Boundary Value Problems
Volume 2009 (2009), Article ID 834158, 12 pages
doi:10.1155/2009/834158
Research Article

Existence of Positive Solutions for Multipoint Boundary Value Problem on the Half-Line with Impulses

Jianli Li1 and Juan J. Nieto2

1Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China
2Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela, 15782 Santiago de Compostela, Spain

Received 7 March 2009; Revised 18 April 2009; Accepted 25 April 2009

Academic Editor: Donal O'Regan

Copyright © 2009 Jianli Li and Juan J. Nieto. 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 consider a multi-point boundary value problem on the half-line with impulses. By using a fixed-point theorem due to Avery and Peterson, the existence of at least three positive solutions is obtained.

1. Introduction

Impulsive differential equations are a basic tool to study evolution processes that are subjected to abrupt changes in their state. For instance, many biological, physical, and engineering applications exhibit impulsive effects (see [1–3]). It should be noted that recent progress in the development of the qualitative theory of impulsive differential equations has been stimulated primarily by a number of interesting applied problems [4–24].

In this paper, we consider the existence of multiple positive solutions of the following impulsive boundary value problem (for short BVP) on a half-line: 𝑢 ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝑓 ( 𝑡 , 𝑢 ) = 0 , 0 < 𝑡 < ∞ , 𝑡 ≠ 𝑡 𝑘 , 𝑡 Δ 𝑢 𝑘 = 𝐼 𝑘 𝑢 𝑡 𝑘 , 𝑘 = 1 , … , 𝑝 , 𝑢 ( 0 ) = 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑢 𝜉 𝑖 , 𝑢 ( ∞ ) = 0 , ( 1 . 1 ) where 𝑢 ( ∞ ) = l i m 𝑡 → + ∞ 𝑢 ( 𝑡 ) ,  0 < 𝜉 1 < 𝜉 2 < ⋯ < 𝜉 𝑚 − 2 < + ∞ , 0 < 𝑡 1 < 𝑡 2 < ⋯ < 𝑡 𝑝 < + ∞ , Δ 𝑢 ( 𝑡 𝑘 ) = 𝑢 ( 𝑡 + 𝑘 ) − 𝑢 ( 𝑡 − 𝑘 ) , and 𝛼 𝑖 , 𝑓 , 𝑞 , and 𝐼 𝑘 satisfy

( 𝐻 1 ) ∑ 0 < 𝑚 − 2 𝑖 = 1 𝛼 𝑖 < 1 ;( 𝐻 2 ) 𝑓 ( 𝑡 , 𝑢 ) ∈ 𝐶 ( [ 0 , ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) ,  𝐼 𝑘 ( 𝑢 ) ∈ 𝐶 ( [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , and when 𝑢 / ( 1 + 𝑡 ) is bounded, 𝑓 ( 𝑡 , 𝑢 ) and 𝐼 𝑘 ( 𝑢 ) are bounded on [ 0 , + ∞ ) ;( 𝐻 3 ) 𝑞 ( 𝑡 ) ∈ 𝐶 ( [ 0 , ∞ ) , [ 0 , + ∞ ) ) and 𝑞 ( 𝑡 ) is not identically zero on any compact subinterval of ( 0 , + ∞ ) . Furthermore 𝑞 ( 𝑡 ) satisfies s u p 𝑡 ∈ [ 0 , + ∞ ) 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑑 𝑠 < + ∞ , ( 1 . 2 ) where 𝐺 ( 𝑡 , 𝑠 ) = 𝑠 , 0 ≤ 𝑠 ≤ 𝑡 < + ∞ , 𝑡 , 0 ≤ 𝑡 ≤ 𝑠 < + ∞ . ( 1 . 3 )

Boundary value problems on the half-line arise quite naturally in the study of radially symmetric solutions of nonlinear elliptic equations and there are many results in this area, see [8, 13, 14, 20, 25–27], for example.

Lian et al. [25] studied the following boundary value problem of second-order differential equation with a 𝑝 -Laplacian operator on a half-line: 𝜑 𝑝 ( 𝑢 ( 𝑡 ) ) + 𝜙 ( 𝑡 ) 𝑓 𝑡 , 𝑢 , 𝑢 = 0 , 0 < 𝑡 < + ∞ , 𝛼 𝑢 ( 0 ) − 𝛽 𝑢 ( 0 ) = 0 , 𝑢 ( ∞ ) = 0 . ( 1 . 4 ) They showed the existence at least three positive solutions for (1.4) by using a fixed point theorem in a cone due to Avery-Peterson [28].

