International Journal of Differential Equations

International Journal of Differential Equations / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 838947 |

R. K. Pandey, A. K. Barnwal, "Multiple Solutions for Nonlinear Doubly Singular Three-Point Boundary Value Problems with Derivative Dependence", International Journal of Differential Equations, vol. 2012, Article ID 838947, 21 pages, 2012.

Multiple Solutions for Nonlinear Doubly Singular Three-Point Boundary Value Problems with Derivative Dependence

Academic Editor: Yuji Liu
Received25 May 2012
Accepted12 Jul 2012
Published29 Aug 2012


We study the existence of multiple nonnegative solutions for the doubly singular three-point boundary value problem with derivative dependent data function βˆ’(𝑝(𝑑)𝑦′(𝑑))β€²=π‘ž(𝑑)𝑓(𝑑,𝑦(𝑑),𝑝(𝑑)𝑦′(𝑑)),0<𝑑<1,𝑦(0)=0,𝑦(1)=𝛼1𝑦(πœ‚). Here, π‘βˆˆπΆ[0,1]∩𝐢1(0,1] with 𝑝(𝑑)>0 on (0,1] and π‘ž(𝑑) is allowed to be discontinuous at 𝑑=0. The fixed point theory in a cone is applied to achieve new and more general results for existence of multiple nonnegative solutions of the problem. The results are illustrated through examples.

1. Introduction

In this paper, we consider the following three-point boundary value problem of Sturm-Liouville type: βˆ’ξ€·π‘(𝑑)π‘¦ξ…žξ€Έ(𝑑)ξ…žξ€·=π‘ž(𝑑)𝑓𝑑,𝑦(𝑑),𝑝(𝑑)π‘¦ξ…žξ€Έ(𝑑),0<𝑑<1,(1.1) with boundary conditions 𝑦(0)=0,𝑦(1)=𝛼1𝑦(πœ‚),(1.2) where 0<𝛼1<(β„Ž(1))/(β„Ž(πœ‚)) and βˆ«β„Ž(𝑑)=𝑑0(1/𝑝(π‘₯))𝑑π‘₯.

Throughout this paper, we assume the following conditions on the functions 𝑝(𝑑), π‘ž(𝑑), and 𝑓(𝑑,𝑦,𝑝𝑦′):(E1)π‘βˆˆπΆ[0,1]∩𝐢1(0,1] with 𝑝(𝑑)>0 on (0,1] and 1/π‘βˆˆπΏ1(0,1);(E2)π‘ž(𝑑)>0, π‘ž is not identically zero on [0,1] and π‘žβˆˆπΏ1(0,1);(E3)π‘“βˆˆπΆ([0,1]Γ—[0,∞)×ℝ,[0,∞)) and 𝑓 is not identically zero.Note that condition (E2) allows π‘ž(𝑑) be discontinuous at 𝑑=0, and if 𝑝(0)=0, then the differential equation (1.1) is called doubly singular [1].

Nonlocal boundary value problem have variety of applications in the area of applied mathematics and physical sciences. The design of a large size bridge with multipoint supports can be considered as an application of these types of boundary value problem [2]. Some more applications can be found in [3–5] and the references therein. Recently, motivated by the wide application of boundary value problems in physical and applied mathematics, the study of multipoint boundary value problems has received increasing interest (see [2, 6–12] and the references therein).

Nonsingular multipoint boundary value problems have been extensively studied in literature, see [13–16] for derivative dependent data function 𝑓(𝑑,𝑦,𝑧) and [8, 10, 12] for derivative independent data function 𝑓(𝑑,𝑦,𝑧)=𝑓(𝑑,𝑦).

Some attention has been devoted to singular multipoint boundary value problems (see [17, 18] and the references therein). When 𝑝(𝑑)=1 and π‘ž(𝑑)𝑓(𝑑,𝑦,𝑧) may have singularity at 𝑑=0,𝑑=1, 𝑦=0 and 𝑧=0, differential equation (1.1) with boundary conditions 𝑦′(0)=0,𝑦(1)=𝛼1𝑦(πœ‚) is considered by Chen et al. [17] and Agarwal et al. [18]. Chen et al. proved the existence of at least one positive solution while Agarwal et al. established that this problem may have at least two positive solutions and also may have no positive solutions under some conditions on π‘ž(𝑑) and 𝑓(𝑑,𝑦,𝑧).

Bai and Ge [19] have generalized the Leggett-Williams fixed point theory and applied to π‘¦ξ…žξ…žξ€·+π‘ž(𝑑)𝑓𝑑,𝑦,π‘¦ξ…žξ€Έ=0,π‘‘βˆˆ(0,1),𝑦(0)=0,𝑦(1)=0,(1.3) to achieve at least three positive solutions of the two-point boundary value problem.

In this work, we consider the problem (1.1)-(1.2) with unbounded coefficient of 𝑦′ along with singularity in the data function 𝑓(𝑑,𝑦,𝑧).

Existence of nonnegative solution(s) of the problem (1.1)-(1.2) may be established either directly or by reducing the problem to π‘¦ξ…žξ…žξ€·+π‘ž(𝑑)𝑓𝑑,𝑦,π‘¦ξ…žξ€Έ=0,(1.4) and applying the existing results. But direct consideration of the problem provides better results, especially as the order of singularity increases. This may be demonstrated by the following simple linear three-point boundary value problem: βˆ’ξ€·π‘‘π‘Ÿπ‘¦ξ…žξ€Έ(𝑑)ξ…ž=π‘‘π‘Ÿβˆ’152(π‘Ÿ+1)π‘‘βˆ’4π‘Ÿξ‚„ξ‚€1,0.5β‰€π‘Ÿ<1,π‘‘βˆˆ(0,1),𝑦(0)=0,𝑦(1)=𝑦4.(1.5) The problem (1.5) can be reduced to the following boundary value problem: βˆ’π‘£ξ…žξ…ž1(π‘₯)=(1βˆ’π‘Ÿ)2π‘₯(2π‘Ÿβˆ’1)/(1βˆ’π‘Ÿ)2(π‘Ÿ+1)π‘₯1/(1βˆ’π‘Ÿ)βˆ’54π‘Ÿξ‚„ξ‚Έξ‚€1,π‘₯∈(0,1),𝑣(0)=0,𝑣(1)=𝑣4(1βˆ’π‘Ÿ)ξ‚Ή,(1.6) by change of variable ∫π‘₯=(1βˆ’π‘Ÿ)𝑑0πœβˆ’π‘Ÿπ‘‘πœ=𝑑1βˆ’π‘Ÿ.

Now we apply the result (Theorem 4.2) of this work to the problem (1.5) and conclude that the problem has at least one nonnegative solution 𝑦(𝑑) with sup[]π‘‘βˆˆ0,1||||≀3𝑦(𝑑)2ξ‚Έ3(1βˆ’π‘Ÿ)4βˆ’4π‘Ÿξ‚Ή(1βˆ’π‘Ÿ)/π‘Ÿξ‚ƒ34ξ‚„.π‘Ÿ+2(1.7)

Further, for 𝑝(𝑑)=1, Theorems 4.2 and 4.3 may be regarded as extension of Theorem 3.1 in [19] for three-point singular boundary value problem. Now applying Theorem 4.2 with 𝑝(𝑑)=1 to the reduced problem (1.6), we get that the problem (1.5) has at least one nonnegative solution 𝑦(𝑑) with sup[]π‘‘βˆˆ0,1||||𝑦(𝑑)=sup[]π‘‘βˆˆ0,1||||≀3𝑣(𝑑)4βˆ’4π‘Ÿξ‚Έ3(1βˆ’π‘Ÿ)4βˆ’4π‘Ÿξ‚Ή(1βˆ’π‘Ÿ)/π‘Ÿξ‚ƒ34ξ‚„.π‘Ÿ+2(1.8)

Now as π‘Ÿ approaches to one, that is, the order of singularity increases, the upper bound for supπ‘‘βˆˆ[0,1]|𝑦(𝑑)| in (1.8) approaches to ∞ while in (1.7) approaches to 4.125, which can be seen from Figure 1. As smaller upper bound for supπ‘‘βˆˆ[0,1]|𝑦(𝑑)| will enable to find nonnegative solution(s) faster and hence will be helpful in constructing efficient numerical algorithms to find multiple nonnegative solutions, thus it is justified to consider the singular problem directly. A detailed working is given in Example 5.1.

