Advances in Difference Equations
Volume 2009 (2009), Article ID 829735, 21 pages
doi:10.1155/2009/829735
Research Article

Positive Solutions for Boundary Value Problems of Second-Order Functional Dynamic Equations on Time Scales

Department of Mathematics, Ege University, Bornova, 35100 Izmir, Turkey

Received 3 February 2009; Accepted 26 February 2009

Academic Editor: Alberto Cabada

Copyright © 2009 Ilkay Yaslan Karaca. 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

Criteria are established for existence of least one or three positive solutions for boundary value problems of second-order functional dynamic equations on time scales. By this purpose, we use a fixed-point index theorem in cones and Leggett-Williams fixed-point theorem.

1. Introduction

In a recent paper [1], by applying a fixed-point index theorem in cones, Jiang and Weng studied the existence of positive solutions for the boundary value problems described by second-order functional differential equations of the form Aykut [2] applied a cone fixed-point index theorem in cones and obtained sufficient conditions for the existence of positive solutions of the boundary value problems of functional difference equations of the form

In this article, we are interested in proving the existence and multiplicity of positive solutions for the boundary value problems of a second-order functional dynamic equation of the form Throughout this paper we let 𝕋 be any time scale (nonempty closed subset of ) and [ 𝑎 , 𝑏 ] be a subset of 𝕋 such that [ 𝑎 , 𝑏 ] = { 𝑡 𝕋 , 𝑎 𝑡 𝑏 } , and for 𝑡 [ 𝜏 1 , 𝑎 ] , 𝑡 is not right scattered and left dense at the same time.

Some preliminary definitions and theorems on time scales can be found in books [3, 4] which are excellent references for calculus of time scales.

We will assume that the following conditions are satisfied.

(H1) 𝑞 ( 𝑡 ) 𝒞 [ 𝑎 , 𝑏 ] , 𝑞 ( 𝑡 ) 0 . (H2) 𝑓 [ 𝑎 , 𝑏 ] × is continuous with respect to 𝑦 and 𝑓 ( 𝑡 , 𝑦 ) 0 for 𝑦 + , where + denotes the set of nonnegative real numbers. (H3) 𝜔 ( 𝑡 ) defined on [ 𝑎 , 𝑏 ] satisfies Let 𝐸 1 = { 𝑡 𝐸 𝑎 𝜔 ( 𝑡 ) 𝜌 ( 𝑏 ) } be nonempty subset of (H4) 𝛼 , 𝛽 , 𝛾 , 𝛿 0 , 𝛼 + 𝛽 > 0 , 𝛾 + 𝛿 > 0 ; if 𝑞 ( 𝑡 ) 0 ( 𝑎 𝑡 𝑏 ) , then 𝛼 + 𝛾 > 0 ; 𝛾 / 𝛿 + for 𝛿 > 0 , where + denotes the set of all positively regressive and rd-continuous functions. (H5) 𝜉 ( 𝑡 ) and 𝜂 ( 𝑡 ) are defined on [ 𝜏 1 , 𝜎 ( 𝑎 ) ] and [ 𝑏 , 𝜎 ( 𝜏 2 ) ] , respectively, where 𝜏 1 = m i n { 𝑎 , 𝑐 } , 𝜏 2 = m a x { 𝑏 , 𝑑 } ; furthermore, 𝜉 ( 𝑎 ) = 𝜂 ( 𝑏 ) = 0 ;

There have been a number of works concerning of at least one and multiple positive solutions for boundary value problems recent years. Some authors have studied the problem for ordinary differential equations, while others have studied the problem for difference equations, while still others have considered the problem for dynamic equations on a time scale [510]. However there are fewer research for the existence of positive solutions of the boundary value problems of functional differential, difference, and dynamic equations [1, 2, 1113].

Our problem is a dynamic analog of the BVPs (1.1) and (1.2). But it is more general than them. Moreover, conditions for the existence of at least one positive solution were studied for the BVPs (1.1) and (1.2). In this paper, we investigate the conditions for the existence of at least one or three positive solutions for the BVP (1.3). The key tools in our approach are the following fixed-point index theorem [14], and Leggett-Williams fixed-point theorem [15].

Theorem 1.1 (see [14]). Let 𝐸 be Banach space and 𝐾 𝐸 be a cone in 𝐸 . Let 𝑟 > 0 , and define Ω 𝑟 = { 𝑥 𝐾 𝑥 < 𝑟 } . Assume 𝐴 Ω 𝑟 𝐾 is a completely continuous operator such that 𝐴 𝑥 𝑥 for 𝑥 𝜕 Ω 𝑟 .
(i) If 𝐴 𝑢 𝑢 for 𝑢 𝜕 Ω 𝑟 , then 𝑖 ( 𝐴 , Ω 𝑟 , 𝐾 ) = 1 . (ii) If 𝐴 𝑢 𝑢 for 𝑢 𝜕 Ω 𝑟 , then 𝑖 ( 𝐴 , Ω 𝑟 , 𝐾 ) = 0 .

