Positive Solutions for (k, n − k) Conjugate Multipoint Boundary Value Problems in Banach Spaces

Abstract

By means of the fixed point index theory of strict-set contraction operator, we study the existence of positive solutions for the multipoint singular boundary value problem (−1)^𝑛−𝑘𝑢^(𝑛)(𝑡)=𝑓(𝑡,𝑢(𝑡)), 0<𝑡<1, 𝑛≥2, 1≤𝑘≤𝑛−1, 𝑢(0)=∑^𝑚−2_𝑖=1𝑎_𝑖𝑢(𝜉_𝑖), 𝑢^(𝑖)(0)=𝑢^(𝑗)(1)=𝜃, 1≤𝑖≤𝑘−1, 0≤𝑗≤𝑛−𝑘−1 in a real Banach space 𝐸, where 𝜃 is the zero element of 𝐸,0<𝜉₁<𝜉₂<⋯<𝜉_𝑚−2<1,𝑎_𝑖∈[0,+∞),𝑖=1,2,…,𝑚−2. As an application, we give two examples to demonstrate our results.

1. Introduction

The theory of ordinary differential equations in Banach spaces has become a new important branch (see, e.g., [1–13] and the references cited therein). In 1988, Guo and Lakshmikantham [4] discussed multiple solutions for two-point boundary value problems of second-order ordinary differential equations in Banach spaces. In [7], Guo obtained the existence of positive solutions for a boundary value problem of nth-order nonlinear impulsive integrodifferential equations in a Banach space by means of fixed point index theory and fixed point theory of completely continuous operators, respectively. Liu et al. in [6] obtained the existence of unbounded nonnegative solutions of a boundary value problem for nth-order impulsive integrodifferential equations on an infinite interval in Banach spaces by means of the Mddotoch fixed point theory in a Banach space. Zhang et al. in [9] dealt with the existence, nonexistence, and multiplicity of positive solutions for a class of nonlinear three-point boundary value problems of nth-order differential equations in Banach spaces. Zhao and Chen in [8, 12] investigated the existence of at least triple positive solutions for nonlinear boundary value problem by upper and low solution methods.

In this paper, the author considers the existence of positive solutions of the following higher-order (𝑘,𝑛−𝑘) conjugate multipoint boundary value problems (BVPs): (−1)^𝑛−𝑘𝑢^(𝑛)(𝑡)((𝑡()𝑢(0)==𝑓𝑡,𝑢)),𝑡∈0,1,𝑛≥2,1≤𝑘≤𝑛−1,_{𝑚−2𝑖=1}𝑎_𝑖𝑢𝜉_𝑖,𝑢^(𝑖)(0)=𝑢^(𝑗)(1)=𝜃,1≤𝑖≤𝑘−1,0≤𝑗≤𝑛−𝑘−1 (1.1) in a real Banach space 𝐸, where 𝜃 is the zero element of 𝐸,0<𝜉₁<𝜉₂<⋯<𝜉_𝑚−2<1,𝑎_𝑖∈[0,+∞),𝑖=1,2,…,𝑚−2.  𝑓∶(0,1)×𝐸→𝐸 is continuous and allowed to be singular at 𝑡=0 and 𝑡=1.

In scalar space, because of the widely applied background in mechanics and engineering, the nonlinear higher-order boundary value problems have received much attention (see Chyan and Henderson [Appl. Math. Letters 15 (2002) 767–774]). In [14], Eloe and Ahmad had solved successfully the existence of positive solution to the following nth-order boundary value problems: 𝑢^(𝑛)(𝑡)(𝑡)𝑓(𝑢)(),𝑢+𝑎=0,𝑡∈0,1_(𝑖)(0)=0,𝑖=0,1,…,𝑛−2,𝑢(1)=𝑎𝑢𝜂. (1.2) Recently, the existence of solutions and positive solutions of nonlinear (𝑘,𝑛−𝑘) focal boundary value problem (−1)^𝑛−𝑘𝑢^(𝑛)(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),…,𝑢^(𝑛−1)(𝑡)(),𝑢,𝑡∈0,1_(𝑖)(0)=0,𝑖=0,1,…,𝑘;𝑢^(𝑗)(1)=0,𝑗=𝑘+1,𝑘+2,…,𝑛−1, (1.3) and its special cases has been studied by many authors (see, e.g., [15–23]).

By using the Krasnoselskii fixed point theorem, Eloe and Henderson in [15], Agarwal and O'Regan in [16], and Kong and Wang in [20] have established the existence of solutions for following the (𝑘,𝑛−𝑘) conjugate boundary value problem: (−1)^𝑛−𝑘𝑢^(𝑛)(𝑡)((𝑡(),𝑢=𝑓𝑡,𝑢)),𝑡∈0,1_(𝑖)(0)=𝑢^(𝑖)(1)=0,0≤𝑖≤𝑘−1,0≤𝑗≤𝑛−𝑘−1. (1.4)

Very recently, by using the fixed point theory in a cone for strict-set contraction operators, Jiang and Zhang in [11] have discussed the existence of positive solutions for the above boundary value problem (1.4) in a Banach space, where the nonlinear term 𝑓(𝑡,𝑢(𝑡))∶𝐼×𝐸→𝐸 is continuous but not allowed to have singularity at 𝑡=0,1, where 𝐼=[0,1].

The organization of this paper is as follows. We shall introduce some lemmas and notations in the rest of this section. The preliminary lemmas are in Section 2. The main results are given in Section 3. Finally, two examples are presented to demonstrate our main results in Section 4.

Let the real Banach space 𝐸 with norm ‖⋅‖ be partially ordered by a cone 𝑃 of 𝐸, that is, 𝑢≤𝑣 if and only if 𝑣−𝑢∈𝑃, and 𝑃^∗ denotes the dual cone of 𝑃, that is, 𝑃^∗={𝜑∣𝜑∈𝐸^∗,𝜑(𝑢)≥0,𝑢∈𝑃}. A cone is called a solid cone if the set of interior points is not empty.

The closed balls in spaces 𝐸 and 𝐶[𝐼,𝐸] are denoted by 𝑃_𝑟={𝑢∈𝐸∶‖𝑢‖≤𝑟}(𝑟>0) and 𝐵_𝑟={𝑢∈𝐶[𝐼,𝐸]∶‖𝑢‖_𝐶≤𝑟}, respectively.

