑠)𝑑𝑠≤𝑆𝑣(𝑡)≤𝑟max_{𝑡∈[1/𝑚,(𝑚−1)/𝑚]}¹₀𝐻₁(𝑡,𝑠)𝑑𝑠≤𝜇𝑟, (2.26) where 𝑙_𝑚=min_{𝑡∈[1/𝑚,(𝑚−1)/𝑚]}∫¹₀𝐻₁(𝑡,𝑠)𝑑𝑠 and 𝜇 is defined in Remark 2.3. Let 𝑀₁1=max{𝑓(𝑡,𝑥,𝑦)∶(𝑡,𝑥,𝑦)∈[𝑚,𝑚−1m𝑙]×[^𝑚𝑟𝑚𝑟,𝜇𝑟]×[−𝑟,−𝑚]}. (2.27) By (𝐻₁)-(𝐻₃), we have sup_{𝑣∈𝜕𝐾𝑟}𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≤sup_{𝑣∈𝜕𝐾𝑟}𝜆_𝐷(𝑚)𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠+sup_{𝑣∈𝜕𝐾𝑟}𝜆^{(𝑚−1)/𝑚}_1/𝑚𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≤1+𝑀₁𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑑𝑠<+∞, (2.28) that is, (2.22) holds. This also implies that 𝑇(𝐵) is uniformly bounded for any bounded set 𝐵⊂𝐾_𝑟,𝑅 from (2.28).
Next we prove that 𝑇 is equicontinuous on 𝐾_𝑟,𝑅. In fact, by (𝐻₃) for any 𝜀>0, there exists a natural number 𝑘 such that
sup_{𝑣∈𝐾𝑟,𝑅}𝜆_𝐷(𝑘)𝐻₂𝜀(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠<4. (2.29) Let 𝑀₂1=max{𝑓(𝑡,𝑥,𝑦)∶(𝑡,𝑥,𝑦)∈[𝑘,𝑘−1𝑘𝑙]×[^𝑘𝑟𝑘𝑟,𝜇𝑟]×[−𝑟,−𝑘]}, (2.30) where 𝑙_𝑘=min_{𝑡∈[1/𝑘,(𝑘−1)/𝑘]}∫¹₀𝐻₁(𝑡,𝑠)𝑑𝑠. Since 𝐻₂(𝑡,𝑠) is uniformly continuous on [0,1]×[0,1], for the above 𝜀>0 and fixed 𝑠∈[1/𝑘,(𝑘−1)/𝑘], there exists 𝛿>0 such that |𝐻₂(𝑡,𝑠)−𝐻₂(𝑡^′,𝑠)|≤(2𝜆𝐿𝑀₂)⁻¹(𝜇+1)𝐻₂(𝑠,𝑠)𝜀 (2.31) for |𝑡−𝑡^′|<𝛿 and 𝑡,𝑡^′∈[0,1]. Consequently, when |𝑡−𝑡^′|<𝛿 and 𝑡,𝑡^′∈[0,1], we have |𝑇𝑣(𝑡)−𝑇𝑣(𝑡^′)|≤2sup_{𝑣∈𝐾𝑟,𝑅}𝜆_𝐷(𝑘)𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠+𝜆sup_{𝑣∈𝐾𝑟,𝑅}^{(𝑘−1)/𝑘}_1/𝑘|𝐻₂(𝑡,𝑠)−𝐻₂(𝑡^′,𝑠)|ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠<𝜀. (2.32) This implies that 𝑇(𝐾_𝑟,𝑅) is equicontinuous. Then by the Arzela-Ascoli theorem 𝑇∶𝐾_𝑟,𝑅→𝐶⁺[0,1] is compact.