Theorem 1.2 (see [15]). Let 𝒫 be a cone in the real Banach space 𝐸 . Set Suppose that 𝐴 𝒫 𝑟 𝒫 𝑟 is a completely continuous operator and 𝜓 is a nonnegative continuous concave functional on 𝒫 with 𝜓 ( 𝑥 ) 𝑥 for all 𝑥 𝒫 𝑟 . If there exists 0 < 𝑝 < 𝑞 < 𝑠 𝑟 such that the following conditions hold:
(i) { 𝑥 𝒫 ( 𝜓 , 𝑞 , 𝑠 ) 𝜓 ( 𝑥 ) > 𝑞 } { } and 𝜓 ( 𝐴 𝑥 ) > 𝑞 for all 𝑥 𝒫 ( 𝜓 , 𝑞 , 𝑠 ) ; (ii) 𝐴 𝑥 < 𝑝 for 𝑥 𝑝 ; (iii) 𝜓 ( 𝐴 𝑥 ) > 𝑞 for 𝑥 𝒫 ( 𝜓 , 𝑞 , 𝑟 ) with 𝐴 𝑥 > 𝑠 . Then 𝐴 has at least three fixed points 𝑥 1 , 𝑥 2 , a n d 𝑥 3 in 𝒫 𝑟 satisfying

2. Preliminaries

First, we give the following definitions of solution and positive solution of BVP (1.3).

Definition 2.1. We say a function 𝑦 ( 𝑡 ) is a solution of BVP (1.3) if it satisfies the following.
(1) 𝑦 ( 𝑡 ) is nonnegative on [ 𝜌 ( 𝜏 1 ) , 𝜎 ( 𝜏 2 ) ] . (2) 𝑦 ( 𝑡 ) = 𝑦 ( 𝜏 1 ; 𝑡 ) as 𝑡 [ 𝜌 ( 𝜏 1 ) , 𝑎 ] , where 𝑦 ( 𝜏 1 ; 𝑡 ) [ 𝜌 ( 𝜏 1 ) , 𝑎 ] [ 0 , ) is defined as (3) 𝑦 ( 𝑡 ) = 𝑦 ( 𝜏 2 ; 𝑡 ) as 𝑡 [ 𝑏 , 𝜎 ( 𝜏 2 ) ] , where 𝑦 ( 𝜏 2 ; 𝑡 ) [ 𝑏 , 𝜎 ( 𝜏 2 ) ] [ 0 , ) is defined as (4) 𝑦 is Δ -differentiable, 𝑦 Δ [ 𝜌 ( 𝑎 ) , 𝑏 ] is -differentiable on [ 𝑎 , 𝑏 ] and ( 𝑦 Δ ) [ 𝑎 , 𝑏 ] is continuous. (5) 𝑦 Δ ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝑦 ( 𝑡 ) = 𝑓 ( 𝑡 , 𝑦 ( 𝜔 ( 𝑡 ) ) ) , for 𝑡 [ 𝑎 , 𝑏 ] .

Furthermore, a solution 𝑦 ( 𝑡 ) of (1.3) is called a positive solution if 𝑦 ( 𝑡 ) > 0 for 𝑡 [ 𝑎 , 𝑏 ] .

Denote by 𝜑 ( 𝑡 ) and 𝜓 ( 𝑡 ) the solutions of the corresponding homogeneous equation under the initial conditions Set Since the Wronskian of any two solutions of (2.3) is independent of 𝑡 , evaluating at 𝑡 = 𝜌 ( 𝑎 ) , 𝑡 = 𝑏 and using the initial conditions (2.4) yield Using the initial conditions (2.4), we can deduce from (2.3) for 𝜑 ( 𝑡 ) and 𝜓 ( 𝑡 ) , the following equations: (See [8].)

