Positive Solutions for Sturm-Liouville Boundary Value Problems in a Banach Space

Abstract

We consider the existence of single and multiple positive solutions for a second-order Sturm-Liouville boundary value problem in a Banach space. The sufficient condition for the existence of positive solution is obtained by the fixed point theorem of strict set contraction operators in the frame of the ODE technique. Our results significantly extend and improve many known results including singular and nonsingular cases.

1. Introduction

Boundary value problems for ordinary differential equations play a very important role in both theoretical study and practical application in many fields. They are used to describe a large number of physical, biological, and chemical phenomena. In this paper, we study the existence of positive solutions for the following second-order nonlinear Sturm-Liouville boundary value problem (BVP) in a Banach Space 𝐸1𝑝(𝑡)𝑝(𝑡)𝑢^′(𝑡)^′(𝑢(𝑡(0)+𝑓))=0,0<𝑡<1,𝛼𝑢−𝛽lim_𝑡→0+𝑝(𝑡)𝑢^′(𝑡)(1)=0,𝛾𝑢+𝛿lim_𝑡→1−𝑝(𝑡)𝑢^′(𝑡)=0, (1.1) where 𝛼,𝛽,𝛿,𝛾≥0 are constants such that 𝐵(𝑡,𝑠)=∫^𝑠_𝑡𝑑𝜏/𝑝(𝜏),𝜔=𝛽𝛾+𝛼𝛾𝐵(0,1)+𝛼𝛿>0, and 𝑝∈𝐶¹((0,1),(0,+∞)). Moreover, 𝑝 may be singular at 𝑡=0 and/or 1.

The BVP (1.1) is often referred to as a model for the deformation of an elastic beam under a variety of boundary conditions [1–12]. We notice that previous work is limited to use the completely continuous operators and the function 𝑓 is required to satisfy some growth condition or assumptions of monotonicity.

The aim of this paper is to consider the existence of positive solutions for the more general Sturm-Liouville boundary value problem (1.1) by using the fixed point theorem of strict set contraction operators. Here we allow 𝑝 to have singularity at 𝑡=0,1. The results obtained in this paper improve and generalize many well-known results.

The rest of the paper is organized as follows. In Section 2, we first present some properties of Green's functions to be used to define a positive operator. Then we approximate the singular second-order boundary value problem by constructing an integral operator. In Section 3, the sufficient condition for the existence of single and multiple positive solutions for the BVP (1.1) is established. In Section 4, we give a example to demonstrate the application of our results.

2. Preliminaries and Lemmas

In this paper, we denote by (𝐸,‖⋅‖₁) a real Banach space. A nonempty closed convex subset 𝑃 in 𝐸 is said to be a cone if 𝜆𝑃∈𝑃 for 𝜆≥0 and 𝑃⋂{−𝑃}={𝜃}, where 𝜃 denotes the zero element of 𝐸. The cone 𝑃 defines a partial ordering in 𝐸 by 𝑥≤𝑦 if and only if 𝑦−𝑥∈𝑃. Recall that the cone 𝑃 is said to be normal if there exists a positive constant 𝜆 such that 0≤𝑥≤𝑦 implies ‖𝑥‖₁≤𝜆‖𝑦‖₁.

In this paper, we assume 𝑃⊆𝐸 is normal, and without loss of generality, we may assume that the normal constant of 𝑃 is 1. Let 𝐽=[0,1], and 𝐶()={(𝑡)},𝐶𝐽,𝐸𝑢∶𝐽→𝐸∣𝑢continuous_𝑖(𝐽,𝐸)=𝑢∶𝐽→𝐸∣𝑢(𝑡)is𝑖-ordercontinuouslydifferentiable,𝑖=1,2,…. (2.1) For 𝑢=𝑢(𝑡)∈𝐶(𝐽,𝐸), let ‖𝑢‖=max_𝑡∈𝐽‖𝑢(𝑡)‖₁, then 𝐶(𝐽,𝐸) is a Banach space with the norm ‖⋅‖.