Yan [20], by using Leray-Schauder theorem and fixed point index theory presents some results on the existence for the boundary value problems on the half-line with impulses and infinite delay.

However to the best knowledge of the authors, there is no paper concerned with the existence of three positive solutions to multipoint boundary value problems of impulsive differential equation on infinite interval so far. Motivated by [20, 25], in this paper, we aim to investigate the existence of triple positive solutions for BVP (1.1). The method chosen in this paper is a fixed point technique due to Avery and Peterson [28].

2. Preliminaries

In this section, we give some definitions and results that we will use in the rest of the paper.

Definition 2.1. Suppose 𝑃 is a cone in a Banach. The map 𝛼 is a nonnegative continuous concave functional on 𝑃 provided 𝛼 ∶ 𝑃 → [ 0 , ∞ ) is continuous and 𝛼 ( 𝑡 𝑥 + ( 1 − 𝑡 ) 𝑦 ) ≥ 𝑡 𝛼 ( 𝑥 ) + ( 1 − 𝑡 ) 𝛼 ( 𝑦 ) ( 2 . 1 ) for all 𝑥 , 𝑦 ∈ 𝑃 , and 𝑡 ∈ [ 0 , 1 ] . Similarly, the map 𝛽 is a nonnegative continuous convex functional on 𝑃 provided 𝛽 ∶ 𝑃 → [ 0 , ∞ ) is continuous and 𝛽 ( 𝑡 𝑥 + ( 1 − 𝑡 ) 𝑦 ) ≤ 𝑡 𝛽 ( 𝑥 ) + ( 1 − 𝑡 ) 𝛽 ( 𝑦 ) ( 2 . 2 ) for all 𝑥 , 𝑦 ∈ 𝑃 , and 𝑡 ∈ [ 0 , 1 ] .

Let 𝛾 , 𝜃 be nonnegative, continuous, convex functionals on 𝑃 and 𝛼 be a nonnegative, continuous, concave functionals on 𝑃 , and 𝜓 be a nonnegative continuous functionals on 𝑃 . Then, for positive real numbers 𝑎 , 𝑏 , 𝑐 , and 𝑑 , we define the convex sets 𝑃 𝑃 ( 𝛾 , 𝑑 ) = { 𝑥 ∈ 𝑃 ∶ 𝛾 ( 𝑥 ) < 𝑑 } , ( 𝛾 , 𝛼 , 𝑏 , 𝑑 ) = { 𝑥 ∈ 𝑃 ∶ 𝑏 ≤ 𝛼 ( 𝑥 ) , 𝛾 ( 𝑥 ) ≤ 𝑑 } , 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) = { 𝑥 ∈ 𝑃 ∶ 𝑏 ≤ 𝛼 ( 𝑥 ) , 𝜃 ( 𝑥 ) ≤ 𝑐 , 𝛾 ( 𝑥 ) ≤ 𝑑 } , ( 2 . 3 ) and the closed set 𝑅 ( 𝛾 , 𝜓 , 𝑎 , 𝑑 ) = { 𝑥 ∈ 𝑃 ∶ 𝑎 ≤ 𝜓 ( 𝑥 ) , 𝛾 ( 𝑥 ) ≤ 𝑑 } . ( 2 . 4 )

To prove our main results, we need the following fixed point theorem due to Avery and Peterson in [28].