The basic space used in this paper is 𝐶[𝐼,𝐸]. For any 𝑢∈𝐶[𝐼,𝐸], evidently, (𝐶[𝐼,𝐸],‖⋅‖_𝑐) is a Banach space with norm ‖𝑢‖_𝐶=max_𝑡∈𝐼‖𝑢(𝑡)‖, and 𝑄={𝑢∈𝐶[𝐼,𝐸]∶𝑢(𝑡)≥𝜃 for 𝑡∈𝐼} is a cone of the Banach space 𝐶[𝐼,𝐸]. A function 𝑢∈𝐶^𝑛[𝐼,𝐸] is called a positive solution of the boundary value problem (1.1) if it satisfies (1.1) and 𝑢∈𝑄,𝑢(𝑡)≢𝜃.

Let 𝑢(𝑡)∶𝐼→𝐸 be continuous, and the abstract generalized integral ∫^𝑎₀𝑢(𝑡)𝑑𝑡(𝑎∈𝐼) can be similarly defined as in the scalar spaces and ∫^𝑎₀𝑢(𝑡)𝑑𝑡∈𝐸. If lim_𝜀→0+∫^𝜀₀𝑢(𝑡)𝑑𝑡 exist, then we say that the abstract integral is convergent, otherwise the abstract integral is divergent.

At the end of this section, we state some definitions and lemmas which will be used in Sections 2 and 3 (for details, see [1–3]).

Definition 1.1 (Kuratovski noncompactness measure). Let 𝐸 be a real Banach space, and 𝑆 is a bounded set in 𝐸. We denote 𝛼(𝑆)=inf{𝛿>0∶𝑆=⋃^𝑚_𝑖=1𝑆_𝑖, all the diameters of 𝑆_𝑖≤𝛿}.

In the following, 𝛼(⋅) denotes the Kuratowski measure of noncompactness in 𝐸 and 𝐶[𝐼,𝐸].

Definition 1.2 (strict-set contraction operator). Let 𝐸₁,𝐸₂ be real Banach spaces, and 𝑆⊂𝐸₁.   𝑇∶𝑆→𝐸₂ is a continuous and bounded operator. If there exists a constant 𝑘, such that 𝛼(𝑇(𝑆))≤𝑘𝛼(𝑆), then 𝑇 is called a 𝑘-set contraction operator. When 𝑘<1,𝑇 is called a strict-set contraction operator.

Lemma 1.3. If 𝐷⊂𝐶[𝐼,𝐸] is bounded and equicontinuous, then 𝛼(𝐷(𝑡)) is continuous on 𝐼 and 𝛼(𝐷)=𝛼(𝐷(𝐼))=sup_𝑡∈𝐼𝛼(𝐷(𝑡)), where 𝐷(𝐼)={𝑢(𝑡)∶𝑢∈𝐷,𝑡∈𝐼},𝐷(𝑡)={𝑢(𝑡)∶𝑢∈𝐷}.

Lemma 1.4. Let 𝑃 be a cone in a real Banach space 𝐸 and let  Ω be a nonempty bounded open convex subset of 𝑃. Suppose that 𝑇∶Ω→𝑃 is a strict-set contraction operator and 𝑇(Ω)⊂Ω, where Ω denotes the closure of in 𝑃. Then the fixed-point index 𝑖(𝑇,Ω,𝑃)=1.

Lemma 1.5. Let 𝑃 be a cone in a real Banach space 𝐸 and let  Ω be a bounded open subset of 𝑃, and suppose that 𝑇∶𝑃∩Ω→𝑃 is a strict-set contraction operator. (i)If 𝜃∈Ω, and 𝑇𝑢≠𝜆𝑢,forall𝑢∈𝜕Ω∩𝑃,𝜆≥1. Then 𝑖(𝑇,Ω∩𝑃,𝑃)=1.(ii)If there exists 𝑢₀∈𝑃⧵{𝜃}, such that 𝑢−𝑇𝑢≠𝜆𝑢,forall𝑢∈𝜕Ω∩𝑃,𝜆≥0. Then 𝑖(𝑇,Ω∩𝑃,𝑃)=0.

2. The Preliminary Lemmas

To prove the main results, we need the following lemmas.

Lemma 2.1 (see [20]). Let 𝑘(𝑡,𝑠) be the Green function for the (𝑘,𝑛−𝑘) conjugate BVP (1.4). Then 𝑘()=1𝑡,𝑠(𝑘−1)!(𝑛−𝑘−1)!^{0𝑡(1−𝑠)}𝑣^𝑘−1(𝑣+𝑠−𝑡)^{𝑛−𝑘−1}∫𝑑𝑣,0≤𝑡≤𝑠≤1,_{0𝑠(1−𝑡)}𝑣^{𝑛−𝑘−1}(𝑣+𝑡−𝑠)^𝑘−1𝑑𝑣,0≤𝑠≤𝑡≤1. (2.1) Obviously, 𝑘(𝑡,𝑠) is continuous on 𝐼×𝐼 and has the following properties. (G₁) There exist nonnegative functions 𝑝(𝑡),𝑚(𝑡),𝑞(𝑡)∈𝐶[0,1] such that 𝑝(𝑡)𝑞(𝑠)≤𝑘(𝑡,𝑠)≤𝑚(𝑡)𝑞(𝑠)≤𝑞(𝑠),∀𝑡,𝑠∈𝐼, (2.2)where 𝑝(𝑡)=𝑡^𝑘(1−𝑡)^𝑛−𝑘(𝑡)=𝑡𝑛−1,𝑚^𝑘−1(1−𝑡)^{𝑛−𝑘−1}{}(𝑠)=𝑠min𝑘,𝑛−𝑘,𝑞^𝑛−𝑘(1−𝑠)^𝐾()!()!.𝑘−1𝑛−𝑘−1 (2.3)(G₂) For any 𝜏∈(0,1/2),𝑘(𝑡,𝑠) satisfies 𝑘(𝑡,𝑠)≥𝜌(𝜏)𝑘(𝑣,𝑠),𝑡∈𝐼_𝜏=[𝜏,1−𝜏],𝑠,𝑣∈𝐼, (2.4)where 𝜌(𝜏)∶=min_{𝑡∈𝐼𝜏}𝑝(𝑡)max_𝑠∈[0,1]𝑚(𝑠). (2.5)

Setting Φ(𝑡)=(𝑛−1)!(𝑘−1)!(𝑛−𝑘−1)!¹_𝑡𝑠^𝑘−1(1−𝑠)^{𝑛−𝑘−1}𝑑𝑠. (2.6) It is obvious that 0≤Φ(𝑡)≤1,𝑡∈[0,1], and by the properties of the Euler integral, we have Φ(0)=1,Φ(1)=0,‖Φ‖=1. (2.7)