Definition 2.1. A function 𝑢(𝑡) is said to be a positive solution of the boundary value problem (1.1) if 𝑢∈𝐶([0,1],𝐸)⋂𝐶¹((0,1),𝐸) satisfies 𝑢(𝑡)>0,𝑡∈(0,1], 𝑝𝑢′∈𝐶¹((0,1),𝐸) and the BVP (1.1).

We notice that if 𝑢(𝑡) is a positive solution of the BVP (1.1) and 𝑝∈𝐶¹(0,1), then 𝑢∈𝐶²(0,1).

Now we denote by 𝐺(𝑡,𝑠) the Green's functions for the following boundary value problem: 1𝑝(𝑡)𝑝(𝑡)𝑢^′(𝑡)^′𝑢(0)=0,0<𝑡<1,−lim_𝑡→0+𝛽𝑝(𝑡)𝑢^′(𝑡)(1)=0,𝛾𝑢+lim_𝑡→1−𝛿𝑝(𝑡)𝑢^′(𝑡)=0. (2.2) It is well known that 𝐺(𝑡,𝑠) can be written by 𝐺()=1𝑡,𝑠𝜔()()((𝛽+𝛼𝐵0,𝑠𝛿+𝛾𝐵𝑡,1,0≤𝑠≤𝑡≤1,𝛽+𝛼𝐵0,𝑡)𝛿+𝛾𝐵𝑠,1),0≤𝑡≤𝑠≤1, (2.3) where 𝐵(𝑡,𝑠)=∫^𝑠_𝑡𝑑𝜏/𝑝(𝜏),𝜔=𝛼𝛿+𝛼𝛾𝐵(0,1)+𝛽𝛾>0.

It is easy to verify the following properties of 𝐺(𝑡,𝑠): (I)𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠)≤(1/𝜔)(𝛽+𝛼𝐵(0,1))(𝛿+𝛾𝐵(0,1))<+∞; (II)𝐺(𝑡,𝑠)≥𝜌𝐺(𝑠,𝑠), for any 𝑡∈[𝑎,𝑏]⊂(0,1),𝑠∈[0,1], where ()𝜌=min𝛿+𝛾𝐵𝑏,1(),()𝛿+𝛾𝐵0,1𝛽+𝛼𝐵0,𝑎().𝛽+𝛼𝐵0,1 (2.4)

Throughout this paper, we adopt the following assumptions:(𝐻₁)𝑝∈𝐶¹((0,1),(0,+∞)) and satisfies 0<¹₀𝑑𝑠𝑝(𝑠)<+∞,0<𝑒=¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠<+∞. (2.5) (𝐻₂)𝑓∶𝑃→𝑃 is a uniformly continuous function and there exists 𝑀>0 such that for any bounded set 𝐵⊂𝐶(𝐽,𝐸), we have 𝛼𝑓(𝐵(𝑡))≤𝑀𝛼(𝐵(𝑡)),𝑡∈𝐽,2𝑀𝑒𝜌<1, (2.6)where 𝛼(⋅) denotes the Kuratowski measure of noncompactness in 𝐸.

The following Lemmas play an important role in this paper (see [13]).

Lemma 2.2. Let 𝐵⊂𝐶(𝐽,𝐸) be bounded and equicontinuous on 𝐽, then 𝛼_𝑐(𝐵)=sup_𝑡∈𝐽𝛼(𝐵(𝑡)).

Lemma 2.3. Let 𝐵⊂𝐶(𝐽,𝐸) be bounded and equicontinuous on 𝐽, then 𝛼(𝐵(𝑡)) is continuous on 𝐽 and 𝛼_𝐽𝑢(𝑡)𝑑𝑡∶𝑢∈𝐵≤_𝐽𝛼(𝐵(𝑡))𝑑𝑡. (2.7)

Lemma 2.4. Let 𝐵⊂𝐶(𝐽,𝐸) be a bounded set on 𝐽. Then 𝛼(𝐵(𝑡))≤2𝛼_𝑐(𝐵).

Now, for the given [𝑎,𝑏]⊂(0,1) and the 𝜌 as in (II), we introduce 𝐾=𝑢∈𝐶(𝐽,𝑃)∶𝑢(𝑡)≥𝜌𝑢(𝑠),𝑡∈[𝑎,𝑏],𝑠∈[0,1]. (2.8) It is easy to check that 𝐾 is a cone in 𝐶(𝐽,𝐸) and for 𝑢(𝑡)∈𝐾,𝑡∈[𝑎,𝑏], we have ‖𝑢(𝑡)‖₁≥𝜌‖𝑢‖.

