Boundary Value Problems
Volume 2009 (2009), Article ID 937064, 19 pages
doi:10.1155/2009/937064
Research Article

Positive Solutions to Singular and Delay Higher-Order Differential Equations on Time Scales

Liang-Gen Hu,1 Ti-Jun Xiao,2 and Jin Liang3

1Department of Mathematics, University of Science and Technology of China, Hefei 230026, China
2School of Mathematical Sciences, Fudan University, Shanghai 200433, China
3Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China

Received 21 March 2009; Accepted 1 July 2009

Academic Editor: Juan José Nieto

Copyright © 2009 Liang-Gen Hu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We are concerned with singular three-point boundary value problems for delay higher-order dynamic equations on time scales. Theorems on the existence of positive solutions are obtained by utilizing the fixed point theorem of cone expansion and compression type. An example is given to illustrate our main result.

1. Introduction

In this paper, we are concerned with the following singular three-point boundary value problem (BVP for short) for delay higher-order dynamic equations on time scales: ( − 1 ) n 𝑢 Δ 2 𝑛 [ ] , [ 𝑢 ( 𝑡 ) = 𝑤 ( 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) , 𝑡 ∈ 𝑎 , 𝑏 𝑢 ( 𝑡 ) = 𝜓 ( 𝑡 ) , 𝑡 ∈ 𝑎 − 𝑐 , 𝑎 ) , Δ 2 𝑖 ( 𝑎 ) − 𝛽 𝑖 + 1 𝑢 Δ 2 𝑖 + 1 ( 𝑎 ) = 𝛼 𝑖 + 1 𝑢 Δ 2 𝑖 ( 𝛾 𝜛 ) , 𝑖 + 1 𝑢 Δ 2 𝑖 ( 𝜛 ) = 𝑢 Δ 2 𝑖 ( 𝑏 ) , 0 ≤ 𝑖 ≤ 𝑛 − 1 , ( 1 . 1 ) where 𝑐 ∈ [ 0 , ( 𝑏 − 𝑎 ) / 2 ] , 𝜛 ∈ ( 𝑎 , 𝑏 ) , 𝛽 𝑖 ≥ 0 , 1 < 𝛾 𝑖 < ( 𝑏 − 𝑎 + 𝛽 𝑖 ) / ( 𝜛 − 𝑎 + 𝛽 𝑖 ) , 0 ≤ 𝛼 𝑖 < ( 𝑏 − 𝛾 𝑖 𝜛 + ( 𝛾 𝑖 − 1 ) ( 𝑎 − 𝛽 𝑖 ) ) / ( 𝑏 − 𝜛 ) , 𝑖 = 1 , 2 , … , 𝑛 and 𝜓 ∈ 𝐶 ( [ 𝑎 − 𝑐 , 𝑎 ] ) . The functional 𝑤 ∶ ( 𝑎 , 𝑏 ) → [ 0 , + ∞ ) is continuous and 𝑓 ∶ [ 𝑎 , 𝑏 ] × ( 0 , + ∞ ) → [ 0 , + ∞ ) is continuous. Our nonlinearity 𝑤 may have singularity at 𝑡 = 𝑎 and/or 𝑡 = 𝑏 , and 𝑓 may have singularity at 𝑢 = 0 .

To understand the notations used in (1.1), we recall the following definitions which can be found in [1, 2].

(a)A time scale 𝕋 is a nonempty closed subset of the real numbers ℝ . 𝕋 has the topology that it inherits from the real numbers with the standard topology. It follows that the jump operators 𝜎 , 𝜌 ∶ 𝕋 → 𝕋 , 𝜎 ( 𝑡 ) = i n f { 𝜏 ∈ 𝕋 ∶ 𝜏 > 𝑡 } , 𝜌 ( 𝑡 ) = s u p { 𝜏 ∈ 𝕋 ∶ 𝜏 < 𝑡 } ( 1 . 2 ) (supplemented by i n f ∅ ∶ = s u p 𝕋 and s u p ∅ ∶ = i n f 𝕋 ) are well defined. The point 𝑡 ∈ 𝕋 is left-dense, left-scattered, right-dense, right-scattered if 𝜌 ( 𝑡 ) = 𝑡 , 𝜌 ( 𝑡 ) < 𝑡 , 𝜎 ( 𝑡 ) = 𝑡 , 𝜎 ( 𝑡 ) < 𝑡 , respectively. If 𝕋 has a left-scattered maximum 𝑡 1 (right-scattered minimum 𝑡 2 ), define 𝕋 𝑘 = 𝕋 − { 𝑡 1 } ( 𝕋 𝑘 = 𝕋 − { 𝑡 2 } ); otherwise, set 𝕋 𝑘 = 𝕋 ( 𝕋 𝑘 = 𝕋 ). By an interval [ 𝑎 , 𝑏 ] we always mean the intersection of the real interval [ 𝑎 , 𝑏 ] with the given time scale, that is, [ 𝑎 , 𝑏 ] ∩ 𝕋 . Other types of intervals are defined similarly. (b) For a function 𝑓 ∶ 𝕋 → ℝ and 𝑡 ∈ 𝕋 𝑘 , the Δ -derivative of 𝑓 at 𝑡 , denoted by 𝑓 Δ ( 𝑡 ) , is the number (provided it exists) with the property that, given any 𝜀 > 0 , there is a neighborhood 𝑈 ⊂ 𝕋 of 𝑡 such that | | 𝑓 ( 𝜎 ( 𝑡 ) ) − 𝑓 ( 𝑠 ) − 𝑓 Δ [ ] | | | | | | ( 𝑡 ) 𝜎 ( 𝑡 ) − 𝑠 ≤ 𝜀 𝜎 ( 𝑡 ) − 𝑠 , ∀ 𝑠 ∈ 𝑈 . ( 1 . 3 ) (c) For a function 𝑓 ∶ 𝕋 → ℝ and 𝑡 ∈ 𝕋 𝑘 , the ∇ -derivative of 𝑓 at 𝑡 , denoted by 𝑓 ∇ ( 𝑡 ) , is the number (provided it exists) with the property that, given any 𝜀 > 0 , there is a neighborhood 𝑈 ⊂ 𝕋 of 𝑡 such that | | 𝑓 ( 𝜌 ( 𝑡 ) ) − 𝑓 ( 𝑠 ) − 𝑓 ∇ [ ] | | | | | | ( 𝑡 ) 𝜌 ( 𝑡 ) − 𝑠 ≤ 𝜀 𝜌 ( 𝑡 ) − 𝑠 , ∀ 𝑠 ∈ 𝑈 . ( 1 . 4 ) (d)If 𝐹 Δ ( 𝑡 ) = 𝑓 ( 𝑡 ) ( Φ ∇ ( 𝑡 ) = 𝑔 ( 𝑡 ) ) , then we define the integral 𝑡 𝑎 𝑓 ( 𝜛 ) Δ 𝜛 = 𝐹 ( 𝑡 ) − 𝐹 ( 𝑎 ) 𝑡 𝑎 . 𝑔 ( 𝜛 ) ∇ 𝜛 = Φ ( 𝑡 ) − Φ ( 𝑎 ) ( 1 . 5 )