Finally, we show that 𝑇∶𝐾_𝑟,𝑅→𝐶⁺[0,1] is continuous. Assume 𝑣_𝑛,𝑣₀∈𝐾_𝑟,𝑅 and ‖𝑣_𝑛−𝑣₀‖→0(𝑛→∞). Then 𝑟≤‖𝑣_𝑛‖≤𝑅 and 𝑟≤‖𝑣₀‖≤𝑅. For any 𝜀>0, by (𝐻₃), there exists a natural number 𝑚>0 such that
sup_{𝑣∈𝐾𝑟,𝑅}𝜆_𝐷(𝑚)𝐻₂𝜀(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠<4. (2.33)
On the other hand, by (2.25), for any 𝑡∈[1/𝑚,(𝑚−1)/𝑚], we have
𝑟𝑚≤𝑣_𝑛𝑙(𝑡)≤𝑅,^𝑚𝑟𝑚≤𝑆𝑣_𝑛(𝑡)≤𝜇𝑅,𝑛=0,1,2,…, (2.34) where 𝑙_𝑚=min_{𝑡∈[1/𝑚,(𝑚−1)/𝑚]}∫¹₀𝐻₁(𝑡,𝑠)𝑑𝑠 and 𝜇 is defined by Remark 2.3.
Since 𝑓(𝑡,𝑥,𝑦) is uniformly continuous in [1/𝑚,(𝑚−1)/𝑚]×[𝑙_𝑚𝑟/𝑚,𝜇𝑅]×[−𝑅,−𝑟/𝑚], we have that
lim_𝑛→+∞|𝑓(𝑠,𝑆𝑣_𝑛(𝑠),−𝑣_𝑛(𝑠))−𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))|=0 (2.35) holds uniformly on 𝑠∈[1/𝑚,(𝑚−1)/𝑚]. Then the Lebesgue dominated convergence theorem yields that ^{(𝑚−1)/𝑚}_1/𝑚𝜆𝐻₂(𝑠,𝑠)ℎ(𝑠)|𝑓(𝑠,𝑆𝑣_𝑛(𝑠),−𝑣_𝑛(𝑠))−𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))|𝑑𝑠→0,as𝑛→∞. (2.36) Thus for above 𝜀>0, there exists a natural number; 𝑁, for 𝑛>𝑁, we have ^{(𝑚−1)/𝑚}_1/𝑚𝜆𝐻₂(𝑠,𝑠)ℎ(𝑠)|𝑓(𝑠,𝑆𝑣_𝑛(𝑠),−𝑣_𝑛(𝑠))−𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀𝜀(𝑠))|𝑑𝑠<2. (2.37) It follows from (2.33) and (2.37) that when 𝑛>𝑁, ‖𝑇𝑣_𝑛−𝑇𝑣₀‖≤^{(𝑚−1)/𝑚}_1/𝑚𝜆𝐻₂(𝑠,𝑠)ℎ(𝑠)|𝑓(𝑠,𝑆𝑣_𝑛(𝑠),−𝑣_𝑛(𝑠))−𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))|𝑑𝑠+2sup_{𝑣∈𝐾𝑟,𝑅𝐷(𝑚)}𝜆𝐻₂<𝜀(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠2𝜀+2×4=𝜀. (2.38) This implies that 𝑇∶𝐾_𝑟,𝑅→𝐶⁺[0,1] is continuous. Thus 𝑇∶𝐾_𝑟,𝑅→𝐶⁺[0,1] is completely continuous. This completes the proof of Lemma 2.6.