Next, we define an operator 𝑇∶𝐾→𝐶(𝐽,𝑃) given by 𝑇𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠,∀𝑢∈𝐾,𝑡∈[0,1]. (2.9) Clearly, 𝑢 is a solution of the BVP (1.1) if and only if 𝑢 is a fixed point of the operator 𝑇.

Through direct calculation, by (II) and for 𝑣∈𝐾,𝑡∈[𝑎,𝑏],𝑠∈𝐽, we have 𝑇𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠≥𝜌¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠=𝜌𝑇𝑢(𝑠). (2.10) So, this implies that 𝑇𝐾⊂𝐾.

Lemma 2.5. Assume that, hold. Then 𝑇∶𝐾→𝐾 is a strict set contraction operator.

Proof. Firstly, The continuity of 𝑇 is easily obtained. In fact, if 𝑣_𝑛,𝑣∈𝐾 and 𝑣_𝑛→𝑣 in the norm in 𝐶(𝐽,𝐸), then for any 𝑡∈𝐽, we get 𝑇𝑣_𝑛(𝑡)−𝑇𝑣(𝑡)₁≤¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑓𝑣_𝑛(𝑠)−𝑓(𝑣(𝑠))₁𝑑𝑠, (2.11) so, by the uniformly continuity of 𝑓, we have 𝑇𝑣_𝑛−𝑇𝑣=sup_𝑡∈𝐽𝑇𝑣_𝑛(𝑡)−𝑇𝑣(𝑡)₁⟶0. (2.12) This implies that 𝑇𝑣_𝑛→𝑇𝑣 in 𝐶(𝐽,𝐸), that is, 𝑇 is continuous.
Now, let 𝐵⊂𝐾 be a bounded set. It follows from that there exists a positive number 𝐿 such that ‖𝑓(𝑣)‖≤𝐿 for any 𝑣∈𝐵. Then, we can get ‖𝑇𝑣(𝑡)‖₁≤𝐿𝑒<∞,∀𝑡∈𝐽,𝑣∈𝐵, (2.13) where 𝑒 is as defined in. So, 𝑇(𝐵)⊂𝐾 is a bounded set in 𝐾.
For any 𝜀>0, by (𝐻₁), there exists a 𝛿′>0 such that ^𝛿′₀𝐺()𝑝(𝑠)≤𝜀𝑠,𝑠,6𝐿¹_1−𝛿^′𝐺()𝑝(𝑠)≤𝜀𝑠,𝑠.6𝐿 (2.14) Let 𝑃=max_{𝑡∈[𝛿′,1−𝛿′]}𝑝(𝑡). It follows from the uniform continuity of 𝐺(𝑡,𝑠) on [0,1]×[0,1] that there exists 𝛿>0 such that 𝐺(𝑡,𝑠)−𝐺𝑡^′≤𝜀,𝑠,3𝑃𝐿𝑡−𝑡^′<𝛿,𝑡,𝑡′∈[0,1],𝑠∈[0,1]. (2.15) Consequently, when |𝑡−𝑡′|<𝛿,𝑡,𝑡′∈[0,1],𝑣∈𝐵, we have 𝑇𝑣(𝑡)−𝑇𝑣𝑡^′₁=¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑝(𝑠)𝑓(𝑣(𝑠))𝑑𝑠₁≤^𝛿′₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑝(𝑠)𝑓(𝑣(𝑠))₁+𝑑𝑠_{1−𝛿′𝛿′}𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑝(𝑠)𝑓(𝑣(𝑠))₁𝑑𝑠+¹_1−𝛿′𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑝(𝑠)𝑓(𝑣(𝑠))₁𝑑𝑠≤2𝐿^𝛿′₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠+2𝐿¹_1−𝛿′𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠+𝑃𝐿¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑑s≤𝜀. (2.16) This