Theoretically, dynamic equations on time scales can build bridges between continuous and discrete mathematics. Practically, dynamic equations have been proposed as models in the study of insect population models, neural networks, and many physical phenomena which include gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems as well as concentration in chemical of biological problems [2]. Hence, two-point and multipoint boundary value problems for dynamic equations on time scales have attracted many researchers' attention (see, e.g., [1–19] and references therein). Moreover, singular boundary value problems have also been treated in many papers (see, e.g., [4, 5, 12–14, 18] and references therein).

In 2004, J. J. DaCunha et al. [13] considered singular second-order three-point boundary value problems on time scales 𝑢 Δ Δ ( ] 𝑢 𝜎 𝑡 ) + 𝑓 ( 𝑡 , 𝑢 ( 𝑡 ) ) = 0 , ( 0 , 1 ∩ 𝕋 , ( 0 ) = 0 , 𝑢 ( 𝑝 ) = 𝑢 2 ( 1 ) ( 1 . 6 ) and obtained the existence of positive solutions by using a fixed point theorem due to Gatica et al. [14], where 𝑓 ∶ ( 0 , 1 ] × ( 0 , ∞ ) → ( 0 , ∞ ) is decreasing in 𝑢 for every 𝑡 ∈ ( 0 , 1 ] and may have singularity at 𝑢 = 0 .

In 2006, Boey and Wong [11] were concerned with higher-order differential equation on time scales of the form ( − 1 ) 𝑛 − 1 𝑦 Δ 𝑛 ( 𝑡 ) = ( − 1 ) 𝑝 + 1 𝐹 𝜎 𝑡 , 𝑦 𝑛 − 1 [ ] , 𝑦 ( 𝑡 ) , 𝑡 ∈ 𝑎 , 𝑏 Δ 𝑖 𝑦 ( 𝑎 ) = 0 , 0 ≤ 𝑖 ≤ 𝑝 − 1 , Δ 𝑖 ( 𝜎 ( 𝑏 ) ) = 0 , 𝑝 ≤ 𝑖 ≤ 𝑛 − 1 , ( 1 . 7 ) where 𝑝 , 𝑛 are fixed integers satisfying 𝑛 ≥ 2 , 1 ≤ 𝑝 ≤ 𝑛 − 1 . They obtained some existence theorems of positive solutions by using Krasnosel'skii fixed point theorem.

Recently, Anderson and Karaca [8] studied higher-order three-point boundary value problems on time scales and obtained criteria for the existence of positive solutions.

The purpose of this paper is to investigate further the singular BVP for delay higher-order dynamic equation (1.1). By the use of the fixed point theorem of cone expansion and compression type, results on the existence of positive solutions to the BVP (1.1) are established.

The paper is organized as follows. In Section 2, we give some lemmas, which will be required in the proof of our main theorem. In Section 3, we prove some theorems on the existence of positive solutions for BVP (1.1). Moreover, we give an example to illustrate our main result.