In this work, we are concerned with existence of multiple nonnegative solutions of the three-point doubly singular boundary value problem (1.1)-(1.2). To achieve this, we use generalized Leggett-Williams fixed point theorem established by Bai and Ge [19].

For this purpose, we first establish certain properties of Green’s function of the corresponding homogeneous boundary value problem. Then fixed point theorem of functional type (generalized Leggett-Williams fixed point theorem) is applied to yield multiple nonnegative solutions for the boundary value problem (1.1)-(1.2).

We organize this work as follows. In Section 2, we present some definitions and basic results required for this work. Section 3 deals with nonnegativity of Green’s function and some basic properties. Section 4 is devoted to existence of at least one and three or odd number of nonnegative solutions. In Section 5, we demonstrate the results through examples.

2. Background and Definitions

The proof of main results is based on fixed point theorem of functional type in a cone given by Bai and Ge [19], which deals with three fixed points of completely continuous nonlinear operators defined in a cone of an ordered Banach space. In this section, we provide some background material from the theory of cone in Banach spaces to make the paper self-contained.

Definition 2.1. A subset 𝐷 of Banach space 𝐸 is said to be retract of 𝐸 if βˆƒ a continuous map π‘ŸβˆΆπΈβ†’π· such that π‘Ÿ(π‘₯)=π‘₯ for every π‘₯∈𝐷.

Corollary 2.2. Every close convex set of a Banach space is a retract of Banach space.

Definition 2.3. Let 𝐸 be a Banach space, π‘ƒβŠ‚πΈ is nonempty convex, closed set, 𝑃 is said to be cone provided that(1)πœ†π‘’βˆˆπ‘ƒ for all πœ†β‰₯0,π‘’βˆˆπ‘ƒ, and(2)π‘’βˆˆπ‘ƒ,βˆ’π‘’βˆˆπ‘ƒ implies 𝑒=0.

Note. From Corollary 2.2, a cone 𝑃 of Banach Space 𝐸 is retract of 𝐸.

Definition 2.4. A subset 𝑅 of Banach space 𝑋 is called relatively compact if 𝑅 (closure of 𝑅) is compact.

Definition 2.5. Consider two Banach spaces 𝑋 and π‘Œ, a subset Ξ© of 𝑋, and a map π‘‡βˆΆΞ©β†’π‘Œ. Then 𝑇 is said to be completely continuous operator if(1)𝑇 is continuous, and(2)𝑇 maps bounded subset of Ξ© into relatively compact sets.

Definition 2.6. The map 𝛼 is said to be a nonnegative continuous convex functional on 𝑃 provided that π›ΌβˆΆπ‘ƒβ†’[0,∞) is continuous and 𝛼(𝑑π‘₯+(1βˆ’π‘‘)𝑦)≀𝑑𝛼(π‘₯)+(1βˆ’π‘‘)𝛼(𝑦),(2.1) for all π‘₯,π‘¦βˆˆπ‘ƒ and 0≀𝑑≀1. Similarly, the map 𝛾 is said to be a nonnegative continuous concave functional on 𝑃 provided that π›ΎβˆΆπ‘ƒβ†’[0,∞) is continuous and 𝛾(𝑑π‘₯+(1βˆ’π‘‘)𝑦)β‰₯𝑑𝛾(π‘₯)+(1βˆ’π‘‘)𝛾(𝑦),(2.2) for all π‘₯,π‘¦βˆˆπ‘ƒ, and 0≀𝑑≀1.

Definition 2.7. Suppose 𝛼,π›½βˆΆπ‘ƒβ†’[0,∞) are two continuous convex functionals satisfying β€–π‘₯‖≀𝑀max{𝛼(π‘₯),𝛽(π‘₯)},forπ‘₯βˆˆπ‘ƒ,(2.3) where 𝑀 is positive constant, and Ξ©={π‘₯βˆˆπ‘ƒβˆΆπ›Ό(π‘₯)<π‘Ÿ,𝛽(π‘₯)<𝐿}β‰ πœ™forπ‘Ÿ>0,𝐿>0.(2.4) From (2.3) and (2.4), Ξ© is a bounded nonempty open subset of 𝑃.

Definition 2.8. Let π‘Ÿ>π‘Ž>0, 𝐿>0 be given constants, 𝛼,π›½βˆΆπ‘ƒβ†’[0,∞) two nonnegative continuous convex functionals satisfying (2.3) and (2.4), and πœ“ a nonnegative continuous concave functional on the cone 𝑃. Then bounded convex sets are defined as π‘ƒξ€·π›Όπ‘Ÿ,𝛽𝐿={π‘₯βˆˆπ‘ƒβˆΆπ›Ό(π‘₯)<π‘Ÿ,𝛽(π‘₯)<𝐿},π‘ƒξ€·π›Όπ‘Ÿ,𝛽𝐿=𝑃𝛼{π‘₯βˆˆπ‘ƒβˆΆπ›Ό(π‘₯)β‰€π‘Ÿ,𝛽(π‘₯)≀𝐿},π‘Ÿ,𝛽𝐿,πœ“π‘Žξ€Έ={π‘₯βˆˆπ‘ƒβˆΆπ›Ό(π‘₯)<π‘Ÿ,𝛽(π‘₯)<𝐿,πœ“(π‘₯)>π‘Ž},π‘ƒξ€·π›Όπ‘Ÿ,𝛽𝐿,πœ“π‘Žξ€Έ={π‘₯βˆˆπ‘ƒβˆΆπ›Ό(π‘₯)β‰€π‘Ÿ,𝛽(π‘₯)≀𝐿,πœ“(π‘₯)β‰₯π‘Ž}.(2.5)

Theorem 2.9 (see [20]). Let 𝑋 be retract of real Banach space 𝐸. Then for every bounded relatively open subset π‘ˆ of 𝑋 and every completely continuous operator π‘‡βˆΆπ‘ˆβ†’π‘‹ which has no fixed point on πœ•π‘ˆ (relative to 𝑋), there exists an integer 𝑖(𝑇,π‘ˆ,𝑋) such that if 𝑖(𝑇,π‘ˆ,𝑋)β‰ 0, then 𝑇 has at least one fixed point in π‘ˆ. Moreover, 𝑖(𝑇,π‘ˆ,𝑋) is uniquely defined.

Theorem 2.10 (see [20]). Let 𝐸 be Banach space, 𝑋 retract of 𝐸, 𝑋1 a bounded convex retract of 𝑋, and π‘ˆβŠ‚π‘‹ nonempty open subset, such that π‘ˆβŠ‚π‘‹1. If π‘‡βˆΆπ‘‹1→𝑋 is completely continuous, 𝑇(𝑋1)βŠ‚π‘‹1, such that there is no fixed point of 𝑇 in 𝑋1β§΅π‘ˆ, then 𝑖(𝑇,π‘ˆ,𝑋)=1.

