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

Abstract

We study the existence of multiple nonnegative solutions for the doubly singular three-point boundary value problem with derivative dependent data function −(𝑝(𝑡)𝑦′(𝑡))′=𝑞(𝑡)𝑓(𝑡,𝑦(𝑡),𝑝(𝑡)𝑦′(𝑡)),0<𝑡<1,𝑦(0)=0,𝑦(1)=𝛼₁𝑦(𝜂). Here, 𝑝∈𝐶[0,1]∩𝐶¹(0,1] with 𝑝(𝑡)>0 on (0,1] and 𝑞(𝑡) is allowed to be discontinuous at 𝑡=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: −𝑝(𝑡)𝑦^(𝑡)^=𝑞(𝑡)𝑓𝑡,𝑦(𝑡),𝑝(𝑡)𝑦^(𝑡),0<𝑡<1,(1.1) with boundary conditions 𝑦(0)=0,𝑦(1)=𝛼₁𝑦(𝜂),(1.2) where 0<𝛼₁<(ℎ(1))/(ℎ(𝜂)) and ℎ(𝑡)=∫^𝑡0(1/𝑝(𝑥))𝑑𝑥.

Throughout this paper, we assume the following conditions on the functions 𝑝(𝑡), 𝑞(𝑡), and 𝑓(𝑡,𝑦,𝑝𝑦′):(E1)𝑝∈𝐶[0,1]∩𝐶¹(0,1] with 𝑝(𝑡)>0 on (0,1] and 1/𝑝∈𝐿¹(0,1);(E2)𝑞(𝑡)>0, 𝑞 is not identically zero on [0,1] and 𝑞∈𝐿¹(0,1);(E3)𝑓∈𝐶([0,1]×[0,∞)×ℝ,[0,∞)) and 𝑓 is not identically zero.Note that condition (E2) allows 𝑞(𝑡) be discontinuous at 𝑡=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(𝑡,𝑦,𝑧) and [8, 10, 12] for derivative independent data function 𝑓(𝑡,𝑦,𝑧)=𝑓(𝑡,𝑦).

Some attention has been devoted to singular multipoint boundary value problems (see [17, 18] and the references therein). When 𝑝(𝑡)=1 and 𝑞(𝑡)𝑓(𝑡,𝑦,𝑧) may have singularity at 𝑡=0,𝑡=1, 𝑦=0 and 𝑧=0, differential equation (1.1) with boundary conditions 𝑦′(0)=0,𝑦(1)=𝛼₁𝑦(𝜂) 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 𝑞(𝑡) and 𝑓(𝑡,𝑦,𝑧).

Bai and Ge [19] have generalized the Leggett-Williams fixed point theory and applied to 𝑦^+𝑞(𝑦𝑡)(𝑓0)=𝑡,𝑦0,,𝑦^=𝑦(01,)=𝑡0∈,(0,1),(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(𝑡,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 𝑦^+𝑞(𝑡)𝑓𝑡,𝑦,𝑦^=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: −𝑡^𝑟𝑦^(𝑡)^=𝑡^𝑟−12(𝑟+1)𝑡−5𝑦(0)=0,𝑦4(1𝑟)=,𝑦0.51≤𝑟<1,𝑡∈(0,1),4.(1.5) The problem (1.5) can be reduced to the following boundary value problem: −𝑣^(𝑥)=1(1−𝑟)²𝑥^{(2𝑟−1)/(1−𝑟)}2(𝑟+1)𝑥^1/(1−𝑟)−5𝑣(0)=0,𝑣(1)=𝑣14𝑟,𝑥∈(0,1),4^(1−𝑟),(1.6) by change of variable 𝑥=(1−𝑟)∫^𝑡0𝜏^−𝑟𝑑𝜏=𝑡^1−𝑟.

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(𝑡) with sup_[]𝑡∈0,1||𝑦(𝑡)||≤3234(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟}34𝑟+2.(1.7)

Further, for 𝑝(𝑡)=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(𝑡)=1 to the reduced problem (1.6), we get that the problem (1.5) has at least one nonnegative solution 𝑦(𝑡) with sup_[]𝑡∈0,1||𝑦(𝑡)||=sup_[]𝑡∈0,1||𝑣(𝑡)||≤34−4^𝑟34(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟}43𝑟+2.(1.8)

Now as 𝑟 approaches to one, that is, the order of singularity increases, the upper bound for sup_𝑡∈[0,1]|𝑦(𝑡)| 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 sup_𝑡∈[0,1]|𝑦(𝑡)| 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.

Figure 1 

Variation of bounds for 𝑦 in both cases.

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 ∃ a continuous map 𝑟∶𝐸→𝐷 such that 𝑟(𝑥)=𝑥 for every 𝑥∈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, 𝑃⊂𝐸 is nonempty convex, closed set, 𝑃 is said to be cone provided that(1)𝜆𝑢∈𝑃 for all 𝜆≥0,𝑢∈𝑃, and(2)𝑢∈𝑃,−𝑢∈𝑃 implies 𝑢=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 𝑌, a subset Ω of 𝑋, and a map 𝑇∶Ω→𝑌. Then 𝑇 is said to be completely continuous operator if(1)𝑇 is continuous, and(2)𝑇 maps bounded subset of Ω into relatively compact sets.

Definition 2.6. The map 𝛼 is said to be a nonnegative continuous convex functional on 𝑃 provided that 𝛼∶𝑃→[0,∞) is continuous and 𝛼(𝑡𝑥+(1−𝑡)𝑦)≤𝑡𝛼(𝑥)+(1−𝑡)𝛼(𝑦),(2.1) for all 𝑥,𝑦∈𝑃 and 0≤𝑡≤1. Similarly, the map 𝛾 is said to be a nonnegative continuous concave functional on 𝑃 provided that 𝛾∶𝑃→[0,∞) is continuous and 𝛾(𝑡𝑥+(1−𝑡)𝑦)≥𝑡𝛾(𝑥)+(1−𝑡)𝛾(𝑦),(2.2) for all 𝑥,𝑦∈𝑃, and 0≤𝑡≤1.

Definition 2.7. Suppose 𝛼,𝛽∶𝑃→[0,∞) are two continuous convex functionals satisfying ‖𝑥‖≤𝑀max{𝛼(𝑥),𝛽(𝑥)},for𝑥∈𝑃,(2.3) where 𝑀 is positive constant, and Ω={𝑥∈𝑃∶𝛼(𝑥)<𝑟,𝛽(𝑥)<𝐿}≠𝜙for𝑟>0,𝐿>0.(2.4) From (2.3) and (2.4), Ω is a bounded nonempty open subset of 𝑃.