Lemma 2.2 (see [8]). Under the conditions (H1) and the first part of (H4) the solutions 𝜓 ( 𝑡 ) and , , 𝜑 𝜑 ( 𝑡 ) 0 , 𝑡 𝜌 ( 𝑎 ) , 𝜎 ( 𝑏 ) , 𝜓 ( 𝑡 ) 0 , 𝑡 𝜌 ( 𝑎 ) , 𝑏 𝜑 ( 𝑡 ) > 0 , 𝑡 𝜌 ( 𝑎 ) , 𝜎 ( 𝑏 ) ] , 𝜓 ( 𝑡 ) > 0 , 𝑡 𝜌 ( 𝑎 ) , 𝑏 Δ ( 𝑡 ) 0 , 𝑡 𝜌 ( 𝑎 ) , 𝑏 , 𝜓 Δ . ( 𝑡 ) 0 , 𝑡 𝜌 ( 𝑎 ) , 𝑏 ( 2 . 9 ) possess the following properties:

Let 𝑦 Δ ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝑦 ( 𝑡 ) = 0 , 𝑡 [ 𝑎 , 𝑏 ] , 𝛼 𝑦 𝜌 ( 𝑎 ) 𝛽 𝑦 Δ 𝜌 ( 𝑎 ) = 0 , 𝛾 𝑦 ( 𝑏 ) + 𝛿 𝑦 Δ ( 𝑏 ) = 0 , ( 2 . 1 0 ) be the Green function for the boundary value problem: given by where 𝜓 ( 𝑡 ) and 𝐷 > 0 are given in (2.7) and (2.8), respectively. It is obvious from (2.6), (H1) and (H4), that 0 𝐺 ( 𝑡 , 𝑠 ) 𝐺 ( 𝑠 , 𝑠 ) holds.

Lemma 2.3. Assume the conditions (H1) and (H4) are satisfied. Then (i) 𝑡 , 𝑠 [ 𝜌 ( 𝑎 ) , 𝑏 ] , for 𝐺 ( 𝑡 , 𝑠 ) Γ 𝐺 ( 𝑠 , 𝑠 ) (ii) 𝑡 [ 𝑎 , 𝜌 ( 𝑏 ) ] for 𝑠 [ 𝜌 ( 𝑎 ) , 𝑏 ] , and 𝐼 Γ = m i n 1 , 𝐼 2 , ( 2 . 1 2 ) where in which