Theorem 2.2. Let 𝑃 be a cone in a real Banach space 𝐸 . Let 𝛾 and 𝜃 be nonnegative continuous convex functionals on a cone 𝑃 , 𝛼 be a nonnegative continuous concave functional on 𝑃 , and 𝜓 be a nonnegative continuous functional on 𝑃 satisfying 𝜓 ( 𝜆 𝑥 ) ≤ 𝜆 𝜓 ( 𝑥 ) for 0 ≤ 𝜆 ≤ 1 , such that for some positive numbers 𝑀 and 𝑑 𝛼 ( 𝑥 ) ≤ 𝜓 ( 𝑥 ) , ‖ 𝑥 ‖ ≤ 𝑀 𝛾 ( 𝑥 ) ( 2 . 5 ) for all 𝑥 ∈ 𝑃 ( 𝛾 , 𝑑 ) . Suppose Φ ∶ 𝑃 ( 𝛾 , 𝑑 ) ⟶ 𝑃 ( 𝛾 , 𝑑 ) ( 2 . 6 ) is completely continuous and there exist positive numbers 𝑎 , 𝑑 , and 𝑐 with 𝑎 < 𝑏 such that
(i) { 𝑥 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) ∶ 𝛼 ( 𝑥 ) > 𝑏 } ≠ ∅ and 𝛼 ( Φ 𝑥 ) > 𝑏 for 𝑥 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) ;(ii) 𝛼 ( Φ 𝑥 ) > 𝑏 for 𝑥 ∈ 𝑃 ( 𝛾 , 𝛼 , 𝑏 , 𝑑 ) with 𝜃 ( Φ 𝑥 ) > 𝑐 ;(iii) 0 ∉ 𝑅 ( 𝛾 , 𝜓 , 𝑎 , 𝑑 ) and 𝜓 ( 𝑇 𝑥 ) < 𝑎 for 𝑥 ∈ 𝑅 ( 𝛾 , 𝜓 , 𝑎 , 𝑑 ) , with 𝜓 ( Φ 𝑥 ) = 𝑎 . Then Φ has at least three fixed points 𝑥 1 , 𝑥 2 , 𝑥 3 ∈ 𝑃 ( 𝛾 , 𝑑 ) such that 𝛾 𝑥 𝑖 𝑥 ≤ 𝑑 , f o r 𝑖 = 1 , 2 , 3 , 𝜓 1 𝑥 < 𝑎 , 𝑎 < 𝜓 2 𝑥 w i t h 𝛼 2 𝑥 < 𝑏 , 𝛼 3 > 𝑏 . ( 2 . 7 )

3. Some Lemmas

Define 𝑃 𝐶 [ 0 , + ∞ ) = { 𝑢 ∶ [ 0 , + ∞ ) → 𝑅 ∣ 𝑢 ( 𝑡 ) is continuous at each 𝑡 ≠ 𝑡 𝑘 , left continuous at 𝑡 = 𝑡 𝑘 , 𝑢 ( 𝑡 + 𝑘 ) exists, 𝑘 = 1 , … , 𝑝 } .

By a solution of (1.1) we mean a function 𝑢 in 𝑃 𝐶 [ 0 , ∞ ) satisfying the relations in (1.1).

Lemma 3.1. 𝑢 ( 𝑡 ) is a solution of (1.1) if and only if 𝑢 ( 𝑡 ) is a solution of the following equation: 𝑢 ( 𝑡 ) = 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝑡 𝐼 𝑘 + ∑ ( 𝑢 ) 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 0 + ∞ 𝐺 𝜉 𝑖 , 𝑠 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐼 𝑘 ( ⎤ ⎥ ⎥ ⎦ 𝑢 ) ∶ = 𝑇 𝑢 ( 𝑡 ) , ( 3 . 1 ) where 𝐺 ( 𝑡 , 𝑠 ) is defined as (1.3).

The proof is similar to Lemma 3 in [9], and here we omit it.

For 𝑡 𝑝 < 𝑎 ∗ < 𝑏 ∗ < + ∞ , let 𝑐 ∗ = m i n { 𝑎 ∗ / ( 1 + 𝑎 ∗ ) , 1 / ( 1 + 𝑏 ∗ ) } . Then 𝐺 ( 𝑡 , 𝑠 ) 1 + 𝑡 ≥ 𝑐 ∗ 𝐺 ( 𝑟 , 𝑠 ) , 1 1 + 𝑟 ≥ 𝑐 1 + 𝑡 ∗ 𝑎 1 + 𝑟 , f o r 𝑡 ∈ ∗ , 𝑏 ∗ [ [ , 𝑟 ∈ 0 , + ∞ ) , 𝑠 ∈ 0 , + ∞ ) . ( 3 . 2 ) It is clear that 0 < 𝑐 ∗ < 1 . Consider the space 𝐸 defined by [ 𝐸 = 𝑢 ∈ 𝑃 𝐶 0 , + ∞ ) ∶ s u p 𝑡 ∈ [ 0 , ∞ ) | | | | 𝑢 ( 𝑡 ) 1 + 𝑡 < + ∞ . ( 3 . 3 )