Definition 2.8. Let 𝑟>𝑎>0, 𝐿>0 be given constants, 𝛼,𝛽∶𝑃→[0,∞) two nonnegative continuous convex functionals satisfying (2.3) and (2.4), and 𝜓 a nonnegative continuous concave functional on the cone 𝑃. Then bounded convex sets are defined as 𝑃𝛼^𝑟,𝛽^𝐿={𝑥∈𝑃∶𝛼(𝑥)<𝑟,𝛽(𝑥)<𝐿},𝑃𝛼𝑃𝛼^𝑟,𝛽^𝐿={𝑥∈𝑃∶𝛼(𝑥)≤𝑟,𝛽(𝑥)≤𝐿},_𝑟,𝛽^𝐿,𝜓_𝑎={𝑥∈𝑃∶𝛼(𝑥)<𝑟,𝛽(𝑥)<𝐿,𝜓(𝑥)>𝑎},𝑃𝛼^𝑟,𝛽^𝐿,𝜓_𝑎={𝑥∈𝑃∶𝛼(𝑥)≤𝑟,𝛽(𝑥)≤𝐿,𝜓(𝑥)≥𝑎}.(2.5)

Theorem 2.9 (see [20]). Let 𝑋 be retract of real Banach space 𝐸. Then for every bounded relatively open subset 𝑈 of 𝑋 and every completely continuous operator 𝑇∶𝑈→𝑋 which has no fixed point on 𝜕𝑈 (relative to 𝑋), there exists an integer 𝑖(𝑇,𝑈,𝑋) such that if 𝑖(𝑇,𝑈,𝑋)≠0, then 𝑇 has at least one fixed point in 𝑈. Moreover, 𝑖(𝑇,𝑈,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 𝑈⊂𝑋 nonempty open subset, such that 𝑈⊂𝑋₁. If 𝑇∶𝑋₁→𝑋 is completely continuous, 𝑇(𝑋₁)⊂𝑋₁, such that there is no fixed point of 𝑇 in 𝑋₁⧵𝑈, then 𝑖(𝑇,𝑈,𝑋)=1.

Theorem 2.11 (see [19]) (fixed point theorem of functional type). Let 𝐸 be Banach space, 𝑃⊂𝐸 a cone, and 𝑐≥𝑏>𝑎>𝑑>0,𝐿₂≥𝐿₁>0 given constants. Assume that 𝛼,𝛽 are nonnegative continuous convex functionals on 𝑃 such that (2.3) and (2.4) are satisfied. 𝜓 is a nonnegative continuous concave functional on 𝑃 such that 𝜓(𝑥)≤𝛼(𝑥) for all 𝑥∈𝑃(𝛼^𝑐,𝛽^𝐿2) and let 𝑇∶𝑃(𝛼^𝑐,𝛽^𝐿2)→𝑃(𝛼^𝑐,𝛽^𝐿2) be a completely continuous operator. Suppose that(1){𝑥∈𝑃(𝛼^𝑏,𝛽^𝐿2,𝜓_𝑎)∶𝜓(𝑥)>𝑎}≠Φ and 𝜓(𝑇𝑥)>𝑎 for 𝑥∈𝑃(𝛼^𝑏,𝛽^𝐿2,𝜓_𝑎),(2)𝛼(𝑇𝑥)<𝑑, 𝛽(𝑇𝑥)<𝐿₁ for all 𝑥∈𝑃(𝛼^𝑑,𝛽^𝐿1),(3)𝜓(𝑇𝑥)>𝑎 for all 𝑥∈𝑃(𝛼^𝑐,𝛽^𝐿2,𝜓_𝑎) with 𝛼(𝑇𝑥)>𝑏.Then 𝑇 has at least three fixed points 𝑥₁,𝑥₂,𝑥₃∈𝑃(𝛼^𝑐,𝛽^𝐿2) such that 𝑥₁∈𝑃𝑥𝛼^𝑑,𝛽^𝐿1,𝑥₂∈𝑥∈𝑃𝛼^𝑐,𝛽^𝐿2,𝜓_𝑎∶𝜓(𝑥)>𝑎₃,∈𝑃𝛼^𝑐,𝛽^𝐿2⧵𝑃𝛼^𝑐,𝛽^𝐿2,𝜓_𝑎∪𝑃𝛼^𝑑,𝛽^𝐿1.(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: 𝐵.𝐶.∶=𝑦(0)=0,−𝑦𝑝(1(𝑡))=𝑦^𝛼(𝑡₁)′=0𝑦(𝜂)⎫⎪⎬⎪⎭,𝑡∈(0,1)(3.1) is given by 𝐺(𝑡,𝑠)=⎧⎪⎨⎪⎩𝐺¹𝐺(𝑡,𝑠),0≤𝑠≤min{𝑡,𝜂}<1;₂𝐺(𝑡,𝑠),0≤𝑡≤𝑠≤𝜂;₃𝐺(𝑡,𝑠),𝜂≤𝑠≤t≤1;₄(𝑡,𝑠),0<max{𝜂,𝑡}≤𝑠≤1.(3.2) Here 𝐺₁(𝑡,𝑠)=ℎ(𝑠)𝛿𝛿−ℎ(𝑡)+𝛼¹𝐺ℎ(𝑡)₂,(𝑡,𝑠)=ℎ(𝑡)𝛿𝛿−ℎ(𝑠)+𝛼¹𝐺ℎ(𝑠)₃,(𝑡,𝑠)=1𝛿ℎ(𝑡)𝛼¹𝐺ℎ(𝜂)−ℎ(𝑠)+𝛿ℎ(𝑠)₄,(𝑡,𝑠)=ℎ(𝑡)𝛿=ℎ(1)−𝛼₁𝛿[ℎ(1)−ℎ(𝑠)],ℎ(𝜂)>0,ℎ(𝑡)=^𝑡01𝑝(𝑥)𝑑𝑥,0<𝛼¹<ℎ(1)ℎ(𝜂).(3.3)