implies that 𝑇(𝐵) is an equicontinuous set on 𝐽. Therefore, by Lemma 2.2, we have 𝛼(𝑇(𝐵))=sup_𝑡∈𝐽𝛼(𝑇(𝐵)(𝑡)). (2.17)
Without loss of generality, by condition, we may assume that 𝑝(𝑡) is singular at 𝑡=0,1. So, there exist {𝑎_𝑛𝑖},{𝑏_𝑛𝑖}⊂(0,1),{𝑛_𝑖}⊂𝑁 with {𝑛_𝑖} being a strictly increasing sequence and lim_𝑖→+∞𝑛_𝑖=+∞ such that 0<⋯<𝑎_𝑛𝑖<⋯<𝑎_𝑛1<𝑏_𝑛1<⋯<𝑏_𝑛𝑖𝑝(𝑡)<⋯<1,≥𝑛_𝑖,𝑡∈0,𝑎_𝑛𝑖∪𝑏_𝑛𝑖,1,𝑝𝑎_𝑛𝑖=𝑝𝑏_𝑛𝑖=𝑛_𝑖,lim_𝑖→+∞𝑎_𝑛𝑖=0,lim_𝑖→+∞𝑏_𝑛𝑖=1. (2.18)(2.19) Next, we let 𝑝_𝑛𝑖(𝑡)=𝑛_𝑖,𝑡∈0,𝑎_𝑛𝑖∪𝑏_𝑛𝑖,𝑝(𝑡)𝑎,1,𝑡∈_𝑛𝑖,𝑏_𝑛𝑖. (2.20) Then, from the above discussion we know that 𝑝_𝑛𝑖 is continuous on 𝐽 for every 𝑖∈𝑁 and 𝑝_𝑛𝑖(𝑡)≥𝑝(𝑡),𝑝_𝑛𝑖(𝑡)⟶𝑝(𝑡),∀𝑡∈(0,1),as𝑖⟶+∞. (2.21)
For any 𝜀>0, by (2.19) and, there exists an 𝑖₀ such that for any 𝑖>𝑖₀, we have 2𝐿^𝑎𝑛𝑖₀𝐺()𝑝(𝑠)𝜀𝑠,𝑠𝑑𝑠<2,2𝐿¹_𝑏𝑛𝑖𝐺()𝑝(𝑠)𝜀𝑠,𝑠𝑑𝑠<2. (2.22) Therefore, for any bounded set 𝐵⊂𝐶(𝐽,𝐸), by Lemmas 2.3 and 2.4,, the above discussion and noting that 𝑝_𝑛𝑖(𝑡)≤𝑝(𝑡),𝑡∈(0,1), as 𝑡∈𝐽,𝑖>𝑖₀, we have that 𝛼(𝑇(𝐵)(𝑡))=𝛼¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑣(𝑠))𝑑𝑠∣𝑣∈𝐵≤𝛼¹₀𝐺(𝑡,𝑠)𝑝(𝑠)−𝑝_𝑛𝑖(𝑠)𝑓(𝑣(𝑠))𝑑𝑠∣𝑣∈𝐵+𝛼¹₀𝐺(𝑡,𝑠)𝑝_𝑛𝑖(𝑠)𝑓(𝑣(𝑠))𝑑𝑠∣𝑣∈𝐵≤2𝐿^𝑎𝑛𝑖₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠+2𝐿¹_𝑏𝑛𝑖𝐺()𝑝(𝑠)+𝑠,𝑠𝑑𝑠₁₀𝛼𝐺(𝑡,𝑠)𝑝_𝑛𝑖(𝑠)𝑓(𝑣(𝑠))∣𝑣∈𝐵𝑑𝑠≤𝜀+𝜌¹₀𝐺()𝑝(𝑠)𝛼𝑓(𝑣(𝑠(𝐵).𝑠,s))∣𝑣∈𝐵𝑑𝑠≤𝜀+2𝑀𝑒𝜌𝛼 (2.23) As 𝜀 is arbitrarily, we get 𝛼(𝑇(𝐵)(𝑡))≤2𝑀𝑒𝜌𝛼_𝑐(𝐵),𝑡∈𝐽. (2.24) So, it follows from (2.17) and (2.24) that for any bounded set 𝐵⊂𝐶(𝐽,𝐸), we have 𝛼_𝑐(𝑇(𝐵))≤2𝑀𝑒𝜌𝛼_𝑐(𝐵). (2.25) And note that 2𝑀𝑒𝜌<1, we have 𝑇∶𝐾→𝐾 is a strict set contraction operator. The proof is completed.