𝐸 is a Banach space, equipped with the norm ‖ 𝑢 ‖ = s u p 0 ≤ 𝑡 < + ∞ ( | 𝑢 ( 𝑡 ) | / ( 1 + 𝑡 ) ) < + ∞ . Define the cone 𝑃 ⊂ 𝐸 by [ 𝑃 = 𝑢 ∈ 𝐸 ∶ 𝑢 ( 𝑡 ) ≥ 0 , 𝑡 ∈ 0 , + ∞ ) , m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑢 ( 𝑡 ) 1 + 𝑡 ≥ 𝑐 ∗ ‖ 𝑢 ‖ . ( 3 . 4 )

Lemma 3.2 (see [20, Theorem 2 . 2 ]). Let 𝑀 ⊂ 𝑃 𝐶 [ 0 , + ∞ ) . Then 𝑀 is compact in 𝑃 𝐶 [ 0 , + ∞ ) , if the following conditions hold:
(a) 𝑀 is bounded in 𝑃 𝐶 [ 0 , + ∞ ) ;(b)the functions belonging to 𝑀 are piecewise equicontinuous on any interval of [ 0 , + ∞ ) ;(c)the functions from 𝑀 are equiconvergent, that is, given 𝜀 > 0 , there corresponds 𝜏 ( 𝜀 ) > 0 such that | 𝑓 ( 𝑡 ) − 𝑓 ( + ∞ ) | < 𝜀 for any 𝑡 ≥ 𝜏 ( 𝜀 ) and 𝑓 ∈ 𝑀 .

Lemma 3.3. 𝑇 ∶ 𝑃 → 𝑃 is completely continuous.