Proof. Consider the following linear differential equation: −𝑝(𝑡)𝑦^(𝑡)^=𝑞(𝑡)𝐹(𝑡),(3.4) where 𝐹∈𝐶([0,1],[0,∞)). Integrating the above differential equation twice first from 𝑡 to 1 and then from 0 to 𝑡, changing the order of integration, and applying the boundary conditions, we get 𝑦(𝑡)=+^𝑡0ℎℎ((𝑠𝑡))𝑞(𝑠)𝐹(𝑠)𝑑𝑠+ℎ(𝑡)^1𝑡𝑞(𝑠)𝐹(𝑠)𝑑𝑠𝛿𝛼¹^𝜂0ℎ(𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠−¹⁰ℎ(𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+𝛼₁ℎ(𝜂)^1𝜂𝑞(𝑠)𝐹(𝑠)𝑑𝑠.(3.5) For 𝑡∈[0,𝜂], 𝑦(𝑡) can be written as 𝑦(𝑡)=^𝑡0ℎ(𝑠)𝛿𝛿−ℎ(𝑡)+𝛼¹+ℎ(𝑡)𝑞(𝑠)𝐹(𝑠)𝑑𝑠_𝜂𝑡ℎ(𝑡)𝛿𝛿−ℎ(𝑠)+𝛼¹ℎ(𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^1𝜂ℎ(𝑡)𝛿[ℎ(1)−ℎ(𝑠)]𝑞(𝑠)𝐹(𝑠)𝑑𝑠,(3.6) or 𝑦(𝑡)=^𝑡0𝐺₁(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^𝜂𝑡𝐺₂(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^1𝜂𝐺₄(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠.(3.7) Similarly, for 𝑡∈[𝜂,1], 𝑦(𝑡) can be written as 𝑦(𝑡)=^𝜂0ℎ(𝑠)𝛿𝛿−ℎ(𝑡)+𝛼¹+ℎ(𝑡)𝑞(𝑠)𝐹(𝑠)𝑑𝑠_𝑡𝜂1𝛿ℎ(𝑡)𝛼¹ℎ(𝜂)−ℎ(𝑠)+𝛿ℎ(𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^1𝑡ℎ(𝑡)𝛿[ℎ(1)−ℎ(𝑠)]𝑞(𝑠)𝐹(𝑠)𝑑𝑠,(3.8) or 𝑦(𝑡)=^𝜂0𝐺₁(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^𝑡𝜂𝐺₃(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠+^1𝑡𝐺₄(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠.(3.9) From (3.7) and (3.9), we may write 𝑦(𝑡)=¹⁰𝐺(𝑡,𝑠)𝑞(𝑠)𝐹(𝑠)𝑑𝑠,(3.10) where 𝐺(𝑡,𝑠) is given in the lemma. It is easy to see that 𝐺(𝑡,𝑠) satisfies all the properties of Green’s function. Hence 𝐺(𝑡,𝑠) is the Green’s function for the boundary value problem (3.1).

Lemma 3.2. The Green’s function 𝐺(𝑡,𝑠) satisfies the following properties:(i)max_𝑡∈[0,1]𝑝(𝑡)(𝜕𝐺(𝑡,𝑠)/𝜕𝑡)<∞,(ii)𝐺(𝑡,𝑠)≥0 for all (𝑡,𝑠)∈{[0,1]×[0,1]},(iii)there exist a constant 𝜆 in (0,1) such that min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)≥𝜆max_𝑡∈[0,1]𝐺(𝑡,𝑠)for𝑠∈[0,1], where 𝜆=⎧⎪⎪⎨⎪⎪⎩min𝛼¹[ℎ(1)−ℎ(𝜂)]ℎ(1)−𝛼₁ℎ(𝜂),𝛼¹ℎ(𝜂)ℎ(1),0<𝛼¹minℎ(1)−𝛼₁≤1,ℎ(𝜂)𝛼₁[ℎ(1)−ℎ(𝜂)],𝛼ℎ₁ℎ(𝜂(1)),1≤𝛼¹<ℎ(1)ℎ(𝜂).(3.11)