Remark 2.6. When 𝐸=𝑅, (2.6) naturally holds. In this case, we may take 𝑀 as 0, consequently, 𝑇∶𝐾→𝐾 is a completely continuous operator. Furthermore, if 𝑝(𝑡)≡1,𝑡∈𝐽, Dalmasso [1] used the following condition: 0<𝑒=¹₀𝑠^𝛼(1−𝑠)^𝛽𝑝(𝑠)𝑑𝑠<+∞,𝛼,𝛽∈[0,1). (2.26) Clearly, our condition is weaker than (2.26).

Our main tool used in this paper is the following fixed point index theorem of cone.

Theorem 2.7 (see [13]). Suppose that 𝐸 is a Banach space, 𝐾⊂𝐸 is a cone, and let the Ω₁,Ω₂ be two bounded open sets of 𝐸 such that 𝜃∈Ω₁,Ω₁⊂Ω₂. Let operator 𝑇∶𝐾∩(Ω₂⧵Ω₁)→𝐾 be strict set contraction. Suppose that one of the following two conditions holds:(i)‖𝑇𝑥‖≤‖𝑥‖, for all 𝑥∈𝐾∩𝜕Ω₁,‖𝑇𝑥‖≥‖𝑥‖, for all 𝑥∈𝐾∩𝜕Ω₂; (ii)‖𝑇𝑥‖≥‖𝑥‖, for all 𝑥∈𝐾∩𝜕Ω₁,‖𝑇𝑥‖≤‖𝑥‖, for all 𝑥∈𝐾∩𝜕Ω₂. Then 𝑇 has at least one fixed point in 𝐾∩(Ω₂⧵Ω₁).

Theorem 2.8 (see [13]). Suppose 𝐸 is a real Banach space, 𝐾⊂𝐸 is a cone. Let Ω_𝑟={𝑢∈𝐾∶‖𝑢‖≤𝑟}, and let the operator 𝑇∶Ω_𝑟→𝐾 be completely continuous and satisfy 𝑇𝑥≠𝑥, for all 𝑥∈𝜕Ω_𝑟. Then(i)If ‖𝑇𝑥‖≤‖𝑥‖, for all 𝑥∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝐾)=1;(ii)If ‖𝑇𝑥‖≥‖𝑥‖, for all 𝑥∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝐾)=0.

3. The Main Results

Denote 𝑓₀=lim_{‖𝑢‖1→0+}𝑓(𝑢)₁‖𝑢‖₁,𝑓_∞=lim_{‖𝑢‖1→∞}𝑓(𝑢)₁‖𝑢‖₁. (3.1) We now present our main results by Theorems 3.1 and 3.2.

Theorem 3.1. Suppose that conditions, hold. Assume that 𝑓 also satisfies(𝐴₁)𝑓(𝑢)≥𝑟𝑢^∗, 𝜌𝑟≤‖𝑢‖₁≤𝑟; (𝐴₂)𝑓(𝑢)≤𝑅𝑢_∗, 0≤‖𝑢‖₁≤𝑅, where 𝑢^∗ and 𝑢_∗ satisfy 𝜌^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑢^∗(𝑠)𝑑𝑠₁≥1,𝑢_∗¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤1. (3.2) Then, the boundary value problem (1.1) has a positive solution 𝑢 such that ‖𝑢‖ is between 𝑟 and 𝑅.