Theorem 2.11 (see [19]) (fixed point theorem of functional type). Let 𝐸 be Banach space, π‘ƒβŠ‚πΈ a cone, and 𝑐β‰₯𝑏>π‘Ž>𝑑>0,𝐿2β‰₯𝐿1>0 given constants. Assume that 𝛼,𝛽 are nonnegative continuous convex functionals on 𝑃 such that (2.3) and (2.4) are satisfied. πœ“ is a nonnegative continuous concave functional on 𝑃 such that πœ“(π‘₯)≀𝛼(π‘₯) for all π‘₯βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2) and let π‘‡βˆΆπ‘ƒ(𝛼𝑐,𝛽𝐿2)→𝑃(𝛼𝑐,𝛽𝐿2) be a completely continuous operator. Suppose that(1){π‘₯βˆˆπ‘ƒ(𝛼𝑏,𝛽𝐿2,πœ“π‘Ž)βˆΆπœ“(π‘₯)>π‘Ž}β‰ Ξ¦ and πœ“(𝑇π‘₯)>π‘Ž for π‘₯βˆˆπ‘ƒ(𝛼𝑏,𝛽𝐿2,πœ“π‘Ž),(2)𝛼(𝑇π‘₯)<𝑑, 𝛽(𝑇π‘₯)<𝐿1 for all π‘₯βˆˆπ‘ƒ(𝛼𝑑,𝛽𝐿1),(3)πœ“(𝑇π‘₯)>π‘Ž for all π‘₯βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2,πœ“π‘Ž) with 𝛼(𝑇π‘₯)>𝑏.Then 𝑇 has at least three fixed points π‘₯1,π‘₯2,π‘₯3βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2) such that π‘₯1ξ€·π›Όβˆˆπ‘ƒπ‘‘,𝛽𝐿1ξ€Έ,π‘₯2βˆˆξ‚†π‘₯βˆˆπ‘ƒξ€·π›Όπ‘,𝛽𝐿2,πœ“π‘Žξ€Έξ‚‡,π‘₯βˆΆπœ“(π‘₯)>π‘Ž3βˆˆπ‘ƒξ€·π›Όπ‘,𝛽𝐿2⧡𝑃𝛼𝑐,𝛽𝐿2,πœ“π‘Žξ€Έβˆͺ𝑃𝛼𝑑,𝛽𝐿1.(2.6)

3. Some Preliminary Results

In this section, we construct the Green’s function and establish some properties, required to establish the main results in Section 4.

Lemma 3.1. The Green’s function for the following boundary value problem: βˆ’ξ€·π‘(𝑑)π‘¦ξ…žξ€Έ(𝑑)β€²=0𝐡.𝐢.∢=𝑦(0)=0,𝑦(1)=𝛼1⎫βŽͺ⎬βŽͺβŽ­π‘¦(πœ‚),π‘‘βˆˆ(0,1)(3.1) is given by ⎧βŽͺ⎨βŽͺ⎩𝐺𝐺(𝑑,𝑠)=1𝐺(𝑑,𝑠),0≀𝑠≀min{𝑑,πœ‚}<1;2𝐺(𝑑,𝑠),0β‰€π‘‘β‰€π‘ β‰€πœ‚;3𝐺(𝑑,𝑠),πœ‚β‰€π‘ β‰€t≀1;4(𝑑,𝑠),0<max{πœ‚,𝑑}≀𝑠≀1.(3.2) Here 𝐺1(𝑑,𝑠)=β„Ž(𝑠)π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑑)+𝛼1ξ€»,πΊβ„Ž(𝑑)2β„Ž(𝑑,𝑠)=(𝑑)π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑠)+𝛼1ξ€»,πΊβ„Ž(𝑠)31(𝑑,𝑠)=π›Ώξ€Ίξ€½π›Όβ„Ž(𝑑)1ξ€Ύξ€»,πΊβ„Ž(πœ‚)βˆ’β„Ž(𝑠)+π›Ώβ„Ž(𝑠)4(𝑑,𝑠)=β„Ž(𝑑)𝛿[],β„Ž(1)βˆ’β„Ž(𝑠)𝛿=β„Ž(1)βˆ’π›Ό1ξ€œβ„Ž(πœ‚)>0,β„Ž(𝑑)=𝑑01𝑝(π‘₯)𝑑π‘₯,0<𝛼1<β„Ž(1).β„Ž(πœ‚)(3.3)

Proof. Consider the following linear differential equation: βˆ’ξ€·π‘(𝑑)π‘¦ξ…žξ€Έ(𝑑)ξ…ž=π‘ž(𝑑)𝐹(𝑑),(3.4) where 𝐹∈𝐢([0,1],[0,∞)). Integrating the above differential equation twice first from 𝑑 to 1 and then from 0 to 𝑑, changing the order of integration, and applying the boundary conditions, we get ξ€œπ‘¦(𝑑)=𝑑0ξ€œβ„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+β„Ž(𝑑)1𝑑+π‘ž(𝑠)𝐹(𝑠)π‘‘π‘ β„Ž(𝑑)𝛿𝛼1ξ€œπœ‚0β„Žξ€œ(𝑠)π‘ž(𝑠)𝐹(𝑠)π‘‘π‘ βˆ’10β„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+𝛼1β„Žξ€œ(πœ‚)1πœ‚π‘žξ‚Ή.(𝑠)𝐹(𝑠)𝑑𝑠(3.5) For π‘‘βˆˆ[0,πœ‚], 𝑦(𝑑) can be written as ξ€œπ‘¦(𝑑)=𝑑0β„Ž(𝑠)π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑑)+𝛼1ξ€»+ξ€œβ„Ž(𝑑)π‘ž(𝑠)𝐹(𝑠)π‘‘π‘ πœ‚π‘‘β„Ž(𝑑)π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑠)+𝛼1ξ€»ξ€œβ„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+1πœ‚β„Ž(𝑑)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠,(3.6) or ξ€œπ‘¦(𝑑)=𝑑0𝐺1ξ€œ(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+πœ‚π‘‘πΊ2ξ€œ(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+1πœ‚πΊ4(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠.(3.7) Similarly, for π‘‘βˆˆ[πœ‚,1], 𝑦(𝑑) can be written as ξ€œπ‘¦(𝑑)=πœ‚0β„Ž(𝑠)π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑑)+𝛼1ξ€»+ξ€œβ„Ž(𝑑)π‘ž(𝑠)𝐹(𝑠)π‘‘π‘ π‘‘πœ‚1π›Ώξ€Ίξ€½π›Όβ„Ž(𝑑)1ξ€Ύξ€»ξ€œβ„Ž(πœ‚)βˆ’β„Ž(𝑠)+π›Ώβ„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+1π‘‘β„Ž(𝑑)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠,(3.8) or ξ€œπ‘¦(𝑑)=πœ‚0𝐺1ξ€œ(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+π‘‘πœ‚πΊ3ξ€œ(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠+1𝑑𝐺4(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠.(3.9) From (3.7) and (3.9), we may write ξ€œπ‘¦(𝑑)=10𝐺(𝑑,𝑠)π‘ž(𝑠)𝐹(𝑠)𝑑𝑠,(3.10) where 𝐺(𝑑,𝑠) is given in the lemma. It is easy to see that 𝐺(𝑑,𝑠) satisfies all the properties of Green’s function. Hence 𝐺(𝑑,𝑠) is the Green’s function for the boundary value problem (3.1).