Proof. (i) 𝑝(𝑡)𝜕𝐺(𝑡,𝑠)𝜕𝑡=⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ℎ(𝑠)𝛿𝛼¹1−1,0≤𝑠≤min{𝑡,𝜂}<1;𝛿𝛿−ℎ(𝑠)−𝛼¹1ℎ(𝑠),0≤𝑡≤𝑠≤𝜂;𝛿𝛼¹1ℎ(𝜂)−ℎ(𝑠),𝜂≤𝑠≤𝑡≤1;𝛿[ℎ(1)−ℎ(𝑠)],0<max{𝜂,𝑡}≤𝑠≤1.(3.12) Since 𝑝(𝑡)(𝜕𝐺(𝑡,𝑠)/𝜕𝑡) is independent of 𝑡, therefore max_𝑡∈[0,1]𝑝(𝑡)(𝜕𝐺(𝑡,𝑠)/𝜕𝑡)<∞.
(ii) For 𝑡<𝜂, 𝛼₁[ℎ(𝑡)−ℎ(𝜂)]>ℎ(1)/ℎ(𝜂)[ℎ(𝑡)−ℎ(𝜂)] and 𝐺₁(𝑡,𝑠)≥1𝛿ℎ(𝑠)ℎ(𝑡)ℎℎ((1𝜂))−1≥0,(3.13) it is easy to show that 𝐺₁(𝑡,𝑠)≥0, for 𝑡≥𝜂.
Next we show 𝐺₂(𝑡,𝑠)≥0, 𝐺₃(𝑡,𝑠)≥0 and 𝐺₄(𝑡,𝑠)≥0 as follows: 𝐺₂(𝑡,𝑠)≥ℎ(𝑡)𝐺𝛿ℎ(𝜂)[ℎ(𝑠)(ℎ(1)−ℎ(𝜂))]≥0;₃(𝑡,𝑠)=1𝛿ℎ(𝑠){ℎ(1)−ℎ(𝑡)}+𝛼¹𝐺ℎ(𝜂){ℎ(𝑡)−ℎ(𝑠)}≥0,₄(𝑡,𝑠)=ℎ(𝑡)𝛿[ℎ(1)−ℎ(𝑠)]≥0.(3.14) Thus 𝐺(𝑡,𝑠)≥0 for all (𝑡,𝑠)∈{[0,1]×[0,1]}.
(iii) We prove the inequality for the following cases:
(a) 𝑠∈[0,𝜂] and (b) 𝑠∈[𝜂,1],
(a) for 𝑠∈[0,𝜂], we further divide this case in two parts as follows.
(1) When 0<𝛼₁≤1, (2) when 1≤𝛼₁<ℎ(1)/ℎ(𝜂).
Case  1 (For 0<𝛼₁≤1). It is easy to see that 𝐺₁(𝑡,𝑠)=ℎ(𝑠)𝛿ℎ(1)−𝛼¹ℎ(𝜂)+ℎ(𝑡)𝛼₁−1(3.15) implies max_[]𝑡∈0,1𝐺₁(𝑡,𝑠)=ℎ(𝑠)𝛿ℎ(1)−𝛼¹ℎ(𝜂),min_{[]𝑡∈𝜂,1}𝐺₁(𝑡,𝑠)=𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)].(3.16) Next,𝐺₂(𝑡,𝑠)=ℎ(𝑡)𝛿ℎ(1)−𝛼¹≤ℎ(𝑠)ℎ(𝜂)+ℎ(𝑠)𝛼₁−1,𝛿ℎ(1)−𝛼¹ℎ(𝜂),asℎ(𝑡)≤ℎ(𝑠).(3.17) Thus for 0<𝛼₁≤1, max_[]𝑡∈0,1𝐺(𝑡,𝑠)=ℎ(𝑠)𝛿ℎ(1)−𝛼¹ℎ(𝜂),min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)=𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)].(3.18)Case  2 (For 1≤𝛼₁<ℎ(1)/ℎ(𝜂)). It is easy to see that max_[]𝑡∈0,1𝐺₁(𝑡,𝑠)=𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],min^{𝑡∈[𝜂,1]}𝐺₁(𝑡,𝑠)=ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)].(3.19) Next, 𝐺₂(𝑡,𝑠)=ℎ(𝑡)𝛿ℎ(1)−𝛼¹≤𝛼ℎ(𝜂)+ℎ(𝑠)𝛼₁−1₁,ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],asℎ(𝑡)≤ℎ(𝑠)≤ℎ(𝜂).(3.20) Thus for 1≤𝛼₁<ℎ(1)/ℎ(𝜂), max_[]𝑡∈0,1𝐺(𝑡,𝑠)=𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],min^{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)=ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)].(3.21) Combining (3.18) and (3.21), we may write for 𝑠∈[0,𝜂], max_[]𝑡∈0,1𝐺(𝑡,𝑠)=⎧⎪⎨⎪⎩ℎ(𝑠)𝛿ℎ(1)−𝛼¹𝛼ℎ(𝜂),0<𝛼₁≤1,₁ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],1≤𝛼¹<ℎ(1)min_{𝑡∈[𝜂,1]}ℎ(𝜂),𝐺(𝑡,𝑠)=⎧⎪⎨⎪⎩𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],0<𝛼¹ℎ(𝑠)≤1,𝛿[ℎ(1)−ℎ(𝜂)],1≤𝛼¹<ℎ(1)ℎ(𝜂).(3.22) (b) For 𝑠∈[𝜂,1]. For this case, 𝐺₃(𝑡,𝑠) and 𝐺₄(𝑡,𝑠) are considered. From (3.2), it can be easily seen that for 𝑠∈[𝜂,1], max_[]𝑡∈0,1𝐺(𝑡,𝑠)=ℎ(1)𝛿[ℎ(1)−ℎ(𝑠)],min^{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)=ℎ(𝜂)𝛿[ℎ(1)−ℎ(𝑠)].(3.23) Thus from (3.22) and (3.23), we get max_[]𝑡∈0,1𝐺(𝑡,𝑠)=⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑠∈[0,𝜂]∶⎧⎪⎪⎨⎪⎪⎩ℎ(𝑠)𝛿ℎ(1)−𝛼¹𝛼ℎ(𝜂),0<𝛼₁≤1,₁ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],1≤𝛼¹<ℎ(1)𝑠∈[𝜂,1]∶ℎ(1)ℎ(𝜂),𝛿[ℎ(1)−ℎ(𝑠)],0<𝛼¹<ℎ(1)min_{𝑡∈[𝜂,1]}ℎ(𝜂).𝐺(𝑡,𝑠)=⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑠∈[0,𝜂]∶⎧⎪⎨⎪⎩𝛼¹ℎ(𝑠)𝛿[ℎ(1)−ℎ(𝜂)],0<𝛼¹ℎ(𝑠)≤1,𝛿[ℎ(1)−ℎ(𝜂)],1≤𝛼¹<ℎ(1)𝑠∈[𝜂,1]∶ℎ(𝜂)ℎ(𝜂),𝛿[ℎ(1)−ℎ(𝑠)],0<𝛼¹<ℎ(1)ℎ(𝜂).(3.24) From (3.24), min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)≥⎧⎪⎪⎨⎪⎪⎩min𝛼¹[ℎ(1)−ℎ(𝜂)]ℎ(1)−𝛼₁ℎ(𝜂),𝛼¹ℎ(𝜂)ℎ(1)max^[]𝑡∈0,1minℎ(1)−𝛼₁𝐺(𝑡,𝑠),0<𝛼₁≤1,ℎ(𝜂)𝛼₁[ℎ(1)−ℎ(𝜂)],𝛼ℎ₁ℎ(𝜂(1))max^[]𝑡∈0,1𝐺(𝑡,𝑠),1≤𝛼₁<ℎ(1)ℎ(𝜂).(3.25) Consequently, setting 𝜆=⎧⎪⎪⎨⎪⎪⎩min𝛼¹[ℎ(1)−ℎ(𝜂)]ℎ(1)−𝛼₁ℎ(𝜂),𝛼¹ℎ(𝜂)ℎ(1),0<𝛼¹minℎ(1)−𝛼₁≤1,ℎ(𝜂)𝛼₁[ℎ(1)−ℎ(𝜂)],𝛼ℎ₁ℎ(𝜂(1)),1≤𝛼¹<ℎ(1)ℎ(𝜂),(3.26) there holds min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)≥𝜆max_[]𝑡∈0,1𝐺(𝑡,𝑠).(3.27) It can be easily seen that 0<𝜆<1. This completes the proof.

4. Existence of Multiple Nonnegative Solutions

Let 𝑋=𝐶[0,1]⋂𝐶²(0,1] be endowed with ordering 𝑥≤𝑦 if 𝑥(𝑡)≤𝑦(𝑡) for all 𝑡∈[0,1] and ‖𝑥‖=max{‖𝑥‖₁,‖𝑥′‖₁}, where ‖𝑥′‖‖𝑥₁‖₁=sup_𝑡∈[0,1]||𝑥(𝑡)||,=sup_𝑡∈(0,1]||𝑝(𝑡)𝑥^(𝑡)||.(4.1) Let 𝐸={𝑥∶𝑥∈𝑋,‖𝑥‖<∞} be bounded subset of 𝑋. 𝐸 is Banach Space.

Now define a cone 𝑃⊂𝐸 as 𝑃=𝑥∈𝐸∶𝑥(𝑡)≥0,min_{𝑡∈[𝜂,1]}𝑥(𝑡)≥𝜆max_[]𝑡∈0,1𝑥(𝑡),𝑝(𝑡)𝑥^(𝑡)^≤0.(4.2) The boundary value problem (1.1)-(1.2) has a solution 𝑦(𝑡) if and only if 𝑦(𝑡) solves the following operator equation: 𝑦(𝑡)=𝑇𝑦(𝑡),(4.3) where the operator 𝑇∶𝑃→𝑃 is given by (𝑇𝑦)(𝑡)=¹⁰𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,0≤𝑡≤1.(4.4) Here 𝐺(𝑡,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 𝑇∶𝑃→𝑃 is well defined and is completely continuous.

