Existence of Solutions for Fractional Integro-Differential Equation with Multipoint Boundary Value Problem in Banach Spaces

Abstract

By means of the fixed-point theorem in the cone of strict-set-contraction operators, we consider the existence of a nonlinear multi-point boundary value problem of fractional integro-differential equation in a Banach space. In addition, an example to illustrate the main results is given.

1. Introduction

The purpose of this paper is to establish the existence results of positive solution to nonlinear fractional boundary value problem 𝐷^𝑞₀₊𝑢(𝑡)+𝑓𝑡,𝑢,𝑢^′,…,𝑢^(𝑛−2)𝑢(0),𝑇𝑢,𝑆𝑢=𝜃,0<𝑡<1,𝑛−1<𝑞≤𝑛,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=𝜃,𝑢^(𝑛−2)(1)=^𝑚−2_𝑖=1𝑎_𝑖𝑢^(𝑛−2)𝜂_𝑖 (1.1) in a Banach space 𝐸, where 𝜃 is the zero element of 𝐸, and 𝑛≥2,0<𝜂₁<⋯<𝜂_𝑚−2<1,  𝑎_𝑖>0(𝑖=1,2,…,𝑚−2),  𝐷^𝛼₀⁺ is Riemann-Liouville fractional derivative, and 𝑇𝑢(𝑡)=^𝑡₀𝐾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠,𝑆𝑢(𝑡)=¹₀𝐻(𝑡,𝑠)𝑢(𝑠)𝑑𝑠, (1.2) where 𝐾∈𝐶[𝐵,𝑅₊],𝐵={(𝑡,𝑠)∈𝐼×𝐼∶𝑡≥𝑠},𝐻∈𝐶[𝐼×𝐼,𝑅₊],  𝐼=[0,1], and 𝑅₊ denotes the set of all nonnegative numbers.

Fractional differential equations have gained importance due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, and probability. For details see [1–3] and the references therein. In recent years, there are some papers dealing with the existence of the solutions of initial value problems or linear boundary value problems for fractional differential equations by means of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, lower and upper solutions method, and so forth), see for example, [4–23].

In [8], by means of the fixed-point theorem for the mixed monotone operator, the authors considers unique existence of positive to singular boundary value problems for fractional differential equation 𝐷^𝑞₀₊𝑢(𝑡)+𝑎(𝑡)𝑓𝑡,𝑢,𝑢^′,…,𝑢^(𝑛−2)𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=𝑢^(𝑛−2)(1)=0, (1.3) where 𝐷^𝑞₀₊ is Riemann-Liouville fractional derivative of order 𝑛−1<𝑞≤𝑛,𝑛≥2.

In [11], El-Shahed and Nieto study the existence of nontrivial solutions for a multi-point boundary value problem for fractional differential equations 𝐷^𝑞₀₊𝑢(𝑡)((𝑡𝑢(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,𝑛−1<𝑞≤𝑛,𝑛∈𝑁,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=0,𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝑢𝜂_𝑖, (1.4) where 𝑛≥2,𝜂_𝑖∈(0,1),  𝑎_𝑖>0(𝑖=1,2,…,𝑚−2), and  𝐷^𝑞₀₊ is Riemann-Liouville fractional derivative. Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of nontrivial solution are obtained by using Leray-Schauder nonlinear alternative. And then, Goodrich [24] was concerned with a partial extension of the problem (1.3) 𝐷^𝑞₀₊𝑢(𝑡)((𝑡𝑢=𝑓𝑡,𝑢)),0<𝑡<1,𝑛−𝑛<𝑞≤𝑛−2,_(𝑖)(0)=0,0≤𝑖≤𝑛−2,𝐷^𝑝₀₊𝑢(1)=0,1≤𝑝≤𝑛−2, (1.5) and the authors derived the Green function for the problem (1.5) and showed that it satisfies certain properties.

By the contraction mapping principle and the Krasnoselskii’s fixed-point theorem, Zhou and Chu [13] discussed the existence and uniqueness results for the following fractional differential equation with multi-point boundary conditions: ^𝐶𝐷^𝑞₀₊𝑢(𝑡)()𝑎+𝑓𝑡,𝑢,𝐾𝑢,𝑆𝑢=0,0<𝑡<1,1<𝑞<2,₁𝑢(0)−𝑏₁𝑢^′(0)=𝑑₁𝑢𝜂₁,𝑎₂𝑢(1)−𝑏₂𝑢^′(1)=𝑑₂𝑢𝜂₂, (1.6) where ^𝐶𝐷^𝑞₀₊ is the Caputo’s fractional derivative, 𝑎₁,𝑎₂,𝑏₁,𝑏₂,𝑑₁, and 𝑑₂ are real numbers, 0<𝜂₁, and 𝜂₂<1.

In [20], Stanĕk has discussed the existence of positive solutions for the singular fractional boundary value problem 𝐷^𝑞𝑢(𝑡)+𝑓𝑡,𝑢,𝑢^′,𝐷^𝑝𝑢𝑢(0)=0,2<𝑞<3,0<𝑝<1,=0,𝑢^′(0)=𝑢^′(1)=0. (1.7)

However, to the best of the author’s knowledge, a few papers can be found in the literature dealing with the existence of solutions to boundary value problems of fractional differential equations in Banach spaces. In [25], Salem investigated the existence of Pseudo solutions for the following nonlinear 𝑚-point boundary value problem of fractional type 𝐷^𝑞₀₊𝑢(𝑡)(𝑡)𝑓((𝑡𝑢(0)+𝑎𝑡,𝑢))=0,0<𝑡<1,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=0,𝑢(1)=^𝑚−2_𝑖=1𝜁_𝑖𝑢𝜂_𝑖 (1.8) in a reflexive Banach space 𝐸, where 𝐷^𝑞₀₊ is the Pseudo fractional differential operator of order 𝑛−1<𝑞≤𝑛,  𝑛≥2.

In [26], by the monotone iterative technique and mönch fixed-point theorem, Lv et al. investigated the existence of solution to the following Cauchy problems for differential equation with fractional order in a real Banach space 𝐸^𝐶𝐷^𝑞𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑢(0)=𝑢₀, (1.9) where ^𝐶𝐷^𝑞𝑢(𝑡) is the Caputo’s derivative order, 0<𝑞<1.