Lemma 3.2. The Green’s function 𝐺(𝑑,𝑠) satisfies the following properties:(i)maxπ‘‘βˆˆ[0,1]𝑝(𝑑)(πœ•πΊ(𝑑,𝑠)/πœ•π‘‘)<∞,(ii)𝐺(𝑑,𝑠)β‰₯0 for all (𝑑,𝑠)∈{[0,1]Γ—[0,1]},(iii)there exist a constant πœ† in (0,1) such that minπ‘‘βˆˆ[πœ‚,1]𝐺(𝑑,𝑠)β‰₯πœ†maxπ‘‘βˆˆ[0,1]𝐺(𝑑,𝑠)forπ‘ βˆˆ[0,1], where ⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©ξ‚»π›Όπœ†=min1[]β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(1)βˆ’π›Ό1,π›Όβ„Ž(πœ‚)1β„Ž(πœ‚)ξ‚Όβ„Ž(1),0<𝛼1≀1,minβ„Ž(1)βˆ’π›Ό1β„Ž(πœ‚)𝛼1[],β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(πœ‚)𝛼1ξ‚Όβ„Ž(1),1≀𝛼1<β„Ž(1).β„Ž(πœ‚)(3.11)

Proof. (i) 𝑝(𝑑)πœ•πΊ(𝑑,𝑠)=⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©πœ•π‘‘β„Ž(𝑠)𝛿𝛼1ξ€»1βˆ’1,0≀𝑠≀min{𝑑,πœ‚}<1;π›Ώξ€Ίπ›Ώβˆ’β„Ž(𝑠)βˆ’π›Ό1ξ€»1β„Ž(𝑠),0β‰€π‘‘β‰€π‘ β‰€πœ‚;𝛿𝛼1ξ€»1β„Ž(πœ‚)βˆ’β„Ž(𝑠),πœ‚β‰€π‘ β‰€π‘‘β‰€1;𝛿[]β„Ž(1)βˆ’β„Ž(𝑠),0<max{πœ‚,𝑑}≀𝑠≀1.(3.12) Since 𝑝(𝑑)(πœ•πΊ(𝑑,𝑠)/πœ•π‘‘) is independent of 𝑑, therefore maxπ‘‘βˆˆ[0,1]𝑝(𝑑)(πœ•πΊ(𝑑,𝑠)/πœ•π‘‘)<∞.
(ii) For 𝑑<πœ‚, 𝛼1[β„Ž(𝑑)βˆ’β„Ž(πœ‚)]>β„Ž(1)/β„Ž(πœ‚)[β„Ž(𝑑)βˆ’β„Ž(πœ‚)] and 𝐺11(𝑑,𝑠)β‰₯π›Ώξ‚Έβ„Ž(𝑠)β„Ž(𝑑)β„Ž(1)ξ‚Ήβ„Ž(πœ‚)βˆ’1β‰₯0,(3.13) it is easy to show that 𝐺1(𝑑,𝑠)β‰₯0, for 𝑑β‰₯πœ‚.
Next we show 𝐺2(𝑑,𝑠)β‰₯0, 𝐺3(𝑑,𝑠)β‰₯0 and 𝐺4(𝑑,𝑠)β‰₯0 as follows: 𝐺2(𝑑,𝑠)β‰₯β„Ž(𝑑)[]πΊπ›Ώβ„Ž(πœ‚)β„Ž(𝑠)(β„Ž(1)βˆ’β„Ž(πœ‚))β‰₯0;3(1𝑑,𝑠)=π›Ώξ€Ίβ„Ž(𝑠){β„Ž(1)βˆ’β„Ž(𝑑)}+𝛼1ξ€»πΊβ„Ž(πœ‚){β„Ž(𝑑)βˆ’β„Ž(𝑠)}β‰₯0,4(𝑑,𝑠)=β„Ž(𝑑)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠)β‰₯0.(3.14) Thus 𝐺(𝑑,𝑠)β‰₯0 for all (𝑑,𝑠)∈{[0,1]Γ—[0,1]}.
(iii) We prove the inequality for the following cases:
(a) π‘ βˆˆ[0,πœ‚] and (b) π‘ βˆˆ[πœ‚,1],
(a) for π‘ βˆˆ[0,πœ‚], we further divide this case in two parts as follows.
(1) When 0<𝛼1≀1, (2) when 1≀𝛼1<β„Ž(1)/β„Ž(πœ‚).
Case  1 (For 0<𝛼1≀1). It is easy to see that 𝐺1(𝑑,𝑠)=β„Ž(𝑠)π›Ώξ€Ίξ€·β„Ž(1)βˆ’π›Ό1ξ€Έξ€·π›Όβ„Ž(πœ‚)+β„Ž(𝑑)1βˆ’1ξ€Έξ€»(3.15) implies max[]π‘‘βˆˆ0,1𝐺1(𝑑,𝑠)=β„Ž(𝑠)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€»β„Ž(πœ‚),min[]π‘‘βˆˆπœ‚,1𝐺1𝛼(𝑑,𝑠)=1β„Ž(𝑠)𝛿[].β„Ž(1)βˆ’β„Ž(πœ‚)(3.16) Next,𝐺2(𝑑,𝑠)=β„Ž(𝑑)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€·π›Όβ„Ž(πœ‚)+β„Ž(𝑠)1,β‰€β„Žβˆ’1ξ€Έξ€»(𝑠)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€»β„Ž(πœ‚),asβ„Ž(𝑑)β‰€β„Ž(𝑠).(3.17) Thus for 0<𝛼1≀1, max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠)=β„Ž(𝑠)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€»β„Ž(πœ‚),minπ‘‘βˆˆ[πœ‚,1]𝛼𝐺(𝑑,𝑠)=1β„Ž(𝑠)𝛿[].β„Ž(1)βˆ’β„Ž(πœ‚)(3.18)Case  2 (For 1≀𝛼1<β„Ž(1)/β„Ž(πœ‚)). It is easy to see that max[]π‘‘βˆˆ0,1𝐺1𝛼(𝑑,𝑠)=1β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),minπ‘‘βˆˆ[πœ‚,1]𝐺1(𝑑,𝑠)=β„Ž(𝑠)𝛿[].β„Ž(1)βˆ’β„Ž(πœ‚)(3.19) Next, 𝐺2(𝑑,𝑠)=β„Ž(𝑑)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€·π›Όβ„Ž(πœ‚)+β„Ž(𝑠)1,β‰€π›Όβˆ’1ξ€Έξ€»1β„Ž(𝑠)𝛿[β„Ž](1)βˆ’β„Ž(πœ‚),asβ„Ž(𝑑)β‰€β„Ž(𝑠)β‰€β„Ž(πœ‚).(3.20) Thus for 1≀𝛼1<β„Ž(1)/β„Ž(πœ‚), max[]π‘‘βˆˆ0,1𝛼𝐺(𝑑,𝑠)=1β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),minπ‘‘βˆˆ[πœ‚,1]𝐺(𝑑,𝑠)=β„Ž(𝑠)𝛿[].β„Ž(1)βˆ’β„Ž(πœ‚)(3.21) Combining (3.18) and (3.21), we may write for π‘ βˆˆ[0,πœ‚], max[]π‘‘βˆˆ0,1𝐺⎧βŽͺ⎨βŽͺ⎩(𝑑,𝑠)=β„Ž(𝑠)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€»β„Ž(πœ‚),0<𝛼1𝛼≀1,1β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),1≀𝛼1<β„Ž(1),β„Ž(πœ‚)minπ‘‘βˆˆ[πœ‚,1]⎧βŽͺ⎨βŽͺβŽ©π›ΌπΊ(𝑑,𝑠)=1β„Ž(𝑠)𝛿[β„Ž](1)βˆ’β„Ž(πœ‚),0<𝛼1≀1,β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),1≀𝛼1<β„Ž(1).β„Ž(πœ‚)(3.22) (b) For π‘ βˆˆ[πœ‚,1]. For this case, 𝐺3(𝑑,𝑠) and 𝐺4(𝑑,𝑠) are considered. From (3.2), it can be easily seen that for π‘ βˆˆ[πœ‚,1], max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠)=β„Ž(1)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠),minπ‘‘βˆˆ[πœ‚,1]𝐺(𝑑,𝑠)=β„Ž(πœ‚)𝛿[].β„Ž(1)βˆ’β„Ž(𝑠)(3.23) Thus from (3.22) and (3.23), we get max[]π‘‘βˆˆ0,1⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩[]∢⎧βŽͺβŽͺ⎨βŽͺβŽͺ⎩𝐺(𝑑,𝑠)=π‘ βˆˆ0,πœ‚β„Ž(𝑠)π›Ώξ€Ίβ„Ž(1)βˆ’π›Ό1ξ€»β„Ž(πœ‚),0<𝛼1𝛼≀1,1β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),1≀𝛼1<β„Ž(1),[]βˆΆξ‚»β„Ž(πœ‚)π‘ βˆˆπœ‚,1β„Ž(1)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠),0<𝛼1<β„Ž(1).β„Ž(πœ‚)minπ‘‘βˆˆ[πœ‚,1]⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩[]∢⎧βŽͺ⎨βŽͺβŽ©π›ΌπΊ(𝑑,𝑠)=π‘ βˆˆ0,πœ‚1β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),0<𝛼1≀1,β„Ž(𝑠)𝛿[]β„Ž(1)βˆ’β„Ž(πœ‚),1≀𝛼1<β„Ž(1),[]βˆΆξ‚»β„Ž(πœ‚)π‘ βˆˆπœ‚,1β„Ž(πœ‚)𝛿[]β„Ž(1)βˆ’β„Ž(𝑠),0<𝛼1<β„Ž(1).β„Ž(πœ‚)(3.24) From (3.24), minπ‘‘βˆˆ[πœ‚,1]⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©ξ‚»π›ΌπΊ(𝑑,𝑠)β‰₯min1[]β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(1)βˆ’π›Ό1,π›Όβ„Ž(πœ‚)1β„Ž(πœ‚)ξ‚Όβ„Ž(1)max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠),0<𝛼1≀1,minβ„Ž(1)βˆ’π›Ό1β„Ž(πœ‚)𝛼1[],β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(πœ‚)𝛼1ξ‚Όβ„Ž(1)max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠),1≀𝛼1<β„Ž(1).β„Ž(πœ‚)(3.25) Consequently, setting ⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©ξ‚»π›Όπœ†=min1[]β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(1)βˆ’π›Ό1,π›Όβ„Ž(πœ‚)1β„Ž(πœ‚)ξ‚Όβ„Ž(1),0<𝛼1≀1,minβ„Ž(1)βˆ’π›Ό1β„Ž(πœ‚)𝛼1[],β„Ž(1)βˆ’β„Ž(πœ‚)β„Ž(πœ‚)𝛼1ξ‚Όβ„Ž(1),1≀𝛼1<β„Ž(1),β„Ž(πœ‚)(3.26) there holds minπ‘‘βˆˆ[πœ‚,1]𝐺(𝑑,𝑠)β‰₯πœ†max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠).(3.27) It can be easily seen that 0<πœ†<1. This completes the proof.