Proof of Theorem 3.1. Without loss of generality, we suppose that 𝑟<𝑅. For any 𝑢∈𝐾, we have ‖𝑢(𝑡)‖₁≥𝜌‖𝑢‖,𝑡∈[𝑎,𝑏]. (3.3) We now define two open subset Ω₁ and Ω₂ of 𝐶(𝐽,𝐸)Ω₁={𝑢∈𝐶(𝐽,𝐸)∶‖𝑢‖<𝑟},Ω₂={𝑢∈𝐶(𝐽,𝐸)∶‖𝑢‖<𝑅}. (3.4) For 𝑢∈𝐾⋂𝜕Ω₁, by (3.3), we have 𝑟=‖𝑢‖≥‖𝑢(𝑡)‖₁≥𝜌‖𝑢‖=𝜌𝑟,𝑡∈[𝑎,𝑏]. (3.5)
For 𝑢∈𝐾∩𝜕Ω₁, if holds, we have ‖𝑇𝑢(𝑡)‖₁=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠₁≥^𝑏_𝑎𝐺(𝑡,𝑠)𝑝(𝑠)𝑢^∗(𝑠)𝑟𝑑𝑠₁≥𝜌𝑟^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑢^∗(𝑠)𝑑𝑠₁≥𝑟=‖𝑢‖,𝑡∈𝐽. (3.6) Therefore, we have ‖𝑇𝑢‖=max_𝑡∈[0,1]‖𝑇𝑢(𝑡)‖₁≥‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₁. (3.7) On the other hand, as 𝑢∈𝐾∩𝜕Ω₂, we have 𝑢(𝑡)≤‖𝑢‖=𝑅 and by, we know ‖𝑇𝑢(𝑡)‖₁=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠₁≤𝑅¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑢_∗(𝑠)𝑑𝑠₁≤𝑅𝑢_∗¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤𝑅=‖𝑢‖. (3.8) Thus ‖𝑇(𝑢)‖=max_𝑡∈[0,1]‖𝑇𝑢(𝑡)‖₁≤‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₂. (3.9) Therefore, by (3.7), (3.9), Theorem 2.7 and 𝑟<𝑅, we have that 𝑇 has a fixed point 𝑢∈𝐾∩(Ω₂⧵Ω₁). Obviously, 𝑢 is a positive solution of the problem (1.1) and 𝑟<‖𝑢‖<𝑅. The proof of Theorem 3.1 is complete.

Theorem 3.2. Suppose that conditions,, and in Theorem 3.1 hold. Assume that 𝑓 also satisfies(𝐴₃)𝑓₀=0;(𝐴₄)𝑓_∞=0. Then, the boundary value problem (1.1) has at least two solutions.

Proof of Theorem 3.2. Firstly, by condition, we can have lim_{‖𝑢‖1→0+}(‖𝑓(𝑢)‖₁/‖𝑢‖₁)=0. Then, there exists an adequately small positive number 𝑚>0 such that 𝑚∫^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤1 and there exists a constant 𝜌_∗∈(0,𝑟) such that 𝑓(𝑢)₁≤𝑚‖𝑢‖₁,0<‖𝑢‖₁≤𝜌_∗,𝑢≠0. (3.10) Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌_∗}, for any 𝑢∈𝜕Ω_𝜌∗, by (3.10), we have 𝑓(𝑢)₁≤𝑚‖𝑢‖₁≤𝑚𝜌_∗. (3.11) For 𝑢∈𝜕Ω_𝜌∗, we have ‖𝑇𝑢(𝑡)‖₁=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠₁≤¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))₁≤^𝑏_𝑎𝐺(𝑡,𝑠)𝑝(𝑠)𝑚𝜌_∗𝑑𝑠≤𝜌_∗𝑚^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤𝜌_∗=‖𝑢‖. (3.12) Therefore, we have ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗. (3.13) Then by Theorem 2.8, we have 𝑖𝑇,Ω_𝜌∗,𝐾=1. (3.14)
Next, by condition (𝐴₄), we have lim_{‖𝑢‖1→∞}(‖𝑓(𝑢)‖₁/‖𝑢‖₁)=0. Then, there exists an adequately small positive number 𝑚>0 such that 𝑚∫^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤1, and there exists a constant 𝜌₀>0 such that 𝑓(𝑢)₁≤𝑚‖𝑢‖₁,‖𝑢‖₁>𝜌₀. (3.15) We choose a constant 𝜌^∗>𝑟, obviously, 𝜌_∗<𝑟<𝜌^∗. Set Ω_𝜌∗={𝑢∈𝐾∶‖𝑢‖<𝜌^∗}, then for any 𝑢∈𝜕Ω_𝜌∗, by (3.15), we have 𝑓(𝑢)₁≤𝑚‖𝑢‖₁≤𝑚𝜌^∗. (3.16) For 𝑢∈𝜕Ω_𝜌∗, we have ‖𝑇𝑢(𝑡)‖₁=¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))𝑑𝑠₁≤¹₀𝐺(𝑡,𝑠)𝑝(𝑠)𝑓(𝑢(𝑠))₁≤^𝑏_𝑎𝐺(𝑡,𝑠)𝑝(𝑠)𝑚𝜌^∗𝑑𝑠≤𝜌^∗𝑚^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤𝜌^∗=‖𝑢‖. (3.17) Therefore, we have ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝜌∗. (3.18) Then by Theorem 2.8, we have 𝑖𝑇,Ω_𝜌∗,𝐾=1. (3.19)
Finally, set Ω_𝑟={𝑢∈𝐾∶‖𝑢‖<𝑟}, For any 𝑢∈𝜕Ω_𝑟, by, Lemma 2.3 and proceeding as for the proof of Theorem 3.1, we have ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝜕Ω_𝑟. (3.20) Then by Theorem 2.8, we have 𝑖𝑇,Ω_𝑟,𝐾=0. (3.21) Therefore, by (3.14), (3.19), (3.21) and 𝜌_∗<𝑅<𝜌^∗, we have 𝑖𝑇,Ω_𝑟⧵Ω_𝜌∗,𝑘=−1,𝑖𝑇,Ω_𝜌∗⧵Ω_𝑟,𝑘=1. (3.22) Then 𝑇 have fixed points 𝑢₁∈Ω_𝑟⧵Ω_𝜌∗ and 𝑢₂∈Ω_𝜌∗⧵Ω_𝑟. Obviously, 𝑢₁ and 𝑢₂ are all positive solutions of the BVP (1.1). The proof of Theorem 3.2 is complete.