By means of Darbo’s fixed-point theorem, Su [27] has established the existence result of solutions to the following boundary value problem of fractional differential equation on unbounded domain [0,+∞)𝐷^𝑞₀₊𝑢(𝑡)((𝑡[)𝑢(0)=𝑓𝑡,𝑢)),𝑡∈0,+∞,1<𝑞≤2,=𝜃,𝐷₀^𝑞−1₊𝑢(∞)=𝑢_∞ (1.10) in a Banach space 𝐸.  𝐷^𝑞₀₊ is the Riemann-Liouville fractional derivative.

Motivated by the above mentioned papers [8, 13, 24, 25, 27, 28] but taking a quite different method from that in [26–29]. By using fixed-point theorem for strict-set-contraction operators and introducing a new cone Ω, we obtain the existence of at least two positive solutions for the BVP (1.1) under certain conditions on the nonlinear term in Banach spaces. Our results are different from those of [8, 13, 24, 25, 28, 30]. Note that the nonlinear term 𝑓 depends on 𝑢 and its derivatives 𝑢^′,𝑢^′′,…,𝑢^(𝑛−2).

2. Preliminaries and Lemmas

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 is said to be normal if there exists a positive constant 𝑁 such that 𝜃≤𝑢≤𝑣 implies ‖𝑢‖≤𝑁‖𝑣‖, where the smallest 𝑁 is called the normal constant of 𝑃. For details on cone theory, see [31].

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=sup_𝑡∈𝐼|𝑢(𝑡)|, and 𝑃={𝑢∈𝐶[𝐼,𝐸]∶𝑢(𝑡)≥𝜃 for 𝑡∈𝐼} is a cone of the Banach space 𝐶[𝐼,𝐸].

Definition 2.1 (see [31]). Let 𝑉 be a bounded set in a real Banach space 𝐸, and 𝛼(𝑉)=inf{𝛿>0∶𝑉=∪^𝑚_𝑖=1𝑉_𝑖, all the diameters of 𝑉_𝑖≤𝛿}. Clearly, 0≤𝛼(𝑉)<∞.  𝛼(𝑉) is called the Kuratovski measure of noncompactness.
We use 𝛼,  𝛼_𝐶 to denote the Kuratowski noncompactness measure of bounded sets in the spaces 𝐸,𝐶(𝐼,𝐸), respectively.

Definition 2.2 (see [31]). Let 𝐸₁,𝐸₂ be real Banach spaces, 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 2.3 (see [31]). If 𝐷⊂𝐶[𝐼,𝐸] is bounded and equicontinuous, then 𝛼(𝐷(𝑡)) is continuous on 𝐼 and 𝛼_𝐶(𝐷)=max_𝑡∈𝐼𝛼(𝐷(𝑡)),𝛼_𝐼𝑢(𝑡)𝑑𝑡∶𝑢∈𝐷≤_𝐼𝛼(𝐷(𝑡))𝑑𝑡, (2.1) where 𝐷(𝑡)={𝑢(𝑡)∶𝑢∈𝐷,𝑡∈𝐼}.

Definition 2.4 (see [2, 3]). The left-sided Riemann-Liouville fractional integral of order 𝑞>0 of a function 𝑦∶𝑅⁰₊→𝑅 is given by 𝐼^𝑞₀₊𝑦(𝑡)=1Γ𝑞^𝑡₀(𝑡−𝑠)^𝑞−1𝑦(𝑠)𝑑𝑠. (2.2)

Definition 2.5 (see [2, 3]). The fractional derivative of order 𝑞>0 of a function 𝑦∶𝑅⁰₊→𝑅 is given by 𝐷^𝑞₀₊𝑦(𝑡)=1Γ𝑑𝑛−𝑞𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝑞−1}𝑦(𝑠)𝑑𝑠, (2.3) where 𝑛=[𝑞]+1,[𝑞] denotes the integer part of number 𝑞, provided that the right side is pointwise defined on 𝑅⁰₊.

Lemma 2.6 (see [2, 3]). Let 𝑞>0. Then the fractional differential equation 𝐷^𝑞₀₊𝑦(𝑡)=0 (2.4) has a unique solution 𝑦(𝑡)=𝑐₁𝑡^𝑞−1+𝑐₂𝑡^𝑞−2+⋯+𝑐_𝑛𝑡^𝑞−𝑛,𝑐_𝑖∈𝑅,𝑖=1,2,…,𝑛; here 𝑛−1<𝑞≤𝑛.

Lemma 2.7 (see [2, 3]). Let 𝑞>0. Then the following equality holds for 𝑦∈𝐿(0,1),𝐷^𝑞₀₊𝑦∈𝐿(0,1), 𝐼^𝑞₀₊𝐷^𝑞₀₊𝑦(𝑡)=𝑦(𝑡)+𝑐₁𝑡^𝑞−1+𝑐₂𝑡^𝑞−2+⋯+𝑐_𝑁𝑡^𝑞−𝑁 (2.5) for some 𝑐_𝑖∈𝑅,  𝑖=1,2,…,𝑁; here 𝑁 is the smallest integer greater than or equal to 𝑞.

Lemma 2.8 (see [31]). Let 𝐾 be a cone in a Banach space 𝐸. Assume that Ω₁,Ω₂ are open subsets of 𝐸 with 0∈Ω₁,  Ω₁⊂Ω₂. If 𝑇∶𝐾∩(Ω₂⧵Ω₁)→𝐾 is a strict-set-contraction operator such that either: (i)‖𝑇𝑥‖≤‖𝑥‖,  𝑥∈𝐾∩𝜕Ω₁, and ‖𝑇𝑥‖≥‖𝑥‖,  𝑥∈𝐾∩𝜕Ω₂, or (ii)‖𝑇𝑥‖≥‖𝑥‖,  𝑥∈𝐾∩𝜕Ω₁, and ‖𝑇𝑥‖≤‖𝑥‖,  𝑥∈𝐾∩𝜕Ω₂,then 𝑇 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