In order to abbreviate our discussion, we give the following assumptions.(C₀)∑^𝑚−2_𝑖=1𝑎_𝑖Φ(𝜉_𝑖)<1. (𝐶₁)𝑓∈𝐶[(0,1)×𝑃,𝑃] and ‖𝑓(𝑡,𝑢)‖≤𝑔(𝑡)‖ℎ(𝑢)‖,𝑡∈(0,1),𝑢∈𝑃, where ℎ∶𝑃→𝑃 is continuous and bounded and 𝑔∶(0,1)→(0,+∞) is continuous and satisfies ∫¹₀𝑔(𝑠)𝑑𝑠<+∞.(𝐶₂) For any 𝑟>0 and [𝑎,𝑏]⊂(0,1),  𝑓(𝑡,𝑢) is uniformly continuous on [𝑎,𝑏]×𝑃_𝑟.(𝐶₃) There exists constant 𝐿≥0 such that for any 𝑡∈(0,1) and the bounded set 𝐷⊂𝑃𝛼𝑓(𝑡,𝐷)≤𝐿𝛼(𝐷), (2.8) where 2𝐿1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖⋅max_𝑠∈𝐼𝑞(𝑠)<1. (2.9)

Lemma 2.2. Let ∑^𝑚−2_𝑖=1𝑎_iΦ(𝜉_𝑖)≠1 and (𝐶₁) be satisfied, then the problem (−1)^𝑛−𝑘𝑢^(𝑛)(𝑡)((𝑡[]𝑢(0)=𝑓𝑡,𝑢)),𝑡∈𝐼=0,1,𝑛≥2,1≤𝑘≤𝑛−1,=^𝑚−2_𝑖=1𝑎_𝑖𝑢𝜉_𝑖,𝑢^(𝑖)(0)=𝑢^(𝑗)(1)=𝜃,1≤𝑖≤𝑘−1,0≤𝑗≤𝑛−𝑘−1 (2.10)(2.11)(2.12) has a unique solution 𝑢(𝑡)=¹₀𝐾(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑡∈𝐼, (2.13) where 𝐾(𝑡,𝑠)=𝑘(𝑡,𝑠)+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖⁻¹Φ(𝑡)^𝑚−2_𝑖=1𝑎_𝑖𝑘𝜉_𝑖,𝑠,0≤𝑡,𝑠≤1. (2.14)

Proof. According to the definitions of generalized integral in abstract space, the proof of this lemma is similar to the proof in scalar spaces, so we omit it.

For 𝑢∈𝐶[𝐼,𝐸], we define an operator 𝑇 by (𝑇𝑢)(𝑡)=¹₀𝐾(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑡∈𝐼. (2.15)

Lemma 2.3. Suppose that (𝐶₀)–(𝐶₃)  hold. Then 𝑇∶𝐶[𝐼,𝐸]∩𝐵_𝑟→𝐶[𝐼,𝐸] is a strict-set contraction operator.

Proof. For 𝑢∈𝑢∈𝐶[𝐼,𝑃], it follows from (𝐶₁) that 𝑇 is well defined and bounded operator. If 𝐵∈𝐶[𝐼,𝑃] is a bounded subset of 𝐶[𝐼,𝑃], then 𝑇𝐵 is bounded.
Next we prove that 𝑇 is continuous on 𝐶[𝐼,𝐸]∩𝐵_𝑟. Let {𝑢_𝑗},{𝑢}⊂𝐶[𝐼,𝐸]∩𝐵_𝑟, and ‖𝑢_𝑗−𝑢‖_𝐶→0(𝑗→∞). Hence {𝑢_𝑗} is a bounded subset of 𝐶[𝐼,𝐸]∩𝐵_𝑟. Thus, there exists 𝑟>0 such that 𝑟=sup_𝑗‖𝑢_𝑗‖_𝐶<∞ and ‖𝑢‖_𝐶≤𝑟.
According to continuity of 𝑓, for all 𝜀>0, there exists 𝐽>0 such that𝑓𝑡,𝑥_𝑗(𝑡)((𝑡<𝜀−𝑓𝑡,𝑥))1+1−∑^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖⁻¹∑^𝑚−2_𝑖=1𝑎_𝑖∫¹₀𝑞(𝑠),𝑑𝑠 (2.16) for 𝑗≥𝐽,forall𝑡∈𝐼.
Then, 𝑇𝑢_𝑗(𝑡)−(𝑇𝑢)(𝑡)≤¹₀𝐾(𝑡,𝑠)𝑓𝑠,𝑥_𝑗(𝑠)((𝑠≤−𝑓𝑠,))𝑑𝑠1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖¹₀𝑞(𝑠)𝑓𝑠,𝑥_𝑗(𝑠)−𝑓(𝑠,𝑥(𝑠))𝑑𝑠<𝜀. (2.17) Therefore, for all 𝜀>0, for any 𝑡∈𝐼 and 𝑗≥𝐽, we get 𝑇𝑢_𝑗(𝑡)−(𝑇𝑢)(𝑡)⟶0. (2.18) This implies 𝑇 is continuous on 𝐶[𝐼,𝐸]∩𝐵_𝑟. By the properties of continuity of 𝐺(𝑡,𝑠), it is easy to see that 𝑇 is equicontinuous on 𝐼.
For any 𝑆⊂𝐶[𝐼,𝐸]∩𝐵_𝑟, it is easy to get that functions 𝑇(𝑆)={𝑇𝑢∣𝑢∈𝑆} are uniformly bounded. By Lemma 1.3, we get 𝛼(𝑇𝑆)=sup_𝑡∈𝐼𝛼((𝑇𝑆)(𝑡)), (2.19) where (𝑇𝑆)(𝑡)={(𝑇𝑢)(𝑡)∶𝑢∈𝑆,𝑡∈𝐼 is fixed}.
Write 𝐷=¹₀𝐾()𝑓((𝑠,𝐷𝑡,𝑠𝑠,𝑢))𝑑𝑠∶𝑢∈𝑆_𝜀=_𝜀^1−𝜀𝐾()𝑓((𝑠,1𝑡,𝑠𝑠,𝑢))𝑑𝑠∶𝑢∈𝑆0<𝜀<2. (2.20)(2.21) By (𝐶₁), for any 𝑢∈𝑆, we have 𝐻_𝜀=_𝜀^1−𝜀𝐾(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠−¹₀𝐾(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≤𝑐₀^𝜀₀𝐾(𝑡,𝑠)𝑔(𝑠)𝑑𝑠+𝑐₀¹_1−𝜀𝐾(𝑡,𝑠)𝑔(𝑠)𝑑𝑠, (2.22) where 𝑐₀∶=max_{𝑢∈𝐵𝑟}‖ℎ(𝑢)‖.
It follows from 𝐻_𝜀 and (𝐶₁) that the Hausdorff metric 𝑑_𝐻(𝐷_𝜀,𝐷)→0,(𝜀→0⁺). Thus lim_𝜀→0+𝛼𝐷_𝜀=𝛼(𝐷). (2.23)
We next shall estimate 𝛼(𝐷_𝜀). For any 𝑢∈𝐶[𝐼,𝐸], by ∫_𝜀^1−𝜀𝑢(𝑡)𝑑𝑡∈(1−2𝜀)𝑐𝑜{𝑢(𝑡)∶𝑡∈𝐼}, then𝛼𝐷_𝜀=𝛼_𝜀^1−𝜀𝐾()𝑓((𝑠≤()𝛼𝑡,𝑠𝑠,𝑢))𝑑𝑠∶𝑢∈𝑆1−2𝜀𝐾()𝑓((𝑠[]𝑘()𝑓((𝑠[]+𝑐𝑜𝑡,𝑠𝑠,𝑢))∶𝑠∈𝜀,1−𝜀,𝑢∈𝑆≤𝛼𝑡,𝑠𝑠,𝑢))∶𝑠∈𝜀,1−𝜀,𝑢∈𝑆1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖⁻¹𝛼^𝑚−2_𝑖=1𝑎_𝑖𝑘𝜉_𝑖𝑓((𝑠[]≤,𝑠𝑠,𝑢))∶𝑠∈𝜀,1−𝜀,𝑢∈𝑆1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1⋅max_𝑠∈𝐼𝑞(𝑠)⋅𝛼𝑓𝐼_𝜀×𝑆𝐼_𝜀, (2.24) where 𝐼_𝜀=[𝜀,1−𝜀],𝑆(𝐼_𝜀)={𝑢(𝑡)∶𝑡∈𝐼_𝜀}.
On the other hand, using a similar method as in the proof of Lemma 2 in [11] we can get that 𝛼(𝑆(𝐼))≤2𝛼(𝑆). (2.25) Therefore, it follows from (2.19), (2.25) that 𝛼𝑓𝐼_𝜀×𝑆𝐼_𝜀=sup_{𝑡∈𝐼𝜀}𝛼𝑓𝑡,𝑆𝐼_𝜀≤𝐿⋅𝛼𝑆𝐼_𝜀≤𝐿⋅𝛼(𝑆(𝐼))≤2𝐿⋅𝛼(𝑆). (2.26) Thus, we have 𝛼𝐷_𝜀≤2𝐿1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖⋅max_𝑠∈𝐼𝑞(𝑠)⋅𝛼(𝑆). (2.27) Combining with (2.24), we get 𝛼(𝐷)≤2𝐿1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖⋅max_𝑠∈𝐼𝑞(𝑠)⋅𝛼(𝑆). (2.28) Therefore, we have 𝛼(𝑇𝑆)≤2𝐿1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖⋅max_𝑠∈𝐼𝑞(𝑠)⋅𝛼(𝑆),∀𝑆∈𝑃∩𝐵_𝑟. (2.29) Notice that by (2.9) we claim that 𝑇∶𝐶[𝐼,𝐸]∩𝐵_𝑟→𝐶[𝐼,𝐸] is a strict-set contraction. The proof is complete.