2. Lemmas

For 1 ≤ 𝑖 ≤ 𝑛 , let 𝐺 𝑖 ( 𝑡 , 𝑠 ) be Green's function of the following three-point boundary value problem: − 𝑢 Δ Δ ( [ ] , 𝑡 ) = 0 , 𝑡 ∈ 𝑎 , 𝑏 𝑢 ( 𝑎 ) − 𝛽 𝑖 𝑢 Δ ( 𝑎 ) = 𝛼 𝑖 𝑢 ( 𝜛 ) , 𝛾 𝑖 𝑢 ( 𝜛 ) = 𝑢 ( 𝑏 ) , ( 2 . 1 ) where 𝜛 ∈ ( 𝑎 , 𝑏 ) and 𝛼 𝑖 , 𝛽 𝑖 , 𝛾 𝑖 satisfy the following condition:

(C) 𝛽 𝑖 ≥ 0 , 1 < 𝛾 𝑖 < 𝑏 − 𝑎 + 𝛽 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 , 0 ≤ 𝛼 𝑖 < 𝑏 − 𝛾 𝑖 𝛾 𝜛 + 𝑖 − 1 𝑎 − 𝛽 𝑖 𝑏 − 𝜛 . ( 2 . 2 ) Throughout the paper, we assume that 𝜎 ( 𝑏 ) = 𝑏 .

From [8], we know that for any ( 𝑡 , 𝑠 ) ∈ [ 𝑎 , 𝑏 ] × [ 𝑎 , 𝑏 ] and 1 ≤ 𝑖 ≤ 𝑛 , 𝐺 𝑖 𝐺 ( 𝑡 , 𝑠 ) = 𝑖 1 [ ] , 𝐺 ( 𝑡 , 𝑠 ) , 𝑠 ∈ 𝑎 , 𝜛 𝑖 2 ( [ ] , 𝑡 , 𝑠 ) , 𝑠 ∈ 𝜛 , 𝑏 ( 2 . 3 ) where 𝐺 𝑖 1 1 ( 𝑡 , 𝑠 ) = 𝑑 𝑖 𝛾 𝑖 ( 𝑡 − 𝜛 ) + 𝑏 − 𝑡 𝜎 ( 𝑠 ) + 𝛽 𝑖 𝛾 − 𝑎 , 𝜎 ( 𝑠 ) ≤ 𝑡 , 𝑖 𝜎 ( 𝑠 ) − 𝜔 + 𝑏 − 𝜎 ( 𝑠 ) 𝑡 + 𝛽 𝑖 − 𝑎 + 𝛼 𝑖 𝐺 ( 𝜛 − 𝑏 ) ( 𝑡 − 𝜎 ( 𝑠 ) ) , 𝑡 ≤ 𝑠 , 𝑖 2 1 ( 𝑡 , 𝑠 ) = 𝑑 𝑖 𝜎 ( 𝑠 ) 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 − 𝑎 ( 𝑏 − 𝑡 ) + 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 𝑡 ( 𝑡 − 𝜎 ( 𝑠 ) ) , 𝜎 ( 𝑠 ) ≤ 𝑡 , 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 𝑑 − 𝑎 ( 𝑏 − 𝜎 ( 𝑠 ) ) , 𝑡 ≤ 𝑠 , 𝑖 = 𝛾 𝑖 − 1 𝑎 − 𝛽 𝑖 + 1 − 𝛼 𝑖 𝛼 𝑏 + 𝜛 𝑖 − 𝛾 𝑖 . ( 2 . 4 )

The following four lemmas can be found in [8].

Lemma 2.1. Suppose that the condition (C) holds. Then the Green function of 𝐺 𝑖 ( 𝑡 , 𝑠 ) in (2.3) satisfies 𝐺 𝑖 ( 𝑡 , 𝑠 ) > 0 , ( 𝑡 , 𝑠 ) ∈ ( 𝑎 , 𝑏 ) × ( 𝑎 , 𝑏 ) . ( 2 . 5 )

Lemma 2.2. Assume that the condition (C) holds. Then Green's function 𝐺 𝑖 ( 𝑡 , 𝑠 ) in (2.3) satisfies 𝐺 𝑖 𝐺 ( 𝑡 , 𝑠 ) ≤ m a x 𝑖 ( 𝑏 , 𝑠 ) , 𝐺 𝑖 , [ ] × [ ] ( 𝜎 ( 𝑠 ) , 𝑠 ) ( 𝑡 , 𝑠 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 . ( 2 . 6 )

Remark 2.3. ( 1 )   If 𝑠 ∈ ( ( 𝛾 𝑖 ( 𝜛 − 𝑎 + 𝛽 𝑖 ) − 𝛼 𝑖 𝜛 − 𝛽 𝑖 + 𝑎 ) / ( 1 − 𝛼 𝑖 ) , 𝑏 ] , 𝑠 ≤ 𝑡 , we know that 𝐺 𝑖 ( 𝑡 , 𝑠 ) is nonincreasing in 𝑡 and 𝐺 𝑖 ( 𝑏 , 𝑠 ) 𝐺 𝑖 = 𝛾 ( 𝜎 ( 𝑠 ) , 𝑠 ) 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 ( 𝑏 − 𝜎 ( 𝑠 ) ) 𝜎 ( 𝑠 ) 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 ( ≥ 𝛾 − 𝑎 𝑏 − 𝜎 ( 𝑠 ) ) 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 𝑏 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 − 𝑎 > 0 . ( 2 . 7 ) Therefore, we have 𝐺 𝑖 ( 𝑏 , 𝑠 ) ≤ 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≤ 𝐺 𝑖 ( 𝜎 ( 𝑠 ) , 𝑠 ) ≤ 𝛿 𝑖 𝐺 𝑖 ( 𝑏 , 𝑠 ) , ( 2 . 8 ) where 𝛿 𝑖 = 𝑏 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 − 𝑎 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 > 1 . ( 2 . 9 ) ( 2 ) If 𝑡 and 𝑠 satisfy the other cases, then we get that 𝐺 𝑖 ( 𝑡 , 𝑠 ) is nondecreasing in 𝑡 and 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≤ 𝐺 𝑖 ( 𝑏 , 𝑠 ) . ( 2 . 1 0 )