3. Main Results

For convenience, we list some following assumptions. (H1) There exist 𝑎∈𝐶[𝐼,𝑅₊] and ℎ∈𝐶[𝑅₊^𝑛+1,𝑅₊] such that 𝑓𝑡,𝑢₁,…,𝑢_𝑛+1≤𝑎(𝑡)ℎ𝑢₁,…,𝑢_𝑛+1,∀𝑡∈𝐼,𝑢_𝑘∈𝑃,𝑘=1,…,𝑛+1. (3.1)(H2)𝑓∶𝐼×𝑃_𝑟^𝑛+1→𝑃, for any 𝑟>0,  𝑓 is uniformly continuous on 𝐼×𝑃_𝑟^𝑛+1 and there exist nonnegative constants 𝐿_𝑘,  𝑘=1,…,𝑛+1, with 2Γ𝑞−𝑛+2𝜂^∗𝑛−2_𝑘=1𝐿_𝑘(𝑛−2−𝑘)!+𝐿_𝑛−1+𝑎^∗()!𝐿𝑛−3^𝑛+𝑏^∗()!𝐿𝑛−3^𝑛+1<1 (3.2)such that 𝛼𝑓𝑡,𝐷₁,𝐷₂,…,𝐷_𝑛+1≤^𝑛+1_𝑘=1𝐿_𝑘𝛼𝐷_𝑘,∀𝑡∈𝐼,boundedsets𝐷_𝑘∈𝑃_𝑟, (3.3) where 𝑎^∗=max{𝐾(𝑡,𝑠)∶(𝑡,𝑠)∈𝐵},𝑏^∗=max{𝐻(𝑡,𝑠)∶(𝑡,𝑠)∈𝐼×𝐼},  𝑃_𝑟={𝑢∈𝑃∶‖𝑢‖≤𝑟},  𝜂^∗=1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1}.

Lemma 3.1. Given 𝑦∈𝐶[𝐼,𝐸] and 1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1}≠0 hold. Then the unique solution of 𝐷₀^{𝑞−𝑛+2}₊𝑥(𝑡)(𝑡)𝑥(0)(1)=+𝑦=0,0<𝑡<1,𝑛−1<𝑞≤𝑛,𝑛≥2,=0,𝑥^𝑚−2_𝑖=1𝑎_𝑖𝑥𝜂_𝑖, (3.4) is 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (3.5) where 𝐺()()+∑𝑡,𝑠=𝑔𝑡,𝑠^{𝑚−2𝑖=1}𝑎_𝑖𝑔𝜂_𝑖,𝑠𝜂^∗𝑡^{𝑞−𝑛+1},𝑔()=(𝑡(𝑡,𝑠1−𝑠))^{𝑞−𝑛+1}−(𝑡−𝑠)^{𝑞−𝑛+1}Γ(𝑡(𝑞−𝑛+2,𝑠≤𝑡,1−𝑠))^{𝑞−𝑛+1}Γ𝑞−𝑛+2,𝑡≤𝑠. (3.6)(3.7)

Proof. Deduced from Lemma 2.7, we have 𝑥(𝑡)=−𝐼^{𝑞−𝑛+2}₀₊𝑦(𝑡)+𝑐₁𝑡^{𝑞−𝑛+1}+𝑐₂𝑡^𝑞−𝑛 (3.8) for some 𝑐₁,𝑐₂∈𝑅. Consequently, the general solution of (3.4) is 𝑥(𝑡)=−^𝑡₀(𝑡−𝑠)^{𝑞−𝑛+1}Γ𝑦(𝑠)𝑞−𝑛+2𝑑𝑠+𝑐₁𝑡^{𝑞−𝑛+1}+𝑐₂𝑡^𝑞−𝑛. (3.9) By boundary value conditions 𝑥(0)=0,𝑥(1)=∑^𝑚−2_𝑖=1𝑎_𝑖𝑥(𝜂_𝑖), there is 𝑐₂=0, and 𝑐₁=11−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+110}(1−𝑠)^{𝑞−𝑛+1}Γ𝑦(𝑠)∑𝑞−𝑛+2𝑑𝑠−^{𝑚−2𝑖=1}𝑎_𝑖1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1𝜂𝑖0}𝜂_𝑖−𝑠^{𝑞−𝑛+1}Γ𝑦(𝑠)𝑞−𝑛+2𝑑𝑠. (3.10) Therefore, the solution of problem (3.4) is 𝑥(𝑡)=−^𝑡₀(𝑡−𝑠)^{𝑞−𝑛+1}Γ𝑦(𝑠)𝑡𝑞−𝑛+2𝑑𝑠+^{𝑞−𝑛+1}1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+110}(1−𝑠)^{𝑞−𝑛+1}Γ𝑦(𝑠)−∑𝑞−𝑛+2𝑑𝑠_{𝑚−2𝑖=1}𝑎_𝑖𝑡^{𝑞−𝑛+1}1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1𝜂𝑖0}𝜂_𝑖−𝑠^{𝑞−𝑛+1}Γ𝑦(𝑠)=𝑞−𝑛+2𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (3.11) The proof is complete.

Moreover, there is one paper [8] in which the following statement has been shown.

Lemma 3.2. The function 𝑔(𝑡,𝑠) defined in (3.7) satisfying the following properties: (1)𝑔(𝑡,𝑠)≥0 is continuous on [0,1]×[0,1], and 𝑔(𝑡,𝑠)≤𝑡^{𝑞−𝑛+1}/Γ(𝑞−𝑛+2),𝑔(𝑡,𝑠)≤𝑔(𝑠,𝑠) for all 0≤𝑡,𝑠≤1;(2)there exists a positive function 𝜌₀∈𝐶(0,1) such that min_{𝛾≤𝑡≤𝛿}𝑔(𝑡,𝑠)≥𝜌₀(𝑠)𝑔(𝑡,𝑠),𝑠∈(0,1), where 0<𝛾<𝛿<1 and 𝜌₀(𝑠)=(𝛿(1−𝑠))^{𝑞−𝑛+1}−(𝛿−𝑠)^{𝑞−𝑛+1}(𝑠(1−𝑠))^{𝑞−𝑛+1}(,𝛾,𝑠∈0,𝜉𝑠^{𝑞−𝑛+1},𝑠∈𝜉,1), (3.12) where 𝛾<𝜉<𝛿 is the solution of (𝛿(1−𝜉))^{𝑞−𝑛+1}−(𝛿−𝜉)^{𝑞−𝑛+1}=𝛾(1−𝜉)^{𝑞−𝑛+1}. (3.13) For our purpose, one assumes that(H3)𝜂^∗=1−∑^𝑚−2_𝑖=1𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1}>0 and 0<𝛾≤min{2𝛿−1,𝛿/2},  2/3≤𝛿<1, where 𝛾,  𝛿 are the constants in (2) of Lemma 3.2.