Further, we construct a cone Ω byΩ=𝑢∈𝑄∶𝑢(𝑡)≥𝜌^∗(𝜏)𝑢(𝑠),∀𝑡∈𝐼_𝜏,∀𝑠∈𝐼, (2.30) where 𝐼_𝜏=[𝜏,1−𝜏], and 𝜌^∗(𝜏)=min𝜌(𝜏),min_{𝑡∈𝐼𝜏}Φ(𝑡), (2.31) where 𝜌(𝜏) is defined in (𝐺₂). It is easy to see that Ω is a closed convex cone of 𝐶[𝐼,𝐸] and Ω⊂𝑄.

Lemma 2.4. Suppose that (𝐶₀)–(𝐶₃)  hold. Then, 𝑇(Ω)⊂Ω.

Proof. From (𝐺₂), (2.6), (2.15), and (2.30), for any 𝑢∈Ω,𝑡∈𝑇_𝜏,𝑡′∈𝐼, we obtain (𝑇𝑢)(𝑡)=¹₀𝐾()𝑓((𝑠(𝜏)𝑡,𝑠𝑠,𝑢))𝑑𝑠≥𝜌¹₀𝑘𝑡^′,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖⁻¹Φ(𝑡)¹₀^𝑚−2_𝑖=1𝑎_𝑖𝑘𝜉_𝑖,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥𝜌^∗(𝜏)¹₀𝑘𝑡^′,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖⁻¹¹₀^𝑚−2_𝑖=1𝑎_𝑖𝑘𝜉_𝑖,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥𝜌^∗(𝜏)¹₀𝐾𝑡^′,𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝜌^∗(𝜏)(𝑇𝑢)𝑡^′,𝑡′∈𝐼. (2.32) Therefore, 𝑇(𝑢)∈Ω, that is, 𝑇(Ω)⊂Ω.

3. The Main Results

Let 𝑀^∗∶=max_𝑠∈𝐼𝑞(𝑠)⋅1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖¹₀𝑔(𝑠)𝑑𝑠. (3.1)

Theorem 3.1. Let (𝐶₀)–(𝐶₃)  hold. In addition, assume that the following conditions are satisfied: (C₄) 𝑑_𝑖∶=𝑀^∗𝑐_𝑖<1,(𝑖=1,2), where 𝑐₁∶=lim_{‖𝑢‖→∞}‖ℎ(𝑢)‖‖𝑢‖,𝑐₂∶=lim_{‖𝑢‖→∞}‖ℎ(𝑢)‖‖𝑢‖, (3.2)(C₅) 𝑃 is a solid cone, and there exist 𝑢₀∈Int(𝑃) and 𝜏∈(0,1/2) such that 𝑓(𝑡,𝑢)≥𝑙(𝑡)𝑢₀,𝑙(𝑡)∈[𝐼_𝜏,𝑅⁺], and 𝑙₀=min_{𝑡∈𝐼𝜏}𝑝(𝑡)_𝜏^1−𝜏𝑞(𝑠)𝑙(𝑠)𝑑𝑠≥1, (3.3)for all 𝑡∈𝐼_𝜏=[𝜏,1−𝜏],𝑢≥𝑢₀.
Then problem (1.1) has at least two positive solutions 𝑢₁,𝑢₂, and 𝑢₁(𝑡)≥𝑙₀𝑢₀, for 𝑡∈𝐼_𝜏,𝑢≥𝑢₀.