Lemma 2.7. One has 𝑇(𝐾_𝑟,𝑅)⊂𝐾.

Proof. For 𝑣∈𝐾_𝑟,𝑅,𝑡∈[0,1], we have (𝑇𝑣)(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑≤𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠. (2.39) Thus ‖𝑇𝑣‖≤𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠. (2.40) On the other hand, by Lemma 2.2 we have (𝑇𝑣)(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≥𝛾𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠 (2.41) which implies min_𝑡∈[0,1](𝑇𝑣)(𝑡)≥𝛾‖𝑇𝑣‖.
In addition, for 𝑣∈𝐾,𝑡∈[0,1], by Remark 2.5, (𝐻₂), and (𝐻₃) we have
(𝑆𝑣)(1−𝑡)=¹₀𝐻₁=(1−𝑡,𝜏)𝑣(𝜏)𝑑𝜏₀₁𝐻₁=(1−𝑡,1−𝜏)𝑣(1−𝜏)𝑑(1−𝜏)₁₀𝐻₁(𝑡,𝜏)𝑣(𝜏)𝑑𝜏=(𝑆𝑣)(𝑡),(𝑇𝑣)(1−𝑡)=𝜆¹₀𝐻₂(1−𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠,=𝜆⁰₁𝐻₂(1−𝑡,1−𝑠)ℎ(1−𝑠)𝑓(1−𝑠,𝑆𝑣(1−𝑠),−𝑣(1−𝑠))𝑑(1−𝑠)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠=(𝑇𝑣)(𝑡), (2.42)(2.43) that is, (𝑇𝑣)(1−𝑡)=(𝑇𝑣)(𝑡), 𝑡∈[0,1]. Therefore, (𝑇𝑣)(𝑡) is symmetric on [0,1]. Obviously, (𝑇𝑣)(𝑡) is concave on [0,1]. Consequently, 𝑇(𝐾_𝑟,𝑅)⊂𝐾. This completes the proof of Lemma 2.7.

3. The Main Results

Our main results of this paper are as follows. For notational convenience, we let

𝑓⁰=limsup_{|𝑥|+|𝑦|→0𝑥>0,𝑦<0}max_𝑡∈[0,1]𝑓(𝑡,𝑥,𝑦)|𝑥|+|𝑦|,𝑓_∞=liminf_{|𝑥|+|𝑦|→+∞𝑥>0,𝑦<0}min_𝑡∈[0,1]𝑓(𝑡,𝑥,𝑦),𝑓|𝑥|+|𝑦|_∞=limsup_{|𝑥|+|𝑦|→+∞𝑥>0,𝑦<0}max_𝑡∈[0,1]𝑓(𝑡,𝑥,𝑦)|𝑥|+|𝑦|,𝑓₀=liminf_{|𝑥|+|𝑦|→0𝑥>0,𝑦<0}min_𝑡∈[0,1]𝑓(𝑡,𝑥,𝑦).|𝑥|+|𝑦| (3.1)

Theorem 3.1. Suppose that conditions (𝐻₁), (𝐻₂), and (𝐻₃) are satisfied. Further assume that the following condition (𝐻₄) holds:(𝐻₄)𝐿𝑓⁰<𝑙𝑓_∞.Then the BVP (1.1) has at least one symmetric positive solution for any 1𝜆∈(𝑙𝑓_∞,1𝐿𝑓⁰), (3.2) where 𝐿 and 𝑙 are defined by Remark 2.3.