Proof. First we show that the operator 𝑇 is well defined. For this, we take 𝑦∈𝑃. From (E2), (E3), and 𝐺(𝑡,𝑠)≥0, it follows that 𝑇𝑦(𝑡)≥0.
Now applying Lemma 3.2, we get min_{𝑡∈[𝜂,1]}𝑇𝑦(𝑡)=min_{𝑡∈[𝜂,1]}¹⁰min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,𝑇𝑦(𝑡)≥=𝜆𝜆m¹⁰axm_[]𝑡∈0,1ax_[]𝑡∈0,1𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(s)𝑦^(𝑠)𝑑𝑠,¹⁰=𝜆max_[]𝑡∈0,1𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,(𝑇𝑦)(𝑡).(4.5) It is easy to show that (𝑝(𝑡)𝑇𝑦′(𝑡))′≤0. Thus 𝑇 is well defined.
We now show that 𝑇 is completely continuous. Let {𝑦_𝑛} be a sequence in 𝑃 and 𝑦₀∈𝑃 with lim_𝑛→∞𝑦_𝑛=𝑦₀. Then, there exists a constant 𝑘₁>0 such that ‖𝑦_𝑛‖<𝑘₁ for all 𝑛∈ℕ∪{0}. Thus ‖𝑦_𝑛−𝑦₀‖→0 as 𝑛→∞ implies sup_𝑡∈[0,1]|(𝑦_𝑛−𝑦₀)(𝑡)|andsup_𝑡∈(0,1]|𝑝(𝑡)(𝑦_𝑛−𝑦₀)′(𝑡)|→0 as 𝑛→∞. So 𝑦_𝑛(𝑡)→𝑦₀(𝑡) and 𝑝(𝑡)𝑦^𝑛(𝑡)→𝑝(𝑡)𝑦₀′(𝑡) as 𝑛→∞.
Since 𝑓 is continuous on {[0,1]×[0,𝑘₁]×[−𝑘₁,𝑘₁]}, so ||𝑇𝑦_𝑛(𝑡)−𝑇𝑦₀(𝑡)||=⟹⟶||||¹⁰0‖‖𝑇𝐺𝑦a(_𝑛s𝑡,𝑛𝑠)⟶𝑞(𝑠)∞𝑓.𝑠,𝑦_𝑛(𝑠),𝑝(𝑠)𝑦^𝑛(𝑠)−𝑓𝑠,𝑦₀(𝑠),𝑝(𝑠)𝑦^0(𝑠)𝑑𝑠||||||𝑝(𝑡)𝑇𝑦_𝑛−𝑇𝑦₀‖‖₁⟶0as𝑛⟶∞.(4.6)(𝑡)−𝑇𝑦^0(𝑡)||=||||𝑝(𝑡)¹⁰𝜕𝐺(𝑡,𝑠)𝜕𝑡𝑞(𝑠)𝑓𝑠,𝑦^𝑛−𝑓𝑠,𝑦₀(𝑠),𝑝(𝑠)𝑦^𝑛(𝑠)⟹⟶‖‖0𝑇a𝑦s_𝑛(𝑛𝑠)⟶,𝑝(𝑠∞)𝑦.^0(𝑠)𝑑𝑠||||−𝑇𝑦₀^‖‖₁⟶0as𝑛⟶∞.(4.7) From (4.6) and (4.7), ‖‖𝑇𝑦_𝑛−𝑇𝑦₀‖‖⟶0as𝑛⟶∞.(4.8) Hence 𝑇∶𝑃→𝑃 is a continuous operator.
Next we prove that 𝑇 maps every bounded subset of 𝑃 into relatively compact set. Let 𝐵={𝑥∈𝑃∶‖𝑥‖≤𝑘₂,𝑘₂isapositiveconstant} be any bounded subset of 𝑃. For 𝑦∈𝐵, 𝑇𝑦(𝑡)=≤m¹⁰a𝐺x_[]𝑡∈0,1(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠¹⁰≤𝐺ma(𝑡x,𝑠)𝑞(𝑠)𝑓_{[]×(𝑠,𝑢,𝑣)∈0,10,𝑘2×−𝑘2,𝑘2}𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠𝑓(𝑠,𝑢,𝑣)max_[]𝑡∈0,1¹⁰⟹𝑇𝑦(𝑡)<∞.𝐺(𝑡,𝑠)𝑞(𝑠)𝑑𝑠(4.9) Therefore 𝑇(𝐵) is uniformly bounded. Further, equicontinuity of 𝑇(𝐵) follows from ||𝑇𝑦𝑡₁−𝑇𝑦𝑡₂||=||||¹⁰⟶0𝐺as(𝑡𝑡,₁𝑠)||^𝑡=𝑡₁−𝐺(𝑡,𝑠)||_𝑡=𝑡2𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠||||||𝑝(𝑡)𝑇𝑦_⟶𝑡₂,𝑡₁−𝑇𝑦^𝑡₂||=≤2||||𝑝m(𝑡a)x¹⁰|||𝜕𝐺𝜕𝐺(𝑡𝜕,(𝑠𝑡𝑡,)𝑠)|||_𝑡=𝑡1−𝜕𝐺𝜕(𝑡𝑡,𝑠)|||_𝑡=𝑡2𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠||||𝜕𝑡|||⋅sup_[]𝑡∈0,1⟶0as𝑡₁𝑝(𝑡)^𝑡2_𝑡1𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,⟶𝑡₂.(4.10) Thus from Arzela-Ascoli Theorem, 𝑇(𝐵) is relatively compact subset of 𝑃 and also 𝑇∶𝑃→𝑃 is completely continuous.

Next, define functionals 𝛼,𝛽,𝜓∶𝑃→[0,∞) such that 𝛽(𝛼𝑦)(𝑦=)=supsu_𝑡∈(0,1]p_𝑡∈[0,1]||𝑦(𝑡)||,𝜓(𝑦)=mi||n𝑝_{𝑡∈[𝜂,1]}(𝑡)𝑦^(𝑡)||,||𝑦(𝑡)||.(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 𝜓 is nonnegative concave functional with 𝜓(𝑦)≤𝛼(𝑦).

Let 𝐶=min_{𝑡∈[𝜂,1]}^1𝜂𝐿=sup_𝑡∈[0,1]𝐺(𝑡,𝑠)𝑞(𝑠)𝑑𝑠,¹⁰𝑁=sup_𝑡∈(0,1]𝐺(𝑡,𝑠)𝑞(𝑠)𝑑𝑠,||||¹⁰𝑝(𝑡)𝜕𝐺(𝑡,𝑠)𝜕𝑡𝑞(𝑠)𝑑𝑠||||.(4.12) Now we state the main results of this work.