Proof. We first show that there exists 𝛽>0, such that ‖𝑣‖≥𝛽,forany𝑣≥𝑢₀. (3.4) If (3.4) is not true, then there exist the sequences {𝑣_𝑛}^∞_𝑛=1 satisfying 𝑣_𝑛≥𝑢₀, and ‖𝑣_𝑛‖<1/𝑛,(𝑛=1,2,…). Thus 𝑢₀≤𝜃, which is contradiction with 𝑢₀∈int(𝑃).
Let 𝑐^′_𝑖=1+𝑐_𝑖𝑀^∗2𝑀^∗,(𝑖=1,2). (3.5) Then 𝑐^′_𝑖>𝑐_𝑖,𝑑^′_𝑖=𝑐^′_𝑖𝑀^∗<1. (3.6) From (𝐶₄) and (3.6), there exist two positive constants 𝑟₁,𝑟₂ with 0<𝑟₁<𝛽,𝑟₂>max𝛽,2𝑢₀ (3.7) such that ‖ℎ(𝑢)‖≤𝑐^′_𝑖‖𝑢‖,𝑢∈𝐵_𝑟1∩𝑃,‖ℎ(𝑢)‖≤𝑐^′_𝑖‖𝑢‖,𝑢∈𝑃⧵𝐵_𝑟2. (3.8) Therefore, for any 𝑢∈𝑃, we have ‖ℎ(𝑢)‖≤𝑐^′_𝑖‖𝑢‖+𝑀, (3.9) where 𝑀=sup{‖ℎ(𝑢)‖∶𝑢∈𝐵_𝑟2}.
Choose 𝑟₃=𝑟₂+1−𝑑^′₂⁻¹𝑀^∗𝑀. (3.10) In the following, let Ω₁=𝑢∈𝑄∶∣‖𝑢‖<𝑟₁,Ω₂=∈𝑄∶∣‖𝑢‖<𝑟₃,Ω₃=𝑢∈𝑄∶∣‖𝑢‖<𝑟₃,if𝑡∈𝐼_𝜏,𝑢(𝑡)≥𝑢₀. (3.11)(3.12)(3.13) It is to see from (3.7) that Ω₃ is nonempty for 2𝑢₀∈Ω₃, which implies Ω₃≠∅. Obviously, Ω_𝑖⊂𝑄  (𝑖=1,2,3) are nonempty, convex, open sets and Ω₁=𝑄∩𝐵_𝑟1,Ω₂=𝑄∩𝐵_𝑟2,Ω₃={𝑢∈Ω₂∶𝑢(𝑡)≥𝑢₀,𝑡∈𝐼_𝜏}. So Ω₁⊂Ω₂,Ω₃⊂Ω₂,Ω₁∩Ω₃=∅. (3.14)
For any 𝑢∈Ω₂, form (3.8) and (3.9), we have ‖(𝑇𝑢)(𝑡)‖≤¹₀𝐾(𝑡,𝑠)𝑔(𝑠)𝑑𝑠𝑐^′₂‖𝑢‖≤+𝑀₁₀𝑔(𝑠)𝑑𝑠⋅𝑐^′₂‖𝑢‖+𝑀⋅1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖max_𝑠∈𝐼𝑞(𝑠)=𝑑^′₂‖𝑢‖+𝑀^∗𝑀<𝑑^′₂𝑟₃+𝑀^∗𝑀<𝑟₃, (3.15) which implies 𝑇Ω₂⊂Ω₂. (3.16) From (2.15) and (3.7), we get 𝑇Ω₁⊂Ω₁. (3.17)
For any 𝑢∈Ω₃, it follows from (3.16) that ‖𝑇𝑢‖<𝑟₃. According to (𝐶₅) and (𝐺₁), we can obtain (𝑇𝑢)(𝑡)≥_𝜏^1−𝜏𝐾()𝑓((𝑠≥𝑡,𝑠𝑠,𝑢))𝑑𝑠_𝜏1−𝜏𝐾(𝑡,𝑠)𝑙(𝑠)𝑢₀𝑑𝑠≥min_{𝑡∈𝐼𝜏}𝑝(𝑡)_𝜏^1−𝜏𝑞(𝑠)𝑙(𝑠)𝑑𝑠⋅𝑢₀=𝑙₀𝑢₀≥𝑢₀,∀𝑡∈𝐼_𝜏, (3.18) which implies 𝑇Ω₃⊂Ω₃. (3.19) Combining (3.16)–(3.19) with Lemma 1.4, we have 𝑖(𝑇,Ω_𝑖,𝑃)=1,𝑖=1,2,3. Furthermore, using the fixed point index theory, we obtain successively 𝑖𝑇,Ω₂⧵Ω₁∪Ω₃,𝑃=𝑖T,Ω₂,𝑃−𝑖𝑇,Ω₁,𝑃−𝑖𝑇,Ω₃,𝑃=−1. (3.20)
Then 𝑇 has at least two fixed points 𝑢₁ and 𝑢₂ which satisfy 𝑢₁∈Ω₃ and 𝑢₂∈Ω₂⧵(Ω₁∪Ω₃). Then Theorem 3.1 is proved.

Theorem 3.2. Assume that (𝐶₀)–(𝐶₃), 𝑑₂<1 hold. Suppose further that (𝐶^′₅)  𝑃 is a cone of the real Banach space 𝐸, and there exist 𝑢₀∈𝑃⧵{𝜃} and 𝜏∈(0,1/2) such that 𝑓(𝑡,𝑢)≥𝑙(𝑡)𝑢₀,𝑙(𝑡)∈[𝐼_𝜏,𝑅⁺], for for all 𝑡∈𝐼_𝜏,𝑢≥𝑢₀.
Then problem (1.1) has at least one positive solutions 𝑢 with 𝑢(𝑡)≥𝑙₀𝑢₀,𝑡∈𝐼_𝜏,𝑢≥𝑢₀, where 𝑙₀ is defined in (3.3).