Proof. Let 𝜆 satisfy (3.2) and at 𝜀>0 be chosen such that 𝑓_∞1−𝜀>0,(𝑓_∞1−𝜀)𝑙≤𝜆≤(𝑓⁰.+𝜀)𝐿 (3.3) Next, by (𝐻₄) there exists 𝑟₀>0 such that 𝑓(𝑡,𝑥,𝑦)≤(𝑓⁰+𝜀)(|𝑥|+|𝑦|),∀𝑡∈[0,1],0<|𝑥|+|𝑦|<𝑟₀,𝑥>0,𝑦<0. (3.4) Take 𝑟=𝑟₀/(𝜇+1). Notice that 0<|𝑆𝑣|+|𝑣|≤(𝜇+1)‖𝑣‖=(𝜇+1)𝑟=𝑟₀,0≤𝑡≤1. (3.5) It follows from (3.4) and (3.5) that, for any 𝑣∈𝜕𝐾_𝑟, ‖𝑇𝑣‖=max_𝑡∈[0,1]𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≤𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≤𝜆(𝑓⁰+𝜀)¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)(|𝑆𝑣(𝑠)|+|𝑣(𝑠)|)𝑑𝑠=𝜆(𝑓⁰+𝜀)(𝜇+1)𝑟¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑑𝑠=𝜆𝐿(𝑓⁰+𝜀)𝑟≤𝑟=‖𝑣‖. (3.6) Thus, ‖𝑇𝑣‖≤‖𝑣‖, for all 𝑣∈𝜕𝐾_𝑟.
On the other hand, for the above 𝜀, by (𝐻₄), there exists 𝑅₀>0 such that
𝑓(𝑡,𝑥,𝑦)>(𝑓_∞−𝜀)(|𝑥|+|𝑦|),𝑡∈[0,1],|𝑥|+|𝑦|≥𝑅₀,𝑥>0,𝑦<0. (3.7) Let 𝑅=max{2𝑟,𝛾⁻¹𝑅₀} and 𝜑(𝑡)≡1, 𝑡∈[0,1]. Then 𝑅>𝑟 and 𝜑(𝑡)∈𝜕𝐾₁.
In the following we show 𝑣≠𝑇𝑣+𝑚𝜑(𝑚>0). Otherwise, there exists 𝑣₀∈𝜕𝐾_𝑅 and 𝑚₀>0 such that 𝑣₀=𝑇𝑣₀+𝑚₀𝜑. Let 𝜉=min{𝑣₀(𝑡)∶𝑡∈[0,1]} and notice that for any 𝑠∈[0,1],
|𝑆𝑣₀(𝑠)|+|𝑣₀(𝑠)|≥min_𝑠∈[0,1][|𝑆𝑣₀(𝑠)|+|𝑣₀(𝑠)|]≥min_𝑠∈[0,1]|𝑣₀(𝑠)|≥𝛾‖𝑣₀‖≥𝑅₀. (3.8) Consequently for any 𝑡∈[0,1], we have 𝑣₀(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))𝑑𝑠+𝑚₀𝜑(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(s))𝑑𝑠+𝑚₀≥𝜆(𝑓_∞−𝜀)¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)(|𝑆𝑣₀(𝑠)|+|𝑣₀(𝑠)|)𝑑𝑠+𝑚₀≥𝜆(𝑓_∞−𝜀)¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑣₀(𝑠)𝑑𝑠+𝑚₀≥𝜆(𝑓_∞−𝜀)𝜉min_𝑡∈[0,1]¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑑𝑠+𝑚₀≥𝜉+𝑚₀>𝜉. (3.9) This implies that 𝜉>𝜉, which is a contradiction. It follows from Lemma 1.1 that 𝑇 has a fixed point 𝑣^∗ with 𝑟<|𝑣^∗|<𝑅. Thus 𝑣^∗ is a symmetric positive solution of the BVP (2.17). Consequently, by Lemma 2.4, one can obtain that BVP (1.1) has a symmetric positive solution. This completes the proof of Theorem 3.1.

Remark 3.2. Since 0<𝑙<𝐿<+∞, we easily obtain 0<1/𝑙𝑓_∞<1, 1/𝐿𝑓⁰>1. Thus 1∈(1/𝑙𝑓_∞,1/𝐿𝑓⁰); so when 𝜆=1, Theorem 3.1 always holds.

Remark 3.3. From Theorem 3.1, we can see that 𝑓(𝑡,𝑥,𝑦) need not be superlinear or sublinear. In fact, Theorem 3.1 still holds, if one of the following conditions is satisfied.(i)If 𝑓_∞=∞, 𝑓⁰>0, then for each 𝜆∈(0,1/𝐿𝑓⁰).(ii)If 𝑓_∞=∞, 𝑓⁰=0, then for each 𝜆∈(0,+∞).(iii)If 𝑓_∞>𝑙⁻¹>0, 𝑓⁰=0, then for each 𝜆∈(1/𝑙𝑓_∞,+∞).

Theorem 3.4. Suppose that conditions (𝐻₁), (𝐻₂), and (𝐻₃) are satisfied. Further assume that the following condition (𝐻₅) holds.(𝐻₅)𝐿𝑓^∞<𝑙𝑓₀.Then the BVP (1.1) has at least one symmetric positive solution for any 1𝜆∈(𝑙𝑓₀,1𝐿𝑓^∞), (3.10) where 𝐿 and 𝑙 are defined by Remark 2.3.