Lemma 2.4. Assume that (C) holds. Then Green's function 𝐺 𝑖 ( 𝑡 , 𝑠 ) in (2.3) verifies the following inequality: 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≥ m i n 𝑡 − 𝑎 , 𝑏 − 𝑎 𝑏 − 𝑡 𝛾 𝑖 𝐺 ( 𝑏 − 𝑎 ) 𝑖 ( 𝑏 , 𝑠 ) ≥ m i n 𝑡 − 𝑎 𝛿 𝑖 ( , 𝑏 − 𝑎 ) 𝑏 − 𝑡 𝛾 𝑖 ( 𝐺 𝑏 − 𝑎 ) m a x 𝑖 ( 𝑏 , 𝑠 ) , 𝐺 𝑖 . ( 𝜎 ( 𝑠 ) , 𝑠 ) ( 2 . 1 1 )

Remark 2.5. If 𝑠 ∈ [ 𝜛 , ( 𝛾 𝑖 ( 𝜛 − 𝑎 + 𝛽 𝑖 ) − 𝛼 𝑖 𝜛 − 𝛽 𝑖 + 𝑎 ) / ( 1 − 𝛼 𝑖 ) ) , 𝑠 ≤ 𝑡 , then we find 𝛾 𝑖 − 1 𝑎 − 𝛽 𝑖 + 1 − 𝛼 𝑖 𝜎 𝛼 ( 𝑠 ) + 𝜛 𝑖 − 𝛾 𝑖 < 0 . ( 2 . 1 2 ) So there exists a misprint on [8, Page 2431, line 23]. From (2.3), it follows that 𝐺 𝑖 ( 𝑡 , 𝑠 ) 𝐺 𝑖 = 𝜎 ( 𝑏 , 𝑠 ) ( 𝑠 ) 1 − 𝛼 𝑖 + 𝛼 𝑖 𝜛 + 𝛽 𝑖 − 𝑎 ( 𝑏 − 𝑡 ) + 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 ( 𝑡 − 𝜎 ( 𝑠 ) ) 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 ( ≥ 𝑏 − 𝜎 ( 𝑠 ) ) 𝜛 + 𝛽 𝑖 − 𝑎 ( 𝑏 − 𝑡 ) + 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 ( 𝑡 − 𝜎 ( 𝑠 ) ) 𝛾 𝑖 𝜛 − 𝑎 + 𝛽 𝑖 ( ≥ 𝑏 − 𝑎 ) 𝑏 − 𝑡 𝛾 𝑖 . ( 𝑏 − 𝑎 ) ( 2 . 1 3 ) Consequently, we get 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≥ 𝑏 − 𝑡 𝛾 𝑖 𝐺 ( 𝑏 − 𝑎 ) 𝑖 ( 𝑏 , 𝑠 ) . ( 2 . 1 4 ) If 𝑠 ∈ ( ( 𝛾 𝑖 ( 𝜛 − 𝑎 + 𝛽 𝑖 ) − 𝛼 𝑖 𝜛 − 𝛽 𝑖 + 𝑎 ) / ( 1 − 𝛼 𝑖 ) , 𝑏 ] , 𝑠 ≤ 𝑡 , then, from (2.8), we obtain 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≥ 𝑡 − 𝑎 𝐺 𝑏 − 𝑎 𝑖 ( 𝑏 , 𝑠 ) ≥ 𝑡 − 𝑎 𝛿 𝑖 𝐺 ( 𝑏 − 𝑎 ) 𝑖 ( 𝜎 ( 𝑠 ) , 𝑠 ) . ( 2 . 1 5 )

Remark 2.6. If we set ℎ 𝑖 ( 𝑡 ) ∶ = m i n { ( 𝑡 − 𝑎 ) / 𝛿 𝑖 ( 𝑏 − 𝑎 ) , ( 𝑏 − 𝑡 ) / 𝛾 𝑖 ( 𝑏 − 𝑎 ) } , then we have 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≥ ℎ 𝑖 𝐺 ( 𝑡 ) m a x 𝑖 ( 𝑏 , 𝑠 ) , 𝐺 𝑖 , [ ] × [ ] ( 𝜎 ( 𝑠 ) , 𝑠 ) ( 𝑡 , 𝑠 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 . ( 2 . 1 6 ) Denote ‖ ‖ 𝐺 𝑖 ‖ ‖ ( ⋅ , 𝑠 ) = m a x 𝑡 ∈ [ 𝑎 , 𝑏 ] | | 𝐺 𝑖 | | [ ] . ( 𝑡 , 𝑠 ) , 𝑠 ∈ 𝑎 , 𝑏 ( 2 . 1 7 ) Thus we have 𝐺 𝑖 ( 𝑡 , 𝑠 ) ≥ ℎ 𝑖 ‖ ‖ 𝐺 ( 𝑡 ) 𝑖 ‖ ‖ [ ] × [ ] . ( ⋅ , 𝑠 ) , ( 𝑡 , 𝑠 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 ( 2 . 1 8 )