Proof . As in the proof of Theorem 3.1, we need only to show that 𝑇 has one positive fixed point 𝑢 with 𝑢(𝑡)≥𝑙₀𝑢₀,𝑡∈𝐼_𝜏,𝑙₀≥1.
Choose 𝑟₀ satisfying 𝑟₀>max{𝛽,2‖𝑢₀‖} and let Ω₀=𝑢∈𝐶[𝐼,𝐸]∶‖𝑢‖<𝑟₀,𝑢(𝑡)≥𝑢₀,∀𝑡∈𝐼_𝜏. (3.21) Obviously, Ω₀ is a bounded closed convex set in 𝐶[𝐼,𝐸].  Ω₀≠∅, for 𝑢^∗(𝑡)≡𝑢₀∈Ω₀. Let 𝑢∈Ω₀. As the proof of (3.19), we have 𝑇(Ω₀)⊂Ω₀, where 𝑇 is given by (2.15). Thus, it follows from Lemmas 2.3 and 1.4 that 𝑇 has a fixed point 𝑢∈Ω₀. The proof is complete.

Theorem 3.3. Assume that (𝐶₀)–(𝐶₃) and the following conditions hold.(𝐶₆) There exists 𝑅₀>0 such that sup_𝑃𝑅0‖ℎ(𝑢)‖<𝑅₀(𝑀^∗)⁻¹, where 𝑃_𝑅0={𝑢∈𝑃∶‖𝑢‖<𝑅₀}.(𝐶₇) There exists 𝜑∈𝑃^∗ such that 𝜑(𝑢)>0 for any 𝑢>𝜃, and lim_{‖𝑢‖→+∞}min_{𝑡∈𝐼𝜏}𝜑𝑓(𝑡,𝑢)𝜑(𝑢)>𝑀₀, (3.22)where 𝜏∈(0,1/2), and 𝑀₀∶=min_{𝑡∈𝐼𝜏}𝑝(𝑡)⋅𝜌^∗(𝜏)_𝜏^1−𝜏𝑞(𝑠)𝑑𝑠⁻¹. (3.23)(𝐶₈) There exists 𝜑∈𝑃^∗ such that 𝜑(𝑢)>0 for any 𝑢>𝜃, and lim_{‖𝑢‖→0}min_{𝑡∈𝐼𝜏}𝜑𝑓(𝑡,𝑢)𝜑(𝑢)>𝑀₀, (3.24) where 𝜏∈(0,1/2), and 𝑀₀ is defined in (3.23). Then problem (1.1) has at least two positive solutions.

Proof. If (𝐶₆) holds, then there exists 𝜀₁ satisfying 0<𝜀₁<𝑅₀(𝑀^∗)⁻¹ such that ‖ℎ(𝑢)‖≤(𝑅₀(𝑀^∗)⁻¹−𝜀₁). For any 𝑢∈Ω∩𝜕𝑃_𝑅0,𝑡∈𝐼_𝜏, it follows from (𝐶₁) and (2.15) that ‖𝑇𝑢‖≤¹₀𝑞(𝑠)𝑔(𝑠)‖ℎ(𝑢)‖𝑑𝑠+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖¹₀𝑞(𝑠)𝑔(𝑠)‖ℎ(𝑢)‖𝑑𝑠≤max_{𝑠∈𝐼𝜏}𝑞(𝑠)1+1−^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖^−1𝑚−2_𝑖=1𝑎_𝑖¹₀𝑔(𝑠)𝑑𝑠⋅𝑅₀𝑀^∗−1−𝜀₁=𝑅₀𝑀^∗−1−𝜀₁𝑀^∗<𝑅₀, (3.25) which implies 𝑇𝑢≠𝜆𝑢,forall𝑢∈Ω∩𝜕𝑃_𝑅0, and 𝜆≥1. From Lemma 1.5 (i), we have 𝑖𝑇,Ω∩𝜕𝑃_𝑅0,Ω=1. (3.26)
According to (𝐶₇), there exist 𝜀₂>0 and 𝑅₂>𝑅₀>0, for any 𝑡∈𝐼_𝜏, such that 𝜑𝑓(𝑡,𝑢)≥𝑀₀+𝜀₂𝜑(𝑢),‖𝑢‖≥𝑅₂. (3.27) Let 𝑈₂={𝑢∈𝐶[𝐼,𝐸]∶‖𝑢‖<𝑅₂}. We need only to show that 𝑢−𝑇𝑢≠𝜆𝑒, for any 𝑢∈Ω∩𝜕𝑈₂, and 𝜆≥0,𝑒∈𝑃 with ‖𝑒‖=1. If it is false, then there exists 𝑢^∗∈Ω∩𝜕𝑈₂ and 𝜆₀≥0, such that 𝑢^∗−𝑇𝑢^∗=𝜆₀𝑒, (3.28) which implies 𝑢^∗(𝑡)≥𝑇𝑢^∗(𝑡)=¹₀𝐾(𝑡,𝑠)𝑓𝑠,𝑢^∗(𝑠)(𝑡)𝑑𝑠≥𝑝_𝜏^1−𝜏𝑞(𝑠)𝑓𝑠,𝑢^∗(𝑠)𝑑𝑠. (3.29) Hence, for any 𝜑∈𝑃^∗, we have 𝜑𝑢^∗(𝑡)≥𝜑𝑇𝑢^∗(𝑡)=¹₀𝐾(𝑡,𝑠)𝜑𝑓𝑠,𝑢^∗(𝑠)≥𝑑𝑠_𝜏1−𝜏𝑝(𝑡)𝑞(𝑠)𝑀₀+𝜀₂𝜑𝑢^∗(𝑠)𝑑𝑠≥min_{𝑡∈𝐼𝜏}𝑝(𝑡)_𝜏^1−𝜏𝑞(𝑠)𝑀₀+𝜀₂𝜑𝑢^∗(𝑠)≥𝑀𝑑𝑠₀+𝜀₂min_{𝑡∈𝐼𝜏}𝑝(𝑡)⋅𝜌^∗(𝜏)_𝜏^1−𝜏𝑞(𝑠)𝑑𝑠⋅𝜑𝑢^∗(𝑡). (3.30)
It is easy to see that 𝜑(𝑢^∗(𝑡))>0. In fact, if 𝜑(𝑢^∗(𝑡))=0, since 𝑢^∗∈Ω∩𝜕𝑈₂, then we have 𝜑(𝑢^∗(𝑡))≥𝜌^∗(𝜏)𝜑(𝑢^∗(𝑠))≥0 and consequently ‖𝑢^∗‖=0, which contradicts with ‖𝑢^∗‖=𝑅₂. By (3.29) and (3.30), we get 1<(𝑀₀+𝜀₂)min_{𝑡∈𝐼𝜏}𝑝(𝑡)𝜌^∗(𝜏)∫_𝜏^1−𝜏𝑞(𝑠)𝑑𝑠≤1, which is a contradiction. It follows from Lemma 1.5 (ii) that 𝑖𝑇,Ω∩𝜕𝑈₂,Ω=0. (3.31)
If (𝐶₈) holds, then there exist 𝜀₃>0 and 𝑅₁>0 with 𝑅₁<𝑅₀, such that 𝜑𝑓(𝑡,𝑢)≥𝑀₀+𝜀₃𝜑(𝑢),‖𝑢‖≥𝑅₁. (3.32) Let 𝑈₁={𝑢∈𝐶[𝐼,𝐸]∶‖𝑢‖<𝑅₁}. As in the proof of (3.31), for any 𝑢∈Ω∩𝜕𝑈₁,𝑡∈𝐼_𝜏, we get 𝑖𝑇,Ω∩𝜕𝑈₁,Ω=0. (3.33) Notice that 𝑈₁⊂𝑃_𝑅0,𝑃_𝑅0⊂𝑈₂. Thus, it follows from (3.26), (3.31), and (3.33) that 𝑖𝑇,Ω∩𝑃_𝑅0⧵𝑈₁,Ω=𝑖𝑇,Ω∩𝑃_𝑅0,Ω−𝑖𝑇,Ω∩𝑈₁𝑖𝑈,Ω=1,𝑇,Ω∩₂⧵𝑃_𝑅0,Ω=𝑖𝑇,Ω∩𝑈₂,Ω−𝑖𝑇,Ω∩𝑃_R0,Ω=−1. (3.34)
Then 𝑇 has at least two fixed points 𝑢₁ and 𝑢₂ which satisfy 𝑢₁∈Ω∩(𝑃_𝑅0⧵𝑈₁) and 𝑢₂∈Ω∩(𝑈₂⧵𝑃_𝑅0). Then Theorem 3.3 is proved.
Similarly to the proofs of Theorem 3.3, we can easily get the following corollaries.