Remark 3.3. We note that if (H3) holds, then the function 𝐺(𝑡,𝑠) defined in (3.4) is satisfying the following properties: (i)𝐺(𝑡,𝑠)≥0   is continuous on [0,1]×[0,1], and 𝐺(𝑡,𝑠)≤Δ𝑡^{𝑞−𝑛+1},   for all 0≤𝑡,  𝑠≤1,   where Δ⁻¹=𝜂^∗Γ(𝑞−𝑛+2);(ii)𝐺(𝑡,𝑠)≤𝐺(𝑠) for all 0≤𝑡,  𝑠≤1, where 𝐺(𝑠)()+∑=𝑔𝑠,𝑠^{𝑚−2𝑖=1}𝑎_𝑖𝑔𝜂_𝑖,𝑠𝜂^∗. (3.14)Indeed, it is obvious from (1) of Lemma 3.2 and (3.6) that 𝐺()()+1𝑡,𝑠≤𝑔𝑠,𝑠𝜂^∗𝑚−2_𝑖=1𝑎_𝑖𝑔𝜂_𝑖(𝑠)≤𝑡,𝑠=𝐺_{𝑞−𝑛+1}Γ+∑𝑞−𝑛+2^{𝑚−2𝑖=1}𝑎_𝑖𝜂_𝑖^{𝑞−𝑛+1}𝜂^∗Γ𝑡𝑞−𝑛+2^{𝑞−𝑛+1}≤Δ𝑡^{𝑞−𝑛+1}. (3.15)

Lemma 3.4. Let 𝑢(𝑡)=𝐼₀^𝑛−2+𝑥(𝑡),  𝑥∈𝐶[𝐼,𝐸]. Then the problem (1.1) can be transformed into the following modified problem: 𝐷₀^{𝑞−𝑛+2}₊𝑥(𝑡)+𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝐼¹₀⁺𝑥(𝑠),𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑥(0)(1)==𝜃,=𝜃,𝑥^𝑚−2_𝑖=1𝑎_𝑖𝑥𝜂_𝑖, (3.16) where 0<𝑡<1,  𝑛−1<𝑞≤𝑛,  𝑛≥2. Moreover, if 𝑥∈𝐶[𝐼,𝐸] is a solutions of problem (3.16), then the function 𝑢(𝑡)=𝐼₀^𝑛−2+𝑥(𝑡) is a solution of (1.1).

The proof follows by routine calculations.

To obtain a positive solution, we construct a cone Ω by 𝑥(𝑡)(𝑡)≥𝜆Ω=∈𝑃∶𝑥3𝑥(𝑠),𝑡∈𝐼^∗,𝑠∈𝐼, (3.17) where 𝑃={𝑥∈𝐶[𝐼,𝐸],𝑥(𝑡)≥𝜃,𝑡∈𝐼},  𝜆=min{min_{𝛾≤𝑡≤𝛿}𝜌(𝑡),𝛾^{𝑞−𝑛+1}},𝐼^∗=[𝛾,𝛿].

Let (𝐴𝑥)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠,0≤𝑡≤1. (3.18)

Lemma 3.5. Assume that (H1)–(H3) hold. Then 𝐴∶Ω→Ω is a strict-set-contraction operator.