Proof. Firstly, for 𝑢 ∈ 𝑃 , from ( 𝐻 1 ) - ( 𝐻 3 ) , it is easy to check that 𝑇 𝑢 is well defined, and 𝑇 𝑢 ( 𝑡 ) ≥ 0 for all 𝑡 ∈ [ 0 , + ∞ ) . For 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 1 1 1 + 𝑡 ( 𝑇 𝑢 ) ( 𝑡 ) = 1 + 𝑡 0 + ∞ 1 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 + 𝑡 𝑝 𝑘 = 1 𝐼 𝑘 + 1 ( 𝑢 ) ∑ 1 + 𝑡 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 0 + ∞ 𝐺 𝜉 𝑖 , 𝑠 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐼 𝑘 ( ⎤ ⎥ ⎥ ⎦ 𝑢 ) ≥ 𝑐 ∗ 0 + ∞ 𝐺 ( 𝑟 , 𝑠 ) 1 1 + 𝑟 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 + 𝑟 𝑝 𝑘 = 1 0 𝑥 0 2 0 0 𝑑 𝐼 𝑘 ( 𝑢 ) + 𝑐 ∗ ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 0 + ∞ 𝐺 𝜉 𝑖 , 𝑠 1 1 + 𝑟 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ( 𝑠 ) ) 𝑑 𝑠 + 1 + 𝑟 0 < 𝑡 𝑘 < 𝜉 𝑖 0 𝑥 0 2 0 0 𝑑 𝐼 𝑘 ⎤ ⎥ ⎥ ⎦ ( 𝑢 ) ≥ 𝑐 ∗ 𝑇 𝑢 ( 𝑟 ) [ 1 + 𝑟 , f o r 𝑟 ∈ 0 , + ∞ ) ( 3 . 5 ) so m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑇 𝑢 ( 𝑡 ) 1 + 𝑡 ≥ 𝑐 ∗ ‖ 𝑇 𝑢 ‖ , ( 3 . 6 ) which shows 𝑇 𝑃 ⊆ 𝑃 .
Now we prove that 𝑇 is continuous and compact, respectively. Let 𝑢 𝑛 → 𝑢 as 𝑛 → + ∞ in 𝑃 . Then there exists 𝑟 0 such that s u p 𝑛 ∈ 𝑁 ⧵ { 0 } ‖ 𝑢 𝑛 ‖ < 𝑟 0 . By ( 𝐻 2 ) we have 𝑓 ( 𝑡 , 𝑢 ) is bounded on [ 0 , + ∞ ) × [ 0 , 𝑟 0 ] . Set 𝐵 0 = s u p { 𝑓 ( 𝑡 , 𝑢 ) ∶ ( 𝑡 , 𝑢 / ( 1 + 𝑡 ) ) ∈ [ 0 , + ∞ ) × [ 0 , 𝑟 0 ] } , and we have 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 | | 𝑓 1 + 𝑡 ( 𝑠 ) 𝑠 , 𝑢 𝑛 | | − 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 ≤ 2 𝐵 0 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 1 + 𝑡 ( 𝑠 ) 𝑑 𝑠 . ( 3 . 7 ) Therefore by the Lebesgue dominated convergence theorem and continuity of 𝑓 and 𝐼 𝑘 , one arrives at ‖ ‖ 𝑇 𝑢 𝑛 ‖ ‖ − 𝑇 𝑢 ≤ s u p 𝑡 ∈ [ 0 , + ∞ ) 1 ⎧ ⎪ ⎨ ⎪ ⎩ 1 + 𝑡 0 + ∞ | | 𝑓 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑠 , 𝑢 𝑛 | | − 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝑡 | | 𝐼 𝑘 𝑢 𝑛 − 𝐼 𝑘 | | + ∑ ( 𝑢 ) 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 × ⎡ ⎢ ⎢ ⎣ 0 + ∞ 𝐺 𝜉 𝑖 | | 𝑓 , 𝑠 𝑞 ( 𝑠 ) 𝑠 , 𝑢 𝑛 | | − 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 | | 𝐼 𝑘 𝑢 𝑛 − 𝐼 𝑘 | | ⎤ ⎥ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭ ( 𝑢 ) ⟶ 0 a s 𝑛 ⟶ + ∞ . ( 3 . 8 ) Therefore 𝑇 is continuous.