4. Existence of Multiple Nonnegative Solutions

Let ⋂𝐢𝑋=𝐢[0,1]2(0,1] be endowed with ordering π‘₯≀𝑦 if π‘₯(𝑑)≀𝑦(𝑑) for all π‘‘βˆˆ[0,1] and β€–π‘₯β€–=max{β€–π‘₯β€–1,β€–π‘₯β€²β€–1}, where β€–π‘₯β€–1=supπ‘‘βˆˆ[0,1]||||,π‘₯(𝑑)β€–π‘₯β€²β€–1=supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘₯ξ…ž||.(𝑑)(4.1) Let 𝐸={π‘₯∢π‘₯βˆˆπ‘‹,β€–π‘₯β€–<∞} be bounded subset of 𝑋. 𝐸 is Banach Space.

Now define a cone π‘ƒβŠ‚πΈ as 𝑃=π‘₯∈𝐸∢π‘₯(𝑑)β‰₯0,minπ‘‘βˆˆ[πœ‚,1]π‘₯(𝑑)β‰₯πœ†max[]π‘‘βˆˆ0,1ξ€·π‘₯(𝑑),𝑝(𝑑)π‘₯ξ…žξ€Έ(𝑑)ξ…žξ‚Όβ‰€0.(4.2) The boundary value problem (1.1)-(1.2) has a solution 𝑦(𝑑) if and only if 𝑦(𝑑) solves the following operator equation: 𝑦(𝑑)=𝑇𝑦(𝑑),(4.3) where the operator π‘‡βˆΆπ‘ƒβ†’π‘ƒ is given by ξ€œ(𝑇𝑦)(𝑑)=10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠,0≀𝑑≀1.(4.4) Here 𝐺(𝑑,𝑠) is the Green’s function of the problem (3.1) defined in Lemma 3.1.

Lemma 4.1. Let (E1)–(E3) hold, then the operator π‘‡βˆΆπ‘ƒβ†’π‘ƒ is well defined and is completely continuous.