Proof. Let 𝜆 satisfy (3.10) and let 𝜀₁>0 be chosen such that 𝐿⁻¹−𝜀₁>0 and 𝜆𝑓^∞<𝐿⁻¹−𝜀₁. By (𝐻₅), there exists 𝜇𝑅^′₀ such that 1𝑓(𝑡,𝑥,𝑦)≤𝜆(𝐿⁻¹−𝜀₁)(|𝑥|+|𝑦|),|𝑥|+|𝑦|≥𝜇𝑅^′₀,𝑥>0,𝑦<0,𝑡∈[0,1]. (3.11)
Let
𝑀₀=sup_{𝑣∈𝜕𝐾𝑅′0}𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠. (3.12) Then 𝑀₀<+∞ by (2.22). Take 𝑅₁>𝑚𝑎𝑥{𝑅^′₀,𝑀₀/𝐿𝜀₁}, then 𝑀₀<𝐿𝑅₁𝜀₁.
Notice that 𝑣∈𝜕𝐾_𝑅′0 implies that
𝑣(𝑡)≤‖𝑣‖=𝑅^′₀,𝑆𝑣(𝑡)≤max_𝑡∈[0,1]¹₀𝐻₁(𝑡,𝑠)𝑑𝑠‖𝑣‖=𝜇𝑅^′₀. (3.13) So for any 𝑣∈𝜕𝐾_𝑅1, let 𝐷(𝑆𝑣,−𝑣)={𝑡∈[0,1]∶(𝑆𝑣,−𝑣)∈[𝜇𝑅^′₀,+∞)×(−∞,−𝑅^′₀]}, (3.14) and then for any 𝑡∈𝐷(𝑆𝑣,−𝑣), clearly 𝜇𝑅^′₀≤|𝑆𝑣|+|𝑣|≤(𝜇+1)‖𝑣‖=(𝜇+1)𝑅₁.
In addition, for any 𝑣∈𝜕𝐾_𝑅1, let 𝑣₁(𝑡)=min{𝑣(𝑡),𝑅^′₀}, then 𝑣₁∈𝜕𝐾_𝑅′0. Thus, for any 𝑣∈𝜕𝐾_𝑅1, we have
‖𝑇𝑣‖=max_𝑡∈[0,1]𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠≤max_𝑡∈[0,1]𝜆_{𝐷(𝑆𝑣,−𝑣)}𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠+𝜆_{[0,1]/𝐷(𝑆𝑣,−𝑣)}𝐻₂≤1(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣(𝑠),−𝑣(𝑠))𝑑𝑠𝜆(𝐿⁻¹−𝜀₁)𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)(|𝑆𝑣(𝑠)|+|𝑣(𝑠)|)𝑑𝑠+𝜆¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣₁(𝑠),−𝑣₁(𝑠))𝑑𝑠≤(𝐿⁻¹−𝜀₁)(𝜇+1)𝑅₁¹₀𝐻₂(𝑠,𝑠)ℎ(𝑠)𝑑𝑠+𝑀₀≤(𝐿⁻¹−𝜀₁)𝐿𝑅₁+𝑀₀<𝑅₁=‖𝑣‖. (3.15) Therefore ‖𝑇𝑣‖≤‖𝑣‖ for any 𝑣∈𝜕𝐾_𝑅1.
Next, let 𝜆 satisfy (3.10). Choose 𝜀₂>0 such that 𝑙⁻¹+𝜀₂<𝜆𝑓₀. Then from (𝐻₅), there exists 0<𝛿<(𝜇+1)𝑅₁ such that
1𝑓(𝑡,𝑥,𝑦)≥𝜆(𝑙⁻¹+𝜀₂)(|𝑥|+|𝑦|),0<|𝑥|+|𝑦|≤𝛿,𝑥>0,𝑦<0,𝑡∈[0,1]. (3.16) Let 𝑟₁=𝛿/(𝜇+1) and 𝜙(𝑡)≡1, 𝑡∈[0,1], then 𝑟₁<𝑅₁ and 𝜙∈𝜕𝐾₁.
Now we prove 𝑣≠𝑇𝑣+𝑚𝜙(𝑚>0). Otherwise, there exists 𝑣₀∈𝜕𝐾_𝑟1 and 𝑚₀>0 such that 𝑣₀=𝑇𝑣₀+𝑚₀𝜙. Let 𝜁=min{𝑣₀(𝑡)∶𝑡∈[0,1]} and apply that |𝑣₀(𝑠)|+|𝑆𝑣₀(𝑠)|<(𝜇+1)𝑟₁=𝛿; then for any 𝑡∈[0,1], we have
𝑣₀(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))𝑑𝑠+𝑚₀𝜙(𝑡)=𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑓(𝑠,𝑆𝑣₀(𝑠),−𝑣₀(𝑠))𝑑𝑠+𝑚₀≥1𝜆(𝑙⁻¹+𝜀₂)𝜆¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)(|𝑆𝑣₀(𝑠)|+|𝑣₀(𝑠)|)𝑑𝑠+𝑚₀≥(𝑙⁻¹+𝜀₂)¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑣₀(𝑠)𝑑𝑠+𝑚₀≥(𝑙⁻¹+𝜀₂)𝜁min_𝑡∈[0,1]¹₀𝐻₂(𝑡,𝑠)ℎ(𝑠)𝑑𝑠+𝑚₀≥(1+𝑙𝜀₂)𝜁+𝑚₀>𝜁. (3.17)(1)(2)(3)(4)(5) This implies that 𝜁>𝜁, which is a contradiction. It follows from Lemma 1.1 that 𝑇 has a fixed point 𝑣^∗∗ with 𝑟₁<|𝑣^∗∗|<𝑅₁. Thus 𝑣^∗∗ is a symmetric positive solution of the BVP (2.17). By Lemma 2.4, BVP (1.1) has a symmetric positive solution.