Let Ω be any bounded subset of 𝑃 . Then there exists 𝑟 > 0 such that ‖ 𝑢 ‖ ≤ 𝑟 for all 𝑢 ∈ Ω . Set 𝐵 1 = s u p { 𝑓 ( 𝑡 , 𝑢 ) ∶ ( 𝑡 , 𝑢 / ( 1 + 𝑡 ) ) ∈ [ 0 , + ∞ ) × [ 0 , 𝑟 ] , 𝐵 2 𝑘 = s u p { 𝐼 𝑘 ( 𝑢 ) ∶ 𝑢 / ( 1 + 𝑡 ) ∈ [ 0 , 𝑟 ] } , then ‖ 𝑇 𝑢 ‖ = s u p 𝑡 ∈ [ 0 , + ∞ ) 1 ⎧ ⎪ ⎨ ⎪ ⎩ 1 + 𝑡 0 + ∞ | | | | 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝑡 | | 𝐼 𝑘 ( | | + ∑ 𝑢 ) 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 0 + ∞ 𝐺 𝜉 𝑖 | | | | , 𝑠 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 | | 𝐼 𝑘 | | ⎤ ⎥ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭ ( 𝑢 ) ≤ 𝐵 1 0 + ∞ ∑ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑑 𝑠 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 0 𝑥 0 2 0 0 𝑑 𝛼 𝑖 0 + ∞ 𝐺 𝜉 𝑖 + ∑ , 𝑠 𝑞 ( 𝑠 ) 𝑑 𝑠 1 + 𝑚 − 2 𝑖 = 1 0 𝑥 0 2 0 0 𝑑 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 0 𝑥 0 2 0 0 𝑑 𝛼 𝑖 𝑝 𝑘 = 1 𝐵 2 𝑘 . ( 3 . 9 ) So 𝑇 Ω is bounded.
Moreover, for any 𝜈 ∈ ( 0 , + ∞ ) and 𝑡 , 𝑡 ∈ ( 𝑡 𝑘 , 𝑡 𝑘 + 1 ] ⊂ [ 0 , 𝜈 ] ( 𝑡 < 𝑡 ) , and 𝑢 ∈ Ω , then | | | | 𝑡 𝑇 𝑢 1 + 𝑡 − 𝑡 𝑇 𝑢 1 + 𝑡 | | | | ≤ ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 𝐵 1 0 + ∞ 𝐺 𝜉 𝑖 𝑞 , 𝑠 ( 𝑠 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐵 2 𝑘 ⎤ ⎥ ⎥ ⎦ | | | 1 1 + 𝑡 − 1 1 + 𝑡 | | | + 𝐵 1 0 + ∞ | | | | 𝐺 𝑡 , 𝑠 1 + 𝑡 − 𝐺 𝑡 , 𝑠 1 + 𝑡 | | | | 𝑞 ( 𝑠 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝑡 ′ 𝐵 2 𝑘 | | | 1 1 + 𝑡 − 1 1 + 𝑡 | | | ⟶ 0 , u n i f o r m l y a s 𝑡 ⟶ 𝑡 . ( 3 . 1 0 ) So 𝑇 Ω is quasi-equicontinuous on any compact interval of [ 0 , + ∞ ) .
Finally, we prove for any 𝜀 , there exists sufficiently large 𝑁 1 > 0 such that | | | | 𝑡 𝑇 𝑢 1 + 𝑡 − 𝑡 𝑇 𝑢 1 + 𝑡 | | | | < 𝜀 , ∀ 𝑡 , 𝑡 ≥ 𝑁 1 , 𝑢 ∈ Ω . ( 3 . 1 1 ) Since ∫ 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑑 𝑠 < + ∞ , we can choose 𝑁 1 > 0 such that ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 𝑁 1 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 𝐵 1 0 + ∞ 𝐺 𝜉 𝑖 𝑞 , 𝑠 ( 𝑠 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐵 2 𝑘 ⎤ ⎥ ⎥ ⎦ < 𝜀 6 , 𝐵 1 ∫ 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑑 𝑠 𝑁 1 < 𝜀 6 , 𝑝 𝑘 = 1 𝐵 2 𝑘 𝑁 1 ≤ 𝜀 6 . ( 3 . 1 2 ) For 𝑡 , 𝑡 ≥ 𝑁 1 , it follows that | | | | 𝑡 ( 𝑇 𝑢 ) 1 + 𝑡 − 𝑡 ( 𝑇 𝑢 ) 1 + 𝑡 | | | | ≤ ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎡ ⎢ ⎢ ⎣ 𝐵 1 0 + ∞ 𝐺 𝜉 𝑖 𝑞 , 𝑠 ( 𝑠 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐵 2 𝑘 ⎤ ⎥ ⎥ ⎦ 1 1 + 𝑡 + 1 1 + 𝑡 + 𝐵 1 0 + ∞ 𝐺 𝑡 , 𝑠 1 + 𝑡 𝑞 ( 𝑠 ) 𝑑 𝑠 + 𝐵 1 0 + ∞ 𝐺 𝑡 , 𝑠 1 + 𝑡 + 𝑞 ( 𝑠 ) 𝑑 𝑠 𝑝 𝑘 = 1 𝐵 2 𝑘 1 1 + 𝑡 + 1 1 + 𝑡 < 𝜀 3 + 𝜀 6 + 𝜀 6 + 𝜀 3 = 𝜀 . ( 3 . 1 3 ) That is (3.11) holds. By Lemma 3.2, 𝑇 Ω is relatively compact. In sum, 𝑇 ∶ 𝑃 → 𝑃 is completely continuous.