Proof. First we show that the operator 𝑇 is well defined. For this, we take π‘¦βˆˆπ‘ƒ. From (E2), (E3), and 𝐺(𝑑,𝑠)β‰₯0, it follows that 𝑇𝑦(𝑑)β‰₯0.
Now applying Lemma 3.2, we get minπ‘‘βˆˆ[πœ‚,1]𝑇𝑦(𝑑)=minπ‘‘βˆˆ[πœ‚,1]ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠,minπ‘‘βˆˆ[πœ‚,1]ξ€œπ‘‡π‘¦(𝑑)β‰₯πœ†10max[]π‘‘βˆˆ0,1𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(s)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠,=πœ†max[]π‘‘βˆˆ0,1ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠,=πœ†max[]π‘‘βˆˆ0,1(𝑇𝑦)(𝑑).(4.5) It is easy to show that (𝑝(𝑑)𝑇𝑦′(𝑑))′≀0. Thus 𝑇 is well defined.
We now show that 𝑇 is completely continuous. Let {𝑦𝑛} be a sequence in 𝑃 and 𝑦0βˆˆπ‘ƒ with limπ‘›β†’βˆžπ‘¦π‘›=𝑦0. Then, there exists a constant π‘˜1>0 such that ‖𝑦𝑛‖<π‘˜1 for all π‘›βˆˆβ„•βˆͺ{0}. Thus β€–π‘¦π‘›βˆ’π‘¦0β€–β†’0 as π‘›β†’βˆž implies supπ‘‘βˆˆ[0,1]|(π‘¦π‘›βˆ’π‘¦0)(𝑑)|andsupπ‘‘βˆˆ(0,1]|𝑝(𝑑)(π‘¦π‘›βˆ’π‘¦0)β€²(𝑑)|β†’0 as π‘›β†’βˆž. So 𝑦𝑛(𝑑)→𝑦0(𝑑) and 𝑝(𝑑)π‘¦ξ…žπ‘›(𝑑)→𝑝(𝑑)𝑦0β€²(𝑑) as π‘›β†’βˆž.
Since 𝑓 is continuous on {[0,1]Γ—[0,π‘˜1]Γ—[βˆ’π‘˜1,π‘˜1]}, so ||𝑇𝑦𝑛(𝑑)βˆ’π‘‡π‘¦0||=||||ξ€œ(𝑑)10𝑓𝐺(𝑑,𝑠)π‘ž(𝑠)𝑠,𝑦𝑛(𝑠),𝑝(𝑠)π‘¦ξ…žπ‘›ξ€Έξ€·(𝑠)βˆ’π‘“π‘ ,𝑦0(𝑠),𝑝(𝑠)π‘¦ξ…ž0||||βŸΉβ€–β€–(𝑠)ξ€Έξ€»π‘‘π‘ βŸΆ0asπ‘›βŸΆβˆž.π‘‡π‘¦π‘›βˆ’π‘‡π‘¦0β€–β€–1||ξ€·βŸΆ0asπ‘›βŸΆβˆž.(4.6)𝑝(𝑑)π‘‡π‘¦ξ…žπ‘›(𝑑)βˆ’π‘‡π‘¦ξ…ž0ξ€Έ||=||||ξ€œ(𝑑)𝑝(𝑑)10πœ•πΊ(𝑑,𝑠)ξ€Ίπ‘“ξ€·πœ•π‘‘π‘ž(𝑠)𝑠,𝑦𝑛(𝑠),𝑝(𝑠)π‘¦ξ…žπ‘›ξ€Έξ€·(𝑠)βˆ’π‘“π‘ ,𝑦0(𝑠),𝑝(𝑠)π‘¦ξ…ž0||||βŸΉβ€–β€–ξ€·(𝑠)ξ€Έξ€»π‘‘π‘ βŸΆ0asπ‘›βŸΆβˆž.π‘‡π‘¦π‘›βˆ’π‘‡π‘¦0ξ€Έξ…žβ€–β€–1⟢0asπ‘›βŸΆβˆž.(4.7) From (4.6) and (4.7), β€–β€–π‘‡π‘¦π‘›βˆ’π‘‡π‘¦0β€–β€–βŸΆ0asπ‘›βŸΆβˆž.(4.8) Hence π‘‡βˆΆπ‘ƒβ†’π‘ƒ is a continuous operator.
Next we prove that 𝑇 maps every bounded subset of 𝑃 into relatively compact set. Let 𝐡={π‘₯βˆˆπ‘ƒβˆΆβ€–π‘₯β€–β‰€π‘˜2,π‘˜2isapositiveconstant} be any bounded subset of 𝑃. For π‘¦βˆˆπ΅, ξ€œπ‘‡π‘¦(𝑑)=10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠≀max[]π‘‘βˆˆ0,1ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ(𝑠)𝑑𝑠≀max[]Γ—ξ€Ί(𝑠,𝑒,𝑣)∈0,10,π‘˜2ξ€»Γ—ξ€Ίβˆ’π‘˜2,π‘˜2𝑓(𝑠,𝑒,𝑣)max[]π‘‘βˆˆ0,1ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)π‘‘π‘ βŸΉπ‘‡π‘¦(𝑑)<∞.(4.9) Therefore 𝑇(𝐡) is uniformly bounded. Further, equicontinuity of 𝑇(𝐡) follows from ||𝑑𝑇𝑦1ξ€Έξ€·π‘‘βˆ’π‘‡π‘¦2ξ€Έ||=||||ξ€œ10||𝐺(𝑑,𝑠)𝑑=𝑑1||βˆ’πΊ(𝑑,𝑠)𝑑=𝑑2ξ‚„ξ€·π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ||||(𝑠)π‘‘π‘ βŸΆ0as𝑑1βŸΆπ‘‘2,||𝑝(𝑑)π‘‡π‘¦ξ…žξ€·π‘‘1ξ€Έβˆ’π‘‡π‘¦ξ…žξ€·π‘‘2||=||||ξ€œξ€Έξ€Έπ‘(𝑑)10ξ‚Έπœ•πΊ(𝑑,𝑠)|||πœ•π‘‘π‘‘=𝑑1βˆ’πœ•πΊ(𝑑,𝑠)|||πœ•π‘‘π‘‘=𝑑2ξ‚Ήξ€·π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ|||||||(𝑠)𝑑𝑠≀2maxπœ•πΊ(𝑑,𝑠)|||πœ•π‘‘β‹…sup[]π‘‘βˆˆ0,1ξ€œπ‘(𝑑)𝑑2𝑑1ξ€·π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…ž(𝑠)𝑑𝑠,⟢0as𝑑1βŸΆπ‘‘2.(4.10) Thus from Arzela-Ascoli Theorem, 𝑇(𝐡) is relatively compact subset of 𝑃 and also π‘‡βˆΆπ‘ƒβ†’π‘ƒ is completely continuous.

Next, define functionals 𝛼,𝛽,πœ“βˆΆπ‘ƒβ†’[0,∞) such that 𝛼(𝑦)=supπ‘‘βˆˆ[0,1]||||,𝑦(𝑑)𝛽(𝑦)=supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘¦ξ…ž||,(𝑑)πœ“(𝑦)=minπ‘‘βˆˆ[πœ‚,1]||||.𝑦(𝑑)(4.11) Clearly, 𝛼,𝛽 are nonnegative continuous convex functionals such that ‖𝑦‖=max{𝛼(𝑦),𝛽(𝑦)} satisfying (2.3) and (2.4), and πœ“ is nonnegative concave functional with πœ“(𝑦)≀𝛼(𝑦).

Let 𝐢=minπ‘‘βˆˆ[πœ‚,1]ξ€œ1πœ‚πΊ(𝑑,𝑠)π‘ž(𝑠)𝑑𝑠,𝐿=supπ‘‘βˆˆ[0,1]ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑑𝑠,𝑁=supπ‘‘βˆˆ(0,1]||||ξ€œ10𝑝(𝑑)πœ•πΊ(𝑑,𝑠)||||.πœ•π‘‘π‘ž(𝑠)𝑑𝑠(4.12) Now we state the main results of this work.

Theorem 4.2. Suppose that (E1)–(E3) are satisfied and 𝑓(𝑑,𝑦,𝑧) satisfies the following condition.(H1) if there exist real constants 𝑑>0 and 𝐿1>0 such that 𝑓(𝑑,𝑦,𝑧)<min{𝑑/𝐿,𝐿1/𝑁} for (𝑑,𝑦,𝑧)∈{[0,1]Γ—[0,𝑑]Γ—[βˆ’πΏ1,𝐿1]},then boundary value problem (1.1)-(1.2) has at least one nonnegative solution 𝑦1 such that supπ‘‘βˆˆ[0,1]|𝑦1(𝑑)|<𝑑 with supπ‘‘βˆˆ(0,1]|𝑝(𝑑)π‘¦ξ…ž1(𝑑)|<𝐿1.

Theorem 4.3. Suppose that (E1)–(E3) are satisfied. There exist real constants π‘Ž,𝑐,𝑑,𝐿1,and𝐿2 with 𝑐β‰₯π‘Ž/πœ†=𝑏>π‘Ž>𝑑>0, 𝐿2β‰₯𝐿1>0 such that π‘Ž/𝐢≀min{𝑐/𝐿,𝐿2/𝑁} and 𝑓(𝑑,𝑦,𝑧) satisfies following conditions.(H1)𝑓(𝑑,𝑦,𝑧)<min{𝑑/𝐿,𝐿1/𝑁} for (𝑑,𝑦,𝑧)∈{[0,1]Γ—[0,𝑑]Γ—[βˆ’πΏ1,𝐿1]},(H2)𝑓(𝑑,𝑦,𝑧)>π‘Ž/𝐢 for (𝑑,𝑦,𝑧)∈{[πœ‚,1]Γ—[π‘Ž,π‘Ž/πœ†]Γ—[βˆ’πΏ2,𝐿2]},(H3)𝑓(𝑑,𝑦,𝑧)≀min{c/L,L2/N} for (𝑑,𝑦,𝑧)∈{[0,1]Γ—[0,𝑐]Γ—[βˆ’πΏ2,𝐿2]}.Then boundary value problem (1.1)-(1.2) has at least three nonnegative solutions 𝑦1, 𝑦2, and 𝑦3 in 𝑃(𝛼𝑐,𝛽𝐿2) such that supπ‘‘βˆˆ[0,1]||𝑦1||(𝑑)<𝑑,supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘¦ξ…ž1||(𝑑)<𝐿1,π‘Ž<minπ‘‘βˆˆ[πœ‚,1]||𝑦2||(𝑑)≀sup[]π‘‘βˆˆ0,1||𝑦2||(𝑑)≀𝑐,supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘¦ξ…ž2(||𝑑)≀𝐿2,sup[]π‘‘βˆˆ0,1||𝑦3(||β‰€π‘Žπ‘‘)πœ†withsup]π‘‘βˆˆ(0,1||𝑝(𝑑)π‘¦ξ…ž3(||𝑑)≀𝐿2.(4.13)