Proof. Let 𝑥∈Ω. Then, it follows from Remark 3.3. that (𝐴𝑥)(𝑡)≤¹₀𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)=𝑑𝑠_𝛾0+^𝛿_𝛾+¹_𝛿𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)𝑑𝑠≤3^𝛿_𝛾𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)𝑑𝑠, (3.19) here, by (H3), we know that 𝛾<𝛿−𝛾 and 𝛿−𝛾>1−𝛿.
From (3.6) and (3.18), we obtain min_{𝑡∈𝛾,𝛿}(𝐴𝑥)(𝑡)=min_{𝑡∈𝛾,𝛿}¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)≥𝑑𝑠_𝛿𝛾𝑔()+𝑡𝑡,𝑠^{𝑞−𝑛+1}𝜂^∗𝑚−2_𝑖=1𝑎_𝑖𝑔𝜂_𝑖,𝑠×𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)≥𝑑𝑠_𝛿𝛾𝜌₀(𝑠)𝑔()+𝛾𝑠,𝑠^{𝑞−𝑛+1}𝜂^∗𝑚−2_𝑖=1𝑎_𝑖𝑔𝜂_𝑖,𝑠×𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)𝑑𝑠≥𝜆^𝛿_𝛾𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)≥𝜆𝑑𝑠3(𝐴𝑥)𝑡^′,𝑡^′∈𝐼, (3.20) which implies that (𝐴𝑥)(𝑡)∈Ω;  that is,  𝐴(Ω)⊂Ω.
Next, we prove that 𝐴 is continuous on Ω. Let {𝑥_𝑗},{𝑥}⊂Ω, and ‖𝑥_𝑗−𝑥‖_Ω→0(𝑗→∞). Hence {𝑥_𝑗} is a bounded subset of Ω. Thus, there exists 𝑟>0 such that 𝑟=sup_𝑗‖𝑥_𝑗‖_Ω<∞ and ‖𝑥‖_Ω≤𝑟. It is clear that 𝐴𝑥_𝑗(𝑡)−(𝐴𝑥)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)−𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠≤Δ𝑡^{𝑞−𝑛+1}𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)−𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠. (3.21) According to the properties of 𝑓,  for all  𝜀>0, there exists 𝐽>0 such that 𝑓𝑠,𝐼₀^𝑛−2+𝑥_𝑗(𝑠),…,𝑥_𝑗(𝑠),𝑇𝐼₀^𝑛−2+𝑥_𝑗(𝑠),𝑆𝐼₀^𝑛−2+𝑥_𝑗(𝑠)−𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)<𝜀Δ, (3.22) for 𝑗≥𝐽,  for all  𝑡∈𝐼.
Therefore,  for all  𝜀>0, for any 𝑡∈𝐼 and 𝑗≥𝐽, we get 𝐴𝑥_𝑗(𝑡)−(𝐴𝑥)(𝑡)<𝑡^{𝑞−𝑛+1}𝜀≤𝜀. (3.23) This implies that 𝐴 is continuous on Ω.
By the properties of continuous of 𝐺(𝑡,𝑠), it is easy to see that 𝐴 is equicontinuous on 𝐼.
Finally, we are going to show that 𝐴 is a strict-set-contraction operator. Let 𝐷⊂Ω be bounded. Then by condition (H1), Lemma 3.1 implies that 𝛼_𝐶(𝐴𝐷)=max_𝑡∈𝐼𝛼((𝐴𝐷)(𝑡)). It follows from (3.18) that 𝛼((𝐴𝐷)(𝑡))≤𝛼𝑐𝑜𝐺(𝑡,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠[]𝑓)∶𝑠∈0,𝑡,𝑡∈𝐼,𝑥∈𝐷≤Δ⋅𝛼𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)[]𝑓𝐼∶𝑠∈0,𝑡,𝑡∈𝐼,𝑥∈𝐷≤Δ⋅𝛼𝐼×₀^𝑛−2+𝐷(𝐼)×⋯×𝐷(𝐼)×𝑇𝐼₀^𝑛−2+𝐷(𝐼)×𝑆𝐼₀^𝑛−2+𝐷(𝐼)≤Δ⋅^𝑛−2_𝑘=1𝐿_𝑘𝛼𝐼₀^{𝑛−1−𝑘+}𝐷(𝐼)+𝐿_𝑛−1𝛼(𝐷(𝐼))+𝑎^∗𝐿_𝑛𝛼𝑇𝐼₀^𝑛−2+𝐷(𝐼)+𝑏^∗𝐿_𝑛+1𝛼𝑆𝐼₀^𝑛−2+𝐷(𝐼), (3.24) which implies 𝛼_𝐶(𝐴𝐷)≤Δ⋅^𝑛−2_𝑘=1𝐿_𝑘𝛼𝐼₀^{𝑛−1−𝑘+}𝐷(𝐼)+𝐿_𝑛−1𝛼(𝐷(𝐼))+𝑎^∗𝐿_𝑛𝛼𝑇𝐼₀^𝑛−2+𝐷(𝐼)+𝑏^∗𝐿_𝑛+1𝛼𝑆𝐼₀^𝑛−2+𝐷(𝐼). (3.25) Obviously, 𝛼𝐼₀^{𝑛−1−𝑘+}𝐷(𝐼)=𝛼^𝑠₀(𝑠−𝜏)^{𝑛−2−𝑘}()!𝑥(𝜏)[]≤1𝑛−2−𝑘𝑑𝜏∶𝜏∈0,𝑠,𝑠∈𝐼,𝑘=1,…,𝑛−2()!𝛼(𝐷(𝐼,𝛼𝑇𝐼𝑛−2−𝑘))_0𝑛−2+𝐷(𝐼)=𝛼^𝑡₀𝐾(𝑡,𝑠)^𝑠₀(𝑠−𝜏)^𝑛−2()!𝑢(𝜏)≤𝑎𝑛−3𝑑𝜏𝑑𝑠∶𝑢∈𝐷,𝑡∈𝐼_∗()!𝛼𝑢(𝑡)≤𝑎𝑛−3({∶𝑡∈𝐼,𝑢∈𝐷})^∗()!𝛼(𝐷(𝐼,𝛼𝑆𝐼𝑛−3))_0𝑛−2+𝐷(𝐼)=𝛼¹₀𝐻(𝑡,𝑠)^𝑠₀(𝑠−𝜏)^𝑛−2()!𝑢(𝜏)≤𝑏𝑛−3𝑑𝜏𝑑𝑠∶𝑢∈𝐷,𝑡∈𝐼_∗()!𝛼𝑢(𝑡)≤𝑏𝑛−3({∶𝑡∈𝐼,𝑢∈𝐷})^∗()!𝛼(𝐷(𝐼.𝑛−3)) (3.26)(3.27)(3.28) Using a similar method as in the proof of Theorem  2.1.1 in [31], we have 𝛼(𝐷(𝐼))≤2𝛼_𝐶(𝐷). (3.29) Therefore, it follows from (3.26)–(3.29) that 𝛼_𝐶(𝐴𝐷)≤2Δ⋅^𝑛−2_𝑘=1𝐿_𝑘(𝑛−2−𝑘)!+𝐿_𝑛−1+𝑎^∗𝐿_𝑛()!+𝑏𝑛−3^∗𝐿_𝑛+1()!𝛼𝑛−3^𝐶(𝐷). (3.30) Noticing that (3.3), we obtain that 𝑇 is a strict-set-contraction operator. The proof is complete.

Theorem 3.6. Let cone 𝑃 be normal and conditions (𝐻1)∼(𝐻3) hold. In addition, assume that the following conditions are satisfied. (H4) There exist 𝑢^∗∈𝑃⧵{𝜃},  𝑐₁∈𝐶[𝐼^∗,𝑅₊] and ℎ₁∈𝐶[𝑃^𝑛+1,𝑅₊] such that 𝑓𝑡,𝑢₁,…,𝑢_𝑛+1≥𝑐₁(𝑡)ℎ₁𝑢₁,…,𝑢_𝑛−1𝑢^∗,∀𝑡∈𝐼^∗,𝑢_𝑘ℎ∈𝑃,₁𝑢₁,…,𝑢_𝑛−1∑^𝑛−1_𝑘=1𝑢_𝑘→∞,as^𝑛−1_𝑘=1𝑢_𝑘→∞,𝑢_𝑘∈𝑃. (3.31)(H5) There exist 𝑢_∗∈𝑃⧵{𝜃},  𝑐₂∈𝐶[𝐼^∗,𝑅₊], and ℎ₂∈𝐶[𝑃^𝑛−1,𝑅₊] such that 𝑓𝑡,𝑢₁,…,𝑢_𝑛+1≥𝑐₂(𝑡)ℎ₂𝑢₁,…,𝑢_𝑛−1𝑢_∗,∀𝑡∈𝐼^∗,𝑢_𝑘ℎ∈𝑃,₂𝑢₁,…,𝑢_𝑛−1∑^𝑛−1_𝑘=1𝑢_𝑖→∞,as^𝑛−1_𝑘=1𝑢_𝑘→0,𝑢_𝑘∈𝑃. (3.32)(H6) There exists a 𝛽>0 such that 𝑁𝑀_𝛽¹₀𝐺(𝑠)𝑎(𝑠)𝑑𝑠<𝛽, (3.33)where 𝑀_𝛽=max_{𝑢𝑘∈𝑃𝛽}{ℎ(‖𝑢₁‖,…,‖𝑢_𝑛+1‖)}. Then problem (1.1) has at least two positive solutions.