4. Existence of Three Positive Solutions

Let the nonnegative continuous concave functional 𝛼 , the nonnegative continuous convex functionals 𝛾 and 𝜃 , and the nonnegative continuous functionals 𝜓 be defined on the cone 𝑃 by 𝛾 ( 𝑢 ) = 𝜓 ( 𝑢 ) = 𝜃 ( 𝑢 ) = s u p 𝑡 ∈ [ 0 , + ∞ ) 𝑢 ( 𝑡 ) 1 + 𝑡 , 𝛼 ( 𝑢 ) = m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑢 ( 𝑡 ) 1 + 𝑡 . ( 4 . 1 ) For notational convenience, we denote by 𝑀 = m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 𝑞 𝑀 1 + 𝑡 ( 𝑠 ) 𝑑 𝑠 , 1 = ∑ 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 0 + ∞ 𝐺 𝜉 𝑖 𝑞 , 𝑠 ( 𝑠 ) 𝑑 𝑠 . ( 4 . 2 ) The main result of this paper is the following.

Theorem 4.1. Assume ( 𝐻 1 ) - ( 𝐻 3 ) hold. Let 𝑎 𝑘 ≥ 0 , 0 < 𝑎 < 𝑏 / 𝑐 ∗ < 𝑐 = 𝑑 , 𝑏 / 𝑀 < 𝑐 ∗ 𝑑 / 2 ( 𝑀 + 𝑀 1 ) and suppose that 𝑓 , 𝐼 𝑘 satisfy the following conditions:
( 𝐴 1 ) 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑐 ∗ 𝑑 / 2 ( 𝑀 + 𝑀 1 ) , 𝐼 𝑘 ( 𝑢 ) ≤ 𝑑 𝑐 ∗ / 2 𝑀 2 f o r ( 𝑡 , 𝑢 / ( 1 + 𝑡 ) ∈ [ 0 , + ∞ ) × [ 0 , 𝑑 ] , ( 𝐴 2 ) 𝑓 ( 𝑡 , 𝑢 ) > 𝑏 / 𝑀 for ( 𝑡 , 𝑢 / ( 1 + 𝑡 ) ) ∈ [ 𝑎 ∗ , 𝑏 ∗ ] × [ 𝑏 , 𝑐 ] ,( 𝐴 3 ) 𝑓 ( 𝑡 , 𝑢 ) < 𝑐 ∗ 𝑎 / 2 ( 𝑀 + 𝑀 1 ) , 𝐼 𝑘 ( 𝑢 ) ≤ 𝑎 𝑐 ∗ 𝑎 𝑘 / 2 𝑀 2 f o r 𝑡 ∈ ( 𝑡 , 𝑢 / ( 1 + 𝑡 ) ) ∈ [ 0 , + ∞ ) × [ 0 , 𝑎 ] , where 𝑀 2 = ∑ 𝑝 𝑘 = 1 𝑎 𝑘 ∑ / ( 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ) . Then (1.1) has at least three positive solutions 𝑢 1 , 𝑢 2 and 𝑢 3 such that 𝛾 𝑢 𝑖 𝑢 ≤ 𝑑 , f o r 𝑖 = 1 , 2 , 3 , 𝜓 1 𝑢 < 𝑎 , 𝑎 < 𝜓 2 𝑢 w i t h 𝛼 2 𝑢 < 𝑏 , 𝛼 3 > 𝑏 . ( 4 . 3 )