Corollary 3.4. Assume that (𝐶₀)–(𝐶₃), 𝑑₁<1 and (𝐶₇) hold. Then problem (1.1) has at least one positive solution.

Corollary 3.5. Assume that (𝐶₀)–(𝐶₃), 𝑑₂<1 and (𝐶₈) hold. Then problem (1.1) has at least one positive solution.

4. Examples

Now we present two examples to illustrate our main results.

Example 4.1. Consider the boundary value problems in 𝐸=𝑅^𝑁 (N-dimensional Euclidean space and ‖𝑢‖=max_{1≤𝑝≤𝑁}|𝑢_𝑝|<+∞) (−1)⁶𝑢_𝑝⁽¹⁰⁾(𝑡)=1+𝑡𝑡(1−𝑡)sin²𝑢_𝑝+1+ln1+𝑢²_𝑝+𝑡(1−𝑡)𝐻𝑢_𝑝𝑢,0<𝑡<1,_𝑝(0)=19𝑢_𝑝14+12𝑢_𝑝12𝑢,𝑝=1,2,3,…_(𝑖)(0)=𝑢^(𝑗)(1)=0,1≤𝑖≤3,0≤𝑗≤5, (4.1) where 𝑛=10,𝑘=4, and 𝐻𝑢_𝑝=23×10⁵𝑢_𝑝+𝑁𝑢₀×10¹²𝑢²_𝑝,𝑢_𝑝∈𝑢0,⁰𝑁,23×10⁵+1×10¹²𝑢_𝑝,𝑢_𝑝∈𝑢₀𝑁,1+𝑟₃𝑢₀+𝑢₀𝑁,23×10⁵𝑢_𝑝+1+𝑟₃𝑢₀+𝑢₀𝑁×10¹²𝑢_𝑝,𝑢_𝑝∈1+𝑟₃𝑢₀+𝑢₀𝑁,+∞, (4.2) where 𝑟₃(⋅) is as defined in (3.10). Then the problem (4.1) can be regarded as a BVP of the form (1.1) in 𝐸. In this situation, 𝑢=(𝑢₁,…,𝑢_𝑃,…,𝑢_𝑁)∈𝑅^𝑁,𝑓=(𝑓₁,…,𝑓_𝑃,…,𝑓_𝑁),𝜃=(0,0,…,0)∈𝑅^𝑁 and 𝑎₁=1/9,𝜉₁=1/4,𝑎₂=𝜉₂=1/2,𝑔=(𝑔₁,…,𝑔_𝑃,…,𝑔_𝑁),ℎ=(ℎ₁,…,ℎ_𝑃,…,ℎ_𝑁), in which 𝑓_𝑝𝑡,𝑢₁,𝑢₂,…,𝑢_𝑁=1+𝑡𝑡(1−𝑡)sin²𝑢_𝑝+1+ln1+𝑢²_𝑝+𝑡(1−𝑡)𝐻𝑢_𝑝,𝑝(𝑡)=𝑡⁴(1−𝑡)⁶9(𝑡)=𝑡,𝑚³(1−𝑡)⁵4(𝑠)=𝑠,𝑞⁶(1−𝑠)⁴.3!⋅5! (4.3) Obviously, 𝑓∶(0,1)×𝐸→𝐸 is continuous. Taking 𝜏=1/4. By direct account, we have ^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖=19Φ14+12Φ12≈0.219<1. (4.4) Now we show that (𝐶₁)–(𝐶₅) hold.

First, it is easy to see that ‖𝑓(𝑡,𝑢)‖≤𝑔(𝑡)‖ℎ(𝑢)‖,𝑡∈(0,1),𝑢∈𝐸, with 𝑔_𝑝(𝑡)=2𝑡(1−𝑡),ℎ_𝑝(𝑢)=sin²𝑢_𝑝+1+ln1+𝑢²_𝑝+12𝐻𝑢_𝑝, (4.5) and ∫¹₀𝑔_𝑝(𝑠)𝑑𝑠=2𝜋<+∞.

For sufficiently large number 𝑅>0, take 𝑃={𝑢=(𝑢₁,…,𝑢_𝑝,…,𝑢_𝑁)∈𝑅^𝑁∶𝑢_𝑝≥0,‖𝑢‖≤𝑅,𝑝=1,2,…,𝑁}.

Then ℎ∶𝑃→𝑃 is continuous and bounded. Therefore, (𝐶₁) is satisfied. In addition, (𝐶₂) and (𝐶₃) are automatically satisfied in 𝑁-dimensional Euclidean space 𝑅^𝑁.