Proof. 𝑡 [ 𝜌 ( 𝑎 ) , 𝜎 ( 𝑏 ) ] for 𝜓 ( 𝑡 ) 0 , and 𝑡 [ 𝜌 ( 𝑎 ) , 𝑏 ] , for 𝜑 ( 𝑡 ) . Besides, 𝜓 ( 𝑡 ) is nondecreasing and 𝑡 [ 𝜌 ( 𝑎 ) , 𝑏 ] is nonincreasing, for 1 0 𝐷 𝜓 ( 𝑡 ) 𝜑 ( 𝑠 ) , i f 𝜌 ( 𝑎 ) 𝑠 𝑡 𝜎 ( 𝑏 ) 𝜓 ( 𝑠 ) 𝜑 ( 𝑡 ) , i f 𝜌 ( 𝑎 ) 𝑡 𝑠 𝜎 ( 𝑏 ) 𝐺 ( 𝑠 , 𝑠 ) . ( 2 . 1 4 ) . Therefore, we have So statement (i) of the lemma is proved. If 𝑠 [ 𝜌 ( 𝑎 ) , 𝑏 ] , for a given 𝑠 [ 𝜌 ( 𝑎 ) , 𝑏 ] then statement (ii) of the lemma is obvious for such values. Now, 𝐺 ( 𝑠 , 𝑠 ) 0 and 𝐺 ( 𝑠 , 𝑠 ) > 0 . Consequently, 𝑠 . , for all such 𝑡 [ 𝑎 , 𝜌 ( 𝑏 ) ] Let us take any 𝑠 [ 𝜌 ( 𝑎 ) , 𝑡 ] . Then we have for 𝐺 ( 𝑡 , 𝑠 ) = 𝐺 ( 𝑠 , 𝑠 ) 𝜓 ( 𝑡 ) 𝜓 𝜓 ( 𝑠 ) 𝜌 ( 𝑏 ) 𝜓 = + 𝜌 ( 𝑎 ) 𝛿 + 𝑏 𝜌 ( 𝑏 ) 𝛾 + 𝛿 𝑞 ( 𝑏 ) 𝑏 𝜌 ( 𝑏 ) 𝛿 + 𝛾 𝑏 𝜌 ( 𝑎 ) 𝑏 𝜌 ( 𝑎 ) 𝑏 𝜏 𝑞 ( 𝑠 ) 𝜓 ( 𝑠 ) 𝑠 Δ 𝜏 1 , ( 2 . 1 5 ) , and we have for 𝐺 ( 𝑡 , 𝑠 ) = 𝐺 ( 𝑠 , 𝑠 ) 𝜑 ( 𝑡 ) 𝜑 ( 𝑠 ) 𝜑 ( 𝑎 ) = + 𝜑 ( 𝑏 ) 𝛽 + 𝛼 𝑎 𝜌 ( 𝑎 ) 𝛽 + 𝛼 𝑏 𝜌 ( 𝑎 ) 𝑏 𝜌 ( 𝑎 ) 𝜏 𝜌 ( 𝑎 ) 𝑞 ( 𝑠 ) 𝜑 ( 𝑠 ) 𝑠 Δ 𝜏 1 . ( 2 . 1 6 ) ,
Let 𝑦 = m a x 𝜌 ( 𝜏 1 ) 𝑡 𝜎 ( 𝜏 2 ) | 𝑦 ( 𝑡 ) | be endowed with maximum norm 𝑦 for 𝐾 , and let 𝐾 = 𝑦 m i n 𝑎 𝑡 𝜌 ( 𝑏 ) 𝑦 ( 𝑡 ) Γ 𝑦 , ( 2 . 1 7 ) be a cone defined by where 𝑦 ( 𝑡 ) is as in (2.12).
Suppose that 𝑦 𝜏 𝑦 ( 𝑡 ) = 1 𝜏 ; 𝑡 , i f 𝜌 1 𝑡 𝑎 , 𝑏 𝜌 ( 𝑎 ) 𝑦 𝜏 𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑦 𝜔 ( 𝑠 ) 𝑠 , i f 𝑎 𝑡 𝑏 , 2 𝜏 ; 𝑡 , i f 𝑏 𝑡 𝜎 2 , ( 2 . 1 8 ) is a solution of (1.3), then it can be written as where
Throughout this paper we assume that 𝑓 0 is the solution of (1.3) with 𝑢 0 ( 𝑡 ) . Clearly, 𝑢 0 𝑢 ( 𝑡 ) = 0 𝜏 1 𝜏 ; 𝑡 , i f 𝜌 1 𝑢 𝑡 𝑎 , 0 , i f 𝑎 𝑡 𝑏 , 0 𝜏 2 𝜏 ; 𝑡 , i f 𝑏 𝑡 𝜎 2 , ( 2 . 2 0 ) can be expressed as follows: where
Let 𝑢 ( 𝑡 ) = 𝑦 ( 𝑡 ) 𝑢 0 ( 𝑡 ) be a solution of (1.3) and 𝑢 ( 𝑡 ) 𝑦 ( 𝑡 ) . Noting that 𝑡 [ 𝑎 , 𝑏 ] for 𝑢 𝜏 𝑢 ( 𝑡 ) = 1 𝜏 ; 𝑡 , i f 𝜌 1 𝑡 𝑎 , 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝑢 𝜏 𝜔 ( 𝑠 ) 𝑠 , i f 𝑎 𝑡 𝑏 , 2 𝜏 ; 𝑡 , i f 𝑏 𝑡 𝜎 2 , ( 2 . 2 2 ) , we have where
Define an operator 𝐴 ( 𝐴 𝑢 ) ( 𝑡 ) = 1 𝑢 𝜏 ( 𝑡 ) , i f 𝜌 1 𝑡 𝑎 , 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝐴 𝜔 ( 𝑠 ) 𝑠 , i f 𝑎 𝑡 𝑏 , 2 𝑢 𝜏 ( 𝑡 ) , i f 𝑏 𝑡 𝜎 2 , ( 2 . 2 4 ) as follows: where
It is easy to derive that 𝑢 = 𝑦 𝑢 0 is a positive solution of BVP (1.3) if and only if 𝑢 is a nontrivial fixed point 𝐴 𝐾 𝐾 of 𝑢 0 , where 𝐴 ( 𝐾 ) 𝐾 . be defined as before.

Lemma 2.4. 𝑢 𝐾

Proof. For 𝐴 𝑢 ( 𝑡 ) 0 , 𝑡 [ 𝜌 ( 𝜏 1 ) , 𝜎 ( 𝜏 2 ) ] , we have 𝐴 . Moreover, we have from definition of 𝐴 𝑢 ( 𝑡 ) 𝐴 𝑢 ( 𝑎 ) that 𝐴 𝑢 ( 𝑡 ) 𝐴 𝑢 ( 𝑏 ) and 𝑡 [ 𝜌 ( 𝜏 1 ) , 𝑎 ] , for 𝑡 [ 𝑏 , 𝜎 ( 𝜏 2 ) ] and 𝐴 𝑢 = 𝐴 𝑢 [ 𝑎 , 𝑏 ] , , respectively. Thus, 𝐴 𝑢 [ 𝑎 , 𝑏 ] = m a x { | 𝐴 𝑢 ( 𝑡 ) | 𝑎 𝑡 𝑏 } where 𝐾 . It follows from the definition m i n 𝑎 𝑡 𝜌 ( 𝑏 ) ( 𝐴 𝑢 ) ( 𝑡 ) = m i n 𝑎 𝑡 𝜌 ( 𝑏 ) 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) 𝑠 Γ 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑠 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) 𝑠 Γ m a x 𝑎 𝑡 𝑏 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑡 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) 𝑠 Γ 𝐴 𝑢 [ 𝑎 , 𝑏 ] = Γ 𝐴 𝑢 , 𝑡 [ 𝑎 , 𝑏 ] . ( 2 . 2 6 ) and Lemma 2.3 that Thus, 𝐴 𝐾 𝐾 .