Lemma 2.7. Assume that condition (C) is satisfied. For 𝐺 𝑖 ( 𝑡 , 𝑠 ) as in (2.3), put 𝐻 1 ( 𝑡 , 𝑠 ) ∶ = 𝐺 1 ( 𝑡 , 𝑠 ) and recursively define 𝐻 𝑗 ( 𝑡 , 𝑠 ) = 𝑏 𝑎 𝐻 𝑗 − 1 ( 𝑡 , 𝑟 ) 𝐺 𝑗 ( 𝑟 , 𝑠 ) Δ 𝑟 ( 2 . 1 9 ) for 2 ≤ 𝑗 ≤ 𝑛 . Then 𝐻 𝑛 ( 𝑡 , 𝑠 ) is Green's function for the homogeneous problem ( − 1 ) 𝑛 𝑢 Δ 2 𝑛 [ ] , 𝑢 ( 𝑡 ) = 0 , 𝑡 ∈ 𝑎 , 𝑏 Δ 2 𝑖 ( 𝑎 ) − 𝛽 𝑖 + 1 𝑢 Δ 2 𝑖 + 1 ( 𝑎 ) = 𝛼 𝑖 + 1 𝑢 Δ 2 𝑖 ( 𝛾 𝜛 ) , 𝑖 + 1 𝑢 Δ 2 𝑖 ( 𝜛 ) = 𝑢 Δ 2 𝑖 ( 𝑏 ) , 0 ≤ 𝑖 ≤ 𝑛 − 1 . ( 2 . 2 0 )

Lemma 2.8. Assume that (C) holds. Denote 𝐾 ∶ = 𝑛 − 1 𝑗 = 1 𝑘 𝑗 , 𝐿 ∶ = 𝑛 − 1 𝑗 = 1 𝑙 𝑗 , ( 2 . 2 1 ) then Green's function 𝐻 𝑛 ( 𝑡 , 𝑠 ) in Lemma 2.7 satisfies ℎ 1 ‖ ‖ 𝐺 ( 𝑡 ) 𝐿 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) ≤ 𝐻 𝑛 ‖ ‖ 𝐺 ( 𝑡 , 𝑠 ) ≤ 𝐾 𝑛 ‖ ‖ [ ] × [ ] , ( ⋅ , 𝑠 ) , ( 𝑡 , 𝑠 ) ∈ 𝑎 , 𝑏 𝑎 , 𝑏 ( 2 . 2 2 ) where 𝑘 𝑗 = 𝑏 𝑎 ‖ ‖ 𝐺 𝑗 ‖ ‖ ( ⋅ , 𝑠 ) Δ 𝑠 > 0 , 𝑙 𝑗 = 𝑏 𝑎 ‖ ‖ 𝐺 𝑗 ‖ ‖ ℎ ( ⋅ , 𝑠 ) 𝑗 + 1 ( 𝑠 ) Δ 𝑠 , 1 ≤ 𝑗 ≤ 𝑛 − 1 . ( 2 . 2 3 )

Proof. We proceed by induction on 𝑛 ≥ 2 . We denote the statement by 𝑃 ( 𝑛 ) . From Lemma 2.7, it follows that ‖ ‖ 𝐻 2 ‖ ‖ = ‖ ‖ ‖ ( 𝑡 , 𝑠 ) 𝑏 𝑎 𝐻 1 ( 𝑡 , 𝑟 ) 𝐺 2 ‖ ‖ ‖ ≤ ( 𝑟 , 𝑠 ) Δ 𝑟 𝑏 𝑎 ‖ ‖ 𝐺 1 ‖ ‖ ‖ ‖ 𝐺 ( ⋅ , 𝑟 ) 2 ‖ ‖ ( ⋅ , 𝑠 ) Δ 𝑟 = 𝑘 1 ‖ ‖ 𝐺 2 ‖ ‖ , ( ⋅ , 𝑠 ) ( 2 . 2 4 ) and from (2.18), we have 𝐻 2 ( 𝑡 , 𝑠 ) = 𝑏 𝑎 𝐻 1 ( 𝑡 , 𝑟 ) 𝐺 2 ≥ ( 𝑟 , 𝑠 ) Δ 𝑟 𝑏 𝑎 ℎ 1 ( ‖ ‖ 𝐺 𝑡 ) 1 ( ‖ ‖ ⋅ , 𝑟 ) × ℎ 2 ( ‖ ‖ 𝐺 𝑟 ) 2 ( ‖ ‖ ⋅ , 𝑠 ) Δ 𝑟 = ℎ 1 ( 𝑡 ) 𝑙 1 ‖ ‖ 𝐺 2 ‖ ‖ . ( ⋅ , 𝑠 ) ( 2 . 2 5 ) So 𝑃 ( 2 ) is true.