Theorem 4.2. Suppose that (E1)–(E3) are satisfied and 𝑓(𝑡,𝑦,𝑧) satisfies the following condition.(H₁) if there exist real constants 𝑑>0 and 𝐿₁>0 such that 𝑓(𝑡,𝑦,𝑧)<min{𝑑/𝐿,𝐿₁/𝑁} for (𝑡,𝑦,𝑧)∈{[0,1]×[0,𝑑]×[−𝐿₁,𝐿₁]},then boundary value problem (1.1)-(1.2) has at least one nonnegative solution 𝑦₁ such that sup_𝑡∈[0,1]|𝑦₁(𝑡)|<𝑑 with sup_𝑡∈(0,1]|𝑝(𝑡)𝑦^1(𝑡)|<𝐿₁.

Theorem 4.3. Suppose that (E1)–(E3) are satisfied. There exist real constants 𝑎,𝑐,𝑑,𝐿₁,and𝐿₂ with 𝑐≥𝑎/𝜆=𝑏>𝑎>𝑑>0, 𝐿₂≥𝐿₁>0 such that 𝑎/𝐶≤min{𝑐/𝐿,𝐿₂/𝑁} and 𝑓(𝑡,𝑦,𝑧) satisfies following conditions.(H₁)𝑓(𝑡,𝑦,𝑧)<min{𝑑/𝐿,𝐿₁/𝑁} for (𝑡,𝑦,𝑧)∈{[0,1]×[0,𝑑]×[−𝐿₁,𝐿₁]},(H₂)𝑓(𝑡,𝑦,𝑧)>𝑎/𝐶 for (𝑡,𝑦,𝑧)∈{[𝜂,1]×[𝑎,𝑎/𝜆]×[−𝐿₂,𝐿₂]},(H₃)𝑓(𝑡,𝑦,𝑧)≤min{c/L,L₂/N} for (𝑡,𝑦,𝑧)∈{[0,1]×[0,𝑐]×[−𝐿₂,𝐿₂]}.Then boundary value problem (1.1)-(1.2) has at least three nonnegative solutions 𝑦₁, 𝑦₂, and 𝑦₃ in 𝑃(𝛼^𝑐,𝛽^𝐿2) such that sup𝑎_𝑡∈[0,1]||𝑦<₁(m𝑡)i||n<_{𝑡∈[𝜂,1]}𝑑,sup_𝑡∈(0,1]||𝑝(𝑡)𝑦^1(𝑡)||<𝐿₁,sup_𝑡∈(0,1]||𝑦₂(𝑡)||≤sup_[]𝑡∈0,1||𝑦₂(𝑡)||≤𝑐,||𝑝(𝑡)𝑦^2(𝑡)||≤𝐿₂,sup_[]𝑡∈0,1||𝑦₃(𝑡)||≤𝑎𝜆withsup_]𝑡∈(0,1||𝑝(𝑡)𝑦^3(𝑡)||≤𝐿₂.(4.13)

Proof of Theorem 4.2. Let 𝑈=𝑃(𝛼^𝑑,𝛽^𝐿1) be open subset of 𝑃. We now show that 𝑇(𝑈)⊂𝑈. For 𝑦∈𝑈, 𝛼(𝑇𝑦)==ssuupp_{[]𝑡∈0,1[]𝑡∈0,1}||𝑇𝑦(𝑡)||¹⁰<𝑑𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,𝐿sup_[]𝑡∈0,1¹⁰𝐺(𝑡,𝑠)𝑞(𝑠)𝑑𝑠<𝑑,fromconditionH₁(4.14) implies that 𝛼(𝑇𝑦)<𝑑.
Consider that 𝛽(𝑇𝑦)==ssuupp_{𝑡∈(0,1]𝑡∈(0,1]}||𝑝(𝑡)𝑇𝑦^(𝑡)||,||||𝑝(𝑡)¹⁰𝜕𝐺(𝑡,𝑠)𝜕𝑡𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^<𝐿(𝑠)₁||||,𝑁sup_]𝑡∈(0,1||||¹⁰𝑝(𝑡)𝜕𝐺(𝑡,𝑠)𝜕𝑡𝑞(𝑠)𝑑𝑠||||,fromconditionH¹,(4.15) implies that 𝛽(𝑇𝑦)<𝐿₁.
Thus 𝑇(𝑈)⊂𝑈. Next, we show that 𝑇 has no fixed point on 𝜕𝑈(=𝑈⧵𝑈). On contrary, suppose there exists a fixed point 𝑦(𝑡) on 𝜕𝑈 such that 𝑇𝑦(𝑡)=𝑦(𝑡). 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 𝜕𝑈 and from Theorem 2.10𝑖(𝑇,𝑈,𝑃)=1. Thus the operator 𝑇 has at least one fixed point in 𝑈 and also the boundary value problem (1.1)-(1.2) has at least one nonnegative solution 𝑦₁ such that sup_𝑡∈[0,1]|𝑦₁(𝑡)|<𝑑 with sup_𝑡∈(0,1]|𝑝(𝑡)𝑦^1(𝑡)|<𝐿₁.