Lemma 2.5. l i m 𝑣 0 + 𝑓 ( 𝑡 , 𝑣 ) 𝑣 = , l i m 𝑣 + 𝑓 ( 𝑡 , 𝑣 ) 𝑣 = , ( 2 . 2 7 ) is completely continuous.

Lemma 2.6. If for all 0 < 𝑟 0 < 𝑅 0 < , then there exist 𝑖 ( 𝐴 , 𝐾 𝑟 , 𝐾 ) = 0 such that 0 < 𝑟 𝑟 0 , for 𝑖 ( 𝐴 , 𝐾 𝑅 , 𝐾 ) = 0 and 𝑅 𝑅 0 , for 𝑀 > 0 .

Proof. Choose Γ 2 𝑀 𝐸 1 𝐺 ( 𝑠 , 𝑠 ) 𝑠 > 1 . ( 2 . 2 8 ) such that By using the first equality of (2.27), we can choose 𝑓 ( 𝑡 , 𝑣 ) 𝑀 𝑣 , 0 𝑣 𝑟 0 . ( 2 . 2 9 ) such that If 𝑡 0 [ 𝑎 , 𝜌 ( 𝑏 ) ] , then for ( 𝐴 𝑢 ) ( 𝑡 0 ) = 𝑏 𝜌 ( 𝑎 ) 𝐺 𝑡 0 𝑓 , 𝑠 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) 𝑠 Γ 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑠 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) 𝑠 Γ 𝐸 1 𝐺 ( 𝑠 , 𝑠 ) 𝑓 𝑠 , 𝑢 𝜔 ( 𝑠 ) 𝑠 Γ 𝑀 𝐸 1 𝐺 ( 𝑠 , 𝑠 ) 𝑢 𝜔 ( 𝑠 ) 𝑠 Γ 2 𝑢 𝑀 𝐸 1 𝐺 ( 𝑠 , 𝑠 ) 𝑠 > 𝑢 . ( 2 . 3 0 ) , we have Therefore we get Thus, we have from Theorem 1.1, 0 < 𝑟 𝑟 0 , for 𝑀 > 0 . On the other hand, the second equality of (2.27) implies for every 𝑅 0 > 𝑟 0 , there is an 𝑓 ( 𝑡 , 𝑣 ) 𝑀 𝑣 , 𝑣 Γ 𝑅 0 . ( 2 . 3 2 ) , such that Here we choose 𝑢 𝜕 𝐾 𝑅 , 𝑅 𝑅 0 satisfying (2.28). For 𝐾 𝑅 , we have definition of 𝑢 ( 𝑡 ) Γ 𝑢 = Γ 𝑅 , 𝑡 [ 𝑎 , 𝜌 ( 𝑏 ) ] . ( 2 . 3 3 ) that It follows from (2.32) that This shows that Thus, by Theorem 1.1, we conclude that 𝑅 𝑅 0 for 𝑓 ( 𝑡 , 𝑣 ) . The proof is therefore complete.

3. Existence of One Positive Solution

In this section, we investigate the conditions for the existence of at least one positive solution of the BVP (1.3).

In the next theorem, we will also assume that the following condition on l i m i n f 𝑣 0 + m i n 𝑡 [ 𝑎 , 𝑏 ] 𝑓 ( 𝑡 , 𝑣 ) 𝑣 > 𝑘 𝜆 1 , l i m s u p 𝑣 + m a x 𝑡 [ 𝑎 , 𝑏 ] 𝑓 ( 𝑡 , 𝑣 ) 𝑣 < 𝑞 𝜆 1 , ( 3 . 1 ) .

(H6): where 𝑘 Γ 𝐸 1 𝜑 1 ( 𝑠 ) 𝑠 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 , ( 3 . 2 ) is large enough such that and Γ 𝜌 ( 𝑏 ) 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 𝑞 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 , ( 3 . 3 ) is small enough such that where 𝜆 1 ( 𝜆 1 > 0 ) is the eigenfunction related to the smallest eigenvalue 𝜑 1 Δ ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝜑 1 ( 𝑡 ) = 𝜆 𝜑 1 ( 𝑡 ) , 𝛼 𝜑 1 𝜌 ( 𝑎 ) 𝛽 𝜑 Δ 1 𝜌 ( 𝑎 ) = 0 , 𝛾 𝜑 1 ( 𝑏 ) + 𝛿 𝜑 Δ 1 ( 𝑏 ) = 0 . ( 3 . 4 ) of the eigenvalue problem:

Theorem 3.1. If (H1)–(H6) are satisfied, then the BVP (1.3) has at least one positive solution.