We now assume that 𝑃 ( 𝑚 ) is true for some positive integer 𝑚 ≥ 2 . From Lemma 2.7, it follows that ‖ ‖ 𝐻 𝑚 + 1 ‖ ‖ = ‖ ‖ ‖ ( 𝑡 , 𝑠 ) 𝑏 𝑎 𝐻 𝑚 ( 𝑡 , 𝑟 ) 𝐺 𝑚 + 1 ‖ ‖ ‖ ≤ ( 𝑟 , 𝑠 ) Δ 𝑟 𝑏 𝑎 𝐻 𝑚 ‖ ‖ 𝐺 ( 𝑡 , 𝑟 ) 𝑚 + 1 ‖ ‖ ≤ ( 𝑟 , 𝑠 ) Δ 𝑟 𝑏 𝑎 𝑚 − 1 𝑗 = 1 𝑘 𝑗 × ‖ ‖ 𝐺 𝑚 ‖ ‖ ‖ ‖ 𝐺 ( ⋅ , 𝑟 ) Δ 𝑟 𝑚 + 1 ‖ ‖ = ( ⋅ , 𝑠 ) 𝑚 𝑗 = 1 𝑘 𝑗 ‖ ‖ 𝐺 𝑚 + 1 ‖ ‖ , 𝐻 ( ⋅ , 𝑠 ) 𝑚 + 1 ( 𝑡 , 𝑠 ) = 𝑏 𝑎 𝐻 𝑚 ( 𝑡 , 𝑟 ) 𝐺 𝑚 + 1 ≥ ( 𝑟 , 𝑠 ) Δ 𝑟 𝑏 𝑎 ℎ 1 ( 𝑡 ) × 𝑚 − 1 𝑗 = 1 𝑙 𝑗 𝐺 𝑚 ( ⋅ , 𝑟 ) ℎ 𝑚 + 1 ‖ ‖ 𝐺 ( 𝑟 ) 𝑚 + 1 ‖ ‖ ( ⋅ , 𝑠 ) Δ 𝑟 = ℎ 1 ( 𝑡 ) 𝑚 𝑗 = 1 𝑙 𝑗 ‖ ‖ 𝐺 𝑚 + 1 ‖ ‖ . ( ⋅ , 𝑠 ) ( 2 . 2 6 ) So 𝑃 ( 𝑚 + 1 ) holds. Thus 𝑃 ( 𝑛 ) is true by induction.

Lemma 2.9 (see [20]). Let ( 𝐸 , ‖ ⋅ ‖ ) be a real Banach space and 𝑃 ⊂ 𝐸 a cone. Assume that 𝑇 ∶ 𝑃 𝜁 , 𝜂 → 𝑃 is completely continuous operator such that
(i) 𝑇 𝑢 ≰ 𝑢 for 𝑢 ∈ 𝜕 𝑃 𝜁 and ̸ 𝑇 𝑢 ≥ 𝑢 for 𝑢 ∈ 𝜕 𝑃 𝜂 , (ii) ̸ 𝑇 𝑢 ≥ 𝑢 for 𝑢 ∈ 𝜕 𝑃 𝜁 and 𝑇 𝑢 ≰ 𝑢 for 𝑢 ∈ 𝜕 𝑃 𝜂 . Then 𝑇 has a fixed point 𝑢 ∗ ∈ 𝑃 with 𝜁 ≤ ‖ 𝑢 ∗ ‖ ≤ 𝜂 .

3. Main Results

We assume that { 𝑎 𝑚 } 𝑚 ≥ 1 and { 𝑏 𝑚 } 𝑚 ≥ 1 are strictly decreasing and strictly increasing sequences, respectively, with l i m 𝑚 → ∞ 𝑎 𝑚 = 𝑎 , l i m 𝑚 → ∞ 𝑏 𝑚 = 𝑏 and 𝑎 1 < 𝑏 1 . A Banach space 𝐸 = 𝐶 ( [ 𝑎 , 𝑏 ] ) is the set of real-valued continuous (in the topology of 𝕋 ) functions 𝑢 ( 𝑡 ) defined on [ 𝑎 , 𝑏 ] with the norm ‖ 𝑢 ‖ = m a x 𝑡 ∈ [ 𝑎 , 𝑏 ] | | | | . 𝑢 ( 𝑡 ) ( 3 . 1 ) Define a cone by ℎ 𝑃 = 𝑢 ∈ 𝐸 ∶ 𝑢 ( 𝑡 ) ≥ 1 ( 𝑡 ) 𝐿 𝐾 [ ] ‖ 𝑢 ‖ , 𝑡 ∈ 𝑎 , 𝑏 . ( 3 . 2 ) Set 𝑃 𝜉 = { 𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ < 𝜉 } , 𝜕 𝑃 𝜉 𝑃 = { 𝑢 ∈ 𝑃 ∶ ‖ 𝑢 ‖ = 𝜉 } , 𝜉 > 0 , 𝜁 , 𝜂 𝑌 = { 𝑢 ∈ 𝑃 ∶ 𝜁 < ‖ 𝑢 ‖ < 𝜂 } , 0 < 𝜁 < 𝜂 , 1 [ ] = { 𝑡 ∈ 𝑎 , 𝑏 ∶ 𝑡 − 𝑐 < 𝑎 } , 𝑌 2 [ ] 𝑌 = { 𝑡 ∈ 𝑎 , 𝑏 ∶ 𝑡 − 𝑐 ≥ 𝑎 } , 𝑚 = 𝑡 ∈ 𝑌 2 ∶ 𝑡 − 𝑐 ∈ 𝑎 , 𝑎 𝑚 ∪ 𝑏 𝑚 . , 𝑏 ( 3 . 3 ) Assume that