4. Application

In order to illustrate the application of our results, we give an example in this section.

Example 4.1. Consider the following singular boundary value problem (SBVP): ₃3√𝑡13³√𝑡𝑢^′(𝑡)+160𝑢^1/2+𝑢^1/3𝑢(0)=𝜃,0<𝑡<1,−lim_𝑡→0+13³√𝑡𝑢^′(𝑡)=𝜃,𝑢(1)+lim_𝑡→1−13³√𝑡𝑢^′(𝑡)=𝜃, (4.1) where (𝑡)=1𝛽=𝛾=𝛿=1,𝛼=3,𝑝3³√𝑡,𝑓(𝑢)=160𝑢^1/2+𝑢^1/3. (4.2) Then obviously, ¹₀1𝑝(𝑡)3𝑑𝑡=2,𝑓_∞=0,𝑓₀=0. (4.3) By computing, we know that the Green's function is 𝐺()=1𝑡,𝑠7()()1+3𝑠)(2−𝑡,0≤𝑠≤𝑡≤1,1+3𝑡)(2−𝑠,0≤𝑡≤𝑠≤1. (4.4) It is easy to note that 0≤𝐺(𝑠,𝑠)≤1 and conditions,,, hold.
Next, by computing, we know that 𝜌=0.44. We choose 𝑟=3,𝑢^∗=104, as 1.32=𝜌𝑟≤‖𝑢‖=max{𝑢(𝑡),𝑡∈𝐽}≤3 and 𝜌‖∫^𝑏_𝑎𝐺(𝑠,𝑠)𝑝(𝑠)𝑢^∗(𝑠)𝑑𝑠‖=1.3>1, because of the monotone increasing of 𝑓(𝑢) on [0,∞), then 𝑓(𝑢)≥𝑓(1.32)=359.3≥312=𝑟𝑢^∗,1.32≤‖𝑢‖≤3. (4.5) Thus condition (𝐴₁) holds. Hence by Theorem 3.2, SBVP (4.1) has at least two positive solutions 𝑢₁,𝑢₂ and 0<‖𝑢₁‖<3<‖𝑢₂‖.

Acknowledgment

The authors were supported financially by Shandong Province Natural Science Foundation (ZR2009AQ004, ZR2010GL013), National Natural Science Foundation of China (11071141).