Proof. Fix 𝑓 1 ( 𝑢 ) = 𝑢 𝑚 + 𝑢 𝑚 1 and let 𝑢 0 for 𝑓 1 ( 𝑢 ) . Then, 𝐴 𝐾 𝐾 satisfies (2.27). Define 𝐴 𝐴 𝑢 ( 𝑡 ) = 1 𝑢 𝜏 ( 𝑡 ) , i f 𝜌 1 𝑡 𝑎 , 𝑏 𝜌 ( 𝑎 ) 𝐺 ( 𝑡 , 𝑠 ) 𝑓 1 𝑢 𝜔 ( 𝑠 ) + 𝑢 0 𝐴 𝜔 ( 𝑠 ) 𝑠 , i f 𝑎 𝑡 𝑏 , 2 𝑥 𝜏 ( 𝑡 ) , i f 𝑏 𝑡 𝜎 2 , ( 3 . 5 ) by
where Then 0 < 𝑟 0 < 𝑅 0 < is a completely continuous operator. One has from Lemma 2.6 that there exist 0 < 𝑟 𝑟 0 i m p l i e s 𝑖 𝐴 , 𝐾 𝑟 , 𝐾 = 0 , ( 3 . 7 ) 𝑅 𝑅 0 i m p l i e s 𝑖 𝐴 , 𝐾 𝑅 , 𝐾 = 0 . ( 3 . 8 ) such that Define 𝐻 by 𝑓 1 then 𝜖 > 0 is a completely continuous operator. By the first equality in (H6) and the definition of 0 < 𝑟 1 𝑟 0 , there are 𝑓 ( 𝑡 , 𝑢 ) 𝑘 𝜆 1 + 𝜖 𝑢 , 𝑡 [ 𝑎 , 𝑏 ] , 0 𝑢 𝑟 1 , 𝑓 1 ( 𝑢 ) 𝑘 𝜆 1 + 𝜖 𝑢 , 0 𝑢 𝑟 1 . ( 3 . 9 ) and 𝐻 ( 𝑡 , 𝑢 ) 𝑢 such that We now prove that 𝑢 𝜕 𝐾 𝑟 1 for all 𝑡 0 [ 0 , 1 ] and 𝑢 1 𝜕 𝐾 𝑟 1 . In fact, if there exists 𝐻 ( 𝑡 0 , 𝑢 1 ) = 𝑢 1 and 𝑢 1 ( 𝑡 ) such that 𝑢 1 Δ ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝑢 1 = ( 𝑡 ) 1 𝑡 0 𝑓 𝑡 , 𝑢 1 𝜔 ( 𝑡 ) + 𝑢 0 𝜔 ( 𝑡 ) + 𝑡 0 𝑓 1 𝑢 1 𝜔 ( 𝑡 ) + 𝑢 0 𝜔 ( 𝑡 ) , 𝑎 𝑡 𝑏 ( 3 . 1 0 ) , then 𝛼 𝑢 1 𝜌 ( 𝑡 ) 𝛽 𝑢 Δ 1 𝜏 𝜌 ( 𝑡 ) = 0 , 𝑡 1 , , 𝑎 𝛾 𝑢 1 ( 𝑡 ) + 𝛿 𝑢 Δ 1 ( 𝑡 ) = 0 , 𝑡 𝑏 , 𝜏 2 . ( 3 . 1 1 ) satisfies the equation and the boundary conditions Multiplying both sides of (3.10) by 𝑏 , then integrating from 𝜆 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑢 1 = ( 𝑠 ) 𝑠 𝑏 𝜌 ( 𝑎 ) 1 𝑡 0 𝑓 𝑠 , 𝑢 1 𝜔 ( 𝑠 ) + 𝑢 0 𝜔 ( 𝑠 ) + 𝑡 0 𝑓 1 𝑢 1 𝜔 ( 𝑠 ) + 𝑢 0 𝜑 𝜔 ( 𝑠 ) 1 = ( 𝑠 ) 𝑠 1 𝑡 0 𝑏 𝜌 ( 𝑎 ) 𝑓 𝑠 , 𝑢 1 𝜔 ( 𝑠 ) + 𝑢 0 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 + 𝑡 0 𝑏 𝜌 ( 𝑎 ) 𝑓 1 𝑢 1 𝜔 ( 𝑠 ) ) + 𝑢 0 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 . ( 3 . 1 2 ) to 𝜆 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑢 1 ( 𝑠 ) 𝑠 1 𝑡 0 𝐸 1 𝑓 𝑠 , 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 + 𝑡 0 𝐸 1 𝑓 1 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 1 𝑡 0 𝑘 𝜆 1 + 𝜖 𝐸 1 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 + 𝑡 0 𝑘 𝜆 1 + 𝜖 𝐸 1 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 = 𝜆 ( 𝑠 ) 𝑠 1 + 𝜖 𝑘 𝑘 𝐸 1 𝜑 1 ( 𝑠 ) 𝑢 1 𝜆 𝜔 ( 𝑠 ) 𝑠 1 + 𝜖 𝑘 𝑢 𝑘 Γ 1 𝐸 1 𝜑 1 𝜆 ( 𝑠 ) 𝑠 1 + 𝜖 𝑘 𝑢 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 . ( 3 . 