(C1) 𝜓 ∶ [ 𝑎 − 𝑐 , 𝑎 ] → ( 0 , ∞ ) is continuous; (C2)we have 0 < 𝐾 𝑞 𝑝 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) Δ 𝑠 , 𝐾 𝑏 𝑎 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) Δ 𝑠 < + ∞ , ( 3 . 4 ) for constants 𝑝 and 𝑞 with 𝑎 + 𝑐 < 𝑝 < 𝑞 < 𝑏 ; (C3)the function 𝑓 ∶ [ 𝑎 , 𝑏 ] × ( 0 , + ∞ ) → ℝ + is continuous and 𝑤 ∶ ( 𝑎 , 𝑏 ) → ℝ + is continuous satisfying l i m 𝑚 → ∞ s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝐾 𝑌 𝑚 ‖ ‖ 𝐺 𝑛 ( ‖ ‖ ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 = 0 , ∀ 0 < 𝜁 < 𝜂 . ( 3 . 5 )

We seek positive solutions 𝑢 ∶ [ 𝑎 , 𝑏 ] → ℝ + , satisfying (1.1). For this end, we transform (1.1) into an integral equation involving the appropriate Green function and seek fixed points of the following integral operator.

Define an operator 𝑇 ∶ 𝐶 + [ 𝑎 , 𝑏 ] → 𝐶 [ 𝑎 , 𝑏 ] by ( 𝑇 𝑢 ) ( 𝑡 ) = 𝑏 𝑎 𝐻 𝑛 ( 𝑡 , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 , ∀ 𝑢 ∈ 𝐶 + ( [ ] 𝑎 , 𝑏 ) , ( 3 . 6 ) where 𝐶 + [ 𝑎 , 𝑏 ] = { 𝑢 ∈ 𝐶 [ 𝑎 , 𝑏 ] ∣ 𝑢 ( 𝑡 ) ≥ 0 , 𝑡 ∈ [ 𝑎 , 𝑏 ] } .

Proposition 3.1. Let (C1), (C2), and (C3) hold, and let 𝜁 , 𝜂 be fixed constants with 0 < 𝜁 < 𝜂 . Then 𝑇 ∶ 𝑃 𝜁 , 𝜂 → 𝑃 is completely continuous.