As far as (𝐶₄) is concerned, we get 𝑀^∗=2𝜋max_[]𝑠∈0,1𝑠⁶(1−𝑠)⁴⋅()3!5!1+1−0.219⁻¹×19+12≈1.86×10⁻⁵,𝑐₁=lim_{‖𝑢‖→∞}‖ℎ(𝑢)‖‖𝑢‖=13×10⁵,𝑐₂=lim_{‖𝑢‖→∞}‖ℎ(𝑢)‖‖𝑢‖=13×10⁵,𝑑_𝑖=𝑀^∗𝑐_𝑖≈0.62<1. (4.6) Then (𝐶₄) is satisfied.

On the other hand, 𝑃 is a solid cone, for all 𝑢₀∈Int(𝑃). For all 𝑡∈𝐼_𝜏=[1/4,3/4], we have 𝑓_𝑝(𝑡,𝑢)≥1×10¹²(1+𝑡)𝑢_𝑝≥𝑙(𝑡)𝑢₀^(𝑝),for𝑢≥𝑢₀, (4.7) where 𝑢₀=(𝑢₀⁽¹⁾,…,𝑢₀^(𝑝),…,𝑢₀^(𝑁)),𝑙(𝑡)=1×10¹²(1+𝑡), and 𝑙₀=min_{[]𝑡∈1/4,3/4}𝑝(𝑡)^3/4_1/4𝑞(𝑠)𝑙(𝑠)𝑑𝑠≈7.055>1. (4.8) So (𝐶₅) is also satisfied. For example, if 𝑢₀={1,1,…,1}, we let 𝛽=1, then 𝑟₃(‖𝑢₀‖)+(‖𝑢₀‖)/𝑁+1≈9.745×10⁷(2𝑁+1). By Theorem 3.1, the problem (4.1) has at least two positive solutions in 𝐸.

Example 4.2. Consider the following boundary value problems in 𝑙^∞: (−1)³𝑢_𝑝⁽⁹⁾(𝑡)=𝜋√𝑡𝑢ln1+^2𝑝𝑝+𝑡(1−𝑡)𝑢_𝑝𝑝sin𝑢_𝑝+1+𝐻𝑢_𝑝𝑢,0<𝑡<1,_𝑝(0)=14𝑢_𝑝34𝑢,𝑝=1,2,3,…_(𝑖)(0)=𝑢^(𝑗)(1)=0,1≤𝑖≤5,0≤𝑗≤2, (4.9) where 𝑛=9,𝑘=6, and 𝐻𝑢_𝑝=8×10⁸𝑢_𝑝,𝑢_𝑝∈10,16×10⁻⁸,100⁴𝑢_𝑝,𝑢_𝑝∈116×10⁻⁸,1×10⁴,8×10⁸𝑢_𝑝+10−8×10¹⁰𝑢_𝑝,𝑢_𝑝∈1×10⁴,+∞. (4.10) Let 𝑃={𝑢=(𝑢₁,…,𝑢_𝑝,…)∈𝑙^∞∶𝑢_𝑝≥0,𝑝=1,2,…},‖𝑢‖=sup_𝑝≥1|𝑢_𝑝|<+∞. Then 𝑙^∞ is a Banach space. Thus the problem (4.9) can be regarded as a BVP of the form (1.1) in 𝑙^∞. In this situation, 𝑢=(𝑢₁,…,𝑢_𝑃,…)∈𝑙^∞,𝑓=(𝑓₁,…,𝑓_𝑃,…),𝜃=(0,0,…)∈𝑙^∞ and 𝑎₁=1/4,𝜉=3/4,𝑔=(𝑔₁,…,𝑔_𝑃,…,),ℎ=(ℎ₁,…,ℎ_𝑃,…,), in which 𝑓_𝑝𝑡,𝑢₁,𝑢₂=𝜋,…,√𝑡𝑢ln1+^2𝑝𝑝+𝑡(1−𝑡)𝑢_𝑝𝑝sin𝑢_𝑝+1+𝐻𝑢_𝑝,𝑝(𝑡)=𝑡⁶(1−𝑡)³8(𝑡)=𝑡,𝑚⁵(1−𝑡)²3(𝑠)=𝑠,𝑞³(1−𝑠)⁶.5!⋅2! (4.11) Obviously, 𝑓∶(0,1)×𝐸→𝐸 is continuous taking 𝜏=1/8. By direct account, we have ^𝑚−2_𝑖=1𝑎_𝑖Φ𝜉_𝑖=14Φ34≈0.0804<1,𝑀^∗=1.099×10⁻⁴,𝜌^∗(𝜏)=min𝜌(𝜏),min_[]1/8,7/8Φ(𝑡)≈min{0.6471,0.0674}=0.0674. (4.12) Let 𝑔_𝑝(𝑡)=𝜋√𝑡,ℎ_𝑝(𝑢)𝑢=ln1+^2𝑝𝑝+𝑢_𝑝2𝑝sin𝑢_𝑝+1+𝐻𝑢_𝑝, (4.13) and ∫¹₀𝑔_𝑝(𝑠)𝑑𝑠=2𝜋<+∞, 𝑀₀=min_{[]𝑡∈1/8,7/8}𝑝(𝑡)⋅𝜌^∗(𝜏)^7/8_1/8𝑞(𝑠)𝑑𝑠⁻¹≈7.0811×10⁸. (4.14) Similar to the proofs of Example 4.1, we can show that (𝐶₁)–(𝐶₃) hold. Taking 𝑅₀=10^8/3, then sup_{𝑢∈𝑃𝑅0}{‖ℎ(𝑢)‖}<𝑀^∗𝑅₀, which implies (𝐶₆), holds.

For 𝑢∈𝑃, choose 𝜑∈𝑃^∗ with 𝜑(𝑢)=∑^+∞_𝑝=1𝑢_𝑝/𝑝². Obviously, 𝜑(𝑢)>0 for 𝑢>𝜃, and lim_{‖𝑢‖→+∞}min_{[]𝑡∈1/8,7/8}𝜑𝑓(𝑡,𝑢)𝜑(𝑢)=lim_{‖𝑢‖→+∞}min_{[]𝑡∈1/8,7/8}𝜋√𝑡⋅∑^+∞_𝑝=11/𝑝²ln1+𝑢²_𝑝/𝑝+𝑢_𝑝/2𝑝sin𝑢_𝑝+1+𝐻𝑢_𝑝∑^+∞_𝑝=1𝑢_𝑝/𝑝²>8×10⁸>𝑀₀, (4.15) which implies (𝐶₇) is satisfied. Similarly, we can show that (𝐶₈) holds. By Theorem 3.3, the problem (4.9) has at least two positive solutions in 𝐸.