1 3 ) , and using integration by parts in the left-hand side two times, we obtain Combining (3.9) and (3.12), we get We also have Equations (3.13) and (3.14) lead to This is impossible. Thus 𝑡 [ 0 , 1 ] for 𝑖 𝐴 , 𝐾 𝑟 1 , 𝐾 = 𝑖 𝐻 ( 0 , ) , 𝐾 𝑟 1 , 𝐾 = 𝑖 𝐻 ( 1 , ) , 𝐾 𝑟 1 , 𝐾 = 𝑖 𝐴 , 𝐾 𝑟 1 , 𝐾 = 0 . ( 3 . 1 6 ) and 𝜖 > 0 . By (3.7) and the homotopy invariance of the fixed-point index (see [11]), we get that On the other hand, according to the second inequality of (H6), there exist 𝑓 ( 𝑡 , 𝑢 ) 𝑞 𝜆 1 𝜖 𝑢 , 𝑢 > 𝑅 , 𝑡 [ 𝑎 , 𝑏 ] . ( 3 . 1 7 ) and 𝐶 = m a x 𝑎 𝑡 𝑏 , 0 𝑢 𝑅 | | 𝑓 ( 𝑡 , 𝑢 ) 𝑞 𝜆 1 𝑢 | | 𝜖 + 1 , ( 3 . 1 8 ) such that We define then it follows that Define 𝐻 1 by 𝑅 1 𝑅 , then 𝐻 1 ( 𝑡 , 𝑢 ) 𝑢 , f o r 𝑡 [ 0 , 1 ] , 𝑢 𝐾 , 𝑢 𝑅 1 . ( 3 . 2 0 ) is a completely continuous operator. We claim that there exists 𝐻 1 ( 𝑡 0 , 𝑢 1 ) = 𝑢 1 such that In fact, if 0 𝑡 0 1 for some 𝜆 1 𝑏 𝜌 ( 𝑎 ) 𝑢 1 ( 𝑠 ) 𝜑 1 ( 𝑠 ) 𝑠 𝑏 𝜌 ( 𝑎 ) 𝑓 𝑠 , 𝑢 1 𝜔 ( 𝑠 ) + 𝑢 0 𝜑 𝜔 ( 𝑠 ) 1 𝜆 ( 𝑠 ) 𝑠 𝑞 1 𝜖 𝑞 𝑢 1 + 𝑢 0 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 + 𝐶 𝑏 𝜌 ( 𝑎 ) 𝜑 1 𝜆 ( 𝑠 ) 𝑠 𝑞 1 𝜖 𝑞 𝑢 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 + 𝐶 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 𝜆 ( 𝑠 ) 𝑠 , ( 3 . 2 1 ) 1 𝑏 𝜌 ( 𝑎 ) 𝑢 1 ( 𝑠 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 𝜌 ( 𝑏 ) 𝜌 ( 𝑎 ) 𝑢 1 ( 𝑠 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 Γ 𝑢 1 𝜌 ( 𝑏 ) 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 𝑞 𝑢 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 , ( 3 . 2 2 ) and 𝐶 1 = 𝑞 ( 𝜆 1 𝜖 / 𝑞 ) 𝑢 0 + 𝐶 . , then where 𝐻 1 ( 𝑡 , 𝑢 ) 𝑢 , f o r 𝑡 [ 0 , 1 ] , 𝑢 𝐾 , 𝑢 𝑅 1 . ( 3 . 2 4 ) Combining (3.21) with (3.22), we have Let Θ Then we get Consequently, by the homotopy invariance of the fixed-point index, we have where ( 𝐾 𝑅 1 𝐾 𝑟 1 ) is zero operator. Use (3.16) and (3.25) to conclude that Hence, 𝑦 ( 𝑡 ) = 𝑢 ( 𝑡 ) has a fixed point in 𝑡 [ 𝑎 , 𝑏 ] .
Let 0 < 𝑟 1 𝑢 = 𝑢 [ 𝑎 , 𝑏 ] = 𝑦 [ 𝑎 , 𝑏 ] 𝑅 1 . Since l i m s u p 𝑣 0 + m a x 𝑡 [ 𝑎 , 𝑏 ] 𝑓 ( 𝑡 , 𝑣 ) 𝑣 < 𝑞 𝜆 1 , l i m i n f 𝑣 + m i n 𝑡 [ 𝑎 , 𝑏 ] 𝑓 ( 𝑡 , 𝑣 ) 𝑣 > 𝑘 𝜆 1 , 𝜉 ( 𝑡 ) 0 , 𝜂 ( 𝑡 ) 0 . ( 3 . 2 7 ) for 𝐻 1 [ 0 , 1 ] × 𝐾 𝐾 and 𝐻 1 ( 𝑡 , 𝑢 ) = 𝑡 𝐴 𝑢 .
(H7)