Remark 3.5. From Theorem 3.4 we can see that the conclusions still hold, if one of the following conditions is satisfied.(i)If 𝑓^∞<𝐿⁻¹, 𝑓₀=∞, then for each 𝜆∈(0,1/𝐿𝑓^∞).(ii)If 𝑓^∞=0, 𝑓₀=+∞, then for each 𝜆∈(0,+∞).(iii)If 𝑓^∞=0, 𝑓₀>𝑙⁻¹>0, then for each 𝜆∈(1/𝑙𝑓₀,+∞).

Remark 3.6. Because of singularity of ℎ, 𝑓, it seems to be difficult to prove our results by using the norm-type expansion and compression theorem. In addition, we need to point out that we not only obtain the existence of symmetric positive solutions of BVP (1.1) but also get the explicit interval about 𝜆, which is different from the previous papers (see [4, 5]).

Remark 3.7. If conditions (𝐻₁), (𝐻₂) and the following condition (𝐻^′₃) are satisfied,(𝐻^′₃)𝑓(𝑡,𝑢)∈𝐶((0,1)×(0,+∞),[0,+∞)), 𝑓((1−𝑡),𝑢)=𝑓(𝑡,𝑢), and for any 0<𝑟<𝑅<+∞,lim_𝑛→+∞sup_{𝑢∈𝐾𝑟,𝑅𝐷(𝑛)}𝐻(𝑠,𝑠)ℎ(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=0, (3.18) where 𝐷(𝑛)=[0,1/𝑛]∪[(𝑛−1)/𝑛,1]. Then our results can be applied to the second-order singular boundary value problem with nonlocal boundary condition: 𝑢^′′(𝑡)+𝜆ℎ(𝑡)𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=𝑢(1)=¹₀𝑎(𝑠)𝑢(𝑠)𝑑𝑠, (3.19) under the corresponding to conditions of Theorems 3.1 and 3.4.

Acknowledgments

The authors are grateful to the referees for their valuable suggestions and comments. This work was supported financially by the National Natural Science Foundation of China (10771117).