Proof. Consider condition (H4), there exists an 𝑟₁>0, such that ℎ₁𝑢₁,…,𝑢_𝑛−1≥3𝑁²∑^𝑛−1_𝑘=1𝑢_𝑘𝜆²∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖,∀𝑢_𝑘∈𝑃,^𝑛−1_𝑘=1𝑢_𝑘≥𝑟₁. (3.34) Therefore, 𝑓𝑡,𝑢₁,⋅,𝑢_𝑛+1≥3𝑁²∑^𝑛−1_𝑘=1𝑢_𝑘𝜆²∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖⋅𝑐₁(𝑡)𝑢^∗,∀𝑢_𝑘∈𝑃,^𝑛−1_𝑘=1𝑢_𝑘≥𝑟₁. (3.35) Take 𝑟₀>max3𝑁𝜆⁻¹𝑟₁,𝛽. (3.36) Then for 𝑡∈[𝛾,𝛿],  ‖𝑥‖_Ω=𝑟₀, we have, by (3.18), ‖𝑥(𝑡≥𝜆)‖‖𝑥‖3𝑁^Ω≥𝜆𝑟3𝑁⁰>𝑟₁. (3.37) Hence, (𝐴𝑥)(𝑡)≥^𝛿_𝛾𝐺(𝑡,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)≥𝑑𝑠3𝑁²𝜆∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖^𝛿𝛾𝐺(𝑠)^𝑛−2_𝑘=1𝐼₀^{𝑛−1−𝑘+}𝑥(𝑠)+‖𝑥(𝑠)‖𝑐₁(𝑠)𝑑𝑠⋅𝑢^∗≥3𝑁²𝜆∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖^𝛿𝛾𝐺(𝑠)𝑐₁(𝑠)‖𝑥(𝑠)‖𝑑𝑠⋅𝑢^∗≥𝑁∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖‖𝑥‖_Ω^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅𝑢^∗=1∫^𝛿_𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠⋅‖𝑢^∗‖^𝛿𝛾𝐺(𝑠)𝑐₁(𝑠)𝑑𝑠𝑢^∗⋅𝑁‖𝑥‖_Ω‖𝑢^∗‖𝑢^∗≥𝑁‖𝑥‖_Ω‖𝑢^∗‖⋅𝑢^∗, (3.38) and consequently, ‖𝐴𝑥‖_Ω≥‖𝑥‖_Ω,∀𝑥∈Ω,‖𝑥‖_Ω=𝑟₀. (3.39) Similarly, by condition (H5), there exists 𝑟₂>0, such that ℎ₂𝑢₁,…,𝑢_𝑛−1≥3𝑁²∑^𝑛−1_𝑘=1𝑢_𝑘𝜆∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗,∀𝑢_𝑘∈𝑃,0<^𝑛−1_𝑘=1𝑢_𝑘≤𝑟₂, (3.40) where 𝜉 is given in (2) of Lemma 3.2. Therefore, 𝑓𝑡,𝑢₁,…,𝑢_𝑛+1≥3𝑁²∑^𝑛−1_𝑘=1𝑢_𝑘𝜆∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗⋅𝑐₂(𝑡)𝑢_∗,∀𝑢_𝑘∈𝑃,0<^𝑛−1_𝑘=1𝑢_𝑘≤𝑟₂. (3.41) Choose 0<𝑟<min^𝑛−2_𝑘=01𝑘!⁻¹𝑟₂,𝛽. (3.42) Then for 𝑡∈[𝛾,𝛿],  𝑥∈Ω,  ‖𝑥‖_Ω=𝑟, we have (𝐴𝑥)(𝜉)=¹₀𝐺(𝜉,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)≥𝑑𝑠_𝛿𝛾𝐺(𝜉,𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)≥𝑑𝑠3𝑁²𝜆∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗^𝛿𝛾𝐺(𝜉,𝑠)^𝑛−2_𝑘=1𝐼₀^{𝑛−1−𝑘+}𝑥(𝑠)+‖𝑥(𝑠)‖𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗≥3𝑁²𝜆∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗^𝛿𝛾𝐺(𝜉,𝑠)‖𝑥(𝑠)‖𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗≥𝑁∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗‖𝑥(𝑠)‖_Ω^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗=1∫^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠⋅𝑢_∗‖𝑥(𝑠)‖_Ω^𝛿_𝛾𝐺(𝜉,𝑠)𝑐₂(𝑠)𝑑𝑠𝑢_∗⋅𝑁‖𝑥(𝑠)‖_Ω𝑢_∗𝑢_∗≥𝑁‖𝑥(𝑠)‖_Ω𝑢_∗⋅𝑢_∗, (3.43) which implies (𝐴𝑥)(𝜉)_Ω≥‖𝑥(𝑠)‖_Ω,∀𝑥∈Ω,‖𝑥‖_Ω=𝑟, (3.44) that is, ‖𝐴𝑥‖_Ω≥‖𝑥(𝑠)‖_Ω,∀𝑥∈Ω,‖𝑥‖_Ω=𝑟. (3.45)
On the other hand, according to (ii) of Remark 3.3 and (3.18), we get (𝐴𝑥)(𝑡)≤¹₀𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠. (3.46) By condition (H1), for 𝑡∈𝐼,  𝑥∈Ω,  ‖𝑥‖_Ω=𝛽, we have 𝑓𝑡,𝑢₁,…,𝑢_𝑛+1≤𝑎(𝑡)ℎ𝑢₁,…,𝑢_𝑛+1≤𝑀_𝛽𝑎(𝑡). (3.47) Therefore, ‖(𝐴𝑥)(𝑡)‖_Ω≤𝑁𝑀_𝛽⋅¹₀𝐺(𝑠)𝑎(𝑠)𝑑𝑠<𝛽=‖𝑥‖_Ω. (3.48)
Applying Lemma 2.7 to (3.39), (3.45), and (3.48) yields that 𝑇 has a fixed-point 𝑥^∗∈Ω_𝑟,𝛽,𝑟≤‖𝑥^∗‖≤𝛽, and a fixed-point 𝑥^∗∗∈Ω_𝛽,𝑟0,𝛽≤‖𝑥^∗∗‖≤𝑟₀. Noticing (3.48), we get ‖𝑥^∗‖≠𝛽 and ‖𝑥^∗∗‖≠𝛽. This and Lemma 3.4 complete the proof.