Proof of Theorem 4.3. It is easy to see that 𝜓(𝑦)≤𝛼(𝑦) for each 𝑦∈𝑃(𝛼^𝑐,𝛽^𝐿2). We now show that 𝑇∶𝑃(𝛼^𝑐,𝛽^𝐿2)→𝑃(𝛼^𝑐,𝛽^𝐿2) is well defined. For 𝑦∈𝑃(𝛼^𝑐,𝛽^𝐿2), 𝛼(𝑇𝑦)==ssuupp_{𝑡∈[0,1][]𝑡∈0,1}||𝑇𝑦(𝑡)||,¹⁰≤𝑐𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠,𝐿sup_[]𝑡∈0,1¹⁰𝛽(𝑇𝑦)=≤𝑐s.up_𝑡∈(0,1]𝐺(𝑡,𝑠)𝑞(𝑠)𝑑𝑠,fromassumptionH₃,(4.16)=sup_𝑡∈(0,1]||𝑝(𝑡)𝑇𝑦^(𝑡)||,||||𝑝(𝑡)¹⁰𝜕𝐺(𝑡,𝑠)𝜕𝑡𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^≤𝐿(𝑠)₂||||,𝑁sup_]𝑡∈(0,1||||¹⁰𝑝(𝑡)𝜕𝐺(𝑡,𝑠)≤𝐿₂𝜕𝑡𝑞(𝑠)𝑑𝑠||||,fromassumptionH³,.(4.17) From (4.16) and (4.17), 𝑇(𝑦)∈𝑃𝛼^𝑐,𝛽^𝐿2.(4.18) Thus, 𝑇∶𝑃(𝛼^𝑐,𝛽^𝐿2)→𝑃(𝛼^𝑐,𝛽^𝐿2) 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(𝑡)=𝑎/𝜆∈𝑃, 0≤𝑡≤1, then 𝛼(𝑦)=sup_𝑡∈[0,1]|𝑦(𝑡)|=𝑎/𝜆,𝛽(𝑦)=sup_𝑡∈(0,1]|𝑝(𝑡)𝑦′(𝑡)|=0<𝐿₂, 𝜓(𝑥)=min_{𝑡∈[𝜂,1]}|𝑦(𝑡)|=𝑎/𝜆>𝑎. Thus, {𝑦∈𝑃(𝛼^{𝑏=𝑎/𝜆},𝛽^𝐿2,Ψ_𝑎)∶Ψ(𝑦)>𝑎}≠Φ. Further if 𝑦∈𝑃(𝛼^𝑎/𝜆,𝛽^𝐿2,Ψ_𝑎), then 𝑎≤𝑦(𝑡)≤𝑎/𝜆 for 𝜂≤𝑡≤1. Then by definition of 𝜓 and assumption (H₂), we have 𝜓(𝑇𝑦)==mmiinn_{𝑡∈[𝜂,1]𝑡∈[𝜂,1]}||𝑇𝑦(𝑡)||¹⁰≥min_{𝑡∈[𝜂,1]}𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠^1𝜂>𝑎𝐺(𝑡,𝑠)𝑞(𝑠)𝑓𝑠,𝑦(𝑠),𝑝(𝑠)𝑦^(𝑠)𝑑𝑠𝐶⋅𝐶=𝑎(4.19) Thus, Condition (1) of Theorem 2.11 is satisfied. We finally show that condition (3) of Theorem 2.11 holds, too. Suppose 𝑦∈𝑃(𝛼^𝑐,𝛽^𝐿2,Ψ_𝑎) with 𝛼(𝑇𝑦)>𝑏. Then by definition of 𝜓 and 𝑇𝑦∈𝑃, we have Ψ(𝑇𝑦)=≥𝜆mmina_{𝑡∈[𝜂,1]}x_[]𝑡∈0,1||𝑇𝑦(𝑡)||≥=𝜆𝑎.𝛼(𝑇𝑦)||𝑇𝑦(𝑡)||(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(𝛼^𝑐,𝛽^𝐿2) such that 𝑎<min_{𝑡∈[𝜂,1]}sup_𝑡∈[0,1]||𝑦₁(𝑡)||<𝑑,sup_𝑡∈(0,1]||𝑝(𝑡)𝑦^1(𝑡)||<𝐿₁;||s𝑦u₂p(𝑡_𝑡∈[0,1])||≤sup_[]𝑡∈0,1||𝑦₂(𝑡)||≤𝑐,sup_]𝑡∈(0,1||𝑝(𝑡)𝑦^2(𝑡)||≤𝐿₂;||𝑦₃(𝑡)||≤𝑎𝜆,sup_]𝑡∈(0,1||𝑝(𝑡)𝑦^3(𝑡)||≤𝐿₂.(4.21)

Corollary 4.4. Suppose that (E1)–(E3) are satisfied. If there exist constants 0<𝑑₁<𝑎₁<𝑎₁/𝜆<𝑑₂<𝑎₂<𝑎₂/𝜆<𝑑₃<⋯<𝑑_𝑛,0<𝐿₁≤𝐿₂≤𝐿₃≤⋯≤𝐿_𝑛𝑛∈ℕ, with 𝑎_𝑖/𝐶<min{𝑑_𝑖+1/𝐿,𝐿_𝑖+1/𝑁},1<𝑖≤𝑛−1 such that 𝑓 satisfies the following conditions:(M1)𝑓(𝑡,𝑦,𝑧)<min{𝑑_𝑖/𝐿,𝐿_𝑖/𝑁},for(𝑡,𝑦,𝑧)∈{[0,1]×[0,𝑑_𝑖]×[−𝐿_𝑖,𝐿_𝑖]},1≤𝑖≤𝑛, (M2)𝑓(𝑡,𝑦,𝑧)>𝑎_𝑖/𝐶,for(𝑡,𝑦,𝑧)∈{[𝜂,1]×[𝑎_𝑖,𝑎_𝑖/𝜆]×[−𝐿_𝑖+1,𝐿_𝑖+1]},1≤𝑖≤𝑛−1, then boundary value problem (1.1)-(1.2) has at least 2𝑛−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,𝑎=𝑎₁). 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𝑡,𝑦,𝑝𝑦^=2(𝑟+1)𝑡−54𝑟,𝑞(𝑡)=𝑡^𝑟−1,max_[]𝑡∈0,1𝑓𝑡,𝑦,𝑝𝑦^=34𝑟+2.(5.1) Following the notations of this work, it is easy to see that 𝐿=3234(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟},𝑁=3(1−𝑟)𝑟(4−4^𝑟).(5.2)
Now for 𝑑≥(3/2)[3(1−𝑟)/4−4^𝑟]^{(1−𝑟)/𝑟}[(3/4)𝑟+2]and𝐿₁≥3(1−𝑟)((3/4)𝑟+1)/𝑟(4−4^𝑟),𝑓(𝑡,𝑦,𝑝𝑦′)≤min{𝑑/𝐿,𝐿₁/𝑁}.
Then from Theorem 4.2, the problem has at least one nonnegative solution 𝑦(𝑡) with sup_𝑡∈[0,1]||𝑦(𝑡)||≤3234(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟}34𝑟+2,sup_]𝑡∈(0,1||𝑝(𝑡)𝑦^(𝑡)||≤3(1−𝑟)𝑟(4−4^𝑟)34𝑟+2.(5.3) Next we reduce the problem and then apply Theorem 4.2 for 𝑝(𝑡)=1. Using the transformation 𝑥=(1−𝑟)∫^𝑡0(1/𝑥^𝑟)𝑑𝑥=𝑡^1−𝑟, the boundary value problem (1.5) can be reduced to regular boundary value problem as −𝑣^(𝑥)=1(1−𝑟)²𝑥^{(2𝑟−1)/(1−𝑟)}2(𝑟+1)𝑥^1/(1−𝑟)−5𝑣(0)=0,𝑣(1)=𝑣14𝑟,𝑥∈(0,1),4^(1−𝑟).(5.4) Here, 𝐹𝑥,𝑣,𝑣^=1(1−𝑟)²𝑡^{(2𝑟−1)/(1−𝑟)}2(𝑟+1)𝑡^1/(1−𝑟)−5max_[]𝑥∈0,14𝑟,𝐹𝑥,𝑣,𝑣^=1(1−𝑟)²2(1+𝑟)−54𝑟.(5.5) Now following the notation of this work for 𝑝(𝑡)=1, 𝐿=3(1−𝑟)²4−4^𝑟𝑟3((41−−4𝑟^𝑟))^{(1−𝑟)/𝑟},𝑁=3(1−𝑟)²𝑟(4−4^𝑟).(5.6) Now for 𝑑≥3/(4−4^𝑟)[3(1−𝑟)/(4−4^𝑟)]^{(1−𝑟)/𝑟}[(3/4)𝑟+2]and𝐿₁≥(3/𝑟(4−4^𝑟))[(3/4)𝑟+2],𝑓(𝑡,𝑣,𝑣′)≤min{𝑑/𝐿,𝐿₁/𝑁}. So the problem has at least one nonnegative solution 𝑣(𝑡) with sup_𝑡∈[0,1]||𝑣(𝑡)||≤34−4^𝑟34(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟}34𝑟+2,sup_]𝑡∈(0,1||𝑣^(𝑡)||≤3𝑟(4−4^𝑟)34𝑟+2.(5.7) Hence the boundary value problem (1.5) has at least one nonnegative solution 𝑦(𝑡) with sup_𝑡∈[0,1]||𝑦(𝑡)||≤34−4^𝑟34(1−−4^𝑟𝑟)^{(1−𝑟)/𝑟}34𝑟+2,sup_]𝑡∈(0,1||𝑦^(𝑡)||≤3𝑟(4−4^𝑟)34𝑟+2.(5.8)
Now in this case it is easy to show that if 𝑟 approaches one, that is, the order of singularity increases, upper bound for sup_𝑡∈[0,1]|𝑦(𝑡)| approaches ∞ while in case of direct solving, upper bound for sup_𝑡∈[0,1]|𝑦(𝑡)| approaches 4.125. As smaller upper bound for sup_𝑡∈[0,1]|𝑦(𝑡)| will enable to find nonnegative solution(s) faster and hence will be helpful in constructing efficient numerical algorithms to find multiple nonnegative solutions.