Proof of Theorem 4.2. Let π‘ˆ=𝑃(𝛼𝑑,𝛽𝐿1) be open subset of 𝑃. We now show that 𝑇(π‘ˆ)βŠ‚π‘ˆ. For π‘¦βˆˆπ‘ˆ, 𝛼(𝑇𝑦)=sup[]π‘‘βˆˆ0,1||||𝑇𝑦(𝑑)=sup[]π‘‘βˆˆ0,1ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έ<𝑑(𝑠)𝑑𝑠,𝐿sup[]π‘‘βˆˆ0,1ξ€œ10ξ€·H𝐺(𝑑,𝑠)π‘ž(𝑠)𝑑𝑠<𝑑,fromcondition1ξ€Έ(4.14) implies that 𝛼(𝑇𝑦)<𝑑.
Consider that 𝛽(𝑇𝑦)=supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘‡π‘¦ξ…ž||,(𝑑)=supπ‘‘βˆˆ(0,1]||||ξ€œπ‘(𝑑)10πœ•πΊ(𝑑,𝑠)ξ€·πœ•π‘‘π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…ž(ξ€Έ||||,<𝐿𝑠)1𝑁sup]π‘‘βˆˆ(0,1||||ξ€œ10𝑝(𝑑)πœ•πΊ(𝑑,𝑠)||||ξ€·Hπœ•π‘‘π‘ž(𝑠)𝑑𝑠,fromcondition1ξ€Έ,(4.15) implies that 𝛽(𝑇𝑦)<𝐿1.
Thus 𝑇(π‘ˆ)βŠ‚π‘ˆ. Next, we show that 𝑇 has no fixed point on πœ•π‘ˆ(=π‘ˆβ§΅π‘ˆ). On contrary, suppose there exists a fixed point 𝑦(𝑑) on πœ•π‘ˆ such that 𝑇𝑦(𝑑)=𝑦(𝑑). Then from (4.14) and (4.15), 𝛼(𝑦)<𝑑 and 𝛽(𝑦)<𝐿1, which are not possible. So the operator 𝑇 has no fixed point on πœ•π‘ˆ and from Theorem 2.10𝑖(𝑇,π‘ˆ,𝑃)=1. Thus the operator 𝑇 has at least one fixed point in π‘ˆ and also the boundary value problem (1.1)-(1.2) has at least one nonnegative solution 𝑦1 such that supπ‘‘βˆˆ[0,1]|𝑦1(𝑑)|<𝑑 with supπ‘‘βˆˆ(0,1]|𝑝(𝑑)π‘¦ξ…ž1(𝑑)|<𝐿1.

Proof of Theorem 4.3. It is easy to see that πœ“(𝑦)≀𝛼(𝑦) for each π‘¦βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2). We now show that π‘‡βˆΆπ‘ƒ(𝛼𝑐,𝛽𝐿2)→𝑃(𝛼𝑐,𝛽𝐿2) is well defined. For π‘¦βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2), 𝛼(𝑇𝑦)=supπ‘‘βˆˆ[0,1]||||,𝑇𝑦(𝑑)=sup[]π‘‘βˆˆ0,1ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…žξ€Έβ‰€π‘(𝑠)𝑑𝑠,𝐿sup[]π‘‘βˆˆ0,1ξ€œ10ξ€·H𝐺(𝑑,𝑠)π‘ž(𝑠)𝑑𝑠,fromassumption3ξ€Έ,≀𝑐.(4.16)𝛽(𝑇𝑦)=supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘‡π‘¦ξ…ž||,(𝑑)=supπ‘‘βˆˆ(0,1]||||ξ€œπ‘(𝑑)10πœ•πΊ(𝑑,𝑠)ξ€·πœ•π‘‘π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…ž(ξ€Έ||||,≀𝐿𝑠)2𝑁sup]π‘‘βˆˆ(0,1||||ξ€œ10𝑝(𝑑)πœ•πΊ(𝑑,𝑠)||||ξ€·Hπœ•π‘‘π‘ž(𝑠)𝑑𝑠,fromassumption3ξ€Έ,≀𝐿2.(4.17) From (4.16) and (4.17), 𝑇(𝑦)βˆˆπ‘ƒξ€·π›Όπ‘,𝛽𝐿2ξ€Έ.(4.18) Thus, π‘‡βˆΆπ‘ƒ(𝛼𝑐,𝛽𝐿2)→𝑃(𝛼𝑐,𝛽𝐿2) is well defined, and by Lemma 4.1, it is completely continuous. Now Condition (2) of Theorem 2.11 can be proved by similar manner. Choose 𝑦(𝑑)=π‘Ž/πœ†βˆˆπ‘ƒ, 0≀𝑑≀1, then 𝛼(𝑦)=supπ‘‘βˆˆ[0,1]|𝑦(𝑑)|=π‘Ž/πœ†,𝛽(𝑦)=supπ‘‘βˆˆ(0,1]|𝑝(𝑑)𝑦′(𝑑)|=0<𝐿2, πœ“(π‘₯)=minπ‘‘βˆˆ[πœ‚,1]|𝑦(𝑑)|=π‘Ž/πœ†>π‘Ž. Thus, {π‘¦βˆˆπ‘ƒ(𝛼𝑏=π‘Ž/πœ†,𝛽𝐿2,Ξ¨π‘Ž)∢Ψ(𝑦)>π‘Ž}β‰ Ξ¦. Further if π‘¦βˆˆπ‘ƒ(π›Όπ‘Ž/πœ†,𝛽𝐿2,Ξ¨π‘Ž), then π‘Žβ‰€π‘¦(𝑑)β‰€π‘Ž/πœ† for πœ‚β‰€π‘‘β‰€1. Then by definition of πœ“ and assumption (H2), we have πœ“(𝑇𝑦)=minπ‘‘βˆˆ[πœ‚,1]||||𝑇𝑦(𝑑)=minπ‘‘βˆˆ[πœ‚,1]ξ€œ10𝐺(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…ž(𝑠)𝑑𝑠β‰₯minπ‘‘βˆˆ[πœ‚,1]ξ€œ1πœ‚ξ€·πΊ(𝑑,𝑠)π‘ž(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)π‘¦ξ…ž(ξ€Έ>π‘Žπ‘ )𝑑𝑠𝐢⋅𝐢=π‘Ž(4.19) Thus, Condition (1) of Theorem 2.11 is satisfied. We finally show that condition (3) of Theorem 2.11 holds, too. Suppose π‘¦βˆˆπ‘ƒ(𝛼𝑐,𝛽𝐿2,Ξ¨π‘Ž) with 𝛼(𝑇𝑦)>𝑏. Then by definition of πœ“ and π‘‡π‘¦βˆˆπ‘ƒ, we have Ξ¨(𝑇𝑦)=minπ‘‘βˆˆ[πœ‚,1]||||𝑇𝑦(𝑑)β‰₯πœ†max[]π‘‘βˆˆ0,1||||𝑇𝑦(𝑑)β‰₯πœ†π›Ό(𝑇𝑦)=π‘Ž.(4.20) So, Condition (3) of Theorem 2.11 is also satisfied. Therefore, Theorem 2.11 yields that boundary value problem (1.1)-(1.2) has at least three nonnegative solutions 𝑦1, 𝑦2, and 𝑦3 in 𝑃(𝛼𝑐,𝛽𝐿2) such that supπ‘‘βˆˆ[0,1]||𝑦1||(𝑑)<𝑑,supπ‘‘βˆˆ(0,1]||𝑝(𝑑)π‘¦ξ…ž1||(𝑑)<𝐿1;π‘Ž<minπ‘‘βˆˆ[πœ‚,1]||𝑦2||(𝑑)≀sup[]π‘‘βˆˆ0,1||𝑦2||(𝑑)≀𝑐,sup]π‘‘βˆˆ(0,1||𝑝(𝑑)π‘¦ξ…ž2||(𝑑)≀𝐿2;supπ‘‘βˆˆ[0,1]||𝑦3||β‰€π‘Ž(𝑑)πœ†,sup]π‘‘βˆˆ(0,1||𝑝(𝑑)π‘¦ξ…ž3||(𝑑)≀𝐿2.(4.21)