Theorem 3.7. Let cone 𝑃 be normal and conditions (𝐻1)∼(𝐻4) hold. In addition, assume that the following condition is satisfied: (H7)ℎ𝑢₁,…,𝑢_𝑛+1∑^𝑛+1_𝑘=1𝑢_𝑘→0,as𝑢_𝑘∈𝑃,^𝑛+1_𝑘=1𝑢_𝑘→0⁺. (3.49)Then problem (1.1) has at least one positive solution.

Proof. By (H4), we can choose 𝑟₀>3𝑁𝜆⁻¹𝑟₁. As in the proof of Theorem 3.6, it is easy to see that (3.39) holds. On the other hand, considering (3.49), there exists 𝑟₃>0 such that ℎ𝑢₁,…,𝑢_𝑛+1≤𝜀₀^𝑛+1_𝑘=1𝑢_𝑘,for𝑡∈𝐼,𝑢_𝑘∈𝑃,0<^𝑛+1_𝑘=1𝑢_𝑘≤𝑟₃, (3.50) where 𝜀₀>0 satisfies 𝜀₀=𝑁^𝑛−1_𝑘=11+𝑎𝑘!^∗+𝑏^∗(𝑛−3)!¹₀𝐺(𝑠)𝑎(𝑠)𝑑𝑠⁻¹. (3.51) Choose 0<𝑟^∗<min{(∑^𝑛−1_𝑘=1(1/𝑘!)+(𝑎^∗+𝑏^∗)/(𝑛−3)!)⁻¹𝑟₃,𝑟₀}. Then for 𝑡∈𝐼,  𝑥∈Ω,  ‖𝑥‖_Ω=𝑟^∗, it follows from (3.46) that ‖(𝐴𝑥)(𝑡)‖≤𝑁¹₀𝐺(𝑠)𝑓𝑠,𝐼₀^𝑛−2+𝑥(𝑠),…,𝑥(𝑠),𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠≤𝑁¹₀𝐺(𝑠)𝑎(𝑠)ℎ𝐼₀^𝑛−2+𝑥(𝑠),…,‖𝑥(𝑠)‖,𝑇𝐼₀^𝑛−2+𝑥(𝑠),𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠≤𝑁𝜀₀¹₀𝐺(𝑠)𝑎(𝑠)^𝑛−1_𝑘=1𝐼₀^{𝑛−1−𝑘+}𝑥(𝑠)+𝑇𝐼₀^𝑛−2+𝑥(𝑠)+𝑆𝐼₀^𝑛−2+𝑥(𝑠)𝑑𝑠≤𝑁𝜀₀^𝑛−1_𝑘=11+𝑎𝑘!^∗+𝑏^∗()!𝑟𝑛−3^∗1₀𝐺(𝑠)𝑎(𝑠)𝑑𝑠=𝑟^∗ (3.52) and consequently, ‖(𝐴𝑥)(𝑡)‖_Ω≤‖𝑥‖_Ω,∀𝑥∈Ω,‖𝑥‖_Ω≤𝑟^∗. (3.53) Since 0<𝑟^∗<𝑟₀, applying Lemma 2.7 to (3.39) and (3.53) yield that 𝑇 has a fixed-point 𝑥^∗∈Ω_{𝑟∗,𝑟0},𝑟^∗≤‖𝑥^∗‖≤𝑟₀. This and Lemma 3.4 complete the proof.

4. An Example

Consider the following system of scalar differential equations of fractional order −𝐷^5/2𝑢_𝑘(𝑡)=(1+𝑡)³960𝑘³𝑢_2𝑘(𝑡)+𝑢^′_3𝑘(𝑡)+^∞_𝑗=1𝑢_2𝑗(𝑡)+^∞_𝑗=1𝑢^′_𝑗(𝑡)³+3𝑢_𝑘(𝑡)+3𝑢^′_𝑘+1(𝑡)+^∞_𝑗=1𝑢_𝑗(𝑡)+^∞_𝑗=1𝑢^′_2𝑗(𝑡)^1/2+1+𝑡³36𝑘^5𝑡0𝑒^{−(1+𝑡)𝑠}𝑢_𝑘(𝑠)𝑑𝑠^2/3+1+𝑡²24𝑘⁴¹⁰𝑒^−𝑠sin²(𝑡−𝑠)𝜋𝑢_2𝑘(𝑠)𝑢𝑑𝑠,𝑡∈𝐼,_𝑘(0)=𝑢^′_𝑘(0)=0,𝑢^′_𝑘(1)=18𝑢^′_𝑘14+12𝑢^′_𝑘49,𝑘=1,2,3,…. (4.1)

Conclusion. The problem (4.1) has at least two positive solutions.