Example 5.2. Consider the following boundary value problem: −𝑡^1/2𝑦^(𝑡)𝑦^(=0)𝑡=0^−1/2𝑓,𝑡,𝑦𝑦((𝑡1)),=𝑡^1/23𝑦^(𝑡),0<𝑡<1,2𝑦31.(5.9)(i) If 𝑓(𝑡,𝑦,𝑧)=⎧⎪⎨⎪⎩sin3𝑡+10𝑦²+8400+30𝑧0²sin𝑡,𝑦≤4;3+2510+30𝑧0²,𝑦≥4;(5.10) then the boundary value problem (5.9) has at least one nonnegative solution.(ii) Further, if 𝑓(𝑡,𝑦,𝑧)=⎧⎪⎪⎨⎪⎪⎩sin𝑡3+10𝑦²+8400+40𝑧00²sin𝑡,𝑦≤43+9𝑦²+24400+40𝑧00²sin𝑡,4≤𝑦≤1913+324803053+40𝑧00²,𝑦≥191,(5.11) then the boundary value problem (5.9) has at least three nonnegative solutions.

Proof. Here, 𝛼₁=3/2 and 𝜂=1/3. After simple calculation, we get 𝐿=5.464, 𝑁=3.732, 𝐶=1.5396, and 𝜆=0.1786.
(i) At least one nonnegative solution: we choose 𝑑=3 and 𝐿₁=10. Here, min{𝑑/𝐿,𝐿₁/𝑁}=0.549; 𝑓(𝑡,𝑦,𝑧)<0.549,for0≤𝑡≤1,0≤𝑦≤3,−10≤𝑧≤10.(5.12) Thus, condition (𝐻₁) is satisfied. Now from Theorem 4.2 the problem has at least one nonnegative solution 𝑦₁ such that sup_𝑡∈[0,1]|𝑦₁(𝑡)|<3 with sup_𝑡∈(0,1]|𝑝(𝑡)𝑦^1(𝑡)|<10.
(ii) At least three nonnegative solutions: we choose constants 𝑑=4,𝑎=34, 𝑐=4800, 𝐿₁=10, and 𝐿₂=3500. Here, min{𝑑/𝐿,𝐿₁/𝑁}=0.732, min{𝑐/𝐿,𝐿₂/𝑁}=878.45, 𝑎/𝜆=190.34, and 𝑎/𝐶=22.084; 𝑓(𝑡,𝑦,𝑓𝑧)(𝑡>,𝑦2,𝑧2).0<840.7f3o2r,1for0≤𝑡≤1,0≤𝑦≤4,−10≤𝑧≤10;𝑓(𝑡,𝑦,𝑧)<878.45for30≤≤𝑡𝑡≤≤1,1,340≤≤𝑦𝑦≤≤149800.03,4,−3−5305000≤≤𝑧𝑧≤≤3530500.0;(5.13) Thus, conditions (𝐻₁), (𝐻₂), and (𝐻₃) are satisfied. Now from Theorem 4.3 the problem has at least three nonnegative solutions 𝑦₁, 𝑦₂, and 𝑦₃ such that 34<min_{𝑡∈[𝜂,1]}sup_𝑡∈[0,1]||𝑦₁(𝑡)||<4,sup_𝑡∈(0,1]||𝑝(𝑡)𝑦₁(𝑡)||<10;su||𝑦p_{2𝑡∈[0,1]}(𝑡)||≤sup_[]𝑡∈0,1||𝑦₂(𝑡)||<4800,sup_]𝑡∈(0,1||𝑝(𝑡)𝑦^2(𝑡)||≤3500,||𝑦₃(𝑡)||≤190.34,sup_]𝑡∈(0,1||𝑝(𝑡)𝑦^3(𝑡)||≤3500.(5.14)

Remark 5.3. For 𝛼₁=1, the problem (1.1)-(1.2) can be regarded as two-point boundary value problem with boundary conditions as 𝑦(0)=0,𝑦^(1)=0,(5.15) in the limiting case 𝜂→1⁻. Thus the results established in this work are also valid for the two-point boundary value problem [21, 22].

Remark 5.4. Theorem 4.3 and Corollary 4.4 extend Theorem 3.1 and Corollary  3.1 of [19] to doubly singular three-point boundary value problem.

Acknowledgments

This work is supported by DST, New Delhi, and UGC, New Delhi, India.