Theorem 3.2. If (H1)–(H5) and (H7) are satisfied, then the BVP (1.3) has at least one positive solution.

Proof. Define 𝜖 > 0 by 0 < 𝑟 1 𝑟 0 , then 𝑓 ( 𝑡 , 𝑣 ) 𝑞 𝜆 1 𝜖 𝑣 , 𝑡 [ 𝑎 , 𝑏 ] , 0 𝑣 𝑟 1 . ( 3 . 2 8 ) is a completely continuous operator. By the first inequality in (H7), there exist 𝐻 1 ( 𝑡 , 𝑢 ) 𝑢 and 0 𝑡 1 such that We claim that 0 𝑡 0 1 for 𝑢 1 𝜕 𝐾 𝑟 1 and 𝐻 1 ( 𝑡 0 , 𝑢 1 ) = 𝑢 1 . In fact, if there exist 𝑢 1 ( 𝑡 ) and 𝜉 ( 𝑡 ) 0 , 𝜂 ( 𝑡 ) 0 such that 𝑢 0 ( 𝑡 ) 0 , then 𝑢 1 Δ ( 𝑡 ) + 𝑞 ( 𝑡 ) 𝑢 1 ( 𝑡 ) = 𝑡 0 𝑓 𝑡 , 𝑢 1 𝜔 ( 𝑡 ) , 𝑎 𝑡 𝑏 . ( 3 . 2 9 ) satisfies the boundary condition (3.11). Since 𝜑 1 ( 𝑡 ) , we have 𝑎 . Then we have Multiplying the last equation by 𝜆 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑢 1 ( 𝑠 ) 𝑠 = 𝑡 0 𝑏 𝜌 ( 𝑎 ) 𝑓 𝑠 , 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 𝑏 𝜌 ( 𝑎 ) 𝑓 𝑠 , 𝑢 1 𝜑 𝜔 ( 𝑠 ) 1 ( 𝑠 ) 𝑠 𝑞 𝜆 1 𝑢 𝜖 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 , ( 3 . 3 0 ) integrating from 𝜆 1 𝑏 𝜌 ( 𝑎 ) 𝑢 1 ( 𝑠 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 𝜌 ( 𝑏 ) 𝜌 ( 𝑎 ) 𝑢 1 ( 𝑠 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 Γ 𝑢 1 𝜌 ( 𝑏 ) 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 𝜆 1 𝑞 𝑢 1 𝑏 𝜌 ( 𝑎 ) 𝜑 1 ( 𝑠 ) 𝑠 . ( 3 . 3 1 ) to 𝜆 1 𝑞 𝜆 𝑞 1 𝜖 . ( 3 . 3 2 ) , by (3.28), we obtain then we have Equations (3.30) and (3.31) lead to This is impossible. By homotopy invariance of the fixed-point index, we get that Define 𝑓 1 by 𝜖 > 0 , then 𝑅 > 𝑅 0 is a completely continuous operator. By the second inequality in (H7), and definition of 𝑓 ( 𝑡 , 𝑢 ) 𝑘 𝜆 1 + 𝜖 𝑢 , 𝑢 > 𝑅 𝑓 , 𝑡 [ 𝑎 , 𝑏 ] , 1 ( 𝑢 ) 𝑘 𝜆 1 + 𝜖 𝑢 , 𝑢 > 𝑅 . ( 3 . 3 4 ) , there exist 𝐶 = m a x 𝑎 𝑡 𝑏 , 0 𝑢 𝑅 | | 𝑓 ( 𝑡 , 𝑢 ) 𝑘 𝜆 1 𝑢 | | + 𝜖 + m a x