Proof. Let 𝐸=𝑙¹={𝑢=(𝑢₁,𝑢₂,…,𝑢_𝑘,…)∶∑^∞_𝑘=1|𝑢_𝑘|<∞} with the norm ‖𝑢‖=∑^∞_𝑘=1|𝑢_𝑘|, and 𝑃={(𝑢₁,…,𝑢_𝑘,…)∶𝑢_𝑘≥0,𝑘=1,2,3,…}. Then 𝑃 is a normal cone in 𝐸 with normal constant 𝑁=1, and system (4.1) can be regarded as a boundary value problem of the form (1.1). In this situation, 𝑞=5/2,𝑛=3,𝑎₁=1/8,𝑎₂=1/2,𝜂₁=1/4,𝜂₂=4/9,𝜂^∗=29/48,𝐾(𝑡,𝑠)=𝑒^{−(1+𝑡)𝑠}𝑒,𝐻(𝑡,𝑠)=_−𝑠sin²(𝑡−𝑠)𝜋,𝑢=(𝑢₁,…,𝑢_𝑘,…),𝑓=(𝑓₁,𝑓₂,…,𝑓_𝑘,…), in which 𝑓_𝑘=()𝑡,𝑢,𝑣,𝑥,𝑦1+𝑡³960𝑘³𝑢_2𝑘+𝑣_3𝑘+^∞_𝑗=1𝑢_2𝑗+^∞_𝑗=1𝑣_𝑗³+3𝑢_𝑘+3𝑣_𝑘+1+^∞_𝑗=1𝑢_𝑗+^∞_𝑗=1𝑣_2𝑗^1/2+1+𝑡³36𝑘⁵𝑥_𝑘^1/5+1+𝑡²24𝑘⁴𝑦_2𝑘. (4.2) Observing the inequality ∑^∞_𝑘=1(1/𝑘³)<3/2, we get, by (4.2), 𝑓𝑡,𝑢,𝑣,𝑥,𝑦=^∞_𝑘=1𝑓_𝑘≤()𝑡,𝑢,𝑣,𝑥,𝑦1+𝑡³21𝑢‖+‖𝑣40(‖‖)³+1𝑢‖+‖𝑣160(‖‖)^1/2+1‖𝑥‖12^1/5+18𝑦. (4.3) Hence (H1) is satisfied for 𝑎(𝑡)=(1+𝑡)³/2 and ℎ=1𝑢,𝑣,𝑥,𝑦()40𝑢+𝑣³+1()160𝑢+𝑣^1/2+1𝑥12^1/5+18𝑦. (4.4)
Now, we check condition (H2). Obviously, 𝑓∶𝐼×𝑃⁴_𝑟→𝑃, for any 𝑟>0, and 𝑓 is uniformly continuous on 𝐼×𝑃⁴_𝑟. Let 𝑓=𝑓⁽¹⁾+𝑓⁽²⁾, where 𝑓⁽¹⁾=(𝑓₁⁽¹⁾,…,𝑓_𝑘⁽¹⁾,…) and 𝑓⁽²⁾=(𝑓₁⁽²⁾,…,𝑓_𝑘⁽²⁾,…), in which 𝑓_𝑘⁽¹⁾=()𝑡,𝑢,𝑣,𝑥,𝑦1+𝑡³960𝑘³𝑢_2𝑘+𝑣_3𝑘+^∞_𝑗=1𝑢_2𝑗+^∞_𝑗=1𝑣_𝑗³+3𝑢_𝑘+3𝑣_𝑘+1+^∞_𝑗=1𝑢_𝑗+^∞_𝑗=1𝑣_2𝑗^1/2+1+𝑡³36𝑘⁵𝑥_𝑘^1/5,(),𝑓𝑘=1,2,3,…_𝑘(2)𝑡,𝑢,𝑣,𝑥,𝑦=1+𝑡²24𝑘⁴𝑦_2𝑘,(𝑘=1,2,3,…). (4.5) For any 𝑡∈𝐼 and bounded subsets 𝐷_𝑖⊂𝐸,𝑖=1,2,3,4, from (4.5) and by the diagonal method, we have 𝛼𝑓⁽¹⁾𝑡,𝐷₁,𝐷₂,𝐷₃,𝐷₄=0,∀𝑡∈𝐼,boundedsets𝐷_𝑖𝛼𝑓⊂𝐸,𝑖=1,2,3,4,₍₂₎𝐼,𝐷₁,𝐷₂,𝐷₃,𝐷₄≤1𝛼𝐷12⁴,∀𝑡∈𝐼,𝐷_𝑖⊂𝐸,𝑖=1,2,3,4. (4.6) It follows from (4.6) that 𝛼𝑓𝐼,𝐷₁,𝐷₂,𝐷₃,𝐷₄≤1𝛼𝐷12⁴,∀𝑡∈𝐼,𝐷_𝑖2⊂𝐸,𝑖=1,2,3,4,Γ𝑞−𝑛+2𝜂^∗𝑛−2_𝑘=1𝐿_𝑘(𝑛−2−𝑘)!+𝐿_𝑛−1+𝑎^∗()!𝐿𝑛−3^𝑛+𝑏^∗()!𝐿𝑛−3^𝑛+1≈0.1565<1. (4.7) that is, condition (H2) holds for 𝐿₁=𝐿₂=𝐿₃=0,𝐿₄=1/12.
On the other hand, take 𝛾=1/4,𝛿=3/4. Then 1/4=𝛾≤min{𝛿/2,2𝛿−1}=3/8,2/3<𝛿, which implies that condition (H3) holds. By (4.2), we have 𝑓_𝑘≥()𝑡,𝑢,𝑣,𝑥,𝑦1+𝑡³960𝑘³(‖𝑢‖+‖𝑣‖)³,∀𝑡∈𝐼^∗(),𝑓,𝑢,𝑣,𝑥,𝑦∈𝑃,𝑘=1,2,3,…_𝑘≥()𝑡,𝑢,𝑣,𝑥,𝑦1+𝑡³960𝑘³‖𝑢‖+‖𝑣‖,∀𝑡∈𝐼^∗,𝑢,𝑣,𝑥,𝑦∈𝑃,(𝑘=1,2,3,…). (4.8) Hence condition (H4) is satisfied for 𝑐₁(𝑡)=(1+𝑡)³960,ℎ_1,𝑘(𝑢,𝑣)=(‖𝑢‖+‖𝑣‖)³,𝑢^∗=11,…,𝑘³,…, (4.9) in this situation, ℎ_1,𝑘=lim_{‖𝑢‖+‖𝑣‖→∞}(‖𝑢‖+‖𝑣‖)³‖𝑢‖+‖𝑣‖=∞. (4.10) And condition (H5) is also satisfied for 𝑐₂(𝑡)=(1+𝑡)³960,ℎ_2,𝑘(𝑢,𝑣)=‖𝑢‖+‖𝑣‖,𝑢_∗=11,…,𝑘³,…, (4.11) in this situation, ℎ_2,𝑘=lim_{‖𝑢‖+‖𝑣‖→0}‖𝑢‖+‖𝑣‖‖𝑢‖+‖𝑣‖=∞. (4.12) Finally, choose 𝛽=1, it is easy to check that condition (H6) is satisfied. In this case, 𝑀_𝛽≈0.4162, and so 𝑁𝑀_𝛽¹₀𝐺(𝑠)𝑎(𝑠)𝑑𝑠≈0.7287<𝛽=1. (4.13) Hence, our conclusion follows from Theorem 3.6.

Acknowledgments

This research is supported by the National Natural Science Foundation of China (11161024, 11271372, and 11201138), Hunan Provincial Natural Science Foundation of China (12JJ2004), NSF of Jiangxi province (2010GZC0115), and Scientific Research Fund of Hunan Provincial Education Department (11C0412).