Proof. Step 1. From the definition 𝛼 , 𝜓 , and 𝛾 , we easily show that 𝛼 ( 𝑢 ) ≤ 𝜓 ( 𝑢 ) , ‖ 𝑢 ‖ ≤ 𝛾 ( 𝑢 ) f o r 𝑢 ∈ 𝑃 ( 𝛾 , 𝑑 ) . ( 4 . 4 ) Next we will show that 𝑇 ∶ 𝑃 ( 𝛾 , 𝑑 ) ⟶ 𝑃 ( 𝛾 , 𝑑 ) . ( 4 . 5 ) In fact, for 𝑢 ∈ 𝑃 ( 𝛾 , 𝑑 ) , then s u p 𝑡 ∈ [ 0 , + ∞ ) 𝑢 ( 𝑡 ) 1 + 𝑡 ≤ 𝑑 . ( 4 . 6 ) From condition ( 𝐴 1 ) , we obtain 𝑓 ( 𝑡 , 𝑢 ) ≤ 𝑑 𝑐 ∗ 2 𝑀 + 𝑀 1 , 𝐼 𝑘 ( 𝑢 ) ≤ 𝑑 𝑐 ∗ 2 𝑀 2 . ( 4 . 7 ) It follows that 𝛾 ( 𝑇 𝑢 ) = s u p 𝑡 ∈ [ 0 , + ∞ ) 𝑇 𝑢 ( 𝑡 ) ≤ 1 1 + 𝑡 𝑐 ∗ m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑇 𝑢 ( 𝑡 ) ≤ 1 1 + 𝑡 𝑐 ∗ m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 1 1 + 𝑡 0 + ∞ 1 𝐺 ( 𝑡 , 𝑠 ) 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 1 + 𝑡 𝑝 𝑘 = 1 𝐼 𝑘 + 1 ( 𝑢 ) ⋅ ∑ 1 + 𝑡 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ⎛ ⎜ ⎜ ⎝ 0 + ∞ 𝐺 𝜉 𝑖 , 𝑠 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 + 0 < 𝑡 𝑘 < 𝜉 𝑖 𝐼 𝑘 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ ≤ 1 ( 𝑢 ) 𝑐 ∗ ⋅ 𝑐 ∗ 𝑑 2 𝑀 + 𝑀 1 m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) ∑ 1 + 𝑡 𝑞 ( 𝑠 ) 𝑑 𝑠 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 0 + ∞ 𝐺 𝜉 𝑖 + 1 , 𝑠 𝑞 ( 𝑠 ) 𝑑 𝑠 𝑐 ∗ ⋅ 𝑐 ∗ 𝑑 ∑ 𝑝 𝑘 = 1 𝑎 𝑘 2 𝑀 2 ∑ 1 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ≤ 𝑑 2 + 𝑑 2 = 𝑑 . ( 4 . 8 ) Thus (4.5) holds.
Step 2. We show that condition (i) in Theorem 2.2 holds. Taking 𝑢 ( 𝑡 ) = ( 1 + 𝑡 ) ( ( 𝑏 + 𝑑 ) / 2 ) , then 𝑢 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) and 𝛼 ( 𝑢 ) > 𝑏 , which shows { 𝑢 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) ∣ 𝛼 ( 𝑢 ) > 𝑏 } ≠ ∅ . Thus for 𝑢 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) , there is 𝑏 ≤ 𝑢 ( 𝑡 ) 𝑎 1 + 𝑡 ≤ f o r 𝑡 ∈ ∗ , 𝑏 ∗ . ( 4 . 9 ) Hence by ( 𝐴 2 ) , we have 𝛼 ( 𝑇 𝑢 ) = m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑇 𝑢 ( 𝑡 ) 1 + 𝑡 ≥ m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) > 𝑏 1 + 𝑡 𝑞 ( 𝑠 ) 𝑓 ( 𝑠 , 𝑢 ) 𝑑 𝑠 𝑀 ⋅ m i n 𝑎 𝑡 ∈ ∗ , 𝑏 ∗ 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) 1 + 𝑡 𝑞 ( 𝑠 ) 𝑑 𝑠 = 𝑏 . ( 4 . 1 0 ) Therefore we have 𝛼 ( 𝑇 𝑢 ) > 𝑏 , ∀ 𝑢 ∈ 𝑃 ( 𝛾 , 𝜃 , 𝛼 , 𝑏 , 𝑐 , 𝑑 ) . ( 4 . 1 1 ) This shows the condition (i) in Theorem 2.2 is satisfied.
Step 3. We now prove (ii) in Theorem 2.2 holds. For 𝑢 ∈ 𝑃 ( 𝛾 , 𝛼 , 𝑏 , 𝑑 ) with 𝜃 ( 𝑇 𝑢 ) > 𝑐 , we have 𝛼 ( 𝑇 𝑢 ) = m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑇 𝑢 ( 𝑡 ) 1 + 𝑡 ≥ 𝑐 ∗ ‖ 𝑇 𝑢 ‖ = 𝑐 ∗ 𝜃 ( 𝑇 𝑢 ) > 𝑐 ∗ 𝑐 > 𝑏 . ( 4 . 1 2 ) Hence, condition (ii) in Theorem 2.2 is satisfied.
Step 4. Finally, we prove (iii) in Theorem 2.2 is satisfied. Since 𝜓 ( 0 ) = 0 < 𝑎 , so 0 ∉ 𝑅 ( 𝛾 , 𝜓 , 𝑎 , 𝑑 ) . Suppose that 𝑢 ∈ 𝑅 ( 𝛾 , 𝜃 , 𝑎 , 𝑑 ) with 𝜓 ( 𝑢 ) = 𝑎 , then 0 ≤ 𝑢 ( 𝑡 ) 1 + 𝑡 ≤ 𝑎 , ( 4 . 1 3 ) by the condition ( 𝐴 3 ) of this theorem, 𝜓 ( 𝑇 𝑢 ) = s u p 𝑡 ∈ [ 0 , + ∞ ) 𝑇 𝑢 ( 𝑡 ) ≤ 1 1 + 𝑡 𝑐 ∗ m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 𝑇 𝑢 ( 𝑡 ) ≤ 1 1 + 𝑡 𝑐 ∗ ⋅ 𝑐 ∗ 𝑎 2 𝑀 + 𝑀 1 m i n 𝑡 ∈ [ 𝑎 ∗ , 𝑏 ∗ ] 0 + ∞ 𝐺 ( 𝑡 , 𝑠 ) ∑ 1 + 𝑡 𝑞 ( 𝑠 ) 𝑑 𝑠 + 𝑚 − 2 𝑖 = 1 𝛼 𝑖 ∑ 1 − 𝑚 − 2 𝑖 = 1 𝛼 𝑖 0 + ∞ 𝐺 𝜉 𝑖 + 1 , 𝑠 𝑞 ( 𝑠 ) 𝑑 𝑠 𝑐 ∗ ⋅ 𝑐