Proof. We separate the proof into four steps.Step 1. For each 𝑢 ∈ 𝑃 𝜁 , 𝜂 , 𝑇 𝑢 is bounded.
By condition (C3), there exists some positive integer 𝑚 0 satisfying s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝐾 𝑌 𝑚 0 ‖ ‖ 𝐺 𝑛 ( ‖ ‖ ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 ≤ 1 , ( 3 . 7 ) where 𝑌 𝑚 0 = 𝑡 ∈ 𝑌 2 ∶ 𝑡 − 𝑐 ∈ 𝑎 , 𝑎 𝑚 0 ∪ 𝑏 𝑚 0 ; , 𝑏 ( 3 . 8 ) here, we used the fact that for each 𝑢 ∈ 𝑃 𝜁 , 𝜂 and 𝑡 ∈ [ 𝑎 𝑚 0 + 𝑐 , 𝑏 𝑚 0 + 𝑐 ] ∩ [ 𝑎 , 𝑏 ] , ℎ 𝜂 ≥ 𝑢 ( 𝑡 − 𝑐 ) ≥ 1 ( 𝑡 − 𝑐 ) 𝐿 𝐾 ‖ ℎ 𝑢 ‖ ≥ 𝜁 m i n 1 𝑎 𝑚 0 𝐿 𝐾 , ℎ 1 𝑏 𝑚 0 𝐿 𝐾 , ℎ 1 ( 𝑏 − 𝑐 ) 𝐿 𝐾 = 𝜁 ℎ > 0 , ( 3 . 9 ) where ℎ ℎ = m i n 1 𝑎 𝑚 0 𝐿 𝐾 , ℎ 1 𝑏 𝑚 0 𝐿 𝐾 , ℎ 1 ( 𝑏 − 𝑐 ) 𝐿 𝐾 . ( 3 . 1 0 )
Set 𝑓 𝐷 ∶ = m a x ( 𝑡 , 𝜓 ( 𝑡 − 𝑐 ) ) ∶ 𝑡 ∈ 𝑌 1 , 𝑄 ∶ = m a x 𝑓 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) ∶ 𝑡 ∈ 𝑌 2 . , 𝜁 ℎ ≤ 𝑢 ( 𝑡 − 𝑐 ) ≤ 𝜂 ( 3 . 1 1 ) Then we obtain 𝑇 𝑢 ( 𝑡 ) ≤ s u p 𝑡 ∈ [ 𝑎 , 𝑏 ] s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝑏 𝑎 𝐻 𝑛 ( 𝑡 , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 ≤ 𝐾 s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝑌 1 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 + s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝐾 𝑌 𝑚 0 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 + s u p 𝑢 ∈ 𝑃 𝜁 , 𝜂 𝐾 𝑌 2 ⧵ 𝑌 𝑚 0 ‖ ‖ 𝐺 𝑛 ( ‖ ‖ ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 ≤ 1 + m a x { 𝐷 , 𝑄 } 𝐾 𝑏 𝑎 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) Δ 𝑠 < + ∞ . ( 3 . 1 2 ) Consequently, 𝑇 𝑢 is bounded and well defined.Step 2. 𝑇 ∶ 𝑃 𝜁 , 𝜂 → 𝑃 . For every 𝑢 ∈ 𝑃 𝜁 , 𝜂 , we get from (2.22) ‖ 𝑇 𝑢 ‖ = s u p 𝑡 ∈ [ 𝑎 , 𝑏 ] 𝑏 𝑎 𝐻 𝑛 ( 𝑡 , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 ≤ 𝐾 𝑏 𝑎 ‖ ‖ 𝐺 𝑛 ‖ ‖ ( ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 . ( 3 . 1 3 ) Then by the above inequality ( 𝑇 𝑢 ) ( 𝑡 ) = 𝑏 𝑎 𝐻 𝑛 ≥ ( 𝑡 , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 𝑏 𝑎 ℎ 1 ( ‖ ‖ 𝐺 𝑡 ) 𝐿 𝑛 ( ‖ ‖ ≥ ℎ ⋅ , 𝑠 ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 1 ( 𝑡 ) 𝐿 𝐾 ‖ 𝑇 𝑢 ‖ . ( 3 . 1 4 ) This leads to 𝑇 𝑢 ∈ 𝑃 .Step 3. We will show that 𝑇 ∶ 𝑃 𝜁 , 𝜂 → 𝑃 is continuous. Let { 𝑢 𝑚 } 𝑚 ≥ 1 be any sequence in 𝑃 𝜁 , 𝜂 such that l i m 𝑚 → ∞ 𝑢 𝑚 = 𝑢 ∈ 𝑃 𝜁 , 𝜂 . Notice also that as 𝑚 → ∞ , 𝜙 𝑚 | | 𝑓 ( 𝑠 ) = 𝑠 , 𝑢 𝑚 | | 𝑤 | | 𝑓 ( 𝑠 − 𝑐 ) − 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) ( 𝑠 ) ⟶ 0 , f o r 𝑠 ∈ ( 𝑎 + 𝑐 , 𝑏 ) , 𝑠 , 𝑢 𝑚 ( | | = | | | | 𝑠 − 𝑐 ) − 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) 𝑤 ( 𝑠 ) 𝑓 ( 𝑠 , 𝜓 ( 𝑠 − 𝑐 ) ) − 𝑓 ( 𝑠 , 𝜓 ( 𝑠 − 𝑐 ) ) 𝑤 ( 𝑠 ) = 0 , f o r 𝑠 ∈ ( 𝑎 , 𝑎 + 𝑐 ) , 𝑌 2 𝐻 𝑛 ( 𝑡 , 𝑠 ) 𝜙 𝑚 ( 𝑠 ) Δ 𝑠 ≤ s u p 𝑥 ∈ 𝑃 𝜁 , 𝜂 2 𝐾 𝑌 2 ‖ ‖ 𝐺 𝑛 ‖ ‖ 𝑤 ( ⋅ , 𝑠 ) ( 𝑠 ) 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) Δ 𝑠 < + ∞ . ( 3 . 1 5 ) Now these together with (C2) and the Lebesgue dominated convergence theorem [10] yield that as 𝑚 → ∞ , ‖ ‖ 𝑇 𝑢 𝑚 ‖ ‖ − 𝑇 𝑢 = s u p [ ] 𝑡 ∈ 𝑎 , 𝑏 𝑏 𝑎 𝐻 𝑛 | | 𝑓 ( 𝑡 , 𝑠 ) 𝑤 ( 𝑠 ) 𝑠 , 𝑢 𝑚 | | ( 𝑠 − 𝑐 ) − 𝑓 ( 𝑠 , 𝑢 ( 𝑠 − 𝑐 ) ) Δ 𝑠 ⟶ 0 . ( 3 . 1 6 ) Step 4. 𝑇 ∶ 𝑃 𝜁 , 𝜂 → 𝑃 is compact.
Define 𝑤 𝑚 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑤 𝑎 ( 𝑡 ) = m i n ( 𝑡 ) , 𝑤 𝑚 , 𝑎 ≤ 𝑡 ≤ 𝑎 𝑚 , 𝑤 ( 𝑡 ) , 𝑎 𝑚 ≤ 𝑡 ≤ 𝑏 𝑚 , 𝑏 m i n 𝑤 ( 𝑡 ) , 𝑤 𝑚 , 𝑏 𝑚 𝑓 ≤ 𝑡 ≤ 𝑏 , 𝑚 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑎 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) = 𝑓 ( 𝑡 , 𝜓 ( 𝑡 − 𝑐 ) ) , 𝑎 ≤ 𝑡 < 𝑎 + 𝑐 , m i n 𝑓 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) , 𝑓 𝑡 , 𝑢 𝑚 , 𝑎 + 𝑐 ≤ 𝑡 ≤ 𝑎 𝑚 𝑎 + 𝑐 , 𝑓 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) , 𝑡 ∈ 𝑚 + 𝑐 , 𝑏 𝑚 ∩ [ ] , 𝑏 + 𝑐 𝑎 , 𝑏 m i n 𝑓 ( 𝑡 , 𝑢 ( 𝑡 − 𝑐 ) ) , 𝑓 𝑡 , 𝑢 𝑚 𝑏 , 𝑡 ∈ 𝑚 ∩ [ ] , + 𝑐 , 𝑏 𝑎 , 𝑏 (