Corollary 4.4. Suppose that (E1)–(E3) are satisfied. If there exist constants 0<𝑑1<π‘Ž1<π‘Ž1/πœ†<𝑑2<π‘Ž2<π‘Ž2/πœ†<𝑑3<β‹―<𝑑𝑛,0<𝐿1≀𝐿2≀𝐿3β‰€β‹―β‰€πΏπ‘›π‘›βˆˆβ„•, with π‘Žπ‘–/𝐢<min{𝑑𝑖+1/𝐿,𝐿𝑖+1/𝑁},1<π‘–β‰€π‘›βˆ’1 such that 𝑓 satisfies the following conditions:(M1)𝑓(𝑑,𝑦,𝑧)<min{𝑑𝑖/𝐿,𝐿𝑖/𝑁},for(𝑑,𝑦,𝑧)∈{[0,1]Γ—[0,𝑑𝑖]Γ—[βˆ’πΏπ‘–,𝐿𝑖]},1≀𝑖≀𝑛, (M2)𝑓(𝑑,𝑦,𝑧)>π‘Žπ‘–/𝐢,for(𝑑,𝑦,𝑧)∈{[πœ‚,1]Γ—[π‘Žπ‘–,π‘Žπ‘–/πœ†]Γ—[βˆ’πΏπ‘–+1,𝐿𝑖+1]},1β‰€π‘–β‰€π‘›βˆ’1, then boundary value problem (1.1)-(1.2) has at least 2π‘›βˆ’1 nonnegative solutions.

Proof. When 𝑛=1, the result follows from Theorem 4.2. When 𝑛=2, it is clear that all the conditions of Theorem 4.3 hold (with 𝑐=𝑑2,𝑑=𝑑1,π‘Ž=π‘Ž1). Thus the boundary value problem (1.1)-(1.2) has at least three positive solutions 𝑦1, 𝑦2, and 𝑦3. Following this way, we complete the proof by induction method.

Finally, we demonstrate these results through examples.

5. Example

In Example 5.1, we demonstrate the detailed working of the boundary value problem (1.5) mentioned in the introduction. Example 5.2 verifies our results.

Example 5.1. Consider the boundary value problem (1.5).
Here, 𝑓𝑑,𝑦,π‘π‘¦ξ…žξ€Έ5=2(π‘Ÿ+1)π‘‘βˆ’4π‘Ÿ,π‘ž(𝑑)=π‘‘π‘Ÿβˆ’1,max[]π‘‘βˆˆ0,1𝑓𝑑,𝑦,π‘π‘¦ξ…žξ€Έ=34π‘Ÿ+2.(5.1) Following the notations of this work, it is easy to see that 3𝐿=2ξ‚Έ3(1βˆ’π‘Ÿ)4βˆ’4π‘Ÿξ‚Ή(1βˆ’π‘Ÿ)/π‘Ÿ,𝑁=3(1βˆ’π‘Ÿ)π‘Ÿ(4βˆ’4π‘Ÿ).(5.2)
Now for 𝑑β‰₯(3/2)[3(1βˆ’π‘Ÿ)/4βˆ’4π‘Ÿ](1βˆ’π‘Ÿ)/π‘Ÿ[(3/4)π‘Ÿ+2]and𝐿1β‰₯3(1βˆ’π‘Ÿ)((3/4)π‘Ÿ+1)/π‘Ÿ(4βˆ’4π‘Ÿ),𝑓(𝑑,𝑦,𝑝𝑦′)≀min{𝑑/𝐿,𝐿1/𝑁}.
Then from Theorem 4.2, the problem has at least one nonnegative solution 𝑦(𝑑) with supπ‘‘βˆˆ[0,1]||||≀3𝑦(𝑑)2ξ‚Έ3(1βˆ’π‘Ÿ)4βˆ’4π‘Ÿξ‚Ή(1βˆ’π‘Ÿ)/π‘Ÿξ‚ƒ34ξ‚„π‘Ÿ+2,sup]π‘‘βˆˆ(0,1||𝑝(𝑑)π‘¦ξ…ž||≀(𝑑)3(1βˆ’π‘Ÿ)π‘Ÿ(4βˆ’4π‘Ÿ)34ξ‚„.π‘Ÿ+2(5.3) Next we reduce the problem and then apply Theorem 4.2 for 𝑝(𝑑)=1. Using the transformation ∫π‘₯=(1βˆ’π‘Ÿ)𝑑0(1/π‘₯π‘Ÿ)𝑑π‘₯=𝑑1βˆ’π‘Ÿ, the boundary value problem (1.5) can be reduced to regular boundary value problem as βˆ’π‘£ξ…žξ…ž1(π‘₯)=(1βˆ’π‘Ÿ)2π‘₯(2π‘Ÿβˆ’1)/(1βˆ’π‘Ÿ)2(π‘Ÿ+1)π‘₯1/(1βˆ’π‘Ÿ)βˆ’54π‘Ÿξ‚„ξ‚Έξ‚€1,π‘₯∈(0,1),𝑣(0)=0,𝑣(1)=𝑣4(1βˆ’π‘Ÿ)ξ‚Ή.(5.4) Here, 𝐹π‘₯,𝑣,π‘£ξ…žξ€Έ=1(1βˆ’π‘Ÿ)2𝑑(2π‘Ÿβˆ’1)/(1βˆ’π‘Ÿ)2(π‘Ÿ+1)𝑑1/(1βˆ’π‘Ÿ)βˆ’54π‘Ÿξ‚„,max[]π‘₯∈0,1𝐹π‘₯,𝑣,π‘£ξ…žξ€Έ=1(1βˆ’π‘Ÿ)225(1+π‘Ÿ)βˆ’4π‘Ÿξ‚„.(5.5) Now following the notation of this work for 𝑝(𝑑)=1, 𝐿=3(1βˆ’π‘Ÿ)24βˆ’4π‘Ÿξ‚Έ3(1βˆ’π‘Ÿ)π‘Ÿ(4βˆ’4π‘Ÿ)ξ‚Ή(1βˆ’π‘Ÿ)/π‘Ÿ,𝑁=3(1βˆ’π‘Ÿ)2π‘Ÿ(4βˆ’4π‘Ÿ).(5.6) Now for 𝑑β‰₯3/(4βˆ’4π‘Ÿ)[3(1βˆ’π‘Ÿ)/(4βˆ’4π‘Ÿ)](1βˆ’π‘Ÿ)/π‘Ÿ[(3/4)π‘Ÿ+2]and𝐿1β‰₯(3/π‘Ÿ(4βˆ’4π‘Ÿ))[(3/4)π‘Ÿ+2],𝑓(𝑑,𝑣,𝑣′)≀min{𝑑/𝐿,𝐿1/𝑁}. So the problem has at least one nonnegative solution 𝑣(𝑑) with supπ‘‘βˆˆ[0,1]||||≀3𝑣(𝑑)4βˆ’4π‘Ÿξ‚Έ3(1βˆ’π‘Ÿ)